Mechanics of Solids Solution Manual - STATICS AND MECHANICS...

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Unformatted text preview: STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS Chapter 1 1-1 Calculate the mass m of a body that weighs 600 lb at the surface of the earth. SOLUTION m 1-2 W g 600 18.63 slug .................................................... Ans. 32.2 Calculate the weight W of a body at the surface of the earth if it has a mass m of 675 kg. SOLUTION W 1-3 mg 675 9.81 6.62 10 3 N 6.62 kN .................................. Ans. If a man weighs 180 lb at sea level, determine the weight W of the man (a) At the top of Mt. McKinley (20,320 ft above sea level). (b) At the top of Mt. Everest (29,028 ft above sea level). SOLUTION W Therefore: where (a) Gme m re2 Wh rh2 Gme m 2.092032 10 7 ft 2 2 W0 r02 r0 rh r0 h 2.090 10 7 Wh W0 r02 rh2 2.090 10 7 ft 2.0320 10 4 180 2.090 10 7 2.092032 10 7 2.9028 10 4 179.7 lb ...................................... Ans. 2.0929028 10 7 ft (b) rh r0 h 2.090 10 7 Wh W0 r02 rh2 180 2.090 10 7 2.0929028 10 7 2 2 179.5 lb .................................... Ans. 1-4 Calculate the weight W of a navigation satellite at a distance of 20,200 km above the earth’s surface if the satellite weighs 9750 N at the earth’s surface. SOLUTION W Therefore: where (a) Gme m re2 Wh rh2 Gme m 26,570 km 2 W0 r02 r0 rh r0 h Wh 6.370 106 m 6370 20, 200 9750 6370 26,570 2 W0 r02 rh2 560 N ........................................... Ans. 1 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 1-5 RILEY, STURGES AND MORRIS Compute the gravitational force acting between two spheres that are touching each other if each sphere weighs 1125 lb and has a diameter of 20 in. SOLUTION F Gm1m2 r2 Gms ms 2 rs rs 8 Gms2 d s2 2 3.439 10 1125 2 32.2 20 1-6 15.11 10 6 lb .................................. Ans. 12 Two spherical bodies have masses of 60 kg and 80 kg, respectively. Determine the gravitational force of attraction between the spheres if the distance from center to center is 600 mm. SOLUTION F 1-7 Gm1m2 r2 6.673 10 11 60 80 2 0.600 0.890 10 6 N ........................ Ans. Determine the weight W of a satellite when it is in orbit 8500 mi above the surface of the earth if the satellite weighs 7600 lb at the earth’s surface. SOLUTION W Therefore: where Gm1m2 r2 Wh rh2 Gme mb W0 r02 r0 rh ro Wh h 2.090 10 7 ft 2.090 10 7 8500(5280) 6.578 10 7 ft Wo ro2 rh2 7600 2.090 107 6.578 107 2 2 768 lb ..................................... Ans. 1-8 Determine the weight W of a satellite when it is in orbit 20.2(106) m above the surface of the earth if the satellite weighs 8450 N at the earth’s surface. SOLUTION W Therefore: where Gm1m2 r2 Wh rh2 Gme mb 26.570 106 m 2 W0 r02 r0 rh ro h Wh 6.370 106 Wo ro2 rh2 6.370 106 m 20.2 106 8450 6.370 106 26.570 10 62 486 N ..................................... Ans. 2 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 1-9 RILEY, STURGES AND MORRIS If a woman weighs 135 lb when standing on the surface of the earth, how much would she weigh when standing on the surface of the moon? SOLUTION W Therefore on the surface of the earth where Gm1m2 r2 4.095 1023 slugs and r 2 me 3960 mi 135 G 4.095 1023 m 3960 5280 7 Gm 1.4413 10 Then on the surface of the moon where lb ft 2 /slug mm 5.037 1021 slugs and r 1080 mi 7 W 1.4413 10 5.037 1021 2 1080 5280 22.33 lb ................................... Ans. 1-10 Determine the weight W of a body that has a mass of 1000 kg (a.) At the surface of the earth. (b.) At the top of Mt. McKinley (6193 m above sea level). (c.) In a satellite at an altitude of 250 km. SOLUTION W (a) Gm1m2 r2 32 W 6.673 10 11 5.976 1024 1000 6370 10 6.673 10 11 9830 N ............................. Ans. (b) W 5.976 1024 1000 3 6370 10 6.673 10 11 6193 2 9810 N ............................. Ans. (c) W 5.976 1024 1000 3 6370 10 250 10 32 9100 N ............................. Ans. 1-11 If a man weighs 210 lb at sea level, determine the weight W of the man (a.) At the top of Mt. Everest (29,028 ft above sea level). (b.) In a satellite at an altitude of 200 mi. SOLUTION W Therefore 210 Gm1m2 r2 Gme m 3960 5280 2 Gme m 9.181 1016 lb ft 2 3 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (a) RILEY, STURGES AND MORRIS 2 W W 9.181 1016 3960 5280 29, 028 9.181 1016 3960 200 5280 2 209.4 lb ....................................... Ans. 190.3 lb ........................................ Ans. (b) 1-12 A space traveler weighs 800 N on earth. A planet having a mass of 5(1025) kg and a diameter of 30(106) m orbits a distant star. Determine the weight W of the traveler on the surface of this planet. SOLUTION W Therefore on the surface of the earth where Gm1m2 r2 5.976 1024 kg and r 2 me 6370 km 800 G 5.976 10 24 m 6370 103 9 Gm 5.432 10 Then on the surface of the planet where N m 2 /kg m 5 1025 kg and r 15 106 m 9 W 5.432 10 5 1025 2 15 106 1207 N ......................................... Ans. 1-13 The planet Jupiter has a mass of 1.302(1026) slug and a visible diameter (top of the cloud layer) of 88,700 mi. Determine the gravitational acceleration g (a.) At a point 100,000 miles above the top of the clouds. (b.) At the top of the cloud layers. SOLUTION W (a) Gm1m2 r2 8 W mg 3.439 10 1.302 10 26 m 5280 2 44,350 100, 000 g 7.71 ft/s 2 ................................................................. Ans. 8 (b) W mg 3.439 10 1.302 1026 m 2 44,350 5280 g 81.7 ft/s 2 ................................................................ Ans. 1-14 The planet Saturn has a mass of 5.67(1026) kg and a visible diameter (top of the cloud layer) of 120,000 km. The weight W of a planetary probe on earth is 4.50 kN. Determine (a.) The weight of the probe when it is 600,000 km above the top of the clouds. (b.) The weight of the probe as it begins its penetration of the cloud layers. SOLUTION W Gm1m2 r2 4 STATICS AND MECHANICS OF MATERIALS, 2nd Edition Therefore on the surface of the earth where RILEY, STURGES AND MORRIS me 5.976 1024 kg and r G 5.976 1024 m 6370 103 8 2 6370 km 4500 Gm 3.055 10 (a) N m 2 /kg 2 W 3.055 10 8 5.67 10 26 103 60, 000 600, 000 3.055 10 8 39.8 N ..................................... Ans. (b) W 5.67 1026 2 60, 000 103 4810 N ...................................... Ans. 1-15 The first U.S. satellite, Explorer 1, had a mass of approximately 1 slug. Determine the force exerted on the satellite by the earth at the low and high points of its orbit, which were 175 mi and 2200 mi, respectively, above the surface of the earth. SOLUTION F F 3.439 10 8 Gm1m2 r2 4.095 10 23 1 5280 2 3960 175 3.439 10 8 29.5 lb ................................... Ans. F 4.095 1023 1 5280 2 3960 2200 13.31 lb .................................. Ans. 1-16 A neutron star has a mass of 2(1030) kg and a diameter of 10 km. Determine the gravitational force of attraction on a 10-kg space probe (a.) When it is 1000 km from the center of the star. (b.) At the instant of impact with the surface of the star. SOLUTION F F 6.673 10 11 Gm1m2 r2 32 2 1030 10 1000 10 6.673 10 11 1.335 109 N ............................... Ans. F 2 1030 10 32 5 10 5.34 1013 N ............................... Ans. 1-17 Determine the weight W, in U.S. customary units, of a 75-kg steel bar under standard conditions (sea level at a latitude of 45 degrees). SOLUTION W mg 75 9.81 735.75 N 0.2248 lb/N 165.4 lb ...................... Ans. 5 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 1-18 Determine the mass m, in SI units, for a 500-lb steel beam under standard conditions (sea level at a latitude of 45 degrees). SOLUTION W m W g 500 32.2 mg 227 kg ........................ Ans. 15.528 slug 14.59 kg/slug 1-19 An automobile has a 440 cubic inch engine displacement. Determine the engine displacement in liters. SOLUTION V 440 in.3 16.39 103 mm3 /in.3 10 1 cm/mm V 3 10 3 L/cm3 7.21 L ................................................................... Ans. 1-20 How many barrels of oil are contained in 100 kL of oil? One barrel (petroleum) equals 42.0 gal. SOLUTION V 100 103 L 0.2642 gal/L 1 barrel 42 gal 629 barrel .................. Ans. 1-21* Express the density, in SI units, of a specimen of material that has a specific weight of 0.025 lb/in.3 SOLUTION g g 0.025 slug 32.2 in.3 14.59 kg slug in. 16.39 103 mm3 3 1000 mm m 3 691 kg/m 3 ............................................................... Ans. 1-22 The viscosity of crude oil under conditions of standard temperature and pressure is 7.13(10-3) N-s/m2. Determine the viscosity of crude oil in U.S. Customary units. SOLUTION Ns 7.13 10 m2 3 0.2248 lb N 0.0929 m 2 ft 2 0.1489 10 3 lb s ........... Ans. ft 2 1-23 One acre equals 43,560 ft2. One gallon equals 231 in.3 Determine the number of liters of water required to cover 2000 acres to a depth of 1 foot. SOLUTION V 2000 acre ft 43,560 ft 2 acre V 12 in ft 3 gal 231 in.3 3.785 L gal 2.47 109 L .............................................................. Ans. 1-24 The stress in a steel bar is 150 MPa. Express the stress in appropriate U.S. Customary units (ksi) by using the values listed in Table 1-6 for length and force as defined values. SOLUTION stress 150 10 N/m stress 6 2 0.2248 lb/N 0.0929 m /ft 2 2 ft 12 in. 2 21.75 103 lb/in.2 21.75 ksi ......................................... Ans. 6 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 1-25 By definition, 1 hp = 33,000 ft-lb/min and 1 W = 1 N-m/s. Verify the conversion factors listed in Table 1-6 for converting power from U.S. Customary units to SI units by using the values listed for length and force as defined values. SOLUTION 1 hp 33, 000 ft lb min 4.448 N lb 0.3048 m ft min 60 s 745.7 Nm ............. Ans. s 1-26 The specific heat of air under standard atmospheric pressure, in SI units, is 1003 N-m/kg- K. Determine the specific heat of air under standard atmospheric pressure in U.S. customary units (ft-lb/slug- R). SOLUTION 1003 N m kg K 0.2248 lb N 3.281 ft m slug R 14.59 kg slug 5K 9R 6000 lb ft .......................................................... Ans. 1-27 Newton’s law of gravitation can be expressed in equation form as F SOLUTION G m1m2 r2 If F is a force, m1 and m2 are masses, and r is a distance, determine the dimensions of G. G Fr 2 m1m2 ML T 2 M M L 2 L3 .......................................... Ans. MT 2 1-28 The elongation of a bar of uniform cross section subjected to an axial force is given by the equation PL EA . What are the dimensions of E if and L are lengths, P is a force, and A is an area? SOLUTION E PL A ML T 2 LL 2 L M LT 2 F ...................................... Ans. L2 T k L g , where T is in seconds, 1-29 The period of oscillation of a simple pendulum is given by the equation L is in feet, g is the acceleration due to gravity, and k is a constant. What are the dimensions of k for dimensional homogeneity? SOLUTION k T gL T L T2 L 12 1 (dimensionless) ............................... Ans. 1-30 An important parameter in fluid flow problems involving thin films is the Weber number (We) which can be expressed in equation form as We v2L where is the density of the fluid, v is a velocity, L is a length, and is the surface tension of the fluid. If the Weber number is dimensionless, what are the dimensions of the surface tension ? SOLUTION v2L We M L3 L T 1 2 L M T2 F ................................ Ans. L 7 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 1-31 In the dimensionally homogeneous equation RILEY, STURGES AND MORRIS P A SOLUTION Mc I is a stress, A is an area, M is a moment of a force, and c is a length. Determine the dimensions of P and I. M LT 2 Therefore P L2 M LT 2 ML2 T 2 I ML T2 L L P I L2 F ............................................... Ans. L4 ...................................................... Ans. ML2 T 2 M LT 2 Pd 1 2 1-32 In the dimensionally homogeneous equation mv 2 1 2 I 2 d is a length, m is a mass, v is a linear velocity, and and I. SOLUTION is an angular velocity. Determine the dimensions of P PL Therefore M L T 2 I 1 T2 P I ML2 T 2 L ML2 T2 T2 ML T2 F ................................................ Ans. ML2 ...................................................... Ans. 1-33 In the dimensionally homogeneous equation Tr J VQ Ib is a stress, T is a torque (moment of a force), V is a force, r and b are lengths and I is a second moment of an area. Determine the dimensions of J and Q. SOLUTION M LT 2 Therefore ML2 T 2 J ML2 T 2 M LT L ML T 2 Q L4 L J L 2 L4 ..................................................... Ans. 8 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS Q M LT 2 ML T 2 L5 L3 ..................................................... Ans. 1-34 In the dimensionally homogeneous equation P Tr AJ is a stress, A is an area, T is a torque (moment of a force), and r is a length. Determine the dimensions of P and J. SOLUTION M LT 2 Therefore P L2 M LT 2 ML2 T 2 J ML T2 L 2 L P J tb L2 F ............................................... Ans. L4 ..................................................... Ans. ML2 T 2 M LT sin at is dimensionally homogeneous. If A is a length and t is time, 1-35 The equation x Ae determine the dimensions of x, a, b, and . SOLUTION x Therefore Le x Tb sin a T L ........................................................... Ans. L11 b T ...................................................................... Ans. a 1 T .................................................................... Ans. 1 (dimensionless) ........................................................ Ans. 1-36 In the dimensionally homogeneous equation dimensions of a, b, and w? SOLUTION If w x3 ax 2 bx a 2 b x , if x is a length, what are the x L , then each term has the dimension L3 . Therefore w a L3 L2 L3 L L3 ..................................................................... Ans. L ............................................................... Ans. b Using the last term as a check, L2 ............................................................... Ans. 9 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS a 2b x y in which A is a length and t is time. SOLUTION bT L2 L bt L2 L3 1-37 Determine the dimensions of a, b, c, and y in the dimensionally homogeneous equation Ae cos 1 a 2 bt c y Therefore Le y cos 1 a2 b T c L11 b L .......................................................... Ans. bT 1 1 T .................................................................... Ans. a 1 (dimensionless) ........................................................ Ans. c 1 (dimensionless) ........................................................ Ans. 1-38 Determine the dimensions of c, , k and P in the differential equation m SOLUTION d 2x dx c kx 2 dt dt P cos t in which m is a mass, x is a length, and t is time. M T Therefore 2 L c L T kL P cos T c k ML T 2 LT ML T 2 L P ML T2 M T2 M ......................................................... Ans. T F .................................................. Ans. L F ............................................................. Ans. 1 T .................................................................... Ans. 1-39 Round off the following numbers to two significant figures. Find the percent difference between each rounded-off number and the original number by using the original number as the reference. (a) 0.015362 (b) 55.33682 (c) 63,746.27 SOLUTION (a) (b) 0.015 0.015362 100 2.36 % ............................................. Ans. 0.015362 55 55.33682 100 0.609 % ............................................... Ans. 55.33682 10 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (c) RILEY, STURGES AND MORRIS 64, 000 63, 746.27 100 63, 746.27 0.398 % ......................................... Ans. 1-40 Round off the following numbers to two significant figures. Find the percent difference between each rounded-off number and the original number by using the original number as the reference. (a) 0.837482 (b) 374.9371 (c) 937,284.9 SOLUTION (a) (b) (c) 0.84 0.837482 100 0.301 % ............................................. Ans. 0.837482 370 374.9371 100 1.317 % ............................................. Ans. 374.9371 940, 000 937, 284.9 100 0.290% ......................................... Ans. 937, 284.9 1-41 Round off the following numbers to three significant figures. Find the percent difference between each rounded-off number and the original number by using the original number as the reference. (a) 0.034739 (b) 26.39473 (c) 55,129.92 SOLUTION (a) (b) (c) 0.0347 0.034739 100 0.1123 % ......................................... Ans. 0.034739 26.4 26.39473 100 0.01997 % .......................................... Ans. 26.39473 55,100 55,129.92 100 0.0543 % ......................................... Ans. 55,129.92 1-42 Round off the following numbers to three significant figures. Find the percent difference between each rounded-off number and the original number by using the original number as the reference. (a) 0.472916 (b) 826.4836 (c) 339,872.8 SOLUTION (a) (b) (c) 0.473 0.472916 100 0.01776 % ......................................... Ans. 0.472916 826 826.4836 100 0.0585 % ............................................ Ans. 826.4836 340, 000 339,872.8 100 0.0374 % ....................................... Ans. 339,872.8 1-43 Round off the following numbers to four significant figures. Find the percent difference between each rounded-off number and the original number by using the original number as the reference. (a) 0.056623 (b) 74.82917 (c) 27,382.84 SOLUTION (a) (b) 0.05662 0.056623 100 5.30 10 3 % ..................................... Ans. 0.056623 74.83 74.82917 100 1.109 10 3 % ...................................... Ans. 74.82917 11 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (c) RILEY, STURGES AND MORRIS 27,380 27,382.84 100 27,382.84 10.37 10 3 % .................................... Ans. 1-44 Round off the following numbers to four significant figures. Find the percent difference between each rounded-off number and the original number by using the original number as the reference. (a) 0.664473 (b) 349.3378 (c) 274,918.2 SOLUTION (a) (b) (c) 0.6645 0.664473 100 4.06 10 3 % ...................................... Ans. 0.664473 349.3 349.3378 100 10.82 10 3 % ...................................... Ans. 349.3378 274,900 274,918.2 100 6.62 10 3 % .................................... Ans. 274,918.2 1-45 The weight of the first Russian satellite, Sputnik I, was 184 lb on the surface of the earth. Determine the force exerted on the satellite by the earth at the low and high points of its orbit which were 149 mi and 597 mi, respectively, above the surface of the earth. SOLUTION F Therefore 184 Gm1m2 r2 Gme m 3960 5280 2 Gme m 8.044 1016 lb ft 2 F F 8.044 1016 3960 149 3960 597 5280 5280 2 170.9 lb ......................................... Ans. 138.9 lb ......................................... Ans. 8.044 1016 2 1-46 The planet Jupiter has a mass of 1.90(1027) kg and a radius of 7.14(107) m. Determine the force of attraction between the earth and Jupiter when the minimum distance between the two planets is 6(1011) m. SOLUTION F F 6.673 10 11 Gm1m2 r2 11 2 5.976 1024 1.90 10 27 6 10 2.10 1018 N .................... Ans. 1-47 Convert 640 acres (1 square mile) to hectares if 1 acre equals 4840 yd2 and 1 hectare equals 104 m2. SOLUTION 640 acres 4840 yd 2 acre 3 ft yd 2 0.0929 m 2 ft 2 hect 104 m 259 hectare ............ Ans. 12 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 1-48 Determine the dimension of c in the dimensionally homogeneous equation v SOLUTION mg 1e c ct m in which v is a velocity, m is a mass, t is time, and g is the gravitational acceleration. L T Therefore M L T2 c ML T 2 LT 1e cT M c As a check, M ......................................................... Ans. T ct m MT T M 1 (dimensionless) 1-49 Develop an expression for the change in gravitational acceleration g between the surface of the earth and a height h when h << Re. SOLUTION W Therefore Gme m r2 mg mg e mg g g ge Gme re re g Gme 4 re h re Gme h re 2 2 2 2 Gme m re2 h 2 2 Gme m re Gme re2 1 re h re2 1 2 re 2re h h 2 2re h h 2 2Gme h .......................................................... Ans. re3 2re h 13 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS Chapter 2 Use the law of sines and the law of cosines, in conjunction with sketches of the force triangles, to solve the following problems. Determine the magnitude of the resultant R and the angle between the x-axis and the line of action of the two forces shown in Fig. P2-1. SOLUTION From the law of cosines 2-1 R2 902 1202 2 90 120 cos 90 R 150.0 lb From the law of sines sin 90 sin 90 R 90sin 90 sin 1 36.87 150.0 R 150 lb 36.87 ....................................................................................................... Ans. Use the law of sines and the law of cosines, in conjunction with sketches of the force triangles, to solve the following problems. Determine the magnitude of the resultant R and the angle between the x-axis and the line of action of the two forces shown in Fig. P2-2. SOLUTION From the law of cosines 2-2 R2 R 2502 2002 2 250 200 cos 50 408.386 N From the law of sines sin 200 sin130 R 200sin130 sin 1 22.03 408.386 R 408 N 36.03 ....................................................................................................... Ans. Use the law of sines and the law of cosines, in conjunction with sketches of the force triangles, to solve the following problems. Determine the magnitude of the resultant R and the angle between the x-axis and the line of action of the two forces shown in Fig. P2-3. SOLUTION From the law of cosines 2-3 R2 R 600 2 8002 2 600 800 cos 75 866.910 lb From the law of sines sin 800 sin 75 R 800sin 75 sin 1 63.05 866.910 R 867 lb 86.95 ....................................................................................................... Ans. 14 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 2-4 RILEY, STURGES AND MORRIS Use the law of sines and the law of cosines, in conjunction with sketches of the force triangles, to solve the following problems. Determine the magnitude of the resultant R and the angle between the x-axis and the line of action of the two forces shown in Fig. P2-4. SOLUTION From the law of cosines R 2 10 2 R From the law of sines 252 2 10 25 cos120 31.225 kN sin 25 sin120 R 25sin120 sin 1 43.90 31.225 R 31.2 kN 19.90 .................................................................................................... Ans. The support ring of the traffic light shown in Fig. P2-5 is acted on by three forces – the weight of the traffic light (220 lb), a force in cable A (280 lb), and a force in cable B (FB). If the resultant of the three forces is zero, determine the magnitude and direction of FB. SOLUTION From the law of cosines 2-5 2 R12 2202 2802 2 220 280 cos 70 R12 sin 12 220 12 290.969 lb sin 70 R12 sin 220sin 70 45.28 290.969 291 lb 25.28 1 From the law of sines R12 But Therefore 2-6 R12 FB FB R 0 291 lb 25.28 .......................................................................................... Ans. Determine the resultant of the three forces shown in Fig. P2-6. Locate the resultant with respect to the x-axis shown. SOLUTION R12 12 32 4 2 tan 1 5 kN 36.870 3 4 Then, using the law of cosines R2 52 82 2 5 8 cos113.130 R 10.974 kN From the law of sines sin 5 sin113.130 R 15 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 5sin113.130 24.77 10.974 R 10.97 kN 5.23 .................................................................................................... Ans. sin 1 A stalled automobile is being pulled by the two forces shown in Fig. P2-7. If the resultant pull is to be 120 lb in the x-direction, determine the magnitude and direction of the force P. SOLUTION From the law of cosines 2-7 P2 P 80 2 1202 2 80 120 cos 20 52.516 lb From the law of sines sin 120 sin 20 P 120sin 20 sin 1 128.60 52.516 P 52.5 lb 31.40 ...................................................................................................... Ans. The eye bolt shown in Fig. P2-8 is subjected to a 2700 N force and an unknown force P. If the resultant pull is 2000 N in the x-direction, determine the magnitude and direction of the force P. SOLUTION From the law of cosines 2-8 P2 P 20002 27002 2 2000 2700 cos 60 2426.932 kN From the law of sines sin 2000 sin 60 P 2000sin 60 sin 1 45.54 2426.932 P 2430 kN 74.46 .................................................................................................. Ans. that will make the vertical Two forces act on the bracket shown in Fig. P2-9. Determine the angle component of the resultant of these two forces zero. SOLUTION From the law of cosines 2-9 1452 1752 R 2 2 175 R cos 50 R 2 224.976 R 9600 0 R 167.747 lb or R 57.229 lb From the law of sines sin 175 sin 50 145 175sin 50 sin 1 145 67.60 or 112.40 ............................................................................................. Ans. 16 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS R 167.7 lb R 57.2 lb when when 67.60 112.40 2-10 A 270-N force and a 400-N force act at point B of the truss shown in Fig. P2-10. A third force F is to be applied at point B so that the resultant of the three forces is zero. Determine the magnitude of F and its orientation with respect to the 400-N force. SOLUTION Using the law of cosines 2 R12 2702 4002 2 270 400 cos 50 R12 sin 12 270 12 306.689 N sin 50 R12 sin 270sin 50 42.41 306.689 307 N 42.41 1 From the law of sines R12 But Therefore R R3 R12 R 3 R12 0 42.41 ........................................................................................ Ans. 307 N 2-11 Three forces are applied to a bracket mounted on a post as shown in Fig. P2-11. Determine (a) The magnitude and direction (angle x) of the resultant R of the three forces. (b) The magnitudes of two other forces Fx and Fy that would have the same resultant. SOLUTION (a) For the system of forces, R R12 F1 F2 F3 R12 F3 From the parallelogram of forces constructed using F1 and F2 1202 1402 2 120 140 cos105 206.631 lb 140sin105 sin 1 40.878 12 206.631 R12 206.631 lb 4.122 For the second parallelogram of forces R 1002 206.6312 2 100 206.631 cos85.878 222.993 lb 206.632sin 85.878 sin 1 67.55 222.993 22.45 ........................................................................................................ Ans. R 223 lb (b) From the definition of sine and cosine Rx Ry 223.993 cos 22.45 223.993 sin 22.45 207.0 lb 85.5 lb ...................................................................... Ans. ........................................................................... Ans. 17 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 2-12 Four forces act on a small airplane in flight, as shown in Fig. P2-12; its weight (25 kN), the thrust provided by the engine (10 kN), the lift provided by the wings (24 kN), and the drag resulting from its motion through the air (3 kN). Determine the resultant of the four forces and its line of action with respect to the axis of the plane. SOLUTION For the system of forces, R F1 F2 F3 F4 R12 F3 F4 R123 F4 From the parallelogram of forces constructed using F1 and F2 R12 32 24 2 2 3 24 cos 90 24.1868 kN 3sin 90 sin 1 7.125 1 24.1868 R12 24.1868 kN 73.875 From the parallelogram of forces constructed using R12 and F3 R123 102 2 R12 2 10 R12 cos 82.875 25.000 kN R sin 82.875 sin 1 12 73.740 2 25.000 R123 25.000 kN 83.740 Finally, from the parallelogram of forces constructed using R123 and F4 R 252 2 R123 2 25 R123 cos 6.260 2.730 kN R sin 6.260 sin 1 123 86.902 3 2.730 3.088 .................................................................................................. Ans. R 2.730 kN 13.088 below the x -axis ......................................................................................... Ans. 2-13 Determine the resultant force R of the three forces applied to the gusset plate shown in Fig. P2-13. SOLUTION For the system of forces, R R12 F1 F2 F3 11, 0002 R12 F3 45002 2 11, 000 4500 cos135 From the parallelogram of forces constructed using F1 and F2 14,534.565 lb 12 sin 1 4500sin135 14,534.565 12.646 77.354 R12 14,534.565 lb 18 STATICS AND MECHANICS OF MATERIALS, 2nd Edition For the second parallelogram of forces RILEY, STURGES AND MORRIS R 90002 2 R12 2 9000 R12 cos137.354 22, 015.69 lb sin R 1 9000sin137.354 22, 015.69 16.078 22, 020 lb 86.57 ............................................................................Ans. 2-14 Determine the magnitude and direction of the resultant of the three forces shown in Fig. P2-14. SOLUTION For the system of forces, R R12 F1 F2 F3 R12 F3 From the parallelogram of forces constructed using F1 and F2 52 7.52 2 5 7.5 cos130 11.3780 kN 7.5sin130 sin 1 30.328 12 11.3780 R12 11.3780 kN 20.328 For the second parallelogram of forces R 102 2 R12 2 10 R12 cos 65.328 11.5961 kN 10sin 65.328 sin 1 51.595 11.5961 71.92 .................................................................................................. Ans. R 11.60 kN 2-15 Two forces A and B are applied to an eyebolt as shown in Fig. P2-15. If the magnitudes of the two forces are A = 50 lb and B = 100 lb, calculate and plot the magnitude of the resultant R as a function of the angle A (0 180 ). Also calculate and plot the angle R that the resultant makes with the force B as a function of A the angle A. When is the resultant a maximum? When is the resultant a minimum? When is the angle R a maximum? Repeat for A = 100 lb and B = 50 lb. SOLUTION From the law of cosines R sin A A2 B2 2 AB cos A From the law of sines B B sin 180 R A sin sin 1 R A A For A = 50 lb and B = 100 lb 19 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS R 502 1002 2 50 100 cos 100 1.25 cos B A A lb sin sin 1 50sin A 100 1.25 cos sin A 2 1.25 cos A 1 A For A = 100 lb and B = 50 lb R 1002 502 2 100 50 cos A A 100 1.25 cos B lb sin sin 1 100sin A 100 1.25 cos sin A 1.25 cos A 1 A 2-16 Two forces A and B are applied to a bracket as shown in Fig. P2-16. If the magnitude of the force B is 325 N, calculate and plot the magnitude of the resultant R as a function of the magnitude of the force A (0 A 900 N). SOLUTION From the law of cosines R2 R 3252 A2 2 325 A cos 75 105, 625 168.2324 A A2 N 20 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 2-17 Determine the magnitudes of the u- and v-components of the 1000-lb force shown in Fig. P2-17. SOLUTION From the law of sines Fu sin 35 Fu Fv 1000 Fv sin100 sin 45 582 lb ........................................................................ Ans. 718 lb ......................................................................... Ans. 2-18 Determine the components of the 3000-N force in the directions of members AB and BC of the truss shown in Fig. P2-18 when = 45 . SOLUTION From the law of sines FAB 3000 FBC sin15 sin 90 sin 75 FAB 776 N ...................................................................... Ans. FBC 2900 N .................................................................... Ans. 2-19 Two cables are used to support a stop light as shown in Fig. P2-19. The resultant R of the cable forces Fu and Fv has a magnitude of 300 lb and its line of action is vertical. Determine the magnitudes of the forces Fu and F v. SOLUTION From the law of sines Fu 300 Fv sin 36.87 sin 98.13 sin 45 Fu 181.8 lb ..................................................................... Ans. Fv 214.3 lb ..................................................................... Ans. 2-20 Two ropes are used to tow a boat upstream as shown in Fig. P2-20. The resultant R of the rope forces Fu and Fv has a magnitude of 1500 N and its line of action is directed along the axis of the boat. Determine the magnitudes of the forces Fu and Fv. SOLUTION From the law of sines Fu 1500 Fv sin 40 sin110 sin 30 Fu 1026 N ....................................................................... Ans. Fv 798 N ......................................................................... Ans. 2-21 Determine the u- and v-components of the 5200-lb force acting on the bracket shown in Fig. P2-21. SOLUTION From the law of sines Fu 5200 Fv sin 22.38 sin135 sin 22.62 Fu 2800 lb ...............................................................Ans. Fv 2830 lb ...............................................................Ans. 21 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 2-22 A 5000-N force acts in the vertical direction on the block shown in Fig. P2-22. Determine the components of the force perpendicular to and parallel to the line AB. SOLUTION From the law of sines Fa sin 60 Fa Fp Fp 5000 sin 90 sin 30 4330 N .................................................................................... Ans. 2500 N .................................................................................... Ans. 2-23 Two forces Fu and Fv are applied to a bracket as shown in Fig. P2-23. If the resultant R of the two forces has a magnitude of 725 lb and a direction as shown on the figure, determine the magnitudes of the forces Fu and F v. SOLUTION From the law of sines Fu 725 Fv sin 47.73 sin 49.40 sin 82.88 Fu 707 lb .................................................................Ans. Fv 948 lb ..................................................................Ans. 2-24 A 20-kN force acts on the member shown in Fig. P2-24. Determine the components of the force in the horizontal and vertical directions (FAC and FBC). SOLUTION From the law of sines FAC 20 FBC sin 53.13 sin 90 sin 36.87 FAC 16.00 N ................................................................................. Ans. FBC 12.00 N ................................................................................. Ans. 2-25 A 500-lb force acts along the line AB shown in Fig. P2-25. Determine the magnitude of the components in the direction of lines AC and AD. SOLUTION From the law of sines FAD 500 FAC sin 40.60 sin108.44 sin 30.96 FAC 271 lb ........................................................Ans. FAD 343 lb ........................................................Ans. 22 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 2-26 Two forces Fu and Fv are applied to a bracket as shown in Fig. P2-26. If the resultant R of the two forces has a magnitude of 375 N and a direction as shown on the figure, determine the magnitudes of the forces Fu and F v. SOLUTION From the law of sines Fu 375 Fv sin 70.35 sin 67.38 sin 42.27 Fu 383 N ...........................................................Ans. Fv 273 N ...........................................................Ans. 2-27 Three forces are applied to a bracket as shown in Fig. P2-27. The magnitude of the resultant R of the three forces is 50 kip. If the force F1 has a magnitude of 30 kip, determine the magnitudes of the forces F2 and F3. SOLUTION For the system of forces, R R23 F1 F2 F3 F1 R 23 From the parallelogram of forces constructed using F1 and R23 302 502 2 30 50 cos 75 51.2205 kip 2 sin R 23 12 38 50sin 75 51.2205 51.2205 kip 1 70.546 27.546 For the second parallelogram of forces 27.546 27.546 39.546 10.454 180 F3 sin F2 51.2205 sin 130 F2 sin 42.6 kip ....................................................................................................................... Ans. F3 12.13 kip ..................................................................................................................... Ans. 2-28 A homogeneous cylinder is subjected to the three forces shown in Fig. P2-28. If the resultant of the three forces is zero, determine the magnitude of N1 and the angle of the surface to which N1 is normal. SOLUTION Since the resultant of the three forces is zero, they must form a closed figure (a triangle). Using the law of sines and the law of cosines gives N1 22202 19402 2 2220 1940 cos15 609.837 N 1940sin15 sin 1 55.42 609.837 N1 610 N ........................................................................ Ans. 55.42 .......................................................................... Ans. 23 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 2-29 Four forces act on the machine component shown in Fig. P2-29. If the resultant of F1, F2, and F3 is horizontal to the left and has a magnitude of 400 lb, determine the magnitudes of the forces F2 and F3. SOLUTION For the system of forces, F1 F2 F3 R23 2 F1 R 23 400 lb 894.427 lb From the force triangle constructed using F1 and R23 4002 8002 tan 1 R 23 400 26.565 800 894.427 lb 63.435 For the second parallelogram of forces F2 894.427 F3 sin 33.435 sin 75 sin 71.565 F2 510 lb .......................................................................................................................... Ans. F3 878 lb ........................................................................................................................... Ans. 2-30 A 140-N light is supported by two cables as shown in Fig. P2-30. If the resultant of the three forces is zero, determine the force T1 and the angle . SOLUTION Since the resultant of the three forces is zero, they must form a closed figure (a triangle). Using the law of sines and the law of cosines gives T1 1402 1552 2 140 155 cos 50 125.411 N sin 40 140 40 sin 1 sin 50 T1 140sin 50 125.411 58.777 T1 125.4 N ........................................................................................................................ Ans. 18.77 ............................................................................................................................ Ans. 2-31 Two forces A and B are applied to an eye bolt using ropes as shown in Fig. P2-31. The resultant R of the two forces has a magnitude R = 4000 lb and makes an angle of 30 with the force A as shown. If both of the forces pull on the eye bolt as shown (ropes cannot push on the eye bolt), what is the range of angles ( min B ) for which this problem has a solution? Calculate and plot the required magnitudes A and B as max functions of the angle angles B ( min B max). Why is the magnitude of B a minimum when B = 90 ? SOLUTION From the law of sines A sin 4000 sin 180 180 B sin 30 B 30 B 30 24 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS A B B B 4000sin sin 180 B 30 B lb 4000sin 30 lb sin 180 B 30 or else A 0 180 or else B 0 The magnitude of B is a minimum when B is vertical because that results in the shortest distance from the end of the 4000 lb force to the horizontal. 2-32 Three forces A, B, and C are applied to an eye bolt using ropes as shown in Fig. P2-32. Force A has a magnitude A = 50 N, and the resultant of the three forces is zero. If all of the forces pull on the eye bolt as shown (ropes cannot push on the eye bolt), what is the range of angles ( min C max) for which this problem has a solution? Calculate and plot the required magnitudes B and C as functions of the angle C ( min C max). Why is the magnitude of C a minimum when C = 90 ? SOLUTION Since the resultant of the three forces is zero, they must form a closed figure (a triangle). Using the law of sines gives B sin 50 sin 180 180 B C C C C sin 40 C 40 40 C C 40 50sin C sin 180 50sin 40 sin 180 kN kN C 40 or else B 180 or else B 0 C The magnitude of C is a minimum when C is vertical because that results in the shortest distance from the end of the 50 kN force to the horizontal. 2-33 Determine the x- and y-components of the 1000-lb force shown in Fig. P2-33. SOLUTION Fx Fy 1000 cos 30 1000sin 30 866 lb ............................................................................................... Ans. 500 lb ............................................................................................... Ans. 2-34 Determine the x- and y-components of the 800-N force shown in Fig. P2-34. SOLUTION Fx Fy 800sin 25 800 cos 25 338 N ................................................................................................. Ans. 725 N ................................................................................................. Ans. 25 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 2-35 Determine the x- and y-components of each force shown in Fig. P2-35. SOLUTION For the 600 lb force RILEY, STURGES AND MORRIS Fx Fy Fx Fy 600sin 60 600 cos 60 800sin 45 800 cos 45 520 lb ................................................................................................. Ans. 300 lb ................................................................................................. Ans. 566 lb ............................................................................................ Ans. 566 lb ................................................................................................. Ans. For the 800 lb force 2-36 Determine the x- and y-components of each force shown in Fig. P2-36. SOLUTION F1x F1 y F2 x F2 y 3 5 950 4 5 950 1 1 570 N ................................................................................................ Ans. 760 N ............................................................................................... Ans. 566 N ...................................................................................... Ans. 566 N ............................................................................................ Ans. 2 800 2 800 2-37 Two forces are applied to a post as shown in Fig. P2-37. Determine (a) The x- and y-components of each force. (b) The x’- and y’-components of each force. SOLUTION (a) F1x F1 y F2 x F2 y 500 cos 70 500sin 70 750 cos 30 750sin 30 500sin 25 500 cos 25 750 cos15 750sin15 171.0 lb ............................................................................................ Ans. 470 lb ................................................................................................ Ans. 650 lb ............................................................................................... Ans. 375 lb .......................................................................................... Ans. 211.3 lb ....................................................................................... Ans. 453 lb ............................................................................................... Ans. 724 lb ............................................................................................... Ans. 194.1 lb ............................................................................................. Ans. (b) F1x F1 y F2 x F2 y 2-38 For the 900-N force shown in Fig. P2-38 (a) Determine the x, y, and z scalar components of the force. (b) Express the force in Cartesian vector form. SOLUTION (a) Fxy Fx Fy Fz 900 cos 35 Fxy cos 60 Fxy sin 60 900sin 35 737.237 N 369 N .................................................................................................. Ans. 638 N ................................................................................................... Ans. 516 N .................................................................................................. Ans. (b) F 369 i 638 j 516 k N .......................................................................................... Ans. 26 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 2-39 As an automobile rounds a curve, a force F of magnitude 600 lb is exerted on one of the tires, as shown in Fig. P2-39. Determine the x, y, and z scalar components of the force. The xy plane is parallel to the roadway. SOLUTION F Fx Fy Fz Fe 600 2 i 2 j 12 k 22 22 122 97.3 i 97.3 j 584 k lb 97.3 lb ....................................................................................................................... Ans. 97.3 lb ......................................................................................................................... Ans. 584 lb .......................................................................................................................... Ans. 2-40 A 50-kN force is applied to an eye bolt as shown in Fig. P2-40. (a) Determine the direction angles x, y, and z. (b) Determine the x, y, and z scalar components of the force. (c) Express the force in Cartesian vector form. SOLUTION (c) F F Fe 50 3i 2 j 2k 32 22 22 36.38 i 24.25 j 24.25 k kN 36.4 i 24.3 j 24.3 k kN .................................................................................. Ans. 36.4 kN ..................................................................................................................... Ans. 24.3 kN ..................................................................................................................... Ans. 24.3 kN ....................................................................................................................... Ans. 36.38 136.69 ............................................................................................ Ans. 50 24.25 cos 1 119.01 ............................................................................................ Ans. 50 24.25 cos 1 60.99 ................................................................................................ Ans. 50 cos 1 (b) Fx Fy Fz (a) x y z 2-41 Two forces are applied at a point on a body as shown in Fig. P2-41. Determine (a) The x- and y- components of each force. (b) The x’- and y’- components of each force. SOLUTION (a) F1x F1 y F2 x F2 y 800sin 40 800 cos 40 1000 cos 20 1000sin 20 800sin 70 800 cos 70 1000 cos 50 1000sin 50 514 lb ........................................................................................... Ans. 613 lb ............................................................................................... Ans. 940 lb ............................................................................................. Ans. 342 lb ............................................................................................. Ans. 752 lb .......................................................................................... Ans. 274 lb ............................................................................................... Ans. 643 lb ............................................................................................. Ans. 766 lb ............................................................................................. Ans. (b) F1x F1 y F2 x F2 y 27 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 2-42 Two forces are applied to an eye bolt as shown in Fig. P2-42. (a) Determine the x, y, and z scalar components of the 30-kN force F1. (b) Express the 30-kN force F1 in Cartesian vector form. (c) Determine the magnitude of the rectangular component of the 30-kN force F1 along the line of action of the 50-kN force F2. (d) Determine the angle between the two forces. SOLUTION (a) F1 F1x F1 y F1z F1e1 30 5i 3 j k 52 32 12 25.355 i 15.213 j 5.071k kN 25.4 kN ...................................................................................................................... Ans. 15.21 kN ................................................................................................................. Ans. 5.07 kN ...................................................................................................................... Ans. 25.4 i 15.21 j 5.07 k kN .................................................................................. Ans. 2i 3 j 2k 2 2 32 F1 e2 22 0.48507 i 0.72761 j 0.48507 k 15.213 0.72761 5.071 0.48507 (b) (c) F1 e2 F12 F12 25.355 0.48507 25.83 kN 25.8 kN ...................................................................................................................... Ans. 25.83 kN 1 (d) F1 e2 cos F1 cos 30.57 ................................................................................................. Ans. 25.83 30 2-43 A wire is stretched between two pylons, one of which is shown in Fig. P2-43. The 250-lb force in the wire is parallel to the xy-plane and makes an angle of 30 with the y-axis. Point A lies in the xz-plane and point C lies in the yz-plane. Determine (a) The magnitude of the rectangular component of the 250-lb force in the direction of member AD. (b) The magnitude of the rectangular component of the 250-lb force in the direction of member CD. (c) The angle between members AD and CD. SOLUTION (a) F e AD eCD FAD FAD Fe 250sin 30 i 250 cos 30 j 125.00 i 216.51 j lb 12 i 18 k 122 182 9 j 18 k 92 182 F e AD F eCD cos 1 0.55470 i 0.83205 k 0.44721 j 0.89443 k 216.51 0 0 0 0.83205 0.89443 125 0.55470 125 0 69.3 lb ....................................................................................................................... Ans. 216.51 0.44721 96.8 lb ....................................................................................................................... Ans. e AD eCD 41.91 .......................................................................................... Ans. (b) FCD FCD (c) 28 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 2-44 The hot-air balloon shown in Fig. P2-44 is tethered with three mooring cables. The force TA in cable AD has a magnitude of 1860 N. (a) Express TA in Cartesian vector form. (b) Determine the magnitude of the rectangular component of the force TA along BD. (c) Determine the angle between cables AD and BD. SOLUTION (a) TA e BD TA / BD TA / BD TAe A 1860 (b) 603 i 905 j 1509 k N .......................... Ans. 202 302 502 16 i 25 j 50 k 0.27517 i 0.42995 j 0.85990 k 162 252 502 TA e BD 603 0.27517 905 0.42995 1509 0.85990 1074 N .................................................................................................................... Ans. 1 20 i 30 j 50 k (c) cos e AD e BD cos 1 TA e BD TA cos 1 1074 1860 54.72 .......................... Ans. 2-45 A system of three cables supports the cylinder shown in Fig. P2-45. The magnitude of the force T1 in cable AO is 2000 lb. Determine (a) The magnitude of the rectangular component of the force T1 along the line OB. (b) The angle between the force T1 and the line OB. SOLUTION (a) T1 eOB TOB T1e1 1024.30 i 768.22 j 1536.44 k lb 42 32 62 4i 6 j 4k 0.48507 i 0.72761 j 0.48507 k 42 62 42 807.39 lb 807 lb ................................................................................. Ans. 2000 1 cos 66.19 .......................................................................................... Ans. 807.39 lb 1 2000 4i 3 j 6k T1 eOB (b) T1 eOB cos 807.39 2000 2-46 A 2000-N force F acts on a machine component as shown in Fig. P2-46. (a) Determine the x, y, and z scalar components of the force. (b) Express the force in Cartesian vector form. (c) Determine the angle between the force and the line AB. SOLUTION (a) F Fx Fy Fz Fe 2000 8 i 14 j 10 k 82 142 102 843.27 i 1475.73 j 1054.09 k N 843 N ........................................................................................................................... Ans. 1476 N ........................................................................................................................ Ans. 1054 N ...................................................................................................................... Ans. 843 i 1476 j 1054 k N ...................................................................................... Ans. (b) F 29 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (c) RILEY, STURGES AND MORRIS e AB FAB 150 i 130 j 1502 1302 F e AB cos 1 0.75569 i 0.65493 j 2000 1 cos 1603.75 N 1603.75 2000 36.69 ............................................................................................. Ans. 2-47 Two flower pots are supported with wires as shown in Fig. P2-47. If = 6 , the tension force T1 has a magnitude of 13 lb, and the tension force T2 has a magnitude of 9 lb, determine the magnitude and orientation of the resultant of the forces T1 and T2. SOLUTION T1 T2 R 13cos 45 i 13sin 45 j 9 cos 6 i 9sin 6 j T1 T2 9.1924 i 9.1924 j lb 8.9507 i 0.94076 j lb 9.1924 0.94076 j 9.1924 8.9507 i 0.2417 i 10.1332 j lb R 10.14 lb 88.63 ..................................................................................................... Ans. 2-48 Two forces are applied to an eye bolt as shown in Fig. P2-48. Determine the magnitude of the resultant R of the two forces and the angle x between the line of action of the resultant and the x-axis. SOLUTION F1 F2 R R 500 cos 60 i 500sin 60 j 375cos 30 i 375sin 30 j F1 F2 846 N 250.00 i 433.01 j N 324.76 i 187.50 j N 433.01 187.50 j 250 324.76 i 574.76 i 620.51 j N 47.19 ........................................................................................................ Ans. 2-49 An automobile stuck in a muddy field is being moved by using a cable fastened to a tree as shown in Fig. P249. The forces T1 and T2 in the two segments of the cable each have a magnitude of 650 lb. If the resultant of the three forces T1, T2, and P is to be zero, determine the magnitude of P. SOLUTION T1 T2 R P 650 cos 5 i 650sin 5 j 650 cos 5 i 650sin 5 j T1 T2 P 113.3 j lb 113.30 j P 647.53 i 56.65 j lb 647.53 i 56.65 j lb 0 P 113.3 lb ......................................................................................................................... Ans. 2-50 Three forces act on the structural member shown in Fig. P2-50. Determine the resultant of the forces and express the result in Cartesian vector form. SOLUTION Rx Ry R 2500 cos 30 900 1265 N 3250 N 2000 2500sin 30 1265 i 3250 j N .................................................................................................... Ans. 30 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 2-51 Express the resultant of the two forces shown in Fig. P2-51 in Cartesian vector form, and determine the angles x, y, and z between the line of action of the resultant and the positive coordinate axes. SOLUTION F1x F1 y F1z F2 xy F2 x F2 y F2 z R R R x 100 cos 44 0 lb 100sin 44 200 cos 53 F2 xy cos 70 F2 xy sin 70 200sin 53 F1 F2 71.934 lb 69.466 lb 120.363 lb 41.167 lb 113.104 lb 159.727 lb 113.101 i 113.104 j 229.193 k lb 113.1i 113.1 j 229 k lb .................................................................................... Ans. 279.49 lb 113.101 cos 1 66.13 ............................................................................................ Ans. 279.49 113.104 cos 1 66.13 ............................................................................................ Ans. 279.49 229.193 cos 1 34.91 ............................................................................................ Ans. 279.49 y z 2-52 Determine the magnitude R of the resultant of the two forces shown in Fig. P2-52. Also determine the angles x, y, and z between the line of action of the resultant and the positive coordinate axes. SOLUTION F1x F1 y F1z F2 x F2 y F2 z R R x 10 cos 67 0 kN 10sin 67 0 kN 20 cos 60 20sin 60 F1 F2 3.9073 kN 9.2051 kN 10.0000 kN 17.3205 kN 3.907 i 10.000 j 26.526 k kN y z 28.616 28.6 kN ....................................................................................................... Ans. 3.907 cos 1 82.15 .............................................................................................. Ans. 28.616 10 cos 1 69.55 ............................................................................................. Ans. 28.616 26.526 cos 1 22.03 ............................................................................................. Ans. 28.616 31 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 2-53 Four forces are applied to the block shown in Fig. P2-53. Determine the magnitude of the resultant and the angle between the resultant and the x-axis. SOLUTION F1x F2 x F3 x F4 x Rx 5 700 29 3 300 34 2 600 29 649.93 lb 154.35 lb 222.83 lb F1 y F2 y F3 y F4 y Ry 2 700 29 5 300 34 259.97 lb 257.25 lb 5 600 557.09 lb 29 1 900 5 1476.80 lb 402.49 lb 2 900 804.98 lb 5 223.54 lb R 1494 lb 81.39 ...................................................................................................... Ans. 2-54 Determine the magnitude and orientation of the resultant of the two forces applied to the eye bolt shown in Fig. P2-54. SOLUTION F1x F1z F2 y R 120 1202 2402 0N 90 90 2 100 44.7214 N F1 y F2 x 240 1202 0N 180 902 1802 2402 100 89.4427 N 180 2 75 33.5410 N F2 z 75 67.0820 N F1 F2 44.721i 122.984 j 67.082k N R 147.054 147.1 N ..................................................................................................... Ans. 44.721 cos 1 72.30 ............................................................................................ Ans. x 147.054 122.984 cos 1 33.25 ............................................................................................ Ans. y 147.054 67.082 cos 1 62.86 ............................................................................................ Ans. z 147.054 2-55 The three forces acting on the machine component shown in Fig. P2-55 have magnitudes F1 = 500 lb, F2 = 300 lb, and F3 = 200 lb. Determine the resultant R of the three forces and express the resultant in Cartesian vector form. Also, determine the angles x, y, and z between the line of action of the resultant and the positive coordinate axes. SOLUTION F1 F2 F3 F1e1 F2 e2 500 300 2 i 35 j 40 k 22 352 402 9 i 22 j 20 k 92 22 2 20 2 18.80 i 329.02 j 376.02 k lb 86.92 i 212.46 j 193.15 k lb 200 j lb 32 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS R R R x F1 F2 F3 105.72 i 83.44 j 569.17 k lb 105.7 i 83.4 j 569 k lb .................................................................................. Ans. 584.89 lb 105.72 cos 1 100.41 ......................................................................................... Ans. 584.89 83.44 cos 1 98.20 ............................................................................................. Ans. 584.89 569.17 cos 1 166.69 ......................................................................................... Ans. 584.89 y z 2-56 Determine the magnitude R of the resultant of the three forces shown in Fig. P 2-56. Also determine the angles x, y, and z between the line of action of the resultant and the positive coordinate axes. SOLUTION F1x F1 y F2 x F2 y F3 x F3 y R R x 10 cos 26 cos 42 10 cos 26 sin 42 16 cos 40 sin 35 16 cos 40 cos 35 24 cos 50 cos 60 24 cos 50 sin 60 F1 F2 F3 6.679 kN 6.014 kN 7.030 kN 10.040 kN 7.713 kN 13.360 kN F1z 10sin 26 4.384 kN F2 z 16sin 40 10.285 kN F3 z 24sin 50 18.385 kN 7.362 i 2.694 j 33.054 k kN y z 33.971 34.0 kN ....................................................................................................... Ans. 7.362 cos 1 77.48 .............................................................................................. Ans. 33.971 2.694 cos 1 94.55 ............................................................................................. Ans. 33.971 33.054 cos 1 13.34 .............................................................................................. Ans. 33.971 2-57 Three forces are applied at the corner of the box shown in Fig. P2-57. Determine the magnitude R of the resultant and the angles x, y, and z between the line of action of of the resultant and the positive coordinate axes. SOLUTION F1 F2 F3 F1e1 10 F2 e2 F3e3 8 i 15 j 20 k 82 152 20 2 16 i 15 j 20 k 16 2 152 20 2 16 i 15 j 5 k 162 152 52 3.0478 i 5.7145 j 7.6194 k kip 10.7811 i 10.1073 j 13.4763 k kip 17.7822 i 16.6708 j 5.5569 k kip 20 25 33 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS R F1 F2 F3 31.611 i 32.493 j 26.653 k kip R 52.587 x 52.6 kip ....................................................................................................... Ans. y z 31.611 53.05 .............................................................................................. Ans. 52.587 32.493 cos 1 128.16 ......................................................................................... Ans. 52.587 26.653 cos 1 59.55 .............................................................................................. Ans. 52.587 cos 1 2-58 The pin A shown in Fig. P2-58 supports a load F of magnitude 1250 N, and is held in place by a wire AD and compression members AB and AC. If the magnitudes of the forces CB and CC in the compression members are 995 N and 700 N, respectively, and the resultant of the four forces is zero, determine the magnitude of the force TD. SOLUTION CB CC TD F Rx Ry Rz TD 995 700 TD 2i 6 j 2k 22 62 22 3i 6 j 2k 300.00 i 900.01 j 300.00 k N 300.00 i 600.00 j 200.00 k N 32 62 22 6 j 3k 0.89443TD j 0.44721TD k N 62 32 1250 k N 0 1500.01 0.89443TD 750.00 0.44721TD 0 0 1677 N ........................................................................................................................ Ans. 2-59 Three forces are applied to an eye bolt as shown in Fig. P2-59. Determine the magnitude R of the resultant of the forces and the angle x between the line of action of the resultant and the x-axis. SOLUTION F1x F2 x F3 x Rx R 2 5000 5 4472.136 lb F1 y F2 y F3 y Ry 1 5000 5 2236.068 lb 1 2000 894.427 lb 5 2 1000 894.427 lb 5 2683.28 lb 5220 lb 2 2000 1788.854 lb 5 1 1000 5 4472.14 lb 447.214 lb 59.04 ...................................................................................................... Ans. 34 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 2-60 Two forces, F1 and F2, are applied to an eye bolt as shown in Fig. P2-60. Determine (a) The magnitude and direction (angle x) of the resultant R of the two forces. (b) The magnitudes of two other forces Fu and Fv (along the axes u and v) that would have the same resultant. SOLUTION (a) F1x F2 x Rx R 80 cos15 60 cos 26 23.3465 N 52.5 N 63.5886 70 45 77.2741 N 53.9276 N F1 y F2 y Ry 80sin15 60sin 26 47.0078 N 20.7055 N 26.3023 N (b) 63.59 ....................................................................................................... Ans. 45 18.5886 115 180 Fu sin Fu Fv 52.4861 sin 46.4114 Fv sin 41.9 N ......................................................................................................................... Ans. 18.46 N ........................................................................................................................ Ans. 2-61 Three forces are applied at a point on a body as shown in Fig. P2-61. Determine the resultant R of the three forces and the angles x, y, and z between the line of action of the resultant and the positive x-, y-, and zcoordinate axes. SOLUTION F1 F2 F3 R R 500 800 700 2 j 2k 22 22 4i 4 j 42 42 2i 2k 22 22 353.553 j 353.553k lb 565.685i 565.685 j lb 494.975i 494.975k lb 1060.660i 919.238 j 848.528k lb F1 F2 F3 1061i 919 j 849k lb ......................................................................................... Ans. R 1640.121 lb 1060.660 cos 1 49.71 .......................................................................................... Ans. x 1640.121 919.238 cos 1 55.91 .......................................................................................... Ans. y 1640.121 848.528 cos 1 58.84 .......................................................................................... Ans. z 1640.121 35 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 2-62 Three forces are applied at a point on a body as shown in Fig. P2-62. Determine the resultant R of the three forces and the angles x, y, and z between the line of action of the resultant and the positive x-, y-, and zcoordinate axes. SOLUTION F1 F2 F3 R R R x 30 20 25 3 i 2 j 3k 32 22 32 3 i 4 j 1.5 k 32 42 1.52 1.5 i 4 j 1.52 4 2 19.1881 i 12.7920 j 19.1881k kN 11.4939 i 15.3252 j 5.7470 k kN 8.7781 i 23.4082 j kN F1 F2 F3 39.4601 i 51.5254 j 24.9351k kN 39.5 i 51.5 j 24.9 k kN ................................................................................. Ans. 69.5250 kN 39.4601 cos 1 124.58 ....................................................................................... Ans. 69.5250 51.5254 cos 1 42.17 ........................................................................................... Ans. 69.5250 24.9351 cos 1 68.98 ............................................................................................ Ans. 69.5250 y z 2-63 Three forces are applied to a stalled automobile as shown in Fig. P2-63. Determine the magnitude of the force F3 and the magnitude of the resultant R if the line of action of the resultant is along the x-axis. SOLUTION F1x F2 x F3 x Rx Ry F3 50 cos 48 30 cos 23 F3 cos 37 33.4565 lb 27.6152 lb 0.79864 F3 lb R 0 lb F1 y F2 y F3 y 50sin 48 30sin 23 F3 sin 37 37.1572 lb 11.7219 lb 0.60182 F3 lb 33.4565 27.6152 0.79864 F3 lb 37.1572 11.7219 0.60182 F3 81.2 lb .......................................................................................................................... Ans. R 125.9 lb ......................................................................................................................... Ans. 2-64 Three cables are used to drag a heavy crate along a horizontal surface as shown in Fig. P2-64. The resultant R of the forces has a magnitude of 2800 N and its line of action is directed along the x-axis. Determine the magnitudes of the forces F1 and F3. SOLUTION F1x F2 x F3 x Rx F1 cos 30 1600 cos10 F3 cos 45 0.86603F1 N 1575.6924 N 0.70711F3 N F1 y F2 y F3 y F1 sin 30 F3 sin 45 0.50000 F1 N 277.8371 N 0.70711F3 N 1600sin10 0.86603F1 1575.6924 0.70711F3 2800 N 36 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS Ry F3 0.50000 F1 277.8371 0.70711F3 0N F1 1100 N ......................................................................................................................... Ans. 385 N ........................................................................................................................... Ans. 2-65 Two forces are applied at a point in a body as shown in Fig. P2-65. Determine (a) The magnitude and direction (angles x, y, and z) of the resultant R of the two forces. (b) The magnitude of the rectangular component of the force F1 along the line of action of the force F2. (c) The angle between forces F1 and F2. SOLUTION (a) F1 F2 x F2 y F2 z R R x F1e1 150 29.4174 i 117.6697 j 88.2523 k lb 1.52 62 4.52 1.5 120 cos 60 36.0000 lb 2.5 2 120 cos 60 48.0000 lb 2.5 120sin 60 103.9231 lb 1.5 i 6 j 4.5 k 65.4174 i 69.6697 j 192.1754 k lb 214.6269 lb 215 lb ................................................................................................ Ans. 65.4174 cos 1 72.25 ......................................................................................... Ans. 214.6269 69.6697 cos 1 71.06 ......................................................................................... Ans. 214.6269 192.1754 cos 1 26.44 ......................................................................................... Ans. 214.6269 F1 F2 4582.333 38.2 lb ................................................................................ Ans. F2 120 y z (b) F1/ 2 (c) cos 1 F1 F2 F2 cos 1 4582.333 150 120 75.25 ............................................................. Ans. 2-66 Three forces are applied with cables to the anchor block shown in Fig. P2-66. Determine (a) The magnitude and direction (angles x, y, and z) of the resultant R of the three forces. (b) The magnitude of the rectangular component of the force F1 along the line of action of the force F2. (c) The angle between forces F1 and F3. SOLUTION (a) F1 136 F2 F3 R 250 325 2.4 i 2.7 j 3.6 k 2.42 2.7 2 3.62 0.6 i 1.8 j 2.7 k 0.62 1.82 2.7 2 3.6 i 1.2 j 0.9 k 3.62 1.22 0.92 64.000 i 72.000 j 96.000 k N 45.455 i 136.364 j 204.545 k N 300.000 i 100.000 j 75.000 k N 409.455 i 164.364 j 375.545 k N 37 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS R (b) 579.399 N 579 N ................................................................................................... Ans. 409.455 cos 1 45.03 ........................................................................................... Ans. x 579.399 164.364 cos 1 106.48 ....................................................................................... Ans. y 579.399 375.545 cos 1 49.60 ............................................................................................ Ans. z 579.399 F1 F2 12, 727.232 F1/ 2 50.9 N ............................................................................ Ans. F2 250 (c) cos 1 F1 F3 F1 F3 cos 1 19, 200 136 325 64.25 ............................................................ Ans. 38 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 3-4 RILEY, STURGES AND MORRIS A 750-kg body is supported by the flexible cable system shown in Fig. P3-4. Determine the tensions in cables AC, BC, and CD. SOLUTION From a free-body diagram for the body Fy 0: 7357.50 0: 0: TCD 750 9.81 0 TCD Fx Fy 7360 N .......................... Ans. TBC cos 60 TAC cos 30 0 From a free-body diagram of the ring, the equations of equilibrium TBC sin 60 1.73205TAC TAC sin 30 7357.50 0 are solved to get TBC TAC TBC 3-5 7360 N ..................................................................................................................................Ans. 12, 740 N ..............................................................................................................................Ans. An 800-lb homogeneous cylinder is supported by two rollers as shown in Fig. P3-5. Determine the forces exerted by the rollers on the cylinder. All surfaces are smooth (frictionless). SOLUTION The equations of equilibrium Fx Fy 0: 0: N A sin 30 N A cos 30 N B sin 30 0 N B cos 30 800 0 are solved to get NA NA NB 3-6 NB 462 lb .............................................................................. Ans. 462 lb .............................................................................. Ans. A worker is using a hoist and cable to lift a 175-kg engine from a car as shown in Fig. P3-6. Determine the forces in the three cables attached to the ring. SOLUTION From a free-body diagram of the ring, the equations of equilibrium Fx Fy T1 T2 3-7 0: 0: T1 cos10 T2 cos10 T2 sin10 T1 sin10 175 9.81 0 0 are solved to get 5.67128T2 317 N ....................................................................................... Ans. T1 1799 N ...................................................................................... Ans. The lightweight collar A shown in Fig. P3-7 is free to slide on the smooth rod BC. Determine the forces exerted on the collar by the cable and by the rod when the 900 lb downward force F is applied to the collar. SOLUTION From a free-body diagram of the collar, the equations of equilibrium 40 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS Fx Fy 0: 0: T cos 20 T sin 20 N sin 30 900 0 0 N cos 30 are solved to get N 1.87939T T 700 lb ............................................................................................... Ans. N 1316 lb ............................................................................................ Ans. An automobile stuck in a muddy field is being moved by using a cable fastened to a tree as shown in Fig. P38. When a 500-N sideways force P is applied to the cable, the cable is pulled 5 to the side as shown. For this position, determine the x- and y-components of the cable force being applied to the automobile. SOLUTION From a free-body diagram of the ring, the equations of equilibrium 3-8 Fx Fy 0: 0: T2 cos 5 500 T1 sin 5 2868.43 N T1 cos 5 T2 sin 5 0 0 are solved to get T1 T2 The x- and y-components of the cable force being applied to the automobile are then T1x T1 y 3-9 T1 cos 5 T1 sin 5 2860 N ..................................................Ans. 250 N .....................................................Ans. Two flower pots are supported with cables as shown in Fig. P3-9. If pot A weighs 10 lb and pot B weighs 8 lb, determine the tension in each of the cables and the slope of cable BC. SOLUTION From a free-body diagram of the upper ring Fx Fy Fx Fy 0: 0: 0: 0: TCD cos 45 TCD sin 45 TBC cos TAB sin 45 TBC cos 0 80 0 (a) (b) TBC sin From a free-body diagram for the lower ring TAB cos 45 TBC sin 0 (c) (d) 10 0 Adding equations (a) and (c) gives TCD cos 45 TCD TAB TAB cos 45 Then adding Eqs. (b) and (d) gives TCD sin 45 TCD TAB TAB sin 45 18 lb 12.7279 12.73 lb ............... Ans. TBC sin TBC cos 1.0000 9.0000 (e) (f) Now Eqs. (b) and (a) can be written Dividing Eq. (e) by Eq. (f) gives 41 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS tan 19 6.340 .................................................... TBC 9.06 lb ........................................................Ans. 3-10 Three smooth homogeneous cylinders A, B, and C are stacked in a box as shown in Fig. P3-10. Each cylinder has a diameter of 250 mm and a mass of 245 kg. Determine (a) The force exerted by cylinder B on cylinder A. (b) The forces exerted on cylinder B by the vertical and horizontal surfaces at D and E. SOLUTION W (a) 245 9.81 Fx Fy 0: 0: N AC Fx Fy 0: 0: NE 2403.45 N N AB cos 40 N AB sin 40 N AC cos 40 0 From a free-body diagram of cylinder A, the equations of equilibrium N AC sin 40 2403.45 0 are solved to get N AB (b) 1869.55 1870 N ............. Ans. ND N AB cos 40 0 From a free-body diagram of cylinder B, the equations of equilibrium N AB sin 40 2403.45 0 are solved to get ND NE 1432 N .....................................................................Ans. 3610 N .....................................................................Ans. 3-11 Three smooth homogeneous cylinders A, B, and C are stacked in a V-shaped trough as shown in Fig. P3-11. Each cylinder weighs 100 lb and has a diameter of 5 in. Determine the minimum angle for equilibrium. SOLUTION From a free-body diagram of cylinder A, the equations of equilibrium Fx Fy 0: 0: N AB sin 30 N AB cos 30 N AC 57.735 lb N AC sin 30 0 N AC cos 30 100 0 are solved to get N AB From a free-body diagram of cylinder B, the equations of equilibrium are Fx Fy 0: 0: FB sin FB cos N BC 57.735sin 30 100 57.735cos 30 0 0 But the contact force between cylinders B and C cannot be negative. Therefore, the minimum angle corresponds to N BC 0 and FB sin FB cos tan 28.868 lb 150.000 lb FB sin FB cos 28.868 150.000 0.19245 10.89 ........................................................................................................................................Ans. 42 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 3-12 A 250-kg body is supported by the flexible cable system shown in Fig. P3-12. Determine the tensions in cables A, B, C, and D. SOLUTION From a free-body diagram of the lower ring, the equations of equilibrium Fx Fy TC TD Fx Fy 0: 0: TC cos 60 TC sin 60 TD 0 0 250 9.81 are solved to get 2831.90 2830 N ..................................................Ans. 1415.95 1416 N ...................................................Ans. 0: 0: TA cos 40 TA sin 40 TB cos 30 TB sin 30 2831.90 cos 60 2831.90sin 60 0 0 From a free-body diagram of the upper ring, the equations of equilibrium are solved to get TA 1507 N .......................................................................Ans. TB 2970 N .......................................................................Ans. 3-13 A 500-lb lawn roller is to be pulled over a curb as shown in Fig. P3-13. Determine the minimum pulling force that must be applied by the man to just start the 3-ft diameter roller over the curb. Also determine the angle which gives the minimum pulling force. Assume that the pulling force is along the handle, which makes an angle of with the horizontal. SOLUTION From a free-body diagram of the roller, the equations of equilibrium are Fx Fy where 0: 0: NF T cos N C sin N C cos T sin 0 (a) (b) 500 0 sin 1 12 18 41.810 The roller just begins to roll over the curb when Therefore, Eqs. (a) and (b) become NF 0. (c) (d) T sin T cos tan 500 0.66667 N C 0.74536 NC Dividing Eq. (c) by Eq. (d) gives 500 0.66667 N C .......................................(e) 0.74536 N C NC 386.968 354.101 333.332 321.553 288.825 T 376.519 373.257 372.679 372.865 375.327 Solving Eqs. (d) and (e) by trial and error for the angle that makes the tension a minimum 40 45 48.19 50 55 Tmin 373 lb @ 48.19 ........................................................................................................... Ans. 43 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS on a rope as 3-14 In order to hold a 130-kg crate in a stationary position, a worker exerts a force P at an angle shown in Fig. P3-14. Determine the force exerted by the worker when = 20 . SOLUTION From a free-body diagram of the ring, the equations of equilibrium Fx Fy 0: 0: TAB cos8 6.75197 P TAB sin 8 P sin 20 P cos 20 130 9.81 0 0 are solved to get TAB P 201.0 N .............................................................................. Ans. TAB 1357 N 3-15 A farmer is extracting a post from the ground using the structure shown in Fig. P3-15. What force must the farmer apply to the cable system if the force required to remove the post is 2000 lb? SOLUTION From a free-body diagram of the right ring, the equations of equilibrium Fx Fy 0: 0: TCE sin15 TCE cos15 TBC 0 2000 0 are solved to get TBC TCE 535.898 lb 2070.552 lb From a free-body diagram of the left ring, the equations of equilibrium Fx Fy 0: 0: 535.898 TAB cos15 TAB sin15 P 0 0 are solved to get TAB 554.803 lb P 143.6 lb ....................................................................................................................................Ans. 3-16 A continuous cable is used to support two blocks as shown in Fig. P3-16. Block A is supported by a small wheel that is free to roll on the cable. Determine the displacement y of block A for equilibrium if the masses of blocks A and B are 22 kg and 34 kg, respectively. SOLUTION From a free-body diagram of the hanging weight, the equations of equilibrium Fy T 0: T 34 9.81 0 are solved to get 333.54 N Since the pulleys are free to rotate, the tension is the same in every part of the rope. From a free-body diagram of the pulley supporting the 22 kg weight, the equations of equilibrium Fx 0: T cos 2 T cos 1 0 44 STATICS AND MECHANICS OF MATERIALS, 2nd Edition Adding Eqs. (a) and (c) together gives RILEY, STURGES AND MORRIS N B sin 60 N B cos 60 NA NB T sin T cos N A sin 30 N A cos 30 0N 1000 N (e) (f) while adding Eqs. (b) and (d) together gives Solving Eqs. (e) and (f) gives 866.03 N 866 N ........................... Ans. 500.00 N 500 N ........................... Ans. 50 N 433.013 N (g) (h) Then Eqs. (b) and (a) can be written Dividing Eq. (g) by Eq. (h) gives tan T 50 433.013 6.587 .......................................................................................................................................Ans. 436 N .......................................................................................................................................Ans. 3-19 Concrete is to be moved from a mixer to the second floor of a building under construction using a container as shown in Fig. P3-19. The container and its contents weigh 3000 lb, and it is supported by three cables equally spaced around the top of the 4-ft diameter container. Determine the force in each cable. SOLUTION The coordinates of points A, B, C, and D are: A: B: x 2 cos 30 1.73205 ft 2sin 30 1.0000 ft y x 0 ft y 2.0000 ft x y 2 cos 30 2sin 30 1.73205 ft 1.0000 ft C: D: where x z y h 0 ft 4.5sin cos 1 2 4.5 63.612 Therefore h TA TC TB Fx 0: TA 4.0311 ft 1.73205 i j 4.0311k 1.732052 12 2 j 4.0311k 2 2 4.03112 4.03112 0.38490TA i 0.22222TA j 0.89581TA k 0.38490TC i 0.22222TC j 0.89581TC k TB 0.44445TB j 0.89581TB k 0.38490TA 0.38490TC 0 Then, the equations of equilibrium 46 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS Fy Fz 0: 0: TA TB 0.22222TA 0.44445TB 0.89581TA 0.89581TB TC 0.22222TC 0 0.89581TC 3000 0 are solved simultaneously to get 1116 lb .................................................................................................................Ans. 3-20 A mass m is to be supported by two cables (A and B) as shown Fig. P3-20. If the maximum force that the cables can withstand is 15 kN, determine the maximum mass m that can be supported. SOLUTION From a free-body diagram of the ring, the equations of equilibrium are Fx Fy If 0: 0: TA max TB mg TB cos 25 TA sin 40 15 kN , then 12.6785 kN TB TA cos 40 mg 0 0 TB sin 25 TA which is okay since 12.6785 kN 15 kN . Then TB sin 25 15 kN TA sin 40 m 1529 kg ....................................................................................................................................Ans. 3-21 The hot-air balloon shown in Fig. P3-21 is tethered with three mooring cables. If the net lift of the balloon is 900 lb, determine the force exerted on the balloon by each of the three cables. SOLUTION TA TA 20 i 30 j 50 k 202 302 502 0.32444TA i 0.48666TA j 0.81111TA k 16 i 25 j 50 k TB TB 162 252 502 0.27517TB i 0.42995TB j 0.85990TB k 25 i 15 j 50 k 252 152 502 0.43193TC i 0.25916TC j 0.86387TC k 0.32444TA 0.27517TB 0.43193TC 0.48666TA 0.42995TB 0.25916TC 0.81111TA 0.85990TB 0.86387TC TA TB TC 0 0 900 0 TC TC Then, the equations of equilibrium Fx Fy Fz 0: 0: 0: are solved simultaneously to get 418 lb ...........................................................................................................................Ans. 205 lb ...........................................................................................................................Ans. 445 lb ...........................................................................................................................Ans. 47 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 3-22 A 100-kg traffic light is supported by a system of cables as shown in Fig. P3-22. Determine the tensions in each of the three cables. SOLUTION TA TB TC Fx Fy Fz 0: 0: 0: TA TB 4 i 8 j 5k 42 82 52 6 i 8 j 5k 0.39036TA i 0.78072TA j 0.48795TA k 0.53666TB i 0.71554TB j 0.44721TB k 62 82 52 8 j 5k 0.84800TC j 0.53000TC k TC 82 52 0.39036TA 0.53666TB 0.78072TA 0.71554TB 0.48795TA 0.44721TB TA TB TC 0.84800TC 0.53000TC 100 9.81 0 0 0 Then, the equations of equilibrium are solved simultaneously to get 603 N ...........................................................................................................................Ans. 439 N ...........................................................................................................................Ans. 925 N ...........................................................................................................................Ans. 3-23 A 250-lb force is applied to the joint at the top of the structure shown in Fig. P3-23. The joint is held in position by the slender members AD, BD, and CD, which can only exert forces that act along the members. If the 250-lb force is in the xy-plane, determine the forces in members AD, BD, and CD. SOLUTION F TAD TBD TCD Fx Fy Fz 0: 0: 0: 250sin 30 i 250 cos 30 j TAD 12 i 18 k 122 182 9 j 18 k 92 182 125 i 216.506 j lb 0.55470TAD i 0.83205TAD k TBD k TCD 0.44721TCD j 0.89443TCD k 0.55470TAD 125 0 0.44721TCD 216.506 0 0 0.83205TAD TBD 0.89443TCD TAD TBD TCD Then, the equations of equilibrium are solved simultaneously to get 225 lb .......................................................................................................................Ans. 621 lb .........................................................................................................................Ans. 484 lb .......................................................................................................................Ans. 48 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 3-24 Three cables are used to support a 250-kg homogeneous plate as shown in Fig. P3-24. Determine the force in each of the three cables. SOLUTION TA TB TC Fx Fy Fz 0: 0: 0: TA TB TC 0.5 i 0.6 j 1.5 k 0.52 0.6 2 1.52 0.5 i 0.6 j 1.5 k 0.29566TA i 0.35479TA j 0.88697TA k 0.29566TB i 0.35479TB j 0.88697TB k 0.52 0.6 2 1.52 0.5 i 1.5 k 0.31623TC i 0.94868TC k 0.52 1.52 0.29566TA 0.29566TB 0.31623TC 0.35479TA 0.35479TB 0 0 0 Then, the equations of equilibrium 0.88697TA 0.88697TB TA TB TC 0.94868TC 250 9.81 are solved simultaneously to get 691 N ............................................................................................................................Ans. 691 N ............................................................................................................................Ans. 1293 N .........................................................................................................................Ans. 3-25 A particle is in equilibrium under the action of four forces as shown on the free-body diagram of Fig. P3-25. Determine the magnitude and the coordinate direction angles of the unknown force F4. SOLUTION F1 F2 F3 Fx Fy Fz 0: 0: 0: 4 5 80 i 150 k lb 3 5 80 j 64 i 48 j lb 173.205 j 100.000 k lb 200 cos 30 j 200sin 30 k 64 F4 x 0 0 0 F4 y Then, the equations of equilibrium 48 173.205 F4 y 150 100 F4 z F4 x F4 x are solved simultaneously to get 64.000 lb 125.205 lb F4 z 50.000 lb 149.239 lb 149.2 lb ...............................................................................................Ans. 64 115.39 ..........................................................................................Ans. 149.239 125.205 cos 1 147.03 .......................................................................................Ans. 149.239 50 cos 1 70.43 ............................................................................................Ans. 149.239 cos 1 y z 49 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 3-26 A pair of steel pipes is stacked in a box as shown in Fig. P3-26. The masses and diameters of the smooth pipes are mA = 5 kg, mB = 20 kg, dA = 100 mm, and dB = 200 mm. Plot the two forces exerted on pipe A (by pipe B and by the side wall) as a function of the distance b between the walls of the box (200 mm b 300 mm). Determine the range of b for which (a) The force at the side wall is less than WA, the weight of pipe A. (b) Neither of the two forces exceed 2WA. (c) Neither of the two forces exceed 4WA. SOLUTION From a free-body diagram of pipe A, the equations of equilibrium are Fx Fy where 0: 0: 1 NA FAB cos b 150 150 FAB sin 5 9.81 0 0 sin Therefore FAB NA (a) (b) (c) 49.05 N ............................................ Ans. cos FAB sin N ........................................ Ans. 50 N for b 100 N for b 200 N for b 260 mm ..................... Ans. 280 mm .................. Ans. 295 mm .................. Ans. NA FAB FAB 3-27 A 75-lb stoplight is suspended between two poles as shown in Fig. P3-27. Neglect the weight of the flexible cables and plot the tension in both cables as a function of the sag distance d (0 d 8 ft). Determine the minimum sag d for which both tensions are less than (a) 100 lb (b) 250 lb (c) 500 lb SOLUTION From a free-body diagram of the ring, the equations of equilibrium are Fx Fy where A 0: 0: TB cos TA sin 1 A B TA cos B A 0 TB sin 75 0 1 tan d 20 B tan d 10 Therefore TB TA TA cos A cos B 75cos B sin A cos B sin B cos A 75cos sin A B B lb ........................................................ Ans. 50 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS TB (a) (b) (c) 75cos sin A A B lb ................................................................................................................ Ans. 5.6 ft ......................... Ans. 2.0 ft ......................... Ans. 1.0 ft ......................... Ans. TB TB TB 100 lb for d 250 lb for d 500 lb for d 3-28 A 50-kg load is suspended from a pulley as shown in Fig. P3-28. The tension in the flexible cable does not change as it passes around the small frictionless pulleys, and the weight of the cable may be neglected. Plot the force P required for equilibrium as a function of the sag distance d (0 d 1 m). Determine the minimum sag d for which P is less than (a) Twice the weight of the load. (b) Four times the weight of the load. (c) Eight times the weight of the load. SOLUTION From a free-body diagram of pulley supporting the load, the equations of equilibrium Fx Fy give A 0: 0: T sin A T cos T sin C C T cos A 0 0 50 9.81 C PT 49.05 N .................................... Ans. 2sin A Since the two angles are equal, the pulley is midway between the supports, and (a) (b) (c) d 1.5 P 1000 N for d 380 mm ................... Ans. P 2000 N for d 185 mm .................... Ans. P 4000 N for d 92 mm ...................... Ans. A tan 1 51 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 3-29 A worker positions a 250-lb crate by pulling on the rope BD as shown in Fig. P3-29. The 3-ft long rope BD is horizontal ( = 0) when the 5-ft long rope AB is vertical ( = 0). (a) What is the maximum distance bmax that the crate can be pulled to the side using this arrangement? (b) Calculate and plot the forces in ropes AB and BD as a function of the distance b for 0 b bmax. (c) How could the worker pull the crate to the side more than the bmax calculated in part a? SOLUTION From a free-body diagram of the ring, the equations of equilibrium are Fx Fy where 0: 0: 1 TAB sin TAB cos TBD cos 0 TBD sin 250 0 sin b 5 5 1 cos sin 1 3 TBD cos sin 250sin cos cos sin sin 250 cos cos cos sin sin lb ............Ans. lb ............Ans. Therefore TAB TBD TAB (a) bmax occurs when TBD goes negative (after it goes to infinity); bmax (c) 3.90 ft ..................................... Ans. To pull further to the side, the worker needs a longer rope to pull on or he needs to attach his rope lower - closer to the crate.................... Ans. 3-30 Two small wheels are connected by a light-weight rigid rod as shown in Fig. P3-30. Plot the angle (between the rod and the horizontal) as a function of the weight W1 (0.25W2 W1 10W2). Determine the weight W1 for which (a) = 50 (b) = 10 (c) = 25 SOLUTION From a free-body diagram of body 1, the equations of equilibrium are Fx Fy Fx Fy 0: 0: 0: 0: T cos N1 cos 30 N1 sin 30 T sin W1 0 0 0 0 (a) (b) From a free-body diagram of body 2, the equations of equilibrium are N 2 sin 60 N 2 cos 60 N1 sin 30 T sin 0 T cos 5 9.81 (c) (d) Adding Eqs. (a) and (c) together gives N 2 sin 60 (e) while adding Eqs. (b) and (d) together gives 52 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS (f) N 2 cos 60 N1 cos 30 W1 49.05 Solving Eqs. (e) and (f) gives N1 1.73205 N 2 N2 W1 49.05 2 T sin T cos tan (a) (b) (c) Then Eqs. (b) and (a) can be written N 2 cos 60 N 2 sin 60 49.05 (g) (h) Dividing Eq. (g) by Eq. (h) gives N 2 cos 60 49.05 N 2 sin 60 50 for W1 14.98 N ....................... Ans. 10 for W1 25 for W1 233 N .............................. Ans. 971 N .............................. Ans. 3-31 An automobile stuck in a muddy field is being moved by using a cable fastened to a tree as shown in Fig. P331. If the side force P has a magnitude of 150 lb, (a) Calculate and plot FC, the force applied to the car as a function of the angle (0 45 ). (b) What is the maximum angle for which this method is effective (that is, for which P FC)? SOLUTION From a free-body diagram of the knot, the equations of equilibrium are Fx Fy (a) If 0: 0: 1 2 T1 cos P T1 sin , then T1 1 1 T2 cos T2 sin FC and 2 2 0 0 T2 FC (b) P P 2sin FC when 150 2sin lb .......................... Ans. 30 .......................................................... Ans. 53 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 3-32 A particle is in equilibrium under the action of four forces as shown on the free-body diagram of Fig. P3-32. Determine the magnitude and the direction angle of the unknown force F4. SOLUTION F1x F2 x F3 x Fx Fy give 500 cos 63 750 cos 30 1000 cos 60 0: 0: F4 x F4 226.995 N 649.519 N 500.000 N F1 y F2 y F3 y 0 0 500sin 63 750sin 30 1000sin 60 445.503 N 375.000 N 866.025 N Then, the equations of equilibrium 226.995 649.519 500 F4 x 445.503 375 866.025 F4 y 1376.514 N 1377 N F4 y 45.522 N 1.89 ........................................................................................................Ans. 3-33 Two 10-in.-diameter pipes and a 6-in.-diameter pipe are supported in a pipe rack as shown in Fig. P3-33. The 10-in.-diameter pipes each weigh 300 lb and the 6-in.-diameter pipe weighs 175 lb. Determine the forces exerted on the pipes by the supports at the contact surfaces A, B, and C. Assume all surfaces to be smooth. SOLUTION AB BC cos 1 6 8 41.410 From a free-body diagram of cylinder B, the equations of equilibrium Fx Fy 0: 0: NB FBC cos 0 FBC sin 453.557 lb 300 0 are solved to get FBC NB 340 lb ................................................. Ans. From a free-body diagram of cylinder C, the equations of equilibrium Fx Fy 0: 0: FBC cos FAB sin 718.132 lb FAB cos FBC sin NC 0 175 0 are solved to get FAB NC 879 lb ................................................. Ans. Finally, from a free-body diagram of cylinder A, the equations of equilibrium Fx Fy 0: 0: NA FAB cos 0 0 N D 300 FAB sin are solved to get NA 539 lb ................................................. Ans. 54 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 3-34 The 250-kg block A of Fig. P3-34 is supported by a small wheel that is free to roll on the continuous cable between supports B and C. If the length of the cable is 42 m, determine the distance x and the tension T in the cable when the system is in equilibrium. SOLUTION Since the pulley is free to rotate, the tension is the same in every part of the rope. From a free-body diagram of the pulley supporting the 250 kg weight, the equations of equilibrium Fx Fy 1 0: 0: T sin 1 T cos T sin 2 2 T cos 1 0 0 250 9.81 are solved to get 2 T 1226.25 N sin 2 (a) From the geometry of the cable 1 1 1 42 m 2 2 (b) cos sin cos sin 40 m 6m (c) (d) Therefore, from Eqs. (b) and (c) 42 cos 40 17.753 T 4020 N .................................................. Ans. Finally, Eq. (d) can be written 1 2 19.6779 m (e) and adding Eqs. (b) and (e) gives 1 30.839 m 1 x cos 29.4 m ................................. Ans. 3-35 A joint in a bridge truss is subjected to the forces shown in Fig. P3-35. Determine the forces C and T required for equilibrium. SOLUTION The equations of equilibrium Fx Fy 0: 0: T 9 T cos 45 T sin 45 9 cos 60 12 0 0 C 9 sin 60 are solved to give 10.61 kip .......................................................................................................................Ans. C 15.29 kip .......................................................................................................................Ans. 55 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 3-36 Four forces act on a small airplane in flight, as shown in Fig. P3-36; the weight W (25 kN), the thrust provided by the engine FT (10 kN), the lift provided by the wings FL, and the drag resulting from motion through the air FD. If the airplane is flying with a constant velocity in the x’ direction, determine the magnitudes of the lift and drag forces. SOLUTION The equations of equilibrium Fx Fy 0: 0: FL FD FD cos10 FL cos10 FL sin10 10 cos10 0 10 sin10 FD sin10 25 0 are solved to give 24.6 kN .......................................................................................................................Ans. 5.66 kN .......................................................................................................................Ans. 3-37 A 500-lb block is supported by three cables as shown in Fig. P3-37. Determine the tensions in the cables AB, AC, and AD. SOLUTION TAB TAC TAD Fx Fy Fz 0: 0: 0: TAB TAC TAD 6 j 12k 0.44721TAB j 0.89443TAB k 62 122 4i 3 j 12k 0.30769TAC i 0.23077TAC j 0.92308TAC k 4 2 32 122 4i 8 j 12k 0.26726TAD i 0.53452TAD j 0.80178TAD k 4 2 82 122 0.30769TAC 0.44721TAB 0.23077TAC 0.26726TAD 0.53452TAD 0 0 Then, the equations of equilibrium 0.89443TAB TAB TAC TAD 0.92308TAC 0.80178TAD 500 0 are solved simultaneously to get 267 lb .........................................................................................................................Ans. 141.3 lb ......................................................................................................................Ans. 162.7 lb ......................................................................................................................Ans. 3-38 A 1250-N force F is supported by a cable AD and by struts AB and AC, as shown in Fig. P3-38. If the struts can transmit only axial tensile or compressive forces, determine the forces in the struts and the tension in the cable. SOLUTION TAD FAB FAC TAD FAB FAC 0.89443TAD j 0.44721TAD k 62 32 2i 6 j 2k 0.30151FAB i 0.90453FAB j 0.30151FAB k 22 62 22 3i 6 j 2k 0.42857 FAC i 0.85714 FAC j 0.28571FAC k 32 62 22 6 j 3k Then, the equations of equilibrium 56 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS Fx Fy Fz 0: 0: 0: 0.89443TAD 0.44721TAD TAD FAB FAC 0.30151FAB 0.90453FAB 0.30151FAB 0.42857 FAC 0.85714 FAC 0 0 0.28571FAC 1250 0 are solved simultaneously to get 1677 N .......................................................................................................................Ans. 995 N .........................................................................................................................Ans. 700 N .........................................................................................................................Ans. 3-39 A 75-lb force F is supported by a tripod as shown in Fig. P3-39. If the legs can transmit only axial tensile or compressive forces, determine the forces in the legs AB, AC, and AD. SOLUTION FAB FAC FAD Fx Fy Fz 0: 0: 0: FAB FAC FAD 25i 20 j 54k 252 202 542 25i 40 j 54k 252 40 2 542 25i 6 j 54k 252 6 2 542 0.39823FAB i 0.31859 FAB j 0.86018 FABk 0.34867 FAC i 0.55787 FAC j 0.75313FAC k 0.41800 FAD i 0.10032 FAD j 0.90289 FAD k 0.41800 FAD 0.10032 FAD 0 0 Then, the equations of equilibrium 0.39823FAB 0.34867 FAC 0.31859 FAB 0.55787 FAC 0.86018FAB FAB FAC FAD 0.75313FAC 0.90289 FAD 75 0 are solved simultaneously to get 33.4 lb ........................................................................................................................Ans. 11.62 lb .....................................................................................................................Ans. 41.5 lb .......................................................................................................................Ans. 3-40 A particle is in equilibrium under the action of four forces as shown on the free-body diagram of Fig. P3-40. Determine the magnitude of the unknown forces F1, F2, and F3. SOLUTION F1 F2 F3 Fx Fy 0: 0: F1 F2 3i 2.5 j 3.5k 32 2.52 3.52 3i 4.5 j 3.5k 0.57208 F1i 0.47673F1 j 0.66742 F1k 0.46569 F2 i 0.69854 F2 j 0.54331F2 k 32 4.52 3.52 3i 5 j 4k 0.42426 F3i 0.70711F3 j 0.56569 F3k F3 32 52 42 0.57208 F1 0.46569 F2 0.42426 F3 0.47673F1 0.69854 F2 0.70711F3 0 0 Then, the equations of equilibrium 57 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS Fz 0: F1 F2 F3 0.66742 F1 0.54331F2 0.56569 F3 50 0 2.34 kN ........................................................................................................................Ans. 43.1 kN ........................................................................................................................Ans. 44.2 kN ........................................................................................................................Ans. are solved simultaneously to get 58 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS Chapter 4 4-1 An aluminum tube with an outside diameter of 1.000 in. will be used to support a 10-kip load. If the axial stress in the member must be limited to 30 ksi T or C, determine the wall thickness required for the tube. SOLUTION P A A di t 4-2 10, 000 A 12 di2 30, 000 psi 10, 000 in.2 30, 000 4 0.75867 in. 1 2t 0.1207 in. .......................................................................................................................... Ans. Three steel bars with 25 15-mm cross sections are welded to a gusset plate as shown in Fig. P4-2. Determine the normal stresses in the bars when the forces shown are being applied to the plate. SOLUTION A B C P A P A P A 40, 000 106.7 106 N/m 2 106.7 MPa ........................................ Ans. 0.025 0.015 50, 000 133.3 106 N/m 2 133.3 MPa ......................................... Ans. 0.025 0.015 20, 000 53.3 106 N/m 2 53.3 MPa ............................................. Ans. 0.025 0.015 4-3 Two ¼-in. diameter steel cables A and B are used to support a 220-lb traffic light as shown in Fig. P4-3. Determine the normal stress in each of the cables. SOLUTION From a free-body diagram of the ring, the equations of equilibrium Fx Fy TB TA A 0: 0: TB cos 25 TA sin 20 TA cos 20 0 TB sin 25 220 0 are solved to get 1.03684TA 281.977 lb TA AA TB AB 281.977 0.25 0.25 2 TB 4 4 292.365 lb 5740 psi ...................................................................................... Ans. 5960 psi ...................................................................................... Ans. 292.365 2 B 4-4 A system of steel pipes is loaded and supported as shown in Fig. P4-4. If the normal stress in each pipe must not exceed 150 MPa, determine the cross-sectional areas required for each of the sections. SOLUTION From free-body diagrams of the pipes, the equations of equilibrium give Fy Fy 0: 0: PA 650 0 PB 650 850 0 PA PB 650 kN 1500 kN 59 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS Fy AA AB AC 4-5 0: PA A PC 650 850 1500 0 4333 10 6 PC m2 6 3000 kN 650 103 150 106 1500 103 150 106 3000 103 150 106 4330 mm 2 ....................................................... Ans. m2 m2 10, 000 mm 2 .............................................. Ans. 20, 000 mm 2 ............................................ Ans. PB B 10, 000 10 20, 000 10 PC C 6 Two 1-in. diameter steel bars are welded to a gusset plate as shown in Fig. P4-5. Determine the normal stresses in the bars when forces F1 = 550 lb and F2 = 750 lb are applied to the plate. SOLUTION A F1 AA F2 AB 550 1 1 2 4 4 700 psi ............................................................................................... Ans. 955 psi ............................................................................................... Ans. 750 2 B 4-6 Two strips of a plastic material are bonded together as shown in Fig. P4-6. The average shearing stress in the glue must be limited to 950 kPa. What length L of splice plate is needed if the axial load carried by the joint is 50 kN? SOLUTION From a free-body diagram of the middle plate, the equations of equilibrium Fx gives 0: 2V 50 103 A 0 V L 4-7 25 103 N 950 10 3 0.3 L 2 0.1754 m 175.4 mm ..................................................................................................... Ans. A coupling is used to connect a 2-in. diameter plastic rod to a 1.5-in. diameter rod, as shown in Fig. P4-7. If the average shearing stress in the adhesive must be limited to 500 psi, determine the minimum lengths L1 and L2 required for the joint if the applied axial load P is 8000 lb. SOLUTION From a free-body diagram of the rod, the equation of equilibrium Fx gives 0: V 8000 0 V 8000 lb 500 500 A 2 L1 1.5 L2 L1 L2 2.55 in. ................................................................Ans. 3.40 in. ................................................................Ans. 60 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 4-8 RILEY, STURGES AND MORRIS Three plates are joined with a 12-mm diameter pin as shown in Fig. P4-8. Determine the maximum load P that can be transmitted by the joint if (a) The maximum normal stress on a cross section at the pin must be limited to 350 MPa. (b) The maximum bearing stress between a plate and the pin must be limited to 650 MPa. (c) The maximum shearing stress on a cross section of the pin must be limited to 240 MPa. (d) The punching shear resistance of the material in the top and bottom plates is 300 MPa. SOLUTION (a) For the outside plates An P An P Pmax (b) 30 12 10 2 An 180 mm 2 180 10 6 6 m2 2 350 106 180 10 325 mm 2 6 126 103 N 6 For the middle plate 25 12 25 An 325 10 m2 350 106 325 10 113.7 103 N Therefore 113.7 kN ......................................................................................................................... Ans. dt Ab dt 2 Ab 12 25 300 mm 2 6 For the bearing stress on the hole in the middle plate Ab P Ab P Pmax (c) 300 10 6 m2 650 106 300 10 12 10 120 mm 2 195 103 N 6 For the outer plates 120 10 6 m2 2 650 106 120 10 156 103 N Therefore 156 kN ............................................................................................................................. Ans. For the shearing stress on the cross section of the pin (which is in double shear) As P (d) d2 4 2 As 12 2 4 6 113.10 mm 2 113.10 10 m2 6 2 240 106 113.10 10 54.3 103 N 54.3 kN ..................................................................................................... Ans. For the punching shear stress in the outside plates As P 2 Lt 2 As 2 18 10 360 10 6 360 mm 2 216 103 N m2 6 2 300 106 360 10 216 kN .......................................................................................................... Ans. 61 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 4-9 RILEY, STURGES AND MORRIS A 100-ton hydraulic punch press is used to punch holes in a 0.50-in.-thick steel plate, as illustrated schematically in Fig. P4-9. If the average punching shear resistance of the steel plate is 40 ksi, determine the maximum diameter hole that can be punched. SOLUTION P As 40 103 dt d 0.5 100 2000 lb d 3.18 in. ............................................................................................. Ans. 4-10 A body with a mass of 250 kg is supported by five 15-mm diameter cables, as shown in Fig. P4-10. Determine the normal stress in each of the cables. SOLUTION The tension in cable E is equal to the weight of the hanging body. The normal stress in cable E is then E TE A 250 9.81 0.015 2 4 13.88 106 N/m 2 13.88 MPa ........................................... Ans. From a free-body diagram of the lower ring, the equations of equilibrium Fx Fy give 0: 0: TC cos 60 TC sin 60 TD 0 0 2 250 9.81 0.015 TC C 2831.90 N C 4 16.03 106 N/m 2 1415.95 N D 16.03 MPa .................................................................................... Ans. 0.015 2 TD D 4 8.01 106 N/m 2 8.01 MPa ........................................................................................ Ans. Then, from a free-body diagram of the upper ring, the equations of equilibrium Fx Fy give 0: 0: TB cos 30 TB sin 30 TA cos 40 TA sin 40 0.015 2 TC cos 60 TC sin 60 4 0 0 TA 1506.83 N A A 8.53 106 N/m 2 2967.85 N B 8.53 MPa ........................................................................................ Ans. 0.015 2 TB B 4 16.79 106 N/m 2 16.79 MPa ................................................................................... Ans. 62 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 4-11 The joint shown in Fig. P4-11 is used in a steel tension member which has a 2 1-in. rectangular cross section. If the allowable normal, bearing, and punching-shearing stresses in the joint are 13.5 ksi, 18.0 ksi, and 6.50 ksi, respectively, determine the maximum load P that can be carried by the joint. SOLUTION For normal stress at the narrowest section An P 0.5 1 An 0.5 in.2 6750 lb 13.5 103 0.5 For bearing stress Ab P 2 38 1 Ab 0.75 in.2 13,500 lb 18.0 103 0.75 For punching shearing stress As P Therefore 2 38 1 As 0.75 in.2 4880 lb 6.5 103 0.75 Pmax 4880 lb ............................................................................................................................ Ans. 4-12 A vertical shaft is supported by a thrust collar and bearing plate, as shown in Fig. P4-12. Determine the maximum axial load that can be applied to the shaft if the average punching shear stress in the collar and the average bearing stress between the collar and the plate are limited to 75 and 100 MPa, respectively. SOLUTION For bearing stress between the collar and the plate Ab P As P Therefore 0.152 0.102 Ab dL As 4 9.8175 10 3 3 m2 100 106 9.8175 10 0.10 0.025 982 103 N 3 For punching shearing stress in the collar 7.854 10 3 m2 75 106 7.854 10 589 103 N Pmax 589 kN ............................................................................................................................ Ans. 4-13 A device for determining the shearing strength of wood is shown in Fig. P4-13. The dimensions of the wood specimen are 6-in. wide by 8-in. high by 2-in. thick. If a load of 16,800 lb is required to fail the specimen, determine the shearing strength of the wood. SOLUTION V As 16,800 1050 psi ................................................................................................... Ans. 82 63 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 4-14 The 75-kg traffic light shown in Fig. P4-14 is supported by three cables of equal diameter. Determine the minimum diameter required if the normal stress in any cable must not exceed 160 MPa. SOLUTION TA TB TC Fx Fy Fz give TA TB 4i 8 j 5k 42 82 52 6i 8 j 5k 0.39036TAi 0.78072TA j 0.48795TAk 0.53666TB i 0.71554TB j 0.44721TB k 6 2 82 52 8 j 5k 0.84800TC j 0.53000TC k TC 82 52 0: 0: 0: 0.39036TA 0.53666TB 0.78072TA 0.71554TB 0.48795TA 0.44721TB 0.53000TC TC 160 106 160 106 160 106 2 dA 4 2 dB 4 2 dC 4 The equations of equilibrium 0 0 0 0.84800TC 75 9.81 TA 1.37478TB TA TB TC Therefore 2.10950TB dA dB dC 0.00190 m 0.00162 m 0.00235 m 452.353 N 329.037 N 694.103 N d min 2.35 mm .......................................................................................................................... Ans. 4-15 A 1000-lb load is securely fastened to a hoisting rope as shown in Fig. P4-15. The force in the weightless flexible cable does not change as it passes around the small frictionless pulley at support C. The sag distance d is 4 ft. Cables AB and BC have the same diameter. The normal stresses in these cables must not exceed 24 ksi, and the shearing stress in pin A (double shear) must not exceed 12 ksi. Determine (a) The minimum diameter of cables AB and BC. (b) The minimum diameter of the pin at A. SOLUTION 4 23.578 10 a 10 cos 1 9.16515 ft 1 sin 1 b 30 a 2 20.83485 ft tan 1 4 10.868 b Then, from a free-body diagram of the ring, the equations of equilibrium Fx Fy give 0: 0: P sin P cos 2 2 TA cos 1 1 0 TA sin 1000 0 64 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS TA 1.07152 P P 1620.35 lb (a) For the normal stress in the cables TA 1736.24 lb 24 103 2 d BC 4 P 1620.35 lb d BC 0.293 in. ........................................................................................................................... Ans. 24 103 2 d AB 4 TA 1736.24 lb d AB (b) 0.304 in. ........................................................................................................................... Ans. 12 103 2 2 dp 4 For the pin at A, which is in double shear, TA 1736.24 lb dp 0.304 in. ............................................................................................................................. Ans. 4-16 A flat steel bar 100 mm wide by 25 mm thick has axial loads applied with 40 mm-diameter pins in double shear at points A, B, C, and D as shown in Fig. P4-16. Determine (a) The axial stress in the bar on a cross section at pin B. (b) The average bearing stress on the bar at pin B. (c) The shearing stress on the pin at A. SOLUTION (a) For axial stress in the bar n P An P Ab V 2 As 350 103 0.06 0.025 250 103 0.04 0.025 350 103 2 0.04 2 233 106 N/m 2 233 MPa .................................................... Ans. (b) For bearing stress at the pin b 250 106 N/m 2 250 MPa ................................................... Ans. (c) For shearing stress on the pin (which is in double shear) 4 139.3 106 N/m2 139.3 MPa ....................................... Ans. 4-17 Two flower pots are supported with steel wires of equal diameter. Pot A weighs 10 lb and pot B weighs 8 lb. Determine the minimum required diameter of the wires if the normal stress in the wires must not exceed 18 ksi. SOLUTION From a free-body diagram of the upper ring Fx Fy Fx Fy 0: 0: 0: 0: TCD cos 45 TCD sin 45 TBC cos TAB sin 45 TBC cos TBC sin 0 80 0 (a) (b) From a free-body diagram for the lower ring TAB cos 45 TBC sin 0 (c) (d) 10 0 Adding equations (a) and (c) gives TCD cos 45 TAB cos 45 65 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS TCD TAB TAB sin 45 12.7279 lb 1.0000 9.0000 19 9.0554 lb P A 12.7279 18 103 psi d2 4 6.340 (e) (f) Then adding Eqs. (b) and (d) gives TCD sin 45 TCD TAB 18 lb Now Eqs. (b) and (a) can be written TBC sin TBC cos tan TBC max Dividing Eq. (e) by Eq. (f) gives d 0.0300 in. ............................................................................................................................ Ans. 4-18 A steel pipe will be used to support a 40-kN load. If the wall thickness of the pipe is 10 percent of the pipe’s outside diameter do, calculate and plot the normal stress in the pipe as a function of the diameter do, (20 mm do 75 mm). If the axial stress in the pipe must be limited to 250 MPa, what is the smallest size standard steel pipe (see Appendix A) that could be used. SOLUTION di d o 2 0.1d o P A P A 0.8d o 141, 471.06 N/m 2 2 do 40 103 d o2 di2 4 40 103 A For standard steel pipe, 250 106 N/m 2 A 160 10 6 m 2 160 mm 2 d 13 mm .................................................... Ans. 4-19 An aluminum tube with an outside diameter do will be used to support a 10-kip load. If the axial stress in the tube must be limited to 30 ksi T or C, calculate and plot the required wall thickness t as a function of do, (0.75 in. do 4 in.). What diameter would be required for a solid aluminum shaft? SOLUTION P A di t do 10, 000 d o2 di2 4 30, 000 psi d o2 0.42441 d o 2t d o2 0.42441 2 For a solid aluminum shaft 66 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS P A d 10, 000 d2 4 30, 000 psi 0.65 in. ................................................................................................................................. Ans. 4-20 The steel pipe column shown in Fig. P4-20a has an outside diameter of 150 mm and a wall thickness of 15 mm. The load imposed on the column by the timber beam is 150 kN. If the bearing stress between the circular steel bearing plate and the timber beam is not to exceed 3.25 MPa, determine the minimum diameter bearing plate that must be used between the column and the beam. Assume that the bearing stress is uniformly distributed over the surface of the plate. If the bearing plate is not rigid, the stress between the bearing plate and the timber beam will not be uniform. If the stress varies as shown in Fig. P4-20b (a uniform value of max above the column and decreasing linearly to max/5 at the outside edge rp of the bearing plate), calculate and plot max versus the radius rp of the bearing plate (75 mm rp 500 mm). Now what minimum diameter bearing plate must be used if the bearing stress must not exceed 3.25 MPa? What is the percent decrease in max for a 400 mmdiameter bearing plate compared with a 150 mm-diameter bearing plate? For a 600 mm-diameter bearing plate compared with a 150 mm-diameter bearing plate? SOLUTION For uniform stress over a rigid bearing plate P A d 150 103 d2 4 3.25 106 N/m 2 0.242 m 242 mm ........................................................................................................... Ans. For the non-rigid bearing plate c1r c2 max where the constants c1 and c2 are chosen such that 1 c1 0.075 0.2 c1rp Therefore, c2 c2 c1 c2 0.8 0.075 rp 1 0.8 0.075 0.075 rp dA 0.075 0 max Integrating the stress to get the axial force F gives 150 103 2 r dr c1 rp 0.075 max c1r c2 2 r dr c2 rp2 2 0.0752 2 2 max 0.0752 2 rp3 3 0.0753 3 67 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 150 103 max 2 When When When 0.075 2 2 c1 r 3 3 p 0.075 3 3 c2 r 2 2 p 0.075 2 2 N/m 2 rp rp rp 0.075 m 0.2 m 0.3 m max max max 8.49 106 N/m 2 8.49 MPa ...................................... Ans. 1.976 MPa (a 76.7% decrease) ................................ Ans. 0.965 MPa (an 88.6% decrease) .............................. Ans. 4-21 The tie rod shown in Fig. P4-21a has a diameter of 1.50 in and is used to resist the lateral pressure against the walls of a grain bin. The force imposed on the wall by the rod is 18,000 lb. If the bearing stress between the washer and the wall must not exceed 400 psi, determine the minimum diameter washer that must be used between the head of the bolt and the grain bin wall. Assume that the bearing stress is uniformly distributed over the surface of the washer. If the washer is not rigid, the stress between the washer and the wall will not be uniform. If the stress varies as shown in Fig. P4-21b (a uniform value of max under the 2.4-in.-diameter restraining nut and decreasing as 1/r to the outside edge rw of the washer), calculate and plot max versus the radius rw of the washer (1 in. rw 8 in.). Now what minimum diameter washer must be used if the bearing stress must not exceed 400 psi? What is the percent decrease in max for an 8-in.-diameter washer compared with a 4-in.diameter washer? For a 12-in.-diameter washer compared with an 8-in.-diameter washer? SOLUTION For uniform stress over a rigid washer P A d 18 103 d 2 1.52 4 400 psi 7.72 in. .................................................. Ans. For the non-rigid washer 1.2 r max Integrating the stress to get the axial force F gives dA 1.2 0.75 max rw 1.2 18 103 2 r dr 1.2 r max 2 r dr max max 1.22 0.752 2.4 rw 1.2 0.87750 2.4 rw 1.2 psi max 18 103 0.87750 2.4 rw 1.2 max When When When When 400 psi 2.4 in. 8 in. d 2rw 13.61 in. ...................................................................... Ans. max max max d d d 2rw 2rw 6529 psi 754 psi 462 psi an 88.5% decrease.................................. Ans. a 38.7% decrease.................................... Ans. 2rw 12 in. 68 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 4-22 The steel bar shown in Fig. P4-22 will be used to carry an axial tensile load of 400 kN. If the thickness of the bar is 45 mm, determine the normal and shearing stresses on plane AB. SOLUTION From a free-body diagram of the bar, the equations of equilibrium Fn Ft give 0: 0: N 400 cos 37 V 400sin 37 0 0 V 240.73 kN N 319.45 kN The areas are related by A 0.075 0.045 Therefore An cos 37 75.6 106 N/m 2 57.0 106 N/m 2 An 4.226 10 3 m2 N An V An 319.45 103 4.226 10 3 240.73 103 4.226 10 3 75.6 MPa ................................................... Ans. 57.0 MPa .................................................... Ans. 4-23 An axial load P is applied to a timber block with a 4 4-in. square cross section, as shown in Fig. P4-23. Determine the normal and shear stresses on the planes of the grain if P = 5000 lb. SOLUTION From a free-body diagram of the bar, the equations of equilibrium Fn Ft give 0: 0: N 5000sin14 V 5000 cos14 0 0 N 1209.61 lb V 4851.48 lb The areas are related by A44 An Therefore An sin14 66.137 in.2 N An V An 1209.61 18.29 psi ............................................................................................... Ans. 66.137 4851.48 66.137 73.4 psi ................................................................................................. Ans. 4-24 A structural steel bar with a 20 25-mm rectangular cross section is subjected to an axial tensile load of 55 kN. Determine the maximum normal and shear stresses in the bar. SOLUTION max P A 55 103 110.0 106 N/m 2 0.02 0.025 max 110.0 MPa ........................................... Ans. P 2A 2 55.0 MPa .................................................................................................... Ans. 69 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 4-25 A steel rod of circular cross section will be used to carry an axial tensile load of 50 kip. The maximum stresses in the rod must be limited to 25 ksi in tension and 15 ksi in shear. Determine the minimum diameter d for the rod. SOLUTION max P A P 2A 50 103 d2 4 50 103 2 d2 4 25 103 psi 15 103 psi d 1.596 in. d 1.457 in. max d min 1.596 in. .......................................................................................................................... Ans. 4-26 Determine the maximum axial load P that can be applied to the wood compression block shown in Fig. P4-26 if specifications require that the shear stress parallel to the grain not exceed 5.25 MPa, the compressive stress perpendicular to the grain not exceed 13.60 MPa, and the maximum shear stress in the block not exceed 8.75 MPa. SOLUTION From a free-body diagram of the bar, the equations of equilibrium are Fn Ft 0: 0: P sin 60 P cos 60 N V 0 0 The areas are related by A 0.2 0.12 Therefore An sin 60 An sin 60 An 0.027713 m 2 P N sin 60 13.60 106 0.027713 sin 60 P 435 103 N 5.25 106 0.027713 V An P P 291 103 N cos 60 cos 60 cos 60 Pmax 291 kN ............................................................................................................................. Ans. 4-27 A steel bar with a 4 1-in. rectangular cross section is being used to transmit an axial tensile load, as shown in Fig. P4-27. Normal and shear stresses on plane AB of the bar are 12 ksi tension and 9 ksi shear. Determine the angle and the applied load P. SOLUTION From a free-body diagram of the bar, the equations of equilibrium are Fn Ft 0: 0: N V P cos P sin 0 0 The areas are related by A 41 Therefore An cos An An 12 103 4 cos 9 103 4 cos An 4 cos in.2 N V Therefore P cos P sin 70 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS tan 9 P sin 12 P cos 36.870 ................................................................................................................................. Ans. 75 103 lb 75 kip ........................................................................................................... Ans. P 4-28 A steel bar with a butt-welded joint, as shown in Fig. P4-28 will be used to carry an axial tensile load of 400 kN. If the normal and shear stresses on the plane of the weld must be limited to 70 MPa and 45 MPa, respectively, determine the minimum thickness t required for the bar. SOLUTION From a free-body diagram of the bar, the equations of equilibrium are Fn Ft 0: 0: N 400sin 57 V 400 cos 57 0 0 The areas are related by A 0.1t Therefore An sin 57 400 103 sin 57 0.11924t 400 103 cos 57 0.11924t An 0.11924t m 2 N An V An tmin 70 106 N/m 2 45 106 N/m 2 t t 0.0402 m 0.0406 m 40.6 mm ........................................................................................................................... Ans. 4-29 The shearing stress on plane AB of the 4 8-in. rectangular block shown in Fig. P4-29 is 2 ksi when the axial load P is applied. If the angle is 35 , determine (a) The load P. (b) The normal stress on plane AB. (c) The maximum normal and shearing stresses in the block. SOLUTION From a free-body diagram of the bar, the equations of equilibrium are Fn Ft 0: 0: P sin 35 P cos 35 N V 0 0 The areas are related by A 48 An (a) An sin 35 55.790 in.2 V cos 35 P sin 35 An cos 35 2 103 55.790 cos 35 An Solving the second equation for P gives P (b) 136, 215 lb 136.2 kip ........................ Ans. Solving the first equation for N gives N (c) 78,129.6 lb 1400 psi ............................................................................................................................... Ans. Then the maximum normal and shearing stresses are given by 71 STATICS AND MECHANICS OF MATERIALS, 2nd Edition max RILEY, STURGES AND MORRIS max P 136, 215 4260 psi ............................................................................................ Ans. 48 A P max 2130 psi ................................................................................................. Ans. 2A 2 4-30 A wood tension member with a 50 100-mm rectangular cross section will be fabricated with an inclined glued joint (45 90 ) at its midsection, as shown in Fig. P4-30. If the allowable stresses for the glue are 5 MPa in tension and 3 MPa in shear, determine (a) The optimum angle for the joint. (b) The maximum safe load P for the member. SOLUTION From a free-body diagram of the bar, the equations of equilibrium are Fn Ft 0: 0: N V P sin P cos 0 0 The areas are related by A 0.05 0.1 An sin An Therefore 0.005 sin P sin P cos P sin P cos m2 An An 5 106 0.005 sin 3 106 0.005 sin 5 1.66667 3 N V and dividing the first equation by the second gives tan 59.04 ................................................................................................................................... Ans. Pmax 34.0 103 N 34.0 kN ............................................................................................... Ans. 4-31 The two parts of the eyebar shown in Fig. P4-31 are connected with two ½-in. diameter bolts (one on each side). Specifications for the bolts require that the axial tensile stress not exceed 12.0 ksi and that the shearing stress not exceed 8.0 ksi. Determine the maximum load P that can be applied to the eyebar without exceeding either specification. SOLUTION From a free-body diagram of the bar, the equations of equilibrium are Fn Ft Ab 0: 0: 2N 2V P cos 30 P sin 30 0 0 The cross sectional areas of the bolts are 0.5 2N cos 30 2 4 0.19635 in.2 2 Ab cos 30 2 12 103 0.19635 cos 30 The first equation gives P 5441 lb while the second equation gives 72 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS P Therefore 2V sin 30 5441 lb 2 Ab sin 30 2 8 103 0.19635 sin 30 6283 lb Pmax 5440 lb ......................................................................................................... Ans. 4-32 The bar shown in Fig. P4-32 has a 200 100-mm rectangular cross section. Determine (a) The normal and shearing stresses on plane a-a. (b) The maximum normal and shearing stresses in the bar. SOLUTION From a free-body diagram of the bar, the equations of equilibrium are Fn Ft 0: 0: N 500 cos 30 V 500sin 30 0 0 The areas are related by A 0.2 0.1 An Therefore An cos 30 3 23.094 10 m2 433.01 kN 250.00 kN An An N V 500 103 cos 30 500 103 sin 30 18.75 106 N/m 2 10.83 106 N/m2 18.75 MPa ...................................................................................... Ans. 10.83 MPa ....................................................................................... Ans. max max P 500 103 25.0 106 N/m 2 25.0 MPa ..................................................... Ans. A 0.2 0.1 P max 12.50 MPa ............................................................................................. Ans. 2A 2 4-33 A steel eyebar with a 4 1-in. rectangular cross section has been designed to transmit an axial tensile load. The length of the eyebar must be increased by welding a new center section in the bar (45 90 ), as shown in Fig. P4-33. The stresses in the weld material must be limited to 12 ksi in tension and 9 ksi in shear. Determine (a) The optimum angle for the joint. (b) The maximum safe load P for the redesigned member. SOLUTION From a free-body diagram of the bar, the equations of equilibrium are Fn Ft 0: 0: N V P sin P cos 0 0 An 12 103 4 sin 9 103 4 sin 4 sin in.2 The areas are related by A 41 Therefore An sin An An N V P sin P cos 73 STATICS AND MECHANICS OF MATERIALS, 2nd Edition and dividing the first equation by the second gives RILEY, STURGES AND MORRIS tan Pmax P sin P cos 12 1.33333 9 53.13 ................................................. Ans. 75, 000 lb ........................................................................................................................ Ans. 4-34 A steel eyebar with a 100 25-mm rectangular cross section has been designed to transmit an axial tensile load P. The length of the eyebar must be increased by welding a new center section in the bar, as shown in Fig. P4-33. If P = 250 kN, calculate and plot the normal stress n and the shear stress n in the weld material for weld angles (30 90 ). If the stresses in the weld material must be limited to 80 MPa in tension and 60 MPa in shear, what ranges of would be acceptable for the joint? Repeat for P = 305 kN and for P = 350 kN. Are weld angles < 30 reasonable? Why or why not? SOLUTION From a free-body diagram of the bar, the equations of equilibrium are Fn Ft 0: 0: N V P sin P cos 0 0 The areas are related by A 0.1 0.025 An 0.0025 sin An sin m2 Therefore N An V An When When When P sin 0.0025 sin P cos 0.0025 sin 250 kN 305 kN 350 kN P sin 2 N/m 2 ............................................................................ Ans. 0.0025 P sin cos 0.0025 0 0 0 N/m 2 ..................................................................... Ans. P P P 64 .................................................................................. Ans. 40 or 50 54 .................................................. Ans. 30 .................................................................................. Ans. 30 are not very practical since the weld gets long quickly Angles less than about as the angle decreases................................................................................................................................. Ans. 4-35 Specifications for the rectangular (3 3 21-in.) block shown in Fig. P4-35 require that the normal and shearing stresses on plane A-A not exceed 800 psi and 500 psi, respectively. If the plane A-A makes an angle = 37 with the horizontal, calculate and plot the ratios / max and / max as a function of the load P (0 P 13 kip). What is the maximum load Pmax that can be applied to the block? Which condition controls what the maximum load can be? Repeat for = 25 . For what angle will the normal stress and the shear stress both reach their limiting values at the same time? SOLUTION From a free-body diagram of the bar, the equations of equilibrium are Fn 0: N P cos 0 74 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS Ft 0: V P sin 0 The areas are related by A 33 An Therefore An cos in.2 P cos 2 9 psi .................................................. Ans. 800 P sin cos 500 9 psi ......... Ans. 9 cos N An max max V An max max When 37 , shear stress controls and Pmax When 9360 lb ............................................. Ans. 25 , normal stress controls and Pmax 8770 lb ............................................. Ans. The normal stress and the shear stress will reach their maximum values at the same time if tan opt opt V N max An max An 500 800 0.62500 32.01 ................................................................................................................................ Ans. 4-36 Compression tests of concrete indicate that concrete fails when the axial compressive strain is 1200 m/m. Determine the maximum change in length that a 200-mm diameter 400-mm long concrete test specimen can tolerate before failure occurs. SOLUTION L 1200 10 6 400 0.480 mm .......................................................................... Ans. 4-37 The 0.5 2 4-in. rubber mounts shown in Fig. P4-37 are used to isolate the vibrational motion of a machine from its supports. Determine the average shearing strain in the rubber mounts if the rigid frame displaces 0.01 in. vertically relative to the support. SOLUTION L 0.01 0.02 in./in. 0.02 rad ................................................................................. Ans. 0.5 4-38 A structural steel bar was loaded in tension to fracture. A 200 mm-length of the bar was marked off in 25mm lengths before loading. After the rod broke, the 25-mm segments were found to have lengthened to 30.0, 30.5, 31.5, 34.0, 44.5, 32.0, 31.0, and 30.0 mm, consecutively. Determine (a) The average strain over the 200-mm length. (b) The maximum average strain over any 50-mm length. SOLUTION Lf L max Li Li 263.5 200 200 0.317 mm/mm ............................................................. Ans. 78.5 50 50 0.570 mm/mm ........................................................................................ Ans. 75 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 4-39 Mutually perpendicular axes in an unstressed member were found to be oriented at 89.92 when the member was stressed. Determine the shearing strain associated with these axes in the stressed member. SOLUTION 2 90 89.92 180 1.396 10 3 rad .......................................................... Ans. 4-40 A thin triangular plate is uniformly deformed as shown in Fig. P4-40. Determine the shearing strain at P associated with the two edges (PQ and PR) that were orthogonal in the undeformed plate. SOLUTION LP Q LMQ LMP 500 mm LMP LMP tan 1 500 cos 45 353.5534 mm 10 363.5534 mm 44.20107 2 180 LMQ LMP 2 2 90 88.40213 3 27.9 10 rad ..................................................................................................................... Ans. 4-41 The sanding-drum mandrel shown in Fig. P4-41 is made for use with a hand drill. The mandrel is made from a rubber-like material which expands when the nut is tightened to secure the sanding drum placed over the outside surface. If the diameter D of the mandrel increases from 2.00 in. to 2.15 in. as the nut is tightened, determine (a) The average normal strain along a diameter of the mandrel. (b) The circumferential strain at the outside surface of the mandrel. SOLUTION (a) D (b) C 2.15 2.00 75.0 10 3 in./in. ................................................................................... Ans. 2.00 2.15 2.00 75.0 10 3 in./in. ..................................................................... Ans. 2.00 4-42 A thin rectangular plate is uniformly deformed as shown by PRSQ in Fig. P4-42. Determine the shearing stain xy at P. SOLUTION 1 tan 2 0.380 0.04354 500 0.200 tan 1 0.04584 250 1 1 2 2 0.08938 1.560 10 3 rad ................................................................................................................... Ans. 76 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 4-43 A steel sleeve is connected to a steel shaft with a flexible rubber insert, as shown in Fig. P4-43. The insert has an inside diameter of 3.3 in. and an outside diameter of 4.3 in. When the unit is subjected to a torque T, the shaft rotates 1.5 with respect to the sleeve. Assume that radial lines in the unstressed state remain straight as the rubber deforms. Determine the shearing strain r in the rubber insert (a) At the inside surface. (b) At the outside surface. SOLUTION b (b) r tan 1 3.3 2 b 0.5 3 1.5 180 4.93774 rad 3 0.0431969 in. 1 86.2 10 (a) 2 1 outside .................................................. Ans. 1.5 6.43774 rad inside 112.4 10 ................................................. Ans. 4-44 A steel rod is subjected to a nonuniform heating that produces an extensional (axial) strain that is proportional to the square of the distance from the unheated end ( = kx2). If the strain is 1250 m/m at the midpoint of a 3.00-m rod, determine (a) The change in length of the rod. (b) The average axial strain over the length L of the rod. (c) The maximum axial strain in the rod. SOLUTION kx 2 d dx (a) 1250 10 kx 2 dx k 3 0 6 k 1.5 2 k 555.556 10 6 d x dx 3 2 kx3 3 3 9k 0 5.00 10 3 m 5.00 mm ................... Ans. (b) (c) avg 0.005 1.667 10 3 3 m/m 1667 m/m ................................................................. Ans. 3 max k3 2 9k 5.00 10 m/m 5000 m/m ................................................ Ans. 4-45 The axial strain in a suspended bar of material of varying cross section due to its own weight, as shown in Fig. P4-45, is given by the expression y/3E, where is the specific weight of the material, y is the distance from the free (bottom) end of the bar, and E is a material constant. Determine, in terms of , L, and E, (a) The change in length of the bar due to its own weight. (b) The average axial strain over the length L of the bar. (c) The maximum axial strain in the bar. SOLUTION y 3E (a) d dy L d 3E 0 y dy y2 3E 2 L 0 L2 .......................................................................... Ans. 6E 77 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (b) (c) avg RILEY, STURGES AND MORRIS L L max L in./in. ................................................................................................................ Ans. 6E L in./in. ............................................................................................................... Ans. 3E 4-46 A steel cable is used to tether an observation balloon. The force exerted on the cable by the balloon is sufficient to produce a uniform strain of 500 m/m in the cable. In addition, at each point in the cable, the weight of the cable reduces the axial strain by an amount that is proportional to the length of the cable between the balloon and the point. When the balloon is directly overhead at an elevation of 300 m, the axial strain at the midlength of the cable is 350 m/m. Determine (a) The total elongation of the cable. (b) The maximum height that the balloon could achieve SOLUTION 500 10 d dy d (a) (b) If L 6 cy 6 350 10 1 10 6 y 6 500 10 6 c 150 c 1 10 6 500 10 10 6 L 0 500 y dy 10 6 500 y y2 2 L 500 L 0.5 L2 0 10 6 300 m 0.1050 m 105.0 mm ..................................................................................................... Ans. If y 500 m , then 0 . If L 1000 m , then 0 . Neither is possible. Therefore ymax 500 m .............................................................................................................................. Ans. 4-47 A steel cable is used to support an elevator cage at the bottom of a 2000 ft-deep mine shaft. A uniform axial strain of 250 in./in. is produced in the cable by the weight of the cage. At each point the weight of the cable produces an additional axial strain that is proportional to the length of the cable below the point. If the total axial strain in the cable at the cable drum (upper end of the cable) is 700 in./in., determine (a) The strain in the cable at a depth of 500 ft. (b) The total elongation of the cable. SOLUTION 250 10 (a) 500 6 6 6 cy 700 10 0.22500 10 6 6 250 10 6 c 2000 c 0.22500 10 6 250 10 587 10 d dy d 1500 in./in. 587 in./in. .................................................................................. Ans. 6 (b) 250 10 10 6 0.22500 10 6 y 250 0.225 y dy 10 0.1125 2000 2 6 2000 0 0.225 y 2 250 y 2 2000 0 250 2000 10 6 0.9500 ft 11.40 in. ......................................................................................................... Ans. 78 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 4-48 At the proportional limit, a 200 mm-gage length of a 15 mm-diameter alloy bar has elongated 0.90 mm and the diameter has been reduced 0.022 mm. The total axial load carried was 62.6 kN. Determine the modulus of elasticity, Poisson’s ratio, and the proportional limit for the material. SOLUTION E 62.6 103 0.015 E 2 4 E 0.90 200 78.7 109 N/m 2 78.7 GPa ...................................................................................... Ans. 0.022 15 t 0.326 ......................................................................................... Ans. 0.9 200 a p P A 62.6 103 0.015 2 4 354 106 N/m 2 354 MPa ............................................... Ans. 4-49 A 1.50 in.-diameter rod 20 ft long elongates 0.48 in. under a load of 53 kip. The diameter of the rod decreases 0.001 in. during the loading. Determine the modulus of elasticity, Poisson’s ratio, and the modulus of rigidity for the material. SOLUTION E 53 103 1.5 2 4 E 0.48 20 12 E 15.00 106 psi 15, 000 ksi ........................................................................................ Ans. t a 0.001 1.5 0.48 20 12 15, 000 2 1 0.333 0.333 .................................................................................. Ans. 5625 ksi .......................................................................... Ans. G E 21 4-50 A tensile test specimen having a diameter of 5.64 mm and a gage length of 50 mm was tested to fracture. Stress and strain values, which were calculated from load and deformation data obtained during the test, are shown in Fig. P4-50. Determine (a) The modulus of elasticity. (b) The proportional limit. (c) The ultimate strength. (d) The yield strength (0.05% offset). (e) The yield strength (0.2% offset). (f) The fracture stress. (g) The true fracture stress if Poisson’s ratio = 0.30 remains constant. (h) The tangent modulus at a stress level of 400 MPa. (i) The secant modulus at a stress level of 400 MPa. SOLUTION From the stress-strain diagram: (a) (b) (c) E pl ult 140 70.0 103 MPa 70.0 GPa ........................................................................... Ans. 0.002 150 MPa ........................................................................................................................... Ans. 450 MPa .......................................................................................................................... Ans. 79 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (d) (e) (f) (g) t .05 .2 f RILEY, STURGES AND MORRIS 220 MPa ............................. Ans. 275 MPa .............................. Ans. 450 MPa .............................. Ans. a 0.3 0.115 d d d 0.19458 mm df 5.64 0.19458 5.4454 mm Pf f Ai 450 5.64 5.4454 2 2 4 tf Af Af 4 483 MPa ...................................................... Ans. (h) (i) Et Es 430 350 1000 MPa 1.000 GPa .......................................................................... Ans. 0.08 0 400 8890 MPa 8.89 GPa ................................................................................... Ans. 0.045 4-51 A tensile test specimen having a diameter of 0.250 in. and a gage length of 2.000 in. was tested to fracture. Stress and strain values, which were calculated from load and deformation data obtained during the test, are shown in Fig. P4-51. Determine (a) The modulus of elasticity. (b) The proportional limit. (c) The ultimate strength. (d) The yield strength (0.05% offset). (e) The yield strength (0.2% offset). (f) The fracture stress. (g) The true fracture stress if the final diameter of the specimen at the location of the fracture was 0.212 in. (h) The tangent modulus at a stress level of 56 ksi. (i) The secant modulus at a stress level of 56 ksi. SOLUTION From the stress-strain diagram: (a) (b) (c) (d) (e) (f) E pl ult .05 .2 f 28 103 28.0 106 psi 28, 000 ksi ........................................................................ Ans. 0.001 34 ksi ................................... Ans. 74 ksi ................................... Ans. 43 ksi ................................... Ans. 43 ksi .................................... Ans. 65 ksi .................................... Ans. Pf f (g) Ai 65 0.25 0.212 2 2 4 tf Af Af 4 (h) Et 90.4 ksi ............................................................................................................................. Ans. 78 35 537 ksi ............................................................................................................ Ans. 0.08 0 80 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (i) RILEY, STURGES AND MORRIS Es 56 1400 ksi ................................................................................................................ Ans. 0.04 4-52 A tensile test specimen having a diameter of 11.28 mm and a gage length of 50 mm was tested to fracture. Determine (a) The modulus of elasticity. (b) The proportional limit. (c) The ultimate strength. (d) The yield strength (0.05% offset). (e) The yield strength (0.2% offset). (f) The fracture stress. (g) The true fracture stress if the final diameter of the specimen at the location of the fracture was 9.50 mm. (h) The tangent modulus at a stress level of 315 MPa. (i) The secant modulus at a stress level of 315 MPa. SOLUTION From the stress-strain diagram: (a) (b) (c) (d) (e) (f) (g) 222.1 185 103 MPa 0.0012 E 185 GPa ...................Ans. 280 MPa ...............Ans. pl E ult .05 .2 f 507 MPa ...............Ans. 300 MPa ...............Ans. 330 MPa ................Ans. 450 MPa ................Ans. Pf 45.1 103 0.0095 2 tf Af 4 636 106 N/m 2 636 MPa .............................................. Ans. (h) (i) Et Es 320 306 17.5 103 MPa 17.5 GPa ........................................................ Ans. 0.0032 0.0024 315 105 103 MPa 105 GPa ............................................................................. Ans. 0.003 4-53 A tensile test specimen having a diameter of 0.505 in. and a gage length of 2.00 in. was tested to fracture. Determine (a) The modulus of elasticity. (b) The proportional limit. (c) The ultimate strength. (d) The yield strength (0.05% offset). (e) The yield strength (0.2% offset). (f) The fracture stress. (g) The true fracture stress if the final diameter of the specimen at the location of the fracture was 0.425 in. (h) The tangent modulus at a stress level of 46,000 psi. (i) The secant modulus at a stress level of 46,000 psi. SOLUTION From the stress-strain diagram: (a) E 32 103 0.0012 26.7 106 psi 26, 700 ksi ........................................................................ Ans. 81 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (b) (c) (d) (e) (f) (g) pl ult .05 .2 f RILEY, STURGES AND MORRIS 40 ksi .....................Ans. 73 ksi .....................Ans. 44 ksi .....................Ans. 47 ksi ......................Ans. 65 ksi ......................Ans. Pf 13 103 0.425 2 tf Af 4 91.6 103 psi 91.6 ksi ..........................................................................Ans. (h) (i) Et Es 47.43 45.93 1875 ksi ........................................................................Ans. 0.004 0.0032 46 14, 370 ksi ..................................................................................Ans. 0.0032 4-54 A cast iron pipe has an inside diameter of 70 mm and an outside diameter of 105 mm. The length of the pipe is 2.5 m. The coefficient of thermal expansion for cast iron is = 12.1(10 6)/ C. Determine the dimension changes caused by (a) An increase in temperature of 70 C. (b) A decrease in temperature of 85 C. SOLUTION (a) L o i 12.1 10 12.1 10 12.1 10 12.1 10 12.1 10 12.1 10 6 6 6 6 6 6 70 2500 70 105 70 70 2.12 mm .......................................................................... Ans. 0.0889 mm ......................................................................... Ans. 0.0593 mm ............................................................................ Ans. 2.57 mm ..................................................................... Ans. 0.108 mm ...................................................................... Ans. 0.072 mm ......................................................................... Ans. (b) L o i 85 2500 85 105 85 70 4-55 A large cement kiln has a length of 225 ft and a diameter of 12 ft. Determine the change in length and diameter of the structural steel shell [ = 6.5(10 6)/ F] caused by an increase in temperature of 250 F. SOLUTION L d 6.5 10 6.5 10 6 6 250 225 12 250 12 12 4.39 in. ....................................................................... Ans. 0.234 in. ........................................................................ Ans. 4-56 An airplane has a wing span of 40 m. Determine the change in length of the aluminum alloy [ = 22.5(10 6)/ C] wing spar if the plane leaves the ground at a temperature of 40 C and climbs to an altitude where the temperature is 40 C. SOLUTION 22.5 10 6 80 40 72.0 10 3 m 72.0 mm ........................................... Ans. 82 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 4-57 Determine the movement of the pointer of Fig. P4-57 with respect to the scale zero when the temperature increases 80 F. The coefficients of thermal expansion are 6.6(10 6)/ F for the steel and 12.5(10 6)/ F for the aluminum. SOLUTION The steel post and the steel scale each stretch the same amount s o 6.6 10 6 6 80 20 0.01056 in. The aluminum post stretches a 12.5 10 80 20 0.02000 in. Therefore, the motion of the pointer relative to the zero on the steel scale is p a s 5 p 1 0.0472 in. ..................................................................................................................... Ans. 4-58 A bronze [ B = 16.9(10 6)/ C] sleeve with an inside diameter of 99.8 mm is to be placed over a solid steel [ S = 11.9(10 6)/ C] cylinder, which has an outside diameter of 100 mm. If the temperatures of the cylinder and sleeve remain equal, how much must the temperature be increased in order for the bronze sleeve to slip over the steel cylinder? SOLUTION d 100 99.8 T 11.9 10 16.9 10 6 6 T 100 T 99.8 403 C ................................................................................................................................ Ans. 4-59 A steel [E = 30,000 ksi and = 6.5(10 6)/ F] surveyor’s tape ½-in. wide 1/32-in. thick is exactly 100 ft long at 72 F and under a pull of 10 lb. What correction should be introduced if the tape is used to make a 100-ft measurement at a temperature of 100 F and under a pull of 25 lb. SOLUTION E T L L L L L 1 L 10 lb , where L is the length at 72 F and zero force. When P 10 1 2 1 32 640 psi , L 100 ft , and 100 640 1 30 106 L L 99.99787 ft Then, when P 0 25 lb , 25 1 2 1 32 1600 psi , and T 28 F 1600 L 1 6.5 10 6 28 6 30 10 L L 100.02140 ft L 0.02140 ft 0.257 in. correction 0.257 in. ............................................................................................................ Ans. 83 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 4-60 A 25-mm diameter aluminum [ = 22.5(10 6)/ C, E = 73 GPa, and = 0.33] rod hangs vertically while suspended from one end. A 2500-kg mass is attached at the other end. After the load is applied, the temperature decreases 50 C. Determine (a) The axial stress in the rod. (b) The axial strain in the rod. (c) The change in diameter of the rod. SOLUTION (a) 2500 9.81 0.025 a 2 4 49.962 106 N/m 2 22.5 10 49.962 106 73 109 6 50.0 MPa ...................................................... Ans. 6 (b) E E t T T 49.962 106 73 109 0.33 50 441 10 6 6 m/m ...................... Ans. 1350.86 10 6 (c) t 22.5 10 50 m/m d d 1350.86 10 25 0.0338 mm ........................................................... Ans. 4-61 The rigid yokes B and C of Fig. P4-61 are securely fastened to the 2-in. square steel (E 30,000 ksi) bar AD. Determine (a) The maximum normal stress in the bar. (b) The change in length of segment AB. (c) The change in length of segment BC. (d) The change in length of the complete bar. SOLUTION (a) max Pmax A PL EA 82 103 22 30 106 20, 500 psi 20.5 ksi ................................................................. Ans. (b) 82 103 8 12 22 AB 0.0656 in. ........... Ans. (c) 12 103 5 12 BC 30 106 22 0.00600 in. .............. Ans. 45 103 4 12 CD 30 106 AB BC 22 CD 0.01800 in. 0.0776 in. ...................................................................................... Ans. (d) total 4-62 A structural tension member of aluminum alloy (E = 70 GPa) has a rectangular cross section 25 75 mm and is 2 m long. Determine the maximum axial load that may be applied if the axial stress is not to exceed 100 MPa and the total elongation is not to exceed 4 mm. SOLUTION PL EA P Pmax A P 2000 70 10 9 0.025 0.075 4 mm 187,500 N P 262,500 N 100 106 0.025 0.075 187.5 kN ......................................................................................................................... Ans. 84 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 4-63 The tension member of Fig. P4-63 consists of a steel (E = 30,000 ksi) pipe A, which has an outside diameter of 6 in. and an inside diameter of 4.5 in., and a solid aluminum alloy (E = 10,600 ksi) bar B, which has a diameter of 4 in. Determine the overall elongation of the member. SOLUTION PA PB A 120 kip 120 103 3 12 B 120 103 4 12 10.6 106 42 4 30 106 6 2 4.52 4 0.01164 0.04324 0.0549 in. ..................................................................................... Ans. 4-64 A steel (E = 200 GPa) rod, which has a diameter of 30 mm and a length of 1.0 m, is attached to the end of a Monel (E = 180 GPa) tube, which has an internal diameter of 40 mm, a wall thickness of 10 mm, and a length of 2.0 m, as shown in Fig. P4-64. Determine the load required to stretch the assembly 3.00 mm. SOLUTION PS PM P P 1000 P 2000 2 9 200 10 P 0.030 4 180 10 9 0.0602 0.0402 4 3.00 mm 212 103 N 212 kN ...................................................................................................... Ans. 4-65 The floor of a warehouse is supported by an air-dried, red-oak, timber column (see Appendix A for properties), as shown in Fig. P4-65. Contents of the warehouse subject the 12 12-in. column to an axial load of 200 kip. If the column is 30-in. long, determine (a) The deformation of the column. (b) The normal stress in the column. (c) The bearing stress between the column and the lower bearing plate. (d) The maximum shearing stress in the column. SOLUTION E 1800 ksi (a) PL EA P A b 200 103 30 1800 103 12 12 200 103 12 12 0.0231 in. ..................................................................... Ans. (b) 1389 psi ................................................................................................ Ans. (c) 200 103 12 12 P 2A 1389 psi ....................................................................................................... Ans. 694 psi .......................................................................................... Ans. (d) max 200 103 2 12 12 4-66 The roof and second floor of a building are supported by the column shown in Fig. P4-66. The column is a structural steel (see Appendix A for properties) section having a cross sectional area of 5700 mm2. The roof and floor subject the column to the axial forces shown. Determine (a) The amount that the first floor will settle. (b) The amount that the roof will settle. SOLUTION E 200 GPa P 1030 kN (C) 1 P2 380 kN (C) 85 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (a) 1 RILEY, STURGES AND MORRIS PL EA 1 1030 103 3500 200 109 5700 10 380 103 3500 6 3.1623 3.16 mm ............................................. Ans. (b) r 200 109 5700 10 6 3.1623 1.1667 4.33 mm ............................... Ans. 4-67 The tension member of Fig. P4-67 consists of a structural steel pipe A, which has an outside diameter of 6 in. and an inside diameter of 4.5 in., and a solid 2014-T4 aluminum alloy bar B, which has a diameter of 4 in. (see Appendix A for properties). Determine (a) The change in length of the steel pipe. (b) The overall deflection of the member. (c) The maximum normal and shearing stresses in the aluminum bar. SOLUTION Est PA (a) A 29, 000 ksi 205 kip (T) PL EA 29 106 120 103 40 Eal PB 205 103 50 62 4.52 10, 600 ksi 120 kip (T) 4 0.0286 in. ...................................................... Ans. (b) B 10.6 106 A B 42 4 0.0360 in. total 0.0646 in. .................................................................................................... Ans. 4 max 2 (c) max P A P 2A 120 103 4 9550 psi ........................................................................................... Ans. max 2 4770 psi ................................................................................................. Ans. 4-68 An aluminum alloy (E = 73 GPa) tube A with an outside diameter of 75 mm is used to support a 25-mm diameter steel (E = 200 GPa) rod B, as shown in Fig. P4-68. Determine the minimum thickness t required for the tube if the maximum deflection of the loaded end of the rod must be limited to 0.40 mm. SOLUTION total A B 0.40 mm 35 103 300 73 109 A 6 35 103 900 200 109 0.025 2 4 A 1817.40 10 752 di2 di t m2 1817.40 mm 2 4 75 2t 57.54 mm 8.73 mm ................................................................................................................................ Ans. 86 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 4-69 A structural steel (see Appendix A for properties) bar of rectangular cross section consists of uniform and tapered sections as shown in Fig. P4-69. The width of the tapered section varies linearly from 2 in. at the bottom to 5 in. at the top. The bar has a constant thickness of ½ in. Determine the elongation of the bar resulting from application of the 30-kip load P. Neglect the weight of the bar. SOLUTION E b 29, 000 ksi 2 3y in. 60 30 103 25 29 106 2 0.5 0 30 103 29 106 3 3 y 60 0 60 in. dy 3y 2 0.5 60 60 0 0.02586 1.03448 10 40 ln 40 y 0.02586 41.37931 10 4.60517 3.68888 0.0638 in. ............................................................................................................................ Ans. 4-70 Determine the change in length of the homogeneous conical bar of Fig. P4-70 due to its own weight. Express the results in terms of L, E, and the specific weight of the material. The taper of the bar is slight enough for the assumption of a uniform axial stress distribution over a cross section to be valid. SOLUTION F 0: AW 0 R2 W y 3 Vol R2 y 3 R ry L P dy EA E dy L 0 y dy 3E y2 3E 2 L 0 L2 ........................................................ Ans. 6E 4-71 The bar shown in Fig. P4-71 is made of annealed bronze (see Appendix A for properties). In addition to its own weight, the bar is subjected to an axial tensile load P of 5000 lb at its lower end. Determine the elongation of the bar due to the combined effects of its weight and the load P. Let r = 4 in. and L = 60 in. SOLUTION E 15, 000 ksi R W F 0: 4 Vol AW 0.320 lb in 3 y 60 in. 15 R 2 y 60 3 5000 0 y 60 60 42 60 3 R2 5000 3 R 2 y 60 960 87 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS P dy EA 5000 225 15 10 6 dy E 60 0 60 0 5000 3 R 2 y 60 E R2 60 960 dy dy y 60 60 2 3 15 10 2 6 0 y 60 dy 960 225 3 15 10 6 dy y 60 60 0 1 0.02387 y 60 60 0.00711 10 0 6 6 y 60 2 2 0 1536.00 10 0.19892 10 0.225 10 3 3 1 y 60 60 0 3 0.03839 10 0.01280 10 3 in. ................................................................................................................... Ans. 4-72 A hollow brass (E = 100 GPa) tube A with an outside diameter of 100 mm and an inside diameter of 50 mm is fastened to a 50-mm diameter steel (E = 200 GPa) rod B, as shown in Fig. P4-72. The supports at the top and bottom of the assembly and the collar C used to apply the 500-kN load P are rigid. Determine (a) The normal stresses in each of the members. (b) The deflection of the collar C. SOLUTION (a) From equilibrium F 0: B TB 4 0.05 PA 500 0 2 A 4 0.12 0.052 500 103 N where B is a tension stress and A is a compressive stress. Then, expressing that the stretch of the rod is equal to the shrink of the tube B A in terms of the stresses gives 2000 200 109 1.5 A B B A B 1500 100 109 A Substituting back into the equilibrium equation gives 56.588 N/m 2 1.5 A A 56.6 MPa .............................................................................. Ans. 84.9 MPa ............................................................................................ Ans. 56.588 106 1500 100 109 0.849 mm ................................................ Ans. (b) C 88 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 4-73 The 7.5 7.5 20-in. oak (E = 1800 ksi) block shown in Fig. P4-73 was reinforced by bolting two 2 7.5 20-in. steel (E = 29,000 ksi) plates to opposite sides of the block. If the stresses in the wood and the steel are to be limited to 4.6 ksi and 22 ksi, respectively, determine (a) The maximum axial compressive load P that can be applied to the reinforced block. (b) The shortening of the block when the load of part (a) is applied. SOLUTION (a) From equilibrium F 0: w Pw 2 Ps 2 P s 0 2 7.5 P 7.5 7.5 s where w and are both compressive stresses. Expressing that s w the shrink of the steel is equal to the shrink of the wood in terms of the stresses gives L 29, 000 103 s s L 1800 103 w w w 16.111 If s 22 ksi , then 1.36552 ksi 4.6 ksi , and 737 kip ....................................... Ans. ............................................................. Ans. P 1.36552 7.5 7.5 (b) 2 22 2 7.5 0.01517 in. 22 103 20 s 29, 000 103 4-74 Five 25-mm diameter steel (E = 200 GPa) reinforcing bars will be used in a 1-m long concrete (E = 31 GPa) pier with a square cross section, as shown in Fig. P4-74. The allowable strengths in compression for steel and concrete are 130 MPa and 9.5 MPa, respectively. Determine the minimum size of pier required to support a 900-kN axial load. SOLUTION From equilibrium F 5 where 0: s s 5Ps 0.025 c 2 Pc 900 0 4 c Ac 900 103 N and are both compressive stresses. Expressing that the s c shrink of the steel rods is equal to the shrink of the concrete in terms of the stresses gives L cL 9 200 10 31 109 6.4516 c s s If c 9.5 MPa , then Ac 78,902 10 b2 5 b s 6 61.29 MPa 130 MPa , and m2 2 78,902 mm 2 25 4 285 mm ................................................................................................................................ Ans. 89 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 4-75 A hollow steel (E = 30,000 ksi) tube A with an outside diameter of 2.5 in. and an inside diameter of 2 in. is fastened to an aluminum (E = 10,000 ksi) bar B that has a 2-in. diameter over one-half of its length and a 1-in. diameter over the other half. The assembly is attached to unyielding supports at the left and right ends and is loaded as shown in Fig. P4-75. Determine (a) The normal stresses in all parts of the bar. (b) The deflection of cross-section a-a. SOLUTION (a) From equilibrium PB where PA 40 kip PC PA 10 kip PA , PB , and PC are all tension forces. Since the total length of the assembly cannot change, the total stretch must 0 . Therefore be zero A B C PA 20 30 103 PA B PA 40 24 4 A PA 10 24 4 10 103 12 4 2.52 22 10 103 22 0 0 kip 0 40 22 0 10 12 4 4 0 ksi ....................................................................... Ans. 12.73 ksi (T) .................................................................................... Ans. 12.73 ksi 12.73 ksi (C) ............................................................ Ans. C (b) 10 103 24 C 10 106 1 2 4 0.03056 in. ............................................................................ Ans. Therefore section a-a moves 0.03056 in. 4-76 A 150-mm diameter 200-mm long polymer (E = 2.10 GPa) cylinder will be attached to a 45-mm diameter 400-mm long brass (E = 100 GPa) rod by using the flange type of connection shown in Fig. P4-76. A 0.15mm clearance exists between the parts as a result of a machining error. If the bolts are inserted and tightened, determine (a) The normal stresses produced in each of the members. (b) The final position of the flange-polymer interface after assembly, with respect to the left support. SOLUTION (a) From equilibrium Tp where Tb Tp and Tb are both tension forces. When the bolts are tightened, the stretch of the two pieces closes the gap, 0.15 mm . Therefore p b Tp 200 2.1 109 Tp Tb 0.15 2 Tb 4 100 109 400 0.045 2 4 0.15 mm 18,976.74 N 90 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 18,976.74 p 0.15 0.045 2 4 2 1.074 106 N/m 2 1.074 MPa (T) ...................................... Ans. 11.93 106 N/m 2 11.93 MPa (T) .................................... Ans. 0.10227 mm 18,976.74 b 4 (b) 18,976.74 200 p 2.1 109 0.15 2 4 Therefore the interface will be located at 200.1023 mm ........................................................... Ans. 4-77 The assembly shown in Fig. P4-77 consists of a steel bar A (Es = 30,000 ksi and As = 1.25 in.2), a rigid bearing plate C that is securely fastened to bar A, and a bronze bar B (EB = 15,000 ksi and AB = 3.75 in.2). A clearance of 0.015 in. exists between the bearing plate C and bar B before the assembly is loaded. After a load P of 95 kip is applied to the bearing plate, determine (a) The normal stresses in bars A and B. (b) The vertical displacement of the bearing plate C. SOLUTION (a) From equilibrium (assuming that the gap is closed and the bearing plate pushes on the brass bar, F 0: 1.25 A TA 3.75 PB 95 0 B or in terms of stresses 95 103 lb in which A is a tension stress and B is a compression stress. If the bearing plate presses against the brass bar, the stretch of the steel rod will exceed the shrink of the brass bar by the 0.015 in. Therefore initial gap, A B 6 12 30 106 A A 2 12 15 106 B B 0.015 in. 23 6250 psi Combining the equilibrium equation and the deformation equation gives A B 18,932 psi 18.93 ksi (T) ................................................................................. Ans. 19, 023 psi 19.02 ksi (C) ................................................................................. Ans. 18,932 6 12 30 106 0.0454 in. ....................................................................... Ans. (b) C A 4-78 A column similar to Fig. P4-74 is being designed to carry a load of 4450 kN. The column, which will have a 500 500 mm square cross section, will be made of concrete (E = 20 GPa) and will be reinforced with 50mm diameter steel (E = 200 GPa) bars. If the allowable stresses are 120 MPa in the steel and 8 MPa in the concrete, determine (a) The number of steel bars required. (b) The stresses in the steel and concrete when the bars of part (a) are used. (c) The change in length of a 3-m long column when the bars of part (a) are used. SOLUTION (a) From equilibrium 91 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS F Pc 0: nPs Pc nPs 4450 0 4450 kN Expressing that the steel rods and the concrete must shrink the same amount, c s , in terms of stresses gives L 20 109 c If c max L 200 109 8 MPa , then s s 10 c s 80 MPa 120 MPa , and 2 8 106 n 17.33 0.52 n 0.05 4 n 80 10 6 0.05 2 4 4450 103 Therefore the required number of rods is n (b) If n 18 .......................................................................... Ans. 18 , then c 0.52 18 0.05 2 4 18 10 c 0.05 2 4 4450 103 c s 7.83 106 N/m 2 10 s c s 7.83 MPa ............................................................................. Ans. 78.3 106 3000 200 109 78.3 MPa .................................................................................................. Ans. 1.175 mm ........................................ Ans. (c) 3000 200 109 4-79 A ½-in. diameter alloy-steel bolt (E = 30,000 ksi) passes through a cold-rolled brass sleeve (E = 15,000 ksi) as shown in Fig. P4-79. The cross-sectional area of the sleeve is 0.375 in.2 Determine the normal stresses produced in the bolt and sleeves by tightening the nut ¼ turn (0.020 in.). SOLUTION From equilibrium Tst Fbr 0.5 2 or in terms of stresses 4 st br 0.375 br st 1.90986 in which st is a tension stress and br is a compression stress. The stretch of the steel bolt, the shrink of the st brass sleeve, and the movement of the nut are related by 0.02 st br . Therefore br 6 30 106 st br st 0.02 6 15 106 br 100, 000 2 Combining the equilibrium equation and the deformation equation gives 25,576 psi 48,847 psi 25.6 ksi (C) .................................................................................. Ans. 48.8 ksi (T) .................................................................................. Ans. 92 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 4-80 The two faces of the clamp shown in Fig. P4-80 are 250 mm apart when the two stainless-steel (Es = 190 GPa) bolts connecting them are unstretched. A force P is applied to separate the faces of the clamp so that an aluminum alloy (Ea = 73 GPa) bar with a length of 251 mm can be inserted as shown. Each of the bolts has a cross-sectional area of 120 mm2 and the bar has a cross-sectional area of 625 mm2. After the load P is removed, determine (a) The axial stresses in the bolts and in the bar. (b) The change in length of the aluminum alloy bar. SOLUTION (a) Expressing the equilibrium equation Fa 2Ts 6 a a a s in terms of stresses gives 625 10 s 2 120 10 6 s 2.60417 s in which is a tension stress and is a compression stress. The stretch of the steel bolts and the shrink of the aluminum bar are related by 1 a Therefore s 330 190 109 s s a 1 251 73 109 a 575.76 106 1.97966 a Combining the equilibrium equation and the deformation equation gives 327 106 N/m 2 125.6 106 N/m 2 125.6 106 251 9 327 MPa ................................................................................ Ans. 125.6 MPa ......................................................................... Ans. 0.432 mm ....................................................................... Ans. (b) a 73 10 4-81 A high-strength steel bolt (Es = 30,000 ksi and As = 0.785 in.2) passes through a brass sleeve (Eb = 15,000 ksi and Ab = 1.767 in.2), as shown in Fig. P4-81. As the nut is tightened, it advances a distance of 0.125 in. along the bolt for each complete turn of the nut. Compute and plot (a) The axial stresses s (in the steel bolt) and b (in the brass sleeve) as functions of the angle of twist of the nut (0 180 ). (b) The elongations s (of the steel bolt) and b (of the brass sleeve) as function of (0 180 ). (c) The distance L between the two washers as a function of (0 180 ). SOLUTION (a) From equilibrium Tst Fbr st or in terms of stresses 0.785 st 1.767 br br 2.25096 st in which is a tension stress and br is a compression stress. The stretch of the steel bolt, the shrink of st the brass sleeve, and the movement of the nut are related by degrees. Therefore 0.125 360 br in which is in 14 30 106 st 0.125 360 12 15 106 br st 744.0476 1.71429 br Combining the equilibrium equation and the deformation equation gives 93 STATICS AND MECHANICS OF MATERIALS, 2nd Edition st br RILEY, STURGES AND MORRIS 422.374 psi (T) ................... Ans. 187.642 psi (C) ................... Ans. st (b) br 14 30 106 br 12 15 106 st 0.19711 10 0.15011 10 3 in. ....................................................................... Ans. in. ...................................................................... Ans. 3 (c) L 12 br 12 0.15011 10 3 in. .................................................................. Ans. 4-82 The short pier shown in Fig. P4-82 is reinforced with nine steel (E = 210 GPa) reinforcing bars. An axial compressive load P is applied to the pier through the rigid capping plate. The axial load carried by the matrix material is a function of R, the percentage of the cross section taken up by the steel reinforcement bars. The load is also a function of the modulus ratio ER/EM where ER and EM are the modulus of elasticity for the reinforcement material and the matrix material, respectively. For the three matrix-reinforcement combinations listed, compute and plot the percentage of the load carried by the matrix as a function of R (0 R 100 %). SOLUTION From equilibrium F 0: PM 9 Ps PM P 9 Ps P 0 Since the steel rods and the pier must shrink the same amount, M s , PM L EM AM Ps L Es As Ps Es As PM EM AM 94 STATICS AND MECHANICS OF MATERIALS, 2nd Edition Combining the equilibrium equation and the deformation equation gives RILEY, STURGES AND MORRIS PM 1 9 Then, since Es As EM AM P R 9 As 100 A AM 9 As 9 As 100 AM 9 As 100 AM 9 As 1 100 1 R R 1 100 R 1 9 As AM and 1 100 R PM P Es R 1 EM 100 R ..............Ans. 4-83 A 3-in. diameter 80-in. long aluminum alloy bar is stress free after being attached to rigid supports, as shown in Fig P4-83. Determine the normal stress in the bar after the temperature drops 100 F. Use E = 10,600 ksi and = 12.5(10 6)/ F. SOLUTION E T 10.6 10 6 12.5 10 6 100 0 13.25 103 psi 13.25 ksi (T) ............................................................................ Ans. 4-84 A 6-m long 50-mm diameter rod of aluminum alloy [E = 70 GPa, = 0.346, and = 22.5(10 6)/ C] is attached at the ends to supports that yield to permit a change in length of 1.00 mm in the rod when stressed. When the temperature is 35 C, there is no stress in the rod. After the temperature of the rod drops to 20 C, determine (a) The normal stress in the rod. (b) The change in diameter of the rod. SOLUTION (a) L E L E TL 6000 70 109 22.5 10 6 55 6000 1 mm 74.958 106 N/m 2 (b) d 75.0 MPa (T) ................................................................... Ans. TL 74.958 106 50 9 55 50 70 10 0.0804 mm 0.0804 mm (shrink) ............................................................... Ans. 0.346 22.5 10 6 95 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 4-85 A bar consists of 3-in. diameter aluminum alloy [E = 10,600 ksi, = 0.33, and = 12.5(10 6)/ F] and 4-in. diameter steel [E = 30,000 ksi, = 0.30, and = 6.6(10 6)/ F] parts, as shown in Fig. P4-85. If end supports are rigid and the bar is stress free at 0 F, determine (a) The normal stress in both parts of the bar at 80 F. (b) The change in diameter of the steel part of the bar. SOLUTION (a) From equilibrium Ps Pa Ps and Pa are compressive forces. in which both Since the end supports are rigid, the total stretch of the rod 0 must be zero s a Ps 20 30 10 6 4 2 4 6.6 10 6 80 20 Pa 30 10.6 10 Ps s 6 3 2 4 12.5 10 6 80 30 0 Pa 89, 449 lb (C) 4 3 2 89, 449 4 4 89, 449 a 2 7118 psi 7.12 ksi ........................................................................ Ans. 12, 654 psi 12.65 ksi ................................................................... Ans. TL 7118 4 30 106 6.6 10 6 (b) ds L E 0.3 80 4 0.00240 in. .......................... Ans. 4-86 A steel tie rod containing a rigid turnbuckle (see Fig. P4-86) has its ends attached to rigid walls. During the summer when the temperature is 30 C, the turnbuckle is tightened to produce a stress in the rod of 15 MPa. Determine the normal stress in the rod in the winter when the temperature is 10 C. Use E = 200 GPa and = 11.9(10 6)/ C. SOLUTION 30 10 L E 9 TL 6 15 106 L 200 10 0 L 11.9 10 200 109 40 L 110.2 106 N/m 2 110.2 MPa ........................................................................... Ans. 4-87 Nine ¾-in. diameter steel (E = 30,000 ksi) reinforcing bars were used when the short concrete (E = 4500 ksi) pier shown in Fig. P4-87 was constructed. After a load P of 150 kip was applied to the pier, the temperature increased 100 F. The coefficients of thermal expansion for steel and concrete are 6.6(10 6)/ F and 6.0(10 6)/ F, respectively. Determine (a) The normal stresses in the concrete and in the steel bars after the temperature increases. 96 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS (b) The change in length of the pier resulting from the combined effects of the temperature change and the load. SOLUTION (a) The equilibrium equation F 0: 100 9 s Pc 9 Ps 150 0 0.75 c 2 is written in terms of stresses 4 c 9 0.75 2 4 s 150 103 lb 24.1504 c 37, 725.62 psi , in which and s are both compressive stresses. Expressing that c s the steel rods and the concrete must stretch the same amount, in terms of stresses gives L 6.0 10 6 100 L 6 4.5 10 6.6667 c 1800.00 psi s c c s L 30 106 s 6.6 10 6 100 L Combining the equilibrium equation and the deformation equation gives 1165.770 psi 1.166 ksi (C) ............................................................................. Ans. 9571.805 psi 9.57 ksi (C) ............................................................................... Ans. 6.6 10 6 (b) 9571.805 24 30 106 100 24 0.00818 in. ............................ Ans. 4-88 The assembly shown in Fig. P4-88 consists of a steel (E = 210 GPa) cylinder A, a rigid bearing plate C, and an aluminum alloy (E = 71 GPa) bar B. Cylinder A has a cross-sectional area of 1850 mm2, and bar B has a cross-sectional area of 2500 mm2. After an axial load of 600 kN is applied, the temperature of cylinder A decreases 50 C and the temperature of bar B increases 25 C. The coefficients of thermal expansion are 11.9(10 6)/ C for the steel and 22.5(10 6)/ C for the aluminum. Determine (a) The normal stresses in the cylinder and in the bar after the load is applied and the temperatures change. (b) The displacement of plate C after the load is applied and the temperatures change. SOLUTION (a) The equilibrium equation F 0: TB TA 600 0 6 A B can be written in terms of stresses 1850 10 A 2500 10 6 B 600 103 N 1.35135 A 324.324 106 N/m 2 in which A B and B are both tension stresses. The total stretch of the assembly must be zero, 0 , therefore A 750 210 109 A 11.9 10 B 6 50 750 600 71 109 B 22.5 10 6 25 600 0 2.36620 30.450 106 N/m 2 217 MPa (T) .................................................................... Ans. 97 Combining the equilibrium equation and the deformation equation gives A 217.5 106 N/m 2 STATICS AND MECHANICS OF MATERIALS, 2nd Edition B RILEY, STURGES AND MORRIS 79.051 106 N/m 2 A 79.1 MPa (C) ............................................................... Ans. 11.9 10 6 (b) 217.5 106 750 C 210 10 9 50 750 C 0.331 mm ......................................................................................................... Ans. 4-89 A high-strength steel [Es = 30,000 ksi, As = 0.785 in.2, and s = 6.6(10 6)/ F] bolt passes through a brass [Eb = 15,000 ksi, Ab = 1.767 in.2, and b = 9.8(10 6)/ F] sleeve, as shown in Fig. P4-89. After the unit is assembled at 40 F, the temperature is increased to 100 F. If the unit is free of stress at 40 F, determine the normal stresses in the bolt and in the sleeve at 100 F. SOLUTION Writing the equilibrium equation Tst Fbr st in terms of stresses gives 0.785 in which st 1.767 br br is a tension stress and is a compressive stress. Expressing that the steel bolt and the brass sleeve must stretch the same amount, st br , in terms of stresses gives 14 6.6 10 6 60 14 6 30 10 14 st 24 br 45,360 psi st st br 12 15 106 br 9.8 10 6 60 12 Combining the equilibrium equation and the deformation equation gives 1839.29 psi 1839 psi (T) ................................................................................ Ans. 817.10 psi 817 psi (C) ................................................................................... Ans. 4-90 The two faces of the clamp of Fig. P4-90 are 250 mm apart when the two stainless-steel [Es = 190 GPa, As = 115 mm2 (each), and s = 17.3(10 6)/ C] bolts connecting them are unstretched. A force P is applied to separate the faces of the clamp so that an aluminum alloy [Ea = 73 GPa, Aa = 625 mm2, and a = 22.5(10 6)/ C] bar with a length of 250.50 mm can be inserted as shown. After the load P is removed, the temperature is raised 100 C. Determine the normal stresses in the bolts and in the bar, and the distance between the faces of the clamps. SOLUTION Writing the equilibrium equation Fa 625 in which s 2Ts a in terms of stresses gives 2 115 s a is a tension stress and is a compressive stress. Expressing that the steel bolts must stretch 0.5 mm more than the aluminum bar must stretch, s 0.5 a , in terms of stresses gives 98 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 330 190 109 s 17.3 10 a 6 100 330 22.5 10 6 0.5 1.73684 s 250.5 73 109 3.43151 a 100 250.5 492.725 106 N/m 2 60.4 MPa (C) ................................................................. Ans. 164.3 MPa (T) ............................................................. Ans. Combining the equilibrium equation and the deformation equation gives a s 60.448 106 N/m 2 164.262 106 N/m 2 Then, the stretch of the aluminum bar is 60.448 106 250.5 a 73 109 250.5 a 22.5 10 6 100 250.5 0.35620 mm and the distance between the faces of the clamp is L 250.856 mm .................................................................................... Ans. 4-91 A prismatic bar [E = 10,000 ksi and = 12.5(10 6)/ F], free of stress at room temperature, is fastened to rigid walls at its ends. One end of the bar is heated 200 F above room temperature while the other end is maintained at room temperature. The change in temperature T along the bar is proportional to the square of the distance from the unheated end. Determine the normal stress in the bar after the change in temperature. SOLUTION d d L E dx E L 0 T dx dx E L3 3 L 0 T 200 x 2 dx L2 0 200x 2 L2 200 L2 0 200 12.5 10 6 10 106 200 E 3 3 8333 psi 8.33 ksi (C) ...................................................................................... Ans. 4-92 The two faces of the clamp shown in Fig. P4-90 are 250 mm apart when the two steel bolts [Es = 190 GPa, As = 115 mm2 (each), and s = 17.3(10 6)/ C] bolts connecting them are unstretched. A force P is applied to separate the faces of the clamp so that an brass bar [Eb = 100 GPa, Ab = 625 mm2, and b = 17.6(10 6)/ C] bar with a length of 250.50 mm can be inserted as shown. After the load P is removed, the temperature of the system is slowly raised. Compute and plot (a) The axial stress s in the steel bolts and the axial stress b in the brass bar as functions of the temperature rise T (0 T 100 C). (b) The elongation s of the steel bolts and the elongation b of the brass bar as functions of the temperature rise T (0 T 100 C). (c) The distance L between the faces of the clamp as a function of the temperature rise T (0 T 100 C). SOLUTION (a) Writing the equilibrium equation 99 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS Pb 625 in which s 2Ts b in terms of stresses gives 2 115 s b is a tension stress and is a compressive stress. Expressing that the steel bolts must stretch 0.5 mm more than the brass bar must stretch, s 0.5 b , in terms of stresses gives 330 190 109 s 17.3 10 b 6 T 330 17.6 10 6 0.5 1.73684 s 250.5 100 109 2.5050 b T 250.5 500 106 1.3002 106 T N/m 2 106 N/m 2 (C) ......................................................... Ans. 106 N/m 2 (T) ........................................................ Ans. Combining the equilibrium equation and the deformation equation gives b s 69.207 0.17997 T 188.063 0.48904 T (b) Then, the elongation of the steel bolts and the brass bar are s 188.063 0.48904 T 190 109 0.32664 4.860 10 3 106 330 17.3 10 6 T 330 T mm ............................................................................. Ans. 106 250.5 3 b 69.207 0.17997 T 100 109 0.17336 4.860 10 17.6 10 Ans. 6 T 250.5 T mm (c) (both stretches). Finally, the distance between the faces of the clamp is L 250.5 b 6 250.32664 4.860 10 T mm .......................................................................... Ans. 100 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 4-93 An aluminum (Eal = 10,000 ksi, al = 12.5 10-6/ F, Aal = 1.4 in.2) bolt passes through a steel (Est = 30,000 ksi, -6 2 st = 6.6 10 / F, Ast = 0.4 in. ) sleeve as shown in Fig. P4-89. Initially, the nut is tightened against the washer at room temperature until the bolt has a tensile force of 3500 lb. Then the temperature of the assembly is slowly raised. Calculate and plot: (a) The stress al in the aluminum bolt and the stress st in the steel sleeve as a function of the temperature increase T (0 F < T < 100 F). (b) The change in length of the aluminum bolt al and the change in length of the steel sleeve st as a function of the temperature increase T (0 F < T < 100 F). SOLUTION (a) Writing the equilibrium equation Tal 1.4 in which Pst al al in terms of stresses gives 0.4 st st is a tension stress and is a compressive stress. The aluminum bolt must stretch more than al st nut the steel sleeve stretches by the amount that the nut moves, . In terms of stresses this gives 6 14 10 106 al 12.5 10 6 T 14 12 30 106 st 6.6 10 T 12 nut When T al 0 the tension in the bolt and the compression in the sleeve are each equal to 3500 lb and 3500 3500 2500 psi 8750 psi st 1.4 0.4 and the initial movement of the nut is nut 0.00700 in. 4 70, 000 958.00 T psi Therefore, the deformation equation can be written 14 al st Combining the equilibrium equation and the deformation equation gives al st 2500 34.214 T psi (T) .............................................................................. Ans. 8750 119.750 T psi (C) ............................................................................ Ans. If the assembly is not welded together, neither stress can be negative. Therefore, the above equations are valid only for T less than about 73 . For T greater than about 73 , the bolt and the sleeve will separate, and both stresses will be zero al st 0 for T 73 .................................................................................. Ans. 101 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (b) Then, the change in length of the aluminum bolt and the steel sleeve are al RILEY, STURGES AND MORRIS st 14 10 106 st 12 30 106 al 12.5 10 6.6 10 6 T 14 in. (stretch) ............................................. Ans. T 12 in. (stretch) ............................................. Ans. 6 4-94 A short standard-weight steel pipe (see Appendix A) is used to support an axial compressive load of 100 kN. If yielding ( y = 250 MPa) should not occur, and the factor of safety is to be 1.6, determine the smallest nominal diameter pipe that may be used to support the load. SOLUTION 100 103 A Use 250 106 1.6 (for which A 640 10 6 m2 d 51 mm A 693.5 mm 2 ) ................................................................ Ans. 4-95 A short column made of structural steel is used to support the floor beams of a building, as shown in Fig. P495. Each floor beam (A and B) transmits a force of 40 kip to the column. The column has the shape of a wide-flange (W) section (see Appendix A). The factor of safety based on failure by yielding is 3.0. Select the lightest wide-flange section that will support the given loads. SOLUTION y 36 ksi FS 3.0 80 103 36 103 A 6.667 in.2 3.0 A W6 25 , W8 24 , W10 30 , W12 30 , etc. all have sufficiently large area. The lightest section is W8 24 .................................................................................................................................... Ans. 4-96 The two structural steel (see Appendix A) rods A and B shown in Fig. P4-96 are used to support a mass m = 2000 kg. If failure is by yielding and a factor of safety of 1.75 is specified, determine the diameters of the rods (to the nearest 1 mm) that must be used to support the mass. Both rods are to have the same diameter. SOLUTION y 250 MPa FS 1.75 From a free body diagram of the connection, the equilibrium equations Fx Fy give 0: 0: TA sin 50 TB cos 30 TB sin 30 250 106 A 1.75 250 106 A 1.75 TA cos 50 2000 9.81 0 0 TA 17, 254 N TB 12,806 N A 120.8 10 A 89.6 10 6 6 m2 m2 If the rods are made from the same material and diameter, then Use d2 A 120.8 mm 2 d 12.4 mm 4 d 13 mm ........................................................................................................................ Ans. 102 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 4-97 The machine component shown in Fig. P4-97 is made of hot-rolled Monel. The forces at B are applied to the component with a rigid collar that is firmly attached to the component. If the mode of failure is yielding and the factor of safety is 1.5, determine the minimum permissible diameter of each segment of the machine component. SOLUTION y 50 ksi 20 103 d2 4 30 103 d2 4 FS 1.5 FAB 20 kip (T) d AB 50 103 1.5 50 103 1.5 0.874 in. ................................................................................................................ Ans. FBC 30 kip (C) d BC 1.070 in. ................................................................................................................. Ans. 4-98 An axial load P = 1000 kN is applied to the rigid steel bearing plate on the top of the short column shown in Fig. P4-98. The outside segment of the column is made of structural steel. The inside core is made of fairly high strength concrete. Both segments are square. The failure modes are yielding for the steel and fracture for the concrete. The factor of safety is to be 1.4. If the area of the concrete is to be 10 times the area of the steel, determine the required dimensions. SOLUTION fc 34 MPa 1.4 ys 250 MPa 10 As FS Ac 1000 kN Ac 1000 103 N Writing the equilibrium equation Fs s Fc As s in terms of stresses gives c in which c and c s are both compressive stresses. Expressing that the steel and the concrete must stretch the s same amount, , in terms of stresses gives s If c fc L cL 9 200 10 31 109 34 MPa , then s 6.4516 c 6.4516 c 219.355 MPa 250 MPa 1000 103 N 10 As 17,877.7 10 6 Therefore 219.355 106 As As 1787.77 10 6 34 106 10 As m2 Ac m2 and the required sizes are concrete 133.7 mm 133.7 mm ............................................................................. Ans. b 140.2 mm 140.2 mm 140.2 mm ............................................................................. Ans. b 2 17,877.7 1787.77 steel 103 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 4-99 Four axial forces are applied to the 1-in. thick, 0.4% C hot-rolled steel bar, as shown in Fig. P4-99. The factor of safety for failure by yielding is 1.75. Determine the minimum width w of the constant crosssectional area bar. SOLUTION y 53 ksi 20 103 1w 10 103 1w 50 103 1w FS 1.75 FAB FBC FBC 20 kip (T) 10 kip (C) 50 kip (T) 53 103 1.75 53 103 1.75 53 103 1.75 w 0.660 in. w 0.330 in. w 1.651 in. w 1.651 in. ..................................................................................................................... Ans. 4-100 The two parts of the eyebar shown in Fig. P4-100 are connected by two bolts (one on each side of the eyebar). The bolts are made of a grade of steel with a tensile yield strength of 1035 MPa and a shear yield strength of 620 MPa. The eyebar is subjected to the forces P = 85 kN. Determine the minimum bolt diameter required to safely support the forces if the mode of failure is yielding and the factor of safety is 1.5. SOLUTION Since there are two bolts, there are two normal forces and two shear forces shown on the free-body diagram. The equilibrium equations gives Fn Ft give 0: 0: 2 N 85 cos 30 2V 85sin 30 0 0 N V d min 85 103 cos 30 2 85 103 sin 30 2 36.806 10 21.25 103 3 1035 106 1.5 620 106 1.5 d2 4 d2 4 d d 0.00824 m 0.00809 m 8.24 mm ............................................................................................................... Ans. 4-101 The two solid rods shown in Fig. P4-101 are pin-connected at the ends and support a weight of 10 kip. The rods are made of SAE 4340 heat-treated steel. The factor of safety for failure by yielding is to be 1.5. For a minimum weight of rod design, determine (a) The optimum angle . (b) The required diameter for the rods. (c) The weight of each rod. Is it reasonable to neglect the weight of the rods in the design? SOLUTION y 132 ksi 0.283 lb/in.3 FS 1.5 (a) From a free body diagram of the pin connection, the equilibrium equations Fx Fy 0: 0: T2 cos T1 sin T1 cos T2 sin 0 10, 000 0 104 STATICS AND MECHANICS OF MATERIALS, 2nd Edition give RILEY, STURGES AND MORRIS T1 T2 10, 000 2sin 132 103 A 1.5 A 56.818 10 sin 3 in.2 The optimum angle is the angle that makes the volume (and therefore the weight and cost) of the rods a minimum. The volume of each rod is V AL 56.818 10 sin 3 25 12 cos 1 17.04545 sin cos 34.0909 sin 2 The minimum volume occurs when sin 2 (b) When 45 .............................................................................................................................. Ans. 45 A d (c) When d 2 56.818 10 3 in.2 4 sin 45 0.31986 in. 0.320 in. .......................................................................................... Ans. 45 V 0.283 34.0909 sin 90 9.65 lb .................................................................... Ans. W Yes, it is safe to neglect the weight of the rods..................................................................... Ans. 4-102 A tension member consists of a 50-mm diameter brass (E = 100 GPa) bar connected to a 32-mm diameter stainless steel (E = 190 GPa) bar, as shown in Fig. P4-102. For an applied load P = 50 kN, determine (a) The normal stresses in each segment of the member. (b) The elongation of the member. SOLUTION FAB (a) AB FBC 50 kN 2 50 103 0.05 4 2 25.5 106 N/m 2 62.2 106 N/m 2 25.5 MPa ............................................... Ans. 62.2 MPa ............................................. Ans. 50 103 BC 0.032 4 (b) 50 103 1500 100 109 0.05 2 50 103 1000 4 190 109 0.032 2 4 0.709 mm ...... Ans. 4-103 An alloy steel (E = 30,000 ksi) bar is loaded and supported as shown in Fig. P4-103. The loading collar at B is free to slide on section BC. The diameters of sections AB, BC, and CD are 2.50 in., 1.50 in., and 1.00 in., respectively. The lengths of all three segments are 15 in. Determine the normal stresses in each section and the overall change in length of the bar. SOLUTION TAB TBC TCD 20 kip 60 kip 10 kip 105 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 20 103 AB 2.5 1.5 1 2 2 4 4 4074 psi 33,950 psi 4.07 ksi (C) ........................................................ Ans. 33.9 ksi (T) ..................................................... Ans. 60 103 BC 2 10 103 CD 4 12, 730 psi 12.73 ksi (T) ..................................................... Ans. 60 103 15 4 30 106 1.5 2 20 103 15 30 106 2.5 2 10 103 15 4 30 106 1 2 4 0.0213 in. .................................................................................................................. Ans. 4-104 The floor beams of a storage shed are supported as shown in Fig. P4-104. Each of the floor beams B and C transmits a 50 kN load to post A. Post A, the baseplate, and the footing have cross-sectional areas of 15,000 mm2, 30,000 mm2, and 260,000 mm2, respectively. Determine (a) The normal stress in post A. (b) The bearing stress between the post and the baseplate. (c) The bearing stress between the baseplate and the footing. (d) The bearing stress between the footing and the ground. SOLUTION (a) n 100 103 15, 000 10 100 103 15, 000 10 100 103 30, 000 10 6 6.67 106 N/m 2 6.67 106 N/m 2 3.33 106 N/m 2 385 103 N/m 2 6.67 MPa ............................................... Ans. 6.67 MPa ............................................... Ans. 3.33 MPa ............................................... Ans. 385 kPa ................................................. Ans. (b) b 6 (c) b 6 (d) b 100 103 260, 000 10 6 4-105 A 1-in.-diameter steel [ = 6.5(10 6)/ F, E = 30,000 ksi, and = 0.30] bar is subjected to a temperature decrease of 150 F. The ends of the bar are supported by two walls that displace a small amount during the temperature change. If the measured strain in the bar is 600 in./in. after the temperature change, determine the load being transmitted to the walls. SOLUTION E 6 T 600 10 6 6 30 10 11, 250 psi 11.25 ksi (T) P A 11.25 1 2 6.5 10 150 600 10 6 4 8.84 kip ................................................................... Ans. 106 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 4-106 A 90-mm diameter brass (E = 100 GPa) bar is securely fastened to a 50-mm diameter steel (E = 200 GPa) bar. The ends of the composite bar are then attached to rigid supports, as shown in Fig. P4-106. Determine the stresses in the brass and the steel after a temperature drop of 70 C occurs. The thermal coefficients of expansion for the brass and the steel are 17.6(10 6)/ C and 11.9(10 6)/ C, respectively. SOLUTION Writing the equilibrium equation Fb Fs 90 2 in terms of stresses gives 4 b 50 2 4 s s 3.24 b in which b s s and b are both tensile stresses. Since the supports are rigid, the total stretch of the bar must be zero, 0 . In terms of stresses 800 100 109 b 17.6 10 s 6 70 800 480 200 109 s 11.9 10 6 70 480 0 8 b 2.4 1.38544 109 N/m 2 87.8 MPa (T) .................................................................... Ans. 285 MPa (T) ..................................................................... Ans. Combining the equilibrium equation and the deformation equation gives b s 87.82 106 N/m 2 284.5 106 N/m 2 4-107 A steel (E = 30,000 ksi) pipe column with an outside diameter of 3 in. and an inside diameter of 2.5 in. is attached to unyielding supports at the top and bottom as shown in Fig. P4-107. A rigid collar C is used to apply a 50-kip load P. Determine (a) The normal stresses in the top and bottom portions of the pipe. (b) The deflection of the collar C. SOLUTION A (a) 32 2.52 4 2.15984 in.2 Writing the equilibrium equation PB TA 50 0 in terms of stresses gives 2.15984 B 2.15984 A 50 103 lb in which A is a tension stress and B is a compressive stress. Since the supports are rigid, the stretch of the top section must be the same as the shrink of the bottom section, A B . In terms of stresses 7 12 30 106 A A B 4 12 30 106 B B 1.75 A Combining the equilibrium equation and the deformation equation gives 8418 psi 8.42 ksi (T) ....................................................................................... Ans. 14, 732 psi 14.73 ksi (C) ................................................................................. Ans. A (b) C 8418 7 12 30 106 0.0236 in. ................................................................ Ans. 107 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 4-108 A 3-mm diameter cord (E = 7 GPa) that is covered with a 0.5-mm thick plastic sheath (E = 14 GPa) is subjected to an axial tensile load P, as shown in Fig. P4-108. The load is transferred to the cord and sheath by rigid blocks attached to the ends of the assembly. The yield strengths for the cord and sheath are 15 MPa and 56 MPa, respectively. Determine the maximum allowable load if a factor of safety of 3 with respect to failure by yielding is specified. SOLUTION Equilibrium gives that the total force is the sum of the force carried by the cord and the force carried by the sheath P PC PS C S The cord and the sheath must both stretch the same amount, . In terms of stresses If C L SL 9 7 10 14 109 15 3 5 MPa , then C max C S S 2 C 2 C 10 MPa 56 18.667 MPa 3 2 Therefore Pmax Pmax 5 106 0.003 4 10 106 0.0042 0.0032 4 90.3 N ................................................................................................................... Ans. 4-109 The assembly shown in Fig. P4-109 consists of a steel (Es = 30,000 ksi, As = 1.25 in.2) bar A, a rigid bearing plate C that is securely fastened to bar A, and a bronze (Eb = 15,000 ksi, Ab = 3.75 in.2) bar B. A clearance of 0.025 in. exists before the assembly is loaded by a force P = 15 kip. Determine, for each segment of the assembly (a) The normal stress. (b) The change in length. SOLUTION (a) Writing the equilibrium equation TA 1.25 PB 15 0 A in terms of stresses gives 3.75 B 15 103 lb in which A is a tension stress and B is a compressive stress. Assuming that the force is sufficient to close the gap, the stretch of bar A must be greater than the shrink of bar B by the amount of the clearance, 0.025 in. In terms of stresses A B 6 12 30 106 A 2 12 15 106 B B 0.025 in. 72 A B A 48 750 103 psi Combining the equilibrium equation and the deformation equation gives 10.7045 103 psi 10.70 ksi (T) ..................................................................... Ans. 431.8 psi 0.432 ksi (C) ................................................................................... Ans. 0.0257 in. (stretch) .............................................. Ans. 10.7045 103 6 12 (b) A 30 106 108 STATICS AND MECHANICS OF MATERIALS, 2nd Edition B RILEY, STURGES AND MORRIS 431.8 2 12 15 106 0.000691 in. (shrink) ......................................................... Ans. 4-110 An 80-kN force P is applied to the 150 180-mm wood block shown in Fig. P4-110. Determine the normal stress perpendicular to the grain of the wood and the shearing stress parallel to the grain of the wood. SOLUTION From a free-body diagram, the equilibrium equations Fn Ft give 0: 0: N V A N 80sin 25 V 80 cos 25 0 0 80sin 25 80 cos 25 n n An An An sin 25 where the areas are related by 0.150 0.180 mm 2 80 103 sin 2 25 Therefore n 0.150 0.180 0.150 0.180 529 103 N/m 2 0.529 MPa ..................................... Ans. 1.135 MPa ....................... Ans. 80 103 sin 25 cos 25 n 1.135 106 N/m 2 4-111 A 4000-lb force P is applied to the square structural steel block shown in Fig. P4-111. Determine (a) The change in length of the block. (b) The maximum normal stress in the block. (c) The maximum shearing stress in the block. (d) The normal stress perpendicular to the plane A-A. (e) The shearing stress parallel to the plane A-A. SOLUTION sin 35 An cos 33 0.322 10 3 cos An 45 11.25 in.2 A 33 (a) PL EA max 4000 21 29 10 6 in. ......................................................... Ans. (b) (c) max P 4000 A 33 P max 2A 2 444 psi .......................................................................................... Ans. 222 psi ....................................... Ans. 0 0 From a free-body diagram, the equilibrium equations Fn Ft give 0: 0: N N 4000 cos V 4000sin 4000 4 5 n 11.25 109 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (d) n RILEY, STURGES AND MORRIS 284 psi ..................................................................................................................... Ans. 4000 3 5 n V (e) n 11.25 213 psi ...................................................................................................................... Ans. 110 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS .Chapter 5 5-1 Two forces are applied to a bracket as shown in Fig. P5-1. Determine (a) The moment of force F1 about point O. (b) The moment of force F2 about point O. SOLUTION (a) (b) 5-2 M1 = 10 (10 ) = 100 lb ⋅ in. M 2 = 25 ( 21) = 525 lb ⋅ in. .........................................................................................Ans. ........................................................................................Ans. Determine the moments of the 225-N force shown in Fig. P5-2 about points A, B, and C. SOLUTION M A = 225 (0.6 ) = 135 N ⋅ m M B = 225 (0.4 ) = 90 N ⋅ m M C = 225 ( 0.8 ) = 180 N ⋅ m 5-3 .....................................................................................Ans. .......................................................................................Ans. .....................................................................................Ans. Two forces are applied at a point in the plane of a rigid steel plate as shown in Fig. P5-3. Determine the moments of (a) The 500-lb force about points A and B. (b) The 300-lb force about points B and C. SOLUTION (a) (b) (c) (d) 5-4 M A = 500 (30 ) = 15, 000 lb ⋅ in. M B = 500 ( 20 ) = 10, 000 lb ⋅ in. M B = 300 (30 ) = 9000 lb ⋅ in. M C = 300 ( 25 ) = 7500 lb ⋅ in. ...............................................................................Ans. ...............................................................................Ans. ..................................................................................Ans. ..................................................................................Ans. Two forces are applied to the bridge truss shown in Fig. P5-4. Determine the moments of (a) The 3.6-kN force about points A and D. (b) The 2.7-kN force about point A. SOLUTION (a) M A = 3.6 (3) = 10.80 kN ⋅ m M D = 3.6 (3) = 10.80 kN ⋅ m M A = 2.7 (3) = 8.1 kN ⋅ m ....................................................................................Ans. ....................................................................................Ans. .........................................................................................Ans. (b) 5-5 A 50-lb force is applied to the handle of a lug wrench, which is being used to tighten the nuts on the rim of an automobile tire as shown in Fig. P5-5. The diameter of the bolt circle is 5.5 in. Determine the moments of the force about the axle of the wheel (point O) and about the point of contact of the wheel with the pavement (point A). SOLUTION M O = (50 cos 20° )( 20 cos 20° ) = 1047 lb ⋅ in. + (50sin 20° )( 2.75 + 20sin 20° ) ...............................................................Ans. 111 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS M A = (50 cos 20° )( 20 cos 20° ) = 1286 lb ⋅ in. 5-6 + (50sin 20° )(16.75 + 20sin 20° ) ....................................................................................................Ans. A 160-N force is applied to the handle of a door as shown in Fig. P5-6. Determine the moments of the force about the hinges A and B. SOLUTION M A = (160 cos 45° )(0.6 ) + (160sin 45° )(0.8 ) = 158.4 N ⋅ m M B = (160sin 45° )(0.8 ) − (160 cos 45° )(0.5 ) = 33.9 N ⋅ m 5-7 Determine the moment of the 425-lb force shown in Fig. P5-7 about point B. SOLUTION ......................Ans. ........................Ans. M B = ( 425cos 35° )(16 ) + ( 425sin 35° )(16 ) = 9470 lb ⋅ in. 5-8 ........................Ans. A pry bar is used to extract a nail from a board as shown in Fig. P5-8. Determine the moment of the 120-N force (a) About point A. (b) About point B. SOLUTION (a) M A = (120 cos 20° )(0.75 ) = 103.5 N ⋅ m + (120sin 20° )(0.46 ) ..........................................Ans. (b) M B = (120 cos 20° )(0.65 ) = 92.2 N ⋅ m + (120sin 20° )(0.460 ) .............................................Ans. 5-9 A man exerts a force P to hold a 60-ft pole in the position shown in Fig. P5-9. If the moment of the force P about point A is 4000 lb ⋅ ft , determine the magnitude of the force P. SOLUTION b = 60 cos 50° = 38.567 ft h = 60sin 50° = 45.963 ft h φ = tan −1 = 23.728° 66 + b M A = ( P cos φ )( 45.963) − ( P sin φ )(38.567 ) = 4000 lb ⋅ ft P = 150.6 lb .......................................................................................................................Ans. 5-10 Two forces act on a truss as shown in Fig. P5-10. Member AC of the truss is perpendicular to members BC and CD. Determine the moment of (a) The 4-kN force about point A. (b) The 3-kN force about point D. SOLUTION (a) (b) M A = 4 ( 2sin 60° ) = 6.93 kN ⋅ m M D = 3 ( 4sin 60° ) = 10.39 kN ⋅ m ............................................................................Ans. ..........................................................................Ans. 112 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 5-11 The crane and boom shown in Fig. P5-11 is lifting a 4000-lb load. Determine the moment of the load about point C. SOLUTION M C = 4000 ( 24 cos 30° − 1 + 1) = 83,100 lb ⋅ ft .....................................................Ans. 5-12 Due to combustion, a compressive force P is exerted on the connecting rod of an automobile engine as shown in Fig. P5-12. The lengths of the crank throw AB and connecting rod BC are 75 mm and 225 mm, respectively. Determine the moment of the force P about the bearing at A in terms of the crank angle θ. SOLUTION From the Law of Sines sin φ sin θ = 75 225 Also sin φ = sin 2 θ 9 sin θ 3 cos φ = 1 − sin 2 φ = 1 − Then M A = ( P cos φ )( 75sin θ ) + ( P sin φ )( 75cos θ ) = 75P (sin θ cos φ + cos θ sin φ ) sin 2 θ sin θ cos θ = 75P sin θ 1 − + 9 3 ..........................................................Ans. 5-13 A 760-lb force acts on a bracket as shown in Fig. P5-13. Determine the moment of the force about point A (a) Using the vector approach. (b) Using the scalar approach. SOLUTION (a) M A = (10 i − 12 j) × (760 cos 40° i + 760sin 40° j) = 10 ( 760sin 40° ) + 12 (760 cos 40° ) k = (11,870 k ) lb ⋅ in. .....................Ans. M A = ( 760 cos 40° )(12 ) + ( 760sin 40° )(10 ) = 11,870 lb ⋅ in. = (11,870 k ) lb ⋅ in. ............................................................Ans. (b) 5-14 Determine the moment of the 675-N force shown in Fig. P5-14 about point O (a) Using the vector approach. (b) Using the scalar approach. SOLUTION (a) M O = ( 0.230 i + 0.250 j) × ( 675cos 22° i − 675sin 22° j) = 0.230 ( −675sin 22° ) − 0.250 ( 675cos 22° ) k = ( −215 k ) N ⋅ m .........Ans. M O = (675cos 22° )( 0.250 ) + ( 675sin 22° )(0.230 ) = 215 N ⋅ m = ( −215 k ) N ⋅ m ......................................................................Ans. (b) 113 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 5-15 Two forces F1 and F2 are applied to a gusset plate as shown in Fig. P5-15. Determine the moment (a) Of force F1 about point A. (b) Of force F2 about point B. SOLUTION (a) M A = (9 i + 4 j) × ( −550sin 45° i + 550 cos 45° j) = 9 (550 cos 45° ) + 4 (550sin 45° ) k = (5060 k ) lb ⋅ in. ............................Ans. M B = ( −7 i + 4 j) × (750sin 60° i + 750 cos 60° j) = −7 ( 750 cos 60° ) − 4 ( 750sin 60° ) k = ( −5220 k ) lb ⋅ in. ......................Ans. (b) 5-16 Two forces F1 and F2 are applied to a bracket as shown in Fig. P5-16. Determine the moment (a) Of force F1 about point O. (b) Of force F2 about point A. SOLUTION (a) M O = ( 0.300 i + 0.500 j) × (5cos 45° i + 5sin 45° j) = 0.300 (5sin 45° ) − 0.500 (5cos 45° ) k = ( −0.707 k ) kN ⋅ m ................Ans. M A = (0.265 i + 0.375 j) × (3cos 45° i − 3sin 45° j) = 0.265 ( −3sin 45° ) − 0.375 (3cos 45° ) k = ( −1.358 k ) kN m .............Ans. (b) 5-17 Determine the moment of the contact force F of the crutch shown in Fig. P5-17 about point A. The length of the crutch is 5 ft and F = 35 lb. SOLUTION M A = (5cos 70° i − 5sin 70° j) × ( −35cos 45° i + 35sin 45° j) = (5cos 70° )(35sin 45° ) − ( −5sin 70° )( −35cos 45° ) k = ( −74.0 k ) lb ⋅ ft ....................................................................................................Ans. 5-18 The moment of the force F shown in Fig. P5-18 about point A is (–2700k) N-mm, and its moment about point C is ( −7500k ) N ⋅ mm . Determine the magnitude and orientation (angle θ) of the force F. SOLUTION M A = (120 i + 75 j) × ( F cosθ i + F sin θ j) = 120 ( F sin θ ) − 75 ( F cos θ ) k = ( −2700 k ) N ⋅ mm M C = (75 j) × ( F cos θ i + F sin θ j) = 75 ( F cos θ ) ( −k ) = ( −7500 k ) N ⋅ mm Therefore 75 F cos θ = 7500 N ⋅ mm 120 F sin θ = 7500 − 2700 = 4800 N ⋅ mm F sin θ 40 tan θ = = F cos θ 100 θ = 21.80° .................................................................................................................Ans. F = 107.7 N ..............................................................................................................Ans. 114 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 5-19 A 970-lb force acts at a point in a body as shown in Fig. P5-19. Determine the moment of the force about point C. SOLUTION F = 970 −25 i + 30 j 252 + 302 = ( −620.98 i + 745.17 j) lb M C = ( −30 j + 20 k ) × ( −620.98 i + 745.17 j) = ( −14,900 i − 12, 420 j − 18, 630 k ) lb ⋅ in. .......................................................Ans. 5-20 A 200-N force is applied to a pipe wrench as shown in Fig. P5-20. Determine the moment of the force about point A. Express the result in Cartesian vector form. SOLUTION M A = ( −0.175 i + 0.580 j + 0.250 k ) × ( −200 k ) = ( −116.0 i − 35.0 j) N ⋅ m ....................................................................................Ans. 5-21 A 650-lb force acts at a point in a body as shown in Fig. P5-21. Determine (a) The moment of the force about point A. (b) The direction angles associated with the moment vector. SOLUTION (a) F = 650 8 i + 16 j + 30 k 82 + 162 + 302 = (148.876 i + 297.751 j + 558.283 k ) lb M A = ( −15 i − 24 j) × (148.876 i + 297.751 j + 558.283 k ) = ( −13,399 i + 8374 j − 893.2 k ) lb ⋅ in. .............................................................Ans. (b) MA = (13,399 ) + (8374 ) + (893.2 ) 2 2 2 = 15,826 lb ⋅ in. θ x = cos −1 θ y = cos −1 θ z = cos −1 −13,399 = 147.85° ......................................................................................Ans. 15,826 8374 = 58.05° ............................................................................................Ans. 15,826 −893.2 = 93.24° ...........................................................................................Ans. 15,826 5-22 A pipe bracket is loaded as shown in Fig. P5-22. Determine the moment of the force F about point O. SOLUTION F = 875 75 i + 150 j + 140 k 752 + 1502 + 1402 = (300.40 i + 600.80 j + 560.75 k ) N M O = ( 0.2 i + 0.25 j + 0.15 k ) × (300.40 i + 600.80 j + 560.75 k ) = (50.1 i − 67.1 j + 45.1k ) N ⋅ m ..........................................................................Ans. 115 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 5-23 The magnitude of the tension T in cable CD of Fig. P5-23 is 150 lb. Determine the moment of T about point B (a) Using a position vector from B to D. (b) Using a position vector from B to C. SOLUTION T = 150 20 i − 22 j + 18 k 202 + 222 + 182 = (86.315 i − 94.947 j + 77.684 k ) lb M B = ( 28 i + 18 k ) × (86.315 i − 94.947 j + 77.684 k ) = (1709 i − 621 j − 2660 k ) lb ⋅ in. ........................................................................Ans. M B = (8 i + 22 j) × (86.315 i − 94.947 j + 77.684 k ) = (1709 i − 621 j − 2660 k ) lb ⋅ in. ........................................................................Ans. 5-24 A 450-N force F acts on a machine component as shown in Fig. P5-24. The direction angles of F are θ x = 70° , θ y = 30° , and θ z = 69° . Determine the moment of the force about point A. Express your answer in Cartesian vector form. SOLUTION F = 450 (cos 70° i + cos 30° j + cos 69° k ) = (153.909 i + 389.711 j + 161.266 k ) N M A = (0.3 i − 0.6 j + 0.4 k ) × (153.909 i + 389.711 j + 161.266 k ) = ( −253 i + 13.18 j + 209 k ) N ⋅ m .......................................................................Ans. 5-25 The magnitude of the force F shown in Fig. P5-25 is 100 lb. Determine the moment of F about the bearing at C. SOLUTION F = 100 −10 i + 7 j + 6 k 102 + 7 2 + 62 = ( −73.521 i + 51.465 j + 44.113 k ) lb M C = ( −9 j − 14 k ) × ( −73.521 i + 51.465 j + 44.113 k ) = (324 i + 1029 j − 662 k ) lb ⋅ in. ..........................................................................Ans. 5-26 Determine the moment of the 800-N force shown in Fig. P5-26 about point D (a) Using a position vector from D to B. (b) Using a position vector from D to E. SOLUTION (a) F = 800 400 i − 400 j − 600 k 4002 + 4002 + 6002 = (388.057 i − 388.057 j − 582.086 k ) N M D = (0.4 i − 0.8 j + 0.6 k ) × (388.057 i − 388.057 j − 582.086 k ) = ( 699 i + 466 j + 155 k ) N ⋅ m .............................................................................Ans. (b) M D = (0.8 i − 1.2 j) × (388.057 i − 388.057 j − 582.086 k ) = ( 699 i + 466 j + 155 k ) N ⋅ m .............................................................................Ans. 116 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 5-27 The magnitudes of the forces F1, F2, and F3 shown in Fig. P5-27 are 550 lb, 300 lb, and 600 lb, respectively. Determine the sum of the moments of the three forces about point B. SOLUTION F1 = 550 −7 i − 11 j + 3 k 7 2 + 112 + 32 = ( −287.763 i − 452.198 j + 123.327 k ) lb M1B = (11 j − 6 k ) × ( −287.763 i − 452.198 j + 123.327 k ) = ( −1357 i + 1727 j + 3165 k ) lb ⋅ ft M 2 B = (11 j − 6 k ) × (300 i ) = ( −1800 j − 3300 k ) lb ⋅ ft M 3 B = ( 6 j − 6 k ) × ( −600 k ) = ( −3600 i ) lb ⋅ ft ΣM B = ( −4960 i − 73 j − 135 k ) lb ⋅ ft .........................................................................Ans. 5-28 If the magnitude of the moment of the cable force T shown in Fig. P5-28 about the hinge at B is M B = 1150 N ⋅ m , determine the magnitude of the force T. SOLUTION CDx = −500 mm CDy = −1400 cos 30 = −1212.436 mm CDz = 750 + 1400sin 30 = 1450 mm T =T −500 i − 1212.436 j + 1450 k 5002 + 1212.4362 + 14502 = ( −0.25574T i − 0.62013T j + 0.74164T k ) N = ( 0.09966T i − 0.70328T j − 0.55369T k ) N ⋅ m MB = M B = 0.5 i − 1.1cos 30° j + ( 0.75 + 1.1sin 30° ) k × T (0.09966T ) + (0.70328T ) + (0.55369T ) 2 2 2 = 0.90062T = 1150 N ⋅ m T = 1277 N .........................................................................................................................Ans. 5-29 The force F in Fig. P5-29 can be expressed in Cartesian vector form as F = (60 i + 100 j + 120 k ) lb . Determine the scalar component of the moment at point B about line BC. SOLUTION M C = (18 i + 32 k ) × (60 i + 100 j + 120 k ) = ( −3200 i − 240 j + 1800 k ) lb ⋅ in. e BC = −9 i + 18 j 92 + 182 = ( −0.44721 i + 0.89443 j) M BC = M C e BC = ( −3200 )( −0.44721) + ( −240 )(0.89443) = 1216 lb ⋅ in. ...........................................................................................................Ans. 117 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 5-30 A 3000-N force is applied to the pipe shown in Fig. P5-30. Determine the scalar component of the moment of the force about the x-axis. SOLUTION F = 3000 280 i − 300 j − 530 k 2802 + 3002 + 5302 = (1253.173 i − 1342.686 j − 2372.078 k ) N M O = ( 0.65 j + 0.53 k ) × (1253.173 i − 1342.686 j − 2372.078 k ) = ( −830.227 i + 664.182 j − 814.562 k ) N ⋅ m M Ox = M O i = −830 N ⋅ m .............................................................................................Ans. 5-31 A force F = ( −30 i + 50 j − 40 k ) lb is applied to the machine component shown in Fig. P5-31. Determine the scalar component of the moment of the force about the z-axis. SOLUTION M O = ( −10 i + 24 j − 16 k ) × ( −30 i + 50 j − 40 k ) = ( −160 i + 80 j + 220 k ) lb ⋅ in. M Oz = M O k = 220 lb ⋅ in. .............................................................................................Ans. 5-32 The magnitude of the force F in Fig. P5-32 is 595 N. Determine the scalar component of the moment at point O about line OC. SOLUTION F = 595 −220 i + 200 k 2202 + 2002 = ( −440.26 i + 400.24 k ) N M O = ( 0.22 i + 0.24 j) × ( −440.26 i + 400.24 k ) = (96.06 i − 88.05 j + 105.66 k ) N ⋅ m eOC = 220 i + 200 k 2202 + 2002 = (0.73994 i + 0.67267 k ) M OC = M O eOC = (96.06 )( 0.73994 ) + (105.66 )( 0.67267 ) = 142.2 N ⋅ m ..........................................................................................................Ans. 5-33 A 40-lb vertical force F is applied to a lug wrench as shown in Fig. P5-33. Determine the magnitude of the component of the moment that would tighten the lug nut. SOLUTION M O = ( 4 i + 8 j − 3 k ) × ( −40 k ) = ( −320 i + 160 j) lb ⋅ in. M Oz = M O i = 320 lb ⋅ in. ............................................................................................Ans. 5-34 If the magnitude of the force T shown in Fig. P5-34 is 1000-N, determine the scalar component of the moment of the force about the line CD. SOLUTION T = 1000 1.8 j + 3.6 k 1.82 + 3.62 = ( 447.21 j + 894.43 k ) N M C = ( −1.2 i + 2.7 j) × ( 447.21 j + 894.43 k ) 118 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS = ( 2414.96 i + 1073.32 j − 536.65 k ) N ⋅ m eCD = −2.4 i − 1.5 j 2.42 + 1.52 = ( −0.84800 i − 0.53000 j) M CD = M C eCD = ( 2414.96 )( −0.84800 ) + (1073.32 )( −0.53000 ) = −2620 N ⋅ m .........................................................................................................Ans. 5-35 A 120-lb force F is applied to a lever-shaft assembly as shown in Fig. P5-35. Determine the moment of the force about each coordinate axis. SOLUTION F = 120 −2 i + 14 j − 16 k 22 + 142 + 162 = ( −11.2390 i + 78.6732 j − 89.9122 k ) lb M O = (13 i + 16 k ) × ( −11.2390 i + 78.6732 j − 89.9122 k ) = ( −1258.77 i + 989.03 j + 1022.75 k ) lb ⋅ in. M Ox = −1259 lb ⋅ in. .........................................................................................................Ans. M Oy = 989 lb ⋅ in. ..............................................................................................................Ans. M Oz = 1023 lb ⋅ in. ............................................................................................................Ans. 5-36 A 650-N force acts on the awning structure shown in Fig. P5-36. Determine the moment of the force about line BC. Express the result in Cartesian vector form. SOLUTION M B = (0.6 i − 0.6 k ) × ( −650 k ) = (390.0 j) N ⋅ m e BC = 1.2 j − 0.9 k 1.22 + 0.92 = (0.8000 j − 0.6000 k ) M BC = M B e BC = (390.0 )(0.8000 ) = 312 N ⋅ m M BC = M BC e BC = 312 ( 0.8000 j − 0.6000 k ) = ( 250 j − 187.2 k ) N ⋅ m .............Ans. 5-37 The magnitude of the force F in Fig. P5-37 is 107 lb. Determine the component of the moment of the force that rotates the door about the axis of the hinges. SOLUTION z A = 34 tan 20° = 12.375 in. F = 107 −36 i − 34 j + 19.625 k 362 + 342 + 19.6252 32 − z A = 19.625 in. = ( −72.318 i − 68.300 j + 39.423 k ) lb M O = ( −36 i + 32 k ) × ( −72.318 i − 68.300 j + 39.423 k ) = ( 2185.60 i − 894.95 j + 2458.80 k ) lb ⋅ in. M Ox = 2190 lb ⋅ in. ...........................................................................................................Ans. 119 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 5-38 A bracket is rigidly attached to a wall at O and is subjected to a 384-N force F as shown in Fig. P5-38. Determine the component of the moment of the force (a) That twists the bracket about the y-axis. (b) That bends the bracket about the x-axis. SOLUTION F = 384 150 i − 250 j + 250 k 1502 + 2502 + 2502 = (149.978 i − 249.963 j + 249.963 k ) N M O = (0.3 i + 0.5 j − 0.2 k ) × (149.978 i − 249.963 j + 249.963 k ) = ( 74.989 i − 104.985 j − 149.978 k ) N ⋅ m (a) (b) M Oy = −105.0 N ⋅ m .........................................................................................................Ans. M Ox = 75.0 N ⋅ m ..............................................................................................................Ans. 5-39 A bar is bent in a circular arc of radius R = 6 ft and is subjected to a 660-lb force F as shown in Fig. P5-39. The force tends to twist and bend the member about the coordinate axes. Determine the twisting and bending moments and state the axes about which each occurs. SOLUTION F = 660 4i − 4k 42 + 42 = ( 466.690 i − 466.690 k ) lb M O = ( 6 j + 6 k ) × ( 466.690 i − 466.690 k ) = ( −2800.14 i + 2800.14 j − 2800.14 k ) lb ⋅ in. Twist: Bend: Bend: 2800 lb ⋅ in. 2800 lb ⋅ in. 2800 lb ⋅ in. about the negative z-axis ..........................................................Ans. about the negative x-axis ..........................................................Ans. about the positive y-axis ...........................................................Ans. 5-40 The magnitude of the force F in Fig. P5-40 is 976 N. Determine (a) The component of the moment at point C parallel to line CE. (b) The component of the moment at point C perpendicular to line CE and the direction angles associated with this moment vector. SOLUTION (a) F = 976 −350 i − 300 j + 160 k 3502 + 3002 + 1602 = ( −700.06 i − 600.06 j + 320.03 k ) N M C = (0.14 i + 0.4 j) × ( −700.06 i − 600.06 j + 320.03 k ) = (128.01 i − 44.80 j + 196.02 k ) N ⋅ m −210 i + 400 j 2102 + 4002 = ( −0.46483 i + 0.88540 j) eCE = M CE = M C eCE = (128.01)( −0.46483) + ( −44.80 )(0.88540 ) = −99.169 N ⋅ m M CE = M CE eCE = ( 46.10 i − 87.80 j) N ⋅ m ................................................................Ans. (b) M ⊥ = M C − M CE = (128.01 i − 44.80 j + 196.02 k ) − ( 46.10 i − 87.80 j) = (81.91 i + 43.00 j + 196.02 k ) N ⋅ m ................................................................Ans. 120 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS M⊥ = (81.91) + ( 43.00 ) + (196.02 ) 2 2 2 = 216.75 N ⋅ m 81.91 = 67.80° ............................................................................................Ans. 216.75 43.00 θ y = cos −1 = 78.56° ...........................................................................................Ans. 216.75 196.02 θ z = cos −1 = 25.26° ............................................................................................Ans. 216.75 θ x = cos −1 5-41 The magnitude of the force F in Fig. P5-41 is 781 lb. Determine (a) The component of the moment at point C parallel to line CD. (b) The component of the moment at point C perpendicular to line CD and the direction angles associated with this moment vector. SOLUTION (a) F = 781 12 i + 10 k 122 + 102 = (599.98 i + 499.98 k ) lb M C = ( −12 i + 25 j) × (599.98 i + 499.98 k ) = (12, 499.60 i + 5999.81 j − 14,999.52 k ) lb ⋅ in. eCD = −12 i + 10 k 122 + 102 = ( −0.76822 i + 0.64018 k ) M CD = M C eCD = (12, 499.60 )( −0.76822 ) + ( −14,999.52 )(0.64018 ) = −19, 204.8 lb ⋅ in. M CD = M CD eCD = (14, 753.5 i − 12, 294.6 k ) lb ⋅ in. .................................................Ans. (b) M ⊥ = M C − M CD = ( −2253.9 i + 5999.8 j − 2704.9 k ) lb ⋅ in. ..............................Ans. M⊥ = ( 2253.9 ) + (5999.8) + ( 2704.9 ) 2 2 2 = 6956.6 lb ⋅ in. −2253.9 = 108.90° .......................................................................................Ans. 6956.6 5999.8 θ y = cos −1 = 30.41° ...........................................................................................Ans. 6956.6 −2704.9 θ z = cos −1 = 112.88° .......................................................................................Ans. 6956.6 θ x = cos −1 5-42 Determine the moment of the couple shown in Fig. P5-42 and the perpendicular distance between the two forces. SOLUTION M = (760 cos 35° )(0.2 ) + (760sin 35° )(0.1) = 168.103 N ⋅ m ...................Ans. 168.103 N ⋅ m = 760d d = 0.221 m = 221 mm ...................................................................................................Ans. 121 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 5-43 A lug wrench is being used to tighten a lug nut on an automobile wheel as shown in Fig. P5-43. Two equal magnitude, parallel forces of opposite sense are applied to the wrench. If the magnitude of each force is 25 lb, determine the couple applied to a lug nut and express the result in Cartesian vector form. SOLUTION M = − ( 25 )(12 ) i = −300 i lb ⋅ in. ..................................................................................Ans. 5-44 To open a valve to a steam line in a power plant requires a couple of magnitude 54 N ⋅ m . Determine the magnitude of each force F shown in Fig. P5-44 required to open the valve. SOLUTION M = 54 N ⋅ m = F (0.3) F = 180 N ..........................................................................................................................Ans. 5-45 A beam is loaded with a system of three couples as shown in Fig. P5-45. Express the resultant of the couple system in Cartesian vector form. SOLUTION M = (500 )(30 ) − ( 250 )(14 ) − (300 )(30 ) = 2500 lb ⋅ in. = ( 2500 k ) lb ⋅ in. ..................................................................Ans. 5-46 The input and output torques (couples) from a gear box are shown in Fig. P5-46. Determine the magnitude and direction of the resultant torque T. SOLUTION T = 750 i + 200 j + 125 k + 150 ( cos 45° j + sin 45° k ) = ( 750 i + 306 j + 231k ) N ⋅ m T= (750 ) + (306 ) + ( 231) 2 2 2 = 842.36 N ⋅ m .......................................................Ans. 750 = 27.08° ............................................................................................Ans. 842.36 306 θ y = cos −1 = 68.70° ............................................................................................Ans. 842.36 231 θ z = cos −1 = 74.08° ............................................................................................Ans. 842.36 θ x = cos −1 5-47 Three couples are applied to a rectangular block as shown in Fig. P5-47. Determine the magnitude of the resultant couple C and the direction angles associated with the resultant couple vector. SOLUTION M = (15 j − 10 k ) × ( −18 i ) + ( −8 i + 10 k ) × ( 20 j) + (8 i + 15 j) × (15 k ) = ( 25 i + 60 j + 110 k ) lb ⋅ in. M= ( 25) + (60 ) + (110 ) 2 2 2 = 127.77 lb ⋅ in. ..........................................................Ans. 25 = 78.72° ............................................................................................Ans. 127.77 60 θ y = cos −1 = 61.99° ............................................................................................Ans. 127.77 θ x = cos −1 122 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS θ z = cos −1 110 = 30.58° ............................................................................................Ans. 127.77 5-48 Three couples are applied to a bent bar as shown in Fig. P5-48. Determine (a) The magnitude of the resultant couple C and the direction angles associated with the resultant couple vector. (b) The scalar component of the resultant couple C about line OA. SOLUTION (a) C = (80 i + 65 j + 95 k ) N ⋅ m C= (80 ) + (65) + (95) 2 2 2 = 140.178 N ⋅ m ............................................................Ans. (b) 80 = 55.20° ..........................................................................................Ans. 140.178 65 θ y = cos −1 = 62.37° .........................................................................................Ans. 140.178 95 θ z = cos −1 = 47.34° ..........................................................................................Ans. 140.178 140 i + 120 j − 80 k eOA = = (0.69653 i + 0.59702 j − 0.39801k ) 1402 + 1202 + 802 θ x = cos −1 COA = C eOA = (80 )( 0.69653) + ( 65 )( 0.59702 ) + (95 )( −0.39801) = 56.7 N ⋅ m ..............................................................................................................Ans. 5-49 Three couples are applied to a rectangular block as shown in Fig. P5-49. Determine (a) The magnitude of the resultant couple C and the direction angles associated with the resultant couple vector. (b) The scalar component of the resultant couple C about line OA. SOLUTION (a) C = (90 i + 125 j + 75 k ) lb ⋅ in. C= (90 ) + (125) + (75) 2 2 2 = 171.318 lb ⋅ in. .........................................................Ans. (b) 90 = 58.31° ..........................................................................................Ans. 171.318 125 θ y = cos −1 = 43.14° .........................................................................................Ans. 171.318 75 θ z = cos −1 = 64.04° ..........................................................................................Ans. 171.318 24 i + 8 j + 32 k eOA = = ( 0.58835 i + 0.19612 j + 0.78446 k ) 242 + 82 + 322 θ x = cos −1 COA = C eOA = (90 )( 0.58835 ) + (125 )(0.19612 ) + (75 )(0.78446 ) = 136.3 lb ⋅ in. ............................................................................................................Ans. 123 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 5-50 Replace the 3-kN force shown in Fig. P5-50 by a force at point A and a couple. SOLUTION FA = 3 kN ↓ ......................................................................................................................Ans. C A = (3000 )(0.15 ) =450 N ⋅ m ...............................................................................Ans. 5-51 Replace the 50-lb force shown in Fig. P5-51 by a force at point A and a couple. Express your answer in Cartesian vector form. SOLUTION FA = (50 j) lb .....................................................................................................................Ans. C A = (50 )(6 )( − k ) = ( −300 k ) lb ⋅ in. ........................................................................Ans. 5-52 Replace the 130-N vertical force shown in Fig. P5-52 by the mechanic’s hand to the wrench by an equivalent force-couple system at the lug nut. SOLUTION FA = 130 N ↓ ....................................................................................................................Ans. C A = (130 )( 0.3) =39.0 N ⋅ m ...................................................................................Ans. 5-53 A gusset plate is riveted to a beam by three rivets as shown in Fig. P5-53. Replace the 2500-lb force by a force-couple system at the top rivet. SOLUTION F = 2500 lb → C = ( 2500 )( 6 ) = 15, 000 lb ⋅ in. 5-54 Four forces are applied to a truss as shown in Fig. P5-54. Determine the magnitude and direction of the resultant of the four forces and the perpendicular distance dR from point A to the line of action of the resultant. SOLUTION R A = (3 i − 12 j) kN = 12.369 kN 75.96° ............................................................Ans. C A = (6 )( 2 ) + ( 4 )( 4 ) + ( 2 )(6 ) = 40 kN ⋅ m = 12.369d d = 3.23 m .........................................................................................................................Ans. 5-55 Three forces are applied to the locked pulley shown in Fig. P5-55. Determine the magnitude and direction of the resultant of the three forces and the perpendicular distance from the axle of the pulley to the line of action of the resultant. SOLUTION R A = ( −160 cos 20° ) i + (160sin 20° ) j − (90 ) j = ( −150.351 i − 35.277 j) lb = 154.43 lb 13.20° ..............................................................................................Ans. C A = (120 )(1) − ( 40 )(1) − (90 )( 2 ) = −100 lb ⋅ ft = −154.43d d = 0.648 ft .......................................................................................................................Ans. 5-56 Determine the resultant of the four forces acting on the bell crank shown in Fig. P5-56, and determine where the resultant intersects the x-axis. SOLUTION R = 175 i + ( 400 cos 45° ) i + ( 400sin 45° ) j + 300 j − 200 j 124 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS = ( 457.84 i + 382.84 j) N = 596.82 N 39.90° ...................................................Ans. C A = ( 400 )(50 ) + ( 200 )(80 ) = 36, 000 N ⋅ mm = 382.84d d = 94.0 mm left of O ....................................................................................................Ans. 5-57 Three 75-lb traffic lights are suspended over a roadway as shown in Fig. P5-57. Determine the resultant of the weights of the traffic lights and locate the resultant with respect to point A. SOLUTION R = (3)(75 ) = 225 lb ↓ ..............................................................................................Ans. C = ( 75 )(8 ) + (75 )(11) + (75 )(14 ) = 2475 lb ⋅ ft = 225d d = 11 ft ..............................................................................................................................Ans. 5-58 Four parallel forces act on a concrete slab as shown in Fig. P5-58. Determine the resultant of the forces and locate the intersection of the line of action of the resultant with the xy-plane. SOLUTION R = ( −4 − 3 − 3 − 2 ) k = ( −12 k ) kN ...........................................................................Ans. C = (3 i + 1.5 j) × ( −4 k ) + (3 i + 3 j) × ( −3 k ) + (1.5 i + 3 j) × ( −3 k ) + ( 4.5 j) × ( −2 k ) = ( −33 i + 25.5 j) kN ⋅ m = ( x i + y j) × ( −12 k ) = ( −12 y i + 12 x j) x = 25.5 12 = 2.12 m ......................................................................................................Ans. y = 33 12 = 2.75 m ..........................................................................................................Ans. 5-59 The forces exerted on the wheels of an airplane by the runway are shown in Fig. P5-59. Determine the resultant of the three forces and locate the intersection of the line of action of the resultant with the xy-plane (the runway). SOLUTION R = (5000 + 5000 + 2500 ) k = (12,500 k ) lb ...........................................................Ans. CO = ( 20 i + 8 j) × (5000 k ) + ( 20 i − 8 j) × (5000 k ) = ( −200, 000 j) lb ⋅ ft = ( x i + y j) × (12,500 k ) = (12,500 y i − 12,500 x j) x = 200, 000 12,500 = 16 ft ..........................................................................................Ans. y = 0 12,500 = 0 ft ..........................................................................................................Ans. 5-60 The magnitude of the force F acting on the casting shown in Fig. P5-60 is 2 kN. Replace the force with a force R through point A and a couple C. SOLUTION R = 2000 3 i + 4 j + 12 k 32 + 42 + 122 = ( 461.538 i + 615.385 j + 1846.154 k ) N .................Ans. C = (0.25 j + 0.075 k ) × ( 461.538 i + 615.385 j + 1846.154 k ) = ( 415.4 i + 34.6 j − 115.4 k ) N ⋅ m ..........................................................................Ans. 125 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 5-61 A force RILEY, STURGES AND MORRIS F = (50i + 50 j − 200k ) lb acts on the wall bracket shown in Fig. P5-61. Replace the force by a force-couple system at rivet B. The eye bolt is small and the force F may be considered as acting at point C. SOLUTION F = (50 i + 50 j − 200 k ) lb ............................................................................................Ans. C = ( 2 i + 4 j − 3 k ) × (50 i + 50 j − 200 k ) = ( −650 i + 250 j − 100 k ) lb ⋅ in. ...............................................................................Ans. 5-62 The homogeneous plate shown in Fig. P5-62 has a mass of 90 kg. The magnitude of the force T in cable BC is 800 N. Replace the weight and cable forces by an equivalent force-couple system at hinge A. SOLUTION W = (90 )(9.81)( − k ) = ( −882.90 k ) N T = 800 −350 i − 610 j + 500 k 3502 + 6102 + 5002 = ( −324.486 i − 565.533 j + 463.552 k ) N R = W + T = ( −324 i − 566 j − 419 k ) N ...................................................................Ans. C = (0.65 i + 0.28 j) × ( −324.486 i − 565.533 j + 463.552 k ) + (0.35 i ) × ( −882.90 k ) = (129.8 i + 7.71 j − 277 k ) N ⋅ m ............................................................................Ans. 5-63 Forces act at A, D, and E of the member shown in Fig. P5-63. If the equivalent force-couple system at B is R = ( −100i − 100 j) lb and a couple C, determine the magnitude of FE, the angle θ, and the required R = ( −100 + FEx ) i + ( FEy − 300 ) j = ( −100 i + 100 j) lb FEx = 0 lb FEy = 200 lb couple C. SOLUTION FE = 200 lb ↑ ...................................................................................................................Ans. C = (100 )(10 ) − (300 )(14 ) + ( 200 )( 20 ) = 800 lb ⋅ in. ..................................Ans. 5-64 Four forces and a couple are applied to a rectangular plate as shown in Fig. P5-64. Determine the magnitude and direction of the resultant of the force-couple system and the distance xR from point O to the intercept of the line of action of the resultant with the x-axis. SOLUTION R = (90 − 60 ) i + (75 − 50 ) j = (30 i + 25 j) N = 39.1 N 39.81° ....................................................................................................Ans. C = 10 − (90 )(0.25 ) − (50 )(0.3) − ( 60 )( 0.25 ) − (75 )( 0.3) = −65.00 N ⋅ m = xR 25 xR = −65.00 25 = −2.6 m ..............................................................................................Ans. 5-65 The magnitude of the resultant R of the four forces acting on the legs of the table shown in Fig. P5-65 is 80 lb. Determine the magnitude of F and the location, xR and yR, of the line of action of R. SOLUTION R = (30 + 20 + 15 + F ) j = (80 j) lb 126 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS F = 15 lb .............................................................................................................................Ans. C = (3 i ) × (30 k ) + (3 j) × (15 k ) + (3 i + 3 j) × ( 20 k ) = (105 i − 150 j) lb ⋅ ft = ( xR i + yR j) × (80 j) = (80 yR i − 80 xR j) xR = 150 80 = 1.875 ft ....................................................................................................Ans. yR = 105 80 = 1.313 ft ....................................................................................................Ans. 5-66 Figure P5-66 shows a crankshaft-flywheel arrangement of a one-cylinder engine. A 1000-N force P is supplied by the connecting rod, and a couple C of magnitude 250 N-m is delivered to the crankshaft by the flywheel. Replace the force and couple by an equivalent force-couple system at the bearing A. SOLUTION P = (1000sin 20° )( − j) + (1000 cos 20° )( − k ) = ( −342.02 j − 939.69 k ) N .......................................................................................Ans. M A = ( −0.225 i − 0.125 j) × ( −342.02 j − 939.69 k ) + ( −250 i ) = ( −132.5 i − 211.4 j + 77.0 k ) N ⋅ m .................................................................Ans. 5-67 A farmer is using the hand winch shown in Fig. P5-67 to raise a 40-lb bucket of water from a well. When the force P and the weight of the bucket are replaced by an equivalent force-couple system at bearing D, the result is R = −10 j − 60k lb and a couple CD Determine the force P applied to the handle of the winch and the couple CD. SOLUTION R = ( Px i + Py j + Pz k ) − 40 k = ( −10 j − 60 k ) lb Px = 0 lb Py = −10 lb Pz = −20 lb ..................................Ans. C D = (30 i + 2.5 j) × ( −40 k ) + ( 61 i − 20 cos 30° j + 20sin 30° k ) × ( −10 j − 20 k ) = (346 i + 2420 j − 610 k ) lb ⋅ in. .........................................................................Ans. 5-68 In order to remove a rusty screw from a steel plate, a worker attaches a screwdriver to the bent bar shown in Fig. P5-68. To hold the screwdriver in place, the worker applies a 30-N force at C; a 50-N force is applied to the handle of the bar in an attempt to remove the screw. Forces Fy and Fz are also applied at C. It is desirable that the screwdriver not bend about the y and z axes. (a) Determine the forces Fy and Fz. (b) Replace the forces at B and C by an equivalent force-couple system at A. SOLUTION (a) M A = (0.45 i ) × ( −30 i + Fy j + Fz k ) + (0.25 i − 0.15 j) × ( −50 k ) Fy = 0 N = 7.5 i + (12.5 − 0.45 Fz ) j + 0.45 Fy k = ( M x i + 0 j + 0 k ) N ⋅ m Fz = 27.78 N ............................................................Ans. R = ( −30 i ) + ( 27.78 k ) + ( −50 k ) = ( −30 i − 22.22 k ) N ....................................Ans. C = (7.5 i ) N ⋅ m ...............................................................................................................Ans. (b) 127 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 5-69 Reduce the forces shown in Fig. P5-69 to a wrench and locate the intersection of the wrench with the xyplane. SOLUTION FB = 420 3i − 5 j + 4k 32 + 52 + 42 = (178.191i − 296.985 j + 237.588 k ) lb R = 60 i + (178.191 i − 296.985 j + 237.588 k ) + 250 j = ( 238.191 i − 46.985 j + 237.588 k ) lb ...............................................................Ans. M O = (3 j) × (60 i ) + (5 j) × (178.191 i − 296.985 j + 237.588 k ) + ( 4 k ) × ( 250 j) = (187.940 i − 1070.955 k ) lb ⋅ ft eR = 238.191 i − 46.985 j + 237.588 k 238.1912 + 46.9852 + 237.5882 = (0.70120 i − 0.13832 j + 0.69942 k ) M R = M O e R = (187.940 )( 0.70120 ) + ( −1070.955 )(0.69942 ) = −617.264 lb ⋅ ft M R = M R e R = ( −432.826 i + 85.380 j − 431.727 k ) lb ⋅ ft ....................................Ans. MO = M R + r × R M O − M R = (620.766 i − 85.380 j − 639.228 k ) lb ⋅ ft r × R = ( xR i + yR j) × ( 238.191 i − 46.985 j + 237.588 k ) = 237.588 yR i − 237.588 xR j − ( 46.985 xR + 238.191 yR ) k xR = 85.380 237.588 = 0.359 ft ..................................................................................Ans. yR = 620.766 237.588 = 2.61 ft ..................................................................................Ans. 5-70 Locate the center of mass for the three particles shown in Fig. P5-70 if mA = 26 kg, mB = 21 kg, and mC = 36 kg. SOLUTION m = 26 + 21 + 36 = 83 kg 83 xG = 26 ( −200 cos 60° ) + 21( −200 cos 30° ) + 36 ( 200 ) = 962.693 83 yG = 26 ( 200sin 60° ) + 21( −200sin 30° ) = 2403.332 xG = 962.693 83 = 11.60 mm .......................................................................................Ans. yG = 2403.332 83 = 29.0 mm ......................................................................................Ans. 5-71 Locate the center of gravity for the four particles shown in Fig. P5-71 if WA = 20 lb, WB = 25 lb, WC = 30 lb, and WD = 40 lb. SOLUTION W = 20 + 25 + 30 + 40 = 115 lb 115 xG = 20 ( 0 ) + 25 ( 0 ) + 30 (8 ) + 40 ( 0 ) = 240 115 yG = 20 (12 ) + 25 (12 ) + 30 (12 ) + 40 (0 ) = 900 115 zG = 20 (10 ) + 25 (0 ) + 30 ( 0 ) + 40 (0 ) = 200 128 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS xG = 240 115 = 2.09 in. ..................................................................................................Ans. yG = 900 115 = 7.83 in. ..................................................................................................Ans. zG = 200 115 = 1.739 in. ................................................................................................Ans. 5-72 Locate the center of mass for the four particles shown in Fig. P5-72 if mA = 16 kg, mB = 24 kg, mC = 14 kg, and mD = 36 kg. SOLUTION m = 16 + 24 + 14 + 36 = 90 kg 90 xG = 16 (300 ) + 24 ( 0 ) + 14 (0 ) + 36 ( 0 ) = 4800 90 yG = 16 ( 0 ) + 24 ( 0 ) + 14 (500 ) + 36 ( 0 ) = 7000 90 zG = 16 (0 ) + 24 (0 ) + 14 (0 ) + 36 ( 400 ) = 14, 400 xG = 4800 90 = 53.3 mm ...............................................................................................Ans. yG = 7000 90 = 77.8 mm ..............................................................................................Ans. zG = 14, 400 90 = 160 mm ............................................................................................Ans. 5-73 Locate the center of gravity for the five particles shown in Fig. P5-73 if WA = 25 lb, WB = 35 lb, WC = 15 lb, WD = 28 lb, and WE = 16 lb. SOLUTION W = 25 + 35 + 15 + 28 + 16 = 119 lb 119 xG = 25 (10 ) + 35 (10 ) + 15 ( 0 ) + 28 (10 ) + 16 ( 0 ) = 880 119 yG = 25 (0 ) + 35 (11) + 15 (16 ) + 28 (16 ) + 16 ( 0 ) = 1073 119 zG = 25 ( 0 ) + 35 (0 ) + 15 (0 ) + 28 (11) + 16 (11) = 484 xG = 880 119 = 7.39 in. ..................................................................................................Ans. yG = 1073 119 = 9.02 in. ................................................................................................Ans. zG = 484 119 = 4.07 in. ..................................................................................................Ans. 5-74 Locate the center of mass for the five particles shown in Fig. P5-74 if mA = 2 kg, mB = 3 kg, mC = 4 kg, mD = 3 kg, and mE = 2 kg. SOLUTION m = 2 + 3 + 4 + 3 + 2 = 14 kg 14 xG = 2 (300 ) + 3 (150 ) + 4 (300 ) + 3 (300 ) + 2 (0 ) = 3150 14 yG = 2 ( 240 ) + 3 ( 400 ) + 4 ( 400 ) + 3 ( 0 ) + 2 ( 200 ) = 3680 14 zG = 2 ( 0 ) + 3 ( 0 ) + 4 ( 270 ) + 3 ( 270 ) + 2 ( 270 ) = 2430 xG = 3150 14 = 225 mm ................................................................................................Ans. yG = 3680 14 = 263 mm ................................................................................................Ans. zG = 2430 14 = 173.6 mm .............................................................................................Ans. 129 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 5-75 Three bodies with masses of 2, 4, and 6 slugs are located at points (2, 3, 4), (3, −4, 5), and (−3, 4, 6), respectively. Locate the mass center of the system if the distances are measured in feet. SOLUTION m = 2 + 4 + 6 = 12 slug 12 xG = 2 ( 2 ) + 4 (3) + 6 ( −3) = −2 12 yG = 2 (3) + 4 ( −4 ) + 6 ( 4 ) = 14 12 zG = 2 ( 4 ) + 4 (5 ) + 6 ( 6 ) = 64 xG = −2 12 = −0.1667 ft ................................................................................................Ans. yG = 14 12 = 1.167 ft ......................................................................................................Ans. zG = 64 12 = 5.33 ft .........................................................................................................Ans. 5-76 Locate the centroid of the shaded area shown in Fig. P5-76 if b = 200 mm and h = 300 mm. SOLUTION dAV = ( hx b ) dx dAH = b − (by h ) dy A = ∫ dAV = ∫ b 0 hx 2 hx bh dx = =2 b 2b 0 b 0 b M y = ∫ xdAV = ∫ My hx3 hx hb 2 x dx = =3 b 3b 0 b hb 2 3 2b xC = = = = 133.3 mm ...........................................................................Ans. A hb 2 3 M x = ∫ y dAH = ∫ yC = h 0 by 2 by 3 by bh 2 y b − dy = − = 3h 0 6 h 2 h M x bh 2 6 h = = = 100.0 mm ............................................................................Ans. A bh 2 3 5-77 Determine the y-coordinate of the centroid of the shaded area shown in Fig. P5-77. SOLUTION dAV = ( x 2 2b ) dx dAH = b − 2by dy A = ∫ dAV = ∫ b 0 x3 x2 b2 dx = = 2b 6b 0 6 b/2 0 b M x = ∫ y dAH = ∫ by 2 y5 / 2 b3 b3 b3 y b − 2by dy = − 2b =−= 5 2 0 8 10 40 2 b/2 130 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS yC = M x b3 40 3b =2 = ...................................................................................................Ans. A b 6 20 5-78 Determine the x-coordinate of the centroid of the shaded area shown in Fig. P5-78. SOLUTION x2 dAV = b 1 − 2 dx a a x2 A = ∫ dAV = ∫ b 1 − 2 dx 0 a b π ab −1 x 2 2 2 = x a − x + a sin a = 4 2a 0 M y = ∫ x dAV = ∫ xC = My A = a 0 a x2 −b x b 1 − 2 dx = a 3a (a 2 −x 23 ) ba 2 = 3 0 a ba 2 3 4a = .................................................................................................Ans. π ab 4 3π 5-79 Locate the centroid of the shaded area shown in Fig. P5-79. SOLUTION x2 dAV = 2 x − dx 25 1 x 2 x2 dM x = yC dAV = 2 x − 2 x − dx 25 25 2 1 4 x3 x 4 = 4x2 − + 2 25 625 A = ∫ dAV = ∫ 50 0 2 x3 x2 2 2 x − dx = x − = 833.33 in. 25 75 0 50 50 x2 M y = ∫ x dAV = ∫ x 2 x − dx 0 25 2 x3 x 4 = − = 20,833.33 in.3 3 100 0 xC = My A = 20,833.33 = 25.0 in. .................................................................................Ans. 833.33 50 50 1 50 4 x3 x 4 1 4 x3 x 4 x5 M x = ∫ dM x = ∫ 4 x 2 − + dx = −+ 20 25 625 2 3 25 3125 0 = 8333.33 in.3 131 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS yC = M x 8333.33 = = 10.00 in. ...................................................................................Ans. 833.33 A 5-80 Determine the y-coordinate of the centroid of the shaded area shown in Fig. P5-80. SOLUTION dAH = ( ay 2 b 2 ) dy b ay 2 ay 3 ab A = ∫ dAH = ∫ 2 dy = 2 = 0 b 3b 0 3 M x = ∫ y dAH = ∫ yC = b 0 b ay 2 ay 4 ab 2 y 2 dy = 2 = 4 b 4b 0 b M x ab 2 4 3b = = ...................................................................................................Ans. 4 A ab 3 5-81 Locate the centroid of the shaded area shown in Fig. P5-81. SOLUTION dAV = ( ax − x dx ) dAH = y − ( y 2 a ) dy A = ∫ dAV = ∫ a 0 ( 0 x3/ 2 x 2 a2 ax − x dx = a − = 3 2 2 0 6 ) a M y = ∫ x dAV = ∫ x xC = My A = a ( x5 / 2 x3 a3 ax − x dx = a − = 5 2 3 0 15 ) a a 3 15 2a = ...................................................................................................Ans. a2 6 5 a M x = ∫ y dAH = ∫ yC = 0 y 2 y3 y 4 a3 y y − dy = − = 3 4a 0 12 a a M x a 3 12 a = 2 = .....................................................................................................Ans. A a6 2 5-82 Determine the y-coordinate of the centroid of the shaded area shown in Fig. P5-82. SOLUTION 4x 5x dAV = x − dx = dx 9 9 5x2 5x 2 A = ∫ dAV = ∫ dx = = 20 mm 3 9 18 3 9 9 4 x 1 5 x 5 x 65 x 2 dM x = yC dAV = + dx = dx 9 2 9 9 162 132 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 9 9 RILEY, STURGES AND MORRIS 65 x 2 65 x3 = 93.889 mm3 M x = ∫ dM x = ∫ yC dAV = ∫ dx = 3 162 486 3 yC = M x 93.889 = = 4.69 mm ....................................................................................Ans. 20 A 5-83 Determine the x-coordinate of the shaded area shown in Fig. P5-83. SOLUTION dAV = ( x 3 2 ) dx 2 x4 x3 15 2 A = ∫ dAV = ∫ dx = = in. −1 2 8 −1 8 x3 x5 33 3 M y = ∫ x dAV = ∫ x dx = = in. −1 2 10 −1 10 2 2 2 xC = My A = 33 10 44 = ≅ 1.760 in. .............................................................................Ans. 15 8 25 5-84 Locate the centroid of the shaded area shown in Fig. P5-84. SOLUTION For the quarter circle (x − r) + ( y − r) 2 2 = r2 2 y = r − r2 − (x − r ) 2 dAV = r − r − r 2 − ( x − r ) dx ( ) = r 2 − ( x − r ) dx = 2rx − x 2 dx 2 A = ∫ dAV = ∫ r 0 2 1 x − r π r 2rx − x dx = ( x − r ) 2rx − x 2 + r 2 sin −1 = 2 4 r 0 2 r r M y = ∫ x dAV = ∫ x 2rx − x 2 dx 0 ( 2rx − x 2 )3/ 2 r π r3 r3 −1 x − r 2 2 − = + ( x − r ) 2rx − x + r sin − = 3 2 4 3 r 0 xC = Since the line r My A (π r 4 ) − (r 3) = r − 4r ........................................................................Ans. = 3 3 π r2 4 3π x = y is an axis of symmetry, yC = xC = r − 4r ...............................................................................................................Ans. 3π 133 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 5-85 Locate the centroid of the volume obtained by revolving the shaded area shown in Fig. P5-85 about the xaxis. SOLUTION x2 dV = π r dx = π 2 − dx 8 x2 x4 = π 4 − + dx 2 64 2 4 x2 x4 V = ∫ dV = ∫ π 4 − + dx 0 2 64 2 x3 x5 = π 4 x − + = 26.808 in.3 6 320 0 M yz = ∫ x dV = ∫ xC = M yz V = 4 0 4 x2 x4 x4 x6 xπ 4 − + dx = π 2 x 2 − + = 33.510 in.4 2 64 8 384 0 4 33.510 = 1.250 in. ....................................................................................Ans. 26.808 Since the x-axis is an axis of symmetry, yC = zC = 0 .........................................................................................................................Ans. 5-86 Locate the centroid of the volume obtained by revolving the shaded area shown in Fig. P5-86 about the xaxis. SOLUTION dV = π r 2 dx = π (50 x ) dx V = ∫ dV = ∫ 50π x dx = 25π x 2 = 25π b 2 0 0 b b M yz = ∫ x dV = ∫ xC = = M yz V = b 0 50π x3 50π b3 x (50π x ) dx = = 3 3 0 b 2 (100 ) ≅ 66.7 mm .......................................Ans. 3 50π b3 3 2b = 25π b 2 3 Since the x-axis is an axis of symmetry, yC = zC = 0 .........................................................................................................................Ans. 5-87 Locate the centroid of the curved homogeneous slender rod shown in Fig. P5-87. SOLUTION x = y2 6 dL = dx 2 + dy 2 = dx dy = y 3 ( dx dy ) + 1 dy = 2 ( y 3) 2 + 1 dy = 12 y + 9 dy 3 134 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS L = ∫ dL = ∫ 12 0 12 12 1 y + 9 dy = y y 2 + 9 + 9 ln y + y 2 + 9 = 27.8807 in. 0 3 6 ( ) 2 12 y 1 M y = ∫ x dL = ∫ y 2 + 9 dy 0 6 3 1 y = 18 4 xC = My L = (y 2 + 9) 3 9y − 8 92 y + 9 − ln y + y 2 + 9 = 304.962 in.2 8 0 2 ( ) 12 304.962 = 10.94 in. ...................................................................................Ans. 27.8807 12 0 M x = ∫ y dA = ∫ yC = y 3 1 1 y + 9 dy = 3 3 2 ( y + 9 ) = 207.278 in.2 0 2 3 12 M x 207.278 = = 7.43 in. .....................................................................................Ans. 27.8807 L 5-88 Locate the centroid of the curved homogeneous slender rod shown in Fig. P5-88 if b = 50 mm. SOLUTION y = x 2 2b dy dx = x b 2 b = 50 mm 2 dL = dx 2 + dy 2 = 1 + ( dy dx ) dx = 1 + ( x 50 ) dx = L = ∫ dL = ∫ = 50 0 1 502 + x 2 dx 50 50 1 1 502 + x 2 dx = x x 2 + 502 + 502 ln x + x 2 + 502 0 50 100 ( ) 1 2 (502 ) + 502 ln 50 + 2 50 − 502 ln 50 = 57.3897 mm 100 50 0 ( ) M y = ∫ x dL = ∫ xC = My L = x 1 1 502 + x 2 dx = 50 50 3 3 ( x2 + 502 ) = 1523.689 mm2 0 50 1523.689 = 26.5 mm ................................................................................Ans. 57.3897 2 50 x 1 M x = ∫ y dL = ∫ 502 + x 2 dx 0 100 50 1 x = 5000 4 (x 2 + 50 23 ) 502 504 − x x 2 + 502 − ln x + x 2 + 502 8 8 0 ( ) 50 = 525.198 mm 2 M 525.198 yC = x = = 9.15 mm ..................................................................................Ans. L 57.3897 5-89 Locate the centroid of the volume of the portion of a right circular cone shown in Fig. P5-89. SOLUTION ˆ r= r (h − z ) h 135 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS dV = ˆ π r2 π r2 2 dz = 2 ( h − z ) dz 4 4h h V = ∫ dV = ∫ 0 π r2 π r2 2 ( h − z ) dz = 2 4h 2 4h h ( h − z )3 π r 2 h − = 3 12 0 h M xy = ∫ z dV = ∫ 0 π r2 2 z ( h − z ) dz 2 4h h π r 2 h 2 z 2 2hz 3 z 4 π r 2 h 2 = 2 − + = 4h 2 3 4 0 48 zC = M xy V = π r 2 h 2 48 h = .............................................................................................Ans. π r 2 h 12 4 ˆ 4r r3 3 dM yz = dV = 3 ( h − z ) dz 3π 3h M yz = ∫ dM yz = ∫ xC = M yz = h 0 r3 r3 3 ( h − z ) dz = 3 3h3 3h ( h − z )4 r 3h − = 4 0 12 h V Since the plane x = y is a plane of symmetry, r 3h 12 r = ...............................................................................................Ans. 2 π r h 12 π yC = xC = r π ....................................................................................................................Ans. 5-90 Locate the mass center of the hemisphere shown in Fig. P5-90 if the density ρ at any point P is proportional to the distance from the xy-plane to the point P. SOLUTION r 2 = y 2 + (r − z ) 2 y 2 = r 2 − ( r − z ) = 2rz − z 2 2 dm = ρ dV = kz (π y 2 ) dz = π k ( 2rz 2 − z 3 ) dz 2rz 3 z 4 5π kr 4 m = ∫ dm = ∫ π k ( 2rz − z ) dz = π k − = 0 4 0 12 3 r 2 3 r M xy = ∫ z dm = ∫ zG = By symmetry, r 0 2rz 4 z 5 3π kr 5 − = zπ k ( 2rz − z ) dz = π k 5 0 10 4 2 3 r M xy m = 3π kr 5 10 18r = .........................................................................................Ans. 5π kr 4 12 25 xG = yG = 0 ........................................................................................................................Ans. 136 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 5-91 Locate the mass center of the right circular cone shown in Fig. P5-91 if the density ρ at any point P is proportional to the distance from the xy-plane to the point P. SOLUTION ˆ r = rz h r2 z2 ˆ dm = ρ dV = kz (π r 2 ) dz = π kz 2 dz h m = ∫ dm = ∫ h 0 π kr 2 3 π kr 2 z 4 π kr 2 h 2 z dz = 2 = h2 h 4 0 4 h 0 h M xy = ∫ z dm = ∫ M xy π kr 2 4 π kr 2 z dz = 2 h2 h z 5 π kr 2 h3 5 = 5 0 h π kr 2 h3 5 4h zG = = = ...........................................................................................Ans. m π kr 2 h 2 4 5 By symmetry, xG = yG = 0 ........................................................................................................................Ans. 5-92 Locate the centroid of the volume of the tetrahedron shown in Fig. P5-92. SOLUTION By similar triangles a (c − z ) c ˆb b = (c − z ) c 1ˆ ab 2 ˆ dV = ab dz = 2 (c − z ) dz 2 2c c ab V = ∫ dV = ∫ (c2 − 2cz + z 2 ) dz 0 2c 2 ˆ a= ab 2 z3 abc 2 = 2 c z − cz + = 2c 3 0 6 M xy = ∫ z dV = ∫ zC = Similarly c 0 c ab 2 ab c 2 z 2 2cz 3 z 4 abc 2 c z − 2cz 2 + z 3 ) dz = 2 − + = ( 2c 2 2c 2 3 4 0 24 c M xy V = abc 2 24 c = ...............................................................................................Ans. 4 abc 6 xC = a .................................................................................................................................Ans. 4 b yC = .................................................................................................................................Ans. 4 137 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS W2 = (0.316 )( −3.5343) = −1.1168 lb V3 = (9 )(9 )(0.5) = 20.2500 in.3 2 4 (9 ) = 3.8197 in. 3π W3 = (0.100 )( 20.2500 ) = 2.0250 lb Wtot = 10.9597 lb Vtot = 48.5243 in.3 xC1 = yC1 = (a) xC = yC = zC = (3.8197 )(31.8086 ) + (3.5)( −3.5343) + (0.25)( 20.2500 ) = 2.35 in. ......Ans. 48.5243 (3.8197 )(31.8086 ) + (3)( −3.5343) + (3)( 20.2500 ) = 3.54 in. ................Ans. 48.5243 (0.25)(31.8086 ) + (0.25)( −3.5343) + (3.5)( 20.2500 ) = 1.606 in. .........Ans. 48.5243 (b) xG = yG = zG = (3.8197 )(10.0515) + (3.5)( −1.1168) + (0.25)( 2.0250 ) = 3.19 in. .........Ans. 10.9597 (3.8197 )(10.0515) + (3)( −1.1168) + (3)( 2.0250 ) = 3.75 in. ...................Ans. 10.9597 (0.25)(10.0515) + (0.25)( −1.1168) + (3.5)( 2.0250 ) = 0.851 in. ............Ans. 10.9597 5-106 A cylinder with a conical cavity and a hemispherical cap is shown in Fig. P5-106. Locate (a) The centroid of the composite volume if R = 200 mm, and h = 250 mm. (b) The center of gravity of the composite volume if the cylinder is made of brass (ρ = 8750 kg/m3) and the cap is made of aluminum (ρ = 2770 kg/m3). SOLUTION Cylinder: V1 = π ( 200 ) ( 250 ) = 31.416 ×106 mm3 2 m1 = (8750 ) (31.416 ×10−3 ) = 274.89 kg Hemisphere: V2 ( 4 3)π ( 200 ) = 2 2 3 = 16.755 ×106 mm3 m2 = ( 2770 ) (16.755 ×10−3 ) = 46.412 kg Cavity: π ( 200 ) ( 250 ) V3 = − = −10.472 × 106 mm3 3 m3 = (8750 ) ( −10.472 × 10−3 ) = −91.630 kg Vtot = 37.699 ×106 mm3 zC 2 = zG 2 = 250 + zC 3 = zG 3 = 3 ( 200 ) = 325 mm 8 mtot = 229.672 kg Totals: 250 = 62.5 mm 4 143 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (a) RILEY, STURGES AND MORRIS zC = 37.699 xC = yC = 0 mm (by symmetry) ..................................................................................Ans. zG = 229.672 xG = yG = 0 mm (by symmetry) ..................................................................................Ans. (125)(31.416 ) + (325)(16.755) + (62.5)( −10.472 ) = 231 mm ...............Ans. (125)( 274.89 ) + (325)( 46.412 ) + (62.5)( −91.630 ) = 190.4 mm ...........Ans. (b) A slender rod is made of two materials; the segment along the x-axis is aluminum (γ = 0.100 lb/in3) and the remainder is made of brass (γ = 0.316 lb/in3). Locate the center of gravity of the slender rod. SOLUTION Note that the curved portion of the wire is a half circle. Then, 5-107 L1 = 16 in. L2 = 14 in. L3 = xG 3 W1 = 0.100 (16 ) = 1.600 lb W2 = 0.316 (14 ) = 4.424 lb xG1 = 8 in. xG 2 = 0 in. yG1 = 0 in. yG 2 = 7 in. zG1 = 0 in. zG 2 = 0 in. πd = π (9.9 ) = 31.102 in. 2 = 0 in. W3 = 0.316 (31.102 ) = 9.828 lb yG 3 = 7 + zG 3 = 7 + 9.9sin (π 2 ) cos 45° = 11.457 in. π2 9.9sin (π 2 ) sin 45° = 11.457 in. π2 Wtot = 15.852 lb xG = yG = zG = 5-108 (8)(1.600 ) + (0 )( 4.424 ) + (0 )(9.828) = 0.807 in. ...........................................Ans. 15.852 (0 )(1.600 ) + (7 )( 4.424 ) + (11.457 )(9.828) = 9.06 in. ..................................Ans. 15.852 (0 )(1.600 ) + (0 )( 4.424 ) + (11.457 )(9.828) = 7.10 in. ...................................Ans. 15.852 The loads acting on a beam are distributed in a triangular manner as shown in Fig. P5-108. Determine and locate the resultant with respect to the left end of the beam. SOLUTION R1 = R2 = ( 2 )(750 ) = 750 N 2 ( 4 )(750 ) = 1500 N 2 R = 750 + 1500 = 2250 N ↓ .........................................................................................Ans. M A = 2250d = 750 ( 2 3)( 2 ) + 1500 2 + ( 4 3) d = 2.67 m .........................................................................................................................Ans. 144 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 5-109 RILEY, STURGES AND MORRIS Determine the resultant of the distributed loads acting on the beam shown in Fig. P5-109, and locate its line of action with respect to the support at A. SOLUTION R1 = ( 200 )(3) = 600 lb R2 = (150 )(3) = 450 lb R3 = (100 )(3) = 300 lb R = 600 + 450 + 300 = 1350 lb ↓ ................................................................................Ans. M A = 1350d = 600 (1.5 ) + 450 ( 4.5 ) + 300 (7.5 ) d = 3.83 ft ..........................................................................................................................Ans. 5-110 A distributed load acts on the beam shown in Fig. P5-110. Determine the resultant of the distributed load and locate its line of action with respect to the support at A. SOLUTION R1 = ( 2.5 )( 4 ) = 10 kN R2 = ( 2.5)( 4 ) = 5 kN 2 R = 10 + 5 = 15 kN ↓ ......................................................................................................Ans. M A = 15d = 10 ( 4 ) + 5 2 + (8 3) d = 4.22 m .........................................................................................................................Ans. 5-111 A distributed load acts on a beam as shown in Fig. P5-111. Determine and locate the resultant of the distributed load with respect to the support at A. SOLUTION R1 = ∫ w dx = ∫ 3 x 2 dx = x 3 = 1000 lb 0 0 10 10 M A1 = R1d1 = ∫ xw dx = ∫ R2 = (300 )(5 ) = 1500 lb 10 0 3x 4 3 x dx = = 7500 lb ⋅ ft 4 0 3 10 R = 1000 + 1500 = 2500 lb ↓ .......................................................................................Ans. M A = 2500d = 7500 + 1500 (10 + 2.5 ) d = 10.50 ft ........................................................................................................................Ans. 5-112 Determine the resultant of the distributed load acting on the beam shown in Fig. P5-112 and locate its line of action with respect to the support. SOLUTION 100 x3 R1 = ∫ w dx = ∫ 100 x dx = = 266.667 N 0 3 0 2 2 2 M A1 = R1d1 = ∫ xw dx = ∫ 100 x 3 dx = 25 x 4 = 400 N ⋅ m 0 0 2 2 145 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS R2 = ( 400 )( 2 ) = 400 N 2 R = 266.667 + 400 = 666.667 ≅ 667 N ↓ ................................................................Ans. M A = 666.667d = 400 + 400 2 + ( 2 3) d = 2.20 m .........................................................................................................................Ans. 5-113 Determine the resultant of the distributed load acting on the beam shown in Fig. P5-113 and locate its line of action with respect to the support. SOLUTION R1 = ( 200 )(6 ) = 600 lb 2 ↓ R2 = ( 200 )( 2 ) = 400 lb ↓ R3 = (100 )(6 ) = 600 lb ↑ R = −600 − 400 + 600 = −400 lb = 400 lb ↓ ............................................................Ans. M A = 400d = 600 ( 4 ) + 400 (7 ) − 600 (3) d = 8.50 ft ..........................................................................................................................Ans. 5-114 A gate is used to hold water as shown in Fig. P5-114. If the width of the gate is 2 m. determine the resultant of the water pressure acting on the gate, and locate its line of action with respect to the bottom of the gate. SOLUTION (9800 )(9 ) (9 )( 2 ) = 793,800 N = 794 kN ....................................................Ans. R= 2 d = 9 3 = 3 m (from bottom) ........................................................................................Ans. 5-115 A flexible cable is used to tether the balloon shown in Fig. P5-115. The cable weighs 1.2 lb/ft along its length. Determine the magnitude of the resultant force and its location with respect to A. SOLUTION y = x 2 10 dL = dy dx = x 5 2 2 ( dx ) + ( dy ) = 1 + ( dy dx ) dx = 1 + ( x 5 ) dx = 2 2 1 5 52 + x 2 dx 15 1 R = ∫ w dL = ∫ 1.2 0 5 52 + x 2 dx 15 1 = 0.24 x x 2 + 52 + 52 ln x + x 2 + 52 = 33.916 lb ...........................Ans. 0 2 ( ) RxR = ∫ xw dL = ∫ 0.24 x 52 + x 2 dx 0 15 1 = 0.24 3 xR = 3 ( x2 + 52 ) = 306.228 lb ⋅ ft 0 15 306.228 = 9.029 ft .................................................................................................Ans. 33.916 2 yR = xR 10 = 9.0292 10 = 8.15 ft ................................................................................Ans. 146 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 5-116 RILEY, STURGES AND MORRIS A loaded rivet causes a force distribution or pressure p on a plate, as shown in Fig. P5-116. The rivet has a diameter of 25 mm, and the plate is 15 mm thick. Determine the magnitude of the resultant force acting on the plate. Let po = 400 N/m2. SOLUTION 0.025 −4 dA = ( r dθ ) t = ( 0.015 ) dθ = (1.875 × 10 ) dθ 2 dRy = ( p dA ) cos θ = ( 400 cos θ ) (1.875 × 10−4 ) dθ cos θ = 0.075cos 2 θ dθ Ry = ∫ dRy = ∫ π /2 −π / 2 0.075cos 2 θ dθ π /2 1 + cos 2θ θ sin 2θ dθ = 0.075 + = 0.075∫ −π / 2 2 4 −π / 2 2 π /2 dRx = ( p dA ) sin θ = ( 400 cos θ ) (1.875 ×10−4 ) dθ sin θ = 0.0375sin 2θ dθ Rx = ∫ dRx = ∫ 5-117 = 0.1178 N ↓ ............................................................................................................Ans. cos 2θ 0.0375sin 2θ dθ = 0.0375 − −π / 2 2 −π / 2 π /2 π /2 = 0 N ............................................................................................................................Ans. The lift on the wing of an airplane due to aerodynamic forces is shown in Fig. P5-117. The lift L may be described by the function L = 25 A. SOLUTION x lb/ft . Determine the moment of the resultant lift force about point dM = x dL = x 25 x dx M = ∫ dM = ∫ 25 x 0 15 3/ 2 ( ) 2 x5 / 2 dx = 25 = 8710 lb ⋅ ft .........................................Ans. 5 0 15 5-118 Determine the moment of the 1650-N force shown in Fig. P5-118 about point O. SOLUTION F = 1650 −180 i + 180 j + 80 k 1802 + 1802 + 802 = ( −1113.055 i + 1113.055 j + 494.691k ) N M O = ( 0.36 i + 0.24 j + 0.4 k ) × ( −1113.055 i + 1113.055 j + 494.691k ) = ( −326 i − 623 j + 668 k ) N ⋅ m ..........................................................................Ans. 5-119 The driving wheel of a truck is subjected to the force-couple system shown in Fig. P5-119. Replace this system by an equivalent single force and determine the point of application of the force along the vertical diameter of the wheel. SOLUTION Rx = 750 lb M O = 600 lb ⋅ ft = 750 y Ry = 1800 − 1800 = 0 lb R = 750 lb → ...................................................................................................................Ans. y = 0.8 ft = 9.60 in. up from the ground ..................................................................Ans. 147 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 5-120 RILEY, STURGES AND MORRIS A 200-N force is applied at corner B of a rectangular plate as shown in Fig. P5-120. Determine the moment of the force (a) About point O. (b) About line OD. SOLUTION (a) F = 200 0.46 i + 0.5 j − 0.9 k 0.462 + 0.52 + 0.92 = (81.5854 i + 88.6798 j − 159.6237 k ) N M O = ( −0.8 i + 2 k ) × (81.5854 i + 88.6798 j − 159.6237 k ) = ( −177.360 i + 35.472 j − 70.944 k ) N ⋅ m .....................................................Ans. (b) eOD = −0.8 i + 0.5 j 0.82 + 0.52 = ( −0.84800 i + 0.53000 j) M OD = M O eOD = ( −177.360 )( −0.84800 ) + (35.472 )(0.53000 ) = 169.2 N ⋅ m ..........................................................................................................Ans. M OD = M OD eOD = 169.2 ( −0.84800 i + 0.53000 j) = ( −143.5 i + 89.7 j) N ⋅ m ..................................................................................Ans. 5-121 A 2500-lb jet engine is suspended from the wing of an airplane as shown in Fig. P5-121. Determine the moment produced by the engine at point A in the wing when the plane is (a) On the ground with the engine not operating. (b) In flight with the engine developing a thrust T of 15,000 lb. SOLUTION (a) (b) M = 2500 (8sin 35° ) = 11, 470 lb ⋅ ft = −86,800 lb ⋅ ft = 86,800 lb ⋅ ft ..................................................................Ans. M = 2500 (8sin 35° ) − 15, 000 (8cos 35° ) ...............................................................Ans. Two forces and a couple act on a beam as shown in Fig. P5-122. Determine the resultant R and its location x. SOLUTION 5-122 R = 1.8 + 1 = 2.8 kN ↓ ...................................................................................................Ans. M O = (1.8 )(1) + (1)( 2 ) + 10 = 13.800 kN ⋅ m = 2.8 x x = 4.92 m .........................................................................................................................Ans. 5-123 Replace the force and couple acting on the wall bracket shown in Fig. P5-123 by a force-couple system at point A. SOLUTION Rx = 0 lb Ry = −40 lb R = 40 lb ↓ .......................................................................................................................Ans. M A = 40 (8 ) + 30 = 350 lb ⋅ in. ..................................................................................Ans. 148 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 5-124 RILEY, STURGES AND MORRIS Determine the resultant of the parallel force system shown in Fig. P5-124 and locate the intersection of its line of action with the xy-plane. SOLUTION R = (125 k ) + ( −80 k ) + (100 k ) + ( −200 k ) + ( −75 k ) = ( −130 k ) N ................Ans. M O = (3 i + 1.5 j) × (125 k ) + (1 i + 1.5 j) × ( −80 k ) + ( 2 i + 3.5 j) × (100 k ) + (3 i + 5.5 j) × ( −200 k ) + (1 i + 5.5 j) × ( −75 k ) = ( −1095 i + 180 j) N ⋅ m = ( x i + y j) × ( −130 k ) = ( −130 y i + 130 x j) x= 5-125 180 = 1.385 m 130 y= 1095 = 8.42 m ..........................................Ans. 130 The concrete floor of a building supports four wood columns as shown in Fig. P5-125. The resultant of the forces transmitted through the columns to the floor is R = 75 kip at the location shown in the figure. Determine the magnitude of the forces F1 and F2. SOLUTION R = 15 + 20 + F1 + F2 = 75 kip ↓ M O = ( 20 i ) × ( −15 k ) + ( 20 i + 30 j) × ( − F1 k ) + (30 j) × ( − F2 k ) = −30 ( F1 + F2 ) i + (300 + 20 F1 ) j = (13 i + 16 j) × ( −75 k ) = ( −1200 i + 975 j) kip ⋅ ft F1 = 33.7 kip ↓ ................................................................................................................Ans. F2 = 6.25 kip ↓ ................................................................................................................Ans. 5-126 A bent rod supports a 450-kN force F as shown in Fig. P5-126. (a) Replace the 450-N force with a force R through the coordinate origin O and a couple C. (b) Determine the twisting moments produced by force F in the three different segments of the rod. SOLUTION xD = 425cos 45° − 750 = −449.48 mm yD = 300 − 150 = 150 mm (a) z D = −425sin 45° = −300.52 mm R = ( 450 k ) N ..................................................................................................................Ans. C = ( −0.44948 i + 0.150 j − 0.30052 k ) × ( 450 k ) = ( 67.5 i + 202.3 j) N ⋅ m ..........................................................................................Ans. M OA = M O j = 202.3 N ⋅ m ............................................................................................Ans. M B = ( −0.75 i − 0.15 j) × ( 450 k ) = ( −67.5 i + 337.5 j) N ⋅ m e AB = (b) ( 425cos 45° ) i − ( 425sin 45° ) k = 425 (0.70711i − 0.70711k ) M AB = M B e AB = ( −67.5 )( 0.70711) = −47.7 N ⋅ m ................................................Ans. M BC = M B i = −67.5 N ⋅ m ...........................................................................................Ans. 149 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 5-127 Locate the centroid of the shaded area shown in Fig. P5-127. SOLUTION RILEY, STURGES AND MORRIS ) dA 2 y − 0.5 y ) dy A = ∫ dA = ∫ ( 2 x − 0.5 x ) dx dAV = H ( =( 2 x − 0.5 x 2 dx 2 2 2 V 0 2 x 3/ 2 x 3 4 = − = in.2 6 0 3 32 M y = ∫ x dAV = ∫ x 0 2 2 ( 2 x5 / 2 x 4 6 2 x − 0.5 x dx = − = in.3 8 0 5 52 2 ) 2 xC = My A = 6 5 18 = = 0.9 in. ......................................................................................Ans. 4 3 20 2 M x = ∫ y dAH = ∫ y 0 ( 2 y5/ 2 y 4 6 2 y − 0.5 y dy = − = in.3 8 0 5 52 2 ) 2 yC = 5-128 M x 6 5 18 = = = 0.9 in. ......................................................................................Ans. A 4 3 20 Two channel sections and a plate are used to form the cross section shown in Fig. P5-128. Each of the channels has a cross-sectional area of 2605 mm2. Locate the y-coordinate of the centroid of the composite section with respect to the top surface of the plate. SOLUTION yC = −737,190 = −64.3 mm ........................Ans. 11, 460 5-129 Locate the centroid of the volume shown in Fig. P5-129 if R = 10 in. and h = 32 in. SOLUTION Ry 2 dV = π r dy = π 2 dy h 2 2 V = ∫ dV = ∫ = h 0 π R2 y4 π R2 y5 dy = 4 h4 h 5 0 h π R2h 5 150 STATICS AND MECHANICS OF MATERIALS, 2nd Edition h 2 h RILEY, STURGES AND MORRIS Ry 2 π R 2 y 6 π R 2h2 M xz = ∫ y dV = ∫ yπ 2 dy = 4 = 0 h 6 0 6 h M π R 2 h 2 6 5h 5 (32 ) yC = xz = = = = 26.7 in. ........................................................Ans. π R2h 5 V 6 6 xC = zC = 0 in. (by symmetry) .....................................................................................Ans. 5-130 Locate the centroid and the mass center of the volume shown in Fig. P5-130, which consists of an aluminum cylinder (ρ= 2770 kg/m3) and a steel (ρ = 7870 kg/m3) cylinder and sphere. SOLUTION V1 = π ( 0.1) ( 0.3) = 9.4248 × 10−3 m3 2 m1 = 2770 (9.4248 ×10−3 ) = 26.1066 kg V2 = π (0.05 ) (0.175 ) = 1.3744 ×10−3 m3 2 m2 = 7870 (1.3744 ×10−3 ) = 10.8169 kg 4π (0.1) V3 = = 4.1888 ×10−3 m3 3 3 m3 = 7870 ( 4.1888 ×10−3 ) = 32.9658 kg mtot = 69.8893 kg 14.9880 Vtot = 14.9880 ×10−3 m3 yC = yG = (0 )(9.4248) + (187.5)(1.3744 ) + (375)( 4.1888) = 122.0 mm .................Ans. (0 )( 26.1066 ) + (187.5)(10.8169 ) + (375)(32.9658) = 205.9 mm .........Ans. 69.8893 xC = xG = zC = zG = 0 mm (by symmetry) ................................................................Ans. 5-131 Determine the resultant R of the system of distributed loads on the beam of Fig. P5-131, and locate its line of action with respect to the left support of the beam. SOLUTION R = ∫ 200 x dx + ( 400 )( 4 ) + 0 4 4 ( 400 )( 4 ) 2 2 = 200 x 3/ 2 + 1600 + 800 = 3466.667 lb ≅ 3470 lb ↓ ....................................Ans. 3 0 4 ( 400 )( 4 ) M O = ∫ x 200 x dx + ( 400 )( 4 ) (6 ) + 8 + ( 4 3) 0 2 ( ) 4 2 = 200 x5 / 2 + 9600 + 7466.667 = 19, 626.667 lb ⋅ ft 5 0 19, 626.667 x= = 5.66 ft ...............................................................................................Ans. 3466.667 151 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS Chapter 6 6-1 Draw a free-body diagram of the angle bracket shown in Fig. P6-1. SOLUTION 6-4 Draw a free-body diagram of the sled shown in Fig. P6-4. SOLUTION 6-5 6-2 Draw a free-body diagram of the diving board shown in Fig. P6-2. The surface at B is smooth. Neglect the weight of the diving board. SOLUTION Draw a free-body diagram of the bracket shown in Fig. P6-5. The contact surfaces between the cylinders and bracket are smooth. SOLUTION 6-6 6-3 Draw a free-body diagram of the lawn mower shown in Fig. P6-3. The lawn mower has a weight W and is resting on a rough surface. SOLUTION Forces P are applied to the handles of the pipe pliers shown in Fig. P6-6. Draw free-body diagrams of each handle. SOLUTION 152 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 6-7 The man shown in Fig. P6-7 has a weight W1; the beam has a weight W2 and a center of gravity G. Draw a free-body diagram of the beam. SOLUTION RILEY, STURGES AND MORRIS 6-10 Draw a free-body diagram of the bent bar shown in Fig. P6-10. The support at A is fixed, and the bar has negligible mass. SOLUTION 6-8 Forces P are applied to the handles of the bolt cutter shown in Fig. P6-8. Draw a free-body diagram of (a) The lower handle. (b) The lower cutter jaw. SOLUTION 6-11 Draw a free-body diagram of the bar shown in Fig. P6-11. The support at A is a journal bearing and the supports at B and C are ball bearings. SOLUTION 6-9 Draw a free-body diagram of the door shown in Fig. P6-9. The homogeneous door has weight W. SOLUTION 6-12 Draw a free-body diagram of the shaft shown in Fig. P6-12. The bearing at A is a thrust bearing, and the bearing at D is a ball bearing. Neglect the weights of the shaft and the levers. SOLUTION 153 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 6-13 A beam is loaded and supported as shown in Fig. P6-13. Determine the reactions at supports A and B. SOLUTION From a free-body diagram of the beam, the equilibrium equations are solved to get the forces → ΣFx = 0 : ΣM A = 0 : Ax = 0 15 B − 3 ( 500 ) − 6 ( 800 ) − 9 ( 700 ) − 12 ( 400 ) = 0 B = 1160 lb ↑ ΣFy = 0 : Ay + B − 500 − 800 − 700 − 400 = 0 Ay = 1240 lb A = 1240 lb ↑ ............................................................................................................................ Ans. B = 1160 lb ↑ ............................................................................................................................ Ans. 6-14 A beam is loaded and supported as shown in Fig. P6-14. Determine the reaction at support A. SOLUTION From a free-body diagram of the beam, the equilibrium equations are solved to get the forces and moment → ΣFx = 0 : ↑ ΣFy = 0 : Ay = 2 kN Ax = 0 Ay − 2 = 0 ΣM A = 0 : M A − 2 ( 4) − 3 = 0 M A = 11 kN ⋅ m A = 2 kN ↑ ................................................................................................................................ Ans. M A = 11 kN ⋅ m ..................................................................................................................... Ans. 6-15 A 30-lb force P is applied to the brake pedal of an automobile as shown in Fig. P6-15. Determine the force Q applied to the brake cylinder and the reaction at support A. SOLUTION From a free-body diagram of the brake pedal, the equilibrium equations are solved to get the forces ΣM A = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : Ay = 15 lb 5.5Q − ( 30 cos 30° )(11) − ( 30sin 30° )( 4 ) = 0 Ax − Q + 30 cos 30° = 0 Ay − 30sin 30° = 0 Q = 62.871 lb Ax = 36.890 lb A = 39.8 lb 22.13° ............................................................ Ans. Q = 62.9 lb ← ........................................................................ Ans. 154 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 6-16 A beam is loaded and supported as shown in Fig. P6-16. Determine the reactions at supports A and B. SOLUTION From a free-body diagram of the beam, the equilibrium equations are solved to get the forces → ΣFx = 0 : ΣM A = 0 : Ax = 0 4.5B − ( 300 )(1.5) ( 0.75 ) 1 − ( 400 )(1.5 ) ( 3 + 0.5) = 0 2 B = 308.333 N ↑ ΣFy = 0 : 1 Ay − ( 300 )(1.5 ) − ( 400 )(1.5 ) + B = 0 2 Ay = 441.667 N A = 442 N ↑ .............................................................................................................................. Ans. B = 308 N ↑ .............................................................................................................................. Ans. 6-17 A rope and pulley system is used to support a body W as shown in Fig. P6-17. Each pulley is free to rotate and the rope is continuous over the pulleys. Determine the tension T in the rope required to hold body W in equilibrium if the weight of body W is 400 lb. Assume that all rope segments are vertical. SOLUTION The pulleys are rigidly connected together and can be treated as a single body for the purpose of drawing a free-body diagram. From a free-body diagram of the pulleys, the equilibrium equations are solved to get the tension ↑ ΣFy = 0 : 4T − 400 = 0 T = 100 lb ................................................................................................ Ans. 6-18 Pulleys 1 and 2 of the rope and pulley system shown in Fig. P6-18 are connected and rotate as a unit. The radii of pulleys 1 and 2 are 100 mm and 300 mm, respectively. Rope A is wrapped around pulley 1 and is fastened to pulley 1 at point A’. Rope B is wrapped around pulley 2 and is fastened to pulley 2 at point B’. Rope C is continuous over pulleys 3 and 4. Determine the tension T in rope C required to hold body W in equilibrium if the mass of body W is 225 kg. SOLUTION W = 225 ( 9.81) = 2207.25 N From a free-body diagram of the pulley 4, the equilibrium equations are solved to get ↑ ΣFy = 0 : 2T + TB − W = 0 TB = W − 2T Next, from a free-body diagram of the pulley 3, the equilibrium equations are solved to get ↑ ΣFy = 0 : TA = 2T TA − 2T = 0 155 STATICS AND MECHANICS OF MATERIALS, 2nd Edition Finally, from a free-body diagram of the pulleys 1 and 2, the equilibrium equations are solved to get RILEY, STURGES AND MORRIS ΣM axle = 0 : 100TA − 300TB = 0 100TA = 300TB 100 ( 2T ) = 300 (W − 2T ) 8T = 3W = 6621.75 N T = 828 N ................................................................................................ Ans. 6-19 Three pipes are supported in a pipe rack as shown in Fig. P6-19. Each pipe weighs 100 lb. Determine the reactions at supports A and B. SOLUTION From a free-body diagram of the pipes and the rack, the equilibrium equations are solved to get the forces ΣM A = 0 : 18 B − (100 cos 30° )( 9 ) − (100 cos 30° )(15 ) B = 249.840 lb − (100 cos 30° )( 21) − 3 (100sin 30° )( 4 ) = 0 → ΣFx = 0 : ↑ ΣFy = 0 : Ax + B sin 30° = 0 Ay − 300 + B cos 30° = 0 Ax = −124.920 lb Ay = 83.632 lb A = 150.3 lb 33.80° ............................................................................................................ Ans. B = 250 lb 60° ..................................................................................................................... Ans. 6-20 The man shown in Fig. P6-20 has a mass of 75 kg; the beam has a mass of 40 kg. The beam is in equilibrium with the man standing at the end and pulling on the cable. Determine the force exerted on the cable by the man and the reaction at support C. SOLUTION From a free-body diagram of the man and the beam, the equilibrium equations are solved to get the forces → ΣFx = 0 : ΣM C = 0 : Cx = 0 75 ( 9.81) ( 3) + 40 ( 9.81) (1.5 ) −3T − 1.5T = 0 T = 621.300 N ↑ ΣFy = 0 : 2T + C y − 75 ( 9.81) − 40 ( 9.81) = 0 C y = −114.450 N C = 114.5 N ↓ ........................................................................................................................... Ans. T = 621 N .................................................................................................................................... Ans. 156 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 6-21 A 75-lb load is supported by an angle bracket, pulley, and cable as shown in Fig. P6-21. Determine (a) The force exerted on the bracket by the pin at C. (b) The reactions at supports A and B of the bracket. (c) The shearing stress on a cross section of the ¼-in.-diameter pin at C, which is in double shear. SOLUTION (a) First, from a free-body diagram of the pulley, the equilibrium equations are solved to get the axle forces → ΣFx = 0 : ↑ ΣFy = 0 : ( 4 5 )( 75 ) − Cx = 0 C y − 75 − ( 3 5)( 75) = 0 63.43° on the pulley 63.43° on the bracket .......................... Ans. C x = 60 lb C y = 120 lb C = 134.2 lb C = 134.2 lb (b) Next, from a free-body diagram of the bracket, the equilibrium equations are solved to get the forces → ΣFx = 0 : ΣM A = 0 : ↑ ΣFy = 0 : Ax + 60 = 0 10 B − 18 ( 60 ) = 0 Ay + B − 120 = 0 11.31° ........................................................................................................... Ans. Ax = −60 lb B = 108 lb Ay = 12 lb A = 61.2 lb B = 108.0 lb ↑ ....................................................................................................................... Ans. (c) Finally, dividing the shear force on the pin by twice its cross-sectional area (since the pin is in double shear) gives the shear stress τC = C 134.2 = = 1367 psi ........................................................................... Ans. 2 As 2 π ( 0.25) 2 4 6-22 A pipe strut BC is loaded and supported as shown in Fig. P6-22. Determine (a) The reactions at supports A and C. (b) The shearing stress on a cross section of the 10-mm-diameter pin at C, which is in double shear. (c) The elongation of cable AB if it is made of aluminum alloy (E = 73 GPa) and has a diameter of 6 mm. SOLUTION (a) From a free-body diagram of the pipe strut, the equilibrium equations are solved to get the forces ΣM C = 0 : → ΣFx = 0 : 1000TAB − 800 ( 750 ) = 0 C x − TAB = 0 TAB = 600 N ...................................................................Ans. Cx = 600 N 157 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ↑ ΣFy = 0 : C y − 750 = 0 51.34° ..................................................................................... Ans. C y = 750 N C = 960.47 N ≅ 960 N (b) Dividing the shear force on the pin by twice its cross-sectional area (since the pin is in double shear) gives the shear stress τC = C 960.47 = = 6.11×106 N/m 2 = 6.11 MPa ................................... Ans. 2 2 As 2 π ( 0.010 ) 4 600 (1600 ) PL = = 0.465 mm ..................................................... Ans. EA ( 73 × 109 ) π ( 0.006 ) 2 4 (c) Finally, the stretch of the cable AB is given by δ AB = 6-23 The lawn mower shown in Fig. P6-23 weighs 35 lb and has a center of gravity at G. Determine (a) The magnitude of the force P required to push the mower at a constant velocity. (b) The forces exerted on the front and rear wheels by the inclined surface. (c) The shearing stresses in the ½-in.-diameter shoulder bolts of the front and rear wheels. SOLUTION (a) From a free-body diagram of the lawn mower, the equilibrium equations are solved to get the forces ΣM O = 0 : ΣFx = 0 : 27 N 2 − ( 35cos15° )(13) + ( 35sin15° )( 4 ) = 0 P cos 30° − 35sin15° = 0 N 2 = 14.936 lb .................................................Ans. P = 10.460 lb ...................................................Ans. ΣFy = 0 : N1 + N 2 − 35cos15° − P sin 30° = 0 N1 = 24.101 lb .................................................Ans. (c) Finally, dividing the shear forces on the pins by their cross-sectional areas gives the shear stresses τ1 = τ2 = N1 24.101 = = 122.7 psi ...................................................................................... Ans. A1s π ( 0.5 )2 4 N2 14.936 = = 76.1 psi ....................................................................................... Ans. A2 s π ( 0.5 )2 4 6-24 The coal wagon shown in Fig. P6-24 is used to haul coal from a mine. If the mass of the coal and wagon is 2000 kg (with its center of mass at G), determine (a) The magnitude of the force P required to move the wagon at a constant velocity. (b) The force exerted on each of the front wheels by the inclined surface. (c) The shearing stress in each of the 25-mm-diameter front axles. SOLUTION W = 2000 ( 9.81) = 19, 620 N (a) From a free-body diagram of the coal wagon, the equilibrium equations are solved to get the forces ΣFx = 0 : P − W sin 30° = 0 158 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS P = 9810 N ....................................................................Ans. ΣM 1 = 0 : ( 2 N 2 )( 3) − P (1) − (W cos 30° )( 2 ) + (W sin 30° )( 2 ) = 0 2 N1 + 2 N 2 − W cos 30° = 0 N 2 = 4028.81 N ............................................................Ans. ΣFy = 0 : N1 = 4466.90 N (c) Finally, dividing the shear force on the front axle by its cross-sectional area gives the shear stress τ2 = N2 4028.81 = = 8.21× 106 N/m 2 = 8.21 MPa ............................................ Ans. A2 s π ( 0.025 )2 4 6-25 A lever is loaded and supported as shown in Fig. P6-25. Determine (a) The reactions at A and C. (b) The normal stress in the ½-in.-diameter rod CD. (c) The shearing stress in the ½-in.-diameter pin at A, which is in double shear. (d) The change in length of rod CD, which is made of a material with a modulus of elasticity of 30(106) psi. SOLUTION θ = tan −1 ( 6 12 ) = 26.565° ΣM A = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : LCD = 62 + 122 = 13.416 in. (a) From a free-body diagram of the lever, the equilibrium equations are solved to get the forces (TCD cosθ )( 6 ) − 40 ( 4 ) − 60 (8) + 80 (12 ) = 0 Ax + 125 + TCD cos θ = 0 Ay − 40 − 60 + 80 − TCD sin θ = 0 5.31° .................................. Ans. TCD = −59.628 lb Ax = −71.667 lb Ay = −6.666 lb A = 72.0 lb TCD = 59.6 lb (C) ......................................... Ans. (b) Dividing the normal force in the rod CD by its cross-sectional area gives the normal stress σ CD = TCD −59.628 = = −304 psi = 304 psi (C) ....................................................... Ans. An π ( 0.5 ) 2 4 (c) Dividing the shear force on the pin by twice its cross-sectional area (since it is in double shear) gives the shear stress τA = VA 72.0 = = 183.3 psi ............................................................................ Ans. 2 As 2 π ( 0.5 )2 4 (d) Finally, the change in length of the rod CD is given by 159 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS δ CD = ( −59.628 )(13.416 ) PL = EA ( 30 × 106 ) π ( 0.5 ) 2 4 = −1.358 ×10−4 in. = 1.358 ×10−4 in. (shrink) ................................................................ Ans. 6-26 The wood plane shown in Fig. P6-26 moves with a constant velocity when subjected to the forces shown. Determine (a) The shearing force of the wood on the plane. (b) The normal force, and its location, of the wood on the plane. SOLUTION From a free-body diagram of the plane, the equilibrium equations are solved to get the forces → ΣFx = 0 : ↑ ΣFy = 0 : ΣM A = 0 : F − 40 cos 70° − 70 cos16° = 0 N − 40 sin 70° − 70sin16° = 0 ( 70 cos16° )( 75) − ( 70sin16° )( 220 ) + ( 40 cos 70° )( 60 ) + Nd = 0 F = 81.0 N ............................................................................................................ Ans. N = 56.882 N ....................................................................................................... Ans. d = −28.5 mm 31.5 mm (from front of plane) ........................................................................ Ans. 6-27 A bracket of negligible weight is used to support the distributed load shown in Fig. P6-27. Determine (a) The reactions at the supports A and B. (b) The shearing stress in the 0.25-in.-diameter pin at A, which is in single shear. (c) The bearing stress between the bracket and the 1-in.×1-in. bearing plate at B. SOLUTION (a) From a free-body diagram of the bracket, the equilibrium equations are solved to get the forces ΣM A = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : 18 B − (12 )(10 ) 2 (12 3) = 0 Ax + B = 0 Ay − (12 )(10 ) 2 = 0 B = 13.333 lb Ax = −13.333 lb Ay = 60.0 lb A = 61.464 lb ≅ 61.5 lb 77.47° ............................................Ans. B = 13.33 lb → ...............................................................................Ans. (b) Dividing the shear force on the pin by its cross-sectional area gives the shear stress τA = VA 61.464 = = 1252 psi ..................................................................................... Ans. As π ( 0.25 )2 4 B 13.333 = = 13.33 psi ............................................................................................. Ans. Ab 1× 1 (c) Finally, the bearing stress is given by σb = 160 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 6-28 Determine the force P required to push the 135-kg cylinder over the small block shown in Fig. P6-28. SOLUTION θ = cos −1 220 − 75 = 48.769° 220 φ = 90° − θ − 20° = 21.231° From a free-body diagram of the cylinder, the equilibrium equations are solved to get the forces ΣM center = 0 : −220 F = 0 F =0 N → ΣFx = 0 : ↑ ΣFy = 0 : P − N cos φ = 0 N sin φ − 135 ( 9.81) = 0 N = 3657.2 N P = 3410 N → ...................................................................................................................... Ans. 6-29 The electric motor shown in Fig. P6-29 weighs 25 lb. Due to friction between the belt and pulley, the belt forces have magnitudes of T1 = 21 lb and T2 = 1 lb. Determine (a) The support reactions at A and B. (b) The shearing stress in the ¼-in.-diameter pin at A, which is in single shear. SOLUTION (a) From a free-body diagram of the motor and support, the equilibrium equations are solved to get the forces → ΣFx = 0 : ΣM A = 0 : ↑ ΣFy = 0 : Ax − 21 − 1 = 0 12 B − 8 ( 25 ) + 5.5 (1) + 10.5 ( 21) = 0 Ay + B − 25 = 0 Ax = 22 lb B = −2.167 lb Ay = 27.167 lb A = 34.958 lb ≅ 35.0 lb 51.00° ................................................................................... Ans. B = 2.17 lb ↓ ......................................................................................................................... Ans. (b) Dividing the shear force on the pin by its cross-sectional area gives the shear stress τA = VA 34.958 = = 712 psi ...................................................................................... Ans. As π ( 0.25 ) 2 4 6-30 Bar AB of Fig. P6-30 has a uniform cross section, a mass of 25 kg, and a length of 1 m. Determine the angle θ for equilibrium. SOLUTION From a free-body diagram of the bar, the equilibrium equations are → ΣFx = 0 : ↑ ΣFy = 0 : ΣM A = 0 : A sin 30° − B sin 45° = 0 A cos 30° + B cos 45° − 25 ( 9.81) = 0 ( B cos 45° )(1cosθ ) + ( B sin 45° )(1sin θ ) − 25 ( 9.81) ( 0.5cosθ ) = 0 161 STATICS AND MECHANICS OF MATERIALS, 2nd Edition Rearranging the first equation gives RILEY, STURGES AND MORRIS A = 1.41421B Substituting this into the second equation gives B = 126.951 N A = 179.535 N Finally, the third equation gives tan θ = 0.366 θ = 20.10° ................................................................................................................................ Ans. 6-31 The wrecker truck of Fig. P6-31 has a weight of 15,000 lb and a center of gravity at G. The force exerted on the rear (drive) wheels by the ground consists of both a normal component By and a tangential component Bx, while the force exerted on the front wheels consists of a normal force Ay only. Determine the maximum pull P that the wrecker can exert when θ = 30° if Bx cannot exceed 0.8By (because of friction considerations) and the wrecker does not tip over backward (the front wheels remain in contact with the ground). SOLUTION From a free-body diagram of the truck, the equilibrium equations are → ΣFx = 0 : ↑ ΣFy = 0 : ΣM A = 0 : P sin 30° − Bx = 0 Ay + By − 15, 000 − P cos 30° = 0 (15, 000 )( 8) − 14.5 Ay − ( P cos 30° )( 5) − ( P sin 30° )(10 ) = 0 Ay = 0 (the front wheels are on the verge of Bx = 0.8 By (the rear wheels are on the verge of slipping). Guessing that the front wheels The fourth equation needed to solve for the four unknowns is either lifting off the ground) or are on the verge of lifting off the ground gives the solution Ay = 0 lb P = 12,861.56 lb By = 26,138.44 lb Bx = 6430.78 lb Since Bx = 6430.78 lb < 0.8 By = 20,911 lb the guess that the front wheels are on the verge of lifting off the ground was the correct guess, and Pmax = 12,860 lb ..................................................................................................................... Ans. 6-32 Pulleys A and B of the chain hoist shown in Fig. P6-32 are connected and rotate as a unit. The chain is continuous, and each of the pulleys contains slots that prevent the chain from slipping. Determine the force F required to hold a 450-kg block W in equilibrium if the radii of pulleys A and B are 90 mm and 100 mm, respectively. SOLUTION From a free-body diagram of the lower pulley, vertical equilibrium equation gives the tension ↑ ΣFy = 0 : 2T − 450 ( 9.81) = 0 T = 2207.25 N Then, from a free-body diagram of the lower pulley, moment equilibrium equation gives the force F 162 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ΣM axle = 0 : 100T − 90T − 100 F = 0 F = 221 N ................................................................................................................................ Ans. 6-33 The crane and boom shown in Fig. P6-33 weigh 12,000 lb and 600 lb, respectively. When the boom is in the position shown, determine (a) The maximum load that can be lifted by the crane. (b) The tension in the cable used to raise and lower the boom when the load being lifted is 3600 lb. (c) The pin reaction at boom support A when the load being lifted is 3600 lb. (d) The required size of pin A ( which is in double shear) if the shearing stress must not exceed 12 ksi. SOLUTION (a) From a free-body diagram of the entire crane, the equilibrium equations are ↑ ΣFy = 0 : ΣM C = 0 : N − 12, 000 − 600 − W = 0 (12, 000 )( 9 ) − Nd − ( 600 )(12 cos 30° − 1) −W ( 24 cos 30° − 1 + 1) = 0 The maximum load that can be lifted corresponds to impending tip; at which point the normal force acts at the front corner C, and d = 0 . At this point W = Wmax = 4930 lb ............................................................................................................... Ans. (b) From a free-body diagram of the pulley, the equilibrium equations give the axle forces → ΣFx = 0 : ↑ ΣFy = 0 : Bx − 3600 cos10° = 0 By − 3600 − 3600 sin10° = 0 Bx = 3545.31 lb By = 4225.13 lb Then, from the free-body diagram of the boom, the equilibrium equations give the forces ΣM A = 0 : 3545.31( 24sin 30° ) − 4225.13 ( 24 cos 30° ) − 600 (12 cos 30° ) + (T cos10° )( 24 sin 30° ) − (T sin10° )( 24 cos 30° ) = 0 T = 6275.13 lb ≅ 6280 lb ....................................... Ans. → ΣFx = 0 : ↑ ΣFy = 0 : Ax − T cos10° − 3545.31 = 0 Ay − T sin10° − 4225.13 − 600 = 0 Ax = 9725.11 lb Ay = 5914.80 lb A = 11, 382.55 lb ≅ 11.38 kip 31.31° ............ Ans. (d) Finally, dividing the shear force on the pin by twice its cross-sectional area (since it is in double shear) gives the shear stress (which must not exceed 12 ksi) τ= VA 11,382.55 = = 12, 000 psi 2 2 As 2 (π d A 4 ) d A = 0.777 in. ......................................................................................................................... Ans. 163 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 6-34 A bar AB of negligible mass is supported in a horizontal position by two cables as shown in Fig. P6-34. Determine (a) The magnitude of force P and the force in each cable. (b) The change in length of the 15-mm-diamerter cable BD, if it is initially 1 m long and is made of steel with a modulus of elasticity of 200 GPa. SOLUTION (a) From a free-body diagram of the bar, the equilibrium equations give the forces ΣM B = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : 1300 L + (TA sin 60° )( L ) = 0 TB cos 68° − TA cos 60° = 0 TA sin 60° + TB sin 68° − 1300 − P = 0 TA = 1501.11 N ≅ 1501 N ........................................ Ans. TB = 2003.58 N ≅ 2004 N ...................................... Ans. P = 1857.69 N ≅ 1858 N ..................................................................................................... Ans. (b) The stretch of wire BD is given by δ= 2003.58 (1000 ) PL = = 0.0567 mm ................................................... Ans. EA ( 200 × 109 ) π ( 0.015 )2 4 6-35 The garage door ABCD shown in Fig. P6-35 is being raised by a cable DE. The one-piece door is a homogeneous rectangular slab weighing 225 lb. Frictionless rollers B and C run in tracks at each side of the door as shown. Determine (a) The force in the cable and the reactions at the rollers when d = 75 in. (b) The shearing stress (single shear) in the 3/8-in.-diameter rods that connect the roller to the door. SOLUTION θ = cos −1 ( 25 48 ) = 58.612° φ = tan −1 → ΣFx = 0 : ↑ ΣFy = 0 : ΣM D = 0 : 12 − 6sin θ = 5.466° 75 − 6 cos θ (a) From a free-body diagram of the door, the equilibrium equations are TD cos φ − 2 B = 0 TD sin φ − 225 − 2C = 0 225 ( 36 cosθ ) − 2 B ( 48sin θ ) + (TD sin φ )( 6 cosθ ) − ( TD cos φ )( 6sin θ ) = 0 The first equation gives B = 0.49773TD Substituting this into the moment equation gives TD = 92.5356 lb ≅ 92.5 lb .................................................................................................... Ans. B = 46.0574 lb ≅ 46.1 lb ..................................................................................................... Ans. C = −108.0926 lb ≅ 108.1 lb ↑ ......................................................................................... Ans. (b) Dividing the shear force on the rods by their cross-sectional areas gives the shear stresses τB = VB 46.0574 = = 417 psi ......................................................................................... Ans. As π ( 3 8 ) 2 4 164 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS τC = VC 108.0926 = = 979 psi ......................................................................................... Ans. As π ( 3 8 ) 2 4 6-36 The lever shown in Fig. P6-36 is formed in a quarter circular arc of radius 450 mm. Determine the angle θ if neither of the reactions at A or B can exceed 200 N. SOLUTION From a free-body diagram of the lever, the equilibrium equations are → ΣFx = 0 : ↑ ΣFy = 0 : ΣM D = 0 : Ax − B sin θ = 0 Ay − 125 + B cos θ = 0 − Ay ( 450 ) = 0 2 A = Ax2 + Ay = 200 N or The fourth needed equation is either B = 200 N . In either case, the moment equation gives Ay = 0 N Now, guessing that B = 200 N gives θ = 51.32° ................................................................................................................................ Ans. Ax = 156.1 N 2 A = Ax2 + Ay = Ax = 156.1 N < 200 N Therefore, the guess was correct. 6-37 A man is slowly raising a 20-ft-long homogeneous pole weighing 150 lb as shown in Fig. P6-37. The lower end of the pole is kept in place by smooth surfaces. Determine the force exerted by the man to hold the pole in the position shown. SOLUTION From a free-body diagram of the pole, the equations of equilibrium give the tension ΣM A = 0 : (T sin 30° )( 20 ) − 150 (10 cos 60° ) = 0 T = 75.0 lb ............................................................................ Ans. 6-38 The lever shown in Fig. P6-36 is a quarter circular arc of radius 450 mm. The 25-mm-diameter pin at A is smooth and frictionless and is in single shear. (a) Plot A, the magnitude of the pin force at A, and B, the force on the smooth support B, as functions of θ (10° ≤ θ ≤ 80°), the angle at which the support is located. (b) Plot τ, the average shear stress in the pin at A, as a function of θ (10° ≤ θ ≤ 80°). SOLUTION (a) From a free-body diagram of the lever, the equilibrium equations are 165 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ΣM D = 0 : Ay = 0 N − Ay ( 450 ) = 0 Ay − 125 + B cos θ = 0 ↑ ΣFy = 0 : → ΣFx = 0 : Therefore B = 125 cos θ N Ax − B sin θ = 0 Ax = B sin θ = 125 tan θ N 2 A = Ax2 + Ay = Ax = 125 tan θ N ...................................................................................... Ans. B = (125 cosθ ) N ................................................................................................................. Ans. (b) Dividing the force in the pin at A by its cross-sectional area gives the shear stress τ= VA 125 tan θ 500 tan θ N/m 2 .................................................................... Ans. = = 2 As π ( 0.025 ) 4 π ( 0.025 )2 6-39 The crane and boom shown in Fig. P6-39 weigh 12,000 lb and 600 lb, respectively. The pulleys at D and E are small and the cables attached to them remain essentially parallel. The 3-in.-diameter pin at A is smooth and frictionless and is in double shear. The distributed force exerted on the treads by the ground is equivalent to a single resultant force N acting at some distance d behind point C. (a) Plot d, the location of the equivalent force N relative to point C, as a function of the boom angle θ (0° ≤ θ ≤ 80°) when the crane is lifting a 3600-lb load. (b) (c) Plot τ, the average shear stress in the pin at A, as a function of θ (0° ≤ θ ≤ 80°) when the crane is lifting a 3600-lb load. It is desired that the resultant force on the tread always be at least 1 ft behind C to ensure that the crane is never in danger of tipping over. Plot Wmax, the maximum load that may be lifted, as a function of θ (0° ≤ θ ≤ 80°). SOLUTION (a) From a free-body diagram of the entire crane, the equilibrium equations are ↑ ΣFy = 0 : ΣM C = 0 : N − 12, 000 − 600 − W = 0 (12, 000 )( 9 ) − Nd − ( 600 )(12 cosθ − 1) −W ( 24 cosθ − 1 + 1) = 0 166 STATICS AND MECHANICS OF MATERIALS, 2nd Edition If RILEY, STURGES AND MORRIS W = 3600 lb , then N = 16, 200 lb d= 108, 600 − 93, 600 cos θ ft ........................................................................................... Ans. 16, 200 (c) If the normal force must be no closer than 1 ft to the front of the tread, then N = 12, 600 + Wmax lb 108, 600 − (12, 600 + Wmax )(1) − 7200 cos θ − 24Wmax cos θ = 0 Wmax = 96, 000 − 7200 cos θ lb ........................................................................................... Ans. 1 + 24 cos θ (b) As the boom is raised, the cable angle φ (assumed to be the same for both cables) and the boom angle θ are related by b = 24 cos θ h = 24sin θ h − 6 24sin θ − 6 tan φ = = b + 9 24 cos θ + 9 Then, from the free-body diagram of the pulley, the equilibrium equations give the forces → ΣFx = 0 : ↑ ΣFy = 0 : Bx − 3600 cos φ = 0 By − 3600 − 3600 sin φ = 0 Bx = 3600 cos φ lb By = 3600 (1 + sin φ ) lb Finally, from the free-body diagram of the boom, the equilibrium equations → ΣFx = 0 : ↑ ΣFy = 0 : ΣM A = 0 : Ax − Bx − TBD cos φ = 0 Ay − By − 600 − TBD sin φ = 0 ( Bx + TBD cos φ )( 24sin θ ) − 600 (12 cosθ ) − ( By + TBD sin φ ) ( 24 cos θ ) = 0 167 STATICS AND MECHANICS OF MATERIALS, 2nd Edition are solved to get the forces RILEY, STURGES AND MORRIS TBD = ( 24 B y 24 cos φ sin θ − 24sin φ cos θ + 7200 ) cos θ − 24 Bx sin θ lb ............Ans. Ax = Bx + TBD cos φ lb Ay = By + 600 + TBD sin φ lb 2 A = Ax2 + Ay Then, dividing the shear force on the pin by twice its cross-sectional area (since it is in double shear) gives the shear stress τ= 2 2 π ( 3) 4 VA = 2 Ax2 + Ay 14.1372 psi ..................................................................................... Ans. Although these graphs appear to be okay, a check of the data reveals that for angles greater than about 66°, the cable tension TBD is negative when the crane is lifting 3600 lb. But the cable cannot push on the boom. Therefore, for angles greater than 66°, the boom would fall back on the cab of the crane and the solution for angles greater than 66° are meaningless. Further analysis would be required to find the critical angle for other weights being lifted. 6-40 An extension ladder arrangement is being raised into position by a hydraulic cylinder as shown in Fig. P6-40. The center of gravity of the 500-kg ladder is at G. The 25-mm-diameter pin at A is smooth, frictionless, and in double shear. (a) Plot σn, the normal stress in the 40-mm-diameter piston rod of the hydraulic cylinder, as a function of the angle θ (10° ≤ θ ≤ 90°). (b) Plot τ, the shear stress in the pin at A, as a function of θ (10° ≤ θ ≤ 90°). SOLUTION W = 500 ( 9.81) = 4905 N (a) From a free-body diagram of the ladder, the equilibrium equations are → ΣFx = 0 : ↑ ΣFy = 0 : ΣM A = 0 : in which Ax + FB sin φ = 0 Ay + FB cos φ − 4905 = 0 ( FB cos φ )( 3cos θ ) − ( FB sin φ )( 3sin θ ) − ( 4905cosθ )( 8 ) + ( 4905sin θ )(1) = 0 3cos θ − 2 φ = tan −1 3sin θ + 1.5 Solving the equilibrium equations gives 168 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS FB = ( 4905cos θ )( 8) − ( 4905sin θ )(1) 3cos φ cos θ − 3sin φ sin θ N Ax = − FB sin φ N Ay = 4905 − FB cos φ N Dividing the piston force by its cross-sectional area gives the normal stress σn = = FB FB = An π ( 0.04 ) 2 4 π ( 0.04 ) 4 FB 2 N/m 2 ...................... Ans. (b) Dividing the force in the pin at A by twice its cross-sectional area (since it is in double shear) gives the shear stress 2 2 Ax2 + Ay 2 Ax2 + Ay VA τ= = = N/m 2 ......................................................... Ans. 2 2 2 As 2 π ( 0.025 ) 4 π ( 0.025) 6-41 The hydraulic cylinder BC is used to tip the box of the dump truck shown in Fig. P6-41. The pins at A (one on each side of the box) each have a diameter of 1.5 in. and are in double shear. If the combined weight of the box and the load is 22,000 lb and acts through the center of gravity G, (a) Plot C, the force in the hydraulic cylinder, as a function of the angle θ (0° ≤ θ ≤ 80°). (b) Plot τ, the shear stress in the pin at A, as a function of θ (0° ≤ θ ≤ 80°). SOLUTION (a) From a free-body diagram of the truck box, the equilibrium equations are → ΣFx = 0 : ↑ ΣFy = 0 : ΣM A = 0 : 2 Ax − FBC cos φ = 0 2 Ay + FBC sin φ − 22, 000 = 0 ( 22, 000 cos θ )(8.5 ) − ( 22, 000sin θ )( 2 ) − FBC sin (φ − θ ) (12.5 ) = 0 in which φ = tan −1 Therefore 12.5sin θ + 0.5 12.5cosθ − 2.0 FBC = 187, 000 cosθ − 44, 000sin θ lb .......................... Ans. 12.5sin (φ − θ ) Ay = ( 22, 000 − FBC sin φ ) 2 lb Ax = ( FBC cos φ ) 2 lb (b) Then, dividing the shear force on the pin by twice its cross-sectional area (since it is in double shear) gives the shear stress τ= 2 2 π (1.5 ) 4 VA = 2 Ax2 + Ay 3.53429 psi .................................................................................. Ans. 169 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 6-42 The wrecker truck of Fig. P6-42 has a mass of 6800 kg and a center of gravity at G. The force exerted on the rear (drive) wheels by the ground consists of both a normal component By and a tangential component Bx, while the force exerted on the front wheels consists of a normal force Ay only. Plot P, the maximum pull that the wrecker can exert, as a function of θ (0° ≤ θ ≤ 90°) if Bx cannot exceed 0.8By (because of friction considerations) and the wrecker does not tip over backward (the front wheels remain in contact with the ground). SOLUTION W = 6800 ( 9.81) = 66, 708 N From a free-body diagram of the truck, the equilibrium equations are → ΣFx = 0 : ↑ ΣFy = 0 : ΣM B = 0 : P sin θ − Bx = 0 Ay + By − 66, 708 − P cos θ = 0 ( 66, 708)( 2.4 ) − 4.4 Ay − ( P cos θ )(1.5 ) − ( P sin θ )( 3) = 0 Ay = 0 (the front wheels are on the verge of Bx = 0.8 By (the rear wheels are on the verge of slipping). Guessing that the front wheels The fourth equation needed to solve for the four unknowns is either lifting off the ground) or are on the verge of lifting off the ground gives the solution Ay = 0 N P= 66, 708 ( 2.4 ) N ...................................................................................................... Ans. 1.5cosθ + 3sin θ Bx = P sin θ N By = 66, 708 + P cos θ N The forces Bx and 0.8 By are plotted on the same P . Since Bx is always less than graph as the force 0.8 By , the guess that the front wheels are on the verge of lifting off the ground was the correct guess, and the solution is valid for all values of θ . 170 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 6-43 The garage door ABCD shown in Fig. P6-43 is being raised by a cable DE. The one-piece door is a homogeneous rectangular slab weighing 225 lb. Frictionless rollers B and C run in tracks at each side of the door as shown. (a) Plot T, the tension in the cable, as a function of d (0 ≤ d ≤ 100 in.). (b) Plot B and C, the forces on the frictionless rollers, as a function of d (0 ≤ d ≤ 100 in.). SOLUTION (a) From a free-body diagram of the door with both rollers on the horizontal track equilibrium equations give ( 0 < d < 40 in.) , the → ΣFx = 0 : T = 0 lb T cos φ = 0 ΣM C = 0 : ↑ ΣFy = 0 : W ( 36 ) − 2 B ( 48 ) + (T sin φ )( 6 ) = 0 T sin φ + 2 B + 2C − W = 0 B = 3W 8 = 675 8 = 84.4 lb C = W 8 = 225 8 = 28.1 lb From a free-body diagram of the door with roller B on the curve and roller C on the horizontal track ( 40 in. < d < 53.5 in., 0° < γ < 90° ) , the equilibrium equations are → ΣFx = 0 : ↑ ΣFy = 0 : ΣM C = 0 : in which T cos φ − 2 B sin γ = 0 T sin φ + 2 B cos γ + 2C − W = 0 W ( 36 cos θ ) − 2 B cos (γ − θ ) ( 48 ) + ( T sin φ )( 6 cosθ ) − (T cos φ )( 6sin θ ) = 0 sin θ = 12 − 12 cos γ 1 − cos γ = 48 4 12 − 6 sin θ d − 6 cos θ d = 100 − 48cosθ − 12 (1 − sin γ ) tan φ = Therefore T= B= 6W cos θ lb 8cos φ cos (γ − θ ) sin γ + sin θ cos φ − cosθ sin φ T cos φ lb 2sin γ C = (W − 2 B cos γ − T sin φ ) 2 lb From a free-body diagram of the door with roller B on the vertical track and roller C on the horizontal track ( 53.5 in. < d < 88 in.) , the equilibrium equations are T cos φ − 2 B = 0 T sin φ + 2C − W = 0 → ΣFx = 0 : ↑ ΣFy = 0 : ΣM C = 0 : W ( 36 cosθ ) − ( 2 B )( 48sin θ ) + (T sin φ )( 6 cosθ ) − (T cos φ )( 6sin θ ) = 0 171 STATICS AND MECHANICS OF MATERIALS, 2nd Edition in which RILEY, STURGES AND MORRIS cos θ = 100 − d 48 12 − 6 sin θ tan φ = d − 6 cos θ Therefore T= 6W cos θ lb 9sin θ cos φ − cos θ sin φ B = T cos φ 2 lb C = (W − T sin φ ) 2 lb Finally, from a free-body diagram of the door with roller B on the vertical track and roller C on the curve ( 88 in. < d < 100 in., 0° < γ < 90° ) , the equilibrium equations are → ΣFx = 0 : ↑ ΣFy = 0 : ΣM C = 0 : in which T cos φ − 2 B − 2C sin γ = 0 T sin φ + 2C cos γ − W = 0 W ( 36 cosθ ) − 2 B ( 48sin θ ) + ( T sin φ )( 6 cosθ ) − (T cos φ )( 6sin θ ) = 0 cos θ = 12 − 12sin γ 1 − sin γ = 48 4 12 − 6 sin θ d − 6 cos θ d = 100 − 12 (1 − sin γ ) tan φ = Therefore T= B= 6W cosθ + 8W sin θ tan γ lb 9 cos φ sin θ − sin φ cosθ + 8sin φ sin θ tan γ T cos φ + T sin φ tan γ − W tan γ lb 2 W − T sin φ C= lb 2 cos γ 172 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 6-44 Determine all forces acting on member ABE of the frame of Fig. P6-44. SOLUTION First, from a free-body diagram of the complete frame, the equilibrium equations give the support forces ΣM A = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : D ( 300 ) − 150 ( 300 ) = 0 Ax + 150 = 0 Ay + D = 0 D = 150 N Ax = −150 N Ay = −150 N Next, from a free-body diagram of member BCD, the equilibrium equations give the pin forces ΣM B = 0 : ΣM C = 0 : 150 ( 300 ) + C y (100 ) = 0 150 ( 200 ) − By (100 ) = 0 C y = −450 N By = 300 N Finally, from a free-body diagram of member ABE, the equilibrium equations give the pin forces ΣM E = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : 300 (100 ) + 150 (100 ) − 150 ( 200 ) − Bx (100 ) = 0 Ex − 150 − 150 = 0 E y − 300 − 150 = 0 Bx = 150 N Ex = 300 N E y = 450 N Therefore, the forces acting on member ABE are: A = 212 N B = 335 N E = 541 N 45° .......................................................................................................... Ans. 63.43° ..................................................................................................... Ans. 56.31° ..................................................................................................... Ans. 6-45 In the linkage of Fig. P6-45, a = 2.0 ft, b = 1.5 ft, θ = 30°, and P = 40 lb. Determine all forces acting on member BCD. SOLUTION Note that member AC is a two-force member. Then, from a free-body diagram of the member BCD, the equilibrium equations give the forces ΣM B = 0 : → ΣFx = 0 : (TAC cos 45° )( 2 ) − ( 40 cos 30° )( 3.5) = 0 Bx + 40 cos 30° − TAC cos 45° = 0 TAC = 85.732 lb Bx = 25.981 lb 173 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ↑ ΣFy = 0 : By − TAC sin 45° + 40 sin 30° = 0 By = 40.622 lb Therefore, the forces acting on member BCD are: B = 48.2 lb 57.40° ................................................................................................... Ans. 45° ...................................................................................................... Ans. TAC = 85.7 lb 6-46 Forces of 5 N are applied to the handles of the paper punch of Fig. P6-46. Determine the force exerted on the paper at D and the force exerted on the pin at B by handle ABC. SOLUTION From a free-body diagram of handle EBD, the equilibrium equations give the forces ΣM B = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : FD ( 40 ) − 5 ( 70 ) = 0 Bx = 0 N 5 − By + FD = 0 FD = 8.75 N By = 13.75 N B = 13.75 N ↓ ................................................................................................................. Ans. The force exerted on the paper is equal and opposite to the force exerted on the handle FD = 8.75 N ↓ ................................................................................................................. Ans. 6-47 Forces of 25 lb are applied to the handles of the pipe pliers shown in Fig. P6-47. Determine the force exerted on the pipe at D and the force exerted on handle DAB by the pin at A. SOLUTION From a free-body diagram of handle DAB, the equilibrium equations give the forces ΣM A = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : FD (1.25 ) − 25 ( 9 ) = 0 FD sin 38° − Ax = 0 Ay − 25 − FD cos 38° = 0 FD = 180 lb Ax = 110.819 lb Ay = 166.842 lb A = 200 lb 56.41° .................................................................................................... Ans. 52° ......................................................................................................... Ans. The force exerted on the pipe at D is equal and opposite to the force exerted on the handle FD = 180 lb 6-48 The jaws and bolts of the wood clamp in Fig. P6-48 are parallel. The bolts pass through swivel mounts so that no moments act on them. The clamp exerts forces of 300 N on each side of the board. Treat the forces on the boards as uniformly distributed over the contact areas and determine the forces in each of the bolts. Show on a sketch all forces acting on the upper jaw of the clamp. SOLUTION The resultant of the distributed force exerted on the upper jaw of the clamp is a force of 300 N at the center of the uniformly distributed load. From a free-body diagram of the upper jaw, the equilibrium equations give the forces 174 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ΣM D = 0 : ΣM C = 0 : 300 ( 225 ) − FC (100 ) = 0 300 (125) − FD (100 ) = 0 FC = 675 N (T) ............................................................. Ans. FD = 375 N (C) ............................................................. Ans. 6-49 Determine all forces acting on member ABCD of the frame of Fig. P6-49. SOLUTION First, from a free-body diagram of the complete frame, the equilibrium equations give the support reactions ΣM A = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : F (12 ) − 100 ( 24 ) = 0 Ax − 100 = 0 Ay + F = 0 F = 200 lb Ax = 100 lb Ay = −200 lb Next, note that member BE is a two-force member. Then, from a free-body diagram of member ABCD, the equilibrium equations give the pin forces ΣM C = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : ( FBE cos 45° )( 6 ) − 100 (18) − 100 ( 6 ) = 0 100 − 100 − FBE cos 45° + C x = 0 C y − 200 + FBE sin 45° = 0 FBE = 565.685 lb C x = 400 lb C y = −200 lb Therefore, the forces acting on member ABCD are: A = 224 lb B = 566 lb C = 447 lb 63.43° .................................................. Ans. 45° ......................................................... Ans. 26.57° .................................................. Ans. 6-50 In Fig. P6-50, a cable is attached to the structure at E, passes around the 0.8-m-diameter, frictionless pulley at A, and then is attached to a 1000-N weight W. Determine (a) The support reaction at G. (b) All forces acting on member ABCD. SOLUTION (a) From a free-body diagram of the complete structure, the equilibrium equations give the support reactions ΣM D = 0 : → ΣFx = 0 : 1000 ( 3.4 ) − G ( 2 ) = 0 Dx − G = 0 G = 1700 N 175 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS Dx = 1700 N ↑ ΣFy = 0 : Dy − 1000 = 0 Dy = 1000 N G = 1700 N ← ............................................................................................................... Ans. (b) Next, from a free-body diagram of the pulley, the equilibrium equations give the pin forces → ΣFx = 0 : ↑ ΣFy = 0 : 1000 − Ax = 0 Ay − 1000 = 0 Ax = 1000 N Ay = 1000 N Then, from a free-body diagram of member ECF, the equilibrium equations give the pin forces ΣM F = 0 : C x (1) + 1000 (1.4 ) = 0 C x = −1400 N Finally, from a free-body diagram of member ABCD, the equilibrium equations give the pin forces ΣM B = 0 : ΣM C = 0 : → ΣFx = 0 : C y (1) + 1000 ( 2 ) + 1000 (1) = 0 1000 ( 2 ) − By (1) + 1000 (1) = 0 1000 + Bx + Cx + 1700 = 0 C y = −3000 N By = 3000 N Bx = −1300 N Therefore the forces on member ABCD are A = 1414 N B = 3270 N C = 3310 N D = 1972 N 45° ........................................................................................................ Ans. 66.57° .................................................................................................. Ans. 64.98° .................................................................................................. Ans. 30.47° ................................................................................................... Ans. 6-51 A pin-connected system of levers and bars is used as a toggle for a press as shown in Fig. P6-51. Three members are joined by pin D, as shown in the insert. Determine (a) The force exerted on the can at A when a force of 1000 lb is applied to the lever at G. (b) All forces that act on member CD. SOLUTION (a) From a free-body diagram of the lever, the moment equilibrium equation gives the force in the link DE ΣM F = 0 : 1000 ( 30 ) − FDE ( 8 ) = 0 FDE = 3750 lb Note that all three links connected to the pin D are two-force members. From the free-body diagram of the pin, the equilibrium equations give the forces in the links CD and BD 176 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS → ΣFx = 0 : ↑ ΣFy = 0 : FCD cos 78° + FBD cos 67° − 3750 = 0 FBD sin 67° − FCD sin 78° = 0 FCD = 6018.19 lb FBD = 6395.05 lb Finally, from a free-body diagram of the plunger, the vertical equilibrium equation gives the crushing force ↑ ΣFy = 0 : FA − 6395.05sin 67° = 0 FA = 5890 lb ↑ on the plunger FA = 5890 lb ↓ on the can ....................................... Ans. (b) Member CD is a two force member in compression; therefore, FCD = 6020 lb FCD = 6020 lb 78° on CD at C ............................................................................. Ans. 78° on CD at D ............................................................................. Ans. 6-52 The front-wheel suspension of an automobile is shown in Fig. P6-52. The pavement exerts a vertical force of 2700 N on the tire. Determine the force in the spring and the forces at A, B, and D. SOLUTION From a free-body diagram of the wheel, the equilibrium equations give the pin forces ΣM B = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : FCD ( 325) 2700 (150 ) = 0 Bx − FCD = 0 2700 − By = 0 FCD = 1246.154 N Bx = 1246.154 N By = 2700 N Next, from a free-body diagram of the control arm, the equilibrium equations give the pin forces ΣM A = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : Fs ( 250 ) − By ( 500 ) − Bx ( 50 ) = 0 Ax − Bx = 0 Ay + By − 5649.23 = 0 Fs = 5649.23 N Ax = 1246.154 N Ay = 2949.23 N Therefore the force on the spring is Fs = 5650 N ↓ ................................................................................................................ Ans. and the forces at A, B, and D are A = 3200 N B = 2970 N 67.09° .................................................................................................. Ans. 65.22° .................................................................................................. Ans. FCD = 1246 N (C) ........................................................................................................... Ans. 177 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 6-53 The fold-down chair of Fig. P6-53 weighs 30 lb and has its center of gravity at G. Determine (a) All forces acting on member ABC. (b) The shearing stress on a cross section of the 3/8-in.-diameter pin at B, which is in single shear. SOLUTION (a) From a free-body diagram of the chair, the equilibrium equations give the support reactions ΣM A = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : 30 ( 3) − E ( 21) = 0 Ax − E = 0 Ay − 30 = 0 E = 4.28571 lb Ax = 4.28571 lb Ay = 30 lb Note that member BD is a two force member which makes an angle of θ = tan −1 (10 12 ) = 39.81° relative to the horizontal. Then, from the free-body diagram of the seat, the equilibrium equations give the pin forces ΣM C = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : 30 ( 3) − (TBD sin θ )(12 ) = 0 TBD cosθ − Cx − 4.28571 = 0 TBD sin θ + C y − 30 = 0 TBD = 11.7154 lb C x = 4.7429 lb C y = 22.500 lb The forces acting on ABC are equal and opposite to those acting on the seat A = 30.3 lb 81.87° ................................................................................................... Ans. FB = 11.72 lb 39.81° ................................................................................................ Ans. C = 22.99 lb 78.10° ................................................................................................. Ans. (b) Dividing the force on the pin at B by its cross-sectional area gives the shear stress τ= FB 11.7154 = = 106.1 psi ................................................................................. Ans. As π ( 3 8 )2 4 6-54 A scissors jack for an automobile is shown in Fig. P6-54. The screw threads exert a force F on the blocks at joints A and B. Determine (a) The force P exerted on the automobile if F = 800 N and θ = 15°, θ = 30°, and θ = 45°. (b) The shearing stress on a cross section of the 10-mm diameter pin at C, which is in single shear. Solve for each angle in part (a). SOLUTION (a) Note that all three members attached to pin A are two-force members. From a free-body diagram of pin A, the equilibrium equations give the forces in members AC and AE ↑ ΣFy = 0 : FAE sin θ − FAC sin θ = 0 178 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS FAE = FAC → ΣFx = 0 : Similarly 800 − FAE cos θ − FAC cosθ = 0 FAE = FAC = ( 800 2 cos θ ) N FBD = FAC = ( 800 2 cos θ ) N Then, from the free-body diagram of the platform, the equilibrium equations give the force P ↑ ΣFy = 0 : If θ FAC sin θ + FBD sin θ − P = 0 P = 2 FAC sin θ = 800 tan θ N = 15° If θ = 30° If θ = 45° P = 214 N ................................... Ans. P = 462 N ................................... Ans. P = 800 N ................................... Ans. (b) Finally, dividing the force on the pin at C by its cross-sectional area gives the shear stress τ= If θ If θ If θ FAC ( 800 2 cos θ ) 5.09296 ×106 = = N/m 2 2 As cos θ π ( 0.010 ) 4 = 15° = 30° = 45° τ = 5.27 × 106 N/m 2 = 5.27 MPa ............................................ Ans. τ = 5.88 × 106 N/m 2 = 5.88 MPa 6 2 ............................................ Ans. τ = 7.20 × 10 N/m = 7.20 MPa ............................................ Ans. 6-55 A force of 20 lb is required to pull the stopper DE in Fig. P6-55. Determine (a) All forces acting on member BCD. (b) The shearing stress on the cross section of the 1/8-in.-diameter pin at B, which is in single shear. (c) The deformation of the 1/8×3/8-in. member AB, which is made of steel with a modulus of elasticity of 30(106) psi. SOLUTION (a) From a free-body diagram of the entire cork puller, the equilibrium equations give the pulling forces ΣM G = 0 : ↑ ΣFy = 0 : E y (1.5) − Dy (1.5 ) = 0 20 − E y − Dy = 0 E y = Dy Dy = E y = 10 lb Note that both links attached to the pin at A are two-force members. Then, from the free-body diagram of pin A, the equilibrium equations give the forces in the links → ΣFx = 0 : ↑ ΣFy = 0 : TAF cos 45° − TAB cos 45° = 0 20 − TAB sin 45° − TAF sin 45° = 0 TAF = TAB TAB = TAF = 14.14214 lb Next, from the free-body diagram of member BCD, the equilibrium equations give the pin forces 179 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ΣM C = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : Dx ( 2 ) − Dy (1.5) − (TAB cos 45° )( 2 ) − (TAB sin 45° )(1) = 0 TAB cos 45° + C x + Dx = 0 TAB sin 45° + C y − Dy = 0 C y = 0 lb Dx = 22.500 lb C x = −32.500 lb FB = 14.14 lb Therefore, the forces acting on member BCD are 45° .................................................................. Ans. C = 32.5 lb ← ............................................................................ Ans. D = 24.6 lb 23.96° ............................................................... Ans. (b) Dividing the force on the pin at B by its cross-sectional area gives the shear stress τ= FB 14.14214 = = 1152 psi .................................................................................. Ans. As π (1 8 )2 4 (c) The change in length of the link AB is given by δ AB = (14.14214 ) 2 PL = = 1.42 × 10−5 in. .......................................... Ans. 6 EA ( 30 ×10 ) (1 8) × ( 3 8) () 6-56 Member BD of the frame shown in Fig. P6-56 is made of structural steel (E = 200 GPa) and has a rectangular cross section 50 mm wide by 15 mm thick. All pins have 15 mm diameters. Determine (a) The axial stress in member BD. (b) The shearing stress on a cross section of pin C if it is loaded in double shear. (c) The change in length of member BD. SOLUTION (a) From a free-body diagram of the entire structure, the equilibrium equations give the support reactions ΣM A = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : E (1.2 ) − ( 2000 )(1.2 ) ( 0.6 ) − 2000 ( 0.7 ) = 0 E = 2366.67 N 2000 − Ax = 0 Ay + E − 2000 (1.2 ) = 0 Ax = 2000 N Ay = 33.333 N Next, from the free-body diagram of member CDE (in which sin θ = 3 5 and cos θ = 4 5 ), the equilibrium equations give the pin forces ΣM C = 0 : → ΣFx = 0 : 2366.667 (1.2 ) − ( 2000 )(1.2 ) ( 0.6 ) − (TBD sin θ )( 0.8 ) = 0 TBD = 2916.67 N C x − 2916.67 cos θ = 0 C x = 2333.33 N 180 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ↑ ΣFy = 0 : C y + 2366.67 − 2000 (1.2 ) − 2916.67 sin θ = 0 C y = 1783.33 N 2 C = C x2 + C y = 2936.78 N Then, dividing the force in the member BD by its cross-sectional area gives the axial (normal) stress σ BD = ( 2916.67 ) = 3.89 ×106 N/m 2 = 3.89 MPa ............................... Ans. TBD = An ( 0.05 × 0.015 ) (b) Dividing the force on the pin at C by twice its cross-sectional area (since it is in double shear) gives the shear stress τ= ( 2936.78 ) = 8.31×106 N/m 2 = 8.31 MPa ............................... Ans. FC = 2 As 2 π ( 0.015 )2 4 ( 2916.67 )(1000 ) = 0.01944 mm ........................................ Ans. PL = EA ( 200 ×109 ) ( 0.05 × 0.015 ) (c) Finally, the change in length of the two-force member BD is given by δ BD = 6-57 The hoist pulley structure of Fig. P6-57 is rigidly attached to the wall at C. A load of sand hangs from the cable that passes around the 1-ft-diameter, frictionless pulley at D. The weight of the sand can be treated as a triangular distributed load with a maximum intensity of 70 lb/ft. Determine (a) All forces acting on member ABC. (b) The shearing stress on the cross section of the ½-in.-diameter pin D, which is in double shear. (c) The change in length of the ¼ × 1-in. member BE [E = 29(106) psi]. SOLUTION T =W = 1 ( 70 )( 2 ) = 70 lb 2 (a) From a free-body diagram of the entire structure, the equilibrium equations give the reaction forces → ΣFx = 0 : ↑ ΣFy = 0 : ΣM D = 0 : T − Cx = 0 Cy − T = 0 Tr − Tr + C y ( 5) − Cx (1.5) − M C = 0 C y = 70 lb M C = 245 lb ⋅ ft C x = 70 lb ΣM A = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : Then, from the free-body diagram of member ABC, the equilibrium equations give the pin forces 70 ( 7 ) − FB ( 4 ) − 245 = 0 Ax − 70 = 0 70 − Ay − FB = 0 FB = 61.25 lb Ax = 70 lb Ay = 8.75 lb 181 STATICS AND MECHANICS OF MATERIALS, 2nd Edition Therefore, the forces acting on member ABC are RILEY, STURGES AND MORRIS A = 70.5 lb 7.13° ...................................................................................................... Ans. B = 61.3 lb ↓ .................................................................................................................. Ans. C = 99.0 lb 45° ......................................................................................................... Ans. M C = 245 lb ⋅ ft .......................................................................................................... Ans. (b) Next, from the free-body diagram of the pulley, the equilibrium equations give the pin forces → ΣFx = 0 : ↑ ΣFy = 0 : 70 − Dx = 0 Dy − 70 = 0 Dx = 70 lb Dy = 70 lb 2 D = Dx2 + Dy = 98.995 lb Dividing the force on the pin at D by twice its cross-sectional area (since it is in double shear) gives the shear stress τ= D 98.995 = = 252 psi ............................................................................... Ans. 2 As 2π ( 0.5 )2 4 (c) The change in length of the two-force member BE is given by δ BE = ( 61.25 )( 3 ×12 ) = 3.04 ×10−4 in. ................................................. Ans. PL = EA ( 29 × 106 ) (1 4 ) × 1 6-58 A pair of vice grip pliers is shown in Fig. P6-58. Determine the force exerted on the bolt by the jaws of the pliers when a force of magnitude 100 N is applied to the handle. Let d = 30 mm. SOLUTION θ = tan −1 ( 30 40 ) = 36.870° ΣM B = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : sin θ = 3 5 cos θ = 4 5 From a free-body diagram of the upper handle, the equilibrium equations give the forces 100 ( 93) − ( FA sin θ )( 38 ) + ( FA cos θ )( 5 ) = 0 FA cos θ − Bx = 0 FA sin θ − 100 − By = 0 FA = 494.681 N Bx = 395.745 N By = 196.809 N Next, from the free-body diagram of the jaw BCD, the moment equilibrium equation is ΣM C = 0 : in which Therefore Da − Bx ( 35 ) − By (12 ) = 0 a = 352 + 152 = 38.079 mm D = 426 N ........................................................................................................................ Ans. 182 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 6-59 Forces of 50 lb are applied to the handles of the bolt cutter of Fig. P6-59. Determine (a) All forces acting on the handle ABC. (b) The force exerted on the bolt at E. (c) (d) The axial stress in the links at D (one on each side) if each has a 1/8 × 3/4-in. rectangular cross section. The change in length of the links at D if they are 4 in. long and made of SAE 4340 heat-treated steel. SOLUTION (a) From a free-body diagram of the jaw CDE, horizontal equilibrium gives C x = 0 lb Then, from a free-body diagram of the handle ABC, the equilibrium equations give the pin forces → ΣFx = 0 : ↑ ΣFy = 0 : ΣM B = 0 : C x − Bx = 0 50 − By + C y = 0 C y (1) − 50 ( 20 ) = 0 By = 1050 lb C y = 1000 lb Bx = 0 lb Therefore, the forces acting on the handle ABC are B = 1050 lb ↓ ................................................................................................................. Ans. C = 1000 lb ↑ ................................................................................................................. Ans. (b) Then, returning to the free-body diagram of the jaw CDE, the equilibrium equations give the force on the bolt ΣM D = 0 : C y ( 3) − E ( 2 ) = 0 E = 1500 lb ↓ on the jaw E = 1500 lb ↑ on the bolt .......................................................................................... Ans. ↑ ΣFy = 0 : TD − 1000 − 1500 = 0 TD = 2500 lb (c) Dividing the link force by twice its cross-sectional area (since there are two links) gives the normal stress σ= TD 1500 = = 13.33 × 103 psi = 13.33 ksi ....................................... Ans. 2 An 2 (1 8 )( 3 4 ) (d) Finally, the change in length of the two-force link is given by δD = ( 2500 2 )( 4 ) PL = = 1.839 ×10−3 in. ............................................ Ans. 6 EA ( 29 × 10 ) (1 8 )( 3 4 ) 6-60 The tower crane of Fig. P6-60 is rigidly attached to the building at F. A cable is attached at D and passes over small frictionless pulleys at A and E. The object suspended from C has a mass of 1530 kg. Determine (a) All forces acting on member ABCD. (b) The support reactions at F. (c) The shearing stress on a cross section of pin A if it has a diameter of 15 mm and is loaded in double shear. SOLUTION W = 1530 ( 9.81) = 15, 009.30 N sin θ = 3 5 φ = tan −1 6 14.4 = 22.620° cos θ = 4 5 183 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS (a) From a free-body diagram of member ABCD and pulley A, the equilibrium equations give the forces ΣM B = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : 8T − ( T sin θ )( 8 ) + ( T sin φ )(14.4 ) − (15, 009.30 )(10 ) = 0 T cos θ + Bx − T cos φ = 0 T sin θ − T + By − 15, 009.30 + T sin φ = 0 T = 17,176.136 N Bx = 2113.97 N By = 15, 273.51 N Therefore, the forces acting on the member ABCD are A = 15,363 N ≅ 15,360 N B = 15, 420 N D = 17,180 N 26.565° ...................................................................... Ans. 82.120° ............................................................................................ Ans. 22.620° ............................................................................................ Ans. C = 15, 010 N ↓ ............................................................................................................. Ans. (b) Next, from the free-body diagram of the entire structure, the equilibrium equations give the support reactions → ΣFx = 0 : ↑ ΣFy = 0 : ΣM F = 0 : Fx = 0 N Fy − 17,176.136 − 15, 009.30 = 0 (17,176.136 )( 8 ) + M F − (15, 009.30 )(10 ) = 0 Fy = 32,185.4 N ≅ 32.2 kN M F = 12, 684 N ⋅ m Therefore, the support reactions are F ≅ 32.2 kN ↑ ................................. Ans. M F ≅ 12.68 kN ⋅ m ..................... Ans. (c) Finally, dividing the force on the pin by twice its cross-sectional area (since it is in double shear) gives the shear stress τ= VA 15,363 = = 43.5 × 106 N/m 2 = 43.5 MPa .............................. Ans. 2 As 2 π ( 0.015 )2 4 6-61 Three bars are connected with smooth pins to form the frame shown in Fig. P6-61. The weights of the bars are negligible. Determine (a) The reactions at supports A and E. (b) The resultant forces at pins B, C, and D. SOLUTION θ = tan −1 (1 2 ) = 26.565° From a free-body diagram of the entire structure, the equilibrium equations give ΣM A = 0 : 26 E y − 675 ( 29.5 ) = 0 184 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS E y = 765.865 lb ↑ ΣFy = 0 : → ΣFx = 0 : E y − Ay − 675 = 0 Ax − Ex = 0 Ay = 90.865 lb Ax = Ex Next, from the free-body diagram of member BD, the equilibrium equations give the pin forces ΣM D = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : ( FB sin θ )(13) − ( 675 )(10 ) = 0 FB cos θ − Dx = 0 Dy − FB sin θ − 675 = 0 FB = 1161.035 lb Dx = 1038.462 lb Dy = 1194.230 lb Finally, from the free-body diagram of member ABC, the equilibrium equations give the pin forces ΣM C = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : 2 (13) Ax + 13 Ay − 6.5 5 FB = 0 Ax − FB cos θ + Cx = 0 − Ay + FB sin θ − C y = 0 C y = 428.365 lb ( ) Ax = Ex = 603.606 lb C x = 434.856 lb (a) Therefore, the reaction forces are A = 610 lb E = 975 lb B = 1161 lb C = 610 lb D = 1583 lb 8.56° ................................................. Ans. 51.76° ............................................... Ans. 26.57° (on ABC ) ......................... Ans. 44.57° (on ABC ) ................................................................................. Ans. 48.99° (on CDE ) ............................................................................... Ans. (b) The forces acting on pins B, C, and D are 6-62 Figure P6-62 is a simplified sketch of the mechanism used to raise the bucket of a bulldozer. The bucket and its contents weigh 10 kN and have a center of gravity at H. Arm ABCD has a weight of 2 kN and a center of gravity at B; arm DEFG has a weight of 1 kN and a center of gravity at E. The weight of the hydraulic cylinders can be ignored. (a) Calculate the force in the horizontal cylinders CJ and EI and all forces acting on arm DEFG for the position shown. (b) Determine the required diameter of the pin at E if the shearing stress cannot exceed 120 MPa. The pin is in double shear. SOLUTION (a) From a free-body diagram of the bucket, the equilibrium equations give the force in cylinder EI ΣM G = 0 : TEI (1.2 cos 30° ) − (10 )( 0.3) = 0 185 STATICS AND MECHANICS OF MATERIALS, 2nd Edition and the pin forces at G RILEY, STURGES AND MORRIS TEI = 2.8868 kN ≅ 2.89 kN .............................................. Ans. → ΣFx = 0 : ↑ ΣFy = 0 : Gx − TEI = 0 G y − 10 = 0 G y = 10 kN Next, from a free-body diagram of member DEFG , the equilibrium equations are Gx = 2.8868 kN ΣM D = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : in which TEI ( 0.6 cos 30° ) − (1)( 0.6sin 30° ) + ( FBF cos φ )(1.2 cos 30° ) + ( FBF sin φ )(1.2sin 30° ) − Gx (1.8cos 30° ) − G y (1.8sin 30° ) = 0 FBF cos φ − Dx + TEI − Gx = 0 FBF sin φ + Dy − 1 − G y = 0 1.8cos 30° − 1.2 cos 30° φ = tan −1 = 19.107° 1.8sin 30° + 1.2sin 30° These equations are solved to give FBF = 10.438 kN Dx = 9.863 kN D = 12.44 kN F = 10.44 kN G = 10.41 kN Dy = 7.583 kN Therefore, the remaining forces acting on the member DEFG are 37.55° ............................................................................................... Ans. 19.11° ................................................................................................ Ans. 73.90° ............................................................................................... Ans. Then, from the free-body diagram of the entire bucket assembly, the moment equilibrium equation gives the force in the cylinder CJ ΣM A = 0 : TCJ (1.8cos 30° ) − ( 2 )( 0.9 cos 30° ) − (1)( 2.4 cos 30° + 0.6 cos 30° ) − (10 )( 2.4 cos 30° + 1.8cos 30° + 0.3) = 0 TCJ = 27.9 kN ..................................................Ans. (b) Finally, dividing the force on pin E by twice its cross-sectional area (since it is in double shear) gives the shear stress VE 2.8868 ×103 τ= = N/m 2 ..................................................................................... Ans. 2 2 As 2 (π d 4 ) Since the shear stress must be no greater than 120 MPa, the minimum diameter of the pin is d= 2 ( 2.8868 × 103 ) π (120 × 10 6 ) = 3.91× 10 −3 m = 3.91 mm ................................................. Ans. 186 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 6-63 The mechanism of Fig. P6-63 is designed to keep its load level while raising it. A pin on the rim of the 4-ftdiameter pulley fits in a slot on arm ABC. Arms ABC and DE are each 4 feet long and the package being lifted weighs 80 lb. The mechanism is raised by pulling on the rope that is wrapped around the pulley. Determine the force P applied to the rope and all forces acting on the arm ABC when the package has been lifted 4 feet, as shown. SOLUTION When h = 4 ft θ = sin −1 ( 2 4 ) = 30° From a free-body diagram of the platform, the equilibrium equations give ΣM C = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : ( FDE cos 30° )( 3) − 80 ( 2 ) = 0 FDE cos 30° − Cx = 0 C y + FDE sin 30° − 80 = 0 C y = 49.208 lb FDE = 61.584 lb Cx = 53.333 lb Next, from the free-body diagram of member ABC, the equilibrium equations give the pin forces ΣM A = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : 2 B − C x ( 4sin 30° ) − C y ( 4 cos 30° ) = 0 C x − Ax − B sin 30° = 0 B cos 30° − C y − Ay = 0 Ay = 70.792 lb B = 138.564 lb Ax = −15.949 lb Finally, from the free-body diagram of the pulley, the moment equilibrium equation gives the rope force ΣM A = 0 : 2P − 2B = 0 P = B = 138.564 lb ≅ 138.6 lb .................................................................................... Ans. Note that the forces Ax and Ay represent the interaction between the arm ABC and the axle of the pulley while the forces FAx and FAy represent the interaction between the pulley support and the axle of the pulley. The forces acting on arm ABC are then A = 72.6 lb B = 138.6 lb C = 72.6 lb 77.30° ................................................................................................... Ans. 60° ....................................................................................................... Ans. 42.70° ................................................................................................... Ans. 6-64 A drum of oil with a mass of 200 kg is supported by a frame (of which there are two) as shown in Fig. P6-64. Determine (a) All forces acting on member ACE. (b) The elongation of the 20-mm diameter wire if it is made of a material with a modulus of elasticity of 200 GPa. SOLUTION (a) From a free-body diagram of the drum, the equilibrium equations give the forces → ΣFx = 0 : ↑ ΣFy = 0 : 2 D cos 45° − 2 E cos 45° = 0 2 D sin 45° + 2 E sin 45° − 200 ( 9.81) = 0 D = E = 693.672 N 187 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS Next, from a free-body diagram of member BCD , the moment equilibrium equation is ΣM B = 0 : ΣM A = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : C y (1) − C x (1) + D 0.8 + 2 = 0 C x (1) + C y (1) − E 0.8 + 2 = 0 T − C x + E cos 45° = 0 A + C y − E sin 45° = 0 ( ) Then, from a free-body diagram of member ACE , the equilibrium equations are ( ) Solving these last four equations gives C x = 1535.937 N Cy = 0 N T = 1045.437 N A = 490.500 N Therefore, the forces acting on the member ACE are A = 491 N ↑ ................................................................................................................... Ans. T = 1045 N → ................................................................................................................ Ans. C = 1536 N ← ................................................................................................................ Ans. E = 694 N 45° ........................................................................................................... Ans. (b) Finally, the stretch of the wire is given by δ= (1045.437 )( 2000 ) = 0.0333 mm ............................................... Ans. PL = EA ( 200 × 109 ) π ( 0.02 )2 4 6-65 The mechanism shown in Fig. P6-65 is designed to keep its load level while raising it. A pin on the rim of the 4-ft-diameter pulley fits in a slot on arm ABC. Arms ABC and DE are each 4 ft long and the package being lifted weighs 80 lb. The mechanism is raised by pulling on the rope that is wrapped around the pulley. (a) Plot P, the force required to hold the platform as a function of the platform height h (0 ≤ h ≤ 5.75 ft). (b) Plot A, C, and E, the magnitudes of the pin reaction forces at A, C, and E as a function of h (0 ≤ h ≤ 5.75 ft). SOLUTION h−2 θ = sin −1 4 (a) From a free-body diagram of the platform, the equilibrium equations give ΣM C = 0 : ( FDE cos θ )( 3) − 80 ( 2 ) = 0 160 lb 3cos θ FDE cos θ − Cx = 0 160 tan θ C y = 80 − lb 3 FDE = → ΣFx = 0 : ↑ ΣFy = 0 : C y + FDE sin θ − 80 = 0 Cx = 53.333 lb 188 STATICS AND MECHANICS OF MATERIALS, 2nd Edition Next, from the free-body diagram of member ABC, the equilibrium equations give the pin forces RILEY, STURGES AND MORRIS ΣM A = 0 : 2 B − C x ( 4sin θ ) − C y ( 4 cos θ ) = 0 C x − Ax − B sin θ = 0 B cos θ − C y − Ay = 0 B = ( 2Cx sin θ + 2C y cos θ ) lb → ΣFx = 0 : ↑ ΣFy = 0 : Ax = ( Cx − B sin θ ) lb Ay = ( B cos θ − C y ) lb Finally, from the free-body diagram of the pulley, the moment equilibrium equation gives the rope force ΣM A = 0 : 2P − 2B = 0 P = B = ( 2C x sin θ + 2C y cos θ ) lb ............................................................................ Ans. Note that the forces Ax and Ay represent the interaction between the arm ABC and the axle of the pulley while the forces FAx and FAy represent the interaction between the pulley support and the axle of the pulley. (b) The magnitudes of the pin forces A, C, and E are 2 A = Ax2 + Ay lb ....................... Ans. 2 C = C x2 + C y lb ....................... Ans. FDE = 160 lb ........................ Ans. 3cos θ 6-66 Forces of P = 100 N are being applied to the handles of the vise grip pliers shown in Fig. P6-66. Plot the force applied on the bolt by the jaws as a function of the distance d (20 mm ≤ d ≤ 30 mm). SOLUTION φ = tan −1 ( d 40 ) ΣM B = 0 : FA = → ΣFx = 0 : ↑ ΣFy = 0 : a = 352 + 152 = 38.079 mm From a free-body diagram of the upper handle, the equilibrium equations give the forces 100 ( 93) − ( FA sin φ )( 38 ) + ( FA cos φ )( 35 − d ) = 0 9300 N 38sin φ − ( 35 − d ) cos φ FA cos φ − Bx = 0 FA sin φ − 100 − By = 0 Bx = FA cos φ N ΣM C = 0 : FD a − Bx ( 35 ) − By (12 ) = 0 By = ( FA sin φ − 100 ) N Next, from the free-body diagram of the jaw BCD, the moment equilibrium equation is 189 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS FD = 35Bx + 12 By 38.079 N ................................................... Ans. 6-67 A load P will be supported by a structure consisting of a rigid bar A, two aluminum alloy (E = 10,600 ksi) bars B, and a stainless steel (E = 28,000 ksi) bar C, as shown in Fig. P6-67. Each bar has a cross sectional area of 2.00 in.2 If the bars are unstressed before the load P is applied, determine the normal stresses in the bars after a 40-kip load is applied. SOLUTION From a free-body diagram of bar A, the equilibrium equations give ΣM C = 0 : ↑ ΣFy = 0 : TB 2 a − TB1a = 0 TB1 + TB 2 + TC − 40 × 103 = 0 TB1 = TB 2 = TB 4σ B + 2σ C = 40, 000 lb Since the force in each of the aluminum bars is the same, they will stretch the same amount and the steel bar must then stretch the same amount as the aluminum bars, δ B = δ C , which in terms of the stresses can be written σ B ( 36 ) σ C ( 72 ) = 10.6 × 106 28 ×106 σ C = 1.32075σ B Combining the equilibrium equation and the deformation equation gives σ B = 6020 psi .................................................................................................................. Ans. σ C = 7950 psi .................................................................................................................. Ans. 6-68 The rigid bar CDE, shown in Fig. P6-68, is horizontal before the load P is applied. Tie rod A is a hot-rolled steel (E = 210 GPa) bar with a length of 450 mm and a cross-sectional area of 300 mm2. Post B is an oak timber (E = 12 GPa) with a length of 375 mm and a cross-sectional area of 4500 mm2. After the 225-kN load P is applied, determine (a) The normal stresses in bar A and post B. (b) The vertical displacement of point D. SOLUTION (a) From a free-body diagram of the bar CDE, the moment equilibrium equation gives ΣM C = 0 : 500TAD + 1350 FBE − 1350 ( 225 ×103 ) = 0 TAD + 2.70 FBE = 607.5 × 103 N 190 STATICS AND MECHANICS OF MATERIALS, 2nd Edition From similar triangles, the stretch of bar AD and the shrink of bar BE are related by RILEY, STURGES AND MORRIS δ AD δ BE = 500 1350 which gives 1350 ( 210 ×10 )( 300 ×10 ) 9 TAD ( 450 ) −6 = 500 (12 ×10 )( 4500 ×10 ) 9 FBE ( 375 ) −6 FBE = 2.77714TAD Combining the equilibrium equation and the deformation equation gives TAD = 71, 485.0 N (T) = σ AD ( 300 × 10−6 ) FBE = 198,523.9 N (C) = σ BE ( 4500 × 10−6 ) σ AD = 238 ×106 N/m 2 = 238 MPa (T) ...................................................................... Ans. σ BE = 44.1×106 N/m 2 = 44.1 MPa (C) .................................................................... Ans. (b) The vertical displacement of point D is the same as the stretch of the tie rod δ AD = ( 210 ×10 )( 300 ×10 ) 9 ( 71, 485.0 )( 450 ) −6 = 0.511 mm ↓ ........................................................ Ans. 6-69 A pin-connected structure is loaded and supported as shown in Fig. P6-69. Member CD is rigid and is horizontal before the load P is applied. Member A is an aluminum alloy bar with a modulus of elasticity of 10,600 ksi and a cross-sectional area of 2.25 in2. Member B is a stainless steel bar with a modulus of elasticity of 28,000 ksi and a cross-sectional area of 1.75 in2. After the load is applied to the structure, determine (a) The normal stresses in bars A and B. (b) The vertical displacement of point D. SOLUTION (a) From a free-body diagram of the bar CD, the moment equilibrium equation gives ΣM C = 0 : 5.5 (10 × 103 ) − 8FA − 3FB = 0 8 FA + 3FB = 55 × 103 lb From similar triangles, the shrink of the two bars and the movement of point D are related by δA δB d = = 8 3 10 which gives 3 (10.6 ×10 ) ( 2.25) 6 FA ( 48 ) =8 ( 28 ×10 ) (1.75) 6 FB ( 36 ) FB = 1.02725FA Combining the equilibrium equation and the deformation equation gives FA = 4963.11 lb (C) = σ A ( 2.25 ) 191 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS FB = 5098.35 lb (C) = σ B (1.75 ) σ A = 2.21×103 psi = 2.21 ksi (C) .............................................................................. Ans. σ B = 2.91×103 psi = 2.91 ksi (C) .............................................................................. Ans. (b) The vertical displacement of point D is then d= 10 ( 4963.11)( 48 ) 10 δA = = 0.01249 in. ↓ ................................................. Ans. 8 8 (10.6 × 106 ) ( 2.25 ) 6-70 Bar B of the pin-connected system of Fig. P6-70 is made of an aluminum alloy [Ea = 70 GPa, Aa = 300 mm2, and αa = 22.5(10-6)/°C] and bar A is made of a hardened carbon steel [Es = 210 GPa, As = 1200 mm2, and αs = 11.9(10-6)/°C]. Bar CDE is to be considered rigid. When the system is unloaded at 40°C, bars A and B are unstressed. After the load P is applied, the temperature of both bars decreases to 15°C. Determine (a) The normal stresses in bars A and B. (b) The vertical displacement (deflection) of pin E. SOLUTION (a) From a free-body diagram of the bar CDE, the moment equilibrium equation gives ΣM C = 0 : 150 FAD + 450 ( FBE − 100 ×103 ) = 0 FAD + 3FBE = 300 ×103 N From similar triangles, the stretch of the two bars are related by δ AD δ BE = 150 450 which gives FAD ( 250 ) + (11.9 ×10−6 ) ( −25 )( 250 ) 450 9 −6 ( 210 × 10 )(1200 ×10 ) FBE ( 500 ) = 150 + ( 22.5 ×10−6 ) ( −25)( 500 ) 9 −6 ( 70 × 10 )( 300 × 10 ) FAD = 8FBE − 19,530 N Combining the equilibrium equation and the deformation equation gives FAD = 212,885 N (T) = σ AD (1200 × 10−6 ) FBE = 29, 048.2 N (T) = σ BE ( 300 × 10−6 ) σ AD = 177.4 ×106 N/m 2 = 177.4 MPa (T) 6 2 ............................................................... Ans. σ BE = 96.8 ×10 N/m = 96.8 MPa (T) .................................................................... Ans. (b) The vertical displacement of pin E is the same as the stretch of bar B ( 29, 048.2 )( 500 ) + ( 22.5 ×10−6 ) ( −25 )( 500 ) δ BE = 9 −6 ( 70 ×10 )( 300 × 10 ) δ BE = 0.410 mm ↓ ......................................................................................................... Ans. 192 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 6-71 A rigid bar CD is loaded and supported as shown in Fig. P6-71. Bars A and B are unstressed before the 30kip load P is applied. Bar A is made of steel (E = 30,000 ksi) and has a cross-sectional area of 2 in2. Bar B is made of brass (E = 15,000 ksi) and has a cross-sectional area of 1.5 in2. Determine (a) The stresses in bars A and B. (b) The vertical displacement (deflection) of pin C. (c) The shearing stress on a cross section of the ¾-in.-diameter pin at D, which is in double shear. SOLUTION (a) From a free-body diagram of the bar CD, the moment equilibrium equation gives ΣM D = 0 : 10 ( 30 × 103 ) − 2 FA − 6 FB = 0 FA + 3FB = 150 × 103 lb From similar triangles, the stretch of the two bars and the movement of point C are related by δA δB c = = 2 6 10 which gives 6 ( 30 ×10 ) ( 2 ) 6 FA ( 8 ) =2 (15 ×10 ) (1.5) 6 FB (15 ) FA = 1.66667 FB Combining the equilibrium equation and the deformation equation gives FA = 53,571 lb (T) = σ A ( 2 ) FB = 32,143 lb (T) = σ B (1.5) σ A = 26.8 ×103 psi = 26.8 ksi (T) .............................................................................. Ans. σ B = 21.4 ×103 psi = 21.4 ksi (T) .............................................................................. Ans. (b) The vertical displacement of point C is then c= 10 ( 53,571)( 8) 10 δA = = 0.0357 in. ↓ ............................................................. Ans. 2 2 ( 30 × 106 ) ( 2 ) (c) Returning to the free-body diagram of the bar CD, the force equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : Dx = 0 FA + FB − Dy − 30 × 103 = 0 Dy = FA + FB − 30 × 103 = 55.714 × 103 lb 2 D = Dx2 + Dy = 55.714 × 103 lb Dividing the pin force by twice its cross-sectional area (since it is in double shear) gives the shear stress τD = VD 55.714 × 103 = = 63.1× 103 psi = 63.1 ksi ....................................... Ans. 2 As 2 π ( 0.75 )2 4 193 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 6-72 A rigid bar CD is loaded and supported as shown in Fig. P6-72. Bars A and B are unstressed before the 150kN load P is applied. Bar A is made of stainless steel (E = 190 GPa) and has a cross-sectional area of 750 mm2. Bar B is made of an aluminum alloy (E = 73 GPa) and has a cross-sectional area of 1250 mm2. Determine (a) The stresses in bars A and B. (b) The vertical displacement (deflection) of pin D. SOLUTION (a) From a free-body diagram of the bar CD, the moment equilibrium equation gives ΣM C = 0 : 0.2 FB + 0.5 ( 4 5 ) FA −0.6 (150 × 103 ) = 0 FB + 2 FA = 450 × 103 N From similar triangles, the stretch of bar B and the vertical movement of pins E and D are related by δB e d = = 0.2 0.5 0.6 When pin E moves down a distance e, bar A stretches δ A = ( 4 5) e . Therefore, δ A = 2δ B −6 and (190 ×10 )( 750 ×10 ) 9 FA (1000 ) =2 ( 73 ×10 )(1250 ×10 ) 9 −6 FB ( 500 ) FA = 1.56164 FB Combining the equilibrium equation and the deformation equation gives FA = 170.432 ×103 N (T) = σ A ( 750 × 10−6 ) FB = 109.136 ×103 N (T) = σ B (1250 × 10−6 ) σ A = 227 ×106 N/m 2 = 227 MPa (T) (b) The vertical displacement of pin D is then 3 ....................................................................... Ans. σ B = 87.3 ×106 N/m 2 = 87.3 MPa (T) ...................................................................... Ans. d = 3δ B (109.136 ×10 ) ( 500 ) = 1.794 mm ↓ .............................................. Ans. =3 ( 73 ×10 )(1250 ×10 ) 9 −6 6-73 A 40-kip load P will be supported by a structure consisting of a rigid bar A, two aluminum alloy [Ea = 10,600 ksi and αa = 12.5(10-6)/°F] bars B, and a stainless steel [Es = 28,000 ksi and αs = 9.6(10-6)/°F] bar C, as shown in Fig. P6-73. The bars are unstressed when the structure is assembled at 72°F. Each bar has a crosssectional area of 2.00 in2. Determine the normal stresses in the bars after the 40-kip load is applied and the temperature is increased to 250°F. SOLUTION From a free-body diagram of bar A, the equilibrium equations give ΣM C = 0 : TB 2 a − TB1a = 0 TB1 = TB 2 = TB 194 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ↑ ΣFy = 0 : TB1 + TB 2 + TC − 40 × 103 = 0 2TB + TC = 40, 000 lb Since stresses are asked for, the equilibrium equation will be written in terms of stresses 4σ B + 2σ C = 40, 000 psi Since the force in each of the aluminum bars is the same (and the temperature change is the same for both bars), they will stretch the same amount and the steel bar must then stretch the same amount as the aluminum bars, δ B = δ C , which in terms of the stresses can be written σ B ( 36 ) σ ( 72 ) + (12.5 × 10−6 ) (168 )( 36 ) = C + ( 9.6 ×10 −6 ) (168 )( 72 ) 6 6 (10.6 ×10 ) ( 28 ×10 ) σ C = 1.32075σ B − 15.7584 ×103 psi Combining the equilibrium equation and the deformation equation gives σ B = 10.77 ×103 psi = 10.77 ksi (T) .......................................................................... Ans. σ C = −1.536 × 103 psi = 1.536 ksi (C) ...................................................................... Ans. 6-74 A pin-connected structure is loaded and supported as shown in Fig. P6-74. Member CD is rigid and is horizontal before the 75-kN load P is applied. Bar A is made of structural steel (E = 200 GPa), and bar B is made of an aluminum alloy (E = 73 GPa). The cross-sectional areas of members A and B are 625 mm2 and 2570 mm2, respectively. Determine (a) The vertical displacement of the pin used to apply the load. (b) The shearing stress on a cross section of the 25-mm diameter pin at C, which is in double shear. SOLUTION θ A = tan −1 ( 3 4 ) = 36.870° θ B = tan −1 ( 3 2 ) = 56.310° (a) From a free-body diagram of the bar CD, the moment equilibrium equation gives LA = 32 + 42 = 5 m LB = 32 + 22 = 13 m ΣM C = 0 : 4 ( FA sin θ A ) + 2 ( FB sin θ B ) −5 ( 75 × 103 ) = 0 2.4 FA + 1.66410 FB = 375 × 103 N From similar triangles, the vertical movement of pins A, B, and D are related by bad == 245 When pin B moves down a distance b, bar B stretches δ B = b sin θ B . Then, when pin A moves down a distance a = 2b , bar A will stretch δ A = a sin θ A = 2b sin θ A = 2δ B sin θ A sin θ B = 1.44222δ B 195 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ( 200 ×10 )( 625 ×10 ) 9 −6 FA ( 5) = 1.44222 ( 73 ×10 )( 2570 ×10 ) 9 −6 FB ( 13 ) FB = 1.44315FA Combining the equilibrium equation and the deformation equation gives FA = 78.0998 × 103 N The vertical displacement of pin D is then FB = 112.7097 × 103 N 3 δB 2.5 (112.7097 × 10 ) 13 d = 2.5b = 2.5 = sin θ B sin 56.310° ( 73 × 109 )( 2570 ×10 −6 ) () d = 0.00651 m = 6.51 mm ↓ ...................................................................................... Ans. (b) Returning to the free-body diagram of the bar CD, the force equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : C x − FA cosθ A − FB cosθ B = 0 FA sin θ A + FB sin θ B − C y − 75 ×103 = 0 C y = 65.6402 ×103 N C x = 124.9997 ×103 N 2 C = Cx2 + C y = 141.186 × 103 N Dividing the pin force by twice its cross-sectional area (since it is in double shear) gives the shear stress τC = VC 141.186 × 103 = = 143.8 × 106 N/m 2 = 143.8 MPa .......................... Ans. 2 As 2 π ( 0.025 ) 4 6-75 Before the 20 kip load P is applied, the arms of the crank C shown in Fig. P6-75 are horizontal and vertical; there is a 0.009-in. gap between the horizontal arm and the brass (Eb = 15,000 ksi and Ab = 12 in2) post B; and the aluminum (Ea = 10,000 ksi and Aa = 2 in2) rod A is horizontal. If the crank C is rigid, determine (a) The stresses in members A and B. (b) The change in length of members A and B. SOLUTION (a) From a free-body diagram of the crank C, the moment equilibrium equation gives ΣM C = 0 : 5 ( 20 × 103 ) − 10TA − 6 FB = 0 5TA + 3FB = 50 × 103 lb From similar triangles, ab = 10 6 in which a = δ A is the stretch of rod A, b = δ B + 0.009 in. , and δ B is the shrink of the post B . Therefore δ A = 5b 3 = ( 5δ B 3) + 0.015 in. 5 FB (15 ) + 0.015 in. = (10 ×106 ) ( 2 ) 3 (15 ×106 ) (12 ) 196 TA ( 50 ) STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS FB = (18TA − 108, 000 ) lb Combining the equilibrium equation and the deformation equation gives TA = 6.3390 × 103 lb = σ A ( 2 ) FB = 6.1017 ×103 lb = σ B (12 ) σ A = 3170 psi (T) ........................................................................................................... Ans. σ B = 508 psi (C) ............................................................................................................. Ans. (b) The change in length of the two members is then 3 δA = δB ( 6.3390 ×10 ) ( 50 ) = 15.85 ×10 (10 ×10 ) ( 2 ) ( 6.1017 ×10 ) (15) = 0.508 ×10 = (15 ×10 ) (12 ) 6 3 6 −3 in. (stretch) ............................................. Ans. −3 in. (shrink) .............................................. Ans. 6-76 The mechanism of Fig. P6-76 consists of a structural steel (E = 200 GPa) rod A with a cross-sectional area of 350 mm2, a cold-rolled brass (E = 100 GPa) rod B with a cross sectional area of 750 mm2, and a rigid bar C. The nuts at the top ends of rods A and B are initially tightened to the point where all slack is removed from the mechanism but the bars remain free of stress. If a nut advances 2.5 mm with each full turn (360°), determine (a) The stresses in the rods when the nut at the top end of rod B rotates 180°. (b) The vertical displacement at the top end of rod A for part (a). SOLUTION (a) From a free-body diagram of the bar C, the moment equilibrium equation gives ΣM C = 0 : 200TA − 125TB = 0 5TB = 8TA From similar triangles, the vertical movement of points A and B are related by a b = 200 125 in which a = δ A is the stretch of rod A, b = 1.25 − δ B mm , and δ B is the stretch of rod B. Therefore 5δ A = 8 (1.25 − δ B ) = 10 − 8δ B ( 200 ×10 )( 350 ×10 ) (100 ×10 )( 750 ×10 ) T = ( 93.33333 ×10 − 1.49333T ) N 9 −6 9 −6 3 5 TA (1500 ) = 10 − 8 B TB (1500 ) A Combining the equilibrium equation and the deformation equation gives TA = 27.537 ×103 N (T) = σ A ( 350 ×10 −6 ) TB = 44.060 × 103 N (T) = σ B ( 750 ×10 −6 ) 197 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS σ A = 78.7 ×106 N/m 2 = 78.7 MPa σ B = 58.7 ×106 N/m 2 = 58.7 MPa 3 ............................................................................ Ans. ............................................................................ Ans. (b) The vertical displacement of the top of rod A is the same as the stretch of rod A δA ( 27.537 ×10 ) (1500 ) = 0.590 mm ↑ ......................................................... Ans. = ( 200 ×10 )( 350 ×10 ) 9 −6 6-77 The pin connected structure shown in Fig. P6-77 consists of a cold-rolled bronze [Eb = 15,000 ksi and αb = 9.4(10-6)/°F] bar A which has a cross-sectional area of 3.00 in2 and two 0.2% C hardened steel [Es = 30,000 ksi and αs = 6.6(10-6)/°F] bars B which have cross-sectional areas of 2.50 in2. If the temperature of bar A decreases 50°F and the temperature of bars B increases 30°F after the 200-kip load is applied, determine (a) The normal stresses in the bars. (b) The displacement of pin C. SOLUTION (a) From a free-body diagram of the pin, the vertical equilibrium equation gives ↑ ΣFy = 0 : 2 ( 4 5 ) TB + TA − 200 × 103 = 0 1.6TB + TA = 200 × 103 lb The stretch of the bronze and the steel bars are related by δ B = δ A cosθ = 4δ A 5 . Therefore TA ( 4 ) TB ( 5) + ( 6.6 × 10−6 ) ( 30 )( 5 ) = 0.8 + ( 9.4 × 10−6 ) ( −50 )( 4 ) 6 6 ( 30 ×10 ) ( 2.50 ) (15 × 10 ) ( 3) TB = (1.06667TA − 37, 410 ) lb TA = 96.006 × 103 lb = σ A ( 3) TB = 64.996 × 103 lb = σ B ( 2.5) Combining the equilibrium equation and the deformation equation gives σ A = 32.0 ×103 psi = 32.0 ksi (T) .............................................................................. Ans. σ B = 26.0 ×103 psi = 26.0 ksi (T) .............................................................................. Ans. (b) The vertical displacement of pin C is the same as the stretch of bar A 3 δA ( 96.006 ×10 ) ( 48) + 9.4 ×10 −50 48 = 0.0798 in. = ( ) ( )( ) (15 ×10 ) ( 3) −6 6 ↓ .................... Ans. 6-78 Three bars are connected by smooth frictionless pins as shown in Fig. P6-78. Bar BCD is rigid, bar AB is aluminum (E = 73 GPa; L = 750 mm; d = 40 mm; σmax = 240 MPa), and bar DE is steel (E = 200 GPa; L = 500 mm; d = 30 mm; σmax = 400 MPa). The 50-mm-diameter pivot pin C is aluminum (τmax = 180 MPa) and is in double shear. After the unit is assembled, the nut D is slowly tightened. (a) Plot σAB, σDE, and τC as functions of the distance that the nut advances (0 ≤ δnut ≤ 2 mm). (b) (c) Plot δAB and δDE as function of δnut (0 ≤ δnut ≤ 2 mm). What is the maximum amount δnut that the nut can be tightened? SOLUTION (a) From a free-body diagram of the bar BCD, the equilibrium equations give 198 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS → ΣFx = 0 : ↑ ΣFy = 0 : ΣM C = 0 : Cx = 0 C y − TAB − TDE = 0 C y = TAB + TDE 200TAB − 125TDE = 0 5TDE = 8TAB From similar triangles, the vertical movement of points B and D are related by b d = 200 125 in which b = δ AB is the stretch of rod AB, d = δ nut − δ DE mm , δ DE is the stretch of rod DE, and δ nut is the movement of the nut along the rod. Therefore 5δ AB = 8 (δ nut − δ DE ) 5 ( 73 ×10 ) π ( 0.04 ) 9 TAB ( 750 ) 2 4 = 8δ nut − 8 ( 200 ×10 ) π ( 0.03) 9 TDE ( 500 ) 2 4 1.44478TAB + TDE = 282, 743.34δ nut N Combining the equilibrium equation and the deformation equation gives TAB = TDE 282, 743.34δ nut = 92,861.66δ nut N 3.04478 = 1.6TAB TAB N/m 2 (T) ...................................................................................... Ans. N/m 2 (T) ...................................................................................... Ans. Dividing the axial forces by the cross-sectional areas gives the normal stresses σ AB = σ DE = π ( 0.04 ) 4 2 π ( 0.03) 4 2 TDE Dividing the force on the pin by twice its cross-sectional area (since it is in double shear) gives the shear stress 2 C x2 + C y VC τC = N/m 2 ............................................................................... Ans. = 2 As 2 π ( 0.05 ) 4 (b) The stretches of the two rods are δ AB = δ DE = ( 73 ×109 ) π ( 0.04 )2 4 ( 200 ×10 ) π ( 0.03) 9 TAB ( 750 ) mm (stretch) ......................................................... Ans. TDE ( 500 ) 2 4 mm (stretch) ...................................................... Ans. (c) Checking the graph reveals that the normal stress in rod DE reaches its limiting value for δ nut ≅ 1.90 mm . The other stresses remain below their limiting values for the entire range graphed. Therefore, 199 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS (δ nut )max = 1.90 mm ....................................................................................................... Ans. 6-79 Three bars are connected by smooth frictionless pins as shown in Fig. P6-79. Bar ABCD is rigid, bar AE is aluminum (E = 10,600 ksi; L = 24 in.; d = ½ in.; σmax = 40 ksi), and bar CF is steel (E = 30,000 ksi; L = 18 in.; d = ¾ in.; σmax = 50 ksi). The ¾-in.-diameter pivot pin B is steel (τmax = 25 ksi) and is in single shear. The holes in bar CF are slightly overdrilled so that pin C moves down 0.06 in. before contact is made with bar CF. (a) Plot σAE, σCF, and τB as functions of P (0 ≤ P ≤ 10 kip). (b) (c) Plot δAE and δCF as functions of P (0 ≤ P ≤ 10 kip). What is the maximum force P that the system can withstand? SOLUTION (a) From a free-body diagram of the bar ABCD, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM B = 0 : Bx = 0 By − TAE + FCF − P = 0 By = P + TAE − FCF 12TAE + 9 FCF − 24 P = 0 4TAE + 3FCF = 8P From similar triangles, the vertical movement of points A, C, and D are related by acd == 12 9 24 in which a = δ AE is the stretch of rod AE, d = δ CF + 0.06 in. , and δ CF is the shrink of rod CF. Therefore 3δ AE = 4δ CF + 0.24 in. 3 (10.6 ×106 ) π ( 0.5)2 4 132,536 + 8 P lb 23.10377 TAE ( 24 ) =4 ( 30 ×106 ) π ( 0.75)2 4 FCF (18 ) + 0.24 in. 6.36792TAE − FCF = 44,178.65 lb Combining the equilibrium equation and the deformation equation gives TAE = 200 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS FCF = 8 P − 4TAE 3 TAE psi (T) ............................................................................................. Ans. psi (C) .......................................................................................... Ans. Dividing the axial forces by the cross-sectional areas gives the normal stresses σ AE = σ CF = π ( 0.5 ) 4 2 π ( 0.75) 4 2 FCF Dividing the force on the pin by its cross-sectional area gives the shear stress 2 Bx2 + By VB τB = psi ........................................................................................... Ans. = As π ( 0.75 )2 4 (b) The stretches of the two rods are δ AE = δ CF = (10.6 ×106 ) π ( 0.5)2 4 ( 30 ×10 ) π ( 0.75) 6 TAE ( 24 ) FCF (18 ) in. (stretch) ........................................................... Ans. 2 4 in. (shrink) ............................................................ Ans. ≅ 6.2 kip . The (c) Checking the graph reveals that the normal stress in rod AE reaches its limiting value for P other stresses remain below their limiting values for the entire range graphed. Therefore, Pmax = 6.2 kip ................................................................................................................... Ans. 6-80 Member ABCD of the pin-connected structure shown in Fig. P6-80 is rigid; bar BF is made of steel [EBF = 210 GPa; ABF = 1200 mm2; αBF = 11.9(10-6)/°C]; and bar CE is made of an aluminum alloy [ECE = 73 GPa; ACE = 900 mm2; αCE = 22.5(10-6)/°C]. As a result of a misalignment of the pin holes at A, B, and C, bar CE must be heated 80°C (after pins A and B are in place) to permit insertion of pin C. Compute and plot (a) The axial stresses σBF (in the steel bar) and σCE (in the aluminum bar) as functions of the temperature decrease (as bar CE cools back down to room temperature) ∆T (0°C ≥ ∆T ≥ −80°C). (b) The elongations δBF (of the steel bar) and δCE (of the aluminum bar) as functions of the temperature decrease ∆T (0°C ≥ ∆T ≥ −80°C). SOLUTION When bar CE is heated 80°C, it stretches 201 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS δ init = ( 22.5 × 10−6 ) ( 80 )( 600 ) = 1.08000 mm (a) From a free-body diagram of the bar ABCD, the moment equilibrium equation gives ΣM A = 0 : 240TCE − 80TBF = 0 TBF = 3TCE From similar triangles, the stretch of bar BF and the movement of pin C are related by δ BF c = 80 240 in which c = (1.08000 − δ CE ) mm is the stretch of bar CE. Therefore and δ CE 3δ BF = (1.0800 − δ CE ) mm TBF (1000 ) = 1.08000 3 9 −6 ( 210 × 10 )(1200 × 10 ) TCE ( 600 ) − + ( 22.5 × 10−6 ) ( 80 + ∆T )( 600 ) 9 −6 ( 73 ×10 )( 900 × 10 ) TCE + 1.30357TBF = −1478.250∆T N in which −80°C ≤ ∆T ≤ 0°C . Combining the equilibrium equation and the deformation equation gives TCE = TBF −1478.250∆T = −301.026∆T N 4.91071 = 3TCE Dividing the axial forces by the cross-sectional areas gives the normal stresses σ BF = σ CE TBF T N/m 2 = BF MPa (T) .............................................................. Ans. −6 1200 × 10 1200 TCE T = N/m 2 = CE MPa (T) .................................................................. Ans. −6 900 × 10 900 (b) The stretches of the two rods are δ BF = ( 210 ×10 )(1200 ×10 ) 9 −6 TBF (1000 ) mm (stretch) ......................................................... Ans. TCE ( 600 ) δ CE = 9 −6 ( 73 ×10 )( 900 × 10 ) + ( 22.5 ×10−6 ) ( 80 + ∆T )( 600 ) mm (stretch) ............................................ Ans. 202 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 6-81 Three bars are connected by smooth frictionless pins as shown in Fig. P6-81. Bar BCD is rigid, bar AC is aluminum [E = 12,000 ksi; L = 36 in.; d = 1 in.; σmax = 20 ksi; α = 12.5(10-6)/°F], and bar DE is steel [E = 30,000 ksi; L = 30 in.; d = ½ in.; σmax = 10 ksi; α = 6.6(10-6)/°F]. The 5/8-in.-diameter pivot pin B is steel (τmax = 5 ksi) and is in single shear. If the temperature drops after the unit is assembled, (a) Plot σAC, σDE, and τB as functions of the temperature drop ∆T (0 ≥ ∆T ≥ −60°F). (b) (c) Plot δAC and δDE as functions of ∆T (0 ≥ ∆T ≥ −60°F). What is the maximum temperature drop that the system can withstand? SOLUTION (a) From a free-body diagram of the bar BCD, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM B = 0 : Bx = 0 By + TAC − TDE = 0 By = TDE − TAC 20TAC − 30TDE = 0 TAC = 1.5TDE From similar triangles, the shrink of bar AC and the stretch of bar DE are related by δ AC δ DE = 20 30 3δ AC = 2δ DE TAC ( 36 ) −3 + (12.5 × 10−6 ) ( ∆T )( 36 ) 2 6 12 × 10 ) π (1) 4 ( TDE ( 30 ) = 2 + ( 6.6 × 10−6 ) ( ∆T )( 30 ) ( 30 × 106 ) π ( 0.5 )2 4 TDE + 1.12500TAC = −171.41315 ∆T lb in which −60°F ≤ ∆T ≤ 0°F . Combining the equilibrium equation and the deformation equation gives TDE = −63.78164 ∆T lb 203 TAC = 1.5TDE STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS Dividing the axial forces by the cross-sectional areas gives the normal stresses σ AC = σ DE = π (1) 4 2 TAC psi (T) ................................................................................................. Ans. psi (T) ............................................................................................. Ans. π ( 0.5 ) 4 2 TDE Dividing the force on the pin by its cross-sectional area gives the shear stress 2 Bx2 + By VB τB = psi ............................................................................................. Ans. = As π ( 5 8 )2 4 (b) The shrink of bar AC and the stretch of bar DE are δ AC TAC ( 36 ) −6 =− + (12.5 × 10 ) ( ∆T )( 36 ) in. (shrink) ............. Ans. (12 × 106 ) π (1)2 4 TDE ( 30 ) −6 = + ( 6.6 × 10 ) ( ∆T )( 30 ) in. (stretch) ............. Ans. ( 30 × 106 ) π ( 0.5 ) 2 4 ∆T ≅ −31 °F . The δ DE (c) Checking the graph reveals that the normal stress in rod DE reaches its limiting value for other stresses remain below their limiting values for the entire range graphed. Therefore, ∆Tmax = −31 °F ................................................................................................................ Ans. 6-82 Use the method of joints and determine the force in each member of the truss shown in Fig. P6-82, if a = 2 m and P = 5 kN. SOLUTION b = 2 tan 30° = 1.15470 m d = b tan 30° = 0.66667 m From a free-body diagram of the entire truss, the equilibrium equations give → ΣFx = 0 : ΣM A = 0 : ↑ ΣFy = 0 : Ax = 0 2.66667C − 2 ( 5) = 0 Ay + C − 5 = 0 204 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS C = 3.7500 kN Ay = 1.2500 kN From a free-body diagram for joint A, the equilibrium equations give ↑ ΣFy = 0 : → ΣFx = 0 : 1.25 + TAB sin 30° = 0 TAD + TAB cos 30° = 0 TAB = −2.500 kN = 2.50 kN (C) ............................... Ans. TAD = 2.1651 kN ≅ 2.17 kN (T) ............................... Ans. From a free-body diagram for joint D, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : TCD − TAD = 0 TBD − 5 = 0 TCD = 2.1651 kN ≅ 2.17 kN (T) .............................................. Ans. TBD = 5.000 kN = 5.00 kN (T) ................................................ Ans. Finally, from a free-body diagram for joint C, the equilibrium equations give ↑ ΣFy = 0 : 3.75 + TBC sin 60° = 0 TBC = −4.3301 kN = 4.33 kN (C) ........................................... Ans. 6-83 A 4000-lb crate is attached by light, inextensible cables to the truss of Fig. P6-83. Determine the force in each member of the truss using the method of joints. SOLUTION From a free-body diagram of the entire truss, the equilibrium equations give ΣM A = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : 6 E − 4 ( 2000 ) − 8 ( 2000 ) = 0 E − Ax = 0 Ay − 4000 = 0 E = 4000 lb Ax = 4000 lb Ay = 4000 lb From a free-body diagram for joint A, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ( 4 5 ) TAB − 4000 = 0 4000 − TAE − ( 3 5 ) TAB = 0 TAB = 5000 lb (T) ................................................................. Ans. TAE = 1000 lb (T) ................................................................. Ans. From a free-body diagram for joint E, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : TDE + ( 4 5 ) TBE + 4000 = 0 TAE + ( 3 5 ) TBE = 0 TBE = −1666.67 lb ≅ 1667 lb (C) .........................................Ans. 205 STATICS AND MECHANICS OF MATERIALS, 2nd Edition From a free-body diagram for joint D, the equilibrium equations give RILEY, STURGES AND MORRIS TDE = −2666.67 lb ≅ 2670 lb (C) ........................................Ans. → ΣFx = 0 : ↑ ΣFy = 0 : TCD − TDE = 0 TBD − 2000 = 0 TCD = −2666.67 lb ≅ 2670 lb (C) ........................................... Ans. TBD = 2000 lb (T) ....................................................................... Ans. Finally, from a free-body diagram for joint C, the equilibrium equations give ↑ ΣFy = 0 : ( 3 5 ) TBC − 2000 = 0 TBC = 3333.33 lb = 3330 lb (T) .............................................. Ans. 6-84 Use the method of joints and determine the force in each member of the truss shown in Fig. P6-84. All members are 3 m long. SOLUTION From a free-body diagram of the entire truss, the equilibrium equations give → ΣFx = 0 : ΣM F = 0 : ↑ ΣFy = 0 : ↑ ΣFy = 0 : → ΣFx = 0 : Ax = 0 3 Ay − 3 ( 3) = 0 F − Ay − 5 − 3 = 0 Ay = 3 kN F = 11 kN From a free-body diagram for joint D, the equilibrium equations give −3 − TDE sin 60° = 0 −TCD − TDE cos 60° = 0 TDE = −3.46410 kN ≅ 3.46 kN (C) ........................................ Ans. TCD = 1.73205 kN ≅ 1.732 kN (T) .......................................... Ans. From a free-body diagram for joint C, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : TCD + TCE cos 60° − TBC cos 60° = 0 −TCE sin 60° − TBC sin 60° − 5 = 0 TBC = −1.15470 kN ≅ 1.155 kN (C) ...................................... Ans. TCE = −4.61880 kN = 4.62 kN (C) ........................................ Ans. From a free-body diagram for joint A, the equilibrium equations give ↑ ΣFy = 0 : → ΣFx = 0 : TAB sin 60° − 3 = 0 TAF + TAB cos 60° = 0 TAB = 3.46410 kN ≅ 3.46 kN (T) ........................................... Ans. TAF = −1.73205 kN ≅ 1.732 kN (C) ...................................... Ans. From a free-body diagram for joint B, the equilibrium equations give ↑ ΣFy = 0 : TBC sin 60° − TBF sin 60° − TAB sin 60° = 0 206 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS TBF = −4.61880 kN ≅ 4.62 kN (C) ........................................ Ans. → ΣFx = 0 : TBE + TBC cos 60° + TBF cos 60° − TAB cos 60° = 0 TBE = 4.61880 kN ≅ 4.62 kN (T) .................................... Ans. Finally, from a free-body diagram for joint F, the equilibrium equations give ↑ ΣFy = 0 : 11 + TBF sin 60° + TEF sin 60° = 0 TEF = −8.08291 kN = 8.08 kN (C) .................................. Ans. 6-85 Each truss member in Fig. P6-85 is 5 ft long. Find the forces in members CD and EF using the method of sections. SOLUTION From a free-body diagram of the entire truss, the equilibrium equations give → ΣFx = 0 : ΣM B = 0 : Ax = 0 800 ( 5 ) + 600 (10 ) + 800 (15 ) +1000 ( 20 ) − 25 Ay = 0 Ay = 1680 lb Cut a section through CD, DE, and EF, and draw a free-body diagram of the left-hand side of the truss. The height of the truss is h = 5sin 60° = 4.33013 ft and the equilibrium equations give ΣM E = 0 : ΣM D = 0 : 1000 ( 5 ) − 1680 (10 ) − TCD ( 4.33013) = 0 800 ( 2.5 ) + 1000 ( 7.5 ) − 1680 (12.5) + TEF ( 4.33013) = 0 TCD = −2725.09 lb ≅ 2730 lb (C) ......................Ans. TEF = 2655.81 lb ≅ 2660 lb (T) .................................................................................. Ans. 6-86 Find the forces in members CJ and KJ of the roof truss shown in Fig. P6-86 using the method of sections. SOLUTION From a free-body diagram of the entire truss, the equilibrium equations give → ΣFx = 0 : ΣM G = 0 : Ax = 0 4 ( 4 ) + 3 ( 8 ) + 5 (12 ) + 4 (16 ) + 3 ( 20 ) − 24 Ay = 0 Ay = 9.33333 kN Cut a section through CD, CJ, and JK, and draw a free-body diagram of the left-hand side of the truss. 207 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS θ CD = tan −1 (10 12 ) = 39.806° θ CJ = tan −1 ( h 4 ) = 59.036° h = ( 2 3)(10 ) = 6.66667 m Then, the equilibrium equations give ΣM A = 0 : ΣM C = 0 : −3 ( 4 ) − 4 ( 8 ) − (TCJ sin θ CJ )(12 ) = 0 3 ( 4 ) − 9.33333 ( 8 ) + TJK ( 6.66667 ) = 0 TCJ = −4.28 kN = 4.28 kN (C) .................................................................................... Ans. TJK = 9.40 kN (T) ........................................................................................................... Ans. 6-87 Find the forces in members BC, BF, and AF, of the stairs truss of Fig. P6-87 using the method of sections. SOLUTION From a free-body diagram of the entire truss, the equilibrium equations give → ΣFx = 0 : ΣM D = 0 : Ax = 0 200 ( 4 ) + 400 ( 8 ) + 600 (12 ) − 12 Ay = 0 Ay = 933.333 lb Cut a section through AF, BF, and BC, and draw a free-body diagram of the upper-portion of the truss. The equilibrium equations give ΣM B = 0 : ΣM F = 0 : ↑ ΣFy = 0 : 600 ( 4 ) − 933.333 ( 4 ) − ( 3 5 ) TAF ( 4 ) = 0 600 ( 4 ) − 933.333 ( 4 ) − ( 4 5 ) TBC ( 3) = 0 TAF = 555.556 lb ≅ 556 lb (T) .......................................Ans. TBC = −555.556 lb ≅ 556 lb (C) ....................................Ans. 933.333 − 600 − ( 3 5 ) TAF − ( 3 5 ) TBC − TBF = 0 TBF = 333.333 lb ≅ 333 lb (T) .......................................Ans. 6-88 The Gambrel truss shown in Fig. P6-88 supports one side of a bridge; an identical truss supports the other side. Floor beams carry vehicle loads to the truss joints. Calculate the forces in members BC, BG, and DE when a truck having a mass of 3500 kg is stopped in the middle of the bridge as shown. The center of gravity of the truck is 1 m in front of the rear wheels. SOLUTION W = 3500 ( 9.81) = 34,335 N From a free-body diagram of the entire bridge (which is consists of two trusses), the equilibrium equations give → ΣFx = 0 : ΣM E = 0 : ΣM A = 0 : 2 Ax = 0 34, 335 ( 4 ) − ( 2 Ay ) (10 ) = 0 ( 2 E )(10 ) − ( 34,335 )( 6 ) = 0 E = 10,300.5 kN Ay = 6867.0 N 208 STATICS AND MECHANICS OF MATERIALS, 2nd Edition Half of the truck’s weight is supported by each of the two trusses. From a free-body diagram of the truck, the equilibrium equations give RILEY, STURGES AND MORRIS ΣM R = 0 : ΣM F = 0 : 34,335 (1) − ( 2 N F )( 4 ) = 0 ( 2 N R )( 4 ) − 34,335 ( 3) = 0 N F = 4291.88 N N R = 12,875.6 N From a free-body diagram of the floor panel between joints H and G, the equilibrium equations give ΣM G = 0 : N F ( 2 ) − 3H = 0 H = 2861.25 N Next, cut a section through BC, BG, and GH, and draw a free-body diagram of the left-hand side of the truss. θ = tan −1 ( 2 3) = 33.690° φ = tan −1 ( 3 3) = 45° Then, the equilibrium equations give ΣM G = 0 : ↑ ΣFy = 0 : 2861.25 ( 3) − 6867 ( 5 ) − ( TBC cos θ )( 5 ) = 0 6867 − 2861.25 + TBC sin θ − TBG sin φ = 0 TBC = −6190 N = 6.19 kN (C) ......................................Ans. TBG = 809.30 N ≅ 0.809 kN (T) ....................................... Ans. Finally, from a free-body diagram for joint E, the equilibrium equations give ↑ ΣFy = 0 : 10,300.5 + TDE cosθ = 0 TDE = −12,380 N = 12.38 kN (C) .......................................Ans. 6-89 A truss is loaded and supported as shown in Fig. P6-89. Determine (a) The normal stress in member CD if it has a diameter of ½ in. (b) The change in length of member CF if it has a diameter of ½ in. and a modulus of elasticity of 29(106) psi. (c) The change in length of member EF if it has the same diameter and modulus of elasticity as member CF. SOLUTION From a free-body diagram of the entire truss, the equilibrium equations give ΣM A = 0 : F (18 ) − 1000 ( 6 ) − 2000 (12 ) −2000 (18) − 1000 ( 24 ) = 0 F = 5000 lb Cut a section through CD, CF, and FG, and draw a free-body diagram of the right-hand side of the truss. The equilibrium equations give 209 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ΣM F = 0 : TCD ( 8) − 1000 ( 6 ) = 0 2 TCD = 750 lb = σ CD π ( 0.5 ) 4 (a) σ CD = 3820 psi (T) ...........................................................Ans. ↑ ΣFy = 0 : ( 4 5 ) TCF + 5000 − 3000 = 0 TCF = −2500 lb (b) δ CF = ( −2500 )(120 ) PL = = −0.0527 in. ............................................... Ans. EA ( 29 × 106 ) π ( 0.5 )2 4 Finally, from a free-body diagram for joint E, the equilibrium equations give ↑ ΣFy = 0 : −1000 − ( 4 5 ) TEF = 0 TEF = −1250 lb (c) δ EF = ( −1250 )(120 ) PL = = −0.0263 in. ............................................... Ans. EA ( 29 × 106 ) π ( 0.5 ) 2 4 6-90 A transmission line truss supports a 5-kN load, as shown in Fig. P6-90. All members of the truss are made of structural steel (see Appendix A for properties). Determine the change in length of members FG and CD if they have cross-sectional areas of 300 mm2 each. SOLUTION Cut a section through CD, DG, and FG, and draw a free-body diagram of the upper-portion of the truss. The equilibrium equations give ΣM G = 0 : ΣM D = 0 : ( 5cos 30° )( 9 ) − ( 5sin 30° )( 4 ) + TCD ( 3) = 0 ( 5cos 30° )( 6 ) − TFG ( 3) = 0 ( −9657.05)( 4000 ) = −0.644 mm .............................................. Ans. PL = EA ( 200 × 109 )( 300 × 10−6 ) ( 8660.25)( 4000 ) = +0.577 mm .............................................. Ans. PL = EA ( 200 × 109 )( 300 × 10−6 ) TCD = −9.65705 kN TFG = 8.66025 kN δ CD = δ FG = 6-91 The truss shown in Fig. P6-91 supports a sign that weighs 3000 lb. The sign is connected to the truss at joints E, G, and H, and the connecting links are adjusted so that each joint carries one-third of the load. All members of the truss are made of structural steel, and each has a cross-sectional area of 0.564 in2. Determine the axial stresses and strains in members CD, CF, CG, and FG of the truss. SOLUTION From a free-body diagram of the entire truss, the equilibrium equations give ΣM A = 0 : TB ( 48 ) − 1000 ( 8) − 1000 ( 24 ) −1000 ( 40 ) = 0 210 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS TB = 1500 lb Cut a section through CD, CF, and FG, and draw a free-body diagram of the right-hand side of the truss. The equilibrium equations give ΣM F = 0 : ΣM C = 0 : ↑ ΣFy = 0 : ↑ ΣFy = 0 : TCD ( 6 ) + 1500 (16 ) − 1000 ( 8 ) = 0 1500 ( 24 ) − 1000 (16 ) − TFG ( 6 ) = 0 1500 − 1000 + ( 3 5) TCF = 0 TCG − 1000 = 0 TCF = −833.333 lb TCG = 1000 lb TCD = −2666.667 lb TFG = 3333.333 lb Next, from a free-body diagram for joint G, the equilibrium equations give Finally, calculating the stresses and strains TCD −2666.667 = = −4728 psi ≅ 4.73 ksi (C) ............................................ Ans. A 0.564 σ −4728 ε CD = = = −163.0 × 10−6 = −163.0 µ in./in. .......................................... Ans. 6 E 29 × 10 −833.333 σ CF = = −1478 psi ≅ 1.478 ksi (C) ........................................................ Ans. 0.564 −1478 ε CF = = −50.9 × 10−6 = −50.9 µ in./in. ........................................................ Ans. 6 29 × 10 +1000 σ CG = = +1773 psi ≅ 1.773 ksi (T) ............................................................... Ans. 0.564 +1773 ε CF = = +61.1×10−6 = +61.1 µ in./in. ........................................................ Ans. 6 29 × 10 +3333.333 σ FG = = +5910 psi ≅ 5.91 ksi (T) ........................................................ Ans. 0.564 +5910 ε CF = = +204 × 10−6 = +204 µ in./in. .......................................................... Ans. 6 29 × 10 σ CD = 6-92 All members of the truss shown in Fig. P6-92 are made of structural steel (E = 200 GPa) and are 25-mm in diameter. Determine the normal stresses in and the change in length of members CD, DI, and HI. SOLUTION From a free-body diagram of the entire truss, the equilibrium equations give → ΣFx = 0 : ΣM G = 0 : Ax = 0 10 ( 6 ) + 18 (12 ) − Ay (12 ) = 0 Ay = 23 kN Cut a section through CD, DI, and HI, and draw a free-body diagram of the right-hand side of the truss. The equilibrium equations give 211 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ΣM I = 0 : ΣM D = 0 : ↑ ΣFy = 0 : 18 ( 3) − 23 ( 3) − TCD ( 3) = 0 18 ( 6 ) + THI ( 3) − 23 ( 6 ) = 0 23 − 18 + TDI sin 45° = 0 TCD = −5 kN THI = 10 kN TDI = −7.07107 kN Finally, calculating the stresses and changes in length σ CD = σ DI = σ HI = δ CD = −5000 π ( 0.025) 4 2 = −10.19 × 106 N/m 2 = 10.19 MPa (C) .............................. Ans. = −14.41× 106 N/m 2 = 14.41 MPa (C) ................................ Ans. = +20.4 ×106 N/m 2 = 20.4 MPa (T) ................................... Ans. −7071.07 π ( 0.025) 4 2 +10, 000 π ( 0.025 ) 4 2 δ DI = δ HI = ( −7071.07 ) ( 3000 2 ) = −0.306 mm .................................................... Ans. 2 200 × 109 ) π ( 0.025 ) 4 ( (10, 000 )( 6000 ) ( 200 ×109 ) π ( 0.025)2 4 = +0.611 mm .................................................... Ans. ( −5000 )( 6000 ) PL = = −0.306 mm ........................................ Ans. EA ( 200 × 109 ) π ( 0.025 )2 4 6-93 Determine the force in member BD of the truss shown in Fig. P6-93. SOLUTION θ = 30° + tan −1 (12 16 ) = 66.870° b= 10 = 11.54701 ft cos 30° φ = sin −1 12 − b sin θ = 6.870° b Cut a section through members BD and CD, and draw a free-body diagram of the left-hand side of the truss. The moment equilibrium equation gives 212 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ΣM A = 0 : (TBD sin φ )( b cosθ ) − 2000 ( b cosθ ) − (TBD cos φ )( b sin θ ) = 0 TBD = −907 lb = 907 lb (C) .......................................................................................... Ans. 6-94 For the simple truss of Fig. P6-94 show that overall equilibrium of the truss is a consequence of equilibrium of all of the pins; hence the equations of overall equilibrium give no new information. (Hint: Write equations of equilibrium for each of the pins and eliminate the unknown member forces from these equations.) SOLUTION From a free-body diagram of the entire truss, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM A = 0 : P − Ax = 0 C − Ay = 0 Ca − Pb = 0 (a) (b) (c) Ax = P → ΣFx = 0 : ↑ ΣFy = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : Ay = C = Pb a (d) (e) From a free-body diagram of joint A, the equilibrium equations give TAC + TAB cos 60° − Ax = 0 TAB sin 60° − Ay = 0 From a free-body diagram of joint B, the equilibrium equations give P + TBC cos 30° − TAB cos 60° = 0 −TBC sin 30° − TAB sin 60° = 0 −TBC cos 30° − TAC = 0 C + TBC sin 30° = 0 (f) (g) From a free-body diagram of joint B, the equilibrium equations give (h) (i) Adding Eqs. (d), (f), and (h) gives Eq. (a) P − Ax = 0 ..................................................................................................................(a) Adding Eqs. (e), (g), and (i) gives Eq. (b) C − Ay = 0 ..................................................................................................................(b) Then, Eqs. (e) and (i) can be rewritten TAB = Ay sin 60° = C =0 sin 60° TBC = − C sin 30° Equation (f) can be rewritten −C C cos 30° cos 60° P = − + cos 30° + cos 60° = C sin 30° sin 60° sin 30° sin 60° sin 60° cos 30° + sin 30° cos 60° sin 90° =C =C sin 30° sin 60° sin 30° sin 60° But sin 90° = 1 , rewritten sin 60° = b d , sin 30° = d a , sin 30° sin 60° = b a , and therefore, Eq. (f) can be 213 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS P= C ................................................................................................................... (c) ba which is equivalent to Eq. (c). 6-95 Show that the overall equilibrium of a truss is a consequence of the equilibrium of the two separate parts generated by the method of sections. That is, section the bridge truss of Fig. P6-95 as indicated, and write the equilibrium equations for each piece. Eliminate the member forces from the resulting six equations and show that the result is equivalent to the equilibrium of the whole truss. SOLUTION From a free-body diagram of the entire truss, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM A = 0 : Ax = 0 Ay − P + D = 0 (a) (b) (c) D ( 3w ) − P ( 2w ) = 0 Cut a section through BC, CF, and EF, and draw a free-body diagram of the left-hand side of the truss. The equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM C = 0 : Ax + TBC + TCF cos θ + TEF = 0 TCF sin θ + Ay = 0 (d) (e) (f) (g) TEF ( h ) + Ax ( h ) − Ay ( 2w ) = 0 ΣM F = 0 : −TBC ( h ) − Ay ( w ) = 0 Cut a section through BC, CF, and EF, and draw a free-body diagram of the right-hand side of the truss. The equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM C = 0 : ΣM F = 0 : −TBC − TCF cos θ − TEF = 0 −TCF sin θ − P + D = 0 D ( w ) − TEF ( h ) = 0 D ( 2w ) + TBC ( h ) − P ( w ) = 0 (h) (i) (j) (k) Adding Eqs. (d) and (h) gives Eq. (a) Ax = 0 ................................................................................................................................ (a) Adding Eqs. (e) and (i) gives Eq. (b) Ay − P + D = 0 ................................................................................................................. (b) Adding two times Eq. (g) to two times Eq. (k) gives 4 Dw − 2 Pw − 2 Ay w = 0 Adding Eq. (f) to Eq. (j) gives Ax h − 2 Ay w + Dw = 0 Subtracting this last equation from the previous (and using Ax = 0 ) gives Eq. (c) 3Dw − 2 Pw = 0 ................................................................................................................ (c) 214 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 6-96 The flat roof of a building is supported by a series of parallel plane trusses spaced 2 m apart (only one such truss is shown in Fig. P6-96). Calculate the forces in all the members of a typical truss when water collects to a depth of 0.2 m as shown. The density of water is 1000 kg/m3. SOLUTION From a free-body diagram of a roof panel, the equilibrium equations give the force carried by the joints at each end of the roof panel as F1 = F2 = W 2 = (1000 )( 9.81)( 2 × 1.2 × 0.2 ) 2 = 2354.4 N Then, from a free-body diagram of the entire truss, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM A = 0 : Ax = 0 Ay + Fy − 6 F1 = 0 Fy ( 3.6 ) − ( 2 F1 )(1.2 ) − ( 2 F1 )( 2.4 ) − ( F1 )( 3.6 ) = 0 Ay = Fy = 3F1 = 7063.20 N From a free-body diagram for joint B, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : TBC = 0 −TAB − F1 = 0 TAB = −2354.4 N ≅ 2.35 kN (C) ............................................. Ans. TBC = 0 kN .................................................................................... Ans. From a free-body diagram for joint A, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : TAH + ( 3 5) TAC = 0 TAB + 7063.2 + ( 4 5 ) TAC = 0 TAC = −5886.0 N ≅ 5.89 kN (C) ............................................. Ans. TAH = 3531.6 N ≅ 3.53 kN (T) ................................................ Ans. From a free-body diagram for joint H, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : TGH − TAH = 0 TCH = 0 TCH = 0 kN ................................................................................... Ans. TGH = 3531.6 N ≅ 3.53 kN (T) ................................................ Ans. From a free-body diagram for joint E, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : TDE = 0 −TEF − F1 = 0 TDE = 0 kN .................................................................................... Ans. TEF = −2354.4 N ≅ 2.35 kN (C) ............................................. Ans. From a free-body diagram for joint F, the equilibrium equations give 215 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS → ΣFx = 0 : ↑ ΣFy = 0 : −TFG − ( 3 5) TDF = 0 TEF + 7063.2 + ( 4 5 ) TDF = 0 TDF = −5886.0 N ≅ 5.89 kN (C) ............................................. Ans. TFG = 3531.6 N ≅ 3.53 kN (T) ................................................ Ans. From a free-body diagram for joint D, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : TDE − TCD + ( 3 5 ) TDF = 0 −2 F1 − TDG − ( 4 5 ) TDF = 0 TCD = −3531.6 N ≅ 3.53 kN (C) ...................................... Ans. TDG = 0 kN ............................................................................ Ans. Finally, from a free-body diagram for joint G, the equilibrium equations give ↑ ΣFy = 0 : TDG + ( 4 5 ) TCG = 0 TCG = 0 kN ............................................................................. Ans. 6-97 Snow on a roof supported by the Howe truss of Fig. P6-97 can be approximated as a distributed load of 20 lb/ft (measured along the roof). Treat the distributed load as you would the weight of the members; that is, replace the total load on each of the upper members as a vertical force, half applied to the joint at each end of the member. Determine the forces in members BC, BG, CG, and GH. SOLUTION From a free-body diagram of a roof panel, the equilibrium equations give the force carried by the joints at each end of the roof panel as F1 = F2 = W 2 = ( 20 ) ( 80 ) 2 = 89.4427 lb Then, from a free-body diagram of the entire truss, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM A = 0 : Ax = 0 Ay + E − 8 F1 = 0 E ( 32 ) − ( 2 F1 )( 8 ) − ( 2 F1 )(16 ) − ( 2 F1 )( 24 ) − ( F1 )( 32 ) = 0 Ay = E = 3F1 = 357.771 lb Next, cut a section through members BC, BG, and GH, and draw a free-body diagram of the left-hand portion of the truss. φ = tan −1 8 16 = 26.565° Then equilibrium equations give ΣM A = 0 : ΣM B = 0 : − ( 2 F1 )( 8 ) − (TBG sin φ )(16 ) = 0 TGH ( 4 ) + F1 ( 8 ) − ( 357.771)( 8 ) = 0 TBG = −200 lb = 200 lb (C) .......................................................................................... Ans. TGH = 536.656 lb ≅ 537 lb (T) .................................................................................... Ans. 216 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ΣM G = 0 : ( 2 F1 )( 8 ) + F1 (16 ) − ( 357.771)(16 ) − (TBC sin φ )(16 ) = 0 TCD cos φ − TBC cos φ = 0 −2 F1 − TCG − TCD sin φ − TBC sin φ = 0 TBC = −400 lb = 400 lb (C) .......................................................................................... Ans. Finally, from a free-body diagram for joint C, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : TCD = −400 lb TCG = 178.9 lb (T) ................................................................ Ans. 6-98 Determine the forces in members CD, DE, and DF of the roof truss shown in Fig. P6-98. Triangle CDF is an equilateral triangle and joints E and G are at the midpoints of their respective sides. SOLUTION LAF = 32 + 42 = 5 m sin φ = 3 5 LAE = LAF = 2.5 m 2 cos φ = 4 5 b = 3 − 2sin 60° = 1.26795 m Then, from a free-body diagram of the entire truss, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM A = 0 : ( 5 + 8 + 10 ) sin φ − Ax = 0 Ay + B − ( 5 + 8 + 10 ) cos φ = 0 B ( 8 ) − 2.5 ( 8 ) − 5 (10 ) = 0 Ay = 9.650 kN B = 8.75 kN Ax = 13.800 kN Next, cut a section through members CD, DF, and EF, and draw a free-body diagram of the right-hand portion of the truss. Then the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM F = 0 : 10sin φ − TEF cos φ − TDF cos 60° − TCD = 0 8.75 − 10cos φ − TEF sin φ − TDF sin 60° = 0 8.75 ( 4 ) − TCD ( 2sin 60° ) = 0 TEF = −32.2765 kN ≅ 32.3 kN (C) TCD = 20.2073 kN ≅ 20.2 kN (T) ............................................................................... Ans. TDF = 23.2279 kN ≅ 23.2 kN (T) .............................................................................. Ans. Finally, from a free-body diagram for joint E, the component of the equilibrium equations normal to members AE and EF gives θ = tan −1 ΣFn = 0 : 1.5 − b = 13.064° 3− 2 8 + TDE sin (φ + θ ) = 0 TDE = −10.4533 kN ≅ 10.45 kN (C) .......................................................................... Ans. 217 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 6-99 Find the forces in members DE, DJ, and JK of the truss of Fig. P6-99. SOLUTION Then, from a free-body diagram of the entire truss, the equilibrium equations give → ΣFx = 0 : ΣM G = 0 : Ax = 0 8000 ( 20 ) + 8000 ( 40 ) +6000 ( 60 ) − Ay ( 80 ) = 0 Ay = 10, 500 lb Next, cut a section through members CD, JK, and JL, and draw a free-body diagram of the left-hand portion of the truss. Then, the equilibrium equations give ΣM J = 0 : ΣM C = 0 : 6000 ( 20 ) − (10,500 )( 40 ) − TCD ( 30 ) = 0 TJK ( 30 ) − (10,500 )( 20 ) = 0 TCD = −10, 000 lb = 10, 000 lb (C) TJK = 7000 lb = 7000 lb (T) ........................... Ans. Finally, from a free-body diagram for joint D, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : TDE − TCD = 0 −8000 − TDJ = 0 TDE = −10, 000 lb = 10, 000 lb (C) .............................................................................. Ans. TDJ = −8000 lb = 8000 lb (C) ...................................................................................... Ans. 6-100 The Gambrel truss shown in Fig. P6-100 supports one side of a bridge; an identical truss supports the other side. A 3400-kg truck is stopped on the bridge at the location shown and floor beams carry the vehicle load to the truss joints. If the center of gravity of the truck is located 1.5 m in front of the rear wheels, plot the force in members BC, BG, and GH as a function of the truck’s location d (0 ≤ d ≤ 20 m) SOLUTION W = 3400 ( 9.81) = 33,354 N From a free-body diagram of the entire truss (one side of the bridge), the equilibrium equations give → ΣFx = 0 : ΣM E = 0 : Ay = Ax = 0 4 PF + 8 PG + 12 PH + 16 PA − 16 Ay = 0 PF + 2 PG + 3PH + 4 PA 4 To determine the joint forces PA, PE, PF, PG, and PH, we need to first consider a free-body diagram of the truck. The equilibrium equations give the forces exerted on the front and rear wheels by the bridge deck panels ΣM R = 0 : ΣM F = 0 : 33,354 (1.5 ) − N F ( 4 ) = 0 N R ( 4 ) − 33,354 ( 2.5 ) = 0 N R = 20,846.25 N 218 N F = 12,507.75 N STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS Next, consider a bridge deck panel. Each of the two trusses supports the left end of the panel with a force F1 and the right end of the panel with a force F2. At the instant shown, one of the truck’s wheels exerts a force N on the panel at a distance b from the right end of the panel. Then the forces transmitted to the trusses are ↑ ΣFy = 0 : ΣM 2 = 0 : 2 F1 + 2 F2 − N = 0 N ( b ) − ( 2 F1 )( 4 ) = 0 F2 = N Nb N ( 4 − b ) − = 2 8 8 F1 = Nb 8 Case 1: As the truck moves across the bridge, there are five different cases to consider: 0 ≤ d ≤ 4 m ; N = NF , b = d PA = PH = PG = 0 PF = F1 = NF b NF d = 8 8 PE = F2 = Case 2: NF ( 4 − d ) 8 4 m ≤ d ≤ 8 m ; b = ( d − 4) m PA = PH = 0 NF b N F ( d − 4) = 8 8 N ( d − 4 ) N F (8 − d ) PF = F1 + F2 = R + 8 8 PG = F1 = Case 3: PE = F2 = N R (8 − d ) 8 8 m ≤ d ≤ 12 m ; b = ( d − 8 ) m PA = PE = 0 N F b N F ( d − 8) = 8 8 N ( d − 8 ) N F (12 − d ) PG = F1 + F2 = R + 8 8 PH = F1 = PF = F2 = Case 4: N R (12 − d ) 8 12 m ≤ d ≤ 16 m ; b = ( d − 12 ) m PE = PF = 0 PA = F1 = N F b N F ( d − 12 ) = 8 8 N ( d − 12 ) N F (16 − d ) PH = F1 + F2 = R + 8 8 219 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS PG = F2 = Case 5: N R (16 − d ) 8 16 m ≤ d ≤ 20 m ; b = ( d − 16 ) m PE = PF = PG = 0 N Rb N R ( d − 16 ) = 8 8 N ( 20 − d ) PH = F2 = R 8 PA = F1 = Next, cut a section through BC, BG, and GH, and draw a free-body diagram of the left-hand side of the truss. θ = tan −1 ( 3 4 ) = 36.870° Then, the equilibrium equations give φ = tan −1 (1 4 ) = 14.036° ΣM G = 0 : ΣM B = 0 : ↑ ΣFy = 0 : 4 PH + 8 ( PA − Ay ) − 4 (TBC cos φ ) = 0 3TGH − 4 ( Ay − PA ) = 0 Ay − PA + TBC sin φ − TBG sin θ = 0 TBC = TGH = TBG = 6-101 PH + 2 PA − 2 Ay cos φ 4 ( Ay − PA ) ..................... Ans. 3 Ay − PA + TBC sin φ sin θ ............................. Ans. ................ Ans. An overhead crane consists of an I-beam supported by a simple truss as shown in Fig. P6-101. If the uniform I-beam weighs 400 lb and is supporting a load of 1000 lb at a distance d from its left end, plot the force in members AB, BC, EF, and FG as a function of the position d (0 ≤ d ≤ 8 ft). SOLUTION From a free-body diagram of the entire truss, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM A = 0 : G − Ax = 0 Ay − PF − PD = 0 3G − 3PF − 9 PD = 0 Ay = PF + PD = 1400 lb G = PF + 3PD Ax = G = PF + 3PD ΣM D = 0 : ΣM F = 0 : in which the joint forces PF and PD are determined from equilibrium of the I-beam 400 ( 3) + 1000 ( 7 − d ) − PF ( 6 ) = 0 6 PD − 400 ( 3) − 1000 ( d − 1) = 0 220 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS PF = 200 + 500 ( 7 − d ) lb 3 500 ( d − 1) PD = 200 + lb 3 TAB cos φ − Ax = 0 1400 − TAG − TAB sin φ = 0 Then, from a free-body diagram for joint A, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : TAG = 1400 − TAB sin φ TAB = in which Ax ....................................................................................................................... Ans. cos φ φ = tan −1 ( 3 9 ) = 18.435° . TFG + TBG cos θ + G = 0 TAG + TBG sin θ = 0 TBG = −TAG sin θ = −G − TBG cos θ ...................................................................................................... Ans. Then, from a free-body diagram for joint G, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : TFG in which θ = tan −1 ( 2 3) = 33.690° . Finally, cut a section through members BC, CF, and EF, and draw a free-body diagram of the right-hand portion of the truss. Then, the equilibrium equations give ΣM F = 0 : ΣM C = 0 : TBC = 6 (TBC sin φ ) − 6 PD = 0 −1TEF − 3PD = 0 PD ......................................... Ans. sin φ TEF = −3PD ......................................... Ans. 6-102 The masses of carton 1, 2, and 3, which rest on the platform shown in Fig. P6-102, are 300 kg, 100 kg, and 200 kg, respectively. The mass of the platform is 500 kg. Determine the tensions in the three cables A, B, and C that support the platform. SOLUTION From a free-body diagram of the platform, the equilibrium equations give ΣM O = 0 : ( i + j) × ( −300 g k ) + ( 2 i + j) × ( −100 g k ) 221 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS + ( 3 i + j) × TB k + (1.5 i + 2 j) × ( −500 g k ) + ( i + 3 j) × ( −200 g k ) + ( i + 4 j) × TC k = 0 x: y: −300 g − 100 g + TB − 1000 g − 600 g + 4TC = 0 +300 g + 200 g − 3TB + 750 g + 200 g − TC = 0 ΣFz = 0 : TA + TB + TC − 1100 g = 0 TA = 340.91g = 3340 N ............................. Ans. TB = 345.45 g = 3390 N ................................................................................................ Ans. TC = 413.64 g = 4060 N ................................................................................................ Ans. 6-103 Determine the reaction at support A of the pipe system shown in Fig. P6-103 when the force applied to the pipe wrench is 50 lb. SOLUTION From a free-body diagram of the platform, the equilibrium equations give ΣM A = 0 : M A + ( −7 i + 23 j + 10 k ) × ( −50 k ) = 0 x: y: M Ax − 1150 = 0 M Ay − 350 = 0 z: ΣFx = 0 : ΣFy = 0 : M Az = 0 Ax = 0 Ay = 0 ΣFz = 0 : Az − 50 = 0 A = ( 50 k ) lb ............................. M A = (1150 i + 350 j) lb ⋅ ft ........................................ Ans. 6-104 The rectangular plate of uniform thickness shown in Fig. P6-104 has a mass of 500 kg. Determine (a) The tensions in the three cables supporting the plate. (b) The deformation of the cable at A, which has a diameter of 10-mm, a length of 1.5 m and a modulus of elasticity of 200 GPa. SOLUTION W = 500 g = 500 ( 9.81) = 4905 N (a) From a free-body diagram of the plate, the equilibrium equations give ΣFz = 0 : ΣM O = 0 : TA + TB + TC − 4905 = 0 ( −1.5 i + 0.9 j) × (TB k ) + ( −0.5 i + 2.5 j) × (TC k ) + ( −0.625 i + 1.25 j) × ( −4905 k ) = 0 0.9TB + 2.5TC − 6131.250 = 0 1.5TB + 0.5TC − 3065.625 = 0 TA = 1560.682 N ≅ 1561 N ........................................................................................... Ans. TB = 1393.466 N ≅ 1393 N .......................................................................................... Ans. x: y: 222 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS TC = 1950.852 N ≅ 1951 N .......................................................................................... Ans. (b) The stretch of wire A is given by δA = 6-105 (1560.862 )(1500 ) = 0.1490 mm .......................................... Ans. PL = EA ( 200 × 109 ) π ( 0.010 ) 2 4 A shaft is loaded through a pulley and a lever (Fig. P6-105) that are fixed to the shaft. Friction between the belt and pulley prevents slipping of the belt. Determine the force P required for equilibrium and the reactions at supports A and B. The support at A is a ball bearing and the support at B is a thrust bearing. The bearings exert only force reactions on the shaft. SOLUTION From a free-body diagram of the shaft, the moment equilibrium equation gives ΣM B = 0 : (14 i + 14 j) × ( 200 j − P k ) + ( −18 j) × ( Ax i + Az k ) + ( −30 j + 6 k ) × ( −500 i ) + ( −30 j − 6 k ) × ( −150 i ) = 0 −14 P − 18 Az = 0 14 P − 2100 = 0 x: y: Az = −116.667 lb P = 150 lb z: ΣFx = 0 : ΣFy = 0 : 18 Ax − 16, 700 = 0 ΣF = 0 gives Ax = 927.778 lb Bx = −277.778 lb By = −200 lb Then, the force equilibrium equation Ax + Bx − 650 = 0 By + 200 = 0 ΣFz = 0 : Az + Bz − P = 0 Bz = 266.667 lb A = ( 928 i − 116.7 k ) lb ................................................................................................ Ans. B = ( −278 i − 200 j + 267 k ) lb ................................................................................... Ans. P = 150 lb ......................................................................................................................... Ans. 6-106 The homogeneous door shown in Fig. P6-106 has a mass of 60 kg, and is held in the position shown by the rod AB. The rod is held in place by smooth horizontal pins at A and B. The hinges at C and D are smooth, and the hinge at C can support thrust along its axis. Determine all forces that act on the door. SOLUTION W = 60 g = 60 ( 9.81) = 588.60 N Note that the rod AB is a two-force member. FAB = ( − FAB cos 60° j + FAB sin 60° k ) Then, from a free-body diagram of the door, the moment equilibrium equation gives = ( −0.5000 FAB j + 0.86603FAB k ) N ΣM C = 0 : ( −950 i ) × ( Dy j + Dz k ) 223 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS + ( −475 i + 475cos 60° j + 475sin 60° k ) × ( −588.60 k ) + ( 950 cos 60° j + 950sin 60° k ) × ( −0.5000 FAB j + 0.86603FAB k ) = 0 x: y: −139, 792.5 + 822.724 FAB = 0 FAB = 169.914 N Dz = 294.300 N Dy = 0 N 950 Dz − 279,585 = 0 −950 Dy = 0 ΣF = 0 gives z: ΣFx = 0 : ΣFy = 0 : Then, the force equilibrium equation Cx = 0 C y + Dy − 0.5000 (169.914 ) = 0 C z + Dz + 0.86603 (169.914 ) − 588.60 = 0 Cx = 0 N C y = 84.957 N ΣFz = 0 : C z = 147.15 N C = ( 85.0 j + 147.2 k ) N ............................................................................................... Ans. D = ( 294 k ) N ................................................................................................................. Ans. FAB = ( −85.0 j + 147.2 k ) N ......................................................................................... Ans. 6-107 A beam is supported by a ball-and-socket joint and two cables as shown in Fig. P6-107. Determine (a) The tensions in the two cables. (b) The reaction at support A (the ball and socket joint). (c) The axial stress and deformation in each of the cables if they are made of steel (E = 30,000 ksi) and have ¼- in. diameters. SOLUTION LBD = 112 + 62 = 12.530 ft TBD = TBD TCD = TCD LCD = 7 2 + 112 + 32 = 13.379 ft −11 j + 6 k = ( −0.87790TBD j + 0.47885TBD k ) 112 + 6 2 −7 i − 11 j + 3 k = ( −0.52320TCD i − 0.82218TCD j + 0.22423TCD k ) 7 2 + 112 + 32 (a) From a free-body diagram of the beam, the moment equilibrium equation gives ΣM A = 0 : (11 j) × ( TBD + TCD + 300 i ) + ( 6 j) × ( −800 k ) = 0 5.26735TBD + 2.46653TCD − 4800 = 0 0=0 x: y: z: 5.75520TCD − 3300 = 0 TBD = 642.77 lb ≅ 643 lb .............................................................................................. Ans. TCD = 573.39 lb ≅ 573 lb .............................................................................................. Ans. (b) Then, the force equilibrium equation ΣF = 0 gives ΣFx = 0 : ΣFy = 0 : Ax − 0.52320 ( 573.39 ) + 300 = 0 Ay − 0.87790 ( 642.77 ) − 0.82218 ( 573.39 ) = 0 Ax = 0 lb Ay = 1035.72 lb 224 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ΣFz = 0 : Az + 0.47885 ( 642.77 ) + 0.22423 ( 573.39 ) − 800 = 0 Az = 363.64 lb A = (1036 j + 364 k ) lb ................................................................................................. Ans. (c) Finally, the stresses and deformation of the two cables is σ BD = σ CD = δ BD = δ CD = 6-108 T 642.77 = = 13.09 × 103 psi = 13.09 ksi ........................................... Ans. A π ( 0.25 )2 4 573.39 π ( 0.25) 4 2 = 11.68 × 103 psi = 11.68 ksi .................................................... Ans. ( 573.39 )(13.379 ×12 ) = 0.0625 in. ........................................................... Ans. ( 30 ×106 ) π ( 0.25)2 4 ( 642.77 )(12.530 ×12 ) = 0.0656 in. ................................................ Ans. TL = EA ( 30 ×106 ) π ( 0.25 ) 2 4 The crankshaft-flywheel arrangement of a one-cylinder engine is shown in Fig. P6-108. A 250-N-m couple C is delivered to the crankshaft by the flywheel, which is rotating at a constant angular velocity. The support at A is a ball bearing and the support at B is a thrust bearing, both of which support only force reactions on the shaft. Determine the magnitudes of the force P and of the resultant bearing reactions at A and B. SOLUTION P = − P sin 20° j − P cos 20° k = −0.34202 P j − 0.93969 P k From a free-body diagram of the crankshaft, the moment equilibrium equation gives ΣM B = 0 : −250 i + ( 0.45 i ) × ( Ay j + Az k ) + ( 0.225 i − 0.125 j) × P = 0 x: y: z: −250 + 0.11746 P = 0 −0.45 Az + 0.21149 P = 0 0.45 Ay − 0.07695 P = 0 P = 2128.38 N Ay = 363.954 N Az = 1000.293 N Then, the force equilibrium equation ΣF = 0 gives ΣFx = 0 : ΣFy = 0 : Bx = 0 Ay + By − 0.34202 ( 2128.38 ) = 0 Az + Bz − 0.93969 ( 2128.38) = 0 Bx = 0 N By = 363.996 N ΣFz = 0 : Bz = 999.728 N A = 3642 + 10002 = 1064 N ...................................................................................... Ans. B = 3642 + 10002 = 1064 N ...................................................................................... Ans. P = 2130 N ...................................................................................................................... Ans. 225 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 6-109 RILEY, STURGES AND MORRIS A farmer is using the hand winch shown in Fig. P6-109 to slowly raise a 40-lb bucket of water from a well. In the position shown, the force P is vertical. The bearings at C and D exert only force reactions on the shaft. Bearing C can support thrust loading; bearing D cannot. Determine the magnitude of force P and the components of the bearing reactions. SOLUTION From a free-body diagram of the winch drum and axle, the moment equilibrium equation gives ΣM C = 0 : ( −20 cos 30° i + 20sin 30° j + 11k ) × ( − P j) + ( 2.5 i − 20 k ) × ( −40 j) + ( −50 k ) × ( Dx i + Dy j) = 0 11P − 800 + 50 Dy = 0 x: y: −50 Dx = 0 20 P cos 30° − 100 = 0 P = 5.7735 lb ≅ 5.77 lb ................................................................................................ Ans. z: Dx = 0 lb ............................................ Dy = 14.7298 lb .............................................. Ans. (b) Then, the force equilibrium equation ΣF = 0 gives ΣFx = 0 : ΣFy = 0 : Cx + Dx = 0 C y + Dy − P − 40 = 0 C x = 0 lb ............................................. Ans. C y = 31.0437 lb ................................ Ans. ΣFz = 0 : 6-110 Cz = 0 C z = 0 lb ............................................. Ans. The block W shown in Fig. P6-110 has a mass of 250 kg. Bar AB rests against a smooth vertical wall at end B and is supported at end A with a ball-and-socket joint. The two cables are attached to a point on the bar midway between the ends. Determine (a) The reactions at supports A and B and the tension in cable CD. (b) The axial stress and deformation in cable CD if it is made of steel (E = 210 GPa) and has an 8-mm diameter. SOLUTION W = 250 g = 250 ( 9.81) = 2452.50 N TCD = TCD −200 i + 200 j + 500 k LCD = 2002 + 2002 + 5002 = 574.456 mm 2002 + 2002 + 5002 = ( −0.34816TCD i + 0.34816TCD j + 0.87039TCD k ) (a) From a free-body diagram of the beam, the moment equilibrium equation gives ΣM A = 0 : ( 200 i − 500 j + 300 k ) × ( TCD − 2452.50 k ) + ( 400 i − 1000 j + 600 k ) × ( By j) = 0 x: y: −539.643TCD + 1, 226, 250 − 600 By = 0 −278.526TCD + 490,500 = 0 −104.448TCD + 400 By = 0 TCD = 1761.056 N By = 459.847 N z: TCD = 1761.056 N ≅ 1761 N ......................................................................................... Ans. 226 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS B = ( 460 j) N .................................................................................................................. Ans. Then, the force equilibrium equation ΣF = 0 gives ΣFx = 0 : ΣFy = 0 : Ax − 0.34816 (1761.056 ) = 0 Ay + By + 0.34816 (1761.056 ) = 0 Az + 0.87039 (1761.056 ) − 2452.50 = 0 Ax = 613 N Ay = −1073 N ΣFz = 0 : Az = 920 N A = ( 613 i − 1073 j + 920 k ) N .................................................................................... Ans. (c) Finally, the stresses and deformation of the cable is σ CD = δ CD = 6-111 1761.056 π ( 0.008) 4 2 = 35.0 × 106 N/m 2 = 35.0 MPa ............................................. Ans. ................................................... Ans. (1761.056 )( 594.456 ) = 0.0958 mm ( 210 ×109 ) π ( 0.008)2 4 A 20-lb piece of electronic equipment is placed on a wooden skid that weighs 10 lb and that rests on a concrete floor (Fig. P6-111). The coefficient of static friction between the skid and the floor is 0.45. (a) Determine the minimum pushing force along the handle necessary to cause the skid to start sliding across the floor (Fig. P6-111a). (b) Determine the minimum force necessary to start the skid in motion if a pulling force instead of a pushing force is applied to the handle (Fig. P6-111b). SOLUTION (a) Impending motion is to the right and friction acts to the left to oppose the impending motion. From a free-body diagram of the skid, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : P cos 30° − 0.45 N = 0 N − 30 − P sin 30° = 0 N = 40.5 lb P = 21.1 lb ....................................................................................................................... Ans. (b) Now, impending motion is to the left and friction acts to the right to oppose the impending motion. From a free-body diagram of the skid, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : 0.45 N − P cos 30° = 0 N − 30 + P sin 30° = 0 N = 23.8 lb P = 12.37 lb ..................................................................................................................... Ans. 6-112 A skier is at the point of experiencing impending motion on a ski slope as illustrated in Fig. P6-112. Determine the coefficient of static friction between a ski and the slope if the angle θ (the angle of repose) is 5°. SOLUTION Impending motion is down the hill and friction acts up the hill to oppose the impending motion. From a free-body diagram of the skier, the equilibrium equations give 227 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ΣFn = 0 : ΣFt = 0 : N − mg cos 5° = 0 N = mg cos 5° µ s N − mg sin 5° = 0 µs = µ s N = mg sin 5° mg sin 5° = tan 5° = 0.0875 ............................................................................... Ans. mg cos 5° 6-113 The block in Fig. P6-113 weighs 500 lb and the coefficient of friction between the block and the inclined plane is 0.2. Determine (a) Whether the system would be in equilibrium for P = 400 lb. (b) The minimum force P to prevent motion. (c) The maximum force P for which the system is in equilibrium. SOLUTION If the impending motion is down the hill, then the friction force would be up the hill. From a free-body diagram of the block, the equilibrium equations are ΣFn = 0 : ΣFt = 0 : (a) If N − 500 cos 30° + P sin 20° = 0 F − 500sin 30° + P cos 20° = 0 P = 400 lb , then the equilibrium equations give N = 296.205 lb F = −125.877 lb Fmax = 0.2 N = 59.24 lb The negative sign on the friction indicates that the direction of impending motion is actually up the hill and the friction is acting down the hill to oppose the motion. However, since the amount of friction required for equilibrium is greater than the maximum available friction, Not in equilibrium. ........................................................................................................ Ans. (b) If F = 0.2 N , then the equilibrium equations give N = 368.872 lb F = 0.2 N = 73.7744 lb P = 187.5 lb = Pmin .......................................................................................................... Ans. (c) If the impending motion is up the hill, then the friction force would be down the hill. From a free-body diagram of the block, the equilibrium equations are ΣFn = 0 : ΣFt = 0 : Now, if N − 500 cos 30° + P sin 20° = 0 − F − 500 sin 30° + P cos 20° = 0 F = 0.2 N , the equilibrium equations give N = 318.812 lb F = 63.7624 lb P = 334 lb = Pmax ............................................................................................................. Ans. 6-114 A 75-kg man starts climbing a 5-m long ladder leaning against a wall (Fig. P6-114). The coefficient of friction is 0.25 at both surfaces. Neglect the weight of the ladder and determine how far up the ladder the man can climb before the ladder starts to slip. SOLUTION Impending motion of the ladder is down and to the left, and friction on the wall and floor acts to oppose the impending motion. From a free-body diagram of the man and the ladder, the equilibrium equations give 228 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS → ΣFx = 0 : ↑ ΣFy = 0 : ΣM 1 = 0 : 0.25 N1 − N 2 = 0 N1 − 75 ( 9.81) + 0.25 N 2 = 0 N 2 ( 5sin 60° ) + ( 0.25 N 2 )( 5cos 60° ) −75 ( 9.81)( d cos 60° ) = 0 N1 = 692.471 N N 2 = 173.118 N d = 2.33 m ..................................................................... Ans. 6-115 A device for lifting rectangular objects such as bricks and concrete blocks is shown in Fig. P6-115. Determine the minimum coefficient of static friction between the contacting surfaces required to make the device work. SOLUTION Impending motion is for the block to slide downward. Friction must act upward on the block to oppose this impending motion. The friction force on the member AB must be equal and opposite to that on the block – it must act downward on the member. From a free-body diagram for member AB, the moment equilibrium equation gives ΣM B = 0 : 6-116 4 N − 12µ s N = 0 µ s = 4 12 = 0.333 ...........................................Ans. A boy is pulling a sled with a box (at constant velocity) up an inclined surface as shown in Fig. P6-116. The mass of the box and sled is 50 kg; the mass of the boy is 40 kg. If the coefficient of kinetic friction between the sled runners and the icy surface is 0.05, determine the minimum coefficient of static friction needed between the boy’s shoes and the icy surface. SOLUTION Impending motion is up the hill and friction acts down the hill to oppose the motion. From a free-body diagram of the sled, the equilibrium equations give ΣFn = 0 : ΣFt = 0 : N1 − 50 ( 9.81) cos15° + P sin 25° = 0 P cos 25 − 50 ( 9.81) sin15° − 0.05 N1 = 0 N1 = 405.142 N P = 162.426 N Impending motion of the boy’s foot is down the hill and friction acts up the hill to oppose the motion. From a free-body diagram of the boy, the equilibrium equations give ΣFn = 0 : ΣFt = 0 : N 2 − 40 ( 9.81) cos15° − P sin 25° = 0 µ s N 2 − 40 ( 9.81) sin15° − P cos 25 = 0 N 2 = 447.673 N µ s N 2 = 248.769 N = F2 µ s = 0.556 ...................................................................... Ans. 229 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 6-117 RILEY, STURGES AND MORRIS A 120-lb girl is walking up a 48-lb uniform beam as shown in Fig P6-117. Determine how far up the beam the girl can walk before the beam starts to slip if (a) The coefficient of friction is 0.20 at all surfaces. (b) The coefficient of friction at the bottom end of the beam is increased to 0.40 by placing a piece of rubber between the beam and the floor. SOLUTION Impending motion of the beam is down and to the right and friction on the wall and floor acts to oppose the impending motion. From a free-body diagram of the girl and the beam, the equilibrium equations are → ΣFx = 0 : ↑ ΣFy = 0 : ΣM 1 = 0 : (a) If ( 3 5) N1 − ( 4 5) F1 − F2 = 0 ( 4 5) N1 + ( 3 5) F1 − 168 + N 2 = 0 48 ( 4 5 )( 6 ) + 120 ( 4 5 ) x − 10 N1 = 0 F1 = 0.2 N1 and F2 = 0.2 N 2 , then → ΣFx = 0 : gives N 2 = 11N1 5 and the other two equations give N1 = 53.846 lb (b) If N 2 = 118.462 lb x = 3.21 ft ......................................................................................................................... Ans. F1 = 0.2 N1 and F2 = 0.4 N 2 , then → ΣFx = 0 : gives N 2 = 11N1 10 and the other two equations give N1 = 83.168 lb 6-118 N 2 = 91.485 lb x = 6.26 ft ........................................................................................................................ Ans. The broom shown in Fig. P6-118 weighs 8 N and is held up by the two cylinders, which are wedged between the broom handle and the side rails. The coefficient of friction between the broom and cylinders and between the cylinders and side rails is 0.30. The side rails are at an angle of θ = 30° to the vertical. The weight of the cylinders may be neglected. Determine whether or not this system is in equilibrium. If the system is in equilibrium, determine the force exerted on the broom handle by the rollers. SOLUTION Impending motion of the broom handle is downward and friction acts upward to oppose the motion. From a freebody diagram of the broom handle, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM center = 0 : N 2 − N1 = 0 F1 + F2 − 8 = 0 F1b − F2b = 0 F1 = F2 = 4 N N1 = N 2 From a free-body diagram of the left-hand roller, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : F3 cos 60° + N 3 cos 30° − N 2 = 0 − F3 sin 60° + N 3 sin 30° − F2 = 0 230 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ΣM center = 0 : F3 r − F2 r = 0 F3 = 4 N N1 = N 2 = N 3 = 14.92820 N Fmax = 0.3N = 4.478 N Since the amount of friction required for equilibrium is less than the maximum amount of friction available at all surfaces, the broom handle Is in equilibrium ............................................................................................................. Ans. N = 14.93 N ..................................................................................................................... Ans. F = 4 N ............................................................................................................................. Ans. 6-119 The device shown in Fig. P6-119 is used to raise boxes and crates between floors of a factory. The frame, which slides on the 4-in.-diameter vertical post, weighs 50 lb. If the coefficient of friction between the post and the frame is 0.10, determine the force P required to raise a 150-lb box. SOLUTION Impending motion of the platform is upward and friction with the post acts to oppose the impending motion. Guessing that the platform rotates slightly clockwise, contact with the post (and the location of the normal and friction forces) is on the top-left and bottom-right sides of the post. Then, from a free-body diagram of the platform and box, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM A = 0 : N A − NB = 0 P − 0.1( N A + N B ) − 200 = 0 26 N B + 4 ( 0.1N B ) + 8P − 4 ( 50 ) − 20 (150 ) = 0 N A = N B = 57.143 lb P = 211 lb ........................................................................................................................ Ans. (Note that the two normal forces are both positive indicating that the guess about how the platform rubs against the post was correct.) 6-120 The automobile shown in Fig. P6-120 has a mass of 1500 kg. The coefficient of friction between the rubber tires and the pavement is 0.70. Determine the maximum incline θ that the automobile can drive up if the automobile has (a) A rear-wheel drive. (b) A front-wheel drive. SOLUTION W = 1500 ( 9.81) = 14, 715 N Impending slip between the tires and the ground is for the bottom of the tires to move backwards and the friction force acts to oppose the slip. From a free-body diagram of the car, the equilibrium equations are ΣFn = 0 : ΣFt = 0 : N1 + N 2 − 14, 715cosθ = 0 F1 + F2 − 14, 715sin θ = 0 1.2 N 2 − 1.7 N1 + 0.85 ( F1 + F2 ) = 0 F2 = 0 ΣM G = 0 : (a) If the vehicle has rear wheel drive, then F1 = 0.7 N1 and and the moment equilibrium equation gives N 2 = 0.92083 N1 231 STATICS AND MECHANICS OF MATERIALS, 2nd Edition and the other two equations give RILEY, STURGES AND MORRIS tan θ = 14, 715sin θ 0.7 N1 = = 0.36443 14, 715 cos θ 1.92083 N1 ........................................................................................................................ Ans. θ = 20.02° (b) If the vehicle has front wheel drive, then F1 = 0 and F2 = 0.7 N 2 and the moment equilibrium equation gives N1 = 1.05588 N 2 and the other two equations give tan θ = 14, 715sin θ 0.7 N 2 = = 0.34049 14, 715 cos θ 2.05588 N 2 θ = 18.80° ......................................................................................................................... Ans. 6-121 When a drawer is pulled by only one of the handles, it tends to twist and rub as shown (highly exaggerated) in Fig. P6-121, which is a top view of the drawer. The weight of the drawer and its contents is 2 lb and is uniformly distributed. The coefficient of friction between the sides of the drawer and the sides of the dresser is 0.6; between the bottom of the drawer and the side rails the drawer rides on, µs = 0.1. Determine the minimum amount of force P necessary to pull the drawer out. SOLUTION F3 = F4 = 0.1(W 2 ) = 0.1 lb Impending slip of the drawer is to the front and friction acts to oppose the impending motion. From a free-body diagram of the drawer, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM 2 = 0 : N1 − N 2 = 0 F1 + F2 + F3 + F4 − P = 0 27 P − 30.375F3 − 0.375 F4 −30.75F1 − 18 N1 = 0 N1 = N 2 = 0.57407 lb F1 = F2 = 0.6 N1 = 0.34444 lb P = 0.889 lb ..................................................................................................................... Ans. 6-122 An ill-fitting window is about 10 mm narrower than its frame (see Fig. P6-122). The window weighs 40 N, and the coefficient of friction between the window and the frame is 0.20. Determine the amount of force P that must be applied at the lower corner to keep the window from lowering. SOLUTION Impending slip of the window is downward and friction acts to oppose the motion. From a free-body diagram of the window, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM 2 = 0 : N1 − N 2 = 0 F1 + F2 − 40 + P = 0 405 ( 40 ) − 600 N1 − 810 F1 = 0 N1 = N 2 = 21.260 N F1 = F2 = 0.2 N1 = 4.252 N P = 31.5 N ....................................................................................................................... Ans. 232 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 6-123 RILEY, STURGES AND MORRIS Determine the minimum coefficient of static friction necessary for the pliers shown in Fig. P6-123 to grip the bolt. SOLUTION Impending motion is for the bolt to slip out of the pliers and friction acts to oppose the motion. From a free-body diagram for bolt, the equilibrium equations give ΣM center = 0 : F1r − F2 r = 0 F1 = F2 = µ s N1 = µ s N 2 N1 = N 2 → ΣFx = 0 : ( N1 + N 2 ) sin17.5° − ( F1 + F2 ) cos17.5° = 0 ( N1 + N 2 ) sin17.5° = µ s ( N1 + N 2 ) cos17.5° µ s = tan17.5° = 0.315 ................................................................................................... Ans. ↑ ΣFy = 0 : , it is also satisfied by the (Note that even though we did not use the third equation of equilibrium values solved above.) 6-124 A homogeneous book of mass 0.46 kg rests in a bookshelf as shown in Fig. P6-104. The thickness of the book is small compared to the other dimensions shown. The coefficient of friction at all surfaces is 0.4. Determine the minimum angle θ for which the book is in equilibrium. SOLUTION W = 0.46 ( 9.81) = 4.5126 N Impending motion of the book is down and to the right. Friction acts to oppose this motion. From a free-body diagram of the book, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM 2 = 0 : 6-125 N1 − 0.4 N 2 = 0 N 2 + 0.4 N1 − 4.5126 = 0 N1 = 1.55607 N tan θ = 1.0500 N 2 = 3.89017 N 4.5126 (130 cosθ ) − N1 ( 260sin θ ) − 0.4 N1 ( 260 cosθ ) = 0 θ = 46.40° ...................................................................... Ans. A wedge is used to raise a 350-lb refrigerator onto a platform (Fig. P6-125). The coefficient of friction is 0.2 at all surfaces. (a) Determine the minimum force P needed to insert the wedge. (b) Determine if the system would still be in equilibrium if P = 0. (c) If the system is not in equilibrium when P = 0, determine the force necessary to keep the wedge in place, or if the system is in equilibrium when P = 0, determine the force necessary to remove the wedge. SOLUTION When the wedge is being pushed to the right, friction acts down and to the left to oppose the motion. From a free-body diagram of the wedge and refrigerator, the equilibrium equations are → ΣFx = 0 : ↑ ΣFy = 0 : (a) If P − N sin15° − F cos15° = 0 N cos15° − F sin15° − 350 = 0 F = 0.2 N , then the equilibrium equations give N = 382.864 lb 233 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS F = 0.2 N = 76.573 lb P = 173.1 lb ...................................................................................................................... Ans. (b) If P = 0 , then the equilibrium equations give N = 390.374 lb F = 104.600 lb However, this solution is not valid since the amount of friction required for equilibrium is greater than the maximum amount of friction available Fmax = 0.2 N = 78.075 lb . Therefore, the wedge will not stay in place if the force P is removed. When the force P is reduced, the wedge wants to slip to the left and friction must act up and to the right to oppose the impending motion. Now from a free-body diagram of the wedge and refrigerator, the equilibrium equations give (using F = 0.2 N ) → ΣFx = 0 : ↑ ΣFy = 0 : P − N sin15° + F cos15° = 0 N cos15° + F sin15° − 350 = 0 N = 343.916 lb F = 0.2 N = 68.783 lb P = 22.6 lb ....................................................................................................................... Ans. 6-126 A 50-kg uniform plank rests on rough supports at A and B (Fig. P6-126). The coefficient of friction is 0.60 at both surfaces. If a man weighing 800 N pulls on the rope with a force of P = 400 N, determine (a) The minimum and maximum angles θmin and θmax for which the system will be in equilibrium. (b) The minimum coefficient of friction that must exist between the man’s shoes and the ground for each of the cases in part (a). SOLUTION W = 50 ( 9.81) = 490.5 N (a) Impending motion of the plank is to the left and friction acts to the right to oppose the motion. From a freebody diagram of the plank, the equilibrium equations are → ΣFx = 0 : ↑ ΣFy = 0 : ΣM B = 0 : If FA + FB − 400 cosθ = 0 N A + N B − 490.5 − 400sin θ = 0 490.5 (1) − N A ( 2 ) − ( 400sin θ )( 2 ) = 0 N A + N B = 490.5 + 400sin θ FA = 0.6 N A and FB = 0.6 N B (impending slip), then the equilibrium equations give 0.6 ( N A + N B ) = 400 cosθ N A + N B = 666.667 cos θ = 490.5 + 400sin θ If θ = 19.920° FA = N A = 0 (impending tip), then the equilibrium equations give θ = sin −1 490.5 = 37.816° 2 ( 400 ) N B = 735.75 N FB = 316.0 N ( FB )max = 0.6 N B = 441.45 N Since the amount of friction required is less than the maximum amount of friction available, the solution is okay and the allowed range of angles is θ min = 19.92° ................. θ max = 37.82° ......................................................................... Ans. 234 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (b) Impending motion of the man’s feet is to the right and friction acts to the left to oppose the motion. From a free-body diagram of the plank, the equilibrium equations are RILEY, STURGES AND MORRIS → ΣFx = 0 : ↑ ΣFy = 0 : If θ 400 cos θ − F = 0 N + 400sin θ − 800 = 0 = 19.920° , the equilibrium equations give F = 376.07 N N = 663.72 N F 376.07 = 0.567 .................................................................................... Ans. ( µ s )min = = N 663.72 If θ = 37.816° , the equilibrium equations give F = 315.99 N N = 554.75 N 315.99 = 0.570 ............................................................................................. Ans. ( µ s )min = 554.75 6-127 The plunger of a door latch is held in place by a spring as shown in Fig. P6-127. Friction on the sides of the plunger may be ignored. If a force of 2 lb is required to just start closing the door and the coefficient of friction between the plunger and the striker plate is 0.25, determine the force exerted on the plunger by the spring. SOLUTION When the door is pushed in the direction of the 2-lb force, friction acts upward and to the left to oppose the motion. From a free-body diagram of the plunger, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : N sin 50° − 0.25 N cos 50° − P = 0 N cos 50° + 0.25 N sin 50° − 2 = 0 N = 2.39722 lb P = 1.451 lb ...................................................................................................................... Ans. 6-128 (a) (b) (c) (d) The 25-kg block in Fig. P6-128 is held against the wall by the brake arm. The coefficient of friction between the wall and the block is 0.20; between the block and the brake arm, 0.50. Neglect the weight of the brake arm. Determine Whether the system would be in equilibrium for P = 230 N. The minimum force P for which the system would be in equilibrium. The maximum force P for which the system would be in equilibrium. The shearing stress on a cross section of the pin at A when impending motion of the block is downward if the pin has a 6-mm diameter and is in double shear. SOLUTION W = 25 ( 9.81) = 245.25 N If the impending motion of the block is downward, friction would act upward on the block (and downward on the brake handle) to oppose the motion. From a free-body diagram of the brake handle, the equilibrium equations are → ΣFx = 0 : ↑ ΣFy = 0 : ΣM A = 0 : N 2 − 140 − Ax = 0 Ay − F2 = 0 (140 − N 2 )( 50 ) + F2 ( 40 ) = 0 235 STATICS AND MECHANICS OF MATERIALS, 2nd Edition From a free-body diagram of the block, the equilibrium equations are RILEY, STURGES AND MORRIS → ΣFx = 0 : ↑ ΣFy = 0 : (a) If N1 − N 2 = 0 P + F1 + F2 − 245.25 = 0 N1 = N 2 F1 + F2 = 15.25 N N 2 = 140 + 0.8F2 P = 230 N , the equilibrium equations give ( F1 + F2 )max = 0.2 N1 + 0.5 N 2 = 0.7 N 2 = 98 + 0.56F2 > 98 N Since the amount of friction required for equilibrium (15.25 N ) is less than the maximum amount of friction available ( at least 98 N ) , the system Is in equilibrium ............................................................................................................. Ans. (b) If F1 = 0.2 N1 and F2 = 0.5 N 2 (impending motion downward), the equilibrium equations give (140 − N 2 )( 50 ) + ( 0.5 N 2 )( 40 ) = 0 F1 = 116.667 N Ax = 93.333 N F2 = 46.667 N Ay = 116.667 N N1 = N 2 = 233.333 N P = 81.9 N = Pmin ............................................................................................................ Ans. 2 A = Ax2 + Ay = 149.406 N (c) If impending motion of the block is upward, the friction forces ( F1 direction and the equilibrium equations give = 0.2 N1 and F2 = 0.5 N 2 ) change (140 − N 2 )( 50 ) − ( 0.5 N 2 )( 40 ) = 0 F1 = 20.00 N F2 = 50.00 N N1 = N 2 = 100.00 N P = 315 N = Pmax ............................................................................................................. Ans. (d) When the impending motion of the block is downward, the force on the pin is A = 149.406 pin force by twice its cross sectional area (since it is in double shear) gives the shear stress N . Dividing the τA = 6-129 VA 149.406 = = 2.64 × 106 N/m 2 = 2.64 MPa ........................... Ans. 2 2 As 2 π ( 0.006 ) 4 A 250-lb weight is suspended from a lightweight rope wrapped around the inner cylinder of a drum (Fig. P6-129). A brake arm is pressed against the outer cylinder of the drum by a hydraulic cylinder. The coefficient of friction between the brake arm and the drum is 0.40. Determine (a) The smallest force in the hydraulic cylinder necessary to prevent motion. (b) The shearing stress on a cross section of the pin at A when motion is impending if the drum weighs 75 lb and the pin has a ½-in. diameter and is in double shear. SOLUTION (a) Impending motion of the drum is a counterclockwise rotation and friction acts to oppose the motion. From a free-body diagram of the drum, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM A = 0 : F − Ax = 0 Ay − 75 − 250 − N = 0 ( 250 )( 6 ) − F (12 ) = 0 236 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS F = 125 lb = 0.4 N N = 312.5 lb Ay = 637.5 lb Ax = 125 lb 2 A = Ax2 + Ay = 649.639 lb From a free-body diagram of the brake arm, the moment equilibrium equation gives ΣM C = 0 : 9 P − 9 F − 27 N = 0 P = 1063 lb ...................................................................................................................... Ans. (b) Dividing the force on the pin by twice its cross sectional area (since it is in double shear) gives the shear stress τA = 6-130 VA 649.639 = = 1654 psi ...................................................................... Ans. 2 As 2 π ( 0.5 )2 4 The brake shown in Fig. P6-130 is used to control the motion of block B. If the mass of block B is 25 kg and the kinetic coefficient of friction between the brake drum and brake pad is 0.30, determine (a) The force P required for a constant-velocity descent. (b) The shearing stress on cross sections of both pins. Each pin has a diameter of 10 mm and is in double shear. SOLUTION W = 25 g = 25 ( 9.81) = 245.25 N (a) Motion of the drum is a clockwise rotation and friction acts to oppose the motion. From a free-body diagram of the drum, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM B = 0 : Bx − F = 0 By − N − 245.25 = 0 F (180 ) − 245.25 ( 90 ) = 0 N = 408.750 N By = 654.000 N F = 122.625 N = 0.3 N Bx = 122.625 N 2 B = Bx2 + By = 665.397 N From a free-body diagram of the brake arm, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM A = 0 : F − Ax = 0 N − Ay − P = 0 200 N + 40 F − 600 P = 0 Ay = 264.325 N Ax = 122.625 N 2 A = Ax2 + Ay = 291.384 N P = 144.4 N ..................................................................................................................... Ans. (b) Dividing the force on the pins by twice their cross sectional areas (since they are in double shear) gives the shear stresses 237 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS VA 291.384 = = 1.855 × 106 N/m 2 = 1.855 MPa .......................... Ans. 2 2 As 2 π ( 0.01) 4 V 665.397 τB = B = = 4.24 ×106 N/m 2 = 4.24 MPa .............................. Ans. 2 2 As 2 π ( 0.01) 4 τA = 6-131 Three identical cylinders are stacked as shown in Fig. P6-131. The cylinders each weigh 22 lb and are 8 in. in diameter. The coefficient of friction is µs = 0.40 at all surfaces. Determine the maximum force P that the cylinders can support without moving. SOLUTION From a free-body diagram for the upper cylinder, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM center = 0 : FAC cos 30° − FAB cos 30° + N AB cos 60° − N AC cos 60° = 0 ( N AB + N AC ) sin 60° + ( FAB + FAC ) sin 30° − P − 22 = 0 FAC r − FAB r = 0 N AC = N AB FAC = FAB P = 2 N AB sin 60° + 2 FAB sin 30° + 22 From a free-body diagram for the lower-left cylinder, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM center = 0 : FB + FAB cos 30° − N AB cos 60° = 0 N B − 22 − FAB sin 30° − N AB sin 60° = 0 FB r − FAB r = 0 FB = FAB = 0.26795 N AB Since the amount of friction required for equilibrium friction available the value of P. Then, ( FAB max = 0.4 N AB ) , the cylinders will not slip at the surface between the cylinders regardless of ( FAB = 0.26795 N AB ) is less than the maximum amount of N B − 22 − FAB sin 30° − ( 3.73205FAB ) sin 60° = 0 FB = FAB = 0.26795 N B − 5.89488 ( FB = 0.26795 N B − 5.89488) is less than the maximum amount of friction available ( FB max = 0.4 N B ) so the cylinders will not slip at the lower surface and again the amount of friction required for equilibrium regardless of the value of P. Therefore, the cylinders will not slip for any value of P and Pmax = ∞ ............................................................................................................................ Ans. 6-132 A 10-kg drum rests on a thin lightweight piece of cardboard as shown in Fig. P6-132. The coefficient of friction is the same at all surfaces. (a) Plot P, the maximum force that may be applied to the cardboard without moving it, as a function of the coefficient of friction µs (0.05 ≤ µs ≤ 0.8). (b) On the same graph, plot (Af)actual and (Bf)actual, the actual amounts of friction force that act at points A and B, and (Af)avail and (Bf)avail, the maximum amounts of friction force available for equilibrium at points A and B. (c) (d) What happens to the system at µs ≅ 0.364? What does the solution for µs > 0.364 mean? 238 STATICS AND MECHANICS OF MATERIALS, 2nd Edition SOLUTION RILEY, STURGES AND MORRIS W = 10 ( 9.81) = 98.1 N Impending motion of the cardboard is downward and friction acts to oppose the motion. From a free-body diagram of the cardboard, the equilibrium equations are ΣFn = 0 : ΣFt = 0 : NC − N B = 0 FB + FC − P = 0 FC = µ s N C From a free-body diagram of the cylinder, the equilibrium equations are ΣFn = 0 : ΣFt = 0 : N B − 98.1cos 20° − FA sin 40° = 0 − FB − 98.1sin 20° − FA cos 40° + N A cos 50° = 0 FA r − FB r = 0 N B = NC FA = µ s N A = FB and ΣM center = 0 : Therefore FA = FB If impending slip at A, then − µ s N A − 98.1sin 20° − µ s N A cos 40° + N A cos 50° = 0 NA = 98.1sin 20° cos 50° − µ s (1 + cos 40° ) N B = 98.1cos 20° + µ s N A sin 40° + N A sin 50° = N C P = FB + FC = FA + FC = µ s ( N A + N C ) = µ s ( N A + N B ) ....................................... Ans. FA = FB = ( FA )max = µ s N A ............................................................................................ Ans. ( FB )max = µ s N B ............................................................................................................... Ans. These five functions are all plotted on the graph below. Note that the force graphed. This verifies the guess that impending slip occurs first at A. (c) When the coefficient of friction is 0.364, the normal force N A goes to infinity. That is, the system locks at FB < ( FB )max for the entire range µ s = 0.364 , and no amount of pulling force will be able to move the cardboard. (d) For coefficients of friction greater than 0.364, the friction and normal forces become negative. The best interpretation of the solution is that a pushing force will be necessary to move the cardboard – therefore, the friction force must be in the opposite direction. 239 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 6-133 RILEY, STURGES AND MORRIS A 150-lb uniformly-loaded file cabinet sits on an inclined surface as shown in Fig. P6-133. A horizontal force P acts on the cabinet, and the coefficient of friction between the cabinet and the inclined surface is 0.4. (a) Determine Pmin and Pmax, the minimum and maximum horizontal forces for which the file cabinet will be in equilibrium. (b) Calculate and plot N and F, the normal and friction forces acting on the bottom of the file cabinet, as functions of P (Pmin ≤ P ≤ Pmax). (c) Calculate and plot d, the distance between the line of action of the normal force and the corner C, as a function of P (Pmin ≤ P ≤ Pmax). SOLUTION Assuming that impending motion is up the incline, friction will act down the incline to oppose the motion. From a free-body diagram of the cabinet, the equilibrium equations are → ΣFx = 0 : ↑ ΣFy = 0 : ΣM C = 0 : P − N sin 25° − F cos 25° = 0 N cos 25° − F sin 25° − 150 = 0 (150 cos 25° )( 9 ) + (150sin 25° )( 24 ) − Nd + ( P sin 25° )(18 ) − ( P cos 25° )( 32 ) = 0 F = 81.382 lb d = −3.306 in. (a) If F = 0.4 N (impending slip up the incline), then N = 203.456 lb P = 159.742 lb But this solution is not possible since the normal force acts at a point outside the body. If d = 0 (impending tip), then N = 190.168 lb F = 52.887 lb P = 128.300 lb = Pmax ..................................................................................................... Ans. If impending slip is down the incline, then the friction force acts up the incline (changes sign in the equilibrium equations above), F = 0.4 N , and the equations of equilibrium give N = 139.489 lb P = 8.383 lb F = 55.796 lb d = 18.393 in. But this solution is also not possible since the normal force again acts at a point outside the body. If d = 18 in. (impending tip), then N = 140.287 lb F = 54.083 lb P = 10.272 lb = Pmin ....................................................................................................... Ans. (b) For 10.272 lb < P < 128.300 lb , the normal and friction forces are F = P cos 25° − 150sin 25° .......................................................................................... Ans. N = P sin 25° + 150 cos 25° .......................................................................................... Ans. 10.272 lb < P < 128.300 lb , the location of the normal force is 2744.9413 − 21.39472P d= ....................................................................................... Ans. N (c) For 240 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 6-134 A 35-kg child is sitting on a swing suspended by a rope that passes over a tree branch (Fig. P6-134). The coefficient of friction between the rope and the branch (which can be modeled as a flat belt over a drum) is 0.5 and the weight of the rope can be ignored. Determine the minimum force the child must exert on the rope to keep suspended. SOLUTION W = 35 ( 9.81) = 343.35 N From a free-body diagram of the child, the vertical equilibrium equation gives ↑ ΣFy = 0 : T1 + T2 − 343.35 = 0 T2 = T1e0.5π = 4.81048T1 which combined with the belt friction equation gives T1 = 59.1 N ....................................................................................................................... Ans. 6-135 A rope attached to a 500-lb block passes over a frictionless pulley and is wrapped for one full turn around a fixed post as shown in Fig. P6-135. If the coefficient of friction between the rope and the post is 0.25, determine (a) The minimum force P that must be used to keep the block from falling. (b) The minimum force P that must be used to begin to raise the block. SOLUTION 500 = Pe 0.25( 2π ) 0.25( 2π ) P = 103.9 lb = Pmin ........................................................ Ans. P = 2410 lb = Pmax ........................................................ Ans. P = 500e 6-136 A rope attached to a 220-kg block passes over a fixed drum (Fig. P6-136) If the coefficient of friction between the rope and the drum is 0.30, determine the minimum force P that must be used to (a) Keep the block from falling. (b) Begin to raise the block. SOLUTION 220 ( 9.81) = Pe 0.3(π 2 ) 0.3(π 2 ) P = 1347 N = Pmin ......................................................... Ans. P = 3460 N = Pmax ........................................................ Ans. P = 220 ( 9.81) e 6-137 A rotating shaft with a belt-type brake is shown in Fig. P6-137. The static coefficient of friction between the brake drum and the brake belt is 0.20. When a 75-lb force P is applied to the brake arm, rotation of the shaft is prevented. Determine the maximum torque that can be resisted by the brake if the shaft is tending to rotate (a) Counterclockwise. (b) Clockwise. 241 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS SOLUTION From a free-body diagram for the pulley, the moment equilibrium equation gives ΣM B = 0 : 6T1 − 6T2 + C = 0 C = 6 (T2 − T1 ) From a free-body diagram for the brake arm, the moment equilibrium equation gives ΣM A = 0 : 9T2 − 3T1 − 27 ( 75 ) = 0 T2 > T1 3T2 − T1 = 675 (a) If the couple C is counterclockwise, then and the belt friction equation gives T2 = T1e0.2π = 1.87446T1 T1 = 145.997 lb T2 = 273.666 lb C = 766 lb ⋅ in. ............................................................................................................ Ans. (a) If T1 > T2 , then the belt friction equation gives T1 = T2 e0.2π = 1.87446T2 T1 = 1124.130 lb 6-138 T2 = 599.710 lb ............................................................................. Ans. C = −3150 lb ⋅ in. = 3150 lb ⋅ in. The scaffolding of Fig. P6-138 is raised using an electric motor that sits on the scaffolding. Frictionless wheels at the ends of the scaffold restrict horizontal motion. The 250-mm-diameter pulley at the top is jammed and will not rotate. The coefficient of friction between the rope and the pulley is 0.25 and the weight of the motor, scaffold, and supplies is 2500 N. Determine the minimum torque that must be supplied by the motor (a) To raise the scaffold at a constant rate. (b) To lower the scaffold at a constant rate. SOLUTION From a free-body diagram for the scaffolding, the vertical equilibrium equation gives ↑ ΣFy = 0 : (a) If TA + TB − 2500 = 0 TA = TB e0.25π = 2.19328TB TA = 1717.106 N TB = 782.894 N Torque = TA r = (1717.106 )( 0.075 ) = 128.8 N ⋅ m (to raise the scaffolding) ...................................................... Ans. TA > TB , then the belt friction equation gives (b) If TB > TA , then the belt friction equation gives TB = TAe0.25π = 2.19328TA TA = 782.894 N TB = 1717.106 N Torque = TA r = ( 782.894 )( 0.075 ) = 58.7 N ⋅ m (to lower the scaffolding) ...................................................... Ans. 242 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 6-139 RILEY, STURGES AND MORRIS A rope is wrapped one full turn around each of two posts as shown in Fig. P6-139. If the coefficient of friction between the rope and the posts is 0.5, determine (a) The ratio of TA to TB. (b) The ratio of TA to TB if only one post is used. SOLUTION (a) TB = TC e 0.5( 2π ) = 23.14069TC 0.5( 2π ) = 23.14069 TAe (b) = 535.492TA TB TA = 535 ...................................................................................................................... Ans. TB = TAe 0.5( 2π ) = 23.14069TA TB TA = 23.1 ..................................................................................................................... Ans. 6-140 A conveyor belt is driven with the 200-mm-diameter multiple-pulley drive shown in Fig. P6-140. Couples CA and CB are applied to the system at pulleys A and B, respectively. The angle of contact between the belt and a pulley is 225° for each 6.5 kg pulley, and the coefficient of friction is 0.30. Determine (a) The maximum force T that can be developed by the drive and the magnitudes of the input couples CA and C B. (b) The shearing stress on a cross section of each 25-mm diameter pin, if each pin is in double shear. SOLUTION W = 6.5 ( 9.81) = 63.765 lb (a) β = ( 225 360 )( 2π ) = 3.92699 rad TC = 2e 0.3( 3.92699 ) 0.3( 3.92699 ) = 6.49638 kN = 21.101 kN ....................................................................................... Ans. T = TC e → ΣFx = 0 : ↑ ΣFy = 0 : ΣM A = 0 : From a free-body diagram for pulley A, the equilibrium equations give 21,101 + 6496.38cos 45° − Ax = 0 Ay − 63.765 + 6496.38sin 45° = 0 C A − 21,101( 0.1) + 6496.38 ( 0.1) = 0 Ay = −4530 N Ax = 25, 695 N 2 A = Ax2 + Ay = 26, 091 N C A = 1460 N ⋅ m .............................................................................................................. Ans. From a free-body diagram for pulley B, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM B = 0 : Bx − 2000 − 6496.38cos 45° = 0 By − 63.765 − 6496.38sin 45° = 0 6496.38 ( 0.1) − 2000 ( 0.1) − CB = 0 By = 4657 N Bx = 6594 N 2 B = Bx2 + By = 8073 N CB = 450 N ⋅ m ................................................................................................................ Ans. 243 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS (b) The shearing stress on the pins is obtained by dividing the forces on the pins by twice their cross sectional areas (since they are in double shear), VA 26, 091 = = 26.6 × 106 N/m 2 = 26.6 MPa ............................ Ans. 2 2 As 2 π ( 0.025) 4 V 8073 τB = B = = 8.22 ×106 N/m 2 = 8.22 MPa ............................ Ans. 2 2 As 2 π ( 0.025 ) 4 τA = 6-141 The electric motor shown in Fig. P6-141 weighs 30 lb and delivers 50 lb-in. of torque to pulley A of a furnace blower by means of a V-belt. The effective diameters of the 36° pulleys are 5 in. The coefficient of friction is 0.30. Determine the minimum distance a to prevent slipping of the belt if the rotation of the motor is clockwise. SOLUTION µ senh = ΣM A = 0 : 0.3 = 0.9708 sin ( 36 2 ) 50 − ( T2 − T1 )( 2.5 ) = 0 From a free-body diagram for the pulley A, the moment equilibrium equation is combined with the belt friction equation T2 = T1e0.9708π = 21.1137T1 to get T1 = 0.99435 lb T2 = 20.9944 lb Then, from a free-body diagram for the motor, the moment equilibrium equation gives ΣM C = 0 : 20.9944 (10.5) + 0.99435 ( 5.5) − 50a = 0 a = 4.52 in. ....................................................................................................................... Ans. 6-142 Complete the derivation of Eq. (6-20). SOLUTION Starting from ∆T cos ( ∆θ 2 ) = 2∆F 2∆P sin (α 2 ) = 2T sin ( ∆θ 2 ) + ∆T sin ( ∆θ 2 ) in the limit as ∆θ → 0 ∆T = 2∆F 2∆P sin (α 2 ) = T ∆θ + ∆T ( ∆θ 2 ) and therefore as ∆θ → 0 2P → 0 ∆F → 0 ∆T → 0 Then, at the point of impending slip, ∆F = µ s ∆P . Divide through by ∆θ and let ∆θ → 0 to get ∆T 2 µ s ∆P µ sT + ( ∆T 2 ) = = ∆θ → 0 ∆θ sin (α 2 ) ∆θ lim 244 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS dT µ sT = dθ sin (α 2 ) ∫ T2 T1 β dT µs = ∫0 dθ sin (α 2 ) T T µs β ln T2 − ln T1 = ln 2 = T1 sin (α 2 ) T2 µβ =e s T1 6-143 sin (α 2 ) ................................................................................................................. Ans. The band wrench of Fig. P6-143 is used to unscrew an oil filter from a car. (The filter acts as if it were a wheel with a resisting torque of 40 lb-ft.) Neglect the weight of the handle and friction between the end of the metal handle and the metal filter case. Determine (a) The minimum coefficient of friction between the band and the filter that will prevent slippage. (b) The shearing stress on a cross section of the pin at B when slipping is impending if the ¼-in.-diameter pin is loaded in double shear. SOLUTION φ = tan −1 ( 2.5 4.5 ) = 29.055° β = 270° + φ = 299.055° = 5.2195 rad (a) From a free-body diagram for the filter, the moment equilibrium equation gives ΣM C = 0 : 40 − (T2 − T1 )( 2.5 12 ) = 0 T2 − T1 = 192.00 lb From a free-body diagram for the wrench, the moment equilibrium equation gives ΣM P = 0 : 9.5T1 − ( T2 sin φ )( 7.5) = 0 T2 = 2.60819T1 T1 = 119.3886 lb T2 = 311.3886 lb Then, the belt friction equation applied to the tensions on the filter gives 311.3886 = 119.3886e5.2195 µs µ s = 0.1837 ...................................................................................................................... Ans. (b) Dividing the force on the pin at B by twice its cross sectional area (since it is in double shear) gives the shear stress τB = 6-144 T2 311.3886 = = 3.17 × 103 psi = 3.17 ksi ....................................... Ans. 2 2 As 2 π ( 0.25 ) 4 A uniform belt 2 m long with a mass of 2 kg hangs over a small fixed peg (Fig. P6-144). If the coefficient of friction between the belt and the peg is 0.40, determine the maximum distance d between the two ends for which the belt will not slip off the peg. SOLUTION T1 = W1 T2 = W2 245 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS W1 + W2 = W = 2 ( 9.81) = 19.62 N L1 + L2 = L = 2 m W2 = W1e0.4π = 3.51359W1 W1 = 4.34688 N W2 = 15.27312 N The weight of each portion of the belt is directly proportional to its length. Therefore W1 19.62 = L1 2 L1 = 0.44311 m 6-145 L2 = 2 − 0.44311 = 1.55689 m d = L2 − L1 = 1.114 m .................................................................................................... Ans. The hand brake of Fig. P6-145 is used to control the rotation of a drum. The coefficient of friction between the belt and the drum is µs = 0.35, and the weight of the handle may be neglected. If a clockwise torque of 350 lb-ft is applied to the drum, determine the minimum force P that must be applied to the handle to prevent motion when a = 4 in., 8 in., and 12 in. SOLUTION From a free-body diagram for the drum, the moment equilibrium equation ΣM E = 0 : (T2 − T1 )( 6 12 ) − 350 = 0 is combined with the belt friction equation T2 = T1e0.35π = 3.00284T1 to get T1 = 349.5043 lb T2 = 1049.504 lb From a free-body diagram for the brake, the moment equilibrium equation gives ΣM B = 0 : P ( 30 − a ) − T2 a = 0 P = 161.5 lb ......................................................................................... Ans. P = 382 lb ............................................................................................ Ans. P = 700 lb ............................................................................................ Ans. a = 4 in. a = 8 in. a = 12 in. 6-146 The structure shown in Fig. P6-146 consists of a circular tie rod AB and a rigid member BC. If the structure is to support a load P = 40 kN, determine the required diameters of the pins at A, B, and C, and the required diameter of the tie rod. The tie rod is made of structural steel and the pins are made of 0.2% C hardened steel. All pins are in double shear. The tie rod is adequately reinforced around the pins so that tensile failure does not occur at the pins. Failure is by yielding, and the factor of safety is 1.3. Take the yield strength in shear to be one-half the yield strength in tension. SOLUTION sin θ = 3 5 structural steel: 0.2% hardened steel: cos θ = 4 5 σ y = 250 MPa σ y = 250 MPa E = 200 GPa Note that the tie rod is a two-force member. Then from a free-body diagram of the rigid member BC, the equilibrium equations give 246 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS → ΣFx = 0 : ↑ ΣFy = 0 : ΣM A = 0 : Cx − TAB cos θ = 0 C y − 40 + TAB sin θ = 0 (TAB sin θ )( 4 ) − 40 ( 3.5 ) = 0 C y = 5.000 kN TAB = 58.333 kN C x = 46.667 kN 2 C = C x2 + C y = 46.934 kN Dividing the force in the tie rod by its cross sectional area gives the normal stress σ AB = TAB 58.333 ×103 σ y 250 ×106 = ≤ = 2 π d AB 4 A FS 1.3 d AB ≥ 0.01965 m = 19.65 mm ...................................................................... Ans. The force on the pins are divided by twice their cross sectional area since they are in double shear V 58.333 ×103 τ y σ y 2 ( 430 × 10 τA = A = ≤ = = 2 FS FS 2 As 1.3 2 π d A 4 6 6 ) 2 58.333 × 103 ( 430 × 10 τB = ≤ 2 1.3 2 π d B 4 d A ≥ 0.01498 m = 14.98 mm ........................................................................ Ans. ) 2 d B ≥ 14.98 mm ................... Ans. dC ≥ 13.44 mm ................... Ans. τC = 6-147 46.934 × 103 ( 430 × 10 ≤ 2 1.3 2 π dC 4 6 ) 2 A rigid handle is used to twist a 2.5-in.-diameter valve, as shown in Fig. P6-147. A pair of oppositely directed parallel forces P = 100 lb is needed to twist the valve. Force is transmitted from the handle to the valve shaft by means of a 1.0-in.-long (into the page) key with a square cross section. The key is to be made of 0.8% C hot-rolled steel. If failure is by fracture, and the factor of safety is to be 2, determine the cross-sectional dimensions of the square key to the nearest 1/64 in. The yield and ultimate strengths in shear are one-half the corresponding values in tension. SOLUTION 0.8% hot-rolled steel: σ f = 122 ksi From a free-body diagram of the handle, the moment equilibrium equation gives ΣM center = 0 : 100 ( 24 ) − V (1.25 ) = 0 3 V = 1920 lb = τ ( b × 1) 1920 τ f σ y 2 (122 × 10 ≤ = = b FS FS 2 b ≥ 0.06295 in. ) 2 ( 4 64 ) < 0.06295 < ( 5 64 ) bmin = 5 64 in. ................................................................................................................. Ans. 247 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 6-148 RILEY, STURGES AND MORRIS Each member of the truss of Fig. P6-148 has a circular cross section and is made of structural steel. If failure is by yielding, and the factor of safety is 2.5, determine the minimum permissible diameter of member EJ. SOLUTION θ = tan −1 ( 3 2 ) = 56.310° structural steel: LEJ = 32 + 2 2 = 3.60555 m σ y = 250 MPa From a free-body diagram of the entire truss, the moment equilibrium equation gives ΣM A = 0 : G ( 8) − 30 ( 8 ) − 30 ( 6 ) − 30 ( 4 ) − 30 ( 2 ) = 0 G = 75 kN Next, divide the truss with a section through members DE, EJ, and HJ, and draw a free-body diagram of the right-hand side. From a free-body diagram of the section, the vertical equilibrium equation gives ↑ ΣFy = 0 : 75 − 30 − 30 − TEJ sin θ = 0 TEJ = 18.0277 kN Dividing the force in the member by its cross-sectional area gives the normal stress σ EJ = TEJ 18.0277 ×103 σ y 250 × 106 = ≤ = πd2 4 A FS 2.5 Ans. d ≥ 0.01515 m = 15.15 mm 6-149 A pin-connected system of levers and bars is used as a toggle for a press to crush cans, as shown in Fig. P6149. The system is designed to provide a crushing force of 550 lb. The handle may be considered rigid. The pin at D is to be made of 2024-T4 wrought aluminum. If failure is by yielding, and the factor of safety is to be 1.5, determine the required diameter of pin D to the nearest 1/32 in. SOLUTION 2024-T4 wrought aluminum: σ y = 48 ksi From a free-body diagram of the plunger, the vertical equilibrium equation gives ↑ ΣFy = 0 : 550 − FBD sin 67° = 0 FBD = 597.498 lb Note that all three members attached to the pin D are two-force members. Therefore, from a free-body diagram of the pin, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : FBD cos 67° + FCD cos 78° − FDE = 0 FBD sin 67° − FCD sin 78° = 0 FCD = 562.287 lb FDE = 350.367 lb FBD = 597.498 lb on the shear surface between BD and CD. 3 Therefore, the maximum shear force on pin D is τ max 597.498 τ y σ y 2 ( 48 × 10 = ≤ = = π d 2 4 FS FS 1.5 ) 2 d ≥ 0.2181 in. 6 32 < 0.2181 < 7 32 d min = 7 32 in. ................................................. Ans. 248 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 6-150 RILEY, STURGES AND MORRIS A pair of vise grip pliers is shown in Fig. P6-150. All members of the pliers, except for the pins and spring, may be treated as rigid. It is anticipated that the largest applied force on the handles will be P = 200 N. If each of the three pins is in double shear and made of 0.4% C hardened steel, determine the required diameters for the pins to the nearest mm. In the analysis, neglect the spring force. Failure is by yielding, and the factor of safety is 1.25. The yield strength in shear is one-half the tensile yield strength. SOLUTION θ = tan −1 ( 30 50 ) = 30.964° 0.4% C hardened steel: φ = tan −1 (15 35) = 23.199° σ y = 360 MPa FAB cos θ − C x = 0 FAB sin θ − 200 − C y = 0 From a free-body diagram of the handle, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM C = 0 : 200 ( 93) − ( FAB sin θ )( 28 ) + ( FAB cos θ )( 5 ) = 0 C y = 745.759 N FAB = 1838.212 N C x = 1576.250 N 2 C = Cx2 + C y = 1743.766 N Next, from a free-body diagram of the jaw, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM D = 0 : N C x − Dx − N sin φ = 0 C y − Dy + N cos φ = 0 ( 352 + 152 − C x ( 35 ) − C y (12 ) = 0 Dy = 2293.426 N ) N = 1683.817 N Dx = 912.951 N 2 D = Dx2 + Dy = 2468.457 N Dividing the force on the pins by twice their cross-sectional areas (since they are in double shear) gives the shear stresses 1838.212 τ y σ y 2 ( 360 × 10 F ≤ = = τ B = AB = 2 2 As 2 π d B 4 FS 1.25 FS C 1743.766 ( 360 × 10 = ≤ τC = 2 2 As 2 π dC 4 1.25 dC ≥ 0.00222 m D 2468.457 ( 360 × 10 = ≤ τD = 2 2 As 2 π d D 4 1.25 d D ≥ 0.00264 m 6 6 ) 2 d B ≥ 0.00285 m 6 ) ) d B min = 3 mm ............................................ Ans. 2 dC min = 3 mm ............................................ Ans. 2 d D min = 3 mm ............................................ Ans. 249 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 6-151 RILEY, STURGES AND MORRIS A device for lifting rectangular objects such as bricks and concrete blocks is shown in Fig. P6-151. The coefficients of friction at all vertical contact surfaces are µs = 0.4 and µk = 0.3. The device is to lift two blocks, each weighing 15 lb. The pin at B is to be made of structural steel with a strength in shear equal to one-half the strength in tension. For failure by yielding and a factor of safety of 3, determine the minimum permissible diameter of the pin at B, which is in double shear. SOLUTION structural steel: σ y = 36 ksi From a free-body diagram of the blocks, the equilibrium equations give ↑ ΣFy = 0 : ΣM G = 0 : FA + FC − 30 = 0 8 FA − 8FC = 0 FA = FC = 15 lb Next, from a free-body diagram of member AB, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM B = 0 : Bx = 45 lb N A − Bx = 0 By − FA = 0 4 N A − 12 FA = 0 By = 15 lb N A = 3FA = 45 lb 2 B = Bx2 + By = 47.434 lb N A = 3FA independent of the weight being lifted. Therefore, as long as the coefficient of friction is greater than FA N A = 1 3 , the device will lift the weight without slipping. Note that Dividing the force on the pin by twice its cross-sectional area (since it is in double shear) gives the shear stress τ σ 2 ( 36 ×10 B 47.434 = ≤ y= y = τB = 2 FS 2 As 2 π d B 4 FS 3 3 ) 2 d B ≥ 0.0709 in. ................................................................................................. Ans. 6-152 The flat roof of a building is supported by a series of parallel plane trusses spaced 2 m apart (only one such truss is shown in Fig. P6-152). Water of density 1000 kg/m3 may collect on the roof to a depth of 0.2 m. All members of the truss are to be the same size. If the truss members are made of structural steel, failure is by yielding, and the factor of safety is 3.0, determine the smallest diameter solid circular rod, to the nearest mm, that can be used. The members are braced so that buckling does not occur. SOLUTION From a free-body diagram of a roof panel, the equilibrium equations give the force carried by the joints at each end of the roof panel as F1 = F2 = W 2 = (1000 )( 9.81)( 2 × 1.2 × 0.2 ) 2 = 2354.4 N Then, from a free-body diagram of the entire truss, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : Ax = 0 Ay + Fy − 6 F1 = 0 250 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ΣM A = 0 : Fy ( 3.6 ) − ( 2 F1 )(1.2 ) − ( 2 F1 )( 2.4 ) − ( F1 )( 3.6 ) = 0 Ay = Fy = 3F1 = 7063.20 N From a free-body diagram for joint B, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : TBC = 0 −TAB − F1 = 0 TAB = −2354.4 N ≅ 2.35 kN (C) ............................................. Ans. TBC = 0 kN .................................................................................... Ans. From a free-body diagram for joint A, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : TAH + ( 3 5) TAC = 0 TAB + 7063.2 + ( 4 5 ) TAC = 0 TAC = −5886.0 N ≅ 5.89 kN (C) ............................................. Ans. TAH = 3531.6 N ≅ 3.53 kN (T) ................................................ Ans. From a free-body diagram for joint H, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : TGH − TAH = 0 TCH = 0 TCH = 0 kN ................................................................................... Ans. TGH = 3531.6 N ≅ 3.53 kN (T) ................................................ Ans. From a free-body diagram for joint E, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : −TDE = 0 −TEF − F1 = 0 TDE = 0 kN .................................................................................... Ans. TEF = −2354.4 N ≅ 2.35 kN (C) ............................................. Ans. From a free-body diagram for joint F, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : −TFG − ( 3 5) TDF = 0 TEF + 7063.2 + ( 4 5 ) TDF = 0 TDF = −5886.0 N ≅ 5.89 kN (C) ............................................. Ans. TFE = 3531.6 N ≅ 3.53 kN (T) ................................................ Ans. From a free-body diagram for joint D, the equilibrium equations give → ΣFx = 0 : ΣFy = 0 : TDE − TCD + ( 3 5) TDF = 0 −2 F1 − TDG − ( 4 5 ) TDF = 0 TCD = −3531.6 N ≅ 3.53 kN (C) ...................................... Ans. TDG = 0 kN ............................................................................ Ans. 251 STATICS AND MECHANICS OF MATERIALS, 2nd Edition Finally, from a free-body diagram for joint G, the equilibrium equations give RILEY, STURGES AND MORRIS ↑ ΣFy = 0 : TDG + ( 4 5 ) TCG = 0 TCG = 0 kN ............................................................................. Ans. Dividing the largest force ( 5886.0 N ) by the cross sectional area of the members gives the normal stress in the member. For structural steel σ y = 250 MPa . Therefore σ max = P 5886.00 σ y 250 × 106 = ≤ = A πd2 4 FS 3.0 d ≥ 0.00948 m = 9.48 mm d min = 10 mm ................................................................................................................... Ans. 6-153 The brake shown in Fig. P6-153 is used to control the motion of block B. The weight of the block is 2200 lb, the coefficient of static friction between the brake arm and drum is 0.35, and the coefficient of kinetic friction is 0.30. A force P is applied to the rigid brake arm such that the block moves downward at a constant rate. Determine the minimum permissible diameter for the pin at the end of the brake arm (to the nearest 1/8-in.) if it is in double shear and is made of 0.2% C hardened steel. Failure is by yielding, and the factor of safety is 2.5. The strength in shear is one-half the strength in tension. SOLUTION 0.2% C hardened steel: σ y = 62 ksi From a free-body diagram of the drum, the moment equilibrium equation gives ΣM C = 0 : 7 F − 3.5 ( 2200 ) = 0 F = 1100 lb = 0.35 N N = 3142.857 lb Next, from a free-body diagram of the brake handle, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM A = 0 : F − Ax = 0 N − Ay − P = 0 8 N + 1.6 F − 24 P = 0 P = 1120.952 lb Ax = 1100 lb Ay = 4263.809 lb 2 A = Ax2 + Ay = 4403.416 lb Dividing the force on the pin by twice its cross-sectional area (since it is in double shear) gives the shear stress 3 VA 4403.416 τ y σ y 2 ( 62 × 10 ) 2 = ≤ = = τA = FS 2 As 2 π d 2 4 FS 2.5 d ≥ 0.47547 in. 3 8 < 0.47547 < 4 8 d min = 4 8 = 1 2 in. ......................................... Ans. 252 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 6-154 RILEY, STURGES AND MORRIS Solid circular bars AB and BC of Fig. P6-154 are pinned at the ends, and the structure is subjected to a load P = 26 kN. The angle θ may vary, but pin A is always directly above pin C. Both bars are made of 6061T6 wrought aluminum alloy. Failure is by yielding, and the factor of safety is 4. In the force analysis assume that the weights of the bars are negligible with respect to the applied loads and that the pins are adequately designed. Determine (a) The angle θ for a minimum weight structure. (b) The diameters of the bars. (c) The weights of the bars. SOLUTION 6061-T6 wrought aluminum: σ y = 270 MPa ρ = 2710 kg/m 3 Note that both AB and BC are two-force members. Then, from a free-body diagram of the pin, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : FBC − TAB cos θ = 0 TAB sin θ − 26 = 0 FBC = TAB cosθ = ( 26 tan θ ) kN d AB ≥ d BC ≥ LBC = 1.8 m 22.1457 × 10−3 m sin θ 22.1457 × 10−3 m tan θ TAB = ( 26 sin θ ) kN Dividing the force in the members by their cross sectional areas gives the normal stresses σ AB = σ BC = TAB 26 × 103 sin θ σ y 270 ×106 = ≤ = 2 A FS 4 π d AB 4 26 × 103 tan θ 270 × 106 ≤ 2 π d BC 4 4 (a) To minimize the weight of the structure, we need to minimize the volume of the bars making up the structure. LAB = (1.8 cosθ ) m V= 2 π d BC πd2 1.8 (1.8 ) + AB 4 4 cos θ 2 1.8π ( 22.1457 × 10−3 ) cos θ 1 = + sin θ sin θ cos θ 4 To minimize the volume, we set dV dθ = 0 −3 2 dV 1.8π ( 22.1457 × 10 = dθ 4 If neither ) sin θ = 0 nor cos θ = 0 , we can multiply through by ( − sin 2 θ cos 2 θ ) to get sin 2 θ cos 2 θ + cos 4 θ + cos 2 θ − sin 2 θ = 0 cos 2 θ 1 1 −1 − − 2+ =0 2 sin θ sin θ cos 2 θ ( cos θ )( sin 2 2 θ + cos 2 θ + 1) − sin 2 θ = 0 2 (1 − sin 2 θ ) − sin 2 θ = 0 2 cos 2 θ − sin 2 θ = 0 2 − 3sin 2 θ = 0 θ = 54.736° ...................................................................................................................... Ans. 253 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS d AB ≥ 24.5 × 10−3 m = 24.5 mm .................................................................................. Ans. d BC ≥ 18.62 ×10 −3 m = 18.62 mm .............................................................................. Ans. wAB wBC π ( 0.0245 )2 1.8 = 2710 ( 9.81) = 39.1 N .................................. Ans. 4 cos 54.736° π ( 0.01862 ) 2 = 2710 ( 9.81) (1.8 ) = 13.03 N ............................................... Ans. 4 6-155 Determine the force P required to pull the 250-lb roller over the step shown in Fig. P6-155. SOLUTION θ = cos −1 ( 9 12 ) = 41.410° When the roller is about to be pulled over the step, it no longer touches the floor and the force exerted on the roller by the floor is zero. From a free-body diagram of the roller, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM center = 0 : P cos 30° − N sin θ − F cos θ = 0 N cosθ − F sin θ + P sin 30° − 250 = 0 Fr = 0 F = 0 lb N = 1.30931P P = 168.7 lb ..................................................................................................................... Ans. 6-156 A machine is activated by force T when a pedal is depressed with a 40-N force, as shown in Fig. P6-156. Determine the magnitude of T and the resultant bearing reaction at B. SOLUTION From a free-body diagram of the pedal, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM B = 0 : Bx = 0 By − T − 40 = 0 T ( 200 ) − (125cos 45° )( 40 ) = 0 By = 57.7 N T = 17.68 N ↓ ........................................................... Ans. Bx = 0 N 6-157 B = 57.7 N ↑ .................................................................................................................. Ans. The 500-lb block shown in Fig. P6-157 is supported by a ball-and-socket joint at A, by a smooth pin at B, and by a cable at C. Determine the components of the reactions at supports A and B and the force in the cable at C. SOLUTION ΣM A = 0 : ( −14 i + 8 k ) × ( By j + Bz k ) + ( −7 i + 9 j + 4 k ) × ( −500 k ) + ( −7 i + 18 j + 8 k ) × (TC k ) + ( −14 i + 18 j + 8 k ) × ( −200 k ) = 0 254 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS x: y: −8 By + 18TC − 8100 = 0 TC = 450 lb ............................................... Ans. Bz = 225 lb ............................................... Ans. By = 0 lb .................................................... Ans. 14 Bz + 7TC − 6300 = 0 −14 By = 0 z: ΣFx = 0 : ΣFy = 0 : Ax = 0 Ay + By + 350 = 0 Ax = 0 lb .................................................... Ans. Ay = −350 lb ............................................ Ans. ΣFz = 0 : 6-158 Az + Bz + TC − 700 = 0 Az = 25 lb .................................................. Ans. The plate shown in Fig. P6-158 has a mass of 80 kg. The brackets at supports A and B exert only force reactions on the plate. Each of the brackets can resist a force along the axis of the pins in one direction only. Determine (a) The reactions at supports A and B and the tension in the cable. (b) The change in length of the cable. (Use E = 200 GPa and d = 2.5 mm.) SOLUTION (a) W = 80 ( 9.81) = 784.800 N ∆y = 1.4 cos 30° = 1.21244 m ∆z = 1.2 + 1.4sin 30° = 1.9000 m LCD = 12 + 1.212442 + 1.90002 = 2.46577 m TC = TC −1i − 1.21244 j + 1.9000 k 12 + 1.212442 + 1.90002 = ( −0.40555TC i − 0.49171TC j + 0.77055TC k ) N From a free-body diagram of the plate, the equilibrium equations give ΣM B = 0 : (1 i + 1.2 k ) × TC + ( 2 i ) × ( Ay j + Az k ) + (1i + 0.60622 j − 0.35 k ) × ( −784.8 k ) = 0 0.59005TC − 475.761 = 0 −1.25721TC − 2 Az + 784.800 = 0 −0.49171TC + 2 Ay = 0 x: y: TC = 806.307 N Az = −114.449 N Ay = 198.235 N z: ΣFx = 0 : ΣFy = 0 : Bx − 0.40555 ( 806.307 ) = 0 Ay + By − 0.49171( 806.307 ) = 0 Az + Bz + 0.77055 ( 806.307 ) − 784.800 = 0 Bx = 326.998 N By = 198.234 N ΣFz = 0 : Bz = 277.949 N A = (198.2 j − 114.4 k ) N ...................... TC = 806 N ................................................ Ans. B = ( 327 i + 198.2 j + 278 k ) N ................................................................................... Ans. (b) The stretch of the cable CD is given by δ CD = ( 806.307 )( 2465.77 ) = 2.03 mm ........................................... Ans. PL = EA ( 200 × 109 ) π ( 0.0025) 2 4 255 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 6-159 RILEY, STURGES AND MORRIS The door to an airplane hangar consists of two uniform sections that are hinged at the middle as shown in Fig. P6-159. The door is raised by means of a cable attached to a bar along the bottom edge of the door. Smooth rollers at the ends of the bar C run in a smooth vertical channel. If the door is 30 ft wide, 15 ft tall, and weighs 1620 lb, determine (a) The force P when the door opening height h = 8 ft. (b) The resultant hinge forces at A and B when h = 8 ft. SOLUTION θ = sin −1 ( 3.5 7.5 ) = 27.818° (a) From a free-body diagram of the door AB, the equilibrium equations are → ΣFx = 0 : ↑ ΣFy = 0 : ΣM A = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : ΣM C = 0 : Ax − Bx = 0 Ay − By − 810 = 0 810 ( 3.75cos θ ) + By ( 7.5cosθ ) − Bx ( 7.5sin θ ) = 0 Bx − N C = 0 By + P − 810 = 0 From a free-body diagram of the door AB, the equilibrium equations are 810 ( 3.75cos θ ) − By ( 7.5cos θ ) − Bx ( 7.5sin θ ) = 0 By = 0 lb Ay = 810 lb Combining the two moment equations gives Bx = 767.562 lb Then the force equations give Ax = 767.562 lb N C = 767.562 lb P = 810 lb ........................................................................................................................ Ans. (b) 2 A = Ax2 + Ay = 1116 lb ............................................................................................... Ans. 2 B = Bx2 + By = 768 lb ................................................................................................. Ans. 6-160 Three bars are connected by smooth frictionless pins as shown in Fig. P6-160. Bar BCDE is rigid, bar AB is aluminum (E = 73 GPa; L = 1.5 m; d = 30 mm) and bar EF is steel (E = 200 GPa; L = 1.2 m; d = 20 mm). The 25-mm-diameter pivot pin D is aluminum and is in double shear. When P = 100 kN determine (a) The deformation of bars AB and EF. (b) The shearing stress on a cross section of the pin at D. SOLUTION (a) From a free-body diagram of member BCDE, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM D = 0 : TEF − Dx + 100 − FAB = 0 Dy = 0 600 FAB + 300TEF − 300 (100 ) = 0 in which TEF is a tensile force and FAB is a compressive force. The stretch of EF and the shrink of AB are related by 256 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS δ AB δ EF = δ AB = 2δ EF 600 300 FAB (1.5 ) TEF (1.2 ) =2 2 9 ( 73 ×10 ) π ( 0.03) 4 ( 200 ×109 ) π ( 0.02 )2 4 FAB = 1.31400TEF Combining the equilibrium equations and the deformation equation gives FAB = 36.2183 kN Dx = 91.3451 kN The deformation of the two rods are 3 TEF = 27.5634 kN Dy = 0 kN δ AB δ EF ( 36.2183 ×10 ) (1500 ) = 1.053 mm (shrink) ............................................ Ans. = ( 73 ×10 ) π ( 0.03) 4 ( 27.5634 ×10 ) (1200 ) = 0.526 mm (stretch) ........................................ Ans. = ( 200 ×10 ) π ( 0.02 ) 4 9 2 3 9 2 (b) Finally, dividing the force on the pin by twice its cross sectional are (since it is in double shear) gives the shear stress τD = 6-161 D 91.3451× 103 = = 93.0 × 106 N/m 2 = 93.0 MPa ............................ Ans. 2 As 2 π ( 0.025 )2 4 A person is holding a 20-lb object as shown in Fig. P6-161. Determine the force T in the biceps muscle and the force F of the humerus against the ulna, in terms of the weight W of the forearm which acts through G. For the position shown, both T and F act vertically. SOLUTION From a free-body diagram of the forearm, the equilibrium equations give ↑ ΣFy = 0 : ΣM F = 0 : T − F − W − 20 = 0 1.5T − 5.5W − 11.5 ( 20 ) = 0 T = 3.667W + 153.33 lb ................................Ans. F = 2.667W + 133.33 lb ...............................Ans. 6-162 Bodies A and B shown in Fig. P6-162 have masses of 2200 kg and 450 kg, respectively. The coefficient of friction between B and the horizontal surface is 0.2, between A and B it is 0.2, and between A and the fixed surface is 0.5. Determine the force P required to cause impending motion of body B to the left. SOLUTION WA = 2200 ( 9.81) = 21,582.0 N WB = 450 ( 9.81) = 4414.5 N For impending motion of block B to the left, the friction forces must act to the right to oppose the motion and FB = µ s N B = 0.2 N B From a free-body diagram for block B, the equilibrium equations are 257 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS → ΣFx = 0 : ↑ ΣFy = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : ΣM A = 0 : Therefore FA + FB − P = 0 N B − N A − 4414.5 = 0 FC − FA = 0 N C + N A − 21,582 = 0 21,582 ( 300 ) − N C (1500 ) = 0 N A = 17, 265.6 N FC ≤ µ s N C = 0.5 N C FA = FC From a free-body diagram for block A, the equilibrium equations are N C = 4316.4 N N B = 21, 680.1 N FA ≤ µ s N A = 0.2 N A FB = 0.2 ( 21, 680.1) = 4336.02 N Guessing that impending slip occurs at C before A gives FC = 0.5 ( 4316.4 ) = 2158.2 N = FA P = FA + FB = 6490 N ................................................................................................... Ans. Since the maximum amount of friction available at A the amount of friction required 6-163 ( FA = 2158.2 N ) , the guess that slip occurs at C first is correct. FA max = 0.2 (17, 265.6 ) = 3453.12 N is greater than Three bars are connected by smooth frictionless pins as shown in Fig. P6-163. Bar BCD is rigid, bar AC is aluminum [E = 12,000 ksi; L = 36 in.; d = 1 in.; α = 12.5(10-6)/°F], and bar DE is steel [E = 30,000 ksi; L = 30 in.; d = ½ in.; α = 6.6(10-6)/°F]. The 5/8-in.-diameter pivot pin B is steel and is in single shear. If the temperature drops 20°F after the unit is assembled, determine (a) The normal stresses in bars AC and DE. (b) The change in length of bars AC and DE. (c) The shearing stress on the cross section of the pin at B. SOLUTION (a) From a free-body diagram of bar BCD, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM A = 0 : Bx = 0 in which Bx = 0 By + TAC − TDE = 0 20TAC − 30TDE = 0 By = TDE − TAC TAC = 1.5TDE TAC and TDE are both tensile forces. The stretch of bar DE and the shrink of bar AC are related by δ DE δ AC = 30 20 δ DE = 1.5δ AC 258 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ( 30 ×10 ) π (.05) 6 TDE ( 30 ) 2 4 + ( 6.6 × 10 −6 ) ( −20 )( 30 ) −TAC ( 36 ) −6 = 1.5 − (12.5 ×10 ) ( −20 )( 36 ) (12 × 106 ) π (1)2 4 5.09296TDE + 5.72958TAC = 9540.0 lb Combining the equilibrium equations and the deformation equation gives TAC = 1045.493 lb (T) TDE = 696.995 lb (T) By = −348.498 lb σ AC = σ DE = (b) TAC 1045.493 = = 1331 psi (T) ...................................................................... Ans. 2 A π (1) 4 696.995 π ( 0.5) 4 2 = 3550 psi (T) ............................................................................... Ans. δ AC = ( −1045.493)( 36 ) − 12.5 ×10−6 −20 36 −TL − α ∆T L = ( ) ( )( ) EA (12 ×106 ) π (1)2 4 0.00501 in. (shrink) .............................................................................................. Ans. ( 696.995)( 30 ) + 6.6 ×10−6 −20 30 TL + α ∆T L = ( ) ( )( ) EA ( 30 ×106 ) π ( 0.5)2 4 0.00751 in. (stretch) ............................................................................................. Ans. 2 Bx2 + By δ DE = (c) 6-164 τB = As = 348.498 π ( 5 8) 4 2 = 1136 psi .................................................................. Ans. An adjustable bracket is held in place by a force P of magnitude 175 N as shown in Fig. P6-164. Determine the minimum coefficient of friction between the bracket and vertical square member if the bracket is to stay in place. SOLUTION From a free-body diagram of the bracket, the equilibrium equations are → ΣFx = 0 : ↑ ΣFy = 0 : ΣM A = 0 : NB − N A = 0 FA + FB − 175 = 0 75FB + 300 N B − 417.5 (175 ) = 0 FB = µ s N B FA = FB = 87.5 N N A = N B = 221.667 N and for impending slip at both A and B FA = µ s N A Therefore NA = NB µs = FA 87.5 = = 0.395 ........................................................................................ Ans. N A 221.667 259 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 6-165 Determine the force in members CD, DE, and FG of the truss shown in Fig. P6-165. SOLUTION sin θ = 3 5 cos θ = 4 5 φ = tan −1 ( 5 20 ) = 14.036° From a free-body diagram of the entire truss, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : ΣM A = 0 : G − Ax − 800sin θ − 600sin θ = 0 Ay − 800 cosθ − 600 cos θ − 500 = 0 G (10 ) − 800 (10 ) − 600 (17.5 ) − 500 ( 20 ) = 0 Ax = 2010 lb Ay = 1620 lb G = 2850 lb N 2 = 447.673 N From a free-body diagram of the joint A, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : TAB cosθ − 2010 = 0 1620 − TAG − TAB sin θ = 0 TAB = 2512.50 lb (T) TAG = 112.50 lb (T) From a free-body diagram of the joint G, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : 2850 + TBG cos β + TFG cos φ = 0 TAG + TBG sin β − TFG sin φ = 0 β = tan −1 ( 4 8 ) = 26.565° TBG = −1229.83 lb ≅ 1230 lb (C) TFG = −1803.86 lb ≅ 1804 lb (C) ................................................... Ans. Finally, from a free-body diagram of the joint D, the equilibrium equations give → ΣFx = 0 : ↑ ΣFy = 0 : −TCD cosθ − TDE cos φ = 0 TCD sin θ + TDE sin φ − 500 = 0 TCD = 1250 lb (T) ............................................................................... Ans. TDE = −1031 lb = 1031 lb (C) ...................................................................................... Ans. 6-166 Determine the change in length of members CF and FG of the truss shown in Fig. P6-166. Each member of the truss is made of structural steel, and is circular with a diameter of 25 mm. SOLUTION structural steel: E = 200 GPa LCF = 4 m = LEF = LFG LDE = 4 cos 30° = 3.46410 m 260 STATICS AND MECHANICS OF MATERIALS, 2nd Edition From a free-body diagram of the upper portion of the truss, the equilibrium equations give RILEY, STURGES AND MORRIS ΣM E = 0 : ΣM C = 0 : 1200 ( 3.46410 ) − TCF ( 3.46410 ) = 0 −1200 ( 3.46410 ) − 300 ( 6.92820 ) − (TFG sin 30° )( 6.92820 ) = 0 TCF = 1200 N TFG = −1800 N δ CF = δ FG = 6-167 ( −1800 )( 4000 ) ( 200 ×109 ) π ( 0.025)2 (1200 )( 4000 ) TL = = 0.0489 mm ........................................ Ans. EA ( 200 × 109 ) π ( 0.025) 2 4 4 = −0.0733 mm ................................................. Ans. The band brake of Fig. P6-167 is used to control the rotation of a drum. The coefficient of friction between the belt and the drum is 0.25, and the weight of the handle is negligible. Determine the force P that must be applied to the end of the handle to resist a maximum torque of 200 lb-in. if the torque is applied (a) Clockwise. (b) Counterclockwise. SOLUTION From a free-body diagram of the drum, the moment equilibrium equation gives ΣM C = 0 : 4TA − 4TB − M = 0 TA − TB = 0.25M From a free-body diagram of the brake handle, the moment equilibrium equation gives ΣM C = 0 : 3 (TA cos 30° ) − 15P = 0 P = 0.2TA cos 30° β = 270° = 3π 2 rad (a) If M = +200 lb ⋅ in. (the drum rotates clockwise), then the tension TA must be bigger than TB and the belt friction equation gives TA = TB e 0.25( 3π 2 ) = 3.24819TB TA = 72.240 lb TB = 22.240 lb TB must be bigger than TA and TB = 72.240 lb P = 12.51 lb ...................................................................................................................... Ans. (b) If M = −200 lb ⋅ in. (the drum rotates counterclockwise), then the tension the belt friction equation gives TB = TAe 0.25( 3π 2 ) = 3.24819TA TA = 22.240 lb P = 3.85 lb ....................................................................................................................... Ans. 261 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS Chapter 7 7-1 (a) (b) For the steel shaft shown in Fig. P7-1 Determine the torques transmitted by cross-sections in intervals AB, BC, CD, and DE of the shaft. Draw a torque diagram for the shaft. SOLUTION (a) Free-body diagrams for parts of the shaft to the left of sections in the intervals AB, BC, CD, and DE of the shaft are shown below. From the free-body diagrams, moment equilibrium about the axis of the shaft gives ΣM axis = 0 : TAB − 80 = 0 TBC − 80 + 100 = 0 TCD − 80 + 100 − 40 = 0 TDE − 80 + 100 − 40 − 25 = 0 TAB = 80 kip ⋅ ft ........................................ Ans. ΣM axis = 0 : ΣM axis = 0 : ΣM axis = 0 : TBC = −20 kip ⋅ ft ..................................... Ans. TCD = 20 kip ⋅ ft ........................................ Ans. TDE = 45 kip ⋅ ft ........................................ Ans. (b) A torque diagram for the shaft is shown below the free-body diagrams............................................. Ans. 7-2 (a) (b) For the steel shaft shown in Fig. P7-2 Determine the maximum torque transmitted by any transverse cross section of the shaft. Draw a torque diagram for the shaft. SOLUTION (a) Free-body diagrams for parts of the shaft to the left of sections in the intervals AB, BC, CD, and DE of the shaft are shown below. From the free-body diagrams, moment equilibrium about the axis of the shaft gives ΣM axis = 0 : TAB = 30 kN ⋅ m ΣM axis = 0 : TBC = −10 kN ⋅ m ΣM axis = 0 : TCD = 5 kN ⋅ m ΣM axis = 0 : TAB − 30 = 0 TBC − 30 + 40 = 0 TCD − 30 + 40 − 15 = 0 TDE − 30 + 40 − 15 − 5 = 0 TDE = 10 kN ⋅ m Tmax = TAB = 30 kN ⋅ m .......................................................................................................... Ans. (b) A torque diagram for the shaft is shown below the free-body diagrams............................................. Ans. 262 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 7-3 (a) (b) RILEY, STURGES AND MORRIS For the steel shaft shown in Fig. P7-3 Determine the maximum torque transmitted by any transverse cross section of the shaft. Draw a torque diagram for the shaft. SOLUTION ΣM axis = 0 : −10 − 15 + T + 15 − 20 = 0 T = 30 kip ⋅ ft (a) Free-body diagrams for parts of the shaft to the left of sections in the intervals AB, BC, CD, and DE of the shaft are shown below. From the free-body diagrams, moment equilibrium about the axis of the shaft gives ΣM axis = 0 : TAB = 10 kip ⋅ ft ΣM axis = 0 : TBC = 25 kip ⋅ ft ΣM axis = 0 : TCD = −5 kip ⋅ ft ΣM axis = 0 : TAB − 10 = 0 TBC − 10 − 15 = 0 TCD − 10 − 15 + 30 = 0 TDE − 10 − 15 + 30 + 15 = 0 TDE = −20 kip ⋅ ft Tmax = TBC = 25 kip ⋅ ft ........................................................................................................... Ans. (b) A torque diagram for the shaft is shown below the free-body diagrams............................................. Ans. 7-4 The motor shown in Fig. P7-4 supplies a torque of 500 N-m to the 25-mm-diameter shaft BCDE. The torques removed at gears C, D, and E are 100 N-m, 150 N-m, and 250 N-m, respectively. (a) Determine the torques transmitted by cross sections in intervals BC, CD, and DE of the shaft. (b) Draw a torque diagram for the shaft. (c) Determine the maximum shearing stress in the shaft. SOLUTION (a) Free-body diagrams for parts of the shaft to the left of sections in the intervals BC, CD, and DE of the shaft are shown below. From the free-body diagrams, moment equilibrium about the axis of the shaft gives ΣM axis = 0 : ΣM axis = 0 : ΣM axis = 0 : TBC − 500 = 0 TCD − 500 + 100 = 0 TDE − 500 + 100 + 150 = 0 TBC = 500 N ⋅ m ......................................................Ans. TCD = 400 N ⋅ m ......................................................Ans. TDE = 250 N ⋅ m ......................................................Ans. (b) A torque diagram for the shaft is shown below the free-body diagrams............................................. Ans. (c) τ max = Tc ( 500 )( 0.0125 ) = = 163.0 × 106 N/m 2 = 163.0 MPa .................................. Ans. 4 J π ( 0.0125 ) 2 263 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 7-5 RILEY, STURGES AND MORRIS A solid circular steel shaft 1-in. in diameter is subjected to a torque of 9000 lb-in. The modulus of rigidity (shear modulus) for the steel is 12,000 ksi. Determine (a) The maximum shearing stress in the shaft. (b) The magnitude of the angle of twist in a 4-ft length. SOLUTION (a) τ= θ= Tc ( 9000 )( 0.5) = = 45,837 psi ≅ 45.8 ksi ............................................................. Ans. 4 J π (1) 32 (b) ( 9000 )( 4 ×12 ) = 0.367 rad ................................................................. Ans. TL = GJ (12 × 106 ) π (1)4 32 7-6 (a) (b) (c) (d) A hollow steel shaft has an outside diameter of 150 mm and an inside diameter of 100 mm. The shaft is subjected to a torque of 35 kN-m. The modulus of rigidity (shear modulus) for the steel is 80 GPa. Determine The shearing stress on a transverse cross section at the outside surface of the shaft. The shearing stress on a transverse cross section at the inside surface of the shaft. The magnitude of the angle of twist in a 2.5-m length. The magnitude of the angle of twist in a 2.5-m long solid shaft that has the same weight as the hollow shaft. SOLUTION J = π ( d o4 − di4 ) 32 = π (1504 − 1004 ) 32 = 39.88 × 106 mm 4 = 39.88 × 10−6 m 4 (a) τ= (b) (c) (d) ( 35 ×103 ) ( 2.5) = 0.0274 rad .......................................................... Ans. TL = θ= GJ ( 80 ×109 )( 39.88 ×10−6 ) A = π (150 2 − 100 2 ) 4 = π d 2 4 4 3 T ρ ( 35 ×10 ) ( 0.050 ) = = 43.88 × 106 N/m 2 ≅ 43.9 MPa .................................. Ans. τ= −6 39.88 ×10 J 3 Tc ( 35 ×10 ) ( 0.075 ) = = 65.82 × 106 N/m 2 ≅ 65.8 MPa .................................... Ans. −6 39.88 ×10 J d = 111.80 mm J = π d 4 32 = π (111.80 ) 32 = 15.338 ×106 mm 4 = 15.338 × 10−6 m 4 ( 35 ×103 ) ( 2.5) = 0.0713 rad ......................................................... Ans. TL = θ= GJ ( 80 ×109 )(15.338 ×10−6 ) 7-7 A torque of 30,000 lb-in. is supplied to the 3-in. diameter factory drive shaft of Fig. P7-7 by a belt that drives pulley A. A torque of 18,000 lb-in. is taken off by pulley B and the remainder by pulley C. Shafts AB and BC are 5 ft and 3.75 ft long, respectively. Both shafts are made of steel (G = 12,000 ksi). Determine (a) The maximum shearing stress in each of the shafts. (b) The magnitude of the angle of twist of pulley B with respect to pulley A. (c) The magnitude of the angle of twist of pulley C with respect to pulley A. SOLUTION J = π d 4 32 = π ( 3) 32 = 7.952 in 4 4 264 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS TAB = 30, 000 lb ⋅ in. TBC = 30,000 − 18, 000 = 12, 000 lb ⋅ in. (a) τ AB = Tc ( 30,000 )(1.5) = J 7.952 = 5659 psi ≅ 5.66 ksi ....................................... Ans. 7.952 TL ( 30, 000 )( 5 ×12 ) = = = 0.018863 rad = 0.01886 rad ................................ Ans. GJ (12 ×106 ) ( 7.952 ) τ BC = (b) (12, 000 )(1.5) = 2264 psi ≅ 2.26 ksi .................................................................... Ans. θB/ A (c) θC / B = (12,000 )( 3.75 ×12 ) = 0.005659 rad (12 106 ) ( 7.952 ) θ C / A = θ B / A + θ C / B = 0.018863 + 0.005659 = 0.0245 rad ............................................. Ans. 7-8 A torque of 10 kN-m is supplied to the steel (G = 80 GPa) factory drive shaft of Fig. P7-8 by a belt that drives pulley A. A torque of 6 kN-m is taken off by pulley B and the remainder by pulley C. Shafts AB and AC are 2.25 m and 1.60 m long, respectively. If the diameter of shaft AB is 80 mm and the diameter of shaft AC is 65 mm, determine (a) The maximum shearing stress in each of the shafts. (b) The angle of twist of pulley B with respect to pulley A. (c) The angle of twist of pulley C with respect to pulley B. SOLUTION J AB = π d 4 32 = π ( 80 ) 32 = 4.021× 106 mm 4 4 J AC = π ( 65 ) 32 = 1.7525 ×106 mm 4 4 TAB = 6 kN ⋅ m TAC = 6 − 10 = −4 kN ⋅ m 3 Tc ( 6 ×10 ) ( 0.040 ) = = = 59.69 × 106 N/m 2 ≅ 59.7 MPa .................................. Ans. −6 4.021×10 J (a) τ AB τ AC = (b) ( 4 ×10 ) ( 0.0325) = 74.18 ×10 3 6 θ A/ B ( 6 ×103 ) ( 2.25) = 0.04197 rad ≅ 0.0420 rad ......................... Ans. TL = = GJ ( 80 ×109 )( 4.021×10−6 ) ( −4 ×103 ) (1.60 ) = −0.04565 rad ≅ −0.0456 rad TL = GJ ( 80 ×109 )(1.7525 ×10−6 ) .................................. Ans. 1.7525 ×10 −6 N/m 2 ≅ 74.2 MPa .......................................... Ans. (c) θC / A = θ C / B = θ C / A + θ A / B = ( −0.04565 ) + ( 0.04197 ) = −0.00368 rad 265 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 7-9 RILEY, STURGES AND MORRIS A solid circular aluminum alloy (G = 4000 ksi) shaft with diameters of 2.5 in. and 1.75 in. is subjected to a torque T, as shown in Fig. P7-9. The shearing stress is limited to 8000 psi, and the angle of twist in the 7-ft length cannot exceed 0.04 rad. Determine the maximum permissible value of T. SOLUTION J AB = π d 4 32 = π ( 2.5 ) 32 = 3.835 in 4 4 J BC = π (1.75 ) 32 = 0.9208 in 4 4 Tc T ( 0.875 ) = ≤ 8000 psi T ≤ 8419 lb ⋅ in. J 0.9208 T ( 3 ×12 ) T ( 4 ×12 ) TL θ =∑ = + ≤ 0.04 rad 6 GJ ( 4 ×10 ) ( 3.835) ( 4 ×106 ) ( 0.9208 ) τ max = τ BC = T ≤ 2601 lb ⋅ in. Tmax = 2601 lb ⋅ in. ≅ 2.60 kip ⋅ in. ....................................................................................... Ans. 7-10 The solid circular steel (G = 80 GPa) shaft of Fig. P7-10 has a diameter of 80 mm. If the gears are spaced at 1.25-m intervals, determine (a) The maximum shearing stress in the shaft. (b) The rotation of a section at D with respect to a section at B. (c) The rotation of a section at E with respect to a section at A. SOLUTION ΣM axis = 0 : −8 + T − 15 − 6 + 14 = 0 4 T = 15 kN ⋅ m J = π d 4 32 = π ( 80 ) 32 = 4.021× 106 mm 4 (a) Tmax = TDE = 14 kN ⋅ m 3 Tc (14 × 10 ) ( 0.040 ) = = 139.27 × 106 N/m 2 ≅ 139.3 MPa ............................ Ans. τ DE = J 4.021×10−6 TL TL θ D / B = θ D / C + θC / B = + GJ CD GJ BC (b) (8 ×10 ) (1.25) + ( −7 ×10 ) (1.25) = (80 ×10 )( 4.021×10 ) (80 ×10 )( 4.021×10 ) 3 3 9 −6 9 −6 (c) = 0.00389 rad ................................................................................................................ Ans. TL θ E / A = ∑θ = ∑ GJ (8 ×10 ) (1.25) + ( −7 ×10 ) (1.25) = (80 ×10 )( 4.021×10 ) (80 ×10 )( 4.021×10 ) (8 ×10 ) (1.25) + (14 ×10 ) (1.25) + (80 ×10 )( 4.021×10 ) (80 ×10 )( 4.021×10 ) 3 3 9 −6 9 −6 3 3 9 −6 9 −6 = 0.0894 rad .................................................................................................................. Ans. 266 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 7-11 RILEY, STURGES AND MORRIS The shaft shown in Fig. P7-11 consists of a brass (G = 5600 ksi) tube AB that is securely connected to a solid stainless steel (G = 12,500 ksi) bar BC. Tube AB has an outside diameter of 5 in. and an inside diameter of 2.5 in. Bar BC has a diameter of 3.5 in. Torques T1 and T2 are 100 kip-in. and 40 kip-in., respectively, in the directions shown. Determine (a) The maximum shearing stress in the shaft. (b) The rotation of a section at C with respect to its no-load position. SOLUTION J AB = π ( 54 − 2.54 ) 32 = 57.52 in 4 J BC = π ( 3.5 ) 32 = 14.732 in 4 4 TAB = 100 − 40 = 60 kip ⋅ in. TBC = −40 kip ⋅ in. (a) (b) Tc ( 60 )( 2.5) = = 2.608 ksi J 57.52 ( 40 )(1.75) = 4.752 ksi = τ ................................................................................. Ans. τ BC = max 14.732 TL θ C / A = ∑θ = ∑ GJ τ AB = = 7-12 ( 60 )( 2 ×12 ) + ( −40 )( 3 ×12 ) = −0.00335 rad ...................................... Ans. ( 5600 )( 57.52 ) (12,500 )(14.732 ) A motor supplies a torque of 5.5 kN-m to the constant diameter steel (G = 80 GPa) line shaft shown in Fig. P7-12. Three machines are driven by gears B, C, and D on the shaft and they require torques of 3 kN-m, 1.5 kN-m, and 1 kN-m, respectively. Determine (a) The minimum diameter required if the maximum shearing stress in the shaft is limited to 100 MPa. (b) The rotation of gear D with respect to the coupling at A if the coupling and gears are spaced at 2-m intervals and the shaft diameter is 75 mm. SOLUTION TAB = 5.5 kN ⋅ m TBC = 5.5 − 3 = 2.5 kN ⋅ m TCD = 5.5 − 3 − 1.5 = 1 kN ⋅ m 3 Tmax c ( 5.5 × 10 ) ( d 2 ) = = = 100 × 106 N/m 2 π d 4 32 J (a) τ max d min = (b) 3 16 ( 5.5 × 103 ) π (100 × 10 4 6 ) = 65.43 × 10−3 m = 65.4 mm ..................................................... Ans. J = π ( 75 ) 32 = 3.106 ×106 mm 4 = 3.106 × 10−6 m 4 θ D / A = ∑θ = ∑ TL GJ 267 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 3 3 3 RILEY, STURGES AND MORRIS ( 5.5 ×10 ) ( 2 ) + ( 2.5 ×10 ) ( 2 ) + (1×10 ) ( 2 ) = 0.0724 rad ........................... Ans. = (80 ×10 )( 3.106 ×10 ) 9 −6 7-13 The hollow circular steel (G = 12,000 ksi) shaft of Fig. P7-13 is in equilibrium under the torques indicated. Determine (a) The maximum shearing stress in the shaft. (b) The rotation of a section at D with respect to a section at A. SOLUTION J = π ( 44 − 2 4 ) 32 = 23.56 in 4 −9 + Q − 21 + 10 = 0 Q = 20 kip ⋅ ft ΣM axis = 0 : TAB = 9 kip ⋅ ft (a) TBC = 9 − 20 = −11 kip ⋅ ft TCD = 9 − 20 + 21 = 10 kip ⋅ ft (b) Tc (11× 12 )( 2 ) = = 11.21 ksi ...................................................................... Ans. J 23.56 TL θ D / A = ∑θ = ∑ GJ (10 ×12 )( 4 ×12 ) + ( −11×12 )( 5 ×12 ) + ( 9 ×12 )( 3 ×12 ) = (12, 000 )( 23.56 ) (12, 000 )( 23.56 ) (12, 000 )( 23.56 ) = 0.00611 rad ................................................................................................................ Ans. Tmax = TBC = 11 kip ⋅ ft τ max = τ BC = 7-14 The solid circular shaft and the hollow tube shown in Fig. P7-14 are both attached to a rigid circular plate at their left ends. A torque TA = 2 kN-m applied to the right end of the shaft is resisted by a torque TB at the right end of the tube. The shaft is made of steel (G = 80 GPa) and the tube is made of an aluminum alloy (G = 28 GPa). If the shaft has a diameter of 50 mm and the tube has an outside diameter of 80 mm, determine (a) The maximum inside diameter that can be used for the tube if the maximum shearing stress in the tube must be limited to 50 MPa. (b) The maximum inside diameter that can be used for the tube if the rotation of the right end of the shaft with respect to the right end of the tube must be limited to 0.25 rad. SOLUTION (a) Tc ( 2 × 10 ) ( 0.040 ) τ= = = 50 ×106 N/m 2 J JB 3 J B = 1.600 ×10 −6 m 4 = 1.600 × 106 mm 4 = π ( 804 − di4 ) 32 di = 70.47 mm ≅ 70.5 mm ................................................................................................. Ans. (b) J A = π ( 50 ) 32 = 0.6136 ×106 mm 4 = 0.6136 × 10−6 m 4 4 θ A/ B J B = 1.6627 × 10−6 m 4 = 1.6627 × 106 mm 4 = π ( 804 − di4 ) 32 ( 2 ×103 ) ( 3.5) ( 2 ×103 ) ( 2.5) = 0.25 rad TL = ∑θ = ∑ = + GJ ( 80 × 109 )( 0.6136 ×10−6 ) ( 28 ×109 ) J B di = 70.01 mm ≅ 70.0 mm .................................................................................................. Ans. 268 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 7-15 RILEY, STURGES AND MORRIS The 4-in. diameter shaft shown in Fig. P7-15 is composed of brass (G = 5000 ksi) and steel (G = 12,000 ksi) sections rigidly connected. Determine the maximum allowable torque, applied as shown, if the maximum shearing stresses in the brass and the steel are not to exceed 7500 psi and 10,000 psi, respectively, and the distance AC, through which the end of the 10-in. pointer AB moves, is not to exceed 0.60 in. SOLUTION J = π ( 4 ) 32 = 25.13 in 4 4 Tc T ( 2 ) = ≤ 10, 000 psi T ≤ 125, 650 lb ⋅ in. J 25.13 Tc T ( 2 ) τB = = ≤ 7500 psi T ≤ 94, 238 lb ⋅ in. J 25.13 T ( 6 × 12 ) T ( 4 × 12 ) s 0.60 θ = θB +θS = + =≤ = 0.060 rad 6 6 ( 5 ×10 ) ( 25.13) (12 ×10 ) ( 25.13) r 10 τS = T ≤ 81,946 lb ⋅ in. Tmax = 81,946 lb ⋅ in. ≅ 81.9 kip ⋅ in. ................................................................................... Ans. 7-16 A stepped steel (G = 80 GPa) shaft has the dimensions and is subjected to the torques shown in Fig. P7-16. Determine (a) The maximum shearing stress on a section 3 m from the left end of the shaft. (b) The rotation of the section at the right end of the shaft with respect to its no-load position. SOLUTION J AC = π (160 ) 32 = 64.34 ×106 mm 4 4 J CD = π (100 ) 32 = 9.817 × 106 mm 4 4 J DE = π ( 50 ) 32 = 0.6136 × 106 mm 4 4 ΣM axis = 0 : TA − 35 + 100 − 20 − 5 = 0 TA = −40 kN ⋅ m TAB = 40 kN ⋅ m TBC = 40 + 35 = 75 kN ⋅ m TCD = 40 + 35 − 100 = −25 kN ⋅ m TDE = 40 + 35 − 100 + 20 = −5 kN ⋅ m (a) 3 Tc ( 25 × 10 ) ( 0.050 ) τ3 = = = 127.33 × 106 N/m 2 ≅ 127.3 MPa .............................. Ans. −6 J 9.817 × 10 (b) θE/ A ( 40 ×103 ) (1) ( 75 ×103 ) (1.5) TL =∑ = + GJ ( 80 × 109 )( 64.34 × 10−6 ) ( 80 ×109 )( 64.34 ×10 −6 ) 3 3 9 −6 9 ( −25 ×10 ) ( 2 ) + ( −5 ×10 ) ( 0.5) + (80 ×10 )( 9.817 ×10 ) (80 ×10 )( 0.6136 ×10 ) −6 = −0.0850 rad ............................................................................................ Ans. 269 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 7-17 RILEY, STURGES AND MORRIS A torque T is applied to the right end of shaft AB of Fig. P7-17. The mean diameter of bevel gear C is twice that of bevel gear B. Both shafts are made of steel (G = 12,000 ksi). Shaft AB has a diameter of 1.5 in., and shaft CD has a diameter of 2.0 in. If the maximum shearing stress in either shaft must not exceed 15 ksi, determine (a) The maximum permissible torque T. (b) The rotation of a section at A relative to its no-load position. SOLUTION J AB = π (1.5 ) 32 = 0.4970 in 4 4 J CD = π ( 2.0 ) 32 = 1.5708 in 4 4 (a) τ AB = τ CD Tc TAB ( 0.75 ) = ≤ 15, 000 psi J 0.4970 Tc TCD (1) = = ≤ 15, 000 psi J 1.5708 TAB ≤ 9940 lb ⋅ in. TCD = 2TAB ≤ 23,562 lb ⋅ in. TAB ≤ 11, 781 lb ⋅ in. T = TAB max = 9940 lb ⋅ in. ≅ 9.94 kip ⋅ in. ........................................................................... Ans. (b) TCD = 2TAB = 2 ( 9940 ) = 19,880 lb ⋅ in. θ A = θ A / B + 2θ C / D = ( 9940 )( 4 ×12 ) + 2 (19,880 )( 3 ×12 ) (12 ×106 ) ( 0.4970 ) (12 ×106 ) (1.5708) = 0.1559 rad ...................................................................................................................... Ans. 7-18 Torque is applied to the steel (G = 80 GPa) shaft shown in Fig. P7-18 through gear C and is removed through gears A and B. If the torque applied to gear C by the motor is 20 kN-m and the torque removed through gear B is 12 kN-m, determine (a) The minimum permissible diameter for each section of the shaft if the maximum shearing stresses must not exceed 125 MPa. (b) The minimum permissible uniform diameter for a shaft with L1 = 1.5 m and L2 = 1.25 m if the rotation of gear A relative to gear C must be less than 0.15 rad. SOLUTION TBC = 20 kN ⋅ m (a) TAB = 20 − 12 = 8 kN ⋅ m d= 3 τ= Tc T ( d 2 ) 16T = = J π d 4 32 π d 3 3 16T πτ d AB = d BC = (b) π (125 × 106 ) π (125 × 106 ) 16 ( 20, 000 ) 16 ( 8000 ) = 0.0688 m = 68.8 mm ............................................................... Ans. 3 = 0.0934 m = 93.4 mm ............................................................... Ans. θ= TL TL 32TL = = GJ G (π d 4 32 ) Gπ d 4 θ A/C = θ A/ B + θ B /C = 32 ( 8 × 103 ) (1.5 ) (80 ×10 )(π d ) 9 4 + 32 ( 20 × 103 ) (1.25) (80 ×10 )(π d ) 9 4 = 0.15 rad d = 0.0749 m = 74.9 mm ............................................................................................. Ans. 270 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 7-19 RILEY, STURGES AND MORRIS A torque of 40 lb-ft is applied through gear A to the left end of the gear train shown in Fig. P7-19. The diameters of gears B and C are 5 in. and 2 in., respectively. If the maximum shearing stresses in the aluminum alloy (G = 3800 ksi) shafts AB and CD are limited to 15 ksi, determine (a) The minimum permissible diameter for shaft AB. (b) The minimum permissible diameter for shaft CD. (c) The maximum length for shaft CD if the rotation of a section at D with respect to a section at C must not exceed 0.5 rad. SOLUTION TAB = 40 lb ⋅ ft (a) TCD = ( 2 5) TAB = 16 lb ⋅ ft d= 3 τ= Tc T ( d 2 ) 16T = = J π d 4 32 π d 3 3 16T πτ d AB = dCD = 16 ( 40 × 12 ) = 0.5462 in. ≅ 0.546 in. ................................................................. Ans. π (15, 000 ) 16 (16 × 12 ) = 0.40246 in. ≅ 0.402 in. ............................................................... Ans. π (15, 000 ) (b) 3 (c) θD/C = (15, 000 )( L ×12 ) = 0.5 rad τL = Gc ( 3.8 × 106 ) ( 0.40246 2 ) L = 2.12 ft ........................................................................................................................ Ans. 7-20 The motor shown in Fig. P7-20 supplies a torque of 45 kN-m to shaft AB. Two machines are powered by gears D and E. The torque delivered by gear E to the machine is 8 kN-m. Shafts AB and DCE are made of steel (G = 80 GPa) and have 150-mm and 80-mm diameters, respectively. If the diameters of gears B and C are 450 mm and 150 mm, respectively, determine (a) The maximum shearing stress in shaft AB. (b) The maximum shearing stress in shaft DCE. (c) The rotation of gear E relative to gear D. SOLUTION J AB = π (150 ) 32 = 49.70 × 106 mm 4 4 J DCE = π ( 80 ) 32 = 4.021× 106 mm 4 4 TC = (1 3) TB = (1 3)( 45) = 15 kN ⋅ m TCD = 15 − 8 = 7 kN ⋅ m TCE = 8 kN ⋅ m (a) τ AB τ CE 3 Tc ( 45 × 10 ) ( 0.075 ) = = = 67.91× 106 N/m 2 ≅ 67.9 MPa ................................ Ans. −6 J 49.70 ×10 (b) (8 ×10 ) ( 0.040 ) = 79.58 ×10 = 3 6 4.021×10−6 N/m 2 ≅ 79.6 MPa ............................................. Ans. 3 (c) θ E / D = θ E /C −θ D /C (8 ×10 ) ( 2.5) − ( 7 ×10 ) (1.5) = (80 ×10 )( 4.021×10 ) (80 ×10 )( 4.021×10 ) 3 9 −6 9 −6 = 0.0295 rad .................................................................................................................. Ans. 271 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 7-21 RILEY, STURGES AND MORRIS The hollow circular tapered shaft of Fig. P7-21 is subjected to a constant torque T. Determine the angle of twist in terms of T, L, G, and r. SOLUTION With the left end of the shaft at x = L: ρ1 = J= 4 π ( ρ14 − ρ 2 ) rx L ρ2 = rx 2L 2 = 15π r 4 x 4 32 L4 dθ = T dx 32TL4 dx = GJ 15π Gr 4 x 4 2L θ = ∫ dθ = L 32TL4 15π Gr 4 ∫ 2L L dx 28TL = ........................................................................... Ans. 4 45π Gr 4 x 7-22 The hollow tapered shaft of Fig. P7-22 has a constant wall thickness t. Determine the angle of twist for a constant torque T in terms of T, L, G, t, and r. Not that when t is small, the approximate expression for the polar second moment of area (J = r2A, where A is the cross-sectional area of the shaft) may be used. SOLUTION With the left end of the shaft at 2 2 x = L: 3 ρ= rx L 33 rx 2π tr x J = ρ A = ρ ( 2πρ t ) = 2π t = L3 L dθ = T dx TL3 dx = 2π tGr 3 x3 GJ 2L θ =∫ 7-23 L TL3 dθ = 2π tGr 3 ∫ 2L L dx 3TL = ........................................................................... Ans. 3 x 16π Gtr 3 The solid cylindrical shaft of Fig. P7-23 is subjected to a uniformly distributed torque of q lb-in. per inch of length. Determine, in terms of q, L, G, and c, the rotation of the left end caused by the applied torque. SOLUTION T = qx dθ = T dx 2qx dx = π Gc 4 GJ L 0 J= π c4 2 θ =∫ 7-24 dθ = 2q π Gc 4 ∫ L 0 x dx = qL2 .................................................................................... Ans. π Gc 4 The solid cylindrical shaft of Fig. P7-24 is subjected to a distributed torque that varies linearly from zero at the left end to q N-m per meter of length at the right end. Determine, in terms of q, L, G, and c, the rotation of the left end caused by the applied torque. SOLUTION T= qx 2 2L L 0 J= q π GLc 4 π c4 2 dθ = T dx qx 2 dx = GJ π GLc 4 θ = ∫ dθ = ∫ L 0 x 2 dx = qL2 .............................................................................. Ans. 3π Gc 4 272 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS τ max = Tmax 7-29 T ( 0.045) ≤ 80 × 106 N/m 2 T ≤ 10, 361 N ⋅ m −6 5.828 ×10 = 9713 N ⋅ m ≅ 9.71 kN ⋅ m ......................................................................................... Ans. A cylindrical tube is fabricated by butt-welding 0.075-in. plate with a spiral seam, as shown in Fig. P7-29. A torque T is applied to the tube through a rigid plate. If the outside diameter of the tube is 1.50 in. and T = 1000 lb-in., determine (a) The normal stress perpendicular to the weld and the shear stress parallel to the weld. (b) The maximum tensile and compressive stresses in the tube. SOLUTION (a) J = π (1.504 − 1.354 ) 32 = 0.17092 in 4 τ xy = Tc − (1000 )( 0.75 ) = = −4388 psi 0.17092 J σ n = 2τ xy sin θ cosθ = 2 ( −4388) sin ( −55° ) cos ( 55° ) = 4123 psi ≅ 4.12 ksi (T) .............................................................................................. Ans. τ nt = τ xy ( cos 2 θ − sin 2 θ ) = ( −4388) cos 2 ( −55° ) − sin 2 ( −55° ) = 1500.8 psi ≅ 1.501 ksi (b) 7-30 ............................................................................................ Ans. σ max = τ max = Tc = 4388 psi (T & C) ............................................................................... Ans. J The hollow circular steel (G = 80 GPa) shaft shown in Fig. P7-30 has an outside diameter of 120 mm and an inside diameter of 60 mm. Determine (a) The maximum compressive stress in the shaft. (b) The maximum compressive stress in the shaft after the inside diameter is increased to 100 mm. SOLUTION T = ( 5000 )(1.5 ) = 7500 N ⋅ m (a) J = π (1204 − 604 ) 32 = 19.085 × 106 mm 4 = 19.085 × 10 −6 m 4 σ max = τ max = (b) J = π (1204 − 1004 ) 32 = 10.540 ×106 mm 4 = 10.540 × 10 −6 m 4 Tc 7500 ( 0.060 ) = = 23.58 × 106 N/m 2 ≅ 23.6 MPa (C) .................. Ans. J 19.085 ×10−6 Tc 7500 ( 0.060 ) = = 42.69 ×106 N/m 2 ≅ 42.7 MPa (C) .................. Ans. J 10.540 ×10−6 σ max = τ max = 7-31 The motor shown in Fig. P7-31 develops a torque T = 1500 lb-ft; the output torques from gears C and D are equal. The mean diameters of gears A and B are 12 in. and 4 in., respectively. If the diameter of the motor shaft is 2 in. and the diameter of the power shaft is 1.25 in., determine (a) The torque in the power shaft between gears B and C. (b) The torque in the power shaft between gears C and D. (c) The maximum tensile and compressive stresses in each shaft. SOLUTION TB = (1 3) TA = (1 3)(1500 ) = 500 lb ⋅ ft 275 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS TC = TD = (1 2 ) TB = (1 2 )( 500 ) = 250 lb ⋅ ft (a) (b) (c) TBC = TB = 500 lb ⋅ ft .......................................Ans. TCD = TB − TC = 500 − 250 = 250 lb ⋅ ft .......Ans. J P = π (1.25 ) 32 = 0.2397 in 4 4 J M = π ( 2 ) 32 = 1.5708 in 4 4 σ max ( motor ) = τ max ( motor ) = Tc (1500 × 12 )(1) = 1.5708 J = 11, 459 psi ≅ 11.46 ksi (T & C) ............................................................. Ans. Tc ( 500 × 12 )( 0.625 ) = 0.2397 J = 15, 644 psi ≅ 15.64 ksi (T & C) ............................................................. Ans. σ max ( power ) = τ max ( power ) = 7-32 Five 600-mm diameter pulleys are keyed to a 40-mm diameter solid steel (G = 76 GPa) shaft, as shown in Fig. P7-32. The pulleys carry belts that are used to drive machinery in a factory. Belt tensions for normal operating conditions are indicated on the figure. Each segment of the shaft is 1.5 m long. Determine (a) The maximum shearing stress in each segment of the shaft. (b) The maximum tensile and compressive stresses in the shaft. (c) The rotation of end E with respect to end A. SOLUTION ΣM axis = 0 : ΣM axis = 0 : ΣM axis = 0 : ΣM axis = 0 : TAB − ( 2500 − 500 )( 0.300 ) = 0 TBC − 600 + ( 2000 − 400 )( 0.300 ) = 0 TCD − 600 + 480 − ( 3000 − 600 )( 0.300 ) = 0 TDE − 600 + 480 − 720 + ( 3000 − 600 )( 0.300 ) = 0 TAB = 600 N ⋅ m TBC = 120 N ⋅ m TCD = 840 N ⋅ m TDE = 120 N ⋅ m J = π ( 40 ) 32 = 0.2513 × 106 mm 4 4 (a) τ AB = τ BC τ CD τ DE Tc 600 ( 0.020 ) = = 47.8 × 106 N/m 2 = 47.8 MPa ........................................... Ans. −6 J 0.2513 ×10 120 ( 0.020 ) = = 9.55 × 106 N/m 2 = 9.55 MPa ..................................................... Ans. −6 0.2513 ×10 840 ( 0.020 ) = = 66.9 × 106 N/m 2 = 66.9 MPa ..................................................... Ans. −6 0.2513 ×10 120 ( 0.020 ) = = 9.55 × 106 N/m 2 = 9.55 MPa ..................................................... Ans. −6 0.2513 ×10 276 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (b) (c) RILEY, STURGES AND MORRIS σ max = τ max = τ CD = 66.9 MPa (T & C) θE/ A = ∑ 120 (1.5 ) 840 (1.5 ) TL = + GJ ( 76 × 109 )( 0.2513 ×10−6 ) ( 76 ×109 )( 0.2513 ×10 −6 ) + ( 76 ×10 )( 0.2513 ×10 ) ( 76 ×10 )( 0.2513 ×10 ) 9 120 (1.5 ) −6 + 600 (1.5 ) 9 −6 = 0.1319 rad ................................................................................................ Ans. 7-33 When the two torques are applied to the steel (G = 12,000 ksi) shaft of Fig. P7-33, point A moves 0.172 in. in the direction indicated by torque T1. Determine (a) The torque T1. (b) The maximum tensile stress in section BC of the shaft. (c) The maximum compressive stress in section CD of the shaft. SOLUTION J BC = π ( 2 ) 32 = 1.5708 in 4 4 J CD = π ( 4 ) 32 = 25.13 in 4 4 (a) θB/ D = ΣM axis = 0 : 0.172 = 0.02867 rad 6 TL − 9000 + T1 = 0 TL = 9000 − T1 lb ⋅ ft θB/ D = ∑ (T1 ×12 )( 24 ) = 0.02867 rad TL − ( 9000 − T1 )(12 )( 48 ) = + 6 GJ (12 ×10 ) ( 25.13) (12 ×106 ) (1.5708) T1 = 2668 lb ⋅ ft = 32, 016 lb ⋅ in. ≅ 32.0 kip ⋅ in. ......................................................... Ans. (b) TBC = T1 = 32, 016 lb ⋅ in. σ max = (c) Tc ( 32, 016 )(1) = = 20,382 psi ≅ 20.38 ksi (T) .............................................. Ans. 1.5708 J Tc ( 75,984 )( 2 ) = = 6047 psi ≅ 6.05 ksi (C) ................................................... Ans. 25.13 J TCD = TL = ( 9000 × 12 ) − 32, 016 = 75,984 lb ⋅ in. σ max = 7-34 A solid circular stepped steel (G = 80 GPa) shaft has the dimensions and is subjected to the torques shown in Fig. P7-34. Determine (a) The maximum tensile stress in section AB of the shaft. (b) The maximum compressive stress in section BC of the shaft. (c) The rotation of a section at C with respect to its no-load position. SOLUTION (a) J AB = π (150 ) 32 = 49.70 × 106 mm 4 = 49.70 ×10 −6 m 4 4 σ max = τ max (b) 3 Tc ( 55 × 10 ) ( 0.075) = = = 83.00 ×106 N/m 2 = 83.0 MPa (T) ........... Ans. −6 J 49.70 ×10 4 J BC = π (100 ) 32 = 9.817 × 106 mm 4 = 9.817 × 10 −6 m 4 277 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS σ max = τ max (c) θC / A ( 55 ×103 ) ( 0.300 ) + (15 ×103 ) ( 0.400 ) TL =∑ = GJ ( 80 × 109 )( 49.70 ×10−6 ) ( 80 ×109 )( 9.817 ×10 −6 ) 3 Tc (15 × 10 ) ( 0.050 ) = = = 76.40 ×106 N/m 2 = 76.4 MPa (C) ........... Ans. J 9.817 ×10−6 = 0.01179 rad .............................................................................................. Ans. 7-35 A 175-lb man climbs a flight of stairs 12 ft high. Determine (a) The work done by the man. (b) The work done on the man by gravity. SOLUTION (a) (b) 7-36 U M = F ( ∆h ) = 175 (12 ) = 2100 lb ⋅ ft .............................................................................. Ans. U G = −W ( ∆h ) = −175 (12 ) = −2100 lb ⋅ ft ...................................................................... Ans. A box with a mass of 600 kg is dragged up an incline 12 m long and 4 m high by using a cable that is parallel to the incline. The force in the cable is 2500 N. Determine the work done on the box (a) By the cable. (b) By gravity. SOLUTION (a) (b) 7-37 U C = Fd = 2500 (12 ) = 30, 000 N ⋅ m = 30.0 kN ⋅ m = 30.0 kJ .................................. Ans. U G = −W ( ∆h ) = − ( 600 × 9.81)( 4 ) = −23,544 N ⋅ m ≅ −23.5 kJ .............................. Ans. A 100-lb block is pushed at constant speed for a distance of 20 ft along a level floor by a force F that makes an angle of 35° with the horizontal, as shown in Fig. P7-37. The coefficient of friction between the block and the floor is 0.35. Determine the work done on the block (a) By the force P. (b) By gravity. (c) By the floor. SOLUTION From a free-body diagram of the block (with the equilibrium equations give AF = µ k AN = 0.35 AN ), → ΣFx = 0 : ↑ ΣFy = 0 : P = 56.60 lb P cos 35° − AF = 0 AN − P sin 35° − 100 = 0 AF = 0.35 (132.47 ) = 46.36 lb AN = 132.47 lb (a) (b) (c) 7-38 U P = ( P cos α ) d = ( 56.60 cos 35° )( 20 ) = 927 lb ⋅ ft .................................................... Ans. U G = W ( ∆h ) = 100 ( 0 ) = 0 lb ⋅ ft ........................................................................................ Ans. U F = − AF d + AN ( y2 − y1 ) = −46.36 ( 20 ) + 132.47 ( 0 ) = −927 lb ⋅ ft ........................ Ans. A block with a mass of 100 kg slides down an inclined surface that is 5 m long and makes an angle of 25° with the horizontal. A man pushes horizontally on the block so that it slides down the incline at a constant speed. The coefficient of friction between the block and the inclined surface is 0.15. Determine the work done on the block (a) By the man. 278 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (b) (c) By gravity. By the inclined surface. RILEY, STURGES AND MORRIS SOLUTION From a free-body diagram of the block (with the equilibrium equations give AF = µ k AN = 0.15 AN ), → ΣFx = 0 : ↑ ΣFy = 0 : P = 290.0 N AN sin 25° − AF cos 25° − P = 0 AN cos 25° + AF sin 25° − 100 ( 9.81) = 0 AN = 1011.7 N AF = 0.15 (1011.7 ) = 151.76 N (a) (b) (c) 7-39 U M = − ( P cos α ) d = − ( 290.0 cos 25° )( 5 ) = −1314 N ⋅ m = −1314 J ..................... Ans. U G = W ( ∆h ) = (100 × 9.81)( 5sin 25° ) = 2073 N ⋅ m = 2073 J .................................. Ans. U S = − AF d = −151.76 ( 5 ) = −758.8 N ⋅ m ≅ −759 J ..................................................... Ans. A constant force acting on a particle can be expressed in Cartesian vector form as particle can be expressed in Cartesian vector form as SOLUTION F = ( 8 i − 6 j + 2 k ) lb . Determine the work done by the force on the particle if the displacement of the s = ( 5 i + 4 j + 6 k ) ft . U = F ⋅ s = ( 8 i − 6 j + 2 k ) ⋅ ( 5 i + 4 j + 6 k ) = 28 lb ⋅ ft .................................................... Ans. 7-40 C = ( 200 i + 300 j + 350 k ) N ⋅ m acts on a rigid body. The unit vector associated with the fixed axis of rotation of the body for an infinitesimal angular displacement d θ is eθ = 0.250i + 0.350 j − 0.903k . Determine the work done on the body by the couple during an angular A constant couple displacement of 1.5 radians. SOLUTION dU = C ⋅ dθ = ( 200 i + 300 j + 350 k ) ⋅ ( 0.250 i + 0.350 j − 0.903 k ) dθ = −161.05 dθ U = ∫ −161.05 dθ = −161.05 [θ ]0 = 242 N ⋅ m = −242 J ..................................... Ans. 1.5 1.5 0 7-41 Determine the horsepower that a 6-in. diameter shaft can transmit at 200 rpm if the maximum shearing stress in the shaft must be limited to 12,000 psi. SOLUTION J = π ( 6 ) 32 = 127.2345 in 4 4 τ max = T ( 3) Tc = = 12,000 psi J 127.2345 3 2π NT 2π ( 200 ) ( 42.4115 × 10 ) = = 1615 hp .................................. Ans. 33, 000 33, 000 T = 508.938 × 103 lb ⋅ in. = 42.4115 ×103 lb ⋅ ft Power = T ω = 279 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 7-42 RILEY, STURGES AND MORRIS A solid circular steel (G = 80 GPa) shaft transmits 200 kW at 180 rpm. Determine the minimum diameter required if the angle of twist in a 2-m length must not exceed 0.025 rad. SOLUTION Power = T ω = 2π NT = 2π (180 60 ) T = 200 × 103 N ⋅ m/s 3 TL (10.61033 ×10 ) ( 2 ) = = 0.025 rad θ= GJ ( 80 × 109 )(π d 4 32 ) T = 10.61033 × 103 N ⋅ m d = 0.1020 m = 102.0 mm .................................................................................................. Ans. 7-43 A steel (G = 12,000 ksi) shaft with a 4-in. diameter must not twist more than 0.06 rad in a 20-ft length. Determine the maximum power that the shaft can transmit at 270 rpm. SOLUTION J = π ( 4 ) 32 = 25.13 in 4 4 θ= T ( 20 × 12 ) TL = = 0.06 rad GJ (12 ×106 ) ( 25.13) 2π NT 2π ( 270 )( 6283) = = 323 hp .................................................... Ans. 33, 000 33, 000 T = 75,390 lb ⋅ in. = 6283 lb ⋅ ft Power = T ω = 7-44 A 3-m-long hollow steel (G = 80 GPa) shaft has an outside diameter of 100 mm and an inside diameter of 60 mm. The maximum shearing stress in the shaft is 80 MPa, and the angular velocity is 200 rpm. Determine (a) The power being transmitted by the shaft. (b) The magnitude of the angle of twist in the shaft. SOLUTION J = π (1004 − 604 ) 32 = 8.545 ×106 mm 4 τ= (a) T ( 0.050 ) Tc = = 80 ×106 N/m 2 −6 J 8.545 × 10 T = 13, 672 N ⋅ m Power = T ω = 2π NT = 2π ( 200 60 )(13, 672 ) (b) 7-45 (80 ×106 ) ( 3) = 0.0600 rad τL = θ= Gc ( 80 × 109 ) ( 0.050 ) = 286.3 ×103 N ⋅ m/s ≅ 286 kw ............................................................................ Ans. ...................................................................... Ans. The hydraulic turbines in a water-power plant rotate at 60 rpm and are rated at 20,000 hp with an overload capacity of 25,000 hp. The 30-in.-diameter shaft between the turbine and the generator is made of steel (G = 12,000 ksi) and is 20 ft long. Determine (a) The maximum shearing stress in the shaft at the rated load. (b) The maximum shearing stress in the shaft at the maximum overload. SOLUTION J = π ( 30 ) 32 = 79.52 × 103 in 4 4 280 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (a) RILEY, STURGES AND MORRIS Power = T ω = 2π NT 2π ( 60 ) TR = = 20, 000 hp 33, 000 33, 000 6 (b) Tc ( 21.01× 10 ) (15) = = 3963 psi ≅ 3.96 ksi ........................................................ Ans. τ= 79.52 × 103 J 2π ( 60 ) TO = 25, 000 hp TO = 2.188 ×106 lb ⋅ ft = 26.26 × 106 lb ⋅ in. Power = 33,000 Tc ( 26.26 × 10 ) (15 ) = = 4953 psi ≅ 4.95 ksi ........................................................ Ans. τ= 79.52 × 103 J 6 TR = 1.7507 ×106 lb ⋅ ft = 21.01×106 lb ⋅ in. 7-46 A solid circular steel (G = 80 GPa) shaft 1.5 m long transmits 200 kW at a speed of 400 rpm. If the allowable shearing stress is 70 MPa and the allowable angle of twist is 0.045 rad, determine (a) The minimum permissible diameter for the shaft. (b) The speed at which this power can be delivered if the shearing stress is not to exceed 50 MPa in a shaft with a diameter of 75 mm. SOLUTION Power = T ω = 2π NT = 2π ( 400 60 ) T = 200 × 103 N ⋅ m/s T = 4775 N ⋅ m (a) τ max = Tc T ( d 2 ) 16T = = J π d 4 32 π d 3 d= 3 d = 3 16T πτ π ( 70 × 106 ) 16 ( 4775 ) = 0.0703 m = 70.3 mm θ= TL TL 32TL = = 4 GJ G (π d 32 ) π d 4 d= 4 d = 4 32TL π Gθ = 0.0671 m = 67.1 mm π ( 80 × 109 ) ( 0.045 ) 32 ( 4775 )(1.5 ) d min = 70.3 mm ...................................................................................................................... Ans. (b) J = π ( 75 ) 32 = 3.106 × 106 mm 4 4 τ= Tc T ( 0.0375) = = 50 × 106 N/m 2 −6 J 3.106 × 10 T = 4141 N ⋅ m Power = Tω = ( 4141) ω = 200 × 103 N ⋅ m/s ω = 48.2975 rad/s = 461 rpm 7-47 .............................................................................. Ans. The motor shown in Fig. P7-47 develops 100 hp at a speed of 360 rpm. Gears A and B deliver 40 hp and 60 hp, respectively, to operating units in a factory. If the maximum shearing stress in the shafts must be limited to 12 ksi, determine (a) The minimum satisfactory diameter for the motor shaft. (b) The minimum satisfactory diameter for the power shaft. 281 STATICS AND MECHANICS OF MATERIALS, 2nd Edition SOLUTION (a) For the motor shaft: RILEY, STURGES AND MORRIS Power = T ω = 2π NT 2π ( 360 ) TR = = 100 hp 33, 000 33, 000 T = 1458.9 lb ⋅ ft = 17,507 lb ⋅ in. τ max = Tc T ( d 2 ) 16T = = J π d 4 32 π d 3 d= 3 d = 3 16T πτ 16 (17,507 ) = 1.951 in. ............................................................................... Ans. π (12, 000 ) (b) For the power shaft: T = (16 96 )(17,507 ) = 2918 lb ⋅ in. d= 7-48 3 16 ( 2918) = 1.074 in. ................................................................................ Ans. π (12, 000 ) A motor delivers 149 kW at 250 rpm to gear B of the factory drive shaft shown in Fig. P7-48. Gears A and C transfer 89 kW and 60 kW, respectively, to operating machinery in the factory. Determine (a) The maximum shearing stress in shaft AB. (b) The maximum shearing stress in shaft BC. (c) The rotation of gear C with respect to gear A if the shafts are both made of steel (G = 80 GPa). SOLUTION Power = T ω = 2π NT (a) T = P 2π N TAB = 89 × 103 = 3399.55 N ⋅ m 2π ( 250 60 ) 4 J AB = π d 4 32 = π ( 0.050 ) 32 = 6.13592 × 10−7 m 4 τ max = (b) Tc ( 3399.55 )( 0.025) = = 138.51× 106 N/m 2 ≅ 138.5 MPa .......................... Ans. −7 J 6.13592 ×10 60 × 103 = 2291.83 N ⋅ m 2π ( 250 60 ) 4 TBC = J BC = π ( 0.075 ) 32 = 3.10631× 10 −6 m 4 τ max = (c) θC / A Tc ( 2291.83)( 0.0375 ) = = 27.67 ×106 N/m 2 = 27.7 MPa ........................... Ans. J 3.10631× 10−6 ( 2291.83)(1.520 ) − ( 3399.55 )( 0.760 ) = θC / B −θ A/ B = (80 ×109 )( 3.10631×10−6 ) (80 ×109 )( 6.13592 ×10−7 ) = −0.0386 rad ................................................................................................................ Ans. 282 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 7-49 RILEY, STURGES AND MORRIS A motor supplies 270 hp at 250 rpm to gear A of the factory drive shaft shown in Fig. P7-49. Gears B and C transfer 170 hp and 100 hp, respectively, to operating machinery in the factory. For an allowable shearing stress of 11 ksi, determine (a) The minimum permissible diameter d1 for shaft AB. (b) The minimum permissible diameter d2 for shaft BC. (c) The rotation of gear C with respect to gear A if both shafts are made of steel (G = 11,600 ksi) and have diameters of 3 in. SOLUTION (a) Power = T ω = 2π NT 2π ( 250 ) TAB = = 270 hp 33, 000 33, 000 TAB = 5672.28 lb ⋅ ft = 68,067 lb ⋅ in. τ max = Tc 68, 067 ( d 2 ) = = 11× 103 psi 4 J π d 32 d1 = 3.16 in. .............................................................................................................. Ans. 2π ( 250 ) TBC = 100 hp 33, 000 (b) Power = TBC = 2100.84 lb ⋅ ft = 25, 210 lb ⋅ in. τ max = 25, 210 ( d 2 ) = 11× 103 psi π d 4 32 d 2 = 2.27 in. .............................................................................................................. Ans. (c) θC / A = θC / B + θ B / A = ( 25, 210 )( 72 ) (11.6 ×106 ) π ( 3)4 32 + ( 68, 067 )( 36 ) (11.6 ×106 ) π ( 3)4 32 = 0.0462 rad .................................................................................................................. Ans. 7-50 A motor provides 180 kW of power at 400 rpm to the drive shafts shown in Fig. P7-50. The maximum shearing stress in the three solid steel (G = 80 GPa) shafts must not exceed 70 MPa. Gears A, B, and C supply 40 kW, 60 kW, and 80 kW, respectively, to operating units in the plant. Determine (a) The minimum satisfactory diameter for shaft D. (b) The minimum satisfactory diameter for shaft E. (c) The minimum satisfactory diameter for shaft F. SOLUTION Power = T ω = 2π NT T = P 2π N d = 3 16T πτ τ max = (a) Tc T ( d 2 ) 16T = = J π d 4 32 π d 3 TD = dD = 40 × 103 = 477.5 N ⋅ m 2π ( 800 60 ) 3 π ( 70 × 106 ) 16 ( 477.5 ) = 0.03263 m ≅ 32.6 mm ................................................................. Ans. (b) TE = 180 × 103 = 4297 N ⋅ m 2π ( 400 60 ) 283 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS dE = (c) 3 π ( 70 × 106 ) 16 ( 4297 ) = 0.06787 m ≅ 67.9 mm ................................................................. Ans. TF = dD = 80 × 103 = 954.9 N ⋅ m 2π ( 800 60 ) 3 π ( 70 × 106 ) 16 ( 954.9 ) = 0.04111 m ≅ 41.1 mm ................................................................. Ans. 7-51 A 3-in. diameter steel drive shaft connects a motor to a machine. The power output of the motor varies with angular speed according to the following data P P ω ω (rpm) (hp) (rpm) (hp) 60 14.3 700 103.3 360 71.3 940 104.7 450 83.5 1120 90.6 600 96.5 1480 22.5 Calculate and plot the shear stress in the shaft τ as a function of the angular speed ω. SOLUTION Power = T ω = T= 2π NT 33, 000 33, 000 ( hp ) 2π N τ max = = 7-52 990.696 ( hp ) ........................................ Ans. N The power output of the motor varies P (kW) 77.2 78.0 67.5 15.0 Tc 33, 000 ( hp ) 1.5 = 4 J 2π N π ( 3) 32 A 75-mm diameter steel drive shaft connects a motor to a machine. with angular speed according to the following data P ω ω (rpm) (kW) (rpm) 50 15.5 750 100 36.0 1000 250 61.7 1250 500 73.3 1500 Calculate and plot the shear stress in the shaft τ as a function of the angular speed ω. SOLUTION Power = T ω = 2π ( N 60 ) T T= 60 ( P ) 2π N (N in revolutions per minute ) τ max 3 Tc 60 ( P ) 2π N ( 0.0375 ) (115.2810 ×10 ) ( P ) = = = .................................... Ans. 4 J N π ( 0.075 ) 32 284 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS Tm = 7.964 kN ⋅ m (a) Ts = 2.036 kN ⋅ m (b) 7-55 Tc ( 2036 )( 0.0625) = = 8.99 × 106 N/m 2 = 8.99 MPa ........................................ Ans. −6 J 14.151×10 ( 7964 )( 0.0875) = 10.23 ×106 N/m 2 = 10.23 MPa ............................................. Ans. τm = 68.11× 10−6 ( 7964 )( 2 ) θ= = 3.60 ×10−3 rad ............................................................... Ans. 9 −6 ( 65 ×10 )( 68.11×10 ) τs = A 3-in. diameter cold-rolled steel (G = 11,600 ksi) shaft, for which the maximum allowable shearing stress is 15 ksi, exhibited severe corrosion in a certain installation. It is proposed to replace the shaft with one in which an aluminum alloy (G = 4000 ksi) tube ¼-in. thick is bonded to the outer surface of the cold-rolled steel shaft to produce a composite shaft. If the shearing stress in the aluminum alloy shell is limited to 12 ksi, determine (a) The maximum torque that the original shaft can transmit. (b) The maximum torque that the replacement shaft can transmit. SOLUTION J s = π ( 3) 32 = 7.952 in 4 4 J a = π ( 3.54 − 34 ) 32 = 6.780 in 4 (a) τs = Tc T (1.5 ) = = 15 ksi J 7.952 Ts = 79.52 kip ⋅ in. ≅ 79.5 kip ⋅ in. ........................................................................ Ans. Ts + Ta − TR = 0 TR = Ts + Ta (b) Equilibrium: ΣM axis = 0 : Deformations: θa = θs = τL Gc τaL τsL + 6 ( 4 ×10 ) (1.75) (11.6 ×106 ) (1.5) τ a = 0.4023τ s τ s = τ max = 15 ksi τs = τ a = 0.4023 (15 ) = 6.035 ksi < τ max = 12 ksi Tc T (1.5 ) = = 15 ksi Ts = 79.52 kip ⋅ in. J 7.952 Tc T (1.75) τa = = = 6.035 ksi Ta = 23.38 kip ⋅ in. J 6.780 TR = 79.52 + 23.38 = 102.9 kip ⋅ in. .................................................................................... Ans. 7-56 The 50-mm diameter steel (G = 80 GPa) shaft shown in Fig. P7-56 is fixed to rigid walls at both ends. When a torque of 4.75 kN-m is applied as shown, determine (a) The maximum shearing stress in the shaft. (b) The angle of rotation of the section where the torque is applied with respect to its no load position. SOLUTION Equilibrium: ΣM axis = 0 : TA + TC − 4.75 = 0 TA + TC = 4.75 kN ⋅ m 286 STATICS AND MECHANICS OF MATERIALS, 2nd Edition Deformations: RILEY, STURGES AND MORRIS θ AB = θ BC = TA ( 0.6 ) TC ( 0.9 ) TL = GJ GJ GJ TA = 2.8500 kN ⋅ m 4 TA = 1.5TC TC = 1.9000 kN ⋅ m (a) J = π ( 0.050 ) 32 = 6.1359 × 10−7 m 4 τ max (b) 7-57 θB ( 2.85 ×10 ) ( 0.6 ) = 0.0348 rad ................................................................. Ans. = (80 ×10 )( 6.1359 ×10 ) 3 9 −7 3 Tc ( 2.8500 × 10 ) ( 0.050 ) = = = 232 ×106 N/m 2 = 232 MPa ........................... Ans. 6.1359 × 10−7 J Two 3-in. diameter solid circular steel (G = 11,600 ksi) and bronze (G = 6500 ksi) shafts are rigidly connected and supported as shown in Fig. P7-57. A torque T is applied at the junction of the two shafts as indicated. The shearing stresses are limited to 18 ksi for the steel and 6 ksi for the bronze. Determine (a) The maximum torque T that can be applied. (b) The angle of rotation of the section where the torque is applied with respect to its no-load position. SOLUTION Equilibrium: ΣM axis = 0 : Deformations: TA + TC − T = 0 T = TA + TC θ AB = θ BC = τL Gc 4 τ b ( 2) τ s (1) = ( 6.5 ×106 ) c (11.6 ×106 ) c τ s = 3.569τ b τ s = τ max = 18 ksi J = π ( 3) 32 = 7.9522 in 4 τ b = τ s 3.569 = 5.0431 ksi < τ max = 6 ksi (a) Tc TA (1.5 ) = = 18 ksi TA = 95.43 kip ⋅ in. = 7.952 kip ⋅ ft J 7.9522 Tc TC (1.5 ) τb = = = 5.0431 ksi TC = 26.74 kip ⋅ in. = 2.228 kip ⋅ ft J 7.9522 T = TA + TC = 7.952 + 2.228 = 10.18 kip ⋅ ft .................................................................... Ans. τs = (b) 7-58 3 τ L (18 ×10 ) (1×12 ) = = 0.01241 rad .................................................................... Ans. θB = Gc (11.6 ×106 ) (1.5) A composite shaft consists of a bronze (G = 45 GPa) shell that has an outside diameter of 100 mm bonded to a solid steel (G = 80 GPa) core. Determine the diameter of the steel core when the torque resisted by the steel core is equal to the torque resisted by the bronze shell. SOLUTION J b = π (1004 − d 4 ) 32 TL GJ J s = π d 4 32 Tb = Ts = T TL (80 ×10 ) π d 4 32 6 θb = θ s = d = 77.5 mm ........................................................................................................................... Ans. ( 45 ×10 ) π (100 6 TL 4 − d 4 ) 32 = 287 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 7-59 RILEY, STURGES AND MORRIS The solid steel (G = 12,000 ksi) shaft shown in Fig. P7-59 is fixed to the wall at C. The bolt holes in the flange at A have an angular misalignment of 0.0010 rad with respect to the holes in the wall. Determine (a) The torque, applied at B, required to align the bolt holes. (b) The maximum shearing stress in the shaft after the bolts are inserted and tightened and the torque at B is removed. (c) The maximum torque that can be applied at section B after the bolts are tightened if the maximum shearing stress in the shaft is not to exceed 10 ksi. SOLUTION (a) J = π ( 6 ) 32 = 127.23 in 4 4 θ A/C +θ B /C = θ B /C = (12 ×10 ) (127.23) 6 T ( 4 ×12 ) = 0.0010 rad TA = 31.8 × 103 lb ⋅ in. = 31.8 kip ⋅ in. ........................... Ans. (b) θ A/C = (12 ×10 ) (127.23) 6 T ( 6 × 12 ) = 0.0010 rad TA = 21.205 × 103 lb ⋅ in. = 21.205 kip ⋅ in. τ max = (c) θB/C Tc ( 21, 205 )( 3) = = 500 psi ................................................................................... Ans. J 127.23 = θ B / A + 0.0010 rad 2τ BC = τ AB + 1.5000 ksi τ BC ( 4 ×12 ) τ ( 2 ×12 ) = AB 6 + 0.0010 rad 6 (12 ×10 ) ( 3) (12 ×10 ) ( 3) τ AB = τ max = 10 ksi τ AB = τ BC = (10 + 1.5000 ) 2 = 5.750 ksi < τ max = 10 ksi Tc TAB ( 3) = = 10 ksi TAB = 424.10 kip ⋅ in. J 127.23 Tc TBC ( 3) τ BC = = = 5.750 ksi TBC = 243.86 kip ⋅ in. J 127.23 T = TAB + TBC = 424.10 + 243.86 = 668 kip ⋅ in. .............................................................. Ans. 7-60 A composite shaft consists of a solid steel (G = 80 GPa) core with an outside diameter of 40 mm covered by a brass (G = 39 GPa) tube with an inside diameter of 40 mm and a wall thickness of 20 mm which is in turn covered by an aluminum alloy (G = 28 GPa) sleeve with an inside diameter of 80 mm and a wall thickness of 10 mm, as shown in Fig. P7-60. The three materials are bonded so that they act as a unit. Determine (a) The maximum shearing stress in each material when the assembly is transmitting a torque of 15 kN-m. (b) The angle of twist in a 3-m length when the assembly is transmitting a torque of 10 kN-m. SOLUTION J s = π ( 0.040 ) 32 = 0.2513 × 10−6 m 4 4 J b = π ( 0.080 4 − 0.0404 ) 32 = 3.770 × 10−6 m 4 J a = π ( 0.1004 − 0.0804 ) 32 = 5.796 × 10 −6 m 4 288 STATICS AND MECHANICS OF MATERIALS, 2nd Edition Equilibrium: RILEY, STURGES AND MORRIS ΣM axis = 0 : Deformations: Ts + Tb + Ta − 15 = 0 Ts + Tb + Ta = 15 kN ⋅ m θ s = θb = θ a = 9 TL GJ Ts L Ta L = −6 9 (80 ×10 )( 0.2513 ×10 ) ( 28 ×10 )( 5.796 ×10−6 ) Tb L Ta L = −6 9 ( 39 ×10 )( 3.770 ×10 ) ( 28 ×10 )( 5.796 ×10−6 ) 9 Ts = 0.12388Ta Tb = 0.90598Ta Tb = 6.695 kN ⋅ m Ta = 7.390 kN ⋅ m 3 Ts = 0.915 kN ⋅ m 6 −6 (a) τs ( 0.915 ×10 ) ( 0.020 ) = 72.8 ×10 = 0.2513 ×10 3 N/m 2 = 72.8 MPa ......................................... Ans. N/m 2 = 71.0 MPa ......................................... Ans. τb = ( 6.695 ×10 ) ( 0.040 ) = 71.0 ×10 3.770 × 10 −6 6 (b) 7-61 3 TL (10 15 ) ( 6.695 ×10 ) ( 3) θb = = = 0.0911 rad ..................................................... Ans. 9 GJ ( 39 ×10 )( 3.770 ×10−6 ) 3 Tc ( 7.390 × 10 ) ( 0.050 ) = 63.8 ×106 N/m 2 = 63.8 MPa .............................. Ans. τa = = 5.796 × 10−6 J The composite shaft shown in Fig. P7-61 is used as a torsional spring. The solid circular polymer (G = 150 ksi) portion of the shaft is encased in and firmly attached to a steel (G = 12,000 ksi) sleeve for part of its length. If a torque T of 1.00 kip-in. is being transmitted by the composite shaft, determine (a) The rotation of a cross section at C. (b) The rotation of a cross section at C if the steel sleeve is assumed to be rigid. (c) The percent error introduced by assuming the steel sleeve to be rigid. SOLUTION J p = π ( 2 ) 32 = 1.5708 in 4 4 J s = π ( 2.254 − 24 ) 32 = 0.9453 in 4 Equilibrium: ΣM axis = 0 : Deformations: 1 − Tp − Ts = 0 Ts + Tp = 1 kip ⋅ in. = 1000 lb ⋅ in. Ts = 48.14Tp θs = θ p = TL GJ Tp L Ts L = (12, 000 )( 0.9453) (150 )(1.5708) Tp = 20.35 lb ⋅ in. (a) Ts = 979.65 lb ⋅ in. θC / A = θC / B + θ B / A = (150 ×10 ) (1.5708) (150 ×10 ) (1.5708) 3 3 1000 ( 4 ) + 20.35 (12 ) = 0.018013 rad ≅ 0.01801 rad ................................................................................... Ans. 289 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (b) If the steel shell is assumed to be rigid, then RILEY, STURGES AND MORRIS θC / A = θC / B = (c) 7-62 (150 ×10 ) (1.5708) 3 1000 ( 4 ) = 0.016976 rad ≅ 0.01698 rad .......................... Ans. Error = 0.018013 − 0.016976 (100 ) = 5.76 % ............................................................... Ans. 0.018013 A disk and two circular shafts are connected and supported between rigid walls, as shown in Fig. P7-62. Shaft AB is made of brass (G = 39 GPa) and has a diameter of 100 mm and a length of 400 mm. Shaft BC is made of Monel (G = 65 GPa) and has a diameter of 80 mm and a length of 600 mm. If a torque of 20 kN-m is applied to the disk, determine (a) The maximum shearing stress in each of the shafts. (b) The angle of rotation of the disk with respect to its no-load position. (c) The maximum tensile and compressive stresses in the shaft. SOLUTION J AB = π (100 ) 32 = 9.817 × 106 mm 4 = 9.817 × 10−6 m 4 4 J BC = π ( 80 ) 32 = 4.021× 106 mm 4 = 4.021× 10 −6 m 4 4 Equilibrium: ΣM axis = 0 : Deformations: TAB + TBC − 20 = 0 TAB + TBC = 20 kN ⋅ m θ AB = θ BC = TAB ( 0.400 ) 9 −6 TL GJ ( 39 ×10 )( 9.817 ×10 ) ( 65 ×10 )( 4.021×10 ) 9 −6 = TBC ( 0.600 ) TAB = 2.197TBC TAB = 13.744 kN ⋅ m (a) TBC = 6.256 kN ⋅ m 6 τ AB τ BC (13.744 ×10 ) ( 0.050 ) = 70.0 ×10 = 3 ( 6.256 ×10 ) ( 0.040 ) = 62.2 ×10 = 3 9.817 × 10 −6 N/m 2 = 70.0 MPa .................................... Ans. N/m 2 = 62.2 MPa ...................................... Ans. 6 (b) (c) 7-63 θB/ A (13.744 ×103 ) ( 0.400 ) = 0.01437 rad .................................................. Ans. TL = = GJ ( 39 × 109 )( 9.817 ×10−6 ) 4.021×10−6 σ max = τ max = τ AB = 70.0 MPa (T & C) ............................................................................ Ans. The steel (G = 12,000 ksi) shaft shown in Fig. P7-63 will be used to transmit a torque of 1000 lb-in. The hollow portion AE of the shaft is connected to the solid portion BF with two pins, at C and D as shown. If the average shearing stress in the pins must be limited to 25 ksi, determine the minimum satisfactory diameters for each of the pins. SOLUTION J h = π ( 24 − 14 ) 32 = 1.4726 in 4 In the interval CD equilibrium gives J s = π (1) 32 = 0.09817 in 4 4 ΣM axis = 0 : Deformations: Th + Ts − 1000 = 0 Th + Ts = 1000 lb ⋅ in. 290 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS θh = θs = TL GJ Th L Ts L = G (1.4726 ) G ( 0.09817 ) Th = 15Ts Th = 937.5 lb ⋅ in. Pin D transmits 937.5 lb ⋅ in. Ts = 62.5 lb ⋅ in. Pin C transmits 62.5 lb ⋅ in. T = (τ A ) d s = ( 25, 000 ) (π d 2 4 ) (1) The pins are in double shear and the moment generated by the shear force (τ A) is dC = dD = 7-64 (a) (b) (c) (d) 4 ( 62.5 ) = 0.0564 in. ......................................................................................... Ans. π ( 25, 000 ) 4 ( 937.5) = 0.219 in. ........................................................................................... Ans. π ( 25, 000 ) A stainless steel (G = 86 GPa) shaft 2.5 m long extends through and is attached to a hollow brass (G = 39 GPa) shaft 1.5 m long, as shown in Fig. P7-64. Both shafts are fixed at the wall. When the two torques shown are applied to the shaft, determine The maximum shearing stress in the steel. The maximum shearing stress in the brass. The maximum compressive stress in the brass. The rotation of the right end of the shaft. SOLUTION J s = π ( 80 ) 32 = 4.021× 106 mm 4 = 4.021× 10−6 m 4 4 J b = π (1204 − 804 ) 32 = 16.336 ×106 mm 4 = 16.336 × 10−6 m 4 Equilibrium: ΣM axis = 0 : (Tb + Ts ) + 36 − 9 = 0 Tb + Ts = −27 kN ⋅ m Deformation in the interval AB: θ s = θb = TL GJ Ts L Tb L = (86 ×109 )( 4.021×10−6 ) ( 39 ×109 )(16.336 ×10−6 ) Ts = 0.5428Tb Ts = 9.499 kN ⋅ m (a) Tb = 17.501 kN ⋅ m 3 Tc ( 9.499 × 10 ) ( 0.040 ) = = 94.5 ×106 N/m 2 = 94.5 MPa ............................... Ans. τs = 4.021× 10−6 J (b) (c) τb (17.501×10 ) ( 0.060 ) = 78.7 ×10 = 3 −6 6 σ max 13.336 × 10 = τ b = 78.7 MPa (C) ................................................................................................... Ans. N/m 2 = 78.7 MPa ....................................... Ans. 291 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 3 RILEY, STURGES AND MORRIS 3 (d) θC / A = θC / B + θ B / A ( −9 ×10 ) (1) + ( 9.499 ×10 ) (1.5) = (86 ×10 )( 4.021×10 ) (86 ×10 )( 4.021×10 ) 9 −6 9 −6 = 0.01518 rad ................................................................................................................ Ans. 7-65 The shaft shown in Fig. P7-65 consists of a 6-ft hollow steel (G = 12,000 ksi) section AB and a 4-ft solid aluminum alloy (G = 4000 ksi) section CD. The torque T of 40 kip-ft is applied initially only to the steel section AB. Section CD is then connected and the torque T is released. When the torque is released, the connection slips 0.010 rad before the aluminum section takes any load. Determine (a) The maximum shearing stress in the aluminum alloy. (b) The maximum shearing stress in the steel after the torque T is released. (c) The final rotation of the collar at B with respect to its no-load position. SOLUTION J AB = π ( 6 4 − 44 ) 32 = 102.10 in 4 J CD = π ( 4 ) 32 = 25.13 in 4 4 θi = Equilibrium: ( 40 ×12 )( 6 ×12 ) = 0.02821 rad TL = GJ (12, 000 )(102.10 ) After the connection is made and the torque is released, TAB = TCD = T f Deformation: θ B / A = θ i − slip − θ C / D T f ( 6 × 12 ) = 0.02821 − 0.010 − T f ( 4 × 12 ) T = 33.96 kip ⋅ in. (a) (b) (c) 7-66 J 25.13 ( 33.96 )( 3) = 0.998 ksi .................................................................................... Ans. τ s = τ AB = 102.10 ( 33.96 )( 6 ×12 ) = 0.01996 rad ............................................................. Ans. TL = θB/ A = GJ (12, 000 )(102.10 ) A torque T of 10 kN-m is applied to the steel (G = 80 GPa) shaft shown in Fig. P7-66 without the brass (G = 40 GPa) shell. The brass shell is then slipped into place and attached to the steel. After the original torque is released, determine (a) The maximum shearing stress in the brass shell. (b) The maximum shearing stress in the steel shaft. (c) The final rotation of the right end of the steel shaft with respect to the left end. SOLUTION τa f (12, 000 )(102.10 ) ( 4000 )( 25.13) Tc ( 33.96 )( 2 ) =τ = = = 2.70 ksi ............................................................................ Ans. CD J s = π ( 80 ) 32 = 4.021× 106 mm 4 4 J b = π (1604 − 1404 ) 32 = 26.62 × 106 mm 4 (10 ×103 ) ( 0.800 ) = 0.02487 rad TL = θi = GJ ( 80 × 109 )( 4.021× 10−6 ) 292 STATICS AND MECHANICS OF MATERIALS, 2nd Edition Equilibrium: After the connection is made and the torque is released, RILEY, STURGES AND MORRIS Ts = Tb = T f Deformation: θ s = θi −θb T f ( 0.800 ) 9 −6 (80 ×10 )( 4.021×10 ) T f = 7680 N ⋅ m (a) = 0.02487 − ( 40 ×10 )( 26.62 ×10 ) 9 −6 T f ( 0.800 ) (b) (c) 7-67 Tc ( 7680 )( 0.080 ) = = 23.1×106 N/m 2 = 23.1 MPa .......................................... Ans. 26.62 × 10−6 J ( 7680 )( 0.040 ) = 76.4 ×106 N/m 2 = 76.4 MPa .................................................... Ans. τs = 4.021×10−6 ( 7680 )( 0.800 ) TL θR/ L = = = 0.01910 rad ................................................... Ans. GJ ( 80 × 109 )( 4.021× 10−6 ) τb = A hollow steel (G = 11,600 ksi) tube with an inside diameter of 2 in., an outside diameter of 2.5 in., and a length of 12 in. is encased with a brass (G = 5600 ksi) tube that has an inside diameter of 2.5 in. and an outside diameter of 3.25 in. The brass and steel tubes, while unstressed, are brazed together at one end. A torque is then applied to the other end of the brass tube, which twists one degree with respect to the steel tube. The tubes are then brazed together in that position. Determine the maximum shearing stresses in the steel tube and in the brass tube after the torque is removed from the brass tube. SOLUTION J s = π ( 2.54 − 24 ) 32 = 2.264 in 4 J b = π ( 3.254 − 2.54 ) 32 = 7.118 in 4 θi = Equilibrium: (1) π 180 = 0.017453 rad Tb = Ts = T After the torque is removed, Deformation: θb = θi −θ s T (12 ) = 0.017453 − T (12 ) T = 23.03 kip ⋅ in. ( 5600 )( 7.118) (11, 600 )( 2.264 ) Tc ( 23.03)(1.25 ) τs = = = 12.72 ksi ................................................................................ Ans. J 2.264 ( 23.03)(1.625) = 5.26 ksi .......................................................................................... Ans. τb = 7.118 7-68 A composite shaft consists of a 2-m long solid circular steel (Gs = 80 GPa) section securely fastened to a 2m long solid circular bronze (Gb = 40 GPa) section, as shown in Fig. P7-68. Both ends of the composite shaft are attached to rigid supports. The maximum shearing stress τmax in the shaft must not exceed 60 MPa, and the rotation of any cross section in the shaft must not exceed 0.04 rad. The ratio of the diameters of the two sections can vary (1/2 ≤ db/ds ≤ 2), but the average diameter (db + ds)/2 must be 100 mm. Compute and plot (a) The maximum allowable torque T as a function of the diameter ratio db/ds (1/2 ≤ db/ds ≤ 2). 293 STATICS AND MECHANICS OF MATERIALS, 2nd Edition When RILEY, STURGES AND MORRIS θ BC > 1° = 17.4533 ×10−3 rad TA + TC = T Equilibrium: Deformation: θ BC = θ AB + 17.4533 ×10−3 rad TC (1.5) 9 −6 (80 ×10 )( 20.36 ×10 ) = (80 ×10 )( 20.36 ×10 ) 9 −6 TA ( 0.75) + 17.4533 × 10−3 TC = 0.5TA + 18,951.96 kN ⋅ m Combining the equilibrium equation and the deformation equation gives TA = 0.66667T − 12,634.64 kN ⋅ m In either case TC = 0.33333T + 12, 634.64 kN ⋅ m TA ( 0.060 ) N/m 2 ................................................... Ans. −6 20.36 × 10 T ( 0.060 ) N/m 2 ................................................... Ans. τ BC = C −6 20.36 × 10 TC (1.5) θ B = θ BC = ................................ Ans. 9 (80 ×10 )( 20.36 ×10−6 ) τ AB = while for the bolts τb = Vb TA db TA 0.3 = = N/m 2 .............................................................................. Ans. Ab Ab 2036 × 10 −6 7-71 A 2-in. diameter solid steel (Gs = 12,000 ksi, τmax = 30 ksi) shaft and a 3-in. diameter hollow aluminum alloy (Ga = 4000 ksi, τmax = 24 ksi) shaft are fastened together with a ½-in. diameter brass (τmax = 36 ksi) pin, as shown in Fig. P7-71. Calculate and plot (a) The maximum shearing stresses, τa in the aluminum shaft and τs in the steel shaft, as a function of the torque T applied to the end of the shaft (0 ≤ T ≤ 60 kip-in.). (b) (c) The average shearing stresses τb on the cross-sectional area of the brass pin at the interface between the shafts as a function of T (0 ≤ T ≤ 60 kip-in.). The rotation of end B with respect to its no-load position θB/D as a function of T (0 ≤ T ≤ 60 kip-in.). 296 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (d) RILEY, STURGES AND MORRIS What is the maximum torque that can be applied to the shaft without exceeding the maximum shearing stresses in either the shaft or the pin? SOLUTION J a = π ( 34 − 24 ) 32 = 6.3814 in 4 Equilibrium: Deformation: J s = π ( 2 ) 32 = 1.5708 in 4 4 Ts + Ta = T θa = θs = TL GJ Ta = 12 (1.5708 ) 4 ( 6.3814 ) Ts = 1.35417Ts Combining the deformation equation with the equilibrium equation gives Ts + 1.35417Ts = T (a) Ts = 0.42478T Ta = 0.57522T (c) T (1) Tc = .................................................................................................................... Ans. J 1.5708 Tc T (1.5) τa = = a .................................................................................................................. Ans. 6.3814 J T (12 ) + T (18 ) TL θB/ D = ∑ =s rad .......................................................................... Ans. GJ (12 × 106 ) (1.5708) τs = (b) τb = Vb Ta d s Ta 2 = = psi ..................................................................................... Ans. 2 Ab Ab π ( 0.5) 4 (d) The stress in the brass pin reaches its limit first when 7-72 T = 24.6 kip ⋅ in. ....................................... Ans. A motor is to transmit 150 kW to a piece of mechanical equipment. The power is transmitted through a solid structural steel shaft. Failure is by yielding and the factor of safety is 1.25. The designer has the freedom to operate the motor at 60 rpm or 6000 rpm. For each case, determine the minimum shaft diameter. Shafts are available with diameters in increments of 5 mm. If weight is important, which speed would be used? SOLUTION Structural steel: σ y = 250 MPa τy = σy = 125 MPa 2 τ all = τ max Power = T ω = 2π NT For 60 rpm: T = P 2π N τ y 125 = = 100 MPa FS 1.25 Tc T ( d 2 ) 16T = = = J π d 4 32 π d 3 297 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 150 ×103 T= = 23.87 ×103 N ⋅ m 2π ( 60 60 ) d= 3 16 ( 23.87 × 103 ) π (100 ×10 6 ) = 1.21569 × 10−3 m3 d = 0.1067 m = 106.7 mm Use d = 110 mm .................................................................................................................... Ans. For 6000 rpm: 150 × 103 T= = 238.7 N ⋅ m 2π ( 6000 60 ) d3 = π (100 ×10 16 ( 238.7 ) 6 ) = 12.1569 × 10−6 m3 d = 0.0230 m = 23.0 mm Use d = 25 mm ...................................................................................................................... Ans. If weight is critical, Use N = 6000 rpm ................................................................................................................ Ans. 7-73 A 3-ft long steel pipe is subjected to a torque of 1200 lb-ft at each end. The pipe is made of 0.2% C hardened steel, failure is by yielding, and the factor of safety is 1.5. Determine the nominal diameter of the lightest standard weight steel pipe that can be used for the shaft. SOLUTION 0.2% C steel: σ y = 36 ksi Tc Tc = J 2I τy = τ all = 2.0-in. pipe: 2.5-in. pipe: σy τ 18 τ all = y = = 18 ksi = 12 ksi FS 1.5 2 I T 1200 × 12 = = = 0.600 in 3 3 c 2τ all 2 (12 × 10 ) S = I c = 0.561 in 3 S = I c = 1.064 in 3 Use d = 2.5 in. ....................................................................................................................... Ans. 7-74 A standard weight structural steel pipe must transmit 150 kW at 60 rpm. The failure mode is yielding and the factor of safety is 1.5. (a) Select the lightest standard weight steel pipe that can be used. (b) If solid structural steel shafts are available with diameters in increments of 10 mm, determine the minimum diameter that can be used. (c) Compare the weights of the two shafts. SOLUTION Structural steel: σ y = 250 MPa ρ = 7870 kg/m3 τ all = T= τy = σy = 125 MPa 2 τ y 125 = = 83.33 MPa FS 1.5 Power = T ω = 2π NT (a) P 150 × 103 = = 23.87 × 103 N ⋅ m 2π N 2π ( 60 60 ) τ all = Tc Tc = J 2I I T 23.87 × 103 = = = 143.22 × 10−6 m3 6 c 2τ all 2 ( 83.33 × 10 ) m = 28.21 kg/m 152-mm pipe: S = I c = 139.16 × 10−6 m 3 298 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 203-mm pipe: RILEY, STURGES AND MORRIS S = I c = 275.67 × 10−6 m3 m = 42.46 kg/m Use d = 203 mm ................................................................................................................... Ans. (b) τ max = 3 Tc ( 23.87 ×10 ) ( d 2 ) = ≤ 83.33 × 106 N/m 2 4 J π d 32 d ≥ 0.1134 m = 113.4 mm Use d = 120 mm .................................................................................................................... Ans. (c) Solid shaft Hollow shaft 7-75 2 m = ρV = 7870 π ( 0.120 ) 4 = 89.0 kg/m ............................................ Ans. m = 42.5 kg/m .................................................................................................. Ans. A shaft is to transmit 100 hp at 200 rpm. The designer has a variety of solid bars and standard steel pipes to select from. Both the bars and the pipes are made of structural steel, the failure mode is yielding, and the factor of safety is 2. (a) Select the lightest standard weight steel pipe that can be used. (b) Select a suitable solid shaft, if they are available with diameters in increments of 1/8 in. (c) If weight is critical, which shaft should be used? SOLUTION Structural steel: σ y = 36 ksi γ = 0.284 lb/in 3 σy τ 18 τ all = y = = 9 ksi = 18 ksi FS 2 2 2π NT 2π ( 200 ) T Power = T ω = = = 100 hp 33, 000 33,000 T = 2626 lb ⋅ ft = 31,513 lb ⋅ in. τy = (a) For a hollow pipe: τ all = 3.0-in. pipe: 3.5-in. pipe: Tc Tc = J 2I I T 31,513 = = = 1.75072 in 3 c 2τ all 2 ( 9 × 103 ) w = 7.58 lb/ft w = 9.11 lb/ft S = I c = 1.724 in 3 S = I c = 2.39 in 3 Tc 31,513 ( d 2 ) = ≤ 9 × 103 psi 4 π d 32 J Use d = 3.5 in. ........................................................................................................................ Ans. (b) τ max = d ≥ 2.613 in. 2 4 8 < 2.613 < 2 5 8 in. Use d = 2 5 8 in. ...................................................................................................................... Ans. (c) Solid shaft Hollow pipe If weight is critical, 2 w = γ V = 0.284 π ( 2.625 ) 4 = 1.537 lb/in. w = 9.11 lb/ft = 0.759 lb/in. Use hollow pipe ..................................................................................................................... Ans. 299 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 7-76 RILEY, STURGES AND MORRIS The motor shown in Fig. P7-76 supplies a torque of 1000 N-m to shaft ABCDE. The torques removed at C, D, and E are 500 N-m, 300 N-m, and 200 N-m, respectively. The shaft is the same diameter throughout and is made of 0.4% C hot-rolled steel. For a factor of safety of 3 and failure by yielding, select a suitable diameter for the shaft, if shafts are available with diameters in increments of 10 mm. SOLUTION 0.4% C hot-rolled steel: σ y = 360 MPa τ y = σ y 2 = 180 MPa τ all = τ y FS = 180 3 = 60 MPa The internal resisting torques are: ΣM axis = 0 : ΣM axis = 0 : ΣM axis = 0 : TBC − 1000 = 0 TCD − 1000 + 500 = 0 TDE − 1000 + 500 + 300 = 0 Tc (1000 )( d 2 ) = ≤ 60 × 106 MPa J π d 4 32 TBC = 1000 N ⋅ m TCD = 500 N ⋅ m TCD = 200 N ⋅ m Tmax = TBC = 1000 N ⋅ m τ max = d ≥ 0.04395 m = 43.95 mm Use d = 50 mm ...................................................................................................................... Ans. 7-77 A torque of 30,000 lb-in. is supplied to the factory drive shaft of Fig. P7-77 by a belt that drives pulley A. A torque of 10,000 lb-in. is removed by pulley B and 20,000 lb-in by pulley C. The shaft is made of structural steel and has a constant diameter over its length. Segment AB of the shaft is 3 ft long, and segment BC is 4 ft long. Failure is by yielding and the factor of safety is 2.25. Select a suitable diameter for the shaft, if shafts are available with diameters in increments of 1/8 in. SOLUTION Structural steel: σ y = 36 ksi τy = σy = 18 ksi 2 τ all = τy 18 = = 8 ksi FS 2.25 The internal resisting torques are: ΣM axis = 0 : ΣM axis = 0 : TBC − 20, 000 = 0 TAB − 20, 000 − 10, 000 = 0 Tmax = TAB = 30, 000 lb ⋅ in. Tc 30, 000 ( d 2 ) = ≤ 8 × 103 psi 4 J π d 32 TBC = 20, 000 lb ⋅ in. TAB = 30, 000 lb ⋅ in. τ max = d ≥ 2.67 in. 2 5 8 < 2.67 < 2 6 8 in. Use d = 2 6 8 = 2 3 4 in. ........................................................................................................... Ans. 7-78 The band brake shown in Fig. P7-78 is part of a hoisting machine. The coefficient of friction between the 500-mm diameter drum and the flat belt is 0.20. The maximum actuating force P that can be applied to the brake arm is 490 N. Rotation of the drum is clockwise. What minimum size shaft should be used to transmit the resisting torque developed by the brake to the machine if the shaft is to be made of 0.4% C hotrolled steel. The factor of safety is 3 for failure by yielding. Circular steel bars are available with diameters in increments of 5 mm. SOLUTION 0.4% C hot-rolled steel: σ y = 360 MPa τ y = σ y 2 = 180 MPa τ all = τ y FS = 180 3 = 60 MPa 300 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ΣM axis = 0 : TB ( 50 ) − 490 ( 650 ) = 0 0.20(1.5π ) TB = 6370 N TA = TB e µs β = 6370e = 16,348 N T = (TA − TB ) R = (16,348 − 6370 )( 0.250 ) = 2495 N ⋅ m τ max = Tc ( 2495)( d 2 ) = ≤ 60 × 106 MPa 4 π d 32 J d ≥ 0.0596 m = 59.6 mm Use d = 60 mm ...................................................................................................................... Ans. 7-79 The motor shown in Fig. P7-79 supplies a torque of 380 lb-ft to shaft BCD. The torques removed at gears C and D are 220 lb-ft and 160 lb-ft, respectively. The shaft BCD has a constant diameter and is made of 0.4% C hot-rolled steel, failure is by yielding, and the factor of safety is 2. Determine (a) The minimum allowable diameter of the shaft, if shafts are available with diameters in increments of 1/8 in. (b) The minimum allowable diameter of the bolts used in the coupling, if eight bolts are used, the material is structural steel, the mode of failure is yielding, and the factor of safety is 1.5. The diameter of the bolt circle is d1 = 3.5 in., and the bolts are available with diameters in increments of 1/16 in. SOLUTION 0.4% C hot-rolled steel: σ y = 53 ksi τy = σy = 26.5 ksi 2 τ all = τ y 18 = = 13.25 ksi FS 2 The internal resisting torques are: ΣM axis = 0 : ΣM axis = 0 : (a) TBC − 380 = 0 TCD − 380 + 220 = 0 Tc ( 380 × 12 )( d 2 ) = ≤ 13.25 × 103 psi 4 J π d 32 TBC = Tmax = 380 lb ⋅ ft TCD = 160 lb ⋅ ft d ≥ 1.206 in. τ max = 1 1 8 < 1.206 < 1 2 8 in. Use d = 1 2 8 = 1 1 4 in. ............................................................................................................. Ans. Structural steel: σ y = 36 ksi τy = ΣM axis = 0 : 8 Fs ( 3.5 2 ) − TBC = 0 < 0.1859 < 316 in. σy τ 18 τ all = y = = 18 ksi = 12 ksi FS 1.5 2 Fs = 325.7 lb = τ As d = 0.1859 in. τ As = (12 × 103 )(π db2 4 ) = 325.7 lb 2 16 Use d = 316 in. ........................................................................................................................ Ans. 7-80 A shaft used to transmit power is constructed by joining two solid segments of shaft with a collar, as shown in Fig. P7-80. The collar has an inside diameter equal to the diameter of the shaft, and both the collar and the shaft are made of the same material. The collar is securely bonded to the shaft segments. Determine the ratio of the diameters of the collar and shaft such that the splice can transmit the same power as the shaft and at the same maximum shearing stress level. Is the solution dependent on the material selected? SOLUTION For the same power, the values of torque are the same in the shaft and the collar. For the shaft: τs = Tc T ( d s 2 ) 16Td s = = π d s4 32 π d s4 J 301 STATICS AND MECHANICS OF MATERIALS, 2nd Edition For the collar: RILEY, STURGES AND MORRIS τc = T ( dc 2 ) Tc 16Td c = = 4 4 J π ( d c − d s ) 32 π ( d c4 − d s4 ) For the same shear stress in the shaft and the collar ds d = 4c4 4 d s dc − d s ( dc ds ) − ( dc ds ) − 1 = 0 4 ( dc d s ) = 1.221 (independent of the type of material) ............................................... Ans. 7-81 A solid circular steel (G = 12,000 ksi) shaft is loaded and supported as shown in Fig. P7-81. Determine (a) The maximum shearing stress in the shaft. (b) The rotation of a section at B with respect to its no-load position. (c) The rotation of end C with respect to its no-load position. SOLUTION J AB = π ( 3) 32 = 7.952 in 4 4 J BC = π ( 2 ) 32 = 1.5708 in 4 4 (a) Tc ( 4000 × 12 )(1.5 ) = = 9054 psi 7.952 J (1000 ×12 )(1) = 7639 psi τ BC = 1.5708 τ max = 9054 psi ≅ 9.05 ksi ................................................................................................... Ans. τ AB = (b) θB/ A = TL ( −4000 × 12 )( 3 × 12 ) = = −0.01811 rad ....................................................... Ans. GJ (12 ×106 ) ( 7.952 ) TL ( −4000 ×12 )( 3 ×12 ) ( −1000 ×12 )( 4 × 12 ) = + GJ (12 ×106 ) ( 7.952 ) (12 ×106 ) (1.5708) (c) θC / A = ∑ = −0.0487 rad ................................................................................................................ Ans. 7-82 A torque of 12.0 kN-m is supplied to the driving gear B of Fig. P7-82 by a motor. Gear A takes off 4.0 kNm of torque, and the remainder is taken off by gear C. Determine the angle of twist of gear A with respect to gear C if both shafts are made of steel (G = 80 GPa) and have diameters of 75 mm. SOLUTION TAB = 4 kN ⋅ m TBC = 4 − 12 = −8 kN ⋅ m J BC = π ( 75 ) 32 = 3.106 × 106 mm 4 4 (c) θ A/C ( 4 ×103 ) (1) ( −8 ×103 ) ( 2 ) TL =∑ = + GJ ( 80 × 109 )( 3.106 ×10−6 ) ( 80 × 109 )( 3.106 × 10−6 ) = −0.0483 rad ................................................................................................................ Ans. 7-83 An aluminum alloy (G = 3800 ksi) tube will be used to transmit a torque in a control mechanism. The tube has an outside diameter of 1.25 in. and a wall thickness of 0.065 in. Because of the tendency of thin sections to buckle, the maximum compressive stress in the tube must be limited to 8000 psi. Determine (a) The maximum torque that can be applied. 302 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (b) SOLUTION RILEY, STURGES AND MORRIS The angle of twist in a 3-ft length when a torque of 1000 lb-in. is applied. J = π (1.254 − 1.124 ) 32 = 0.08520 in 4 (a) σ max = τ max = Tc T (1.25 2 ) = ≤ 8000 psi 0.08520 J T ≤ 1091 lb ⋅ in. ................................................................................................................ Ans. (b) 7-84 θ= (1000 )( 3 ×12 ) = 0.1112 rad ................................................................ Ans. TL = GJ ( 3.8 × 106 ) ( 0.08520 ) A 2-m long hollow brass (G = 39 GPa) shaft has an outside diameter of 50 mm. The maximum shearing stress in the shaft must be limited to 50 MPa, and the angle of twist in the 2-m length must be limited to 2°. If the shaft will rotate at 1500 rpm and transmit 50 kW, determine the maximum inside diameter that can be used for the shaft. SOLUTION Power = T ω = 2π NT = 2π (1500 60 ) T = 50 × 103 N ⋅ m/s T = 318.3 N ⋅ m To satisfy the stress specification: τ max = θ= Tc 318.3 ( 0.025 ) = ≤ 50 × 106 N/m 2 J J J ≥ 0.15915 × 10−6 m 4 To satisfy the deformation specification: TL ( 318.3)( 2 ) = ≤ 2° = 0.03491 rad GJ ( 39 ×109 ) J J ≥ 0.46758 × 10−6 m 4 J min = π ( 50 4 − d i4 ) 32 = 0.46758 × 106 mm 4 di = 34.9 mm .......................................................................................................................... Ans. 7-85 The inner surface of the aluminum alloy (G = 4000 ksi) sleeve A and the outer surface of the steel (G = 12,000 ksi) shaft B of Fig. P7-85 are smooth. Both the sleeve and the shaft are rigidly fixed to the wall at D. The 0.500-in. diameter pin C fills a hole drilled completely through a diameter of the sleeve and shaft. If the average shearing stress on the cross-sectional area of the pin at the interface between the shaft and the sleeve must not exceed 5000 psi, determine (a) The maximum torque T that can be applied to the right end of the steel shaft B. (b) The maximum shearing stress in the aluminum alloy sleeve A when the torque of part (a) is applied. (c) The rotation of the right end of the shaft when the maximum torque T is applied. SOLUTION J a = π ( 34 − 24 ) 32 = 6.3814 in 4 J s = π ( 2 ) 32 = 1.5708 in 4 4 τp = Equilibrium: Deformation: Vp Ap = π ( 0.5 ) 4 2 Ta 2 = 5000 psi Ta = 1963.5 lb ⋅ in. Ts + Ta = T θa = θs = TL GJ Ts = 12 (1.5708 ) 4 ( 6.3814 ) Ta = 0.73846Ta = 1450.0 lb ⋅ in. 303 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (a) (b) RILEY, STURGES AND MORRIS T = Ts + Ta = 3413.5 lb ⋅ in. ≅ 3410 lb ⋅ in. ........................................................................ Ans. Tc 1963.5 (1.5) = = 462 psi ....................................................................................... Ans. J 6.3814 TL 1450.0 (12 ) + 3413.5 (18) θB/ D = ∑ = = 0.00418 rad .......................................... Ans. GJ (12 ×106 ) (1.5708) τa = (c) 7-86 The 100-mm diameter segment ABC of the shaft shown in Fig. P7-86 is initially not connected to the 60mm diameter segment CD. Torque TB = 15 kN-m is applied at section B, and a secure connection between the two segments is then made at C, after which the torque TB is removed. Determine the resulting maximum shearing stress in segment CD after torque TB is removed. The moduli of rigidity are 40 GPa for ABC and 80 GPa for CD. SOLUTION J AB = J BC = π (100 ) 32 = 9.817 × 106 mm 4 4 J CD = π ( 60 ) 32 = 1.2723 × 106 mm 4 4 With the torque TB = 15 kN ⋅ m applied: TL GJ 3 (15 ×10 ) (1.2 ) 9 θ Bi = θ B / A = θ C / A = = ( 40 ×10 )( 9.817 ×10 ) −6 = 0.04584 rad After the connection is made and TB is removed: TAC = TCD = T From the deformation of the shaft: θ C / A + θ C / D = θ Bi ( 40 ×10 )( 9.817 ×10 ) (80 ×10 )(1.2723 ×10 ) 9 −6 9 −6 T ( 2.2 ) + T (1.6 ) = 0.04584 rad TAC = TCD = T = 2149.9 N ⋅ m τ CD = 7-87 Tc 2149.9 ( 0.030 ) = = 50.7 × 106 N/m 2 = 50.7 MPa ......................................... Ans. −6 1.2723 × 10 J The motor shown in Fig. P7-87 delivers 200 hp at 300 rpm to a piece of equipment at B. The shaft is made from 0.4% C hot-rolled steel, which is available with diameters in increments of 1/8 in. Determine the shaft diameter required if failure is by yielding and the factor of safety is 1.5. SOLUTION 0.4% C hot-rolled steel: σ y = 53 ksi τy = σy = 26.5 ksi 2 τ all = τ y 26.5 = = 17.667 ksi FS 1.5 Power = T ω = 2π NT 2π ( 300 ) T = = 200 hp 33, 000 33, 000 T = 3501 lb ⋅ ft = 42.01 kip ⋅ in. 304 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS τ max 3 Tc ( 42.01× 10 ) ( d 2 ) = = ≤ 17.667 × 103 psi π d 4 32 J d ≥ 2.296 in. 2 2 8 < 2.296 < 2 3 8 in. Use d = 2 3 8 in. ...................................................................................................................... Ans. 7-88 The 160-mm diameter steel (G = 80 GPa) shaft shown in Fig. P7-88 has a 100-mm diameter bronze (G = 40 GPa) core inserted in 3 m of the right end and securely bonded to the steel. Determine (a) The maximum shearing stress in each of the materials. (b) The rotation of the free end of the shaft. SOLUTION J ABs = π (160 ) 32 = 64.34 × 106 mm 4 = 64.34 × 10 −6 m 4 4 J BCs = π (1604 − 1004 ) 32 = 54.52 × 106 mm 4 = 54.52 × 10−6 m 4 J BCb = π (100 ) 32 = 9.817 ×106 mm 4 = 9.817 ×10 −6 m 4 4 (a) In the interval AB: 3 Tc ( 85 × 10 ) ( 0.080 ) = = 105.69 ×106 N/m 2 ≅ 105.7 MPa −6 J 64.34 ×10 In the interval BC: Tb + Ts = 75 kN ⋅ m TA = 160 − 75 = 85 kN ⋅ m τs = θb = θ s = TL GJ Tb L Ts L = −6 9 ( 40 ×10 )( 9.817 ×10 ) (80 ×10 )( 54.52 ×10−6 ) 9 Ts = 11.107Tb Tb = 6.195 kN ⋅ m 3 Ts = 68.805 kN ⋅ m 3 Tc ( 6.195 × 10 ) ( 0.050 ) τb = = = 31.55 × 106 N/m 2 ≅ 31.6 MPa −6 J 9.817 × 10 For the steel: τ max = 105.7 MPa ............................................................................................. Ans. Tc ( 68.805 × 10 ) ( 0.080 ) τs = = = 100.96 × 106 N/m 2 ≅ 101.0 MPa J 54.52 ×10 −6 For the bronze: (b) τ max = 31.6 MPa ............................................................................................... Ans. θD/ A (85 ×103 ) ( 2 ) − ( 68.805 ×103 ) (1.5) TL = θs = ∑ = GJ ( 80 × 109 )( 64.34 × 10−6 ) ( 80 × 109 )( 54.52 ×10 −6 ) = 0.00936 rad ......................................................................................................................... Ans. 305 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS Chapter 8 8-1 (a) Determine the second moment of the shaded area shown in Fig. P8-1 with respect to The x-axis. (b) The y-axis. SOLUTION (a) 6 (8) 4 ( 6 ) Ix = − = 736 in 4 ............................................................................................. Ans. 3 3 3 3 (b) 8-2 2 ( 6) 6 ( 4) 6 ( 2) Iy = + − = 256 in 4 ............................................................................. Ans. 3 3 3 3 3 3 Determine the second moments of the shaded area shown in Fig. P8-2 with respect to x- (horizontal) and y(vertical) axes through the centroid of the area. SOLUTION A = 200 ( 60 ) + 60 (160 ) + 200 ( 60 ) = 33, 600 mm 2 yC = 100 200 ( 60 ) + 30 60 (160 ) + 100 200 ( 60 ) 33, 600 3 3 3 = 80 mm 60 ( 200 ) 160 ( 60 ) 60 ( 200 ) Ix = + + = 331.52 ×106 mm 4 3 3 3 2 I xC = I x − AyC = 331.52 × 106 − 33, 600 ( 80 ) = 116.5 ×106 mm 4 .............................. Ans. 2 I yC 8-3 200 ( 280 ) 140 (160 ) = − = 318 × 106 mm 4 ........................................................... Ans. 12 12 3 3 Determine the second moments of the shaded area shown in Fig. P8-3 with respect to x- (horizontal) and y(vertical) axes through the centroid of the area. SOLUTION I xC I yC 8-4 4 ( 6) 3( 4) = − = 56.0 in 4 .......................................................................................... Ans. 12 12 3 3 1( 4 ) 4 (1) 1( 4 ) = + + = 11.00 in 4 .......................................................................... Ans. 12 12 12 3 3 3 Four C305×45 channels are welded together to form the cross section shown in Fig. P8-4. Determine the second moments of the area with respect to x- (horizontal) and y- (vertical) axes through the centroid of the area. SOLUTION From Table A-6 for a C305 × 45 channel: d = 304.8 mm I x = 67.4 × 106 mm 4 tw = 13.0 mm xC = 17.1 mm A = 5690 mm 2 I y = 2.14 ×106 mm 4 2 I xC = 2 ( 67.4 ×106 ) + 2 ( 2.14 ×106 ) + (148.4 ) ( 5690 ) = 390 ×106 mm 4 ............. Ans. 2 I yC = 2 ( 67.4 ×106 ) + 2 ( 2.14 ×106 ) + (17.1) ( 5690 ) = 142.4 ×106 mm 4 ............. Ans. 306 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 8-5 RILEY, STURGES AND MORRIS Two 10×1-in. steel plates are welded to the flanges of an S18×70 I-beam as shown in Fig. P8-5. Determine the second moments of the area with respect to x- (horizontal) and y- (vertical) axes through the centroid of the area. SOLUTION From Table A-3 for an S18 × 70 beam: I x = 926 in 4 I y = 24.1 in 4 d = 18.00 in. I xC I yC 8-6 (a) 10 (1)3 2 = 926 + 2 + ( 9.5 ) (10 × 1) = 2733 in 4 ≅ 2730 in 4 .................................. Ans. 12 1(10 )3 = 24.1 + 2 = 190.8 in 4 ................................................................................... Ans. 12 Determine the second moment of the shaded area shown in Fig. P8-6 with respect to The x-axis. (b) The y-axis. SOLUTION (a) π (120 ) 120 ( 30 ) 60 ( 60 ) Ix = + − = 37.5 ×106 mm 4 16 3 3 4 3 3 4 3 3 .............................................. Ans. (b) 8-7 (a) (b) π (120 ) 30 (120 ) 60 ( 60 ) Iy = + − = 53.7 ×106 mm 4 .............................................. Ans. 16 3 3 Determine the second moments of the shaded area shown in Fig. P8-7 with respect to The x- and y-axes shown on the figure. The x- and y-axes through the centroid of the area. SOLUTION xC 3 = 20 − (a) 4 ( 3) 4R = 20 − = 18.727 in. 3π 3π 3 4 2 π ( 3) 4 20 (10 ) π ( 3) 2 2 π ( 3) 2 Ix = − + ( 5) π ( 3 ) − + ( 5) 3 2 4 8 = 6667 − 770 − 385 = 5512 in 4 ≅ 5510 in 4 ................................................................ Ans. 3 4 2 π ( 3)4 8 ( 3)4 10 ( 20 ) π ( 3) 2 2 2 π ( 3) Iy = − + ( 5 ) π ( 3) − − + (18.727 ) 3 9π 2 4 8 = 26, 667 − 770 − 4967 = 20,930 in 4 ≅ 20,900 in 4 ................................................. Ans. (b) π ( 3) A = 20 (10 ) − π ( 3) − = 157.59 in 2 2 2 2 2 2 10 20 (10 ) − 5 π ( 3) − (18.727 ) π ( 3) 2 = 10.114 in. xC = 157.59 2 I xC = I x − yC A = 5512 − ( 5) (157.59 ) = 1572 in 4 ......................................................... Ans. 2 2 I yC = I y − xC A = 20, 930 − (10.114 ) (157.59 ) = 4810 in 4 .......................................... Ans. 2 307 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 8-8 (a) (b) RILEY, STURGES AND MORRIS Determine the second moments of the shaded area shown in Fig. P8-8 with respect to The x- and y-axes shown on the figure. The x- and y-axes through the centroid of the area. SOLUTION xC 2 = 160 + 4 ( 60 ) 4R = 160 + = 185.46 mm 3π 3π 3 4 2 π ( 30 )4 160 (120 ) π ( 60 ) 2 π ( 60 ) 2 2 (a) I x = + + ( 60 ) + ( 60 ) π ( 30 ) − 12 2 4 8 = 37.68 × 106 mm 4 ≅ 37.7 × 106 mm 4 .......................................................................... Ans. 120 (160 )3 2 (160 )(120 ) Iy = + ( 320 3) 36 2 2 π ( 60 )4 8 ( 60 )4 π ( 30 )4 2 π ( 60 ) 2 2 − − + (185.46 ) + (160 ) π ( 30 ) − 9π 2 4 8 = 245.78 × 106 mm 4 ≅ 246 × 106 mm 4 ........................................................................ Ans. (b) 160 (120 ) π ( 60 ) 2 A= − − π ( 30 ) = 12, 427 mm 2 2 2 2 xC = yC = ( 320 3) 160 (120 ) ( 40 ) 160 (120 ) 2 2 2 + (185.46 ) π ( 60 ) 2 − (160 ) π ( 30 ) = 130.39 mm 12, 427 2 2 2 + ( 60 ) π ( 60 ) 2 − ( 60 ) π ( 30 ) = 44.55 mm 12, 427 2 2 I xC = I x − yC A = 37.68 ×106 − ( 44.55 ) (12, 427 ) = 13.02 × 106 mm 4 ...................... Ans. 2 I yC = I y − xC A = 245.78 × 106 − (130.39 ) (12, 427 ) = 34.5 × 106 mm 4 .................... Ans. 2 8-9 Determine the second moments of the shaded area shown in Fig. P8-9 with respect to the x- and y-axes. SOLUTION yC 2 = 10 − 4 ( 4) 4R = 10 − = 8.302 in. 3π 3π 2 A2 = A3 = π R 2 2 = π ( 4 ) 2 = 25.13 in 2 Ix = 3 4 4 π ( 4 )4 20 (10 ) π ( 4 ) 8 ( 4 ) 2 − − + ( 8.302 ) ( 25.13) − 3 9π 8 8 = 6667 − 1760 − 101 = 4806 in 4 ≅ 4810 in 4 ............................................................... Ans. 3 4 π ( 4 )4 10 ( 20 ) π ( 4 ) 2 2 Iy = − + ( 5 ) ( 25.13) − + (15 ) ( 25.13) 3 8 8 = 26, 667 − 729 − 5755 = 20,183 in 4 ≅ 20, 200 in 4 .................................................. Ans. 308 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 8-10 RILEY, STURGES AND MORRIS The maximum flexural stress on a transverse plane in a beam with a rectangular cross section 150 mm wide × 300 mm deep is 15 MPa. Determine the resisting moment Mr transmitted by the plane. SOLUTION 150 ( 300 ) I= = 337.5 × 106 mm 4 = 337.5 × 10−6 m 4 12 M c M ( 0.150 ) σ max = r = r = 15 × 106 N/m 2 −6 337.5 ×10 I 3 M r = 33.8 ×103 N ⋅ m = 33.8 kN ⋅ m ................................................................................. Ans. 8-11 The maximum flexural stress on a transverse plane in a beam with a rectangular cross section 4 in. wide × 8 in. deep is 1500 psi. Determine the resisting moment Mr transmitted by the plane. SOLUTION 4 ( 8) I= = 170.67 in 4 12 3 σ max = M r c M r ( 4) = = 1500 psi 170.67 I M r = 64.0 ×103 lb ⋅ in. = 64.0 kip ⋅ in. ............................................................................... Ans. 8-12 Determine the maximum flexural stress required to produce a resisting moment Mr of –15 kN-m if the beam has the cross section shown in Fig. P8-12. SOLUTION yC = 100 50 ( 200 ) + 25 50 ( 200 ) + 100 50 ( 200 ) 50 ( 200 ) + 50 ( 200 ) + 50 ( 200 ) 3 3 3 = 75 mm 100 (125 ) 300 ( 75 ) 200 ( 25) I= + − 3 3 3 = 106.25 × 106 mm 4 = 106.25 × 10−6 m 4 σ =− 8-13 ( −15 ×103 ) ( 0.125) = 17.65 ×106 N/m2 = 17.65 MPa (T) ............. Ans. Mr y =− 106.25 ×10−6 I Determine the maximum flexural stress required to produce a resisting moment Mr of +5000 lb-ft if the beam has the cross section shown in Fig. P8-13. SOLUTION 1 6 ( 2 ) + 5 6 ( 2 ) = 3 in. yC = 6 ( 2) + 6 ( 2) 2 ( 5 ) 6 ( 3) 4 (1) I= + − = 136 in 4 3 3 3 ( 5000 ×12 )( 5 ) = −2206 lb/in.2 ≅ 2210 psi (C) ................................ Ans. Mc σ =− r =− 136 I 3 3 3 8-14 A timber beam consists of three 50×200-mm planks fastened together to form an I-beam 200 mm wide × 300 mm deep, as shown in Fig. P8-14. If the flexural stress at point A of the cross section is 7.5 MPa (C), determine the flexural stress (a) At point B of the cross section. (b) At point C of the cross section (c) At point D of the cross section. 309 STATICS AND MECHANICS OF MATERIALS, 2nd Edition SOLUTION 3 3 RILEY, STURGES AND MORRIS 200 ( 300 ) 150 ( 200 ) I= − = 350 × 106 mm 4 = 350 × 10−6 m 4 12 12 M ( 0.150 ) My σA = − r A = − r = −7.5 ×106 N/m 2 M r = 17.5 ×103 N ⋅ m 350 ×10−6 I (a) (17.5 ×103 ) ( −0.100 ) = 5.00 ×106 N/m2 = 5.00 MPa (T) ......... Ans. M r yB =− σB = − 350 × 10−6 I σC = − (b) (c) 8-15 (17.5 ×103 ) ( −0.125) = 6.25 ×106 N/m2 = 6.25 MPa (T) ......... Ans. M r yD =− σD = − 350 × 10−6 I A timber beam is made of three 2×6-in. planks fastened together to form an I-beam 6 in. wide × 10 in. deep. If the maximum flexural stress must not exceed 1200 psi, determine the maximum resisting moment Mr that the beam can support. SOLUTION (17.5 ×103 ) ( 0.050 ) = −2.50 ×106 N/m2 = 2.50 MPa (C) ......... Ans. M r yC =− 350 ×10−6 I I= 6 (10 ) 4 ( 6 ) − = 428 in 4 12 12 3 3 σ max = M r c M r ( 5) = = 1200 psi 428 I M r = 102.7 × 103 lb ⋅ in. = 102.7 kip ⋅ in. ........................................................................... Ans. 8-16 The beam of Fig. P8-16 is made of material that has a yield strength of 250 MPa. Determine the maximum resisting moment that the beam can support if yielding must be avoided. SOLUTION yC = 25 100 ( 50 ) + 150 200 ( 37.5 ) 100 ( 50 ) + 200 ( 37.5) 3 3 = 100 mm 3 100 (100 ) 62.5 ( 50 ) 37.5 (150 ) I= − + = 72.92 ×106 mm 4 3 3 3 M c M ( 0.150 ) σ max = r = r = 250 × 106 N/m 2 −6 72.92 ×10 I M r = 121.5 ×103 N ⋅ m = 121.5 kN ⋅ m ............................................................................. Ans. 8-17 The maximum flexural stress on a transverse cross section of a beam with a 2×4-in. rectangular cross section must not exceed 10 ksi. Determine the maximum resisting moment Mr that the beam can support if the neutral axis is (a) Parallel to the 4-in. side. (b) Parallel to the 2-in. side. SOLUTION (a) 4 ( 2) I= = 2.667 in 4 12 3 σ max = M r c M r (1) = = 10 ksi 2.667 I M r = 26.7 kip ⋅ in. .................................................................................................................. Ans. 310 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 3 RILEY, STURGES AND MORRIS (b) 2 ( 4) I= = 10.667 in 4 12 M c M ( 2) σ max = r = r = 10 ksi I 10.667 M r = 53.3 kip ⋅ in. .................................................................................................................. Ans. A hardened steel (E = 210 GPa) bar with a 50-mm square cross section is subjected to a flexural form of loading that produces a flexural strain of +1200 µm/m at a point on the top surface of the beam. Determine (a) The maximum flexural stress at the point. (b) The resisting moment Mr developed in the beam on a transverse cross section through the point. SOLUTION 8-18 (a) σ = Eε = ( 210 ×109 )(1200 × 10−6 ) = 252.0 × 106 N/m 2 = 252 MPa ............................................................ Ans. (b) 50 ( 50 ) I= = 0.5208 ×106 mm 4 = 0.5208 ×10−6 m 4 12 M ( 0.025 ) My σ =− r =− r = 252.0 ×106 N/m 2 −6 I 0.5208 ×10 3 M r = −5.25 ×103 N ⋅ m = −5.25 kN ⋅ m ............................................................................ Ans. 8-19 A steel (E = 29,000 ksi) bar with a rectangular cross section is bent over a rigid mandrel (R = 10 in.) as shown in Fig. P8-19. If the maximum flexural stress in the bar is not to exceed the yield strength (σy = 36 ksi) of the steel, determine the maximum allowable thickness h for the bar. SOLUTION ρ = R+ h= 8-20 h h = 10 + in. 2 2 σ x = Eε x = E E ( h 2) c = ρ 10 + ( h 2 ) 20 ( 36 ) 20σ x = = 0.0249 in. ............................................................................ Ans. E − σ x 29, 000 − 36 An aluminum alloy (E = 73 GPa) bar with a rectangular cross section is bent over a rigid mandrel as shown in Fig. P8-19. The thickness h of the bar is 25 mm. If the maximum flexural stress in the bar must be limited to 200 MPa, determine the minimum allowable radius R for the mandrel. SOLUTION ρ = R+ R= 8-21 h 25 = R+ = R + 12.5 mm 2 2 σ x = Eε x = E E (12.5 ) c = ρ R + (12.5 ) E (12.5 ) 73, 000 (12.5 ) − 12.5 = − 12.5 = 4550 mm = 4.55 m ............................ Ans. σx 200 The load carrying capacity of an S24×80 American Standard beam (see Appendix A for dimensions) is to be increased by fastening two 8 × ¾-in. plates to the flanges of the beam, as shown in Fig. P8-21. The maximum flexural stress in both the original and modified beams must be limited to 15 ksi. Determine (a) The maximum resisting moment that the original beam can support. (b) The maximum resisting moment that the modified beam can support. SOLUTION 311 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (a) From Table A-3 for an RILEY, STURGES AND MORRIS S 24 × 80 section: d = 24.00 in. I = 2100 in 4 σ max = M r c M r (12.00 ) = = 15 ksi I 2100 M r = 2625 kip ⋅ in. ≅ 2630 kip ⋅ in. .................................................................................... Ans. 8 ( 0.75 )3 2 I = 2100 + 2 + (12.375 ) ( 8 × 0.75 ) = 3938 in 4 12 (b) σ max = M r c M r (12.75 ) = = 15 ksi I 3938 M r = 4633 kip ⋅ in. ≅ 4630 kip ⋅ in. .................................................................................... Ans. 8-22 Two L102×102×12.7-mm structural steel angles (see Appendix A for dimensions) are attached back to back to form a T-section, as shown in Fig. P8-22. Determine the maximum resisting moment Mr that can be supported by the beam if the maximum flexural stress must be limited to 125 MPa. SOLUTION From Table A-8 for an L102 × 102 ×12.7 angle: S = 32.3 × 103 mm 3 = 32.3 × 10−6 m3 σ max = Mr Mr = = 125 × 106 N/m 2 S 2 ( 32.3 × 10 −6 ) M r = 8.08 ×103 N ⋅ m = 8.08 kN ⋅ m .................................................................................. Ans. 8-23 An I-beam is fabricated by welding two 16×2 in. flange plates to a 24×1-in. web plate. The beam is loaded in the plane of symmetry parallel to the web. On a section where the resisting moment Mr = 1400 kip-ft, determine the maximum flexural stress. SOLUTION 16 ( 28 ) 15 ( 24 ) I= − = 11,989 in 4 12 12 (1400 ×12 )(14 ) = −19.62 kip/in.2 = 19.62 ksi (C) Mc σ top = − r = − 11,989 I 3 3 σ bottom = − (1400 ×12 )( −14 ) = +19.62 kip/in.2 = 19.62 ksi (T) M rc =− 11,989 I σ max = 19.62 ksi (T on bottom & C on top) ................................................................... Ans. 8-24 Determine the maximum tensile and compressive flexural stresses on a section where the resisting moment Mr = −4 kN-m if the beam has the cross section shown in Fig. P8-24. The beam is loaded in the plane of symmetry parallel to the web. SOLUTION yC = 237.5 100 ( 25 ) + 125 200 ( 25 ) + 12.5 200 ( 25 ) 100 ( 25) + 200 ( 25 ) + 200 ( 25) 3 3 3 = 102.50 mm 3 100 (147.5 ) 75 (122.5) 200 (102.5) 175 ( 77.5 ) − + − I= 3 3 3 3 6 4 4 −6 = 105.65 × 10 mm = 105.65 ×10 m 312 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS σ top σ bot 8-25 ( −4 ×103 ) ( 0.1475) = +5.58 ×106 N/m2 = 5.58 MPa (T) .......... Ans. Mr y =− =− I 105.65 ×10−6 ( −4 ×103 ) ( −0.1025) = −3.88 ×106 N/m2 = 3.88 MPa (C) ....... Ans. Mr y =− =− I 105.65 ×10−6 Determine the maximum tensile and compressive flexural stresses on a section where the resisting moment Mr = +30,000 lb-in. if the beam has the cross section shown in Fig. P8-25. The beam is loaded in the vertical plane of symmetry. SOLUTION yC = 4R 4 ( 2) = = 0.8488 in. 3π 3π 4 4 π R 4 8R 4 π ( 2 ) 8 ( 2 ) − = − = 1.7561 in 4 I= 8 9π 8 9π ( 30 )(1.1512 ) = −19.67 kip/in.2 = 19.67 ksi (C) .......................... Ans. Mc σ top = − r = − 1.7561 I ( 30 )( −0.8488 ) = +14.50 kip/in.2 = 14.50 ksi (T) ................... Ans. Mc σ bottom = − r = − 1.7564 I 8-26 Determine the maximum resisting moment Mr that can be supported by a beam having the cross section shown in Fig. P8-26 if the maximum flexural stress must be limited to 120 MPa. SOLUTION 50 ( 50 ) π (10 ) − = 0.5130 × 106 mm 4 = 0.5130 × 10−6 m 4 I= 12 4 M r ( 0.025) Mc σ max = r = = 120 × 106 N/m 2 0.5130 ×10−6 I 3 4 M r = 2.46 × 103 N ⋅ m = 2.46 kN ⋅ m ................................................................................. Ans. 8-27 A beam has the cross section shown in Fig. P8-27. On a section where the resisting moment is –30 kip-ft, determine (a) The maximum tensile flexural stress. (b) The maximum compressive flexural stress. SOLUTION yC = 0.5 8 (1) + 3 4 (1) + 7 π ( 22 − 1.52 ) 8 (1) + 4 (1) + π ( 22 − 1.52 ) = 3.113 in. 8 (1)3 1( 2.113)3 1(1.887 )3 2 + ( 2.613) ( 8 × 1) + + I = 3 3 12 π ( 24 − 1.54 ) 2 + + ( 3.887 ) π ( 22 − 1.52 ) = 152.33 in 4 4 σ top = − ( −30 ×12 )( 5.887 ) = +13.91 kip/in.2 = 13.91 ksi (T) ................... Ans. Mrc =− I 152.33 313 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS σ bottom = − 8-28 ( −30 ×12 )( −3.113) = −7.36 kip/in.2 = 7.36 ksi (C) ............... Ans. M rc =− I 152.33 Determine the maximum flexural stress on a section where the resisting moment Mr = +100 kN-m if the beam has the cross section shown in Fig. P8-28. The beam is loaded in the vertical plane of symmetry. SOLUTION A = 250 ( 25 ) + 150 ( 25 ) + π ( 50 ) = 17,854 mm 2 2 2 12.5 ( 250 ( 25 ) ) + 100 150 ( 25) + 225 π ( 50 ) = 124.36 mm yC = 17,854 250 ( 25 )3 25 ( 99.36 )3 25 ( 50.64 )3 2 + (111.86 ) ( 250 × 25 ) + + I = 3 3 12 π ( 50 )4 2 + + (100.64 ) π ( 502 ) = 172.24 × 106 mm 4 = 172.24 × 10−6 m 4 4 3 6 σ bot 8-29 ( −100 ×10 ) ( −0.15064 ) = +87.5 ×10 =− 172.24 × 10 −6 N/m 2 = 87.5 MPa (T) .................. Ans. The cantilever beam shown in Fig. P8-29 is subjected to a moment M = 20,000 lb-in. at its free end. The beam has a 2×2-in. cross section. Determine the maximum tensile flexural stress in the beam. SOLUTION 2 ( 2) I= = 1.3333 in 4 12 ( 20 )( −1) = +15.00 kip/in.2 = 15.00 ksi (T) .............................. Ans. Mc σ bottom = − r = − I 1.3333 3 8-30 The cantilever beam shown in Fig. P8-30a is subjected to a moment M at its free end. The cross section of the beam is shown in Fig. P8-30b. If the allowable stresses are 110 MPa (T) and 170 MPa (C), determine the maximum moment that can be applied to the beam. SOLUTION yC = 140 120 ( 40 ) + 60 120 ( 40 ) 120 ( 40 ) + 120 ( 40 ) = 100 mm 120 ( 60 )3 40 (120 )3 2 2 + ( 40 ) (120 × 60 ) + + ( 40 ) ( 60 × 120 ) I = 12 12 −6 6 4 4 = 21.76 × 10 mm = 21.76 × 10 m At the top of the beam, σ = 110 MPa (T) σ top = − M ( 0.060 ) Mr y =− r ≤ 110 × 106 N/m 2 −6 I 21.76 × 10 M r ≤ 39.89 ×103 N ⋅ m ≅ 39.9 kN ⋅ m At the bottom of the beam, σ = 170 MPa (C) 314 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS σ bot = − M ( −0.100 ) Mr y =− r ≥ −170 × 106 N/m 2 −6 I 21.76 × 10 M r ≤ 36.99 ×103 N ⋅ m ≅ 37.0 kN ⋅ m M max = 37.0 kN ⋅ m ............................................................................................................... Ans. 8-31 A steel pipe with an outside diameter of 4 in. and an inside diameter of 3 in. is simply supported at its ends and carries two concentrated loads, as shown in Fig. P8-31. On a section 5 ft from the right support, determine the flexural stress (a) At point A (top, outside) on the cross section. (b) At point B (bottom, inside) on the cross section. SOLUTION ΣM C = 0 : ΣM cut = 0 : RD ( 20 ) − 2000 (18) − 1000 ( 3) = 0 1950 ( 5) − 2000 ( 3) − M = 0 RD = 1950 lb M = 3750 lb ⋅ ft I= π ( 24 − 1.54 ) 4 = 8.590 in 4 (a) σA = − ( 3750 ×12 )( 2 ) Mr y =− 8.590 I ( 3750 ×12 )( −1.5 ) = +7858 lb/in.2 ≅ 7.86 ksi (T) ......................... Ans. Mr y =− I 8.590 = −10, 477 lb/in.2 ≅ 10.48 ksi (C) .............................................................................. Ans. (b) 8-32 σB = − A beam has the cross section shown in Fig. P8-32 and is loaded in the vertical plane of symmetry. On a section where the resisting moment Mr = +50 kN-m, determine (a) The maximum tensile flexural stress. (b) The maximum compressive flexural stress. SOLUTION AO = π ( 502 − 402 ) = 2827 mm 2 IO = yC = π ( 504 − 404 ) 4 = 2.898 × 106 mm 4 = 52.82 mm 3 ( 2827 ) + 30 (135 ) + 30 (150 ) 82.5 30 (135 ) + 200 ( 2827 ) 150 ( 30 )3 2 2 + ( 52.82 ) ( 30 × 150 ) I = 2 ( 2.898 × 106 ) + ( 52.82 ) ( 2827 ) + 12 30 (135 )3 2 2 + + ( 29.68 ) ( 30 × 135 ) + ( 2.898 ×106 ) + (147.18) ( 2827 ) 12 315 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 8-40 RILEY, STURGES AND MORRIS A beam is loaded and supported as shown in Fig. P8-40. Using the coordinate axes shown, write equations for the shear force V and the bending moment M for any section of the beam in the interval 0 < x < 4 m. SOLUTION From overall equilibrium: ΣM B = 0 : ↑ ΣFy = 0 : ΣM cut = 0 : 20 ( 2 ) + 15 ( 4 ) ( 6 ) − RA ( 8 ) = 0 0 m ≤ x ≤ 4 m: RA = 50 kN For the initial section of the beam, RA − 15 x − V = 0 M + (15 x )( x 2 ) − RA x = 0 V = ( −15 x + 50 ) kN ............................................................................................................. Ans. M = ( −7.5 x 2 + 50 x ) kN ⋅ m ................................................................................................ Ans. 8-41 A beam is loaded and supported as shown in Fig. P8-41. Using the coordinate axes shown, write equations for the shear force V and the bending moment M for any section of the beam in the interval 0 < x < 10 ft. SOLUTION From overall equilibrium: ΣM B = 0 : ↑ ΣFy = 0 : ΣM cut = 0 : 3000 + 200 (10 ) ( 5 ) − 500 ( 2 ) − RA (10 ) = 0 0 ft ≤ x ≤ 10 ft : RA = 1200 lb For the middle section of the beam, RA − 200 x − V = 0 M + 3000 + 200 x ( x 2 ) − RA x = 0 V = ( −200 x + 1200 ) lb ......................................................................................................... Ans. M = 1200 x − 3000 − 100 x 2 = ( −100 x 2 + 1200 x − 3000 ) lb ⋅ ft ................................... Ans. 8-42 A beam is loaded and supported as shown in Fig. P8-42. Using the coordinate axes shown, write equations for the shear force V and the bending moment M for any section of the beam in the interval 0 < x < L. SOLUTION From overall equilibrium: ↑ ΣFy = 0 : ΣM A = 0 : RA + wL − 2wL = 0 M R + ( wL )( L 2 ) + ( 3wL2 2 ) − ( 2wL )( 3L 2 ) = 0 RA = wL For the initial section of the beam, M R = wL2 0≤ x≤ L: ↑ ΣFy = 0 : ΣM cut = 0 : RA + wx − V = 0 M − ( wx )( x 2 ) − RA x + M R = 0 V = wL + wx = w ( L + x ) N ................................................................................................. Ans. M = ( wx 2 2 ) + wLx − wL2 = ( w 2 ) ( x 2 + 2 Lx − 2 L2 ) N ⋅ m ........................................ Ans. 8-43 The beam shown in Fig. P8-39 has a solid rectangular cross section that is 3 in. wide and 8 in. deep. Determine the maximum tensile flexural stress on a section at x = 10ft. SOLUTION 319 STATICS AND MECHANICS OF MATERIALS, 2nd Edition From overall equilibrium: RILEY, STURGES AND MORRIS ΣM A = 0 : At RB (12 ) − 2000 ( 4 ) − 5000 ( 8) = 0 RB ( 2 ) − M 10 = 0 3 RB = 4000 lb M 10 = 8000 lb ⋅ ft x = 10 ft (using the right-hand segment of the beam), ΣM cut = 0 : I = 3 ( 8 ) 12 = 128 in 4 On the bottom of the beam, y = −4 in. σ =− 8-44 ( 8000 ×12 )( −4 ) = +3000 lb/in.2 = 3000 psi (T) .............................. Ans. Mr y =− 128 I The beam shown in Fig. P8-40 is an S254×52 steel section. On a section at x = 5 m, determine the maximum tensile and compressive flexural stresses. SOLUTION From overall equilibrium: ΣM B = 0 : At 20 ( 2 ) + 15 ( 4 ) ( 6 ) − RA ( 8 ) = 0 M 5 + (15 )( 4 ) ( 3) − RA ( 5 ) = 0 S 254 × 52 section: y = −127.0 mm 3 6 −6 RA = 50 kN x = 5 m (using the left-hand segment of the beam), ΣM cut = 0 : M 5 = 70 kN ⋅ m From Table A-4 for an d = 254.00 mm I = 61.2 × 106 mm 4 At the bottom of the beam, 61.2 × 10 At the top of the beam, y = +127.0 mm 3 σ bot ( 70 ×10 ) ( −0.1270 ) = +145.26 ×10 =− N/m 2 ≅ 145.3 MPa (T) .................. Ans. σ top 8-45 (a) (b) (c) (d) ( 70 ×10 ) ( +0.1270 ) = −145.26 ×10 =− 61.2 × 10−6 6 N/m 2 ≅ 145.3 MPa (C) .................. Ans. A beam is loaded and supported as shown in Fig. P8-45. Using the coordinate axes shown, write equations for the shear force V and the bending moment M for any section of the beam In the interval –3 ft < x < 0. In the interval 0 < x < 2 ft. In the interval 2 ft < x < 8 ft. In the interval 8 ft < x < 10 ft. SOLUTION From overall equilibrium: ΣM A = 0 : ΣM B = 0 : (a) 2000 ( 3) + RB (10 ) − 1000 ( 6 ) ( 5 ) − 2000 ( 8 ) = 0 2000 (13) − RA (10 ) + 1000 ( 6 ) ( 5) + 2000 ( 2 ) = 0 RB = 4000 lb RA = 6000 lb V = −2000 lb .......................................................................................................................... Ans. M = −2000 ( x + 3) = ( −2000 x − 6000 ) lb ⋅ ft ................................................................. Ans. (b) V = RA − 2000 = 6000 − 2000 = 4000 lb ......................................................................... Ans. 320 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ΣM cut = 0 : M + 2000 ( x + 3) − RA x = 0 M = RA x − 2000 ( x + 3) = ( 4000 x − 6000 ) lb ⋅ ft ........................................................... Ans. (c) V = RA − 2000 − 1000 ( x − 2 ) = ( −1000 x + 6000 ) lb .................................................... Ans. ΣM cut = 0 : x−2 M + 2000 ( x + 3) − RA x + 1000 ( x − 2 ) 2 =0 x−2 M = RA x − 2000 ( x + 3) − 1000 ( x − 2 ) 2 = ( −500 x 2 + 6000 x − 8000 ) lb ⋅ ft ................................................................................ Ans. (d) V = − RB = −4000 lb .............................................................................................................. Ans. M = RB (10 − x ) = ( −4000 x + 40, 000 ) lb ⋅ ft .................................................................. Ans. A beam is loaded and supported as shown in Fig. P8-46. Using the coordinate axes shown, write equations for the shear force V and the bending moment M for any section of the beam In the interval –2 m < x < 0. In the interval 0 < x < 4 m. In the interval 4 m < x < 6 m. In the interval 6 m < x < 10 m. 8-46 (a) (b) (c) (d) SOLUTION From overall equilibrium: ΣM A = 0 : ΣM B = 0 : (a) 6 + RB (10 ) − 12 ( 6 ) (1) − 24 ( 6 ) = 0 6 + 12 ( 6 ) ( 9 ) + 24 ( 4 ) − RA (10 ) = 0 RB = 21 kN RA = 75 kN V = −12 ( x + 2 ) = ( −12 x − 24 ) kN .................................................................................... Ans. x+2 2 M = −6 − 12 ( x + 2 ) 2 = ( −6 x − 24 x − 30 ) kN ⋅ m ........................................ Ans. (b) V = −12 ( x + 2 ) + RA = ( −12 x + 51) kN ........................................................................... Ans. x+2 2 M = −6 − 12 ( x + 2 ) 2 + RA x = ( −6 x + 51x − 30 ) kN ⋅ m ............................ Ans. (c) V = 24 − RB = 3 kN ............................................................................................................... Ans. M = RB (10 − x ) − 24 ( 6 − x ) = ( 3x + 66 ) kN ⋅ m ............................................................ Ans. V = − RB = −21 kN ................................................................................................................. Ans. M = RB (10 − x ) = ( −21x + 210 ) kN ⋅ m .......................................................................... Ans. (d) 8-47 A beam is loaded and supported as shown in Fig. P8-47. Using the coordinate axes shown, (a) Write equations for the shear force V and the bending moment M for any section of the beam. (b) Determine the magnitudes and locations of the maximum shear force and the maximum bending moment in the beam. SOLUTION From overall equilibrium: 321 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ΣM B = 0 : (a) 1500 (12 ) 2 (12 3) − RA (12 ) = 0 RA = 3000 lb V = RA − (1500 x 12 ) x = 2 ( −62.5 x 2 + 3000 ) lb .............................................................. Ans. (1500 x 12 ) x x 3 M = RA x − = ( −20.833 x + 3000 x ) lb ⋅ ft ................................. Ans. 2 3 (b) The shear force is monotonically decreasing on the interval will occur either at the beginning or the end of the interval. ( dV dx = −125 x ) . Therefore, the maximum Vx =0 = 3000 lb Vx =12 = −62.5 (12 ) + 3000 = −6000 lb = Vmax ................................................................. Ans. 2 The maximum bending moment in the interval will occur when dM dx = −62.5 x 2 + 3000 = 0 M x =6.928 = 13,858 lb ⋅ ft Checking the ends of the interval x = 3000 62.5 = 6.928 ft M x = 0 = 0 lb ⋅ ft Therefore M x =12 = 0 lb ⋅ ft M max = M x =6.928 = 13,858 lb ⋅ ft ≅ 13.86 kip ⋅ ft ................................................. Ans. 8-48 A beam is loaded and supported as shown in Fig. P8-48. Using the coordinate axes shown, (a) Write equations for the shear force V and the bending moment M for any section of the beam. (b) Determine the magnitudes and locations of the maximum shear force and the maximum bending moment in the beam. SOLUTION From overall equilibrium: ΣM A = 0 : (a) 25 ( 5 ) 2 ( 5 3) − RB ( 5 ) = 0 RB = 20.833 kN V= 1 25 ( 5 − x ) 2 ( 5 − x ) − RB = ( 2.5 x − 25 x + 41.667 ) kN ................................... Ans. 2 5 25 ( 5 − x ) 2 ( 5 − x ) M = RB ( 5 − x ) − 10 3 = ( 0.8333x3 − 12.5 x 2 + 41.667 x ) kN ⋅ m ................................................................... Ans. (b) The shear force is monotonically decreasing on the interval ( dV dx = 5 x − 25 < 0 ) . Therefore, the maximum will occur either at the beginning or the end of the interval. Vx =0 = 41.667 kN ≅ 41.7 kN = Vmax .................................................................................. Ans. Vx =5 = −20.833 kN The maximum bending moment in the interval will occur when dM dx = 2.5 x 2 − 25 x + 41.667 = 0 M x = 2.113 = 40.1 kN ⋅ m x = 2.113 m, 7.89 m (The second root, 7.89 m is not in the interval and is not a valid solution.) Checking the ends of the interval 322 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS M x =0 = 0 kN ⋅ m Therefore M x =5 = 0 kN ⋅ m M x = 2.113 = 40.1 kN ⋅ m = M max .............................................................................. Ans. 8-49 An S8×23 steel beam (see Appendix A) is loaded and supported as shown in Fig. P8-49. (a) Using the coordinate axes shown, write equations for the shear force V and the bending moment M for any section of the beam in the interval 6 ft < x < 8 ft. (b) Determine the flexural stress at a point 1 in. below the top of the beam on a section at x = 3 ft. (c) Determine the maximum flexural stress on a section at x = 3 ft. SOLUTION From overall equilibrium: ΣM B = 0 : (a) 500 ( 6 ) ( 7 ) + 3000 ( 2 ) − RA (10 ) − 800 ( 5 ) ( 2.5 ) = 0 RA = 1700 lb V = RA − 500 ( 6 ) = −1300 lb ............................................................................................... Ans. M = RA x − 500 ( 6 ) ( x − 3) = ( −1300 x + 9000 ) lb ⋅ ft ................................................ Ans. (b) From Table A-3 for an S8 × 23 section: d = 8.00 in. S = 16.2 in 2 I = 64.9 in 4 M 3 = RA ( 3) − 500 ( 3) (1.5 ) = 2850 lb ⋅ ft (c) ( 2850 ×12 )( 3) = −1581 lb/in.2 = 1581 psi (C) .................................... Ans. My =− I 64.9 ( 2850 ×12 ) = 2111 lb/in.2 ≅ 2110 psi (C top & T bottom) ................. Ans. M σ= =− S 16.2 σ =− 8-50 An S178×30 steel beam (see Appendix A) is loaded and supported as shown in Fig. P8-50. (a) Using the coordinate axes shown, write equations for the shear force V and the bending moment M for any section of the beam in the interval 0.5 m < x < 2.5 m. (b) Determine the flexural stress at a point 15 mm above the bottom of the beam on a section at x = 1.5 m. (c) Determine the maximum flexural stress on a section at x = 1.5 m. SOLUTION From overall equilibrium: ΣM B = 0 : (a) 9 ( 4 ) + 15 ( 2 ) (1.5 ) − RA ( 3) = 0 RA = 27 kN V = −9 + RA − 15 ( x − 0.5) = ( −15 x + 25.5 ) kN .............................................................. Ans. x − 0.5 M = RA x − 9 ( x + 1) − 15 ( x − 0.5 ) 2 = ( −7.5 x 2 + 25.5 x − 10.88 ) kN ⋅ m ............................................................................... Ans. S178 × 30 section: 2 (b) From Table A-4 for an d = 177.8 mm I = 17.6 × 106 mm 4 S = 198 × 103 mm3 M 1.5 = −7.5 (1.5 ) + 25.5 (1.5 ) − 10.875 = 10.500 kN ⋅ m y = − (177.8 2 ) + 15 = −73.9 mm 323 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS (10.5 ×103 ) ( −0.0739 ) = +44.1×106 N/m2 = 44.1 MPa (T) ............ Ans. My =− σ =− 17.6 × 10−6 I (c) 8-51 3 M (10.5 × 10 ) = = 53.0 × 106 N/m 2 = 53.0 MPa (C top & T bottom) ........... Ans. σ= −6 S 198 ×10 A beam is loaded and supported as shown in Fig. P8-51. On a section 6 ft to the right of support A, determine the maximum tensile and compressive flexural stresses if the load P = 3000 lb. SOLUTION From overall equilibrium: ΣM B = 0 : yC = 3000 ( 6 ) − RA (10 ) = 0 0.5 4 (1) + 5 1( 8) 4 (1) + 1( 8 ) = 3.500 in. RA = 1800 lb M 6 = RA ( 6 ) − 3000 ( 2 ) = 4800 lb ⋅ ft At the top of the section, 4 (1)3 1 ( 8 )3 2 2 + ( 3) ( 4 ×1) + + (1.5 ) (1× 8) = 97.00 in 4 I = 12 12 y = +5.5 in. ( 4800 ×12 )( 5.5 ) = −3266 lb/in.2 ≅ 3.27 ksi (C) ................................ Ans. My =− I 97.00 At the bottom of the section, y = −3.5 in. σ =− σ =− 8-52 ( 4800 ×12 )( −3.5) = +2078 lb/in.2 ≅ 2.08 ksi (T) ............................. Ans. My =− I 97.00 A beam is loaded and supported as shown in Fig. P8-52. On a section 3 m to the right of support A, determine the maximum tensile and compressive flexural stresses if w = 3.5 kN/m. SOLUTION From overall equilibrium: ΣM A = 0 : yC = RB ( 4 ) − 3500 ( 2 ) (1) = 0 RB = 1750 N 140 120 ( 40 ) + 60 40 (120 ) 120 ( 40 ) + 40 (120 ) = 100 mm 120 ( 40 )3 40 (120 )3 2 2 + ( 40 ) (120 × 40 ) + + ( 40 ) ( 40 × 120 ) I = 12 12 6 4 4 −6 = 21.76 × 10 mm = 21.76 × 10 m M 3 = RB (1) = 1750 N ⋅ m At the top of the section, y = +60 mm : (1750 )( 0.060 ) = −4.83 ×106 N/m 2 = 4.83 MPa (C) ........................ Ans. My =− I 21.76 × 10−6 At the bottom of the section, y = −100 mm σ =− 324 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS σ =− (1750 )( −0.100 ) = +8.04 ×106 N/m 2 = 8.04 MPa (T) ...................... Ans. My =− I 21.76 × 10−6 8-53 A beam is loaded and supported as shown in Fig. P8-53. Using the coordinate axes shown, (a) Write equations for the shear force V and the bending moment M for any section of the beam. (b) Determine the magnitudes and locations of the maximum shear force and the maximum bending moment in the beam. SOLUTION RA = RB = 0.5∫ w ds = 0.5∫ 1000sin (π s 10 ) ds 0 0 10 10 5000 πs 10, 000 =− lb cos 10 = π π 0 (a) 10 V = RA − ∫ = x 0 x 10, 000 10, 000 10, 000 π s πs − ∫ 1000sin ds = + w ds = cos 0 π π π 10 0 10 x 10, 000 πx πx cos lb ≅ 3.18cos kip ......................................................................... Ans. π 10 10 x x 10, 000 x πs − ∫ 1000 ( x − s ) sin ds M = RA x − ∫ w ( x − s ) ds = 0 0 π 10 10, 000 x 10, 000 x π s 100, 000 π s 10, 000 π s = + cos 10 + π 2 sin 10 − π s cos 10 π π 0 0 0 = (b) x x x Vmax 100, 000 πx πx sin lb ⋅ ft ≅ 10.13sin kip ⋅ ft ....................................................... Ans. 2 π 10 10 = Vx = 0 = Vx =10 = 10, 000 π lb ≅ 3.18 kip .................................................................. Ans. 2 M max = M x =5 = 100, 000 π lb ⋅ ft ≅ 10.13 kip ⋅ ft ......................................................... Ans. 8-54 A beam is loaded and supported as shown in Fig. P8-54. Using the coordinate axes shown, (a) Write equations for the shear force V and the bending moment M for any section of the beam. (b) Determine the magnitudes and locations of the maximum shear force and the maximum bending moment in the beam. SOLUTION From overall equilibrium: ΣM B = 0 : − RA ( 4 ) + ∫ w ( 4 − s ) ds = 0 0 4 4 RA = 0.25∫ ( 4 − s ) 25 cos (π s 8 ) ds 0 4 4 4 200 400 50 s 400 = sin (π s 8 ) 0 − 2 cos (π s 8 ) 0 − sin (π s 8 ) 0 = 2 kN π π π π (a) V = RA − ∫ = x 0 x 400 400 200 π s πs w ds = 2 − ∫ 25cos ds = 2 − sin 0 π π π 8 0 8 x 400 200 sin (π x 8 ) kN ............................................................................................. Ans. − π2 π 325 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS M = RA x − ∫ w ( x − s ) ds = 0 x x 400 x πs − ∫ 25 ( x − s ) cos ds 2 0 π 8 x x x 400 x 200 x π s 1600 π s 200 π s = 2− sin 8 + π 2 cos 8 + π s sin 8 π π 0 0 0 = (b) 400 1600 x − 4 ) + 2 cos (π x 8 ) kN ⋅ m ................................................................... Ans. 2( π π Vmax = Vx = 0 = 400 π 2 kN ≅ 40.5 kN ................................................................................. Ans. The maximum bending moment in the interval will occur when dM 400 200 πx sin (π x 8 ) = 0 = 2− = 0.6901 rad x = 1.7573 m dx π π 8 M max = M x =1.7573 = 34.1 kN ⋅ m ............................................................................................ Ans. 8-55 A beam is loaded and supported as shown in Fig. P8-55. Using the coordinate axes shown, (a) Write equations for the shear force V and bending moment M for any section of the beam. (b) Determine the magnitudes and locations of the maximum shear force and the maximum bending moment in the beam. SOLUTION From overall equilibrium: ΣM B = 0 : − RA (10 ) + ∫ w (10 − s ) ds = 0 0 10 10 10 s 3 s 4 1 10 RA = ∫ (10 s 2 ) (10 − s ) ds = − = 833.3 lb 10 0 4 0 3 (a) V = RA − ∫ w ds = 833.3 − ∫ 10 s 2 ds ≅ ( −3.333 x3 + 833.3) lb .................................. Ans. x x 0 0 M = RA x − ∫ w ( x − s ) ds = 833.3x − ∫ 10s 2 ( x − s ) ds 0 0 3 x x ≅ ( −0.8333 x 4 + 833 x ) lb ⋅ ft ........................................................................................ Ans. (b) Vmax = Vx =10 = 833.3 − 3.333 (10 ) = −2500 lb ................................................................. Ans. The maximum bending moment in the interval will occur when dM dx = −3.333 x 3 + 833.3 = 0 x = 6.300 ft M max = M x =6.300 = 3937 lb ⋅ ft ≅ 3.94 kip ⋅ ft .................................................................... Ans. 8-56 A beam is loaded and supported as shown in Fig. P8-56. Using the coordinate axes shown, (a) Write equations for the shear force V and the bending moment M for any section of the beam. (b) Determine the magnitudes and locations of the maximum shear force and the maximum bending moment in the beam. SOLUTION From overall equilibrium: ΣM B = 0 : − RA ( 8) + ∫ w ( 8 − s ) ds = 0 0 8 326 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS RA = 8 18 ( 64 − s 2 ) (8 − s ) ds = 1 ∫0 ( s3 − 8s 2 − 64s + 512 ) ds 8 ∫0 8 8 1 s 4 8s 3 =− − 32s 2 + 512s = 213.3 kN 8 4 3 0 (a) V = RA − ∫ w ds = 213.3 − ∫ ( 64 − s 2 ) ds = 213.3 − 64 x + x x 0 0 ≅ ( 0.333 x3 − 64.0 x + 213 ) kN ........................................................................................ Ans. x3 3 M = RA x − ∫ w ( x − s ) ds = 213.3x − ∫ ( 64 − s 2 ) ( x − s ) ds x x 0 0 = 213.3x − ≅ ( 0.0833 x 4 − 32.0 x 2 + 213 x ) kN ⋅ m ....................................................................... Ans. (b) The maximum shear force in the interval will occur when x x + + 32 x 2 − 64 x 2 43 4 4 dV dx = x 2 − 64.0 = 0 Checking the beginning of the interval x=8 m Vx =8 = −128.0 kN Vx =0 = 213.3 kN ≅ 213 kN = Vmax ...................................................................................... Ans. The maximum bending moment in the interval will occur when dM dx = 0.3333 x3 − 64.0 x + 213.3 = 0 8-57 x = 3.570 m M max = M x =3.570 = 367 kN ⋅ m .............................................................................................. Ans. Two C10×15.3 steel channels (see Appendix A) are placed back to back to form a 10-in. deep beam, as shown in Fig. P8-57. The beam is 10 ft long and is simply supported at its ends. The beam carries a uniformly distributed load of 1000 lb/ft over its entire length and a concentrated load P at the center of the span, as shown in Fig. P8-57. If the maximum permissible flexural stress at the center of the span is 16,000 psi, determine the maximum permissible value of the concentrated load P. SOLUTION From overall equilibrium: ΣM B = 0 : 1000 (10 ) ( 5 ) + P ( 5 ) − RA (10 ) = 0 RA = 5000 + 0.5 P lb M P = RA ( 5 ) − 1000 ( 5 ) ( 2.5 ) = (12,500 + 2.5 P ) lb ⋅ ft From Table A-5 for a C10 ×15.3 section: S = 13.5 in 3 σ max = 8-58 M (12,500 + 2.5 P )(12 ) = ≤ 16 × 103 psi 2 × 13.5 S P ≤ 9400 lb ............................................................................................................................. Ans. Two 50×200-mm structural timbers are used to fabricate a beam with an inverted T cross section, as shown in Fig. P8-58. The beam is simply supported at the ends and is 4 m long. If the maximum flexural stress must be limited to 30 MPa, determine (a) The maximum moment that can be resisted by the beam. (b) The largest concentrated load that can be supported at the center of the span. (c) The largest uniformly-distributed load (over the entire span) that can be supported by the beam. 327 STATICS AND MECHANICS OF MATERIALS, 2nd Edition SOLUTION RILEY, STURGES AND MORRIS yC = 25 200 ( 50 ) + 150 50 ( 200 ) 200 ( 50 ) + 50 ( 200 ) = 87.5 mm 200 ( 50 )3 50 ( 200 )3 2 2 I = + ( 62.5 ) ( 200 × 50 ) + + ( 62.5 ) ( 50 × 200 ) 12 12 = 113.54 × 106 mm 4 = 113.54 × 10 −6 m 4 (a) σ max = M ( 0.1625 ) Mc =− = 30 ×106 N/m 2 −6 113.54 × 10 I M = 20.96 × 103 N ⋅ m ≅ 21.0 kN ⋅ m ................................................................................ Ans. For a concentrated load P: RA = RB = P 2 M max = M 2 = RA ( 2 ) = P P = M max = 21.0 kN ............................................................................................................. Ans. For a distributed load w: RA = RB = 4w 2 = 2w M max = M 2 = RA ( 2 ) − ( 2w )(1) = 2w w = M max 2 = 10.48 kN/m .................................................................................................. Ans. 8-59 The supports for the beam shown in Fig. P8-59 are symmetrically located. If the distance d from the supports to the ends of the beam is adjustable, compute and plot (a) The bending moment M(x) in the beam for d = 2 ft, 3 ft, and 4 ft as a function of x (0 ft ≤ x ≤ 15 ft). (b) The maximum bending moments (Mmax)AB in segment AB and (Mmax)BC in segment BC as functions of d (0 ft ≤ d ≤ 6 ft). (c) What value of d gives the smallest Mmax in the beam? SOLUTION RB = RC = wL 2 = 1200 (15) 2 = 9000 lb (a) For the initial section of the beam, 0 ft ≤ x ≤ d ft : M = −1200 x ( x 2 ) = −600 x 2 lb ⋅ ft ................................................................................... Ans. For the middle section of the beam, d ft ≤ x ≤ (15 − d ) ft : M = RA ( x − d ) − 1200 x ( x 2 ) = ( −600 x 2 + 9000 x − 9000d ) lb ⋅ ft .......................... Ans. For the last section of the beam, (15 − d ) ft ≤ x ≤ 15 ft : M x = 7.5 = ( 33, 750 − 9000d ) lb ⋅ ft d = 3.107 ft M = −1200 (15 − x ) (15 − x ) 2 = ( −600 x 2 + 18, 000 x − 135, 000 ) lb ⋅ ft .............. Ans. (b) M x = d = −600d 2 lb ⋅ ft d ≤ 3.107 ft : M x = d = M x = 7.5 = −600d 2 = ( 33, 750 − 9000d ) When 328 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ( M max ) AB = M x =d = −600d 2 lb ⋅ ft ..................................................................................... Ans. ( M max ) BC = M x=7.5 = ( 33,750 − 9000d ) lb ⋅ ft ................................................................ Ans. When 3.107 ft ≤ d ≤ 6 ft : ( M max ) AB = ( M max ) BC = M x= d = −600d 2 lb ⋅ ft ............................................................... Ans. The smallest maximum moment in the beam occurs when d = 3.107 ft = 3.11 ft ............................................................................................................ Ans. 8-60 An overhead crane consists of a carriage which moves along a beam as shown in Fig. P8-60. If the carriage moves slowly along the beam, compute and plot (a) The bending moment M(x) in the beam for b = 1.2 m, 2 m, and 2.8 as a function of x (0 m ≤ x ≤ 5 m). (b) (c) The bending moments MB under the left wheel and MC under the right wheel as functions of b (0.3 m ≤ b ≤ 4.7 m). What is the largest bending moment in the beam? When and where does it occur? SOLUTION From overall equilibrium: ΣM A = 0 : ΣM D = 0 : RD ( 5 ) − 2500b = 0 2500 ( 5 − b ) − RA ( 5 ) = 0 0 m ≤ x ≤ ( b − 0.3) m : RD = 500b N RA = 500 ( 5 − b ) N (a) For the initial section of the beam, M = RA x = 500 ( 5 − b ) x = ( 2500 x − 500bx ) N ⋅ m For the middle section of the beam, ( b − 0.3) m ≤ x ≤ ( b + 0.3) m : M = RA x − 1250 ( x − b + 0.3) = 500 ( 5 − b ) x − 1250 ( x − b + 0.3) = (1250 x − 500bx + 1250b − 375 ) N ⋅ m ( b + 0.3) m ≤ x ≤ 5 m : M = RD ( 5 − x ) = 500b ( 5 − x ) = ( 2500b − 500bx ) N ⋅ m 329 For the last section of the beam, STATICS AND MECHANICS OF MATERIALS, 2nd Edition (b) RILEY, STURGES AND MORRIS M B = 500 ( 5 − b )( b − 0.3) = ( −500b 2 + 2650b − 750 ) N ⋅ m M C = 500b ( 4.7 − b ) = ( 2350b − 500b 2 ) N ⋅ m b ≅ 2.35 m and under the left wheel when (c) The maximum bending moment occurs under the right wheel when b ≅ 2.65 m : M max ≅ 2.76 kN ⋅ m ............................................................................................................... Ans. 8-61 Draw complete shear force and bending moment diagrams for the beam shown in Fig. P8-61. SOLUTION From overall equilibrium: ΣM E = 0 : 2000 (12 ) + 4000 ( 8 ) +6000 ( 4 ) − RA (16 ) = 0 RA = 5000 lb ↑ ΣM A = 0 : RE (16 ) − 2000 ( 4 ) −4000 ( 8) − 6000 (12 ) = 0 RE = 7000 lb ↑ 8-62 Draw complete shear force and bending moment diagrams for the beam shown in Fig. P8-62. SOLUTION From overall equilibrium: ΣM A = 0 : 15 ( 7.5 ) + 20 ( 2 ) − RC ( 5 ) = 0 RC = 30.5 kN ↑ ΣM C = 0 : RA ( 5 ) + 15 ( 2.5 ) − 20 ( 3) = 0 RA = 4.5 kN ↑ 330 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 8-63 Draw complete shear force and bending moment diagrams for the beam shown in Fig. P8-63. SOLUTION From overall equilibrium: ΣM C = 0 : 2000 (12 ) (10 ) − 8000 ( 8 ) − RA (16 ) = 0 RA = 11, 000 lb ↑ ΣM A = 0 : RC (16 ) − 2000 (12 ) ( 6 ) − 8000 ( 24 ) = 0 RC = 21, 000 lb ↑ 8-64 Draw complete shear force and bending moment diagrams for the beam shown in Fig. P8-64. SOLUTION From overall equilibrium: ΣM D = 0 : 10 ( 7 ) + 20 ( 3) − 15 ( 5 ) − RA (10 ) = 0 RA = 5.5 kN ↑ ΣM A = 0 : RD (10 ) − 10 ( 3) − 20 ( 7 ) − 15 (15 ) = 0 RD = 39.5 kN ↑ 8-65 Draw complete shear force and bending moment diagrams for the beam shown in Fig. P8-65. SOLUTION From overall equilibrium: ΣM E = 0 : 300 ( 5 ) (13.5 ) + 2000 ( 7 ) + 3000 ( 4 ) − 400 ( 5 ) ( 2.5 ) − RB (11) = 0 RB = 3750 lb ↑ ΣM B = 0 : RE (11) + 300 ( 5 ) ( 2.5 ) − 2000 ( 4 ) −3000 ( 7 ) − 400 ( 5 ) (13.5 ) = 0 RE = 4750 lb ↑ 331 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 8-66 Draw complete shear force and bending moment diagrams for the beam shown in Fig. P8-66. SOLUTION From overall equilibrium: ΣM D = 0 : 20 ( 6 ) + 30 ( 4 ) ( 2 ) − RA ( 8 ) = 0 RA = 45 kN ↑ ΣM A = 0 : RD ( 8 ) − 20 ( 2 ) − 30 ( 4 ) ( 6 ) = 0 RD = 95 kN ↑ 8-67 Draw complete shear force and bending moment diagrams for the beam shown in Fig. P8-67. SOLUTION From overall equilibrium: ΣM D = 0 : 1500 (15 ) + 250 (10 ) ( 5 ) +1000 − RA ( 20 ) = 0 RA = 1800 lb ↑ ΣM A = 0 : RD ( 20 ) + 1000 − 1500 ( 5 ) − 250 (10 ) (15 ) = 0 RD = 2200 lb ↑ 8-68 Draw complete shear force and bending moment diagrams for the beam shown in Fig. P8-68. SOLUTION From overall equilibrium: ΣM E = 0 : 9 + 18 ( 6 ) ( 9 ) + 36 ( 4 ) − RB (10 ) = 0 RB = 112.5 kN ↑ ΣM B = 0 : RE (10 ) + 9 − 18 ( 6 ) (1) − 36 ( 6 ) = 0 RE = 31.5 kN ↑ 332 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 8-69 RILEY, STURGES AND MORRIS The beam shown in Fig. P8-69a has the cross section shown in Fig. P8-69b. Determine the maximum tensile and compressive flexural stresses in the beam. SOLUTION From overall equilibrium: ΣM C = 0 : 6000 + 400 (10 ) ( 5 ) − 1000 ( 2 ) − RB (10 ) = 0 RB = 2400 lb ↑ ΣM B = 0 : 6000 − 400 (10 ) ( 5 ) + RC (10 ) − 1000 (12 ) = 0 RC = 2600 lb ↑ 7 6 ( 2 ) + 3 2 ( 6 ) = 5 in. yC = 6 ( 2) + 2 ( 6) 6 ( 2 )3 2 ( 6 )3 2 2 I = + ( 2 ) ( 6 × 2 ) + + ( 2 ) ( 2 × 6 ) = 136.0 in 4 12 12 From the moment diagram, M max = −6000 lb ⋅ ft At the top of the beam, M max = +1200 lb ⋅ ft y = +3.00 in. ( −6000 ×12 )( 3.00 ) = +1588 lb/in.2 = 1588 psi (T) ...................... Ans. My =− 136.0 I At the bottom of the beam, y = −5.00 in. σ top = − ( −6000 ×12 )( −5.00 ) = −2647 lb/in.2 ≅ 2650 psi (C) .................. Ans. My =− 136.0 I At the section where M = +1200 lb ⋅ ft , the tensile and compressive stresses are both less than these values. σ bot = − A W102 × 19 wide-flange beam is loaded and supported as shown in Fig. P8-70. Determine the maximum tensile and compressive flexural stresses in the beam. SOLUTION From overall equilibrium: 8-70 ΣM E = 0 : 3 ( 2 ) ( 7 ) + 6 + 5 ( 4 ) ( 2 ) +3 ( 2 ) − 2 − RB ( 6 ) = 0 RB = 15.333 kN ↑ ΣM B = 0 : RE ( 6 ) − 2 + 3 ( 2 ) (1) + 6 − 5 ( 4 ) ( 4 ) − 3 ( 4 ) = 0 RE = 13.667 kN ↑ From the moment diagram 333 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS M max = +17.38 kN ⋅ m For a W102 × 19 section: S = 89.5 × 103 mm 4 M 17.38 × 103 = = 194.19 × 106 N/m 2 ≅ 194.2 MPa (T bottom & C top) ............. Ans. S 89.5 × 10−6 At the section where M = −4 kN ⋅ m , the tensile and compressive stresses are both less than these values. σ= A beam is loaded and supported as shown in Fig. P8-71a. Two 10 × 1-in. steel plates are welded to the flanges of an S18 × 70 American standard I-beam to form the cross section shown in Fig. P8-71b. Determine the maximum tensile and compressive flexural stresses in the beam. SOLUTION From overall equilibrium: 8-71 ΣM D = 0 : 3 (17.5) + 10 ( 7.5 ) − 5 ( 7.5) − RA ( 27.5) = 0 RA = 3.273 kip ↑ ΣM A = 0 : RD ( 27.5 ) − 3 (10 ) − 10 ( 20 ) − 5 ( 35 ) = 0 RD = 14.727 kip ↑ From the moment diagram, M max = −37.5 kip ⋅ ft For an S18 × 70 section: d = 18.00 in. I XX = 926 in 4 10 (1)3 2 I = I C + I P = 926 + 2 + ( 9.5) (10 × 1) = 2733 in 4 12 At the top of the beam, y = (18 2 ) + 1.00 = 10.00 in. ( −37.5 ×12 )(10.00 ) = +1.647 kip/in.2 = 1.647 ksi (T) ................ Ans. My =− 2733 I At the bottom of the beam, y = −10.00 in. σ top = − σ bot = − 8-72 ( −37.5 ×12 )( −10.00 ) = −1.647 kip/in.2 = 1.647 ksi (C) ............. Ans. My =− 2733 I A beam is loaded and supported as shown in Fig. P8-72a. Two 250 × 25-mm steel plates and two C254 × 45 channels are welded together to form the cross section shown in Fig. P8-72b. Determine the maximum tensile and compressive flexural stresses in the beam. SOLUTION For a C 254 × 45 channel: 6 d = 254.0 mm I XX = 42.9 × 106 mm 4 250 ( 25 )3 2 I = I C + I P = 2 ( 42.9 × 10 ) + 2 + (139.5 ) ( 250 × 25 ) 12 334 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS = 329.4 × 106 mm 4 = 329.4 ×10 −6 m 4 From overall equilibrium: ΣM D = 0 : ΣM A = 0 : 40 ( 2 ) + 30 ( 4 ) ( 6 ) − RA ( 8 ) = 0 RA = 100 kN ↑ RD ( 8 ) − 30 ( 4 ) ( 2 ) − 40 ( 6 ) = 0 RD = 60 kN ↑ From the moment diagram M max = +166.667 kN ⋅ m At the top of the beam, y = ( 254 2 ) + 25 = +152.0 mm (166.667 ×103 ) ( 0.152 ) = −76.91 N/m2 ≅ 76.9 MPa (C) ............. Ans. My σ top = − =− I 329.4 × 10−6 At the bottom of the beam, y = −152.0 mm σ bot 8-73 (166.667 ×10 ) ( −0.152 ) = +76.91 N/m2 ≅ 76.9 MPa (T) .......... Ans. My =− =− I 329.4 × 10−6 3 A WT8 × 25 structural steel T-section is loaded and supported as shown in Fig. P8-73. Determine the maximum tensile and compressive flexural stresses in the beam. SOLUTION From overall equilibrium: ΣM B = 0 : 6000 ( 6 ) + 1000 (12 ) ( 6 ) − RA (12 ) = 0 RA = 9000 = 9.00 kip ↑ From the moment diagram, M max = 36.0 kip ⋅ ft For a WT 8 × 25 section: d = 8.130 in. ytop = yc = 1.89 in. I XX = 42.3 in 4 ybot = − ( d − yc ) = − ( 8.130 − 1.89 ) = −6.24 in. σ top = − σ bot 8-74 ( 36.0 ×12 )(1.89 ) = −19.302 kip/in.2 ≅ 19.30 ksi (C) .................. Ans. My =− 42.3 I ( 36.0 ×12 )( −6.24 ) = +63.73 kip/in.2 ≅ 63.7 ksi (T) .................... Ans. My =− =− 42.3 I A 203 × 203-mm nominal size structural timber (see Appendix A) is supported by two brick columns, as shown in Fig. P8-74. Assume that the brick columns transmit only vertical forces to the timber beam. The beam supports the roof of a building through three timber columns. Columns A and C each transmit forces of 8 kN to the beam; column B transmits a force of 10 kN. (a) Draw complete shear force and bending moment diagrams for the beam. (b) Determine the maximum tensile and compressive flexural stresses in the beam. SOLUTION (a) From overall equilibrium: 335 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ΣM C = 0 : 10 ( 3) + 8 ( 6 ) − RA ( 6 ) = 0 RA = 13 kN ↑ (b) From the moment diagram M max = +15.00 kN ⋅ m For a 203 × 203 − mm timber: S = 1150 ×103 mm3 = 1150 × 10−6 m3 σ= M 15.00 ×103 = = 13.04 × 106 N/m 2 −6 S 1150 × 10 = 13.04 MPa (T bottom & C top) ................................................................................. Ans. 8-75 Draw complete shear force and bending moment diagrams for the beam shown in Fig. P8-75. SOLUTION From overall equilibrium: ΣM D = 0 : 1000 ( 8 ) − 2000 − RB ( 6 ) + 500 ( 4 ) ( 4 ) + 2000 ( 2 ) +4000 − 1000 ( 2 ) = 0 RB = 3333 lb ↑ ΣM B = 0 : RD ( 6 ) − 2000 + 1000 ( 2 ) − 500 ( 4 ) ( 2 ) − 2000 ( 4 ) +4000 − 1000 ( 8 ) = 0 RD = 2667 lb ↑ 8-76 An S457 × 81 American standard steel beam (see Appendix A) is loaded and supported as shown in Fig. P8-76. The two segments of the beam are connected with a smooth pin at D. (a) Draw complete shear force and bending moment diagrams for the beam. (b) Determine the maximum tensile and compressive flexural stresses in the beam. SOLUTION (a) From equilibrium for the right-hand segment of the beam: ΣM D = 0 : ↑ ΣFy = 0 : RE (1.5 ) − 40 (1.5 ) 2 ( 2.5 ) = 0 RE = 50 kN ↑ RD + 50 − 40 (1.5 ) 2 = 0 RD = −20 kN = 20 kN ↓ For the left-hand segment of the beam: ΣM C = 0 : 80 (1.5 ) ( 2.75 ) + 40 ( 2 ) (1) − RB ( 2 ) + 20 (1) = 0 336 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS RB = 215 kN ↑ ΣM B = 0 : RC ( 2 ) + 80 (1.5 ) ( 0.75 ) − 40 ( 2 ) (1) + 20 ( 3) = 0 RC = −35 kN = 35 kN ↓ (b) From the moment diagram For an M max = −90 kN ⋅ m S = 1465 × 103 mm3 = 1465 ×10 −6 m3 S 457 × 81 section: σ= 8-77 M 90.0 × 103 = = 61.4 × 106 N/m 2 ≅ 61.4 MPa (T top & C bottom) ........... Ans. −6 S 1465 × 10 An S15 × 50 American standard steel beam (see Appendix A) is loaded and supported as shown in Fig. P877. The total length of the beam is 15 ft. If the flexural stress is limited to 15 ksi, determine the maximum permissible value for the distributed load w. SOLUTION From overall equilibrium: ΣM C = 0 : w ( 3L ) 2 ( L ) − RB ( 2 L ) = 0 RB = ( 3wL 4 ) lb ↑ ΣM B = 0 : RC ( 2 L ) − w ( 3L ) 2 ( L ) = 0 RC = ( 3wL 4 ) lb ↑ The maximum bending moment between the supports occurs where the shear force is zero, V= 3wL ( wx 3L ) x − =0 4 2 2 ≅ 2.121L x = 3L M max = M 2.121L For an ( w 3L )( 2.121L ) 2.121L ≅ 0.3107 wL2 lb ⋅ ft 3wL = (1.121L ) − 4 2 3 2 S15 × 50 section: S = 64.8 in 3 2 M 0.3107 wL2 0.3107 w ( 5 ×12 ) σ= = = = 15, 000 lb/in.2 S S 64.8 w = 869 lb/in. ≅ 10.43 kip/ft ............................................................................................... Ans. 8-78 Draw complete shear force and bending moment diagrams for segments AB and CD of the structure shown in Fig. P8-78. SOLUTION For the complete structure: ΣM D = 0 : ↑ ΣFy = 0 : RC ( 3) − 3 (1.5 ) − 3 (1.5 ) (1.5 ) = 0 RC = 3.75 kN ↑ RD − 3 − 3 (1.5 ) + 3.75 = 0 RD = 3.75 kN ↑ 337 STATICS AND MECHANICS OF MATERIALS, 2nd Edition For member AB: RILEY, STURGES AND MORRIS ↑ ΣFy = 0 : −3 − VB = 0 VB = −3 kN = 3 kN ↑ ΣM B = 0 : M B + 3 (1.5) = 0 = 4.5 kN ⋅ m M B = −4.5 kN ⋅ m For member CD: ↑ ΣFy = 0 : ΣM C = 0 : 3.75 − 1.5 ( 3) − VC = 0 VC = −0.75 kN = 0.75 kN ↑ M C + 1.5 ( 3) (1.5 ) − 3.75 ( 3) = 0 M C = 4.5 kN ⋅ m = 4.5 kN ⋅ m 8-79 (a) (b) (c) (d) Three members are connected with smooth pins to form the frame shown in Fig. P8-79. The weights of the members are negligible. Member CD has a 3 × 3-in. cross section, and member BE has a 1 × 1-in. cross section. The pin at A has a ½-in diameter and is in double shear. Draw complete shear force and bending moment diagrams for member CD. Determine the maximum tensile and compressive flexural stresses in the member CD. Determine the normal stress in member BE. Determine the shearing stress in the pin at A. SOLUTION (a) For the complete frame: → ΣFx = 0 : ↑ ΣFy = 0 : ΣM A = 0 : Ax + 500 = 0 Ay − 400 (12 ) + RD = 0 RD (12 ) − 500 ( 7 ) − 400 (12 ) ( 6 ) = 0 RD = 2692 lb ↑ Ax = −500 lb = 500 lb ← Ay = 2108 lb ↑ For member CD: ↑ ΣFy = 0 : C y + E y − 400 (12 ) + 2692 = 0 ΣM C = 0 : E y ( 8 ) + 2692 (12 ) − 400 (12 ) ( 6 ) = 0 E y = −438 lb = 438 lb ↓ C y = 2546 lb ↑ 338 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (b) From the moment diagram, RILEY, STURGES AND MORRIS and M max = 8103 lb ⋅ ft I = 3 ( 3) 12 = 6.75 in 4 3 ( 8103 ×12 )(1.50 ) = −21, 608 lb/in.2 ≅ 21.6 ksi (C) ..................... Ans. My =− I 6.75 My ( 8103 × 12 )( −1.50 ) σ bot = − = = +21, 608 lb/in.2 ≅ 21.6 ksi (T) ....................... Ans. I 6.75 438 (c) For member BE: E y = ( 3 5 ) FBE FBE = = 730 lb 35 σ top = − σ BE = (d) For pin A: FBE 730 = = 730 psi (T) .......................................................................................... Ans. A 1× 1 2 FA = Ax2 + Ay = 5002 + 21082 = 2166 lb τA = 8-80 VA 2166 = = 5516 psi ≅ 5.52 ksi ......................................................... Ans. 2 As 2 π ( 0.5) 2 4 A body with a mass of 1500 kg is supported by a roller on an I-beam, as shown in Fig. P8-80. The roller moves slowly along the beam, thereby causing the shear force V and the bending moment M to be functions of the position b. (a) Draw complete shear force and bending moment diagrams when the roller is at position b. (b) Determine the position of the roller when the bending moment is maximum. SOLUTION (a) P = mg = 1500 ( 9.81) = 14,715 N = 14.715 kN ΣM B = 0 : ΣM A = 0 : P (10 − x ) − A (10 ) = 0 A = ( P − 0.1Px ) kN B (10 ) − Px = 0 B = ( 0.1Px ) kN (b) M P = Ax = ( Px − 0.1Px 2 ) kN ⋅ m dM P = P (1 − 0.2 x ) = 0 dx x = 5.00 m ........................................................................................................................ Ans. M max = 14.715 ( 5 ) − 0.1(14.715 )( 5 ) = 36.8 kN ⋅ m 2 8-81 Member AB supports a 55-lb sign, as shown in Fig. P8-81. Determine (a) The maximum tensile flexural stress in the ½-in. nominal diameter standard steel pipe AB (see Appendix A). (b) The normal stress in the 3/16-in. diameter wire BC. (c) The shearing stress in the ¼-in. diameter pin at A, which is in double shear. SOLUTION (a) For pipe AB: ↑ ΣFy = 0 : Ay + By − 2 ( 27.5 ) = 0 339 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS " ΣM A = 0 : By ( 64 ) − 27.5 (16 ) − 27.5 ( 48) = 0 Ay = By = 27.5 lb ↑ From the moment diagram: For a ½-in. diameter pipe: Therefore, M max = 440 lb ⋅ in. S = 0.041 in 3 σ= M 440 = = 10, 732 lb/in.2 S 0.041 ≅ 10.73 ksi (T bottom & C top) .................Ans. (b) θ = tan −1 ( 35 64 ) = 28.67° FBC = By sin θ = 27.5 = 57.32 lb = 57.32 lb (T) sin 28.67° 2 π ( 3 16 ) ABC = = 0.02761 in 2 4 F 57.32 σ BC = BC = = 2076 psi ≅ 2.08 ksi (T) ........................................................... Ans. 0.02761 A (c) For the pin A: → ΣFx = 0 : Ax − 57.32 cos 28.67° = 0 2 Ax2 + Ay Ax = 50.29 lb → τA = 8-82 2 As = 50.292 + 27.52 = 584 psi ................................................................. Ans. 2 2 ªπ ( 0.25 ) 4º ¬ ¼ Two beams AD and EH are spliced, as shown in Fig. P8-82. Draw complete shear force and bending moment diagrams for the beams AD, CF, and EH. SOLUTION For the system of beams: " ΣM B = 0 : " ΣM A = 0 : For beam ABD: 44 ( 6.5 ) − RA (11) = 0 RH (11) − 44 ( 4.5 ) = 0 RA = 26.0 kN ↑ RH = 18.00 kN ↑ ↑ ΣFy = 0 : 26.0 − By − 44 + Dy = 0 Dy = 96.0 kN ↑ By = 78.0 kN ↓ " ΣM D = 0 : − ( 26.0 )( 4.5 ) + By (1.5 ) = 0 For beam EGH: 340 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ↑ ΣFy = 0 : 18.00 − G + E y = 0 E y = 36.0 kN ↑ G y = 54.0 kN ↓ " ΣM E = 0 : 18.00 ( 4.5) − Gy (1.5) = 0 8-83 A tractor is moving slowly over a bridge, as shown in Fig. P8-83. The forces exerted on one beam of the bridge by the tractor are 4050 lb by the rear wheels and 1010 lb by the front wheels. Determine the position b of the tractor for which the bending moment in the beam is maximum. SOLUTION ΣM B = 0 : 1010 ( 40 − b − 9.5 ) +4050 ( 40 − b ) − A ( 40 ) = 0 A = ( 4820 − 126.5b ) lb ↑ ΣM A = 0 : B ( 40 ) − 4050b − 1010 ( b + 9.5 ) = 0 B = ( 239.9 + 126.5b ) lb ↑ The moment under the left wheel M C = Ax = ( −126.5b 2 + 4820b ) lb ⋅ ft is a maximum when dM C dx = −253b + 4820 = 0 b = 4820 253 = 19.05 ft ( M C )max = −126.5 (19.05 ) The moment under the right wheel 2 + 4820 (19.05 ) = 45,914 lb ⋅ ft M D = B ( 30.5 − b ) = ( 239.9 + 126.5b )( 30.5 − b ) = ( 7316.95 + 3618.35b − 126.5b 2 ) lb ⋅ ft is a maximum when dM D dx = 3618.35 − 253b = 0 = 33,191 lb ⋅ ft b = 3618.35 253 = 14.30 ft 2 ( M D )max = 7316.95 + 3618.35 (14.30 ) − 126.5 (14.30 ) 341 STATICS AND MECHANICS OF MATERIALS, 2nd Edition Therefore, the maximum moment occurs when RILEY, STURGES AND MORRIS b = 19.05 ft .............................................................................................................................. Ans. 8-84 A timber beam is loaded and supported as shown in Fig. P8-84. At section A-A of the beam, determine the shearing stresses on horizontal planes that pass through points a, b, and c of the cross section. SOLUTION From overall equilibrium ΣM C = 0 : 10 ( 2 ) − RB ( 4 ) = 0 RB = 5.00 kN ↑ For the left-hand portion of the beam ↑ ΣFy = 0 : 3 5 − VA = 0 VA = +5.00 kN = 5.00 kN ↓ 150 ( 200 ) = 100.0 × 106 mm 4 = 100.0 × 10−6 m 4 12 Qa = 0 mm3 I= For point a: τ a = 0 MPa .......................................................................................... Ans. For point b: Qb = yC A = 75 (150 × 50 ) = 562.5 ×103 mm3 3 −6 VQ ( 5 × 10 )( 562.5 × 10 ) = = 187.5 × 103 N/m 2 = 187.5 kPa ........................ Ans. τb = −6 It (100.0 ×10 ) ( 0.150 ) For point c: Qc = yC A = 50 (150 × 100 ) = 750.0 ×103 mm 3 3 −6 VQ ( 5 × 10 )( 750.0 × 10 ) = = 250 ×103 N/m 2 = 250 kPa .............................. Ans. τc = −6 It (100.0 ×10 ) ( 0.150 ) 8-85 The transverse shear V at a certain section of a timber beam is 7500 lb. If the beam has the cross section shown in Fig. P8-85, determine (a) The horizontal shearing stress in the glued joint 2 in. below the top of the beam. (b) The transverse shearing stress at a point 3 in. below the top of the beam. (c) The magnitude and location of the maximum transverse shearing stress on the cross section. SOLUTION 3 3 8 (12 ) 4 ( 8 ) I= − = 981.3 in 4 12 12 (a) For the glue joint, y = 4 in. , Q4 = yC A = 5 ( 8 × 2 ) = 80 in 3 VQ 7500 ( 80 ) = = 152.9 lb/in 2 = 152.9 psi ......................................................... Ans. It ( 981.3 )( 4 ) y = 3 in. , τ4 = (b) For a point 3-in. below the top of the beam, Q3 = yC A = 5 ( 8 × 2 ) + 3.5 ( 4 × 1) = 94 in 3 342 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS τ3 = VQ 7500 ( 94 ) = = 179.6 lb/in 2 = 179.6 psi ......................................................... Ans. It ( 981.3)( 4 ) y = 0 in. , (c) The maximum shear stress occurs at the neutral axis, QNA = yC A = 5 ( 8 × 2 ) + 2 ( 4 × 4 ) = 112 in 3 τ max = 8-86 VQ 7500 (112 ) = = 214 lb/in 2 = 214 psi (at NA) ............................................ Ans. It ( 981.3)( 4 ) A W254 × 89 structural steel wide-flange beam (see Appendix A) is loaded and supported as shown in Fig. P8-86. Determine the maximum transverse shearing stress at section A-A of the beam (a) Using Eq. (8-15). (b) Using Eq. (8-17). SOLUTION From overall equilibrium ΣM C = 0 : 25 (1) + 20 ( 3) − RB ( 5) = 0 RB = 17.00 kN ↑ For the left-hand portion of the beam ↑ ΣFy = 0 : For a 17 − 20 − VA = 0 VA = −3.00 kN = 3.00 kN ↑ d = 260 mm w f = 256 mm W 254 × 89 section, I = 142 × 106 mm 4 tw = 10.7 mm t f = 17.3 mm (a) QNA = yC A = 121.35 ( 256 × 17.3) + 56.35 (10.7 ×112.7 ) = 605.4 × 103 mm3 τ max = τ NA VQ ( 3.00 × 10 )( 605.4 × 10 = = It (142 ×10−6 ) ( 0.0107 ) 3 −6 ) = 1.195 × 106 N/m 2 = 1.195 MPa ................................................ Ans. (b) Aw = 260 − 2 (17.3) (10.7 ) = 2412 mm 2 τ avg = 8-87 V 3.00 × 103 = = 1.244 × 106 N/m 2 = 1.244 MPa ......................................... Ans. −6 Aw 2412 × 10 A WT7 × 34 structural steel wide-flange beam (see Appendix A) is loaded and supported as shown in Fig. P8-87. Determine the maximum transverse shearing stress at section A-A of the beam. SOLUTION From overall equilibrium ΣM C = 0 : 500 (10 ) ( 5 ) − RB (10 ) = 0 RB = 2500 lb ↑ For the left-hand portion of the beam ↑ ΣFy = 0 : 2500 − 500 ( 3) − VA = 0 343 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS VA = +1000 lb = 1000 lb ↓ For a WT 7 × 34 section, I = 32.6 in 4 d = 7.020 in. t s = 0.415 in. yC = 1.29 in. QNA = yC A = ( 5.73 2 )( 5.73 × 0.415 ) = 6.813 in 3 τ max = τ NA = 8-88 VQ 1000 ( 6.813) = = 504 lb/in 2 = 504 psi (at NA) ........................... Ans. It ( 32.6 )( 0.415 ) A timber beam 4 m long is simply supported at its ends and carries a uniformly distributed load w of 8 kN/m over its entire length. If the beam has the cross section shown in Fig. P8-88, determine (a) The maximum horizontal shearing stress in the glued joints between the web and flanges of the beam. (b) The maximum horizontal shearing stress in the beam. SOLUTION I NA QJ = yC A = 80 (180 × 40 ) 180 ( 200 ) 140 (120 ) = − 12 12 6 4 = 99.84 × 10 mm = 99.84 × 10−6 m 4 3 3 QNA = 80 (180 × 40 ) + 30 ( 40 × 60 ) = 576 × 103 mm 3 = 576 × 10−6 m3 = 648 ×103 mm3 = 648 × 10−6 m 3 Vmax = RA = RB = ( wL 2 ) = 8 ( 4 ) 2 = 16.00 kN (a) 3 −6 VQ (16 × 10 )( 576 × 10 ) = τJ = It ( 99.84 × 10−6 ) ( 0.040 ) = 2.31× 106 N/m 2 = 2.31 MPa ................................................. Ans. (b) τ max = τ NA 3 −6 VQ (16 ×10 )( 648 × 10 ) = = It ( 99.84 × 10−6 ) ( 0.040 ) = 2.60 × 106 N/m 2 = 2.60 MPa ................................................................................. Ans. 8-89 A W10 × 30 structural steel wide-flange beam (see Appendix A) is loaded and supported as shown in Fig. P8-89. At section A-A of the beam, determine (a) The maximum transverse shearing stress due to the 5000-lb load. (b) The maximum transverse shearing stress due to the 5000-lb load plus the weight of the beam. SOLUTION From overall equilibrium ΣM C = 0 : − RB (10 ) − 5000 ( 4 ) = 0 RB = −2000 lb = 2000 lb ↓ For the left-hand portion of the beam ↑ ΣFy = 0 : −2000 − VA = 0 344 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS VA = −2000 lb = 2000 lb ↑ For a W10 × 30 section, I = 170 in 4 d = 10.47 in. w f = 5.810 in. tw = 0.300 in. t f = 0.510 in. (a) QNA = yC A = 4.980 ( 5.810 × 0.510 ) + 2.3625 ( 0.300 × 4.725 ) = 18.105 in 3 τ max = τ NA = VQ 2000 (18.105 ) = = 710 lb/in 2 = 710 psi (at NA) ........................... Ans. It (170 )( 0.300 ) w = 30 lb/ft (b) With the weight of the beam included, ΣM C = 0 : ↑ ΣFy = 0 : 3 30 (14 ) − RB (10 ) − 5000 ( 4 ) = 0 RB = −1874 lb = 1874 lb ↓ −1874 − 30 ( 5 ) − VA = 0 VA = −2024 lb = 2024 lb ↑ 2024 (18.105 ) τ max = 8-90 (170 )( 0.300 ) = 719 lb/in 2 = 719 psi (at NA) .................................................. Ans. A W203 × 60 structural steel wide-flange section (see Appendix A) is used for the cantilever beam shown in Fig. P8-90. Determine the maximum flexural and transverse shearing stresses in the beam and state where they occur. SOLUTION From overall equilibrium ΣM A = 0 : ↑ ΣFy = 0 : For a − M A − 30 ( 2.5 ) = 0 M A = −75.0 kN ⋅ m = 75.0 kN ⋅ m VA − 30 = 0 VA = +30 kN = 30 kN ↑ W 203 × 60 section, d = 210 mm w f = 205 mm I = 60.8 × 106 mm 4 S = 582 × 103 mm 3 tw = 9.1 mm t f = 14.2 mm From the shear force and bending moment diagrams, Vmax = 30.0 kN (over the full length of the beam) M max = −75.0 kN ⋅ m (at the wall) M 75 × 103 = = 128.9 × 106 N/m 2 = 128.9 MPa (T top & C bottom) ........... Ans. S 582 × 10−6 VQ ( 30.0 × 10 )( 322.5 × 10 = It ( 60.8 ×10−6 ) ( 0.0091) 3 −6 σ max = QNA = yC A = 97.9 ( 205 × 14.2 ) + 45.4 ( 9.1× 90.8 ) = 322.5 × 103 mm3 τ max = τ NA = ) = 17.49 ×10 6 N/m 2 = 17.49 MPa ......... Ans. 345 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 8-91 RILEY, STURGES AND MORRIS The beam shown in Fig. P8-91a is composed of two 1 × 6-in. and two 1 × 3-in. hard maple boards that are glued together as shown in Fig. P8-91b. Determine the magnitude and location of (a) The maximum tensile flexural stress in the beam. (b) The maximum horizontal shearing stress in the beam. SOLUTION From shear force and bending moment diagrams for the beam Vmax = 1000 lb M max = 1800 lb ⋅ ft 2 ( 6 ) 2 ( 3) = + = 40.5 in 4 12 12 3 3 I NA (a) At the bottom of the beam, 2 ft from the left support, σ max = − (1800 ×12 )( −3) Mc =− I 40.5 = +1600 lb/in 2 = 1600 psi (T) ................................................................................... Ans. (b) QNA = yC A = 0.75 ( 2 × 1.5) + 1.5 ( 2 × 3) = 11.25 in 3 At the left support and from the end of the distributed load to the right support, τ NA = VQ 1000 (11.25 ) = = 69.4 lb/in 2 = 69.4 psi It ( 40.5)( 4 ) 1000 ( 6.75 ) Q1.5 = 2.25 ( 2 × 1.5) = 6.75 in 3 τ 1.5 = ( 40.5 )( 4 ) = 83.3 lb/in 2 = 83.3 psi ............................................. Ans. τ max = τ1.5 = 83.3 psi (1.5 in. above and below the NA) 8-92 The beam shown in Fig. P8-92a is composed of three pieces of timber that are glued together as shown in Fig. P8-92b. Determine (a) The maximum horizontal shearing stress in the glued joints. (b) The maximum horizontal shearing stress in the wood. (c) The maximum tensile and compressive flexural stresses in the beam. SOLUTION From overall equilibrium ΣM B = 0 : ↑ ΣFy = 0 : 9 + 6 ( 0.5) + 6 ( 2.5) − RA ( 3) = 0 RA = 9.00 kN ↑ RA + RB − 6 − 6 = 0 RB = 3.00 kN ↑ From the shear force and bending moment diagrams, Vmax = 9.00 kN 346 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS M max = 10.5 kN ⋅ m 100 ( 240 ) I= = 115.2 × 106 mm 4 = 115.2 × 10−6 m 4 12 3 (a) QJ = yC A = 95 (100 × 50 ) = 475 × 103 mm3 = 475 × 10−6 m3 3 −6 VQ ( 9.00 × 10 )( 475 × 10 ) = = 371× 103 N/m 2 = 371 kPa .................................... Ans. τJ = −6 It (115.2 ×10 ) ( 0.100 ) (b) QNA = yC A = 60 (100 × 120 ) = 720 × 103 mm3 = 720 × 10−6 m3 τ NA 3 −6 VQ ( 9.00 × 10 )( 720 × 10 ) = = = 563 ×103 N/m 2 = 563 kPa ................................. Ans. −6 It (115.2 ×10 ) ( 0.100 ) (c) σ max 3 Mc (10.5 × 10 ) ( 0.120 ) = = I 115.2 ×10−6 = 10.94 × 106 N/m 2 = 10.94 MPa (T bottom & C top) ..................................... Ans. 8-93 The lintel beam AB shown in Fig. P8-93a has a rectangular cross section as shown in Fig. P8-93b, and is used to support a brick wall over a door opening. The brick wall is assumed to produce a triangular load distribution. The total load carried by the beam is 500 lb. If the beam is simply supported at the ends, determine the maximum flexural and transverse shearing stresses in the beam and state where they occur. SOLUTION From overall equilibrium ΣM B = 0 : 250 ( 4 ) 2 ( 2 ) − RA ( 4 ) = 0 RA = 250 lb ↑= RB From the shear force and bending moment diagrams, Vmax = 250 lb (at ends of the beam) M max = 333.3 lb ⋅ ft (at midspan) 8 ( 0.5 ) I= = 0.08333 in 4 12 Mc ( 333.3 × 12 )( 0.25 ) = σ max = I 0.08333 3 = 11,999 lb/in 2 ≅ 12.00 ksi (T bottom & C top) .................................................. Ans. QNA = yC A = 0.125 ( 8 × 0.25 ) = 0.25 in 3 250 ( 0.25 ) VQ = = 93.8 lb/in 2 = 93.8 psi (at NA) ..................................... Ans. It ( 0.08333)( 8 ) are worthless since τ max = The results for 8-94 τ max w d = 16 and Eq. 8-15 gives useful results only when w d < 1 . A timber beam is simply supported and carries a uniformly distributed load of 4 kN/m over the full length of the beam. If the beam has the cross section shown in Fig. P8-94 and a span of 6 m, determine (a) The horizontal shearing stress in the glued joint 50 mm below the top of the beam and 1 m from the left support. 347 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (b) (c) (d) RILEY, STURGES AND MORRIS The horizontal shearing stress in the glued joint 50 mm above the bottom of the beam and ½ m from the left support. The maximum horizontal shearing stress in the beam. The maximum tensile flexural stress in the beam. SOLUTION From overall equilibrium " ΣM B = 0 : ↑ ΣFy = 0 : ª 4 ( 6 ) º ( 3 ) − RA ( 6 ) = 0 ¬ ¼ RA = 12.00 kN ↑ RA + RB − 4 ( 6 ) = 0 RB = 12.00 kN ↑ From the shear force and bending moment diagrams, Vmax = 12.00 kN M max = +18.00 kN ⋅ m yC = V1.0 = 12.00 − 4 (1) = 8.00 kN V0.5 = 12.00 − 4 ( 0.5 ) = 10.00 kN = 143.18 mm 25 ª150 ( 50 ) º + 125 ª50 (150 ) º + 225 ª 250 ( 50 ) º ¬ ¼ ¬ ¼ ¬ ¼ 150 ( 50 ) + 50 (150 ) + 250 ( 50 ) ª150 ( 50 )3 º 50 ( 93.18 )3 2 I =« + (118.18 ) (150 × 50 ) » + 3 « 12 » ¬ ¼ ª 250 ( 50 )3 º 2 + +« + ( 81.82 ) ( 250 × 50 ) » 3 « 12 » ¬ ¼ 6 4 4 −6 = 209.14 × 10 mm = 209.14 ×10 m 3 50 ( 56.82 ) (a) QTJ = yC A = 81.82 ( 250 × 50 ) = 1022.8 ×103 mm3 = 1022.8 ×10−6 m3 τ TJ (b) 3 −6 VQ ( 8.00 × 10 )(1022.8 × 10 ) = = = 783 × 103 N/m 2 = 783 kPa ..................... Ans. −6 It ( 209.1×10 ) ( 0.050 ) QBJ = 118.18 (150 × 50 ) = 886.4 ×103 mm3 = 886.4 ×10−6 m3 τ BJ = (c) (10.00 ×10 )(886.4 ×10 ) = 848 ×10 ( 209.1×10 ) ( 0.050 ) 3 −6 −6 3 −6 3 N/m 2 = 848 kPa ................................ Ans. QNA = 81.82 ( 250 × 50 ) + 28.41( 50 × 56.82 ) = 1103.5 ×103 mm 3 = 1103.5 ×10−6 m3 τ NA = σ max (d) (12.00 ×10 )(1103.5 ×10 ) = 1267 ×10 ( 209.1×10 ) ( 0.050 ) (18.00 ×10 ) ( −0.14318) My =− =− −6 3 3 N/m 2 = 1267 kPa .......................... Ans. I 209.1× 10−6 = +12.33 ×106 N/m 2 = 12.33 MPa (T) .................................................................. Ans. 348 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 8-95 RILEY, STURGES AND MORRIS The timber beam shown in Fig. P8-95a is fabricated by gluing two 1 × 5-in. and two 1 × 4-in. boards together as shown in Fig. P8-95b. Determine (a) The maximum horizontal shearing stress in the glued joints. (b) The maximum horizontal shearing stress in the wood. (c) The maximum tensile and compressive flexural stresses in the beam. SOLUTION From overall equilibrium " ΣM B = 0 : ª 200 ( 6 ) º ( 6 ) + 600 ( 3) ¬ ¼ −600 ( 3) − RA ( 9 ) = 0 RA = 800 lb ↑ ↑ ΣFy = 0 : RA + RB − 200 ( 6 ) − 600 − 600 = 0 RB = 1600 lb ↑ From the shear force and bending moment diagrams, Vmax = −1000 lb M max = −1800 lb ⋅ ft 5 ( 6) 3( 4) I= − = 74.0 in 4 12 12 3 3 (a) QJ = yC A = 2.5 ( 5 ×1) = 12.5 in 3 τJ = (b) VQ 1000 (12.5 ) = = 84.5 lb/in 2 = 84.5 psi ................. Ans. It ( 74.0 )( 2 ) 1000 (16.5 ) QNA = 2.5 ( 5 × 1) + 2 ª1(1× 2 ) º = 16.5 in 3 ¬ ¼ τ NA = (c) 8-96 σ max ( 74.0 )( 2 ) (1800 ×12 )( 3) = 876 lb/in 2 ≅ 876 psi (T top & C bottom) ........... Ans. Mc = =− I 74.0 = 111.5 lb/in 2 = 111.5 psi .................................................................. Ans. A cantilever beam is used to support a concentrated load of 20 kN at the end of the beam. The beam is fabricated by bolting two C457 × 86 steel channels (see Appendix A) back-to-back to form the H-section shown in Fig. P8-96. If the pairs of bolts are spaced at 300-mm intervals along the beam, determine (a) The shear force carried by each of the bolts. (b) The bolt diameter required if the shear and bearing stresses for the bolts must be limited to 60 MPa and 125 MPa, respectively. SOLUTION For a C 457 × 86 channel, A = 11, 030 mm 2 I = 7.41×106 mm 4 xC = 21.9 mm tw = 17.8 mm For the beam 349 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 2 I NA = 2 ( 7.41×106 ) + ( 21.9 ) (11, 030 ) = 25.40 ×106 mm 4 = 25.40 ×10−6 m 4 (a) QNA = yC A = 21.9 (11, 030 ) = 241.6 ×103 mm3 = 241.6 ×10−6 m3 V = 20 kN (entire length of beam) 3 −6 VQ ( 20.0 ×10 )( 241.6 ×10 ) = = = 416.1× 103 N/m 2 = 416.1 kPa −6 It ( 25.40 ×10 ) ( 0.4572 ) 3 τ NA τ A ( 416.1×10 FB = NA s = 2 (b) ) ( 0.4572 × 0.300 ) = 28.54 ×10 2 3 N ≅ 28.5 kN ............... Ans. τ= FB 28.54 ×103 = ≤ 60 × 106 N/m 2 AB πd2 4 FB 28.54 × 103 = ≤ 125 × 106 N/m 2 dtw d ( 0.0178 ) d ≥ 0.02461 m d ≥ 0.01283 m σb = d min = 0.02461 m ≅ 24.6 mm ............................................................................................. Ans. 8-97 A timber beam is fabricated from one 2 × 8-in. and two 2 × 6-in. pieces of lumber to form the cross section shown in Fig. P8-97. The flanges of the beam are fastened to the web with nails that can safely transmit a shear force of 100 lb. If the beam is simply supported and carries a 1000-lb load at the center of a 12-ft span, determine (a) The shear force transferred by the nails from the flange to the web in a 12-in. length of the beam. (b) The spacing required for the nails. SOLUTION (a) RA = RB = 1000 2 = 500 lb ↑ V = 500 lb (over entire length of the beam) I NA 6 (12 ) 4 ( 8 ) = − = 693.3 in 4 12 12 3 3 QJ = yC A = 5 ( 6 × 2 ) = 60.0 in 3 τJ = VQ 500 ( 60.0 ) = = 7.212 lb/in 2 = 7.212 psi It ( 693.3)( 6 ) FJ = τ J A = 7.212 (12 × 6 ) = 519 lb .................................................................................... Ans. (b) N = FJ FN = 519.3 100 = 5.19 nails s = 12 5.19 = 2.31 in. Use 6 nails spaced 2 in. apart in a 12-in. length .......................................................... Ans. 8-98 A box beam will be fabricated by bolting two 15 × 260-mm steel plates to two C305 × 45 steel channels (see Appendix A), as shown in Fig. P8-98. The beam will be simply supported at the ends and will carry a concentrated load of 125 kN at the center of a 5-m span. Determine the bolt spacing required if the bolts have a diameter of 20 mm and an allowable shearing stress of 150 MPa. SOLUTION 350 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS RA = RB = 125 2 = 62.5 kN ↑ V = 62.5 kN (over entire length of the beam) For a C305 × 45 channel, I = 67.4 × 106 mm 4 d = 304.8 mm w f = 80.5 mm 260 (15 )3 2 I NA = 2 ( 67.4 ×106 ) + 2 + (159.9 ) ( 260 × 15) 12 = 334.4 × 106 mm 4 QJ = yC A = 152.4 ( 260 × 15 ) = 594.4 × 103 mm3 3 −6 VQ ( 62.5 ×10 )( 594.4 ×10 ) = = 690.0 × 103 N/m 2 = 690.0 kPa τJ = −6 It ( 334.4 ×10 ) ( 2 × 0.0805) FJ = τ J A = ( 690.0 × 103 ) ( 2 × 0.0805 × s ) = (111, 087 s ) N 2 FJ = FB = τ B AB = 2 (150 × 106 ) π ( 0.020 ) 4 = 94, 248 N s = 94, 248 111, 087 = 0.848 m = 848 mm ..................................................................... Ans. 8-99 A W21 × 101 structural steel wide-flange section (see Appendix A) is simply supported at its ends and carries a concentrated load at the center of a 20-ft span. The concentrated load must be increased to 125 kip, which requires that the beam be strengthened. It has been decided that two 3/4 × 16-in. steel plates will be bolted to the flanges, as shown in Fig. P8-99. Determine the bolt spacing required if the bolts have a diameter of ¾-in. and an allowable shearing stress of 17.5 ksi. SOLUTION RA = RB = 125 2 = 62.5 kip ↑ V = 62.5 kip (over entire length of the beam) For a W 21× 101 section, I = 2420 in 4 d = 21.36 in. w f = 12.290 in. I NA 16 ( 0.75)3 2 2420 + 2 = + (11.055) (16 × 0.75 ) 12 = 5354 in 4 QJ = yC A = 10.68 (16 × 0.75) = 128.160 in 3 3 VQ ( 62.5 ××10 ) (128.16 ) = = 121.73 lb/in 2 = 121.73 psi τJ = It ( 5354 )(12.290 ) FJ = τ J A = 121.73 (12.290 × s ) = (1496.1 s ) lb 2 FJ = FB = τ B AB = 2 (17.5 ×103 ) π ( 0.75 ) 4 = 15, 463 lb s = 15, 463 1496.1 = 10.34 in. ............................................................................................. Ans. 351 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 8-100 RILEY, STURGES AND MORRIS A W356 × 122 structural steel wide-flange section (see Appendix A) has a C381 × 74 channel bolted to the top flange, as shown in Fig. P8-100. The beam is simply supported at its ends and carries a concentrated load of 96 kN at the center of an 8-m span. If the pairs of bolts are spaced at 500-mm intervals along the beam, determine (a) The shear force carried by each of the bolts. (b) The bolt diameter required if the shear and bearing stresses for the bolts must be limited to 60 MPa and 125 MPa, respectively. SOLUTION RA = RB = 96 2 = 48 kN ↑ V = 48 kN (over entire length of the beam) For a C381× 74 channel, W 356 ×122 section, I = 4.58 × 106 mm 4 A = 9485 mm 2 xC = 20.3 mm For a tw = 18.2 mm w f = 257 mm I = 367 × 106 mm 4 d = 363 mm For the beam A = 15,550 mm 2 yC = 360.9 ( 9485) + 181.5 (15,550 ) = 249.5 mm 9485 + 15,550 2 I NA = ( 367 × 106 ) + ( 68 ) (15,550 ) 2 + ( 4.58 ×106 ) + (111.4 ) ( 9485 ) = 561.2 ×106 mm 4 QJ = yC A = 111.4 ( 9485) = 1056.6 ×103 mm 3 3 −6 VQ ( 48 × 10 )(1056.6 ×10 ) τJ = = = 351.6 ×103 N/m 2 = 351.6 kPa −6 It ( 561.2 ×10 ) ( 0.257 ) FB = τ J A 2 = ( 351.6 × 103 ) ( 0.257 × 0.500 ) 2 = 22.6 × 103 N = 22.6 kN ............... Ans. 8-101 A timber beam is simply supported and carries a uniformly distributed load w of 360 lb/ft over its entire 18ft span (Fig. P8-101a). If the beam has the cross section shown in Fig. P8-101b, compute and plot the vertical shearing stress τ as a function of distance y from the neutral axis for a cross section 2 ft from the left end of the beam. SOLUTION RA = RB = ( 360 )(18) 2 = 3240 lb ↑ V2 = 3240 − 360 ( 2 ) = 2520 lb 1 6 ( 2 ) + 2 1( 4 ) + 2 1( 4 ) = 1.400 in. yC = 6 ( 2 ) + 1 ( 4 ) + 1( 4 ) I NA 6 ( 2 )3 1 ( 4 )3 2 2 = + ( 0.4 ) ( 6 × 2 ) + 2 + ( 0.6 ) (1× 4 ) = 19.467 in 4 12 12 352 STATICS AND MECHANICS OF MATERIALS, 2nd Edition For RILEY, STURGES AND MORRIS −1.4 in. ≤ y ≤ 0.6 in. Q = yC A = t = 8 in. 2 VQ 2520 ( 4 ) (1.9600 − y ) = = 64.72 (1.9600 − y 2 ) psi τ= It (19.467 )( 8 ) 1.40 + y ( 8 )(1.40 − y ) = 4 (1.9600 − y 2 ) in 3 2 For 0.6 in. ≤ y ≤ 2.6 in. t = 2 in. 2.6 + y Q = yC A = 2 (1)( 2.6 − y ) 2 = ( 6.7600 − y 2 ) in 3 2 VQ 2520 ( 6.7600 − y ) = τ= It (19.467 )( 2 ) = 64.72 ( 6.7600 − y 2 ) psi 8-102 The transverse shear force V at a certain section of a timber beam is 18 kN. If the beam has the cross section shown in Fig. P8-102, compute and plot the vertical shearing stress τ as a function of distance y (−150 mm < y < 150 mm) from the neutral axis. SOLUTION 3 3 200 ( 300 ) 100 ( 200 ) I= − = 383.3 × 106 mm 4 12 12 For −100 mm ≤ y ≤ 100 mm t = 100 mm Q = yC A = 125 ( 200 × 50 ) + 100 + y (100 )(100 − y ) 2 = 1.250 × 106 + 50 (1002 − y 2 ) mm3 3 6 2 2 −9 VQ (18 × 10 ) 1.250 × 10 + 50 (100 − y ) (10 ) τ= = It ( 383.3 ×10−6 ) ( 0.100 ) = 0.46997 1.250 × 106 + 50 (1002 − y 2 ) N/m 2 t = 100 mm For −150 mm ≤ y ≤ −100 mm and 100 mm ≤ y ≤ 150 mm Q = yC A = 150 + y ( 200 )(150 − y ) 2 3 2 2 −9 VQ (18 × 10 ) 100 (150 − y ) (10 ) τ= = It ( 383.3 ×10−6 ) ( 0.200 ) = 100 (1502 − y 2 ) mm3 = 23.48 (150 2 − y 2 ) N/m 2 353 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 8-103 RILEY, STURGES AND MORRIS The transverse shear force V at a certain section of a timber beam is 2500 lb. If the beam has the cross section shown in Fig. P8-103, compute and plot the vertical shearing stress τ as a function of distance y (−3 in. < y < 3 in.) from the neutral axis. SOLUTION 3 3 2 ( 6 ) 2 ( 3) + = 40.500 in 4 12 12 For −1.5 in. ≤ y ≤ 1.5 in. t = 4 in. I NA = Q = yC A = 2.25 ( 2 × 1.5 ) + 1.50 + y ( 4 )(1.50 − y ) = 6.75 + 2 ( 2.2500 − y 2 ) in 3 2 2 VQ 2500 6.75 + 2 ( 2.2500 − y ) = 104.167 + 30.8642 2.2500 − y 2 psi τ= = ( ) It 40.5)( 4 ) ( −3.0 in. ≤ y ≤ −1.5 in. and 1.5 in. ≤ y ≤ 3.0 in. For t = 2 in. 3.0 + y Q = yC A = ( 2 )( 3.0 − y ) 2 = ( 9 − y 2 ) in 3 2 VQ 2500 ( 9 − y ) τ= = It ( 40.5)( 2 ) = 30.8642 ( 9 − y 2 ) psi 8-104 The beam shown in Fig. 8-104a is fabricated by gluing two pieces of timber together to form the cross section shown in Fig. 8-104b. Compute and plot the vertical shearing stress τ as a function of distance y from the neutral axis for a cross section 0.5 m from the left end of the beam. SOLUTION From overall equilibrium ΣM B = 0 : ↑ ΣFy = 0 : 3 ( 2 ) ( 3 ) − RA ( 4 ) = 0 RA = 4.50 kN ↑ RA + RB − 3 ( 2 ) = 0 RB = 1.50 kN ↑ V0.5 = 4.50 − 3 ( 0.5) = 3.0 kN yC = 100 25 ( 200 ) + 212.5 100 ( 25 ) 25 ( 200 ) + 100 ( 25 ) = 137.5 mm 25 ( 200 )3 100 ( 25)3 2 2 + ( 37.5 ) ( 25 × 200 ) + + ( 75.0 ) (100 × 25 ) I = 12 12 −6 6 4 4 = 37.89 × 10 mm = 37.89 × 10 m 354 STATICS AND MECHANICS OF MATERIALS, 2nd Edition For RILEY, STURGES AND MORRIS −137.5 mm ≤ y ≤ 62.5 mm Q = yC A = t = 25 mm 137.5 + y ( 50 )(137.5 − y ) = 12.5 (137.52 − y 2 ) mm3 2 3 2 2 −9 VQ ( 3 × 10 ) 12.5 (137.5 − y ) (10 ) τ= = = 39.59 (137.52 − y 2 ) N/m 2 −6 It ( 37.89 ×10 ) ( 0.025) For 62.5 mm ≤ y ≤ 87.5 mm Q = yC A = t = 100 mm 87.5 + y (100 )(87.5 − y ) 2 3 2 2 −9 VQ ( 3 × 10 ) 50 ( 87.5 − y ) (10 ) τ= = It ( 37.89 ×10−6 ) ( 0.100 ) = 50 ( 87.52 − y 2 ) mm3 = 39.59 ( 87.52 − y 2 ) N/m 2 8-105 A 20-ft long simply supported timber beam is loaded with 1800-lb concentrated loads applied 6 ft from each support. The allowable flexural stress is 1900 psi, and the allowable shearing stress is 90 psi. Select the lightest standard structural timber that can be used to support the loading. SOLUTION From overall equilibrium ΣM A = 0 : ΣM B = 0 : B ( 20 ) − 1800 ( 6 ) − 1800 (14 ) = 0 B = 1800 lb ↑ 1800 ( 6 ) + 1800 (14 ) − A ( 20 ) = 0 A = 1800 lb ↑ M 10,800 ×12 = ≤ 1900 lb/in 2 S S 10,800 × 12 S≥ = 68.21 in 3 1900 Try an 8 × 8-in. timber with σ max = S = 70.3 in 3 I = 264 in 4 2 A = 56.3 in 2 S≥ w = 15.6 lb/ft With the weight of the beam included, the maximum moment is 11,580 × 12 = 73.14 in 3 1900 which is bigger than the section modulus of the timber. Next, try a 4 ×12-in. timber with S = 79.9 in 3 I = 459 in 4 2 15.6 ( 20 ) M = 10,800 + = 11,580 lb ⋅ ft 8 A = 41.7 in 2 S≥ w = 11.6 lb/ft 11,380 × 12 = 71.87 in 3 1900 With the weight of the beam included, the maximum moment is M = 10,800 + 11.6 ( 20 ) = 11,380 lb ⋅ ft 8 355 STATICS AND MECHANICS OF MATERIALS, 2nd Edition which is okay. Next, check the shear stress, RILEY, STURGES AND MORRIS Vmax = Vload + Vweight = 1800 + 11.6 ( 20 ) = 1916 lb 2 Vmax 1.5 (1916 ) = = 68.92 lb/in 2 A 41.7 which is less than the allowable shear stress of 90 psi . Therefore, this design is okay. τ max = 1.5 Use a 4 × 12-in. timber ......................................................................................................... Ans. 8-106 A simply supported timber beam 5 m long is loaded with 6700 N concentrated loads applied 2 m from each support. If the allowable flexural stress is 9 MPa and the allowable shearing stress is 0.6 MPa, select the lightest standard structural timber that can be used to support the loading. SOLUTION From overall equilibrium ΣM A = 0 : ΣM B = 0 : B ( 5 ) − 6700 ( 2 ) − 6700 ( 3) = 0 B = 6700 N ↑ 6700 ( 2 ) + 6700 ( 3) − A ( 5 ) = 0 A = 6700 N ↑ σ max = M 13, 400 = ≤ 9 × 106 N/m 2 S S 13, 400 = 1488.9 × 10−6 m3 S≥ 6 9 × 10 = 1488.9 × 103 mm 3 Try a 203 × 254-mm timber with S = 1850 ×103 mm3 I = 223 × 106 mm 4 A = 46, 000 mm 2 2 m = 29.4 kg/m With the weight of the beam included, the maximum moment is M = M load + M weight S≥ ( 29.4 × 9.81)( 5 ) = 13, 400 + 8 = 14,301 N ⋅ m 14,301 = 1589 × 10 −6 m 3 = 1589 × 103 mm 3 6 9 × 10 which is okay. Next, check the shear stress, Vmax = Vload + Vweight = 6700 + ( 29.4 × 9.81)( 5 ) = 7421 N 2 Vmax 1.5 ( 7421) = = 242 × 103 N/m 2 A 46 ×10−3 which is much less than the allowable shear stress of 600 kPa . Therefore, this design is okay. Use a 203 × 254-mm timber ............................................................................................... Ans. τ max = 1.5 8-107 The lever shown in Fig. P8-107 is used to lift a 600 lb rock. Select a standard steel pipe to perform the task. The allowable flexural stress is 20 ksi. Neglect the effects of shear. SOLUTION 356 STATICS AND MECHANICS OF MATERIALS, 2nd Edition From overall equilibrium RILEY, STURGES AND MORRIS ΣM A = 0 : ΣM B = 0 : B ( 5 ) − 600 ( 5.8 ) = 0 B = 696 lb ↑ P ( 5 ) − 600 ( 0.8 ) = 0 P = 96 lb ↓ M max = M B = 96 ( 5 ) = 480 lb ⋅ ft σ max = Use a d = 1.5-in. pipe .............. ( S = 0.326 in 3 ) ............................................................. Ans. 8-108 A structural steel beam is subjected to the loading shown in Fig. P8-108. The allowable flexural stress is 152 MPa, and the allowable shearing stress is 100 MPa. Select the lightest American Standard beam that can be used to support the loading. SOLUTION From overall equilibrium M 480 × 12 = ≤ 20 × 103 lb/in 2 S S S≥ 480 ×12 = 0.288 in 3 20 × 103 ΣM A = 0 : B ( 5.5 ) − 22 (1.25 ) − 40 ( 2.75 ) − 22 ( 4.25 ) = 0 B = 42 kN ↑ ΣM B = 0 : 22 ( 4.25 ) + 40 ( 2.75) + 22 (1.25 ) − A ( 5.5 ) = 0 A = 42 kN ↑ σ max = S≥ M 82.5 × 103 = ≤ 152 × 106 N/m 2 S S 82.5 × 103 = 542.8 ×10−6 m 3 = 542.8 ×103 mm3 6 152 × 10 Try an S305 × 47 section with S = 596 × 103 mm 3 tw = 8.9 mm d = 304.8 mm 2 m = 47 kg/m With the weight of the beam included, the maximum moment is M = M load + M weight = 82.5 ×10 S≥ 3 ( 47 × 9.81)( 5.5 ) + 8 = 84.24 ×103 N ⋅ m 84.24 × 103 = 554.2 ×10−6 m 3 = 554.2 × 103 mm3 152 × 106 which is okay. Next, check the shear stress, Vmax = Vload + Vweight = 42 ×103 + ( 47 × 9.81)( 5.5 ) = 43.27 ×103 N 2 τ max = Vmax 43.27 × 10 = = 15.95 × 106 N/m 2 Aw 0.0089 × 0.3048 3 357 STATICS AND MECHANICS OF MATERIALS, 2nd Edition which is much less than the allowable shear stress of 100 RILEY, STURGES AND MORRIS MPa . Therefore, this design is okay. Use an S305 × 47 section .................................................................................................... Ans. 8-109 A 16-ft long simply supported beam is loaded with a uniform load of 4000 lb/ft over its entire length. If the allowable flexural stress is 22 ksi and the allowable shearing stress is 14.5 ksi, select the lightest structural steel wide-flange beam that can be used to support the loading. SOLUTION From overall equilibrium ΣM A = 0 : ΣM B = 0 : B (16 ) − 4 (16 ) ( 8 ) = 0 B = 32 kip ↑ 4 (16 ) ( 8 ) − A (16 ) = 0 A = 32 kip ↑ M 128 × 12 = ≤ 22 ksi S S 128 × 12 S≥ = 69.82 in 3 22 Try a W18 × 60 section with σ max = S = 108 in 3 2 tw = 0.415 in. d = 18.24 in. w = 60 lb/ft With the weight of the beam included, the maximum moment is 0.060 (16 ) M = 128 + = 129.92 kip ⋅ ft 8 which is still okay. Next, check the shear stress, S≥ 129.92 × 12 = 70.87 in 3 22 Vmax = Vload + Vweight = 32 + 0.060 (16 ) = 32.48 kip 2 τ max = Vmax 32.48 = = 4.291 kip/in 2 = 4.291 ksi Aw 0.415 ×18.24 14.5 ksi . Therefore, this design is okay. which is less than the allowable shear stress of Use a W18 × 60 section ....................................................................................................... Ans. 8-110 Select the lightest wide-flange beam that can be used to support the loading shown in Fig. P8-110. The allowable flexural stress is 152 MPa, and the allowable shearing stress is 100 MPa. SOLUTION From overall equilibrium ΣM A = 0 : ΣM B = 0 : B ( 7.5) − 15 ( 2.5 ) − 15 ( 5 ) = 0 B = 15 kN ↑ 15 ( 5 ) + 15 ( 2.5 ) − A ( 7.5) = 0 A = 15 kN ↑ σ max M 37.5 × 103 = = ≤ 152 × 106 N/m 2 S S 358 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 37.5 × 103 = 246.7 ×10−6 m 3 = 246.7 × 103 mm 3 6 152 × 10 Try a W 254 × 33 section with S≥ S = 380 × 103 mm 3 tw = 6.1 mm d = 258 mm 2 m = 33 kg/m With the weight of the beam included, the maximum moment is M = M load + M weight = 37.5 × 10 S≥ 3 ( 33 × 9.81)( 7.5 ) + 8 = 39.78 × 103 N ⋅ m 39.78 × 103 = 261.7 ×10−6 m 3 = 261.7 × 103 mm 3 152 × 106 which is still okay. Next, check the shear stress, Vmax = Vload + Vweight = 15 × 103 + ( 33 × 9.81)( 7.5 ) = 16.21×103 N 2 τ max = Vmax 16.21× 10 = = 10.30 × 106 N/m 2 Aw 0.0061× 0.258 3 which is much less than the allowable shear stress of 100 MPa . Therefore, this design is okay. Use a W 254 × 33 section .................................................................................................... Ans. 8-111 The floor-framing plan for a residential dwelling is shown in Fig. P8-111. The floor decking is to be supported by 2-in. nominal width joists spaced 16 in. apart. Each joist is to span 12 ft and is simply supported at the ends. The floor decking is subjected to a uniform loading of 60 lb/ft2, which includes the live load plus an allowance for the dead load of the flooring system. The joists are made of construction grade Douglas fir with an allowable flexural stress of 1200 psi and an allowable shearing stress of 120 psi. Determine the required nominal depth of the joists. SOLUTION A joist is a simply supported beam subjected to a uniformly distributed load on a 12-ft span. Since the joists are spaced 16 in. apart, the uniformly distributed load is w = 60 (16 12 ) = 80 lb/ft From overall equilibrium ΣM A = 0 : B (12 ) − 80 (12 ) ( 6 ) = 0 B = 480 lb ↑ ΣM B = 0 : 80 (12 ) ( 6 ) − A (12 ) = 0 A = 480 lb ↑ M 1440 × 12 = ≤ 1200 lb/in 2 S S 1440 ×12 S≥ = 14.40 in 3 1200 Try a 2 × 8-in. timber with σ max = S = 15.3 in 3 I = 57.1 in 4 A = 12.2 in 2 w = 3.39 lb/ft With the weight of the beam included, the maximum moment is 359 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 3.39 (12 ) M = 1440 + = 1501 lb ⋅ ft 8 2 S≥ 1501× 12 = 15.01 in 3 1200 which is still okay. Next, check the shear stress, Vmax = Vload + Vweight = 480 + 3.39 (12 ) = 500.3 lb 2 Vmax 1.5 ( 500.3) = = 61.51 lb/in 2 A 12.2 which is less than the allowable shear stress of 120 psi . Therefore, this design is okay. τ max = 1.5 Use a 2 × 8-in. timber section ............................................................................................ Ans. 8-112 A 15-kN load is supported by a roller on an I-beam, as shown in Fig. P8-112. The roller moves slowly along the beam, thereby causing the shear force and bending moment to be functions of b. Select the lightest permissible American Standard beam to support the loading. The allowable flexural stress is 152 MPa, and the allowable shearing stress is 100 MPa. Neglect the weight of the beam. SOLUTION From overall equilibrium ΣM A = 0 : ΣM B = 0 : B ( 8 ) − 15 x = 0 B = (1.875 x ) kN ↑ 15 ( 8 − x ) − A ( 8 ) = 0 A = (15 − 1.875 x ) kN ↑ M C = Ax = 15 x − 1.875 x 2 x = 4.00 m The moment under the load is The maximum moment occurs under the load when dM C dx = 15 − 3.75 x = 0 M max = 30 kN ⋅ m σ max = S≥ M 30 × 103 = ≤ 152 × 106 N/m 2 S S 30 × 103 = 197.4 ×10−6 m3 = 197.4 ×103 mm3 6 152 × 10 Try an S 203 × 27 section with S = 236 × 103 mm 3 Next, check the shear stress, t w = 6.9 mm d = 203.2 mm m = 27 kg/m τ max Vmax 15 × 103 = = = 10.70 × 106 N/m 2 Aw 0.0069 × 0.2032 MPa . Therefore, this design is okay. which is much less than the allowable shear stress of 100 Use an S 203 × 27 section ................................................................................................... Ans. 8-113 A carriage moves slowly along a simply supported I-beam, as shown in Fig. P8-113. Select the lightest permissible wide-flange beam to support the loading, if the allowable flexural stress is 24 ksi and the allowable shearing stress is 14.5 ksi. Neglect the weight of the beam. Note that the shear force and bending moment are functions of b, the position of the left hand wheel. SOLUTION 360 STATICS AND MECHANICS OF MATERIALS, 2nd Edition From overall equilibrium RILEY, STURGES AND MORRIS ΣM A = 0 : ↑ ΣFy = 0 : D ( 20 ) − 2000 x − 2000 ( x + 5) = 0 D = ( 200 x + 500 ) lb ↑ A + D − 4000 = 0 A = ( −200 x + 3500 ) lb ↑ The moment under the left wheel M B = Ax = ( −200 x 2 + 3500 x ) lb ⋅ ft dM B = −400 x + 3500 = 0 dx x = 8.75 ft is a maximum when M B max = 15,313 lb ⋅ ft The moment under the right wheel M C = D (15 − x ) = ( −200 x 2 + 2500 x + 7500 ) lb ⋅ ft dM C = −400 x + 2500 = 0 dx x = 6.25 ft is a maximum when M C max = 15,313 lb ⋅ ft Therefore, the maximum bending moment occurs under the wheel closest to the center of the beam, M max = 15,313 lb ⋅ ft σ max = Try a M 15,313 × 12 = ≤ 24 × 103 lb/in 2 S S S≥ 15, 313 × 12 = 7.657 in 3 24 × 103 w = 15 lb/ft W 8 × 15 section with S = 11.8 in 3 tw = 0.245 in. d = 8.11 in. Next, check the shear stress, τ max = Vmax 3500 = = 1762 lb/in 2 Aw 0.245 × 8.11 ksi . Therefore, this design is okay. which is much less than the allowable shear stress of 14.5 Use a W 8 × 15 section ......................................................................................................... Ans. 8-114 A beam has the cross section shown in Fig. P8-114. If the flexural stress at point A is 14 MPa (T), determine (a) The maximum flexural stress on the section. (b) The resisting moment Mr at the section. SOLUTION yC = 25 200 ( 50 ) + 100 50 ( 200 ) + 100 50 ( 200 ) 200 ( 50 ) + 50 ( 200 ) + 50 ( 200 ) = 75.00 mm 200 ( 50 )3 50 ( 200 )3 2 2 + ( 50 ) ( 200 × 50 ) + 2 + ( 25) ( 50 × 200 ) = 106.25 ×106 mm 4 I = 12 12 361 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (a) RILEY, STURGES AND MORRIS σ max = c 125 σA = (14 ) = −70 MPa = 70 MPa (C) ................................................... Ans. yA −25 (b) σA = − M ( −0.025) M r yA =− r = 14 × 106 −6 I 106.25 ×10 M r = 59.5 ×103 N ⋅ m = 59.5 kN ⋅ m ................................................................................. Ans. 8-115 A T-beam has the cross section shown in Fig. P8-115. Determine the maximum tensile and compressive flexural stresses on a cross section of the beam where the resisting moment being transmitted is 74 kip-ft. SOLUTION 4 1.5 ( 8) + 9 4 ( 2 ) = 6.00 in. yC = 1.5 ( 8) + 4 ( 2 ) 1.5 ( 8 )3 4 ( 2 )3 2 2 + ( 2 ) (1.5 × 8 ) + + ( 3) ( 4 × 2 ) = 186.667 in 4 I = 12 12 σ bot = − σ top 8-116 ( 74 ×12 )( −6 ) = +28.5 kip/in.2 = 28.5 ksi (T) ................................ Ans. My =− I 186.667 ( 74 ×12 )( 4 ) = −19.03 kip/in.2 = 19.03 ksi (C) .............................................. Ans. =− 186.667 Determine the percentage of the resisting moment Mr carried by the flanges of a W838 × 226 wide-flange beam (see Appendix A for dimensions). SOLUTION From Table A-2 for a W 838 × 226 section: d = 851 mm I = 3395 × 106 mm 4 t f = 26.8 mm σ max = For the web Mc I M= tw = 16.1 mm 3 σ max I σ max ( 3395 ) = = ( 7.9788σ max ) kN ⋅ m c 425.5 d w = 851 − 2 ( 26.8 ) = 797.4 mm 16.1( 797.4 ) Iw = = 680.3 × 106 mm 4 12 398.7 σ max ( web ) = (σ max ) = 0.9370 σ max 425.5 Mw = Mf = 8-117 σ max ( web ) I w ( 0.9370 σ max )( 680.3) = = (1.5988σ max ) kN ⋅ m cw 398.7 7.9788 − 1.5988 (100 ) = 80.0 % ............................................................................. Ans. 7.9788 The maximum flexural stress on the cross section of the beam shown in Fig. P8-117 is 8000 psi (C). Determine (a) The resisting moment being transmitted by the section. (b) The magnitude of the flexural force carried by the flange. SOLUTION 362 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 3 2 ( 6 ) + 7 6 ( 2 ) = 5.00 in. yC = 2 ( 6) + 6 ( 2) 2 ( 6 )3 6 ( 2 )3 2 2 + ( 2) ( 2 × 6) + + ( 2 ) ( 6 × 2 ) = 136 in 4 I = 12 12 (a) σ =− M ( −5 ) Mr y =− r = −8 ×103 I 136 M r yCf I 3 M r = −217.6 × 103 lb ⋅ ft ≅ −218 kip ⋅ in. .......................................................................... Ans. (b) σ Cf = − ( −217.6 ×10 ) ( 2 ) = 3200 lb/in =− 136 2 Ff = σ Cf Af = 3200 ( 6 × 2 ) = 38, 400 lb = 38.4 kip (T) ................................................ Ans. 8-118 A beam is loaded and supported as shown in Fig. P8-118. (a) Draw complete shear force and bending moment diagrams for the beam. (b) Using the coordinate axes shown, write equations for the shear force and bending moment for any section of the beam in the interval 0 < x < 4 m. SOLUTION From overall equilibrium: ΣM E = 0 : 20 ( 2 ) + 8 ( 6 ) ( 7 ) +10 (12 ) − RB ( 8 ) = 0 RB = 62 kN For 0m≤x≤4m V = −10 − 8 ( x + 2 ) + RB = ( −8 x + 36 ) kN ................................ Ans. M = −10 ( x + 4 ) + RB x − 8 ( x + 2 ) x+2 2 = ( −4 x 2 + 36 x − 56 ) kN ⋅ m ............ Ans. 8-119 A beam is loaded and supported as shown in Fig. P8-119. (a) Draw complete shear force and bending moment diagrams for the beam. (b) Using the coordinate axes shown, write equations for the shear force and bending moment for any section of the beam in the interval 0 < x < 10 ft. SOLUTION For 0 ft ≤ x ≤ 10 ft V = 200 ( 6 ) + 760 − 300 x = ( −300 x + 1960 ) lb .................................... Ans. M = 200 ( 6 ) ( x + 3) + 760 x − 300 x ( x 2 ) = ( −150 x 2 + 1960 x + 3600 ) lb ⋅ ft ............ Ans. 363 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 8-120 RILEY, STURGES AND MORRIS Select the lightest pair of structural-steel angles (see Appendix A) that may be used for the beam of Fig. P8-120 if the maximum flexural stress must be limited to 60 MPa. The angles will be fastened back-toback to form a T-section. SOLUTION From the moment diagram, M max = 7.5 kN ⋅ m σ max = S= M 7.5 × 103 = = 60 × 106 S S 7.5 × 103 = 125 ×10−6 m3 = 125 × 103 mm 3 60 × 106 For each angle S= 125 × 103 = 62.5 ×103 mm3 2 Use two L178 ×102 × 9.5-mm angles .............................................................................. Ans. 8-121 A timber beam is loaded and supported as shown in Fig. P8-121a. The beam has the cross section shown in Fig. P8-121b. Determine (a) The flexural stresses at points A and B on a transverse cross section 1 ft from the left end of the beam. (b) The maximum tensile and compressive flexural stresses in the beam. SOLUTION I NA 8 (10 ) 6 ( 6 ) = − = 558.7 in 4 12 12 3 3 M 1 = −5600 + 200 (1) ( 0.5 ) = −5500 lb ⋅ ft (a) σA = − ( −5500 ×12 )( −3) My =− I 558.7 ( −5500 ×12 )( 5) My =− I 558.7 = −354.4 lb/in 2 ≅ 354 psi (C) ............Ans. σB = − = +590.7 lb/in 2 ≅ 591 psi (T) ............Ans. (b) From the moment diagram, M max = −5600 lb ⋅ ft σ top = − ( −5600 ×12 )( 5 ) 558.7 = +601.4 lb/in 2 ≅ 601 psi (T) ............Ans. σ bot = − ( −5500 ×12 )( −5) 558.7 = −601.4 lb/in 2 ≅ 601 psi (C) ............Ans. 364 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 8-122 RILEY, STURGES AND MORRIS Select the lightest steel wide-flange or American standard beam (see Appendix A) that may be used for the beam of Fig. P8-122 if the maximum flexural stress must be limited to 75 MPa. SOLUTION From the moment diagram, M max = 20 kN ⋅ m σ max = S= M 20 × 103 = = 75 × 106 S S 20 × 103 = 266.7 × 10−6 m3 75 × 106 = 266.7 × 103 mm3 Use a W 254 × 33 section ....................Ans. 8-123 The beam shown in Fig. P8-123a has the cross section shown in Fig. P8-113b. Determine the maximum tensile and compressive flexural stresses in the beam. SOLUTION 4 1( 8 ) + 8.5 4 (1) = 5.50 in. yC = 1( 8) + 4 (1) 1 ( 8 )3 4 (1)3 2 2 I = + (1.5) (1× 8 ) + + ( 3) ( 4 × 1) = 97 in 4 12 12 From the moment diagram, M max = 14, 742 lb ⋅ ft σ top = − (14, 742 ×12 )( 3.5 ) My =− I 97 (14, 742 ×12 )( −5.5) 97 = −6380 lb/in 2 = 6.38 ksi (C) σ bot = − = +10, 030 lb/in 2 = 10.03 ksi (T) Therefore σ max T = 10.03 ksi (T) ............................................................................................................ Ans. σ max C = 6.38 ksi (C) 8-124 (a) (b) (c) (d) ............................................................................................................. Ans. A WT305 × 70 structural tee (see Appendix A) is loaded and supported as a beam (with the flange on top) as shown in Fig. P8-124. Determine The maximum tensile flexural stress in the beam. The maximum compressive flexural stress in the beam. The maximum vertical shearing stress in the beam. The vertical shearing stress at a point in the stem just below the flange on the cross section where the maximum vertical shearing stress occurs. SOLUTION 365 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ΣM E = 0 : ΣM B = 0 : For a 10 (1) + 15 ( 2 ) ( 2 ) + 10 ( 5 ) − RB ( 4 ) = 0 RB = 30 kN 10 (1) − 15 ( 2 ) ( 2 ) − 10 ( 3) + RE ( 4 ) = 0 RE = 30 kN WT 305 × 70 section: I = 77.4 × 106 mm 4 t f = 22.2 mm d = 308.7 mm w f = 230.3 mm tw = 13.1 mm yC = 75.9 mm (a) At the bottom of the beam 2.33 m from the left support: σ max T = − ( 23.33 ×103 ) ( −0.2328) My =− 77.4 ×10−6 I = +70.17 ×106 N/m 2 ≅ 70.2 MPa (T) ................................................................. Ans. (b) At the bottom of the beam at the left support: σ max C ( −10.00 ×103 ) ( −0.2328) My =− =− 77.4 × 10−6 I = −30.08 ×106 N/m 2 ≅ 30.1 MPa (C) ................................................................. Ans. (c) QNA = yC A = 116.4 (13.1× 232.8 ) = 355.0 × 103 mm3 τ max (d) 3 −6 VQ ( 20 × 10 )( 355.0 × 10 ) = = = 7.00 × 106 N/m 2 = 7.00 MPa .................... Ans. −6 It ( 77.4 ×10 ) ( 0.0131) QJ = 64.8 ( 22.2 × 230.3) = 331.3 × 103 mm 3 τJ 8-125 (a) (b) (c) (d) ( 20 ×10 )( 331.3 ×10 ) = 6.53 ×10 = ( 77.4 ×10 ) ( 0.0131) 3 −6 −6 6 N/m 2 = 6.53 MPa ................................... Ans. A WT12 × 52 structural tee (see Appendix A) is loaded and supported as a beam (with the flange on the bottom), as shown in Fig. P8-125. Determine The maximum tensile flexural stress in the beam. The maximum compressive flexural stress in the beam. The maximum vertical shearing stress in the beam. The vertical shearing stress at a point in the stem just above the flange on the cross section where the maximum vertical shearing stress occurs. SOLUTION ΣM E = 0 : ΣM B = 0 : 7 + 1( 8 ) ( 7 ) + 5 (16 ) − 1( 3) 2 ( 2 ) − RB (14 ) = 0 RB = 10 kip 7 − 1( 8 ) ( 7 ) + 5 ( 2 ) − 1( 3) 2 (16 ) + RE (14 ) = 0 RE = 4.50 kip 366 STATICS AND MECHANICS OF MATERIALS, 2nd Edition For a RILEY, STURGES AND MORRIS WT12 × 52 section: I = 189 in 4 t f = 0.750 in. d = 12.030 in. w f = 12.75 in. tw = 0.500 in. yC = 2.59 in. (a) At the top of the beam at the left support: σ max T = − ( −17.00 ×12 )( 9.44 ) My =− 189 I (10.50 ×12 )( 9.44 ) 189 = +10.19 kip/in 2 = 10.19 ksi (T) ............Ans. (b) At the top of the beam 8 ft from the left support: σ max C = − = −6.29 kip/in 2 = 6.29 ksi (C) .............................................................................. Ans. (c) QNA = yC A = 4.72 ( 0.5 × 9.44 ) = 22.28 in 3 τ max = (d) VQ 5000 ( 22.28 ) = = 1179 lb/in 2 = 1179 psi .................................................... Ans. It 189 ( 0.500 ) QJ = 2.215 (12.75 × 0.750 ) = 21.18 in 3 τJ = 8-126 5000 ( 21.18 ) 189 ( 0.500 ) = 1121 lb/in 2 = 1121 psi ................................................................... Ans. An extruded aluminum alloy beam has the cross section shown in Fig. P8-126. All parts of the section are 5 mm thick. When a constant shear force V = 4450 N is being supported by the beam, determine (a) The shearing stresses at points A and B of the cross section. (b) The maximum horizontal shearing stress in the beam. SOLUTION A = 5 ( 50 + 100 + 50 + 100 + 100 ) = 2000 mm 2 yC = 50 5 (100 ) + 100 100 ( 5) 2000 = 37.5 mm 5 ( 50 )3 100 ( 5)3 2 + + ( 37.5 ) ( 2 × 5 × 50 + 100 × 5) I = 2 12 12 5 (100 )3 100 ( 5 )3 2 2 + + (12.5 ) ( 5 ×100 ) + + ( 62.5) (100 × 5) = 3.960 × 106 mm 4 12 12 (a) QA = yC A = 62.5 (100 × 5 ) = 31.25 × 103 mm3 4450 ( 31.25 ×10−6 ) VQ = = 7.02 × 106 N/m 2 = 7.02 MPa ........................... Ans. τA = −6 It ( 3.960 × 10 ) ( 0.005 ) QB = 62.5 (100 × 5 ) + 12.5 ( 5 × 100 ) = 37.50 ×103 mm3 367 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS τB = (b) ( 3.960 ×10 ) ( 0.005) −6 4450 ( 37.50 ×10−6 ) = 8.43 × 106 N/m 2 = 8.43 MPa ....................................... Ans. QC = 62.5 (100 × 5 ) + 31.25 ( 5 × 62.5) = 41.02 × 103 mm3 τC = ( 3.960 ×10 ) ( 0.005) −6 4450 ( 41.02 × 10−6 ) = 9.22 × 106 N/m 2 = 9.22 MPa ....................................... Ans. 368 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS Chapter 9 9-1 (a) (b) (c) (a) A beam is loaded and supported as shown in Fig. P9-1. Determine The equation of the elastic curve. Use the designated axes. The deflection at the left end of the beam. The slope at the left end of the beam. SOLUTION EIy′′ = − Px EIy′ = − Px 2 + C1 2 At x = L, y′ = 0 : y = 0: PL2 C1 = 2 − PL3 C2 = 3 Px3 EIy = − + C1 x + C2 At x = L, 6 P y′ = ( − x2 + L2 ) 2 EI P y= ( − x3 + 3L2 x − 2L3 ) .................................Ans. 6 EI (b) δ A = yx=0 P − PL3 PL3 3 = ( 0 + 0 − 2 L ) = 3EI = 3EI ↓ ............................................................. Ans. 6 EI P + PL2 PL2 2 ( 0 + L ) = 2 EI = 2 EI 2 EI ..................................................................... Ans. (c) 9-2 (a) (b) (c) θ A = y′ =0 = x A beam is loaded and supported as shown in Fig. P9-2. Determine The equation of the elastic curve. Use the designated axes. The deflection at the right end of the beam. The slope at the right end of the beam. SOLUTION ↑ ΣFy = 0 : VA − wL = 0 VA = + wL = wL ↑ − M A − wL ( L 2 ) = 0 wL2 wx 2 − 2 2 At ΣM A = 0 : M A = − wL2 2 = wL2 2 (a) EIy′′ = wLx − EIy′ = EIy = wLx 2 wL2 x wx3 − − + C1 2 2 6 x = 0, x = 0, y′ = 0 : y = 0: C1 = 0 C2 = 0 wLx 3 wL2 x 2 wx 4 − − + C1 x + C2 6 4 24 w y′ = ( − x3 + 3Lx2 − 3L2 x ) 6 EI At 369 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS y= (b) w ( − x 4 + 4Lx3 − 6 L2 x2 ) ......................................................................................... Ans. 24 EI 4 w wL4 ( − L4 + 4 L4 − 6L4 ) = −8wL = 8EI ↓ ................................................. Ans. EI 24 EI 3 w wL3 ( − L3 + 3L3 − 3L3 ) = −6wL = 6EI EI 6 EI δ B = yx = L = θ B = y′ = L = x (a) (b) (c) (c) 9-3 ................................................... Ans. A beam is loaded and supported as shown in Fig. P9-3. Determine The equation of the elastic curve. Use the designated axes. The deflection midway between the supports. The slope at the left end of the beam. SOLUTION (a) EIy′′ = EIy′ = wLx wx 2 − 2 2 wLx 2 wx 3 − + C1 4 6 L At x = , 2 y′ = 0 : − wL3 C1 = 24 wLx 3 wx 4 EIy = − + C1 x + C2 At x = 0, y = 0 : C2 = 0 12 24 w y′ = ( −4 x3 + 6 Lx2 − L3 ) 24 EI w y= ( − x 4 + 2Lx3 − L3 x ) .............................................................................................. Ans. 24 EI (b) δ M = yx= L 2 = θ A = y′ =0 = x (a) (b) (c) w L4 L4 L4 −5wL4 5wL4 − + − = = ↓ ........................................ Ans. 24 EI 16 4 2 384 EI 384 EI .......................................................... Ans. (c) 9-4 w − wL3 wL3 ( 0 + 0 − L3 ) = 24 EI = 24 EI 24 EI A beam is loaded and supported as shown in Fig. P9-4. Determine The equation of the elastic curve. Use the designated axes. The deflection at the left end of the beam. The slope at the left end of the beam. SOLUTION w ( x ) = wx L ( wx L )( x ) x wx3 =− M ( x) = − 2 6L 3 (a) EIy′′ = − wx 3 6L 370 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS At EIy′ = − EIy = − wx 4 + C1 24 L x = L, y′ = 0 : C1 = wL3 24 wx5 − wL4 + C1 x + C2 At x = L, y = 0 : C2 = 120 L 30 w y′ = ( − x4 + L4 ) 24 EIL w y= ( − x5 + 5L4 x − 4 L5 ) .......................................................................................... Ans. 120 EIL w − wL4 wL4 0 + 0 − 4 L5 ) = = ↓ .................................................... Ans. ( 120 EIL 30 EI 30 EI w + wL3 wL3 = 0 + L4 ) = ( 24 EIL 24 EI 24 EI ............................................................... Ans. (b) δ A = yx=0 = θ A = y′ =0 = x (a) (b) (c) (c) 9-5 A beam is loaded and supported as shown in Fig. P9-5. Determine The equation of the elastic curve. Use the designated axes. The deflection midway between the supports. The slope at the right end of the beam. SOLUTION ΣM B = 0 : ( wL 2 )( L 3) − RA L = 0 RA = wL 6 w ( x ) = wx L ( wx L )( x ) x wLx wx 3 − M ( x ) = RA x − = 2 6 6L 3 (a) EIy′′ = EIy′ = EIy = wLx wx3 − 6 6L wLx 2 wx 4 − + C1 12 24 L At x = 0, y = 0: C2 = 0 wLx 3 wx 5 −7 wL3 − + C1 x + C2 At x = L, y = 0 : C1 = 36 120 L 360 w y′ = ( −15 x4 + 30 L2 x2 − 7 L4 ) 360 EIL w y= ( −3x5 + 10 L2 x3 − 7 L4 x ) .................................................................................. Ans. 360 EIL w 3L5 10 L5 7 L5 −5wL4 5wL4 − + − = = ↓ .......................... Ans. 360 EIL 32 8 2 768 EI 768EI ................................... Ans. (b) δ M = yx= L 2 = θ B = y′ = L = x (c) w + wL3 wL3 −15L4 + 30 L4 − 7 L4 ) = = ( 360 EIL 45 EI 45EI 371 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 9-6 RILEY, STURGES AND MORRIS For the steel beam [E = 200 GPa and I = 32.0(106) mm4] shown in Fig. P9-6, determine the deflection at a section midway between the supports. SOLUTION RA = RB = 10 kN M ( x ) = RA x − 10 ( x + 1) = −10 kN ⋅ m EIy′′ = −10 kN ⋅ m EIy′ = ( −10 x + C1 ) kN ⋅ m 2 EIy = ( −5 x 2 + C1 x + C2 ) kN ⋅ m3 EIy = ( −5 x 2 + 10 x ) kN ⋅ m3 At At x = 0 m, x = 2 m, y = 0: y = 0: C2 = 0 C1 = 10 kN ⋅ m3 δM 9-7 −5 (1)2 + 10 (1) ×103 = yx =1 = = +0.000781 m = 0.781 mm ↑ ...................... Ans. 9 ( 200 ×10 )( 32.0 ×10−6 ) For the steel beam (E = 30,000 ksi and I = 32.1 in.4) shown in Fig. P9-7, determine the deflection at a section midway between the supports. SOLUTION EIy′′ = 7500 lb ⋅ ft EIy′ = ( 7500 x + C1 ) lb ⋅ ft 2 EIy = ( 3750 x 2 + C1 x + C2 ) lb ⋅ ft 3 EIy = ( 3750 x 2 − 45, 000 x ) lb ⋅ ft 3 At At x = 0 ft, x = 12 ft, y = 0: y = 0: C2 = 0 C1 = −45, 000 lb ⋅ ft 3 δ M = yx=6 9-8 3750 ( 6 )2 − 45, 000 ( 6 ) (12 )3 = = −0.242 in. = 0.242 in. ↓ .................. Ans. 6 ( 30 ×10 ) ( 32.1) A cantilever beam is fixed at the left end and carries a uniformly distributed load w over the full length of the beam. In addition, the right end is subjected to a moment of +3wL2/8, as shown in Fig. P9-8. Determine the maximum deflection in the beam if I = 2.5(106) mm4, E = 210 GPa, L = 3 m, and w = 1500 N/m. SOLUTION ↑ ΣFy = 0 : ΣM A = 0 : 2 VA − wL = 0 VA = wL = wL ↑ M A = − wL2 8 = wL2 8 ( 3wL 8) − ( wL )( L 2 ) − M wL2 wx 2 − 8 2 A =0 wx x M ( x ) = VA x + M A − 2 2 = wLx − 372 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS EIy′′ = wLx − EIy′ = EIy = wL2 wx 2 − 8 2 At wLx 2 wL2 x wx3 − − + C1 2 8 6 x = 0, y′ = 0 : y = 0: C1 = 0 C2 = 0 wLx 3 wL2 x 2 wx 4 − − + C1 x + C2 At x = 0, 6 16 24 w y′ = ( −4 x3 + 12 Lx2 − 3L2 x ) 24 EI w y= ( −2 x4 + 8Lx3 − 3L2 x2 ) 48 EI The maximum deflection occurs either when y ′ = 0 or at x = L y′ = w ( −4 x3 + 12 Lx 2 − 3L2 x ) = 0 24 EI x = 0, 0.2753L, 2.725 L δ 1 = yx =0.2753 L = −0.001379wL4 EI δ B = y x = L = + wL4 16 EI = +0.06250wL4 EI 1500 ( 3) wL4 = δB = = = +0.01446 m = 14.46 mm ↑ ....... Ans. 16 EI 16 ( 210 × 109 )( 2.5 × 10−6 ) 4 δ max 9-9 (a) (b) (c) (a) A beam is loaded and supported as shown in Fig. P9-9. Determine The equation of the elastic curve. Use the designated axes. The slope at the left end of the beam. The deflection midway between the supports. SOLUTION EIy′′ = Px 2 EIy′ = EIy = Px 2 + C1 4 Px 3 + C1 x + C2 12 At x = L 2, At y′ = 0 : y = 0: C1 = − PL2 16 x = 0, C2 = 0 P ( 4 x 2 − L2 ) 16 EI P y= ( 4 x3 − 3L2 x ) ............................................ Ans. 48 EI y′ = (b) θ A = y′ = 0 = x P − PL2 PL2 ( 0 − L2 ) = 16 EI = 16EI 16 EI ................................................................. Ans. (c) δ M = yx= L 2 = P 4 L3 3L3 − PL3 PL3 − = = ↓ ..................................................... Ans. 48EI 8 2 48EI 48EI 373 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 9-10 RILEY, STURGES AND MORRIS A 100 × 300-mm timber having a modulus of elasticity of 8 GPa is loaded and supported as shown in Fig. P9-10. Determine (a) The deflection at the 7-kN load. (b) The deflection at the free end of the beam. SOLUTION 100 ( 300 ) = 225 × 106 mm 4 = 225 × 10−6 m 4 I= 12 3 RA = 7 kN ↑ M ( x ) = RA x − M A = ( 7 x − 14 ) kN ⋅ m (a) M A = 14 kN ⋅ m 0m≤x≤2m At EIy′′ = ( 7 x − 14 ) kN ⋅ m 7x2 EIy′ = − 14 x + C1 kN ⋅ m 2 2 7 x3 EIy = − 7 x 2 + C1 x + C2 kN ⋅ m3 6 7 x2 EIy′ = − 14 x kN ⋅ m 2 2 7 x3 EIy = − 7 x 2 kN ⋅ m3 6 x = 0, x = 0, y′ = 0 : y = 0: C1 = 0 C2 = 0 At 0m≤x≤2m 0m≤x≤2m δ B = yx=2 θ B = y′ =2 x 7 ( 2 )3 − 42 ( 2 )2 ×103 = = −0.01037 m = 10.37 mm ↓ .......................... Ans. 9 6 ( 8 × 10 )( 225 × 10−6 ) 7 ( 2 )2 − 28 ( 2 ) ×103 = = −0.007778 rad = 0.007778 rad 9 2 ( 8 × 10 )( 225 × 10−6 ) (b) δ C = δ B + θ B (1.5 ) = −0.01037 + ( −0.007778 )(1.5) = −0.0220 m = 22.0 mm ↓ .......................................................................................... Ans. 9-11 A 130-lb boy is standing on a 1.6 × 12-in. wood (E = 1400 ksi) diving board, as shown in Fig. P9-11. If length AB is 2 ft and length BC is 5 ft, determine the maximum deflection in the diving board. SOLUTION ΣM A = 0 : ↑ ΣFy = 0 : RB L − P ( 3.5L ) = 0 RA + RB − P = 0 RB = 3.5P = 3.5P ↑ RA = −2.5P = 2.5 P ↓ For the interval 0 ≤ x≤L ′′ EIy1 = −2.5 Px 374 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ′ EIy1 = − EIy1 = − 2.5Px 2 + C1 2 At x = 0, x = L, y1 = 0 : y1 = 0 : C2 = 0 C1 = 2.5PL2 6 2.5Px 3 + C1 x + C2 6 2.5 P ′ y1 = ( −3x 2 + L2 ) 6 EI ′ y1 = 2.5 P ( −3x2 + L2 ) = 0 6 EI At y1 = 2.5 P ( − x3 + L2 x ) 6 EI The maximum deflection in the interval 0 ≤ x ≤ L occurs when x = 0.5774 L δ max = yx =0.5774 L = +0.16038 PL3 EI = 0.16038 PL3 EI ↑ For the interval L ≤ x ≤ 3.5 L 2 ′′ EIy2 = −2.5 Px + 3.5P ( x − L ) 2.5Px 2 3.5P ( x − L ) ′ + + C3 EIy2 = − 2 2 2.5Px3 3.5P ( x − L ) EIy2 = − + + C3 x + C4 6 6 The two solutions must match at x = L, 3 ′ ′ EIy1 = EIy2 : EIy1 = EIy2 : 3 − − 2.5PL2 2.5PL2 2.5 PL2 + =− + 0 + C3 2 6 2 2.5PL3 2.5PL3 + C1 L = − + 0 + C3 L + C4 6 6 C3 = 2.5 PL2 6 C4 = 0 12 (1.6 ) I= = 4.096 in 4 12 The maximum deflection in the interval L ≤ x ≤ 3.5 L occurs at C 3 δ max = yx =3.5 L −7.292 PL3 −7.292 (130 )( 2 ×12 ) = = EI (1.4 ×106 ) ( 4.096 ) = −2.29 in. = 2.29 in. ↓ ............................................................................................... Ans. 9-12 A timber beam 150 mm wide × 300 mm deep is loaded and supported as shown in Fig. P9-12. The modulus of elasticity of the timber is 10 GPa. A pointer is attached to the right end of the beam. Determine (a) The deflection of the right end of the pointer. (b) The maximum deflection of the beam. SOLUTION For the interval (a) 0≤ x≤ L 2 EIy′′ = Px 2 375 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS EIy′ = EIy = Px 2 + C1 4 At x = L 2, At y′ = 0 : y = 0: C1 = − PL2 16 Px 3 + C1 x + C2 12 P y′ = ( 4 x 2 − L2 ) 16 EI 3 x = 0, C2 = 0 y= P ( x3 − 3L2 x ) 48 EI I= 150 ( 300 ) = 337.5 × 106 mm 4 = 337.5 × 10−6 m 4 12 P − PL2 PL2 = 0 − L2 ) = ( 16 EI 16 EI 16 EI As a result of symmetry, θ B = −θ A θ A = y′ = 0 = x δ P = θ B LP = (b) 16 (10 × 109 )( 337.5 ×10−6 ) 8900 ( 5.5) 2 (1.5 ) = +0.00748 m = 7.48 mm ↑ ........... Ans. δ max = yx = L 2 = P 4 L3 3L3 − PL3 − = 48 EI 8 2 48EI 3 −8900 ( 5.5 ) − PL3 = = = −0.00914 m = 9.14 mm ↓ ................ Ans. 48EI 48 (10 × 109 )( 337.5 × 10−6 ) 9-13 A beam is loaded and supported as shown in Fig. P9-13. Determine (a) The equation of the elastic curve. Use the designated axes. (b) The slope at the right end of the beam. SOLUTION ΣM B = 0 : M ( x) = PL − RA L − ( PL 3) = 0 PL PL Px + RA x − Px = − 3 3 3 R A = +2 P 3 = 2 P 3 ↑ (a) EIy′′ = PL Px − 3 3 EIy′ = EIy = PLx Px 2 − + C1 3 6 PLx 2 Px3 − + C1 x + C2 6 18 At x = 0, x = L, y = 0: y = 0: C2 = 0 C1 = − PL2 9 At P ( −3x2 + 6 Lx − 2 L2 ) 18 EI P y= ( − x3 + 3Lx 2 − 2 L2 x ) ............................................................................................ Ans. 18 EI y′ = 376 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ............................................. Ans. θ B = y′ = L = x P + PL2 PL2 −3L2 + 6 L2 − 2 L2 ) = = ( 18EI 18EI 18EI 9-14 A beam is loaded and supported as shown in Fig. P9-14. Determine (a) The equation of the elastic curve. Use the designated axes. (b) The deflection at the left end of the beam. SOLUTION − ( kx 2 ) x x −kx 4 M ( x) = = 3 4 12 EIy′′ = − EIy′ = − EIy = − y′ = kx 4 12 kx 5 + C1 60 At x = L, x = L, y′ = 0 : y = 0: C1 = C2 = kL5 60 −5kL6 360 kx 6 + C1 x + C2 360 y= At k ( − x5 + L5 ) 60 EI k ( − x6 + 6 L5 x − 5L6 ) 360 EI (a) At x = L , w = kL2 . Therefore, k = w L2 y= w − x 6 + 6 L5 x − 5 L6 ) ........................................................................................ Ans. 2( 360 EIL (b) 9-15 δ A = yx=0 −5wL4 wL4 = = ↓ ............................................................................................. Ans. 360 EI 72 EI Determine the deflection midway between the supports for beam AB of Fig. P9-15. Segment BC of the beam is rigid. SOLUTION ΣM B = 0 : ( wL )( L 2 ) + ( wL 2 )( L 2 ) − RA L = 0 3wLx wx 2 − 4 2 3wLx 2 wx 3 − + C1 8 6 At RA = 3wL 4 = 3wL 4 ↑ EIy′′ = EIy′ = EIy = x = 0, x = L, y = 0: y = 0: C2 = 0 C1 = − wL3 12 3wLx3 wx 4 − + C1 x + C2 24 24 w y′ = ( −4 x3 + 9 Lx2 − 2 L3 ) 24 EI At y= w ( − x 4 + 3Lx3 − 2L3 x ) 24 EI δ max = yx = L 2 = −11wL4 11wL4 w L4 3L4 −+ − L4 = = ↓ .................................. Ans. 24 EI 16 8 384 EI 384 EI 377 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 9-16 RILEY, STURGES AND MORRIS The beam shown in Fig. P9-16 is a W203 × 60 structural steel (E = 200 GPa) wide-flange section (see Appendix A). Determine (a) The equation of the elastic curve for the region of the beam between the supports. Use the designated axes. (b) The deflection midway between the supports if w = 3.5 kN/m and L = 5 m. SOLUTION ΣM B = 0 : ( wL 2 ) − ( wL 2 )( L 4 ) − R L = 0 2 M A − M B − RA L = 0 A RA = 3wL 8 ↑ (a) EIy′′ = EIy′ = EIy = 3wLx wL2 − 8 2 3wLx 2 wL2 x − + C1 16 2 At x = 0, x = L, y = 0: y = 0: C2 = 0 C1 = 3wL3 16 3wLx 3 wL2 x 2 − + C1 x + C2 48 4 w y′ = ( 3Lx2 − 8L2 x + 3L3 ) 16 EI At y= w ( Lx3 − 4L2 x2 + 3L3 x ) ................... Ans. 16 EI 4 (b) For a W 203 × 60 wide-flange section: I = 60.8 × 106 mm 4 δ max = yx = L 2 5 ( 3.5 × 103 ) ( 5 ) w L4 4 L4 3L4 5wL4 = + = − = 16 EI 8 4 2 128 EI 128 ( 200 × 109 )( 60.8 × 10−6 ) = +7.03 × 10−3 m = 7.03 mm ↑ .................................................................................. Ans. 9-17 A beam AB is loaded and supported as shown in Fig. P9-17. The load P is applied through a collar which can be positioned on the load bar DE at any location in the interval L/4 < a < 3L/4. Determine (a) The equation of the elastic curve for beam AB. (b) The location of load P for maximum deflection at end B. (c) The location of load P for zero deflection at end B. SOLUTION ↑ ΣFy = 0 : ΣM A = 0 : (a) VA − P = 0 − M A − Pa = 0 VA = P M A = − Pa EIy′′ = Px − Pa EIy′ = EIy = Px 2 − Pax + C1 2 At x = 0, x = 0, y′ = 0 : y = 0: C1 = 0 C2 = 0 Px 3 Pax 2 − + C1 x + C2 6 2 P y′ = ( x 2 − 2ax ) 2 EI At y= P ( x3 − 3ax2 ) ....................................... Ans. 6 EI 378 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (b) RILEY, STURGES AND MORRIS δ B = yx= L = For P ( L3 − 3aL2 ) 6 EI a=L 4 a = 3L 4 δB = δB = P 3 3L3 + PL3 PL3 = = ↑ L− 6 EI 4 24 EI 24 EI P 3 9 L3 −5PL3 5PL3 = ↓ L − = 6 EI 4 24 EI 24 EI a = 3L 4 ............................................................. Ans. a = L 3 .............................................................. Ans. For The maximum deflection at B occurs when δB = 0 9-18 when L3 − 3aL2 = 0 The cantilever beam ABC shown in Fig. P9-18 has a second moment of area IAB = 2I in the interval AB and a second moment of area IBC = I in the interval BC. Determine (a) The deflection at section B. (b) The deflection at section C. SOLUTION ↑ ΣFy = 0 : ΣM A = 0 : For the interval VA − 2 P = 0 −M A − P ( L ) − P ( 2L ) = 0 VA = 2 P M A = −3PL 0≤ x≤ L ′′ 2 EIy1 = 2 Px − 3PL ′ 2 EIy1 = Px 2 − 3PLx + C1 2 EIy1 = ′ y1 = At At x = 0, x = 0, ′ y1 = 0 : y1 = 0 : C1 = 0 C2 = 0 Px3 3PLx 2 − + C1 x + C2 3 2 y1 = P ( x2 − 3Lx ) 2 EI For the interval L ≤ x ≤ 2 L P ( 2 x3 − 9Lx2 ) 12 EI ′′ EIy2 = 2 Px − 3PL − P ( x − L ) P ( x − L) ′ + C3 EIy2 = Px − 3PLx − 2 2 2 Px3 3PLx 2 P ( x − L ) EIy2 = − − + C3 x + C4 3 2 6 3 The two solutions must match at x = L, C3 = PL2 ′ ′ EIy1 = EIy2 : EIy1 = EIy2 : y2 = PL2 3PL2 − = PL2 − 3PL2 + 0 + C3 2 2 PL3 3PL3 PL3 3PL3 − = − + 0 + C3 L + C4 6 4 3 2 C4 = −5PL3 12 P3 3 4 x − 18 Lx 2 − 2 ( x − L ) + 12 L2 x − 5 L3 12 EI 379 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (a) RILEY, STURGES AND MORRIS δ B = ( y1 ) x = L = ( y2 ) x = L = δ C = ( y2 ) x = 2 L = −7 PL3 7 PL3 = ↓ ...................................................................... Ans. 12 EI 12 EI (b) 9-19 −23PL3 23PL3 = ↓ .................................................................................. Ans. 12 EI 12 EI The simply supported beam ABCD shown in Fig. P9-19 has a second moment of area IBC = 2I in the center section BC and a second moment of area IAB = ICD = I in the other two sections near the supports. Determine (a) The deflection at section B. (b) The maximum deflection in the beam. SOLUTION For the interval 0≤ x≤ L ′′ EIy1 = Px 2 ′ EIy1 = EIy1 = Px 2 + C1 4 At Px 3 + C1 x + C2 12 For the interval L ≤ x ≤ 3L 2 ′′ 2 EIy2 = Px 2 ′ 2 EIy2 = 2 EIy2 = Px 2 + C3 4 At x = 0, y1 = 0 : C2 = 0 x = 3L 2, ′ y2 = 0 : C3 = −9 PL2 16 Px3 + C3 x + C4 12 The two solutions must match at x = L, ′ ′ EIy1 = EIy2 : EIy1 = EIy2 : y1 = (a) PL2 PL2 1 −9 PL2 + C1 = + 4 8 2 16 PL3 PL3 1 −9 PL2 1 + C1 L = + L + C4 12 24 2 16 2 y2 = P ( 4 x3 − 27 L2 x − 8L3 ) 96 EI C1 = C4 = −13PL2 32 − PL3 6 P (8 x3 − 39L2 x ) 96 EI δ B = ( y1 ) x = L = ( y2 ) x = L = δ max = ( y2 ) x =3 L 2 = −31PL3 31PL3 = ↓ .................................................................. Ans. 96 EI 96 EI (b) 9-20 −35PL3 35PL3 = ↓ ............................................................................ Ans. 96 EI 96 EI The cantilever beam ABC shown in Fig. P9-20 has a second moment of area IAB = 4I in the interval AB and a second moment of area IBC = I in the interval BC. Determine (a) The deflection at section B. (b) The deflection at section C. SOLUTION 380 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ↑ ΣFy = 0 : ΣM A = 0 : For the interval VA − 2wL − wL = 0 − M A − 2wL ( L 2 ) − wL ( 2 L ) = 0 0≤ x≤ L VA = 3wL M A = −3wL2 ′′ 4 EIy1 = 3wLx − 3wL2 − wx 2 ′ 4 EIy1 = 4 EIy1 = ′ y1 = 3wLx 2 wx 3 − 3wL2 x − + C1 2 3 wLx3 3wL2 x 2 wx 4 − − + C1 x + C2 2 2 12 y1 = 2 At x = 0, x = 0, ′ y1 = 0 : C1 = 0 y1 = 0 : C2 = 0 At w ( 9 Lx2 − 18L2 x − 2 x3 ) 24 EI For the interval L ≤ x ≤ 2 L ′′ EIy2 = 3wLx − 3wL2 − wx 2 + w ( x − L ) w ( 6 Lx3 − 18L2 x2 − x4 ) 48 EI 3wLx 2 wx3 w ( x − L ) ′ − 3wL2 x − + + C3 EIy2 = 2 3 3 3 wLx3 3wL2 x 2 wx 4 w ( x − L ) EIy2 = − − + + C3 x + C4 2 2 12 12 The two solutions must match at x = L, 4 ′ ′ EIy1 = EIy2 : EIy1 = EIy2 : y2 = (a) 9wL3 2wL3 wL3 3wL3 wL3 − − = − 3wL3 − + 0 + C3 24 3 12 2 3 wL4 3wL4 wL4 wL4 3wL4 wL4 − − = − − + 0 + C3 L + C4 8 8 48 2 2 12 C3 = C4 = 33wL3 24 −27 wL4 48 w 4 −4 x 4 + 24 Lx3 − 72 L2 x 2 + 4 ( x − L ) + 66 L3 x − 27 L4 48 EI δ B = ( y1 ) x = L = ( y2 ) x = L = δ C = ( y2 ) x = 2 L = −13wL4 13wL4 = ↓ ................................................................. Ans. 48EI 48 EI (b) 9-21 −51wL4 51wL4 = ↓ .................................................................................. Ans. 48EI 48EI A beam is loaded and supported as shown in Fig. P9-21. Determine the deflection midway between the supports. SOLUTION ↑ ΣFy = 0 : ΣM B = 0 : For the interval RA + RB − wa = 0 wa ( 3a 2 ) − RA ( 3a ) = 0 RA = RB = wa 2 0≤ x≤a ′′ EIy1 = wax 2 381 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ′ EIy1 = EIy1 = wax 2 + C1 4 At wax3 + C1 x + C2 12 For the interval a ≤ x ≤ 2a wax w ( x − a ) ′′ − EIy2 = 2 2 2 x = 0, y1 = 0 : C2 = 0 wax 2 w ( x − a ) ′ − + C3 EIy2 = 4 6 3 At x = 3a 2, ′ y2 = 0 : C3 = −13wa 3 24 wax3 w ( x − a ) EIy2 = − + C3 x + C4 12 24 The two solutions must match at x = a, 4 ′ ′ EIy1 = EIy2 : EIy1 = EIy2 : y2 = wa3 wa3 13wa 3 − 0 + C1 = −0− 4 4 24 −13wa3 wa 4 wa 4 − 0 + C1a = −0+ a + C4 12 12 24 C1 = −13wa 3 24 C4 = 0 w 4 2ax3 − ( x − a ) − 13a 3 x 24 EI δ M = ( y2 ) x = 3 a 2 = 9-22 w 27a 4 a 4 39a 4 −205wa 4 205wa 4 −− = = ↓ .................. Ans. 24 EI 4 16 2 384 EI 384 EI A beam is loaded and supported as shown in Fig. P9-22. Determine the deflection at the left end of the distributed load. SOLUTION ΣM B = 0 : ↑ ΣFy = 0 : wL ( L 2 ) − RA ( 2 L ) = 0 RA + RB − wL = 0 RA = wL 4 RB = 3wL 4 For the interval 0≤ x≤ L L ≤ x ≤ 2L ′′ EIy1 = wLx 4 wLx 2 ′ EIy1 = + C1 8 wLx 3 EIy1 = + C1 x + C2 24 The boundary conditions give: wLx w ( x − L ) ′′ − EIy2 = 4 2 ′ EIy2 = 2 wLx 2 w ( x − L ) − + C3 8 6 3 4 wLx3 w ( x − L ) EIy2 = − + C3 x + C4 24 24 382 STATICS AND MECHANICS OF MATERIALS, 2nd Edition At At RILEY, STURGES AND MORRIS x = 0, y1 = 0 : y2 = 0 : x = L, C2 = 0 2 LC3 + C4 = −7 wL4 24 x = 2 L, The two solutions must match at ′ ′ EIy1 = EIy2 : EIy1 = EIy2 : Therefore, wL3 wL3 + C1 = − 0 + C3 8 8 wL4 wLx 4 + C1 L = − 0 + C3 L + C4 24 24 C2 = C 4 = 0 y2 = w 4 2 Lx3 − ( x − a ) − 7 L3 x 48 EI C1 = C3 = −7 wL3 48 w 2 Lx3 − 7 L3 x 48 EI y1 = δ M = ( y1 ) x = L = ( y2 ) x = L = −5wL4 5wL4 = ↓ .................................................................... Ans. 48EI 48EI 9-23 A beam is loaded and supported as shown in Fig. P9-23. Determine (a) The slope at the left end of the beam. (b) The maximum deflection between the supports. SOLUTION ΣM B = 0 : P ( L ) + P ( 2 L ) − RA ( 3L ) = 0 RA = P ↑ ΣFy = 0 : RB = P For the interval RA + RB − P − P = 0 0≤ x≤ L L ≤ x ≤ 2L ′′ EIy1 = Px Px 2 ′ EIy1 = + C1 2 Px 3 EIy1 = + C1 x + C2 6 The boundary conditions give: At At ′′ EIy2 = Px − P ( x − L ) Px 2 P ( x − L ) ′ − + C3 EIy2 = 2 2 2 Px 3 P ( x − L ) EIy2 = − + C3 x + C4 6 6 3 x = 0, y1 = 0 : ′ y2 = 0 : x = L, C2 = 0 C3 = − PL2 x = 3L 2, The two solutions must match at ′ ′ EIy1 = EIy2 : EIy1 = EIy2 : PL2 PL2 + C1 = − 0 + C3 2 2 PL3 PL3 + C1 L = − 0 + C3 L + C4 6 6 383 C1 = C3 = − PL2 C4 = C 2 = 0 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (a) RILEY, STURGES AND MORRIS y1 = P ( x3 − 6 L2 x ) 6 EI ′ y1 = P ( x2 − 2L2 ) 2 EI θ A = ( y1′ ) x =0 = (b) − PL2 PL2 = EI EI ............................................................................................ Ans. y2 = P3 3 x − ( x − L ) − 6 L2 x 6 EI δ M = ( y2 ) x = 3 L 2 = 9-24 −23PL3 23PL3 = ↓ .............................................................................. Ans. 24 EI 24 EI The cantilever beam ABC shown in Fig. P9-24 has a second moment of area IAB = 4I in the interval AB and a second moment of area IBC = I in the interval BC. Determine (a) The deflection at section B. (b) The deflection at section C. SOLUTION ↑ ΣFy = 0 : 0≤ x≤ L VA − 2wL − wL = 0 VA = 3wL M A = −5wL2 2 ΣM A = 0 : − M A − 2 wL ( L 2 ) − wL ( 3L 2 ) = 0 For the interval 4 EIy1′′ = 3wLx − ′ 4 EIy1 = 5wL2 − wx 2 2 At 3wLx 2 5wL2 x wx3 − − + C1 2 2 3 x = 0, ′ y1 = 0 : C1 = 0 wLx3 5wL2 x 2 wx 4 − − + C1 x + C2 At x = 0, y1 = 0 : C2 = 0 4 EIy1 = 2 4 12 w w ′ y1 = y1 = ( 9 Lx 2 − 15L2 x − 2 x3 ) ( 6 Lx3 − 15L2 x 2 − x4 ) 24 EI 48 EI For the interval L ≤ x ≤ 2 L w( x − L) 5wL2 ′′ − wx 2 + EIy2 = 3wLx − 2 2 2 3wLx 2 5wL2 x wx 3 w ( x − L ) ′ − − + + C3 EIy2 = 2 2 3 6 3 wLx3 5wL2 x 2 wx 4 w ( x − L ) EIy2 = − − + + C3 x + C4 2 4 12 24 The two solutions must match at x = L, 4 ′ ′ EIy1 = EIy2 : EIy1 = EIy2 : 3wL3 5wL3 wL3 3wL3 5wL3 wL3 − − = − − + 0 + C3 8 8 12 2 2 3 wL4 5wL4 wL4 wL4 5wL4 wL4 − − = − − + 0 + C3 L + C4 8 16 48 2 4 12 C3 = wL3 C4 = −3wL4 8 384 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS y2 = (a) w 4 −2 x 4 + 12 Lx 3 − 30 L2 x 2 + ( x − L ) + 24 L3 x − 9 L4 24 EI δ B = ( y1 ) x = L = ( y2 ) x = L = δ C = ( y2 ) x = 2 L = −5wL4 5wL4 = ↓ ..................................................................... Ans. 24 EI 24 EI (b) 9-25 −3wL4 3wL4 = ↓ ...................................................................................... Ans. 2 EI 2 EI A 160-lb diver walks slowly onto a diving board. The diving board is a wood (E = 1800 ksi) plank 10 ft long, 18 in. wide, and 2 in. thick, and is modeled as the cantilever beam shown in Fig. P9-25. For the diver at positions, a = nL 5 ( n = 1, 2, ,5) , compute and plot the deflection curve for the diving board (plot y as a function of x for 0 ≤ x ≤ 10 ft). SOLUTION ↑ ΣFy = 0 : ΣM A = 0 : For the interval VA − 160 = 0 VA = 160 lb ↑ − M A − 160a = 0 0≤ x≤a x = 0, x = 0, M A = −160a = 160a lb ⋅ ft ′′ EIy1 = (160 x − 160a ) lb ⋅ ft ′ EIy1 = ( 80 x 2 − 160ax + C1 ) lb ⋅ ft 2 At At ′ y1 = 0 : y1 = 0 : C1 = 0 C2 = 0 80 x3 EIy1 = − 80ax 2 + C1 x + C2 lb ⋅ ft 3 3 For the interval a ≤ x ≤ L ′′ EIy2 = 160 x − 160a − 160 ( x − a ) lb ⋅ ft 2 ′ EIy2 = 80 x 2 − 160ax − 80 ( x − a ) + C3 lb ⋅ ft 2 3 80 x 3 80 ( x − a ) 2 EIy2 = − − 80ax − + C3 x + C4 lb ⋅ ft 3 3 3 The two solutions must match at x = a, ′ ′ EIy1 = EIy2 : EIy1 = EIy2 : 80a 2 − 160a 2 = 80a 2 − 160a 2 − 0 + C3 C3 = 0 C4 = 0 80a3 80a3 − 80a 3 = − 80a3 − 0 + C3 a + C4 3 3 3 3 18 ( 2 ) EI = (1800 × 10 ) = 21.6 × 106 lb ⋅ in 2 12 Therefore: 80 x 3 y1 = − 80ax 2 in. ........................................ 0 ≤ x ≤ a ........................ Ans. 6 21.6 × 10 3 (12 ) 3 385 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 3 80 x 3 80 ( x − a ) 2 y2 = − 80ax − in. ............... a ≤ x ≤ 10 ft ................. Ans. 21.6 ×106 3 3 3 (12 ) 9-26 A diving board consists of a wood (E = 12 GPa) plank which is pinned at the left end and rests on a movable support, as shown in Fig. P9-26. The board is 3 m long, 500 mm wide, and 80 mm thick. If a 70kg diver stands at the end of the board, (a) Compute and plot the deflection curve for the beam (plot y as a function of x for 0 ≤ x ≤ 3 m) for the right support at positions b = 0.5 m, 1.0 m, and 1.5 m. (b) If the stiffness of the board is defined as the ratio of the diver’s weight to the deflection at the end of the board SOLUTION k = (W y ) , compute and plot the stiffness of the board as a function of b for 0 ≤ b ≤ 1.5 m. Does the stiffness depend on the weight of the diver? W = 70 ( 9.81) = 686.7 N ΣM A = 0 : ΣM B = 0 : Bb − WL = 0 B = WL b ↑ Ab − W ( L − b ) = 0 0≤ x≤b A = −W ( L − b ) b = W ( L − b ) b ↓ (a) For the interval ′′ EIy1 = − W ( L − b) x b At W ( L − b) x2 ′ EIy1 = − + C1 2b W ( L − b ) x3 + C1 x + C2 6b For the interval b ≤ x ≤ L EIy1 = − x = 0, x = b, y1 = 0 : y1 = 0 : C2 = 0 C1 = W ( L − b) b 6 At ′′ EIy2 = − W ( L − b ) x WL ( x − b ) + b b 2 W ( L − b ) x 2 WL ( x − b ) ′ EIy2 = − + + C3 2b 2b 386 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS EIy′′′ = − wx + C1 EIy′′ = − EIy′ = − EIy = − At At (a) wx 2 + C1 x + C2 2 wx3 C1 x 2 + + C2 x + C3 6 2 wx 4 C1 x3 C2 x 2 + + + C3 x + C4 24 6 2 C1 = 0 C2 = 0 At At Boundary conditions: x = 0, x = 0, y= V = EIy′′′ = 0 : M = EIy′′ = 0 : x = L, x = L, y′ = 0 : y = 0: C3 = wL3 6 C4 = − wL4 8 w ( − x4 + 4 L3 x − 3L4 ) .............................................................................................. Ans. 24 EI (b) δ A = yx=0 = − wL4 wL4 = ↓ .................................................................................................. Ans. 8 EI 8EI 9-28 For the beam and loading shown in Fig. P9-28, determine (a) The equation of the elastic curve. (b) The deflection midway between the supports. SOLUTION EIy′′′′ = − wx L EIy′′′ = − EIy′′ = − EIy′ = − EIy = − At At (a) wx 2 + C1 2L wx3 + C1 x + C2 6L wx 4 C1 x 2 + + C2 x + C3 24 L 2 wx5 C1 x3 C2 x 2 + + + C3 x + C4 120 L 6 2 C2 = 0 C4 = 0 At At Boundary conditions: x = 0, x = 0, y= M = EIy′′ = 0 : y = 0: x = L, x = L, M = EIy′′ = 0 : y = 0: C1 = wL 6 C3 = −7 wL3 360 w ( −3x5 + 10 L2 x3 − 7 L4 x ) .................................................................................. Ans. 360 EIL (b) δ M = yx= L 2 = w 3L5 10 L5 7 L5 −5wL4 5wL4 + − = ↓ .......................... Ans. − = 360 EIL 32 8 2 768 EI 768 EI 9-29 A beam is loaded and supported as shown in Fig. P9-29. Determine (a) The equation of the elastic curve. (b) The deflection at the left end of the beam. (c) The support reactions VB and MB. 388 STATICS AND MECHANICS OF MATERIALS, 2nd Edition SOLUTION RILEY, STURGES AND MORRIS EIy′′′′ = − wx3 L3 EIy′′′ = − EIy′′ = − EIy′ = − EIy = − x = 0, x = 0, y= wx 4 + C1 4 L3 wx5 + C1 x + C2 20 L3 wx 6 C1 x 2 + + C2 x + C3 120 L3 2 wx 7 C x3 C x 2 + 1 + 2 + C3 x + C4 840 L3 6 2 C1 = 0 C2 = 0 At At Boundary conditions: At At (a) V = EIy′′′ = 0 : M = EIy′′ = 0 : x = L, x = L, y′ = 0 : y = 0: C3 = wL3 120 C4 = − wL4 140 w − x 7 + 7 L6 x − 6 L7 ) ........................................................................................ Ans. 3( 840 EIL (b) (c) (d) δ A = yx =0 = − wL4 wL4 = ↓ ............................................................................................ Ans. 140 EI 140 EI .............................................................................. Ans. VB = EIy′′′= L = − wL 4 = wL 4 ↑ .......................................................................................... Ans. x M B = EIy′′= L = − wL2 20 = wL2 20 x 9-30 A beam is loaded and supported as shown in Fig. P9-30. Determine (a) The equation of the elastic curve. (b) The deflection midway between the supports. (c) The maximum deflection of the beam. (d) The support reactions RA and RB. SOLUTION EIy′′′′ = − wx3 L3 EIy′′′ = − EIy′′ = − EIy′ = − wx 4 + C1 4 L3 wx5 + C1 x + C2 20 L3 wx 6 C1 x 2 + + C2 x + C3 120 L3 2 wx 7 C1 x3 C2 x 2 EIy = − + + + C3 x + C4 840 L3 6 2 Boundary conditions: At At x = 0, x = L, M = EIy′′ = 0 : M = EIy′′ = 0 : C2 = 0 C1 = wL 20 At At x = 0, x = L, y = 0: y = 0: C4 = 0 C3 = − wL3 140 389 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (a) RILEY, STURGES AND MORRIS y= w − x 7 + 7 L4 x3 − 6 L6 x ) .................................................................................... Ans. 3( 840 EIL (b) δ M = yx= L 2 = −13wL4 13wL4 w L7 7 L7 − + − 3L7 = = ↓ ....................... Ans. 840 EIL3 128 8 5120 EI 5120 EI x = 0.5424 L (c) The maximum deflection occurs where y′ = w ( −7 x6 + 21L4 x2 − 6 L6 ) = 0 840 EIL3 δ max = yx =0.5424 L = (d) −0.00256wL4 0.00256 wL4 = ↓= 1.008δ M ..................................... Ans. EI EI RA = EIy′′′=0 = + wL 20 = wL 20 ↑ .................................................................................... Ans. x wL wL wL RB = − EIy′′′= L = − − + ↑ .......................................................................... Ans. x = 20 5 4 9-31 A beam is loaded and supported as shown in Fig. P9-31. Determine (a) The equation of the elastic curve. (b) The deflection at the left end of the beam. (c) The support reactions VB and MB. SOLUTION πx 2L 2 wL πx sin EIy′′′ = − + C1 2L π EIy′′′′ = − w cos EIy′′ = EIy′ = 4wL2 πx + C1 x + C2 cos 2 π 2L 8wL3 π x C1 x 2 + + C2 x + C3 sin π3 2L 2 16wL4 π x C1 x3 C2 x 2 cos + + + C3 x + C4 π4 2L 6 2 C1 = 0 C2 = −4wL2 π 2 C3 = C4 = 4wL3 (π − 2 ) π3 2wL4 (4 −π ) π3 EIy = − At At At Boundary conditions: x = 0, x = 0, x = L, x = L, y= V = EIy′′′ = 0 : M = EIy′′ = 0 : y′ = 0 : y = 0: w 2π 4 EI At (a) πx 4 222 3 4 −32 L cos 2 L − 4π L x + 8 (π − 2 ) π L x + 4π ( 4 − π ) L .................. Ans. w −0.1089wL4 0.1089 wL4 −32 L4 + 4π ( 4 − π ) L4 = = ↓ ......... Ans. EI EI 2π 4 EI 390 (b) δ A = yx=0 = STATICS AND MECHANICS OF MATERIALS, 2nd Edition (c) (d) RILEY, STURGES AND MORRIS VB = EIy′′′= L = −2 wL π = 2 wL π ↑ .................................................................................... Ans. x M B = EIy′′= L = −4 wL2 π 2 = 4 wL2 π 2 x ........................................................................ Ans. 9-32 A beam is loaded and supported as shown in Fig. P9-32. Determine (a) The equation of the elastic curve. (b) The deflection midway between the supports. (c) The maximum deflection of the beam. (d) The slope at the left end of the beam. (e) The support reactions RA and RB. SOLUTION πx 2L 2 wL πx cos EIy′′′ = + C1 2L π EIy′′′′ = − w sin EIy′′ = 4wL2 πx + C1 x + C2 sin 2 π 2L 8wL3 π x C1 x 2 + + C2 x + C3 cos π3 2L 2 EIy′ = − EIy = − At At At At (a) 16wL4 π x C1 x3 C2 x 2 sin + + + C3 x + C4 π4 2L 6 2 C2 = 0 C1 = −4wL π2 C4 = 0 Boundary conditions: x = 0, x = L, x = 0, x = L, y= M = EIy′′ = 0 : M = EIy′′ = 0 : y = 0: y = 0: 2w 3π 4 EI C3 = 2wL3 ( 24 + π 2 ) 3π 4 πx 4 2 3 2 3 −24 L sin 2 L − π Lx + ( 24 + π ) L x ..................................................... Ans. 2w π π 2 L4 24 + π 2 4 4 + −24 L sin − L 3π 4 EI 4 8 2 (b) δ M = yx= L 2 = = −0.00869wL4 0.00869 wL4 = ↓ .............................................................................. Ans. EI EI (c) The maximum deflection occurs where y′ = 2w 3π 4 EI πx −12π L3 cos − 3π 2 Lx 2 + ( 24 + π 2 ) L3 = 0 2L x = 0.5154 L δ max = yx =0.5154 L = −0.00870wL4 = 1.001δ M .................................................................... Ans. EI 391 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (d) RILEY, STURGES AND MORRIS θ A = y′ =0 = x (e) 2w −0.0262wL3 0.0262wL3 −12π L3 + ( 24 + π 2 ) L3 = = ....... Ans. EI EI 3π 4 EI 2wL 4wL 2wL (π − 2 ) RA = EIy′′′=0 = − 2= ↑ ................................................................... Ans. x π π π2 −4 wL 4 wL RB = − EIy′′′= L = − x = 2 ↑ ................................................................................ Ans. 2 π π 9-33 A cantilever beam is loaded and supported as shown in Fig. P9-33. Use singularity functions to determine the deflection (a) At a distance x = L from the support. (b) At the right end of the beam. SOLUTION ↑ ΣFy = 0 : VA = 0 ΣM A = 0 : VA + P − P = 0 PL − P ( 2 L ) − M A = 0 M A = − PL = PL EIy′′ = − PL + P x − L 1 P x−L EIy′ = − PLx + 2 2 + C1 3 At x = 0, x = 0, y′ = 0 : y = 0: C1 = 0 C2 = 0 PLx 2 P x − L EIy = − + + C1 x + C2 2 6 P 3 y= −3Lx 2 + x − L 6 EI (a) (b) 9-34 At δ B = yx = L = − PL3 2 EI = PL3 2 EI ↓ ................................................................................. Ans. δ C = yx=2 L = P −11PL3 11PL3 2 3 −3 L ( 2 L ) + ( L ) = = ↓ ...................................... Ans. 6 EI 6 EI 6 EI A cantilever beam is loaded and supported as shown in Fig. P9-34. Use singularity functions to determine the deflection (a) At a distance x = L from the support. (b) At the right end of the beam. SOLUTION ↑ ΣFy = 0 : VA = 2 P ΣM A = 0 : VA − P − P = 0 PL − P ( L ) − P ( 2 L ) − M A = 0 M A = −2 PL = 2 PL 392 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS EIy′′ = 2 Px − 2 PL − PL x − L − P x − L 0 1 P x−L EIy′ = Px − 2 PLx − PL x − L − 2 2 1 2 + C1 3 PL x − L Px 3 EIy = − PLx 2 − 3 2 Boundary conditions: At At 2 P x−L − 6 + C1 x + C2 x = 0, x = 0, y′ = 0 : y = 0: C1 = 0 C2 = 0 y= (a) P 3 2 3 2 x − 6 Lx 2 − 3L x − L − x − L 6 EI δ B = yx = L = P −2 PL3 2 PL3 2 L3 − 6 L3 = = ↓ .......................................................... Ans. 6 EI 3EI 3EI P −2 PL3 2 PL3 3 2 2 3 2 ( 2 L ) − 6 L ( 2 L ) − 3L ( L ) − ( L ) = = ↓ ........ Ans. EI EI 6 EI (b) 9-35 δ C = yx=2 L = A beam is loaded and supported as shown in Fig. P9-35. Use singularity functions to determine the deflection (a) At a distance x = L from the left support. (b) At the middle of the span. SOLUTION ↑ ΣFy = 0 : ΣM D = 0 : RA = RB = P RA + RD − P − P = 0 P ( L ) + P ( 2 L ) − RA ( 3 L ) = 0 EIy′′ = Px − P x − L − P x − 2 L 1 1 Px 2 P x − L EIy′ = − 2 2 Px 3 P x − L EIy = − 6 6 Boundary conditions: At At 2 P x − 2L − 2 P x − 2L − 6 3 2 + C1 + C1 x + C2 C2 = 0 C1 = − PL2 3 x = 0, y = 0: y = 0: x = 3 L, y= (a) P3 3 3 x − x − L − x − 2 L − 6 L2 x 6 EI δ B = yx= L = P −5PL3 5PL3 L3 − 6 L3 = = ↓ ............................................................. Ans. 6 EI 6 EI 6 EI 393 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (b) 9-36 RILEY, STURGES AND MORRIS δ M = y x =3 L 2 = P 6 EI 27 L3 L3 −23PL3 23PL3 − − 9 L3 = = ↓ ..................................... Ans. 8 8 24 EI 24 EI A beam is loaded and supported as shown in Fig. P9-36. Use singularity functions to determine the deflection (a) At the right end of the beam. (b) At a section midway between the supports. SOLUTION ΣM B = 0 : ↑ ΣFy = 0 : RB = 3P ↑ − RA ( L 2 ) − P ( L ) = 0 RA + RB − P = 0 RA = −2 P = 2 P ↓ EIy′′ = −2 Px + 3P x − EIy′ = − Px 2 + EIy = − y= (a) L 2 1 3P L x− 2 2 2 + C1 3 At x = 0, y = 0: y = 0: C2 = 0 C1 = PL2 12 Px3 3P L + x− 3 6 2 + C1 x + C2 3 At x = L 2, P 12 EI L 3 −4 x + 6 x − 2 + L2 x δ C = y x =3 L 2 = δ M = yx= L 4 = P 12 EI P 12 EI 108L3 3L3 − PL3 PL3 − + 6 L3 + = = ↓ ..................................... Ans. 8 2 2 EI 2 EI L3 L3 + PL3 PL3 − +0+ = = ↑ ................................................ Ans. 16 4 64 EI 64 EI (b) 9-37 A cantilever beam is loaded and supported as shown in Fig. P9-37. Use singularity functions to determine the deflection (a) At the point of application of the concentrated load 3P. (b) At the right end of the beam. SOLUTION ↑ ΣFy = 0 : VA = 3 P ΣM A = 0 : VA − 3P = 0 2 PL − 3P ( L ) − M A = 0 M A = − PL = PL EIy′′ = 3Px − PL − 3P x − L 1 394 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 3P x − L 3Px 2 EIy′ = − PLx − 2 2 Px 3 PLx 2 P x − L EIy = − − 2 2 2 Boundary conditions: At At 3 2 + C1 + C1 x + C2 C1 = 0 C2 = 0 x = 0, x = 0, y′ = 0 : y = 0: (a) P3 3 x − Lx 2 − x − L 2 EI P L3 − L3 = 0 ............................................................................................ Ans. δ B = yx= L = 2 EI y= (b) 9-38 δ C = y x =3 L 2 P 27 L3 9 L3 L3 + PL3 PL3 = − − = = ↑ ............................................. Ans. 2 EI 8 4 8 2 EI 2 EI A beam is loaded and supported as shown in Fig. P9-38. Use singularity functions to determine the deflection (a) At the left end of the beam. (b) At a point midway between the supports. SOLUTION ΣM B = 0 : 2 PL 4L =0 P − RA ( L ) − 3 3 2P L x− 3 3 1 RA = 2 P 3 ↑ EIy′′ = − Px + EIy′ = − EIy = − + 2 2 PL 5L x− 3 6 2 PL 5L x− 3 6 PL 5L x− 3 6 2 0 Px 2 P L + x− 2 3 3 Px3 P L + x− 6 9 3 y = 0: y = 0: 1 + 3 + C1 + C1 x + C2 + Boundary conditions: At At x = L 3, C1 = 7 PL2 36 C2 = −19 PL3 324 3 x = 4 L 3, P 324 EI y= (a) L 3 −54 x + 36 x − 3 + 108 L x − 5L 6 2 + 63L2 x − 19 L3 δ L = yx =0 −19 PL3 19 PL3 = = ↓ ......................................................................................... Ans. 324 EI 324 EI + PL3 PL3 = = ↑ .......................................................................................... Ans. 48EI 48 EI (b) δ M = y x =5 L 6 395 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 9-39 RILEY, STURGES AND MORRIS Use singularity functions to determine the deflection at the left end of the cantilever beam shown in Fig. P9-39. SOLUTION 4 wLx w 3L EIy′′ = − − x− 25 2 2 2 wLx 2 w 3L EIy′ = − − x− 25 6 2 2 wLx3 w 3L EIy = − − x− 75 24 2 Boundary conditions: At At 2 3 + C1 4 + C1 x + C2 x = 5L 2, x = 5L 2, w 600 EI y′ = 0 : y = 0: 4 C1 = 2 wL3 3 C2 = −29wL4 24 + 400 L3 x − 725 L4 y= 3L 3 −16 Lx − 25 x − 2 δ A = yx=0 = 9-40 −725wL4 29wL4 = ↓ ...................................................................................... Ans. 600 EI 24 EI A beam is loaded and supported as shown in Fig. P9-40. Use singularity functions to determine the deflection (a) At the left end of the distributed load. (b) At a section midway between the supports. SOLUTION ΣM B = 0 : w ( 2 L ) ( L ) − RA ( 3L ) = 0 RA = 2wL 3 ↑ 2wLx w x − L EIy′′ = − 3 2 wLx 2 w x − L EIy′ = − 3 6 2 3 + C1 4 At x = 0, y = 0: y = 0: C2 = 0 −7 wL3 C1 = 9 wLx 3 w x − L EIy = − + C1 x + C2 9 24 w 4 8 Lx 3 − 3 x − L − 56 L3 x y= 72 EI (a) At x = 3 L, δ L = yx= L = w −48wL4 2 wL4 8L4 − 56 L4 = = ↓ ................................................... Ans. 72 EI 72 EI 3EI w 72 EI (b) δ M = y x =3 L 2 = −305wL4 305wL4 3L4 27 L4 − − 84 L4 = = ↓ ........................ Ans. 16 384 EI 384 EI 396 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 9-41 A beam is loaded and supported as shown in Fig. P9-41. Use singularity functions to determine (a) The deflection midway between the supports (b) The maximum deflection of the beam. SOLUTION ΣM B = 0 : w ( L ) ( L 2 ) + ( 2 wL )( 3L 2 ) − RA ( 5 L 2 ) = 0 2 RA = 7 wL 5 ↑ 7 wLx 3L w 1 EIy′′ = − 2 wL x − L − x− 5 2 2 7 wLx 2 w 3L 2 EIy′ = − wL x − L − x− 10 6 2 7 wLx3 wL w 3L 3 EIy = − x−L − x− 30 3 24 2 Boundary conditions: At At 3 + C1 4 + C1 x + C2 x = 0, y = 0: y = 0: C2 = 0 C1 = −119wL3 120 4 x = 5L 2, w 3L 3 3 y= 28 Lx − 40 L x − L − 5 x − 120 EI 2 (a) − 119 L3 x δ M = y x =5 L 4 = w 875L4 10 L4 595L4 −101wL4 101wL4 − − = = ↓ ................ Ans. 120 EI 16 16 4 128EI 128EI 3 (b) The maximum deflection occurs when y′ = 3L w 2 2 84 Lx − 120 L x − L − 20 x − 120 EI 2 − 119 L3 = 0 x = 1.2186 L δ max = yx =1.2186 L = = w 50.669 L4 − 0.418 L4 − 145.013L4 120 EI −0.7897 wL4 0.7897 wL4 = ↓ = 1.001δ M ............................................................. Ans. EI EI 9-42 A beam is loaded and supported as shown in Fig. P9-42. Use singularity functions to determine (a) The deflection at the right end of the beam. (b) The deflection midway between the supports. SOLUTION ΣM B = 0 : ↑ ΣFy = 0 : ( wL )( 3L 2 ) − w ( L 2 ) ( L 4 ) − RA L = 0 RA + RB − wL − w ( L 2 ) = 0 RA = 11wL 8 ↑ RB = wL 8 ↑ 397 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 2 L EIy′′ = − wL x + 2 wL L EIy′ = − x+ 2 2 wL L EIy = − x+ 6 2 Boundary conditions: At At 1 + 2 11wL wL w 1 1 x−0 + x−L − x−L 8 8 2 + 3 11wL wL w 2 2 3 x−0 + x−L − x − L + C1 16 16 6 + 11wL wL w 3 3 4 x−0 + x−L − x − L + C1 x + C2 48 48 24 x = 0, x = L, y = 0: y = 0: 3 C2 = wL4 48 C1 = 5wL3 16 3 3 4 + 11L x − 0 + L x − L − 2 x − L + 15 L3 x + L4 w L y= −8 L x + 48 EI 2 (a) δ C = y x =3 L 2 = δ M = yx= L 2 = −9wL4 9 wL4 = ↓ ....................................................................................... Ans. 128EI 128EI w 48EI (b) 9-43 +5wL4 5wL4 11L4 15L4 −8 L4 + + + L4 = = ↑ ...................... Ans. 8 2 128EI 128EI Use singularity functions to determine the deflection at the right end of the cantilever beam shown in Fig. P9-43. SOLUTION ΣM A = 0 : ↑ ΣFy = 0 : ( wL 8)( 2 L ) − [ wL ]( L 2 ) − M A = 0 VA − wL + ( wL 8 ) = 0 M A = − wL2 4 = wL2 4 VA = 7 wL 8 ↑ EIy′′ = 7 wLx wL2 wx 2 w x − L − − + 8 4 2 2 2 7 wLx 2 wL2 x wx3 w x − L EIy′ = − − + 16 4 6 6 7 wLx3 wL2 x 2 wx 4 w x − L EIy = − − + 48 8 24 24 Boundary conditions: At At 3 + C1 4 + C1 x + C2 x = 0, x = 0, y′ = 0 : y = 0: C1 = 0 C2 = 0 y= w 4 7 Lx3 − 2 x 4 − 6 L2 x 2 + 2 x − L 48 EI δ C = yx=2 L = w + wL4 wL4 56 L4 − 32 L4 − 24 L4 + 2 L4 = = 24 EI 24 EI ↑ .......................... Ans. 48EI 398 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 9-44 RILEY, STURGES AND MORRIS A beam is loaded and supported as shown in Fig. P9-44. Use singularity functions to determine the deflection (a) At the left end of the distributed load. (b) At a section midway between the supports. SOLUTION ↑ ΣFy = 0 : ΣM B = 0 : RA + RB − wa = 0 ( wa )( 3a 2 ) − RA ( 3a ) = 0 2 RA = RB = wa 2 ↑ wax w x − a EIy′′ = − 2 2 wax 2 w x − a EIy′ = − 4 6 w x − 2a − 2 w x − 2a − 6 2 3 3 + C1 4 At x = 0, y = 0: y = 0: C2 = 0 C1 = −13wa 3 24 w x − 2a wax3 w x − a EIy = − − + C1 x + C2 12 24 24 w 4 4 y= 2ax3 − x − a + x − 2a − 13a3 x 24 EI (a) 4 At x = 3a, δ L = yx =a = δ M = y x =3a 2 w −11wa 4 11wa 4 2a 4 − 0 + 0 − 13a 4 = = ↓ .................................... Ans. 24 EI 24 EI 24 EI w 54a 4 a 4 39a 4 −205wa 4 205wa 4 = − +0− = = ↓ ................. Ans. 24 EI 8 16 2 384 EI 384 EI (b) 9-45 A beam is loaded and supported as shown in Fig. P9-45. Use singularity functions to determine (a) The deflection midway between the supports. (b) The maximum deflection of the beam. SOLUTION ΣM C = 0 : ( wL 2 )( L 3) + ( wL )( L ) − RA ( 2 L ) = 0 RA = 7 wL 12 ↑ w x−L 7 wLx 1 − wL x − L − EIy′′ = 12 6L EIy′ = 7 wLx 2 wL x − L − 24 2 2 3 − 3 w x−L 24 L 5 4 + C1 + C1 x + C2 C2 = 0 C1 = −217 wL3 720 7 wLx3 wL x − L − EIy = 72 6 Boundary conditions: At At w x−L − 120 L x = 0, y = 0: y = 0: x = 2 L, 399 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS y= (a) w 3 5 70 L2 x 3 − 120 L2 x − L − 6 x − L − 217 L4 x 720 EIL δ M = yx= L = y′ = w −147 wL4 147 wL4 70 L5 − 0 − 0 − 217 L5 = = ↓ .................... Ans. 720 EIL 720 EI 720 EI x = 1.0168 L (b) The maximum deflection occurs when w 2 4 210 L2 x 2 − 360 L2 x − L − 30 x − L − 217 L4 720 EIL δ max = yx =1.0168 L = −0.2043wL4 0.2043wL4 = ↓ = 1.0004δ M ....................................... Ans. EI EI 9-46 A beam is loaded and supported as shown in Fig. P9-46. Use singularity functions to determine (a) The deflection at the middle of the span. (b) The maximum deflection in the beam. SOLUTION ΣM B = 0 : ( wL 2 )( 4 L 3) − RA ( 2 L ) = 0 2 RA = wL 3 ↑ wLx wx 3 w x − L EIy′′ = − + 3 6L 2 wLx 2 wx 4 w x − L − + EIy′ = 6 24 L 6 wLx 3 wx5 w x − L − + EIy = 18 120 L 24 Boundary conditions: At At w x−L + 6L 3 3 w x−L + 24 L w x−L + 120 L 4 + C1 5 4 + C1 x + C2 x = 0, y = 0: y = 0: C2 = 0 C1 = −41wL3 360 x = 2 L, y= (a) w 4 5 20 L2 x 3 − 3 x 5 + 15 L x − L + 3 x − L − 41L4 x 360 EIL δ M = yx= L = y′ = w −24wL4 wL4 20 L5 − 3L5 + 0 + 0 − 41L5 = 360 EI = 15 EI ↓ .................... Ans. 360 EIL x = 0.9352 L (b) The maximum deflection occurs when w 3 4 60 L2 x 2 − 15 x 4 + 60 L x − L + 15 x − L − 41L4 = 0 360 EIL δ max = yx =0.9352 L = 9-47 −0.06703wL4 0.0670 wL4 ≅ ↓ = 1.005δ M ...................................... Ans. EI EI A 160-lb diver walks slowly onto a diving board. The diving board is a wood (E = 1800 ksi) plank 10 ft long, 18 in. wide, and 2 in. thick, and is modeled as the cantilever beam shown in Fig. P9-47. For the diver at positions, a = nL 5 ( n = 1, 2, ,5) , compute and plot the deflection curve for the diving board (plot y as a function of x for 0 ≤ x ≤ 10 ft). SOLUTION 400 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ↑ ΣFy = 0 : ΣM A = 0 : VA − 160 = 0 VA = 160 lb ↑ − M A − 160a = 0 M A = −160a = 160a lb ⋅ ft 1 EIy′′ = 160 x − 160a − 160 x − a lb ⋅ ft 2 EIy′ = 80 x 2 − 160ax − 80 x − a + C1 lb ⋅ ft 2 80 x 3 80 x − a EIy = − 80ax 2 − 3 3 Boundary conditions: At At 3 + C1 x + C2 lb ⋅ ft 3 x = 0, x = 0, y′ = 0 : y = 0: 3 3 C1 = 0 C2 = 0 18 ( 2 ) EI = (1800 × 10 ) 12 6 = 21.6 × 10 lb ⋅ in 2 Therefore: 80 x3 80 x − a y= − 80ax 2 − 6 21.6 × 10 3 3 3 (12 ) 3 in. ............................................................ Ans. 9-48 A diving board consists of a wood (E = 12 GPa) plank which is pinned at the left end and rests on a movable support, as shown in Fig. P9-48. The board is 3 m long, 500 mm wide, and 80 mm thick. If a 70kg diver stands at the end of the board, (a) Compute and plot the deflection curve for the beam (plot y as a function of x for 0 ≤ x ≤ 3 m) for the right support at positions b = 0.5 m, 1.0 m, and 1.5 m. (b) If the stiffness of the board is defined as the ratio of the diver’s weight to the deflection at the end of the board SOLUTION k = (W y ) , compute and plot the stiffness of the board as a function of b for 0 ≤ b ≤ 1.5 m. W = 70 ( 9.81) = 686.7 N Does the stiffness depend on the weight of the diver? ΣM A = 0 : ΣM B = 0 : B = WL b ↑ Bb − WL = 0 − Ab − W ( L − b ) = 0 A = −W ( L − b ) b = W ( L − b ) b ↓ 500 ( 80 ) I= = 21.33 × 106 mm 4 12 3 401 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ΣM A = 0 : D (15 ) − 210a − 210 ( a + 4.5 ) − 40 (15 ) ( 7.5 ) = 0 D = ( 363 + 28a ) lb ↑ ↑ ΣFy = 0 : A + D − 210 − 210 − 40 (15 ) = 0 A = ( 657 − 28a ) lb ↑ 1 1 EIy′′ = ( 657 − 28a ) x − 20 x 2 − 210 x − a − 210 x − a − 4.5 lb ⋅ ft ( 657 − 28a ) x 2 20 x3 2 2 − − 105 x − a − 105 x − a − 4.5 + C1 lb ⋅ ft 2 EIy′ = 2 3 ( 657 − 28a ) x3 5 x 4 3 3 EIy = − − 35 x − a − 35 x − a − 4.5 + C1 x + C2 lb ⋅ ft 3 6 3 Boundary conditions: At At x = 0 ft, x = 15 ft, y = 0: y = 0: C2 = 0 7 15 − a C1 = 1050a − 19, 012.5 + 3 3 3 7 15 − a − 4.5 + 3 3 ( 657 − 28a ) x3 5 x 4 3 EIy = − − 35 x − a − 35 x − a − 4.5 6 3 3 3 7 15 − a x 7 15 − a − 4.5 x +1050ax − 19, 012.5 x + + lb ⋅ ft 3 3 3 (a) The deflection is maximum at the middle of the span when the cart is at the middle of the bridge. Therefore, x = 7.5 ft a = 5.25 ft y = −2 in. E = 1800 ksi I = 25.109 in 4 = 4h3 12 h = 4.22 in. .............................................................................................................................. Ans. (b) When the depth of the beam is h = 5 in. 4 ( 5) I= = 41.667 in 4 12 3 EI = (1800 × 103 ) ( 41.667 ) = 75.00 × 106 lb ⋅ in 2 403 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 9-50 RILEY, STURGES AND MORRIS The gangplank between a fishing boat and a dock consists of a wood (E = 12 GPa) plank 3 m long, 300 mm wide, and 50 mm thick. If the plank is modeled as the simply supported beam shown in Fig. P9-50, compute and plot the deflection curve for the gangplank (plot y as a function of x for 0 ≤ x ≤ 3 m) as a 75kg man walks across the plank. Plot curves for the man at positions a = nL 7 ( n = 1, 2, ,5) . SOLUTION ΣM A = 0 : ΣM C = 0 : CL − Wa = 0 C = WL a ↑ W ( L − a ) − AL = 0 A = W ( L − a) L ↑ W = 85 ( 9.81) = 833.85 N 300 ( 50 ) I= = 3.125 × 106 mm 4 12 3 EI = (12 × 109 )( 3.125 × 10 −6 ) = 37.50 × 103 N ⋅ m 2 EIy′′ = EIy′ = W ( L − a) x −W x − a L 1 W ( L − a ) x2 W x − a − 2L 2 2 + C1 3 W ( L − a ) x3 W x − a EIy = − 6L 6 Boundary conditions: At At Therefore: + C1 x + C2 C2 = 0 W ( L − a) L W ( L − a) + C1 = − 6 6L 3 x = 0, x = L, y = 0: y = 0: y= ( L − a ) x3 − 3 x − a 3 − ( 3 − a )( 3)2 x + ( 3 − a )3 x mm ..... Ans. 6 ( 37.5 × 10 ) ( 3) 3 833.85 (103 ) 404 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 9-51 Determine the deflection at the right end of the cantilever beam shown in Fig. P9-51. SOLUTION Using the solutions for cases 1 and 2 in Table A-19 with P = wL : δ C = δ w + δ P = δ Bw + θ Bw + δ CP w ( 2L ) w ( 2L ) =− − 8EI 6 EI 4 3 ( L) ( wL )( 3L ) − 3EI 3 = −37 wL4 37 wL4 = ↓ ................................................................................................... Ans. 3EI 3EI 9-52 Determine the deflection at the right end of the cantilever beam shown in Fig. P9-52. SOLUTION Using the solutions for cases 1, 2, and 4 in Table A-19 with P = wL and M = wL2 2 : δB = δw +δP +δM 3 2 wL4 ( wL )( L ) ( wL 2 ) ( L ) =− + + 8EI 3EI 2 EI 2 = +11wL4 11wL4 = ↑ ..................................................................................................... Ans. 24 EI 24 EI 9-53 For the cantilever beam shown in Fig. P9-53, determine (a) The slope and deflection at section B. (b) The slope and deflection at section C. SOLUTION Using the solutions for cases 1 and 4 in Table A-19 with M = PL : (a) θB = θP +θM = ( 2 P ) L2 − ( PL )( L ) =− 2 EI EI ........................................................................................................ Ans. 2 −2 PL2 2 PL2 = EI EI 3EI ( 2 P ) L3 − ( PL )( L ) δB = δP +δM = − 2 EI 2 −7 PL3 7 PL3 = = ↓ ........................................ Ans. 6 EI 6 EI ................................................ Ans. (b) θC = θ B + θC / B 2 PL2 P ( L ) −5PL2 5PL2 =− − = = EI 2 EI 2 EI 2 EI 3 P ( L) 7 PL3 2 PL2 −7 PL3 7 PL3 δC = δ B +θBL + δ P = − − = = ↓ .................. Ans. ( L) − 6 EI 3EI 2 EI 2 EI EI 405 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 9-54 RILEY, STURGES AND MORRIS Determine the deflection at a point midway between the supports of the beam shown in Fig. P9-54 when P = 13.5 kN, L = 3 m, I = 80(106) mm4, and E = 200 GPa. SOLUTION Using the solution for case 5 in Table A-19 with P = 2 P and P2 = P : 1 δ = δ P1 + δ P2 = − ( 2 P )( 3L 4 ) 3 ( 9 L 4 ) 48 EI 2 2 − 4 ( 3L 4 ) 2 2 P ( L 2 ) 3 ( 9 L 4 ) − 4 ( L 2 ) −53PL3 53PL3 = − = ↓ 48 EI 96 EI 96 EI δ= 96 ( 200 × 109 )( 80 × 10 −6 ) −53 (13.5 × 103 ) ( 3) 3 = −12.58 m = 12.58 mm ↓ ................................................. Ans. 9-55 Determine the deflection at the right end of the cantilever beam shown in Fig. P9-55. SOLUTION Using the solutions for cases 1 and 2 in Table A-19 with P = wL : δ D = δ w + δ P = [δ Cw + θ Cw L ] + δ BP + θ BP ( 2 L ) w ( 2L ) w ( 2L ) =− − 8 EI 6 EI 4 3 ( L) ( wL )( L ) − 3EI 3 ( wL )( L ) − 2 EI 2 ( 2L ) −14wL4 14wL4 = = ↓ ............................................................................................................. Ans. 3EI 3EI 9-56 For the beam shown in Fig. P9-56, determine (a) The deflection at a point midway between the supports. (b) The deflection at the right end of the beam. SOLUTION Using the solutions for cases 6 and 7 in Table A-19 with P = wL : (a) δ B = δ Bw + δ BP = 5wL4 ( wL )( L ) =− − 384 EI 48EI 3 −13wL4 13wL4 = ↓ ................................................................................................... Ans. 384 EI 384 EI (b) wL3 ( wL )( L )2 δ D = θ C ( L 3) = [θ Cw + θ CP ]( L 3) = + ( L 3) 16 EI 24 EI = +5wL4 5wL4 = ↑ ...................................................................................................... Ans. 144 EI 144 EI 406 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 9-57 RILEY, STURGES AND MORRIS Member AB of Fig. P9-57 is the flexural member of a scale that is used to weigh food in a microwave oven. Determine the deflection of point C when W = 5 lb, L = 2 in. and EI = 100 lb-in.2 SOLUTION Using the solutions for cases 1 and 4 in Table A-19 with P = W and M = WL 2 : δ C = δ B − θ B ( L 2 ) = [δ BP + δ BM ] − [θ BP + θ BM ]( L 2 ) 2 2 WL3 (WL 2 )( L ) W ( L ) (WL 2 )( L ) =− + − − + ( L 2) 3EI 2 EI EI 2 EI 3 −WL = 12 EI With W = 5 lb , L = 2 in. , and EI = 100 lb ⋅ in 2 : δC = − ( 5 )( 2 ) 3 12 (100 ) = −0.0333 in. = 0.0333 in. ↓ ................................................................ Ans. 9-58 Determine the deflection at a point midway between the supports of the beam shown in Fig. P9-58. SOLUTION Using the solutions for cases 7 and 8 in Table A-19 with M = wL2 3 : δ = δM +δw ( wL 3) ( L ) =− 2 2 16 EI 5wL4 − 384 EI −13wL 13wL4 = = ↓ ................................................................................................... Ans. 384 EI 384 EI 4 9-59 Determine the deflection at a point midway between the supports of the beam shown in Fig. P9-59. SOLUTION Using the solutions for cases 7 and 8 in Table A-19 with M B = wL2 8 and M C = wL2 4 : δ = δ w + δ M B + δ MC 2 2 5wL4 ( wL 8 ) ( L ) ( wL 4 ) ( L ) =− + + 384 EI 16 EI 16 EI 2 2 = + wL4 wL4 = ↑ ........................................................................................................... Ans. 96 EI 96 EI 407 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 9-60 RILEY, STURGES AND MORRIS Determine the deflection at the free end of the cantilever beam shown in Fig. P9-60 when w = 7 kN/m, L = 1.8 m, I = 130(10-6) m4, and E = 200 GPa. SOLUTION Using the solutions for cases 1 and 2 in Table A-19 with VB = 2 wL and M B = wL2 − ( 2 w )( L ) ( L 2 ) = 0 : δ C = δ Cw + δ B + θ B L =− 8 EI −23wL4 = 12 EI ( 2w ) L4 − ( 2wL )( L ) 3EI 3 − ( 2wL )( L ) 2 EI 2 ( L) δC = 12 ( 200 ×109 )(130 × 10−6 ) −23 ( 7 × 103 ) (1.8 ) 4 = −5.42 × 10−3 m = 5.42 mm ↓ .............................. Ans. 9-61 Determine the deflection at the right end of the cantilever beam shown in Fig. P9-61. SOLUTION Using the solutions for cases 1, 2, and 4 in Table A-19 with VB = wL and M B = − wL2 2 : δ C = δ B + θ B L + δ Cw = (δ BM + δ BV ) + (θ BM + θ BV ) L + δ Cw ( wL2 2 ) ( L )2 ( wL )( L )3 ( wL2 2 ) ( L ) ( wL )( L )2 wL4 − + − ( L) − = − − 2 EI 3EI 2 EI 8 EI EI wL4 −41wL4 41wL4 −7 wL4 wL3 = − = = ↓ .................................................. Ans. ( L) − 12 EI 8EI 24 EI 24 EI EI 9-62 Determine the deflection at the right end of the cantilever beam shown in Fig. P9-62. SOLUTION Using the solutions for cases 1 and 2 in Table A-19 with P = wL 2 : δ C = δ BP + θ BP L + δ Cw ( wL 2 ) L3 − ( wL 2 )( L ) =− 3EI 2 EI = 2 w ( 2L ) ( L) − 8EI 4 −29wL4 29wL4 = ↓ ........................................................................................................... Ans. 12 EI 12 EI 408 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 9-63 Determine the midspan deflection of beam AC of Fig. P9-63 if both beams have the same flexural rigidity. SOLUTION Using the solutions for cases 4, 5, and 6 in Table A-19 with P = P 2 and C M C = PL 2 : ( P 2 ) L 3 ( 4 L )2 − 4 ( L )2 3 3 = −11PL = 11PL ↓ δ F = −2 48 EI 12 EI 12 EI δC = δ F − δC / F = − 2 11PL3 ( PL 2 ) L −2 PL3 2 PL3 + = = ↓ 12 EI 2 EI 3EI 3EI 3 δ B = δC + δ B/C 2 PL3 P ( 2 L ) −5PL3 5PL3 =− − = = ↓ .............................................. Ans. 3EI 48EI 6 EI 6 EI 9-64 Determine the deflection at the right end of the beam shown in Fig. P9-64. SOLUTION Using the solution for case 8 in Table A-19 with M B = 2wL2 : δ D = θ C ( 3L 4 ) ( 2wL ) ( L ) 3L = − wL =− 2 4 6 EI 4 4 EI = wL4 ↓ .............................................. Ans. 4 EI 9-65 Determine the deflection at the free end of the cantilever beam shown in Fig. P9-65. SOLUTION Using the solutions for cases 1, 2, and 4 in Table A-19 with VB = wL and M B = wL2 2 : δ D = δ B + θ B L + δ Cw + θ C L = (δ BM + δ BV ) + (θ BM + θ BV ) L + δ Cw + (θ BM + θ BV + θ Cw ) L ( wL2 2 ) ( L )2 ( wL )( L )3 ( wL2 2 ) ( L ) ( wL )( L )2 wL4 + − ( L) − = − − − 2 EI 3EI 2 EI 8 EI EI ( wL2 2 ) ( L ) ( wL )( L )2 wL3 ( L) − + + EI 2 EI 6 EI = −69wL4 23wL4 = ↓ ................................................................................................... Ans. 24 EI 8EI 409 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 9-66 RILEY, STURGES AND MORRIS Determine the deflection at the free end of the cantilever beam shown in Fig. P9-66 when w = 7.5 kN/m, L = 3 m, I = 180(106) mm4, and E = 200 GPa. SOLUTION Using the solutions for cases 1, 2, and 4 in Table A-19 with VB = 2 wL and M B = 3wL2 2 − ( 2 w )( L ) ( L 2 ) = wL2 2 : δ C = δ Cw + δ B + θ B L = δ Cw + (δ Bw + δ BM − δ BV ) + (θ Bw + θ BM − θ BV ) L ( 2w ) L4 + wL4 − ( wL2 =− 8 EI 8 EI 2) ( L) 2 2 EI − ( 2wL )( L ) 3EI 3 δC = 8 ( 200 × 10 −7 ( 7.5 × 103 ) ( 3) 9 wL3 ( wL2 2 ) ( L ) ( 2wL )( L )2 −7 wL4 ( L) = + + − 2 EI 8EI EI 6 EI 4 −6 )(180 ×10 ) = −14.77 × 10−3 m = 14.77 mm ↓ ............................ Ans. 9-67 Determine the deflection at the free end of the cantilever beam shown in Fig. P9-67. SOLUTION Using the solution for case 3 in Table A-19: δ C = δ C 2 w + 2 δ Bw + θ Bw ( L 2 ) 4 3 ( 2w ) L4 + 2 w ( L 2 ) + w ( L 2 ) L =− 30 EI 24 EI 2 30 EI =− wL4 wL4 wL4 −11wL4 11wL4 + + = = ↓ ....................................................... Ans. 15EI 240 EI 192 EI 192 EI 192 EI 9-68 Determine the deflection at the right end of the cantilever beam shown in Fig. P9-68. SOLUTION Using the solution for case 3 in Table A-19: δ C = δ C 2 w + δ Bw + θ Bw ( L ) ( 2w )( 2 L ) =− 30 EI =− 4 4 w( L) wL3 + + ( L) 30 EI 24 EI 4 16wL wL4 wL4 −119wL4 119 wL4 + + = = ↓ .................................................... Ans. 15EI 30 EI 24 EI 120 EI 120 EI 410 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 9-69 A beam is loaded and supported as shown in Fig. P9-69. Determine (a) The reactions at supports A and B. (b) The deflection at the middle of the span. SOLUTION (a) RILEY, STURGES AND MORRIS EIy′′ = RA x − EIy′ = EIy = At At wx 2 2 RA x 2 wx3 − + C1 2 6 RA x 3 wx 4 − + C1 x + C2 6 24 y = 0: y′ = 0 : y = 0: Boundary conditions: x = 0, x = L, x = L, C2 = 0 RA L2 wL3 − + C1 = 0 2 6 RA L3 wL4 − + C1 L + C2 = 0 6 24 C1 = − wL3 48 and At Solving simultaneously yields: RA = +3wL 8 = 3wL 8 ↑ ...................................................................................................... Ans. Then, overall equilibrium gives: ↑ ΣFy = 0 : RA − wL − VB = 0 VB = −5wL 8 = 5wL 8 ↑ ............................................................................................... Ans. wL2 − RA L = 0 2 ............................................................................................. Ans. ΣM B = 0 : MB + M B = − wL2 8 = wL2 8 (b) y= w ( 3Lx3 − 2 x4 − L3 x ) 48 EI δ M = yx= L 2 9-70 w 3L4 L4 L4 − wL4 wL4 = − − = = ↓ .......................................... Ans. 48EI 8 8 2 192 EI 192 EI When a moment M is applied to the left end of the beam shown in Fig. P9-70, the slope at the left end of the beam is zero. Determine the magnitude of the moment M. SOLUTION EIy′′ = − M + RA x − EIy′ = − Mx + EIy = − wx 2 2 RA x 2 wx3 − + C1 2 6 Mx 2 RA x 3 wx 4 + − + C1 x + C2 2 6 24 411 STATICS AND MECHANICS OF MATERIALS, 2nd Edition Boundary conditions: At At At RILEY, STURGES AND MORRIS x = 0, x = 0, x = L, x = L, y = 0: y′ = 0 : y′ = 0 : y = 0: C2 = 0 C1 = 0 − ML + − RA L2 wL3 − =0 2 6 At ML2 RA L3 wL4 + − =0 2 6 24 Solving simultaneously yields: RA = + wL 2 = wL 2 ↑ M = + wL2 12 = wL2 12 9-71 ................................................................................................. Ans. A beam is loaded and supported as shown in Fig. P9-71. Determine the magnitude of the moment M required to make the slope at the left end of the beam zero. SOLUTION EIy′′ = M − EIy′ = Mx − EIy = At At wx 2 2 wx 3 + C1 6 Mx 2 wx 4 − + C1 x + C2 2 24 y′ = 0 : y′ = 0 : Boundary conditions: x = 0, x = L, C1 = 0 ML − wL3 =0 6 M = + wL2 6 = wL2 6 9-72 ..................................................................................................... Ans. When a moment M is applied to the left end of the cantilever beam shown in Fig. P9-72, the slope at the left end of the beam is zero. Determine the magnitude of the moment M. SOLUTION EIy′′ = Px − M EIy′ = EIy = At At Px 2 − Mx + C1 2 Px 3 Mx 2 − + C1 x + C2 6 2 y′ = 0 : y′ = 0 : Boundary conditions: x = 0, x = L, C1 = 0 PL2 − ML = 0 2 ......................................................................................................... Ans. M = + PL 2 = PL 2 412 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 9-73 RILEY, STURGES AND MORRIS When a load P is applied to the right end of the cantilever beam shown in Fig. P9-73, the deflection at the right end of the beam is zero. Determine the magnitude of the load P. SOLUTION EIy′′ = Px − EIy′ = wx 2 2 Px 2 wx 3 − + C1 2 6 Px 3 wx 4 EIy = − + C1 x + C2 6 24 Boundary conditions: At At x = 0, x = L, x = L, y = 0: y′ = 0 : y = 0: C2 = 0 PL2 wL3 − + C1 = 0 2 6 PL3 wL4 − + C1 L = 0 6 24 C1 = − wL3 48 and At Solving simultaneously yields: P = +3wL 8 = 3wL 8 ↑ ........................................................................................................ Ans. 9-74 A beam is loaded and supported as shown in Fig. P9-74. Determine (a) The reactions at supports A and B. (b) The maximum deflection in the beam. SOLUTION (a) EIy′′ = RA x − EIy′ = EIy = At At wx3 6L RA x 2 wx 4 − + C1 2 24 L RA x 3 wx 5 − + C1 x + C2 6 120 L y = 0: y′ = 0 : y = 0: Boundary conditions: x = 0, x = L, x = L, C2 = 0 RA L2 wL3 − + C1 = 0 2 24 RA L3 wL4 − + C1 L = 0 6 120 C1 = − wL3 120 and At Solving simultaneously yields: RA = + wL 10 = wL 10 ↑ ...................................................................................................... Ans. Then, overall equilibrium gives: ↑ ΣFy = 0 : RA − ( wL 2 ) − VB = 0 413 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS VB = −2wL 5 = 2wL 5 ↑ ............................................................................................... Ans. ΣM B = 0 : wL2 MB + − RA L = 0 6 ........................................................................................ Ans. M B = − wL2 15 = wL2 15 (b) The maximum deflection occurs where w ( 6 L2 x2 − 5x4 − L4 ) = 0 120 EIL w y= ( 2L2 x3 − x5 − L4 x ) 120 EIL y′ = x = 0.4472 L δ max = yx =0.4472 L = −0.00239 wL4 0.00239wL4 = ↓ ........................................................ Ans. EI EI 9-75 A beam is loaded and supported as shown in Fig. P9-75. Determine (a) The reactions at supports A and B. (b) The maximum deflection in the beam. SOLUTION w ( x ) = kx 2 (a) At x = L: w( x) = w Therefore, w ( x ) = wx 2 2 EIy′′ = RA x − EIy′ = EIy = At At wx 4 12 L2 RA x 2 wx5 − + C1 2 60 L2 RA x 3 wx 6 − + C1 x + C2 6 360 L2 y = 0: y′ = 0 : y = 0: Boundary conditions: x = 0, x = L, x = L, C2 = 0 RA L2 wL3 − + C1 = 0 2 60 RA L3 wL4 − + C1 L = 0 6 360 C1 = − wL3 240 and At Solving simultaneously yields: RA = + wL 24 = wL 24 ↑ ..................................................................................................... Ans. Then, overall equilibrium gives: ↑ ΣFy = 0 : RA − ( wL 3) − VB = 0 VB = −7 wL 24 = 7 wL 24 ↑ .......................................................................................... Ans. wL L − RA L = 0 3 4 ....................................................................................... Ans. ΣM B = 0 : MB + M B = − wL2 24 = wL2 24 (b) The maximum deflection occurs where 414 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS w 5 L3 x 2 − 4 x5 − L5 ) = 0 2( 240 EIL w y= 5 L3 x 3 − 2 x 6 − 3L5 x ) 2( 720 EIL y′ = x = 0.4666 L δ max = yx =0.4666 L = −0.001267 wL4 0.001267 wL4 = ↓ ................................................... Ans. EI EI 9-76 A beam is loaded and supported as shown in Fig. P9-76. Determine the reactions at supports A and B. SOLUTION EIy′′′′ = − w sin (π x 2 L ) EIy′′′ = 2 wL πx + C1 cos 2L π EIy′′ = 4wL2 πx + C1 x + C2 sin 2 π 2L 8wL3 π x C1 x 2 cos + + C2 x + C3 π3 2L 2 EIy′ = − EIy = − 16wL4 π x C1 x3 C2 x 2 + + + C3 x + C4 sin π4 2L 6 2 Boundary conditions: At At At x = 0, x = 0, x = L, x = L, y = 0: M = EIy′′ = 0 : y′ = 0 : y = 0: C4 = 0 C2 = 0 C1 L2 + C3 = 0 2 − 16wL4 C1 L3 + + C3 L = 0 π4 6 and At Solving simultaneously yields: C1 = −48wL π 4 C3 = 24 wL3 π 4 2wL π ( 0 ) 48wL 2wL 48wL cos − 4 = − 4 ↑ .................................... Ans. VA = EIy′′′= 0 = x 2L π π π π π 48wL −48wL 48wL 2 wL VB = EIy′′′= L = cos − = = ↑ .......................................... Ans. x 2 π4 π4 π4 π 4 wL2 48wL2 4wL2 48wL2 M B = EIy′′= L = 2 sin − = − x 2 π π4 π2 π4 9-77 A beam is loaded and supported as shown in Fig. P9-77. Determine (a) The reactions at supports A and B. (b) The deflection at the middle of the span. SOLUTION (a) .................................. Ans. EIy′′ = RA x − P x − L 1 415 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS R x2 P x − L EIy′ = A − 2 2 R x3 P x − L EIy = A − 6 6 Boundary conditions: At At At 2 + C1 3 + C1 x + C2 x = 0, y = 0: y′ = 0 : y = 0: x = 2 L, x = 2 L, 2 RA L2 − ( PL2 2 ) + C1 = 0 C2 = 0 4 RA L3 PL3 − + 2C1 L = 0 3 6 C1 = − PL2 8 and Solving simultaneously yields: RA = +5P 16 = 5P 16 ↑ ....................................................................................................... Ans. Then, overall equilibrium gives: ↑ ΣFy = 0 : ΣM B = 0 : RA − P − VB = 0 M B + PL − RA ( 2 L ) = 0 ............................................................................................ Ans. VB = −11P 16 = 11P 16 ↑ ............................................................................................. Ans. M B = −3PL 8 = 3PL 8 (b) y= P 3 5 x 3 − 16 x − L − 12 L2 x 96 EI ( ) δ M = yx= L P −7 PL3 7 PL3 3 3 = ( 5L − 0 − 12 L ) = 96EI = 96EI ↓ ............................................... Ans. 96 EI 9-78 A beam is loaded and supported as shown in Fig. P9-78. Determine (a) The reactions at supports A and B. (b) The deflection at the middle of the span. SOLUTION (a) From symmetry: VA = + P 2 = P 2 ↑ ................................. VB = − P 2 = P 2 ↑ ......................................... Ans. Then for the interval 0≤ x≤ L 2 EIy′′ = Px + MA 2 EIy′ = EIy = At Px 2 + M A x + C1 4 Px 3 M A x 2 + + C1 x + C2 12 2 y′ = 0 : Boundary conditions: x = 0, C1 = 0 416 STATICS AND MECHANICS OF MATERIALS, 2nd Edition At At RILEY, STURGES AND MORRIS x = 0, y = 0: y′ = 0 : C2 = 0 PL2 M A L + =0 16 2 ....................................................................................................... Ans. x = L 2, M A = − PL 8 = PL 8 Then, from symmetry, (b) M B = − PL 8 = PL 8 .............................................. Ans. y= P ( 4 x3 − 3Lx2 ) 48 EI 0≤ x≤ L 2 δ M = yx= L 2 = P L3 3L3 − PL3 PL3 − = = ↓ ................................................... Ans. 48EI 2 4 192 EI 192 EI 9-79 A beam is loaded and supported as shown in Fig. P9-79. Determine the reactions at supports B, C, and D. SOLUTION EIy′′ = RD x + RC x − L + RB x − 2 L 1 1 EIy′ = RD x 2 RC x − L + 2 2 2 + 3 RB x − 2 L 2 3 2 + C1 + C1 x + C2 C2 = 0 R x3 R x − L EIy = D + C 6 6 Boundary conditions: At At At R x − 2L +B 6 x = 0, x = L, y = 0: y = 0: y = 0: (R ( 4R D D L3 6 ) + C1 L = 0 x = 2 L, L3 3 ) + ( RC L3 6 ) + 2C1 L = 0 Solving simultaneously yields: Then, overall equilibrium gives: C1 = RD L2 6 and RD = − RC 6 ↑ ΣFy = 0 : ΣM B = 0 : RB = − P + RB + RC + RD = 0 P ( L 2 ) + RC ( L ) + RD ( 2 L ) = 0 +13P 13P −3 P 3 P +P P = ↑ .......... RC = = ↓ .............. RD = = ↑ ......................... Ans. 8 8 8 8 4 4 9-80 A beam is loaded and supported as shown in Fig. P9-80. Determine (a) The reactions at supports A, B, and C. (b) The bending moment over the middle support. (c) The deflection at the middle of span BC. SOLUTION (a) EIy′′ = RA x − w x−L wx 2 1 + RB x − L + 2 2 2 2 R x 2 wx 3 RB x − L EIy′ = A − + 2 6 2 w x−L + 6 3 + C1 417 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS R x 3 wx 4 RB x − L EIy = A − + 6 24 6 Boundary conditions: At At 3 w x−L + 24 4 + C1 x + C2 C2 = 0 x = 0, x = L, y = 0: y = 0: y = 0: RA L3 wL4 − + C1 L = 0 6 24 4 RA L3 2wL4 RB L3 wL4 − + + + 2C1 L = 0 3 3 6 24 RA + RB + RC − wL = 0 At x = 2 L, Overall equilibrium gives: ↑ ΣFy = 0 : ΣM C = 0 : Solving simultaneously yields: ( wL )( 3 L 2 ) − RB ( L ) − RA ( 2 L ) = 0 C1 = − wL3 32 and RA = (b) (c) +7 wL 7 wL +5wL 5wL − wL wL = ↑ ........ RB = = ↑ ........ RC = = ↓ .................... Ans. 16 16 16 16 8 8 M B = RA L − ( wL )( L 2 ) = y= 7 wL wL2 − wL2 L− = ......................................................... Ans. 16 2 16 w 3 4 7 Lx3 − 4 x 4 + 10 L x − L + 4 x − L − 3L3 x 96 EI ( ) δ MBC = yx =3 L 2 = = w 189 L4 81L4 10 L4 L4 9 L4 − + +− 96 EI 8 4 8 4 2 + wL4 wL4 = ↑ .................................................................................................. Ans. 256 EI 256 EI 9-81 A beam is loaded and supported as shown in Fig. P9-81. Determine (a) The reactions at supports A, C and D. (b) The deflection at the middle of span AC. SOLUTION (a) EIy′′ = RA x − P x − L + RC x − 2 L 1 1 R x2 P x − L EIy′ = A − 2 2 R x3 P x − L EIy = A − 6 6 Boundary conditions: At At 2 R x − 2L +C 2 R x − 2L +C 6 3 2 + C1 + C1 x + C2 3 x = 0, y = 0: y = 0: C2 = 0 4 RA L3 PL3 − + 2C1 L = 0 3 6 x = 2 L, 418 STATICS AND MECHANICS OF MATERIALS, 2nd Edition At RILEY, STURGES AND MORRIS x = 3 L, x = 3 L, y′ = 0 : y = 0: 9 RA L2 4 PL2 RC L2 − + + C1 = 0 2 2 2 9 RA L3 4 PL3 RC L3 − + + 3C1 L = 0 2 3 6 C1 = −7 PL2 44 and At Solving simultaneously yields: RA = +4 P 4 P +23P 23P = ↑ ............................ RC = = ↑ ...................................... Ans. 11 11 22 22 Then, overall equilibrium gives: ↑ ΣFy = 0 : 4P 23P −P+ − VD = 0 11 22 23P 4P ΣM D = 0 : M D + P ( 2 L ) − ( L ) − ( 3L ) = 0 22 11 +9 P 9 P +3PL 3PL VD = = ↓ ............................ M D = = 22 22 22 22 .................................... Ans. (b) x−L P 2 x3 y= − 6 EI 33 3 23 x − 2 L + 132 3 − 7 L2 x 44 δ B = yx= L = P 2 L3 7 L3 −13PL3 13PL3 −0+0− = ↓ ......................................... Ans. = 44 132 EI 132 EI EI 33 9-82 A beam is loaded and supported as shown in Fig. P9-82. Determine (a) The reactions at supports A, B, and C. (b) The bending moment over the center support. (c) The maximum bending moment in the beam. SOLUTION (a) EIy′′ = RA x − wx 2 + RB x − L 2 1 R x 2 wx 3 RB x − L + EIy′ = A − 2 6 2 R x 3 wx 4 RB x − L EIy = A − + 6 24 6 Boundary conditions: At At 2 + C1 3 + C1 x + C2 C2 = 0 RA L3 wL4 − + C1 L = 0 6 24 4 wL3 RB L2 2 RA L − + + C1 = 0 3 2 2 x = 0, x = L, y = 0: y = 0: y′ = 0 : y = 0: At x = 2 L, x = 2 L, At 4 RA L3 2wL4 RB L3 − + + 2C1 L = 0 3 3 6 419 STATICS AND MECHANICS OF MATERIALS, 2nd Edition Solving simultaneously yields: RILEY, STURGES AND MORRIS C1 = − wL3 42 and RA = +11wL 11wL +8wL 8wL = ↑ .................. RB = = ↑ ..................................... Ans. 28 28 7 7 Overall equilibrium gives: ↑ ΣFy = 0 : VC = RA + RB − VC − 2 wL = 0 −13wL 13wL = ↑ ................................................................................................. Ans. 28 28 ΣM C = 0 : ( w2 L )( L ) − RB ( L ) − RA ( 2 L ) = 0 ................................................................................................... Ans. − wL2 wL2 MC = = 14 14 (b) 11wL wL2 −3wL2 M B = RA L − ( wL )( L 2 ) = L− = ..................................................... Ans. 28 2 28 11wLx wx 2 8wL x − L − + M ( x ) = EIy′′ = 28 2 7 8wL x − L 11wL dM dx = − wx + 28 7 0 1 (c) The bending moment is a maximum where =0 x = 11L 28 : x = 43L 28 : x = L: x = 2L : M = 121wL2 1568 ≅ 0.0772wL2 M = 57 wL2 1568 ≅ 0.0364wL2 M = −3wL2 28 ≅ −0.1071wL2 M = − wL2 14 ≅ −0.0714wL2 M max = −3wL2 28 .................................................................................................................. Ans. 9-83 A beam is loaded and supported as shown in Fig. P9-83. Determine (a) The reactions at supports A, B, and C. (b) The bending moment over the center support. SOLUTION (a) EIy′′ = VA x + M A − wx 2 + RB x − L 2 1 V x2 wx 3 RB x − L + EIy′ = A + M A x − 2 6 2 V x3 M x 2 wx 4 RB x − L EIy = A + A − + 6 2 24 6 Boundary conditions: At At At 2 + C1 3 + C1 x + C2 C1 = 0 C2 = 0 x = 0, x = 0, x = L, y′ = 0 : y = 0: y′ = 0 : VA L2 wL3 + M AL − =0 2 6 420 STATICS AND MECHANICS OF MATERIALS, 2nd Edition At RILEY, STURGES AND MORRIS x = L, y = 0: VA L3 M A L2 wL4 + − =0 6 2 24 ........................ Ans. Solving simultaneously yields: VA = + wL 2 = wL 2 ↑ .................... M A = − wL2 12 = wL2 12 From symmetry: VC = − wL 2 = wL 2 ↑ .................... Ans. M C = − wL2 12 = wL2 12 Then, overall equilibrium gives: ......... Ans. ↑ ΣFy = 0 : VA + RB − VC − 2 wL = 0 RB = wL = wL ↑ .............................................................................................................. Ans. wL wL2 wL2 − wL2 L− − = ..................................... Ans. 2 12 2 12 (b) M B = VA L + M A − ( wL )( L 2 ) = 9-84 A beam is loaded and supported as shown in Fig. P9-84. Determine (a) The reactions at supports A and D. (b) The deflection at B if E = 200 GPa and I = 350(106) mm4. SOLUTION (a) 1 2 EIy′′ = RA x − 36 x − 3 − 3 x − 6 kN ⋅ m R x2 2 3 ′ = A − 18 x − 3 − x − 6 + C1 kN ⋅ m 2 EIy 2 R x3 x−6 3 EIy = A − 6 x − 3 − 4 6 Boundary conditions: At At At 4 + C1 x + C2 kN ⋅ m3 x = 0 m, y = 0: y′ = 0 : y = 0: C2 = 0 72 RA − 1674 + C1 = 0 288RA − 4698 + 12C1 = 0 C1 = −249.75 kN ⋅ m 2 and x = 12 m, x = 12 m, Solving simultaneously yields: RA = +26.72 kN ≅ +26.7 kN ↑ ................................................................................... Ans. Then, overall equilibrium gives: ↑ ΣFy = 0 : RA − 36 − 6 ( 6 ) − VD = 0 VD = −45.3 kN = 45.3 kN ↑ ......................................................................................... Ans. ΣM D = 0 : M D + 6 ( 6 ) ( 3 ) + 36 ( 9 ) − RA (12 ) = 0 M D = −111.4 kN ⋅ m = 111.4 kN ⋅ m (b) .................................................................... Ans. EIy = 1 3 4 53.44 x3 − 72 x − 3 − 3 x − 6 − 2997 x kN ⋅ m 12 421 ( ) STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS δ B = y x =3 = −629 −629 ×103 = = −8.99 × 10−3 m = 8.99 mm ↓ ......... Ans. 9 −6 EI ( 200 ×10 )( 350 ×10 ) 9-85 A beam is loaded and supported as shown in Fig. P9-85. Determine (a) The reactions at supports A, B, and D. (b) The deflection at C if E = 30,000 ksi and I = 26 in.4 SOLUTION (a) EIy′′ = RA x − 600 x 2 + RB x − 4 2 1 + RD x − 10 lb ⋅ ft 1 +600 x − 4 − 3000 x − 6 1 R x2 R x−4 EIy′ = A − 200 x 3 + B 2 2 2 2 + 200 x − 4 2 3 R x − 10 −1500 x − 6 + D 2 + C1 lb ⋅ ft 2 4 3 R x3 RB x − 4 4 A EIy = − 50 x + + 50 x − 4 6 6 3 3 RD x − 10 −500 x − 6 + + C1 x + C2 lb ⋅ ft 3 6 Boundary conditions: At At At x = 0 ft, x = 4 ft, y = 0: y = 0: y = 0: C2 = 0 ( 32 RA 3) − 12,800 + 4C1 = 0 ( 500 RA 3) − 467, 200 + 36 RB + 10C1 = 0 RA + RB + RD − 1200 ( 4 ) − 3000 − 2000 = 0 x = 10 ft, Overall equilibrium gives: ↑ ΣFy = 0 : ΣM D = 0 : − RA (10 ) + 1200 ( 4 ) ( 8 ) − RB ( 6 ) + 3000 ( 4 ) − 2000 ( 2 ) = 0 Solving simultaneously yields: C1 = −2027 lb ⋅ ft 2 and RA = 1960 lb ↑ ................ RB = 4467 lb ↑ ............... RD = 3373 lb ↑ .......................... Ans. (b) EIy = 1 3 1960 x3 − 300 x 4 + 4467 x − 4 + 300 x − 4 6 3 3 4 −3000 x − 6 + 3373 x − 10 − 12,162 x lb ⋅ ft 3 δ C = yx=6 354 (12 ) 354 = = = 0.000784 in. = 0.000784 in. ↑ ........................ Ans. EI ( 30 × 106 ) ( 26 ) 3 422 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 9-92 RILEY, STURGES AND MORRIS A beam is loaded and supported as shown in Fig. P9-92. Determine the magnitude of the load P required to make the slope at the left end of the beam zero. SOLUTION Using the solutions for cases 1 and 2 in Table A-19: P ( L) w( L) θ A = θP +θw = − =0 2 EI 6 EI wL wL P= = ↑ ...................................................................................................................... Ans. 3 3 2 3 9-93 A beam is loaded and supported as shown in Fig. P9-93. Determine the reactions at supports A and B. SOLUTION Using the solutions for cases 1 and 4 in Table A-19: δ A = δ RA + δ M + θ M L RA ( 2 L ) ML2 ML = − − ( L) = 0 3EI 2 EI EI 3 RA = +9 M 9 M = ↑ ......................... Ans. 16 L 16 L Then, overall equilibrium gives: ↑ ΣFy = 0 : ΣM B = 0 : RA − VB = 0 M + M B − RA ( 2 L ) = 0 VB = +9 M 9 M = ↓ ......................... Ans. 16 L 16 L MM ............................ Ans. MB = = 8 8 9-94 A beam is loaded and supported as shown in Fig. P9-94. Determine the reactions at supports A and B. SOLUTION Using the solutions for cases 1 and 2 in Table A-19: δ B = δ w + δ RB = wL4 RB L3 + =0 8EI 3EI RB = +3wL 8 = 3wL 8 ↑ ............................................................................................... Ans. Then, overall equilibrium gives: ↑ ΣFy = 0 : VA + RB − wL = 0 VA = +5wL 5wL = ↑ ....................... Ans. 8 8 ΣM A = 0 : RB L − M A − ( wL )( L 2 ) = 0 MA = − wL2 wL2 = 8 8 .................... Ans. 9-95 A beam is loaded and supported as shown in Fig. P9-95. Determine the reactions at supports A and B. SOLUTION Using the solutions for cases 1 and 4 in Table A-19 with M B = − PL 2 and VB = P : δ B = δ BVB + δ BM B + δ BRB 431 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS PL3 ( PL 2 )( L ) R L3 =− − + B =0 3EI 2 EI 3EI 2 RB = +7 P 7 P = ↑ ............................ Ans. 4 4 Then, overall equilibrium gives: ↑ ΣFy = 0 : ΣM A = 0 : VA + RB − P = 0 RB ( L ) − M A − P ( 3L 2 ) = 0 VA = −3P 4 = 3P 4 ↓ ...................... Ans. M A = + PL 4 = PL 4 ................. Ans. 9-96 A beam is loaded and supported as shown in Fig. P9-96. Determine the reactions at supports A, B, and C. SOLUTION Using the solutions for cases 1, 2, 7, and 8 in Table A-19 with and M Bw = wL2 2 M BRC = RC L : δ C = δ Cw + θ B L + δ CRC 3 2 wL4 w ( 2 L ) ( wL 2 ) ( 2 L ) ( RC L )( 2 L ) R L3 ( L) + C = 0 =− + − + 8EI 24 EI 3EI 3EI 3EI + wL wL RC = = ↑ ................................................................................................................. Ans. 8 8 Then, overall equilibrium gives: ΣM A = 0 : ↑ ΣFy = 0 : 9-97 RB ( 2 L ) + RC ( 3L ) − ( 3wL )( 3L 2 ) = 0 RA + RB + RC − 3wL = 0 RB = +33wL 16 = 33wL 16 ↑ ...................................................................................... Ans. RA = +13wL 16 = 13wL 16 ↑ ...................................................................................... Ans. A beam is loaded and supported as shown in Fig. P9-97. When the load P is applied, the slope at the right end of the beam is zero. Determine (a) The magnitude of the load P. (b) The reactions at supports A and B. SOLUTION Using the solutions for cases 1, 2, and 4 in Table A-19 with M B = PL 4 and (a) VB = P : θ C = θ CP + θ Cw + θ CRB P ( 5L 4 ) w ( L ) RB L2 =− − + =0 2 EI 6 EI 2 EI 2 3 −75P + 48RB − 16wL = 0 δ B = δ Bw + δ BVB + δ BM B + δ BRB 432 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 3 2 RILEY, STURGES AND MORRIS wL4 P ( L ) ( PL 4 )( L ) RB L3 =− − − + =0 8EI 3EI 2 EI 3EI −11P + 8RB − 3wL = 0 Solving simultaneously gives: RB = +49 wL 49 wL +2 wL 2 wL = ↑ ........................ P = = ↓ ....................................... Ans. 72 72 9 9 (b) Then, overall equilibrium gives: ↑ ΣFy = 0 : VA = VA − wL + RB − P = 0 +13wL 13wL = ↑ ................................................................................................. Ans. 24 24 ΣM A = 0 : RB ( L ) − P ( 5L 4 ) − ( wL )( L 2 ) − M A = 0 −7 wL2 7 wL2 = 72 72 .............................................................................................. Ans. MA = 9-98 A beam is loaded and supported as shown in Fig. P9-98. Determine (a) The reactions at supports A and C. (b) The deflection at the right end of the distributed load. SOLUTION Using the solutions for cases 1, 2, and 4 in Table A-19: (a) δ C = δ Bw + θ Bw ( L ) + δ RC =− R ( 2L ) wL4 w ( L ) − =0 ( L) + C 8EI 6 EI 3EI +7 wL 7 wL RC = = ↑ ..................................................................................................... Ans. 64 64 3 3 Then, overall equilibrium gives: ↑ ΣFy = 0 : VA = RC − wL + VA = 0 +57 wL 57 wL = ↑ ................................................................................................. Ans. 64 64 ΣM A = 0 : RC ( 2 L ) − ( wL )( L 2 ) − M A = 0 .............................................................................................. Ans. −9wL2 9wL2 MA = = 32 32 (b) δ B = δ BV + δ BM + δ Bw ( 7 wL 64 )( L ) = 3EI 4 3 ( 7wL + 4 2 64 ) ( L ) 2 2 EI − wL4 8 EI = −13wL 13wL = ↓ ................................................................................................... Ans. 384 EI 384 EI 433 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 9-99 RILEY, STURGES AND MORRIS A beam is loaded and supported as shown in Fig. P9-99. Determine the magnitude of the moment M required to make (a) The slope at the right end of the beam zero. (b) The deflection at the right end of the beam zero. SOLUTION Using the solutions for cases 4 and 8 in Table A-19 with M A = PL 3 and MB = M : (a) θ C = θ B + θ CM = θ BM A + θ BM B + θ CM ( PL 3)( L ) + ML + M ( L 3) = 0 =− EI 6 EI 3EI + PL PL M= = ........................................................................................................ Ans. 12 12 δ C = θ B ( L 3) + δ CM = θ BM A + θ BM B ( L 3) + δ CM M ( L 3) ( PL 3)( L ) ML = − + =0 ( L 3) + 6 EI 3EI 2 EI + PL PL M= = ........................................................................................................ Ans. 9 9 2 (b) 9-100 Draw complete shear force and bending moment diagrams for the beam shown in Fig. P9-100. SOLUTION Using the solutions for cases 1 and 4 in Table A-19: δ C = δ BM + θ BM L + δ CRC M ( L ) ML R ( 2L ) =− − =0 ( L) + C EI 2 EI 3EI +9 M 9 M RC = = ↑ ..................................................................................................................... Ans. 16 L 16 L 2 3 Then, overall equilibrium gives: ↑ ΣFy = 0 : ΣM A = 0 : VA + RC = 0 RC ( 2 L ) − M A − M = 0 VA = −9 M 9 M = ↓ ................................ Ans. 16 L 16 L +M M MA = = ................................. Ans. 8 8 9-101 Draw complete shear force and bending moment diagrams for the beam shown in Fig. P9-101. SOLUTION Using the solutions for cases 1, 3, and 4 in Table A-19: 434 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS δ A = δ AM + δ AV + δ Aw M ( L ) VA ( L ) wL4 =A + − =0 2 EI 3EI 30 EI 2 3 15M A + 10VA L = wL2 θ A = θ AM + θ AV + θ Aw =− M A L VA ( L ) w ( L ) − + =0 EI 2 EI 24 EI 2 3 24M A + 12VA L = wL2 Solving simultaneously gives: VA = +3wL 3wL − wL2 wL2 = ↑ .............................. M A = = 20 20 30 30 ................................... Ans. Then, overall equilibrium gives: ↑ ΣFy = 0 : VB = VA − ( wL 2 ) − VB = 0 −7 wL 7 wL = ↑ ..................................................................................................... Ans. 20 20 wL L MB + − M A − VA ( L ) = 0 2 3 ΣM B = 0 : MB = 9-102 − wL2 wL2 = 20 20 ................................................................................................... Ans. Two beams are loaded and supported as shown in Fig. P9-102. Determine the reactions at supports A, B, and D. Both beams have the same flexural rigidity. SOLUTION Using the solution for case 1 in Table A-19: δB A = δB D δ BRB = δ CP + θ CP ( L 2 ) + δ BRB P ( L 2) P ( L 2) R L3 − B =− − 3EI 3EI 2 EI RB = 5P 32 3 2 ( L 2) + RB L3 3EI RB = 5P 32 ↓ .................................................. Ans. For beam AB : ↑ ΣFy = 0 : ΣM A = 0 : For beam VA − RB = 0 − M A − RB L = 0 RB − P − VD = 0 M D + P ( L 2 ) − RB L = 0 VA = +5P 32 = 5 P 32 ↑ ................................ Ans. M A = −5PL 32 = 5 PL 32 ...................... Ans. CD : RB = 5P 32 ↑ .................................................. Ans. VD = −27 P 32 = 27 P 32 ↑ .......................... Ans. M D = −11PL 32 = 11PL 32 .................. Ans. ↑ ΣFy = 0 : ΣM D = 0 : 435 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 9-103 RILEY, STURGES AND MORRIS A steel (E = 29,000 ksi and I = 120 in.4) beam is loaded and supported as shown in Fig. P9-103. The post BD is a 6 × 6-in. timber (E = 1500 ksi) that is braced to prevent buckling. Determine the load carried by the post if it is unstressed before the 530-lb/ft distributed load is applied. SOLUTION AP = ( 6 × 6 ) = 36 in 2 The amount that the center of the beam settles is equal to the amount that the post shrinks: δ beam = δ Bw + δ BRB = δ post Using the solutions for cases 6 and 7 in Table A-19: − 5wL4 R L3 RL + B =− B P AP EP 384 EI 48 EI RB ( 20 ×12 ) 3 48 ( 29 ×106 ) (120 ) 9-104 + ( 36 ) (1.5 ×106 ) RB ( 20 ×12 ) = 5 ( 530 12 )( 20 ×12 ) 4 384 ( 29 ×106 ) (120 ) RB = 6287 lb ≅ 6.29 kip ...................................................................................................... Ans. A uniformly distributed load of 7 kN/m is supported by two 100 × 100-mm timber (E = 8.5 GPa) beams arranged as shown in Fig. P9-104. Beam AB is fixed at the wall and beam CD is simply supported. Before the load is applied, the beams are in contact at B, but the reaction at B is zero. After the 7-kN/m distributed load is applied, determine (a) The maximum flexural stress in each beam. (b) The maximum longitudinal shearing stress in each beam. SOLUTION Using the solutions for cases 1, 2, and 6 in Table A-19: δB A = δB D δ Bw + δ BRB = δ BRB R ( 2L ) wL4 RB L3 − + =− B 8EI 3EI 48EI RB = wL 4 For beam 3 AB : RB = wL 4 ↑ VA − wL + RB = 0 RB L − M A − ( wL )( L 2 ) = 0 ↑ ΣFy = 0 : ΣM A = 0 : VA = +3wL 4 = 3wL 4 ↑ M A = − wL2 4 = wL2 4 From the shear force and bending moment diagrams: Vmax = 7.875 kN M max = −3.938 kN ⋅ m 436 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 100 (100 ) I= = 8.333 ×106 mm 4 12 3 Q = yC A = 25 (100 × 50 ) = 125.00 × 103 mm3 (a) σ max 3 Mc ( 3.938 ×10 ) ( 0.050 ) = = 8.333 ×10−6 I = 23.6 × 106 N/m 2 = 23.6 MPa (T bottom & C top) ........................................... Ans. (b) τ max 3 −6 VQ ( 7.875 ×10 )(125.00 ×10 ) = = It (8.333 ×10−6 ) ( 0.100 ) = 1.181×106 N/m 2 = 1.181 MPa (at neutral surface) ......................................... Ans. For beam CD : RB = wL 4 ↓ RC = RD = wL 8 ↑ From symmetry: From the shear force and bending moment diagrams: Vmax = 1.3125 kN M max = 1.9688 kN ⋅ m Mc (1.9688 ×10 ) ( 0.050 ) = = 8.333 ×10−6 I 3 (a) σ max = 11.81×106 N/m 2 = 11.81 MPa (T bottom & C top) ........................................ Ans. (b) τ max VQ (1.3125 ×10 )(125.00 ×10 = = It (8.333 ×10−6 ) ( 0.100 ) 3 −6 ) = 0.1969 ×106 N/m 2 = 0.1969 MPa (at neutral surface) ................................... Ans. 9-105 The 4-in. wide × 6-in. deep timber (E = 1200 ksi) beam shown in Fig. P9-105 is fixed at the left end and supported at the right end with a tie rod that has a cross-sectional area of 0.125 in.2 Determine the tension in the tie rod if it is unstressed before the load is applied to the timber beam and (a) The tie rod is made of steel (E = 30,000 ksi). (b) The tie rod is made of aluminum alloy (E = 10,000 ksi). SOLUTION I NA 4 ( 6) = = 72 in 4 12 3 The amount that the end of the beam settles is equal to the amount that the rod shrinks: δ beam = δ Bw + δ BRB = δ rod Using the solutions for cases 1 and 2 in Table A-19: wL4 RB L3 RB LR − = 8EI 3EI AR ER 437 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (a) For the steel RILEY, STURGES AND MORRIS ( E = 30, 000 ksi ) rod: 3 (120 12 )(10 ×12 ) + = 6 6 3 (1.2 ×10 ) ( 72 ) ( 0.125 ) ( 30 ×10 ) 8 (1.2 ×106 ) ( 72 ) (b) For the aluminum alloy RB (10 ×12 ) RB ( 30 ×12 ) 4 RB = 444 lb ............................................................................................................... Ans. ( E = 10, 000 ksi ) rod: RB ( 30 ×12 ) 4 (120 12 )(10 ×12 ) + = 6 6 3 (1.2 ×10 ) ( 72 ) ( 0.125 ) (10 ×10 ) 8 (1.2 ×106 ) ( 72 ) 3 RB (10 ×12 ) RB = 431 lb ............................................................................................................... Ans. 9-106 In Fig. P9-106, the aluminum alloy tie rod passes through a hole in the aluminum alloy cantilever beam and through the coil spring positioned on the end of the tie rod. Before loading, there is a clearance of 2.5 mm between the bottom of the beam and the top of the spring. The cross-sectional area of the tie rod is 100 mm2, the second moment of area of the cross section of the beam with respect to the neutral axis is 40(106) mm4, the modulus of elasticity of the aluminum alloy is 70 GPa, and the spring modulus is 1000 kN/m.. Determine the axial stress in the tie rod when M = 9 kN-m and w = 90 kN/m. SOLUTION Using the solutions for cases 1, 2, and 4 in Table A-19: δ Bw + δ BM + δ BR = δ rod + δ spring + clearance − wL4 ML2 RL3 RL3 R b b b r − + =− − −C 8Eb I b 2 Eb I b 48 Eb I b Ar Er k 3 4 3 2 3 9 ( 90 ×10 ) (1.25) − ( 9 ×10 ) (1.25) + R (1.25 ) − 8 ( 70 × 10 )( 40 × 10 ) 2 ( 70 × 10 )( 40 ×10 ) 3 ( 70 × 10 )( 40 × 10 ) −6 9 −6 9 −6 =− R = 6.959 × 103 N ( 70 ×10 )(100 ×10 ) 9 R (1.25 ) −6 − R − 2.5 ×10−3 3 1000 ×10 R 6.959 ×103 = = 69.6 N/m 2 ≅ 69.6 MPa (T) .................................................... Ans. σ= −6 Ar 100 ×10 9-107 Two steel (E = 29,000 ksi) beams support a 1600-lb concentrated load, as shown in Fig. P9-107. In the unloaded condition, beam AB touches but exerts no force on beam CD. Beam AB is an S4 × 9.5 American standard section, and beam CD is an S5 × 14.75 American standard section (see Appendix A). Determine (a) The maximum flexural stress in each beam. (b) The maximum transverse shearing stress in each beam. SOLUTION For an S 4 × 9.5 section: I = 6.79 in 4 S = 3.39 in 3 I = 15.2 in 4 S = 6.09 in 3 w f = 2.796 in. d = 4.00 in. w f = 3.284 in. d = 5.00 in. t f = 0.293 in. ( Beam For an AB ) CD ) tw = 0.326 in. t f = 0.326 in. S5 × 14.75 section: ( Beam tw = 0.494 in. 438 STATICS AND MECHANICS OF MATERIALS, 2nd Edition Using the solutions for cases 1 and 4 in Table A-19: RILEY, STURGES AND MORRIS δE A = δE D δ EP + δ EM + δ ER = δ ER ( PL 2 ) L2 + RL3 = − R ( 2L ) PL3 − − 3EI AB 2 EI AB 3EI AB 48EI CD R= 3 7 (15.2 )(1600 ) 7 I CD P = = 2289 lb 2 I AB + 4 I CD 2 ( 6.79 ) + 4 (15.2 ) R = 2289 lb ↑ For beam AB : ↑ ΣFy = 0 : ΣM B = 0 : −VB + R − 1600 = 0 VB = +689 lb = 689 lb ↓ M B − R ( 3) + 1600 ( 4.5) = 0 M B = −333 lb ⋅ ft = 333 lb ⋅ ft From the shear force and bending moment diagrams: Vmax = 1600 lb M max = 2400 lb ⋅ ft QNA = yC A = 1.8535 ( 2.796 × 0.293) +0.8535 ( 0.326 ×1.707 ) = 1.9936 in 3 (a) σ max = M 2400 × 12 = = 8496 lb/in 2 ≅ 8.50 ksi (T top & C bottom) ..................... Ans. S 3.39 VQ 1600 (1.9936 ) = = 1441 lb/in 2 = 1441 psi (at neutral surface) ............... Ans. It 6.79 ( 0.326 ) R = 2289 lb ↓ RC = RD = 1122.5 lb ↑ (b) τ max = For beam AB : From symmetry From the shear force and bending moment diagrams: Vmax = 1144.5 lb M max = 3434 lb ⋅ ft QNA = yC A = 2.337 ( 3.284 × 0.326 ) +1.087 ( 0.494 × 2.174 ) = 3.669 in 3 (a) σ max = τ max = M 3434 × 12 = = 6767 lb/in 2 ≅ 6.77 ksi (T bottom & C top) ..................... Ans. S 6.09 VQ 1144.5 ( 3.669 ) = = 559 lb/in 2 = 559 psi (at neutral surface) ................ Ans. It 15.2 ( 0.494 ) (b) 439 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 9-108 A beam is loaded and supported as shown in Fig. P9-108. (a) Determine the reactions at supports A and C. (b) Draw complete shear force and bending moment diagrams for the beam. SOLUTION (a) Using the solutions for cases 6 and 8 in Table A-19 with M C = PL 2 and VC = P : θ A = θ A 4 P + θ AM A + θ AM C MA = − PL PL = 2 2 ( 4 P ) L2 − M A L + ( PL 2 ) L = 0 =− 16 EI 3EI 6 EI ...................................................................................................... Ans. Then, overall equilibrium gives: ΣM A = 0 : RC ( L ) − ( 4 P )( L 2 ) − P ( 3L 2 ) − M A = 0 RC = +3P = 3P ↑ ............................. Ans. ↑ ΣFy = 0 : VA − 4 P + RC − P = 0 VA = +2 P = 2 P ↑ .............................. Ans. 9-109 A beam is loaded and supported as shown in Fig. P9-109. (a) Determine the reactions at supports A, B, and C. (b) Draw complete shear force and bending moment diagrams for the beam. SOLUTION (a) Using the solutions for cases 6, 7, and 8 in Table A-19: θ A = θ Aw + θ AM A + θ ARB w ( 2 L ) M A ( 2 L ) RB ( 2 L ) =− − + =0 24 EI 3EI 16 EI 3 2 gives and −8M A + 3RB − 4wL2 = 0 δ B = δ Bw + δ BM A + δ BRB 5 ( w )( 2 L ) M A ( 2 L ) R ( 2L ) =− − +B =0 384 EI 16 EI 48EI 4 2 3 gives Solving simultaneously gives −6M A + 4 RB − 5wL2 = 0 RB = +8wL 8wL = ↑ ............................. Ans. 7 7 MA = − wL2 wL2 = 14 14 ........................... Ans. 440 STATICS AND MECHANICS OF MATERIALS, 2nd Edition Then, overall equilibrium gives: RILEY, STURGES AND MORRIS ΣM A = 0 : RC ( 2 L ) + RB ( L ) − w ( 2 L ) ( L ) − M A = 0 +11wL 11wL = ↑ .......................... Ans. 28 28 +13wL 13wL VA = = ↑ .......................... Ans. 28 28 RC = ↑ ΣFy = 0 : 9-110 VA − w ( 2 L ) + RB + RC = 0 A 3-m simply supported beam is loaded with a uniformly distributed load of 2.6 kN/m over its entire length. The beam is made of air-dried Douglas fir (E = 13 GPa) with an allowable flexural stress of 8 MPa and an allowable shearing stress of 0.7 MPa. The maximum deflection at the center of the span must not exceed 10 mm. Select the lightest standard structural timber that can be used for the beam. SOLUTION ΣM A = 0 : ΣM B = 0 : BL − wL ( L 2 ) = 0 wL ( L 2 ) − AL = 0 A = B = + wL 2 = ( wL 2 ) kN ↑ From the shear force and bending moment diagrams: Vmax = 3.9 kN M max = 2.925 kN ⋅ m The properties required by the three specifications are: σ max = τ max = M max : S 1.5Vmax : A S= A= M max 2.925 × 103 = = ( 365.6 × 10−6 ) m3 = ( 365.6 × 103 ) mm3 σ max 8 × 106 1.5Vmax 1.5 ( 3.9 ×10 = τ max 0.7 × 106 3 )= (8357 ×10 ) m = (8357 ) mm −6 2 2 δ max 5wL4 = : 384 EI 5 ( 2.6 × 103 ) ( 3) 5wL4 I= = 384 Eδ max 384 (13 × 109 ) ( 0.010 ) 4 = ( 21.09 ×10 −6 ) m 4 = ( 21.09 × 106 ) mm 4 Try a 51× 254 − mm timber with m = 6.38 kg/m I = 48.3 × 106 mm 4 A = 9880 mm 2 S = 400 × 103 mm 3 With the weight of the beam included: w = 2.6 × 103 + 6.38 ( 9.81) = ( 2.663 × 103 ) N/m = 2.663 kN/m Vmax = M max wL 2.663 ( 3) = = 3.995 kN 2 2 2 wL2 2.663 ( 3) = = = 2.996 kN ⋅ m 8 8 Then, the maximum normal stress, shear stress, and deflection are: 441 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS σ max = τ max δ max Therefore: 9-111 3 1.5Vmax 1.5 ( 3.995 × 10 ) = = = ( 0.607 × 106 ) N/m 2 < 0.7 MPa = τ all A 9880 × 10−6 4 M max 2.996 × 103 = = ( 7.490 ×106 ) N/m 2 < 8 MPa = σ all −6 S 400 × 10 5 ( 2.663 ×103 ) ( 3) 5wL4 = = = ( 4.47 × 10−3 ) m < 10 mm = δ all 384 EI 384 (13 × 109 )( 48.3 ×10 −6 ) Use a 51× 254 − mm timber section ....................................................................... Ans. A portion of a pedestrian walkway along the side of a bridge is shown in Fig. P9-111. Cantilever beams support the loading, one of which is shown. The beams are select structural eastern Hemlock (E = 1200 ksi) with allowable flexural and shearing stresses of 1300 psi and 80 psi, respectively. The allowable deflection is 0.2 in. Select the lightest standard structural timber that can be used for the beams. SOLUTION ↑ ΣFy = 0 : ΣM A = 0 : VA − 600 ( 4 ) − 500 = 0 VA = +2900 lb = 2900 lb ↑ M A − 600 ( 4 ) ( 2 ) − 500 ( 4 ) = 0 M A = +6800 lb ⋅ ft = 6800 lb ⋅ ft The properties required by the three specifications are: σ max = τ max = M max : S 1.5Vmax : A S= M max 6800 × 12 = = 62.77 in 3 1300 σ max A= 1.5Vmax 1.5 ( 2900 ) = = 54.38 in 2 τ max 80 3 4 δ max PL3 wL4 = + : 3EI 8EI 8PL3 + 3wL4 8 ( 500 )( 4 × 12 ) + 3 ( 600 12 )( 4 × 12 ) = = 215.0 in 4 I= 6 24 Eδ max 24 (1.2 × 10 ) ( 0.2 ) w = 15.6 lb/ft I = 264 in 4 A = 56.3 in 2 S = 70.3 in 3 Try an 8 × 8 − in. timber with With the weight of the beam included: w = 600 + 15.6 = 615.6 lb/ft Vmax = P + wL = 500 + 615.5 ( 4 ) = 2962 lb M max = PL + 615.6 ( 4 ) wL2 = 500 ( 4 ) + = 6925 lb ⋅ ft 2 2 2 Then, the maximum normal stress, shear stress, and deflection are: σ max = τ max M max 6925 (12 ) = = 1182 psi < 1300 psi = σ all S 70.3 1.5Vmax 1.5 ( 2962 ) = = = 78.9 psi < 80 psi = τ all A 56.3 442 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 3 RILEY, STURGES AND MORRIS 4 δ max Therefore: 9-112 500 ( 4 × 12 ) ( 615.6 12 )( 4 ×12 ) = 0.1656 in. < 0.2 in. = δ PL3 wL4 = + = + all 6 3EI 8 EI 3 (1.2 × 10 ) ( 264 ) 8 (1.2 × 106 ) ( 264 ) Use an 8 × 8 − in. timber section ............................................................................... Ans. A solid circular shaft made of ASTM A36 steel (E = 200 GPa) is supported by bearings spaced 1.5 m apart. The shaft is to support a 4-kN load perpendicular to the shaft; the load may be placed at any point between the bearings. The allowable flexural stress is 152 MPa, the allowable shearing stress is 100 MPa, and the allowable deflection is 5 mm. If shafts are available with diameters in increments of 5 mm, determine the smallest diameter shaft that can be used to support the load. Neglect the weight of the shaft. SOLUTION ΣM A = 0 : ↑ ΣFy = 0 : B (1.5 ) − 4 x = 0 A+ B−4 = 0 B = ( 2.667 x ) kN ↑ A = ( 4 − 2.667 x ) kN ↑ The maximum bending moment occurs at C. M C = Ax = ( 4 − 2.667 x ) x The biggest bending moment at C occurs when dM C dx = 4 − 5.333x = 0 x = 0.750 m M max = 1.500 kN ⋅ m Vmax = 4 kN ( when x ≅ 0 m or when x ≅ 1.5 m ) To satisfy the bending stress condition: σ max = 3 M max c M max c 4 M max =4= πc 4 π c3 I 3 4M max 4 (1.50 × 10 ) = = (12.565 × 10−6 ) m3 c= πσ max π (152 ×106 ) c = ( 23.25 × 10 −3 ) m = 23.25 mm d = 2c = 46.50 mm ≅ 50 mm Checking the maximum shear stress and the maximum deflection: 2 4r π r 2 4 ( 25 ) π ( 25 ) Q = yC A = = 10, 417 mm 3 = 3π 2 3π 2 π r 4 π ( 25) = = ( 306.8 × 103 ) mm 4 I= 4 4 4 τ max δ max Therefore: 3 −6 Vmax Q ( 4 ×10 )(10.417 ×10 ) = = = ( 2.716 ×106 ) N/m 2 < 100 MPa = τ all −9 It ( 306.8 ×10 ) ( 0.050 ) ( 4 ×103 ) (1.5) PL3 = = = ( 4.584 × 10−3 ) m < 5 mm = δ all −9 9 48 EI 48 ( 200 × 10 )( 306.8 × 10 ) 3 Use a 50 − mm diameter shaft ................................................................................. Ans. 443 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 9-113 RILEY, STURGES AND MORRIS A simply supported structural steel (E = 29,000 ksi) beam has a span of 24 ft and carries a uniformly distributed load of 1200 lb/ft. The beam has an allowable flexural stress of 24 ksi, an allowable shearing stress of 14 ksi, and an allowable deflection of 1/360 of the span. Select the lightest American standard (Sshape) beam that can be used to support the loading. SOLUTION ΣM A = 0 : ΣM B = 0 : B ( 24 ) − 1200 ( 24 ) (12 ) = 0 1200 ( 24 ) (12 ) − A ( 24 ) = 0 A = B = 14.400 × 103 lb = 14.40 kip Vmax = A = 14.40 kip M max = M x =12 = A (12 ) − 1200 (12 ) ( 6 ) = 86.40 ×103 lb ⋅ ft = 86.40 kip ⋅ ft δ max = ( 24 ×12 ) 360 = 0.800 in. The properties required by the three specifications are: σ max = τ max = M max : S Vmax : Aw S= M max 86.4 ×12 = = 43.2 in 3 24 σ max Vmax 14.40 = = 1.0286 in 2 14 τ max 4 Aw = δ max = 5wL4 : 384 EI I= 5 (1200 12 )( 24 × 12 ) 5wL4 = = 386.1 in 4 6 384 Eδ max 384 ( 29 × 10 ) ( 0.800 ) d = 15.00 in. I = 447 in 4 Try an S15 × 42.9 section with tw = 0.411 in. S = 59.6 in 3 With the weight of the beam included: Vmax = wL 2 = 1242.9 ( 24 ) 2 = (14.915 × 103 ) lb = 14.915 kip w = 1200 + 42.9 = 1242.9 lb/ft 2 M max = wL2 8 = 1242.9 ( 24 ) 8 = ( 89.49 ×103 ) lb ⋅ ft = 89.49 kip ⋅ ft Then, the maximum normal stress, shear stress, and deflection are: σ max = τ max M max 89.49 (12 ) = = 18.02 ksi < 24 ksi = σ all S 59.6 V 14.915 = max = = 2.42 ksi < 14 ksi = τ all Aw 15.00 × 0.411 5wL4 5 (1242.9 12 )( 24 × 12 ) = = = 0.716 in. < 0.800 in. = δ all 384 EI 384 ( 29 × 106 ) ( 447 ) 4 δ max Therefore: Use an S15 × 42.9 section .......................................................................................... Ans. 444 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 9-114 RILEY, STURGES AND MORRIS The simply supported structural steel (E = 200 GPa) beam shown in Fig. P9-114 has an allowable flexural stress of 165 MPa, an allowable shearing stress of 100 MPa, and an allowable deflection of 1/360 of the span. Select the lightest wide-flange beam that can be used to support the loading shown in the figure. SOLUTION ΣM A = 0 : ΣM E = 0 : E ( 9 ) − 70 (1.5 ) − 70 ( 4.5) − 70 ( 7.5) = 0 70 (1.5 ) + 70 ( 4.5) + 70 ( 7.5 ) − A ( 9 ) = 0 A = E = 105 kN ↑ From the shear force and bending moment diagrams: Vmax = 105 kN M max = 262.5 kN ⋅ m The properties required by the three specifications are: σ max = M max : S S= M max 262.5 ×103 = 165 × 106 σ max = (1591× 10−6 ) m3 = 1591 mm3 τ max = Vmax : Aw Aw = Vmax 105 × 103 = τ max 100 ×106 = (1050 ×10 −6 ) m 2 = 1050 mm 2 δ max = 2 Pb ( 3L2 − 4b 2 ) 48EI 48 Eδ max + 9 PL3 = = 0.02500 m = 25.00 mm 48EI 360 2 2 2 ( 70 × 103 ) (1.5 ) 3 ( 9 ) − 4 (1.5 ) + ( 70 × 103 ) ( 9 ) = 9 48 ( 200 × 10 ) ( 0.025 ) I= 2 Pb ( 3L2 − 4b 2 ) + PL3 = ( 417.4 ×10 −6 ) m 4 = ( 417.4 × 106 ) mm 4 Try a W 533 × 92 section with I = 554 × 106 mm 4 w = 92 ( 9.81) = 902.5 N/m S = 2080 × 103 mm 3 d = 533 mm tw = 10.2 mm With the weight of the beam included: Vmax = 105 ×103 + M max 902.5 ( 9 ) wL = 105 × 103 + = (109.06 × 103 ) N 2 2 2 3 902.5 ( 9 ) wL2 = 262.5 × 10 + = 262.5 × 103 + = ( 271.6 × 103 ) N ⋅ m 8 8 M max 271.6 × 103 = = = (130.58 × 106 ) N/m 2 < 165 MPa = σ all −6 S 2080 × 10 Then, the maximum normal stress, shear stress, and deflection are: σ max τ max Vmax 109.06 × 103 = = = ( 20.06 × 106 ) N/m 2 < 100 MPa = τ all Aw 0.533 × 0.0102 445 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS δ max = 2 Pb ( 3L2 − 4b 2 ) 48EI + PL3 5wL4 + 48EI 384 EI 2 2 2 ( 70 × 103 ) (1.5) 3 ( 9 ) − 4 (1.5 ) ( 70 ×103 ) ( 9 )3 + = 48 ( 200 × 109 )( 554 × 10−6 ) 48 ( 200 × 109 )( 554 × 10−6 ) + Therefore: 9-115 384 ( 200 × 10 5 ( 902.2 )( 9 ) 9 4 −6 )( 554 ×10 ) = (19.53 ×10−3 ) m < 25 mm = δ all Use a W 533 × 92 section ........................................................................................... Ans. The simply supported beam shown in Fig. P9-115 is made of air-dried Douglas fir (E = 1900 ksi) with an allowable flexural stress of 1900 psi, an allowable shearing stress of 85 psi, and an allowable deflection of 1/360 of the span. Select the lightest standard structural timber that can be used to support the loads shown in the figure. SOLUTION ΣM A = 0 : ΣM D = 0 : D ( 24 ) − 4200 ( 8) − 4200 (16 ) = 0 4200 ( 8 ) + 4200 (16 ) − A ( 24 ) = 0 A = D = 4200 lb ↑ Vmax = 4200 lb M max = 33, 600 lb ⋅ ft = 33.6 kip ⋅ ft The properties required by the three specifications are: σ max = τ max = M max : S 1.5Vmax : A S= A= = M max ( 33.6 × 10 ) × 12 = = 212.2 in 3 σ max 1900 3 1.5Vmax 1.5 ( 4200 ) = = 74.12 in 2 τ max 85 δ max = 2 Pb ( 3L2 − 4b 2 ) 48EI I= 48 Eδ max 24 × 12 = 0.800 in. 360 2 2 2 ( 4200 )( 8 × 12 ) 3 ( 24 × 12 ) − 4 ( 8 × 12 ) = 2343 in 4 = 3 48 (1900 × 10 ) ( 0.800 ) 2 Pb ( 3L2 − 4b 2 ) Try a 10 × 16 − in. timber with w = 40.9 lb/ft I = 2948 in 4 A = 147 in 2 S = 380 in 3 With the weight of the beam included: Vmax = 4200 + M max 40.9 ( 24 ) wL = 4200 + = 4691 lb 2 2 2 40.9 ( 24 ) wL2 = 33, 600 + = 33, 600 + = ( 36.54 × 103 ) lb ⋅ ft = 36.54 kip ⋅ ft 8 8 446 STATICS AND MECHANICS OF MATERIALS, 2nd Edition Then, the maximum normal stress, shear stress, and deflection are: 3 RILEY, STURGES AND MORRIS ( 36.54 ×10 ) (12 ) = 1154 psi < 1900 psi = σ M σ max = max = all 380 S 1.5Vmax 1.5 ( 4691) τ max = = = 47.9 psi < 85 psi = τ all A 147 δ max = 2 Pb ( 3L2 − 4b 2 ) 48EI + 5wL4 384 EI 2 2 2 ( 4200 )( 8 × 12 ) 3 ( 24 ×12 ) − 4 ( 8 ×12 ) 5 ( 40.9 12 )( 24 × 12 )4 + = 3 48 (1900 × 10 ) ( 2948 ) 384 (1900 × 103 ) ( 2948) = 0.690 in. < 0.800 in. = δ all Therefore: 9-116 Use a 10 × 16 − in. timber section ............................................................................. Ans. The boards for a concrete form are to be bent to a circular curve of 5-m radius. What maximum thickness can be used if the stress is not to exceed 15 MPa? The modulus of elasticity for the wood is 10 GPa. SOLUTION From Eq. 9-2: 1M = ρ EI or M= EI ρ σ max = Mc ( EI ρ ) c Ec E ( t 2 ) = = = I I ρ ρ 6 2σ max ρ 2 (15 × 10 ) ( 5 ) = = (15.00 × 10−3 ) m = 15.00 mm ................................... Ans. t= 9 E 10 × 10 9-117 During fabrication of a laminated timber arch, one of the 10-in wide × 1-in. thick Douglas fir (E = 1900 ksi) planks is bent to a radius of 12 ft. Determine the maximum flexural stress developed in the plank. SOLUTION From Eq. 9-2: 1M = ρ EI or M= EI ρ σ max = Mc ( EI ρ ) c Ec 1900 (1 2 ) = = = = 6.60 ksi (T & C) .................................. Ans. I I ρ 12 × 12 9-118 A beam is loaded and supported as shown in Fig. P9-118. Determine (a) The equation of the elastic curve, using the designated axes. (b) The slope at the right end of the beam. (c) The deflection at the right end of the beam. SOLUTION ↑ ΣFy = 0 : ΣM A = 0 : VA = wL 2 VA − wL 2 = 0 − M A − ( wL 2 )( 2 L 3) = 0 M A = − wL2 3 M ( x ) = VA x + M A − ( wx 2 2 L ) ( x 3) 447 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS EIy′′ = EIy′ = EIy = (a) (b) wLx wL2 wx 3 − − 2 3 6L wLx 2 wL2 x wx 4 − − + C1 4 3 24 L At x = 0, y′ = 0 : C1 = 0 wLx 3 wL2 x 2 wx5 − − + C1 x + C2 At x = 0, y = 0 : C2 = 0 12 6 120 L w y= ( − x5 + 10L2 x3 − 20L3 x2 ) ................................................................................ Ans. 120 EIL w y′ = ( − x4 + 6 L2 x2 − 8L3 x ) 24 EIL y′ = L = x w − wL3 wL3 − L4 + 6 L4 − 8L4 ) = = ( 24 EIL 8 EI 8 EI ........................................................ Ans. (c) 9-119 yx = L = w −11wL4 11wL4 − L5 + 10 L5 − 20 L5 ) = = ↓ .......................................... Ans. ( 120 EIL 120 EI 120 EI Select the lightest structural steel (E = 29,000 ksi) wide flange or American standard beam (Appendix A) that can be used for the beam shown in Fig. P9-119 if the maximum flexural stress must not exceed 10 ksi and the maximum deflection must not exceed 0.200 in. when L = 8 ft and w = 2000 lb/ft. SOLUTION ΣM C = 0 : ( wL )( L 2 ) − RA ( 2 L ) = 0 2 RA = wL 4 wLx w x − L EIy′′ = − 4 2 wLx 2 w x − L EIy′ = − 8 6 3 + C1 4 At x = 0, y = 0: y = 0: C2 = 0 −7 wL3 C1 = 48 wLx 3 w x − L EIy = − + C1 x + C2 24 24 w 4 y= 2 Lx 3 − 2 x − L − 7 L3 x 48 EI At x = 2 L, ( ) ) x = 1.0804 L The maximum deflection occurs when y′ = w 3 6 Lx 2 − 8 x − L − 7 L3 = 0 48 EIL ( δ max = yx =1.0804 L = I min Smin −0.1050wL4 0.1050 wL4 = ↓ EI EI 4 0.1050wL4 0.1050 ( 2000 12 )( 8 ×12 ) = = = 256.3 in 4 Eδ max 29 × 106 ) ( 0.200 ) ( 9wL2 32 9 ( 2000 12 )( 8 ×12 ) 32 M = max = = = 43.2 in 3 10 × 103 σ max σ max 4 448 STATICS AND MECHANICS OF MATERIALS, 2nd Edition The lightest beam satisfying both requirements is a 9-120 RILEY, STURGES AND MORRIS W16 × 40 wide flange section. ............................. Ans. A cantilever beam is fabricated by bolting two structural steel (E = 200 GPa) C127 × 10 channel sections together, as shown in Fig. P9-120. Determine (a) The deflection at the right end of the beam. (b) The maximum tensile flexural stress in the beam. SOLUTION For a C127 × 10 channel section: A = 1270 mm 2 x = 12.3 mm IYY = ( 0.199 × 106 ) mm 4 w f = 44.5 mm cBC = w f − x = 44.5 − 12.3 = 32.2 mm 2 I AB = 2 ( 0.199 ×106 ) + 1270 (12.3) = ( 0.782 × 106 ) mm 4 (a) Using the solutions for cases 1 and 4 in Table A-19: δ B = δ BP + δ BM = − PL3AB ML2 AB − 3EI AB 2 EI AB 3 2 500 (1) (1) =− − 9 9 −6 3 ( 200 ×10 )( 0.782 ×10 ) 2 ( 200 ×10 )( 0.782 ×10−6 ) 500 (1) = ( −2.664 × 10−3 ) m = 2.664 mm ↓ θ B = θ BP + θ BM =− PL2 MLAB AB =− − 2 EI AB EI AB 500 (1) 9 2 −6 2 ( 200 ×10 = ( −4.795 ×10−3 )( 0.782 ×10 ) ) rad = ( 4.795 ×10 ) rad −3 − 500 (1) (1) 9 ( 200 ×10 )( 0.782 ×10−6 ) δ C = δ B + θ B LBC + δ C / B = δ B + θ B LBC − = −2.664 ×10 − ( 4.795 ×10 −3 −3 PL3 BC 3EI BC 500 (1) 9 3 −6 ) (1) − 3 200 ×10 0.199 ×10 ( )( ) = ( −11.65 × 10−3 ) m = 11.65 mm ↓ ........................................................................... Ans. (b) σA = − σB = − ( −500 × 2 )( 0.0445 ) = +56.9 ×106 N/m 2 M Ac A =− ( ) I AB 0.782 ×10−6 ( −500 ×1)( 0.0322 ) = +80.9 ×106 N/m 2 M B cB =− ( ) I BC 0.199 ×10−6 σ max = σ B = +80.9 ×106 N/m 2 = 80.9 MPa (T) ............................................................. Ans. 449 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 9-121 RILEY, STURGES AND MORRIS An aluminum beam (E = 10,000 ksi and I = 48 in.4) is loaded and supported as shown in Fig. P9-121. Determine (a) The deflection at the left end of the beam. (b) The deflection at the middle of span BD. SOLUTION ΣM D = 0 : ΣM B = 0 : 2200 (13) − RB (10 ) − 15, 000 = 0 2200 ( 3) − RD (10 ) − 15, 000 = 0 RB = 1360 lb ↑ RD = 840 lb ↑ With the origin of coordinates at the right end of the beam, 1 0 1 EIy′′ = 840 x − 0 − 15, 000 x − 5 + 1360 x − 10 lb ⋅ ft 2 1 2 EIy′ = 420 x − 0 − 15, 000 x − 5 + 680 x − 10 + C1 lb ⋅ ft 2 3 2 3 EIy = 140 x − 0 − 7500 x − 5 + 226.67 x − 10 + C1 x + C2 lb ⋅ ft 3 Boundary conditions: At At x = 0 ft, x = 10 ft, y = 0: y = 0: C2 = 0 C1 = 4750 lb ⋅ ft 2 3 2 3 EIy = 140 x − 0 − 7500 x − 5 + 226.67 x − 10 + 4750 x lb ⋅ ft 3 (a) At A ( x = 13 ft ) 140 (13)3 − 7500 ( 8 ) 2 + 226.67 ( 3)3 + 4750 (13) (10 ×10 ) ( 48) 6 δA = (12 ) 3 = −0.376 in. = 0.376 in. ↓ ............................................................................................ Ans. (b) At C ( x = 5 ft ) 140 ( 5 )3 − 0 + 0 + 4750 ( 5 ) (10 ×10 ) ( 48) 6 δC = (12 ) 3 = +0.1485 in. = 0.1485 in. ↑ ........................................................................................ Ans. 9-122 The steel (E = 200 GPa) beam AB of Fig. P9-122 is fixed at ends A and B and supported at the center by the pin connected timber (E = 10 GPa) struts CD and CE. The cross-sectional area of each strut is 6400 mm2, and the second moment of the cross-sectional area of the beam with respect to its neutral axis is 25(106) mm4. Determine the force in each strut after the 6-kN/m distributed load is applied to the beam. SOLUTION As a result of symmetry, for a beam with fixed ends and a uniformly distributed load w: VA = VB = wL 2 MA = MB = M Using the solutions for cases 6, 7, and 8 in Table A-19: θ A = θw +θMA +θMB = − wL3 ML ML − − =0 24 EI 3EI 6 EI M= − wL2 12 450 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS δw = − 5wL4 ML2 ML2 5wL4 wL4 wL4 wL4 + + =− + + =− 384 EI 16 EI 16 EI 384 EI 192 EI 192 EI 384 EI As a result of symmetry, for a beam with fixed ends and a concentrated load P at the center: VA = VB = P 2 MA = MB = M − PL 8 = ( −2.664 × 10−3 ) m = 2.664 mm ↓ θ A = θP +θMA +θMB PL2 ML ML =− − − =0 16 EI 3EI 6 EI M= PL3 ML2 ML2 PL3 PL3 PL3 PL3 δP = − + + =− + + =− 48EI 16 EI 16 EI 48 EI 128 EI 128 EI 192 EI For the system: RC = ( 3 5 ) Pstrut + ( 3 5) Pstrut = ( 6 5 ) Pstrut = ( 6 5 ) Ps δ beam = δ Cw + δ CRB = ( 5 3) δ strut wL4 RC L3 5 PL − + =− s s 384 EI 192 EI 3 As Es RC L3 5 Ps Ls wL4 + = 192 EI 3 As Es 384 EI Ps ( 2.5 ) ( 6 ×103 ) ( 4 )4 5 = + 192 ( 200 × 109 )( 25 × 10−6 ) 3 ( 6.4 × 10−3 )(10 × 109 ) 384 ( 200 ×109 )( 25 ×10−6 ) ( 6 Ps 5)( 4 ) 3 Ps = ( 5.51×103 ) N = 5.51 kN ............................................................................................ Ans. 9-123 In Fig. P9-123, beam AC is made of brass and beam BC is made of steel. The second moment of the crosssectional area of beam BC with respect to its neutral axis is twice that of beam AC, and the modulus of elasticity of the steel is twice that of the brass. The reaction at C is zero before the load w is applied. Determine (a) The reaction at C on beam BC. (b) The reactions at supports A and B. SOLUTION Using the solutions for cases 1 and 2 in Table A-19 with I BC = 2 I AC and EBC = 2 E AC : δC / A = δC / B δ CRC = δ Cw + δ CRC − RC ( L 2 ) 3EI 3 wL4 RC L3 =− + 8 ( 2 E )( 2 I ) 3 ( 2 E )( 2 I ) RC = + wL wL = ↑ ..........................................Ans. 4 4 451 STATICS AND MECHANICS OF MATERIALS, 2nd Edition For beam RILEY, STURGES AND MORRIS AC : ↑ ΣFy = 0 : ΣM A = 0 : VA − RC = 0 − M A − RC ( L 2 ) = 0 ............................................................................................. Ans. VA = + wL 4 = wL 4 ↑ ................................................................................................... Ans. M A = − wL2 8 = wL2 8 For beam BC : ↑ ΣFy = 0 : ΣM B = 0 : RC − wL − VB = 0 M B + ( wL )( L 2 ) − RC L = 0 ............................................................................................ Ans. VB = −3wL 4 = 3wL 4 ↑ ............................................................................................... Ans. M B = − wL2 4 = wL2 4 9-124 A beam is loaded and supported as shown in Fig. P9-124. Determine (a) The reactions at supports A and D. (b) The deflection at B if E = 200 GPa and I = 350(106) mm4. SOLUTION (a) 1 2 EIy′′ = RA x − 36 x − 3 − 3 x − 6 kN ⋅ m R x2 2 3 EIy′ = A − 18 x − 3 − x − 6 + C1 kN ⋅ m 2 2 R x3 x−6 3 EIy = A − 6 x − 3 − 4 6 Boundary conditions: At At At 4 + C1 x + C2 kN ⋅ m3 x = 0 m, y = 0: y′ = 0 : y = 0: C2 = 0 72 RA − 1674 + C1 = 0 288RA − 4698 + 12C1 = 0 and x = 12 m, x = 12 m, Solving simultaneously gives C1 = −249.75 N ⋅ m 2 RA = +26.72 kN ≅ 26.7 kN ↑ ............................................................................................. Ans. Then, overall equilibrium gives: ↑ ΣFy = 0 : ΣM D = 0 : RA − 36 − 6 ( 6 ) − VD = 0 36 ( 9 ) + 6 ( 6 ) ( 3) − RA (12 ) − M D = 0 ......................................................................... Ans. VD = −45.28 kN ≅ 45.3 kN ↑ ............................................................................................. Ans. M D = −111.36 kN ⋅ m ≅ 111.4 kN ⋅ m (b) EIy = 1 3 4 53.44 x3 − 72 x − 3 − 3 x − 6 − 2997 x kN ⋅ m3 12 ( ) y x =3 629 629 ×103 =− =− = −8.99 × 10−3 m = 8.99 mm ↓ ............... Ans. −6 9 EI ( 200 ×10 )( 350 ×10 ) 452 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 9-125 A beam is loaded and supported as shown in Fig. P9-125. Determine (a) The reactions at supports A, B, and D. (b) The deflection at C if E = 30,000 ksi and I = 26 in.4 SOLUTION RILEY, STURGES AND MORRIS (a) 1 2 1 1 EIy′′ = RA x − 600 x 2 + RB x − 4 + 600 x − 4 − 3000 x − 6 + RD x − 10 lb ⋅ ft R x2 R x−4 EIy′ = A − 200 x 3 + B 2 2 2 R x − 10 + 200 x − 4 − 1500 x − 6 + D 2 3 2 2 + C1 lb ⋅ ft 2 3 3 R x3 RB x − 4 RD x − 10 4 3 4 A EIy = − 50 x + + 50 x − 4 − 500 x − 6 + + C1 x + C2 lb ⋅ ft 3 6 6 6 Boundary conditions: At At At x = 0 ft, x = 4 ft, x = 10 ft, y = 0: y = 0: y = 0: C2 = 0 ( 32 RA 3) − 12,800 + 4C1 = 0 ( 500 RA 3) − 467, 200 + 36 RB + 10C1 = 0 RA + RB + RD − 1200 ( 4 ) − 3000 − 2000 = 0 From equilibrium: ↑ ΣFy = 0 : ΣM D = 0 : Solving simultaneously gives 1200 ( 4 ) ( 8 ) + 3000 ( 4 ) − 2000 ( 2 ) − RA (10 ) − RB ( 6 ) = 0 C1 = −2027 lb ⋅ ft 2 and RA = 1960 lb ↑ .................... RB = 4467 lb ↑ ................... RD = 3373 lb ↑ ................... Ans. (b) EIy = 1 3 1960 x3 − 300 x 4 + 4467 x − 4 + 300 x − 4 6 3 3 4 −3000 x − 6 + 3373 x − 10 − 12,162 x lb ⋅ ft 3 yx =6 354 (12 ) 354 = = = +0.000784 in. = 0.000784 in. ↑ ............................... Ans. EI ( 30 ×106 ) ( 26 ) 3 453 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 9-126 A beam is loaded and supported as shown in Fig. P9-126. Determine (a) The reactions at supports A, B, and C. (b) The deflection midway between supports B and C. SOLUTION (a) RILEY, STURGES AND MORRIS EIy′′ = RA x − wx 2 + RB x − L 2 1 R x 2 wx 3 RB x − L EIy′ = A − + 2 6 2 RA x 3 wx 4 RB x − L EIy = − + 6 24 6 Boundary conditions: At At At From equilibrium: 2 + C1 3 + C1 x + C2 C2 = 0 4 RA L2 − wL3 + 24 LC1 = 0 108RA L2 − 81wL3 + 32 RB L2 + 72 LC1 = 0 RA + RB + RC − w ( 3L ) = 0 w ( 3L ) ( 3L 2 ) − RA ( 3L ) − RB ( 2 L ) = 0 x = 0, x = L, y = 0: y = 0: y = 0: x = 3 L, ↑ ΣFy = 0 : ΣM D = 0 : Solving simultaneously gives C1 = wL3 48 and + wL wL = ↑ ................................................................................................................. Ans. 8 8 +33wL 33wL RB = = ↑ ....................................................................................................... Ans. 16 16 +13wL 13wL RC = = ↑ ........................................................................................................ Ans. 16 16 RA = (b) y= w 3 2 Lx3 − 4 x 4 + 33L x − L + 2 L3 x 96 EI w yx =2 L = (16 L4 − 64 L4 + 33L4 + 4 L4 ) 96 EI ( ) = −11wL4 11wL4 = ↓ ................................................................................................. Ans. 96 EI 96 EI 454 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS Chapter 10 10-1 At a point in a thin plate, there are normal stresses of 20 ksi (T) on a vertical plane and 10 ksi (C) on a horizontal plane, as shown in Fig. P10-1. Determine the normal and shear stresses at this point on the inclined plane AB shown in the figure. SOLUTION ΣFn = 0 : σ dA + (10dA cos 64° )( cos 64° ) − ( 20dA sin 64° )( sin 64° ) = 0 σ = +14.23 ksi = 14.23 ksi (T) ................................. Ans. ΣFt = 0 : τ dA − (10dA cos 64° )( sin 64° ) − ( 20dA sin 64° )( cos 64° ) = 0 τ = +11.82 ksi ................................................................................................................. Ans. 10-2 At a point in a stressed body, there are normal stresses of 95 MPa (T) on a vertical plane and 125 MPa (T) on a horizontal plane, as shown in Fig. P10-2. Determine the normal and shear stresses at this point on the inclined plane AB shown in the figure. SOLUTION ΣFn = 0 : σ dA − ( 95dA sin 20° )( sin 20° ) − (125dA cos 20° )( cos 20° ) = 0 σ = +121.5 MPa = 121.5 MPa (T) .......................... Ans. ΣFt = 0 : τ dA + ( 95dA sin 20° )( cos 20° ) − (125dA cos 20° )( sin 20° ) = 0 τ = +9.64 MPa ................................................................................................................ Ans. 10-3 The stresses shown in Fig. P10-3 act at a point on the surface of a circular shaft which is subjected to a twisting moment M as shown. Determine the normal and shear stresses at this point on the inclined plane AB shown in the figure. SOLUTION ΣFn = 0 : σ dA − (15dA cos 60° )( sin 60° ) − (15dA sin 60° )( cos 60° ) = 0 σ = +12.99 ksi = 12.99 ksi (T) ................................. Ans. ΣFt = 0 : τ dA + (15dA cos 60° )( cos 60° ) − (15dA sin 60° )( sin 60° ) = 0 ................................................................................................................... Ans. τ = +7.50 ksi 10-4 The stresses shown in Fig. P10-4 act at a point on the surface of a cantilever beam. Determine the normal and shear stresses at this point on the inclined plane AB shown in the figure. SOLUTION 455 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ΣFn = 0 : σ dA + (170dA sin 55° )( sin 55° ) − ( 70dA sin 55° )( cos 55° ) − ( 70dA cos 55° )( sin 55° ) = 0 σ = −48.3 MPa = 48.3 MPa (C) .............................. Ans. ΣFt = 0 : τ dA − (170dA sin 55° )( cos 55° ) − ( 70dA sin 55° )( sin 55° ) + ( 70dA cos 55° )( cos 55° ) = 0 ............................................................................................................. Ans. τ = +103.8 MPa 10-5 The stresses shown in Fig. P10-5 act at a point on the surface of a thin-walled pressure vessel that is subjected to an internal pressure, an axial load P, and a torque T. Determine the normal and shear stresses at this point on the inclined plane AB shown in the figure. SOLUTION ΣFn = 0 : σ dA − (18dA sin 65° )( sin 65° ) − ( 6dA cos 65° )( cos 65° ) + (15dA sin 65° )( cos 65° ) + (15dA cos 65° )( sin 65° ) = 0 σ = +4.37 ksi = 4.37 ksi (T) ..................................... Ans. ΣFt = 0 : τ dA − (18dA sin 65° )( cos 65° ) + ( 6dA cos 65° )( sin 65° ) − (15dA sin 65° )( sin 65° ) + (15dA cos 65° )( cos 65° ) = 0 τ = +14.24 ksi ................................................................................................................. Ans. 10-6 The stresses shown in Fig. P10-6 act at a point in a stressed body. Determine the normal and shear stresses at this point on the inclined plane AB shown in the figure. SOLUTION ΣFn = 0 : σ dA + ( 65dA sin 55° )( sin 55° ) + (125dA cos 55° )( cos 55° ) + ( 75dA sin 55° )( cos 55° ) + ( 75dA cos 55° )( sin 55° ) = 0 σ = −155.2 MPa = 155.2 MPa (C) .......................... Ans. ΣFt = 0 : τ dA + ( 65dA sin 55° )( cos 55° ) − (125dA cos 55° )( sin 55° ) − ( 75dA sin 55° )( sin 55° ) + ( 75dA cos 55° )( cos 55° ) = 0 τ = +53.8 MPa ................................................................................................................ Ans. 10-7 At a point in a machine component, the normal and shear stresses on an inclined plane are 4800 psi (T) and 1500 psi, respectively, as shown in Fig. P10-7. The normal stress on a vertical plane through the point is zero. Determine (a) The shear stresses on horizontal and vertical planes. (b) The normal stress on a horizontal plane. 456 STATICS AND MECHANICS OF MATERIALS, 2nd Edition SOLUTION RILEY, STURGES AND MORRIS sin θ = 3 5 → ΣFx = 0 : (a) cos θ = 4 5 + (1500dA ) cos θ = 0 τ xy ( dA cosθ ) + ( 4800dA ) sin θ τ xy = −5100 psi = −5.10 ksi ....................................... Ans. ↑ ΣFy = 0 : (b) 10-8 σ y ( dA cosθ ) − ( 4800dA ) cos θ + (1500dA ) sin θ − τ xy ( dA sin θ ) = 0 σ y = −150.0 psi = 150.0 psi (C) ................................................................................. Ans. The stresses on horizontal and vertical planes at a point on the outside surface of a solid circular bar subjected to an axial load P and a torsional load T are shown in Fig. P10-8. The normal stress on the inclined plane AB is 15 MPa (T). Determine (a) The normal stress σx on the vertical plane. (b) The magnitude and direction of the shear stress on the inclined plane AB. SOLUTION sin θ = 5 13 ΣFn = 0 : cos θ = 12 13 15dA − (σ x dA sin θ )( sin θ ) + ( 25dA sin θ )( cos θ ) + ( 25dA cosθ )( sin θ ) = 0 (a) σ x = +221.4 MPa = 221.4 MPa (T) ........................ Ans. ΣFt = 0 : (b) 10-9 τ = +61.0 MPa (as shown on sketch) ....................................................................... Ans. The thin-walled cylindrical pressure vessel shown in Fig. P10-9 was constructed by wrapping a thin steel plate into a helix that forms an angle θ = 35° with respect to a transverse plane through the cylinder and butt-welding the resulting seam. In a thin-walled cylindrical pressure vessel, the normal stress σy on a horizontal plane through a point on the surface of the vessel is twice as large as the normal stress σx on a vertical plane through the point, and the shear stresses on both the horizontal and vertical planes are zero. If the stresses in the weld material on the plane of the weld must be limited to 10 ksi in tension and 7 ksi in shear, determine the maximum normal stress σx permitted in the vessel. SOLUTION τ dA − (σ x dA sin θ )( cosθ ) − ( 25dA sin θ )( sin θ ) + ( 25dA cos θ )( cosθ ) = 0 For σ = 10 ksi (T): ΣFn = 0 : 10dA − (σ x dA cos 35° )( cos 35° ) − ( 2σ x dA sin 35° )( sin 35° ) = 0 σ x = +7.52 ksi For τ = 7 ksi: ΣFt = 0 : 7 dA + (σ x dA cos 35° )( sin 35° ) − ( 2σ x dA sin 35° )( cos 35° ) = 0 ........................................................................................ Ans. σ x = +14.90 ksi Therefore: σ x ( max ) = 7.52 ksi 457 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 10-10 RILEY, STURGES AND MORRIS The thin-walled cylindrical pressure vessel shown in Fig. P10-9 was constructed by wrapping a thin steel plate into a helix that forms an angle θ with respect to a transverse plane through the cylinder and buttwelding the resulting seam. The normal stresses σx and σy on vertical and horizontal planes through a point on the surface of the vessel are 50 MPa and 100 MPa, respectively, and the shear stresses are zero. Prepare a plot showing the variation of normal and shear stresses on the plane of the weld as the angle θ is varied from 0° to 180°. SOLUTION ΣFn = 0 : σ n dA − ( 50dA cosθ )( cosθ ) − (100dA sin θ )( sin θ ) = 0 σ n = ( 50 cos 2 θ + 100 sin 2 θ ) MPa .................................................................................... Ans. ΣFt = 0 : τ nt dA + ( 50dA cosθ )( sin θ ) − (100dA sin θ )( cos θ ) = 0 τ nt = ( 50sin θ cosθ ) MPa .................... Ans. 10-11 In a solid circular shaft subjected to a torsional form of loading, the normal stresses σx and σy on vertical and horizontal planes through a point on the surface of the vessel are zero and the shear stresses are 15 ksi, as shown in Fig. P10-11. Prepare a plot showing the variation of normal and shear stresses on plane AA through the point as the angle θ is varied from 0° to 180°. SOLUTION ΣFn = 0 : ΣFt = 0 : σ n dA − (15dA cosθ )( sin θ ) − (15dA sin θ )( cos θ ) = 0 .................................................................. Ans. σ n = ( 30sin θ cosθ ) ksi τ nt dA − (15dA cos θ )( cosθ ) + (15dA sin θ )( sin θ ) = 0 ............................................................................................... Ans. τ nt = 15 ( cos 2 θ − sin 2 θ ) ksi 458 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 10-12 RILEY, STURGES AND MORRIS Use the stress transformation equations for plane stress (Eqs. 10-1 and 10-2) to solve for the normal and shear stresses on the inclined plane AB at the point shown in Fig. P10-12. SOLUTION The given values are: σ x = +20 MPa σ y = +80 MPa τ xy = 0 MPa θ = tan −1 ( 3 4 ) = 36.870° σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cosθ = 20 cos 2 ( 36.870° ) + 80sin 2 ( 36.870° ) + 2 ( 0 ) sin ( 36.870° ) cos ( 36.870° ) = +41.6 MPa = 41.6 MPa (T) ...................................................................................... Ans. τ nt = − (σ x − σ y ) sin θ cosθ + τ xy ( cos 2 θ − sin 2 θ ) = − ( 20 − 80 ) sin ( 36.870° ) cos ( 36.870° ) + ( 0 ) cos 2 ( 36.870° ) − sin 2 ( 36.870° ) = +28.8 MPa ..................................................................................................................... Ans. 10-13 Use the stress transformation equations for plane stress (Eqs. 10-1 and 10-2) to solve for the normal and shear stresses on the inclined plane AB at the point shown in Fig. P10-13. SOLUTION The given values are: σ x = +25 ksi σ y = +7 ksi τ xy = +12 ksi θ = 90° − 60° = 30° σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ = 25cos 2 ( 30° ) + 7 sin 2 ( 30° ) + 2 (12 ) sin ( 30° ) cos ( 30° ) = +30.9 ksi = 30.9 ksi (T) ............................................................................................. Ans. τ nt = − (σ x − σ y ) sin θ cosθ + τ xy ( cos 2 θ − sin 2 θ ) = − ( 25 − 7 ) sin ( 30° ) cos ( 30° ) + 12 cos 2 ( 30° ) − sin 2 ( 30° ) = −1.794 ksi ...................................................................................................................... Ans. 10-14 Use the stress transformation equations for plane stress (Eqs. 10-1 and 10-2) to solve for the normal and shear stresses on the inclined plane AB at the point shown in Fig. P10-14. SOLUTION The given values are: σ x = −65 MPa σ y = −125 MPa τ xy = +75 MPa θ = 55° − 90° = −35° σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cosθ = ( −65 ) cos 2 ( −35° ) + ( −125) sin 2 ( −35° ) + 2 ( 75 ) sin ( −35° ) cos ( −35° ) = −155.2 MPa = 155.2 MPa (C) .................................................................................. Ans. τ nt = − (σ x − σ y ) sin θ cosθ + τ xy ( cos 2 θ − sin 2 θ ) = − ( −65 ) − ( −125 ) sin ( −35° ) cos ( −35° ) + 75 cos 2 ( −35° ) − sin 2 ( −35° ) = +53.8 MPa ..................................................................................................................... Ans. 459 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 10-15 RILEY, STURGES AND MORRIS Use the stress transformation equations for plane stress (Eqs. 10-1 and 10-2) to solve for the normal and shear stresses on the inclined plane AB at the point shown in Fig. P10-15. SOLUTION The given values are: σ x = +25 ksi σ y = +12 ksi τ xy = −10 ksi θ = 90° − 52° = 38° σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ = 25cos 2 ( 38° ) + 12sin 2 ( 38° ) + 2 ( −10 ) sin ( 38° ) cos ( 38° ) = +10.37 ksi = 10.37 ksi (T) ......................................................................................... Ans. τ nt = − (σ x − σ y ) sin θ cosθ + τ xy ( cos 2 θ − sin 2 θ ) = − ( 25 − 12 ) sin ( 38° ) cos ( 38° ) + ( −10 ) cos 2 ( 38° ) − sin 2 ( 38° ) = −8.73 ksi ......................................................................................................................... Ans. 10-16 Use the stress transformation equations for plane stress (Eqs. 10-1 and 10-2) to solve for the normal and shear stresses on the inclined plane AB at the point shown in Fig. P10-16. SOLUTION The given values are: σ x = +20 MPa σ y = +120 MPa τ xy = −80 MPa θ = −64° σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cosθ = 20 cos 2 ( −64° ) + 120 sin 2 ( −64° ) + 2 ( −80 ) sin ( −64° ) cos ( −64° ) = +163.8 MPa = 163.8 MPa (T) .................................................................................. Ans. τ nt = − (σ x − σ y ) sin θ cosθ + τ xy ( cos 2 θ − sin 2 θ ) = − ( 20 − 120 ) sin ( −64° ) cos ( −64° ) + ( −80 ) cos 2 ( −64° ) − sin 2 ( −64° ) = +9.85 MPa ..................................................................................................................... Ans. 10-17 The stresses shown in Fig. P10-17a act at a point on the free surface of a stressed body. Determine (a) The normal and shear stresses at this point on the inclined plane AB shown in the figure. (b) The normal stresses σn and σt and the shear stress τnt at this point if they act on the rotated stress element shown in Fig. P10-17b. SOLUTION (a) The given values are: σ x = +18 ksi σ y = +13 ksi τ xy = +6 ksi θ = 71° − 90° = −19° σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cosθ = 18cos 2 ( −19° ) + 13sin 2 ( −19° ) + 2 ( 6 ) sin ( −19° ) cos ( −19° ) = +13.78 ksi = 13.78 ksi (T) ......................................................................................... Ans. τ nt = − (σ x − σ y ) sin θ cosθ + τ xy ( cos 2 θ − sin 2 θ ) = − (18 − 13) sin ( −19° ) cos ( −19° ) + 6 cos 2 ( −19° ) − sin 2 ( −19° ) = +6.27 ksi ......................................................................................................................... Ans. 460 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (b) The given values are: RILEY, STURGES AND MORRIS σ x = +18 ksi σ y = +13 ksi τ xy = +6 ksi θ n = 26° θ t = 116° σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cosθ = 18cos 2 ( 26° ) + 13sin 2 ( 26° ) + 2 ( 6 ) sin ( 26° ) cos ( 26° ) = +21.8 ksi = 21.8 ksi (T) ............................................................................................. Ans. σ t = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cosθ = 18cos 2 (116° ) + 13sin 2 (116° ) + 2 ( 6 ) sin (116° ) cos (116° ) = +9.23 ksi = 9.23 ksi (T) .............................................................................................. Ans. τ nt = − (σ x − σ y ) sin θ cosθ + τ xy ( cos 2 θ − sin 2 θ ) = − (18 − 13) sin ( 26° ) cos ( 26° ) + 6 cos 2 ( 26° ) − sin 2 ( 26° ) = +1.724 ksi ...................................................................................................................... Ans. 10-18 The stresses shown in Fig. P10-18a act at a point on the free surface of a stressed body. Determine (a) The normal and shear stresses at this point on the inclined plane AB shown in the figure. (b) The normal stresses σn and σt and the shear stress τnt at this point if they act on the rotated stress element shown in Fig. P10-18b. SOLUTION The given values are: (a) σ x = −10 MPa σ y = −70 MPa τ xy = +40 MPa θ = 62° − 90° = −28° σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cosθ = ( −10 ) cos 2 ( −28° ) + ( −70 ) sin 2 ( −28° ) + 2 ( 40 ) sin ( −28° ) cos ( −28° ) = −56.4 MPa = 56.4 MPa (C) ...................................................................................... Ans. τ nt = − (σ x − σ y ) sin θ cosθ + τ xy ( cos 2 θ − sin 2 θ ) = +47.2 MPa ..................................................................................................................... Ans. (b) = − ( −10 ) − ( −70 ) sin ( −28° ) cos ( −28° ) + 40 cos 2 ( −28° ) − sin 2 ( −28° ) θ n = 15° θ t = 105° σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cosθ = ( −10 ) cos 2 (15° ) + ( −70 ) sin 2 (15° ) + 2 ( 40 ) sin (15° ) cos (15° ) = +5.98 MPa = 5.98 MPa (T) ....................................................................................... Ans. σ t = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cosθ = ( −10 ) cos 2 (105° ) + ( −70 ) sin 2 (105° ) + 2 ( 40 ) sin (105° ) cos (105° ) = −86.0 MPa = 86.0 MPa (C) ...................................................................................... Ans. τ nt = − (σ x − σ y ) sin θ cosθ + τ xy ( cos 2 θ − sin 2 θ ) = − ( −10 ) − ( −70 ) sin (15° ) cos (15° ) + 40 cos 2 (15° ) − sin 2 (15° ) = +19.64 MPa ................................................................................................................... Ans. 461 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 10-19 RILEY, STURGES AND MORRIS The stresses shown in Fig. P10-19 act at a point in a structural member. Determine the normal stresses σn and σt and the shear stress τnt at this point. SOLUTION The given values are: σ x = +18 ksi σ y = +10 ksi τ xy = +9 ksi θ n = 15° θ t = 105° σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cos θ = 18cos 2 (15° ) + 10sin 2 (15° ) + 2 ( 9 ) sin (15° ) cos (15° ) = +22.0 ksi = 22.0 ksi (T) ............................................................................................. Ans. σ t = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cosθ = 18cos 2 (105° ) + 10 sin 2 (105° ) + 2 ( 9 ) sin (105° ) cos (105° ) = +6.04 ksi = 6.04 ksi (T) ............................................................................................. Ans. τ nt = − (σ x − σ y ) sin θ cosθ + τ xy ( cos 2 θ − sin 2 θ ) = +5.79 ksi ......................................................................................................................... Ans. The stresses shown in Fig. P10-20 act at a point in a structural member. Determine the normal stresses σn and σt and the shear stress τnt at this point. SOLUTION The given values are: 10-20 = − (18 − 10 ) sin (15° ) cos (15° ) + 9 cos 2 (15° ) − sin 2 (15° ) σ x = +170 MPa 2 σ y = +50 MPa 2 τ xy = +30 MPa θ n = −19° θ t = 71° σ n = σ x cos θ + σ y sin θ + 2τ xy sin θ cosθ = 170 cos 2 ( −19° ) + 50sin 2 ( −19° ) + 2 ( 30 ) sin ( −19° ) cos ( −19° ) = +138.8 MPa = 138.8 MPa (T) .................................................................................. Ans. σ t = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cosθ = 170 cos 2 ( 71° ) + 50sin 2 ( 71° ) + 2 ( 30 ) sin ( 71° ) cos ( 71° ) = +81.2 MPa = 81.2 MPa (T) ...................................................................................... Ans. τ nt = − (σ x − σ y ) sin θ cosθ + τ xy ( cos 2 θ − sin 2 θ ) = +60.6 MPa ..................................................................................................................... Ans. 10-21 The stresses on horizontal and vertical planes at a point on the outside surface of a solid circular shaft subjected to an axial load P and a torque T are shown in Fig. P10-21. The normal stress on plane AA at this point is 8000 psi (T). Determine (a) The magnitude of the shearing stresses τh and τv. (b) The magnitude and direction of the shearing stress on the inclined plane AA. = − (170 − 50 ) sin ( −19° ) cos ( −19° ) + 30 cos 2 ( −19° ) − sin 2 ( −19° ) SOLUTION (a) The given values are: σ x = +8 ksi σ y = 0 ksi σ n = +8 ksi θ = − tan −1 ( 4 3) = −53.130° σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cosθ 462 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 8 = 8cos 2 ( −53.130° ) + ( 0 ) sin 2 ( −53.130° ) + 2τ v sin ( −53.130° ) cos ( −53.130° ) τ v = −5.333 ksi τ v = τ h ≅ 5.33 ksi ................................................................................................................ Ans. (b) τ nt = − (σ x − σ y ) sin θ cosθ + τ xy ( cos 2 θ − sin 2 θ ) = − ( 8 − 0 ) sin ( −53.130° ) cos ( −53.130° ) + τ v cos 2 ( −53.130° ) − sin 2 ( −53.130° ) = +5.33 ksi ......................................................................................................................... Ans. 10-22 The stresses on horizontal and vertical planes at a point are shown in Fig. P10-22. The normal stress on plane AB is 15 MPa (T). Determine (a) The normal stress σx on the vertical plane. (b) The magnitude and direction of the shearing stress on the inclined plane AB. SOLUTION The given values are: σ y = 0 MPa (a) σ n = +15 MPa τ xy = +25 MPa θ = − tan −1 (12 5 ) = −67.38° σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cosθ 15 = σ x cos 2 ( −67.38° ) + ( 0 ) sin 2 ( −67.38° ) + 2 ( 25 ) sin ( −67.38° ) cos ( −67.38° ) σ x = +221.4 MPa ≅ 221 MPa (T) ............................................................................... Ans. (b) τ nt = − (σ x − σ y ) sin θ cosθ + τ xy ( cos 2 θ − sin 2 θ ) = +61.0 MPa ..................................................................................................................... Ans. 10-23 = − ( 221.4 − 0 ) sin ( −67.38° ) cos ( −67.38° ) + 25 cos 2 ( −67.38° ) − sin 2 ( −67.38° ) At a point in a structural member, there are stresses on horizontal and vertical planes, as shown in Fig. P1023. The magnitude of the compressive stress σc is three times the magnitude of the tensile stress σt. Specifications require that the shearing stress on plane AB not exceed 4000 psi, and the normal stress on plane AB not exceed 7800 psi. Determine the maximum value of stress σc that will satisfy the specifications. SOLUTION The given values are: σ x = σt σ y = −3σ t τ xy = 0 psi θ = tan −1 ( 4 3) = 53.130° σ n = ±7800 psi σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cosθ τ nt = ±4000 psi σ t = −0.6410σ n = −0.6410 ( −7800 ) = +5000 psi = σ t cos 2 ( 53.130° ) + ( −3σ t ) sin 2 ( 53.130° ) + 2 ( 0 ) sin ( 53.130° ) cos ( 53.130° ) τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ ) σ t = 0.5208τ nt = 0.5208 ( +4000 ) = +2083 psi Therefore: = − σ t − ( −3σ t ) sin ( 53.13° ) cos ( 53.13° ) + ( 0 ) cos 2 ( 53.13° ) − sin 2 ( 53.13° ) σ c ( max ) = −3σ t ( max ) = −3 ( 2083) = −6249 psi ≅ 6.25 ksi (C) ............................. Ans. 463 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 10-24 RILEY, STURGES AND MORRIS At a point in a stressed body, the stresses on two perpendicular planes are as shown in Fig. P10-24. Determine (a) The stresses on plane aa. (b) The stresses on horizontal and vertical planes at the point. SOLUTION (a) The given values are: σ n = +200 MPa σ t = +50 MPa τ nt = 0 MPa θ = 135° σ aa = σ n cos 2 θ + σ t sin 2 θ + 2τ nt sin θ cosθ = 200 cos 2 (135° ) + 50sin 2 (135° ) + 2 ( 0 ) sin (135° ) cos (135° ) = +125.0 MPa = 125.0 MPa (T) .................................................................................. Ans. τ aa = − (σ n − σ t ) sin θ cos θ + τ nt ( cos 2 θ − sin 2 θ ) = − ( 200 − 50 ) sin (135° ) cos (135° ) + ( 0 ) cos 2 (135° ) − sin 2 (135° ) = +75.0 MPa ..................................................................................................................... Ans. (b) The given values are: σ n = +200 MPa σ t = +50 MPa τ nt = 0 MPa θ x = −18° θ y = 72° σ x = σ n cos 2 θ + σ t sin 2 θ + 2τ nt sin θ cos θ = 200 cos 2 ( −18° ) + 50sin 2 ( −18° ) + 2 ( 0 ) sin ( −18° ) cos ( −18° ) = +185.7 MPa = 185.7 MPa (T) .................................................................................. Ans. σ y = σ n cos 2 θ + σ t sin 2 θ + 2τ nt sin θ cosθ = 200 cos 2 ( 72° ) + 50 sin 2 ( 72° ) + 2 ( 0 ) sin ( 72° ) cos ( 72° ) = +64.3 MPa = 64.3 MPa (T) ....................................................................................... Ans. τ xy = − (σ n − σ t ) sin θ cosθ + τ nt ( cos 2 θ − sin 2 θ ) = − ( 200 − 50 ) sin ( −18° ) cos ( −18° ) + ( 0 ) cos 2 ( −18° ) − sin 2 ( −18° ) = +44.1 MPa ..................................................................................................................... Ans. 10-25 At a point in a stressed body, the stresses on two perpendicular planes are as shown in Fig. P10-25. Determine (a) The stresses on plane aa. (b) The stresses on horizontal and vertical planes at the point. SOLUTION (a) The given values are: σ n = −20 ksi σ t = −3 ksi τ nt = 0 ksi θ = 135° σ aa = σ n cos 2 θ + σ t sin 2 θ + 2τ nt sin θ cosθ = −20 cos 2 (135° ) + ( −3) sin 2 (135° ) + 2 ( 0 ) sin (135° ) cos (135° ) = −11.50 ksi = 11.50 ksi (C) ......................................................................................... Ans. 464 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS τ aa = − (σ n − σ t ) sin θ cosθ + τ nt ( cos 2 θ − sin 2 θ ) = − ( −20 ) − ( −3) sin (135° ) cos (135° ) + ( 0 ) cos 2 (135° ) − sin 2 (135° ) = −8.50 ksi ......................................................................................................................... Ans. (b) The given values are: σ n = −20 ksi σ t = −3 ksi τ nt = 0 ksi θ x = 28° θ y = 118° σ x = σ n cos 2 θ + σ t sin 2 θ + 2τ nt sin θ cosθ = ( −20 ) cos 2 ( 28° ) + ( −3) sin 2 ( 28° ) + 2 ( 0 ) sin ( 28° ) cos ( 28° ) = −16.25 MPa = 16.25 ksi (C) ..................................................................................... Ans. σ y = σ n cos 2 θ + σ t sin 2 θ + 2τ nt sin θ cosθ = ( −20 ) cos 2 (118° ) + ( −3) sin 2 (118° ) + 2 ( 0 ) sin (118° ) cos (118° ) = −6.75 ksi = 6.75 ksi (C) ............................................................................................. Ans. τ xy = − (σ n − σ t ) sin θ cosθ + τ nt ( cos 2 θ − sin 2 θ ) = − ( −20 ) − ( −3) sin ( 28° ) cos ( 28° ) + ( 0 ) cos 2 ( 28° ) − sin 2 ( 28° ) = +7.05 ksi ......................................................................................................................... Ans. 10-26 The stresses shown in Fig. P10-26 act at a point on the free surface of a stressed body. (a) Calculate and graph the normal stress σ n and the shearing stress τ nt on the inclined plane AB as a function of the angle (b) (c) θ ( 0° ≤ θ ≤ 180° ) . For what angle θ is the normal stress a maximum? A minimum? What is the value of the shear stress on these planes? For what angle θ is the shear stress a maximum? A minimum? What is the value of the normal stress on these planes? SOLUTION (a) σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cosθ τ nt = − (σ x − σ y ) sin θ cosθ + τ xy ( cos 2 θ − sin 2 θ ) = 40 cos 2 θ + ( −80 ) sin 2 θ + 2 ( 80 ) sin θ cos θ MPa ............................................ Ans. = − 40 − ( −80 ) sin θ cosθ + 80 ( cos 2 θ − sin 2 θ ) MPa ..................................... Ans. (b) At { } θ = 26.565° σ max = 80 MPa ........................................ Ans. τ = 0 MPa ................................................. Ans. At θ = 116.565° σ min = −120 MPa .................................... Ans. (c) τ = 0 MPa ................................................. Ans. At θ = 71.565° τ min = −100 MPa .................................... Ans. 465 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS σ = −20 MPa .......................................................................................................................... Ans. At θ = 161.565° τ max = 100 MPa ...................................................................................................................... Ans. σ = −20 MPa .......................................................................................................................... Ans. 10-27 The stresses shown in Fig. P10-27 act at a point on the free surface of a stressed body. Calculate the normal stress σ n and the shearing stress τ nt on the inclined plane AB as a function of the angle θ ( 0° ≤ θ ≤ 180° ) . For each angle θ , graph the negative of the shearing stress ( −τ nt , vertical axis) as a function of the normal stress ( σ n , horizontal axis). On your graph, clearly identify the points associated with the angles θ SOLUTION = 0 , 30 , 45 , 60 , 90 , 120 ,135 , 150 , and 180 . σ n = σ x cos 2 θ + σ y sin 2 θ +2τ xy sin θ cos θ = ( 0 ) cos 2 θ + 60sin 2 θ +2 ( 40 ) sin θ cos θ ............. Ans. τ nt = − (σ x − σ y ) sin θ cosθ +τ xy ( cos 2 θ − sin 2 θ ) = − ( 0 − 60 ) sin θ cos θ +40 ( cos 2 θ − sin 2 θ ) ......... Ans. 10-28 Normal and shear stresses on horizontal and vertical planes through a point in a structural member subjected to plane stress. Determine, and show on a sketch, the principal stresses and the maximum shear stress at the point. SOLUTION The given values are: σ x = +50 MPa 2 σ y = +20 MPa τ xy = +40 MPa σ p1, p 2 σ +σ y σ −σ y 2 =x ±x + τ xy 2 2 50 + 20 2 50 − 20 = ± + ( 40 ) 2 2 2 σ p1 = 35 + 42.72 = 77.7 MPa = 77.7 MPa (T) σ p 3 = σ z = 0 MPa ........................... τ p .............................................................. Ans. σ p 2 = 35 − 42.72 = −7.72 MPa = 7.72 MPa (C) ........................................................... Ans. 2τ xy 2 ( 40 ) 1 1 θ p = tan −1 = tan −1 = 34.72° .................................................... Ans. 2 σ x −σ y 2 ( 50 ) − ( 20 ) 466 = τ max = 42.7 MPa ........................................... Ans. STATICS AND MECHANICS OF MATERIALS, 2nd Edition 10-29 RILEY, STURGES AND MORRIS Normal and shear stresses on horizontal and vertical planes through a point in a structural member subjected to plane stress. Determine, and show on a sketch, the principal stresses and the maximum shear stress at the point. SOLUTION The given values are: σ x = +12 ksi 2 σ y = −4 ksi τ xy = −6 ksi σ p1, p 2 σ x +σ y σ x −σ y 2 = ± + τ xy 2 2 = 12 + ( −4 ) 2 12 − ( −4 ) 2 ± + ( −6 ) 2 2 σ p1 = 4 + 10.00 = 14.00 ksi = 14.00 ksi (T) .................................................................... Ans. σ p 2 = 4 − 10.00 = −6.00 ksi = 6.00 ksi (C) ..................................................................... Ans. σ p 3 = σ z = 0 ksi ............................... τ p = τ max = 10.00 ksi ............................................. Ans. 2τ xy 2 ( −6 ) 1 1 = tan −1 = −18.43° θ p = tan −1 2 σ x −σ y 2 (12 ) − ( −4 ) 10-30 ................................................. Ans. Normal and shear stresses on horizontal and vertical planes through a point in a structural member subjected to plane stress. Determine, and show on a sketch, the principal stresses and the maximum shear stress at the point. SOLUTION The given values are: σ x = +75 MPa 2 σ y = −25 MPa τ xy = −35 MPa σ p1, p 2 σ x +σ y σ x −σ y 2 = ± + τ xy 2 2 = 75 + ( −25 ) 2 75 − ( −25 ) 2 ± + ( −35 ) 2 2 σ p1 = 25 + 61.03 = 86.0 MPa = 86.0 MPa (T) ............................................................... Ans. σ p 2 = 25 − 61.03 = −36.0 MPa = 36.0 MPa (C) σ p 3 = 0 MPa ....................... τ p ........................................................... Ans. = τ max = 61.0 MPa ......................................................... Ans. 2τ xy 2 ( −35 ) 1 1 = tan −1 = −17.50° ............................................... Ans. θ p = tan −1 2 σ x −σ y 2 ( 75 ) − ( −25 ) 10-31 Normal and shear stresses on horizontal and vertical planes through a point in a structural member subjected to plane stress. Determine, and show on a sketch, the principal stresses and the maximum shear stress at the point. SOLUTION The given values are: σ x = −15 ksi τ xy = +8 ksi σ y = +10 ksi 467 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS σ p1, p 2 σ x +σ y σ x −σ y 2 = ± + τ xy 2 2 2 ( −15 ) + 10 ± = 2 ( −15 ) − 10 2 + (8) 2 2 σ p1 = −2.50 + 14.84 = 12.34 ksi = 12.34 ksi (T) ............................................................ Ans. σ p 2 = −2.50 − 14.84 = −17.34 ksi = 17.34 ksi (C) ........................................................ Ans. σ p 3 = σ z = 0 ksi ................. τ p = τ max = 14.84 ksi ........................................................... Ans. 2τ xy 2 (8) 1 1 = tan −1 = −16.31° ............................................... Ans. θ p = tan −1 2 σ x −σ y 2 ( −15) − (10 ) 10-32 Normal and shear stresses on horizontal and vertical planes through a point in a structural member subjected to plane stress. Determine, and show on a sketch, the principal stresses and the maximum shear stress at the point. SOLUTION The given values are: σ x = −30 MPa τ xy = −10 MPa σ p1, p 2 σ y = +40 MPa σ x +σ y σ x −σ y 2 = ± + τ xy 2 2 2 ( −30 ) + 40 ± = 2 ( −30 ) − 40 2 + ( −10 ) 2 2 σ p1 = 5 + 36.4 = 41.4 MPa = 41.4 MPa (T) ................................................................... Ans. σ p 2 = 5 − 36.4 = −31.4 MPa = 31.4 MPa (C) ................................................................ Ans. σ p 3 = σ z = 0 MPa ........................... τ p = τ max = 36.4 MPa ........................................... Ans. 2τ xy 2 ( −10 ) 1 1 = tan −1 = +7.97° ................................................. Ans. θ p = tan −1 2 σ x −σ y 2 ( −30 ) − ( 40 ) 10-33 Normal and shear stresses on horizontal and vertical planes through a point in a structural member subjected to plane stress. Determine, and show on a sketch, the principal stresses and the maximum shear stress at the point. SOLUTION The given values are: σ x = +25 ksi τ xy = +12 ksi σ y = +7 ksi σ p1, p 2 σ +σ y σ −σ y 2 =x ±x + τ xy 2 2 468 2 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS = 25 + 7 2 25 − 7 ± + (12 ) 2 2 2 σ p1 = 16.00 + 15.00 = 31.0 ksi = 31.0 ksi (T) ................................................................. Ans. σ p 2 = 16.00 − 15.00 = +1.000 ksi = 1.000 ksi (T) ......................................................... Ans. σ p 3 = σ z = 0 ksi ............................... τ p = 15.00 ksi ......................................................... Ans. σ − σ min 31 − 0 τ max = max = = 15.50 ksi ............................................................................ Ans. 2 2 2τ xy 2 (12 ) 1 1 θ p = tan −1 = tan −1 = +26.57° .................................................... Ans. 2 σ x −σ y 2 ( 25 ) − ( 7 ) 10-34 Normal and shear stresses on horizontal and vertical planes through a point in a structural member subjected to plane stress. Determine, and show on a sketch, the principal stresses and the maximum shear stress at the point. SOLUTION The given values are: σ x = +36 MPa τ xy = +12 MPa σ y = +26 MPa σ p1, p 2 σ +σ y σ −σ y 2 =x ±x + τ xy 2 2 36 + 26 2 36 − 26 = ± + (12 ) 2 2 2 2 σ p1 = 31 + 13.00 = 44.0 MPa = 44.0 MPa (T) ............................................................... Ans. σ p 2 = 31 − 13.00 = +18.00 MPa = 18.00 MPa (T) ........................................................ Ans. σ p 3 = σ z = 0 MPa ........................... τ p = 13.00 MPa ..................................................... Ans. σ − σ min 44 − 0 τ max = max = = 22.0 MPa .......................................................................... Ans. 2 2 2τ xy 2 (12 ) 1 1 θ p = tan −1 = tan −1 = +33.69° ................................................. Ans. 2 σ x −σ y 2 ( 36 ) − ( 26 ) 10-35 Normal and shear stresses on horizontal and vertical planes through a point in a structural member subjected to plane stress. Determine, and show on a sketch, the principal stresses and the maximum shear stress at the point. SOLUTION The given values are: σ x = −2 ksi τ xy = −8 ksi σ y = −14 ksi σ p1, p 2 σ +σ y σ −σ y 2 =x ±x + τ xy 2 2 2 469 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ( −2 ) + ( −14 ) ± = 2 ( −2 ) − ( −14 ) 2 + ( −8 ) 2 2 σ p1 = −8 + 10.00 = +2.00 ksi = 2.00 ksi (T) ................................................................... Ans. σ p 2 = −8 − 10.00 = −18.00 ksi = 18.00 ksi (C) .............................................................. Ans. σ p 3 = 0 ksi ......................................... τ p = τ max = 10.00 ksi ............................................. Ans. 2τ xy 2 ( −8 ) 1 1 θ p = tan −1 = tan −1 = −26.57° 2 σ x −σ y 2 ( −2 ) − ( −14 ) 10-36 .............................................. Ans. Normal and shear stresses on horizontal and vertical planes through a point in a structural member subjected to plane stress. Determine, and show on a sketch, the principal stresses and the maximum shear stress at the point. SOLUTION The given values are: σ x = +72 MPa τ xy = −24 MPa σ y = +36 MPa σ p1, p 2 σ +σ y σ −σ y 2 =x ±x + τ xy 2 2 72 + 36 2 72 − 36 = ± + ( −24 ) 2 2 2 2 σ p1 = 54 + 30.00 = 84.0 MPa = 84.0 MPa (T) ............................................................... Ans. σ p 2 = 54 − 30.00 = +24.0 MPa = 24.0 MPa (T) ........................................................... Ans. σ p 3 = σ z = 0 MPa ........................... τ p = 30.00 MPa ..................................................... Ans. σ − σ min 84 − 0 τ max = max = = 42.0 MPa .......................................................................... Ans. 2 2 2τ xy 2 ( −24 ) 1 1 θ p = tan −1 = tan −1 = −26.57° ................................................. Ans. 2 σ x −σ y 2 ( 72 ) − ( 36 ) 10-37 Normal and shear stresses on horizontal and vertical planes through a point in a structural member subjected to plane stress. Determine, and show on a sketch, the principal stresses and the maximum shear stress at the point. SOLUTION The given values are: σ x = +13 ksi τ xy = +6 ksi σ y = +18 ksi σ p1, p 2 σ +σ y σ −σ y 2 =x ±x + τ xy 2 2 2 470 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS = 13 + 18 2 13 − 18 ± + (6) 2 2 ............................................................... Ans. 2 σ p1 = 15.50 + 6.50 = 22.0 ksi = 22.0 ksi (T) .................................................................. Ans. σ p 2 = 15.50 − 6.50 = +9.00 ksi = 9.00 ksi (T) σ p 3 = σ z = 0 ksi ............................... τ p = 6.50 ksi ........................................................... Ans. σ max − σ min 22 − 0 = = 11.00 ksi ........................................................................... Ans. 2 2 2τ xy 2 ( 6) 1 1 θ p = tan −1 = tan −1 = −33.69° ................................................... Ans. 2 σ x −σ y 2 (13) − (18 ) τ max = 10-38 The stresses shown in Fig. P10-38 act at a point on the free surface of a stressed body. The magnitude of the maximum shear stress at the point is 125 MPa. Determine the principal stresses, and the magnitude of the unknown shear stress on the vertical plane, and the angle θp between the x-axis and the maximum tensile stress at the point. SOLUTION Since σx and σy have opposite signs, τ max = τ p = 125 MPa σ p1 + σ p 2 = σ x + σ y = 80 + ( −120 ) = −40 MPa Therefore: σ p1 = 105.0 MPa = 105.0 MPa (T) .................................................................... Ans. σ p 2 = −145.0 MPa = 145.0 MPa (C) ................................................................. Ans. 80 − ( −120 ) 2 τp = ± τ xy = ±75.0 MPa + τ xy = 125 MPa 2 Therefore (as shown on Fig. P10-38): τ xy = −75.0 MPa ........................................................... Ans. 2 σ p1 − σ p 2 = 2τ max = 2 (125 ) = 250 MPa 2τ xy 2 ( −75 ) 1 1 θ p = tan −1 = tan −1 = +18.43° 2 σ x −σ y 2 ( 80 ) − ( −120 ) ............................................ Ans. 10-39 The principal compressive stress on a vertical plane through a point in a wooden block is equal to four times the principal compressive stress on a horizontal plane, as shown in Fig. P10-39. The plane of the grain is 30° clockwise from the vertical plane. If the normal and shear stresses on the plane of the grain must not exceed 300 psi (C) and 125 psi shear, determine the maximum allowable compressive stress on the horizontal plane. SOLUTION σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cosθ = 4σ c cos 2 ( −30° ) + σ c sin 2 ( −30° ) + 0 > −300 psi τ nt = − (σ x − σ y ) sin θ cosθ + τ xy ( cos 2 θ sin 2 θ ) σ c > −92.3 psi = 92.3 psi (C) = − ( 4σ c − σ c ) sin ( −30° ) cos ( −30° ) + 0 > −125 psi Therefore σ c ( max ) = 92.3 psi (C) ......................................................................................... Ans. 471 σ c > −96.2 psi = 96.2 psi (C) STATICS AND MECHANICS OF MATERIALS, 2nd Edition 10-40 RILEY, STURGES AND MORRIS At a point on the free surface of a stressed body, a normal stress of 64 MPa (C) and an unknown positive shear stress exist on a horizontal plane. One principal stress at the point is 8 MPa (C). The maximum shear stress at the point has a magnitude of 95 MPa. Determine the unknown stresses on the horizontal and vertical planes shown in Fig. P10-40 and the unknown principal stress at the point. SOLUTION The principal stress σ p2 must be compressive with a magnitude greater than Then 64.0 MPa and the principal stress σ p 3 = σ z = 0 MPa (since it is a free surface). τ max = σ p2 σ max − σ min 0 − σ p 2 = = 95.0 MPa 2 2 = −190.0 MPa = 190.0 MPa (C) ............................................................................... Ans. σ p1 + σ p 2 = σ x + σ y ( −8 ) + ( −190 ) = σ x + ( −64 ) σ x = −134.0 MPa = 134.0 MPa (C) ................................................................................. Ans. τp = σ p1 − σ p 2 ( −8) − ( −190 ) = = 91.0 MPa 2 2 2 ( −134 ) − ( −64 ) 2 τp = ± + τ xy = 91.0 MPa 2 τ xy = 84.0 MPa ....................................................................................................................... Ans. 10-41 At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a horizontal plane, as shown in Fig. P10-41. An unknown negative shear stress acts on the horizontal and vertical planes. The maximum shear stress at the point has a magnitude of 32 ksi. Determine the principal stresses and the shear stress on the vertical plane at the point. SOLUTION Since σx and σ y have opposite signs, 2 τ max = τ p = 32 ksi 20 − ( −30 ) 2 τp = ± + τ xy = 32 ksi 2 τ xy = −19.97 ksi ..................................................................................................................... Ans. σ p1 + σ p 2 = σ x + σ y = 20 + ( −30 ) = −10 ksi σ p1 − σ p 2 = 2τ p = 2 ( 32 ) = 64 ksi σ p1 = +27.0 ksi = 27.0 ksi (T) ........................................................................................... Ans. σ p 2 = −37.0 ksi = 37.0 ksi (C) .......................................................................................... Ans. σ p 3 = −0 ksi ............................................................................................................................ Ans. 472 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 10-42 RILEY, STURGES AND MORRIS At a point on the free surface of a stressed body, a normal stress of 75 MPa (T) and an unknown negative shear stress exist on a horizontal plane, as shown in Fig. P10-42. One principal stress at the point is 200 MPa (T). The maximum in-plane shear stress at the point has a magnitude of 85 MPa. Determine the unknown stresses on the vertical plane, the unknown principal stress, and the maximum shear stress at the point. SOLUTION σ p1 − σ p 2 200 − σ p 2 = = 85.0 MPa 2 2 σ p 2 = +30.0 MPa = 30.0 MPa (T) .................................................................................... Ans. σ − σ min 200 − 0 τ max = max = = 100.0 MPa ..................................................................... Ans. 2 2 σ p1 + σ p 2 = σ x + σ y 200 + 30 = σ x + 75 τp = σ x = +155.0 MPa = 155.0 MPa (T) ................................................................................. Ans. 155 − 75 2 τp = ± + τ xy = 85.0 MPa 2 τ xy = −75.0 MPa .................................................................................................................... Ans. 10-43 The stresses shown in Fig. P10-43 act at a point on the free surface of a stressed body. Use Mohr’s circle to determine the normal and shear stresses at this point on the inclined plane AB shown in the figure. SOLUTION The given values for use in drawing Mohr’s circle are: 2 σ x = σ p1 = +12 ksi σ y = σ p 2 = −16 ksi σ z = σ p 3 = 0 ksi 12 + ( −16 ) = −2 ksi 2 16 + 12 R= = 14 ksi 2 σ n = −2 + 14cos 52° = +6.62 ksi = 6.62 ksi (T) ............................................................ Ans. a= τ nt = 14sin 52° = 11.03 ksi 10-44 = +11.03 ksi .................................................................... Ans. The stresses shown in Fig. P10-44 act at a point on the free surface of a stressed body. Use Mohr’s circle to determine the normal and shear stresses at this point on the inclined plane AB shown in the figure. SOLUTION The given values for use in drawing Mohr’s circle are: σ x = σ p 2 = −27 MPa σ y = σ p1 = +45 MPa σ z = σ p 3 = 0 MPa a= 2 27 + 45 R= = 36 MPa 2 ( −27 ) + 45 = 9 MPa 473 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS σ n = 9 + 36 cos 66° = 23.6 MPa = 23.6 MPa (T) .......................................................... Ans. τ nt = 36sin 66° = 32.9 MPa 10-45 = −32.9 MPa ................................................................. Ans. At a point in a structural member subjected to plane stress there are normal and shear stresses on horizontal and vertical planes through the point, as shown in Fig. P10-45. Use Mohr’s circle to determine the principal stresses and the maximum shear stress at the point. SOLUTION The given values for use in drawing Mohr’s circle are: σ x = +25 ksi τ xy = +12 ksi a= σ y = +7 ksi σ z = σ p 3 = 0 ksi 25 + 7 = 16 ksi 2 R = 92 + 122 = 15 ksi σ p1 = 16 + 15 = +31.0 ksi = 31.0 ksi (T) .......................................................................... Ans. σ p 2 = 16 − 15 = +1.00 ksi = 1.00 ksi (T) .......................................................................... Ans. σ p 3 = 0 ksi ........................... τ p = R = 15.00 ksi ............................................................... Ans. σ − σ min 31 − 0 τ max = max = = 15.50 ksi ............................................................................ Ans. 2 2 1 12 θ p1 = tan −1 = 26.57° ..................................................................................................... Ans. 2 9 10-46 At a point in a structural member subjected to plane stress there are normal and shear stresses on horizontal and vertical planes through the point, as shown in Fig. P10-46. Use Mohr’s circle to determine the principal stresses and the maximum shear stress at the point. SOLUTION The given values for use in drawing Mohr’s circle are: σ x = +50 MPa τ xy = +40 MPa a= σ y = +20 MPa σ z = σ p 3 = 0 MPa 50 + 20 = 35 MPa 2 R = 152 + 402 = 42.72 MPa σ p1 = 35 + 42.72 = +77.7 MPa = 77.7 MPa (T) ............................................................ Ans. σ p 2 = 35 − 42.72 = −7.72 MPa = 7.72 MPa (C) ........................................................... Ans. σ p 3 = 0 MPa ....................... τ max = τ p = R = 42.7 MPa ................................................. Ans. 1 40 θ p1 = tan −1 = 34.72° ..................................................................................................... Ans. 2 15 474 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 10-47 RILEY, STURGES AND MORRIS At a point in a structural member subjected to plane stress there are normal and shear stresses on horizontal and vertical planes through the point, as shown in Fig. P10-47. Use Mohr’s circle to determine the principal stresses and the maximum shear stress at the point. SOLUTION The given values for use in drawing Mohr’s circle are: σ x = −15 ksi τ xy = +8 ksi a= σ y = +10 ksi σ z = σ p 3 = 0 ksi ( −15 ) + 10 = −2.5 ksi 2 R = 12.52 + 82 = 14.841 ksi σ p1 = −2.5 + 14.841 = +12.34 ksi = 12.34 ksi (T) ......................................................... Ans. σ p 2 = −2.5 − 14.841 = −17.34 ksi = 17.34 ksi (C) ......................................................... Ans. σ p 3 = 0 ksi ........................... τ max = τ p = R = 14.84 ksi ................................................... Ans. 1 8 θ p1 = − tan −1 = −16.31° ............................................................................................ Ans. 2 12.5 10-48 At a point in a structural member subjected to plane stress there are normal and shear stresses on horizontal and vertical planes through the point, as shown in Fig. P10-48. Use Mohr’s circle to determine the principal stresses and the maximum shear stress at the point. SOLUTION The given values for use in drawing Mohr’s circle are: σ x = −65 MPa τ xy = +75 MPa a= σ y = −125 MPa σ z = σ p 3 = 0 MPa ( −65 ) + ( −125) = −95 MPa 2 R = 302 + 752 = 80.78 MPa σ p1 = −95 + 80.78 = −14.22 MPa = 14.22 MPa (C) ..................................................... Ans. σ p 2 = −95 − 80.78 = −175.8 MPa = 175.8 MPa (C) ..................................................... Ans. σ p 3 = 0 MPa ....................... τ p = R = 80.8 MPa .............................................................. Ans. σ max − σ min 0 − ( −175.8 ) = = 87.9 MPa .............................................................. Ans. 2 2 1 75 θ p1 = tan −1 = 26.87° ..................................................................................................... Ans. 2 55 τ max = 475 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 10-49 RILEY, STURGES AND MORRIS At a point in a structural member subjected to plane stress there are normal and shear stresses on horizontal and vertical planes through the point, as shown in Fig. P10-49. Use Mohr’s circle to determine (a) The principal stresses and the maximum shear stress at the point. (b) The normal and shear stresses on the inclined plane AB shown in the figure. SOLUTION The given values for use in drawing Mohr’s circle are: σ x = +12 ksi τ xy = +10 ksi a= σ y = −6 ksi σ z = σ p 3 = 0 ksi 12 + ( −6 ) = 3 ksi 2 R = 92 + 102 = 13.454 ksi (a) σ p1 = 3 + 13.454 = +16.45 ksi = 16.45 ksi (T) ............................................................... Ans. σ p 2 = 3 − 13.454 = −10.45 ksi = 10.45 ksi (C) ............................................................... Ans. σ p 3 = 0 ksi ........................... τ max = τ p = R = 13.45 ksi ................................................... Ans. (b) σ n = 3 − 13.454 cos 71.99° = −1.160 ksi = 1.160 ksi (C) τ nt = 13.454sin 71.99° = 12.79 ksi ............................................ Ans. = +12.79 ksi ...................................................... Ans. 10-50 At a point in a structural member subjected to plane stress there are normal and shear stresses on horizontal and vertical planes through the point, as shown in Fig. P10-50. Use Mohr’s circle to determine (a) The principal stresses and the maximum shear stress at the point. (b) The normal and shear stresses on the inclined plane AB shown in the figure. SOLUTION The given values for use in drawing Mohr’s circle are: σ x = +80 MPa τ xy = −60 MPa a= σ y = −100 MPa σ z = σ p 3 = 0 MPa 80 + ( −100 ) = −10 MPa 2 R = 902 + 602 = 108.17 MPa (a) σ p1 = −10 + 108.17 = +98.2 MPa = 98.2 MPa (T) ....................................................... Ans. σ p 2 = −10 − 108.17 = −118.2 MPa = 118.2 MPa (C) .................................................. Ans. σ p 3 = 0 MPa ....................... τ max = τ p = R = 108.2 MPa ............................................... Ans. (b) σ n = −10 − 108.17 cos 62.31° = −60.3 MPa = 60.3 MPa (C) ..................................... Ans. τ nt = 108.17 sin 62.31° = 95.8 MPa = −95.8 MPa ................................................... Ans. 476 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 10-51 RILEY, STURGES AND MORRIS At a point in a structural member subjected to plane stress there are normal and shear stresses on horizontal and vertical planes through the point, as shown in Fig. P10-51. Use Mohr’s circle to determine (a) The principal stresses and the maximum shear stress at the point. (b) The normal and shear stresses on the inclined plane AB shown in the figure. SOLUTION The given values for use in drawing Mohr’s circle are: σ x = +25 ksi τ xy = −10 ksi a= σ y = +12 ksi σ z = σ p 3 = 0 ksi 25 + 12 = 18.50 ksi 2 R = 6.52 + 102 = 11.927 ksi (a) σ p1 = 18.5 + 11.927 = +30.4 ksi = 30.4 ksi (T) .............................................................. Ans. σ p 2 = 18.5 − 11.927 = +6.57 ksi = 6.57 ksi (T) ............................................................. Ans. σ p 3 = 0 ksi ........................... τ p = R = 11.93 ksi τ max = ............................................................... Ans. (b) σ max − σ min 30.42 − 0 = = 15.21 ksi ........................................................................ Ans. 2 2 σ n = 18.5 − 11.927 cos 47.02° = +10.37 ksi = 10.37 ksi (T) ....................................... Ans. = −8.73 ksi ........................................................... Ans. τ nt = 11.927 sin 47.02° = 8.73 ksi 10-52 The thin rectangular plate shown in Fig. P10-52 is uniformly deformed such that ε x = −2000 µ m/m , ε y = −1500 µ m/m , and γ xy = +1250 µ rad . Determine the normal strain ε n in the plate. SOLUTION The given values are: ε x = −2000 µ m/m ε y = −1500 µ m/m γ xy = +1250 µ rad θ n = −35° ε n = ε x cos 2 θ n + ε y sin 2 θ n + γ xy sin θ n cos θ n = ( −2000 ) cos 2 ( −35° ) + ( −1500 ) sin 2 ( −35° ) + 1250 sin ( −35° ) cos ( −35° ) = −2413 µ m/m ≅ −2410 µ m/m ................................................................................... Ans. 10-53 The thin rectangular plate shown in Fig. P10-53 is uniformly deformed such that ε x = +880 µin./in. , along diagonal AC of ε y = +960 µ in./in. , and γ xy = −750 µ rad . the plate. SOLUTION The given values are: Determine the normal strain ε AC ε x = +880 µin./in. ε y = +960 µ in./in. γ xy = −750 µ rad θ AC = tan −1 ( 2 4 ) = 26.57° ε AC = ε x cos 2 θ n + ε y sin 2 θ n + γ xy sin θ n cosθ n = 880 cos 2 ( 26.57° ) + 960sin 2 ( 26.57° ) + ( −750 ) sin ( 26.57° ) cos ( 26.57° ) = +596 µ in./in. .............................................................................................................. Ans. 477 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 10-54 RILEY, STURGES AND MORRIS The thin rectangular plate shown in Fig. P10-54 is uniformly deformed such that ε x = +1500 µ m/m , ε y = −1250 µ m/m , and γ xy = +1000 µ rad . Determine the normal strain ε BD along diagonal BD of the plate. SOLUTION The given values are: ε x = +1500 µ m/m ε y = −1250 µ m/m γ xy = +1000 µ rad θ BD = − tan −1 (150 200 ) = −36.87° ε BD = ε x cos 2 θ n + ε y sin 2 θ n + γ xy sin θ n cosθ n = 1500 cos 2 ( −36.87° ) + ( −1250 ) sin 2 ( −36.87° ) + 1000 sin ( −36.87° ) cos ( −36.87° ) = +30.0 µ m/m .................................................................................................................. Ans. 10-55 The strain components for a point in a body subjected to plane strain are P10-55. SOLUTION The given values are: ε x = −800 µin./in. , ε y = +640 µ in./in. , and γ xy = −960 µ rad . Determine the strain components ε n , ε t , and γ nt . at the point if the nt-axes are rotated with respect to the xy-axes by θ = 42° in the direction indicated in Fig. ε x = −800 µin./in. ε y = +640 µ in./in. γ xy = −960 µ rad θ n = 42° θ t = 132° ε n = ε x cos 2 θ n + ε y sin 2 θ n + γ xy sin θ n cos θ n = ( −800 ) cos 2 ( 42° ) + 640sin 2 ( 42° ) + ( −960 ) sin ( 42° ) cos ( 42° ) = −633 µ in./in. .................................................................................................................. Ans. ε t = ε x cos 2 θ t + ε y sin 2 θ t + γ xy sin θ t cosθ t = ( −800 ) cos 2 (132° ) + 640sin 2 (132° ) + ( −960 ) sin (132° ) cos (132° ) = +473 µ in./in. .................................................................................................................. Ans. γ nt = −2 ( ε x − ε y ) sin θ n cos θ n + γ xy ( cos 2 θ n − sin 2 θ n ) = +1332 µ rad .................................................................................................................... Ans. 10-56 The strain components for a point in a body subjected to plane strain are = −2 ( −800 ) − 640 sin ( 42° ) cos ( 42° ) + ( −960 ) cos 2 ( 42° ) − sin 2 ( 42° ) P10-56. SOLUTION The given values are: ε x = +720 µ m/m , ε y = −480 µ m/m , and γ xy = +360 µ rad . Determine the strain components ε n , ε t , and γ nt . at the point if the nt-axes are rotated with respect to the xy-axes by θ = 30° in the direction indicated in Fig. ε x = +720 µ m/m ε y = −480 µ m/m γ xy = +360 µ rad θ n = −30° θ t = 60° ε n = ε x cos 2 θ n + ε y sin 2 θ n + γ xy sin θ n cos θ n = 720 cos 2 ( −30° ) + ( −480 ) sin 2 ( −30° ) + 360 sin ( −30° ) cos ( −30° ) = +264 µ m/m ................................................................................................................... Ans. 478 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ε t = ε x cos 2 θ t + ε y sin 2 θ t + γ xy sin θ t cosθ t = 720 cos 2 ( 60° ) + ( −480 ) sin 2 ( 60° ) + 360 sin ( 60° ) cos ( 60° ) = −24.1 µ m/m ................................................................................................................... Ans. γ nt = −2 ( ε x − ε y ) sin θ n cosθ n + γ xy ( cos 2 θ n − sin 2 θ n ) = −2 720 − ( −480 ) sin ( −30° ) cos ( −30° ) + 360 cos 2 ( −30° ) − sin 2 ( −30° ) = +1219 µ rad .................................................................................................................... Ans. 10-57 The strain components for a point in a body subjected to plane strain are P10-55. SOLUTION The given values are: ε x = +900 µin./in. , ε y = +650 µ in./in. , and γ xy = −300 µ rad . Determine the strain components ε n , ε t , and γ nt . at the point if the nt-axes are rotated with respect to the xy-axes by θ = 32° in the direction indicated in Fig. ε x = +900 µin./in. ε y = +650 µ in./in. γ xy = −300 µ rad θ n = 32° θ t = 122° ε n = ε x cos 2 θ n + ε y sin 2 θ n + γ xy sin θ n cos θ n = 900 cos 2 ( 32° ) + 650 sin 2 ( 32° ) + ( −300 ) sin ( 32° ) cos ( 32° ) = +695 µ in./in. .................................................................................................................. Ans. ε t = ε x cos 2 θ t + ε y sin 2 θ t + γ xy sin θ t cosθ t = 900 cos 2 (122° ) + 650 sin 2 (122° ) + ( −300 ) sin (122° ) cos (122° ) = +855 µ in./in. .................................................................................................................. Ans. γ nt = −2 ( ε x − ε y ) sin θ n cosθ n + γ xy ( cos 2 θ n − sin 2 θ n ) = −356 µ rad ...................................................................................................................... Ans. 10-58 The strain components for a point in a body subjected to plane strain are = −2 ( 900 − 650 ) sin ( 32° ) cos ( 32° ) + ( −300 ) cos 2 ( 32° ) − sin 2 ( 32° ) P10-56. SOLUTION The given values are: ε x = +800 µ m/m , ε y = −950 µ m/m , and γ xy = −800 µ rad . Determine the strain components ε n , ε t , and γ nt . at the point if the nt-axes are rotated with respect to the xy-axes by θ = 42° in the direction indicated in Fig. ε x = +800 µ m/m ε y = −950 µ m/m γ xy = −800 µ rad θ n = −42° θ t = 48° ε n = ε x cos 2 θ n + ε y sin 2 θ n + γ xy sin θ n cos θ n = 800 cos 2 ( −42° ) + ( −950 ) sin 2 ( −42° ) + ( −800 ) sin ( −42° ) cos ( −42° ) = +414 µ m/m ................................................................................................................... Ans. ε t = ε x cos 2 θ t + ε y sin 2 θ t + γ xy sin θ t cosθ t = 800 cos 2 ( 48° ) + ( −950 ) sin 2 ( 48° ) + ( −800 ) sin ( 48° ) cos ( 48° ) = −564 µ m/m ................................................................................................................... Ans. 479 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS γ nt = −2 ( ε x − ε y ) sin θ n cos θ n + γ xy ( cos 2 θ n − sin 2 θ n ) = −2 800 − ( −950 ) sin ( −42° ) cos ( −42° ) + ( −800 ) cos 2 ( −42° ) − sin 2 ( −42° ) = +1657 µ rad .................................................................................................................... Ans. 10-59 The strain components for a point in a body subjected to plane strain are P10-55. SOLUTION The given values are: ε x = +750 µin./in. , ε y = +360 µ in./in. , and γ xy = −300 µ rad . Determine the strain components ε n , ε t , and γ nt . at the point if the nt-axes are rotated with respect to the xy-axes by θ = 52° in the direction indicated in Fig. ε x = +750 µin./in. ε y = +360 µ in./in. γ xy = −300 µ rad θ n = 52° θ t = 142° ε n = ε x cos 2 θ n + ε y sin 2 θ n + γ xy sin θ n cos θ n = 750 cos 2 ( 52° ) + 360 sin 2 ( 52° ) + ( −300 ) sin ( 52° ) cos ( 52° ) = +362 µ in./in. .................................................................................................................. Ans. ε t = ε x cos 2 θ t + ε y sin 2 θ t + γ xy sin θ t cosθ t = 750 cos 2 (142° ) + 360 sin 2 (142° ) + ( −300 ) sin (142° ) cos (142° ) = +748 µ in./in. .................................................................................................................. Ans. γ nt = −2 ( ε x − ε y ) sin θ n cosθ n + γ xy ( cos 2 θ n − sin 2 θ n ) = −306 µ rad ...................................................................................................................... Ans. 10-60 The strain components for a point in a body subjected to plane strain are = −2 ( 750 − 360 ) sin ( 52° ) cos ( 52° ) + ( −300 ) cos 2 ( 52° ) − sin 2 ( 52° ) P10-56. SOLUTION The given values are: ε x = −100 µ m/m , ε y = −700 µ m/m , and γ xy = +400 µ rad . Determine the strain components ε n , ε t , and γ nt . at the point if the nt-axes are rotated with respect to the xy-axes by θ = 28° in the direction indicated in Fig. ε x = −100 µ m/m ε y = −700 µ m/m γ xy = +400 µ rad θ n = −28° θ t = 62° ε n = ε x cos 2 θ n + ε y sin 2 θ n + γ xy sin θ n cos θ n = ( −100 ) cos 2 ( −28° ) + ( −700 ) sin 2 ( −28° ) + 400sin ( −28° ) cos ( −28° ) = −398 µ m/m ................................................................................................................... Ans. ε t = ε x cos 2 θ t + ε y sin 2 θ t + γ xy sin θ t cosθ t = ( −100 ) cos 2 ( 62° ) + ( −700 ) sin 2 ( 62° ) + 400sin ( 62° ) cos ( 62° ) = −402 µ m/m ................................................................................................................... Ans. γ nt = −2 ( ε x − ε y ) sin θ n cos θ n + γ xy ( cos 2 θ n − sin 2 θ n ) = +721 µ rad ....................................................................................................................... Ans. 480 = −2 ( −100 ) − ( −700 ) sin ( −28° ) cos ( −28° ) + 400 cos 2 ( −28° ) − sin 2 ( −28° ) STATICS AND MECHANICS OF MATERIALS, 2nd Edition 10-61 RILEY, STURGES AND MORRIS The strain components for a point in a body subjected to plane strain are P10-55. SOLUTION The given values are: ε x = +850 µin./in. , ε y = +450 µ in./in. , and γ xy = +600 µ rad . Determine the strain components ε n , ε t , and γ nt . at the point if the nt-axes are rotated with respect to the xy-axes by θ = 20° in the direction indicated in Fig. ε x = +850 µin./in. ε y = +450 µ in./in. γ xy = +600 µ rad θ n = 20° θ t = 110° ε n = ε x cos 2 θ n + ε y sin 2 θ n + γ xy sin θ n cos θ n = 850 cos 2 ( 20° ) + 450sin 2 ( 20° ) + 600sin ( 20° ) cos ( 20° ) = +996 µ in./in. .................................................................................................................. Ans. ε t = ε x cos 2 θ t + ε y sin 2 θ t + γ xy sin θ t cosθ t = 850 cos 2 (110° ) + 450sin 2 (110° ) + 600sin (110° ) cos (110° ) = +304 µ in./in. .................................................................................................................. Ans. γ nt = −2 ( ε x − ε y ) sin θ n cos θ n + γ xy ( cos 2 θ n − sin 2 θ n ) = −2 ( 850 − 450 ) sin ( 20° ) cos ( 20° ) + 600 cos 2 ( 20° ) − sin 2 ( 20° ) = +203 µ rad ...................................................................................................................... Ans. 10-62 The strain components for a point in a body subjected to plane strain are P10-56. SOLUTION The given values are: ε x = +1200 µ m/m , ε y = +200 µ m/m , and γ xy = −1600 µ rad . Determine the strain components ε n , ε t , and γ nt . at the point if the nt-axes are rotated with respect to the xy-axes by θ = 64° in the direction indicated in Fig. ε x = +1200 µ m/m ε y = +200 µ m/m γ xy = −1600 µ rad θ n = −64° θ t = 26° ε n = ε x cos 2 θ n + ε y sin 2 θ n + γ xy sin θ n cos θ n = 1200 cos 2 ( −64° ) + 200sin 2 ( −64° ) + ( −1600 ) sin ( −64° ) cos ( −64° ) = +1023 µ m/m .................................................................................................................. Ans. ε t = ε x cos 2 θ t + ε y sin 2 θ t + γ xy sin θ t cosθ t = 1200 cos 2 ( 26° ) + 200 sin 2 ( 26° ) + ( −1600 ) sin ( 26° ) cos ( 26° ) = +377 µ m/m ................................................................................................................... Ans. γ nt = −2 ( ε x − ε y ) sin θ n cosθ n + γ xy ( cos 2 θ n − sin 2 θ n ) = −2 (1200 − 200 ) sin ( −64° ) cos ( −64° ) + ( −1600 ) cos 2 ( −64° ) − sin 2 ( −64° ) = +1773 µ rad .................................................................................................................... Ans. 481 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 10-63 RILEY, STURGES AND MORRIS The thin rectangular plate shown in Fig. P10-63 is uniformly deformed such that ε n = +1575 µin./in. , (a) (b) ε t = +1350 µ in./in. , and ε x = +1250 µin./in. . The shearing strain γ nt in the plate. The normal strain Determine εy in the plate. SOLUTION The given values are: ε n = +1575 µin./in. θ n = 45° (a) ε t = +1350 µ in./in. θ x / n = −45° ε x = +1250 µin./in. θ y / n = 45° ε x = ε n cos 2 θ x / n + ε t sin 2 θ x / n + γ nt sin θ x / n cos θ x / n 1250 = 1575cos 2 ( −45° ) + 1350sin 2 ( −45° ) + γ nt sin ( −45° ) cos ( −45° ) γ nt = +425 µ rad (b) ..................................................................................................................... Ans. ε y = ε n cos 2 θ y / n + ε t sin 2 θ y / n + γ nt sin θ y / n cosθ y / n = 1575cos 2 ( 45° ) + 1350sin 2 ( 45° ) + γ nt sin ( 45° ) cos ( 45° ) = +1675 µ in./in. ............................................................................................................... Ans. 10-64 The thin rectangular plate shown in Fig. P10-64 is uniformly deformed such that ε BD = +750 µ m/m , and ε AD γ xy in the plate. SOLUTION The given values are: ε AB = −1200 µ m/m , = −600 µ m/m . Determine the normal strain ε y and the shearing strain ε AB = −1200 µ m/m ε BD = +750 µ m/m ε AD = −600 µ m/m θ AB = tan −1 ( 240 200 ) = 50.19° θ BD = − tan −1 ( 240 200 ) = −50.19° ε AB = ε x cos 2 θ AB + ε y sin 2 θ AB + γ xy sin θ AB cosθ AB −1200 = ( −600 ) cos 2 ( 50.19° ) + ε y sin 2 ( 50.19° ) + γ xy sin ( 50.19° ) cos ( 50.19° ) 0.5901ε y + 0.4918γ xy = −954.1 (a) ε BD = ε x cos 2 θ BD + ε y sin 2 θ BD + γ xy sin θ BD cosθ BD 750 = ( −600 ) cos 2 ( −50.19° ) + ε y sin 2 ( −50.19° ) + γ xy sin ( −50.19° ) cos ( −50.19° ) 0.5901ε y − 0.4918γ xy = +995.9 Solving Eqs. (a) and (b) gives: (b) ε y = +35.4 µ m/m ............................................................................................................. Ans. γ xy = −1983 µ rad ............................................................................................................. Ans. 482 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 10-65 RILEY, STURGES AND MORRIS The strain components for a point in a body subjected to plane strain are ε x = +600 µin./in. , ε y = −200 µ in./in. , and γ xy = +500 µ rad . Determine the principal strains and the maximum shearing strain at the point. Show the principal strain deformations and the maximum shearing strain distortion on a sketch. SOLUTION The given values are: ε x = +600 µin./in. γ xy = +500 µ rad ε p1, p 2 ε y = −200 µ in./in. εx +εy ε x − ε y γ xy = ± + 2 2 2 = 600 + ( −200 ) 2 2 2 600 − ( −200 ) 500 ± + 2 2 2 2 ε p1 = 200 + 471.7 = 671.7 µ in./in. ≅ 672 µ in./in. .......................................................... Ans. ε p 2 = 200 − 471.7 = −271.7 µ in./in. ≅ −272 µ in./in. .................................................... Ans. ε p 3 = ε z = 0 µ in./in. .................. γ p = γ max = 2 ( 471.7 ) = 943 µ rad ............................. Ans. γ 1 1 500 θ p = tan −1 xy = tan −1 = 16.00° 2 600 − ( −200 ) εx −εy 2 10-66 .................................................. Ans. The strain components for a point in a body subjected to plane strain are ε x = +960 µ m/m , ε y = −320 µ m/m , and γ xy = −480 µ rad . Determine the principal strains and the maximum shearing strain at the point. Show the principal strain deformations and the maximum shearing strain distortion on a sketch. SOLUTION The given values are: ε x = +960 µ m/m ε y = −320 µ m/m 2 2 γ xy = −480 µ rad ε p1, p 2 εx +εy ε x − ε y γ xy = ± + 2 2 2 = 960 + ( −320 ) 2 960 − ( −320 ) −480 ± + 2 2 2 2 ε p1 = 320 + 683.5 = 1003.5 µ m/m ≅ 1004 µ m/m ........................................................... Ans. ε p 2 = 320 − 683.5 = −363.5 µ m/m ≅ −364 µ m/m ......................................................... Ans. ε p 3 = ε z = 0 µ m/m .................... γ p = γ max = 2 ( 683.5 ) = 1367 µ rad ........................... Ans. γ 1 1 −480 = −10.28° θ p = tan −1 xy = tan −1 2 960 − ( −320 ) εx − ε y 2 ............................................... Ans. 483 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 10-67 RILEY, STURGES AND MORRIS The strain components for a point in a body subjected to plane strain are ε x = +900 µin./in. , ε y = −300 µ in./in. , and γ xy = −420 µ rad . Determine the principal strains and the maximum shearing strain at the point. Show the principal strain deformations and the maximum shearing strain distortion on a sketch. SOLUTION The given values are: ε x = +900 µin./in. 2 2 ε y = −300 µ in./in. γ xy = −420 µ rad ε p1, p 2 ε +εy ε − ε y γ xy =x ±x + 2 2 2 900 − ( −300 ) −420 900 + ( −300 ) = ± + 2 2 2 2 2 ε p1 = 300 + 635.7 = 935.7 µ in./in. ≅ 936 µ in./in. .......................................................... Ans. ε p 2 = 300 − 635.7 = −335.7 µ in./in. ≅ −336 µ in./in. .................................................... Ans. ε p 3 = ε z = 0 µ in./in. .................. γ p = γ max = 2 ( 635.7 ) = 1271 µ rad ........................... Ans. γ 1 1 −420 = −9.65° .................................................. Ans. θ p = tan −1 xy = tan −1 2 900 − ( −300 ) εx − ε y 2 10-68 The strain components for a point in a body subjected to plane strain are ε x = −900 µ m/m , ε y = +600 µ m/m , and γ xy = +480 µ rad . Determine the principal strains and the maximum shearing strain at the point. Show the principal strain deformations and the maximum shearing strain distortion on a sketch. SOLUTION The given values are: ε x = −900 µ m/m 2 2 ε y = +600 µ m/m γ xy = +480 µ rad ε p1, p 2 ε +εy ε − ε y γ xy =x ±x + 2 2 2 ( −900 ) + 600 ± = 2 ( −900 ) − 600 480 + 2 2 2 2 ε p1 = −150 + 787.4 = 637.4 µ m/m ≅ 637 µ m/m ........................................................... Ans. ..................................................... Ans. ε p 2 = −150 − 787.4 = −937.4 µ m/m ≅ −937 µ m/m ε p 3 = ε z = 0 µ m/m ................................................................................................................. Ans. γ p = γ max = 2 ( 787.4 ) = 1575 µ rad .................................................................................... Ans. γ 1 1 480 θ p = tan −1 xy = tan −1 = −8.87° 2 εx −εy 2 ( −900 ) − 600 ................................................. Ans. 484 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 10-69 RILEY, STURGES AND MORRIS The strain components for a point in a body subjected to plane strain are ε x = −750 µin./in. , ε y = +1000 µ in./in. , and γ xy = −250 µ rad . Determine the principal strains and the maximum shearing strain at the point. Show the principal strain deformations and the maximum shearing strain distortion on a sketch. SOLUTION The given values are: ε x = −750 µin./in. 2 ε y = +1000 µ in./in. 2 γ xy = −250 µ rad ε p1, p 2 εx +εy ε x − ε y γ xy = ± + 2 2 2 ( −750 ) + 1000 ± = 2 ( −750 ) − 1000 −250 + 2 2 2 2 ε p1 = 125 + 883.9 = 1008.9 µ in./in. ≅ 1009 µ in./in. ....................................................... Ans. ε p 2 = 125 − 883.9 = −758.9 µ in./in. ≅ −759 µ in./in. ..................................................... Ans. ε p 3 = ε z = 0 µ in./in. .................. γ p = γ max = 2 ( 883.9 ) = 1768 µ rad ........................... Ans. γ 1 1 −250 = 4.07° .................................................. Ans. θ p = tan −1 xy = tan −1 2 εx −εy 2 ( −750 ) − 1000 10-70 The strain components for a point in a body subjected to plane strain are ε x = +750 µ m/m , ε y = −410 µ m/m , and γ xy = +360 µ rad . Determine the principal strains and the maximum shearing strain at the point. Show the principal strain deformations and the maximum shearing strain distortion on a sketch. SOLUTION The given values are: ε x = +750 µ m/m 2 ε y = −410 µ m/m 2 γ xy = +360 µ rad ε p1, p 2 εx +εy ε x − ε y γ xy = ± + 2 2 2 = 750 + ( −410 ) 2 750 − ( −410 ) 360 ± + 2 2 2 2 ε p1 = 170 + 607.3 = 777.3 µ m/m ≅ 777 µ m/m ............................................................... Ans. ε p 2 = 170 − 607.3 = −437.3 µ m/m ≅ −437 µ m/m ......................................................... Ans. ε p 3 = ε z = 0 µ m/m .................... γ p = γ max = 2 ( 607.3) = 1215 µ rad ........................... Ans. γ 1 1 360 θ p = tan −1 xy = tan −1 = 8.62° ..................................................... Ans. 2 750 − ( −410 ) εx −εy 2 485 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 10-71 RILEY, STURGES AND MORRIS The strain components for a point in a body subjected to plane strain are ε x = +720 µin./in. , ε y = +520 µ in./in. , and γ xy = +480 µ rad . Determine the principal strains and the maximum shearing strain at the point. Show the principal strain deformations and the maximum shearing strain distortion on a sketch. SOLUTION The given values are: ε x = +720 µin./in. ε y = +520 µ in./in. γ xy = +480 µ rad ε +εy ε − ε y γ xy =x ±x + 2 2 2 2 2 2 ε p1, p 2 720 + 520 720 − 520 480 = ± + 2 2 2 2 ε p1 = 620 + 260.0 = 880 µ in./in. ........................................................................................ Ans. ε p 2 = 620 − 260.0 = 360 µ in./in. ........................................................................................ Ans. ε p 3 = ε z = 0 µ in./in. .................. γ p = 2 ( 260.0 ) = 520 µ rad γ 1 1 480 θ p = tan −1 xy = tan −1 = 33.69° 2 720 − 520 εx −εy 2 10-72 ......................................... Ans. γ max = ( ε max − ε min ) = ( 880.0 − 0 ) = 880 µ rad ................................................................. Ans. ........................................................ Ans. The strain components for a point in a body subjected to plane strain are ε x = −540 µ m/m , ε y = −980 µ m/m , and γ xy = −560 µ rad . Determine the principal strains and the maximum shearing strain at the point. Show the principal strain deformations and the maximum shearing strain distortion on a sketch. SOLUTION The given values are: ε x = −540 µ m/m ε y = −980 µ m/m γ xy = −560 µ rad εx +εy ε x − ε y γ xy = ± + 2 2 2 2 2 ε p1, p 2 ( −540 ) + ( −980 ) ± = 2 ( −540 ) − ( −980 ) −560 + 2 2 2 2 ε p1 = −760 + 356.1 = −403.9 µ m/m ≅ −404 µ m/m ....................................................... Ans. ε p 2 = −760 − 356.1 = −1116.1 µ m/m ≅ −1116 µ m/m ................................................. Ans. ε p 3 = ε z = 0 µ m/m .................... γ p = 2 ( 356.1) ≅ 712 µ rad .......................................... Ans. γ 1 1 −560 = −25.92° ......................................... Ans. θ p = tan −1 xy = tan −1 2 εx −εy 2 ( −540 ) − ( −980 ) γ max = ( ε max − ε min ) = 0 − ( −1116 ) = 1116 µ rad ......................................................... Ans. 486 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 10-73 RILEY, STURGES AND MORRIS The strain components for a point in a body subjected to plane strain are ε x = +864 µin./in. , ε y = +432 µ in./in. , and γ xy = −288 µ rad . Determine the principal strains and the maximum shearing strain at the point. Show the principal strain deformations and the maximum shearing strain distortion on a sketch. SOLUTION The given values are: ε x = +864 µin./in. ε y = +432 µ in./in. γ xy = −288 µ rad ε +εy ε − ε y γ xy =x ±x + 2 2 2 2 2 2 ε p1, p 2 864 + 432 864 − 432 −288 = ± + 2 2 2 2 ε p1 = 648 + 259.6 = 907.6 µ in./in. ≅ 908 µ in./in. ...................Ans. ε p 2 = 648 − 259.6 = 388.4 µ in./in. ≅ 388 µ in./in. ...................Ans. ε p 3 = ε z = 0 µ in./in. .................. γ p = 2 ( 259.6 ) ≅ 519 µ rad ......................................... Ans. γ max = ( ε max − ε min ) = ( 908 − 0 ) = 908 µ rad .................................................................... Ans. γ 1 1 −288 = −16.85° ...................................................... Ans. θ p = tan −1 xy = tan −1 2 864 − 432 εx −εy 2 10-74 The strain components for a point in a body subjected to plane strain are ε x = +650 µ m/m , ε y = +900 µ m/m , and γ xy = +300 µ rad . Determine the principal strains and the maximum shearing strain at the point. Show the principal strain deformations and the maximum shearing strain distortion on a sketch. SOLUTION The given values are: ε x = +650 µ m/m γ xy = +300 µ rad 2 2 ε y = +900 µ m/m ε p1, p 2 ε +εy ε − ε y γ xy =x ±x + 2 2 2 2 650 + 900 650 − 900 300 = ± + 2 2 2 2 ε p1 = 775 + 195.3 = 970.3 µ m/m ≅ 970 µ m/m ............................................................... Ans. ε p 2 = 775 − 195.3 = 579.7 µ m/m ≅ 580 µ m/m .............................................................. Ans. .......................................... Ans. γ max = ( ε max − ε min ) = ( 970 − 0 ) = 970 µ rad .................................................................... Ans. γ 1 1 300 θ p = tan −1 xy = tan −1 = −25.09° ...................................................... Ans. 2 650 − 900 εx −εy 2 ε p 3 = ε z = 0 µ m/m .................... γ p = 2 (195.3) ≅ 391 µ rad 487 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 10-75 RILEY, STURGES AND MORRIS The strain components for a point in a body subjected to plane strain are ε x = −325 µ in./in. , ε y = −625 µ in./in. , and γ xy = +380 µ rad . Determine the principal strains and the maximum shearing strain at the point. Show the principal strain deformations and the maximum shearing strain distortion on a sketch. SOLUTION The given values are: ε x = −325 µ in./in. ε y = −625 µ in./in. γ xy = +380 µ rad εx +εy ε x − ε y γ xy = ± + 2 2 2 2 2 ε p1, p 2 ( −325 ) + ( −625 ) ± = 2 ( −325 ) − ( −625 ) 380 + 2 2 2 2 ε p1 = −475 + 242.1 = −232.9 µin./in. ≅ −233 µin./in. ................................................... Ans. ε p 2 = −475 − 242.1 = −717.1 µ in./in. ≅ −717 µ in./in. .................................................. Ans. ε p 3 = ε z = 0 µ in./in. .................. γ p = 2 ( 242.1) ≅ 484 µ rad .......................................... Ans. γ 1 1 380 θ p = tan −1 xy = tan −1 = 25.85° ............................................ Ans. 2 εx − ε y 2 ( −325 ) − ( −625 ) 10-76 The strain components for a point in a body subjected to plane strain are γ max = ( ε max − ε min ) = 0 − ( −717 ) = 717 µ rad ............................................................. Ans. ε x = +900 µ m/m , ε y = +650 µ m/m , and γ xy = +600 µ rad . Determine the principal strains and the maximum shearing strain at the point. Show the principal strain deformations and the maximum shearing strain distortion on a sketch. SOLUTION The given values are: ε x = +900 µ m/m ε y = +650 µ m/m γ xy = +600 µ rad ε +εy ε − ε y γ xy =x ±x + 2 2 2 2 2 2 ε p1, p 2 900 + 650 900 − 650 600 = ± + 2 2 2 2 ε p1 = 775 + 325 = 1100 µ m/m ............................................................................................ Ans. ε p 2 = 775 − 325 = 450 µ m/m .............................................................................................. Ans. ε p 3 = ε z = 0 µ m/m .................... γ p = 2 ( 325) ≅ 650 µ rad ............................................. Ans. γ max = ( ε max − ε min ) = (1100 − 0 ) = 1100 µ rad ................................................................ Ans. ........................................................ Ans. γ 1 1 600 θ p = tan −1 xy = tan −1 = 33.69° 2 900 − 650 εx −εy 2 488 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 10-77 Given that RILEY, STURGES AND MORRIS at a point in a body ε p1 = +600 µ in./in. , ε p 2 = −400 µ in./in. , and θ p = +18.43° subjected to plane strain, use Mohr’s circle to determine showing the angle ε x , ε y , γ xy , γ p , and γ max , and prepare a sketch θ p , the principal strain deformations, and the maximum shearing strain distortions. SOLUTION The given values for use in drawing Mohr’s circle are: ε p1 = +600 µ in./in. a= ε p 2 = −400 µ in./in. ε p 3 = ε z = 0 µ in./in. R= θ p = 18.43° 600 + ( −400 ) = 100 µ in./in. 2 600 − ( −400 ) = 500 µ in./in. 2 ε x = 100 + 500cos ( 36.86° ) = +500 µin./in. .................................................................... Ans. ε y = 100 − 500 cos ( 36.86° ) = −300 µ in./in. ................................................................... Ans. γ xy = 2 ( 500 ) sin ( 36.86° ) = 600 µ rad = +600 µ rad ................................................. Ans. γ max = γ p = 2 ( 500 ) = 1000 µ rad ........................................................................................ Ans. 10-78 Given that ε p1 = +785 µ m/m , ε p 2 = −945 µ m/m , and θ p = +16.85° at a point in a body subjected to plane strain, use Mohr’s circle to determine showing the angle ε x , ε y , γ xy , γ p , and γ max , and prepare a sketch θ p , the principal strain deformations, and the maximum shearing strain distortions. SOLUTION The given values for use in drawing Mohr’s circle are: ε p1 = +785 µ m/m ε p 2 = −945 µ m/m ε p 3 = ε z = 0 µ m/m θ p = +16.85° a= 785 + ( −945 ) = −80 µ m/m 2 785 − ( −945) R= = 865 µ m/m 2 489 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ε x = −80 + 865cos ( 33.70° ) = +640 µ m/m .......... Ans. ε y = −80 − 865cos ( 33.70° ) = −800 µ m/m .......... Ans. γ xy = 2 ( 865) sin ( 33.70° ) = +960 µ rad ................. Ans. γ max = γ p = 2 ( 865 ) = 1730 µ rad 10-79 Given that ............................. Ans. ε p1 = +708 µ in./in. , ε p 2 = −104 µ in./in. , and θ p = −34.10° at a point in a body subjected to plane strain, use Mohr’s circle to determine showing the angle ε x , ε y , γ xy , γ p , and γ max , and prepare a sketch θ p , the principal strain deformations, and the maximum shearing strain distortions. SOLUTION The given values for use in drawing Mohr’s circle are: ε p1 = +708 µ in./in. ε p 3 = ε z = 0 µ in./in. a= ε p 2 = −104 µ in./in. θ p = −34.10° 708 + ( −104 ) = 302 µ in./in. 2 708 − ( −104 ) R= = 406 µ in./in. 2 ε x = 302 + 406 cos ( −68.20° ) = +453 µ in./in. ...............Ans. ε y = 302 − 406 cos ( −68.20° ) = +151.2 µ in./in. ...........Ans. γ xy = 2 ( 406 ) sin ( 68.20° ) = 754 µ rad = −754 µ rad ........................................Ans. ........................................Ans. γ max = γ p = 2 ( 406 ) = 812 µ rad 10-80 Given that ε p1 = −114 µ m/m , ε p 2 = −903 µ m/m , and θ p = +19.26° at a point in a body subjected to plane strain, use Mohr’s circle to determine showing the angle ε x , ε y , γ xy , γ p , and γ max , and prepare a sketch θ p , the principal strain deformations, and the maximum shearing strain distortions. SOLUTION The given values for use in drawing Mohr’s circle are: ε p1 = −114 µ m/m a= ε p 2 = −903 µ m/m ε p 3 = ε z = 0 µ m/m R= θ p = +19.26° ( −114 ) + ( −903) = −508.5 µ m/m 2 ( −114 ) − ( −903) = 394.5 µ m/m 2 ........................................................... Ans. ε x = −508.5 + 394.5cos ( 38.52° ) = −199.8 µ m/m ........................................................ Ans. ε y = −508.5 − 394.5cos ( 38.52° ) = −817 µ m/m 490 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS γ xy = 2 ( 394.5 ) sin ( 38.52° ) = 491 µ rad = +491 µ rad .............................................. Ans. γ p = 2 ( 394.5 ) = 789 µ rad ................................................................................................... Ans. γ max = ( ε max − ε min ) = 0 − ( −903) = 903 µ rad .............................................................. Ans. 10-81 Given that ε x = +950 µin./in. , ε y = −225 µ in./in. , and γ xy = +275 µ rad at a point in a body subjected to plane strain, use Mohr’s circle to determine sketch showing the angle distortions. SOLUTION The given values for use in drawing Mohr’s circle are: ε p1 , ε p 2 , γ p , γ max , and θ p , and prepare a θ p , the principal strain deformations, and the maximum shearing strain ε x = +950 µin./in. a= ε y = −225 µ in./in. γ xy = +275 µ rad 950 + ( −225) 2 2 R = ( 587.5 ) + (137.5) = 603.4 µ in./in. = 362.5 µ in./in. 2 ε p1 = 362.5 + 603.4 = +965.9 µ in./in. ≅ +966 µ in./in. ................................................. Ans. ε p 2 = 362.5 − 603.4 = −240.9 µ in./in. ≅ −241 µ in./in. ................................................. Ans. ε p 3 = 0 µ in./in. ............ γ max = γ p = 2 ( 603.4 ) ≅ 1207 µ rad .......................................... Ans. 1 137.5 θ p = tan −1 = 6.59° 2 587.5 .................................................................................................. Ans. 491 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 10-82 Given that RILEY, STURGES AND MORRIS at a point in a body ε x = +900 µ m/m , ε y = −333 µ m/m , and γ xy = +982 µ rad subjected to plane strain, use Mohr’s circle to determine sketch showing the angle distortions. SOLUTION The given values for use in drawing Mohr’s circle are: ε p1 , ε p 2 , γ p , γ max , and θ p , and prepare a θ p , the principal strain deformations, and the maximum shearing strain ε x = +900 µ m/m ε y = −333 µ m/m γ xy = +982 µ rad a= R= 900 + ( −333) = 283.5 µ m/m 2 ( 616.5) + ( 491) 2 2 = 788.1 µ m/m ................................................ Ans. ε p1 = 283.5 + 788.1 = +1071.6 µ m/m ≅ +1072 µ m/m ε p 2 = 283.5 − 788.1 = −504.6 µ m/m ≅ −505 µ m/m ..................................................... Ans. ε p 3 = 0 µ m/m ............... γ max = γ p = 2 ( 788.1) ≅ 1576 µ rad .......................................... Ans. 1 491 θ p = tan −1 = 19.27° 2 616.5 ................................................................................................ Ans. 10-83 Given that ε x = −750 µin./in. , ε y = −390 µ in./in. , and γ xy = +900 µ rad at a point in a body subjected to plane strain, use Mohr’s circle to determine sketch showing the angle distortions. SOLUTION The given values for use in drawing Mohr’s circle are: ε p1 , ε p 2 , γ p , γ max , and θ p , and prepare a θ p , the principal strain deformations, and the maximum shearing strain ε x = −750 µin./in. a= ε y = −390 µ in./in. γ xy = +900 µ rad 2 ε p1 = −570 + 484.7 = −85.3 µ in./in. .................................................................................. Ans. ( −750 ) + ( −390 ) = −570 µin./in. R = 1802 + 4502 = 484.7 µ in./in. ε p 2 = −570 − 484.7 = −1054.7 µ in./in. ≅ −1055 µ in./in. ............................................ Ans. ε p 3 = 0 µ in./in. ............ γ p = 2 ( 484.7 ) ≅ 969 µ rad ........................................................ Ans. 492 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS γ max = ( ε max − ε min ) = 0 − ( −1055 ) = 1055 µ rad ......................................................... Ans. 1 −450 = −34.10° .............................................................................................. Ans. θ p = tan −1 2 180 10-84 Given that ε x = +600 µ m/m , ε y = +480 µ m/m , and γ xy = −480 µ rad at a point in a body subjected to plane strain, use Mohr’s circle to determine sketch showing the angle distortions. SOLUTION The given values for use in drawing Mohr’s circle are: ε p1 , ε p 2 , γ p , γ max , and θ p , and prepare a θ p , the principal strain deformations, and the maximum shearing strain ε x = +600 µ m/m a= ε y = +480 µ m/m γ xy = −480 µ rad 600 + 480 = 540 µ m/m 2 R = 602 + 2402 = 247.4 µ m/m ε p1 = 540 + 247.4 = +787.4 µ m/m ≅ +787 µ m/m ........................................................ Ans. ε p 2 = 540 − 247.4 = +292.6 µ m/m ≅ +293 µ m/m ........................................................ Ans. ε p 3 = 0 µ m/m ............................. γ p = 2 ( 247.4 ) ≅ 495 µ rad .......................................... Ans. γ max = ( ε max − ε min ) = ( 787 − 0 ) = 787 µ rad .................................................................... Ans. 1 −480 = −37.98° .............................................................................................. Ans. θ p = tan −1 2 120 493 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 10-85 Given that RILEY, STURGES AND MORRIS ε x = −680 µin./in. , ε y = +320 µ in./in. , and ε p1 = +414 µ in./in. at a point in a body γ xy , ε p 2 , γ p , γ max , and θ p , and prepare a θ p , the principal strain deformations, and the maximum shearing strain subjected to plane strain, use Mohr’s circle to determine sketch showing the angle distortions. SOLUTION The given values for use in drawing Mohr’s circle are: ε x = −680 µin./in. a= 2 ε y = +320 µ in./in. ε p1 = +414 µ in./in. ( −680 ) + 320 = −180 µin./in. γ xy = 2 R 2 − ( ε x − a ) 2 R = ε p1 − a = 414 − ( −180 ) = 594 µ in./in. = 2 5942 − 320 − ( −180 ) = ±641 µ in./in. ......................................................... Ans. 2 ε p 2 = −180 − 594 = −774 µ in./in. ...................................................................................... Ans. ε p 3 = 0 µ in./in. .......................... γ max = γ p = 2 ( 594 ) ≅ 1188 µ rad ............................... Ans. 1 ±321 = ±16.35° θ p = tan −1 2 500 .............................................................................................. Ans. 10-86 Given that ε x = +450 µ m/m , ε y = +150 µ m/m , and ε p1 = +780 µ m/m at a point in a body subjected to plane strain, use Mohr’s circle to determine sketch showing the angle distortions. SOLUTION The given values for use in drawing Mohr’s circle are: γ xy , ε p 2 , γ p , γ max , and θ p , and prepare a θ p , the principal strain deformations, and the maximum shearing strain ε x = +450 µ m/m a= ε y = +150 µ m/m ε p1 = +780 µ m/m 450 + 150 = 300 µ m/m 2 2 R = ε p1 − a = 780 − 300 = 480 µ m/m 2 γ xy = 2 R 2 − ( ε x − a ) = 2 4802 − ( 450 − 300 ) = ±911.9 µ m/m ≅ ±912 µ m/m .................................................................................. Ans. 494 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ε p 2 = 300 − 480 = −180 µ m/m ........................................................................................... Ans. ε p 3 = 0 µ m/m ............................. γ max = γ p = 2 ( 480 ) ≅ 960 µ rad ................................. Ans. 1 ±456 = ±35.89° .............................................................................................. Ans. θ p = tan −1 2 150 10-87 Given that ε x = +360 µin./in. , ε y = +750 µ in./in. , and ε p 2 = +197 µin./in. at a point in a body subjected to plane strain, use Mohr’s circle to determine sketch showing the angle distortions. SOLUTION The given values for use in drawing Mohr’s circle are: γ xy , ε p1 , γ p , γ max , and θ p , and prepare a θ p , the principal strain deformations, and the maximum shearing strain ε x = +360 µin./in. ε y = +750 µ in./in. ε p 2 = +197 µin./in. a= 360 + 750 = +555 µ in./in. 2 R = a − ε p 2 = 555 − 197 = 358 µ in./in. 2 2 γ xy = 2 R 2 − ( ε y − a ) = 2 3582 − ( 750 − 555 ) = ±600 µ in./in. .......................................................Ans. ε p1 = 555 + 358 = +913 µ in./in. .................................Ans. ε p 3 = 0 µ in./in. ..............................................................Ans. γ p = 2 R = 2 ( 358) = 716 µ rad ...................................Ans. γ max = ε max − ε min = ( 913 − 0 ) = 913 µ rad ...............Ans. 1 ±300 = ±28.49° .....................................Ans. θ p = tan −1 2 195 495 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 10-88 Given that RILEY, STURGES AND MORRIS at a point in a body ε x = −300 µ m/m , ε y = +600 µ m/m , and ε p 2 = −522 µ m/m subjected to plane strain, use Mohr’s circle to determine sketch showing the angle distortions. SOLUTION The given values for use in drawing Mohr’s circle are: γ xy , ε p1 , γ p , γ max , and θ p , and prepare a θ p , the principal strain deformations, and the maximum shearing strain ε x = −300 µ m/m a= 2 ε y = +600 µ m/m ε p 2 = −522 µ m/m ( −300 ) + 600 = 150 µ m/m 2 R = a − ε p 2 = 150 − ( −522 ) = 672 µ m/m 2 γ xy = 2 R 2 − ( ε y − a ) = 2 6722 − ( 600 − 150 ) = ±998.2 µ m/m ≅ ±998 µ m/m .................................................................................. Ans. ε p1 = 150 + 672 = +822 µ m/m ............................................................................................ Ans. ε p 3 = 0 µ m/m ............................. γ max = γ p = 2 ( 672 ) = 1344 µ rad .............................. Ans. 1 ±499 = ±23.98° .............................................................................................. Ans. θ p = tan −1 2 450 10-89 At a point on the surface of an aluminum alloy (E = 10,000 ksi and G = 3800 ksi) machine part subjected to a biaxial state of stress, the measured strains were ε x = +900 µin./in. , ε y = −300 µ in./in. , and γ xy = −400 µ rad . SOLUTION The given values are: Determine the stresses σ x , σ y , and τ xy at the point. ε x = +900 µin./in. G = 3800 ksi ε y = −300 µ in./in. E = 10, 000 ksi = 2 (1 + ν ) G γ xy = −400 µ rad ν = 0.3158 σx = E 10, 000 ε + νε y ) = 900 + 0.3158 ( −300 ) (10−6 ) 2( x 2 1 −ν 1 − 0.3158 = +8.94 ksi = 8.94 ksi (T) ............................................................................................. Ans. 496 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS σy = E 10, 000 ε + νε x ) = −300 ) + 0.3158 ( 900 ) (10−6 ) 2( y 2 ( 1 −ν 1 − 0.3158 = −0.1753 ksi = 0.1753 ksi (C) .................................................................................... Ans. τ xy = Gγ xy = 3800 ( −400 × 10−6 ) = −1.520 ksi ................................................................ Ans. 10-90 At a point on the surface of an structural steel (E = 200 GPa and G = 76 GPa) machine part subjected to a biaxial state of stress, the measured strains were ε x = +750 µ m/m , ε y = +350 µ m/m , and γ xy = −560 µ rad . SOLUTION The given values are: Determine the stresses σ x , σ y , and τ xy at the point. ε x = +750 µ m/m G = 76 GPa ε y = +350 µ m/m E = 200 GPa = 2 (1 + ν ) G γ xy = −560 µ rad ν = 0.3158 σx = E 200 × 109 ε x + νε y ) = 750 + 0.3158 ( 350 ) (10 −6 ) 2( 2 1 −ν 1 − 0.3158 E 200 × 109 ε y + νε x ) = 350 + 0.3158 ( 750 ) (10 −6 ) 2( 2 1 −ν 1 − 0.3158 .......................... Ans. = +191.2 N/m 2 = 191.2 MPa (T) ................................................................................. Ans. σy = = +130.4 N/m 2 = 130.4 MPa (T) ................................................................................. Ans. τ xy = Gγ xy = ( 76 × 109 )( −560 × 10−6 ) = −42.6 N/m 2 = −42.6 MPa 10-91 At a point on the surface of a titanium alloy (E = 14,000 ksi and G = 5300 ksi) machine part subjected to a biaxial state of stress, the measured strains were ε x = +1250 µ in./in. , ε y = +600 µ in./in. , and γ xy = +650 µ rad . SOLUTION The given values are: Determine the stresses σ x , σ y , and τ xy at the point. ε x = +1250 µin./in. G = 5300 ksi ε y = +600 µ in./in. E = 14, 000 ksi = 2 (1 + ν ) G γ xy = +650 µ rad ν = 0.3208 σx = E 14, 000 ε + νε y ) = 1250 + 0.3208 ( 600 ) (10−6 ) 2( x 1 −ν 1 − 0.32082 = +22.5 ksi = 22.5 ksi (T) ............................................................................................. Ans. E 14, 000 ε + νε x ) = 600 + 0.3208 (1250 ) (10−6 ) 2( y 2 1 −ν 1 − 0.3208 = +15.62 ksi = 15.62 ksi (T) ......................................................................................... Ans. σy = τ xy = Gγ xy = 5300 ( 650 × 10−6 ) = +3.45 ksi ..................................................................... Ans. 497 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 10-92 RILEY, STURGES AND MORRIS At a point on the surface of a stainless steel (E = 190 GPa and G = 76 GPa) machine part subjected to a biaxial state of stress, the measured strains were ε x = +1175 µ m/m , ε y = −1250 µ m/m , and γ xy = +850 µ rad . SOLUTION The given values are: Determine the stresses σ x , σ y , and τ xy at the point. ε x = +1175 µ m/m G = 76 GPa ε y = −1250 µ m/m E = 190 GPa = 2 (1 + ν ) G γ xy = +850 µ rad ν = 0.250 σx = E 190 × 109 ε x + νε y ) = 1175 + 0.250 ( −1250 ) (10 −6 ) ( 1 −ν 2 1 − 0.2502 E 190 × 109 ε y + νε x ) = −1250 ) + 0.250 (1175 ) (10−6 ) 2( 2 ( 1 −ν 1 − 0.250 = +174.8 N/m 2 = 174.8 MPa (T) ................................................................................. Ans. σy = = −193.8 N/m 2 = 193.8 MPa (C) ................................................................................. Ans. τ xy = Gγ xy = ( 76 × 109 )( 850 ×10−6 ) = +64.6 N/m 2 = +64.6 MPa ............................. Ans. 10-93 Determine the state of strain that corresponds to the following state of stress at a point in a steel (E = 30,000 ksi and ν = 0.30 ) machine part: σ x = 15, 000 psi , σ y = 5000 psi , σ z = 7500 psi , τ xy = 5500 psi , τ yz = 4750 psi , and τ zx = 3200 psi . SOLUTION The given values are: σ x = 15.0 ksi τ xy = 5.5 ksi εx = σ y = 5.0 ksi τ yz = 4.75 ksi σ z = 7.5 ksi τ zx = 3.2 ksi γ xy = E = 30, 000 ksi ν = 0.30 1 σ x − ν (σ y + σ z ) E 2 (1 + ν ) 1 τ xy = τ xy G E εx = εy = εz = 1 15.0 − 0.30 ( 5.0 + 7.5 ) = +375 × 10−6 = +375 µ in./in. ........................ Ans. 30, 000 1 5.0 − 0.30 ( 7.5 + 15.0 ) = −58.3 × 10−6 = −58.3 µ in./in. ..................... Ans. 30, 000 1 7.5 − 0.30 (15.0 + 5.0 ) = +50.0 × 10−6 = +50.0 µ in./in. ..................... Ans. 30, 000 2 (1 + 0.30 ) ( 5.5) = +477 ×10−6 = +477 µ rad ....................................................... Ans. 30, 000 2 (1 + 0.30 ) ( 4.75) = +412 ×10−6 = +412 µ rad .................................................... Ans. 30, 000 2 (1 + 0.30 ) ( 3.20 ) = +277 ×10−6 ≅ +277 µ rad .................................................... Ans. 30, 000 γ xy = γ yz = γ zx = 498 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 10-94 RILEY, STURGES AND MORRIS Determine the state of strain that corresponds to the following state of stress at a point in an aluminum alloy (E = 73 GPa and ν = 0.33 ) machine part: σ x = 120 MPa , σ y = −85 MPa , σ z = 45 MPa , τ xy = 35 MPa , τ yz = 48 MPa , and τ zx = 76 MPa . SOLUTION The given values are: σ x = 120 MPa τ xy = 35 MPa εx = σ y = −85 MPa τ yz = 48 MPa σ z = 45 MPa τ zx = 76 MPa γ xy = E = 73 GPa ν = 0.33 1 σ x − ν (σ y + σ z ) E 2 (1 + ν ) 1 τ xy = τ xy G E εx = εy = εz = 1 120 − 0.33 ( −85 + 45 ) = +1825 ×10−6 = +1825 µ m/m ...................... Ans. 73, 000 1 −85 − 0.33 (120 + 45) = −1910 ×10−6 = −1910 µ m/m ...................... Ans. 73, 000 1 45 − 0.33 (120 − 85) = +458 ×10−6 = +458 µ m/m .............................. Ans. 73, 000 2 (1 + 0.33) ( 35 ) = +1275 ×10−6 = +1275 µ rad ................................................... Ans. 73, 000 2 (1 + 0.33) ( 48 ) = +1749 ×10−6 = +1749 µ rad ................................................... Ans. 73, 000 2 (1 + 0.33) ( 76 ) = +2769 ×10−6 ≅ +2770 µ rad ................................................... Ans. 73, 000 γ xy = γ yz = γ zx = 10-95 Given that ε x = +900 µ in./in. , ε y = −300 µ in./in. , and γ xy = −400 µ rad at a point on the free surface of a machine component ( E = 10, 000 ksi and ν = 0.30 ), determine the principal stresses and the maximum shear stress at the point. Locate the planes on which these stresses act and show the stresses on a complete sketch. SOLUTION The given values are: ε x = +900 µ in./in. ν = 0.30 σx = ε y = −300 µ in./in. E = 10, 000 ksi = 2 (1 + ν ) G γ xy = −400 µ rad G = 3846 ksi τ xy = Gγ xy = 3846 ( −400 × 10−6 ) = −1.5384 ksi 2 E 10, 000 ε + νε y ) = 900 + 0.30 ( −300 ) (10−6 ) = +8.901 ksi 2( x 1 −ν 1 − 0.302 E 10, 000 σy = ε + νε x ) = −300 ) + 0.30 ( 900 ) (10−6 ) = −0.3297 ksi 2( y 2 ( 1 −ν 1 − 0.30 σ p1, p 2 2 σ x +σ y σ x −σ y 8.901 − 0.3297 8.901 + 0.3297 2 2 = ± ± + τ xy = + 1.5384 2 2 2 2 499 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS σ p1 = 4.286 + 4.865 = +9.151 ksi ≅ 9.15 ksi (T) ........................................................... Ans. σ p 2 = 4.286 − 4.865 = −0.579 ksi ≅ 0.579 ksi (C) τ max = τ p = 4.865 ksi ≅ 4.87 ksi ....................................................... Ans. σ p 3 = 0 ksi ............................................................................................................................... Ans. ........................................................................................ Ans. 2τ xy 2 ( −1.5384 ) 1 1 = tan −1 = −9.22° ....................................... Ans. θ p = tan −1 2 8.901 − ( −0.3297 ) σ x −σ y 2 10-96 Given that maximum shear stress at the point. Locate the planes on which these stresses act and show the stresses on a complete sketch. SOLUTION The given values are: ε x = +750 µ m/m , ε y = +350 µ m/m , and γ xy = −560 µ rad at a point on the free surface of a machine component ( E = 200 GPa and ν = 0.30 ), determine the principal stresses and the ε x = +750 µ m/m ν = 0.30 ε y = +350 µ m/m E = 200 GPa = 2 (1 + ν ) G γ xy = −560 µ rad G = 76.92 GPa σx = E 200 × 109 ε x + νε y ) = 750 + 0.30 ( 350 ) (10−6 ) 2( 2 1 −ν 1 − 0.30 = +187.91×106 N/m 2 = 187.91 MPa (T) E 200 × 109 σy = ε + νε x ) = 350 + 0.30 ( 750 ) (10−6 ) 2( y 2 1 −ν 1 − 0.30 = +126.37 ×106 N/m 2 = 126.37 MPa (T) τ xy = Gγ xy = ( 76.92 × 109 )( −560 × 10−6 ) = −43.08 × 106 N/m 2 = −43.08 MPa σ p1, p 2 σ +σ y σ −σ y 2 =x ±x + τ xy 2 2 187.91 + 126.37 187.91 − 126.37 2 = ± + 43.08 2 2 2 2 σ p1 = 157.14 + 52.94 = +210.08 MPa ≅ 210 MPa (T) ................................................ Ans. σ p 2 = 157.14 − 52.94 = +104.20 MPa ≅ 104.2 MPa (T) ............................................. Ans. σ p 3 = 0 ksi ............................................................................................................................... Ans. τ p = 52.94 MPa ≅ 52.9 MPa .............................................................................................. Ans. σ max − σ min 210.08 − 0 = = 105.0 MPa ≅ 105.0 MPa ...................................... Ans. 2 2 2τ xy 2 ( −43.08 ) 1 1 = tan −1 = −27.25° ......................................... Ans. θ p = tan −1 2 187.91 − 126.37 σ x −σ y 2 τ max = 500 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 10-97 Given that RILEY, STURGES AND MORRIS ε x = +1250 µ in./in. , ε y = +600 µ in./in. , and γ xy = +650 µ rad at a point on the free surface of a machine component ( E = 14, 000 ksi and ν = 0.32 ), determine the principal stresses and the maximum shear stress at the point. Locate the planes on which these stresses act and show the stresses on a complete sketch. SOLUTION The given values are: ε x = +1250 µ in./in. ν = 0.32 σx = ε y = +600 µ in./in. E = 14, 000 ksi = 2 (1 + ν ) G γ xy = +650 µ rad G = 5303 ksi τ xy = Gγ xy = 5303 ( 650 × 10−6 ) = +3.447 ksi E 14, 000 ε + νε y ) = 1250 + 0.32 ( 600 ) (10−6 ) = +22.49 ksi 2( x 2 1 −ν 1 − 0.32 E 14, 000 σy = ε + νε x ) = 600 + 0.32 (1250 ) (10 −6 ) = +15.597 ksi 2( y 2 1 −ν 1 − 0.32 σ p1, p 2 σ +σ y σ −σ y 2 =x ±x + τ xy 2 2 = 22.49 + 15.597 22.49 − 15.597 2 ± + 3.447 2 2 ........................................................ Ans. ................................................... Ans. ........................................... Ans. 2 2 σ p1 = 19.044 + 4.874 = +23.92 ksi ≅ 23.9 ksi (T) σ p 2 = 19.044 − 4.874 = +14.170 ksi ≅ 14.17 ksi (T) σ p 3 = 0 ksi .................................. τ p = 4.874 ksi ≅ 4.87 ksi τ max = σ max − σ min 23.92 − 0 = = 11.96 ksi ..................................................................... Ans. 2 2 2τ xy 2 ( 3.447 ) 1 1 = tan −1 = 22.50° ............................................. Ans. θ p = tan −1 2 22.49 − 15.597 σ x −σ y 2 10-98 At a point on the free surface of an aluminum alloy (E = 73 GPa and ν = 0.33 ) machine part, the strain rosette shown in Fig. P10-98 was used to obtain the following normal strain data: ε a = +780 µ m/m , (a) (b) ε b = +345 µ m/m , and ε c = −332 µ m/m . Determine The strain components ε x , ε y , and γ xy at the point. The principal strains and the maximum shearing strain at the point. SOLUTION The given values are: ε a = ε x = +780 µ m/m (a) ε b = ε −60 = +345 µ m/m ε c = ε +60 = −332 µ m/m ε n = ε x cos 2 θ n + ε y sin 2 θ n + γ xy sin θ n cosθ n ε b = 780 cos 2 ( −60° ) + ε y sin 2 ( −60° ) + γ xy sin ( −60° ) cos ( −60° ) = +345 µ m/m ε c = 780 cos 2 ( 60° ) + ε y sin 2 ( 60° ) + γ xy sin ( 60° ) cos ( 60° ) = −332 µ m/m 501 STATICS AND MECHANICS OF MATERIALS, 2nd Edition Therefore RILEY, STURGES AND MORRIS ........................ Ans. ε x = +780 µ m/m ......... ε y = −251 µ m/m ......... γ xy = −782 µ rad εx +εy ε x − ε y γ xy = ± + 2 2 2 = 780 + ( −251) 2 2 2 (b) ε p1, p 2 780 − ( −251) −782 ± + 2 2 2 2 ε p1 = 264.5 + 647 = +911.5 µ m/m ≅ +912 µ m/m ......................................................... Ans. ε p 2 = 264.5 − 647 = −382.5 µ m/m ≅ −383 µ m/m ........................................................ Ans. ε p3 = ε z = − γ p = γ max = 2 ( 647 ) = 1294 µ rad ν 0.33 (ε x + ε y ) = − 1 − 0.33 ( 780 − 251) = −261 µ m/m ........................... Ans. 1 −ν ....................................................................................... Ans. γ 1 1 −782 θ p = tan −1 xy = tan −1 = −18.59° ................................................ Ans. εx −εy 2 2 780 − ( −251) 10-99 At a point on the free surface of a steel (E = 30,000 ksi and ν = 0.30 ) machine part, the strain rosette shown in Fig. P10-99 was used to obtain the following normal strain data: ε a = +750 µ in./in. , (a) (b) ε b = −125 µ in./in. , and ε c = −250 µ in./in. . Determine The strain components ε x , ε y , and γ xy at the point. The principal strains and the maximum shearing strain at the point. SOLUTION The given values are: ε a = ε x = +750 µ in./in. (a) ε b = ε +45 = −125 µ in./in. ε c = ε y = −250 µ in./in. ε n = ε x cos 2 θ n + ε y sin 2 θ n + γ xy sin θ n cos θ n ε b = 750 cos 2 ( 45° ) + ( −250 ) sin 2 ( 45° ) + γ xy sin ( 45° ) cos ( 45° ) = −125 µ in./in. Therefore ε x = +750 µ in./in. ....... ε y = −250 µ in./in. ....... γ xy = −750 µ rad 2 2 ........................ Ans. (b) ε p1, p 2 2 2 εx + ε y ε x − ε y γ xy 750 − 250 750 + 250 −750 = ± ± + = + 2 2 2 2 2 2 ε p1 = 250 + 625 = +875 µ in./in. .......................................................................................... Ans. ε p 2 = 250 − 625 = −375 µ in./in. ......................................................................................... Ans. ε p3 = ε z = − ν 0.30 (ε x + ε y ) = − 1 − 0.30 ( 750 − 250 ) = −214 µin./in. ........................ Ans. 1 −ν γ p = γ max = 2 ( 625) = 1250 µ rad ........................................................................................ Ans. γ 1 1 −750 θ p = tan −1 xy = tan −1 = −18.43° εx −εy 2 2 750 − ( −250 ) 502 ............................................... Ans. STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 10-100 At a point on the free surface of a steel (E = 200 GPa and ν = 0.30 ) machine part, the strain rosette shown in Fig. P10-100 was used to obtain the following normal strain data: ε a = −555 µ m/m , (a) (b) ε b = +925 µ m/m , and ε c = +740 µ m/m . Determine The strain components ε x , ε y , and γ xy at the point. The principal strains and the maximum shearing strain at the point. SOLUTION The given values are: ε a = ε x = −555 µ m/m (a) ε b = ε −60 = +925 µ m/m ε c = ε 60 = +740 µ m/m ε n = ε x cos 2 θ n + ε y sin 2 θ n + γ xy sin θ n cos θ n ε b = −555cos 2 ( −60° ) + ε y sin 2 ( −60° ) + γ xy sin ( −60° ) cos ( −60° ) = +925 µ m/m ε c = −555cos 2 ( 60° ) + ε y sin 2 ( 60° ) + γ xy sin ( 60° ) cos ( 60° ) = +740 µ m/m Therefore ε x = −555 µ m/m ......... ε y = +1295 µ m/m εx +εy ε x − ε y γ xy = ± + 2 2 2 2 2 ...... γ xy = −214 µ rad ........................ Ans. (b) ε p1, p 2 ( −555 ) + 1295 ± = 2 ( −555 ) − 1295 −214 + 2 2 2 2 ε p1 = 370 + 931 = +1301 µ m/m .......................................................................................... Ans. ε p 2 = 370 − 931 = −561 µ m/m ............................................................................................ Ans. ε p3 = ε z = − ν 0.30 (ε x + ε y ) = − 1 − 0.30 ( −555 + 1295 ) = −317 µ m/m ...................... Ans. 1 −ν γ p = γ max = 2 ( 931) = 1862 µ rad ........................................................................................ Ans. γ 1 1 −214 θ p = tan −1 xy = tan −1 = 3.30° εx −εy 2 2 −555 − 1295 ...................................................... Ans. 10-101 The strain rosette shown in Fig. P10-101 was used to obtain normal strain data at a point on the free surface of a structural steel machine part. The gage readings were ε a = −350 µ in./in. , ε b = +1000 µ in./in. , and (a) (b) (c) ε c = +550 µ in./in. . Determine The strain components ε x , ε y , and γ xy The stress components at the point. at the point. σ x , σ y , and τ xy The principal stresses and the maximum shearing stress at the point. Express the stresses in U.S. customary units. SOLUTION The given values are: ε a = ε x = −350 µ in./in. (a) ε b = ε 60 = +1000 µ in./in. ε c = ε −60 = +550 µ in./in. ε n = ε x cos 2 θ n + ε y sin 2 θ n + γ xy sin θ n cos θ n 503 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS ε b = −350 cos 2 ( 60° ) + ε y sin 2 ( 60° ) + γ xy sin ( 60° ) cos ( 60° ) = +1000 µ in./in. ε c = −350 cos 2 ( −60° ) + ε y sin 2 ( −60° ) + γ xy sin ( −60° ) cos ( −60° ) = +550 µ in./in. Therefore ε x = −350 µ in./in. ....... ε y = +1150 µ in./in. .... γ xy = +520 µ rad G = 11, 000 ksi ........................ Ans. (b) For structural steel: E = 29, 000 ksi = 2 (1 + ν ) G ν = 0.3182 σx = E 29, 000 ε + νε y ) = −350 + 0.3182 (1150 ) (10−6 ) 2( x 2 1 −ν 1 − 0.3182 = +0.514 ksi = 0.514 ksi (T) ......................................................................................... Ans. E 29, 000 ε + νε x ) = 1150 + 0.3182 ( −350 ) (10−6 ) 2( y 2 1 −ν 1 − 0.3182 = +33.514 ksi ≅ 33.5 ksi (T) ......................................................................................... Ans. σy = τ xy = Gγ xy = 11, 000 ( 520 ×10 −6 ) = +5.72 ksi .................................................................. Ans. (c) σ p1, p 2 σ +σ y σ −σ y 2 =x ±x + τ xy 2 2 = 0.514 + 33.514 0.514 − 33.514 2 ± + 5.72 2 2 .................................................... Ans. 2 2 σ p1 = 17.012 + 17.463 = +34.475 ksi ≅ 34.5 ksi (T) ..................................................... Ans. σ p 2 = 17.012 − 17.463 = −0.451 ksi ≅ 0.451 ksi (C) σ p 3 = σ z = 0 ksi ........................ τ max = τ p = 17.46 ksi .................................................... Ans. 2τ xy 2 ( 5.72 ) 1 1 θ p = tan −1 = tan −1 = −9.56° ............................................. Ans. 2 σ x −σ y 2 0.514 − 33.514 10-102 The strain rosette shown in Fig. P10-102 was used to obtain normal strain data at a point on the free surface of a 2024-T4 aluminum alloy machine part. The gage readings were ε a = +525 µ m/m , (a) (b) (c) ε b = +450 µ m/m , and ε c = +1425 µ m/m . Determine The strain components ε x , ε y , and γ xy at the point. The stress components σ x , σ y , and τ xy at the point. The principal stresses and the maximum shearing stress at the point. Express the stresses in SI units. SOLUTION The given values are: ε a = ε x = +525 µ m/m (a) ε b = ε 45 = +450 µ m/m ε c = ε −45 = +1425 µ m/m ε n = ε x cos 2 θ n + ε y sin 2 θ n + γ xy sin θ n cos θ n ε b = 525cos 2 ( 45° ) + ε y sin 2 ( 45° ) + γ xy sin ( 45° ) cos ( 45° ) = +450 µ m/m ε c = 525cos 2 ( −45° ) + ε y sin 2 ( −45° ) + γ xy sin ( −45° ) cos ( −45° ) = +1425 µ m/m Therefore ε x = +525 µ m/m ......... ε y = +1350 µ m/m 504 ...... γ xy = −975 µ rad ........................ Ans. STATICS AND MECHANICS OF MATERIALS, 2nd Edition (b) For 2024-T4 aluminum alloy: RILEY, STURGES AND MORRIS G = 28 GPa E = 73 GPa = 2 (1 + ν ) G ν = 0.3036 σx = E 73 × 109 (ε x +νε y ) = 1 − 0.30362 525 + 0.3036 (1350 ) (10−6 ) 1 −ν 2 E 73 ×109 ε x + νε y ) = 1350 + 0.3036 ( 525) (10−6 ) 2( 2 1 −ν 1 − 0.3036 .............. Ans. = +75.17 ×106 N/m 2 ≅ 75.2 MPa (T) ........................................................................ Ans. σy = = +121.37 ×106 N/m 2 ≅ 121.37 MPa (T) ................................................................. Ans. τ xy = Gγ xy = ( 28 × 109 )( −975 × 10−6 ) = −27.30 × 106 N/m 2 ≅ −27.3 MPa 2 (c) σ p1, p 2 2 σ x +σ y σ x −σ y 75.17 + 121.37 2 75.17 − 121.37 2 = ± + τ xy = ± + ( 27.30 ) 2 2 2 2 σ p1 = 98.27 + 35.76 = +134.03 MPa ≅ 134.0 MPa (T) ............................................... Ans. σ p 2 = 98.27 − 35.76 = +62.51 MPa ≅ 62.5 MPa (T) ................................................... Ans. σ p 3 = σ z = 0 MPa .................... τ p = 35.76 MPa ≅ 35.8 MPa .................................... Ans. 1 1 (σ max − σ min ) = (134.03 − 0 ) ≅ 67.0 MPa ..................................................... Ans. 2 2 2τ xy 2 ( −27.30 ) 1 1 θ p = tan −1 = tan −1 = 24.88° ............................................. Ans. 2 σ x −σ y 2 75.17 − 121.37 τ max = 10-103 The strain rosette shown in Fig. P10-103 was used to obtain normal strain data at a point on the free surface of an 18-8 cold-rolled stainless steel machine part. The gage readings were ε a = +665 µ in./in. , (a) (b) (c) ε b = +390 µ in./in. , and ε c = +870 µ in./in. . Determine The strain components ε x , ε y , and γ xy at the point. The stress components σ x , σ y , and τ xy at the point. The principal stresses and the maximum shearing stress at the point. Express the stresses in U.S. customary units. SOLUTION The given values are: ε a = ε x = +665 µ in./in. (a) ε b = ε y = +390 µ in./in. ε c = ε −60 = +870 µ in./in. ε n = ε x cos 2 θ n + ε y sin 2 θ n + γ xy sin θ n cos θ n ε c = 665cos 2 ( −60° ) + 390sin 2 ( −60° ) + γ xy sin ( −60° ) cos ( −60° ) = +870 µ in./in. Therefore ε x = +665 µ in./in. ....... ε y = +390 µ in./in. ....... γ xy = −950 µ rad G = 12,500 ksi ........................ Ans. (b) For 18-8 stainless steel: E = 28, 000 ksi = 2 (1 + ν ) G ν = 0.12 σx = E 28, 000 ε + νε y ) = 665 + 0.12 ( 390 ) (10−6 ) 2( x 2 1 −ν 1 − 0.12 = +20.22 ksi ≅ 20.2 ksi (T) ........................................................................................... Ans. 505 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS σy = E 28, 000 ε + νε x ) = 390 + 0.12 ( 665 ) (10−6 ) 2( y 2 1 −ν 1 − 0.12 = +13.347 ksi ≅ 13.35 ksi (T) ....................................................................................... Ans. τ xy = Gγ xy = 12,500 ( −950 × 10−6 ) = −11.875 ksi ≅ −11.88 ksi ................................. Ans. (c) σ p1, p 2 σ +σ y σ −σ y 2 =x ±x + τ xy 2 2 20.22 + 13.347 20.22 − 13.347 2 = ± + 11.875 2 2 2 2 σ p1 = 16.784 + 12.362 = +29.146 ksi ≅ 29.1 ksi (T) .................................................... Ans. σ p 2 = 16.784 − 12.362 = +4.422 ksi ≅ 4.42 ksi (T) ...................................................... Ans. σ p 3 = σ z = 0 ksi ........................ τ p = 12.362 ksi ≅ 12.36 ksi ........................................ Ans. 1 1 (σ max − σ min ) = ( 29.146 − 0 ) ≅ 14.57 ksi ...................................................... Ans. 2 2 2τ xy 2 ( −11.875 ) 1 1 θ p = tan −1 = tan −1 = −36.93° ........................................... Ans. 2 σ x −σ y 2 20.22 − 13.347 τ max = 10-104 At a point on the free surface of an aluminum alloy (E = 73 GPa and ν = 0.33 ) machine part, the strain rosette shown in Fig. P10-104 was used to obtain the following normal strain data: ε a = +875 µ m/m , (a) (b) (c) ε b = +700 µ m/m , and ε c = −650 µ m/m . Determine The stress components σ x , σ y , and τ xy at the point. The principal strains and the maximum shearing strain at the point. Prepare a sketch showing all of these strains. The principal stresses and the maximum shearing stress at the point. Prepare a sketch showing all of these stresses. SOLUTION The given values are: ε a = ε x = +875 µ m/m (a) ε b = ε −60 = +700 µ m/m ε c = ε 60 = −650 µ m/m ε n = ε x cos 2 θ n + ε y sin 2 θ n + γ xy sin θ n cos θ n ε b = 875 cos 2 ( −60° ) + ε y sin 2 ( −60° ) + γ xy sin ( −60° ) cos ( −60° ) = +700 µ m/m ε c = 875 cos 2 ( 60° ) + ε y sin 2 ( 60° ) + γ xy sin ( 60° ) cos ( 60° ) = −650 µ m/m Therefore ε x = +875 µ m/m ν = 0.33 ε y = −258 µ m/m γ xy = −1559 µ rad G = 27.44 GPa For aluminum alloy: E = 73 GPa = 2 (1 + ν ) G σx = E × 109 (ε x +νε y ) = 173 0.332 875 + 0.33 ( −258) (10−6 ) 1 −ν 2 − 506 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS = +64.71×106 N/m 2 ≅ 64.7 MPa (T) ......................................................................... Ans. E 73 × 109 σy = ε + νε y ) = −258 + 0.33 ( 875) (10−6 ) 2( x 2 1 −ν 1 − 0.33 = +2.519 ×106 N/m 2 ≅ 2.52 MPa (T) ........................................................................ Ans. τ xy = Gγ xy = ( 27.44 ×109 )( −1559 ×10 −6 ) = −42.78 × 106 N/m 2 ≅ −42.8 MPa ........................................................................... Ans. (b) ε p1, p 2 εx +εy ε x − ε y γ xy = ± + 2 2 2 = 875 + ( −258 ) 2 2 2 875 − ( −258 ) −1559 ± + 2 2 2 2 ε p1 = 308.5 + 963.6 = +1272.1 µ m/m ≅ +1272 µ m/m ................................................. Ans. ε p 2 = 308.5 − 963.6 = −655.1 µ m/m ≅ −655 µ m/m ..................................................... Ans. ε p3 = ε z = − ν 0.33 (ε x + ε y ) = − 1 − 0.33 (875 − 258 ) = −304 µ m/m ........................... Ans. 1 −ν γ p = γ max = 2 ( 963.6 ) = 1927.2 µ rad ≅ 1927 µ rad ........................................................ Ans. γ 1 1 −1559 θ p = tan −1 xy = tan −1 = −27.00° ................................................ Ans. εx −εy 2 2 875 − ( −258 ) (c) σ p1 = E 73 × 109 ε p1 + νε p 2 ) = 1272.1 + 0.33 ( −655.1) (10−6 ) 2( 2 1 −ν 1 − 0.33 E 73 × 109 ε p1 + νε p 2 ) = −655.1 + 0.33 (1272.1) (10−6 ) 2( 2 1 −ν 1 − 0.33 = +86.50 ×106 N/m 2 ≅ 86.5 MPa (T) ....................................................................... Ans. σ p2 = = −19.277 ×106 N/m 2 ≅ 19.28 MPa (C) .................................................................. Ans. σ p 3 = σ z = 0 MPa ................................................................................................................. Ans. τ max = Gγ max = ( 27.44 × 109 )(1927.2 ×10 −6 ) = 52.89 × 106 N/m 2 ≅ 52.9 MPa ............................................................................... Ans. 10-105 At a point on the free surface of an aluminum alloy (E = 10,600ksi and ν = 0.33 ) machine part, the strain rosette shown in Fig. P10-105 was used to obtain the following normal strain data: ε a = +800 µ in./in. , (a) (b) (c) ε b = +950 µ in./in. , and ε c = +600 µ in./in. . Determine The stress components σ x , σ y , and τ xy at the point. The principal strains and the maximum shearing strain at the point. Prepare a sketch showing all of these strains. The principal stresses and the maximum shearing stress at the point. Prepare a sketch showing all of these stresses. 507 STATICS AND MECHANICS OF MATERIALS, 2nd Edition SOLUTION The given values are: RILEY, STURGES AND MORRIS ε a = ε x = +800 µ in./in. (a) 2 2 ε b = ε 36.870 = +950 µ in./in. ε c = ε y = +600 µ in./in. ε n = ε x cos 2 θ n + ε y sin 2 θ n + γ xy sin θ n cos θ n ε b = 800 ( 4 5 ) + 600 ( 3 5 ) + γ xy ( 3 5 )( 4 5 ) = +950 µin./in. Therefore ε x = +800 µ in./in. ν = 0.33 ε y = +600 µ in./in. γ xy = +462.5 µ rad G = 3985 ksi For aluminum alloy: E = 10, 600 ksi = 2 (1 + ν ) G σx = E 10, 600 ε + νε y ) = 800 + 0.33 ( 600 ) (10−6 ) 2( x 1 −ν 1 − 0.332 = +11.872 ksi ≅ 11.87 ksi (T) ....................................................................................... Ans. E 10, 600 ε + νε x ) = 600 + 0.33 ( 800 ) (10−6 ) 2( y 2 1 −ν 1 − 0.33 = +10.278 ksi ≅ 10.28 ksi (T) ....................................................................................... Ans. .......................................... Ans. σy = τ xy = Gγ xy = 3985 ( 462.5 × 10−6 ) = 1.8431 ksi ≅ 1.843 ksi 2 2 (b) ε p1, p 2 2 2 εx + ε y ε x − ε y γ xy 800 + 600 800 − 600 462.5 = ± ± + = + 2 2 2 2 2 2 ε p1 = 700 + 252 = +952 µ in./in. ......................................................................................... Ans. ε p 2 = 700 − 252 = +448 µ in./in. ......................................................................................... Ans. ε p3 = ε z = − ν 0.33 (ε x + ε y ) = − 1 − 0.33 (800 + 600 ) = −690 µin./in. ......................... Ans. 1 −ν γ p = 2 ( 252 ) = 504 µ rad ...................................................................................................... Ans. γ max = ( ε max − ε min ) = 952 − ( −690 ) = 1642 µ rad ........................................................... Ans. γ 1 1 462.5 θ p = tan −1 xy = tan −1 = 33.31° ......................................................... Ans. 2 800 − 600 εx −εy 2 (c) σ p1 = E 10, 600 ε + νε p 2 ) = 952 + 0.33 ( 448 ) (10 −6 ) 2 ( p1 2 1 −ν 1 − 0.33 = +13.083 ksi ≅ 13.08 ksi (T) ....................................................................................... Ans. E 10, 600 ε + νε p1 ) = 448 + 0.33 ( 952 ) (10−6 ) 2 ( p2 2 1 −ν 1 − 0.33 = +9.066 ksi ≅ 9.07 ksi (T) ........................................................................................... Ans. σ p2 = σ p 3 = σ z = 0 ksi ..................................................................................................................... Ans. τ p = Gγ p = 3985 ( 504 × 10−6 ) = 2.008 ksi ≅ 2.01 ksi ................................................... Ans. τ max = Gγ max = 3985 (1642 × 10−6 ) = 6.543 ksi ≅ 6.54 ksi ........................................... Ans. 508 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 10-106 At a point on the free surface of an steel (E = 200 GPa and ν = 0.30 ) machine part, the strain rosette shown in Fig. P10-106 was used to obtain the following normal strain data: ε a = +875 µ m/m , (a) (b) (c) ε b = +700 µ m/m , and ε c = +350 µ m/m . Determine The stress components σ x , σ y , and τ xy at the point. The principal strains and the maximum shearing strain at the point. Prepare a sketch showing all of these strains. The principal stresses and the maximum shearing stress at the point. Prepare a sketch showing all of these stresses. SOLUTION The given values are: ε a = ε x = +875 µ m/m (a) 2 2 ε b = ε 53.130 = +700 µ m/m ε c = ε y = +350 µ m/m ε n = ε x cos 2 θ n + ε y sin 2 θ n + γ xy sin θ n cos θ n ε b = 875 ( 3 5 ) + 350 ( 4 5 ) + γ xy ( 4 5 )( 3 5 ) = +700 µ m/m Therefore ε x = +875 µ m/m ν = 0.30 ε y = +350 µ m/m E = 200 GPa = 2 (1 + ν ) G γ xy = +335.4 µ rad G = 76.92 GPa For steel: σx = E × 109 (ε x +νε y ) = 2000.302 875 + 0.30 ( 350 ) (10−6 ) 1 −ν 2 1− E × 109 (ε x +νε y ) = 2000.302 350 + 0.30 (875) (10−6 ) 1 −ν 2 1− = +215.4 ×106 N/m 2 ≅ 215 MPa (T) .......................................................................... Ans. σy = = +134.62 ×106 N/m 2 ≅ 134.6 MPa (T) .................................................................... Ans. τ xy = Gγ xy = ( 76.92 × 109 )( 335.4 × 10 −6 ) = +25.80 × 106 N/m 2 ≅ +25.8 MPa ........................................................................... Ans. 2 2 εx + ε y ε x − ε y γ xy 875 + 350 875 − 350 335.4 = ± + = ± + 2 2 2 2 2 2 2 2 (b) ε p1, p 2 ε p1 = 612.5 + 311.5 = +924 µ m/m .................................................................................... Ans. ε p 2 = 612.5 − 311.5 = +301 µ m/m .................................................................................... Ans. ε p3 = ε z = − ν 0.30 (ε x + ε y ) = − 1 − 0.30 (875 + 350 ) = −525 µ m/m ........................... Ans. 1 −ν γ p = 2 ( 311.5 ) = 623 µ rad ................................................................................................... Ans. γ max = ( ε max − ε min ) = 924 − ( −525 ) = 1449 µ rad ........................................................... Ans. γ 1 1 335.4 θ p = tan −1 xy = tan −1 = 16.29° ......................................................... Ans. 2 875 − 350 εx −εy 2 509 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (c) RILEY, STURGES AND MORRIS σ p1 = E 200 × 109 ε p1 + νε p 2 ) = 924 + 0.30 ( 301) (10−6 ) 2( 2 1 −ν 1 − 0.30 E 200 × 109 + νε p 2 ) = ε 301 + 0.30 ( 924 ) (10−6 ) 2 ( p1 2 1 −ν 1 − 0.30 = +222.9 ×106 N/m 2 ≅ 223 MPa (T) ........................................................................ Ans. σ p2 = = +127.08 ×106 N/m 2 ≅ 127.1 MPa (T) ..................................................................... Ans. σ p 3 = σ z = 0 MPa ................................................................................................................. Ans. τ max = Gγ max = ( 76.92 ×109 )( 623 × 10−6 ) τ max = Gγ max = ( 76.92 ×109 )(1449 × 10−6 ) = 47.92 × 106 N/m 2 ≅ 47.9 MPa ................................................................................ Ans. = 111.46 × 106 N/m 2 ≅ 111.5 MPa ............................................................................. Ans. 10-107 A spherical gas storage tank, similar to the one shown in Fig. 10-33, has a diameter of 35 ft and a wall thickness of 7/8 in. Determine the maximum normal stress in the tank if the gas pressure is 100 psi. SOLUTION σ= pr (100 ) 210 − ( 7 8 ) = 11,950 psi = 11.95 ksi ............................................. Ans. = 2t 2 ( 7 8) 10-108 Determine the maximum normal stress in a 300-mm diameter basketball that has a 2-mm wall thickness after it has been inflated to a pressure of 100 kPa. SOLUTION 3 pr (100 ×10 ) ( 0.148 ) σ= = = 3.70 × 106 N/m 2 = 3.70 MPa ................................... Ans. 2t 2 ( 0.002 ) 10-109 A steel pipe with an inside diameter of 10 in. will be used to transmit steam under a pressure of 800 psi. If the hoop stress in the pipe must be limited to 10 ksi because of a longitudinal weld in the pipe, determine the minimum satisfactory thickness for the pipe. SOLUTION σh = pr : t t= pr ( 800 )( 5 ) = = 0.400 in. ...................................................................... Ans. 10, 000 σh 10-110 A cylindrical propane tank, similar to the ones shown in Fig. 10-35, has an outside diameter of 3.25 m and a wall thickness of 22 mm. If the allowable hoop stress is 100 MPa and the allowable axial stress is 45 MPa, determine the maximum internal pressure that can be applied to the tank. SOLUTION σa = σh = pr : 2t pr : t p= p= 2tσ h 2 ( 0.022 )( 45 ) = = 1.2352 MPa ri 1.603 tσ h ( 0.022 )(100 ) = = 1.3724 MPa ri 1.603 Therefore: pmax = 1.2352 MPa ≅ 1.235 MPa ................................................................ Ans. 510 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 10-111 A cylindrical boiler with hemispherical ends has an outside diameter of 6 ft. If the hoop and axial stresses in the boiler must be limited to 15 ksi and 8 ksi, respectively, determine the thickness of steel plate required if the internal pressure in the boiler will be 200 psi. SOLUTION pr ( 200 )( 36 − t ) t = 0.4444 in. = = 8000 psi 2t 2t pr ( 200 )( 36 − t ) σh = t = 0.4737 in. = = 15, 000 psi t t Therefore tmin = 0.4737 in. ≅ 0.474 in. .......................................................................... Ans. σa = 10-112 The cylindrical tank shown in Fig. P10-112 is 20 m in diameter, is made of structural steel, and will be used to store stove oil stress is 80 MPa. SOLUTION ( ρ = 850 kg/m ) . Determine the minimum wall thickness required if the allowable 3 p = ρ gh = 850 ( 9.81)( 6 ) = 50.03 × 103 N/m 2 = 50.03 kPa 3 pr ( 50.03 ×10 ) (10 − t ) σh = = = 80 × 106 N/m 2 t t t = 0.00625 m = 6.25 mm ................................................................................................... Ans. 10-113 A standpipe 12 ft in diameter and 50 ft tall is being constructed for use as a storage tank for water 3 (γ = 62.4 lb/ft ) . Determine the minimum thickness of steel plate that can be used if the hoop stress in p = γ h = 62.4 ( 50 ) = 3120 lb/ft 2 = 21.67 psi the standpipe must be limited to 5000 psi. SOLUTION σh = pr ( 21.67 )( 72 − t ) = = 5000 psi t t t = 0.311 in. ............................................................................................................................. Ans. 10-114 A cylindrical pressure vessel is fabricated by butt-welding 20-mm plate with a spiral seam, as shown in Fig. P10-114. The pressure in the tank is 2800 kPa. Determine (a) The normal stress perpendicular to the weld. (b) The shearing stress parallel to the weld. (c) The maximum shearing stress at a point on the outside surface of the vessel. (d) The maximum shearing stress at a point on the inside surface of the vessel. SOLUTION 3 pr ( 2800 × 10 ) ( 0.600 ) σa = = 2t 2 ( 0.020 ) 3 pr ( 2800 × 10 ) ( 0.600 ) σh = = t ( 0.020 ) = 42.00 × 106 N/m 2 = 42.00 MPa = 84.00 × 106 N/m 2 = 84.00 MPa 511 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS θ = tan −1 ( 3 4 ) = 36.870° (a) σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cosθ = 42.00 cos 2 ( 36.870° ) + 84.00sin 2 ( 36.870° ) + 0 = 57.1 MPa .......................... Ans. (b) τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ ) = − ( 42.00 − 84.00 ) sin ( 36.870° ) cos ( 36.870° ) + 0 = 20.2 MPa ....................... Ans. (c) τ max = (d) τ max σ max − σ min σ h − 0 84.00 − 0 = = = 42.0 MPa ................................................... Ans. 2 2 2 σ − σ min σ h − ( − p ) 84.00 − ( −2.800 ) = max = = = 43.4 MPa ............................. Ans. 2 2 2 10-115 A steel boiler 3 ft in diameter is welded using a spiral seam that makes an angle of 30° with respect to a transverse plane of the boiler, as shown in Fig. P10-115. For an internal pressure of 140 psi and a wall thickness of 2 in., determine (a) The normal stress perpendicular to the weld. (b) The shearing stress parallel to the weld. (c) The maximum shearing stress in the boiler. SOLUTION σa = σh = pr (140 )(18 ) = = 630 psi 2t 2 ( 2) pr (140 )(18 ) = = 1260 psi t ( 2) (a) σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cosθ = 630 cos 2 ( −30° ) + 1260sin 2 ( −30° ) + 0 = 787 psi ............................................... Ans. (b) τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ ) = − ( 630 − 1260 ) sin ( −30° ) cos ( −30° ) + 0 = −273 psi .......................................... Ans. (c) τ max = σ max − σ min σ h − ( − p ) 1260 − ( −140 ) = = = 700 psi ....................................... Ans. 2 2 2 10-116 A spherical pressure vessel 3 m in diameter is being designed to withstand a maximum internal pressure of 500 kPa. The material being used in its construction has an allowable stress of 105 MPa, a modulus of elasticity of 210 GPa, and a Poisson’s ratio of 0.20. Determine (a) The minimum satisfactory wall thickness. (b) The circumferential normal strain at maximum pressure when the wall thickness of part (a) is used. (c) The change in diameter of the pressure vessel at maximum pressure when the wall thickness of part (a) is used. SOLUTION (a) σ= pr ( 0.500 )(1.5 − t ) = = 105 MPa 2t 2t 512 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS t = 3.571× 10 −3 m ≅ 3.57 mm ............................................................................................. Ans. (b) (c) ε= ∆C = ε C = επ D = ( 350 × 10−6 ) (π )( 3) = 3300 × 10−6 m σ −νσ 105 − 0.30 (105) = = 350 × 10−6 m/m = 350 µ m/m E 210 × 103 ................................. Ans. C + ∆C = π ( D + ∆D ) = C + π∆D ∆D = ∆C 3300 × 10−6 = = 1050.4 × 10−6 m ≅ 1.050 mm ............................................. Ans. π π 10-117 A cylindrical tank, similar to the one shown in Fig. 10-35, has an outside diameter of 8 ft and a wall thickness of 3/4 in. The tank is subjected to an internal pressure of 150 psi. (a) Determine the axial and hoop stresses in the tank. (b) Prepare a plot, similar to Fig. 4-19, showing the variation of normal and shearing stresses on planes through a point in the wall of the tank as the angle θ, measured counterclockwise from a transverse plane, varies from 0° to 90°. SOLUTION (a) σa = σh = pr (150 )( 47.25 ) = = 4725 psi ≅ 4.73 ksi ............................................................ Ans. 2t 2 (3 4) pr (150 )( 47.25 ) = = 9450 psi ≅ 9.45 ksi ............................................................ Ans. t (3 4) (b) σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cosθ = ( 4725cos 2 θ + 9450sin 2 θ + 0 ) psi τ nt = − (σ x − σ y ) sin θ cosθ + τ xy ( cos 2 θ − sin 2 θ ) = − ( 4725 − 9450 ) sin θ cos θ + 0 = ( 4725sin θ cosθ ) psi 10-118 The strains measured on the outside surface of the cylindrical pressure vessel shown in Fig. P10-118 are ε1 = +619 µ m/m and ε 2 = +330 µ m/m . The angle θ = 30°, the outside diameter of the vessel is 510 mm, and the wall thickness is 3 mm. The vessel is made of 0.4 percent carbon hot-rolled steel (see Appendix A for properties). Determine (a) The stresses σ 1 and σ 2 in the vessel. (b) The internal pressure applied to the vessel. SOLUTION For 0.4 percent carbon hot-rolled steel: G = 80 GPa (a) E = 210 GPa = 2 (1 + ν ) G ν = 0.3125 σ1 = E 210 × 109 ε1 + νε 2 ) = 619 + 0.3125 ( 330 ) ×10−6 2( 2 1 −ν 1 − 0.3125 = 168.1× 106 N/m 2 = 168.1 MPa ................................................................................ Ans. 513 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS σ2 = E 210 × 109 330 + 0.3125 ( 619 ) ×10−6 ( ε 2 + νε1 ) = 2 2 1 −ν 1 − 0.3125 pr p ( 255 ) = = ( 42.5 p ) MPa 2t 2 ( 3) pr p ( 255 ) = = ( 85.0 p ) MPa t ( 3) = 121.82 × 106 N/m 2 ≅ 121.8 MPa ............................................................................. Ans. (b) σa = σh = σ 2 = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cosθ = ( 42.5 p ) cos 2 ( 30° ) + ( 85.0 p ) sin 2 ( 30° ) + 0 = 121.82 MPa p = 2.29 MPa ......................................................................................................................... Ans. 10-119 A 4-in. diameter shaft is subjected to both a torque of 30 kip⋅in. and an axial tensile load of 50 kip, as shown in Fig. P10-119. Determine the principal stresses and the maximum shearing stress at point A on the surface of the shaft. SOLUTION A= π d 2 π ( 4) = = 12.566 in 2 4 4 2 4 σx = P 50 = = 3.979 ksi A 12.566 π d 4 π ( 4) J= = = 25.13 in 4 32 32 σ p1, p 2 σ +σ y σ −σ y 2 =x ±x + τ xy 2 2 3.979 + 0 3.979 − 0 2 = ± + 2.388 2 2 2 2 τ xy = Tc 30 ( 2 ) = = 2.388 ksi 25.13 J σ p1 = 1.9895 + 3.1082 = +5.0977 ksi ≅ 5.10 ksi (T) .................................................... Ans. σ p 2 = 1.9895 − 3.1082 = −1.1187 ksi ≅ 1.119 ksi (C) 2τ xy 2 ( 2.388 ) 1 1 θ p = tan −1 = tan −1 = 25.10° 2 σ x −σ y 2 3.979 − 0 ................................................. Ans. σ p 3 = σ z = 0 ksi ........................ τ max = τ p = 3.1082 ksi ≅ 3.11 ksi .............................. Ans. ....................................................... Ans. 10-120 A hollow shaft with an outside diameter of 400 mm and an inside diameter of 300 mm is subjected to both a torque of 350 kN⋅m and an axial tensile load of 1500 kN, as shown in Fig. P10-120. Determine the principal stresses and the maximum shearing stress at a point on the outside surface of the shaft. SOLUTION A= J= π ( 4002 − 3002 ) 4 4 = 54.98 × 103 mm 2 = 54.98 × 10−3 m 2 = 1718.1× 106 mm 4 = 1718.1× 10−6 m 4 π ( 400 − 3004 ) 32 514 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS σx = 3 −3 Tc ( 350 ×10 )( 200 × 10 ) τ xy = = = 40.74 × 106 N/m 2 = 40.74 MPa J 1718.1×10−6 P 1500 × 103 = = 27.28 × 106 N/m 2 = 27.28 MPa −3 A 54.978 ×10 σ p1, p 2 σ +σ y σ −σ y 2 =x ±x + τ xy 2 2 27.28 + 0 27.28 − 0 2 = ± + 40.74 2 2 2 2 σ p1 = 13.64 + 42.96 = +56.60 MPa ≅ 56.6 MPa (T) .................................................... Ans. σ p 2 = 13.64 − 42.96 = −29.32 MPa ≅ 29.3 MPa (C) ................................................... Ans. σ p 3 = σ z = 0 MPa .................... τ max = τ p = 42.96 ksi ≅ 43.0 ksi ............................... Ans. 2τ xy 2 ( 40.74 ) 1 1 θ p = tan −1 = tan −1 = 35.74° ........................................................ Ans. 2 σ x −σ y 2 27.28 − 0 10-121 A 2-in. diameter shaft is used in an aircraft engine to transmit 360 hp at 1500 rpm to a propeller that develops a thrust of 2800 lb. Determine the principal stresses and the maximum shearing stress produced at any point on the outside surface of the shaft. SOLUTION T= 33, 000 ( hp ) 2π N = 2 33, 000 ( 360 ) 2π (1500 ) = 1260.5 lb ⋅ ft π d 2 π ( 2) A= = = 3.142 in 2 4 4 π d 4 π ( 2) J= = = 1.5708 in 4 32 32 4 σx = P 2800 = = 891.2 psi A 3.142 τ xy = 2 Tc (1260.5 × 12 )(1) = = 9629 psi 1.5708 J σ p1, p 2 σ +σ y σ −σ y 2 =x ±x + τ xy 2 2 = 891.2 + 0 891.2 − 0 2 ± + 9629 2 2 ................................................... Ans. 2 σ p1 = 445.6 + 9639 = +10, 084.6 psi ≅ 10.08 ksi (T) σ p 2 = 445.6 − 9639 = −9193.4 psi ≅ 9.19 ksi (C) ......................................................... Ans. σ p 3 = σ z = 0 ksi ..................................................................................................................... Ans. τ max = τ p = 9639 psi ≅ 9.64 ksi .......................................................................................... Ans. 2τ xy 2 ( 9629 ) 1 1 θ p = tan −1 = tan −1 = 43.68° ........................................................ Ans. 2 σ x −σ y 2 891.2 − 0 515 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 10-122 A 60-mm diameter shaft must transmit a torque of unknown magnitude while it is supporting an axial tensile load of 150 kN. Determine the maximum allowable value for the torque if the tensile principal stress at a point on the outside surface of the shaft must not exceed 125 MPa. SOLUTION π d 2 π ( 60 ) A= = = 2827 mm 2 4 4 2 π d 4 π ( 60 ) J= = = 1.2723 × 106 mm 4 32 32 4 σx = P 150 × 103 = = 53.06 ×106 N/m 2 = 53.06 MPa (T) −6 A 2827 ×10 2 2 σx +σ y σ x −σ y 53.06 + 0 53.06 − 0 2 2 + + σ p1 = + τ xy = + τ xy = 125 MPa 2 2 2 2 τ xy = 94.83 MPa τ xy = Tc J = 94.83 MPa 6 −6 τ xy J ( 94.83 ×10 )(1.2723 ×10 ) T= = c 0.030 = 4.022 ×103 N ⋅ m ≅ 4.02 kN ⋅ m ................................................................................ Ans. 10-123 The T-section shown in Fig. P10-123 is used as a short post to support a compressive load P = 150 kip. The load is applied on the centerline of the stem at a distance e = 2 in. from the centroid of the cross section. Determine the normal stresses at points C and D on section AB. SOLUTION A = 2 ( 20 × 6 ) = 24 in 2 xC = 3( 2 × 6) + 7 ( 2 × 6) = 5 in. 24 P Mc 150 (150 × 2 )( 5) + =− + = +4.779 ksi ≅ 4.78 ksi (T) ....................... Ans. AI 24 136 150 (150 × 2 )( 3) P Mc σD = − − =− − = −12.868 ksi ≅ 12.87 ksi (C) .................. Ans. 24 136 AI 2 ( 6 )3 6 ( 2 )3 2 2 Iy = + ( 2 × 6 )( 2 ) + + ( 2 × 6 )( 2 ) = 136 in 4 12 12 σC = − 10-124 The cross section of the straight vertical portion of the coil-loading hook shown in Fig. P10-124a is shown in Fig. P10-124b. The horizontal distance from the line of action of the applied load to the inside face CD of the cross section is 600 mm. Determine the maximum tensile and compressive stresses on section CDEF for a 40-kN load. SOLUTION A = (100 × 50 ) + ( 30 × 80 ) + ( 60 × 20 ) = 8600 mm 2 xC = 25 (100 × 50 ) + 90 ( 30 × 80 ) + 140 ( 60 × 20 ) = 59.19 mm 8600 100 ( 50 )3 2 Iy = + (100 × 50 )( 34.19 ) 12 516 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 30 ( 80 )3 60 ( 20 )3 2 2 + + ( 30 × 80 )( 30.81) + + ( 60 × 20 )( 80.81) 12 12 M = ( 40 × 103 )( 659.19 × 10−3 ) = 26.37 × 103 N ⋅ m = 26.37 kN ⋅ m = 18.321× 106 mm 4 σ CD ( 26.37 ×103 ) ( 0.05919 ) P Mc 40 × 103 =+ = + AI 8600 × 10−6 18.321×10−6 = 89.845 ×106 N/m 2 ≅ 89.8 MPa (T) ..................................................................... Ans. σ EF ( 26.37 ×103 ) ( 0.09081) P Mc 40 × 103 =− = − AI 8600 × 10−6 18.321×10−6 = −126.05 ×106 N/m 2 ≅ 126.1 MPa (C) ................................................................. Ans. 10-125 The estimated maximum total force P to be exerted on the C-clamp shown in Fig. P10-125 is 450 lb. If the normal stress on section A–A is not to exceed 16,000 psi, determine the minimum allowable value for the dimension h of the cross section. SOLUTION A = 0.5 × h = ( h 2 ) in 2 M = 450 ( 3) = 1350 lb ⋅ in I= ( 0.5 ) h3 = 12 (h 3 24 ) in 4 σ= P Mc 450 (1350 × 2 )( h 2 ) + = + < 16, 000 psi AI h2 h3 24 16, 000h 2 − 900h − 16, 200 = 0 h = 1.035 in. ............................................................................................................................ Ans. 10-126 A 150-mm diameter shaft will be used to support the axial load and torques shown in Fig. P10-126. Determine the principal stresses and the maximum shearing stress at point A on the surface of the shaft. SOLUTION π d 2 π (150 ) A= = = 17.671× 103 mm 2 4 4 2 π d 4 π (150 ) J= = = 49.70 ×106 mm 4 32 32 4 σx = 3 Tc (19.6 × 10 ) ( 0.075) τ xy = = = 29.58 × 106 N/m 2 = 29.58 MPa −6 J 49.70 ×10 P 780 × 103 = = 44.14 ×106 N/m 2 = 44.14 MPa A 17.671×10−3 σ p1, p 2 σ +σ y σ −σ y 2 =x ±x + τ xy 2 2 44.14 + 0 44.14 − 0 2 = ± + 29.58 2 2 2 2 σ p1 = 22.07 + 36.91 = +58.98 MPa ≅ 59.0 MPa (T) .................................................... Ans. 517 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS σ p 2 = 22.07 − 36.91 = −14.84 MPa ≅ 14.84 MPa (C) ................................................. Ans. σ p 3 = σ z = 0 MPa .................... τ max = τ p = 36.91 MPa ≅ 36.9 MPa ........................ Ans. ....................................................... Ans. 2τ xy 2 ( 29.58 ) 1 1 θ p = tan −1 = tan −1 = 26.64° 2 44.14 − 0 σ x −σ y 2 10-127 The thin-walled cylindrical pressure vessel shown in Fig. P10-127 has an inside diameter of 24 in. and a wall thickness of 1/2 in. The vessel is subjected to an internal pressure of 250 psi. In addition, a torque of 150 kip⋅ft is applied to the vessel through rigid plates on the ends of the vessel. Determine the maximum normal and shearing stresses at a point on the outside surface of the vessel. SOLUTION J= π ( 254 − 244 ) 32 = 5777.5 in 4 σx =σa = σ y =σh = pr 250 (12 ) = = 3000 psi = 3.000 ksi 2t 2 (1 2 ) τ xy = Tc (150 × 12 )(12.5 ) = = 3.894 ksi 5777.5 J 2 pr 250 (12 ) = = 6000 psi = 6.000 ksi t (1 2 ) σ p1, p 2 2 σ +σ y σ −σ y 3+ 6 3 − 6 2 2 =x ±x + τ xy = ± + 3.894 2 2 2 2 σ p1 = 4.5 + 4.173 = +8.673 ksi ≅ 8.67 ksi (T) ............................................................... Ans. ............................................................ Ans. σ p 2 = 4.5 − 4.173 = +0.327 ksi ≅ 0.327 ksi (T) σ p 3 = σ z = 0 ksi ..................................................................................................................... Ans. σ − σ min 8.673 − 0 τ max = max = = 4.337 ksi ≅ 4.34 ksi ................................................. Ans. 2 2 10-128 A steel shaft is loaded and supported as shown in Fig. P10-128. If the maximum shearing stress in the shaft must not exceed 55 MPa and the maximum tensile stress in the shaft must not exceed 83 MPa, determine the maximum torque T that can be applied to the shaft. SOLUTION A100 J100 π d 2 π (100 ) = = = 7.854 × 103 mm 2 4 4 2 A150 J150 π d 2 π (150 ) = = = 17.671× 103 mm 2 4 4 2 π d 4 π (100 ) = = = 9.817 × 106 mm 4 32 32 4 π d 4 π (150 ) = = = 49.70 × 106 mm 4 32 32 4 σ 100 = P 550 × 103 = A 7.854 × 10 −3 = 70.03 × 106 N/m 2 = 70.03 MPa σ 150 = τ100 = T ( 0.050 ) Tc = = 5093T N/m 2 −6 J 9.817 ×10 P 550 ×103 = A 17.671×10 −3 = 31.12 × 106 N/m 2 = 31.12 MPa τ150 = Tc 3T ( 0.075 ) = = 4527T N/m 2 −6 J 49.70 ×10 518 STATICS AND MECHANICS OF MATERIALS, 2nd Edition Since both RILEY, STURGES AND MORRIS σ 100 and τ100 are larger than σ 150 2 and τ150 , the maximum stresses will occur in the 100-mm section. 2 σx +σ y σ x −σ y 70.03 + 0 70.03 − 0 2 2 + + σ p1 = + τ xy = + τ xy ≤ 83 MPa 2 2 2 2 τ xy ≤ 32.67 MPa τ max σ −σ y 70.03 − 0 2 2 =τp = x + τ xy = + τ xy ≤ 55 MPa 2 2 2 2 τ xy ≤ 42.41 MPa Therefore, τ xy ( max ) = 32.67 MPa and Tmax 6 −6 τ xy J100 ( 32.67 ×10 )( 9.817 10 ) = = = 6410 N ⋅ m .......................................... Ans. c100 0.050 10-129 A 30-lb force P is applied to the brake pedal of an automobile as shown in Fig. P10-129. Force Q is applied to the brake cylinder. Determine the maximum tensile and compressive normal stresses on section a–a, which is midway between points A and B. Section a–a may be modeled as a 3/16×1-in. rectangle. SOLUTION ΣM A = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : − ( 30 cos 30° )(11) − ( 30sin 30° )( 4 ) + Q ( 5.5 ) = 0 Q = 62.87 lb = 62.87 lb ← Ax + 30 cos 30° − Q = 0 Ay − 30sin 30° = 0 Ax = 37.89 lb = 36.89 lb → Ay = 15.00 lb = 15.00 lb ↑ θ = tan −1 ( 5.5 4 ) = 53.97° ΣFn = 0 : 36.89 cos 53.97° + 15.00sin 53.97° − N = 0 ΣM O = 0 : N = 33.83 lb M O + 36.89 ( 2.75 ) − 15.00 ( 2 ) = 0 M O = −71.45 lb ⋅ in. = 71.45 lb ⋅ in. A = ( 3 16 )(1) = 0.1875 in 2 ( 3 16 )(1) I= 12 3 = 0.015625 in 4 σ top = − σ bot N M Oc 33.83 71.45 ( 0.500 ) + =− + = +2106 psi ≅ 2110 psi (T) ......... Ans. A I 0.1875 0.015625 N Mc 33.83 71.45 ( 0.500 ) =− − O =− − = −2467 psi ≅ 2470 psi (C) ......... Ans. A I 0.1875 0.015625 519 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 10-130 An automobile engine with a mass of 360 kg is supported by an engine hoist, as shown in Fig. P10-130. Determine the maximum tensile and compressive normal stresses on section a–a if member ABC is a hollow square 100×100-mm section with a wall thickness of 20-mm. SOLUTION W = mg = 360 ( 9.81) = 3532 N a = 1430 cos16° = 1374.6 mm b = 1430sin16° = 394.2 mm 1374.6 − 890 = 20.53° φ = tan −1 394.2 + 900 ΣM A = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : ( B cos 20.53° )(1374.6 ) − ( B sin 20.53° )( 394.2 ) − 3532 ( 2700 cos16° ) = 0 Ax + 7978sin 20.53 = 0 Ay + 7978cos 20.53° − 3532 = 0 B = 7978 N Ax = −2798 N = 2798 N ← Ay = −3939 N = 3939 N ↓ ΣFn = 0 : P − 2798cos16° − 3939sin16° = 0 ΣM O = 0 : M O + 3939 ( 530 cos16° ) − 2798 ( 530sin16° ) = 0 P = 3775 N M O = −1.5980 × 106 N ⋅ mm = 1.5980 × 103 N ⋅ m A = 100 2 − 602 = 6400 mm 2 100 (100 ) 60 ( 60 ) I= − = 7.253 × 106 mm 4 12 12 3 3 σ top (1.5980 ×10 ) ( 0.050 ) 3775 P Mc = + O= + −6 6400 × 10 7.253 ×10 −6 A I 3 = +11.606 ×106 N/m 2 ≅ 11.61 MPa (T) ................................................................. Ans. σ bot (1.5980 ×10 ) ( 0.050 ) P Mc 3775 = − O= − −6 A I 6400 × 10 7.253 × 10−6 3 = −10.426 × 106 N/m 2 ≅ 10.43 MPa (C) ................................................................. Ans. 10-131 A short post supports a vertical force P = 9600 lb and a horizontal force H = 800 lb, as shown in Fig. P10131. Determine the vertical normal stresses at corners A, B, C, and D of the post. Neglect stress concentrations. SOLUTION A = ( 4 × 6 ) = 24 in 2 Ix ( 6 )( 4 ) = 12 3 = 32 in 4 Iy ( 4 )( 6 ) = 12 3 = 72 in 4 520 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS M x = 9600 (1) = 9600 lb ⋅ in. M y = 800 ( 24 ) = 19, 200 lb ⋅ in. P 9600 = = 400 psi (C) 24 A σP = σ Mx = σMy = M x c 9600 ( 2 ) = = 600 psi (T & C) Ix 32 M yc Iy = 19, 200 ( 3) 72 = 800 psi (T & C) σ A = −400 + 600 + 800 = +1000 psi = 1000 psi (T) σ B = −400 − 600 + 800 = −200 psi = 200 psi (C) ..................................................... Ans. ......................................................... Ans. σ C = −400 − 600 − 800 = −1800 psi = 1800 psi (C) ..................................................... Ans. σ D = −400 + 600 − 800 = −600 psi = 600 psi (C) ......................................................... Ans. 10-132 Determine the vertical normal stresses at points A, B, C, and D of the rectangular post shown in Fig. P10132. Neglect stress concentrations. SOLUTION A = ( 200 ×150 ) = 30 × 103 mm 2 Ix ( 200 )(150 ) = 12 12 3 = 56.25 ×106 mm 4 3 Iy = (150 )( 200 ) = 100.0 ×106 mm 4 M x = ( 75 × 103 ) ( 0.075 ) = 5625 N ⋅ m M y = ( 75 × 103 ) ( 0.050 ) = 3750 N ⋅ m σP = M x P 75 ×103 = = 2.50 ×106 N/m 2 = 2.50 MPa (C) A 30 ×10−3 M c 5625 ( 0.075 ) = x= = 7.50 ×106 N/m 2 = 7.50 MPa (T & C) Ix 56.25 ×10−6 M yc Iy = 3750 ( 0.100 ) 100.0 × 10 −6 σMy = = 3.75 ×106 N/m 2 = 3.75 MPa (T & C) σ A = −2.50 + 7.50 + 3.75 = +8.75 MPa = 8.75 MPa (T) ............................................. Ans. σ B = −2.50 + 7.50 − 3.75 = +1.25 MPa = 1.25 MPa (T) σ D = −2.50 − 7.50 + 3.75 = −6.25 MPa = 6.25 MPa (C) ............................................. Ans. σ C = −2.50 − 7.50 − 3.75 = −13.75 MPa = 13.75 MPa (C) ........................................ Ans. ............................................ Ans. 521 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 10-133 Determine the principal stresses and the maximum shearing stress at points on the top and bottom of section a–a of the pipe system shown in Fig. P10-133. The pipe has an outside diameter of 1 in. and a wall thickness of 1/8 in. SOLUTION π (1) π ( 0.75 ) I= − = 0.03356 in 4 64 64 4 4 π (1) π ( 0.75 ) J= − = 0.06711 in 4 32 32 4 4 At Section a-a: ΣFx = 0 : ΣFy = 0 : ΣFz = 0 : ΣM x = 0 : ΣM y = 0 : Vx = 0 N =0 Vz − 50 = 0 M x − 50 (18 ) = 0 T − 50 ( 7 ) = 0 Mz = 0 Vz = 50 lb M x = 900 lb ⋅ in. T = 350 lb ⋅ in. ΣM z = 0 : M x c 900 ( 0.5 ) = = 13, 409 psi (T on top & C on bottom) I 0.03356 Tc 350 ( 0.5) τ= = = 2608 psi J 0.06711 σ= On the top of the pipe: σ p1, p 2 13, 409 + 0 13, 409 − 0 2 = ± + 2608 = 6705 ± 7194 psi 2 2 ........................................................ Ans. ............................................................. Ans. 2 σ p1 = 6705 + 7194 = +13,899 psi ≅ 13.90 ksi (T) σ p 2 = 6705 − 7194 = −489 psi ≅ 0.489 ksi (C) Since σ p 3 = σ z = 0 ksi ..................................................................................................................... Ans. σ p1 and σ p2 have opposite signs: τ max = τ p = 7194 psi ≅ 7.19 ksi .......................................................................................... Ans. On the bottom of the pipe: σ p1, p 2 −13, 409 + 0 −13, 409 − 0 2 = ± + 2608 = −6705 ± 7194 psi 2 2 ........................................................... Ans. 2 σ p1 = −6705 + 7194 = +489 psi ≅ 0.489 ksi (T) σ p 2 = −6705 − 7194 = −13,899 psi ≅ 13.90 ksi (C) ..................................................... Ans. σ p 3 = σ z = 0 ksi ..................................................................................................................... Ans. Since σ p1 and σ p2 have opposite signs: τ max = τ p = 7194 psi ≅ 7.19 ksi .......................................................................................... Ans. 522 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS 10-134 The output from a strain gage located on the bottom surface of the hat-section shown in Fig. P10-134 will be used to indicate the magnitude of the load P applied to the section. The hat-section is made of aluminum alloy (E = 73 GPa and ν = 1 3 ) and is 24 mm wide. When the maximum load P = 500 N is applied to the section, the strain gage should read ε = +1000 µ m/m . Plot a curve showing the combinations of thickness t and height h that will satisfy the specification. Limit the range of h from 0 to 50 mm. SOLUTION A = 24t mm 2 Ix = 500h ( t 2 ) (×10−6 ) P Mc 500 = + = ( 73 ×109 )(1000 ×10−6 ) N/m 2 σ P = Eε = + −12 3 AI 24t ×10−6 2t (×10 ) gives 24t 3 = 2t 3 mm 4 12 1752t 2 − 500t = 3000h h (mm) 0 10 20 30 40 t (mm) 0.285 4.283 5.996 7.311 8.420 10-135 A solid shaft 4-in. in diameter is acted on by forces P and Q, as shown in Fig. P10-135. Determine the principal stresses and the maximum shearing stress at point A on the surface of the shaft. SOLUTION π ( 4) A= = 12.566 in 2 4 2 π ( 4) I= = 12.566 in 4 64 4 π ( 4) J= = 25.133 in 4 32 4 At Point A: ΣFx = 0 : ΣFy = 0 : Vx − 2.25 = 0 N − 18 = 0 Vx = 2.25 kip N = 18 kip ΣFz = 0 : ΣM x = 0 : ΣM y = 0 : Vz = 0 Mx = 0 T − 2.25 ( 24 ) 0 M z − 2.25 ( 36 ) = 0 T = 54 kip ⋅ in. ΣM z = 0 : M z = 81 kip ⋅ in. ( 81)( 2 ) = 11.459 ksi (T) N Mc 18 + =− + A I 12.566 12.566 Tc 54 ( 2 ) τ= = = 4.297 ksi J 25.133 σ =− 523 STATICS AND MECHANICS OF MATERIALS, 2nd Edition 2 RILEY, STURGES AND MORRIS σ p1, p 2 = 11.459 + 0 11.459 − 0 2 ± + 4.297 = 5.579 ± 7.042 ksi 2 2 σ p1 = 5.579 + 7.042 = +12.621 ksi ≅ 12.62 ksi (T) ....................................................... Ans. σ p 2 = 5.579 − 7.042 = −1.463 ksi ≅ 1.463 ksi (C) ........................................................ Ans. σ p 3 = σ z = 0 ksi ..................................................................................................................... Ans. Since σ p1 and σ p2 have opposite signs: ........................................................................................ Ans. τ max = τ p = 7.042 ksi ≅ 7.04 ksi 10-136 A thin-walled cylindrical pressure vessel with an inside diameter of 1.2 m is fabricated by butt-welding 15mm plate with a spiral seam as shown in Fig. P10-136. The pressure in the tank is 2500 kPa. Additional loads are applied to the cylinder through a rigid end plate as shown in Fig. P10-136. Determine (a) The normal and shearing stresses on the plane of the weld at a point on the outside surface of the tank. (b) The principal stresses and the maximum shearing stress at a point on the inside surface of the tank. SOLUTION A= J= π (12302 − 12002 ) 4 3 = 57.256 × 103 mm 2 = 21.134 × 109 mm 4 θ = − tan −1 ( 4 3) = −53.130° π (12304 − 12004 ) 32 3 pr ( 2500 ×10 ) ( 0.600 ) = σx = 0.015 t = 100.0 ×106 N/m 2 = 100.0 MPa (T) 3 125 ×103 pr P ( 2500 × 10 ) ( 0.600 ) −= − σy = 2t A 2 ( 0.015 ) 57.256 × 10−3 = 47.817 ×106 N/m 2 = 47.817 MPa (T) 3 Tc ( 750 ×10 ) ( 0.615 ) = = 21.825 ×106 N/m 2 = 21.825 MPa τ xy ( outside ) = −3 21.134 ×10 J 3 Tr ( 750 ×10 ) ( 0.600 ) = 21.293 ×106 N/m 2 = 21.293 MPa τ xy ( inside ) = = 21.134 ×10−3 J (a) σ n = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cosθ +2 ( −21.83) sin ( −53.130° ) cos ( −53.130° ) = 87.56 MPa ≅ 87.6 MPa (T) ..................................................................................... Ans. = 100 cos 2 ( −53.130° ) + 47.82sin 2 ( −53.130° ) τ nt = − (σ x − σ y ) sin θ cos θ + τ xy ( cos 2 θ − sin 2 θ ) = − (100 − 47.82 ) sin ( −53.130° ) cos ( −53.130° ) + ( −21.83 ) cos 2 ( −53.130° ) − sin 2 ( −53.130° ) = 31.16 MPa ≅ 31.2 MPa ............................................................................................ Ans. 524 STATICS AND MECHANICS OF MATERIALS, 2nd Edition (b) RILEY, STURGES AND MORRIS σ p1, p 2 σ +σ y σ −σ y 2 =x ±x + τ xy 2 2 100 + 47.82 100 − 47.82 2 = ± + 21.29 2 2 2 2 σ p1 = 73.91 + 33.67 = +107.58 MPa ≅ 107.6 MPa (T) ............................................... Ans. σ p 2 = 73.91 − 33.67 = +40.24 MPa ≅ 40.2 MPa (T) ................................................... Ans. σ p 3 = − p = −2.50 ×106 N/m 2 = 2.50 MPa (C) .............................................................. Ans. τ max = σ max − σ min 107.58 − ( −2.5 ) = = 55.04 MPa ≅ 55.0 MPa 2 2 .............................. Ans. 10-137 A 1-in. diameter steel (E = 30,000 ksi and ν = 0.30 ) bar is subjected to a tensile load P and a torque T, as shown in Fig. P10-137. Determine the axial load P and the torque T if the strains indicated by gages a and b on the bar are ε a = +1084 µ in./in. and ε b = −754 µ in./in. . SOLUTION σ 45 = σ −45 σ 45 = σ x cos 2 ( 45° ) + σ y sin 2 ( 45° ) + 2τ xy sin ( 45° ) cos ( 45° ) For a shaft subjected to an axial load P and a torque T: Therefore: E 30, 000 ε + νε b ) = 1084 + 0.30 ( −754 ) ×10 −6 = 28.279 ksi 2( a 1 −ν 1 − 0.302 E 30, 000 = ε + νε a ) = −754 ) + 0.30 (1084 ) × 10−6 = −14.136 ksi 2( b 2 ( 1 −ν 1 − 0.30 σ −45 = σ x cos 2 ( −45° ) + σ y sin 2 ( −45° ) + 2τ xy sin ( −45° ) cos ( −45° ) σ y = 0 ksi 0.50σ x + τ xy = 28.279 ksi 0.50σ x − τ xy = −14.136 ksi Solving yields: Then, 2 σ x = 14.143 ksi τ xy = 21.2008 ksi 4 π (1) π (1) A= = 0.7854 in 2 J= = 0.09817 in 4 4 32 P = σ x A = 14.143 ( 0.7854 ) = 11.11 kip ........................................................................... Ans. T= τ xy J 21.208 ( 0.09817 ) = = 4.16 kip ⋅ in. ................................................................. Ans. c 0.50 10-138 A bag of potatoes is resting on a chair, as shown in Fig. P10-138. The force exerted by the p0tatoes on the frame at one side of the chair is equivalent to horizontal and vertical forces of 24 N and 84 N, respectively, at E and a force of 28 N perpendicular to member BH at G. Determine the maximum tensile and compressive normal stresses on a section midway between pins C and F. Member BH has a 10 × 30-mm cross section; the 30-mm dimension lies in the plane of the page. SOLUTION 525 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS θ = tan −1 ( 3 5 ) = 30.96° For the entire chair, the equilibrium equations: ↑ ΣFy = 0 : ΣM B = 0 : A + B − 84 − 28sin θ = 0 0.2 ( 84 ) − 0.5 ( 24 ) − 0.4A + 0.3 + ( 0.5 cos θ ) ( 28 ) = 0 B = 24.58 N give A = 73.82 N For member DF the equilibrium equations give: ΣM D = 0 : → ΣFx = 0 : ↑ ΣFy = 0 : 0.4 ( 84 ) − 0.5Fy = 0 Fy = 67.2 N Dx − Fx + 24 = 0 Dy + Fy − 84 = 0 Dx = ( Fx − 24 ) N For member BH the equilibrium equations give: Dy = 16.80 N ΣM C = 0 : 0.3 + ( 0.1667 sin θ ) ( 28 ) + 0.1333 ( 24.58 ) +0.1667 ( 67.2 ) − 0.2777 Fx = 0 Fx = 115.1 N → ΣFx = 0 : ↑ ΣFy = 0 : Fx + Cx − 28cosθ = 0 24.58 + C y − 67.2 − 28sin θ = 0 C y = 57.0 N 400 500 = a b b = 277.8 mm d = 166.7 mm Cx = −91.0 N a + b = 500 mm Therefore: On a section midway between pins C and F: a = 222.2 mm c = 133.3 mm b 2 = 138.9 mm d 2 = 83.4 mm ΣFn = 0 : 24.6 cos 30.96° + 57.0 cos 30.96° + 91.0sin 30.96° − P = 0 ΣM O = 0 : 24.6 ( 216.7 ) + 57.0 ( 83.4 ) − 91.0 (138.9 ) − M = 0 M = −2555 N ⋅ mm = 2.555 N ⋅ m 3 2 P = 116.79 N 10 ( 30 ) A = 10 × 30 = 300 mm I= = 22.5 ×103 mm 4 12 2.555 ( 0.015 ) 116.79 P Mc σ top = − + =− + 300 ×10−6 22.5 ×10−9 AI = +1.3140 ×106 N/m 2 ≅ 1.314 MPa (T) ................................................................. Ans. 526 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS σ bot = − 2.555 ( 0.015) 116.79 P Mc − =− − −6 300 × 10 22.5 × 10−3 AI = −2.093 ×106 N/m 2 ≅ 2.09 MPa (C) ..................................................................... Ans. 10-139 Three strain gages are mounted on a 1-in. diameter aluminum alloy (E = 10,000 ksi and ν = 1 3 ) rod, as shown in Fig. P10-139. When loads P and Q are applied to the rod, they produce longitudinal strains ε A = +550 µ in./in. , ε B = +400 µ in./in. , and ε C = −300 µ in./in. . Determine the magnitudes of loads P and Q and the location x of load P. SOLUTION π (1) A= = 0.7854 in 2 4 2 σ A = Eε A = (10 × 106 )( 550 × 10−6 ) = 5500 psi σ B = Eε B = (10 × 106 )( 400 × 10−6 ) = 4000 psi σ C = Eε C = (10 × 106 )( −300 × 10−6 ) = −3000 psi Q P ( x + 3) c = Q + P ( x + 3)( 0.50 ) = 5500 psi + A I 0.7854 0.04909 π (1) Ix = = 0.04909 in 4 64 4 σA = σB = (a) Px ( 0.50 ) Q ( Px ) c Q + = + = 4000 psi (b) A I 0.7854 0.04909 Px ( 0.50 ) Q ( Px ) c Q σC = − = − = −3000 psi (c) A I 0.7854 0.04909 From Eqs. (b) and (c): Q = 392.7 lb ≅ 393 lb ................................................................ Ans. From Eq. (b): From Eq. (a): Solving yields: Px = 343.63 lb P ( x + 3) = 490.90 lb P = 49.09 lb ≅ 49.1 lb .................... x = 7.00 in. ................................. Ans. 10-140 A steel shaft 120 mm in diameter is supported in flexible bearings at its ends. Two pulleys, each 500 mm in diameter are keyed to the shaft. The pulleys carry belts that produce the forces shown in Fig. P10-140. Determine the principal stresses and the maximum shearing stress at point A on the top surface of the shaft. SOLUTION π (120 ) = 10.179 × 106 mm 4 I= 64 4 π (120 ) = 20.358 × 106 mm 4 J= 32 4 (120 ) Q= 12 ΣM B = 0 : ΣFz = 0 : 3 = 144 × 103 mm3 RC ( 2800 ) − 35 ( 800 ) = 0 RB − 35 + RC = 0 TA = 30 ( 0.250 ) − 5 ( 0.250 ) = 6.25 kN ⋅ m RC = 10.00 kN RB = 25.00 kN For the horizontal plane: 527 STATICS AND MECHANICS OF MATERIALS, 2nd Edition For the vertical plane: RILEY, STURGES AND MORRIS ΣM B = 0 : ΣFz = 0 : RC ( 2800 ) − 35 ( 2000 ) = 0 RB − 35 + RC = 0 RC = 25.00 kN RB = 10.00 kN From the shear force and bending moment diagrams: M A = 10 (1.200 ) = 12 kN ⋅ m VA = 35 − 25 = 10 kN 3 3 −6 Tc VQ ( 6.25 ×10 ) ( 0.060 ) (10 ×10 )(144 ×10 ) τ xy = + = + J It 20.358 ×10−6 (10.179 ×10−6 ) ( 0.120 ) (12 ×103 ) ( 0.060 ) = −70.73 ×106 N/m2 = 70.73 MPa (C) Mc σx = − =− I 10.179 × 10−6 = 19.60 ×106 N/m 2 = 19.60 MPa σ p1, p 2 = −70.73 + 0 −73.70 − 0 2 ± + 19.60 = −35.365 ± 40.433 MPa 2 2 2 σ p1 = −35.365 + 40.433 = +5.068 MPa ≅ 5.07 MPa (T) ............................................ Ans. σ p 2 = −35.365 − 40.433 = −75.798 MPa ≅ 75.8 MPa (C) ......................................... Ans. σ p 3 = 0 MPa Since ........................................................................................................................... Ans. σ p1 and σ p2 have opposite signs: τ max = τ p = 40.433 MPa ≅ 40.4 MPa ............................................................................... Ans. 10-141 A 120-lb girl is walking up a uniform 2 × 12-in. beam of negligible weight, as shown in Fig. P10-141. The coefficient of friction is 0.20 at all surfaces. When the beam begins to slip, determine the maximum tensile and compressive normal stresses on a section at a distance x/2 from the right end of the beam. SOLUTION Af = µ An = 0.2 An B f = µ Bn = 0.2 Bn 528 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS → ΣFh = 0 : ↑ ΣFv = 0 : Solving yields: Bn ( 3 5 ) − 0.2 Bn ( 4 5) − 0.2 An = 0 Bn ( 4 5) + 0.2 Bn ( 3 5 ) + An − 120 = 0 An = 84.62 lb Bn = 38.46 lb 84.62 ( 3 5 ) + 0.2 ( 84.62 )( 4 5 ) − P = 0 120 x ( 4 5) − 38.46 (10 ) = 0 P = 64.31 lb ΣFn = 0 : ΣM A = 0 : x = 4.006 ft ΣM O = 0 : 84.62 ( 4 5 )( 2.003) − M − 0.2 ( 84.62 )( 3 5)( 2.003) = 0 3 M = 115.25 lb ⋅ ft A = 2 ×12 = 24.0 in 2 σ top σ bot 12 ( 2 ) = 8.00 in 4 12 P Mc 64.31 (115.25 ×12 )(1) =− − =− − = −175.56 psi ≅ 175.6 psi (C) ....... Ans. AI 24.0 8.00 P Mc 64.31 (115.25 ×12 )(1) =− + =− + = +170.19 psi ≅ 170.2 psi (T) ....... Ans. AI 24.0 8.00 I= 10-142 Four strain gages are mounted at 90° intervals around the circumference of a 100-mm diameter steel (E = 200 GPa and ν = 0.30 ) shaft, as shown in Fig. P10-142. All of the gages are oriented at an angle of 45° with respect to a line on the surface of the shaft that is parallel to the axis of the shaft. Determine the torque T, shear force V, and bending moment M at the cross section where the gages are located if ε A = +450 µ m/m , ε B = +325 µ m/m , ε C = +550 µ m/m , and ε D = +675 µ m/m . SOLUTION π (100 ) = 7854 mm 2 A= 4 2 4 G= E 200 = = 76.92 GPa 2 (1 + ν ) 2.6 4 π (100 ) π (100 ) = 4.909 × 106 mm 4 = 9.817 × 106 mm 4 I= J= 64 32 2 4 ( 50 ) π (100 ) Q = yC A = = 83.33 × 103 mm3 3π 8 At point A: τ xy = Tc VQ − = σ 45 = −σ −45 J It At point C: 1 1 + ν Tc VQ (σ 45 −νσ −45 ) = − = εA E E J It Tc VQ τ xy = + = σ 45 = −σ −45 J It ε 45 = ε 45 = Therefore: 1 1 + ν Tc VQ (σ 45 −νσ −45 ) = + = εC E E J It 1 + ν Tc ε A + εC = 2 E J 529 STATICS AND MECHANICS OF MATERIALS, 2nd Edition from which: RILEY, STURGES AND MORRIS ( 200 ×109 )( 9.817 ×10−6 ) 450 + 550 ×10−6 EJ T= (ε A + ε C ) = ( ) 2 (1 + ν ) c 2 (1 + 0.30 )( 0.050 ) = 15.103 ×103 N ⋅ m ≅ 15.10 kN ⋅ m ............................................................................. Ans. 1 + ν VQ εC − ε A = 2 E It Similarly: V= ( 200 ×109 )( 4.909 ×10−6 ) ( 0.100 ) 550 − 450 ×10−6 EIt εC − ε A ) = ( ( ) 2 (1 + ν ) Q 2 (1 + 0.30 ) ( 83.33 × 10−6 ) 3 At point B: Tc (15.103 ×10 ) ( 0.050 ) = = 76.92 ×106 N/m 2 −6 J 9.817 ×10 M ( 0.050 ) Mc σx = − =− = − M (10.185 × 103 ) N/m 2 −6 4.909 ×10 I τ σ νσ ε B = x cos 2 45° − x sin 2 45° + xy sin 45° cos 45° E E G = 45.32 × 103 N ≅ 45.3 kN ............................................................................................. Ans. τ xy = = from which: (1 − 0.30 ) ( −10.185 ×103 ) M 200 × 109 ( 0.5) + 76.92 ×106 ( 0.50 ) = 325 ×10−6 76.92 ×109 M = 9.818 × 103 N ⋅ m ≅ 9.82 kN ⋅ m ........................................................................... Ans. 10-143 A 4-in. diameter solid circular steel shaft is loaded and supported as shown in Fig. P10-143. Because of discontinuities at points A and B, the normal and shearing stresses at these points must be limited to 13 ksi (T) and 8.5 ksi, respectively. Determine the maximum permissible value for the load P. SOLUTION A= π ( 4) = 12.566 in 2 4 2 4 QB = yC A = 4 2 4 ( 2 ) π ( 4) = 5.333 in 3 3π 8 π ( 4) Ix = = 12.566 in 4 64 At point A: π ( 4) = 25.13 in 4 J= 32 (16 P )( 2 ) = 3.183P F Mc 8P + = + 12.566 12.566 A I Tc ( 8P )( 2 ) τ xy = = = 0.6367 P 25.13 J σx = At point B: σx = ( 64 P )( 2 ) = 10.823P F Mc 8P + = + 12.566 12.566 A I 530 STATICS AND MECHANICS OF MATERIALS, 2nd Edition RILEY, STURGES AND MORRIS τ xy = Tc VQ ( 8 P )( 2 ) P ( 5.333) + = + = 0.7428 P 25.13 12.566 ( 4 ) J It 2 Therefore, maximum stresses will occur at point B. 10.823P + 0 2 10.823P − 0 σ p1 = + + ( 0.7428 P ) < 13 ksi 2 2 gives P < 1.1955 kip 2 2 10.823P − 0 τ p = τ max = + ( 0.7428 P ) < 8.5 ksi 2 gives P < 1.5561 kip Therefore: Pmax = 1.1955 kip ≅ 1.196 kip ................................................................ Ans. 10-144 The frame shown in Fig. P10-144 is constructed of 100 × 100-mm timbers. Determine and show on sketches the principal and maximum shearing stresses at points G and H. SOLUTION ΣM C = 0 : 6 (1.5 ) + 3 (1.5 ) − Ay ( 3) = 0 Ay = 4.50 kN At the section containing G and H: ΣF = 0 : ΣF = 0 : ΣM O = 0 : 4.50 ( 4 5 ) − P = 0 4.50 ( 3 5 ) − V = 0 P = 3.60 kN V = 2.70 kN M − 4.50 ( 3 5 ) (1) = 0 M = 2.7 kN ⋅ m 3 A = 100 ×100 = 10 ×103 mm 2 100 (100 ) I= = 8.333 × 106 mm 4 12 QH = yC A = 37.5 (100 × 25 ) = 93.75 × 103 mm3 At point G: 3 P Mc 3.6 × 103 ( 2.7 × 10 ) ( 0.050 ) σ =− − =− − = −16.561× 106 N/m 2 −3 −6 AI 10 × 10 8.333 ×10 Therefore: σ p1 = 0 MPa ............. σ p 2 = −16.561×106 N/m 2 ≅ 16.56 MPa (C) ................ Ans. σ p 3 = 0 MPa .................... τ max = σ p1 − σ p 2 = 8.28 MPa 2 .................................... Ans. At point H: 3 P My 3.6 × 103 ( 2.7 × 10 ) ( 0.025 ) σ =− + =− + = +7.740 ×106 N/m 2 −3 −6 AI 10 × 10 8.333 ×10 531 STATICS AND MECHANICS OF MATERIALS, 2nd Edition Maximum-Distortion-Energy: RILEY, STURGES AND MORRIS σ e = σ 21 − σ p1σ p 3 + σ 2 3 = 302 − ( 30 )( −50 ) + 502 = 70 ksi p p σ e = 70 ksi > σ f = 60 ksi (Failure) .................................................................................. Ans. 10-150 The state of stress at a point on the surface of a machine component is shown in Fig. P10-150. If the yield strength of the material is 250 MPa, determine which, if any, of the theories will predict failure by yielding for this state of stress. SOLUTION σ x = +100 MPa σ p1, p 2 σ y = +80 MPa 2 τ xy = −40 MPa σ +σ y σ +σ y 2 =x ±x + τ xy 2 2 100 + 80 2 100 − 80 = ± + ( −40 ) = 90 ± 41.23 MPa 2 2 2 σ p1 = 90 + 41.23 = +131.23 MPa ≅ 131.2 MPa (T) σ p 2 = 90 − 41.23 = +48.77 MPa ≅ 48.8 MPa (T) σ p 3 = σ z = 0 MPa Maximum-Normal-Stress: σ f = 250 MPa σ max = σ p1 = 131.2 MPa < σ f = 250 MPa (No failure) .............................................. Ans. Maximum-Shear-Stress: 1 (σ p1 − σ p3 ) = 1 (131.2 − 0 ) = 65.6 MPa 2 2 1 1 τ f = σ f = ( 250 ) = 125 MPa 2 2 τ max = 65.6 MPa < τ f = 125 MPa (No failure) .............................................................. Ans. τ max = Maximum-Distortion-Energy: σ e = σ 21 − σ p1σ p 3 + σ 2 3 = 131.22 − (131.2 )( 48.8 ) + 48.82 = 114.9 MPa p p σ e = 114.9 MPa < σ f = 250 MPa (No failure) ............................................................. Ans. 10-151 At a point on