EM 319 - 1st Homework

EM 319 - 1st Homework - 1.2-1 A hollow circular post ABC...

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Unformatted text preview: 1.2-1 A hollow circular post ABC (see figure) supports a load P. = l?00lb acting at the top. A second load P2 is uniformly distributed around the cap plate at B. The diam— eters and thicknesses of the upper and lower parts of the post are aim = 1.25 111.. L15 = 0.5 in.. dm- = 2.25 in.. and rm- --—- 0.375 in.. respectively. ta) Calculate the normal stress 0",”; in the upper part of the post. {b} If it is desired that the lower part of the post have the same compressive stress as the upper part. what should be the magnitude of the load P3? (c) If PI remains at 1700 lb and P3 is now set at 2260 lb, what new thickness of HC will result in the same compres- sive stress in both part s‘? PRDB. 1.2-1 CHAPTER 1 Problems 1.2-2 A force F of 70 N is applied by a rider to the front hand brake of a bicycle (P is the resultant of an evenly distributed pressure]. As the hand brake pivots at A, a tension Tdevelops in the 460—1nm long brake cable (Ar = [.025 Inma) which elongates by 5 = 0.2 [4 mm. Find normal stress or and strain .9 in the brake cable. Brake cable. L = 460 mm Hand brake pivot A P (Resultant of distributed pressure) - . H I ‘ “700 “HR /¢- Uniform hand “he. 0303.. .r brake pressure PHDE. 1.2—2 CHAPTER 1 Tension, Compression, and Shear 1.2-3 A bicycle rider would like to compare the effective— ness of cantilever hand brakes [see figure part (all versus V brakes [figure part (btl. (a) Calculate the braking force RH at the wheel rims for each of the bicycle brake systems shown. Assume that all forces act in the plane of the figure and that cable tension 1" = 45 lbs. Also. what is the average compressive normal stress or. on the brake pad (A — 0.035 ml)? anchored to frame ta) Cantilever brakes FHCIB. 1.2v3 1.2-4 A circular aluminum tube of length L = 400 mm is loaded in compression by forces P (see figure). The outside and inside diameters are 60 mm and 50 mm. respectively. A strain gage is placed on the outside of the bar to measure normal strains in the longitudinal direction. (a) if the measured strain is E = 550 X IO '5. what is the shortenng 5 of the bar? {b} If the compressive stress in the bar is intended to he 40 MPa. what should be the load I"? (b) For each braking system. what is the stress in the brake cable (assume effective cross—sectional area of 0.00 l 6? if}? (HINT: Because of symmetry. you only need to use the right halfof each figure in your analysis.) —- PIVOII porntS—_-_ anchored to frame - [b] V brakes Strain gage /. u— — ——L=4(imm —’i PRIDE. 1.2—4 1.2-5 The cross section of a concrete corner column that is loaded uniformly in compression is shown in the figure. CHAPTER 1 Problems 3:25 (a) Determine the average compressive stress rrf. in the ,8 = 48°. Both wires have a diameter of 30 mils. (Wire (lia- concrete if the load is equal to 3200 k. meters are often expressed in mils: one mil equals 0.001 in.) (b) Determine the coordinates xr and v. of the point Determine the tensile stresses or. and 0'; in the two wires. where the resultant load must act in order to produce unilorrn normal stress in the column. y 4— 24 in.—»lv 20 lit—v PRIJB. 1.2-5 PRDB. 1.2-7 1.2-6 A car weighing 130 kN when fully loaded is pulled slowly up a steep inclined track by a steel cable (see figure). The cable has an effective cross-sectional area of 490 mmz. 1.2-8 A long retaining wall is braced by wood shores set at and the angle a of the incline is 30°. an angle of 30° and supported by concrete thrust blocks. as Calculate the tensile stress Uf in the cable. shown in the first part of the figure. The shores are evenly spaced, 3 m apart. For analysis purposes, the wall and shores are ideal- ized as shown in the second part of the figure. Note that the base of the wall and both ends of the shores are assumed to be pinned. The pressure of the soil against the wall is assumed to be triangularly distributed. and the resultant force acting on a 3—meter length of the wall is F = 190 kN. If each shore has a 150 mm X 150 mm square cross section. what is the compressive stress (rt in the shores? Soil Retaining wall Concrete Shore thrust / PRIJB. 1.2-6 K block 1.2-? Two steel wires support a moveablc overhead camera weighing W = '25 ll) (see figure) used for close-up viewing of field action at sporting events. At some instant. wire 1 is at an angle a = 20° to the horizontal and wire 2 is at an angle PRDB. 1.2-8 51} CHAPTER 1 Tension, Compression, and Shear 1.2-9 A pickup truck tailgate supports a crate (WC = 1501b), as shown in the figure. The tailgate weighs IMF-=60 lb and is supported by two cables (only one is shown in the figure). Each cable has an effective cross—sectional area A. = 0.017 inz. (a) Find the tensile force T and normal stress 0' in each cable. (b) If each cable clongales 5: 0.0] in. due to the weight of both the crate and the tailgate, what is the average strain in the cable'? PROBS.1.2-9 and 12-10 1.2-10 Solve the preceding problem if the mass of the tail gate is MT = 27 kg and that ofthe crate is MC = 68 kg. Use dimen- sions H = 305 mm, L = 406 mm. dc = 460 mm, and (if: 350 mm. The cable cross-sectional area is AC, = l l .0 mmz. (a) Find the tensile force T and normal stress 0' in each cable. (b) If each cable elongates 5 = 0.25 mm due to the weight of both the crate and the tailgate. what is the average strain in the cable? L = 406 mm PRDB. 1.240 1"1.2-" An L-shaped reinforced concrete slab [2 ft X 12 ft (but with a 6 ft X 6 ft cutout) and thickness t = 9.0 in. is lifted by three cables attached at 0, B and D, as shown in the figure. The cables are combined at point Q, which is 7.0 ft above the top of the slab and directly above the center of mass at C. Each cable has an effective cross~sectional area 01°14.e = 0.12 inz. (a) Find the tensile force T,- (i = l, 2, 3) in each cable due to the weight W of the concrete slab [ignore weight of cables). (b) Find the average stress o,- in each cable. (See Table 1-1—1 in Appendix H for the weight density of reinforced concrete.) (0‘ U, ) _ l‘ 6 ft -L B(IZ,0,0) lb Concrete slab ’y = 150 E Thickness t, cg at (5 ft, 5 ft. 0) PHOB. 1.2-11 5-2? CHAPTER 1 Tension, Compression, and Shear 1.3-3 Three different materials. designated A, B, and C, are tested in tension using test specimens having diameters of 0.505 in. and gage lengths of2.0 in. (see figure). At failure. the distances between the gage marks are found to be 2.13, 2.48. and 2.78 in.. respectively. Also. at the failure cross sections the diameters are found to be 0.484. 0.398. and 0.253 in.. respectively. Determine the percent elongation and percent reduction in area of each specimen. and then. using your Own judg- ment, classify each material as hrittle or ductile. Gage PRIJB. 1.3-3 1.3-4 The .s'tr'engrli-to-n‘cight ratio of a structural material is defined as its load-carrying capacity divided by its weight. For materials in tension. we may use a charac— teristic tensile stress (as obtained from a stress-strain curve) as a measure of strength. For instance. either the yield stress or the ultimate stress could he used. depending upon the particular application. Titus. the strength-to— weight ratio REM for a material in tension is defined as 0' Rs'xw : _ 1’ in which 0' is the characteristic stress and 'y is the weight density. Note that the ratio has units of length. Using the ultimate stress or“ as the strength parameter. calculate the strength—to—weight ratio (in units of meters) for each of the following materials: alumintlm alloy 60bl—T6. Douglas fir (in bending). nylon, structural steel ASTM-A572. and a titanium alloy. (Obtain the material properties from Tables H—1 and 1-1-3 of Appendix H. When a range of values is given in a table. use the average value.) 1.3-5 A symmetrical framework consisting of three pin— connected bars is loaded by a force P (see figure). The angle between the inclined bars and the horizontal is a = 48°. The axial strain in the middle bar is measured as 0.0? [3. Determine the tensile stress in the outer bars if they are constructed of aluminum alloy having the stress-strain diagram shown in Fig. 1-13. (Express the stress in USCS units.) PRDB. 1.3-5 1.3-6 A specimen of a methacrylate plastic is tested in tension at room temperature (see figure), producing the stress-strain data listed in the accompanying table (see the next page). Plot the stress—strain curve and determine the propor- tional limit. modulus of elasticity tie. the slope of the initial part of the stress—strain curve), and yield stress at 0.2% offset. Is the material ductile or brittle? P Md.th ( gteTH-i PROB. 1.3-5 STRESS-STRAIN DATA FOR PROBLEM 1.3-6 Stress (Ml-’21) Strain 8.0 0.0032 17.5 0.0073 25.6 0.01 1 1 31.1 0.0129 39.8 0.0163 44.0 0.0184 48.2 0.0209 53.9 0.0260 58.1 0.0331 62.0 0.0429 62.1 Fracture its 7 The data shown in the accompanying table were _i"|'ll'i from a tensile test of high-strength steel. The test gs en had a diameter of 0.505 in. and a gage length of in. {see figure for Prob. 1.3-3). At fracture. the elonga— _a between the gage marks was 0. [2 in. and the minimum in was 0.42 in. Plot the conventional stress-strain curve for the steel determine the proportional limit, modulus of elasticity ., the slope of the initial part of the stress—strain curve), "_f--.I stress 310.1% offset. ultimate stress, percent elonga— _-:: in 2.00 in., and percent reduction in area. TENSILE-TEST DATA FOR PROBLEM 1.3-7 Load (lb) Elongation (in) l ,000 0.0002 2,000 0.0006 6,000 0.00 19 10.000 0.0033 12.000 00039 12,900 0.0043 I 3 ,400 0.004? I 3 .600 0.0054 1 3 .800 0.0063 I 4,000 0.0090 14.400 0.0102 1 5 .200 0.0 I 30 16.8 00 0.02 30 18,400 0.0336 20,000 0.050? 22,400 0.1 108 22.600 Fracture -- 'city and Plasticity .- 'I A bar made of structural steel having the stress-strain am shown in the figure has a length of 48 in. The yield 5 of the steel is 42 ksi and the slope of the initial linear _ of the stress—strain curve (modulus of elasticity) is X 103 ksi. The bar is loaded axially until it elongates .320 in., and then the load is removed. CHAPTER 1 Problems 63 How does the final length of the bar compare with its original length? (Him: Use the concepts illustrated in Fig. l—le.) 0(ksi] 60 ‘7 40 20 0 . 0 0.002 0.004 0.006 e PHUB. 1.4-1 1.4-2 A bar of length 2.0 m is made of a structural steel having the stress-strain diagram shown in the figure. The yield stress of the steel is 250 MPa and the slope of the initial linear pan of the stress-strain curve (modulus of elasticity) is 200 (Spa. The bar is loaded axially until it clongates 6.5 mm. and then the load is removed. How does the final length of the bar compare with its original length? (Hint: Use the concepts illustrated in Fig. 1-18b.) 0'( MPa) 300 200 100 0 0.002 0.004 0.006 PRUB. 1.4~2 64 CHAPTER 1 Tension, Compression, and Shear 1.4-3 An aluminum bar has length l. = 5 ft and diameter d = 1.25 in. The stress—strain curve for the aluminum is shown in Fig. 1—13 of Section 1.3. The initial straightw line part of the curve has a slope (modulus of elasticity) of 10 X 10‘I psi. The bar is loaded by tensile forces P = 39 k and then unloaded. (a) What is the permanent set of the bar? (b) If the bar is reloaded. what is the proportional limit? (Him: Use the concepts illustrated in Figs. 1—18b and 1—19.) 1.4-4 A circular bar of magnesium alloy is 750 mm long. The stress—strain diagram for the material is shown in the figure. The bar is loaded in tension to an elongation of 6.0 mm. and then the load is removed. (a) What is the permanent set ofthe bar? (b) lfthe bar is reloaded, what is the proportional limit? (Him: Use the concepts illustrated in Figs. 1-18b and 1-19.] 200 rI(MPa) 100 l1 0 0.005 0.010 P9035. 1.4-3 and 1.4-4 in1.4-5 A wire of length L = 4 ft and diameter d = 0.125 in. is stretched by tensile forces P = 600 lb. The wire is made of a copper alloy having a stress-strain relationship that may be described mathematically by the following equation: 18.0005 0 003 k _ = — E S I = . ‘7 1 + 3005 E (a 51) in which 6 is nondimensional and a" has units of kips per square inch {ksi}. (a) Construct a stress—strain diagram for the material. (b) Determine the elongation of the wire due to the forces P. (c) If the forces are removed, what is the permanent set of the bar? (d) If the forces are applied again, what is the propor— tional limit? Roche’s Law and Poisson‘s Basia When solving the problems for Secrirm l .5. assume that the material behaves linearly elastically. 1.5-1 A high-strength steel bar used in a large crane has diameter d = 2.00 in. {see figure). The steel has modulus of elasticity E = 29 X 10" psi and Poisson's ratio V = 0.29. Because of clearance requirements. the diameter of the bar is limited to 2.001 in. when it is compressed by axial forces. What is the largest compressive load Pm.“1x that is permitted? PRUB. 1.5—1 1.5-2 A round bar of 10 mm diameter is made of aluminum alloy 7075—T6 (see figure}. When the bar is stretched by axial forces P, its diameter decreases by 0.016 mm. Find the magnitude of the load P. (Obtain the material properties from Appendix H.) TOYS—T6 PROB. 1.5-2 1.5-3 A polyethylene bar having diameter d1 = 4.0 in. is placed inside a steel tube having inner diameter d: = 4.01 in. (see figure). The polyethylene bar is then compressed by an axial force P. At what value of the force P will the space between 'the polyethylene bar and the steel tube be closed? (For polyethylene, assume E = 200 ksi and v = 04.) Steel tube Polyethylene bar PHDB. 1.5-3 1.5-4 A prismatic bar with a circular cross section is loaded by tensile forces P = 65 kN (see figure). The bar has length L = 1.75 m and diameter d = 32 mm. It is made of aluminum alloy with modulus of elasticity E = 75 GPa and Poisson’s ratio U = U3. Find the increase in length of the bar and the percent decrease in its cross—sectional area. P8083. 1.54 and 1.5-5 1.5-5 A bar of monel metal as in the figure (length L = 9 in., diameter d = 0.225 in.) is loaded axially by a tensile force P. If the bar elongates by 0.0195 in., what is the decrease in diameter a"? What is the magnitude of the load P? Use the data in Table 1-1-2, Appendix H. 1.5-5 A tensile test is peformed on a brass specimen 10 mm in diameter using a gage length of 50 mm (see figure]. When the tensile load P reaches a value of 20 kN, the distance between the gage marks has increased by 0.122 mm. CHAPTER 1 Problems 55 (a) What is the modulus of elasticity E of the brass? (b) If the diameter decreases by 0.00830 mm, what is Poisson’s ratio? PRDB. 1.5-6 1.5-7 A hollow, brass circular pipe ABC (see figure) supports a load P. = 26.5 kips acting at the top. A second load P2 = 22.0 kips is unifonnly distributed around the cap plate at B. The diameters and thicknesses of the upper and lower parts of the pipe are LEM = 1.25 in., .5”,- = 0.5 in.. (lac = 2.25 in., and rm; = 0.375 in., respectively. The modulus of elasticity is [4,000 ksi. When both loads are fully applied, the wall thick- ness of pipe BC increases by 200 X 10 h in. (a) Find the increase in the inner diameter of pipe segment BC. (b) Find Poisson‘s ratio for the brass. (c) Find the increase in the wall thickness of pipe segment AB and the increase in the inner diameter ofAB. PRUB. 1.5-? 1.2-1 1.2-2 1.2-3 1.2-4 1 .2-5 1.2-6 1.2-? 1 .2-8 1.2-9 1.2-10 1.2-“ 1.2-“! 1.2-13 1.2-14 1.3-1 1.3-2 1.3-3 1.3-4 1.3-5 1.3-6 1.3-? CHAPTER 1 (1003”, = 1443 psi; (b) Pg = 1487.5 lbs; (c) rm; = 0.5 in. (11):)" = 65 MPa; (b) a: = 4.652 X 10 4 (a) Rh. = 127.3 lb {cantilever}, 191.3 lb (V—brakes); t. = 204 psi (cantilever), 306 psi (V—brakes); (b) (b) (rum.e = 26,946 psi (both) (a15 = 0.220 mm; (b) P = 34.6 kN (a) 05 = 2.128 ksi: IC = 19.22 in., ya = 19.22 in. or, = 133 MPa or. = 25.5 ksi: U2 = 35.8 ksi: cr,_. = 5.21 MPa (3) T: 184 1b, 0' = 10.8 ksi; (b) em“, = 5 x 10'4 (a) T = 819 N, 0' = 74.5 MPa; (b) em“, = 4.923 >< 10‘4 (a) T. = 5877 lb, T2 = 46791b. T3 = 7159 lb; (610. = 49 k51,0'2 = 39 ksi. 03 = 60 ksi (a) (er = yw2(L2 - x2)7'2g: (b) Um“ = yw2L272g ((1)7113 = 16201b, TM: = 1536 lb, To) = 164016 (b) (IN, = 13,501 psi, 17m; = 12.799 psi, (rt-p = 13.667 psi (a) TAQ = TBQ = 50.5 kN; (b} 0' = 166 MPa (a) Lmax = 11,800 ft; (6)12”, = 13,500 ft (a) Lmax = 7900 m; (b) Lnm = 8330 m % elongation = 6.5, 24.0, 39.0; % reduction = 8.1, 37.9, 74.9: Brittle. ductile, ductile 11.9 X103m;12.7 ><103m;6.1><103 m; 6.5 X 103 m; 23.9 X 103 m rr H”: 31 ksi Up. = 47 MPa, Slope % 2.4 GPa. 01' = 53 MPa; Brittle up. =-- 65,000 psi, Slope 4: 30 x 10“ psi, cry ‘-= 69,000 psi, 0'” 2 113,000 psi; Elongation = 6%, Reduction = 31% 0.13 in. longer Answers to Problems 1.4-2 1.4-3 1.4-4 1.4-5 1.5-1 1.5-2 1.5-3 1.5-4 1.5-5 1.5-6 1.5-7 1 .5-3 1.6-1 1.6-2 1.6-3 1.6—4 1.6-5 1.6-6 1.8-7 1.6-8 1.6-9 1.6-"! 1.6-“ 1.8-12 1.6-13 4.0 mm longer (a) 2.809 in.; (b) 31.8 ksi (a) 2.966 mm; (b) 180 MPa (1)) 0.71 in.: (1:10.58 ir1.; {(1)49 ksi P11m = 157 k P = 27.4 kN (tension) P = - 15.708 kips AL = 1.886 mm; 92: decrease in .r—sec area = 0.072% Ad 2 —1.56 x 10“1 in., P = 2.154 kips (a) E = 104 GPa; (b) v = 0.34 (a) AdflCinner : 8 X 10—4 in- 16) 9W, 2 0.34 {0)AL13 = 2.?32 x 10416, Admmr = 1.366 >< 10‘4 in. M = 9789 mm3 0;, = 7.04 ksi, “rave = 10.756 ksi 01:, = 139.86 MPa; Pun = 144.45 kN (a) 1' = 12.732 ksi; (b) any: = 20 ksi: 0b,, = 26.667 ksi (a) Ax = 254.6 N, A}. = 1072 N, 3,, = —254.6 N (b) Aresullam 2 1101-8 N (c) T = 5.48 MPa, 0!, = 6.886 MPa {3) “rm, = 2979 psi; 1b) orbmx = 936 psi T] = 13.176 kN, T3 = 10.772 RN, TIM, = 25.888 MPa, 72m = 21.166 MPa, orb. = 7.32 MPa, 0b; = 5.985 MPa (6) Resultant = 1097 lb; (b) 0;», = 4999 psi (1:) Tm“ = 2793 psi, 7p] 2 609 psi G = 2.5 MPa (at) ‘yaver = 0.004; (b) V = 89.6 k (a) yam, = 0.50; (b) 5 = 4.50 mm (a) rave, = 6050 psi; (b) 0;, = 9500 psi 7;..." = 429 MPa (a) Ax = 0, 14).: 17016114,, = 4585 in.—lb 1MB,r = 253.6 lb, B). = 1601b, Bras = 299.8 lb, ct = _B.'( 995 ...
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This note was uploaded on 09/18/2011 for the course EM 319 taught by Professor Staff during the Spring '11 term at Oklahoma State.

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EM 319 - 1st Homework - 1.2-1 A hollow circular post ABC...

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