solutions_Chapter03
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solutions_Chapter03

Course Number: CIVL 3322, Spring 2011

College/University: U. Memphis

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3.1 At the proportional limit, a 12-inch gage length of a 0.75-in.-diameter alloy bar has elongated 0.0325 in. and the diameter has been reduced 0.0006 in. The total tension force on the bar was 17.5 kips. Determine the following properties of the material: (a) the modulus of elasticity. (b) Poissons ratio. (c) the proportional limit. Solution (a) The bar cross-sectional area is D2 = (0.75 in.) 2 = 0.441786...

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At 3.1 the proportional limit, a 12-inch gage length of a 0.75-in.-diameter alloy bar has elongated 0.0325 in. and the diameter has been reduced 0.0006 in. The total tension force on the bar was 17.5 kips. Determine the following properties of the material: (a) the modulus of elasticity. (b) Poissons ratio. (c) the proportional limit. Solution (a) The bar cross-sectional area is D2 = (0.75 in.) 2 = 0.441786 in.2 4 4 and thus, the normal stress corresponding to the 17.5-kip force is 17.5 kips = = 39.611897 ksi 0.441786 in.2 The strain in the bar is e 0.0325 in. = = = 0.002708 in./in. L 12 in. The modulus of elasticity is therefore 39.611897 ksi = 14, 626 ksi = 14,630 ksi E= = 0.002708 in./in. A= Ans. (b) The longitudinal strain in the bar was calculated previously as long = 0.002708 in./in. The lateral strain can be determined from the reduction of the diameter: D 0.0006 in. = = 0.000800 in./in. lat = 0.75 in. D Poissons ratio for this specimen is therefore 0.000800 in./in. = lat = = 0.295421 = 0.295 long 0.002708 in./in. Ans. (c) Based on the problem statement, the stress in the bar is equal to the proportional limit; therefore, PL = 39.6 ksi Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.2 At the proportional limit, a 20 mm thick 75 mm wide bar elongates 6.8 mm under an axial load of 480 kN. The bar is 1.6-m long. If Poissons ratio is 0.32 for the material, determine: (a) the modulus of elasticity. (b) the proportional limit (c) the change in each lateral dimension. Solution (a) The bar cross-sectional area is A = (20 mm)(75 mm) = 1,500 mm 2 and thus, the normal stress corresponding to the 480-kN axial load is (480 kN)(1,000 N/kN) = = 320 MPa 1,500 mm 2 The strain in the bar is e 6.8 mm = = = 0.004250 mm/mm L (1.6 m)(1,000 mm/m) The modulus of elasticity is therefore 320 MPa E= = = 75, 294 MPa = 75.3 GPa 0.004250 mm/mm Ans. (b) Based on the problem statement, the stress in the bar is equal to the proportional limit; therefore, PL = 320 MPa Ans. (c) Poissont ratio is given as = 0.32. The longitudinal strain in the bar was calculated previously as long = 0.004250 mm/mm The corresponding lateral strain can be determined from Poissons ratio: lat = long = (0.32)(0.004250 mm/mm) = 0.001360 mm/mm Using this lateral strain, the change in bar width is width = lat (width) = ( 0.001360 mm/mm)(75 mm) = 0.1020 mm and the change in bar thickness is thickness = lat (thickness) = (0.001360 mm/mm)(20 mm) = 0.0272 mm Ans. Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.3 At an axial load of 22 kN, a 15 mm thick 45 mm wide polyimide polymer bar elongates 3.0 mm while the bar width contracts 0.25 mm. The bar is 200 mm long. At the 22-kN load, the stress in the polymer bar is less than its proportional limit. Determine: (a) the modulus of elasticity. (b) Poissons ratio (c) the change in the bar thickness. Solution (a) The bar cross-sectional area is A = (15 mm)(45 mm) = 675 mm 2 and thus, the normal stress corresponding to the 22-kN axial load is (22 kN)(1,000 N/kN) = = 32.592593 MPa 675 mm 2 The strain in the bar is e 3.0 mm = = = 0.0150 mm/mm L 200 mm The modulus of elasticity is therefore 32.592593 MPa E= = = 2,173 MPa = 2.17 GPa 0.0150 mm/mm (b) The longitudinal strain in the bar was calculated previously as long = 0.0150 mm/mm The lateral strain can be determined from the reduction of the bar width: width 0.25 mm = = 0.005556 mm/mm lat = width 45 mm Poissons ratio for this specimen is therefore 0.005556 mm/mm = 0.370370 = 0.370 = lat = 0.0150 mm/mm long (c) The change in bar thickness can be found from the lateral strain: thickness = lat (thickness) = (0.005556 mm/mm)(15 mm) = 0.0833 mm Ans. Ans. Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.4 A 0.75-in.-thick rectangular alloy bar is subjected to a tensile load P by pins at A and B. The width of the bar is w = 3.0 in. Strain gages bonded to the specimen measure the following strains in the longitudinal (x) and transverse (y) directions: x = 2,136 and y = 673 . (a) Determine Poissons ratio for this specimen. (b) If the measured strains were produced by an axial load of P = 50 kips, what is the modulus of elasticity for this specimen? Fig. P3.4 Solution (a) Poissons ratio for this specimen is 673 = 0.315 = lat = y = 2,136 long x (b) The bar area is A = (3.0 in.)(0.75 in.) = 2.25 in.2 and so the normal stress for an axial load of P = 50 kips is 50 kips = = 22.222222 ksi 2.25 in.2 The modulus of elasticity is thus 22.222222 ksi = 10, 404 ksi = 10, 400 ksi E= = 1 in./in. (2,136 ) 1,000,000 Ans. Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.5 A 6-mm-thick rectangular alloy bar is subjected to a tensile load P by pins at A and B. The width of the bar is w = 30 mm. Strain gages bonded to the specimen measure the following strains in the longitudinal (x) and transverse (y) directions: x = 1,525 and y = 540 . (a) Determine Poissons ratio for this specimen. (b) If the measured strains were produced by an axial load of P = 27.5 kN, what is the modulus of elasticity for this specimen? Fig. P3.5 Solution (a) Poissons ratio for this specimen is 540 = 0.354 = lat = y = 1,525 long x (b) The bar area is A = (30 mm)(6 mm) = 180 mm 2 and so the normal stress for an axial load of P = 50 kips is (27.5 kN)(1,000 N/kN) = = 152.777778 MPa 180 mm 2 The modulus of elasticity is thus 152.777778 MPa = 100,182 MPa = 100.2 GPa E= = 1 in./in. (1,525 ) 1,000,000 Ans. Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.6 A copper rod [E = 110 GPa] originally 600-mm long is pulled in tension with a normal stress of 275 MPa. If the deformation is entirely elastic, what is the resulting elongation? Solution Since the deformation is elastic, the strain in the rod can be determined from Hookes Law, 275 MPa = 0.002500 mm/mm = = E 110, 000 MPa The elongation in the rod is thus e = L = (0.002500 mm/mm)(600 mm) = 1.500 mm Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.7 A 6061-T6 aluminum tube [E = 10,000 ksi; = 0.33] has an outside diameter of 4.000 in. and a wall thickness of 0.065 in. (a) Determine the tension force that must be applied to the tube to cause its outside diameter to contract by 0.005 in. (b) If the tube is 84-in. long, what is the overall elongation? Solution (a) The strain associated with the given diameter contraction is 0.005 in. lat = = 0.001250 in./in. 4.000 in. From the given Poissons ratio, the longitudinal strain in the tube must be 0.001250 in./in. long = lat = = 0.003788 in./in. 0.33 and from Hookes Law, the normal stress can be calculated as = E = (10, 000 ksi)(0.003788 in./in.) = 37.878788 ksi The area of the tube is needed to determine the tension force. Given that the outside diameter of the tube is 4.000 in. and the wall thickness is 0.065 in., the inside diameter of the tube is 3.870 in. The tube cross-sectional area is thus (4.000 in.) 2 (3.870 in.) 2 = 0.803541 in.2 4 and the force applied to the tube is F = A = (37.878788 ksi)(0.803541 in.2 ) = 30.437154 kips = 30.4 kips Ans. (b) The strain was calculated previously. Use the longitudinal strain to determine the overall elongation: e = L = (0.003788 in./in.)(84 in.) = 0.318 in. Ans. A= Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.8 A metal specimen with an original diameter of 0.500 in. and a gage length of 2.000 in. is tested in tension until fracture occurs. At the point of fracture, the diameter of the specimen is 0.260 in. and the fractured gage length is 3.08 in. Calculate the ductility in terms of percent elongation and percent reduction in area. Solution Percent elongation is simply the longitudinal strain at fracture: e (3.08 in. 2.000 in.) 1.08 in. = = 0.54 in./in. = = L 2.000 in. 2.000 in. percent elongation = 54% Ans. The initial cross-sectional area of the specimen is D2 = (0.500 in.) 2 = 0.196350 in.2 4 4 The final cross-sectional area at the fracture location is A0 = D2 = (0.260 in.) 2 = 0.053093 in.2 4 4 The percent reduction in area is A0 Af (0.196350 in.2 0.053093 in.2 ) percent reduction of area = (100%) = (100%) = 73.0% Ans. A0 0.196350 in.2 Af = Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.9 A portion of the stress-strain curve for a stainless steel alloy is shown in Fig. P3.9. A 350mm-long bar is loaded in tension until it elongates 2.0 mm and then the load is removed. (a) What is the permanent set in the bar? (b) What is the length of the unloaded bar? (c) If the bar is reloaded, what will be the proportional limit? Fig. P3.9 Solution (a) The normal strain in the specimen is e 2.0 mm = = = 0.005714 mm/mm L 350 mm Construct a line parallel to the elastic modulus line that passes through the data curve at a strain of = 0.005714 mm/mm. The strain value at which this modulus line intersects the strain axis is the permanent set: permanent set = 0.0035 mm/mm Ans. (b) The length of the unloaded bar is therefore: e = L = (0.0035 mm/mm)(350 mm) = 1.225 mm L f = 350 mm + 2.225 mm = 352.225 mm (c) From the stress-strain curve, the reload proportional limit is 444 MPa . Ans. Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.10 The 16 22 25 mm rubber blocks shown in Fig. P3.10 are used in a double U shear mount to isolate the vibration of a machine from its supports. An applied load of P = 285 N causes the upper frame to be deflected downward by 5 mm. Determine the shear modulus G of the rubber blocks. Fig. P3.10 Solution Consider a FBD of the upper U frame. The downward force P is resisted by two upward shear forces V; therefore, V = 285 N / 2 = 142.5 N. Next, consider a FBD of one of the rubber blocks. The shear force acting on one rubber block is V = 142.5 N. The area of the rubber block that is parallel to the direction of V is AV = (22 mm)(25 mm) = 550 mm 2 Consequently, the shear stress in one rubber block is V 142.5 N = = = 0.259091 MPa AV 550 mm 2 The shear strain associated with the 5-mm downward displacement of the rubber blocks is given by: 5 mm = 0.312500 = 0.302885 rad tan = 16 mm From Hookes Law for shear stress and shear strain, the shear modulus G can be computed: 0.259091 MPa = G G = = = 0.855411 MPa = 0.855 MPa 0.302885 rad Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.11 Two hard rubber blocks are used in an anti-vibration mount to support a small machine, as shown in Fig. P3.11. An applied load of P = 150 lb causes a downward deflection of 0.25 in. Determine the shear modulus of the rubber blocks. Assume a = 0.5 in., b = 1.0 in., and c = 2.5 in. Fig. P3.11 Solution Determine the shear strain from the angle formed by the downward deflection and the block thickness a: 0.250 in. tan = = 0.500 = 0.463648 rad 0.50 in. Note: The small angle approximation tan is not applicable in this instance. Determine the shear stress from the applied load P and the block area. Note that this is a double shear configuration; therefore, the shear force V acting on a single rubber block is half of the total load: V = P/2 = 75 lb. To determine the area needed here, consider the surface that is bonded to the plate. This area has dimensions of bc. The shear stress acting on a single block is therefore: V 75 lb = = = 30 psi A (1.0 in.)(2.5 in.) The shear modulus G can be calculated from Hookes law for shear stress and strain: 30 psi G = = = 64.704 psi = 64.7 psi = G 0.463648 rad Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.12 Two hard rubber blocks [G = 350 kPa] are used in an antivibration mount to support a small machine, as shown in Fig. P3.12. Determine the downward deflection that will occur for an applied load of P = 900 N. Assume a = 20 mm, b = 50 mm, and c = 80 mm. Fig. P3.12 Solution Note that this is a double shear configuration; therefore, the shear force V acting on a single rubber block is half of the total load: V = P/2 = 450 N. Determine the shear stress from the shear force V and the block area. To determine the area needed here, consider the surface that is bonded to the plate. This area has dimensions of bc. The shear stress acting on a single block is therefore: V 450 N = = = 0.112500 MPa A (50 mm)(80 mm) Since the shear modulus G is given, the shear strain can be calculated from Hookes law for shear stress and shear strain: 0.112500 MPa = G = = = 0.321429 rad G 0.350 MPa From the angle and the block thickness a, the downward deflection of the block can be determined from: tan = a = a tan = (20 mm) tan(0.321429 rad) = 6.659512 mm = 6.66 mm Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.13 A load test on a 6 mm diameter by 150 mm long magnesium alloy rod found that a tension load of 780 N caused an elastic elongation of 0.55 mm in the rod. Using this result, determine the elastic elongation that would be expected for a 19-mm-diameter rod of the same material if the rod were 1.2 m long and subjected to a tension force of 2.6 kN. Solution The area of the 6-mm-diameter rod is D2 = (6 mm) 2 = 28.274334 mm 2 4 4 Thus, the normal stress in the rod due to a 780-N load is F 780 N = = = 27.586857 MPa A 28.274334 mm 2 The strain in the 150-mm long rod associated with a 0.55-mm elongation is e 0.55 mm = = = 0.003667 mm/mm L 150 mm Therefore, the elastic modulus of the magnesium alloy is 27.586857 MPa E= = = 7,523.688217 MPa 0.003667 mm/mm A= The area of the 19-mm-diameter rod is A= D2 = (19 mm) 2 = 283.528737 mm 2 4 4 Thus, the normal stress in the 19-mm-diameter rod due to a 2.6-kN load is F (2.6 kN)(1,000 N/kN) = = = 9.170146 MPa A 283.528737 mm 2 From Hookes Law, the strain that would be expected is 9.170146 MPa = = = 0.001219 mm/mm E 7,523.688217 MPa Since the 19-mm-diameter rod is 1.2-m long, the expected elongation is e = L = (0.001219 mm/mm)(1.2 m)(1,000 mm/m) = 1.462604 mm = 1.463 mm Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.14 The stress-strain diagram for a particular stainless steel alloy is shown in Fig. P3.14. A rod made from this material is initially 800 mm long at a temperature of 20C. After a tension force is applied to the rod and the temperature is increased by 200C, the length of the rod is 804 mm. Determine the stress in the rod and state whether the elongation in the rod is elastic or inelastic. Assume the coefficient of thermal expansion for this material is 18 106/C. Fig. P3.14 Solution The 4-mm total elongation of the rod is due to a combination of load and temperature increase. The 200C temperature increase causes a normal strain of: T = T = (18 106 / C )(200C ) = 0.003600 mm/mm which means that the rod elongates eT = T L = (0.003600 mm/mm)(800 mm) = 2.8800 mm The portion of the 4-mm total elongation due to load is therefore e = e eT = 4 mm 2.8800 mm = 1.1200 mm The strain corresponding to this elongation is e 1.1200 mm = = = 0.001400 mm/mm L 800 mm By inspection of the stress-strain curve, this strain is clearly in the linear region. Therefore, the rod is elastic in this instance. For the linear region, the elastic modulus can be determined from the lower scale plot: (400 MPa 0) E= = = 200,000 MPa (0.002 mm/mm 0) Using Hookes Law (or directly from the - diagram), the stress corresponding to the 0.001400 mm/mm strain is = E = (200, 000 MPa)(0.001400 mm/mm) = 280 MPa Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.15 In Fig. P3.15, rigid bar ABC is supported by axial member (1), which has a cross-sectional area of 400 mm2, an elastic modulus of E = 70 GPa, and a coefficient of thermal expansion of = 22.5 106 /C. After load P is applied to the rigid bar and the temperature rises 40C, a strain gage affixed to member (1) measures a strain increase of 1,650 . Determine: (a) the normal stress in member (1). (b) the magnitude of applied load P. (c) the deflection of the rigid bar at C. Fig. P3.15 Solution (a) The strain measured in member (1) is due to both the internal force in the member and the temperature change. The strain caused by the temperature change is T = T = (22.5 106 / C )(40C ) = 0.000900 mm/mm Since the total strain is = 1,650 = 0.001650 mm/mm, the strain caused by the internal force in member (1) must be = T = 0.001650 mm/mm 0.000900 mm/mm = 0.000750 mm/mm The elastic modulus of member (1) is E = 70 GPa; thus, from Hookes Law, the stress in the member is: 1 = E = (70, 000 MPa)(0.000750 mm/mm) = 52.5 MPa Ans. (b) If the normal stress in member (1) is 52.5 MPa, the axial force in the member is F1 = 1 A1 = (52.5 N/mm 2 )(400 mm 2 ) = 21, 000 N Consider moment equilibrium of rigid bar ABC about joint A to determine the magnitude of P: M A = (1.4 m)(21,000 N) (2.4 m)P = 0 P = 12, 250 N = 12.25 kN Ans. (c) The strain in member (1) was measured as = 1,650 = 0.001650 mm/mm; therefore, the total elongation of member (1) is e1 = 1 L1 = (0.001650 mm/mm)(2,500 mm) = 4.125 mm The deflection of the rigid bar at B is equal to this elongation; therefore, vB = e1 = 4.125 mm (downward). By similar triangles, the deflection of the rigid bar at C is given by: v vB =C 1.4 m 2.4 m 2.4 m 2.4 Ans. vC = vB = (4.125 mm) = 7.071429 mm = 7.07 mm 1.4 m 1.4 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.16 A tensile test specimen of 1045 hot-rolled steel having a diameter of 0.505 in. and a gage length of 2.00 in. was tested to fracture. Stress and strain data obtained during the test are shown in Fig. P3.16. Determine (a) the modulus of elasticity. (b) the proportional limit. (c) the ultimate strength. (d) the yield strength (0.20% offset). (e) the fracture stress. (f) the true fracture stress if the final diameter of the specimen at the location of the fracture was 0.392 in. Fig. P3.16 Solution From the stress-strain curve, the proportional limit will be taken as = 60 ksi at a strain of = 0.0019. (Obviously, there can be quite a bit of leeway in pulling numbers from such a limited plot.) (a) The modulus of elasticity is 60 ksi E= = = 31, 600 ksi 0.0019 in./in. (b) From the diagram, the proportional limit is taken as PL = 60 ksi (c) The ultimate strength is ult = 105 ksi (d) The yield strength is Y = 68 ksi (e) The fracture stress is fracture = 98 ksi Ans. Ans. Ans. Ans. Ans. (f) The original cross-sectional area of the specimen is Af = D2 = (0.505 in.) 2 = 0.200296 in.2 4 4 The area of the specimen at the fracture location is A0 = D2 = (0.392 in.) 2 = 0.120687 in.2 4 4 The true fracture stress is therefore 0.200296 in.2 true fracture = (98 ksi) = 162.6 ksi 0.120687 in.2 Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.17 A tensile test specimen of stainless steel alloy having a diameter of 0.495 in. and a gage length 2.00 of in. was tested to fracture. Stress and strain data obtained during the test are shown in Fig. P3.17. Determine: (a) the modulus of elasticity. (b) the proportional limit. (c) the ultimate strength. (d) the yield strength (0.20% offset). (e) the fracture stress. (f) the true fracture stress if the final diameter of the specimen at the location of the fracture was 0.350 in. Fig. P3.17 Solution From the stress-strain curve, the proportional limit will be taken as = 60 ksi at a strain of = 0.002. (Obviously, there can be quite a bit of leeway in pulling numbers from such a limited plot.) (a) The modulus of elasticity is 60 ksi E= = = 30, 000 ksi 0.002 in./in. (b) From the diagram, the proportional limit is taken as PL = 60 ksi (c) The ultimate strength is ult = 159 ksi (d) The yield strength is Y = 80 ksi (e) The fracture stress is fracture = 135 ksi Ans. Ans. Ans. Ans. Ans. (f) The original cross-sectional area of the specimen is D2 = (0.495 in.) 2 = 0.192442 in.2 4 4 The area of the specimen at the fracture location is A0 = D2 = (0.350 in.) 2 = 0.096211 in.2 4 4 The true fracture stress is therefore 0.192442 in.2 true fracture = (135 ksi) = 270 ksi 0.096211 in.2 Af = Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.18 A bronze alloy specimen having a diameter of 12.8 mm and a gage length of 50 mm was tested to fracture. Stress and strain data obtained during the test are shown in Fig. P3.18. Determine: (a) the modulus of elasticity. (b) the proportional limit. (c) the ultimate strength. (d) the yield strength (0.20% offset). (e) the fracture stress. (f) the true fracture stress if the final diameter of the specimen at the location of the fracture was 10.5 mm. Fig. P3.18 Solution From the stress-strain curve, the proportional limit will be taken as = 210 MPa at a strain of = 0.002. (Obviously, there can be quite a bit of leeway in pulling numbers from such a limited plot.) (a) The modulus of elasticity is 210 MPa E= = = 105, 000 MPa 0.002 in./in. (b) From the diagram, the proportional limit is taken as PL = 210 MPa (c) The ultimate strength is ult = 380 MPa (d) The yield strength is Y = 290 MPa (e) The fracture stress is fracture = 320 MPa Ans. Ans. Ans. Ans. Ans. (f) The original cross-sectional area of the specimen is A0 = D2 = (12.8 mm) 2 = 128.679635 mm 2 4 4 The area of the specimen at the fracture location is D2 = (10.5 mm) 2 = 86.590148 mm 2 4 4 The true fracture stress is therefore 128.679635 mm 2 true fracture = (320 MPa) = 476 MPa 86.590148 mm 2 Af = Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.19 An alloy specimen having a diameter of 12.8 mm and a gage length of 50 mm was tested to fracture. Load and deformation data obtained during the test are given. 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 (0.20% offset). (f) the fracture stress. (g) the true fracture stress if the final diameter of the specimen at the location of the fracture was 11.3 mm. (kN) Change in Length (mm) (kN) Change in Length (mm) 0 7.6 14.9 22.2 28.5 29.9 30.6 32.0 33.0 33.3 36.8 41.0 0 0.02 0.04 0.06 0.08 0.10 0.12 0.16 0.20 0.24 0.50 1.00 43.8 45.8 48.3 49.7 50.4 50.7 50.4 50.0 49.7 47.9 45.1 1.50 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 fracture Load Load Solution The plot of the stress-strain data is shown below. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. (a) The modulus of elasticity is 224.976 MPa E= = = 140, 600 MPa 0.0016 mm/mm Ans. (b) From the diagram, the proportional limit is taken as PL = 234 MPa Ans. (c) The ultimate strength is ult = 400 MPa Ans. (d) The yield strength by the 0.05% offset method is Y = 239 MPa Ans. (e) The yield strength by the 0.2% offset method is Y = 259 MPa Ans. (f) The fracture stress is fracture = 356 MPa Ans. (f) The original cross-sectional area of the specimen is D2 = (12.8 mm) 2 = 128.679635 mm 2 4 4 The area of the specimen at the fracture location is A0 = D2 = (11.3 mm) 2 = 100.287492 mm 2 4 4 The true fracture stress is therefore 128.679635 mm 2 true fracture = (356 MPa) = 457 MPa 100.287492 mm 2 Af = Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.20 A 1035 hot-rolled steel specimen with a diameter of 0.500 in. and a 2.0-in. gage length was tested to fracture. Load and deformation data obtained during the test are given. 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 (0.20% offset). (f) the fracture stress. (g) the true fracture stress if the final diameter of the specimen at the location of the fracture was 0.387 in. (lb) Change in Length (in.) 0 2,690 5,670 8,360 11,050 12,540 13,150 13,140 12,530 12,540 12,840 12,840 0 0.0009 0.0018 0.0028 0.0037 0.0042 0.0046 0.0060 0.0079 0.0098 0.0121 0.0139 Load (lb) Change in Length (in.) 12,540 12,540 14,930 17,020 18,220 18,820 19,110 19,110 18,520 17,620 16,730 16,130 15,900 0.0209 0.0255 0.0487 0.0835 0.1252 0.1809 0.2551 0.2968 0.3107 0.3246 0.3339 0.3385 fracture Load Solution The plot of the stress-strain data is shown below. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. (a) The modulus of elasticity is 63.849 ksi E= = = 30, 400 ksi 0.0021 in./in. Ans. (b) From the diagram, the proportional limit is taken as PL = 63.8 ksi Ans. (c) The ultimate strength is ult = 97.3 ksi Ans. (d) The yield strength using the 0.05% offset method is Y = 65.4 ksi Ans. (e) The yield strength using the 0.2% offset method is Y = 63.8 ksi Ans. (f) The fracture stress is fracture = 82.1 ksi Ans. (g) The original cross-sectional area of the specimen is D2 = (0.500 in.) 2 = 0.196350 in.2 4 4 The area of the specimen at the fracture location is A0 = D2 = (0.387 in.) 2 = 0.117628 in.2 4 4 The true fracture stress is therefore 0.196350 in.2 true fracture = (82.1 ksi) = 137.0 ksi 0.117628 in.2 Af = Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.21 A 2024-T4 aluminum test specimen with a diameter of 0.505 in. and a 2.0-in. gage length was tested to fracture. Load and deformation data obtained during the test are given. 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 (0.20% offset). (f) the fracture stress. (g) the true fracture stress if the final diameter of the specimen at the location of the fracture was 0.452 in. Load (lb) 0 1,300 2,390 3,470 4,560 5,640 6,720 7,380 8,240 8,890 9,330 9,980 10,200 10,630 Change in Length (in.) 0.0000 0.0014 0.0023 0.0032 0.0042 0.0051 0.0060 0.0070 0.0079 0.0088 0.0097 0.0107 0.0116 0.0125 Load (lb) 11,060 11,500 12,360 12,580 12,800 13,020 13,230 13,450 13,670 13,880 14,100 14,100 14,100 14,100 14,100 Change in Length (in.) 0.0139 0.0162 0.0278 0.0394 0.0603 0.0788 0.0974 0.1159 0.1391 0.1623 0.1994 0.2551 0.3200 0.3246 fracture Solution The plot of the stress-strain data is shown below. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. (a) The modulus of elasticity is 33.55 ksi E= = = 11,180 ksi 0.003 in./in. Ans. (b) From the diagram, the proportional limit is taken as PL = 33.6 ksi Ans. (c) The ultimate strength is ult = 70.4 ksi Ans. (d) The yield strength using the 0.05% offset method is Y = 44.4 ksi Ans. (e) The yield strength using the 0.2% offset method is Y = 54.5 ksi Ans. (f) The fracture stress is fracture = 70.4 ksi Ans. (g) The original cross-sectional area of the specimen is A0 = D2 = (0.505 in.) 2 = 0.200296 in.2 4 4 The area of the specimen at the fracture location is Af = D2 = (0.452 in.) 2 = 0.160460 in.2 4 4 The true fracture stress is therefore 0.200296 in.2 true fracture = (70.4 ksi) = 87.9 ksi 0.160460 in.2 Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.22 A 1045 hot-rolled steel tension test specimen has a diameter of 6.00 mm and a gage length of 25 mm. In a test to fracture, the stress and strain data below were obtained. 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 (0.20% offset). (f) the fracture stress. (g) the true fracture stress if the final diameter of the specimen at the location of the fracture was 4.65 mm. (kN) Change in Length (mm) 0.00 2.94 5.58 8.52 11.16 12.63 13.02 13.16 13.22 13.22 13.25 13.22 0.00 0.01 0.02 0.03 0.05 0.05 0.06 0.08 0.08 0.10 0.14 0.17 Load (kN) Change in Length (mm) 13.22 16.15 18.50 20.27 20.56 20.67 20.72 20.61 20.27 19.97 19.68 19.09 18.72 0.29 0.61 1.04 1.80 2.26 2.78 3.36 3.83 3.94 4.00 4.06 4.12 fracture Load Solution The plot of the stress-strain data is shown below. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. (a) The modulus of elasticity is 394.765 MPa E= = = 247,000 MPa 0.0016 mm/mm Ans. (b) From the diagram, the proportional limit is taken as PL = 400 MPa Ans. (c) The ultimate strength is ult = 732 MPa Ans. (d) The yield strength by the 0.05% offset method is Y = 465 MPa Ans. (e) The yield strength by the 0.2% offset method is Y = 465 MPa Ans. (f) The fracture stress is fracture = 675 MPa Ans. (f) The original cross-sectional area of the specimen is D2 = (6 mm) 2 = 28.274334 mm 2 4 4 The area of the specimen at the fracture location is A0 = D2 = (4.65 mm) 2 = 16.982272 mm 2 4 4 The true fracture stress is therefore 28.274334 mm 2 true fracture = (675 MPa) = 1,124 MPa 16.982272 mm 2 Af = Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.23 Rigid bar BCD in Fig. P3.23 is supported by a pin at C and by aluminum rod (1). A concentrated load P is applied to the lower end of aluminum rod (2), which is attached to the rigid bar at D. The crosssectional area of each rod is A = 0.20 in.2 and the elastic modulus of the aluminum material is E = 10,000 ksi. After the load P is applied at E, the strain in rod (1) is measured as 900 (tension). (a) Determine the magnitude of load P. (b) Determine the total deflection of point E relative to its initial position. Fig. P3.23 Solution (a) From the measured strain, the stress in rod (1) is 1 = E11 = (10, 000 ksi)(900 106 in./in.) = 9 ksi and thus, the force in rod (1) is F1 = 1 A1 = (9 ksi)(0.20 in.2 ) = 1.8 kips (T) Consider the equilibrium of the rigid bar, and write a moment equilibrium equation about C to determine the magnitude of load P: M C = (20 in.)(1.8 kips) (30 in.)P = 0 P = 1.2 kips Ans. (b) From the measured strain, the elongation of rod (1) is e1 = 1 L1 = (900 106 in./in.)(50 in.) = 0.0450 in. From similar triangles, the deflection of the rigid bar at D can be expressed in terms of the deflection at B: vB v =D 20 in. 30 in. 30 in. 30 in. v D = vB = (0.045 in.) = 0.0675 in. 20 in. 20 in. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. The elongation of rod (2) due to the 1.2-kip load must be determined. The stress in rod (2) is F 1.2 kips 2 = 2 = = 6 ksi A2 0.2 in.2 and consequently, the strain in rod (2) is 6 ksi 2 = 2 = = 0.000600 in./in. E2 10,000 ksi From the strain, the elongation in rod (2) can be computed: e2 = 2 L2 = (0.000600 in./in.)(100 in.) = 0.06 in. The deflection of joint E is the sum of the rigid bar deflection at D and the elongation in rod (2): vE = vD + e2 = 0.0675 in. + 0.06 in. = 0.1275 in. Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.24 The rigid bar AC in Fig. P3.24 is supported by two axial bars (1) and (2). Both axial bars are made of bronze [E = 100 GPa; = 18 106 mm/mm/C]. The cross-sectional area of bar (1) is A1 = 240 mm2 and the cross-sectional area of bar (2) is A2 = 360 mm2. After load P has been applied and the temperature of the entire assembly has increased by 20C, the total strain in bar (2) is measured as 800 (elongation). Determine: (a) the magnitude of load P. (b) the vertical displacement of pin A. Fig. P3.24 Solution (a) The total strain in bar (2) is caused partly by the axial force in the bar and partly by the increase in temperature. The strain caused by the 20C temperature increase is: T = T = (18 106 mm/mm/ C)(20C) = 0.000360 mm/mm The strain caused by the axial force in the bar is thus: 2, = 2 2,T = 0.000800 mm/mm 0.000360 mm/mm = 0.000440 mm/mm The stress in bar (2) is 2 = E2 2, = (100, 000 MPa)(0.000440 mm/mm) = 44 MPa and the force in bar (2) is F2 = 2 A2 = (44 N/mm 2 )(360 mm 2 ) = 15,840 N Next, consider a FBD of the rigid bar AC. Equilibrium equations for this FBD are: Fy = F1 + F2 P = 0 M A = (1, 400 mm)F2 (500 mm)P = 0 which can be solve simultaneously to give: 500 mm F2 = P = 0.357143P 1,400 mm and F1 = 0.642857 P The applied load P can be expressed in terms of F2 as 1 P= F2 = 2.8 F2 0.357143 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. and so the magnitude of load P is P = 2.8F2 = 2.8(15,840 N) = 44,352 N = 44.4 kN Ans. (b) The force in bar (1) is F1 = 0.642857 P = (0.642857)(44,352 N) = 28,512 N Thus, the stress in bar (1) is F 28,512 N 1 = 1 = = 118.800 MPa A1 240 mm 2 The normal strain due to the axial force in bar (1) is 118.800 MPa 1, = 1 = = 0.001188 mm/mm E1 100, 000 MPa The normal strain caused by the 20C temperature increase is: 1,T = T = (18 106 mm/mm/ C)(20C) = 0.000360 mm/mm Therefore, the total strain in bar (1) is 1 = 1, + 1,T = 0.001188 mm/mm + 0.000360 mm/mm = 0.001548 mm/mm and the elongation in bar (1) is e1 = 1, L1 = (0.001548 mm/mm)(1,300 mm) = 2.012400 mm Since rigid bar ABC is connected to bar (1) (with a perfect connection), joint A displaces downward by an amount equal to the elongation of bar (1); therefore, vA = e1 = 2.01 mm Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.25 The rigid bar in Fig. P3.25 is supported by axial bar (1) and by a pin connection at C. Axial bar (1) has a cross-sectional area of A1 = 275 mm2, an elastic modulus of E = 200 GPa, and a coefficient of thermal expansion of = 11.9 106 mm/mm/C. The pin at C has a diameter of 25 mm. After load P has been applied and the temperature of the entire assembly has been increased by 20C, the total strain in bar (1) is measured as 675 (elongation). Determine: (a) the magnitude of load P. (b) the shear stress in pin C. Fig. P3.25 Solution The total strain in bar (1) consists of thermal strain as well as normal strain caused by normal stress: = + T The normal strain due to the increase in temperature is: T = T = (11.9 106 mm/mm/C)(20C) = 0.000238 mm/mm Therefore, the normal stress in bar (1) causes a normal strain of: = T = 0.000675 mm/mm 0.000238 mm/mm = 0.000437 mm/mm From Hookes law, the normal stress in bar (1) can be calculated as: 1 = E = (200, 000 MPa)(0.000437 mm/mm) = 87.4 MPa and thus the axial force in bar (1) must be: F1 = 1 A1 = (87.4 N/mm 2 )(275 mm 2 ) = 24, 035 N Next, consider a free-body diagram of the rigid bar. Write a moment equilibrium equation about pin C: M C = (200 mm)F1 (380 mm)P = (200 mm)(24, 035 N) (380 mm)P = 0 P = 12, 650 N = 12.65 kN Ans. Now that P is known, the horizontal and vertical reactions at C can be calculated: Fx = Cx F1 = 0 Cx = F1 = 24, 035 N Fy = C y P = 0 C y = P = 12, 650 N The resultant force acting on pin C is: 2 C = Cx2 + C y = (24, 035 N) 2 + (12, 650 N) 2 = 27,160.702 N Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Since the pin at C is a double shear connection, the shear force acting on one shear plane is half of the resultant force: V = 13,580.351 N. The area of one shear plane of the 25-mm-diameter pin at C (in other words, the cross-sectional area of the pin) is: Apin = (25 mm)2 = 490.874 mm 2 4 and thus the shear stress in pin C is: V 13,580.351 N C = = = 27.666 N/mm 2 = 27.7 MPa 2 AV 490.874 mm Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.26 The rigid bar in Fig. P3.26 is supported by axial bar (1) and by a pin connection at C. Axial bar (1) has a crosssectional area of A1 = 275 mm2, an elastic modulus of E = 200 GPa, and a coefficient of thermal expansion of = 11.9 106 mm/mm/C. The pin at C has a diameter of 25 mm. After load P has been applied and the temperature of the entire assembly has been decreased by 30C, the total strain in bar (1) is measured as 675 (elongation). Determine: (a) the magnitude of load P. (b) the shear stress in pin C. Fig. P3.26 Solution The total strain in bar (1) consists of thermal strain as well as normal strain caused by normal stress: = + T The normal strain due to the decrease in temperature is: T = T = (11.9 106 mm/mm/ C)( 30C) = 0.000357 mm/mm Therefore, the normal stress in bar (1) causes a normal strain of: = T = 0.000675 mm/mm (0.000357 mm/mm) = 0.001032 mm/mm From Hookes law, the normal stress in bar (1) can be calculated as: 1 = E = (200, 000 MPa)(0.001032 mm/mm) = 206.4 MPa and thus the axial force in bar (1) must be: F1 = 1 A1 = (206.4 N/mm 2 )(275 mm 2 ) = 56, 760 N Next, consider a free-body diagram of the rigid bar. Write a moment equilibrium equation about pin C: M C = (200 mm)F1 (380 mm)P = (200 mm)(56, 760 N) (380 mm)P = 0 P = 29,874 N = 29.9 kN Ans. Now that P is known, the horizontal and vertical reactions at C can be calculated: Fx = Cx F1 = 0 Cx = F1 = 56, 760 N Fy = C y P = 0 C y = P = 29,874 N The resultant force acting on pin C is: 2 C = Cx2 + C y = (56, 760 N) 2 + (29,874 N) 2 = 64,142 N Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Since the pin at C is a double shear connection, the shear force acting on one shear plane is half of the resultant force: V = 32,071 N. The area of one shear plane of the 25-mm-diameter pin at C (in other words, the cross-sectional area of the pin) is: Apin = (25 mm)2 = 490.874 mm 2 4 and thus the shear stress in pin C is: V 32, 071 N C = = = 65.3345 N/mm 2 = 65.3 MPa 2 AV 490.874 mm Ans. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

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U. Memphis - CIVL - 3322
4.1 A stainless steel alloy bar 25 mm wide by16 mm thick is subjected to an axial load ofP = 145 kN. Using the stress-strain diagramgiven in Fig. P4.1, determine:(a) the factor of safety with respect to theyield strength defined by the 0.20% offsetm
U. Memphis - CIVL - 3322
5.1 A steel [E = 200 GPa] rod with a circular cross section is 6-m long. Determine the minimumdiameter D required if the rod must transmit a tensile force of 30 kN without exceeding an allowablestress of 180 MPa or stretching more than 5 mm.SolutionIf
U. Memphis - CIVL - 3322
5.14 Rigid bar ABCD is loaded and supported asshown in Fig. P5.14. Steel [E = 30,000 ksi] bars(1) and (2) are unstressed before the load P isapplied. Bar (1) has a cross-sectional area of0.625 in.2 and bar (2) has a cross-sectional areaof 1.25 in.2.
U. Memphis - CIVL - 3322
5.23 The 200 200 1,200-mm oak [E = 12 GPa] block (2)shown in Fig. P5.23 was reinforced by bolting two 6 200 1,200mm steel [E = 200 GPa] plates (1) to opposite sides of the block. Aconcentrated load of 360 kN is applied to a rigid cap. Determine:(a) th
U. Memphis - CIVL - 3322
5.44 A 20-mm-diameter by 3.5-m-long steel rod (1)is stress free after being attached to rigid supports,as shown in Fig. P5.44. At A, a 25-mm-diameterbolt is used to connect the rod to the support.Determine the normal stress in steel rod (1) and thesh
U. Memphis - CIVL - 3322
5.62 The machine part shown in Fig. P5.62is in. thick and is made of SAE 4340 heattreated steel (see Appendix D for properties).Determine the maximum safe load P if afactor of safety of 2 with respect to failure byyield is specified.Fig. P5.62Soluti
U. Memphis - CIVL - 3322
6.1 A solid circular steel shaft having an outside diameter of D = 1.25 in. is subjected to a pure torque ofT = 1,900 lb-in. Determine the maximum shear stress in the shaft.SolutionThe polar moment of inertia for the shaft isD4 =(1.25 in.) 4 = 0.2396
U. Memphis - CIVL - 3322
6.26 A torque of TA = 525 lb-ft is applied to gear Aof the gear train shown in Fig. P6.26. The bearingsshown allow the shafts to rotate freely.(a) Determine the torque TD required for equilibriumof the system.(b) Assume shafts (1) and (2) are solid 1
U. Memphis - CIVL - 3322
6.42 The driveshaft of an automobile is being designed to transmit 280 hp at 3,500 rpm. Determine theminimum diameter required for a solid steel shaft if the allowable shear stress in the shaft is not toexceed 4,000 psi.SolutionThe torque in the drive
U. Memphis - CIVL - 3322
6.67 A hollow circular cold-rolled bronze [G1 = 6,500 ksi] tube (1) with an outside diameter of 2.00 in.and an inside diameter of 1.25 in. is securely bonded to a solid 1.25-in.-diameter cold-rolled stainlesssteel [G2 = 12,500 ksi] core (2) as shown in
U. Memphis - CIVL - 3322
6.79 The torsional assembly of Fig. P6.79 consistsof a cold-rolled stainless steel tube connected to asolid cold-rolled brass segment at flange C. Theassembly is securely fastened to rigid supports at Aand D. Stainless steel tube (1) and (2) has anou
U. Memphis - CIVL - 3322
6.94 A fillet with a radius of 0.15 in. is used at the junction in a stepped shaft where the diameter isreduced from 4.00 in. to 3.00 in. Determine the maximum shear stress in the fillet when the shaft istransmitting a torque of 4,000 lb-ft.SolutionFi
U. Memphis - CIVL - 3322
6.101 A torque of magnitude T = 1.5kip-in. is applied to each of the barsshown in Fig. P6.101. If the allowableshear stress is specified as allow = 8 ksi,determine the minimum requireddimension b for each bar.Fig. P6.101Solution(a) Circular Sectio
U. Memphis - CIVL - 3322
7.1 For the cantilever beam and loading shown,(a) Derive equations for the shear force V and thebending moment M for any location in the beam.(Place the origin at point A.)(b) Plot the shear-force and bending-momentdiagrams for the beam using the der
U. Memphis - CIVL - 3322
7.7 For the simply supported beam subjected to theloading shown,(a) Derive equations for the shear force V and thebending moment M for any location in the beam.(Place the origin at point A.)(b) Plot the shear-force and bending-momentdiagrams for the
U. Memphis - CIVL - 3322
7.24 Use the graphical method to construct theshear-force and bending-moment diagrams for thebeams shown. Label all significant points on eachdiagram and identify the maximum moments (bothpositive and negative) along with their respectivelocations. C
U. Memphis - CIVL - 3322
7.31 Draw the shear-force and bending-momentdiagram for the beam shown in Fig. P7.31. Assumethe upward reaction provided by the ground to beuniformly distributed. Label all significant pointson each diagram. Determine the maximum value of(a) the inte
U. Memphis - CIVL - 3322
8.1 During fabrication of a laminated timber arch, one of the 10 in. wide by 1 in. thick Douglas fir [E =1,900 ksi] planks is bent to a radius of curvature of 12 ft. Determine the maximum bending stressdeveloped in the plank.SolutionFrom Eq. (8.3):E
U. Memphis - CIVL - 3322
8.19 A WT230 26 standard steel shape is used to support the loads shown on the beam in Fig. P8.19a.The dimensions from the top and bottom of the shape to the centroidal axis are shown on the sketch ofthe cross section (Fig. P8.19b). Consider the entire
U. Memphis - CIVL - 3322
8.31 A solid steel shaft supports loadsPA = 200 lb and PD = 300 lb as shownin Fig. P8.31. Assume L1 = 6 in.,L2 = 20 in., and L3 = 10 in. The bearingat B can be idealized as a roller supportand the bearing at C can be idealizedas a pin support. If th
U. Memphis - CIVL - 3322
8.42 A composite beam is fabricated by bolting two 3 in. wide 12 in. deep timber planks to the sidesof a 0.50 in. 12 in. steel plate (Fig. P8.42b). The moduli of elasticity of the timber and the steel are1,800 ksi and 30,000 ksi, respectively. The simpl
U. Memphis - CIVL - 3322
8.52 A steel pipe assembly supports aconcentrated load of 17 kN as shown in Fig.P8.52. The outside diameter of the pipe is 142mm and the wall thickness is 6.5 mm.Determine the normal stresses produced atpoints H and K.Fig. P8.52SolutionSection pro
U. Memphis - CIVL - 3322
8.65 A beam with a box cross section is subjected toa resultant moment magnitude of 2,100 N-m actingat the angle shown in Fig. P8.65. Determine:(a) the maximum tension and the maximumcompression bending stresses in the beam.(b) the orientation of the
U. Memphis - CIVL - 3322
9.1 For the following problems, a beam segment subjected to internal bending moments at sections Aand B is shown along with a sketch of the cross-sectional dimensions. For each problem:(a) Sketch a side view of the beam segment and plot the distribution
U. Memphis - CIVL - 3322
9.11 A 1.6-m long cantilever beam supports a concentrated load of 7.2 kN, as shown below. The beamis made of a rectangular timber having a width of 120 mm and a depth of 280 mm. Calculate themaximum horizontal shear stresses at points located 35 mm, 70
U. Memphis - CIVL - 3322
9.18 A 50-mm-diameter solid steel shaftsupports loads PA = 1.5 kN and PC = 3.0 kN,as shown in Fig. P9.18. Assume L1 = 150mm, L2 = 300 mm, and L3 = 225 mm. Thebearing at B can be idealized as a rollersupport and the bearing at D can be idealizedas a
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9.38 A wooden beam is fabricated from one 2 8 and two 2 4 piecesof dimension lumber to form the I-beam cross section shown in Fig.P9.38. The flanges of the beam are fastened to the web with nails thatcan safely transmit a force of 100 lb in direct shea
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10.1 For the loading shown, use the doubleintegration method to determine (a) theequation of the elastic curve for thecantilever beam, (b) the deflection at the freeend, and (c) the slope at the free end.Assume that EI is constant for each beam.Fig.
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10.21 For the beam and loading shown inFig. P10.21, integrate the load distribution todetermine (a) the equation of the elasticcurve for the beam, and (b) the maximumdeflection for the beam. Assume that EI isconstant for the beam.Fig. P10.21Solutio
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10.29a For the beams and loadings shownbelow, determine the beam deflection atpoint H. Assume that EI = 8 104 kN-m2 isconstant for each beam.Fig. P10.29aSolutionDetermine beam slope at A.[Appendix C, SS beam with concentrated moment.]Relevant equa
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10.47 The simply supported beam shown inFig. P10.47 consists of a W410 60structural steel wide-flange shape [E = 200GPa; I = 216 106 mm4]. For the loadingshown, determine:(a) the beam deflection at point B.(b) the beam deflection at point C.(c) the
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11.1 A beam is loaded and supported asshown in Fig. P11.1. Use the doubleintegration method to determine themagnitude of the moment M0 required tomake the slope at the left end of the beamzero.Fig. P11.1SolutionMoment equation:xM a a = M ( x) + w
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11.18a For the beams and loadings shownbelow, assume that EI = 3.0 104 kN-m2 isconstant for each beam.(a) For the beam in Fig. P11.18a, determinethe concentrated upward force P required tomake the total beam deflection at B equal tozero (i.e., vB =
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11.26 For the beam and loading shownbelow, derive an expression for the reactionsat supports A and B. Assume that EI isconstant for the beam.Fig. P11.26SolutionChoose the reaction force at A as the redundant; therefore, the released beam is a cantil
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11.35 The beam shown in Fig. P11.35consists of a W360 79 structural steelwide-flange shape [E = 200 GPa; I = 225 106 mm4]. For the loading shown,determine:(a) the reactions at A, B, and C.(b) the magnitude of the maximum bendingstress in the beam.
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11.47 A W530 92 structural steel wideflange shape [E = 200 GPa; I = 554 106mm4] is loaded and supported as shown inFig. P11.47. Determine:(a) the force and moment reactions atsupports A and C.(b) the maximum bending stress in thebeam.(c) the deflec
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12.1 The stresses shown in the figure act at a point in a stressedbody. Using the equilibrium equation approach, determine thenormal and shear stresses at this point on the inclined plane shown.Fig. P12.1SolutionFn = n dA (215 MPa) cos 25(dA cos 25)
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12.9 The stresses shown in the figure act at a point in a stressed body.Determine the normal and shear stresses at this point on the inclinedplane shown.Fig. P12.9SolutionThe given stress values are: x = 4, 200 psi, y = 1,800 psi, xy = 0 psi, = +50
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12.25 Consider a point in a structural member that is subjected to planestress. Normal and shear stresses acting on horizontal and vertical planesat the point are shown.(a) Determine the principal stresses and the maximum in-plane shear stressacting a
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12.47 Mohrs circle is shown for a point in aphysical object that is subjected to plane stress.(a) Determine the stresses x, y, and xy and showthem on a stress element.(b) Determine the principal stresses and themaximum in-plane shear stress acting at
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12.75 At a point in a stressed body, the known stresses are x = 40 MPa (T), y = 20 MPa (C), z = 20MPa (T), xy = +40 MPa, yz = 0, and zx = +30 MPa. Determine:(a) the normal and shear stresses on a plane whose outward normal is oriented at angles of 40, 7
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13.1 The thin rectangular plate shown in Fig. P13.1 isuniformly deformed such that x = +890 , y = 510 ,and xy = +680 rad.(a) Determine the normal strain AC along diagonal AC ofthe plate.(b) Determine the normal strain BD along diagonal BD ofthe plat
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In Problems 13.23 through 13.26, the principal strains are given for a point in a body subjected to planestrain. Construct Mohrs circle and use it to(a) determine the strains x, y, and xy. (Assume x > y)(b) determine the maximum in-plane shear strain a
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13.39 The strain rosette shown in the figure was used to obtain normalstrain data at a point on the free surface of a machine part.(a) Determine the strain components x, y, and xy at the point.(b) Determine the principal strains and the maximum in-plan
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13.49 A 10-mm-thick aluminum [E = 70 GPa; = 0.33]plate is subjected to biaxial stress with x = 120 MPa andy = 60 MPa. The plate dimensions are b = 100 mm and h= 50 mm (see Fig. P13.49).(a) Determine the change in length of edges AB and AD.(b) Determi
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14.1 Determine the normal stress in a ball, which has an outside diameterof 220 mm and a wall thickness of 3 mm, when the ball is inflated to agage pressure of 110 kPa.Fig. P14.1SolutionD = 220 mmt = 3 mmd = 220 mm 2(3 mm) = 214 mma =pd (0.110 MP
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15.1 A 3-in.-diameter solid shaft is subjected toboth a torque of T = 25 kip-in. and an axialtension load of P = 40 kips, as shown in Fig.P15.1.(a) Determine the principal stresses and themaximum shear stress at point H on thesurface of the shaft.(
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15.19 A tee-shaped flexural member (Fig. P15.19b) is subjected to an internal axial force of P = 1,000lb, an internal shear force of V = 600 lb, and an internal bending moment of M = 1,500 lb-ft, as shown inFig. P15.19a. Determine the principal stresses
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15.29 For the vertical flexural member shown,determine the principal stresses and the maximumshear stress acting at points H and K, as shown onFigs. P15-29a and P15-29b. Show these stresses onan appropriate sketch for each point.Fig. P15.29aFig. P15
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15.37 A short rectangular post supports a compressive load of P = 120 kN as shown in Fig. P15.37a. Atop view of the post showing the location where load P is applied to the top of the post is shown in Fig.P15.37b. Determine the vertical normal stresses
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15.43 A 2.5-in.-diameter solid aluminum post issubjected to a horizontal force of V = 3 kips, a verticalforce of P = 7 kips, and a concentrated torque of T = 11kip-in., acting in the directions shown in Fig. P15.43.Assume L = 3.5 in. Determine the nor
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15.53 A steel shaft with an outside diameter of1.25 in. is supported in flexible bearings at itsends. Two pulleys are keyed to the shaft, andthe pulleys carry belt tensions as shown in Fig.P15.53.(a) Determine the normal and shear stresses onthe top
FSU - ENC - 1101
How To Write A Paper (3rd draft) By Colton BatehamMe: Heyy. Carl: Yo dude Me: Whatcha up to bro? Carl: nm, just ate. U? Me: Dangg. Tryin to procrastinate on this paper I gotta write. Carl: oh, well, I already told some friends I was down to longboard man
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Crots Paper Process Memo Colton BatehamNothing directly inspired me to write about procrastinating in my crots paper. I started texting my friends, and I noticed I was putting off my work by texting. Anything was allowed to be in the essay, so, I put my
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Colton Bateham p.61-p.81 - = New slide - = big dash * = smaller (sub) dash-btw Rachael, make a title page from the chapter title and the list on page 62. -Google vs. the Library -Google: *Convenient *Only about 50% of results were considered "good" -the
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Research Paper Process Memo Colton BatehamThe research paper was the least enjoyable of the three essays for me. The topic idea came to me easily though. After thinking about what interests me, my paper topic became an obvious choice. I am obsessed with
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Colton Bateham Professor Burnett ENC 1105 27 September 2010 Florida State on the Upswing In College Football there are a handful of programs known as the elite. These teams dominate their opponents year after year and create a dynasty for their school. Fl
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Radical Remediation Process Memo Colton BatehamFor my radical remediation or, radical revision, I chose to create a post secret postcard. The postcard was modeled as if it were created by the Narrator in my 3rd paper, the short story paper. My Narrator i
FSU - AMH - 2097
Exam 3 (Final Exam) Study GuideItalians I. Stereotypes of Italians A. Organized Crime B. Food Pasta & Pizza C. Guido D. Family-Oriented II. Italy A. No such thing as "Italy" A.1. City-states are controlled by families. A.2. 1859-1919 Unification A.2.a. S
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Notes for Race and Ethnicity in the United States Test 1 Terms Race the assumption of differences based on real or imagined physical characteristics Racism assuming that someone is inferior or superior to you based on race Ethnicity similar cultural
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Notes for test 2 V. Forced Labor Movement Atlantic/ Triangle Slave Trade Involvement in the trade: 1500-1880 (this class focuses on the English involvement) Types of slaves traded: o 75% male o 75% are adults (18-26 years old) Passage to the New World (Fo