ch7-4 - 466 CHAPTER 7 Analysis of Stress and Strain Problem...

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Problem 7.5-8 A brass cube 50 mm on each edge is compressed in two perpendicular directions by forces P 5 175 kN (see figure). Calculate the change D V in the volume of the cube and the strain energy U stored in the cube, assuming E 5 100 GPa and n 5 0.34. Solution 7.5-8 Biaxial stress-cube 466 CHAPTER 7 Analysis of Stress and Strain P = 175 kN P = 175 kN Side b 5 50 mm P 5 175 kN E 5 100 GPa n5 0.34 (Brass) s x 5 s y 52 P b 2 (175 kN) (50 mm) 2 70.0 MPa C HANGE IN VOLUME Eq. (7-47): V 0 5 b 3 5 (50 mm) 3 5 125 3 10 3 mm 3 D V 5 eV 0 56 mm 3 (Decrease in volume) S TRAIN ENERGY Eq. (7-50): 5 0.03234 MPa U 5 uV 0 5 (0.03234 MPa)(125 3 10 3 mm 3 ) 5 4.04 J u 5 1 2 E ( s x 2 1 s y 2 2 2 ns x s y ) e 5 1 2 2 n E ( s x 1 s y ) 448 3 10 2 6 P P Problem 7.5-9 A 4.0-inch cube of concrete ( E 5 3.0 3 10 6 psi, n 5 0.1) is compressed in biaxial stress by means of a framework that is loaded as shown in the figure. Assuming that each load F equals 20 k, determine the change D V in the volume of the cube and the strain energy U stored in the cube. Solution 7.5-9 Biaxial stress – concrete cube F F b 5 4 in. E 5 3.0 3 10 6 psi 0.1 F 5 20 kips Joint A : 5 28.28 kips s x 5 s y P b 2 1768 psi P 5 F Ï 2 C HANGE IN VOLUME Eq. (7-47): V 0 5 b 3 5 (4 in.) 3 5 64 in. 3 D V 5 eV 0 0.0603 in. 3 (Decrease in volume) S TRAIN ENERGY Eq. (7-50): 5 0.9377 psi U 5 uV 0 5 60.0 in.-lb u 5 1 2 E ( s x 2 1 s y 2 2 2 ns x s y ) e 5 1 2 2 n E ( s x 1 s y ) 0.0009429 F F A F F P A
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Problem 7.5-10 A square plate of width b and thickness t is loaded by normal forces P x and P y , and by shear forces V , as shown in the figure. These forces produce uniformly distributed stresses acting on the side faces of the plate. Calculate the change D V in the volume of the plate and the strain energy U stored in the plate if the dimensions are b 5 600 mm and t 5 40 mm, the plate is made of magnesium with E 5 45 GPa and n 5 0.35, and the forces are P x 5 480 kN, P y 5 180 kN, and V 5 120 kN. Solution 7.5-10 Square plate in plane stress SECTION 7.5 Hooke’s Law for Plane Stress 467 P y P y P x P x y t b b V V V V x O b 5 600 mm t 5 40 mm E 5 45 GPa n5 0.35 (magnesium) P x 5 480 kN P y 5 180 kN V 5 120 kN C HANGE IN VOLUME Eq. (7-47): V 0 5 b 2 t 5 14.4 3 10 6 mm 3 D V 5 eV 0 5 2640 mm 3 (Increase in volume) e 5 1 2 2 n E ( s x 1 s y ) 5 183.33 3 10 2 6 t xy 5 V bt 5 5.0 MPa s y 5 P y bt 5 7.5 MPa s x 5 P x bt 5 20.0 MPa S TRAIN ENERGY Eq. (7-50): Substitute numerical values: u 5 4653 Pa U 5 uV 0 5 67.0 N . m 5 67.0 J G 5 E 2(1 1 n ) 5 16.667 GPa u 5 1 2 E ( s x 2 1 s y 2 2 2 ns x s y ) 1 t 2 xy 2 G Problem 7.5-11 Solve the preceding problem for an aluminum plate with b 5 12 in., t 5 1.0 in., E 5 10,600 ksi, n 5 0.33, P x 5 90 k, P y 5 20 k, and V 5 15 k. Solution 7.5-11 Square plate in plane stress b 5 12.0 in. t 5 1.0 in. E 5 10,600 ksi 0.33 (aluminum) P x 5 90 k P y 5 20 k V 5 15 k C HANGE IN VOLUME Eq. (7-47): V 0 5 b 2 t 5 144 in. 3 D V 5 eV 0 5 0.0423 in. 3 (Increase in volume) e 5 1 2 2 n E ( s x 1 s y ) 5 294 3 10 2 6 t xy 5 V bt 5 1250 psi s y 5 P y bt 5 1667 psi s x 5 P x bt 5 7500 psi S TRAIN ENERGY Eq. (7-50): Substitute numerical values: u 5 2.591 psi U 5 uV 0 5 373 in.-lb G 5 E 1 n ) 5 3985 ksi u 5 1 2 E ( s x 2 1 s y 2 2 2 ns x s y ) 1 t xy 2 2 G Probs. 7.5-10 and 7.5-11
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Problem 7.5-12 A circle of diameter d 5 200 mm is etched on a brass plate (see figure). The plate has dimensions 400 3 400 3 20 mm.
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ch7-4 - 466 CHAPTER 7 Analysis of Stress and Strain Problem...

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