617_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

617_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: 07Ch07.qxd 9/27/08 1:20 PM Page 611 611 SECTION 7.5 Hooke’s Law for Plane Stress Problem 7.5-7 Solve the preceding problem for an aluminum plate with sx (compression), dimensions 20 30 0.5 in., E 106 psi, and 10.5 12,000 psi (tension), sy 0.33. 3,000 psi Solution 7.5-7 Biaxial stress sx E 12,000 psi sy 3,000 psi 6 10.5 * 10 psi (b) CHANGE IN THICKNESS 0.33 Eq. (7-39c): âz Dimensions of Plate: 20 in. * 30 in. * 0.5 in. Shear Modulus (Eq. 7-38): G E 2(1 + ) ¢t 3.9474 * 106 psi 6 âz t 6 141 * 10 ; in. (Decrease in thickness) Principal stresses: s1 (c) CHANGE IN VOLUME 12,000 psi s2 From Eq. (7-47): ¢ V 3,000 psi s2 s1 2 Eq. (7-35): gmax (sx + sy) 282.9 * 10 (a) MAXIMUM IN-PLANE SHEAR STRAIN Eq. (7-26): tmax E tmax G V0 7,500 psi 1,900 * 10 6 ; 1 ‹ ¢V 2 E 1 2 E b (sx + sy) 300 in.3 (20)(30)(0.5) Also, a V0 a b (sx + sy) 291.4 * 10 (300 in.3)(291.4 * 10 3 0.0874 in. 6 6 ) ; (Increase in volume) Problem 7.5-8 A brass cube 50 mm on each edge is compressed P = 175 kN in two perpendicular directions by forces P 175 kN (see figure). Calculate the change V in the volume of the cube and the strain energy U stored in the cube, assuming E 100 GPa and 0.34. P = 175 kN Solution 7.5-8 Biaxial stress-cube sx P sy (175 kN) 2 (50 mm)2 b 70.0 MPa CHANGE IN VOLUME Eq. (7-47): e 1 V0 E 50 mm P 100 GPa 175 kN 0.34 ( Brass) b3 ¢V Side b 2 E eV0 (Decrease in volume) (sx + sy) (50 mm)3 56 mm3 448 * 10 125 * 103mm3 ; 6 ...
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This note was uploaded on 12/22/2011 for the course MEEG 310 taught by Professor Staff during the Fall '11 term at University of Delaware.

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