617_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

# 617_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online