626_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

626_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 9/27/08 620 1:22 PM Page 620 CHAPTER 7 Analysis of Stress and Strain Solution 7.6-8 Block of rubber (b) DILATATION 1 Eq. (7-56): e 2 (sx + sy + sz) E 1 2 E ( p p0) Substitute for p: p sy p0 âx sz 0 ây Z 0 âz Z 0 sx E OR 0 E ( p0) p 2 )p0 ; 0 (a) LATERAL PRESSURE Eq. (7-53 a): âx (1 + )(1 E e sx (c) STRAIN ENERGY DENSITY Eq. (7-57b): u (sy + sz) ‹p p0 ; 1 (s2 + s2 + s2) x y z 2E v (sx sy + sx sz + sy sz) E Substitute for sx, sy, sz, and p: u )p 2 0 2 (1 ; 2E Problem 7.6-9 A solid spherical ball of brass (E 15 * 106 psi 0.34) is lowered into the ocean to a depth of 10,000 ft. The diameter of the ball is 11.0 in. Determine the decrease ¢ d in diameter, the decrease ¢ V in volume, and the strain energy U of the ball. Solution 7.6-9 E 15 * 10 Brass sphere 6 psi DECREASE IN VOLUME 0.34 Lowered in the ocean to depth h Diameter d 11.0 in. 63.8 lb/ft Pressure: s0 gh 638,000 lb/ft2 4431 psi DECREASE IN DIAMETER Eq. (7-59): â0 ¢d s0 (1 E ¢V eV0 283.6 * 10 43 pr 3 V0 3 Sea water: g 3â0 Eq. (7-60): e 10,000 ft 0.198 in.3 (decrease) 6 4 11.0 in. 3 (p) a b 3 2 ; STRAIN ENERGY 2) 94.53 * 10 â0d 1.04 * 10 (decrease) 3 in. ; 6 Use Eq. (7-57 b) with sx u U 3(1 uV0 2 2E )s2 0 sy 0.6283 psi 438 in.-lb ; sz s0: 696.9 in.3 ...
View Full Document

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.

Ask a homework question - tutors are online