626_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

626_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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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 ...
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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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