625_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

625_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: 07Ch07.qxd 9/27/08 1:22 PM Page 619 619 SECTION 7.6 Triaxial Stress Problem 7.6-7 A rubber cylinder R of length L and cross-sectional F F area A is compressed inside a steel cylinder S by a force F that applies a uniformly distributed pressure to the rubber (see figure). (a) Derive a formula for the lateral pressure p between the rubber and the steel. (Disregard friction between the rubber and the steel, and assume that the steel cylinder is rigid when compared to the rubber.) (b) Derive a formula for the shortening d of the rubber cylinder. R S L S Solution 7.6-7 Rubber cylinder Solve for p: p v 1 F ab vA ; (b) SHORTENING Eq. (7-53 b): ây F A p sy sx sz p âz âx (sz + sx) E E ( 2p) Substitute for p and simplify: 0 F (1 + )( EA 1 1+2) (Positive ây represents an increase in strain, that is, elongation.) (a) LATERAL PRESSURE sx E d E p âyL (sy + sz) d or 0 E F EA ây Eq. (7-53 a): âx sy a F A pb (1 + )(1 2 ) FL a b (1 ) EA ; (Positive d represents a shortening of the rubber cylinder.) Problem 7.6-8 A block R of rubber is confined between plane F F parallel walls of a steel block S (see figure). A uniformly distributed pressure p0 is applied to the top of the rubber block by a force F. (a) Derive a formula for the lateral pressure p between the rubber and the steel. (Disregard friction between the rubber and the steel, and assume that the steel block is rigid when compared to the rubber.) (b) Derive a formula for the dilatation e of the rubber. (c) Derive a formula for the strain-energy density u of the rubber. S S R ...
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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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