615_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

# 615_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Problem 7.5-3 Assume that the normal strains and for an element in plane stress (see figure) are measured with strain gages. (a) Obtain a formula for the normal strain in the direction in terms of , and Poisson’s ratio . (b) Obtain a formula for the dilatation in terms of , and Poisson’s ratio . ± P x , P y e ± P x , P y z P z P y P x y x z O s x t xy s y Solution 7.5-3 Plane stress Given: (a) N ORMAL STRAIN Eq. (7-34c): Eq. (7-36a): Eq. (7-36b): Substitute and into the first equation and simplify: â z ²³ ± 1 ³± ( â x + â y ) ; s y s x s y ² E (1 ³± 2 ) ( â y + ± â x ) s x ² E (1 ³± 2 ) ( â x + ± â y ) â z ²³ ± E ( s x + s y ) â z â x , â y , ± (b) D ILATATION Eq. (7-47): Substitute and from above and simplify: e ² 1 ³ 2 ± 1 ³± ( â x + â y ) ; s y s x e ² 1 ³ 2 ± E ( s x + s y ) Problem 7.5-4 A magnesium plate in biaxial stress is subjected to tensile stresses s x ² 24 MPa and s y ² 12 MPa (see figure). The corresponding strains in the plate are
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Unformatted text preview: x 440 10 6 and y 80 10 6 . Determine Poissons ratio and the modulus of elasticity E for the material. s y s x y x O Solution 7.5-4 Biaxial stress P OISSON S RATION AND MODULUS OF ELASTICITY Eq. (7-39a): Eq. (7-39b): y 1 E ( s y s x ) x 1 E ( s x s y ) x 440 * 10 6 y 80 * 10 6 s x 24 MPa s y 12 MPa Substitute numerical values: Solve simultaneously: 0.35 E 45 GPa ; E (80 * 10 6 ) 12 MPa (24 MPa) E (440 * 10 6 ) 24 MPa (12 MPa) SECTION 7.5 Hookes Law for Plane Stress 609 Probs. 7.5-4 through 7.5-7 07Ch07.qxd 9/27/08 1:20 PM Page 609...
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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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