148_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

# 148_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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142 CHAPTER 2 Axially Loaded Members ± under redundant R A A ² 1 ³ 2 so and now use statics to find R S and solve for allowable load P lower value of P controls P Aa ³ 1713 N P Sa ³ 1504 N P allow is controlled by steel wires (b) P allow if load P at x ³ a/2 where 1 ³ displ. at x ³ a Use 2 from (a) above so equating 1 & 2 , solve for R A same as in (a) ± stress in left steel wire exceeds that in right steel wire R SL ³ 3P 4 + W 2 ´ (P + W) E A A A E A A A + 2E S A S 2 R SL ³ 3P 4 + W 2 ´ R A 2 R A ³ ( P + W) E A A A E A A A + 2E S A S d 2 ³ R A a L 2E S A S + L E A A A b d 1 ³ P + W 2 a L E S A S
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Unformatted text preview: b d 1 ³ d 1L + d 1R 2 d 1R ³ a P 4 + W 2 ba L E S A S b d 1L ³ a 3P 4 + W 2 ba L E S A S b ; P Sa ³ s Sa a E A A A + 2E S A S E S b ´ W P Aa ³ s Aa a E A A A + 2E S A S E A b ´ W s Sa ³ ( P + W) E S E A A A + 2E S A S s Aa ³ ( P + W) E A E A A A + 2E S A S s Sa ³ R S A S R S ³ (P + W) E S A S E A A A + 2E S A S R S ³ P + W ´ (P + W) E A A A E A A A + 2E S A S 2 R S ³ P + W ´ R A 2 s Aa ³ R A A A R A ³ ( P + W) E A A A E A A A + 2E S A S R A ³ P + W 2 a L E S A S b L 2E S A S + L E A A A d 2 ³ R A a L 2E S A S + L E A A A b Sec_2.4.qxd 9/25/08 11:38 AM Page 142...
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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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