153_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

153_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: Sec_2.4.qxd 9/25/08 11:38 AM Page 147 147 SECTION 2.4 Statically Indeterminate Structures Solution 2.4-15 Bar supported by two wires EQUATION OF COMPATIBILITY dc dD c d FORCE-DISPLACEMENT RELATIONS dC TCh EA (Eq. 2) TD(2h) EA dD (Eqs. 3, 4) SOLUTION OF EQUATIONS Substitute (3) and (4) into Eq. (2): TD(2h) TCh TC 2TD or cEA dEA c d TENSILE FORCES IN THE WIRES h 18 in. 2h 36 in. c 20 in. TC d 50 in. L 66 in. E 30 TENSILE STRESSES IN THE WIRES TC 2cPL sC A A(2c2 + d2) A 0.0272 in.2 P 340 lb (Eq. 5) Solve simultaneously Eqs. (1) and (5): 106 psi 2cPL 2 2c + d TD A sD FREE-BODY DIAGRAM 2 dPL TD (Eqs. 6, 7) 2 2c + d2 (Eq. 8) dPL (Eq. 9) A(2c2 + d2) DISPLACEMENT AT END OF BAR dB L dD a b d 2hPL2 2hTD L ab EA d EA(2c2 + d 2) (Eq. 10) SUBSTITUTE NUMERICAL VALUES 2c2 d2 (a) sC 2(20 in.)2 2 A(2c + d ) sD (50 in.)(340 lb)(66 in.) dPL 2 2 A(2c + d ) 12,500 psi (b) dB (0.0272 in.2)(3300 in.2) ; 10,000 psi DISPLACEMENT DIAGRAM 3300 in.2 2(20 in.)(340 lb)(66 in.) 2cPL 2 (50 in.)2 (0.0272 in.2)(3300 in.2) ; 2hPL2 EA(2c2 + d2) 2(18 in.)(340 lb)(66 in.)2 (30 * 106 psi)(0.0272 in.2)(3300 in.2) EQUATION OF EQUILIBRIUM MA 0 TC (c) TD(d) PL (Eq. 1) 0.0198 in. ; ...
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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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