149_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

149_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 143 SECTION 2.4 Statically Indeterminate Structures PEAAA + 6PESAS + 4WESAS 4EAAA + 8ESAS RSL 143 PEAAA + 6PESAS + 4WESAS 1 ab 4EAAA + 8ESAS AS s Sa solve for Pallow based on allowable stresses in steel & alum. PSa (4WESAS) sSa(4ASEAAA + 8ESAS2 ) EAAA + 6ESAS PSa 820 N ; PAa 1713 N same as in (a) steel controls (c) Pallow if wires are switched as shown & x a/2 select RA as the redundant statics on the two released structures (1) cut alum. wire - apply P & W, compute forces in left & right steel wires, then compute displacements at each steel wire P 2 RSL RSR L P a b 2 ESAS d1L by geometry, P +W 2 a d1R L P + Wb a b 2 ESAS at alum. wire location at far right is d1 a P L b + 2W b a 2 ESAS (2) next apply redundant RA at right wire, compute wire force & displ. at alum. wire RSL RA RSR (3) compatibility equate a d2 2RA 1, P L + 2Wb a b 2 ES AS 2 RA a 5L L + b ESAS EAAA and solve for RA then Pallow for alum. wire EAAAP + 4EAAAW 10EAAA + 2ESAS EAP + 4EAW 10EAAA + 2ESAS PAa 5L L + ESAS EAAA RA sAa RA sAa(10EAAA + 2ESAS) EA s Aa RA AA 4EAW PAa 1713 N (4) statics or superposition - find forces in steel wires then Pallow for steel wires RSL P + RA 2 RSL EAAAP + 4EAAAW P + 2 10EAAA + 2ESAS RSL 6EAAAP + PESAS + 4EAAAW 10EAAA + 2ESAS larger than RSR below so use in allow. stress calcs ...
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