253_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

253_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: Sec_2.8-2.12.qxd 9/25/08 11:44 AM Page 247 SECTION 2.12 Elastoplastic Analysis BAR d (b) PLASTIC LOAD PP 1.5 in. FT 2 AB pd 4 PP 1.7671 in.2 YAT FT EAT c 21,827 lb L Let 1 shortening of tube FB a BP P ( 19,330 psi c 1 1 Y 0.010 in. Y(AT AB) ; L b EAB c TP P1 YAB shortening of bar BP Initially, the tube supports all of the load. Let P1 load required to close the clearance s1 FB 104,300 lb INITIAL SHORTENING OF TUBE T P1 AT FB sYL E 0.028621 in. 0.02862 in. TP 0.018621 in. ; (c) LOAD-DISPLACEMENT DIAGRAM OK) (a) YIELD LOAD PY Because the tube and bar are made of the same material, and because the strain in the tube is larger than the strain in the bar, the tube will yield first. FT TY s TY Y BY YAT 40,644 lb shortening of tube at the yield stress sYL E FTL EAT TY 0.018621 in. 0.018621 in. ; shortening of bar TY c PY P1 FB EAB d L BY 29,453 lb PY FT 40,644 lb FB 70,097 lb PY 70,100 lb 29,453 lb 3.21 dY d1 1.86 PP PY 0.008621 in. 1.49 dP dY 1.54 0 P P1: slope 2180 k/in. P1 ; P PY: slope 5600 k/in. PY P PP: slope 3420 k/in. 247 ...
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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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