147_Mechanics Homework Mechanics of Materials Solution

147_Mechanics Homework Mechanics of Materials Solution -...

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Unformatted text preview: PROBLEM 2 114 2.113 Bar AB has a cross-sectional area of 1200 mm2 and is made of a steel that is ' assumed to be elastoplastic with E = 200 GPa and a, = 250 MPa. Knowing that the force F increases from 0 to 520 kN and then decreases to zero, determine (a) the permanent deflection of point C, (b) the residual stress in the bar. 2.114 Solve Prob. 2.113, assuming that a = 180 mm. a = 120 mm ‘ 440 mm ' SOLUTION R = IQOO hum" = 12003110“ MIL Force 4-»: yi‘elu For‘HM AC3 fie. = A5} = (IQOOKIOJLCZS'tDflo‘j = 300 we to Fur equipn‘brwm F + 'PGB —- PM = O 9:: é‘l__._____—._.""*F :. F” Pa; 7 Pk -' F” : 300x103.— 520 #03 =' - 210 W03 N _ Peel-.58 2 (420xt09(omo-o.rso) EA (200 XIO")(J200 x10") 3 ._. -M : - 123.333 MO“ 194. I200 no“ = o. zssssswo“ m I l . __~_:_Pcal.es = (F’Pm)Lcn . R‘(%+L§c_z): FLCB EA EA " F Lee. (519 “031(Ofi'40 «9.130) L +L ’ WW"- = 307.27EXI03N kc- t-ll - EA FAQ—F = 307.273xio'5—s-2w 103 = —2:2.727>(103 N I = (307.273xlo’)(o.:§9_l__ (ZOOXJO' )(1200 #10") Pa: _ 307.273 K10 3 _ T ‘ 1200 Kurt? Lo; = - 2:2.727Kto‘ A taooxm“ = $6 — S" = 0.132333 #63— 0.230%910'3 0.00783 x104 m = 0.00788 Mm 4 = 0.230455v/o'3 m 256-0“)!10‘ Pa = — 117.273xto‘? 250 >402 256.061xlo‘ = — 6.06 xzo‘Pa 1' -6.06 MP9, 4 messes xlo‘ + :77.st XID‘ = - Log xto‘ Pa; ...
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