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144_Mechanics Homework Mechanics of Materials Solution

144_Mechanics Homework Mechanics of Materials Solution -...

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Unformatted text preview: i—l m‘i—lfll—n L 2.11] Two tempered-steel bars, each % -in. thick, are bonded to a % -in. mild-steel bar. This composite bar is subjected as shown to a centric axial load of magnitude P. Both steels are elastoplastic with E = 29 x 10" psi and with yield strengths equal to 100 ksi and 50 ksi, respectively, for the tempered and mild steel. The load P is gradually increased from zero until the defamation of the bar reaches a maximum value 6,” = 0,014 in. and then decreased back to zero. Determine (a) the maximum value of P, (b) the maximum stress in the tempered-steel bars, (c) the permanent set after the load is removed. PROBLEM 2. 1 ll SOLUTION FD!" 'H\Q MVIJ S‘l’eej A, =‘ (Fixld ‘ LOO {“1 g ._ LGw _ (”Kts'oxroa) Y‘ ' E ‘ W For ”we ramper‘ea’ 5+9.er A2 = 20,34 )(2) "-' 0.75 {-1.1 _ L6}: __ (H)Cl00x.'03) _. . 3“. ._E.. a _zq¥_..__l03 _ 0.0Ll827é. m. To‘l‘af QV'veGL‘. A = And-A1 '= l-75rnz 5w * 3m < 5w. Thelma” sl'ee} yr‘e/Js. TempereJ She) is dash}. (a) Forces P. = A, 5}, = (1.00)(50xto3}= 5'0 X103 [5... 122; was» = camoacO-vsxO-ow : a-” w 1%. L 14 P: P.+ a = HQJIIWOS .919 1/2.} hips q :: 0. 02.9138 in. (b) Siresses 6] = EL = 6;. v 50m)” PS: ~= ED ks: 6; = I? = €229;qu —_~ 82.363105f9n‘ 82.86 ks: L t 2. ( Unflodinj 5’ = _'E>__L_ ‘ (“ZULU“) ) Hi :- 0.0309”! EA ’ (R‘ixlo‘ )(rns) l‘hi LC.) Permaneni' so‘l‘ 0.0‘1 __ 0.08099 - ='0,00‘?06 in. ...
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