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# mat33hw4 - Mat. 33 Homework # 4 2/16/09 1) 2) delta l =...

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Mat. 33 Homework # 4 2/16/09 1) delta l = (Fl 0 )/(AE) = (65250*8*10 -3 )/(4.540*10 -6 *207*10 9 ) = .00555m = 5.55mm 2) aluminum alloy: delta l = (Fl 0 )/(AE) = (29000*.5)/(6.334*10 -5 *70*10 9 ) = .00327m = 3.27mm brass alloy: delta l = (Fl 0 )/(AE) = (29000*.5)/(6.334*10 -5 *100*10 9 ) = .00229m = 2.229mm copper: delta l = (Fl 0 )/(AE) = (29000*.5)/(6.334*10 -5 *110*10 9 ) = .00208m = 2.08mm steel alloy: delta l = (Fl 0 )/(AE) = (29000*.5)/(6.334*10 -5 *207*10 9 ) = .00111m = 1.11mm According to my calculations the only possible candidate is the steel alloy. 3) (delta l*A*E)/(l 0 ) = F = (.00225*1.134*10 -5 *207*10 9 )/(.375) = 14084.28 N stress = F/A = 14084.28/(1.134*10 -5 ) = 1242000000 Pa = 1242 MPa according to the value for stress the value of strain according to the chart in the book is approximately = .0075 4) In edge dislocation an extra half plane of atoms exists which defines the dislocation line. Plastic dislocation is the result of many edge dislocations, which are the result of shear stress. Sheared stress is applied perpendicular to the

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## mat33hw4 - Mat. 33 Homework # 4 2/16/09 1) 2) delta l =...

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