Question

2 slabs, slab A and slab B, are stacked on top of each other in a parallel configuration, each of negligible width and Length=3 in. If the temperature is increase 30°C, how does the bars deform? Note: The area of slab A is .6 the area of slab B. da = 13 x 10-600-1 EA = 20 x 106 psi AB = 19 x 10-6°C-1 ER = 25 x 106 psi

Answer 1

STI ko SRA SRB . 3 in k 1 3 in As slab B has higher x, it will try to deform mare than slab A, but as they are stacked together, the final total deformation will be in between. Thus a compressive force will be generated in B, pushing it back and a tensile force in a pulling it forward From the fig at top sit 81,1+ SpA = $1,2- SRB Thermal VERSAL deformation due to thermal deformalim KL = xp LaTT RL AREA B AB EA also, Br RLS Foto ] = (XB-4A) 2 AT AA= D.CAB 4 ABAn [ 1 - Mo] (06-09) LA7 06 ? pL Г A ACAT - 016X207 - (11-13) X10 X 3x30 Team 5 n ak be * - 6*3*30 x love 13500 x 10 s I Now, da St, 17 RL LALAT RL = (13* 3 430 + 13 500)xion - 2.2084 10-3 eins total deformation of stack is 2.2088103 in . . So .