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Stress, strain and deflections due to temperature change

  1. Nov 24, 2009 #1
    in the following problem
    Capture.JPG

    i know that ΔT=120

    now the maximum deflection is 0.5mm so i looked for the total deflection had there been no restrictions

    ΔL(aluminium)=300*(23e-6*120)= 0.8280
    ΔL(st steel)=250*(18e-6*120)= 0.5400

    this is clearly more than the maximum deflection of 0.5- there are 0.868 "extra" which cause the stress.

    now putting this all together is where i get stumped.


    F=ε*E*A
    and ΔL=ε*L

    F(aluminium)=ε(al)*(70e9)*(2000)
    F(steel)=ε(s)*(190e9)*(800)
    ε(al)*300+ε(s)*250=0.5

    now the force in the aluminium and in the steel must be equal so i have 3 equation system to solve, after solving i get

    F= 1.3201e+011
    ε(al)= 0.942928e-3
    ε(s)=0.868486e-3

    now simply using

    σ=ε*E or σ=F/A

    σ=ε*E
    =0.942928e-3*70e9
    =66004960

    but the correct answer is -114.6MPa

    i can see where this might be wrong, nowhere in my stress calculations do i take into account the amount that each material expands. but i have no idea how to fix it
     
  2. jcsd
  3. Nov 25, 2009 #2
    think i got it,

    F(aluminium)=ε(al)*(70e9)*(2000)
    F(steel)=ε(s)*(190e9)*(800)
    ε(al)*300+ε(s)*250=0.868
     
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