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Transposing a formula

  1. Dec 4, 2013 #1
    Need to find stress in a bar

    (α1L1ΔT1+x1)+(α2L2ΔT2+x2)=0

    (α1L1ΔT1+ε1L1)+(α2L2ΔT2+ε2L2)=0

    (α1L1ΔT1+σ1/E1*L)+(α2L2ΔT2+σ2/E2*L)=0

    (α1L1ΔT1+α2L2ΔT2)+σ1/E1*L1+σ2/E2*L2=0

    αΔT(L1+L2)+1/E(σ1L1+σ2L2)=0

    I can work out most of this formula but i don't understand how we got the highlighted line 1/E(σ1L1+σ2L2)=0
     
  2. jcsd
  3. Dec 4, 2013 #2

    haruspex

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    You don't mean the '=0' there, right?
    I presume it should be (σ1/E1)*L1+(σ2/E2)*L2 and (1/E)(σ1L1+σ2L2), which are the same if E=E1=E2. With no context, I cannot tell whether E1=E2.
     
  4. Dec 4, 2013 #3
    No=0 fine. It 1/E where did that come from.
     
  5. Dec 4, 2013 #4

    haruspex

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    You have (σ1/E1)*L1+(σ2/E2)*L2. If E1=E2=E then that reduces to (σ1/E)*L1+(σ2/E)*L2 = (1/E)(σ1)*L1+(1/E)(σ2)*L2 = (1/E)(σ1L1+σ2L2).
     
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