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Question about stress and strain of a fiber

  1. Feb 18, 2017 #1
    1. The problem statement, all variables and given/known data
    2vbmvm1.jpg
    i can not find the N.
    what i can only found is:
    D0=0.02mm;V=0.25;E=10G Pa;A=π(0.02mm/2)2
    ΔD=εshear×D0
    εshearaxial×V
    εaxialaxial / E
    σaxial=N/A

    2. Relevant equations


    3. The attempt at a solution
     
  2. jcsd
  3. Feb 19, 2017 #2

    haruspex

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    I don't see anything in the question about a tension being applied, so why should it change diameter?
     
  4. Feb 19, 2017 #3
    stressShear= E×I
    R=0.01mm,E=10GPa, I0= (π/4)×(0.01×10(-3))4=7.85 zm4
    Stressshear=(10×109×7.85×10-21)
    =78.5p Pa

    Shear modulus=E/[2(1+V)] = 10G/[2(1+0.25)]=4GPa
    Shear modulus= Stressshearshear
    εshear=78.5p/4G=19.625z

    ΔD=εshear×D0
    =19.625z×0.02m
    =0.3925y m
    Is it right?
     
  5. Feb 19, 2017 #4

    haruspex

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    You seem to be using the given Young's modulus as though it is the applied stress. They happen to have the same dimension, pressure, but they are not the same thing. Specifically, the modulus is that stress which would, in theory, double the wire's length.
    If the problem statement does not specify an applied load then there is no way to answer it.

    Even then, I am unable to follow the rest of your calculation. Not sure what all those ys and zs mean. It makes no sense that the change in diameter can be 20 times the initial diameter.
     
  6. Feb 19, 2017 #5
    Stressshear=E×I=F/A
    V=strainshear/strainaxial
    ⇒strainshear=Vχstrainaxial=ΔD/D0
    ⇒ΔD=D0χVχStrainaxial
    =D0χVχ(stressaxial/E)
    =D0χVχ(1/E)χ(FχA)
    F in shear and F in axial are same?
     
  7. Feb 19, 2017 #6
    However, I calculate out 3.925*10^-26 again
    I think that's wrong
     
  8. Feb 19, 2017 #7

    haruspex

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    Am I getting through?
     
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