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Question about stress and strain of a fiber

  • Thread starter billy722
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  • #1
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Homework Statement


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

Homework Equations




The Attempt at a Solution

 

Answers and Replies

  • #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?
 
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  • #3
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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?
 
  • #4
haruspex
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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?
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.
 
  • #5
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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?
 
  • #6
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However, I calculate out 3.925*10^-26 again
I think that's wrong
 
  • #7
haruspex
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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?
If the problem statement does not specify an applied load then there is no way to answer it.
If the problem statement does not specify an applied load then there is no way to answer it.
If the problem statement does not specify an applied load then there is no way to answer it.
If the problem statement does not specify an applied load then there is no way to answer it.
If the problem statement does not specify an applied load then there is no way to answer it.
Am I getting through?
 

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