Question about stress and strain of a fiber

AI Thread Summary
The discussion revolves around calculating stress and strain in a fiber, focusing on the relationship between diameter change and applied stress. The original poster struggles to find the necessary tension (N) and questions the validity of their calculations, particularly regarding the change in diameter. Respondents clarify that Young's modulus is not the same as applied stress and emphasize that without a specified load, the problem cannot be solved. There is confusion about the calculations involving shear modulus and the meaning of certain variables, leading to repeated assertions that an applied load is essential for a proper solution. The conversation highlights the importance of clear problem statements in mechanics-related queries.
billy722
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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

 
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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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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?
 
billy722 said:
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.
 
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?
 
However, I calculate out 3.925*10^-26 again
I think that's wrong
 
billy722 said:
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?
haruspex said:
If the problem statement does not specify an applied load then there is no way to answer it.
haruspex said:
If the problem statement does not specify an applied load then there is no way to answer it.
haruspex said:
If the problem statement does not specify an applied load then there is no way to answer it.
haruspex said:
If the problem statement does not specify an applied load then there is no way to answer it.
haruspex said:
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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