# Question about stress and strain of a fiber

1. Feb 18, 2017

### billy722

1. The problem statement, all variables and given/known data

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. Feb 19, 2017

### haruspex

I don't see anything in the question about a tension being applied, so why should it change diameter?

3. Feb 19, 2017

### billy722

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. Feb 19, 2017

### haruspex

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. Feb 19, 2017

### billy722

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. Feb 19, 2017

### billy722

However, I calculate out 3.925*10^-26 again
I think that's wrong

7. Feb 19, 2017

### haruspex

Am I getting through?