- #1

- 19

- 0

## Homework Statement

i can not find the N.

what i can only found is:

D

_{0}=0.02mm;V=0.25;E=10G Pa;A=π(0.02mm/2)

^{2}

ΔD=ε

_{shear}×D

_{0}

ε

_{shear}=ε

_{axial}×V

ε

_{axial}=σ

_{axial}/ E

σ

_{axial}=N/A

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- Thread starter billy722
- Start date

- #1

- 19

- 0

i can not find the N.

what i can only found is:

D

ΔD=ε

ε

ε

σ

- #2

- 36,204

- 6,818

- #3

- 19

- 0

R=0.01mm,E=10GPa, I

Stress

=78.5p Pa

Shear modulus=E/[2(1+V)] = 10G/[2(1+0.25)]=4GPa

Shear modulus= Stress

ε

ΔD=ε

=19.625z×0.02m

=0.3925y m

Is it right?

- #4

- 36,204

- 6,818

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._{Shear}= E×I

R=0.01mm,E=10GPa, I_{0}= (π/4)×(0.01×10^{(-3)})^{4}=7.85 zm^{4}

Stress_{shear}=(10×10^{9}×7.85×10^{-21})

=78.5p Pa

Shear modulus=E/[2(1+V)] = 10G/[2(1+0.25)]=4GPa

Shear modulus= Stress_{shear}/ε_{shear}

ε_{shear}=78.5p/4G=19.625z

ΔD=ε_{shear}×D_{0}

=19.625z×0.02m

=0.3925y m

Is it right?

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

- 19

- 0

V=strain

⇒strain

⇒ΔD=D

=D

=D

F in shear and F in axial are same?

- #6

- 19

- 0

However, I calculate out 3.925*10^-26 again

I think that's wrong

I think that's wrong

- #7

- 36,204

- 6,818

_{shear}=E×I=F/A

V=strain_{shear}/strain_{axial}

⇒strain_{shear}=Vχstrain_{axial}=ΔD/D_{0}

⇒ΔD=D_{0}χVχStrain_{axial}

=D_{0}χVχ(stress_{axial}/E)

=D_{0}χVχ(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.

Am I getting through?If the problem statement does not specify an applied load then there is no way to answer it.

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