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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
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.stress_{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?
Stress_{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.