billy722
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Homework Statement
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
εshear=εaxial×V
εaxial=σaxial / E
σaxial=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.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= Stressshear/εshear
εshear=78.5p/4G=19.625z
ΔD=εshear×D0
=19.625z×0.02m
=0.3925y m
Is it right?
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.
Am I getting through?haruspex said:If the problem statement does not specify an applied load then there is no way to answer it.