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?
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
Homework EquationsThe Attempt at a Solution
i think that's wrong, it should be right
Tex= g/(10^2m) ; Mass per length= g/m = (10^2g)/(10^2m) = (10^2g) / (m) Tex = (1.5*10^2g) / m = 150 Tex ; Stress= 100/150= 2/3 N/Tex
Den= g/ (9*10^3m) ; Mass per length= g/m = (9*10^3g)/(9*10^3m) = (9*10^3g)/ (m) den = (1.5*9*10^3g)/ m =13500 Den = 1/135...
So,the stress=force/mass per length
Mass per length=(1.5*10^-2)/(1*10^-3)(1*10^-3)(100*10^-3)=150000 g/m
=150000/1000 g/tex=150 g/tex
=150000/9000 g/den=16.6667 g/ den
Stress=100/150
Stress=100/1.6667
I don't sure