Recent content by billy722

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

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?

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?

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

Thank you,you help me learn it,not just ans

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...

Then, 1.5g/m=1.5/1000 tex=1.5*10^-3 tex 1.5/9000 deg=1.6667*10^-4 den Stress=100/(1.5*10^-3)=66666.67 N/tex =100/(1.6667*10^-4)=599988 N/den?

(L/(Cm)^3)*(mm)^2= (L/(10^-2m)^3)*(10^-3m)^2= (L/M^3)*(M^2) =>1.5*1 g/m?

Ok, mass per length=1.5*(10^12) g/m

1.5 g/cm^3 =1.5*(10^-6) g/m^3 1 mm^2=1*(10^-6) m^2 Mass per length=1.5*10^(-12) g/m

L^2=cross section area (1*1 mm^2)?

M/L=(M/L^3)*L^2?

Density per unit length=density per unit volume(1.5g/cm^3)/total length(100mm)

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

1. I need it to find the mass which in tex unit? 2.i just now stress=force/cross section area 3.no,it will same?