Answer:
Explanation:
Using equation of pure torsion
[tex]\frac{T}{I_{polar} }=\frac{t}{r}[/tex]
where
T is the applied Torque
[tex]I_{polar}[/tex] is polar moment of inertia of the shaft
t is the shear stress at a distance r from the center
r is distance from center
For a shaft with
[tex]D_{0} =[/tex] Outer Diameter
[tex]D_{i} =[/tex] Inner Diameter
[tex]I_{polar}=\frac{\pi (D_{o} ^{4}-D_{in} ^{4}) }{32}[/tex]
Applying values in the above equation we get
[tex]I_{polar} =\frac{\pi(0.042^{4}-(0.042-.008)^{4})}{32}\\
I_{polar}= 1.74[/tex] x [tex]10^{-7} m^{4}[/tex]
Thus from the equation of torsion we get
[tex]T=\frac{I_{polar} t}{r}[/tex]
Applying values we get
[tex]T=\frac{1.74X10^{-7}X100X10^{6} }{.021}[/tex]
T =829.97Nm
