Answer:
[tex]\tau _{max}=\dfrac{4}{3}\tau _{avg}[/tex]
Explanation:
Lets take
d= diameter of rod
T=Applied torque on the rod
τ=Shear stress on the rod
We know that shear stress in the rod is varying with radius r of the rod.Shear stress is zero at center of the rod and maximum at the outer most section of the rod.
From Torque-Stress
[tex]\dfrac{T}{J}=\dfrac{\tau }{r}[/tex]
So we can say that shear stress is varying linearly with radius of rod.
We know that [tex]J=\dfrac{\pi d^4}{32}[/tex]
And for maximum shear stress r=R
So we can say that shear stress due to torque
[tex]\tau _{max}=\dfrac{16T}{\pi d^3}[/tex]
If we consider that only shear force acting on the rod then maximum shear stress is 4/3 times more than average shear stress.
[tex]\tau _{max}=\dfrac{4}{3}\tau _{avg}[/tex]
