Answer:
The correct answer is 231 Mpa i.e option a.
Explanation:
using the equation of torsion we Have
[tex]\frac{T}{I_{p}}=\frac{\tau }{r}\\\\\therefore \tau =\frac{T}{I_{p}}\times r[/tex]
where,
[tex]\tau[/tex]= shear stress at a distance 'r' from the center
T = is the applied torque
[tex]I_{p}[/tex] = polar moment of inertia of the section
r = radial distance from the center
Thus we can see that if a point is located at center i.e r = 0 there will be no shearing stresses at the center due to torque.
We know that in case of a circular section the maximum shearing stresses due to a shear force occurs at the center and equals
[tex]\tau _{max}=\frac{4}{3}\times \frac{V}{A}[/tex]
Applying values we get
[tex]\tau _{max}=\frac{4}{3}\times \frac{85\times 10^{3}}{0.25\times \pi \times (25\times 10^{-3})^{2}}\\\\\therefore \tau _{max}=230.88Mpa\approx 231Mpa[/tex]
