Question 2.12: A solid shaft is subjected to a maximum torque of 25 kN m. F...

Let the shaft diameter be d mm. Therefore, maximum shear stress is

\tau_{\max }=\frac{16 T}{\pi d^3}

Substituting the given values, we get

80=\frac{16 \times 25 \times 10^6}{\pi d^3}

Therefore d = 116.8 mm. The angle of twist is

\theta=\frac{T L}{G J}

Substituting the given values, we get

\frac{\pi}{80}=\frac{25 \times 10^6 \times 20 \times d}{80 \times 10^3 \times\left(\pi d^4 / 32\right)}

Therefore, d = 153.9 mm. The required shaft diameter which will satisfy both the requirements is the maximum of the two values, that is 153.9 mm.
Therefore, the required shaft diameter is d = 153.9 mm ≈ 154 mm.