Given: For solid shaft, d = 110 mm , R = 55 mm, T = 12.5 kN-m = $12.5 \times 10^{6} N-mm$ , L = 2.5m = 2500 mm, G = 80 GPa = $80 \times 10^{3} N/mm^{2}$

Solution:

$Polar M.I = I_{p} = \frac{\pi}{32} d^{4} = \frac{\pi}{32} \times 110^{4} = 14.37 \times 10^{6} mm^{4}$

1) Using the relation:

$\frac{I}{I_{p}} = \frac{q_{max}}{R}$

$\therefore q_{max} = \frac{I}{i_{p}} \times R = \frac{12.5 \times 10^{6} \times 55}{14.37 \times 10^{6}}$

$q_{max} = 47.84 N/mm^{2}$

2) Using the relation,

$\theta = \frac{T}{I_{p}} \times \frac{L}{G} = \frac{12.5 \times 10^{6} \times 2500}{14.37 \times 10^{6} 80 \times 10^{3}}$

$\theta = 0.02718$ radians