Data: P = 200 HP, N = 180 rpm,  = 80 N/mm2 =1, L = 3 m C = 0.82 x 105 N/mm2

Find d

Case – I

Diameter of shaft based on Shear Strength Criteria:

$\mathrm{P}=\left(\frac{2 \pi N T}{4500}\right) \mathrm{HP}$

$200=\left(\frac{2 \pi \times 180 \times T}{4500}\right)$

$\mathrm{T}=795.775 \mathrm{kg}-\mathrm{m}$

$\mathrm{T}=795.775 \times 9.81 \mathrm{N}-\mathrm{m}$

$\mathrm{T}=7806.5499 \times 10^{3} \mathrm{N}-\mathrm{mm}$

By using the relation

$\mathrm{T}=\frac{\pi}{16} \times \tau \times d^{3}$

$7806.5499 \times 10^{3}=\frac{\pi}{16} \times 80 \times d^{3}$

d = 79.21mm

Case – II

Diameter of shaft based on Rigidity Criteria:

By using the relation:

$\frac{T}{J}=\frac{C \theta}{L}$

$\frac{7806.5499 \times 10^{3}}{\frac{\pi}{32} d^{4}}=\frac{0.82 \times 10^{5} \times\left(1^{0} \times \frac{\pi}{180}\right)}{3 \times 10^{3}}$

$\mathrm{d}=113.63 \mathrm{mm}$

Choose the diameter of solid circular shaft equal to 114mm to satisfy the given conditions.