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Select a diameter for a solid circular shaft to transmit 200 HP at 180 rpm. The allowable shear stress is $80 N/mm^{2}$ and the allowable angle is twist is $1^{o}$ in a length of 3m.$C=0.82x10^{5}$
written 4.7 years ago by gravatar for teamques10 teamques10 ★ 65k modified 2.0 years ago by gravatar for RakeshBhuse RakeshBhuse • 3.2k
strength of materials
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written 4.7 years ago by gravatar for teamques10 teamques10 ★ 65k

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:

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

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

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

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