A solid circular shaft is to transmit power, P kW when turning N revolutions per min. For a given maximum shear stress, the shaft diameter will be pro

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A solid circular shaft is to transmit power, P kW when turning N revolutions per min. For a given maximum shear stress, the shaft diameter will be proportional to
1. (P/N)0.31
2. (P/N)0.33
3. (P/N)0.67
4. (P/N)0.3
5. (P/N)0.5

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Correct Answer - Option 2 : (P/N)0.33

Concept:

For a solid circular shaft,

Power transmitted in kW, $P = \frac{{2\pi NT}}{{60}}$  where,

T = Torque

N = Revolutions per minute

From equation of pure torsion,

$\frac{T}{{{I_P}}} = \frac{τ }{r} = \frac{{Gθ }}{L}$ ⇒ $T = \frac{{\tau \times {I_P}}}{r}$

Calculation:

$P = \frac{{2\pi NT}}{{60}}$

${I_P} = \frac{π }{{32}}{D^4}$

$\begin{array}{l} P = \frac{{2\pi N}}{{60}} \times \frac{{τ \times {I_P}}}{r}\\ τ = \frac{{60}}{{2\pi }}\frac{{\frac{D}{2}}}{{\frac{{\pi {D^4}}}{{32}}}}\frac{P}{N} \end{array}$

${D^3} = \frac{1}{{48.63 \times \tau }} \times \frac{P}{N}$

Since, τ is constant (given) and neglecting other constant terms,

${D^3} \propto \frac{P}{N} \Rightarrow D \propto {\left( {\frac{P}{N}} \right)^{0.33}}$

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