Question

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

Answer 1

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,

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