Question

The power transmitted by a 75 mm diameter shaft at 140 r.p.m., subjected to a maximum shear stress of 60 N/mm², is nearly
1. 68 kW
2. 70 kW
3. 73 kW
4. 76 kW

Answer 1

Correct Answer - Option 3 : 73 kW

Concept:

Power transmitted by shaft subjected to torque ‘T’ is given by,

\(P = \frac{{2\pi NT}}{{60}}\;Watt\)

Where, N = Number of revolutions per minute, T = Applied torque N-m

Maximum shear stress induced in shaft is given by,

\(\frac{{{τ _{max}}}}{R} = \frac{T}{{{I_P}}}\)

Where, R = Radius of shaft (m), I_P = Polar moment of inertia (m⁴) and τ_max = Maximum shear stress N/m²

Polar moment of inertia for solid circular shaft,

\({I_p} = \frac{\pi }{{32}} \times {D^4}\)

Calculation:

Diameter of shaft, D = 75 mm

Number of revolutions per minute, N = 140 rpm

Maximum shear stress = 60 MPa

Maximum torque that can be applied on the shaft ‘T’ is given by,

\(T = {τ _{max}} \times {Z_P}\)

\( = 60 \times \frac{\pi }{{16}} \times {D^3} = 60 \times \frac{\pi }{{16}} \times {75^3}\)

\( = 4967578.13\;Nmm \approx 4967.578\;Nm\)

Power Transmitted by shaft when subjected to torque ‘T’ is given by,

\(P = \frac{{2\pi NT}}{{60}}\)

\( = \frac{{2\pi \times 140 \times 4967.578}}{{60}}\)

\( = 72791.57\;W \approx 72.791\;kW\)