# The power transmitted by a shaft 60 mm diameter at 180 RPM, if the permissible stress is 85 N/mm2

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The power transmitted by a shaft 60 mm diameter at 180 RPM, if the permissible stress is 85 N/mm2

1. 68 kW
2. 650 kW
3. 1200 kW
4. 7 kW

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Correct Answer - Option 1 : 68 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 the shaft is given by,

$\frac{{{τ _{max}}}}{R} = \frac{T}{{{I_P}}}$

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

Polar moment of inertia for solid circular shaft,

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

Calculation:

The diameter of the shaft, D = 60 mm

Number of revolutions per minute, N = 180 rpm

permissible shear stress = 85 N/mm2

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

$T = {τ _{max}} \times {Z_P}$

$= 85 \times \frac{\pi }{{16}} \times {D^3} = 85 \times \frac{\pi }{{16}} \times {60^3}$

$= 3604977.56\;Nmm \approx 3604.977\;Nm$

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

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

$= \frac{{2\pi \times 180 \times 3604.977}}{{60}}$

$= 67952.21\;W \approx 67.95\;kW$