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Find the power that can be transmitted by a shaft of 40 mm dia. rotating at 200 RPM, if maximum shear stress is not to exceed 85 MPa.
somd-2 • 132  views
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Given:

For solid circular shaft d = 40mm, N = 200 rpm , $q_{max} = 85 N/mm^{2}$

Solution:

For solid shaft, $I_{p}$ =$ \frac{\pi}{32} \times d^{4}$ = $\frac{\pi}{32} \times 40^{4}$ = $2.51 \times 10^{5} mm^{4}$

R = d/2 = 40/2 = 20 mm

Using the relation, $\frac{T}{I_{p}}$ = $\frac{q_{max}}{R}$

$\therefore T = \frac{q_{max} \times I_{p}}{R} = \frac{85 \times 2.51 \times 10^{5}}{20} = 1.07 \times 10^{6} N-mm$

$T = 1.07 \times 10^{3} N-mm$

Assuming $T_{max} = T_{avg} = 1.07 \times 10^{3} N-mm$

Power = $\frac{2 \pi T_{avg}}{60}$ = $\frac{2 \pi \times 200 \times 1.07 \times 10^{3}}{60}$

$= 22410 watts$

$= 22.41 kW$

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