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December 9th, 2016, 07:02 AM   #1
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Shear stress, angle of twist

A solid shaft of 100mm diameter, transmits 75kW at 150rpm, determine the value of the maximum shear stress set up in the shaft and the angle of twist in degrees per metre of the shaft length if G = 80GN/m^2
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December 9th, 2016, 12:41 PM   #2
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Have you had a go yourself? Do you know what shear stress is? And what the angle of twist represents?
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December 13th, 2016, 07:58 PM   #3
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Knowns..

$G = 80GN/m^2$

$\omega = 150rpm = \dfrac{150 * 2\pi}{60} = 5\pi \ rad/s$

$P = 75kW$

$d = 100mm = 0.1m$

$c = \dfrac{d}{2} = 50mm = 0.05m$

Relevant equations..

Polar moment of inertia ($m^4$): $J = \dfrac{1}{2} \pi c^4$

Angle of twist (in radians): $\phi = \sum\limits_i\dfrac{T_i L_i}{J_i G} = \dfrac{T L}{J G} $ (since shaft is continuous).

Power, torque, speed relation: $P = T \omega \implies T = \dfrac{P}{\omega}$

Max shear stress (Pascals): $\tau_{max} = \dfrac{Tc}{J}$

Now basically plug and play. To find angle of twist we need to know the torque and inertia,

$T = \dfrac{P}{\omega} = \dfrac{75000}{5\pi} \approx 4.77kNm$

$J = \dfrac{1}{2} \pi (0.05)^4 = 9.82 * 10^{-6} m^4$

$\tau_{max} = \dfrac{Tc}{J} = \dfrac{4770 * 0.05}{9.82 * 10^{-6}} \approx 24.3MPa$

$\phi$ (degrees per length) $= \dfrac{T}{JG}L = \dfrac{4770}{9.82 * 10^{-6} * 80 * 10^9}L = 0.0061 * L = 0.348L$
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December 14th, 2016, 09:02 AM   #4
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Thank you, i got the same give or take a couple of rounding errors, it was the angle that was troubling me, just wondered where that final L value Comes from to give 0.347?
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December 14th, 2016, 04:28 PM   #5
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$0.0061 rad \approx 0.347^{\circ}$. L is an arbitrary length in meters. The angle of twist is related to the length of the shaft multiplied by 0.347.
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