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Find the minimum diameter of a solid shaft required to transmit 250 kW of power at 250 RPM if the maximum allowable shear stress is 95 MPa.?
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- Anonymous3 years agoFavorite Answer
torque
τ = P / ω = 250e3W / (250rev/min * 2π rad/rev * 1min/60s) = 9549 N·m
Now it's a little iffy -- I think shear stress
σ = τ*r / J
where J is the second moment of area, = πr⁴/2 for a circular section
then σ = τ*r / (πr⁴/2) = 2τ / πr³
and so
r³ = 2τ / πσ = 2*9549N·m / π*95e6N/m² = 6.4e-5 m³
r = 0.040 m = 4.0 cm
diameter d = 2*r = 8.0 cm
Source(s): https://en.wikipedia.org/wiki/List_of_second_momen... https://en.wikipedia.org/wiki/Torsion_(mechanics)#...
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