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# Answer to Question #80131 in Other Engineering for Chris

Question #80131
The axle of a heavy duty tractor 120 K W at 600 rpm it acts as a cantilever supporting a load of 280 N located 60 cm from the fixed end. The allowable normal stress Is 137 MPa and the allowable shear stress is 62 MPa. If the axial load is negligible, find the axle diameter. A:52 mm B:50 mm C:56 mm D:54 mm
1
2020-01-12T10:19:27-0500

Normal stress is given by

@$\sigma = \frac{Mr}{I}, @$

where @$M=Fx=280N\cdot60cm=168Nm -@$ the maximum bending moment,

@$r=\frac{d}{2} - @$ the radius of the axle,

@$I=\frac{\pi d^4}{64}-@$ the moment of inertia.

Substitute

@$\sigma = \frac{Md}{2\cdot\frac{\pi d^4}{64}}=\frac{32M}{\pi d^3},@$

@$d=\sqrt[3]{\frac{32M}{\pi\sigma}}=\sqrt[3]{\frac{32\cdot168}{137\cdot10^6\pi}}=0.0232m=23.2mm.@$

Shear stress is given by

@$\tau = \frac{Tr}{J}, @$

where @$T=\frac{P}{2\pi n}=\frac{120\cdot10^3}{2\pi\cdot\frac{600}{60}}=1909.86Nm-@$ torque,

@$J=\frac{\pi d^4}{32}-@$ polar moment of inertia.

Substitute

@$\tau = \frac{Td}{2\cdot\frac{\pi d^4}{32}}=\frac{16T}{\pi d^3},@$

@$d=\sqrt[3]{\frac{16T}{\pi\tau}}=\sqrt[3]{\frac{16\cdot1909.86}{62\cdot10^6\pi}}=0.0539m=53.9mm.@$

Thus, the axle diameter is 54 mm (Option D).

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