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
Expert's answer
2020-01-12T10:19:27-0500

Normal stress is given by


σ=MrI,\sigma = \frac{Mr}{I},

where M=Fx=280N60cm=168NmM=Fx=280N\cdot60cm=168Nm - the maximum bending moment,

r=d2r=\frac{d}{2} - the radius of the axle,

I=πd464I=\frac{\pi d^4}{64}- the moment of inertia.

Substitute


σ=Md2πd464=32Mπd3,\sigma = \frac{Md}{2\cdot\frac{\pi d^4}{64}}=\frac{32M}{\pi d^3},

d=32Mπσ3=32168137106π3=0.0232m=23.2mm.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


τ=TrJ,\tau = \frac{Tr}{J},

where T=P2πn=1201032π60060=1909.86NmT=\frac{P}{2\pi n}=\frac{120\cdot10^3}{2\pi\cdot\frac{600}{60}}=1909.86Nm- torque,

J=πd432J=\frac{\pi d^4}{32}- polar moment of inertia.

Substitute


τ=Td2πd432=16Tπd3,\tau = \frac{Td}{2\cdot\frac{\pi d^4}{32}}=\frac{16T}{\pi d^3},

d=16Tπτ3=161909.8662106π3=0.0539m=53.9mm.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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