What interference fit is required in order to transmit 18 kN m torque between a 150-mmdiameter steel shaft and a 300-mm-outside-diameter cast iron hub that is 250 mm long? (Take Young’s modulus for the steel and cast iron to be 200 and 100GPa respectively,and assume the coefficient of friction is

Question

What interference fit is required in order to transmit 18 kN m torque between a 150-mmdiameter steel shaft and a 300-mm-outside-diameter cast iron hub that is 250 mm long? (Take Young’s modulus for the steel and cast iron to be 200 and 100 GPa respectively, and assume the coefficient of friction is 0.12 and that Poisson’s ratio is 0.3 for both materials).

Accepted Answer

The torque is given by [latex]T=2 f p \pi b^{2} L[/latex] and this equation can be rearranged to evaluate the pressure required to transmit the torque:

[latex]p=\frac{T}{2 f \pi b^{2} L}=\frac{18 imes 10^{3}}{2 imes 0.12 \pi(0.075)^{2} imes 0.25}=16.97 imes 10^{6} N / m ^{2} .[/latex]

The interference required to generate this pressure is given by

[latex]\delta=2 b p\left[\frac{1}{E_{o}}\left(\frac{c^{2}+b^{2}}{c^{2}-b^{2}}+\mu_{o} ight)+\frac{1}{E_{i}}\left(1-\mu_{i} ight) ight][/latex]

[latex]=2 imes 0.075 imes 16.97 imes 10^{6}\left[\frac{1}{100 imes 10^{9}}\left(\frac{0.15^{2}+0.075^{2}}{0.15^{2}-0.075^{2}}+0.3 ight)+\frac{1}{200 imes 10^{9}}(1-0.3) ight][/latex]

[latex]=5.89 imes 10^{-5} m[/latex]

So the required interference fit is 0.059 mm. This could be specified using the H7/s6 tolerance band with the hole diameter dimension as [latex]150.000 \begin{array}{c}+0.04 \+0\end{array}[/latex] mm and the shaft diameter as [latex]150.000 \begin{array}{l}+0.125 \+0.100\end{array}[/latex] mm.

Question 19.3: What interference fit is required in order to transmit 18 kN...

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