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

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).

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The torque is given by T=2fpπb2LT=2 f p \pi b^{2} L and this equation can be rearranged to evaluate the pressure required to transmit the torque:

p=T2fπb2L=18×1032×0.12π(0.075)2×0.25=16.97×106N/m2.p=\frac{T}{2 f \pi b^{2} L}=\frac{18 \times 10^{3}}{2 \times 0.12 \pi(0.075)^{2} \times 0.25}=16.97 \times 10^{6} N / m ^{2} .

The interference required to generate this pressure is given by

δ=2bp[1Eo(c2+b2c2−b2+μo)+1Ei(1−μi)]\delta=2 b p\left[\frac{1}{E_{o}}\left(\frac{c^{2}+b^{2}}{c^{2}-b^{2}}+\mu_{o}\right)+\frac{1}{E_{i}}\left(1-\mu_{i}\right)\right]

 

=2×0.075×16.97×106[1100×109(0.152+0.07520.152−0.0752+0.3)+1200×109(1−0.3)]=2 \times 0.075 \times 16.97 \times 10^{6}\left[\frac{1}{100 \times 10^{9}}\left(\frac{0.15^{2}+0.075^{2}}{0.15^{2}-0.075^{2}}+0.3\right)+\frac{1}{200 \times 10^{9}}(1-0.3)\right]

 

=5.89×10−5m=5.89 \times 10^{-5} m

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 150.000+0.04+0150.000 \begin{array}{c}+0.04 \\+0\end{array} mm and the shaft diameter as 150.000+0.125+0.100150.000 \begin{array}{l}+0.125 \\+0.100\end{array} mm.

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