Two equal-length steel cylindrical parts are press-fitted as shown. E = 200 GPa, ν = 0.33, radial interference δ = 0.05 mm, interface radius R = 25 mm and outer radius R₀ = 40 mm. Find contact pressure in MPa.
For the solid inner shaft, radial contraction at R is $pR(1-\nu)/E$. The outer hub expands by $pR[(R_0^2+R^2)/(R_0^2-R^2)+\nu]/E$. Their sum equals radial interference; ν cancels for identical materials. Therefore $p=\delta E/[R\{1+(R_0^2+R^2)/(R_0^2-R^2)\}]=0.05(200000)/[25(1+2225/975)]=121.875$ MPa. **121.9 MPa, C**.
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