Consider a press-fit connection consisting of a shaft and a nave. Let the radii $r_2=2r_1=1\text{ m}.$ The inner radius of the nave is less than the one of the shaft by the interference $\Delta r_{1}.$ Determine the contact pressure developed between the two parts.

$E=2\times10^5\text{ MPa},$ $\nu=0.3.$

Axisymmetric–plane stress.

$\Delta r_1=0.6667\times10^{-3}\text{ m}.$

• Quadrilaterals, PRESS7K1

In the absence of external forces, we have no definite tolerance for the residual vector criterion, which, therefore, cannot be used directly for this and similar self-equilibrium problems. We may, however, control the magnitude of the out-of-balance vector in the .oL file manually until it reduces to a ‘reasonable small’ value. To this end, we first execute the solver with default tolerance values on the RP line which will cause the program to stop after $\mathtt{NITER}$ iterations have been performed. When satisfied, we re-execute the solver with the convergence criteria disabled by setting $\mathtt{RTOL}=\mathtt{XTOL}=-1$ and the solution is obtained in the next iteration step.

The relation between the contact pressure $p_c$ and the interference $\Delta r_{1}$ can be expressed as $$p_c=E\frac{\Delta r_{1}}{r_{1}\left(1+\frac{r_2^2+r_1^2}{r_2^2-r_1^2}\right)} =2\times10^5\frac{0.6667\times10^{-3}}{0.5\left(1+\frac{1+0.5^2}{1-0.5^2}\right)} =100\text{ MPa}.$$

The computed distribution of radial and hoop stresses in the shaft and nave with the penalty parameter set to $\xi=10^{15}\text{ N/m}^3$ is shown below. The analytical solution is plotted by a solid line.