Calculating the pressure between a housing and a shaft in an interference fit

Question

I'm trying to calculate the pressure between a journal/shaft and a housing bore in a case of interference fit.

Considere a 2D axisymmetric (effectively 1D) journal bearing with interference fit:

Where $D_1 > D_2$ and after fitting we should have $D_1>D_4>D_2$ and I think also $D_5>D_3$. Writing the continuum equation of equilibrium for the 1D problem:

$$ \frac{d \sigma_r}{d r} +\frac{\sigma_r-\sigma_\theta}{r}=0 \tag{1}$$

and the elements of the strain tensor can be calculated from radial displacement:

$$\left\{ \begin{matrix} \epsilon_r=\frac{d u}{d r}\\ \epsilon_\theta=\frac{u}{r} \end{matrix} \right. \tag{2}$$

where $u$ is the radial displacement. And the Hooke's law including the Poisson's ratio for stress-strain tensors :

$$ \left\{ \begin{matrix} \sigma_r=\frac{\lambda}{\nu}\left( \left(1-\nu \right)\epsilon_r +\nu \epsilon_\theta \right)\\ \sigma_\theta=\frac{\lambda}{\nu}\left( \left(1-\nu \right)\epsilon_\theta +\nu \epsilon_r \right) \end{matrix} \right. \tag{3}$$

Where $\lambda=\frac{\nu E}{\left(\nu+1 \right)\left(1-2 \nu \right)}$ is the Lame’s elastic constant. Combining 1, 2 and 3 yields:

$$ u_{rr}+\frac{u_r}{r}-\frac{u}{r^2}=0 \tag{4}$$

Which is a simple second order Cauchy-Euler ODE and has an analytic solution of:

$$u=c_1 r +\frac{c_2}{r}\tag{5}$$

The issue is that this solution is singular at $r=0$, whereas based on the boundary condition we should have $u=0$ there. This doesn't make sense. I'm probably making some mistakes. I would appreciate if you could help me know what is the problem and how I can solve it.

P.S.1. The final goal is to answer my other question over here.

P.S.2. The assumption of $c_2=0$ also does not work. stress can't be constant.

P.S.3. I borrowed the equation 3 from this lecture notes of "Structural Mechanics in Nuclear Power Technology" MIT course. But it doesn't makes sense because it is exactly the same as the Cartesian one.

Answer 1

Following this Lecture notes from "Mechanics of Materials Laboratory, University of Washington" and also considering the planar version of equations 2 and 3 as:

$$ \left\{ \begin{matrix} \sigma_r=\frac{E}{1-\nu^2}\left( u_r +\nu \frac{u}{r} \right)\\ \sigma_\theta= \frac{E}{1-\nu^2}\left( \nu u_r +\frac{u}{r} \right) \end{matrix} \right. \tag{6}$$

Combining with 5:

$$ \left\{ \begin{matrix} \sigma_r=\frac{E}{1-\nu^2}\left( c_1 \left( 1+\nu \right) +c_2 \left( \frac{1-\nu}{r^2} \right) \right)\\ \sigma_\theta= \frac{E}{1-\nu^2}\left( c_1 \left( 1+\nu \right) -c_2 \left( \frac{1-\nu}{r^2} \right) \right) \end{matrix} \right. \tag{7}$$

If we use the hypothetical cylinder proposed in the reference with the pressure boundary conditions of $P_o$ at $r_o$ and $P_i$ at $r_i$ then:

$$ \left\{ \begin{matrix} c_1=\frac{1-\nu}{E}\left( \frac{r_i^2 P_i -r_o^2 P_o}{r_o^2-r_i^2} \right)\\ c_2=\frac{1-\nu}{E}\left( \frac{r_i^2 r_o^2 \left( P_i - P_o\right)}{r_o^2-r_i^2} \right) \end{matrix} \right. \tag{8} $$

Which if applied for our problem implies:

$$ \left\{ \begin{matrix} c_{11}=\frac{1-\nu_1}{E_1}P=A_1 P\\ c_{21}=0 \end{matrix} \right. \tag{9} $$

For the shaft and

$$ \left\{ \begin{matrix} c_{12}=\frac{1-\nu_2}{E_2}\left( \frac{r_2^2 }{r_3^2-r_2^2} \right)P=A_2 P\\ c_{22}=\frac{1-\nu_2}{E_2}\left( \frac{r_2^2 r_3^2 }{r_3^2-r_2^2} \right)P=A_3P \end{matrix} \right. \tag{10} $$

For the housing, Where in both 9 and 10 $P$ is the contact pressure between the two. Also considering another boundary condition

$$r_2 +\delta_2 =r_1-\delta_1=r_4 \tag{11}$$

Where $\delta_1$ and $\delta_2$ are the displacements of journal and housing at contact point respectively. Which can be calculated as:

$$ \left\{ \begin{matrix} \delta_1=c_{11}r_1\\ \delta_2=c_{12}r_2+\frac{c_{22}}{r_2} \end{matrix} \right. \tag{12} $$

Now if we combine equations 9, 10, 11 and 12 we can calculate interference pressure as:

$$P=\frac{r_1-r_2}{A_1 r_1 +A_2 r_2 +\frac{A_3}{r_2}} \tag{13}$$

This gives the solution I was looking for but the paradox I raised above is still there. Displacement at the center must be zero.