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:

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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.

  • $\begingroup$ You are assuming plane strain, correct? Are you assuming separate displacement relationships for the two regions, correct? $\endgroup$ Mar 2, 2018 at 13:17
  • $\begingroup$ @ChesterMiller If I understood you correctly you are asking if I'm assuming the displacement of the two is equal at the contact point, which the answer is no. only the radial stress is equal there. $\endgroup$
    – Foad
    Mar 2, 2018 at 13:30
  • $\begingroup$ OK. That is correct. Are you assuming that the strain in the z direction is zero? $\endgroup$ Mar 2, 2018 at 14:51
  • $\begingroup$ @ChesterMiller That's the assumption I have not considered and That probably will change both my question and answer below. depending on either I consider $\epsilon_z=0$ or $\sigma_z=0$ there will be different results. The former is basically a planar problem. $\endgroup$
    – Foad
    Mar 2, 2018 at 14:53
  • 1
    $\begingroup$ In my judgment, zero axial strain is more realistic. $\endgroup$ Mar 2, 2018 at 16:52

1 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.

  • $\begingroup$ @ChesterMiller here I think I have achieved the parameter I was looking for (if not making any silly mistakes as always) but still the paradox creeps my mind! $\endgroup$
    – Foad
    Mar 2, 2018 at 14:55
  • 1
    $\begingroup$ For the shaft, if u is proportional to r, it means that the radial stress is equal to the hoop stress, and the radial stress is constant. So the state of stress is homogeneous and transversely isotropic in the shaft. $\endgroup$ Mar 2, 2018 at 17:00
  • $\begingroup$ @ChesterMiller true. And this I think is the key to resolving the paradox above. basically if the internal diameter of a thick cylinder is zero then the radial pressure is constant and displacement is of course zero at the center. $\endgroup$
    – Foad
    Mar 2, 2018 at 17:04
  • 1
    $\begingroup$ I totally agree. $\endgroup$ Mar 2, 2018 at 17:12

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