Lame's equations for thick cylinders

Assumptions

Material is homogeneous, isotropic and linear elastic; the cylinder is long, and the section remains plane (axial strain uniform); the stress is symmetric about the axis, so $\sigma_r$ and $\sigma_\theta$ depend on $r$ only.

Equilibrium

Consider an element between radii $r$ and $r+dr$ subtending $d\theta$. Radial force balance:

$$(\sigma_r + d\sigma_r)(r+dr)\,d\theta - \sigma_r\,r\,d\theta - 2\sigma_\theta\,dr\,\sin\frac{d\theta}{2} = 0$$

For small angles ($\sin(d\theta/2) \approx d\theta/2$) and neglecting higher-order terms:

$$\frac{d\sigma_r}{dr} + \frac{\sigma_r - \sigma_\theta}{r} = 0 \quad\text{or}\quad \sigma_\theta = \sigma_r + r\frac{d\sigma_r}{dr}$$

Compatibility

With radial displacement $u$: $\varepsilon_\theta = u/r$ and $\varepsilon_r = du/dr$, so

$$\frac{d\varepsilon_\theta}{dr} = \frac{\varepsilon_r - \varepsilon_\theta}{r}$$

With Hooke's law and a constant axial stress $\sigma_z$ (equivalently constant $\varepsilon_z$), $\varepsilon_\theta = \dfrac1E[\sigma_\theta - \nu(\sigma_r + \sigma_z)]$ and $\varepsilon_r = \dfrac1E[\sigma_r - \nu(\sigma_\theta + \sigma_z)]$. Substituting:

$$\frac{d}{dr}(\sigma_\theta - \nu\sigma_r) + \frac{1+\nu}{r}(\sigma_\theta - \sigma_r) = 0$$

Put $\sigma_\theta = \sigma_r + r\sigma_r'$ from equilibrium:

$$\sigma_r'' + \frac{3}{r}\sigma_r' = 0 \ \Rightarrow\ \sigma_r' = \frac{2B}{r^3}$$

Integrating:

$$\boxed{\sigma_r = A - \frac{B}{r^2},\qquad \sigma_\theta = A + \frac{B}{r^2}}$$

Note that $\sigma_r + \sigma_\theta = 2A$ is constant across the wall.

Boundary conditions

At $r = a$ (inner): $\sigma_r = -p_i$. At $r = b$ (outer): $\sigma_r = -p_o$.

$$A = \frac{p_ia^2 - p_ob^2}{b^2 - a^2},\qquad B = \frac{a^2b^2(p_i - p_o)}{b^2 - a^2}$$

For $p_o = 0$:

$$\sigma_r = \frac{p_ia^2}{b^2 - a^2}\left(1 - \frac{b^2}{r^2}\right),\qquad \sigma_\theta = \frac{p_ia^2}{b^2 - a^2}\left(1 + \frac{b^2}{r^2}\right)$$

Distribution

 stress
   ^ sigma_theta (max at r=a, tension)
   |\
   | \_____
   |       ----__ sigma_theta (at r=b)
 --+-------------------> r
   | ____----
   |/  sigma_r (compressive; -p_i at r=a, 0 at r=b)
   a                   b

The hoop stress is highest at the inner surface, $\sigma_{\theta,max} = p_i\dfrac{b^2 + a^2}{b^2 - a^2}$, and always exceeds $p_i$, however thick the wall.

Axial stress

• Closed ends: $\sigma_z = \dfrac{p_ia^2}{b^2 - a^2}$ (uniform, from end force balance).
• Open ends (no end load): $\sigma_z = 0$.