Thick- Wall 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): $\sigma_z = 0$.

Data

$a = 80$ mm, $b = 120$ mm, $p_i = 40$ MPa, $p_o = 0$.

$$\frac{p_ia^2}{b^2 - a^2} = \frac{40\times6400}{14400 - 6400} = 32\ \text{MPa}$$

(a) Lame's stresses

$$\sigma_\theta = 32\left(1 + \frac{14400}{r^2}\right),\quad \sigma_r = 32\left(1 - \frac{14400}{r^2}\right)$$

Axial stress for closed ends: $\sigma_z = p_ia^2/(b^2 - a^2) = 32.0$ MPa.

At the inner surface: $\sigma_\theta = 104$, $\sigma_r = -40$, $\sigma_z = 32$ MPa; check $\sigma_\theta + \sigma_r = 64 = 2A$ and equals $\sigma_\theta+\sigma_r$ at the outer surface $(64 + 0)$. $\checkmark$

(b) Failure criteria

Maximum shear stress (Tresca): $\sigma_1 = \sigma_\theta = 104$, $\sigma_3 = \sigma_r = -40$ MPa.

$$\sigma_1 - \sigma_3 = 104 - (-40) = 144\ \text{MPa},\qquad \tau_{max} = 72\ \text{MPa}$$ $$FS = \frac{\sigma_y}{\sigma_1 - \sigma_3} = \frac{350}{144} = 2.43$$

von Mises (distortion energy):

$$\begin{aligned} \sigma_{vm} &= \sqrt{\tfrac12\left[(\sigma_\theta - \sigma_r)^2 + (\sigma_r - \sigma_z)^2 + (\sigma_z - \sigma_\theta)^2\right]} \\ &= \sqrt{\tfrac12\left[144^2 + 72^2 + 72^2\right]} = 124.7\ \text{MPa} \end{aligned}$$ $$FS = \frac{350}{124.7} = 2.81$$

Tresca is more conservative, which is why it is commonly used for thick cylinders.

(c) Strains at the inner surface

$$\begin{aligned} \varepsilon_\theta &= \frac{1}{E}\left[\sigma_\theta - \nu(\sigma_r + \sigma_z)\right] = \frac{104 - 0.3(-40 + 32)}{200\times10^3} = 5.320e-04 \\ \varepsilon_z &= \frac{1}{E}\left[\sigma_z - \nu(\sigma_\theta + \sigma_r)\right] = \frac{32 - 0.3(64)}{200\times10^3} = 6.40e-05 \\ \varepsilon_r &= \frac{1}{E}\left[\sigma_r - \nu(\sigma_\theta + \sigma_z)\right] = \frac{-40 - 0.3(136)}{200\times10^3} = -4.040e-04 \end{aligned}$$

Volumetric strain:

$$\varepsilon_v = \varepsilon_\theta + \varepsilon_r + \varepsilon_z = \frac{1-2\nu}{E}(\sigma_\theta + \sigma_r + \sigma_z) = \frac{0.4\times96}{200\times10^3} = 1.92e-04$$

The inner radius increases by $u = \varepsilon_\theta\,a = 5.320e-04\times80 = 0.0426$ mm.

Answer: (a) At $r = 80$: $\sigma_\theta = 104$ MPa, $\sigma_r = -40$ MPa; at $r = 120$: $\sigma_\theta = 64$ MPa, $\sigma_r = 0$; $\sigma_z = 32$ MPa. (b) FS $= 2.43$ (Tresca), $2.81$ (von Mises). (c) $\varepsilon_\theta = 5.32\times10^{-4}$, $\varepsilon_z = 6.0\times10^{-5}$, $\varepsilon_v = 1.92\times10^{-4}$.

Data

Inner tube: $a = 30$ mm, $r = 50$ mm. Outer tube: $r = 50$ mm, $b = 70$ mm. Radial interference $\delta = 0.06/2 = 0.03$ mm. Both tubes are steel, $E = 200\times10^3$ MPa.

        outer tube (50 to 70 mm)
     .--------------------.
    /   .--------------.   \
   |   /  inner tube    \   |
   |  |   (30 to 50)     |  |
   |   \                /   |
    \   '--------------'   /
     '--------------------'
        contact pressure p at r = 50

Contact pressure

After assembly, a pressure $p$ acts on the contact surface. The inner tube has external pressure $p$ only, the outer tube has internal pressure $p$ only. The radial interference equals the sum of the outward movement of the outer tube's inner surface and the inward movement of the inner tube's outer surface.

For a tube with internal pressure $p$ (inner radius $r$, outer radius $b$): hoop stress at inner surface $= p\dfrac{b^2 + r^2}{b^2 - r^2}$, so

$$u_{outer} = \frac{p\,r}{E}\left[\frac{b^2 + r^2}{b^2 - r^2} + \nu\right]$$

For a tube with external pressure $p$ only (inner radius $a$, outer radius $r$): hoop stress at outer surface $= -p\dfrac{r^2 + a^2}{r^2 - a^2}$, so the inward movement is

$$u_{inner} = \frac{p\,r}{E}\left[\frac{r^2 + a^2}{r^2 - a^2} - \nu\right]$$

Sum (the Poisson terms cancel for the same material):

$$\delta = \frac{p\,r}{E}\left[\frac{b^2 + r^2}{b^2 - r^2} + \frac{r^2 + a^2}{r^2 - a^2}\right]$$ $$\frac{b^2 + r^2}{b^2 - r^2} = \frac{4900 + 2500}{2400} = 3.0833,\qquad \frac{r^2 + a^2}{r^2 - a^2} = \frac{2500 + 900}{1600} = 2.1250$$ $$p = \frac{\delta E}{r\,(3.0833 + 2.1250)} = \frac{0.03\times200\times10^3}{50\times5.2083} = 23.04\ \text{MPa}$$

(b) Hoop stresses

Outer tube (internal pressure $p$):

$$\sigma_\theta(r=50) = p\,\frac{b^2 + r^2}{b^2 - r^2} = 23.04\times3.0833 = 71.04\ \text{MPa (tension)}$$ $$\sigma_\theta(r=70) = \frac{2pr^2}{b^2 - r^2} = \frac{2\times23.04\times2500}{2400} = 48.00\ \text{MPa (tension)}$$

Inner tube (external pressure $p$):

$$\sigma_\theta(r=50) = -p\,\frac{r^2 + a^2}{r^2 - a^2} = -23.04\times2.125 = -48.96\ \text{MPa}$$ $$\sigma_\theta(r=30) = -\frac{2pr^2}{r^2 - a^2} = -\frac{2\times23.04\times2500}{1600} = -72.00\ \text{MPa (compression)}$$

The radial stress is $-p = -23.04$ MPa at the contact surface and zero on the free surfaces.

(c) Heating of the outer tube

To slip the outer tube over, its bore must expand by the diametral interference $0.06$ mm (a small extra clearance is used in practice). The bore of 100 mm diameter expands by $\alpha\,d\,\Delta T$:

$$\Delta T = \frac{0.06}{12\times10^{-6}\times100} = 50\ ^\circ\text{C}$$

Answer: Contact pressure $= 23.04$ MPa; hoop stresses: outer tube $71.0$ MPa (bore) and $48.0$ MPa (outside); inner tube $-49.0$ MPa (outside) and $-72.0$ MPa (bore); outer tube to be heated by at least $50^\circ$C above the inner tube.