Thick cylinder stresses and failure criteria

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.

Check: $\sigma_\theta + \sigma_r = 2A = 64$ MPa is the same at every radius, for example $104 - 40 = 64$ at the inner surface and $64 + 0 = 64$ at the outer surface. $\checkmark$

At the inner surface: $\sigma_\theta = 104$, $\sigma_r = -40$, $\sigma_z = 32$ MPa.

(b) Failure criteria

Maximum shear stress (Tresca): $\sigma_1 = \sigma_\theta = 104$, $\sigma_3 = \sigma_r = -40$ MPa (the axial stress, 32 MPa, lies between them).

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

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

$$\begin{aligned} \varepsilon_\theta &= \frac{1}{E}\left[\sigma_\theta - \nu(\sigma_r + \sigma_z)\right] = \frac{104 - 0.3(-40 + 32)}{200\times10^3} = 5.32\times10^{-4} \\ \varepsilon_z &= \frac{1}{E}\left[\sigma_z - \nu(\sigma_\theta + \sigma_r)\right] = \frac{32 - 0.3(64)}{200\times10^3} = 6.4\times10^{-5} \\ \varepsilon_r &= \frac{1}{E}\left[\sigma_r - \nu(\sigma_\theta + \sigma_z)\right] = \frac{-40 - 0.3(136)}{200\times10^3} = -4.04\times10^{-4} \end{aligned}$$

Volumetric strain:

$$\varepsilon_v = \varepsilon_\theta + \varepsilon_r + \varepsilon_z = \frac{1-2\nu}{E}(\sigma_\theta + \sigma_r + \sigma_z) = \frac{0.4\times96}{200\times10^3} = 1.92\times10^{-4}$$

The inner radius increases by $u = \varepsilon_\theta\,a = 5.32\times10^{-4}\times80 = 0.0426$ mm.

Answer: (a) At $r = 80$: $\sigma_\theta = 104$ MPa, $\sigma_r = -40$ MPa; at $r = 100$: $\sigma_\theta = 78.1$ MPa, $\sigma_r = -14.1$ 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) $\varepsilon_\theta = 5.32\times10^{-4}$, $\varepsilon_z = 6.4\times10^{-5}$, $\varepsilon_v = 1.92\times10^{-4}$.