8.5 Thin Cylinders, Spheres & Thick-Walled Pressure Vessels

Why Pressure Vessels Matter in a Coal Context

A Management Trainee in Coal India meets pressure vessels constantly: compressed-air receivers driving pneumatic drills, hydraulic cylinders on shovels and draglines, boiler drums in captive power plants, and methane drainage pipelines. The stress analysis is short, the formulae are few, and they are examined almost every year in mechanical PSU papers.

The Thin-Cylinder Assumption

A cylinder is classed as thin when

$\frac{t}{d} < \frac{1}{20}, \qquad \text{equivalently } d > 20t$

Under this condition the stress is taken as uniform through the wall thickness, and the radial stress — which varies from $-p$ at the bore to zero at the outside — is small enough to neglect against the hoop stress. That single simplification is what makes the thin-cylinder formulae so simple.

Circumferential (Hoop) Stress

Consider a cylinder of internal diameter $d$, wall thickness $t$ and length $l$, under internal gauge pressure $p$. Cut it along a longitudinal plane through the axis. The bursting force on the projected area is $p,d,l$, and it is resisted by the stress acting on two wall sections each of area $t,l$:

$p\,d\,l = \sigma_h \times 2\,t\,l$

$\boxed{\sigma_h = \frac{p\,d}{2t}}$

Longitudinal (Axial) Stress

Now cut the cylinder on a transverse plane. The force on the end closure is $p,\dfrac{\pi d^2}{4}$, resisted by the annular wall area $\pi d t$:

$p\,\frac{\pi d^2}{4} = \sigma_l\,\pi d t$

$\boxed{\sigma_l = \frac{p\,d}{4t}}$

The consequence

$\sigma_h = 2\,\sigma_l$

The hoop stress is always twice the longitudinal stress. A cylindrical vessel therefore fails by splitting along a longitudinal line, not by parting circumferentially — which is why longitudinal welded seams in a boiler drum are designed to a higher joint efficiency than circumferential ones. This ratio is the most examined single fact in the topic.

Maximum shear stress

Since the third principal stress (radial) is neglected, the in-plane principal stresses are $\sigma_h$ and $\sigma_l$, and

$\tau_{\max} = \frac{\sigma_h - \sigma_l}{2} = \frac{pd}{8t}$

If the radial stress at the bore, $-p$, is retained, the absolute maximum shear stress becomes $(\sigma_h + p)/2$.

Thin Spherical Shell

Cutting a sphere on any diametral plane gives the same equilibrium in every direction:

$p\,\frac{\pi d^2}{4} = \sigma\,\pi d t \quad\Longrightarrow\quad \sigma_1 = \sigma_2 = \frac{p\,d}{4t}$

A sphere therefore carries half the maximum stress of a cylinder of the same diameter, thickness and pressure. For a given duty a spherical vessel needs half the wall thickness, which is why large LPG and cryogenic storage vessels are spherical despite the higher fabrication cost.

Strains and Dimensional Changes

Using the two-dimensional form of Hooke's law with Young's modulus $E$ and Poisson's ratio $\mu$:

Circumferential (hoop) strain, which equals the diametral strain:

$\epsilon_h = \frac{\sigma_h}{E} - \mu\frac{\sigma_l}{E} = \frac{pd}{2tE}\left(1 - \frac{\mu}{2}\right) = \frac{\delta d}{d}$

Longitudinal strain:

$\epsilon_l = \frac{\sigma_l}{E} - \mu\frac{\sigma_h}{E} = \frac{pd}{4tE}\left(1 - 2\mu\right) = \frac{\delta l}{l}$

Volumetric strain is the sum of one longitudinal and two circumferential strains:

$\frac{\delta V}{V} = \epsilon_l + 2\epsilon_h = \frac{pd}{4tE}\left(5 - 4\mu\right)$

For a sphere, the volumetric strain is

$\frac{\delta V}{V} = 3\epsilon = \frac{3pd}{4tE}(1-\mu)$

Note a consequence of the longitudinal strain expression: if $\mu = 0.5$, the longitudinal strain is zero. Real metals have $\mu$ near 0.3, so the cylinder does lengthen, but only modestly.

Worked example

A compressed-air receiver of internal diameter 800 mm and wall thickness 10 mm operates at 2 MPa. Take $E = 200$ GPa and $\mu = 0.3$.

$\sigma_h = \frac{2 \times 800}{2\times10} = 80 \text{ MPa}, \qquad \sigma_l = 40 \text{ MPa}$

Check thinness: $t/d = 10/800 = 0.0125 < 0.05$, so the thin assumption is valid.

Diametral change:

$\delta d = \frac{pd^2}{2tE}\left(1-\frac{\mu}{2}\right) = \frac{2\times800^2}{2\times10\times200000}(0.85) = 0.272 \text{ mm}$

Thick Cylinders: Lame's Equations

When $t/d$ exceeds about one twentieth, stress varies significantly through the wall and the radial stress can no longer be ignored. Lame's equations give, at radius $r$,

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

with $A$ and $B$ fixed by boundary conditions. For a cylinder of inner radius $r_i$ and outer radius $r_o$ under internal pressure $p$ only:

$A = \frac{p\,r_i^2}{r_o^2 - r_i^2}, \qquad B = \frac{p\,r_i^2 r_o^2}{r_o^2 - r_i^2}$

Key results:

• Hoop stress is maximum at the bore and falls towards the outside. Adding wall thickness therefore gives diminishing returns, because the extra material is at the low-stress radius.
• At the inner surface, $\sigma_r = -p$ (compressive) and

$\sigma_{h,\max} = p\,\frac{r_o^2 + r_i^2}{r_o^2 - r_i^2}$

• The sum $\sigma_h + \sigma_r = 2A$ is constant through the wall, a useful check.

Compound Cylinders

Because hoop stress peaks at the bore, a single thick cylinder wastes material. Shrink-fitting an outer cylinder over an inner one puts the inner cylinder into initial compression and the outer into initial tension. When internal pressure is later applied, it must first overcome that compression, so the peak resultant hoop stress is reduced and the stress distribution is far more uniform.

This is the principle behind gun barrels, high-pressure extrusion containers and heavy hydraulic cylinders. Autofrettage achieves the same end differently, by over-pressurising a single cylinder until the bore yields, so that on release the plastically stretched inner layers are held in residual compression by the still-elastic outer layers.

Design Note: Joint Efficiency

For a riveted or welded vessel, the wall must carry the stress across the joint, so the design formulae become

$\sigma_h = \frac{pd}{2t\eta_l}, \qquad \sigma_l = \frac{pd}{4t\eta_c}$

where $\eta_l$ and $\eta_c$ are the efficiencies of the longitudinal and circumferential joints respectively. Because hoop stress is the larger, it is the longitudinal joint efficiency that governs the design.