Thin-walled pressure vessels

By the end of this topic you will be able to

State when the thin-wall assumption is valid.
Calculate hoop and longitudinal stress in a cylinder and the stress in a sphere.
Find the wall thickness needed for a given pressure and allowable stress.

Pressure vessels

Boilers, gas cylinders, pipelines, tanks and aircraft cabins contain fluid under pressure. The pressure tries to burst the wall, which carries it as tension in the wall, like a hoop around a barrel.

A vessel is thin-walled when the wall thickness is small compared with the radius, roughly $d/t \ge 20$ . Then the stress is nearly uniform through the thickness, and simple equilibrium gives the formulas below. (For thicker walls, the stress varies and other equations such as Lamé's are needed.)

Cylinder: two stresses

Cut the cylinder along its length. The pressure pushes the two halves apart with a force $p\,d\,L$ on the projected area, resisted by the wall's two cut edges, $2tL\sigma_h$ :

\sigma_h = \frac{pd}{2t} \quad \text{(hoop or circumferential stress)}

Now cut across the cylinder. The pressure on the end $p\pi d^2/4$ is balanced by the wall's ring area $\pi d t$ :

\sigma_L = \frac{pd}{4t} \quad \text{(longitudinal or axial stress)}

The hoop stress is twice the longitudinal stress, which is why an overcooked sausage splits lengthwise and a pressurised cylinder is most likely to rupture along its length.

Sphere

A sphere has the same stress in every direction, and it is the more efficient shape:

\sigma = \frac{pd}{4t}

(half the cylinder's hoop stress for the same $d$ , $t$ and $p$ .)

Interactive modelChange the pressure, diameter and wall thickness for a cylinder or sphere.

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Worked example

A cylindrical air receiver has an inside diameter of 1.0 m and a wall 10 mm thick. The pressure is 2 MPa. Find the stresses and the minimum thickness if the allowable stress is 150 MPa.

Thin-wall check. $d/t = 1000/10 = 100 \ge 20$ ✓

Stresses. $\sigma_h = \dfrac{2\times1000}{2\times10} = 100$ MPa and $\sigma_L = 50$ MPa.

Required thickness (hoop governs): $t = \dfrac{pd}{2\sigma_{\text{allow}}} = \dfrac{2\times1000}{2\times150} = 6.67$ mm. The 10 mm wall gives a factor of 1.5 on thickness.

Practice problems

A spherical tank of inside diameter 6 m and wall 15 mm holds gas at 0.5 MPa. Find the stress.

Show answer

$\sigma = pd/4t = 0.5\times6000/(4\times15) = 50$ MPa.
For the worked-example cylinder, what thickness is required if it were a sphere instead?

Show answer

$t = pd/(4\sigma) = 2\times1000/(4\times150) = 3.33$ mm, half the cylinder's.
A boiler drum has $d = 1.2$ m, $t = 12$ mm, $p = 1.5$ MPa. Find $\sigma_h$ and $\sigma_L$ .

Show answer

$\sigma_h = 1.5\times1200/24 = 75$ MPa; $\sigma_L = 37.5$ MPa.
Why must a pressure-vessel's longitudinal weld be stronger than the circumferential weld?

Show answer

The longitudinal weld carries the hoop stress, which is twice the longitudinal stress carried across the circumferential weld.

Key formulae

Cylinder, hoop stress: $\sigma_h = \frac{pd}{2t}$
Cylinder, longitudinal stress: $\sigma_L = \frac{pd}{4t}$
Sphere (any direction): $\sigma = \frac{pd}{4t}$
Thin-wall condition: $\frac{d}{t} \ge 20$

Lecturer notes: suggested plan, activity and assessment

Suggested plan (35 min). 5 min: demonstrate with a long balloon (it bulges and splits along its length); 10 min: derive both formulas from free bodies; 10 min: worked example; 10 min: model and sphere vs cylinder.

Activity. Ask which dimension has the biggest effect on stress (the diameter-to-thickness ratio) and verify in the model.

Discussion. Why are LPG tanks cylinders with domed ends rather than flat ends?

Assessment idea. Thickness design for given pressure, diameter and allowable stress, including the weld efficiency.