Module 7 · Lesson 7.4

Pressure vessels and shared load paths

Two places where direct stress alone answers the whole question.

Why this matters

Direct stress looks like the simplest topic in the course, and it settles two problems that look much harder than they are: how thin-walled tanks and pipes carry pressure, and how load divides between members that share it. The second is your first encounter with an indeterminate structure, and it is solved by exactly the idea that later powers the whole of Module 16.

By the end of this lesson you should be able to

Calculate hoop and longitudinal stress in a thin cylinder
Explain why hoop stress is twice the longitudinal stress
Calculate membrane stress in a thin sphere
Share a load between parallel members using compatibility

What you should already know

Direct stress and strain (this module, lesson 1)
Axial extension δ = PL/AE (this module, lesson 2)
Equilibrium and free bodies (Module 2)

A thin-walled cylinder is one whose wall is thin enough — conventionally less than about a twentieth of the radius — that the stress can be taken as uniform through the thickness. That single assumption turns a hard elasticity problem into two free-body diagrams.

Cut the cylinder along its length. The pressure pushes the two halves apart. Over a length L the pressure acts on a projected area of 2rL, and it is resisted by the wall on both cut edges, of area 2tL. Balancing:

p(2rL) = σ_h(2tL), so σ_h = pr/t

Now cut it across. The pressure acts on the circular end of area πr², resisted by the ring of wall material of area 2πrt:

p(πr²) = σ_l(2πrt), so σ_l = pr/2t

The hoop stress is twice the longitudinal stress. That is why an overpressurised pipe splits along its length rather than snapping in two, and why the seam of a welded tank runs longitudinally at the most critical location.

Thin cylinder under internal pressure

σ_{h} = \frac{p r}{t}, σ_{l} = \frac{p r}{2 t}

What it calculates: hoop and longitudinal membrane stresses

σ_h: hoop (circumferential) stress (N/mm²)
σ_l: longitudinal stress (N/mm²)
p: internal gauge pressure (N/mm²)
r: internal radius (mm)
t: wall thickness (mm)

This assumes

Thin wall: t is small compared with r, conventionally r/t greater than about 20
Stress uniform through the thickness
Away from ends, joints and supports, where local bending occurs

In plain terms: Both stresses are tensile and act at right angles to each other, so a point in the wall is in a two-dimensional stress state — which is exactly the situation Module 14 analyses with Mohr's circle.

A sphere is the efficient shape. Cut it through any diameter and the same argument gives σ = pr/2t in every direction — half the hoop stress of a cylinder of the same radius and thickness. That is why pressure vessels for high pressures are spherical, and why the ends of a cylindrical tank are domed rather than flat.

Predict first

An overpressurised pipe fails. Which way does the crack run?

Now the second problem. Two or more members share a load in parallel — a column encased in concrete, a bolt group, a hanger with two rods of different sizes. Statics gives you one equation, ΣF = 0, and you have as many unknowns as members. It is statically indeterminate.

The missing equations come from compatibility: the members are connected, so they must all deform by the same amount.

If each member has stiffness k = AE/L and they all shorten by the same δ, then each carries P_i = k_i δ. Adding them up gives δ = P/Σk, and therefore:

Load sharing between parallel members

P_{i} = P \frac{k _{i}}{\sum k}, k_{i} = \frac{A _{i} E _{i}}{L _{i}}

What it calculates: how a load divides between members that deform together

P_i: force in member i (N)
P: total applied load (N)
k_i: axial stiffness of member i (N/mm)

This assumes

All members deform by the same amount — they are rigidly connected
Linearly elastic behaviour
The members are the same length, or their differing lengths are in k

In plain terms: Load follows stiffness. This is the same principle as the distribution factors in moment distribution, and the same principle that makes an accidentally stiff element attract force nobody intended.

Worked example

A pressure vessel, and a column sharing load with its casing

Given

A thin cylinder: internal radius 600 mm, wall thickness 10 mm, internal pressure 1.6 N/mm²
Separately: a steel core (A = 2000 mm², E = 205 000) encased in concrete (A = 60 000 mm², E = 30 000), same length, carrying 1500 kN together

Find

The hoop and longitudinal stresses, and how the 1500 kN divides.

Practice

A thin cylinder has an internal radius of 600 mm and a wall thickness of 10.0 mm, under an internal pressure of 1.60 N/mm². What is the hoop stress, in N/mm²?

Practice

For the same cylinder, what is the longitudinal stress, in N/mm²?

Practice

A steel core (A = 2000 mm², E = 205 000 N/mm²) is encased in concrete (A = 60 000 mm², E = 30 000 N/mm²) of the same length. Together they carry 1500 kN. What force does the steel carry, in kN?

Summary

Thin cylinder: hoop stress pr/t, longitudinal stress pr/2t — the hoop is always twice
That is why pressurised pipes split lengthwise
A thin sphere carries pr/2t in every direction, so it is the efficient shape
Parallel members that deform together share load in proportion to stiffness AE/L
Load sharing is the simplest indeterminate problem, solved by compatibility

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This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint