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Structural Analysis Fundamentals / Stress and strain

Pressure vessels and shared load paths

Membrane pressure stresses and compatible axial load sharing.

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This lesson has received a full-context editorial review and a check of its technical explanations. Selected arithmetic has been checked separately. This is not independent engineering sign-off.

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Why this matters

Thin-wall membrane theory estimates pressure stresses away from local disturbances. Parallel-member load sharing introduces compatibility: equilibrium alone does not determine each member force.

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)

For a thin cylindrical wall, membrane stresses can be approximated as uniform through its thickness. This lesson uses r/t greater than about 20 as a conservative teaching criterion; the acceptable approximation depends on the required accuracy.

A longitudinal cut through a length L gives p(2rL) = σh(2tL), hence σh = pr/t. For closed ends whose pressure thrust is carried by the wall, a transverse cut gives pπr² = σl(2πrt), hence σl = pr/(2t).

The hoop stress is twice the longitudinal membrane stress in this model. Actual fracture also depends on welds, defects, material behaviour and other loads.

Thin cylinder under internal pressure
σh=prt,σl=pr2t\sigma_h = \frac{pr}{t}, \qquad \sigma_l = \frac{pr}{2t}

What it calculates: hoop and longitudinal membrane stresses

σh
hoop (circumferential) stress (N/mm²)
σl
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
  • Membrane stress approximately uniform through the thickness; closed ends transmit pressure thrust through the wall
  • 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 thin spherical shell under internal pressure has membrane stress pr/(2t) in each tangential direction. This is half the cylindrical hoop stress for the same radius, thickness and pressure. Vessel selection also depends on manufacture, capacity, supports and local stresses.

Predict first

For the ideal closed-end thin cylinder, which crack orientation would the larger tensile membrane stress tend to open?

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 Pi = ki δ. Adding them up gives δ = P/Σk, and therefore:

Load sharing between parallel members
Pi=P ki∑k,ki=AiEiLiP_i = P\,\frac{k_i}{\sum k}, \qquad k_i = \frac{A_iE_i}{L_i}

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

Pi
force in member i (N)
P
total applied load (N)
ki
axial stiffness of member i (N/mm)

This assumes

  • Equal axial displacement is imposed at the member ends, without end-plate rotation, gaps or differing initial strains
  • 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
    • The larger hoop membrane stress tends to open longitudinal cracks; real failure modes require further assessment
    • 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