Thin-Walled Composite Tube and Pressure Vessel Theory

How a thin-walled composite tube carries pressure, axial force, torque, bending and a temperature change: membrane statics on the mean radius, the laminate solve with the wall curvature held at zero, the reaction moments of an unsymmetric wall, and shell buckling with knock-down factors.

Thin-walled composite tube theory treats the wall of a laminated cylinder as a membrane on its mean radius. Whatever the tube carries, the wall sees it as three force resultants N_x, N_y and N_xy in N/mm. The laminate is then solved with its wall curvature held at zero (κ = 0)¹, and three shell buckling checks, with their interaction, run on the same resultants².

This is the method under the Tube & Pressure Vessel tool. The tube and pressure vessel reference covers the inputs, what each result means and a quick-reference table of the formulas. This page derives those formulas: the cuts through the wall that give the resultants, why the wall stays at zero curvature and what that costs an unsymmetric layup, where the thermal terms enter, and how the buckling checks and their knock-down factors are built.

The wall as a membrane on the mean radius

The wall of thickness t has inner radius r_i, outer radius r_o and mean radius r̄, all in mm:

Every load acts on r̄. This is the thin-wall form, the one behind the boiler formula. Exact equilibrium of the wall would put p_i on r_i and p_o on r_o, and the two differ by a term of order t/r̄. That difference is the first reason the model asks for a thin wall.

The axes are the laminate axes. x runs along the tube, y around the hoop (the direction θ), and xy is the in-plane shear xθ, so a 0° ply runs along the tube and a 90° ply around it. z points radially outward from the mid-surface of the wall, and ply 1, the top ply with the most positive z, is the outer surface. Internal pressure p_i and external pressure p_o are in MPa and never negative: p_i pushes outward, p_o inward. The axial force F (N) is positive in tension, and the torque T (N·mm) is positive by the right-hand rule about +x. The sign of the bending moment M_b (N·mm) does not matter, for a reason given under bending.

The same tube under internal pressure with two kinds of end. Left, closed: the heads collect the pressure and pass it into the wall as an axial load. Right, open: the pressure loads the wall in the hoop direction only.

Statics: from loads to force resultants

The resultants superpose, so each load can be taken on its own and the results added. Every term comes from cutting the tube and writing equilibrium for the piece that comes away. Two cuts give the pressure terms.

Left, a cut across the axis of a closed-end tube: the pressure on the end cap, p_iπr̄², is carried by N_x around the circumference 2πr̄. Right, a half-ring of unit length cut along a generator: the pressure over the projected width 2r̄ is carried by N_y at the two cut edges.

Pressure and axial force, closed and open ends

In a closed-end thin-walled pressure vessel the hoop resultant is twice the axial one, whatever the layup. Cut across the axis, the pressure on a closed end, πr̄²(p_i − p_o), and the axial force F are carried by the wall around its circumference 2πr̄:

N_x is in N/mm, with the pressures in MPa, r̄ in mm and F in N. An open end is a pipe whose end load goes to a support or a sliding seal, the case NASA SP-8007 calls lateral pressure². Cut along a generator, the pressure on a half-ring of unit length acts over the projected width 2r̄ and is carried by N_y at its two edges:

For an internal pressure p on a closed end, dividing by t gives the membrane stresses in MPa²:

So N_y = 2N_x. Statics alone sets that ratio. The layup only decides how the plies share the 2:1 state, which the worked example below shows.

Torque

A torque T is carried by a uniform shear flow N_xy around the circumference. All of it acts at lever arm r̄, so T = N_xy · 2πr̄ · r̄, and

with N_xy in N/mm and T in N·mm.

Bending and the two sides of the section

A thin circular section has second moment of area I = πr̄³t in mm⁴. A bending moment M_b gives the axial stress σ = M_br̄/I at the extreme fibres, and the extreme-fibre axial resultant is σt:

This is the relation M_cr = πr²N_x that NASA SP-8007 uses for a cylinder in bending². Bending adds ΔN_x to the axial resultant on one side of the section and subtracts it on the other:

N_y and N_xy are the same on both sides. The sign of M_b only names which side is which, so the analysis takes |M_b|, checks both sides, and the side with the lower first-ply safety factor governs. With M_b = 0 there is one state and no side.

Worked example: a ±55° vessel under internal pressure

Take an E-glass/epoxy ply with E₁ = 41000 MPa, E₂ = 10400 MPa, G₁₂ = 4300 MPa and ν₁₂ = 0.28. Lay it up [55/−55/−55/55] in 0.125 mm plies, so t = 0.5 mm, on r̄ = 50 mm with closed ends, and apply p_i = 1 MPa. The two cuts give N_x = 25 N/mm and N_y = 50 N/mm, with N_xy = 0. The layup is symmetric, so the zero-curvature solve below gives the ordinary laminate result, and every ply carries the same stresses apart from the sign of the shear:

Every ply sits at the same angle to the hoop, so σ_x and σ_y come out at the average wall stress N/t. The shear stresses of the +55° and −55° plies cancel, which is how N_xy stays zero. In the ply axes the picture is clearer: the fibres carry 121.3 MPa, more than four times the 28.7 MPa across them, and the in-plane shear stays under 10 MPa. A ±55° winding turns the 2:1 membrane state into mostly fibre load.

Why the wall is solved with zero curvature

The laminate law, with the mechanical and thermal resultants summed on the left¹, is

N (N/mm) and M (N·mm/mm) are the mechanical resultants, N^T and M^T the thermal ones at the temperature change ΔT¹. ε⁰ holds the dimensionless mid-surface strains and κ the curvatures in 1/mm. A, B and D are the blocks of the ABD matrix, in N/mm, N and N·mm.

A flat laminate with B ≠ 0 curls when it is loaded in plane, because B couples membrane force into curvature. A closed circular tube cannot curl its wall into a flat-plate curvature. The hoop continuity of the section holds κ = 0, and the membrane strains come from the A block alone:

With κ = 0 the strain ε(z) = ε⁰ + zκ is the same at every height in the wall¹. The top, middle and bottom of a ply carry the same strain and, through σ = Q̄(ε − αΔT), the same stress¹. Here σ and the transformed ply stiffness Q̄ are in MPa, the thermal expansion coefficients α in 1/°C and ΔT in °C. Plies at different angles still carry different stresses, because Q̄ and α turn with the ply. Those differences are what the ply check ranks.

Left, a free strip of an unsymmetric laminate curls under in-plane load. Right, the same laminate closed into a tube: the end view stays round and the side view stays straight, so κ = 0, and the wall carries the reaction moments M_x and M_y instead. The reactions oppose the curl. Their signs depend on the layup.

The moments are reactions, not loads

The moment rows of the laminate law, at κ = 0, give the mechanical moment the wall has to carry to stay at zero curvature:

M is in N·mm/mm. Nobody applies it. The closed section supplies it, the way a support supplies the reaction on a beam.

A symmetric layup has B = 0 and M^T = 0. It needs no reaction, and the zero-curvature solve gives the same strains as the free-plate solve with M = 0. An unsymmetric layup needs a non-zero reaction, and there the two solves differ: the free plate would curl and spread its strain through the thickness, while the tube wall keeps one strain at every height. Put the M above and the same ΔT back into the full ABD solve and it returns κ = 0 and the same ε⁰, by construction. That makes a quick check if you reproduce the numbers in your own laminate solver.

Worked example: an unsymmetric [0/45/90] tube

Take three T300/5208 plies, each 0.125 mm thick (t = 0.375 mm), with E₁ = 181000 MPa, E₂ = 10300 MPa, G₁₂ = 7170 MPa and ν₁₂ = 0.28. The 0° ply is on the inner surface, the 45° ply in the middle and the 90° ply outermost, as ply 1. On r̄ = 50 mm with closed ends and p_i = 1 MPa, the resultants are the same as for any closed tube of that radius, N = [25, 50, 0] N/mm. The zero-curvature solve gives:

To stay round and straight, the wall carries a hoop bending moment M_y of 4.732 N·mm/mm and an axial one, M_x, of −2.062 N·mm/mm. A flat [0/45/90] plate under the same N would take that as curvature instead. The shear strain is not zero either. The single 45° ply leaves the layup unbalanced, so the tube twists under pressure alone, at dφ/dx = γ⁰_xy/r̄ = −3.57 × 10⁻⁵ rad/mm (see twist rate).

Thermal terms

A temperature change ΔT enters through the thermal force resultant N^T (N/mm) and moment resultant M^T (N·mm/mm)¹, built ply by ply as described under hygrothermal effects. Both go into the zero-curvature solve as written above, so ΔT moves the membrane strains and the reaction moment alike.

A symmetric tube has M^T = 0. An unsymmetric one does not, and then a temperature change alone needs a reaction. Cool the [0/45/90] tube above by ΔT = −100 °C with no mechanical load and it carries M = [0.835, −0.835, 0] N·mm/mm. Add p_i = 1 MPa and the reaction becomes M = [−1.227, 3.896, 0] N·mm/mm. That is the sum of the two cases to rounding, −2.062 + 0.835 and 4.732 − 0.835, because the solve is linear.

A station at height z in the wall, in mm, sits at radius r and moves outward by Δr, both in mm:

ε⁰_y is the total hoop strain of the zero-curvature solve, free thermal expansion included. No radial correction is made to the strain, because that would be a step toward a thick-wall solution.

The buckling checks below read the mechanical resultants only. A free tube's thermal expansion puts no net force on the wall, so ΔT does not enter them.

Twist rate, torsional stiffness and average wall stress

Under pure torque the zero-curvature solve gives γ⁰_xy = (A⁻¹)₆₆N_xy. The twist per unit length is the shear strain over the radius, and the torsional stiffness is GJ = T/(dφ/dx):

dφ/dx is in rad/mm, GJ in N·mm² and (A⁻¹)₆₆ in mm/N. For a symmetric layup the in-plane shear modulus is G_xy = 1/(t(A⁻¹)₆₆) in MPa, and GJ becomes the familiar 2πr̄³tG_xy. The ±55° vessel above has G_xy = 10768 MPa, which gives GJ = 4.23 × 10⁹ N·mm². The twist rate takes γ⁰_xy from the combined solve. An unbalanced layup, with A₁₆ or A₂₆ non-zero, therefore reports the twist it picks up under pressure alone, as the [0/45/90] tube did.

The average wall stress, in MPa, is each resultant over the thickness, per side when M_b ≠ 0:

Shell buckling with knock-down factors

Three checks run on the mechanical resultants, each against one load: external pressure, axial compression (which also covers the compression side of bending) and torsion. When more than one runs, an interaction check combines them². Each check needs the length L between the supports, in mm, and each gives the same three numbers. The classical value is the buckling load of a perfect tube. The knock-down factor γ is the empirical factor of NASA SP-8007 that brings it down to what tested tubes carry. The design value is γ times the classical value, and the safety factor of the check is the design value over the applied value. How the results appear in the tool is under Buckling on the tube reference.

Classical buckling of a laminated tube

SP-8007 gives the classical loads of an orthotropic cylinder as the minimum, over the wave numbers, of a determinant built from the wall stiffnesses². Those closed forms assume a specially orthotropic wall, A₁₆ = A₂₆ = D₁₆ = D₂₆ = 0, and one buckle shape per pair of wave numbers. A laminate with angle plies has D₁₆ and D₂₆ non-zero, and an unbalanced one A₁₆ and A₂₆ as well. Those terms lower the critical loads, skew the buckles and make the torsion load depend on its sign. The one-term form shows none of that. So the analysis solves the same shell equations with the full A and D matrices by the Rayleigh-Ritz method.

The kinematics are Sanders' thin-shell theory³, with x along the axis, y = r̄θ around it, and u, v, w the axial, hoop and outward radial displacements of the mid-surface:

SP-8007 writes its orthotropic equations in Donnell's form, which drops the v/r̄ terms. That is accurate when the tube buckles into many circumferential waves, but it puts the buckling pressure of a long tube at 4D/r̄³ instead of 3D/r̄³², so the Sanders terms stay. External pressure is applied as a follower load that stays normal to the deformed wall. A pressure that kept its direction would also give 4D/r̄³.

Both ends are simply supported: no radial or hoop displacement, free to rotate and free to move axially². For one circumferential wave number n, u, v and w are sine and cosine series along the tube, each with a cos nθ and a sin nθ part. A₁₆, A₂₆, D₁₆, D₂₆ and N_xy couple those parts. The strain energy of the wall and the work of the prebuckling resultants then give the eigenproblem

K is the stiffness from the A and D energy, K_G the geometric stiffness of a unit load pattern, and the critical load factor λ the lowest positive root, minimised over n ≥ 0. The unit patterns are N_x = −1 for axial compression, p = 1 with N_y = −r̄ for external pressure, and N_xy = ±1 for torsion, with the sign of the applied torque. Axial compression is reported as the stress σ_cr = N_x,cr/t in MPa and torsion as the torque T_cr = 2πr̄²N_xy,cr in N·mm. The buckle shape is the eigenvector: n full waves around the tube, and m, the axial half-wave count that carries the largest share of w.

The solution has no closed form. For an isotropic wall the SP-8007 equations, σ_cr = 0.605Et/r̄ in axial compression and p_cr = 3D/r̄³ for a long tube under external pressure, are asymptotes of it², so far from their range the Ritz value is the more accurate one. The energy assumes B = 0. An unsymmetric wall gets the same checks on its A and D, and the tool marks them approximate. The ply check does take B.

External pressure

The check runs when the net external pressure, in MPa, is positive:

The knock-down factor is γ_p = 0.75 at every length, the factor SP-8007 puts on the lateral pressure of orthotropic cylinders². The safety factor is SF_p = γ_pp_cr/p.

Axial compression and the bending compression side

When M_b ≠ 0 the axial resultant differs between the two sides of the section. The check takes the peak compressive membrane stress over both sides, in MPa, and runs when it is positive:

One check covers a compressive F, the closed-end axial load of external pressure and the compression side of bending, because SP-8007 treats the peak bending stress of a cylinder by the axial compression procedure². The safety factor is SF_c = γ_cσ_cr/σ_c, with the knock-down factor γ_c below. Internal pressure on a closed end puts the wall in axial tension, so the check runs only when the other loads overcome it. Internal pressure alone never triggers any of the three checks.

Torsion

The check runs when T ≠ 0. The knock-down factor is γ_t = 0.67, the factor SP-8007 puts on the torsion load of moderately long orthotropic cylinders². The safety factor is SF_t = γ_tT_cr/|T|. T_cr is solved for the sign of the applied torque, because the coupling terms of angle plies can make a tube buckle at a different torque in each direction.

Combined loads

When two or more checks run, the loads act together. SP-8007 carries the interaction equations of the isotropic cylinder over to orthotropic ones: the load ratios add up to one at buckling².

Each ratio is the applied value over the design value of its own check, so R = 1/SF for that check. Scaling every load by the same factor scales every ratio by it, so the combined safety factor over the checks that ran is

With one check it equals that check's SF. With two or more it is below every single one. SP-8007 notes that the straight-line sum is conservative for compression with torsion². SF_comb is the buckling safety factor of the tube, and the load with the largest ratio governs.

Why the knock-down factor is not optional

A knock-down factor scales the classical buckling load of a perfect shell down to what real cylinders reach in test. The classical value is a bifurcation load, and initial geometric imperfections make thin cylinders buckle well below it, so SP-8007 multiplies it by an empirical factor drawn from test data². For axial compression the factor falls as the wall gets thinner relative to its radius. For a layered orthotropic wall SP-8007 keeps the form of the isotropic factor and replaces the thickness with the geometric mean of the axial and hoop radii of gyration²:

SP-8007 prints φ with the constant 29.8, which is 16 · 12^1/4 = 29.78 rounded. The analysis keeps the unrounded form above, so an isotropic wall, where D/A = t²/12 and t_eq = t, gets the isotropic factor exactly. At r̄/t_eq = 50, φ = √50/16 = 0.442 and γ_c = 0.678. Double the ratio to 100 and γ_c drops to 0.581.

The pressure and torsion factors are the √γ = 0.75 and γ^3/4 = 0.67 of the isotropic equations, with the correlation factor γ = 0.5625 for pressure². For very long isotropic cylinders SP-8007 allows more, 0.90 under external pressure and 0.80 in torsion². The analysis keeps one factor per check at every length. That is conservative on a long tube, and it keeps the design pressure continuous as the buckle shape moves to n = 2 waves.

The factors come from tests on moderately long cylinders. SP-8007 states them above a value of γZ, where Z is the curvature parameter: 2.85 in axial compression, 100 under external pressure and 50 in torsion². For a laminate the analysis takes Z on the equivalent thickness:

A shorter tube buckles more like a curved plate. It is less sensitive to imperfections and carries more than γ times its classical value, so its design value is conservative, and the tool says so. The classical value holds at every length.

For pure bending SP-8007 gives a separate, higher factor, 1 − 0.731(1 − e^−φ)². Using the axial compression factor on the bending compression side as well is conservative.

What the model leaves out

• Away from the ends. No domed head, end closure, head junction or support is modelled. Near them the wall bends, and the stresses differ from the membrane result.
• No radial stress. The pressures on the inner and outer surface are not applied to the plies as a through-thickness stress.
• Bending moment only. The model takes no transverse shear force from bending.
• No moisture. Only the temperature change enters the hygrothermal terms.
• Linear elastic plies, small strains. Thick walls are out of scope.
• Buckling. Simply supported ends and B = 0. The prebuckling state is membrane only, so internal pressure does not raise the axial buckling load, which is conservative². A wall whose A or D matrix is not positive definite has no buckling result.

A tube treated as a beam, with deflection under a transverse load and column buckling, is covered by the beam reference. The other structural tools are listed on the structural analysis overview.

Author: Rick Schrijver

References

1. Jones, R.M. (1999). Mechanics of Composite Materials, 2nd ed. Taylor & Francis. Chapter 4. ISBN: 978-1560327127. (Eq. 4.13 for the strain through the thickness, Eq. 4.22-4.24 for the laminate stiffnesses, Eq. 4.103-4.107 for the thermal stresses, the thermal force and moment resultants and the laminate law with thermal terms)
2. NASA Langley Research Center. (2020). Buckling of Thin-Walled Circular Cylinders. NASA/SP-8007-2020/REV 2, the second revision of NASA SP-8007 (1965, revised 1968). (Section 4.1.1: Eq. 4 for the curvature parameter, Eq. 7-10 for axial compression and its knock-down factor, Eq. 13-14 for bending, Eq. 17-18 for the pressure membrane stresses, Eq. 23-29 for external pressure and its correlation factors, Eq. 33-38 for torsion, Eq. 41-47 for combined loads; Section 4.1.2: Eq. 52-59 for orthotropic axial compression, Eq. 60-63 for the knock-down factors, Eq. 64-66 and the 0.75 factor for lateral pressure, Eq. 69 and the 0.67 factor for torsion, Section 4.1.2.5 for combined loads) ntrs.nasa.gov
3. Sanders, J.L., Jr. (1963). Nonlinear theories for thin shells. Quarterly of Applied Mathematics, 21, 21-36. (Strain-displacement relations and rotations of the shell; the cylindrical form is also in Brush, D.O. and Almroth, B.O. (1975), Buckling of Bars, Plates, and Shells, McGraw-Hill, Chapter 5)

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