CivilSolve

Statics & Mechanics of Materials

Thin-Walled Pressure Vessel

Thin-wall stresses in a pressurized cylinder (hoop σh = p·r/t, longitudinal σl = p·r/2t) or sphere (σ = p·r/2t) — or, given an allowable stress instead of a thickness, the minimum wall thickness. Warns when r/t < 10 (thin-wall theory becomes unreliable). Use for: 'find the hoop stress in the tank', boiler/pipe wall stress, 'what wall thickness is required'. Not for: thick-walled cylinders (Lamé) or external pressure buckling.

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Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

SI cylinder: p = 1.2 MPa, D = 1 m, t = 10 mm

Given

internal pressure
1.2 MPa
diameter
1 m
wall thickness
10 mm

Assumptions

  • Thin-wall membrane theory: uniform stress through the wall, valid for r/t of about ten or more; p is gage pressure.

Solution steps

  1. Hoop and longitudinal stresses

    The hoop direction carries twice the longitudinal stress — cylinders split along their length, not around it.

    σh = p·r/t; σl = p·r/(2t)

    r/t = 50; σh = 60 MPa; σl = 30 MPa

    = 60 MPa

Results

Hoop (circumferential) stress σh

60MPa

Longitudinal (axial) stress σl

30MPa

r/t ratio

50

Where this answer was checked
source
σh = p·r/t, σl = p·r/2t (NCEES FE Reference Handbook) — hoop and longitudinal stresses
verified by
hand-recomputed
derivation
r = 0.5 m. σh = 1.2e6·0.5/0.010 = 60 MPa; σl = 30 MPa; r/t = 50.

Example 2

US cylinder: p = 200 psi, r = 18 in, t = 0.5 in

Given

internal pressure
200 psi
radius
18 in
wall thickness
0.5 in

Assumptions

  • Thin-wall membrane theory: uniform stress through the wall, valid for r/t of about ten or more; p is gage pressure.

Solution steps

  1. Hoop and longitudinal stresses

    The hoop direction carries twice the longitudinal stress — cylinders split along their length, not around it.

    σh = p·r/t; σl = p·r/(2t)

    r/t = 36; σh = 7.2 ksi; σl = 3.6 ksi

    = 7.2 ksi

Results

Hoop (circumferential) stress σh

7.2ksi

Longitudinal (axial) stress σl

3.6ksi

r/t ratio

36

Where this answer was checked
source
σh = p·r/t in US units — tank wall stress
verified by
hand-recomputed
derivation
σh = 200·18/0.5 = 7200 psi = 7.2 ksi; σl = 3.6 ksi.

Example 3

size the wall: p = 0.8 MPa, r = 0.6 m, allowable 80 MPa

Given

internal pressure
0.8 MPa
radius
0.6 m
allowable stress
80 MPa

Assumptions

  • Thin-wall membrane theory: uniform stress through the wall, valid for r/t of about ten or more; p is gage pressure.

Solution steps

  1. Required wall thickness

    The hoop stress governs a cylinder, so size the wall from σh = p·r/t.

    t = p·r/σ_allow

    t = 6 mm (r/t = 100)

    = 6 mm

Results

Minimum wall thickness t

0.006m

Resulting r/t ratio

100

Notes

  • This is the theoretical minimum — real designs round UP to available plate and apply code factors (joint efficiency, corrosion allowance).
Where this answer was checked
source
t = p·r/σ_allow (hoop governs) — required thickness
verified by
hand-recomputed
derivation
t = 0.8e6·0.6/80e6 = 6 mm; r/t = 100.