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Calculations with context

Thin-wall cylinder: internal pressure

Calculate hoop, axial and equivalent membrane stresses from inner diameter, end condition and nonnegative internal pressure differential within a strict thin-wall scope.

STRUCTURAL MECHANICS · 1.0.2

Thin-wall cylinder under internal pressure

Derive hoop and axial membrane stresses from two cuts through a cylinder. Compare a closed shell carrying its end thrust with a wall carrying no axial end load.

Internal gauge differential pressure only, INNER diameter, uniform thin wall, far from discontinuities. Thick-wall, external-pressure and buckling problems are rejected or excluded.

Define the model

Use a decimal point or comma; no thousands separators. Changing a unit reinterprets the displayed number, so enter the equivalent value when converting units.

Measured on the fluid side. Outer or mean diameter must not be entered as D.
Uniform actual wall thickness for this ideal model; no corrosion or weld allowance is inferred.
p = p_inside − p_outside ≥ 0. Use consistent pressure references. Negative differential pressure is unsupported.

Calculated response

The static lesson and example are available below. Interactive controls are enabled when the module loads.

Runs locally in your browser. Inputs are not uploaded or stored. CSV contains the current inputs, SI conversions, equations, assumptions, sources and results.

From the free body to the equation

A pressure is a force per projected area. Cut a unit-length cylinder along its axis and keep one half. Pressure pushes on a projected rectangle of width D. Two cut wall strips resist with 2tσh, so 2tσh = pD. This gives the hoop membrane stress σh = pD/(2t). The length cancels.

For a closed end whose thrust is carried by the cylindrical shell, pressure acts over πD²/4. In this inner-diameter thin-wall approximation the resisting wall area is πDt. Equating the two gives σz = pD/(4t), half the hoop stress. The exact annular area differs; this calculator deliberately keeps the stated first-order approximation.

An open or independently restrained end does not automatically guarantee zero axial stress. The zero-axial mode requires that no end pressure thrust, support restraint, thermal restraint or other axial load enters the wall. If a separate frame carries a sealed end’s pressure thrust, that frame must carry pπD²/4; the tool reports this equivalent diverted thrust. For a truly uncapped pipe, no attached cap exists; the displayed pA is a reference capped-end thrust, not a calculated anchor load.

The model uses the inner radius ri = D/2 throughout. It requires t/ri < 0.1 as a teaching-model gate; thinning the wall increases membrane stress and does not prove adequacy. Inner, mean and outer diameter conventions must not be interchanged in formulas, especially near the edge of a thin-wall approximation.

Tensile membrane stress is positive. Radial stress is compressive at the inner wall and follows the pressure boundary condition; with an exterior gauge reference its surface values are −p and 0. Plane stress neglects this smaller component. The displayed t/ri quantifies |σr,inner|/σh when p > 0; it does not assert that radial stress is literally absent.

The optional-equivalent interpretation is purely algebraic: for principal stresses (σh, σz, 0), σeq = √[((σh−σz)² + σz² + σh²)/2]. There is no in-plane shear here. This scalar is not a stress direction, material strength, allowable pressure or verdict. No material database, yield strength or factor of safety is invented.

Symbols and SI units
QuantityValueUnit
pinternal minus external gauge pressurePa
D; riINNER diameter; INNER radius D/2m
t; ℓwall thickness; arbitrary cut lengthm
σh; σzhoop; longitudinal membrane stress, tension positivePa
σrradial normal stressPa
σeqprincipal plane-stress equivalent stressPa
Fendpressure force pπD²/4N
t/rirelative wall thickness; radial/hoop magnitude ratio for p>01

Worked maritime-context example

Synthetic training case: a straight section of a shipboard compressed-air receiver is idealised far from its heads, with inner D = 800 mm, t = 8 mm and internal differential p = 0.6 MPa. Closed ends transfer pressure thrust into the shell. This is not an actual receiver specification or an operating-pressure recommendation.

Calculated response

Closed ends · shell carries the end thrust

Calculated response
QuantityValueUnit
Hoop membrane stress σh30MPa
Axial membrane stress σz15MPa
Plane-stress equivalent σeq25.9807621MPa
Inner radial boundary stress (gauge)-0.6MPa
Outer radial boundary stress (gauge)0MPa
ri/t501
t/ri0.021
|σr,inner| / σh0.021
Pressure end thrust301.592895kN
End thrust carried by wall301.592895kN
Equivalent end thrust outside wall0kN

Displayed results are rounded. Validation and CSV retain the full converted floating-point precision.

σeq assumes principal plane stress (σh, σz, 0) and zero shear. Radial stress is neglected. No strength or safety comparison is made.

Cylinder free bodies: hoop cut and end thrustLeft: pressure resultant p D ell on a longitudinal half-cylinder is balanced by two wall forces sigma_h t ell. Right: axial pressure resultant p pi D squared over four is carried by the approximate annular wall area pi D t for closed ends, or diverted away from the wall in the zero-axial mode.p D ℓσh t ℓσh t ℓD (inner)2 σh t ℓ = p D ℓp πD²/4σz πDt = p πD²/4σz πDt
Force arrows show the free-body balances, not wall thickness or deformation to scale. D is inner diameter. The axial resisting area πDt is the stated thin-wall approximation.

Active analytic equations

  • ri=D/2; t/ri<0.1
  • 2 sigma_h t ell=p D ell => sigma_h=pD/(2t)
  • closed: sigma_z (pi D t)=p(pi D^2/4) => sigma_z=pD/(4t)
  • open / externally carried end thrust: sigma_z=0
  • sigma_eq=sqrt(((sigma_h-sigma_z)^2+sigma_z^2+sigma_h^2)/2)
  • sigma_r(inner)=-p; sigma_r(outer)=0 relative to exterior gauge reference; plane-stress model takes sigma_r approximately 0

Substituted calculation

  1. D = 800 × 0.001 = 0.8 m; t = 8 × 0.001 = 0.008 m
  2. p = 0.6 × 1,000,000 = 600,000 Pa
  3. ri = D/2 ≈ 0.4 m; t/ri ≈ 0.008/0.4 ≈ 0.02
  4. The geometric gate uses unrounded converted inputs. t < D/20: 0.008 m < 0.04 m.
  5. σh = 600,000 × 0.8 / (2 × 0.008) = 30 MPa
  6. σz = 600,000 × 0.8 / (4 × 0.008) = 15 MPa
  7. σeq = √[30² − (30 × 15) + 15²] = 25.9807621 MPa
  8. Fend = 600,000 × π × 0.8² / 4 = 301.592895 kN
Equilibrium and compatibility residuals
QuantityValueUnit
2*t*sigma_h-p*D0N/m
sigma_z*pi*D*t+external_thrust-p*pi*D^2/40N

Residuals compare the pressure resultants with the membrane-force resultants. They verify the chosen approximation’s equilibrium, not vessel adequacy.

Current input snapshot
Current input snapshot
QuantityValue
Pressure modeInternal differential pressure
Diameter conventionInner diameter
Where does pressure end thrust go?Closed ends · shell carries the end thrust
Inner diameter D800
Inner diameter D · Unitmm
Wall thickness t8
Wall thickness t · Unitmm
Internal gauge differential pressure p0.6
Internal gauge differential pressure p · UnitMPa

Assumptions and boundaries

  • Uniform circular membrane region, away from ends, nozzles, changes of thickness, weld details and supports. End-cap geometry and local stresses are not solved.
  • Only internal p ≥ 0, and ri/t > 10. The strict geometric gate is a simplified educational applicability convention, not an acceptance or safety boundary.
  • The gauge differential model neglects a common ambient hydrostatic compression. High external confining pressure, subsea pressure hulls, external-pressure collapse and vacuum service require another model.
  • Hoop and axial values are first-order membrane estimates. Through-thickness variation and radial stress are neglected in the equivalent-stress calculation. It is not a thick-cylinder or full three-dimensional stress result.
  • No yield comparison, allowable stress, corrosion or weld efficiency, fatigue, creep, temperature dependence, stability or pressure-vessel code certification.
  • Numerical implementation bounds: SI D 10⁻⁶–10⁴ m, t 10⁻⁹–10² m, p zero or 10⁻¹²–10¹⁰ Pa, hoop stress ≤10¹⁶ Pa. These broad bounds are not physical operating limits.

Teaching calculation only. No capacity, safety or regulatory approval is implied.

Check your understanding

Why is the closed-end hoop stress twice the axial stress?

The two free bodies have different projected pressure areas and resisting wall areas. Hoop equilibrium uses two lengthwise strips; axial equilibrium uses an approximately circular wall strip.

Why does the model reject a negative p?

Negative internal-minus-external pressure is an external-pressure problem. A simple sign reversal would omit instability and collapse, so this model does not apply the internal-pressure formulas.

Can I enter the outside diameter?

No. Convert a measured outer diameter to inner diameter D = Do − 2t first and use consistent units. This tool fixes the inner-radius convention for every equation.

Is σeq a safe operating limit?

No. It is a derived scalar for the assumed plane-stress state. No material condition, allowable strength, code rule or safety factor is supplied.

Sources and curriculum connection

Curriculum connection: the cylinder is a bounded application of the combined-stress and stress-analysis topics in İTÜ MUK204. The course page does not claim this specific calculator; the cylinder equations are supported separately by the primary mechanics references.

Primary source content checked 9 October 2026. Historical mechanics publications support the equations; they are not current design codes.

Related context

The method explanation and worked example are on this page. The articles below provide additional context.

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