In a thin-walled cylinder under internal pressure P, the hoop stress is P times the inner radius r divided by the wall thickness t, and the longitudinal stress is exactly half of that, which is why such a cylinder splits along its length. The formula gives only the membrane stress in a perfect seamless tube, so the thickness it returns is a lower bound that welds, openings, corrosion, vacuum and fatigue all push upward, and the real design thickness must come from a pressure-vessel code.
Part of Pressure Vessel Theory
The statics behind the two wall stresses in a pressurised cylinder, what the simple formula leaves out, and why it is used to understand a vessel rather than to size one.
Both Wall Stresses Fall Out Of One Force Balance
Cut a cylinder of inner radius r, wall thickness t and length L in half along a plane through its axis. The pressure P presses on the inside of each half. The part of that force that tries to separate the halves acts over the projected area of the cut, which is a rectangle 2r wide and L long, so it is P × 2r × L. The wall resists with two strips, one on each side of the cut, each t × L in cross-section, carrying a stress σ_h. Balance the forces:
σ_h × 2 × t × L = P × 2r × L
so σ_h = P × r / t (hoop, or circumferential, stress).
Now cut the cylinder across, near the middle. The pressure on the end cap pushes with force P × π r². The wall ring that resists it has a cross-section of about 2π r t (circumference times thickness). Balance:
σ_l × 2π r t = P × π r²
so σ_l = P × r / (2t) (longitudinal, or axial, stress).
The hoop stress is therefore twice the longitudinal stress at every point of a thin-walled cylinder. A crack runs where the stress is highest, and it runs perpendicular to that stress, so it opens along the length of the shell. This matches the carrier article’s statement that cylinders fail along their length first.
A sphere or a hemispherical end cap balances against P × π r² over a circumference of 2π r, giving σ = P × r / (2t): half the cylinder’s hoop stress for the same thickness. That is the whole reason a domed head can be thinner than the shell it closes, and why a flat head cannot (see End Caps (Heads)).
Units And Validity
Use consistent units. With P in megapascals (1 bar = 0.1 MPa), r and t in millimetres, the stress comes out in MPa, because MPa is the same as N/mm².
The derivation assumes a thin wall, commonly stated as a radius at least ten times the thickness (recalled from textbook statics; flagged in the verification list). The assumption also ignores everything that is not the plain cylinder: the seam, any opening, any change of section and any load other than internal pressure.
Worked Example: The Carrier’s Number, Checked And Put In Its Place
The Autoclave Construction article works one example: 2 bar gauge, inner radius 200 mm, a safety factor of 4 and a steel strength of 250 MPa. Its arithmetic is correct:
- Convert: 2 bar = 0.2 MPa.
- Required thickness: t = P × r × SF / σ = 0.2 × 200 × 4 / 250 = 160 / 250 = 0.64 mm.
- Check the thin-wall assumption: r / t = 200 / 0.64, about 310, comfortably above ten.
The 250 MPa is the carrier’s assumed strength for mild steel. It is not a property of any particular plate, and an unknown scavenged plate has no known strength at all.
The carrier then specifies a wall many times thicker and says why: welding allowance and corrosion margin. Run the same formula backwards at the carrier’s chosen wall of 8 mm, which is the carrier’s figure and not one to copy, and the membrane hoop stress is:
σ_h = 0.2 × 200 / 8 = 5 MPa, which is 5 / 250 = 2% of the assumed strength.
So at the carrier’s own wall, the pressure term in the formula uses up a small fraction of the steel. What the extra thickness actually pays for are the things the formula cannot see:
- The weld. A seam does not carry the load as well as plate. Pressure-vessel codes handle this with a joint-efficiency factor below one applied to the weld, and a seam that has been radiographed or tested earns a better factor than one that has not (the principle is recalled; the values belong to the code).
- Corrosion. Stress scales with 1/t, so halving the remaining wall doubles the stress: 5 MPa becomes 10 MPa, and local pitting thins the wall far more than average loss does.
- Openings. Every hole for a gauge, valve, drain or door concentrates stress at its edge to several times the nominal value, which is why the code requires reinforcement around openings and why an ill-placed hole is a design error.
- External pressure. As an autoclave cools, steam condenses and the inside pressure can fall below the outside. A thin shell that is strong against internal pressure can collapse (buckle) under a vacuum that is far smaller, and that is a separate calculation with a much harsher dependence on thickness (recalled principle, not computed here). A vacuum breaker is part of the design.
- Fatigue. Each pressure cycle flexes the wall and welds slightly; the vessel is cycled every day it is used.
- Handling and shape. A very thin shell cannot be rolled round, welded or lifted without distorting, and distortion concentrates stress.
What The Formula Is For
- Scaling. Stress is proportional to P and to r, and inversely proportional to t. A 10% rise in operating or relief pressure raises wall stress by 10%; doubling the diameter at the same thickness doubles it. Anyone proposing to raise the set pressure of an existing vessel has changed its stress by exactly that proportion and needs the rating changed by the code, not by wishing.
- Comparing shapes. It shows why a sphere or dome beats a flat plate, and why a long thin cylinder is not made safer by shortening it (length does not appear).
- Judging an inspection finding. A measured wall loss translates directly into a stress rise.
It is not for choosing the thickness of a vessel that will hold steam beside people. The safety factors the corpus quotes disagree: the carrier asks for at least 4 for amateur welding and 6 if the steel or welding is uncertain, while Safety Factors tabulates 5-8 for boilers and pressure vessels. They cannot both be the right single answer. The safer reading is the higher figure, and modern codes do not use a single multiplier on strength at all: they set an allowable stress per material and add joint efficiency, corrosion allowance and testing rules.
Common Mistakes
- Using the result as the thickness. The 0.64 mm above would not survive being rolled. It is a floor in an equation.
- Mixing units. Putting pressure in bar and strength in MPa with no conversion gives an answer wrong by a factor of ten.
- Forgetting the weld. Treating a seamed shell as seamless plate overstates what it can carry.
- Ignoring vacuum. The cooling phase is where a poorly vented vessel can crush inward.
Safety
A calculation does not make an unsourced vessel safe
A wall thickness that satisfies this formula has not been shown safe: it has been shown not to fail in the one way the formula models. Do not build or operate a pressure vessel from this formula. Take the design from a recognised pressure-vessel code or copy an existing certified vessel, prove it with water, and keep a relief device that the operator cannot adjust or block.
What’s Next
- Learn next: Failure Modes — what the corrosion, fatigue and flaw terms look like when they win
- Combine with: End Caps (Heads) — the hemisphere result applied to real head shapes