What is the formula for mawp and required thickness for a cylindrical shell?

Circumferential stress (longitudinal seam): P = S·E·t / (R + 0.6t); Longitudinal stress (circumferential seam): P = 2·S·E·t / (R − 0.4t); Sphere or hemispherical head: P = 2·S·E·t / (R + 0.2t); Required thickness from OD: t = P·D_o / (2·(S·E + 0.4P)); t_c = t·(1 − undertolerance) − CA ; ordering wall = (t_req + CA)/(1 − undertolerance); R = inside radius in the corroded condition = D_o/2 − t_c; Validity: cylinder t ≤ 0.5R and P ≤ 0.385·S·E ; sphere t ≤ 0.356R and P ≤ 0.665·S·E. Calculates the maximum allowable working pressure and the required thickness of a cylindrical shell or sphere from ASME VIII-1 UG-27, with the thin-wall validity checks.

What should I check before relying on the result?

Use the measured thickness with undertolerance at 0, or the nominal thickness with the specification undertolerance (12.5 % on most seamless pipe) - never nominal with 0. Allowable stress must be taken at the design temperature. For carbon steel it falls steeply above about 350 degC. Joint efficiency of 1.00 requires a fully radiographed double-welded butt joint. Do not assume it. This covers the shell only. Nozzles, heads, cones and reinforcement are separate calculations and often govern. For an in-service assessment against measured metal loss, use the API 579 routes rather than a bare UG-27 calculation. Verify against the governing edition: ASME BPVC Section VIII Division 1, UG-27(c)(1) and (c)(2) and UG-27(d); joint efficiency per UW-12.

MAWP and required thickness for a cylindrical shell

A thin-walled cylinder under internal pressure carries two membrane stresses. The circumferential (hoop) stress is roughly PR/t and is resisted by the longitudinal weld seam. The longitudinal stress is roughly PR/2t — half as large — and is resisted by the circumferential seam. Because the hoop stress is twice the longitudinal, the longitudinal seam governs almost every cylindrical shell, which is why a longitudinal weld is normally radiographed to a higher standard than a girth weld and why the joint efficiency E in the hoop equation matters so much.

ASME VIII-1 UG-27 writes the two cases as P = S·E·t/(R + 0.6t) for circumferential stress and P = 2·S·E·t/(R − 0.4t) for longitudinal stress, with R the inside radius in the corroded condition. The 0.6t and 0.4t terms are the Lamé correction that keeps the thin-wall formula honest as the wall gets thicker; they are not arbitrary. For a sphere or a hemispherical head the equation is P = 2·S·E·t/(R + 0.2t), which is why a sphere needs roughly half the wall of a cylinder of the same diameter and pressure.

Two validity limits apply. The UG-27 equations are only valid while t ≤ 0.5R and P ≤ 0.385·S·E for the circumferential case; beyond that the through-wall stress gradient is too steep for the thin-wall approximation and Appendix 1-2 thick-wall equations are required. Every ultrasonic thickness survey should be checked against these limits before the numbers are used.

Getting the inputs right matters more than the arithmetic. Thickness must be the measured thickness less any corrosion allowance you intend to retain, and less mill undertolerance on new construction (12.5 % on seamless pipe). Allowable stress S is from the code stress tables at the design temperature, not at ambient, and drops steeply above about 350 °C for carbon steel. Joint efficiency E comes from UW-12 and depends on the joint type and the extent of radiography — it is 1.0 only for a full-radiographed double-welded butt joint. Using E = 1.0 by default on a spot-radiographed seam over-states the MAWP by 15 %.

Worked example

Outside diameter	508 mm
Measured or nominal thickness	12.7 mm
Corrosion allowance to retain	1.5 mm
Allowable stress at design temperature	118 MPa
Mill undertolerance	0 %
Component	cylinder
Design pressure	5 MPa
Weld joint efficiency	1
Thickness less undertolerance and corrosion allowance	11.2 mm
Inside radius (corroded)	242.8 mm
MAWP - circumferential stress	5.297 MPa
MAWP - longitudinal stress	11.091 MPa
MAWP	5.297 MPa
Required thickness for the design pressure	10.583 mm
Nominal thickness to order	12.083 mm

Measured wall, so undertolerance = 0 and t_c = 12.7 x 1.0 - 1.5 = 11.2 mm; R = 254 - 11.2 = 242.8 mm. Circumferential: P = 118 x 1.0 x 11.2/(242.8 + 0.6x11.2) = 1321.6/249.52 = 5.2966 -> 5.297 MPa. Longitudinal: P = 2 x 118 x 11.2/(242.8 - 0.4x11.2) = 2643.2/238.32 = 11.0910 -> 11.091 MPa. The circumferential case governs, as it always does on a cylinder, so MAWP = 5.297 MPa. Required thickness for 5.0 MPa from the OD form: t = 5.0 x 508/(2(118 + 0.4x5.0)) = 2540/240 = 10.5833 -> 10.583 mm, or 12.083 mm with the 1.5 mm corrosion allowance. Validity: t_c/R = 0.046, well under 0.5, and 0.385 S E = 45.4 MPa, well above the MAWP.

Use at your own risk — verify before you act

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