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ASME B31.1 Pipe Bend Wall Thickness Calculator

Returns the minimum wall a finished pipe bend needs at the intrados, the extrados and the sidewall under ASME B31.1-2026 ¶104.2.1(c), eqs. (7), (9) and (10) — ¶102.4.5(b), eqs. (3), (5) and (6) in B31.1-2024. The intrados governs: it needs more wall than straight pipe.

Method last updated (calculation changelog) · fixture-verified on every build — most recently 2026-11-12.

Built and fixture-verified by Matthew Norris, P.E. — active P.E. licensure in Arizona, California, Kansas, Missouri, North Carolina, Texas.

A bend is a piece of a torus, and pressure does not load a torus evenly. The inside of the bend carries more hoop stress than the straight pipe it was made from and the outside carries less, which is the opposite of what bending does to the metal: the intrados thickens and the extrados thins. B31.1 handles both facts with one factor, I, applied to the allowable stress in the straight-pipe equation. This calculator returns I and the minimum wall tm at the three stations the Code names, so you can write the after-bending thickness requirements on a bend data sheet for a main steam, hot reheat or feedwater bend, or check a fabricator's thickness survey against them. S, E, W and y are typed inputs from your governing edition.

Pipe bend with intrados, extrados and centerline A ninety-degree pipe bend showing the bend radius R to the centerline, the intrados on the inside of the bend and the extrados on the outside. R extrados — I < 1 outside of the bend, stretched thin intrados — I > 1 inside of the bend, governs t_m sidewall on the centerline — I = 1.0 t_m = P·D_o / 2(S·E·W / I + P·y) + A
A pipe bend in its finished form: the intrados (inside of the bend) carries I above 1 and needs more wall than straight pipe, the extrados (outside) carries I below 1 and may be thinner, and the sidewall on the bend centerline keeps I = 1.0. R is the bend radius.
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Method

Minimum required thickness of a bend, after bending, in its finished form — B31.1-2026 ¶104.2.1(c), eq. (7); ¶102.4.5(b), eq. (3) in B31.1-2024:

tm = P·Do / ( 2 (S·E·W / I + P·y) ) + A

with the factor I set by position on the cross-section — eqs. (9) and (10) in 2026, eqs. (5) and (6) in 2024:

intrados: I = ( 4(R/Do) − 1 ) / ( 4(R/Do) − 2 )

extrados: I = ( 4(R/Do) + 1 ) / ( 4(R/Do) + 2 )

sidewall on the bend centerline: I = 1.0

R is the bend radius to the pipe centerline. Dividing the allowable by I is what makes the intrados wall thicker and the extrados wall thinner than straight pipe; at I = 1.0 the equation is exactly ¶104.1.2 eq. (3), so the sidewall result doubles as the straight-pipe minimum for the tangents. The tighter the bend, the further the two factors move from 1: a long-radius induction bend barely notices, a short-radius bend notices a great deal. W is the weld strength reduction factor of ¶102.4.7; ¶104.2.1(c) sets it to 1 for seamless pipe, for seam-welded pipe below the creep range, and for the parts of a bend that do not contain a weld. The thickness requirements apply at mid-arc of the bend and at all three stations, with gradual transitions between them, and the ends of the bend must still meet ¶104.1.2 for straight pipe. The equations moved in the 2026 edition; the arithmetic did not.

Inputs
PInternal design gauge pressurepsi
DoPipe outside diameterin
RBend radius to the pipe centerlinein
SMaximum allowable stress (Mandatory Appendix A — user-supplied; the tabulated value includes E)psi
EJoint efficiency factor (1.0 when the Appendix A value is entered as S)—
WWeld strength reduction factor (¶102.4.7 — user-supplied, default 1.0)—
yCoefficient (Table 104.1.2-1 — user-supplied)—
AAdditional thicknessin
Outputs
IFactor at intrados and at extrados—
tm intradosMinimum wall at the inside of the bend (governs)in
tm extradosMinimum wall at the outside of the bendin
tm sidewallMinimum wall on the bend centerline (I = 1.0, straight-pipe value)in

Limitations — what this calculator is not

Worked example — fixture-verified

Main steam lead to the turbine: NPS 12 (12.75 in OD) seamless 9Cr–1Mo–V pipe, 2,500 psig at 1,000°F, with a 5D induction bend: centerline radius R = 5 × 12.75 = 63.75 in. ASSUMED inputs for this example: SE = 14,000 psi, y = 0.7, W = 1.0 (seamless), A = 0 (welded ends, steam service). Read SE and y for your own material and temperature from Mandatory Appendix A and Table 104.1.2-1.

Given
Design pressure P2,500psig
Outside diameter Do12.75in
Bend radius R (5D)63.75in
Allowable SE (ASSUMED)14,000psi
Weld strength reduction factor W1.0—
Coefficient y (ASSUMED)0.7—
Additional thickness A0in

Step by step

  1. Radius ratio: R/Do = 63.75 / 12.75 = 5, so 4(R/Do) = 20.
  2. Intrados factor, eq. (9): I = (20 − 1) / (20 − 2) = 19 / 18 = 1.05556.
  3. Extrados factor, eq. (10): I = (20 + 1) / (20 + 2) = 21 / 22 = 0.95455.
  4. Common terms: P·Do = 2,500 × 12.75 = 31,875 lb/in; P·y = 2,500 × 0.7 = 1,750 psi.
  5. Intrados, eq. (7): tm = 31,875 / (2 × (14,000 / 1.05556 + 1,750)) = 31,875 / (2 × 15,013.2) = 31,875 / 30,026.3 = 1.06157 in.
  6. Extrados: tm = 31,875 / (2 × (14,000 / 0.95455 + 1,750)) = 31,875 / (2 × 16,416.7) = 31,875 / 32,833.3 = 0.97081 in.
  7. Sidewall, I = 1.0: tm = 31,875 / (2 × 15,750) = 31,875 / 31,500 = 1.01190 in — the straight-pipe minimum for the tangents.
Result COMPUTED
tm at intrados — governs1.06157in
tm at extrados0.97081in
tm at sidewall (straight-pipe value)1.01190in
I at intrados / extrados1.05556 / 0.95455—

On a 5D bend the intrados needs 4.9% more wall than straight pipe and the extrados may be 4.1% thinner. Grossing the straight-pipe minimum up for a 12.5% mill tolerance gives 1.156 in, so Sch 160 (1.312 in nominal, 1.148 in at the bottom of the tolerance) is the first standard wall that covers the tangents. On that wall the extrados can lose about 15% in bending before it reaches 0.97081 in. Compare that margin with the thinning the bender's procedure predicts before the pipe is ordered, not after the thickness survey comes back.

Why you can trust these numbers: this exact case is fixture bend-thickness.json — case “B31.1 ¶104.2.1(c) — main steam NPS 12 seamless 5D bend, R=63.75 in, P=2500, SE=14000, W=1, y=0.7, A=0” (tolerance 0.0001) — in the calc-core release gate. It re-runs on every commit; a red fixture blocks deployment. See the validation methodology.

Sources & citations

Per the source & citation policy, allowable-stress and factor table values are inputs — typed from your governing edition, or autofilled on paid tiers from the embedded datasets enumerated there, always cited by standard, edition, table and line. Where a page does reproduce specific ASME data (the B16.5 ratings, the quick-reference tables), it states the source table and conditions inline.

FAQ

Why does the inside of a bend need more wall than straight pipe?

Because of the shape. A straight pipe under pressure is a cylinder, and every strip of wall carries the same hoop load. A bend is a segment of a torus. On the inside of the bend a short strip of wall has to hold back the pressure acting on a wider wedge of bore, and on the outside a long strip holds back a narrower wedge, so the hoop stress rises at the intrados and falls at the extrados. The factor I in ¶104.2.1(c) is that stress ratio written in terms of R/Do, and dividing the allowable by it puts the thick wall where the stress is. Bending happens to move metal the helpful way, thickening the intrados and thinning the extrados, which is why a well-made bend from adequately heavy pipe usually satisfies both requirements at once. The trouble comes with tight radii, where the intrados factor climbs fast: at R/Do = 1.5 the intrados factor is 1.25, and no bending process hands you a quarter more wall for free.

What changed for bends between B31.1-2024 and B31.1-2026?

The location and the ordering guidance, not the equations. In B31.1-2024 the bend equations sit in ¶102.4.5(b) as eqs. (3) through (6), in the chapter on allowances, and ¶102.4.5(a) offers the alternative of holding the straight-pipe minimum of ¶104.1.2(a) at every point of the completed bend, with a table of suggested pre-bend walls as a guide for ordering. In B31.1-2026 the equations live with the rest of the pressure design rules in ¶104.2.1(c) as eqs. (7) through (10). ¶102.4.5 is now a short pointer to ¶104.2.1(a) plus the statement of where the thickness requirements apply, and for ordering wall it refers the designer to PFI ES-24. Either edition gives the same required walls from the same inputs, so a bend calculated to one is not invalidated by the other. What a reviewer will check is that the paragraph and equation numbers on the calculation match the edition in the design basis, which is why this page gives both.

Do I have to use the I-factor equation at all?

No. ¶104.2.1(a) of the 2026 edition says the minimum wall of a bend shall meet the requirements of (c) or of ¶104.1.2(a), and in both cases the fabrication requirements of ¶129.1. So there are two acceptable ways to qualify a bend for pressure. The first holds the straight-pipe minimum everywhere on the finished bend. It is simple to specify and simple to inspect, and it wastes nothing on gentle bends, but it means the extrados, which is the part that thins, may not go below the straight-pipe value. The second uses the I-factor profile: more wall than straight pipe at the intrados, less allowed at the extrados. That profile matches what bending does to the metal, so it lets you accept extrados thinning that the first method would reject, at the price of a thickness requirement at the intrados that has to be surveyed. Say on the bend data sheet which basis applies. A survey taken against one and judged against the other is how good bends get rejected.

How do I choose the wall to order for a pipe that will be bent?

Work back from the two finished-bend requirements. After bending, the extrados must still be at least the extrados tm, and the intrados at least the intrados tm. Before bending, the pipe is at worst nominal × (1 − mill tolerance) all the way round. So the ordered wall has to satisfy two inequalities: the mill-minimum wall less the extrados thinning the process produces must clear the extrados requirement, and the mill-minimum wall plus the intrados thickening must clear the intrados requirement. The second one is the trap. Induction bending thickens the intrados noticeably; some cold-bending methods barely do, and then the intrados requirement is met only by starting with heavier pipe. Ask the fabricator for the predicted thinning and thickening for the size, radius and process before the pipe is bought. ¶102.4.5(a) of the 2026 edition points to PFI ES-24 for that guidance, and ¶129.1 makes its manufacturing limits the default where the design sets none.

Does the weld strength reduction factor W apply in a bend?

Only where there is a weld. ¶104.2.1(c) states that W equals 1 for seamless pipe, for seam-welded pipe operating below the creep range, and for the parts of the bend that do not contain a weld. So a seamless bend never sees it. A bend made from longitudinally welded pipe in the creep range takes the Table 102.4.7-1 factor at the seam and 1.0 everywhere else, which makes the seam's position on the cross-section part of the pressure design: put the seam at the intrados and the lower W lands on the station that already needs the most wall. That is why seam orientation belongs on the bend drawing for hot service. Run the calculator twice in that case, once with W = 1.0 for the stations away from the seam and once with the tabulated W for the station that carries it, and take each requirement where it applies. Carbon steel is exempt from ¶102.4.7 altogether, per Note (7) of the table.

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