ASME B31.1 Pipe Bend Wall Thickness Calculator — Sample Report

1 · Design inputs

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.

2 · 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:

t_m = P·D_o / ( 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/D_o) − 1 ) / ( 4(R/D_o) − 2 )

extrados: I = ( 4(R/D_o) + 1 ) / ( 4(R/D_o) + 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.

3 · Calculation

1. Radius ratio: R/D_o = 63.75 / 12.75 = 5, so 4(R/D_o) = 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·D_o = 2,500 × 12.75 = 31,875 lb/in; P·y = 2,500 × 0.7 = 1,750 psi.
5. Intrados, eq. (7): t_m = 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: t_m = 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: t_m = 31,875 / (2 × 15,750) = 31,875 / 31,500 = 1.01190 in — the straight-pipe minimum for the tangents.

4 · Results COMPUTED

5 · Verification statement

The result values above are locked as 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 continuous verification gate (see https://pipingtoolset.com/trust/validation.html). Allowable-stress and factor data are user-supplied inputs; citations: ASME B31.1-2026, Power Piping — ¶104.2.1(c), eq. (7) (outside-diameter form; eq. (8) is the inside-diameter form), eq. (9) intrados, eq. (10) extrados, I = 1.0 at the sidewall on the bend centerline; Figure 104.2.1-1 nomenclature. In B31.1-2024 the same equations are ¶102.4.5(b), eqs. (3) through (6). · ASME B31.1-2026 ¶104.2.1(a) and (b) — the minimum wall shall meet (c) or ¶104.1.2(a) and the fabrication requirements of ¶129.1; flattening and buckling limits by design, otherwise ¶129.1. · ASME B31.1-2026 ¶102.4.5 — thickness requirements apply at mid-arc, intrados, extrados and bend centerline; PFI ES-24 named as guidance on thinning due to bending. ¶129.1 — bending by any hot or cold method to a surface free of cracks; PFI ES-24 manufacturing limits where the design sets none. · ASME B31.1-2026 Mandatory Appendix A (SE), Table 104.1.2-1 (y), ¶102.4.7 and Table 102.4.7-1 (W) — typed from your governing edition.