Local Strain Limit Calculator

Free elastic–plastic local strain limit check per ASME Section VIII Division 2, paragraph 5.3.3.1 — protection against local failure. Enter the principal stresses and equivalent plastic strain from your elastic–plastic FE model to get the limiting triaxial strain (Table 5.7, Eq. 5.7) and the Eq. 5.8 acceptance check, with a detailed step-by-step report. Jump to how it works ↓

Identification

Component / Analysis Point

Load Case

Prepared By

Step 1 — Stress State at the Point Eq. 5.1

Stress Units

Stress Input Method

Eq. (5.7) uses only stress ratios, so the result is unit-independent — the unit selection labels the report. Take all values at the point from the elastic–plastic analysis performed at the Table 5.5 local-criterion load combination.

Principal Stress σ₁

psi

Principal Stress σ₂

psi

Principal Stress σ₃

psi

Equivalent Stress σₑ (from FE)

psi

Total Equivalent Plastic Strain, ε_peq

Step 2 — Limiting Triaxial Strain Table 5.7 · Eq. 5.7

Material Category (Table 5.7)

Analysis Temperature (optional)

Override ε_Lu, m₂ and α_sl directly (bypass the Table 5.7 lookup)

Yield-to-Tensile Ratio, R = S_y / S_u (Eq. 3-D.11)

R Input Method

Ratio R

Yield Strength, S_y

psi

Tensile Strength, S_u

psi

Engineering yield and ultimate tensile strengths, both at the analysis temperature (Annex 3-D). Not a reduction-of-area value.

Elongation / Reduction of Area (Table 5.7 Columns 4 & 5 — optional)

Elongation, E

Reduction of Area, RA

Per Table 5.7 Note (1): leave both blank to use ε_Lu = m₂. If E and/or RA are given in the material specification, ε_Lu is the maximum of Columns 3, 4 and 5 as applicable.

Step 3 — Forming Strain Part 6

Heat treated in accordance with Part 6 (forming strain taken as zero)

Forming Strain, ε_cf

Enter your data and click Calculate to see the strain limit check and the step-by-step solution.

How This Local Strain Limit Calculator Works

ASME Section VIII Division 2, paragraph 5.3.3 protects against local failure, a ductile-rupture failure mode driven by accumulated plastic strain under a triaxial stress state. It is a separate check from protection against plastic collapse (5.2.4): a component can have ample collapse margin and still fail locally at a highly constrained point such as a nozzle or head-to-shell junction. This calculator implements the single-load-case procedure of 5.3.3.1 at one point of your elastic–plastic FE model. Calculations run on a protected server; only your inputs are sent and only the results and worked solution come back.

The Assessment Procedure

Step 1 — Stress State (Eq. 5.1)

From the FE results at the point, take σ₁, σ₂, σ₃ and the total equivalent plastic strain ε_peq. The von Mises equivalent stress σₑ and the triaxiality term (σ₁+σ₂+σ₃)/(3σₑ) are computed.

Step 2 — Limiting Strain (Eq. 5.7)

ε_Lu, m₂ and α_sl come from Table 5.7 for the material category. The uniaxial limit is reduced exponentially as triaxiality rises above 1/3 (uniaxial tension).

Step 3 — Forming Strain (Part 6)

The forming strain ε_cf is based on the material and fabrication method per Part 6. If the part is heat treated per Part 6, ε_cf may be taken as zero.

Step 4 — Acceptance (Eq. 5.8)

The point is acceptable for the load case when ε_peq + ε_cf ≤ ε_L. The report also gives the utilization ratio (ε_peq + ε_cf)/ε_L.

Governing Equations

\[ \sigma_e = \frac{1}{\sqrt{2}}\left[(\sigma_1-\sigma_2)^2+(\sigma_2-\sigma_3)^2+(\sigma_3-\sigma_1)^2\right]^{0.5} \qquad (5.1) \]

\[ \varepsilon_L = \varepsilon_{Lu} \cdot \exp\left\{-\left(\frac{\alpha_{sl}}{1+m_2}\right)\left[\left(\frac{\sigma_1+\sigma_2+\sigma_3}{3\sigma_e}\right)-\frac{1}{3}\right]\right\} \qquad (5.7) \]

\[ \varepsilon_{peq} + \varepsilon_{cf} \le \varepsilon_L \qquad (5.8) \]

Table 5.7 — Uniaxial Strain Limit for the Multiaxial Strain Limit Criterion

Material	Max. Temperature	m₂ (Col. 3)	Elongation Specified (Col. 4)	RA Specified (Col. 5)	α_sl
Ferritic steel	480°C (900°F)	0.60(1.00 − R)	2·ln[1 + E/100]	ln[100/(100 − RA)]	2.2
Stainless steel and nickel base alloys	480°C (900°F)	0.75(1.00 − R)	3·ln[1 + E/100]	ln[100/(100 − RA)]	0.6
Duplex stainless steel	480°C (900°F)	0.70(0.95 − R)	2·ln[1 + E/100]	ln[100/(100 − RA)]	2.2
Precipitation-hardening nickel-based austenitic alloys	540°C (1,000°F)	1.09(0.93 − R)	ln[1 + E/100]	ln[100/(100 − RA)]	2.2
Aluminum	120°C (250°F)	0.52(0.98 − R)	1.3·ln[1 + E/100]	ln[100/(100 − RA)]	2.2
Copper	65°C (150°F)	0.50(1.00 − R)	2·ln[1 + E/100]	ln[100/(100 − RA)]	2.2
Titanium and zirconium	260°C (500°F)	0.50(0.98 − R)	1.3·ln[1 + E/100]	ln[100/(100 − RA)]	2.2

Notes: (1) If elongation and reduction of area are not specified, ε_Lu = m₂; if either is specified, ε_Lu is the maximum of Columns 3, 4 and 5 as applicable. (2) R = S_y/S_u per Eq. (3-D.11). (3) E is the % elongation and RA the % reduction of area from the material specification. (4) Ferritic steel includes carbon, low alloy and alloy steels, and ferritic, martensitic and iron-based age-hardening stainless steels. Source: ASME BPVC.VIII.2-2025, Table 5.7.

Verified Example — PTB-3-2022, Example E5.3.3

Top head-to-shell junction of a vessel under 420 psig internal pressure, analysed at the factored local-criterion load of 1.7 × 420 = 714 psig. Ferritic steel, R = 0.5429. Use Load Verified Example above to reproduce it.

Quantity	Value	Source
σ₁ / σ₂ / σ₃	45,095 / 34,603 / 1,118 psi	FE result
σₑ (von Mises)	39,783 psi	Eq. (5.1)
m₂ = ε_Lu	0.2743	Table 5.7
ε_L	0.1515	Eq. (5.7)
ε_peq	0.002468	FE result
(ε_peq + ε_cf)/ε_L	0.0163 — acceptable	Eq. (5.8)

⚠ This tool implements the single-load-case procedure of 5.3.3.1 only. The strain-limit damage procedure of 5.3.3.2 for a specific loading sequence (Eqs. 5.9–5.12) is not implemented. Always verify Code data against your controlled copy of ASME BPVC Section VIII Division 2 before use in a certified deliverable.