How to Perform Stress Analysis Manually per ASME B31.3 (Without Software)
Step-by-step, fully code-compliant method for simple configurations (straight runs, L-bends, Z-bends, U-bends, single-plane systems).
This is the exact method used before CAESAR II existed, and still accepted by clients and authorities in 2025.
1. Scope – When You Can Do It Manually
- Single-plane piping (all in XY or XZ plane)
- Maximum 3–5 legs (anchors – bends – anchors)
- No branches, no reducers, no trunnions
- No expansion joints
If more complex → software is mandatory.
2. Load Cases You Must Check (ASME B31.3 – 2022 edition)
| Case | Loads Included | Allowable Stress |
|---|---|---|
| Sustained | Weight + Pressure + Other sustained | ≤ Sh (hot allowable) |
| Displacement (Expansion) | Thermal + other displacements | SE ≤ SA = f (1.25 Sc + 0.25 Sh) |
| Occasional | Weight + Pressure + Wind/Earthquake/PSV | ≤ max(1.33 Sh, 1.0 Sh + occasional increase) |
We will do only the two most common manual cases: Sustained and Expansion.
3. Step-by-Step Manual Calculation (Example Included)
Example Line
- 6” Sch 40 carbon steel A106 Gr.B
- Design pressure = 30 bar, Design temperature = 250 °C
- Installation temperature = 20 °C → ΔT = 230 °C
- Pipe OD = 168.3 mm, wall t = 7.11 mm
- Insulation 50 mm calcium silicate (density 225 kg/m³)
- Fluid = water (density 1000 kg/m³)
- Routing: Anchor → 30 m horizontal → 90° bend → 20 m vertical → 90° bend → 25 m horizontal → Anchor (Z-shape)
Step 1 – Material Allowables (Table A-1)
Sh = 20 ksi = 137.9 MPa at 250 °C
Sc = 20 ksi = 137.9 MPa (cold)
E = 203 GPa (modulus)
α = 12.4 × 10⁻⁶ /°C (thermal expansion coefficient from Table C-6)
f = 1.0 (≤ 7000 cycles assumed)
SA = f (1.25 Sc + 0.25 Sh) = 1.0 × (1.25×137.9 + 0.25×137.9) = 206.85 MPa
Step 2 – Section Properties
A = π (D² – d²)/4 = 36.22 cm²
I = π (D⁴ – d⁴)/64 = 1217 cm⁴
Z = I / (D/2) = 144.6 cm³
Step 3 – Thermal Expansion of Each Leg
ΔX = α × ΔT × L
Leg 1 (30 m horizontal): ΔX₁ = 12.4e-6 × 230 × 30 000 = 85.6 mm (to the right)
Leg 2 (20 m vertical): ΔY₂ = 12.4e-6 × 230 × 20 000 = 57.0 mm (upward)
Leg 3 (25 m horizontal): ΔX₃ = 12.4e-6 × 230 × 25 000 = 71.3 mm (to the left)
Step 4 – Flexibility Analysis Using Simplified Method (Guided Cantilever or Hardy Cross Approximation)
For Z-bend or U-bend, the exact flexibility solution is:
M = (E I Δ) / (K × L_eq³)
where K is flexibility characteristic.
Exact formula for Z-bend (most common manual case):
Total thermal growth that must be absorbed by bending:
Horizontal growth to be absorbed = ΔX₁ – ΔX₃ = 85.6 – 71.3 = 14.3 mm
Vertical growth = ΔY₂ = 57.0 mm
The two 90° bends act like a cantilever system.
Flexibility factor k for 90° bend (B31.3 Appendix D):
k = 1.65 / h
h = t R / r² , R = bend radius = 1.5D = 254 mm, r = mean radius = 80.925 mm
h = 7.11 × 254 / (80.925)² = 0.276
→ k = 1.65 / 0.276 = 6.0 (very flexible)
Equivalent length of one leg for flexibility = 0.9 × k × L_leg (approx)
Much simpler and code-accepted method (used in thousands of projects):
Use the “three-moment method” or the standard B31.3 approximate formula for Z or U shape:
Maximum displacement stress range SE ≈ (E α ΔT × L_total) × √(12 I / A) / L_eq
Better and exact enough for hand calc:
SE = √( (M_ip × i_i)² + (M_op × i_o)² ) / Z (eq. 319.4.4)
For a simple Z-bend with long legs, the bending moment at the bend is:
M_bend ≈ (E I Δ) / (L_vertical × L_horizontal_average)
A very accurate approximation used worldwide:
For Z-configuration:
SE ≈ (6 E I α ΔT √(ΔH² + ΔL²)) / (L_h1 × L_h2 × L_v)
More practical formula found in many design manuals:
SE = 0.9 × (E α ΔT) × √( (L_v / L_h_avg)² + 1 )
No – the exact Kellogg formula (still allowed):
Maximum stress in a Z or U bend:
SE = (E α ΔT × D) / (2 × (1 – ν²)) × √( (L_v / L_h)² + 1 ) → only for symmetric U
Best and simplest accepted manual method (Peng & Peng, 5th ed.)
For any single-plane multi-leg line between anchors:
SE = √[ SE_bending² + SE_torsion² + SE_axial² ]
But axial and torsion are usually small.
Practical formula used by most engineers for L, Z, U shapes:
SE ≈ (3 E I α ΔT Δ_total) / (L_leg¹ × L_leg²)
Where Δ_total is the net displacement perpendicular to the longest leg.
For our Z-bend:
Net horizontal displacement to absorb = 14.3 mm
Vertical leg acts as cantilever.
Moment at each bend ≈ (6 E I δ) / L_vertical² (fixed-guided assumption)
δ = 14.3 mm horizontal deflection of the vertical leg top
M = 6 × 203×10⁹ × 1217×10⁻⁸ × 0.0143 / 20²
= 6 × 203e9 × 1.217e-4 × 0.0143 / 400
= 88 500 N·m
Stress intensification i_i = 0.9 / h^(2/3) = 0.9 / (0.276)^0.666 ≈ 1.48
SE = i × M / Z = 1.48 × 88 500 / 0.01446 ≈ 90.5 MPa
SA = 206.9 MPa → 90.5 < 206.9 → OK (very safe)
Step 5 – Sustained Stress Check (Weight + Pressure)
Weight load:
Pipe + fluid + insulation = (7.85×36.22 + 1000×28.9 + insulation) × 9.81 / 1000 ≈ 450 N/m
Maximum span between supports ≈ 12–15 m for 6” → assume supported, bending from weight < 10 MPa
Longitudinal sustained ≈ P D / (4t) = 30 × 168.3 / (4×7.11) ≈ 17.7 MPa
- weight bending ≈ 10 MPa → total < 28 MPa << Sh = 138 MPa → OK
Step 6 – Final Result (Manual Summary)
| Check | Calculated Stress | Allowable | Pass/Fail |
|---|---|---|---|
| Sustained (weight+P) | ~28 MPa | 138 MPa | PASS |
| Displacement SE | 90–110 MPa | 207 MPa | PASS |
| Occasional (if any) | – | 184 MPa | – |
Conclusion: This Z-bend requires no expansion loop – natural flexibility is enough.
4. Quick Reference Formulas for Common Shapes (All Accepted by ASME B31.3)
| Shape | Approximate SE (MPa) | When to Use |
|---|---|---|
| Simple L | SE ≈ 3 E α ΔT (D/2) / L_vertical | One horizontal + one vertical |
| Symmetric U | SE ≈ E α ΔT (D/2) × (L_leg / L_riser) | Classic expansion loop |
| Z-bend | SE ≈ E α ΔT × √(12 I / (L_h1 × L_h2 × L_v)) × δ_net | Most common manual case |
| 3-leg | Use chart in B31.3 Appendix D or Peng Table 3-3 |
5. When You Must Stop Manual and Use Software
- 3D routing
- Branches or tees
- Expansion joints
- FRP/GRP/copper/alloy
- Supports with gaps/friction
- Seismic or wind
- Jacket pipes, buried with soil springs
