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Truss Analysis: worked example
A pin-jointed steel truss solved for member forces and support reactions, with member checks.
EN 1993-1-1:2005 — Eurocode 3: Design of steel structures — Part 1-1: General rules and rules for buildings · SI units · Calculator v1.0
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Inputs
| Section family / default size | HEA / HEA160 |
| Steel grade | S235 |
| Nodes | 8 |
| Members | 13 |
| Geometry valid / solved | Yes |
Design checks All checks pass
| Check | Required | Provided | Utilisation | Result |
|---|---|---|---|---|
| M1 | 200.0 kN (tension) | 911.1 kN | 22.0% | PASS |
| M2 | 200.0 kN (tension) | 911.1 kN | 22.0% | PASS |
| M3 | 200.0 kN (tension) | 911.1 kN | 22.0% | PASS |
| M4 | 200.0 kN (tension) | 911.1 kN | 22.0% | PASS |
| M5 | 266.7 kN (compression) | 456.7 kN | 58.4% | PASS |
| M6 | 266.7 kN (compression) | 456.7 kN | 58.4% | PASS |
| M7 | 100.0 kN (tension) | 911.1 kN | 11.0% | PASS |
| M8 | 0.0 kN (zero) | 911.1 kN | 0.0% | PASS |
| M9 | 100.0 kN (tension) | 911.1 kN | 11.0% | PASS |
| M10 | 250.0 kN (compression) | 340.8 kN | 73.3% | PASS |
| M11 | 250.0 kN (compression) | 340.8 kN | 73.3% | PASS |
| M12 | 83.3 kN (tension) | 911.1 kN | 9.1% | PASS |
| M13 | 83.3 kN (tension) | 911.1 kN | 9.1% | PASS |
Node coordinates, supports and applied nodal loads
| Node | X (m) | Y (m) | Support | Fx (kN) | Fy (kN) |
|---|---|---|---|---|---|
| B0 | 0.00 | 0.00 | pinned | 0.00 | 0.00 |
| B1 | 4.00 | 0.00 | free | 0.00 | -100.00 |
| B2 | 8.00 | 0.00 | free | 0.00 | -100.00 |
| B3 | 12.00 | 0.00 | free | 0.00 | -100.00 |
| B4 | 16.00 | 0.00 | rollerY | 0.00 | 0.00 |
| T1 | 4.00 | 3.00 | free | 0.00 | 0.00 |
| T2 | 8.00 | 3.00 | free | 0.00 | 0.00 |
| T3 | 12.00 | 3.00 | free | 0.00 | 0.00 |
Members
| Member | From | To | Section |
|---|---|---|---|
| M1 | B0 | B1 | Default (HEA160) |
| M2 | B1 | B2 | Default (HEA160) |
| M3 | B2 | B3 | Default (HEA160) |
| M4 | B3 | B4 | Default (HEA160) |
| M5 | T1 | T2 | Default (HEA160) |
| M6 | T2 | T3 | Default (HEA160) |
| M7 | B1 | T1 | Default (HEA160) |
| M8 | B2 | T2 | Default (HEA160) |
| M9 | B3 | T3 | Default (HEA160) |
| M10 | B0 | T1 | Default (HEA160) |
| M11 | B4 | T3 | Default (HEA160) |
| M12 | T1 | B2 | Default (HEA160) |
| M13 | T3 | B2 | Default (HEA160) |
Calculation steps
Geometry & determinacy (2 steps)
—
Model summary
—
8 joints, 13 members, 3 reaction components
8 joints, 13 members, 3 reactions
Statics
Determinacy count (necessary, not sufficient — the stiffness solve below is the rigorous check)
m + r = 2j ?
13 + 3 vs 2×8 = 16
m+r=16 → statically determinate
Stiffness-method solve (2 steps)
—
Bar-element assembly & solve
Kff · Uf = Pf (2 DOF/joint, axial-only elements)
16 total DOF, 13 free
Solved — joint displacements recovered
—
Global equilibrium check
ΣFx = ΣFy = 0 (applied loads + reactions)
ΣFx=-0.000 kN, ΣFy=0.000 kN
Satisfied
Member axial forces (13 steps)
—
Member M1 (HEA160)
N = (EA/L)·Δ, tension +ve
L=4.00 m
N = 200.0 kN (tension)
—
Member M2 (HEA160)
N = (EA/L)·Δ, tension +ve
L=4.00 m
N = 200.0 kN (tension)
—
Member M3 (HEA160)
N = (EA/L)·Δ, tension +ve
L=4.00 m
N = 200.0 kN (tension)
—
Member M4 (HEA160)
N = (EA/L)·Δ, tension +ve
L=4.00 m
N = 200.0 kN (tension)
—
Member M5 (HEA160)
N = (EA/L)·Δ, tension +ve
L=4.00 m
N = -266.7 kN (compression)
—
Member M6 (HEA160)
N = (EA/L)·Δ, tension +ve
L=4.00 m
N = -266.7 kN (compression)
—
Member M7 (HEA160)
N = (EA/L)·Δ, tension +ve
L=3.00 m
N = 100.0 kN (tension)
—
Member M8 (HEA160)
N = (EA/L)·Δ, tension +ve
L=3.00 m
N = 0.0 kN (zero)
—
Member M9 (HEA160)
N = (EA/L)·Δ, tension +ve
L=3.00 m
N = 100.0 kN (tension)
—
Member M10 (HEA160)
N = (EA/L)·Δ, tension +ve
L=5.00 m
N = -250.0 kN (compression)
—
Member M11 (HEA160)
N = (EA/L)·Δ, tension +ve
L=5.00 m
N = -250.0 kN (compression)
—
Member M12 (HEA160)
N = (EA/L)·Δ, tension +ve
L=5.00 m
N = 83.3 kN (tension)
—
Member M13 (HEA160)
N = (EA/L)·Δ, tension +ve
L=5.00 m
N = 83.3 kN (tension)
Support reactions (2 steps)
—
Reaction at B0
—
—
Rx = -0.0 kN, Ry = 150.0 kN
—
Reaction at B4
—
—
Rx = 0.0 kN, Ry = 150.0 kN
Governing member M11 — compression (9 steps)
Table 5.2
Material coefficient
ε = √(235 / fy)
ε = √(235 / 235)
ε = 1.000
Table 5.2
Flange (outstand in compression)
c = (b − tw − 2r)/2; limits 9ε | 10ε | 14ε
c/tf = 62.0/9 = 6.89 vs 9.00 | 10.00 | 14.00
Flange: Class 1
Table 5.2
Web (internal, in bending)
c = h − 2tf − 2r; limits 72ε | 83ε | 124ε
c/tw = 104.0/6 = 17.33 vs 72.0 | 83.0 | 124.0
Web: Class 1
§5.5.2(6)
Section class = least favourable element class
max(flange, web)
max(1, 1)
Section: Class 1
EN 1993-1-1 Table 6.2
Buckling curve selection (h/b=0.95, tf=9.0 mm)
—
—
curve y-y = b, curve z-z = c
EN 1993-1-1 §6.3.1.3
Elastic critical buckling force, both axes
Ncr = π²·E·I/Lcr²
Lcr,y=5000 mm, Lcr,z=5000 mm
Ncr,y=1387.0 kN, Ncr,z=510.7 kN
EN 1993-1-1 (6.50)
Non-dimensional slenderness
λ̄ = √(Npl,Rd/Ncr)
Npl,Rd=911.1 kN
λ̄y=0.810, λ̄z=1.336
EN 1993-1-1 (6.49)
Reduction factor χ (imperfection factor α from the curve)
χ = 1/(Φ+√(Φ²−λ̄²)) ≤ 1
αy=0.34, αz=0.49
χy=0.718, χz=0.374
EN 1993-1-1 (6.47)
Buckling resistance, both axes — governing axis controls
Nb,Rd = χ·A·fy/γM1
γM1=1
Nb,Rd,y=654.1 kN, Nb,Rd,z=340.8 kN → governs z-z
Design basis and assumptions
- National Annex: none applied as a whole — the EN recommended values are used, except where noted below. There is no National Annex selector; check the values against the annex for your project.
- γM0 = 1.0 and γM1 = 1.0 (EN 1993-1-1 §6.1 recommended values).
- Models an idealized pin-jointed truss — every member carries pure axial force only, and loads can only be applied at nodes, not along a member’s length.
- Stiffness-method solve with 2-DOF/joint bar elements — mathematically equivalent to the method of joints for a statically determinate truss; a linear-elastic, small-deflection (first-order) analysis only.
- Net-section rupture at connections is not checked — verify separately at any bolted/welded joint.
- A distributed load must be approximated as statically-equivalent point loads at the nearest joints.
- Geometry and results are in SI units (m, kN, kN·m) only.
- If the structure actually has rigid, moment-resisting joints, use Frame Analysis instead.
Example values for illustration only. Results are checked against a published validation suite; for real projects, have a licensed engineer review the results.