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Frame Analysis: worked example
A rigid steel frame solved by the stiffness method for member forces, moments and 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 / HEA200 |
| Steel grade | S235 |
| Nodes | 4 |
| Members | 3 |
| Geometry valid / solved | Yes |
Design checks All checks pass
| Check | Required | Provided | Utilisation | Result |
|---|---|---|---|---|
| M1 — bending | 17.7 kN·m | 100.9 kN·m | 17.6% | PASS |
| M1 — shear | 2.6 kN | 245.3 kN | 1.1% | PASS |
| M1 — axial | 39.7 kN | 1265.0 kN | 3.1% | PASS |
| M1 — interaction | — | — | 20.7% | PASS |
| M2 — bending | 72.3 kN·m | 100.9 kN·m | 71.6% | PASS |
| M2 — shear | 50.3 kN | 245.3 kN | 20.5% | PASS |
| M2 — axial | 22.6 kN | 1265.0 kN | 1.8% | PASS |
| M2 — interaction | — | — | 73.4% | PASS |
| M3 — bending | 49.7 kN·m | 100.9 kN·m | 49.2% | PASS |
| M3 — shear | 22.6 kN | 245.3 kN | 9.2% | PASS |
| M3 — axial | 50.3 kN | 1265.0 kN | 4.0% | PASS |
| M3 — interaction | — | — | 53.2% | PASS |
Node coordinates, supports and applied nodal loads/moments
| Node | X (m) | Y (m) | Support | Fx (kN) | Fy (kN) | Mz (kN·m) |
|---|---|---|---|---|---|---|
| C0 | 0.00 | 0.00 | fixed | 0.00 | 0.00 | 0.00 |
| C1 | 0.00 | 4.00 | free | 20.00 | 0.00 | 0.00 |
| C2 | 6.00 | 4.00 | free | 0.00 | 0.00 | 0.00 |
| C3 | 6.00 | 0.00 | fixed | 0.00 | 0.00 | 0.00 |
Members and member loads
| Member | From | To | Section | Loads |
|---|---|---|---|---|
| M1 | C0 | C1 | Default (HEA200) | — |
| M2 | C1 | C2 | Default (HEA200) | w = 15.00 kN/m |
| M3 | C2 | C3 | Default (HEA200) | — |
Calculation steps
Stiffness-method solve (2 steps)
—
Beam-column assembly & solve
Kff · Uf = Pf (3 DOF/node: u, v, θ)
12 total DOF, 6 free
Solved — nodal displacements/rotations recovered
—
Global equilibrium check
ΣFx = ΣFy = 0 (applied + member self-loads + reactions)
ΣFx=0.000 kN, ΣFy=-0.000 kN
Satisfied
Member end forces & governing demand (3 steps)
—
Member M1 (HEA200)
Recovered via q = k·(T·u) + FEF
L=4.00 m
Mmax=17.7 kN·m, Vmax=2.6 kN, N=-39.7 kN
—
Member M2 (HEA200)
Recovered via q = k·(T·u) + FEF
L=6.00 m
Mmax=72.3 kN·m, Vmax=50.3 kN, N=-22.6 kN
—
Member M3 (HEA200)
Recovered via q = k·(T·u) + FEF
L=4.00 m
Mmax=49.7 kN·m, Vmax=22.6 kN, N=-50.3 kN
Support reactions (2 steps)
—
Reaction at C0
—
—
Rx=2.6 kN, Ry=39.7 kN, Mz=7.2 kN·m
—
Reaction at C3
—
—
Rx=-22.6 kN, Ry=50.3 kN, Mz=40.8 kN·m
Governing member M2 (17 steps)
§6.2.5(2)
Section modulus for Class 1
Class 1/2 → plastic modulus
Wpl,y = 429.5 × 10³ mm³
Wpl,y = 429500 mm³
§6.2.5(2)
Design resistance for bending
Mc,Rd = Wpl,y · fy / γM0
Mc,Rd = 429500 × 235 / 1
Mc,Rd = 100.9 kN·m
§6.2.5(1)
Cross-section bending check
MEd / Mc,Rd ≤ 1.0
72.3 / 100.9
Ratio = 0.716 → OK
§6.2.6(3)a
Shear area, rolled I-section
Av = A − 2b·tf + (tw + 2r)·tf ≥ η·hw·tw
Av = 5383 − 2×200×10 + (6.5 + 2×18)×10 = 1808; min 1326
Av = 1808 mm²
§6.2.6(2)
Plastic shear resistance
Vpl,Rd = Av · (fy/√3) / γM0
Vpl,Rd = 1808 × (235/√3) / 1
Vpl,Rd = 245.3 kN
§6.2.6(6)
Web shear buckling check
hw/tw ≤ 72ε/η, else EN 1993-1-5 §5 governs
26.2 vs 60.0
OK — no shear buckling check needed
§6.2.6(1)
Shear check
VEd / Vpl,Rd ≤ 1.0
50.3 / 245.3
Ratio = 0.205 → OK
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 = 78.8/10 = 7.88 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 = 134.0/6.5 = 20.62 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=10.0 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=6000 mm, Lcr,z=6000 mm
Ncr,y=2125.6 kN, Ncr,z=769.2 kN
EN 1993-1-1 (6.50)
Non-dimensional slenderness
λ̄ = √(Npl,Rd/Ncr)
Npl,Rd=1265.0 kN
λ̄y=0.771, λ̄z=1.282
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=1.000 (λ̄≤0.2, buckling ignored per §6.3.1.2(4)), χz=1.000 (λ̄≤0.2, buckling ignored)
EN 1993-1-1 (6.47)
Buckling resistance, both axes — governing axis controls
Nb,Rd = χ·A·fy/γM1
γM1=1
Nb,Rd,y=1265.0 kN, Nb,Rd,z=1265.0 kN → governs y-y
EN 1993-1-1 (disclosed simplification)
Combined axial + bending — linear interaction
N/Nb,Rd + My/Mc,Rd ≤ 1.0
22.6/1265.0 + 72.3/100.9
Ratio = 0.734 → OK
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).
- First-order elastic analysis only — no P-Δ (second-order) effects.
- No lateral-torsional buckling check on frame members.
- EC3 combined axial+bending uses a disclosed conservative linear interaction (N/Nb,Rd + My/Mc,Rd ≤ 1), not the full §6.3.3 sway-buckling interaction.
- A member’s own UDL is automatically resolved into along-member and perpendicular components regardless of the member’s angle.
- A concentrated moment at a node transfers into every member framing into that node, since joints are modeled as fully rigid.
- Geometry and results are in SI units (m, kN, kN·m) only.
Example values for illustration only. Results are checked against a published validation suite; for real projects, have a licensed engineer review the results.