Worked example · 2D portal frame
Portal frame analysis: reactions, shear, moment and deflection, step by step
A 4 m × 3 m steel portal, solved twice. First with pinned bases under gravity alone, where the answer can be checked against the classical closed-form solution — the two agree to within 0.04 %. Then with fixed bases and a lateral load, the case that has no textbook formula, where equilibrium becomes the proof.
The frame
Both cases share the same geometry and the same section, so the two sets of results are directly comparable. Only the base fixity and the load change.
| Span L (centre to centre) | 4.0 m |
| Eaves height h | 3.0 m |
| Section (all members) | IPE 180 |
| Steel grade | S235 (E = 210 GPa) |
| Second moment of area I | 1 317 cm⁴ = 1.317 × 10⁻⁵ m⁴ |
| Area A | 23.9 cm² = 2.39 × 10⁻³ m² |
| Analysis | First-order linear elastic, in-plane (2D) |
| Mesh | Rafter split at midspan (see FAQ) |
Case 1 — pinned bases, gravity only: checked against the closed form
With both bases pinned and only the 5 kN/m rafter load acting, the frame is symmetric and indeterminate to the first degree. The classical two-hinged portal formulas give an exact answer to compare against.
k = (I_beam · h) / (I_column · L) = (1 · 3) / (1 · 4) = 0.75
H = w·L² / (4h(2k + 3)) = 5 · 4² / (4 · 3 · 4.5) = 1.4815 kN
V = w·L / 2 = 5 · 4 / 2 = 10.000 kN
M_eaves = H · h = 1.4815 · 3 = 4.4444 kN·m
M_mid = w·L²/8 − H·h = 10.000 − 4.4444 = 5.5556 kN·m| Quantity | Closed form | FERS | Difference |
|---|---|---|---|
| Vertical reaction, each baseV | 10.000 kN | 10.000 kN | 0.00 % |
| Horizontal thrust, each baseH | 1.4815 kN | 1.4809 kN | −0.04 % |
| Bending moment at the eavesM_eaves | 4.4444 kN·m | 4.4426 kN·m | −0.04 % |
| Sagging moment at midspanM_mid | 5.5556 kN·m | 5.5574 kN·m | +0.03 % |
| Rafter sag, relative to the eavesδ | 2.8124 mm | 2.8135 mm | +0.04 % |
The vertical reactions match exactly because they follow from statics alone. The small residual on the thrust, the moments and the sag is the axial flexibility that the closed form neglects and the finite-element model includes — see the FAQ below.
Case 2 — fixed bases with a lateral load: equilibrium as the proof
Fix both bases and add a 5 kN horizontal load at the left eaves, and the frame becomes indeterminate to the third degree. There is no clean textbook formula for this combination, so the check that remains is equilibrium — and it has to close exactly.
| Support | Horizontal | Vertical | Moment |
|---|---|---|---|
| Left base (load side) | −0.087 kN | 8.468 kN | 2.034 kN·m |
| Right base (far side) | −4.913 kN | 11.532 kN | 6.837 kN·m |
ΣFx : −0.087 + (−4.913) = −5.000 kN balances the applied 5 kN
ΣFy : 8.468 + 11.532 = 20.000 kN = w·L = 5 · 4
ΣM : about the left base, from the applied loads,
the right-base reaction and both fixing
moments = 0.000 kN·mNote how unevenly the lateral load splits the vertical reactions: 8.47 kN on the near base against 11.53 kN on the far one, even though the gravity load is perfectly symmetric. A 5 kN horizontal load acting 3 m above the bases has to be balanced by a couple across the 4 m span, which adds 1.53 kN to one base and takes it off the other.
| Location | Moment | Note |
|---|---|---|
| Left column base | 2.034 kN·m | Hogging, outside face in tension |
| Left eaves joint | 1.774 kN·m | Column and rafter balance exactly |
| Rafter, 1.5 m from the left eaves | 5.303 kN·m | Peak sagging moment in the rafter |
| Rafter midspan (node 5) | 5.161 kN·m | Sagging |
| Right eaves joint | 7.903 kN·m | Governs the frame |
| Right column base | 6.837 kN·m | Hogging |
Bending at the far eaves governs this frame, at more than three times the moment carried by the near column base. Sway and rafter sag are of the same order — around 3 mm — so on a real frame the serviceability limit would be worth checking alongside strength. The levers available are the usual ones: change the section, change the base fixity, or brace the frame so that it is no longer relying on frame action alone to resist the lateral load.
Frequently asked questions
- Why is a portal frame statically indeterminate?
- A portal with two pinned bases has four reaction components against the three equations of planar equilibrium, so it is indeterminate to the first degree: the horizontal thrust H cannot be found from statics alone and depends on the relative stiffness of rafter and columns. Fixing both bases adds two more reaction components and makes it indeterminate to the third degree. That is where a stiffness-method solver earns its keep. For a simply supported beam you can check the answer with a single formula; for the frame you need the compatibility conditions the solver enforces.
- How is the closed-form thrust derived?
- For a two-hinged (pinned-base) symmetric portal carrying a uniformly distributed load w on the rafter, the standard result is H = wL squared / (4h(2k+3)), with the stiffness ratio k = (I_beam times h) / (I_column times L). With equal sections, L = 4 m and h = 3 m, k = 0.75 and H = 1.4815 kN. Two limiting cases confirm the formula. As the columns become infinitely stiff (k tends to 0) it reduces to H = wL squared / 12h, which is the fixed-end beam moment wL squared / 12 divided by the height. As the columns become very flexible (k tends to infinity) H tends to zero and the rafter behaves as a simply supported beam with M = wL squared / 8.
- Why do FERS and the closed form differ by 0.04 %?
- The classical portal formulas assume axially rigid members: the rafter cannot stretch and the columns cannot shorten. The finite-element solution includes axial flexibility, so the frame spreads a fraction of a millimetre at the eaves and sheds a correspondingly small amount of thrust. The residual is a genuine difference between the two models, not solver error. The vertical reactions, which do not depend on that assumption, agree to every digit shown.
- Why is the rafter split at midspan?
- Support reactions and internal forces are exact at the nodes however the rafter is meshed, so none of the forces in this article change if you model it as a single member. The deflected shape does change. FERS returns the deflected shape by interpolating the nodal displacements and rotations with a cubic Hermite polynomial, which carries no element-interior term for a distributed load, so an undivided member under a UDL under-reports its own sag. For this frame the difference is 1.67 mm undivided against 2.61 mm split. Splitting any span that carries a distributed load into at least two elements removes the discrepancy, and gives a smoother diagram in any case.
- What changes when the bases are fixed instead of pinned?
- Fixing the bases draws moment into the foundations and relieves the rafter. In case 2 the base moments reach 2.03 and 6.84 kN·m, moments the pinned frame of case 1 cannot develop at all. The frame also becomes markedly stiffer against sway, which is the reason fixed bases get chosen where lateral drift governs, at the cost of a foundation that has to be designed for that moment.
- Can I reproduce these numbers myself?
- Yes. Open the free 2D frame calculator, choose the Portal frame example and press Solve. It is the same model solved by the same engine that produced this page. Reactions and the N, V and M diagrams are free and need no account; deflections are unlocked with a free account.
Keep going
- Free 2D frame calculator — this exact frame, ready to edit: move a node, change the load, re-solve.
- Worked Eurocode 3 steel beam check — take a member like these through an EN 1993-1-1 resistance check.
- Benchmarks — more FERS results compared against analytical solutions.
- Sign conventions — how FERS orients local axes and signs the diagrams above.
From one portal to the whole building
The full FERS Cloud app takes the same solver to 3D frames and trusses, load combinations and Eurocode EC3 steel checks across every member.