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    Structural / frame

    Portal frame analysis with P-Δ

    Member magnification checks one column at a time. A frame carries its gravity load through sway, so the columns soften together: this solver assembles the elastic and geometric stiffness of the whole frame, iterates the second-order displacements, and finds the load factor at which the frame buckles.

    BasesFixedShape2 baysGravity300 kN
    Eaves sway
    5.57 mm
    Second order · 5.54 mm first order
    Drift ratio
    1 / 718
    0.139% of the storey height
    Critical load factor
    λ 188
    1.866e+4% reserve before buckling
    Largest end moment
    126.8 kN·m
    P-Δ amplification 1.005
    Member end forces
    Second-order axial, shear and bending at each member end. Moments are magnitudes about the member axis.
    5 members
    Member end forces
    MemberAxial (kN)Shear (kN)Moment i (kN·m)Moment j (kN·m)
    C152.713.145.47.3
    C2168.335.875.367.8
    C379.041.582.883.2
    B116.952.77.3126.8
    B211.471.059.083.1

    Positive axial is compression. Column moments at the base govern the foundation design.

    Inspect the model
    Choose a quick result view, a scaled comparison, or the complete diagnostic set.
    Sway as gravity approaches buckling
    Gravity load as a share of the 4692 kN that would buckle the frame. At 90% the sway has grown from 5.6 to 57 mm.
    Sway as gravity approaches buckling0.00012.6925.3938.0850.7763.470%0%11%11%23%23%34%34%45%45%56%56%68%68%79%79%90%90%Second orderFirst order
    End moments by member
    Absolute moments at each end; the base of a column usually governs.
    End moments by member0.00029.9259.8489.75119.67149.5945.407.30C1C175.3567.81C2C282.8483.16C3C37.31126.77B1B159.0183.13B2B2End iEnd j
    Canonical equations
    The exact relationships implemented by this engine.

    K_e = Σ Tᵀ k_e T · 3 DOF per node: ux, uy, θ.

    K_g = (P/L) · g(θ, L) · Compression reduces lateral stiffness; 6/5 term included.

    (K_e − K_g) u = F · Iterated because K_g depends on the axial forces.

    det(K_e − λ K_g) = 0 · Smallest λ from the negative-pivot count.

    engineering methodology

    A clear path from inputs to results.

    A per-member magnification check treats each column on its own. A real frame leans as a whole, and that lean feeds back into the moments: the P-Δ effect. This solver assembles the elastic and geometric stiffness of a complete portal frame, iterates the second-order displacements, and reports the load factor at which the frame would buckle.

    Step-by-step methodWorked exampleAssumptions explained

    How to use

    1. Set the number of bays, the span and the eaves height.
    2. Choose fixed or pinned bases, and whether to include second-order effects.
    3. Enter the section properties for the columns and the beams.
    4. Apply the beam gravity load and the lateral load at each eave.
    5. Read the sway, the critical factor and the largest end moment.

    The formula

    Keu=F  ;  (Ke−Kg)u=F  ;  det⁡(Ke−λKg)=0K_e u = F \;;\; (K_e - K_g)u = F \;;\; \det(K_e - \lambda K_g) = 0

    Reference: Direct stiffness method for 2D frames; consistent geometric stiffness (Cook et al., Concepts and Applications of Finite Element Analysis, ch. 11).

    Worked example

    Two-bay portal, fixed bases, 25 kN/m and 30 kN per eave

    Inputs
    • Bays 2 × 6 m, eaves 4 m.
    • E 200 GPa, columns 20,000 cm⁴ and 150 cm², beams 30,000 cm⁴ and 120 cm².
    • Beam gravity 25 kN/m, lateral 30 kN at each of the three eaves.
    Calculation path
    1. 1

      The elastic solve gives 5.54 mm of sway before the geometric stiffness is added.

    2. 2

      The gravity load puts 300 kN down the columns, so K_g softens the frame slightly: 5.57 mm.

    3. 3

      The critical factor is λ = 187.6, so the P-Δ amplification is 1/(1 − 1/187.6) = 1.005.

    4. 4

      The largest end moment is 126.8 kN·m, at the base of the leeward column.

    Result

    5.57 mm sway · drift 1/718 · λ 187.6 · 126.8 kN·m peak moment.

    Interpreting the result

    The drift ratio (sway over height) is the serviceability number; 1/718 is well inside a typical 1/300 limit. The critical factor is the stability number: a λ of 187 means the frame could carry 187 times the applied load before elastic buckling, so the P-Δ amplification is negligible here. Shorten the columns or raise the gravity load and λ falls fast.

    Assumptions & limitations

    • Euler–Bernoulli frame elements with 3 degrees of freedom per node.
    • Two sub-elements per member; buckling converges from above as the mesh is refined.
    • Elastic material, prismatic members, loads applied at the nodes and as uniform beam loads.
    • No joint eccentricity, no shear deformation and no residual stresses.

    Frequently asked questions

    What is the difference between P-δ and P-Δ?

    P-δ is the moment from curvature within a member; P-Δ is the moment from the relative displacement of the member ends. A frame solver captures P-Δ directly and P-δ through the geometric stiffness.

    Does a pinned base help or hurt?

    It reduces the base moment but increases sway and lowers λ, because the frame has less rotational restraint.

    Why does the critical factor matter if the frame is stable?

    It sets the amplification. At λ = 5 the second-order moments are 25% larger than first order; at λ = 1.2 they are five times larger.

    Can I model a braced frame?

    Add a member between the eaves and the base; its axial stiffness will carry the lateral load and the frame will behave as braced.

    Methodology last reviewed 2026-10-21. Results are informational and not financial or engineering advice — see our disclaimer.