Portal Frame Under Lateral Load — Fixed vs Pinned Bases Worked Example
The simplest moment frame there is, solved by hand with slope-deflection and then by the stiffness method, with fixed and then pinned bases. The comparison shows why base fixity is the first question in any frame model: pinning the bases here multiplies the drift by 4.3.
The frame
| Geometry | One bay, L = 24 ft, h = 12 ft |
| Members | Columns Ic = 300 in&sup4;, A = 11.7 in² · Beam Ib = 500 in&sup4;, A = 10.3 in² · E = 29,000 ksi |
| Load | H = 10 kip lateral at the beam level (service wind) |
| Bases | Case A fixed, case B pinned |
Build it in the frame tool: 4 joints, 3 members, a "W" joint load of 10 kip in +X, 2D mode on.
Open 3D frame analysis →
Two unknowns: joint rotation θ and sway Δ
Column: Mtop = 2kc(2θ − 3ψ), Mbase = 2kc(θ − 3ψ), ψ = Δ/h
Beam (antisymmetric, both joints rotate θ): M = 2kb(2θ + θ) = 6kbθ
Joint equilibrium + story shear
Shear: H = 2 × 2kc(6ψ − 3θ) / h → ψ = Hh / [4kc(6 − 2.0)]
Δ = 0.2145 in (h/671), θ = 0.000993 rad
|Mbase| = 35.0 kip-ft, |Mtop| = 25.0 kip-ft (inflection point 7.0 ft above the base)
For comparison, a perfectly rigid beam gives Δ = Hh³/24EIc = 0.143 in with 30 kip-ft at each end. A real beam lets the joints rotate, so the moment migrates toward the base and the drift grows 50%.
Same answer, plus what the hand method ignored
| Quantity | Slope-deflection | Frame tool, left column | Frame tool, right column |
|---|---|---|---|
| Sway at beam level (in) | 0.2145 | 0.2173 | 0.2125 |
| Base moment (kip-ft) | 35.0 | 35.35 | 34.70 |
| Top moment (kip-ft) | 25.0 | 25.12 | 24.83 |
| Column shear (kip) | 5.0 | 5.04 | 4.96 |
| Column axial (kip) | — | ±2.08 | ∓2.08 |
The averages match the hand solution exactly. The small split comes from the beam shortening axially (it pushes the far column), which slope-deflection assumes away. The ±2.08 kip column axial force is the overturning couple: the 120 kip-ft overturning moment (10 kip × 12 ft), less the 70.05 kip-ft the fixed bases take, divided by the 24 ft bay = 2.08 kip. With a rigid beam it rises to 2.5 kip, and with pinned bases to 5.0 kip, which the anchorage and footing have to resist.
Same frame, half the restraint
Mtop = Hh/2 = 60 kip-ft in each column and at each beam end
The frame tool gives 0.920 / 0.915 in and 60.09 / 59.91 kip-ft. Pinning the bases cut the base moment to zero but put 2.4 times the moment in the beam-column joint and 4.3 times the drift on the frame.
Tools used in this example
3D frame analysis (2D mode), and the continuous beam tool for the gravity design of the beam. See the frame modeling guide for supports, releases and local axes.
Frequently asked questions
How do you calculate the lateral drift of a portal frame?
For a single-bay portal with fixed bases, write the slope-deflection equations for the two columns and the beam, using the joint rotation and the sway as unknowns. Solve the joint moment equilibrium and the story shear equation together. With a rigid beam, the drift reduces to Hh³/24EI for two fixed-base columns. With pinned bases, it is (Hh³/6EI)(1 + IcL/(2Ibh)).
Where is the inflection point in a fixed-base portal frame column?
With a rigid beam, it is at mid-height. With a flexible beam, the joints rotate and the inflection point moves up toward the beam, so the base moment exceeds the top moment. In this example, the base carries 35 kip-ft and the top 25 kip-ft, with inflection 7 ft above a 12 ft column's base.
Are column bases pinned or fixed?
Real base plates are partially fixed. Typical designs assume pinned bases for gravity-only columns and for moment frames on small footings, and fixed bases only where the base plate, anchor rods and footing are detailed for the moment. Model both bounds when drift or the beam-column joint moment governs.