2-D frame analysis
A plane frame of any shape (portal frames, multi-bay frames, gable frames, braced frames, trusses) analysed by the direct stiffness method for each ASCE 7-22 load combination: nodal displacements, support reactions, member end forces and the axial force, shear and moment along every member.
Inputs of the frame analysis #
- Sections: up to three AISC shapes (Shapes Database v16.0); a member uses its area A and strong-axis Ix. A member can instead use an A and I entered in its row (any material).
- Nodes: numbered 1, 2, ... in table order, at (x, y) with x to the right and y up. A node's support is free, fixed, pin, a roller holding Y (on a horizontal surface) or a roller holding X (on a vertical surface).
- Members: from node i to node j, the section, E (29,000 ksi for steel, AISC 360-22 B4, prefilled; enter the modulus of your material) and the end conditions: rigid, pinned at i, pinned at j or pinned both ends (a truss member). A node where every member is pinned is a pin joint; it cannot carry a nodal moment.
- Nodal loads: Fx (right positive), Fy (up positive) and Mz (counter-clockwise positive) per load type.
- Member loads: a uniform load over the whole member or a point load at a distance a from end i, acting by gravity (downward; per unit length of the member for an inclined member), along global X, or perpendicular to the member along its local y axis (90° counter-clockwise from the direction i to j).
- Load combinations: the ASCE 7-22 2.3.1 (LRFD) or 2.4.1 (ASD) combinations of the load types present, chosen by the method switch (W and E in both directions), or one load set: the unfactored sum of the loads.
Results and checks of the frame analysis #
| Item | Basis |
|---|---|
| Equilibrium of the frame | ΣFx, ΣFy and ΣM about the origin of the applied loads and the reactions, for every combination; the largest residual is reported and must be below 10⁻⁶ of the applied load |
| Analysis basis | Direct stiffness method: 6 × 6 member stiffness with axial and bending deformation, member loads as fixed-end forces, end releases by static condensation |
| Second-order effects | A note: P-Δ and P-δ are not included (AISC 360-22 C1, C2, Appendix 8) |
| Reactions and member forces | Largest and smallest reaction components at each support, and the axial force, shear and moment of each member, over all combinations |
| Largest nodal displacement | The largest x or y displacement and its combination |
The drawings (on screen, in the report and as a PDF / DXF drawing sheet) show the frame elevation on the member centre lines with node numbers, member marks (M1, M2, ...; the sections are listed by mark in the sheet notes), supports, moment releases (small circles), the nodal and member loads by load type with their values, and the bay and story dimensions; and, for the combination with the largest moment, the moment (on the tension side), shear and axial force (tension +) diagrams drawn on the frame with the largest and smallest value of each member labelled (the units are in the scale line). Uniform loads stacked on a sloping member share one label, listed from the outer band in. The largest displacement is in the design summary; the deformed shape is not drawn. Member design, connections and bases are not part of the calculation. Member moments are positive with tension on the member's local −y side (the bottom of a member drawn left to right).
Limitations #
- First-order linear elastic analysis: second-order effects (P-Δ, P-δ), geometric and material nonlinearity are out of scope. For stability design amplify the forces (AISC 360-22 Appendix 8) or use a second-order analysis.
- No shear deformation, rigid end zones, semi-rigid connections, springs, support settlement, temperature, inclined supports or partial uniform loads (use point loads or split the member).
- No pattern live load: enter the live load arrangements as separate load cases or calculations if they govern.
- No member design: use the forces in the member's material calculation.
Verification examples #
Closed-form solutions (hand calculation), one unfactored load set, E = 29,000 ksi. The portal frame is solved by slope-deflection with axially rigid members (A = 10,000 in² in the model); the derivation is in the test file (backend/tests/test_verify_frame_2d.py).
| Example | What it checks | Closed form | ILX Calcs |
|---|---|---|---|
| FR-1 | Fixed-fixed beam, w = 1 kip/ft, 20 ft | wL²/12 = 33.33, wL²/24 = 16.67 kip-ft | 33.33, 16.67 |
| FR-2 | Propped cantilever | wL²/8 = 50.0 kip-ft; 3wL/8 = 7.5 kips; 9wL²/128 = 28.13 kip-ft | 50.0; 7.5; 28.13 |
| FR-3 | Two equal continuous spans | wL²/8 = 50.0 kip-ft; 1.25wL = 25.0 kips; 0.375wL = 7.5 kips | 50.0; 25.0; 7.5 |
| FR-4 | Fixed-base portal, h = L = 12 ft, H = 10 kips | base 0.2857Hh = 34.29, top 0.2143Hh = 25.71 kip-ft; Δ = Hh³/(16.8EI) = 0.0613 in | 34.29, 25.71; 0.0613 |
| FR-5 | Three-bar truss, P = 10 kips at the apex | diagonals −P/(2 sin α) = −8.33 kips; chord +6.67 kips | −8.33; +6.67 |
Published examples (backend/tests/test_verify_frame_2d.py): F. Udoeyo, Structural Analysis (Temple University Press, 2020, open textbook), Chapters 11 (slope-deflection) and 12 (moment distribution), and AISC Steel Construction Manual, 16th Ed., Table 3-22 beam cases modelled as frames. End moments are compared as magnitudes.
| Example | What it checks | Published | ILX Calcs |
|---|---|---|---|
| UD-11.5 | Fixed column with 2 kips/ft lateral load, fixed-end beam with 20 kips (Example 11.5) | M = 17.09, 15.83, 14.58 kip-ft; Ax = 10.13, Ay = 10.20, Cy = 9.80, Cx = 9.87 kips | 17.08, 15.83, 14.58; 10.13, 10.21, 9.79, 9.88 |
| UD-12.3 | Beam over a column with a propped end, moment distribution (Example 12.3) | M = 12.48, 60.37, 75.31, 14.94, 7.47 kip-ft | 12.53, 60.27, 75.36, 15.09*, 7.55* |
| UD-12.5 | Column fixed at both ends with a pinned 65 kN/m beam, SI input (Example 12.5) | M = 13.17, 26.33, 53.39 kN·m | 13.30, 26.59, 53.18 |
| AISC-17F | Table 3-22 case 17 (p. 3-220) | M1 = −28.13, M2 = −9.38, Ma = 14.06 kip-ft; R1 = 8.44 kips | same |
| AISC-39F | Table 3-22 case 39 (p. 3-229) | R = 8.0 / 22.0 kips; MB = −40.0; M+ = 32.0 / 10.0 kip-ft | 8.0 / 22.0; −40.0; 31.94 / 10.0 |
\* Documented deviation UDOEYO-12.3-MD: the textbook stops the moment distribution after two cycles with distribution factors rounded to 0.01; the app gives the exact elastic solution (1 % apart). In Example 11.5 the textbook writes every slope-deflection equation with the same K = I/L for the 10 ft column and the 6 ft beam; the test gives the column I × 10/6 so both members have that K (with equal I the app gives 16.98 / 16.04 / 14.48 kip-ft). Examples 11.7 and 12.6 of the textbook are not used: their printed moments violate the zero column-top shear of a frame whose beam ends on a roller.