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3D Frame Analysis

A lightweight structural analysis program in your browser, in the spirit of RISA-3D and SAP2000. Generate a beam, portal frame or multi-story building, or type joints and members into the tables. Assign sections, releases and loads, then read displacements, reactions and moment, shear and axial diagrams for every ASCE 7 combination. P-Δ and modal analysis (periods and mass participation) are one click away. Nothing is uploaded; the model stays on your computer unless you share a link.

Changing a generator input rebuilds the joints, members and loads (sections, materials and combinations are kept). Choose "Custom" first to edit the tables freely.

Drag to rotate · Shift-drag to pan · Wheel to zoom · Click a member for its diagrams
Status—
Max displacement— in
Max |Mz|— kip-ft
Max |My|— kip-ft
Max |V|— kip
Max tension— kip
Max compression— kip
Max torsion— kip-ft
ΣRy / applied— kip
Max story drift—
Worst member L/Δ—
Period T1— s
#XY (up)ZSupport

Coordinates in ft. Fixed = all 6 restrained; pinned = 3 translations; roller = Y only; "pin+Rx" also restrains rotation about X.

#ijSectionEnd releasesβ (deg)L

A pinned end releases My and Mz at that end. β rotates the local y/z axes about the member axis.

#NameShaped1d2d3d4MaterialAIz (strong)IyJ

#NameEGUnit weight

JointCaseFxFyFzMxMyMz

Forces in kip, moments in kip-ft, global axes (Y up, so gravity is −Fy).

MemberCaseDirectionw

Uniform load over the full member length, in kip/ft of member length. Global directions are signed (gravity = GY with a negative w); local y / z follow the member axes.

NameDLW

Factors on the three load cases. Without P-Δ the results superpose; with P-Δ each combination is solved separately, as it must be.

JointRxRyRzMxMyMz
MemN iN jmax |Vy|max |Vz|max |T|Mz iMz max +Mz jmax |My|L/Δ

Tension positive. Mz positive = tension on the −y face (sagging for beams). Values in kip and kip-ft. L/Δ is the deflection relative to the member chord.

JointΔxΔyΔzθxθyθz

Translations in in, rotations in radians.

ModePeriod T (s)Frequency (Hz)Mass XMass YMass ZΣ XΣ YΣ Z

The share link stores the whole model in the URL fragment, which is never sent to the server.

$$ \mathbf{k}_e^{(xy)} = \frac{EI_z}{L^3}\begin{bmatrix} 12 & 6L & -12 & 6L\\ 6L & 4L^2 & -6L & 2L^2\\ -12 & -6L & 12 & -6L\\ 6L & 2L^2 & -6L & 4L^2\end{bmatrix},\quad \mathbf{k}_G = \frac{N}{L}\begin{bmatrix}1&-1\\-1&1\end{bmatrix},\quad \mathbf K\boldsymbol\phi = \omega^2\mathbf M\boldsymbol\phi $$
k(xy) bending block for vi, θzi, vj, θzj (the xz block uses Iy) · kG P-Δ geometric stiffness on lateral translations, N tension positive · M lumped translational mass · end releases by static condensation of the released rotations.

How the solver works

Every member is a straight, prismatic, Euler-Bernoulli beam-column with six degrees of freedom at each end: three translations and three rotations. The page forms each 12 × 12 member stiffness matrix in local axes, condenses out any released end moments, rotates it to global axes, and assembles the global stiffness matrix. It renumbers the joints by reverse Cuthill–McKee to keep the matrix banded, then factors it as LDLᵀ. A 10-story, 5 × 5-bay building (2,000+ equations) solves in well under a second.

Uniform member loads become fixed-end forces. After the solve, each member's end forces are recovered and the internal forces are computed at 40 stations from statics, using the end forces plus the distributed load. Deflections along each member come from integrating the curvature M/EI twice between the computed end displacements, so they include the effect of end releases exactly. The page checks equilibrium of every combination by comparing the sum of the vertical reactions with the total applied load.

Instability. If a translational pivot vanishes during factorization, the structure is a mechanism, and the page names the joint and direction. If a rotational pivot vanishes, for example at a joint where every member is pinned, that rotation has no stiffness and does not affect the answer. The page restrains it and lists it as a warning.

Rigid floor diaphragms. With the option on, the joints at each floor level (the same Y, not restrained against X, Z or rotation about Y) are tied to a master point at the level's mass centroid: ux = uxm + θy·Δz, uz = uzm − θy·Δx, θy = θym. The floor then translates and twists as a rigid body in plan while each joint keeps its own vertical displacement and bending rotations, which is how a concrete slab or a composite deck is normally idealized. Because the master sits at the mass centroid, the floor mass and its polar moment of inertia stay uncoupled, and torsional modes come out correctly. Without diaphragms, each frame line moves on its own, and an unbraced interior frame can dominate the first mode.

P-Δ. With P-Δ on, each combination is solved, the member axial forces are taken from that solution, and the geometric stiffness N/L is added to both lateral translations of every member. This repeats until the largest displacement changes by less than 0.01%. A negative pivot means the frame has buckled under that combination. Only the chord (P-Δ) effect is modelled, not member curvature (P-δ).

Modal analysis. Mass is lumped at the joints in all three translations: half of each member's self-weight at each end, plus, optionally, the dead-load joint and member loads converted to mass. Rotational inertia is neglected. The lowest modes come from subspace iteration (Bathe), with Jacobi rotations for the reduced eigenproblem. Each mode is reported with its period, frequency and effective modal mass ratio in X, Y and Z.

Verification problems

Closed-form checks (all reproduced by the automated test suite)
ProblemExactHow to set it up
Cantilever, tip load PΔ = PL³/3EI, M = PLCantilever, any segments
Simple beam, uniform wM = wL²/8, Δ = 5wL⁴/384EIContinuous beam, 1 span, pinned
Fixed–fixed beam, uniform wMend = wL²/12, Mmid = wL²/24Continuous beam, 1 span, fixed
Two equal spans, uniform wMB = −wL²/8, RB = 1.25wLContinuous beam, 2 spans
Cantilever, tip torque Tθ = TL/GJCantilever, tip torque
Column, axial P + lateral H, P-Δ, 1 elementΔ = H / (3EI/L³ − P/L)Vertical cantilever, 1 segment, P-Δ on
Simple beam, first modef₁ = (π/2)√(EI/(m̄L⁴))1 span, pinned, many segments

Worked example — fixed-base portal frame under lateral load

One bay, 20 ft wide, 12 ft tall, rigid beam, H = 10 kip at the roof

Given: fixed bases, columns with EIc, and a beam so stiff that the joints do not rotate.
Each fixed–fixed column with no end rotation has lateral stiffness k = 12EIc/h³; two columns share H equally, so each takes a 5 kip shear.
Column end moments: M = V h / 2 = 5 × 12 / 2 = 30 kip-ft at the top and bottom of each column (inflection at mid-height).
Sway: Δ = H h³ / (24 EIc). With EIc = 29,000 ksi × 500 in⁴ and h = 144 in: Δ = 10 × 144³ / (24 × 29,000 × 500) = 0.0858 in.
To reproduce it, set the beam section's Iz about 1000 times the column's. The page then gives 30.0 kip-ft and 0.086 in. With a real beam, the joints rotate, the moments shift toward the base and the drift grows, which is why a frame analysis is needed.

Limits

References: McGuire, W., Gallagher, R.H. and Ziemian, R.D., Matrix Structural Analysis, 2nd ed. (2000), Ch. 4–5 (3D frame element, transformation, condensation of releases) and Ch. 9 (geometric stiffness). Przemieniecki, J.S., Theory of Matrix Structural Analysis (1968). Bathe, K.-J., Finite Element Procedures, §11.6 (subspace iteration). Gibbs, N.E., Poole, W.G. and Stockmeyer, P.K., "An algorithm for reducing the bandwidth and profile of a sparse matrix," SIAM J. Numer. Anal. 13(2), 1976. ASCE/SEI 7-22 §2.3 and §2.4 (combinations) and §12.9.1.1 (90% modal mass participation). ANSI/AISC 360-22 Chapter C and Appendix 7.

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