Free worked example · complete calculation
See the complete reactions and force diagrams first, then work through the direct stiffness calculation for this 6 m × 4 m fixed-base frame.
Answer at a glance
Portal frame fundamentals
A portal frame is a structural system formed by columns connected to a beam or rafters through rigid joints. The frame carries vertical and horizontal loads through a combination of axial force, shear force and bending moment, while the moment-resisting joints allow the columns and roof member to act together as one continuous frame.
Portal frame analysis calculates the horizontal, vertical and moment support reactions; the translations and rotations of the joints; and the axial-force, shear-force and bending-moment response of each member. A fixed-base rigid portal frame is statically indeterminate, so this worked example uses the matrix direct stiffness method to enforce both equilibrium and displacement compatibility and to calculate frame sway and deflection.
For a rigid, statically indeterminate portal frame, the direct stiffness method is commonly used. Each beam and column contributes an element stiffness matrix; the matrices and equivalent nodal loads are transformed and assembled; support restraints are applied; and the resulting joint displacements are solved before reactions and member end forces are recovered.
The analysis calculates horizontal and vertical support reactions, base moments, joint translations, joint rotations, frame sway, member axial force, shear force and bending moment. These values are presented in reaction, deflected-shape, axial-force, shear-force and bending-moment diagrams.
Horizontal loading, unsymmetrical gravity loading or unequal stiffness can translate the roof joints laterally. In this example, the 12 kN horizontal load causes rightward sway; the rigid roof beam couples the columns so both columns bend and participate in resisting the lateral shear.
No. Analysis determines reactions, internal forces and displacements for the idealized frame. Design must additionally verify member strength, buckling, stability, connections, base plates, bracing, serviceability, load combinations and the requirements of the applicable structural code.
01 · Model inputs
A single-bay portal frame spans 6 m and rises 4 m. Both column bases are fixed and all three members meet at rigid beam-to-column joints. The roof carries an 8 kN/m downward UDL and node N3 carries a 12 kN rightward load.
| Node | X | Y | Support | Restrained DOFs |
|---|---|---|---|---|
| N1 | 0.0 | 0.0 | Fixed | UX, UY, RZ |
| N2 | 0.0 | 4.0 | Free | — |
| N3 | 6.0 | 4.0 | Free | — |
| N4 | 6.0 | 0.0 | Fixed | UX, UY, RZ |
| Load | Target | Definition |
|---|---|---|
| L1 | N3 | 12 kN in global +X |
| L2 | M2 | 8 kN/m downward over the full 6 m beam |
02 · Solved answer
The 8 kN/m roof load and 12 kN horizontal load act together as the one worked loading condition. The diagrams below are exported by the Optimal Beam calculator from this exact model.
| Support | Horizontal Rx (kN) | Vertical Ry (kN) | Moment Mz (kN·m) |
|---|---|---|---|
| N1 | 0.743 kN | 20.811 kN | 5.455 kN·m |
| N4 | -12.743 kN | 27.189 kN | 23.413 kN·m |
| Node | UX (mm) | UY (mm) | RZ (mrad) |
|---|---|---|---|
| N2 | 1.289 | -0.083 | -0.694 |
| N3 | 1.285 | -0.109 | 0.207 |
| Member | Axial (kN) | Vi / Vj (kN) | Mi / Mj (kN·m) |
|---|---|---|---|
| M1 | -20.811 | -0.743 / -0.743 | -5.455 / -8.426 |
| M2 | -0.743 | 20.811 / -27.189 | -8.426 / -27.558 |
| M3 | -27.189 | 12.743 / 12.743 | -27.558 / 23.413 |
03 · Hand calculation
This is the numerical path from the applied loads to the six reactions. The same model values are used by the calculator.
Each member is written in its own local axes, transformed to global X/Y, and assembled by matching common joint degrees of freedom. The fixed bases are then removed from the unknown set, leaving the six-by-six system shown below.
Global +X is right, +Y is up and +θ is counter-clockwise. Each member local x-axis runs from its i-node to its j-node.Start with every joint translation and rotation, then identify which values are known.
{d} = {u₁, v₁, θ₁, u₂, v₂, θ₂, u₃, v₃, θ₃, u₄, v₄, θ₄}ᵀ
{dc} = {u₁, v₁, θ₁, u₄, v₄, θ₄}ᵀ = {0, 0, 0, 0, 0, 0}ᵀ
{df} = {u₂, v₂, θ₂, u₃, v₃, θ₃}ᵀ
The complete frame has twelve global degrees of freedom. The fixed bases set six of them to zero, leaving the six roof-joint translations and rotations as the unknown vector.
| Joint | Global DOFs | Status |
|---|---|---|
| N1 | u₁, v₁, θ₁ | Restrained |
| N2 | u₂, v₂, θ₂ | Free |
| N3 | u₃, v₃, θ₃ | Free |
| N4 | u₄, v₄, θ₄ | Restrained |
Ds = 3m + r − 3j → Ds = 3(3) + 6 − 3(4) = 3
The frame is stable but statically indeterminate to degree three, so equilibrium alone cannot determine all reactions and member end actions.
This is the complete axial-and-bending matrix used for each unreleased member.
{d′e} = {uᵢ, vᵢ, θᵢ, uⱼ, vⱼ, θⱼ}ᵀ
{f′e} = [k′e]{d′e}
| uᵢ | vᵢ | θᵢ | uⱼ | vⱼ | θⱼ | |
|---|---|---|---|---|---|---|
| uᵢ | a | 0 | 0 | −a | 0 | 0 |
| vᵢ | 0 | b | c | 0 | −b | c |
| θᵢ | 0 | c | d | 0 | −c | e |
| uⱼ | −a | 0 | 0 | a | 0 | 0 |
| vⱼ | 0 | −b | −c | 0 | b | −c |
| θⱼ | 0 | c | e | 0 | −c | d |
The DOF order is [uᵢ, vᵢ, θᵢ, uⱼ, vⱼ, θⱼ]. Because translations are in metres and rotations in radians, the matrix contains compatible kN/m, kN and kN·m terms.
E = 200 GPa, A = 5000 mm², I = 200,000,000 mm⁴
E = 200,000,000 kN/m²; A = 0.005 m²; I = 0.0002 m⁴; EA = 1,000,000 kN; EI = 40,000 kN·m²
| Coefficient | Definition | Purpose |
|---|---|---|
| a | EA/L | Axial stiffness |
| b | 12EI/L³ | Transverse stiffness |
| c | 6EI/L² | Shear–rotation coupling |
| d | 4EI/L | Near-end rotational stiffness |
| e | 2EI/L | Far-end rotational coupling |
| Term | 4 m column | 6 m beam |
|---|---|---|
| EA/L | 250,000.000 kN/m | 166,666.667 kN/m |
| 12EI/L³ | 7,500.000 kN/m | 2,222.222 kN/m |
| 6EI/L² | 15,000.000 kN | 6,666.667 kN |
| 4EI/L | 40,000.000 kN·m | 26,666.667 kN·m |
| 2EI/L | 20,000.000 kN·m | 13,333.333 kN·m |
| uᵢ | vᵢ | θᵢ | uⱼ | vⱼ | θⱼ | |
|---|---|---|---|---|---|---|
| uᵢ | 250,000.000 | 0.000 | 0.000 | -250,000.000 | 0.000 | 0.000 |
| vᵢ | 0.000 | 7,500.000 | 15,000.000 | 0.000 | -7,500.000 | 15,000.000 |
| θᵢ | 0.000 | 15,000.000 | 40,000.000 | 0.000 | -15,000.000 | 20,000.000 |
| uⱼ | -250,000.000 | 0.000 | 0.000 | 250,000.000 | 0.000 | 0.000 |
| vⱼ | 0.000 | -7,500.000 | -15,000.000 | 0.000 | 7,500.000 | -15,000.000 |
| θⱼ | 0.000 | 15,000.000 | 20,000.000 | 0.000 | -15,000.000 | 40,000.000 |
| uᵢ | vᵢ | θᵢ | uⱼ | vⱼ | θⱼ | |
|---|---|---|---|---|---|---|
| uᵢ | 166,666.667 | 0.000 | 0.000 | -166,666.667 | 0.000 | 0.000 |
| vᵢ | 0.000 | 2,222.222 | 6,666.667 | 0.000 | -2,222.222 | 6,666.667 |
| θᵢ | 0.000 | 6,666.667 | 26,666.667 | 0.000 | -6,666.667 | 13,333.333 |
| uⱼ | -166,666.667 | 0.000 | 0.000 | 166,666.667 | 0.000 | 0.000 |
| vⱼ | 0.000 | -2,222.222 | -6,666.667 | 0.000 | 2,222.222 | -6,666.667 |
| θⱼ | 0.000 | 6,666.667 | 13,333.333 | 0.000 | -6,666.667 | 26,666.667 |
Direction cosines connect the member’s local axial/transverse directions to the frame axes.
[ke] = [Te]ᵀ[k′e][Te]The beam is already aligned with global X. The left column local x-axis points upward, while the right column local x-axis points downward because M3 is defined from N3 to N4.
| 0 | 1 | 0 | 0 | 0 | 0 |
| -1 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 0 | 0 |
| 0 | 0 | 0 | 0 | 1 | 0 |
| 0 | 0 | 0 | -1 | 0 | 0 |
| 0 | 0 | 0 | 0 | 0 | 1 |
| 1 | 0 | 0 | 0 | 0 | 0 |
| 0 | 1 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 0 | 0 |
| 0 | 0 | 0 | 1 | 0 | 0 |
| 0 | 0 | 0 | 0 | 1 | 0 |
| 0 | 0 | 0 | 0 | 0 | 1 |
| 0 | -1 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 0 | 0 |
| 0 | 0 | 0 | 0 | -1 | 0 |
| 0 | 0 | 0 | 1 | 0 | 0 |
| 0 | 0 | 0 | 0 | 0 | 1 |
The member load must enter the same joint-DOF system as the 12 kN nodal load.
V = wL/2; M = wL²/12
V = 8(6)/2 = 24.000 kN; M = 8(6²)/12 = 24.000 kN·m
{pM2} = {0, −24, −24, 0, −24, 24}ᵀ
{Ff} = {0, −24, −24, 12, −24, 24}ᵀ
The beam consistent-load vector is first placed at the M2 element DOFs. Adding the +12 kN nodal force at u₃ produces the complete free-DOF load vector.
| Free DOF | Applied action |
|---|---|
| u2 | 0.000 kN |
| v2 | -24.000 kN |
| θ2 | -24.000 kN·m |
| u3 | 12.000 kN |
| v3 | -24.000 kN |
| θ3 | 24.000 kN·m |
Common joint DOFs land in the same rows and columns, so their stiffness terms add directly.
[K] = Σ[Ae]ᵀ[ke][Ae]
[Kff] = [Kff]M1 + [Kff]M2 + [Kff]M3
Each transformed member matrix is placed into the global rows and columns belonging to its end nodes. The tables show the actual contribution of each member after retaining only the six free roof DOFs.
| Member | Element DOFs | Global placement |
|---|---|---|
| M1 | 1–6 | u₁, v₁, θ₁, u₂, v₂, θ₂ |
| M2 | 1–6 | u₂, v₂, θ₂, u₃, v₃, θ₃ |
| M3 | 1–6 | u₃, v₃, θ₃, u₄, v₄, θ₄ |
| u₂ | v₂ | θ₂ | u₃ | v₃ | θ₃ | |
|---|---|---|---|---|---|---|
| u₂ | 7,500.000 | 0.000 | 15,000.000 | 0.000 | 0.000 | 0.000 |
| v₂ | 0.000 | 250,000.000 | 0.000 | 0.000 | 0.000 | 0.000 |
| θ₂ | 15,000.000 | 0.000 | 40,000.000 | 0.000 | 0.000 | 0.000 |
| u₃ | 0.000 | 0.000 | 0.000 | 0.000 | 0.000 | 0.000 |
| v₃ | 0.000 | 0.000 | 0.000 | 0.000 | 0.000 | 0.000 |
| θ₃ | 0.000 | 0.000 | 0.000 | 0.000 | 0.000 | 0.000 |
| u₂ | v₂ | θ₂ | u₃ | v₃ | θ₃ | |
|---|---|---|---|---|---|---|
| u₂ | 166,666.667 | 0.000 | 0.000 | -166,666.667 | 0.000 | 0.000 |
| v₂ | 0.000 | 2,222.222 | 6,666.667 | 0.000 | -2,222.222 | 6,666.667 |
| θ₂ | 0.000 | 6,666.667 | 26,666.667 | 0.000 | -6,666.667 | 13,333.333 |
| u₃ | -166,666.667 | 0.000 | 0.000 | 166,666.667 | 0.000 | 0.000 |
| v₃ | 0.000 | -2,222.222 | -6,666.667 | 0.000 | 2,222.222 | -6,666.667 |
| θ₃ | 0.000 | 6,666.667 | 13,333.333 | 0.000 | -6,666.667 | 26,666.667 |
| u₂ | v₂ | θ₂ | u₃ | v₃ | θ₃ | |
|---|---|---|---|---|---|---|
| u₂ | 0.000 | 0.000 | 0.000 | 0.000 | 0.000 | 0.000 |
| v₂ | 0.000 | 0.000 | 0.000 | 0.000 | 0.000 | 0.000 |
| θ₂ | 0.000 | 0.000 | 0.000 | 0.000 | 0.000 | 0.000 |
| u₃ | 0.000 | 0.000 | 0.000 | 7,500.000 | 0.000 | 15,000.000 |
| v₃ | 0.000 | 0.000 | 0.000 | 0.000 | 250,000.000 | 0.000 |
| θ₃ | 0.000 | 0.000 | 0.000 | 15,000.000 | 0.000 | 40,000.000 |
| DOF | u₂ | v₂ | θ₂ | u₃ | v₃ | θ₃ |
|---|---|---|---|---|---|---|
| u₂ | 174,166.667 | 0.000 | 15,000.000 | -166,666.667 | 0.000 | 0.000 |
| v₂ | 0.000 | 252,222.222 | 6,666.667 | 0.000 | -2,222.222 | 6,666.667 |
| θ₂ | 15,000.000 | 6,666.667 | 66,666.667 | 0.000 | -6,666.667 | 13,333.333 |
| u₃ | -166,666.667 | 0.000 | 0.000 | 174,166.667 | 0.000 | 15,000.000 |
| v₃ | 0.000 | -2,222.222 | -6,666.667 | 0.000 | 252,222.222 | -6,666.667 |
| θ₃ | 0.000 | 6,666.667 | 13,333.333 | 15,000.000 | -6,666.667 | 66,666.667 |
Partitioning removes the six known zero displacements without discarding their reaction rows.
[[Kcc, Kcf], [Kfc, Kff]] {dc, df}ᵀ = {Fc, Ff}ᵀ
{dc} = 0 ⇒ [Kff]{df} = {Ff}
[Kff]{df} = {Ff}
The transformed member matrices and consistent member-load vector are assembled into the global system. Applying the six base restraints leaves this 6 × 6 roof-joint equation.
| DOF | Value | Unit |
|---|---|---|
| u₂ | 0.001289030548 | m |
| v₂ | -0.000083245350 | m |
| θ₂ | -0.000694027884 | rad |
| u₃ | 0.001284574413 | m |
| v₃ | -0.000108754650 | m |
| θ₃ | 0.000207225404 | rad |
| DOF | Value | Unit |
|---|---|---|
| u2 | 1.289031 | mm |
| v2 | -0.083245 | mm |
| θ2 | -0.694028 | mrad |
| u3 | 1.284574 | mm |
| v3 | -0.108755 | mm |
| θ3 | 0.207225 | mrad |
174166.667(0.001289031) + 15000(−0.000694028) − 166666.667(0.001284574) = 0.000000 kN
−166666.667(0.001289031) + 174166.667(0.001284574) + 15000(0.000207225) = 12.000000 kN
| Row | [Kff]{df} | Applied Ff | Residual |
|---|---|---|---|
| u₂ | 0.000000 kN | 0.000000 kN | 0.000000000 |
| v₂ | -24.000000 kN | -24.000000 kN | 0.000000000 |
| θ₂ | -24.000000 kN·m | -24.000000 kN·m | 0.000000000 |
| u₃ | 12.000000 kN | 12.000000 kN | 0.000000000 |
| v₃ | -24.000000 kN | -24.000000 kN | 0.000000000 |
| θ₃ | 24.000000 kN·m | 24.000000 kN·m | 0.000000000 |
The restrained rows were not solved for displacement; they are now used to recover all six base actions.
{Rc} = [Kcf]{df} − {Fc}Recovering the restrained rows gives every fixed-base component, not only the vertical reactions.
| -7,500.000 | 0.000 | -15,000.000 |
| 0.000 | -250,000.000 | 0.000 |
| 15,000.000 | 0.000 | 20,000.000 |
[R] = [Kcf] {0.001289030548, -0.000083245350, -0.000694027884}ᵀ
{0.742689, 20.811337, 5.454901}ᵀ
Rx kN · Ry kN · Mz kN·m
[R] = [Kcf] {0.001284574413, -0.000108754650, 0.000207225404}ᵀ
{-12.742689, 27.188663, 23.413124}ᵀ
Rx kN · Ry kN · Mz kN·m
| Support | Rx (kN) | Ry (kN) | Mz (kN·m) |
|---|---|---|---|
| N1 | 0.742689 | 20.811337 | 5.454901 |
| N4 | -12.742689 | 27.188663 | 23.413124 |
The beam displacement vector is returned to its local stiffness equation, including the UDL fixed-end actions.
{f′M2} = [k′M2]{d′M2} − {pM2}
{f′M2} = {0.742689, 20.811337, 8.425657, -0.742689, 27.188663, -27.557632}ᵀ
The recovered vector contains the actions exerted by M2 at its two nodes. The plotted internal-force signs follow the corresponding member cut-face convention.
| M2 component | [k′M2]{d′M2} | −{pM2} | Recovered end action |
|---|---|---|---|
| Nᵢ | 0.742689 kN | 0.000000 kN | 0.742689 kN |
| Vᵢ | -3.188663 kN | 24.000000 kN | 20.811337 kN |
| Mᵢ | -15.574343 kN·m | 24.000000 kN·m | 8.425657 kN·m |
| Nⱼ | -0.742689 kN | 0.000000 kN | -0.742689 kN |
| Vⱼ | 3.188663 kN | 24.000000 kN | 27.188663 kN |
| Mⱼ | -3.557632 kN·m | -24.000000 kN·m | -27.557632 kN·m |
ΣFx = 0.742689 − 12.742689 + 12.000000 = 0.000000 kNΣFy = 20.811337 + 27.188663 − 48.000000 = 0.000000 kNΣMN1 = 5.454901 + 23.413124 + 6(27.188663) − 48(3) − 12(4) = 0.000000 kN·m 04 · Hand-versus-solver verification
Independent values from the calculation above are compared directly with the stored calculator result.
| Check | Hand calculation | Optimal Beam | Difference | Status |
|---|---|---|---|---|
| Horizontal reaction at N4 | -12.742689 kN | -12.742689 kN | 0.000000% | Pass |
| Vertical reaction at N1 | 20.811337 kN | 20.811337 kN | 0.000000% | Pass |
| Vertical reaction at N4 | 27.188663 kN | 27.188663 kN | 0.000000% | Pass |
| Base moment at N1 | 5.454901 kN·m | 5.454901 kN·m | 0.000000% | Pass |
| Base moment at N4 | 23.413124 kN·m | 23.413124 kN·m | 0.000000% | Pass |
| Roof sway at N2 | 1.289031 mm | 1.289031 mm | 0.000000% | Pass |
| Rotation at N2 | -0.694028 mrad | -0.694028 mrad | 0.000000% | Pass |
| Beam end moment at N3 | -27.557632 kN·m | -27.557632 kN·m | 0.000000% | Pass |
| Right-column compression | -27.188663 kN | -27.188663 kN | 0.000000% | Pass |
05 · Analysis scope
This is a first-order structural analysis example, not a completed member or connection design.
Optimal Beam currently performs first-order linear-elastic analysis and does not include P-Delta effects. Use an appropriate second-order method when axial compression acting through sway may materially amplify moments or drift.
06 · Common modeling mistakes
Using pinned beam-to-column joints when the intended frame is moment-resisting.
Leaving either base rotation unrestrained when the idealization requires a fixed base.
Applying the roof UDL in global X or local X instead of transverse to the horizontal beam.
Reporting only vertical reactions and omitting the horizontal and moment reactions at fixed bases.
Reading a local member sign as a global direction without checking the member i-to-j axis.
Treating first-order sway as a P-Delta result or using the analysis as a code design check.
Continue with the exact model
The model, diagrams, result tables and calculation report come from this same calculator analysis.