Solution, reactions and equilibrium

Full and reduced systems

Original equation:

K u = F + R

Here R denotes support reactions. Partition rows and columns into free f and constrained c:

In FEM, a free degree of freedom is a nodal translation or rotation whose value is not prescribed by boundary conditions and must be determined by analysis. The term refers to an individual displacement component, not to the entire node. It does not imply motion without resistance: elements connected to the node may deform and develop elastic forces. For example, the unsupported end of a cantilever frame has free X and Y translations and a free rotation, but the cantilever stiffness resists these motions.

[ Kff  Kfc ] [ uf ] = [ Ff ] + [  0 ]
[ Kcf  Kcc ] [ uc ]   [ Fc ]   [ Rc ]

For the implemented homogeneous supports, u_c = 0. The first block row becomes a system for the unknown free displacements:

K_ff u_f = F_f

Loads applied to constrained DOFs do not enter the reduced right-hand side but remain in the full F and are included in reaction recovery.

Checks before solving

validateReducedSystem checks coefficient finiteness and positive definiteness of the symmetrized K_ff using Cholesky factorization. Failure usually means one of two things:

• insufficient supports allow rigid-body motion;
• an internal mechanism remains, for example a node can move transversely to every connected truss member without extending them in the linear approximation.

For a unique solution to the linear static problem considered here, the supports and element stiffness must exclude rigid-body motions and internal mechanisms — possible motions without elastic resistance. Free DOFs are allowed: the corresponding unknown displacements are determined from the equilibrium equations K_ffu_f = F_f, and reactions are calculated after solving. A fully constrained model is a valid special case: with zero prescribed displacements, there are no free DOFs, all displacements are zero, and reactions balance the applied loads.

Expanding to the full vector

After solving the reduced system, a zero vector of length n is created and the computed components are inserted at indices freeDOFs:

u = zeros(numberOfDOFs, 1);
u(freeDOFs) = K(freeDOFs, freeDOFs) \ F(freeDOFs);

The full vector is needed to recover element strains, reactions and results at any node. Constrained components remain exactly zero.

Reactions and global equilibrium

After expanding the solution, reactions are calculated from the full equation:

R = K u − F

In an ideal solution, components of R at free DOFs are zero. At constrained DOFs, they give the forces and moments exerted by the supports on the model.

A small residual vector Ku − F − R alone is insufficient: it checks algebra but not the signs and positions of external forces. Therefore calculateStaticEquilibrium independently sums:

ΣF_x = 0,   ΣF_y = 0,   ΣM_z = 0

Moments include the lever arms of nodal forces xF_y − yF_x and the applied nodal moments M_z of frames.

Two analytical checks

Axially loaded member

For a member of length L, area A and Young’s modulus E subjected to axial force F:

u = FL/(EA),   R_x = −F

This problem checks axial stiffness, DOF mapping, reduction, reaction sign and global balance.

Euler–Bernoulli cantilever

For a cantilever of length L subjected to transverse force P at the free end:

v = PL³/(3EI),   θ = PL²/(2EI)

R_y = −P,   M_z = −PL

One frame element reproduces this solution exactly for a nodal end force. The check is in tests/run_verification_tests.m.

Postprocessing in the offline HTML/SVG viewer

After analysis, save the model and results to a compact JSON file and open it in the MKE-F offline postprocessor. The viewer works directly through file://: no local server, npm or internet connection is required.

1. Run the analysis and export the result

Start GNU Octave from the project root and run:

addpath(pwd);
setup;

problem = StructFEProblem(fullfile('examples', 'cases', 'ANSYSBeamStatic01.txt'));
model = problem.GetAnalysisModel();
result = problem.RunStatic();

exportOptions = struct( ...
    'title', 'Static frame', ...
    'lengthUnit', 'm', ...
    'forceUnit', 'N', ...
    'momentUnit', 'N*m', ...
    'stressUnit', 'Pa', ...
    'prettyPrint', true);

exportPostprocessorData(model, result, fullfile('output', 'static-frame.json'), exportOptions);

exportPostprocessorData checks dimensions and finiteness, excludes global and element matrices, and writes versioned UTF-8 JSON. Unit fields are labels only: numerical values are not converted. For this SI export, the postprocessor displays coordinates and displacements in meters, forces, loads and reactions in newtons, moments in N·m, and stresses in Pa.

2. Open and adjust the view

1. Open postprocessor/index.html in a modern Edge, Chromium or Firefox browser.
2. Click Open JSON or drag output/static-frame.json into the dashed area. A supplied example is also available in examples/results/static-frame.json.
3. Toggle original and deformed geometry, supports, loads, reactions, and node or element numbers in the Layers.
4. Leave Scale = Automatic for automatic displacement magnification, or enter a positive factor manually. Use the mouse wheel to zoom and drag to pan; Reset view fits the model to the window.
5. Click a node or element to select it and view its ID and basic data.

3. Select a static result

• For type 112 truss elements, the Overlay list displays axial force or stress. Red indicates tension; blue indicates compression.
• For type 113 frame elements, the Overlay list draws diagrams of N, V or M. Without distributed loads, N = −f₁ is positive in tension, V = −f₂, and the moment varies linearly from −f₃ to f₆, where f denotes local end forces. Under a uniform transverse load, version-2 JSON produces a parabolic moment diagram. Positive values are plotted toward the local +Y axis.
• Deformation, arrow and diagram scales are independent and shown in the legend. An empty or unsupported diagram is replaced with a clear error message.

4. Save the illustration

Export SVG saves a scalable vector drawing with embedded styles and analysis context. For a raster image, select 1×, 2× or 4× resolution and click Export PNG. The export includes the result area and, for transient analysis, the current chart, but excludes buttons and the sidebar.

Verification: three-member portal under a uniform load

M. F. Astakhov, A. V. Karavaltsev, S. Ya. Makarov — Handbook of Aircraft Strength Calculations, p. 327, Table III.35. Identical columns AB and DC, beam BC, and rigid corners B and C. Both translations are restrained at bases A and D, with free rotations (support type 4). There are exactly three type-113 elements and four nodes. The model uses linear Euler–Bernoulli theory without shear deformation or geometric nonlinearity.

Selected data: l = 6 m, h = 3 m, q = 10 kN/m, E = 200 GPa, A₁ = A₂ = 0.02 m², J₁ = 8·10⁻⁵ m⁴, J₂ = 1.6·10⁻⁴ m⁴. The program denotes second moments of area by I. Density 7800 kg/m³ is required input but does not affect statics; self-weight is not added automatically. Global X points right and Y up. Only member 2 is loaded: 20,2,2,0,-10000.

Drawing reference and axial deformation

With positive downward load intensity q and sagging-positive beam moments:

k = (J₂/J₁)(h/l) = 1;   V_A = V_D = ql/2
H₀ = ql² / [4h(2k + 3)];   H_A = +H₀, H_D = −H₀
M_B = M_C = −H₀h;   M(l/2) = ql²/8 − H₀h

The drawing formulas neglect axial deformation. Element 113 includes finite EA. An independent force-method solution, releasing the horizontal restraint at D, gives:

H [2h³/(3EJ₁) + h²l/(EJ₂) + l/(EA₂)] = ql³h/(12EJ₂)
H = ql² / {4h[2k + 3 + 3J₂/(A₂h²)]}

Substituting this H in the moment expressions gives the exact reference for the finite-EA model. Equal column shortening does not affect horizontal thrust but is included in the vertical displacements at B and C. The horizontal reaction differs from the drawing by approximately 0.0533%.

Running the example and checking results

Load CasePinnedPortalUniform.txt with Open TXT in the preprocessor and select Run analysis. Automated verification is in example_08_pinned_portal_uniform.m:

addpath(pwd);
setup;
addpath(fullfile(pwd, 'reference', 'examples'));
r = example_08_pinned_portal_uniform();
exportPostprocessorData(r.model, r.analysis, 'output/pinned-portal-uniform.json');

The example checks all reactions, 18 member end forces, 12 nodal displacements, axial forces, beam section moments, equilibrium, and convergence toward the drawing formula when the beam area increases by a factor of 100. Force tolerance is 10⁻⁷ + 10⁻⁹|reference| in N or N·m; displacement tolerance is 10⁻¹² + 10⁻⁹|reference| in m or rad. It runs as part of test_reference_examples.

Moments in localEndForces act on the element, so for BC: M_B = -f(3) and M_C = f(6). Include the load when recovering the beam diagram: M(x) = −Hh + qx(l−x)/2. Linear interpolation between end moments would miss the positive maximum. Three elements suffice for these forces and nodal displacements; deflection inside the loaded beam additionally requires the particular solution for q or mesh refinement.

Open the supplied JSON result in the postprocessor with Open JSON, then select Overlay → M for the moment diagram.

The original formulas come from the user-supplied fragment titled “Rectangular frame with equal-section columns and pinned bases”. No bibliographic details appear in the fragment. Dimensions and section properties were selected for this example.

Laboratory 03. Axial member extension: displacement and support reaction

The aim is to obtain displacement from equilibrium and explain how one unknown becomes a full vector of four displacements.

Input data and notation

Length L = 2 m, cross-sectional area A = 0.01 m², modulus E = 200 GPa, force F = 1000 N. Positive X points right. Full-vector order: [u₁x; u₁y; u₂x; u₂y].

Before running: hand calculations

1. Identify the free displacement in the diagram and write its global number.
2. Calculate the stiffness k = EA/L in N/m. Write the equation k·u₂x = F and find u₂x = FL/(EA).
3. Use equilibrium of the entire member to find reaction R₁x. Construct the full displacement vector, inserting zeros at constrained DOFs.

Run in Octave

Run from the project root. The supplied function prints the results; you do not need to edit its code for this exercise.

addpath(pwd);
setup;
addpath(fullfile(pwd, "reference", "examples"));
r = example_03_static_bar();

Working with the results

1. In the lines Fixed DOFs and Free DOFs find the numbers from your table. Number 3 denotes horizontal displacement of node 2.
2. Compare Reduced stiffness Kff [N/m] with your EA/L, and Reduced load Ff [N] with force F. Divide the second number by the first: the result should equal the third component of Complete displacement vector [m].
3. Compare the full vector with your prediction and the first component of Reaction vector [N] with reaction −F. Check R₁x + F = 0.
4. Use the formula to predict the displacement change when separately doubling L, A and E. Explain why the reaction stays the same at a fixed force.

What to include in the report

The diagram with DOF numbering, the hand-derived equation, a quantity/hand calculation/Octave/units table for stiffness, displacement and reaction; explain zero components and predict the three parameter changes.

Checking distributed loads

Consistent element loads enter the full F before solving and calculating R=Ku−F. For a uniform load, the resultant is qL and acts at the midpoint of the original element; global equilibrium is also checked using these original resultants. For a cantilever, |v(L)|=|q|L⁴/(8EI), |θ(L)|=|q|L³/(6EI); free-end forces are zero.

MPC

MPC analysis solves Kr q = Fr and reconstructs full u = Tq. Forces are recovered from the original system, including loads on dependent and fixed DOFs. Global static equilibrium includes both support reactions and MPC forces.

Complete MPC guide · JSON v3 · CaseMPCFrame.txt · CaseMPCTruss.txt