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Validation/Structural mechanics

Cantilever beam

The first published-reference case for the structural backend — and the one that found the peak stress at a re-entrant corner does not converge at all, which is why this page validates a deflection and says plainly why it does not validate a stress.

VerificationDeflection converges to +2.76 % of Euler-Bernoulli; −1.34 % once shear is added

Why this case

A cantilever under a distributed load is the textbook problem every strength-of-materials course opens with: fix one end, load the beam along its length, and elementary theory gives the tip deflection in closed form. Like the shock tube and the plate with a hole, the reference is an equation this project can solve itself, not a measurement with its own uncertainty to argue about.

It is also the first published-reference case for the structural (code_aster) backend: seven analysis types ship, and until this page had a comparison behind none of them.

PhysicsLinear elastic, small strain, static
Solvercode_aster, MECA_STATIQUE, tetrahedral quadratic (TET10)
ReferenceEuler-Bernoulli beam theory; Timoshenko & Goodier, Theory of Elasticity
AgreementDeflection converges to +2.76 % of Euler-Bernoulli, −1.34 % once shear is added
ReproduceImport the bar → Structural → Linear static → Netgen mesh → Run

The problem

A steel bar, one end clamped, a uniform pressure on the top face bending it downward — the standard idealisation of a cantilever under a distributed load.

Geometry

  clamped                                    p = 1e5 Pa
    ┃█████████████████████████████████████│  ↓↓↓↓↓↓↓↓↓↓
    ┃                                       ▔▔▔▔▔▔▔▔▔▔▔▔
    ┃─────────────── L = 0.1 m ────────────▶
  x = 0                                  free tip

  cross-section 0.02 m × 0.02 m (y, z), L/h = 5

Steel: E = 210 GPa, ν = 0.3. The load is a uniform pressure of 1×10⁵ Pa on the top face, which over the 0.1 m span is a distributed load w = pressure × width = 2000 N/m.

The exact solution

For a uniformly distributed load on a cantilever, elementary beam theory gives

    delta = w L^4 / (8 E I),   I = b h^3 / 12

which for w = 2000 N/m, L = 0.1 m, E = 210 GPa and I = 1.3333×10⁻⁸ m⁴ gives δ = 8.9286×10⁻⁶ m.

Results

The mesh coloured by axial displacement, zero at the clamped root and rising toward the free end
The finest mesh, coloured by axial (X) displacement — zero at the clamped root, rising toward the free end. The surface shown sits off the neutral axis, so this component reads the bending rotation as well as the transverse deflection it accompanies. From the application's own Post-processing image export.
Tip deflection against mesh refinement, converging above the Euler-Bernoulli line and settling close to the Timoshenko-corrected one
Tip deflection on five meshes from 19 to 62,797 tetrahedra, against both references.
Tip deflection, across mesh refinement
MeshTetsSHD Sim δ (m)vs Euler-Bernoullivs Timoshenko
Default198.5941 × 10⁻⁶−3.75 %−7.59 %
maxh 0.0087499.0991 × 10⁻⁶+1.91 %−2.16 %
maxh 0.0042 1369.1546 × 10⁻⁶+2.53 %−1.56 %
maxh 0.00220 0019.1715 × 10⁻⁶+2.72 %−1.38 %
maxh 0.001562 7979.1754 × 10⁻⁶+2.76 %−1.34 %
The deflection converges cleanly: the last refinement, from 20 001 to 62 797 tetrahedra, moves the answer by only 0.04 percentage points. This is a settled number, not one still finding its way to an answer.

The finest mesh sits +2.76 % above the theory that ignores shear and −1.34 % against the theory that includes it — closer to the more complete physics, which is the expected direction: a finite-element solution has no reason to reproduce a simplification’s own error.

What the stress does not show

The same five meshes, read for peak von Mises stress instead of deflection:

Peak von Mises stress, across mesh refinement
MeshSHD Sim peak von Mises (Pa)Euler-Bernoulli σ at the root (Pa)
Default5.31 × 10⁶7.50 × 10⁶
maxh 0.0086.95 × 10⁶7.50 × 10⁶
maxh 0.0048.30 × 10⁶7.50 × 10⁶
maxh 0.0021.107 × 10⁷7.50 × 10⁶
maxh 0.00151.217 × 10⁷7.50 × 10⁶

What this case settles

Before this study, the only number connected to this set-up was a peak displacement sitting in the application’s own end-to-end test suite since the structural module shipped, with nothing to check it against. This page is what checks it: on the same coarse default mesh the closed form is 3.75 % away, squarely inside what a 19-element mesh should cost a bending problem, and refinement carries the answer to within a per cent and a half of the more complete shear-corrected theory.

It also settles what this set-up cannot answer: a stress read at a re-entrant corner, which is a statement about the idealisation, not about the solver.

Reproducing this

  1. Import a long, thin bar; choose Structural → Linear static, backend code_aster
  2. Material: Steel (mild) from the catalogue
  3. Restraints: fix all three translations on the clamped end face
  4. Loads: a Pressure load of 1×10⁵ Pa on the top face
  5. Mesh with Netgen — the comparison above used the default sizing and three levels of refinement
  6. Solve; peak displacement and peak von Mises are read from the Results pane

Repeat this yourself

Every case here is set up from the worked examples in the product, with no hand-editing of solver files — so you can run it, and get the same numbers. The free tier runs real cases up to 250,000 cells of fluids, or 100,000 nodes of solid, with no account needed to download and no time limit.

All validation cases · Written by the team building SHD Sim.