Validation / 2 of 10

Bar under end pressure

The axial bar again, now compressed by a uniform pressure on its end face. Pressure pushes along the inward normal, so a positive pressure must shorten the bar. The example checks the sign of pressure loads and that a pressure is treated as a force per area, not as a total force.

Tip displacement u; reference u = −p·L / E
Reference	KoFEM	Difference
−4.7619 × 10⁻³ mm	−4.7358 × 10⁻³ mm	−0.55 % within ±2 %

Open in KoFEMOpens the solved model in the app.

The problem

Prismatic bar of length L along x, W × H section. Sketch, not to scale; thin lines show the element mesh.

$L$	1000 mm
$W \times H$	100 × 100 mm
$E$	210 GPa
$ν$	0.3
$p$	1 MPa

Supports	x = 0 face fixed (Ux, Uy, Uz).
Load	Pressure p on the x = L face.
Measured	Mean x-displacement of the x = L face.

Analytical solution

On the end face the outward normal is $n = e_{x}$ , so the traction is $t = - p n = - p e_{x}$ . The bar carries an axial force $N = - p A$ and a uniform compressive stress $σ_{x x} = - p$ . Following the axial bar derivation,

u (L) = \frac{N L}{E A} = - \frac{p L}{E} .

The cross-section cancels: an end pressure shortens a bar by the same amount whatever its size. With $p = 1 MPa$ , the tip moves by $u (L) = - 4.7619 \times 10^{- 3} mm$ .

Finite element model

Mesh	20 × 4 × 4 grid: 320 linear 8-node hexahedra (P1)
Unknowns	525 nodes, 1,575 degrees of freedom
Material	Linear elastic, isotropic
Solver	KoFEM engine (MFEM), the WebAssembly build the app runs: engine-5ceb743373dd

The pressure is integrated over the end face with the same boundary integrator as the axial bar's force, with the traction $t = - p n$ built from each face's normal.

Results

Quantity	tip displacement u
Reference	u = −p·L / E = −4.7619 × 10⁻³ mm (Gere & Goodno, Mechanics of Materials)
KoFEM	−4.7358 × 10⁻³ mm
Difference	−0.55 % within ±2 %

Discussion

The difference, −0.55 %, matches the axial bar in tension (−0.55 %). That is what linearity predicts: the resultant here is $p A = 10 kN$ instead of $P = 1000 kN$ , and every displacement scales with the load. An identical relative difference shows that the pressure path builds the same load vector as the force path with $t = - p n$ , so both its sign and its area scaling are right.

The two ways this could go wrong are easy to spot. A flipped sign would stretch the bar instead (about −200 %). Treating $p$ as a total force instead of a force per area would be off by a factor of $A = 1.0000 \times 10^{4} {mm}^{2}$ .

Reproduce

In the app: open the model, then Studies → Static 1 → Run static solve, select the result plot and set Field to Ux. The Result summary shows Min Ux -4.736e-3 mm.

From a clone (Node.js 20+, bash, curl), this solves the model with the same engine and prints its line:

git clone https://github.com/mkofler96/KoFEM.git && cd KoFEM
bash scripts/fetch-wasm-engine.sh
node examples/validation/run.mjs --case surface-pressure

Case                                  Quantity                                  FE   Reference     Err%  Tol%  Status
Bar under end pressure                tip displacement u                -4.7358e-3  -4.7619e-3    -0.55     2  PASS

References

J. M. Gere and B. J. Goodno, Mechanics of Materials, Cengage Learning: axially loaded members, torsion, deflections of beams.

KoFEM is research software and is not certified for safety-critical design. It solves linear-static, linear-elastic problems. Check results that matter against your own references.