Validation / 1 of 10

Axial bar in tension

0.00 mm0.474 mm

Open in KoFEMOpens the solved model in the app.

The problem

Analytical solution

Cut the bar at a section $x$. Equilibrium of the part to the right of the cut needs an internal axial force $N(x)=P$, the same at every section, because nothing loads the bar between its ends. Assume that this force is carried by a normal stress that is uniform over the cross-section $A=WH$:

Hooke's law relates the stress to the axial strain, and the strain is the rate of change of the axial displacement $u(x)$:

With the clamp, $u(0)=0$, the displacement grows linearly along the bar,

and the loaded end moves by

With the inputs above, $A=1.0000\times {10}^4{\text{ mm}}^2$, the stress is $\sigma =100\text{ MPa}$ and $\delta =0.47619\text{ mm}$.

Two simplifications hide in this one-dimensional model. The stress is uniform over every section, and the lateral (Poisson) contraction is free everywhere. A 3-D solid meets neither exactly near its ends: the clamp prevents the contraction at $x=0$, and at the loaded face the stress follows however the load is applied. By Saint-Venant's principle these disturbances die out within about one section width, so in a bar ten times longer than it is wide they change $\delta$ only slightly.

Finite element model

The force is applied the way the app applies a force to a face: as a uniform traction $t=P/A$, integrated over the 16 element faces of the end face. In the finite element method a traction enters through the work it does, so the force on node $i$ is

where $N_i$ is the shape function of that node. On linear quadrilateral faces of size $h\times h$, a node collects $t{h}^2/4$ from each face it touches: an interior node of the end face gets $t{h}^2$, an edge node $t{h}^2/2$ and a corner node $t{h}^2/4$. The forces add up to $tA=P$. Splitting $P$ equally over the nodes instead would get the total right but overload the edges and corners.

Results

Discussion

KoFEM's extension differs from $PL/(EA)$ by −0.55 %. Linear elements represent a linearly varying displacement exactly, so the uniform state inside the bar carries no discretisation error. What remains is the clamp. The x = 0 face is fully fixed, so near it the bar cannot contract sideways. That material is in a multiaxial state and stiffer than the 1-D model assumes, so the end moves slightly less. It is a difference between the 3-D body and the 1-D theory, so a finer mesh does not remove it.

The same bar under an end pressure comes out −0.55 %: the pressure path builds the same load vector.

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 Max Ux 4.736e-1 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 axial-bar

Case                                  Quantity                                  FE   Reference     Err%  Tol%  Status
Axial bar in tension                  tip extension δ                      0.47358     0.47619    -0.55     2  PASS

References

1. J. M. Gere and B. J. Goodno, Mechanics of Materials, Cengage Learning: axially loaded members, torsion, deflections of beams.
2. S. P. Timoshenko and J. N. Goodier, Theory of Elasticity, 3rd ed., McGraw-Hill, 1970.
3. T. J. R. Hughes, The Finite Element Method: Linear Static and Dynamic Finite Element Analysis, Dover, 2000.

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.