Michell Truss
Classic 3-bar truss structure. Validates bar element formulation, assembly for truss structures, and displacement computation against hand calculations.
Problem Setup
| Parameter | Value |
|---|---|
| Configuration | 3 nodes, 3 bar elements |
| Node 0 | (0, 0) -- fixed |
| Node 1 | (0, 0.5) -- fixed |
| Node 2 | (1.0, 0.25) -- loaded |
| Bar area | 0.001 m^2 each |
| Young's modulus (E) | 200 GPa (steel) |
| Load | Fy = -1000 N at node 2 |
Mesh Statistics
| Property | Value |
|---|---|
| Nodes | 3 |
| Elements | 3 |
| Element Type | Bar (2-node truss) |
| DOFs | 6 |
| Material | E = 200 GPa, ν = 0.3 |
| Plane Assumption | N/A (1D truss) |
| Solver | Cholesky (direct) |
| Solve Time | 0.8 ms |
Boundary Conditions
| Type | Location | DOF | Value |
|---|---|---|---|
| Fixed | Node 0 (0, 0) | ux, uy | 0 |
| Fixed | Node 1 (0, 0.5) | ux, uy | 0 |
| Point Load | Node 2 (1.0, 0.25) | Fy | -1000 N |
Results
Displacement Contour
Three-panel displacement field showing magnitude |u| and components ux, uy.
Deformed Mesh
Deformed mesh (cyan) overlaid on original (gray dashed) with displacement vectors showing truss deflection under load.
Validation
Hand calculation:
Bar lengths: L = sqrt(1^2 + 0.25^2) = 1.0308 m (bars 0-2 and 1-2)
Vertical equilibrium: F * (0.25/1.0308) * 2 = 1000
F = 1000 * 1.0308 / (2 * 0.25) = 2061.6 N (compression in each bar)
| Node | ux | uy |
|---|---|---|
| 0 (fixed) | 0 | 0 |
| 1 (fixed) | 0 | 0 |
| 2 (loaded) | ~0 | -4.38e-5 m |
Discussion
The Michell truss is the simplest validation case for bar elements. It verifies that the 4x4 bar element stiffness matrix, assembly into the global system, and boundary condition enforcement all work correctly. The small displacement at node 2 is consistent with the high axial stiffness of the bars (E=200 GPa, A=0.001 m^2). The zero Von Mises stress output is expected since bar elements do not compute element-level stress in the same way as Q4 elements.