Case 11 — Cantilever Column with Combined Axial and Lateral Load (P-Delta Effect)
A vertical cantilever column, fixed at the base and free at the top, is loaded simultaneously by a lateral (horizontal) force and an axial (vertical) compressive force applied at the free end. The axial force amplifies the lateral deflection through second-order (P-Delta) effects, producing a top displacement larger than the first-order (linear) solution would predict. This verification case validates RodX's geometric nonlinearity (P-Delta) implementation.
Description
A cantilever column of length $L$, fixed at the base (all three degrees of freedom restrained) and free at the top. A horizontal force $H$ and a vertical (axial, compressive) force $P$ are applied simultaneously at the free end.
Determine (accounting for second-order (P-Delta) effects):
- Lateral displacement at the top of the column
- Bending moment at the base of the column
Structural scheme
Geometry, boundary conditions, and load applications used in the verification model.
Model parameters
| Parameter | Value |
|---|---|
| Units | m, kN |
| Element type | Beam-column element (geometric nonlinearity enabled) |
| Material | Steel, $E = 2.1 \times 10^{8}$ kN/m² |
| Section properties | $I = 8.014 \times 10^{-7}$ m⁴ |
| Length | $L = 4$ m |
| Boundary conditions | Base node: fixed (all 3 DOF restrained); top node: free |
| Loads | Horizontal load $H = 2$ kN and vertical (axial) load $P = 5$ kN at top node |
Analytical solution
The column is governed by the beam-column differential equation:
$$EI\,v''(x) + P\,v(x) = M_0(x)$$
where $M_0(x) = H(L - x)$ is the first-order moment from the lateral load, and $P$ enters through the parameter $k = \sqrt{\dfrac{P}{EI}}$.
Applying the boundary conditions of a fixed base ($v(0) = 0$, $v'(0) = 0$) and evaluating the solution at the free end ($x = L$) gives the closed-form top displacement:
$$\Delta = \frac{HL^3}{3EI}\left[\frac{3(\tan kL - kL)}{(kL)^3}\right]$$
The bending moment at the base includes the second-order contribution from the axial force acting through the lateral displacement:
$$M_{\text{base}} = HL + P\Delta$$
For the parameters above:
$$kL = 0.6895$$
$$\Delta_{\text{theoretical}} = 313.219 \text{ mm}$$
$$M_{\text{base}} = -9.566\,\mathrm{kN{\cdot}m}$$
Numerical results
Displacements
Bending moment diagram
Comparison
| Parameter | Theoretical | RodX | Midas Civil |
|---|---|---|---|
| Δx(B), mm | 313.219 | 313.066 | 313.123 |
| M, kN·m | −9.566 | −9.565 | −9.566 |
The numerical results obtained with RodX are in agreement with the analytical solution and reference FEA results.