Stability of Finite-DOF Systems

When solving a stability problem, the first question is: how many independent parameters are needed to describe the new post-buckling displacement configuration? This number is the system's buckling DOF.

The DOF depends on constraints and rigidity assumptions. For an elastic column the deflection curve needs infinitely many parameters — it is an infinite-DOF problem (discussed in §11-3); whereas in simplified engineering models the column is often taken as a rigid bar with deformation concentrated at individual hinges, giving a finite buckling DOF.

① Infinite DOF

Any cross-section of an elastic column may deflect laterally — the deflection curve $y(x)$ is continuous and needs infinitely many independent parameters; this is an infinite-DOF problem.

② Simplified finite-DOF model

When the column itself is treated as rigid and deformation is concentrated only at elastic supports or hinges, the buckling configuration is described by finitely many independent parameters. Below, a column formed by three rigid bars connected by two elastic hinges has buckling DOF $n = 2$ or $3$, depending on the number of hinges.

The static method (equilibrium method) rests on the idea: near the original equilibrium, assume a small new displacement configuration and write the equilibrium equations in that new configuration; if the equations admit a non-trivial solution, the new configuration can coexist with equilibrium — the corresponding load is the critical load.

① Equilibrium in the new configuration

Take Animation 11.2.2-a's single-DOF model: a rigid column $AC$ fixed at the bottom with a horizontal support at the top ($A$'s rotational stiffness $k_\theta = 3EI/l$). After buckling, the rotation $\theta$ at $A$ is the only DOF. Writing the moment equilibrium for the rigid bar:

② Non-trivial solution of the homogeneous equation

Rearranging gives a homogeneous equation in $\theta$:

Setting the coefficient to zero: $F_{\mathrm P} l - 3EI/l = 0$, giving the critical load

③ Generalization to n DOF · Characteristic equation

For a system with $n$ DOF, set up $n$ independent equilibrium equations in the new configuration, giving a homogeneous linear system in the $n$ independent parameters. The condition for a non-trivial solution is that the coefficient determinant vanishes:

In the system of Animation 11.2.3-a, bars $AB$, $BC$, $CD$ are all rigid; at nodes $B$ and $C$ there are elastic supports (stiffness $k$); at the $D$ end there is an axial compression $F_{\mathrm P}$. Find the critical load and the corresponding mode shape.

① DOF · New displacement configuration

At buckling, the new configuration is fully determined by the two vertical displacements $y_1, y_2$ of hinges $B, C$ — DOF $n = 2$. The elastic-support reactions are $k y_1$ and $k y_2$.

② Free-body equilibrium

Overall moment equilibrium about $A$ ($\sum M_A = 0$) gives the vertical reaction at $D$:

Taking moments to the right of hinges $B$ and $C$ ($\sum M_B = 0,\ \sum M_C = 0$):

③ Reduce to a homogeneous system

Eliminating $F_{yD}$ gives a homogeneous system in $y_1, y_2$:

④ Stability equation · Eigenvalues

Non-trivial solution requires the coefficient determinant to vanish:

Expanding: $3F_{\mathrm P}^2 - 4klF_{\mathrm P} + (kl)^2 = 0$, giving two eigenvalues:

⑤ Take the smaller · Critical load

The two eigenvalues $F_{\mathrm{P1}} = kl/3$ and $F_{\mathrm{P2}} = kl$ of Ex. 11-1 correspond to two independent buckling mode shapes. Back-substitution into the homogeneous system yields the ratio $y_2 / y_1$ at the two hinges:

The energy method starts from the stationary-potential principle: a system is in equilibrium if and only if, for any admissible displacement / deformation, the total potential $E_{\mathrm P}$ is stationary, i.e. the first variation vanishes:

The animation below shows the stationary-potential principle intuitively — a perturbed system remains in equilibrium only if the first variation of its potential is zero:

The sign of the second variation $\delta^{2} E_{\mathrm P}$ near the stationary point classifies the stability. Intuitively, the three equilibria of a ball on a surface capture this:

Total potential of a deformable system · Two parts

For a deformable system, the total potential consists of strain energy $U$ and load potential $U_{\mathrm P}$:

Two equivalent ways to assess stability via the potential: (i) stationary potential $\delta E_{\mathrm P} = 0$ with a non-trivial displacement — i.e. an eigenvalue problem; (ii) second variation vanishing $\delta^{2} E_{\mathrm P} = 0$ — the critical-state criterion. Both are equivalent; the first is used in practice.

Using the single-DOF rigid column of the figure, we re-derive $F_{\mathrm{Pcr}}$ by the energy method and compare with the static-method result. Before buckling the total energy is zero; we look at the energy change due to rotation $\theta$ at $A$.

Stationary potential → critical load

Total potential:

Stationary condition $\mathrm d E_{\mathrm P}/\mathrm d \theta = 0$:

For an $n$-DOF system, let $a_1, a_2, \ldots, a_n$ be $n$ independent generalized coordinates. The total potential after buckling is a function of these coordinates:

① Stationary-potential condition

From $\delta E_{\mathrm P} = 0$ and the arbitrariness of $\delta a_1, \ldots, \delta a_n$, we require all first-order partial derivatives of $E_{\mathrm P}$ with respect to $a_i$ to vanish:

② Homogeneous linear system

Since $E_{\mathrm P}$ is a quadratic form in $a_i$, its first derivatives are linear in $a_i$, so (11-5) is a homogeneous linear algebraic system in $a_1, \ldots, a_n$. $a_1 = \cdots = a_n = 0$ is the original equilibrium.

③ Stability equation · Eigenvalues

Buckling requires a non-trivial solution, so the coefficient determinant must vanish:

$D = 0$ gives $n$ eigenvalues — $n$ possible critical loads. The smallest is the actual critical load, and the corresponding $a_i$ ratios give the buckling mode.

Re-solve Ex. 11-1 (see Animation 11.2.8-a) by the energy method and compare with the static-method result $F_{\mathrm{Pcr}} = kl/3$.

① Geometry and animation

The buckling configuration is still described by two parameters $y_1, y_2$. The rigid bars store no energy; all strain energy is in the elastic supports at $B$ and $C$.

② Strain energy U

Sum of strain energies in the two elastic supports:

③ Load potential · Horizontal displacement at D

Using (11-4) approximation $l(1-\cos\theta) \approx l\theta^{2}/2$ and summing over the rigid segments gives the horizontal displacement at $D$:

④ Total potential · Stationary condition

Setting $\partial E_{\mathrm P}/\partial y_1 = 0$ and $\partial E_{\mathrm P}/\partial y_2 = 0$ and rearranging:

⑤ Characteristic equation · Result

Vanishing of the determinant:

Expanding: $3 F_{\mathrm P}^{2} - 4 kl\, F_{\mathrm P} + (kl)^{2} = 0$ — exactly the same characteristic equation as Ex. 11-1 by the static method.

With the finite-DOF system as vehicle, this section has developed two solution paths — the static method and the energy method — and verified via a single example (Animation 11.2.9-a, two DOF) that both yield the same characteristic equation and the same critical load, exposing their deep equivalence.