Stability of Rigid Frames

A frame under uniformly distributed vertical loads begins to sway immediately, and its members are in bending equilibrium. The axial-force-induced additional moment increases with sway, which grows nonlinearly — at the critical load a maximum point appears (see Animation 11.6.1-a), i.e. Type-II (limit-point) buckling.

Stiffness change in axially loaded members

The larger the axial compression $F_{\mathrm P}$, the smaller the member's rotational stiffness and lateral stiffness — when $F_{\mathrm P}$ reaches the critical value, the stiffness drops to zero. Without axial force, a cantilever member in has rotational stiffness $S_{BA} = 3i = 3EI/l$ and lateral stiffness $k = 3i/l^{2} = 3EI/l^{3}$. Once either stiffness vanishes, the rest of the structure can no longer support the member.

① Equilibrium of an axially loaded member

Consider the member in the figure: rotations $\theta_{A}, \theta_{B}$ at the ends, relative lateral displacement $\Delta$; end moments and shears as in the figure.

The third term in parentheses captures the effect of axial force on bending moment. Let

So equation becomes $y'' + (u/l)^{2}\, y = -(M_{AB} + F_{\mathrm Q} x)/EI$; general solution

② Fixed-fixed member · Slope-deflection equation

Applying BCs $y=0, y' = \theta_{A}$ at $x = 0$; $y = \Delta, y' = \theta_{B}$ at $x = l$, we obtain

Here $i = EI/l$ is the linear stiffness; $\xi_{1},\xi_{2},\eta_{1},\eta_{2}$ are the axial-force-modified stiffness coefficients (functions of $u$; see Table 11-2).

③ Fixed-pinned case

If end $A$ is pinned, use $M_{AB}=0$ to eliminate $\theta_{A}$:

Table 11-2 also lists the end moments and shears induced by unit displacements.

④ Displacement-method stability equation · $D = 0$

For a frame with $n$ joint displacements $Z_{1}, \ldots, Z_{n}$, assembly gives the displacement-method canonical system:

$Z_{i}=0$ is the trivial solution (frame in its original equilibrium); buckling requires a non-trivial solution, so the determinant must vanish:

Example 11-7 · Critical load of Animation 11.6.2-b

The frame has two rotational DOF at joints $B$ and $D$ as basic unknowns $Z_{1}, Z_{2}$. Because both columns have the same size and axial force:

Drawing unit moment diagrams $\overline{M}_{1}, \overline{M}_{2}$ (Animation 11.6.2-b,d):

Stability equation:

Substituting:

Roots $\xi_{1}(u) = -1.307, -3.443$; from Table 11-2 the smaller-$u$ root gives $u = 5.46$. Hence

In Ch. 8's matrix displacement method, the element force $\boldsymbol{F}^{e}$ and end displacement $\boldsymbol{\Delta}^{e}$ are related by the element stiffness equation:

For stability analysis of a frame, incorporating the axial-force effect on element stiffness modifies this to:

Structural stiffness equation

Assembly using the same procedure as in the displacement method gives the global form:

The RHS is zero, as the "free" terms in the displacement-method canonical equations all vanish.

Stability equation (FEM form)

At the critical state, Eq. (11-39) has a non-trivial $\boldsymbol{\Delta}$, which requires

Because $\boldsymbol{K}_{G}$ scales linearly with the axial forces (which scale with $F_{\mathrm P}$), Eq. (11-40) is a generalized eigenvalue problem — the smallest eigenvalue is the critical-load factor. This approach is easily programmed for arbitrarily complex frames and is the standard modern stability-analysis procedure.