Chapter 11 · Section 3

Critical Load of Elastic Columns by the Static Method

Extending the static method of §11-2 to an infinite-DOF elastic column: starting from the deflection ODE, combined with displacement boundary conditions, we get the stability equation. For symmetric structures, sym./antisym. modes simplify analysis; the length factor $\mu$ compactly encodes the result $F_{\mathrm{Pcr}} = \pi^{2} EI/(\mu l)^{2}$. Key worked examples: frame sway and stepped cantilever column.

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11.3.1

From finite to infinite DOF

Motivation

§11-2 used the rigid-bar + elastic-hinge finite-DOF model as "the simplest example" — its displacement is described by a few independent parameters. However, for real elastic columns, every cross-section can deflect slightly; the deflection curve is a continuous function $y(x)$ requiring infinitely many parameters to describe completely.

New tool

Deflection differential equation

For a continuous deflection $y(x)$, we can no longer use "algebraic systems + determinants"; instead we write a deflection ODE:

$$ EI\, y'' \;=\; -\, M(x)$$

(11-11)

$EI$ = flexural rigidity; $M(x)$ = section moment; sign convention depends on coordinates

Expressing $M(x)$ in terms of support reactions and $y(x)$ yields a homogeneous linear ODE in $y(x)$.

Anim. 11.3.1-a

A unified static-method procedure

Whether in finite DOF (algebraic) or infinite DOF (differential form), the static method has the same three steps: ① write the equations in a perturbed configuration → ② apply boundary conditions to get a homogeneous system → ③ the non-trivial condition (determinant $D=0$) gives the characteristic equation. The smallest positive $F_{\mathrm P}$ is the critical load.

11.3.2

model · Deflection ODE

Governing ODE

Using Animation 11.3.2-a — a prismatic axially compressed column fixed at the bottom with a horizontal support at the top (length $l$, flexural rigidity $EI$) — we illustrate the static-method procedure. At $F_{\mathrm P} = F_{\mathrm{Pcr}}$ the column bifurcates from straight equilibrium to bent equilibrium.

Anim. 11.3.2-a

① Moment in the perturbed configuration

Choose coordinates: $x$ along the axis, $y$ lateral. From the free-body moment equilibrium (see Animation 11.3.2-a):

$$ M(x) \;=\; F_{\mathrm P}\, y \;+\; F_{\mathrm R}\, (l - x) $$

(11.3-1)

$F_{\mathrm R}$ = horizontal-support reaction at the top; $y$ = lateral deflection at $x$

② Introducing α² to simplify

Substitute into $EIy'' = -M$ and let

$$ \alpha^{2} \;=\; \dfrac{F_{\mathrm P}}{EI}$$

(11-12)

to obtain a 2nd-order constant-coefficient non-homogeneous linear ODE:

$$ y'' \;+\; \alpha^{2}\, y \;=\; -\, \dfrac{F_{\mathrm R}}{EI}\,(l - x) $$

(11.3-2)

11.3.3

Boundary conditions · Stability equation

Characteristic Equation

① General solution

$$ y \;=\; A\cos\alpha x \;+\; B\sin\alpha x \;-\; \dfrac{F_{\mathrm R}}{F_{\mathrm P}}\,(l - x) $$

(11.3-3)

Differentiating:

$$ y' \;=\; -\,\alpha A\sin\alpha x \;+\; \alpha B\cos\alpha x \;+\; \dfrac{F_{\mathrm R}}{F_{\mathrm P}} $$

(11.3-4)

② Displacement boundary conditions · three constraints

Fixed at $x = 0$: $y = 0,\ y' = 0$
Horizontal support at $x = l$: $y = 0$

Substituting into the general solution gives a homogeneous linear system in $A, B, F_{\mathrm R}/F_{\mathrm P}$.

③ Non-trivial condition → stability equation

Setting the coefficient determinant to zero — the stability equation:

$$ \tan \alpha l \;=\; \alpha l $$

(11.3-5)

a transcendental equation in $\alpha l$

The smallest positive root is $\alpha l = 4.493$; substituting back into $\alpha^2 = F_{\mathrm P}/EI$ gives the critical load:

$$ F_{\mathrm{Pcr}} \;=\; \alpha^{2}\, EI \;=\; \dfrac{(4.493)^{2}\, EI}{l^{2}} \;\approx\; 20.19\,\dfrac{EI}{l^{2}} \;=\; \dfrac{\pi^{2} EI}{(0.7\,l)^{2}} $$

(11.3-6)

Equivalent to a length factor $\mu l = 0.7\,l$ — matching a fixed-pinned column

Anim. 11.3.3-a

11.3.4

Symmetric / antisymmetric buckling of symmetric structures

Symmetric / Antisymmetric Modes

For symmetric structures under symmetric loads, the buckling mode is necessarily symmetric or antisymmetric. This property simplifies analysis — decompose the original problem, analyze the symmetric and antisymmetric half-structures, and take the smaller critical load as the true $F_{\mathrm{Pcr}}$.

Anim. 11.3.4-a

Symmetric mode Symmetric Mode

Anim. 11.3.4-b

Mode shape is symmetric about the axis of symmetry — both sides move the same way;
at the axis of symmetry: rotation $= 0$, shear $= 0$;
on the symmetric half-structure, apply a sliding support at the axis.

Antisymmetric mode Antisymmetric Mode

Anim. 11.3.4-c

Mode shape is antisymmetric about the axis — the two sides move oppositely;
at the axis: deflection $= 0$, moment $= 0$;
on the antisymmetric half-structure, apply a pin at the axis.

Typical modes of the two paths

Possible symmetric buckling

Symmetric buckling of a symmetric structure under symmetric loading — usually the higher critical load.

Anim. 11.3.4-d

Possible antisymmetric buckling

Antisymmetric buckling — usually the lower critical load, i.e. the true $F_{\mathrm{Pcr}}$.

Anim. 11.3.4-e

Value of the symmetry simplification

For complex symmetric structures (frames, trusses, arches), splitting the problem in half halves the DOF and the number of unknowns — the transcendental equation's order is also halved, dramatically easing solution. This is a key tool in §11-6 frame stability analysis.

11.3.5

Length factor $\mu$ and Euler's formula

Effective Length

For elastic columns with different boundary conditions, the critical load admits the unified Euler-type expression:

$$ F_{\mathrm{Pcr}} \;=\; \dfrac{\pi^{2}\, EI}{(\mu l)^{2}} $$

(11.3-7)

$\mu$ = length factor; $\mu l$ = effective length — the "equivalent length" when converted to a pinned-pinned column (distance between two consecutive inflection points of the buckled shape)

Anim. 11.3.5-e

Four classical boundary conditions

Pinned-pinned · $\mu = 1$

$F_{\mathrm{Pcr}} = \pi^{2} EI/l^{2}$ — one half-wave spans the bar; the ends are the inflection points.

Anim. 11.3.5-a

Fixed-free · $\mu = 2$

$F_{\mathrm{Pcr}} = \pi^{2} EI/(2l)^{2}$ — a half-wave occupies only the physical bar; the mirrored effective length is $2l$.

Anim. 11.3.5-b

Fixed-fixed · $\mu = 0.5$

$F_{\mathrm{Pcr}} = \pi^{2} EI/(0.5\,l)^{2}$ — inflection points $0.5\,l$ apart; capacity 4× higher.

Anim. 11.3.5-c

Fixed-pinned · $\mu \approx 0.7$

$F_{\mathrm{Pcr}} = \pi^{2} EI/(0.7\,l)^{2}$ — corresponds to the $\tan\alpha l = \alpha l$ case in 11.3.2.

Anim. 11.3.5-d

Slenderness ratio λ and critical stress

Define the column's slenderness ratio $\lambda$:

$$ \lambda \;=\; \dfrac{\mu l}{i}$$

(11-9)

$i = \sqrt{I/A}$ = radius of gyration

Dividing $F_{\mathrm{Pcr}}$ by the cross-sectional area $A$ gives the critical stress

$$ \sigma_{\mathrm{cr}} \;=\; \dfrac{\pi^{2} E}{\lambda^{2}}$$

(11-10)

general formula for Euler's critical stress

Anim. 11.3.5-f

Stronger restraint, smaller $\mu$, higher capacity

Fixed-fixed ($\mu = 0.5$) has 4× the capacity of pinned-pinned; fixed-free ($\mu = 2$) has only $1/4$. This governs the engineering choice of end restraints.

11.3.6

Comparison of frame buckling modes

Frame Failure Modes

A multi-column frame (see Animation 11.3.6-a) has many more buckling forms than a single column. The two main classes are single-member buckling (e.g. the $AB$ member alone) and overall-sway buckling (all columns sway together). The stiffness ratio $EI_{1}/EI$ between columns and beams, plus the boundary restraints, determine which mode governs.

Anim. 11.3.6-a

Single-member buckling

Buckling confined to a single bar; other members stay put;
A "local Euler problem" under the current boundary conditions;
the restraint stiffness provided by the other members fixes $\mu$.

Overall sway buckling

All columns sway together;
typical when the beam is soft, or the column bases are pinned or on springs;
often the actual failure mode — capacity well below any single-bar value.

Bounds on the critical load

For the antisymmetric mode of the figure, the critical load can be bracketed between two bounds:

$$ \dfrac{\pi^{2} EI}{(2h)^{2}} \;\le\; F_{\mathrm{Pcr}} \;\le\; \dfrac{\pi^{2} EI}{h^{2}} $$

(11.3-8)

Lower bound: column top is a cantilever ($EI_{1}\to 0$); upper bound: column top is a sliding support ($EI_{1}\to \infty$)

This shows the strong influence of beam stiffness — stiffer beams raise the capacity substantially.

11.3.7

Example 11-3 · Frame sway (static method)

Worked Example

In the portal frame of Animation 11.3.7-a, both columns have the same $EI$ and the beam has $EA = \infty$. Both column tops carry axial loads $F_{\mathrm P}$. Use the static method to compute the critical load for sway buckling (see Animation 11.3.7-a).

Anim. 11.3.7-a

① Reduce the frame to a column with an elastic support

Take column $CD$ as the computation object; the column $EF$ plus the beam act as an elastic support restraining its lateral motion. Treating the sway $\Delta$ as an elastic displacement, the support reaction is

$$ F_{\mathrm R} \;=\; k\,\Delta \;=\; \dfrac{3EI}{l^{3}}\,\Delta $$

(11.3-9)

$k = 3EI/l^{3}$ = lateral stiffness at the top of $EF$ (as a cantilever)

② Deflection ODE of CD

Free-body moment equilibrium:

$$ EI\, y'' \;=\; F_{\mathrm P}\,(\Delta - y) \;-\; F_{\mathrm R}\,(l - x) \;+\; F_{\mathrm{NDB}}\,(l - x) $$

(11.3-10)

Here $F_{\mathrm{NDB}} = (\Delta/l)\, F_{\mathrm P}$ is the horizontal reaction at $D$ (induced by the load at $B$). Applying the displacement boundary conditions $y(0) = 0,\ y'(0) = 0,\ y(l) = \Delta$ gives a homogeneous system in $A, B, \Delta$.

③ Stability equation · smallest positive root

The determinant-zero condition expands to a transcendental equation in $\alpha l$:

$$ \left[\dfrac{3}{(\alpha l)^{2}} - 1\right]\tan \alpha l \;+\; \alpha l\left[2 - \dfrac{3}{(\alpha l)^{2}}\right] \;=\; 0 $$

(11.3-11)

Numerical search: smallest positive root $\alpha l = 1.645$; substitute into $\alpha^{2} = F_{\mathrm P}/EI$:

$$ F_{\mathrm{Pcr}} \;=\; (1.645)^{2}\,\dfrac{EI}{l^{2}} \;\approx\; 2.706\,\dfrac{EI}{l^{2}} \;=\; \dfrac{\pi^{2} EI}{(1.91\, l)^{2}} $$

(11.3-12)

Length factor $\mu \approx 1.91$ — slightly less than the cantilever value $\mu = 2$ (elastic support provides partial restraint)

11.3.8

Example 11-4 · Stepped cantilever column

Stepped Column

Animation 11.3.8-a: a stepped cantilever column with lower segment length $l_{1}$ and rigidity $EI_{1}$, upper segment length $l_{2}$ and rigidity $EI_{2}$, under the two loads $F_{\mathrm{P1}}$ and $F_{\mathrm{P2}}$. Use the static method to set up the stability equation.

Anim. 11.3.8-a

① Piecewise deflection ODEs

Let $y_{1}, y_{2}$ denote the deflections of the lower and upper segments (see Animation 11.3.8-a). The lower carries $F_{\mathrm{P1}} + F_{\mathrm{P2}}$, the upper only $F_{\mathrm{P2}}$:

$$ \left\{\begin{aligned} EI_{1}\, y_{1}'' &= F_{\mathrm{P2}}(\Delta_{2} - y_{2}) \\[2pt] EI_{2}\, y_{2}'' &= F_{\mathrm{P2}}(\Delta_{2} - y_{2}) + F_{\mathrm{P1}}(\Delta_{1} - y_{1}) \end{aligned}\right. $$

(11.3-13)

Define

$$ \alpha_{1} \;=\; \sqrt{\dfrac{F_{\mathrm{P1}} + F_{\mathrm{P2}}}{EI_{1}}},\qquad \alpha_{2} \;=\; \sqrt{\dfrac{F_{\mathrm{P2}}}{EI_{2}}} $$

(11.3-14)

② Boundary and continuity conditions

Fixed at $x = 0$: $y_{1} = 0,\ y_{1}' = 0$
Section change at $x = l_{1}$: $y_{1} = y_{2} = \Delta_{1},\ y_{1}' = y_{2}'$ (continuity of displacement and rotation)

Six unknowns ($A_{1}, B_{1}, A_{2}, B_{2}, \Delta_{1}, \Delta_{2}$) in a homogeneous system.

③ Stability equation

Setting the $6\times 6$ determinant to zero gives

$$ \tan \alpha_{1} l_{1}\,\cdot\, \tan \alpha_{2} l_{2} \;=\; \dfrac{\alpha_{2}}{\alpha_{1}}\,\left(\dfrac{F_{\mathrm{P1}} + F_{\mathrm{P2}}}{F_{\mathrm{P2}}}\right) $$

(11.3-15)

Special case: $EI_{1} = 2EI_{2}$, $F_{\mathrm{P1}} = F_{\mathrm{P2}} = F_{\mathrm P}$, $l_{1} = l_{2} = l/2$ gives

$$ \tan^{2}\!\left(\dfrac{\alpha_{2}\, l}{2}\right) \;=\; 2 $$

(11.3-16)

smallest positive root $\alpha_{2}\, l = 1.91$, so

$$ F_{\mathrm{Pcr}} \;=\; 3.65\,\dfrac{EI_{2}}{l^{2}} $$

(11.3-17)

Effective length $\mu l = \pi l/1.91 \approx 1.64\,l$

Section summary

For infinite-DOF columns, the deflection ODE $EIy'' = -M$ replaces algebraic equations;
Displacement BCs → homogeneous system → $D = 0$ → a transcendental stability equation in $\alpha l$;
For symmetric structures under symmetric loads, modes are symmetric or antisymmetric — half-structure analysis is a major simplifier;
The four classical column cases fit the unified $F_{\mathrm{Pcr}} = \pi^{2} EI/(\mu l)^{2}$, $\mu \in \{0.5, 0.7, 1, 2\}$;
Frame buckling has two broad kinds — single-member vs overall sway — pick the weaker one;
Problems with elastic supports (frames) or stepped sections are handled by piecewise ODE + continuity conditions;
Next, §11-4 introduces the energy method — more convenient for variable coefficients.

§11-4 Energy Method →