Buckling load, effective length, Euler hinged

Buckling load

The buckling (crippling or critical) load is the axial compressive load at which a long, slender column just begins to buckle (deflect sideways). It is the highest load the column can carry in its straight, stable form.

Effective length

The effective length $L_e$ is the length of an equivalent column with both ends hinged that has the same buckling load. It equals the distance between the points of zero bending moment (inflection points) in the buckled shape:

For any end condition, $P_{cr} = \dfrac{\pi^2EI}{L_e^2}$.

Euler's formula for both ends hinged

Assumptions

Assumptions: the column is perfectly straight, of uniform section and homogeneous material, the load is perfectly axial, the material obeys Hooke's law, deflections are small, and the effect of shear and axial shortening is ignored.

Derivation

Let the column AB of length $L$ be hinged at both ends, with the load $P$ producing a small deflection $y$ at distance $x$ from A.

   P
   |  B
   |  /\
   |  \ y
   |  /
   A /
   P
 (A origin, x along column)

Bending moment at the section: $M = -Py$ (taking sagging moment as positive, the moment due to $P$ tends to reduce the curvature, so it is negative). Using the differential equation of the elastic curve:

$$EI\frac{d^2y}{dx^2} = -Py \;\Rightarrow\; \frac{d^2y}{dx^2} + k^2 y = 0,\qquad k^2 = \frac{P}{EI}$$

The solution is

$$y = C_1\cos kx + C_2\sin kx$$

Boundary conditions:

• At $x = 0$, $y = 0 \Rightarrow C_1 = 0$.
• At $x = L$, $y = 0 \Rightarrow C_2\sin kL = 0$.

For a buckled shape $C_2 \ne 0$, so $\sin kL = 0$ and $kL = n\pi$. The smallest non-zero load is for $n = 1$:

$$k L = \pi \;\Rightarrow\; \frac{PL^2}{EI} = \pi^2$$ $$P_{cr} = \frac{\pi^2 EI}{L^2}$$

Here $I$ is the least moment of inertia, since the column buckles about the axis of least resistance. The shape is a half sine wave, $y = C_2\sin(\pi x/L)$, with maximum deflection at mid-span.