Euler's formula: both ends fixed

Euler's crippling load is the smallest axial load at which an initially straight, slender column buckles. For both ends fixed, the ends are held in position and direction.

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

When the column buckles, the fixed ends develop an end moment $M_0$ (fixing moment) while the load $P$ stays axial. Take the origin at the lower fixed end A, with $y$ the lateral deflection at distance $x$.

   P
   |  B  (fixed, M0)
   |  /\
   | (  y
   |  \/
   A  fixed, M0
   P

The moment at the section is $M = M_0 - Py$, so

$$EI\frac{d^2y}{dx^2} = M_0 - Py \;\Rightarrow\; \frac{d^2y}{dx^2} + k^2y = \frac{M_0}{EI},\qquad k^2 = \frac{P}{EI}$$

The solution is

$$y = C_1\cos kx + C_2\sin kx + \frac{M_0}{P}$$

Boundary conditions:

• At $x=0$: $y = 0 \Rightarrow C_1 = -\dfrac{M_0}{P}$.
• At $x=0$: $\dfrac{dy}{dx} = 0 \Rightarrow C_2 k = 0 \Rightarrow C_2 = 0$.

So

$$y = \frac{M_0}{P}\,(1 - \cos kx)$$

• At $x = L$: $y = 0$ (the top end is at the same lateral position) gives $\cos kL = 1$.

The smallest non-zero solution is $kL = 2\pi$:

$$\frac{PL^2}{EI} = 4\pi^2$$ $$P_{cr} = \frac{4\pi^2EI}{L^2} = \frac{\pi^2EI}{(L/2)^2}$$

The effective length is $L_e = L/2$, so a fixed-fixed column carries 4 times the load of the same hinged column. The buckled shape has points of inflection at $L/4$ from each end (at $x = L/4$ and $x = 3L/4$).