Euler's formula: one end fixed, other hinged

Euler's critical load for a column fixed at one end and hinged at the other is derived below.

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 have its top end B hinged and its bottom end A fixed, length $L$. When it buckles, the fixed end develops a moment, and a horizontal reaction $H$ appears at the hinge to keep B in position. Take the origin at the hinged end B, with $x$ measured downward and $y$ the lateral deflection.

   P
   |  B  hinged (H)
   |  )
   |  ) y
   |  )
   A  fixed
   P

Moment at the section: $M = -Py + Hx$, so

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

The solution is

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

Boundary conditions:

• At $x=0$: $y = 0 \Rightarrow C_1 = 0$.
• At $x=L$ (fixed end): $y = 0 \Rightarrow C_2\sin kL + \dfrac{HL}{P} = 0$
• At $x=L$: $\dfrac{dy}{dx} = 0 \Rightarrow C_2k\cos kL + \dfrac{H}{P} = 0$

Eliminate $C_2$ and $H$ between the last two equations:

$$\tan kL = kL$$

This transcendental equation is solved by trial. The smallest non-zero root is

$$kL = 4.493 \;\Rightarrow\; \frac{PL^2}{EI} = 4.493^2 = 20.19$$ $$P_{cr} = \frac{20.19\,EI}{L^2} \approx \frac{2\pi^2EI}{L^2} = \frac{\pi^2EI}{(0.7L)^2}$$

The effective length is $L_e \approx 0.7L$. The bending moment is zero where $\sin kx = 0$, so the inflection point lies at $x = \pi/k \approx 0.7L$ from the hinged end, that is, about $0.3L$ from the fixed end.