---
title: "Euler's formula: one end fixed, other hinged"
subject: "Strength of Materials"
chapter: 8
key: "semester_3/strength_of_materials:euler-one-fixed-one-hinged"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/euler-one-fixed-one-hinged
site: BCT Engineering
---

# Euler's formula: one end fixed, other hinged

Strength of Materials · Chapter 8: Column Theory

- Asked in: 2080 Baisakh, 8 marks; 2078 Kartik, 8 marks (asked 2 times)
- Video: [COLUMN WITH ONE END FIXED OTHER END HINGED:- EXPRESSION FOR BUCKLING LOAD](https://www.youtube.com/watch?v=opNy4xO0Mec) — TECHNICAL CLASSES
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/euler-one-fixed-one-hinged
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-8

## Question

Derive the Euler's formula for the end support condition one end fixed and another end hinged.

## Short answer

The fixed-hinged column equation yields tan(kL) = kL, with smallest root kL ≈ 4.493. Critical load is P_cr ≈ 20.19EI/L² = π²EI/(0.7L)², effective length about 0.7L.

## Answer

**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.

```text
   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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why does a horizontal reaction H appear at the hinge?
   A. Because load P is horizontal
   B. To keep the hinged end in position while the fixed end resists rotation
   C. To balance self weight
   D. H is zero by symmetry
   Answer: B. The fixed-hinged column needs H, giving M = -Py + Hx. (Key point: Reaction at hinge)

2. What condition emerges after eliminating C2 and H?
   A. sin kL = 0
   B. cos kL = 0
   C. tan kL = kL
   D. cos kL = 1
   Answer: C. Both conditions at the fixed end combine to tan kL = kL. (Key point: Governing equation)

3. What is the smallest non-zero root and the resulting load?
   A. kL = pi, P_cr = pi^2EI/L^2
   B. kL = pi/2
   C. kL = 2 pi
   D. kL = 4.493, P_cr = 20.19 EI/L^2
   Answer: D. 4.493^2 = 20.19. (Key point: Root)

4. What effective length results?
   A. About 0.7 L
   B. L
   C. 2 L
   D. 0.5 L
   Answer: A. 20.19 EI/L^2 is approximately pi^2 EI/(0.7L)^2. (Key point: Effective length)

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