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

# Euler's formula: fixed-hinged strut with limitation

Strength of Materials · Chapter 8: Column Theory

- Asked in: 2076 Asoj, 6 marks; 2066 Bhadra (old course), 8 marks (asked 2 times)
- Video: [Module 5: Euler's formula, Assumption & limitations, equivalent length, buckling stress and load.](https://www.youtube.com/watch?v=ARgoXXgKpMY) — jeevan ealias
- 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-limitation
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-8

## Question

Derive the Euler's formula for critical load for a strut with one end fixed another hinged. Also mention the limitation for using this formula.

## Short answer

Derivation gives P_cr ≈ 20.19EI/L², effective length ~0.7L. Limitations: valid only for slenderness ratio above ~80-100, for perfectly straight elastic columns with axial load; it overestimates for short columns and ignores imperfections.

## 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.

#### Limitations of Euler's formula
1. It applies only to **long (slender) columns**. In terms of stress, $\sigma_{cr} = \dfrac{\pi^2E}{\lambda^2}$ must not exceed the proportional limit, so $\lambda \ge \pi\sqrt{E/\sigma_p}$ (about 80 to 100 for mild steel). For short and intermediate columns it over-estimates the load.
2. It assumes a perfectly straight column, a perfectly axial load and homogeneous material. Real columns have initial crookedness and eccentricity, so they fail at a lower load.
3. It is valid only within the elastic range (Hooke's law), so $\sigma_{cr}$ cannot exceed the proportional limit.
4. It ignores the effect of direct compression, shear deformation and imperfect end conditions.
5. For very slender columns it is accurate, and for intermediate columns empirical formulas (Rankine, Johnson, Secant, straight-line) are used instead.

## Check yourself

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

1. 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)

2. 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)

3. 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)

4. What is the main limitation of Euler's formula?
   A. It holds for short columns
   B. It holds only for long, slender columns where critical stress is below the proportional limit
   C. It overestimates nothing for short columns
   D. It holds beyond the elastic range
   Answer: B. For short and intermediate columns it overestimates the load, so empirical formulas are used. (Key point: Limitation)

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