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

# Euler's formula: fixed-free column with limitations

Strength of Materials · Chapter 8: Column Theory

- Asked in: 2075 Chaitra, 6 marks
- Video: [Euler's derivation for column having one end fixed and other free ! SOM ! By Ramashish Sah](https://www.youtube.com/watch?v=IOLkrYdfvrI) — RS EDUCATION
- 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-fixed-free-limitations
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-8

## Question

Derive Euler critical buckling load formula for a column having one end fixed and the other end free. Discuss the limitations of Euler buckling formula.

## Short answer

Derivation gives P_cr = π²EI/(4L²) = π²EI/(2L)². Limitations: valid only for long, slender columns with stress below the proportional limit; assumes perfect straightness, axial load, homogeneous elastic material; ignores imperfections, shear, and direct compression.

## Answer

**Euler's critical load** for a column fixed at one end and free 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 AB, length $L$, have its lower end A fixed and its upper end B free. Under the load $P$, the free end deflects by $\delta$. Take the origin at the fixed end A, with $y$ measured from the original axis.

```text
      P
      v
   B  )  <- free end, deflection delta
      )
      ) y
      )
   A  fixed
```

At a section distance $x$ from A, the lever arm of $P$ is $(\delta - y)$, so $M = P(\delta - y)$:

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

The solution is

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

Boundary conditions:
- At $x=0$: $y=0 \Rightarrow C_1 = -\delta$.
- At $x=0$: $\dfrac{dy}{dx} = 0 \Rightarrow C_2 = 0$.

So $y = \delta\,(1 - \cos kx)$.

- At $x=L$: $y = \delta \Rightarrow \delta = \delta(1 - \cos kL) \Rightarrow \cos kL = 0$

The smallest solution is $kL = \dfrac{\pi}{2}$:

$$
\frac{PL^2}{EI} = \frac{\pi^2}{4}
$$

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

The effective length is $L_e = 2L$, so this column carries one quarter of the load of a hinged-hinged column of the same length.

#### 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 is the bending moment at a section?
   A. M = -Py
   B. M = M0 - Py
   C. M = Py
   D. M = P(delta - y)
   Answer: D. Lever arm is delta - y with the free end deflected by delta. (Key point: Moment)

2. What boundary condition gives cos kL = 0?
   A. y = delta at x = L
   B. y = 0 at x = L
   C. slope zero at L
   D. y = 0 at x = 0 only
   Answer: A. delta = delta(1 - cos kL) leads to cos kL = 0. (Key point: Condition at free end)

3. What is P_cr and Le?
   A. pi^2EI/L^2, Le = L
   B. pi^2EI/(4L^2), Le = 2L
   C. 4pi^2EI/L^2, Le = L/2
   D. pi^2EI/(0.7L)^2
   Answer: B. kL = pi/2. (Key point: Result)

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

## Related in Strength of Materials

- [Euler's formula: one end fixed, other hinged](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aeuler-one-fixed-one-hinged) — Strength of Materials, chapter 8, asked 2×
- [Euler's formula both ends fixed, with limitations](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aeuler-both-ends-fixed-limitation) — Strength of Materials, chapter 8
- [Euler's formula: fixed-hinged strut with limitation](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aeuler-one-fixed-one-hinged-limitation) — Strength of Materials, chapter 8, asked 2×
- [Round bar fixed-free: buckling load from lateral deflection](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Around-bar-fixed-free-buckling-3m) — Strength of Materials, chapter 8, asked 2×
- [Round bar fixed-free (effective length 2 m): buckling load](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Around-bar-clamped-free-buckling-2m) — Strength of Materials, chapter 8, asked 2×
- [Buckling load, effective length, Euler hinged](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aeuler-both-ends-hinged-define-buckling) — Strength of Materials, chapter 8

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