---
title: "Euler's buckling load for columns"
subject: "Strength of Materials"
chapter: 10
key: "semester_4/strength_of_materials_mechanical:euler-crippling-load-pinned-column"
source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/euler-crippling-load-pinned-column
site: BCT Engineering
---

# Euler's buckling load for columns

Strength of Materials · Chapter 10: Columns

- Asked in: year not given, 8 marks
- Video: [EULER'S FORMULA OF COLUMN /Euler's Theory Buckling Load (for COLUMN BOTH END HINGED)](https://www.youtube.com/watch?v=p5S_v-_sFo4) — TECHNICAL CLASSES
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/euler-crippling-load-pinned-column
- Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-10

## Question

Derive Euler's expression for the crippling load of a column with both ends hinged. State the assumptions and limitations. Give the expressions for the crippling load for other end conditions in terms of effective length, and define slenderness ratio.

## Answer

#### Assumptions
1. The column is initially perfectly straight and the load is purely axial.
2. The material is homogeneous, isotropic and obeys Hooke's law (stresses within the proportional limit).
3. The cross-section is uniform and the column bends in the plane of least resistance.
4. The length is large compared with the lateral dimensions; the effect of direct compression is neglected.
5. Self weight is neglected; the end conditions are ideal.

#### Derivation (both ends hinged)
```text
    P (down)
     |
     v
     o A (hinge)        x measured from A
     |
     |  y (lateral deflection)
     )
     |
     o B (hinge)
     ^
     |
    P (up)
```

Column of length $L$ with load $P$, buckled into a slightly bent shape. At a section distance $x$ from A the lateral deflection is $y$, so the bending moment is $M = -Py$ (the load tends to increase the deflection):

$$
EI\frac{d^2y}{dx^2} = -Py \;\Rightarrow\; \frac{d^2y}{dx^2} + \alpha^2y = 0, \qquad \alpha^2 = \frac{P}{EI}
$$

Solution: $y = A\cos\alpha x + B\sin\alpha x$.

- $x = 0,\ y = 0 \Rightarrow A = 0$.
- $x = L,\ y = 0 \Rightarrow B\sin\alpha L = 0$. For a buckled shape $B\neq0$, so $\sin\alpha L = 0$, $\alpha L = n\pi$.

The smallest non-trivial value is $n = 1$: $\alpha^2 = \pi^2/L^2$:

$$
\boxed{P_{cr} = \frac{\pi^2EI}{L^2}}
$$

$I$ is the **least** second moment of area, $I_{min} = Ak^2$, $k$ being the least radius of gyration. The buckled shape is a half sine wave.

#### Other end conditions
$$
P_{cr} = \frac{\pi^2EI}{L_e^2}
$$

| End conditions | Effective length $L_e$ |
|---|---|
| Both ends hinged | $L$ |
| Both ends fixed | $L/2$ |
| One end fixed, other hinged | $L/\sqrt2$ (about $0.7L$) |
| One end fixed, other free | $2L$ |

#### Slenderness ratio and limitation
Slenderness ratio $\lambda = L_e/k$. The Euler stress is

$$
\sigma_{cr} = \frac{P_{cr}}{A} = \frac{\pi^2E}{\lambda^2}
$$

Euler's formula holds only for **long** columns, for which $\sigma_{cr}$ is below the proportional limit. For mild steel ($E = 200$ GPa, $\sigma_p \approx 250$ MPa) this needs $\lambda \gtrsim 90$ to $100$. For short and intermediate columns Euler's formula overestimates the strength, and Rankine's or Johnson's formulas are used.

## Check yourself

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

1. What equation is obtained for the pinned column?
   A. EI y'' = Py
   B. EI y' = -Py
   C. EI y'' = -Px
   D. EI y'' + Py = 0, from M = -Py
   Answer: D. The load increases deflection, giving d²y/dx² + α²y = 0 with α² = P/EI. (Key point: Differential equation)

2. How is P found?
   A. y = 0 at x = L requires sin αL = 0, and n = 1 gives P = π²EI/L²
   B. cos αL = 0, giving P = π²EI/4L²
   C. y = 0 at x = L requires B = 0, so no buckling
   D. Slope zero at x = L, giving P = 2π²EI/L²
   Answer: A. Non-trivial buckled shape requires sin αL = 0; smallest n = 1. (Key point: Boundary conditions)

3. Which effective lengths are correct?
   A. Both fixed 2L; fixed-free L/2
   B. Both fixed L/2; fixed-free 2L
   C. Both fixed L/2; fixed-free L
   D. Both fixed L; fixed-free L/2
   Answer: B. Pcr = π²EI/Le², with Le = L/2 for fixed-fixed and 2L for cantilever. (Key point: Effective length)

4. When is Euler's formula valid?
   A. For short columns with small λ
   B. For all columns
   C. For long columns where σcr is below the proportional limit (λ above about 90-100 for mild steel)
   D. Only when eccentricity is large
   Answer: C. For short and intermediate columns, Euler overestimates; Rankine or Johnson formulas are used. (Key point: Limitation)

## Related in Strength of Materials

- [Tubular column load and eccentric loading](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Atube-column-and-eccentric-short-column) — Strength of Materials, chapter 10

## Related in other subjects

- [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
- [Euler's formula both ends hinged, with limitations](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aeuler-both-ends-hinged-limitation) — Strength of Materials, chapter 8
- [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: both ends hinged](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aeuler-both-ends-hinged) — Strength of Materials, chapter 8, asked 4×