---
title: "Buckling load, effective length, Euler hinged"
subject: "Strength of Materials"
chapter: 8
key: "semester_3/strength_of_materials:euler-both-ends-hinged-define-buckling"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/euler-both-ends-hinged-define-buckling
site: BCT Engineering
---

# Buckling load, effective length, Euler hinged

Strength of Materials · Chapter 8: Column Theory

- Asked in: 2069 Chaitra, 2+6 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_3/strength_of_materials/euler-both-ends-hinged-define-buckling
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-8

## Question

Define buckling load and effective length of column and derive a Euler's formula for crippling load of a column of length L with its both ends hinged condition.

## Short answer

Buckling (crippling) load is the axial compressive load at which a long, slender column suddenly deflects laterally. Effective length L_e is the length of an equivalent hinged-hinged column with the same buckling load, equal to the distance between inflection points. For both ends hinged, P_cr = π²EI/L².

## Answer

#### Buckling load
The **buckling (crippling or critical) load** is the axial compressive load at which a long, slender column just begins to buckle (deflect sideways). It is the highest load the column can carry in its straight, stable form.

#### Effective length
The **effective length** $L_e$ is the length of an equivalent column with both ends hinged that has the same buckling load. It equals the distance between the points of zero bending moment (inflection points) in the buckled shape:

| End conditions | $L_e$ |
|---|---|
| Both ends hinged | $L$ |
| Both ends fixed | $L/2$ |
| One fixed, one hinged | $0.7L$ |
| One fixed, one free | $2L$ |

For any end condition, $P_{cr} = \dfrac{\pi^2EI}{L_e^2}$.

#### Euler's formula for both ends hinged

#### 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 of length $L$ be hinged at both ends, with the load $P$ producing a small deflection $y$ at distance $x$ from A.

```text
   P
   |  B
   |  /\
   |  \ y
   |  /
   A /
   P
 (A origin, x along column)
```

Bending moment at the section: $M = -Py$ (taking sagging moment as positive, the moment due to $P$ tends to reduce the curvature, so it is negative). Using the differential equation of the elastic curve:

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

The solution is

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

Boundary conditions:
- At $x = 0$, $y = 0 \Rightarrow C_1 = 0$.
- At $x = L$, $y = 0 \Rightarrow C_2\sin kL = 0$.

For a buckled shape $C_2 \ne 0$, so $\sin kL = 0$ and $kL = n\pi$. The smallest non-zero load is for $n = 1$:

$$
k L = \pi \;\Rightarrow\; \frac{PL^2}{EI} = \pi^2
$$

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

Here $I$ is the least moment of inertia, since the column buckles about the axis of least resistance. The shape is a half sine wave, $y = C_2\sin(\pi x/L)$, with maximum deflection at mid-span.

## Check yourself

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

1. What is the buckling (crippling) load of a column?
   A. The axial compressive load at which a long slender column just begins to deflect sideways
   B. The load at which the material of the column crushes
   C. The transverse load that causes the maximum bending stress
   D. The load at which the column shortens by 1%
   Answer: A. Buckling load is the axial compressive load at which a long, slender column just begins to buckle; it is the highest load it can carry while staying straight and stable. (Key point: Buckling load)

2. The effective length of a column is best described as:
   A. The actual length of the column
   B. The length of an equivalent both-ends-hinged column with the same buckling load
   C. The length of the column minus the end fixing depth
   D. The length at which slenderness ratio equals 1
   Answer: B. Effective length is the length of an equivalent hinged-hinged column with the same buckling load, equal to the distance between inflection points of the buckled shape. (Key point: Effective length)

3. What is the effective length of a column with one end fixed and the other free, of actual length L?
   A. L/2
   B. 0.7L
   C. L
   D. 2L
   Answer: D. For fixed-free, Le = 2L, giving the lowest buckling load of the standard end conditions. (Key point: Effective length values)

4. In the hinged-hinged derivation, y = C1 cos kx + C2 sin kx. Applying y = 0 at x = 0 and x = L gives:
   A. C2 = 0 and cos kL = 0
   B. C1 = 0 and sin kL = 0 for non-zero C2
   C. C1 = C2 and kL = pi/2
   D. C1 = 0 and kL = 2pi always
   Answer: B. y(0)=0 gives C1=0; then y(L)=0 requires sin kL = 0 for a buckled (C2 not 0) shape, so kL = n pi. (Key point: Boundary conditions in derivation)

5. Why is n = 1 taken in kL = n pi to get Euler's load?
   A. Because n = 1 gives the smallest non-zero load at which buckling can occur
   B. Because n = 1 gives the largest load the column can carry
   C. Because higher n makes the column straight
   D. Because n = 0 gives the buckling load
   Answer: A. The column buckles at the lowest non-zero load; n=0 gives zero load (no buckling) and larger n give higher loads, so n=1 gives Pcr = pi^2 EI/L^2. (Key point: Smallest critical load)

6. Which moment of inertia is used in Euler's formula?
   A. The greatest moment of inertia
   B. The polar moment of inertia
   C. The least moment of inertia, since the column buckles about the axis of least resistance
   D. The average of Ixx and Iyy
   Answer: C. The column buckles about the axis of least resistance, so the least I is used. (Key point: Axis for I)

## Related in Strength of Materials

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

## Related in other subjects

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- [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