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

# Euler's formula both ends hinged, with limitations

Strength of Materials · Chapter 8: Column Theory

- Asked in: 2074 Chaitra, 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_3/strength_of_materials/euler-both-ends-hinged-limitation
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-8

## Question

Derive the expression for the Euler's formula for crippling load on a column with both ends hinged condition. Explain the limitation of Euler's Formula also.

## Short answer

Euler's critical load for both ends hinged is P_cr = π²EI/L², derived from EI y'' = -Py. Limitations: applies only to slender columns deforming elastically; neglects initial crookedness, load eccentricity, and shear; it overestimates the load for short or intermediate columns.

## Answer

**Euler's critical (crippling) load** is the smallest axial load at which an initially straight, slender column becomes unstable and buckles sideways.

#### 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.

#### 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 equation governs the hinged column?
   A. EI y'' = -Py
   B. EI y'' = P(delta - y)
   C. EI y'' = M0 - Py
   D. EI y'' = -Px
   Answer: A. Moment is P y. (Key point: Differential equation)

2. What is P_cr?
   A. 4pi^2EI/L^2
   B. pi^2EI/L^2 with n = 1
   C. pi^2EI/4L^2
   D. 2pi^2EI/L^2
   Answer: B. kL = pi. (Key point: Critical load)

3. What is the buckled shape?
   A. Full sine wave
   B. Parabola
   C. Half sine wave, maximum at mid span
   D. Straight line
   Answer: C. y = C2 sin(pi x/L). (Key point: Buckled shape)

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 beyond the elastic range
   D. It holds only for long, slender columns where critical stress is below the proportional limit
   Answer: D. For short and intermediate columns it overestimates the load, so empirical formulas are used. (Key point: Limitation)

## Related in Strength of Materials

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- [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×
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- [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

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