---
title: "Straight bar buckling: maximum central deflection"
subject: "Strength of Materials"
chapter: 8
key: "semester_3/strength_of_materials:euler-straight-bar-central-deflection"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/euler-straight-bar-central-deflection
site: BCT Engineering
---

# Straight bar buckling: maximum central deflection

Strength of Materials · Chapter 8: Column Theory

- Asked in: 2080 Bhadra, 8 marks
- Video: [BUCKLING - Column Stability in UNDER 10 Minutes](https://www.youtube.com/watch?v=3FNEV6yNUFY) — Less Boring Lectures
- 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-straight-bar-central-deflection
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-8

## Question

A straight bar 1.5 m long and $12\ \text{mm}\times5\ \text{mm}$ in section is mounted on a test machine and is loaded axially till it buckles. Assuming Euler's formula for both ends hinged, estimate maximum central deflection before it attains the yield point at $290\ \text{N/mm}^2$. Take $E = 2\times10^5\ \text{MPa}$.

## Short answer

Euler buckling load is 109.7 N. At yield stress 290 N/mm², maximum central deflection is the sum of direct and bending stresses, giving about 131 mm.

## Answer

**Given:** $L = 1.5$ m = 1500 mm, section $12\times5$ mm, hinged ends, $E = 2\times10^5\ \text{N/mm}^2$, yield stress $290\ \text{N/mm}^2$.

The bar buckles about the axis of least resistance (thickness 5 mm).

$$
I_{min} = \frac{12\times5^3}{12} = 125\ \text{mm}^4,\qquad A = 60\ \text{mm}^2,\qquad Z = \frac{I}{y} = \frac{125}{2.5} = 50\ \text{mm}^3
$$

#### Euler's buckling load
$$
P_{cr} = \frac{\pi^2EI}{L^2} = \frac{\pi^2\times2\times10^5\times125}{1500^2} = 109.66\ \text{N}
$$

Direct stress at this load:

$$
\sigma_d = \frac{P_{cr}}{A} = \frac{109.66}{60} = 1.83\ \text{N/mm}^2
$$

#### Central deflection at yield
After buckling, the load remains at $P_{cr}$ while the deflection $\delta$ grows. The maximum stress at mid-span is direct stress plus bending stress:

$$
\sigma_{max} = \frac{P}{A} + \frac{P\,\delta}{Z}
$$

$$
290 = 1.83 + \frac{109.66\,\delta}{50}
$$

$$
\delta = \frac{(290 - 1.83)\times50}{109.66} = 131.4\ \text{mm}
$$

**Answer: Euler load = 109.7 N; maximum central deflection before yielding $\approx 131$ mm.** This is an estimate, because the small-deflection theory is used.

## Check yourself

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

1. Which I is used for the 12x5 mm bar?
   A. Least, I = 12x5^3/12 = 125 mm^4
   B. 12^3 x 5/12 = 720 mm^4
   C. 60 mm^4
   D. Average
   Answer: A. Bar buckles about the axis of least resistance. (Key point: Axis of buckling)

2. What is the Euler load?
   A. 1097 N
   B. pi^2EI/L^2 = 109.66 N
   C. 10.97 N
   D. 29.6 kN
   Answer: B. pi^2 x 2e5 x 125/1500^2 = 109.66 N. (Key point: Euler load)

3. How is the maximum stress expressed?
   A. P/A only
   B. P delta/A
   C. sigma = P/A + P delta/Z
   D. P/Z + delta/A
   Answer: C. Direct plus bending stress from moment P delta. (Key point: Stress at mid span)

4. What is the deflection at yield?
   A. About 1.31 mm
   B. About 13 mm
   C. About 290 mm
   D. About 131 mm
   Answer: D. (290 - 1.83) x 50/109.66 = 131.4 mm. (Key point: Deflection)

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