---
title: "Round bar fixed-free (effective length 2 m): buckling load"
subject: "Strength of Materials"
chapter: 8
key: "semester_3/strength_of_materials:round-bar-clamped-free-buckling-2m"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/round-bar-clamped-free-buckling-2m
site: BCT Engineering
---

# Round bar fixed-free (effective length 2 m): buckling load

Strength of Materials · Chapter 8: Column Theory

- Asked in: 2069 Asar, 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/round-bar-clamped-free-buckling-2m
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-8

## Question

A round bar is clamped at bottom and free at top. Its effective length is 2m. A horizontal force of 300N at top produces a horizontal deflection of 20mm. Determine the buckling load for the bar if the load is applied axially on top.

## Short answer

Effective length 2 m implies actual length 1 m. From cantilever deflection, EI = 300 × 1000³ / (3 × 20) = 5×10⁹ N·mm². Buckling load P_cr = π²EI / (2000)² = 12.34 kN.

## Answer

**Given:** the bar is clamped at the bottom and free at the top (cantilever column). The effective length is $L_e = 2$ m, so the actual length is $L = L_e/2 = 1$ m. A horizontal force $H = 300$ N at the top produces a deflection $\delta = 20$ mm.

#### Step 1: find $EI$ from the lateral load
Acting as a cantilever:

$$
\delta = \frac{HL^3}{3EI} \;\Rightarrow\; EI = \frac{300\times(1000)^3}{3\times20} = 5\times10^{9}\ \text{N mm}^2
$$

#### Step 2: buckling load

$$
P_{cr} = \frac{\pi^2EI}{L_e^2} = \frac{\pi^2\times5\times10^9}{(2000)^2} = 12{,}337\ \text{N}
$$

**Answer: buckling load $\approx 12.34$ kN.**

*If the 2 m were taken as the actual length (so $L_e = 4$ m), $EI = 4\times10^{10}\ \text{N mm}^2$ and $P_{cr} = 24.67$ kN; the usual reading of "effective length" gives 12.34 kN.*

## Check yourself

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

1. What is the actual length?
   A. 2 m
   B. 4 m
   C. 1 m, since Le = 2L = 2 m
   D. 0.5 m
   Answer: C. Clamped-free Le = 2L. (Key point: Actual length)

2. What is EI?
   A. 4x10^10
   B. 5x10^6
   C. 1.5x10^9
   D. HL^3/(3 delta) = 5x10^9 N mm^2
   Answer: D. 300x1000^3/(60). (Key point: Find EI)

3. Which Le is in P_cr?
   A. Le = 2000 mm, giving 12.34 kN
   B. L = 1000, 49.3 kN
   C. 4000, 3.08 kN
   D. Le = 1000
   Answer: A. pi^2 x 5e9/2000^2 = 12337 N. (Key point: Buckling load)

4. Why is the other reading different?
   A. It is identical
   B. Using 2 m as actual length gives Le = 4 m and EI = 4e10 so 24.67 kN
   C. It gives 12.34
   D. It cannot be defined
   Answer: B. The reading changes both EI and Le. (Key point: Interpretation)

## Close variants set in other exams

- [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×

## Related in Strength of Materials

- [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
- [Beam deflection used for column buckling load](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ass-beam-deflection-to-column-buckling) — Strength of Materials, chapter 8

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