---
title: "Round bar fixed-free: buckling load from lateral deflection"
subject: "Strength of Materials"
chapter: 8
key: "semester_3/strength_of_materials:round-bar-fixed-free-buckling-3m"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/round-bar-fixed-free-buckling-3m
site: BCT Engineering
---

# Round bar fixed-free: buckling load from lateral deflection

Strength of Materials · Chapter 8: Column Theory

- Asked in: 2079 Bhadra, 8 marks
- Video: [Understanding Buckling](https://www.youtube.com/watch?v=21G7LA2DcGQ) — The Efficient Engineer
- 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-fixed-free-buckling-3m
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-8

## Question

A round bar fixed at bottom and free at top has a length of 3 m. Determine the buckling load for the bar if the load is applied axially on top. If a horizontal force of 15 kN at top can produces a horizontal deflection of 35 mm.

## Short answer

From the cantilever test, EI = 15×10³ × 3000³ / (3 × 35) = 3.857×10¹² N·mm². For fixed-free, effective length is 2L = 6000 mm, giving buckling load P_cr = π²EI / (6000)² = 1.057×10⁶ N ≈ 1057 kN.

## Answer

**Given:** round bar fixed at the bottom, free at the top, $L = 3$ m = 3000 mm. A horizontal force $H = 15$ kN at the top gives a horizontal deflection $\delta = 35$ mm.

#### Step 1: find $EI$ from the lateral load test
The bar acts as a cantilever under $H$:

$$
\delta = \frac{HL^3}{3EI} \;\Rightarrow\; EI = \frac{HL^3}{3\delta} = \frac{15\times10^3\times3000^3}{3\times35} = 3.857\times10^{12}\ \text{N mm}^2
$$

#### Step 2: buckling load (fixed-free)
Effective length $L_e = 2L = 6000$ mm.

$$
P_{cr} = \frac{\pi^2EI}{L_e^2} = \frac{\pi^2\times3.857\times10^{12}}{(6000)^2} = 1.057\times10^6\ \text{N}
$$

**Answer: buckling load $P_{cr} = 1057$ kN (about 1.06 MN).**

## Check yourself

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

1. How is EI found?
   A. From delta = HL^3/(48EI)
   B. From cantilever deflection delta = HL^3/(3EI)
   C. From delta = HL/AE
   D. From Euler formula
   Answer: B. The lateral test treats the bar as a cantilever. (Key point: Finding EI)

2. What is EI?
   A. 3.857x10^9
   B. 1.2x10^13
   C. 3.857x10^12 N mm^2
   D. 4.5x10^12
   Answer: C. 15000x3000^3/(3x35). (Key point: EI value)

3. What is Le for fixed-free?
   A. L
   B. 0.7L
   C. 0.5L
   D. 2L = 6000 mm
   Answer: D. Fixed-free column has effective length twice its length. (Key point: Effective length)

4. What is P_cr?
   A. About 1057 kN
   B. About 4228 kN
   C. About 264 kN
   D. About 15 kN
   Answer: A. pi^2 x 3.857e12/6000^2 = 1.057e6 N. (Key point: Buckling load)

## Close variants set in other exams

- [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×

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- [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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