---
title: "Beam deflection used for column buckling load"
subject: "Strength of Materials"
chapter: 8
key: "semester_3/strength_of_materials:ss-beam-deflection-to-column-buckling"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/ss-beam-deflection-to-column-buckling
site: BCT Engineering
---

# Beam deflection used for column buckling load

Strength of Materials · Chapter 8: Column Theory

- Asked in: 2067 Asar (old course), 10 marks
- Video: [Mechanics of Materials: Lesson 67 - Beam Column Buckling Example](https://www.youtube.com/watch?v=H_B4f-MjNKM) — Jeff Hanson
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/ss-beam-deflection-to-column-buckling
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-8

## Question

A 5m long simply supported beam has 30mm maximum deflection due to 800N transverse load applied at the center of the beam. Determine the buckling load if the same beam of 5m length is used as a column with one end fixed and other end hinged.

## Short answer

EI from simply supported beam central deflection: EI = 800 × 5000³ / (48 × 30) = 6.944×10¹⁰ N·mm². As a column fixed at one end and hinged at the other, P_cr = 2π²EI/L² ≈ 54.8 kN.

## Answer

**Given:** simply supported beam, $L = 5$ m = 5000 mm, central load $W = 800$ N, maximum (central) deflection $\delta = 30$ mm.

#### Step 1: find $EI$
For a simply supported beam with a central point load:

$$
\delta = \frac{WL^3}{48EI} \;\Rightarrow\; EI = \frac{WL^3}{48\delta} = \frac{800\times5000^3}{48\times30} = 6.944\times10^{10}\ \text{N mm}^2
$$

#### Step 2: buckling load as a column
For one end fixed and the other hinged, the effective length is $L_e = 0.7L$, which corresponds to the standard textbook form $P_{cr} = 2\pi^2EI/L^2$:

$$
P_{cr} = \frac{2\pi^2EI}{L^2} = \frac{2\pi^2\times6.944\times10^{10}}{5000^2} = 54{,}831\ \text{N}
$$

(Using the exact root of $\tan kL = kL$, that is $P_{cr} = 20.19EI/L^2$, gives about 56 kN.)

**Answer: buckling load $\approx 54.8$ kN.**

## Check yourself

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

1. Which relation is used to find EI from the beam data (central load W, span L, deflection delta)?
   A. delta = WL^3/(3EI)
   B. delta = WL^3/(48EI)
   C. delta = 5WL^4/(384EI)
   D. delta = WL^2/(8EI)
   Answer: B. A simply supported beam with a central point load has maximum deflection WL^3/(48EI), so EI = WL^3/(48 delta). (Key point: Finding EI from beam)

2. With W = 800 N, L = 5000 mm and delta = 30 mm, EI equals about:
   A. 6.94 x 10^10 N mm^2
   B. 6.94 x 10^8 N mm^2
   C. 2.08 x 10^11 N mm^2
   D. 1.67 x 10^9 N mm^2
   Answer: A. EI = 800 x 5000^3/(48 x 30) = 1 x 10^14/1440 = 6.944 x 10^10 N mm^2. (Key point: Value of EI)

3. For the column with one end fixed and the other hinged, the buckling load used is:
   A. pi^2 EI/L^2
   B. pi^2 EI/(4L^2)
   C. 2 pi^2 EI/L^2 (Le about 0.7L)
   D. 4 pi^2 EI/L^2
   Answer: C. Fixed-hinged has Le = 0.7L, so Pcr is taken as 2 pi^2 EI/L^2. (Key point: End condition formula)

4. The buckling load of the column is approximately:
   A. 27.4 kN
   B. 54.8 kN
   C. 109.7 kN
   D. 13.7 kN
   Answer: B. Pcr = 2 pi^2 x 6.944 x 10^10/5000^2 = 54,831 N, about 54.8 kN. (Key point: Final result)

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