---
title: "Hollow steel strut: crippling and safe load"
subject: "Strength of Materials"
chapter: 8
key: "semester_3/strength_of_materials:hollow-steel-strut-5m-fixed-hinged"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/hollow-steel-strut-5m-fixed-hinged
site: BCT Engineering
---

# Hollow steel strut: crippling and safe load

Strength of Materials · Chapter 8: Column Theory

- Asked in: 2066 Jestha (old course), 8 marks
- Video: [Euler's Crippling load for tube/Hollow circular section/ Column/ Strength of Materials/ structures](https://www.youtube.com/watch?v=FjFigjvVy74) — Prof.Dr.Pravin Patil
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/hollow-steel-strut-5m-fixed-hinged
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-8

## Question

A hollow 5m long mild steel tube has the internal diameter of 4cm and thickness of 6mm. It is used as a strut with one end fixed and the other hinged. Find crippling load, crippling stress and the safe compressive load for the member if $E = 2\times10^5\ \text{N/mm}^2$ and factor of safety is 3.

## Short answer

Using effective length 0.7 × 5 = 3.5 m and slenderness ratio 213, Euler’s formula applies. Crippling load = 36.83 kN, crippling stress = 42.48 N/mm², safe load with factor of safety 3 = 12.28 kN.

## Answer

**Given:** $L = 5$ m = 5000 mm, $d_i = 40$ mm, $t = 6$ mm so $d_o = 52$ mm. One end fixed, other hinged: $P_{cr} = \dfrac{2\pi^2EI}{L^2}$ ($L_e = 0.7L$). $E = 2\times10^5\ \text{N/mm}^2$, FOS = 3.

#### Section properties
$$
I = \frac{\pi}{64}(52^4 - 40^4) = 2.332\times10^5\ \text{mm}^4
$$

$$
A = \frac{\pi}{4}(52^2 - 40^2) = 867.1\ \text{mm}^2
$$

$r = \sqrt{I/A} = 16.4$ mm, and $L_e/r = 3500/16.4 = 213$, so the strut is slender and Euler's formula is valid.

#### Crippling load
$$
P_{cr} = \frac{2\pi^2\times2\times10^5\times2.332\times10^5}{5000^2} = 36{,}832\ \text{N} = 36.83\ \text{kN}
$$

#### Crippling stress
$$
\sigma_{cr} = \frac{P_{cr}}{A} = \frac{36{,}832}{867.1} = 42.48\ \text{N/mm}^2
$$

#### Safe load
$$
P_{safe} = \frac{36.83}{3} = 12.28\ \text{kN}
$$

**Answer: crippling load = 36.83 kN; crippling stress = 42.48 N/mm²; safe load = 12.28 kN.**

## Check yourself

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

1. For a tube with internal diameter 40 mm and thickness 6 mm, the outer diameter is:
   A. 46 mm
   B. 52 mm
   C. 44 mm
   D. 58 mm
   Answer: B. d_o = d_i + 2t = 40 + 12 = 52 mm; thickness is on both sides. (Key point: Outer diameter)

2. Which expression gives the moment of inertia of the hollow tube?
   A. (pi/64)(52^4 - 40^4)
   B. (pi/64)(52 - 40)^4
   C. (pi/32)(52^4 - 40^4)
   D. (pi/64)(52^2 - 40^2)
   Answer: A. For a hollow circular section I = (pi/64)(D^4 - d^4), giving 2.332 x 10^5 mm^4. (Key point: Section properties)

3. Why is Euler's formula valid here?
   A. Because the strut is hollow
   B. Because the slenderness ratio Le/r is about 213, so the strut is slender
   C. Because the factor of safety is 3
   D. Because both ends are partly fixed
   Answer: B. Euler's formula applies to long, slender struts; Le/r = 3500/16.4 = 213 is large enough. (Key point: Validity of Euler)

4. The crippling load for the fixed-hinged strut is about:
   A. 18.4 kN
   B. 36.8 kN
   C. 73.7 kN
   D. 12.3 kN
   Answer: B. Pcr = 2 pi^2 E I/L^2 = 2 pi^2 x 2x10^5 x 2.332x10^5/5000^2, about 36.83 kN. (Key point: Crippling load)

5. How is the safe load obtained and what is it?
   A. Pcr x 3 = 110.5 kN
   B. Pcr / 3 = 12.28 kN
   C. Pcr / 2 = 18.4 kN
   D. sigma_cr x 3 = 127 N
   Answer: B. Safe load = crippling load / factor of safety = 36.83/3 = 12.28 kN. (Key point: Safe load)

## Related in Strength of Materials

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