---
title: "Hollow steel tube strut: critical, safe load"
subject: "Strength of Materials"
chapter: 8
key: "semester_3/strength_of_materials:hollow-steel-strut-both-hinged-safe"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/hollow-steel-strut-both-hinged-safe
site: BCT Engineering
---

# Hollow steel tube strut: critical, safe load

Strength of Materials · Chapter 8: Column Theory

- Asked in: 2078 Bhadra, 6 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-both-hinged-safe
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-8

## Question

A hollow mild steel tube is 5 m long and 4 cm internal diameter. Thickness of tube is 8 mm and it is used as a strut with both ends hinged. Determine critical load and safe load on the strut. Take $E = 2.1\times10^5\ \text{N/mm}^2$ F.O.S = 3

## Short answer

With both ends hinged, L_e = 5 m. I = 3.571×10⁵ mm⁴, critical load P_cr = π²EI/L² = 29.60 kN. Safe load with factor of safety 3 is 9.87 kN.

## Answer

**Given:** $L = 5$ m = 5000 mm, $d_i = 40$ mm, $t = 8$ mm, so $d_o = 40 + 2\times8 = 56$ mm. Both ends hinged ($L_e = L$), $E = 2.1\times10^5\ \text{N/mm}^2$, FOS = 3.

#### Section properties
$$
I = \frac{\pi}{64}(56^4 - 40^4) = \frac{\pi}{64}(9834496 - 2560000) = 3.571\times10^5\ \text{mm}^4
$$

$$
A = \frac{\pi}{4}(56^2 - 40^2) = 1206.4\ \text{mm}^2
$$

Radius of gyration $r = \sqrt{I/A} = 17.2$ mm, so $\lambda = 5000/17.2 = 290.6$, which is very slender, so Euler's formula applies.

#### Critical load
$$
P_{cr} = \frac{\pi^2EI}{L^2} = \frac{\pi^2\times2.1\times10^5\times3.571\times10^5}{5000^2} = 29{,}604\ \text{N} = 29.60\ \text{kN}
$$

#### Safe load
$$
P_{safe} = \frac{P_{cr}}{FOS} = \frac{29.60}{3} = 9.87\ \text{kN}
$$

**Answer: critical load = 29.60 kN; safe load = 9.87 kN.**

## Check yourself

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

1. What is the outer diameter?
   A. 48 mm
   B. 40 mm
   C. 56 mm (40 + 2x8)
   D. 64 mm
   Answer: C. Thickness is added on both sides. (Key point: Outer diameter)

2. How is I found?
   A. pi/64 (d_o - d_i)^4
   B. pi/32 (d_o^4 - d_i^4)
   C. pi/4 (d_o^2 - d_i^2)
   D. pi/64 (d_o^4 - d_i^4)
   Answer: D. Hollow circular section. (Key point: Moment of inertia)

3. What is P_cr for hinged ends?
   A. pi^2EI/L^2 = 29.60 kN
   B. pi^2EI/(2L)^2
   C. 118 kN
   D. 9.87 kN
   Answer: A. Le = L for hinged ends. (Key point: Critical load)

4. How is safe load found?
   A. P_cr x FOS
   B. P_cr/FOS = 9.87 kN
   C. P_cr
   D. P_cr - 3
   Answer: B. Divide by factor of safety 3. (Key point: Safe load)

## Related in Strength of Materials

- [Hollow steel strut: crippling and safe load](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ahollow-steel-strut-5m-fixed-hinged) — Strength of Materials, chapter 8
- [Euler's formula: both ends hinged](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aeuler-both-ends-hinged) — Strength of Materials, chapter 8, asked 4×
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- [Bar fixed at both ends: load shares, displacements](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Acircular-bar-fixed-both-ends-50kn) — Strength of Materials, chapter 3
- [Euler's formula: fixed-hinged strut with limitation](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aeuler-one-fixed-one-hinged-limitation) — Strength of Materials, chapter 8, asked 2×

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