---
title: "Hollow section replacing solid compression member"
subject: "Strength of Materials"
chapter: 6
key: "semester_3/strength_of_materials:hollow-replace-solid-compression-member"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/hollow-replace-solid-compression-member
site: BCT Engineering
---

# Hollow section replacing solid compression member

Strength of Materials · Chapter 6: Torsion

- Asked in: 2068 Baisakh (old course), 8 marks
- 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-replace-solid-compression-member
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-6

## Question

A solid circular compression member 50mm in diameter is to be replaced by a hollow circular section of the same material. Find the size of the hollow section, if internal diameter is 0.8 times the external diameter.

## Short answer

Equating cross-sectional areas gives outer diameter 83.3 mm and inner diameter 66.7 mm.

## Answer

**Condition:** the hollow member must carry the same axial load as the solid one with the same material and the same stress, so the **cross-sectional area must be equal**. Let the hollow section have outer diameter $D$ and inner diameter $d_i = 0.8D$.

$$
A_{solid} = \frac{\pi}{4}(50)^2 = 1963.5\ \text{mm}^2
$$

$$
A_{hollow} = \frac{\pi}{4}(D^2 - d_i^2) = \frac{\pi}{4}D^2(1-0.8^2) = \frac{\pi}{4}D^2(0.36)
$$

Equating areas:

$$
0.36D^2 = 50^2 \Rightarrow D = \frac{50}{0.6} = 83.33\ \text{mm}, \qquad d_i = 0.8D = 66.67\ \text{mm}
$$

Wall thickness $t = \dfrac{D - d_i}{2} = 8.33$ mm.

#### Comparison (why the hollow section is better in compression)
| Property | Solid | Hollow |
|---|---|---|
| Area (mm²) | 1963.5 | 1963.5 |
| $I$ (mm⁴) | $3.0680\times10^{5}$ | $1.3976\times10^{6}$ |
| $r = \sqrt{I/A}$ (mm) | 12.50 | 26.68 |

For the same weight, the hollow section has a much larger moment of inertia and radius of gyration, so its slenderness ratio is smaller and it resists buckling better.

**Answer: outer diameter $\approx 83.3$ mm and inner diameter $\approx 66.7$ mm.**

## Check yourself

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

1. What condition must the hollow member satisfy to replace the solid one in compression?
   A. Same outer diameter
   B. Same cross-sectional area (same load, same stress)
   C. Same moment of inertia
   D. Same radius of gyration
   Answer: B. For the same material, load and stress, the area must be equal. (Key point: Equivalence condition)

2. With d_i = 0.8D, what is the hollow area in terms of D?
   A. (pi/4) D^2 (0.36)
   B. (pi/4) D^2 (0.2)
   C. (pi/4) D^2 (0.8)
   D. (pi/4) D^2 (0.64)
   Answer: A. 1 - 0.8^2 = 0.36, so A = (pi/4)D^2(0.36). (Key point: Area equation)

3. What outer and inner diameters result?
   A. D = 50 mm, d = 40 mm
   B. D = 62.5 mm, d = 50 mm
   C. D = 83.3 mm, d = 66.7 mm
   D. D = 138.9 mm, d = 111.1 mm
   Answer: C. 0.36 D^2 = 50^2 gives D = 50/0.6 = 83.33 mm and d = 66.67 mm. (Key point: Sizes)

4. Why is the hollow section better as a compression member of the same area?
   A. It is stronger in tension only
   B. It has a larger area
   C. It has lower stress
   D. It has larger I and radius of gyration, so lower slenderness ratio and better buckling resistance
   Answer: D. I rises from 3.07e5 to 1.40e6 mm^4 and r from 12.5 to 26.7 mm, reducing slenderness. (Key point: Why hollow is better)

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