---
title: "Reinforced concrete column stresses in steel and concrete"
subject: "Strength of Materials"
chapter: 3
key: "semester_3/strength_of_materials:rc-column-8-bars"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/rc-column-8-bars
site: BCT Engineering
---

# Reinforced concrete column stresses in steel and concrete

Strength of Materials · Chapter 3: Simple Stress and Strain

- Asked in: 2080 Bhadra, 6 marks
- Video: [4-31 Determine stress in concrete & steel | Axial Loading | Mechanics of Materials by R.C Hibbeler](https://www.youtube.com/watch?v=rwt8-cZNP_4) — Engr. Adnan Rasheed Mechanical
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/rc-column-8-bars
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-3

## Question

A reinforced concrete column of size $230\ \text{mm}\times400\ \text{mm}$ has 8 steel bars of 12 mm diameter and subjected to an axial compression of 600 kN, find the stresses developed in steel and concrete, take modular ratio as 18.67.

## Short answer

Using modular ratio m=18.67, equal strain gives σs = m σc. Equilibrium of total load P = σc Ac + σs As yields σc = 5.56 N/mm² and σs = 103.73 N/mm² (both compressive).

## Answer

The steel and concrete shorten by the same amount, so the strain is equal and $\sigma_s = m\,\sigma_c$ with the modular ratio $m = E_s/E_c = 18.67$.

#### Areas
$$
\begin{aligned}
A_s &= 8\times\frac{\pi}{4}(12)^2 = 904.78\ \text{mm}^2\\
A_g &= 230\times400 = 92000\ \text{mm}^2\\
A_c &= A_g - A_s = 91095.22\ \text{mm}^2
\end{aligned}
$$

#### Equilibrium
$$
P = \sigma_c A_c + \sigma_s A_s = \sigma_c\,(A_c + mA_s)
$$
$$
600\times10^3 = \sigma_c\,(91095.22 + 18.67\times904.78) = \sigma_c\,(107987.44)
$$
$$
\sigma_c = 5.556\ \text{N/mm}^2,\qquad \sigma_s = m\sigma_c = 18.67\times5.556 = 103.73\ \text{N/mm}^2
$$

Check: load carried by concrete 506.14 kN + steel 93.86 kN = 600.00 kN.

**Answer: stress in concrete = 5.56 N/mm² and stress in steel = 103.73 N/mm² (both compressive).**

## Check yourself

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

1. Why is sigma_s = m * sigma_c in the RC column?
   A. Steel and concrete shorten by the same amount, so their strains are equal
   B. Steel and concrete carry equal loads
   C. Steel and concrete have equal areas
   D. Steel and concrete have the same modulus
   Answer: A. Both materials are bonded and shorten equally, so sigma_s/E_s = sigma_c/E_c, giving sigma_s = m sigma_c. (Key point: Compatibility)

2. How is the net concrete area found?
   A. Gross area plus steel area
   B. Gross area minus steel area
   C. Gross area times modular ratio
   D. Steel area times modular ratio
   Answer: B. The steel bars occupy part of the section, so A_c = A_g - A_s = 91095 mm^2. (Key point: Concrete area)

3. Which equation gives the concrete stress?
   A. P = sigma_c A_c only
   B. P = sigma_c (A_c - m A_s)
   C. P = sigma_c (A_c + m A_s)
   D. P = sigma_c A_g / m
   Answer: C. Load is shared: P = sigma_c A_c + sigma_s A_s, and substituting sigma_s = m sigma_c gives sigma_c (A_c + m A_s). (Key point: Equilibrium)

4. Which pair of stresses is obtained?
   A. Concrete 103.73 N/mm^2, steel 5.56 N/mm^2
   B. Both 6.67 N/mm^2
   C. Concrete 5.56 N/mm^2, steel 5.56 N/mm^2
   D. Concrete 5.56 N/mm^2, steel 103.73 N/mm^2
   Answer: D. sigma_c = 600000/107987 = 5.556 and sigma_s = 18.67 x 5.556 = 103.73 N/mm^2. (Key point: Result)

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