---
title: "WSM tensile steel for 230x425 beam"
subject: "Design of RCC Structure"
chapter: 2
key: "semester_7/design_of_rcc_structure:wsm-tensile-steel-for-230x425-beam"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/wsm-tensile-steel-for-230x425-beam
site: BCT Engineering
---

# WSM tensile steel for 230x425 beam

Design of RCC Structure · Chapter 2: Working Stress Method of Design

- Asked in: 2076 Chaitra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/wsm-tensile-steel-for-230x425-beam
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-2

## Question

Calculate the tensile reinforcement required for a rectangular RC beam of size 230 mm x 425 mm (overall) if it has to carry a moment of 64 kNm at service condition. Use M20 grade concrete mix and Fe500 grade steel in working stress method.

## Short answer

Balanced moment is 27.68 kN·m, so the section is doubly reinforced with M2=36.32 kN·m. Required tension steel is 668 mm2 and compression steel is 1342 mm2.

## Answer

**Assumptions:** effective cover $d'=40$ mm (top and bottom), so $d=425-40=385$ mm. M20: $\sigma_{cbc}=7$ N/mm$^2$, $m=13.33$. Fe500: $\sigma_{st}=275$ N/mm$^2$ (IS 456 Table 22 / Annex B). $M=64$ kN·m.

**Step 1: Balanced moment of the singly reinforced section**

$$
\begin{aligned}
k&=\frac{13.33\times7}{13.33\times7+275}=0.2533,\quad j=1-\frac{k}{3}=0.9156\\
Q&=\tfrac12\sigma_{cbc}kj=0.8118\ \text{N/mm}^2\\
M_{bal}&=Qbd^2=0.8118\times230\times385^2=27.68\ \text{kN·m}
\end{aligned}
$$

Since $M=64>M_{bal}=27.68$ kN·m, a singly reinforced section is not enough. The beam must be **doubly reinforced**.

**Step 2: Moment shared by compression steel**

$$
M_2=M-M_{bal}=64-27.68=36.32\ \text{kN·m}
$$

**Step 3: Compression steel**

Depth of neutral axis (balanced) $x_c=kd=97.54$ mm. Stress in concrete at the compression steel level:

$$
\sigma_c'=\sigma_{cbc}\frac{x_c-d'}{x_c}=7\times\frac{97.54-40}{97.54}=4.13\ \text{N/mm}^2
$$

$$
A_{sc}=\frac{M_2}{(1.5m-1)\,\sigma_c'\,(d-d')}=\frac{36.32\times10^6}{(1.5\times13.33-1)\times4.13\times(385-40)}=1342\ \text{mm}^2
$$

Stress in compression steel $\sigma_{sc}=1.5m\sigma_c'=82.57$ N/mm$^2$, below the permissible value, OK.

**Step 4: Tension steel**

$$
A_{st}=\frac{M_{bal}}{\sigma_{st}jd}+\frac{M_2}{\sigma_{st}(d-d')}=\frac{27.68\times10^6}{275\times0.9156\times385}+\frac{36.32\times10^6}{275\times(385-40)}
$$

$$
A_{st}=286+383=668\ \text{mm}^2
$$

**Answer:** tensile reinforcement $A_{st}=668$ mm$^2$ (e.g. 3-20 mm bars); compression reinforcement $A_{sc}=1342$ mm$^2$ (e.g. 7-16 mm bars).

## Check yourself

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

1. Why is the beam doubly reinforced?
   A. M = 64 kN.m exceeds M_bal = 27.68 kN.m
   B. M is less than M_bal
   C. Fe500 cannot be used singly
   D. Cover is too small
   Answer: A. The singly reinforced balanced section cannot carry 64 kN.m. (Key point: Check for doubly reinforced)

2. How is the moment for compression steel found?
   A. M2 = M - M_bal = 36.32 kN.m
   B. M2 = M + M_bal
   C. M2 = M_bal / 2
   D. M2 = M x 1.5
   Answer: A. The excess over the balanced moment is taken by the compression steel and extra tension steel. (Key point: Moment split)

3. Which expression gives Asc?
   A. M2 / ((1.5m-1) sigma_c' (d-d'))
   B. M2 / (sigma_st (d-d'))
   C. M_bal/(sigma_st j d)
   D. M2 / (sigma_cbc b d)
   Answer: A. Asc = 36.32e6/((1.5x13.33-1) x 4.13 x 345) = 1342 mm2. (Key point: Compression steel formula)

4. What are the two parts of Ast?
   A. M_bal/(sigma_st j d) and M2/(sigma_st(d-d'))
   B. M/(sigma_st d) and Asc
   C. M_bal/(sigma_cbc b d) and M2/d
   D. Only M/(sigma_st j d)
   Answer: A. 286 + 383 = 668 mm2. (Key point: Tension steel)

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