---
title: "WSM steel area and resistance, M15"
subject: "Design of RCC Structure"
chapter: 2
key: "semester_7/design_of_rcc_structure:wsm-steel-area-and-resistance-m15"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/wsm-steel-area-and-resistance-m15
site: BCT Engineering
---

# WSM steel area and resistance, M15

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

- Asked in: 2080 Bhadra, 8 marks
- Video: [Design of Singly Reinforced Beam | Working Stress Method | #WSM | Moment of Resistance | Rcc](https://www.youtube.com/watch?v=DFwdx2_tC8E) — GATE HUB
- 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-steel-area-and-resistance-m15
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-2

## Question

Determine the area of steel required and the moment of resistance for a RC beam 250 mm x 450 mm. Consider M15 grade concrete and mild steel. Use the working stress method.

## Short answer

For M15 and mild steel, balanced constants are k=0.400, j=0.867, Q=0.867 N/mm2. With b=250 mm and d=450 mm, moment of resistance is 43.88 kN·m. Required steel area is 803.7 mm2, say 4-16 mm bars.

## Answer

**Assumption:** the section is given as $b\times d$, so $b=250$ mm and effective depth $d=450$ mm. For M15 concrete $\sigma_{cbc}=5$ N/mm$^2$, $m=\dfrac{280}{3\times5}=18.67$; for mild steel $\sigma_{st}=140$ N/mm$^2$.

**Step 1: Balanced section constants**

$$
\begin{aligned}
k&=\frac{m\sigma_{cbc}}{m\sigma_{cbc}+\sigma_{st}}=\frac{18.67\times5}{18.67\times5+140}=0.400\\
j&=1-\frac{k}{3}=0.867\\
Q&=\frac12\sigma_{cbc}\,k\,j=\frac12\times5\times0.400\times0.867=0.867\ \text{N/mm}^2
\end{aligned}
$$

**Step 2: Moment of resistance (balanced section, maximum for singly reinforced)**

$$
M_r=Q\,b\,d^2=0.867\times250\times450^2=43.88\ \text{kN·m}
$$

**Step 3: Area of steel**

$$
A_{st}=\frac{M_r}{\sigma_{st}\,j\,d}=\frac{43.88\times10^6}{140\times0.867\times450}=803.7\ \text{mm}^2
$$

Minimum steel check (IS 456, cl. 26.5.1.1): $0.85bd/f_y=0.85\times250\times450/250=382$ mm$^2$ $<A_{st}$, OK.

Provide 4-16 mm dia bars ($804$ mm$^2$).

**Answer:** $A_{st}=803.7$ mm$^2$ (say 4-16 mm bars) and $M_r=43.88$ kN·m.

## Check yourself

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

1. For M15 with m = 18.67, sigma_cbc = 5 and sigma_st = 140, what is k?
   A. 0.289
   B. 0.400
   C. 0.867
   D. 0.253
   Answer: B. k = 18.67x5/(18.67x5+140) = 0.400. (Key point: Balanced constant k)

2. How is the balanced moment of resistance written?
   A. M_r = Q b d^2 with Q = 0.5 sigma_cbc k j
   B. M_r = sigma_st Ast d
   C. M_r = 0.36 fck b xu (d - 0.416 xu)
   D. M_r = Ast / (b d)
   Answer: A. Q = 0.5 x 5 x 0.400 x 0.867 = 0.867, so M_r = 43.88 kN.m. (Key point: Moment of resistance)

3. Which formula gives the steel area?
   A. Ast = M_r / (sigma_st j d)
   B. Ast = M_r / (sigma_cbc b d)
   C. Ast = 0.85 b d / fy
   D. Ast = M_r / (0.87 fy d)
   Answer: A. Ast = Mr/(sigma_st j d) = 803.7 mm2 with j = 1 - k/3 = 0.867. (Key point: Area of steel)

4. What is the purpose of the 0.85bd/fy check?
   A. It gives the maximum steel allowed
   B. It ensures the provided Ast is not below the code minimum
   C. It finds the modular ratio
   D. It sets the neutral axis depth
   Answer: B. IS 456 cl. 26.5.1.1 sets a minimum; 382 mm2 < 803.7 mm2, OK. (Key point: Minimum steel check)

## Related in Design of RCC Structure

- [WSM moment of resistance, 250x450 beam](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Awsm-moment-of-resistance-250x450-beam) — Design of RCC Structure, chapter 2
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