---
title: "WSM moment of resistance, 4.2 m beam"
subject: "Design of RCC Structure"
chapter: 2
key: "semester_7/design_of_rcc_structure:wsm-moment-of-resistance-4-2"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/wsm-moment-of-resistance-4-2
site: BCT Engineering
---

# WSM moment of resistance, 4.2 m beam

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

- Asked in: 2074 Asoj, 6 marks
- Video: [How to find Moment of resistance of beam by WSM](https://www.youtube.com/watch?v=OnGUkxGnoxI) — MOHD SHARIQUE AHMAD
- 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-moment-of-resistance-4-2
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-2

## Question

A simply supported rectangular RC beam of effective span 4.2 m and overall dimensions 230 mm x 450 mm is reinforced with 4-20 mm dia. bars in tension. Determine the moment of resistance. Take permissible stresses for M20 concrete and Fe415 grade steel.

## Short answer

Moment of resistance is 52.29 kN·m. Over-reinforced, concrete governs.

## Answer

**Assumption:** clear cover 25 mm, so $d=450-25-\dfrac{20}{2}=415$ mm. $b=230$ mm, $A_{st}=4\times\frac{\pi}{4}(20)^2=1256.6$ mm$^2$. Permissible stresses: M20 $\sigma_{cbc}=7$ N/mm$^2$, Fe415 $\sigma_{st}=230$ N/mm$^2$, $m=13.33$. (The span 4.2 m is not needed for $M_r$.)

**Step 1: Constants** (M20: $\sigma_{cbc}=7$, Fe415: $\sigma_{st}=230$ N/mm$^2$, $m=13.33$)

$$
\begin{aligned}
k &= \frac{m\sigma_{cbc}}{m\sigma_{cbc}+\sigma_{st}} = \frac{13.33\times 7}{13.33\times 7 + 230} = 0.289\\
x_c &= k d = 0.289\times 415 = 119.77\ \text{mm}
\end{aligned}
$$

**Step 2: Actual neutral axis** (taking moments of areas about the neutral axis, transformed section)

$$
\frac{b x^2}{2} = m A_{st}(d-x)\ \Rightarrow\ \frac{230 x^2}{2} = 13.33\times 1256.6\,(415-x)
$$

$$
x = 183.59\ \text{mm}
$$

Since $x = 183.59 > x_c = 119.77$ mm, the section is **over-reinforced**: concrete reaches $\sigma_{cbc}=7$ N/mm$^2$ first, so concrete governs.

**Step 3: Moment of resistance**

$$
\begin{aligned}
M_r &= \frac12\sigma_{cbc}\,b\,x\left(d-\frac{x}{3}\right) = 0.5\times 7\times 230\times 183.59\times\left(415-\frac{183.59}{3}\right)\\
&= 52.29\ \text{kN·m}
\end{aligned}
$$

Stress in steel at this moment: $\sigma_{st} = \dfrac{m\,\sigma_{cbc}(d-x)}{x} = 117.61$ N/mm$^2$ $<$ 230 N/mm$^2$.

**Answer: $M_r=52.29$ kN·m.**

## Check yourself

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

1. Is the span 4.2 m needed for the moment of resistance?
   A. Yes, M_r = wL^2/8
   B. No, M_r depends only on section and material properties
   C. Yes, to find d
   D. Yes, to find m
   Answer: B. Span is needed only when converting to load. (Key point: Irrelevant data)

2. D = 450, clear cover 25, 20 mm bars. d equals:
   A. 415 mm
   B. 425 mm
   C. 450 mm
   D. 405 mm
   Answer: A. d = 450 - 25 - 10 = 415 mm. (Key point: Effective depth)

3. x = 183.59 mm > x_c = 119.77 mm. Which M_r formula applies?
   A. sigma_st Ast (d - x/3)
   B. 0.5 sigma_cbc b x (d - x/3)
   C. Q b d^2
   D. 0.87 fy Ast (d - 0.416x)
   Answer: B. Over-reinforced, concrete governs: 52.29 kN.m. (Key point: Classification and formula)

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