---
title: "WSM moment of resistance, 230x350 beam"
subject: "Design of RCC Structure"
chapter: 2
key: "semester_7/design_of_rcc_structure:wsm-moment-of-resistance-230x350-beam"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/wsm-moment-of-resistance-230x350-beam
site: BCT Engineering
---

# WSM moment of resistance, 230x350 beam

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

- Asked in: 2075 Asoj, 7 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-moment-of-resistance-230x350-beam
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-2

## Question

A rectangular R.C beam of size 230 x 350 mm overall is reinforced with 4-16 mm dia bars at tension zone in bottom. Determine the moment of resistance of that beam section if the permissible stresses in concrete and steel do not exceed 7.0 MPa and 140 MPa. Take nominal cover to re-bar as 25 mm and m = 13.33.

## Short answer

Moment of resistance is 28.92 kN·m. Over-reinforced section, concrete governs.

## Answer

**Given:** $b=230$ mm, $D=350$ mm, nominal cover 25 mm, so $d=350-25-\dfrac{16}{2}=317$ mm. $A_{st}=4\times\frac{\pi}{4}(16)^2=804.2$ mm$^2$. $\sigma_{cbc}=7$, $\sigma_{st}=140$ N/mm$^2$, $m=13.33$.

**Step 1: Constants** ($\sigma_{cbc}=7$, $\sigma_{st}=140$ N/mm$^2$, $m=13.33$)

$$
\begin{aligned}
k &= \frac{m\sigma_{cbc}}{m\sigma_{cbc}+\sigma_{st}} = \frac{13.33\times 7}{13.33\times 7 + 140} = 0.400\\
x_c &= k d = 0.400\times 317 = 126.78\ \text{mm}
\end{aligned}
$$

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

$$
\frac{b x^2}{2} = m A_{st}(d-x)\ \Rightarrow\ \frac{230 x^2}{2} = 13.33\times 804.2\,(317-x)
$$

$$
x = 131.50\ \text{mm}
$$

Since $x = 131.50 > x_c = 126.78$ mm, the section is **over-reinforced**: concrete reaches $\sigma_{cbc}=7$ N/mm$^2$ first, so concrete governs.

**Step 3: Moment of resistance**

$$
\begin{aligned}
M_r &= \frac12\sigma_{cbc}\,b\,x\left(d-\frac{x}{3}\right) = 0.5\times 7\times 230\times 131.50\times\left(317-\frac{131.50}{3}\right)\\
&= 28.92\ \text{kN·m}
\end{aligned}
$$

Stress in steel at this moment: $\sigma_{st} = \dfrac{m\,\sigma_{cbc}(d-x)}{x} = 131.62$ N/mm$^2$ $<$ 140 N/mm$^2$.

**Answer: $M_r=28.92$ kN·m.**

## Check yourself

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

1. D = 350 mm, nominal cover 25 mm, 16 mm bars. Find d.
   A. 317 mm
   B. 325 mm
   C. 350 mm
   D. 309 mm
   Answer: A. d = 350 - 25 - 8 = 317 mm. (Key point: Effective depth with nominal cover)

2. With sigma_st = 140, sigma_cbc = 7, m = 13.33, k equals:
   A. 0.289
   B. 0.400
   C. 0.253
   D. 0.867
   Answer: B. k = 93.31/(93.31+140) = 0.400. (Key point: Balanced k for steel 140)

3. x = 131.50 mm vs x_c = 126.78 mm. Which governs?
   A. Steel
   B. Concrete (over-reinforced)
   C. Both equally
   D. Neither
   Answer: B. x > x_c so concrete limits: M_r = 28.92 kN.m. (Key point: Governing material)

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