---
title: "WSM moment of resistance, 300x550 beam"
subject: "Design of RCC Structure"
chapter: 2
key: "semester_7/design_of_rcc_structure:wsm-moment-of-resistance-300x550-beam"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/wsm-moment-of-resistance-300x550-beam
site: BCT Engineering
---

# WSM moment of resistance, 300x550 beam

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

- Asked in: 2081 Bhadra, 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-moment-of-resistance-300x550-beam
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-2

## Question

A rectangular beam is 300 mm wide and has 550 mm effective depth. It is reinforced with 4 bars of 12 mm diameter. Determine the moment of resistance using the working stress method of analysis. Use M20 concrete and Fe415 steel.

## Answer

**Given:** $b=300$ mm, $d=550$ mm, 4-12 mm dia, $A_{st}=4\times\frac{\pi}{4}(12)^2=452.4$ mm$^2$. M20: $\sigma_{cbc}=7$ N/mm$^2$, $m=13.33$; Fe415: $\sigma_{st}=230$ N/mm$^2$.

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

$$
\begin{aligned}
k &= \frac{m\sigma_{cbc}}{m\sigma_{cbc}+\sigma_{st}} = \frac{13.33\times 7}{13.33\times 7 + 230} = 0.289\\
x_c &= k d = 0.289\times 550 = 158.73\ \text{mm}
\end{aligned}
$$

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

$$
\frac{b x^2}{2} = m A_{st}(d-x)\ \Rightarrow\ \frac{300 x^2}{2} = 13.33\times 452.4\,(550-x)
$$

$$
x = 129.95\ \text{mm}
$$

Since $x = 129.95 < x_c = 158.73$ mm, the section is **under-reinforced**: steel reaches $\sigma_{st}=230$ N/mm$^2$ first, so steel governs.

**Step 3: Moment of resistance**

$$
\begin{aligned}
M_r &= \sigma_{st}A_{st}\left(d-\frac{x}{3}\right) = 230\times 452.4\times\left(550-\frac{129.95}{3}\right)\\
&= 52.72\ \text{kN·m}
\end{aligned}
$$

Stress in concrete at this moment: $\sigma_{c} = \dfrac{\sigma_{st}\,x}{m(d-x)} = 5.34$ N/mm$^2$ $<$ 7 N/mm$^2$ (safe).

**Answer: $M_r = 52.72$ kN·m** (under-reinforced section).

## Check yourself

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

1. How is m obtained for M20 here?
   A. m = 280/(3 x 7) = 13.33
   B. m = 230/7
   C. m = 7/230
   D. m = 0.87 x 415
   Answer: A. m = 280/(3 sigma_cbc) with sigma_cbc = 7 N/mm2 gives 13.33. (Key point: Modular ratio)

2. x = 129.95 mm and x_c = 158.73 mm. What type of section is it?
   A. Over-reinforced
   B. Under-reinforced, steel reaches sigma_st first
   C. Balanced
   D. Doubly reinforced
   Answer: B. x < x_c: steel reaches its permissible stress first. (Key point: Classification)

3. Which expression gives M_r for the under-reinforced section?
   A. 0.5 sigma_cbc b x (d - x/3)
   B. sigma_st Ast (d - x/3)
   C. sigma_st Ast d
   D. 0.87 fy Ast (d - 0.416 x)
   Answer: B. Steel governs, so M_r = sigma_st Ast (d - x/3) = 52.72 kN.m. (Key point: Moment formula)

4. After finding M_r, why is the concrete stress computed (5.34 N/mm2)?
   A. To confirm it is below the permissible 7 N/mm2, so the section is safe
   B. To replace the steel stress
   C. To find the modular ratio
   D. To compute the self-weight
   Answer: A. For an under-reinforced section concrete stress must stay below sigma_cbc; 5.34 < 7. (Key point: Concrete stress check)

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