---
title: "WSM design of 300x450 section"
subject: "Design of RCC Structure"
chapter: 2
key: "semester_7/design_of_rcc_structure:wsm-design-of-300x450-section"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/wsm-design-of-300x450-section
site: BCT Engineering
---

# WSM design of 300x450 section

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

- Asked in: 2071 Chaitra, 4 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-design-of-300x450-section
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-2

## Question

Using working stress method, design a rectangular section 300 mm width and 450 mm height carrying 30 kN/m load in the effective span 3.6 m. Use mild steel and M20 grade of concrete.

## Short answer

Provide 5-16 mm bars at effective depth 400 mm. Required steel area is 1001 mm^2, provided 1005 mm^2; section is under-reinforced and safe.

## Answer

**Assumptions:** the 30 kN/m includes self weight; effective cover 50 mm so $d=450-50=400$ mm. M20: $\sigma_{cbc}=7$, $m=13.33$; mild steel: $\sigma_{st}=140$ N/mm$^2$.

**Step 1: Bending moment**

$$
M=\frac{wL^2}{8}=\frac{30\times3.6^2}{8}=48.60\ \text{kN·m}
$$

**Step 2: Balanced section capacity**

$$
k=\frac{13.33\times7}{13.33\times7+140}=0.400,\quad j=0.867,\quad Q=1.213
$$

$$
M_{bal}=Qbd^2=1.213\times300\times400^2=58.23\ \text{kN·m}>M
$$

So the section is under-reinforced and singly reinforced steel is enough.

**Step 3: Area of steel**

$$
A_{st}=\frac{M}{\sigma_{st}\,j\,d}=\frac{48.60\times10^6}{140\times0.867\times400}=1001\ \text{mm}^2
$$

(Using $j$ of the balanced section is slightly conservative.) Minimum steel $0.85bd/f_y=408$ mm$^2$ is satisfied.

**Provide 5-16 mm dia bars**, $A_{st}=1005$ mm$^2$.

**Check:** neutral axis $x=149.57$ mm $<x_c=kd=159.98$ mm; $M_r=49.28$ kN·m $>$ 48.60 kN·m, safe.

**Answer:** Section 300 x 450 mm with 5-16 mm bars at effective depth 400 mm (plus nominal 2-10 mm hanger bars and stirrups).

## Check yourself

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

1. How is the bending moment found for the simply supported beam?
   A. M = wL^2/8 = 48.60 kN.m
   B. M = wL = 108 kN.m
   C. M = wL^2/2
   D. M = wL^2/12
   Answer: A. 30 x 3.6^2/8 = 48.60 kN.m. (Key point: Design moment)

2. Why is singly reinforced steel enough?
   A. M_bal = 58.23 kN.m exceeds M = 48.60 kN.m
   B. Because mild steel is used
   C. Because span is short
   D. Because M_bal is less than M
   Answer: A. The balanced section capacity exceeds the design moment, so no compression steel is needed. (Key point: Section check)

3. Which formula gives Ast?
   A. M/(sigma_st j d)
   B. M/(sigma_cbc b d)
   C. M/(0.87 fy d)
   D. M/(Q b)
   Answer: A. 48.60e6/(140 x 0.867 x 400) = 1001 mm2. (Key point: Area of steel)

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