---
title: "WSM udl capacity of 25x60 cm beam"
subject: "Design of RCC Structure"
chapter: 2
key: "semester_7/design_of_rcc_structure:wsm-udl-capacity-of-25x60-cm"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/wsm-udl-capacity-of-25x60-cm
site: BCT Engineering
---

# WSM udl capacity of 25x60 cm beam

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

- Asked in: 2078 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-udl-capacity-of-25x60-cm
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-2

## Question

An R.C.C beam 25 cm wide and 60 cm deep has 4 bars of 20 mm dia. as tension reinforcement. The centre of bars being 5 cm above the bottom of the beam. Determine the uniformly distributed load the beam can carry over a simply supported effective span of 6.10 m. The permissible stresses in concrete and steel are taken as 7 MPa and 230 MPa respectively. Use modular ratio.

## Short answer

With d=550 mm and Ast=1256.6 mm2, actual neutral axis is 212.63 mm, greater than balanced 158.73 mm, so over-reinforced. Moment of resistance is 89.14 kN·m. Safe total UDL is 19.16 kN/m including self weight, giving 15.41 kN/m superimposed load.

## Answer

**Given:** $b=250$ mm, $D=600$ mm, $d=600-50=550$ mm, 4-20 mm dia, $A_{st}=4\times\frac{\pi}{4}(20)^2=1256.6$ mm$^2$, span $L=6.10$ m, $\sigma_{cbc}=7$, $\sigma_{st}=230$ N/mm$^2$, $m=13.33$ (modular ratio for M20).

**Step 1: Constants** ($\sigma_{cbc}=7$, $\sigma_{st}=230$ 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)

$$
\frac{b x^2}{2} = m A_{st}(d-x)\ \Rightarrow\ \frac{250 x^2}{2} = 13.33\times 1256.6\,(550-x)
$$

$$
x = 212.63\ \text{mm}
$$

Since $x = 212.63 > x_c = 158.73$ mm, the section is **over-reinforced**: concrete reaches $\sigma_{cbc}=7$ N/mm$^2$ first, so concrete governs.

**Step 3: Moment of resistance**

$$
\begin{aligned}
M_r &= \frac12\sigma_{cbc}\,b\,x\left(d-\frac{x}{3}\right) = 0.5\times 7\times 250\times 212.63\times\left(550-\frac{212.63}{3}\right)\\
&= 89.14\ \text{kN·m}
\end{aligned}
$$

Stress in steel at this moment: $\sigma_{st} = \dfrac{m\,\sigma_{cbc}(d-x)}{x} = 148.05$ N/mm$^2$ $<$ 230 N/mm$^2$.

**Step 4: Safe UDL on the simply supported span**

$$
M_r=\frac{wL^2}{8}\ \Rightarrow\ w=\frac{8M_r}{L^2}=\frac{8\times89.14}{6.10^2}=19.16\ \text{kN/m}
$$

Self weight $=0.25\times0.60\times25=3.75$ kN/m, so the superimposed load is $19.16-3.75=15.41$ kN/m.

**Answer:** total safe UDL $=19.16$ kN/m (including self weight), i.e. superimposed load $=15.41$ kN/m.

## Check yourself

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

1. Overall depth 60 cm with bar centres 5 cm above the bottom gives d equal to:
   A. 500 mm
   B. 550 mm
   C. 600 mm
   D. 650 mm
   Answer: B. d = 600 - 50 = 550 mm. (Key point: Effective depth)

2. x = 212.63 mm > x_c = 158.73 mm. What governs the moment of resistance?
   A. Steel, at 230 N/mm2
   B. Concrete, at 7 N/mm2
   C. Shear
   D. Self weight
   Answer: B. Over-reinforced: concrete reaches sigma_cbc first. (Key point: Governing material)

3. Which formula converts M_r into the total UDL on a simply supported span?
   A. w = M_r / L
   B. w = 8 M_r / L^2
   C. w = M_r L^2 / 8
   D. w = 4 M_r / L
   Answer: B. M = wL^2/8, so w = 8x89.14/6.10^2 = 19.16 kN/m. (Key point: UDL from moment)

4. How is the superimposed load found from the total load?
   A. Add self weight
   B. Subtract self weight 3.75 kN/m, giving 15.41 kN/m
   C. Multiply by 1.5
   D. Divide by span
   Answer: B. Self weight = 0.25 x 0.60 x 25 = 3.75 kN/m, so 19.16 - 3.75 = 15.41 kN/m. (Key point: Superimposed load)

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