---
title: "WSM stresses under 50 kNm moment"
subject: "Design of RCC Structure"
chapter: 2
key: "semester_7/design_of_rcc_structure:wsm-stresses-under-50-knm-moment"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/wsm-stresses-under-50-knm-moment
site: BCT Engineering
---

# WSM stresses under 50 kNm moment

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

- Asked in: 2081 Baisakh, 9 marks
- Video: [Design of Singly Reinforced Beam | Working Stress Method | #WSM | Moment of Resistance | Rcc](https://www.youtube.com/watch?v=DFwdx2_tC8E) — GATE HUB
- 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-stresses-under-50-knm-moment
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-2

## Question

A rectangular beam of size 300 mm x 550 mm effective depth is reinforced with 4-16 mm diameter tension bars. Determine the stress induced in the top compression fibre of the concrete and tension steel when it is subjected to a moment of 50 kN-m. Take M20 concrete and Fe 415 grade steel with the permissible stress in concrete and steel not exceeding 5 MPa and 140 MPa respectively.

## Short answer

With neutral axis 165.72 mm and lever arm 494.76 mm, the extreme concrete stress is 4.07 N/mm2 and tension steel stress is 125.66 N/mm2, both below the permissible limits, so the section is safe.

## Answer

**Given:** $b=300$ mm, $d=550$ mm, 4-16 mm dia, $A_{st}=4\times\frac{\pi}{4}(16)^2=804.2$ mm$^2$, $M=50$ kN·m. Permissible: $\sigma_{cbc}=5$, $\sigma_{st}=140$ N/mm$^2$. For M20, modular ratio $m=13.33$ (IS 456, Annex B-1.3).

**Step 1: Neutral axis** (elastic cracked section)

$$
\frac{b x^2}{2}=m A_{st}(d-x)\ \Rightarrow\ \frac{300x^2}{2}=13.33\times 804.2(550-x)
$$

$$
x=165.72\ \text{mm}
$$

**Step 2: Lever arm**

$$
z=d-\frac{x}{3}=550-\frac{165.72}{3}=494.76\ \text{mm}
$$

**Step 3: Stresses**

$$
\begin{aligned}
\sigma_{c}&=\frac{2M}{b\,x\,z}=\frac{2\times 50\times10^6}{300\times 165.72\times 494.76}=4.07\ \text{N/mm}^2\\
\sigma_{s}&=\frac{M}{A_{st}\,z}=\frac{50\times10^6}{804.2\times 494.76}=125.66\ \text{N/mm}^2
\end{aligned}
$$

Both are less than the permissible values (5 and 140 N/mm$^2$), so the section is safe.

**Answer:** stress in extreme compression concrete $=4.07$ N/mm$^2$; stress in tension steel $=125.66$ N/mm$^2$.

## Check yourself

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

1. How is the neutral axis of the cracked elastic section found?
   A. b x^2/2 = m Ast (d - x)
   B. b x = m Ast
   C. x = d/3
   D. b x^2/2 = Ast (d - x)
   Answer: A. Take first moments of the compression concrete area and the transformed steel area about the neutral axis. (Key point: Neutral axis equation)

2. What is the lever arm z in this analysis?
   A. z = d
   B. z = d - x/3
   C. z = d - x/2
   D. z = x/3
   Answer: B. Compression resultant acts at x/3 from the top, so z = d - x/3 = 494.76 mm. (Key point: Lever arm)

3. Which formula gives the stress in the extreme compression concrete?
   A. M/(Ast z)
   B. 2M/(b x z)
   C. M/(b x z)
   D. m M/(b x z)
   Answer: B. Compression force is 0.5 sigma_c b x, so sigma_c = 2M/(b x z) = 4.07 N/mm2. (Key point: Concrete stress formula)

4. The computed stresses are 4.07 and 125.66 N/mm2. What is concluded?
   A. Unsafe, concrete exceeds 5
   B. Unsafe, steel exceeds 140
   C. Safe, both are below permissible 5 and 140 N/mm2
   D. Cannot be judged without the span
   Answer: C. Both are below the permissible values. (Key point: Safety conclusion)

## Related in Design of RCC Structure

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