---
title: "Beam design, 250x475 section"
subject: "Design of RCC Structure"
chapter: 4
key: "semester_7/design_of_rcc_structure:beam-design-250x475-section"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/beam-design-250x475-section
site: BCT Engineering
---

# Beam design, 250x475 section

Design of RCC Structure · Chapter 4: Design of beams: Behavior in Flexure

- Asked in: 2079 Bhadra, 10 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/beam-design-250x475-section
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-4

## Question

A simply supported RCC beam of effective span 4 m and overall dimensions 250 mm x 475 mm is subjected to superimposed load of 45 kN/m excluding its self-weight with point load 75 kN at midspan. Design the beam for limit state of collapse in flexure. Also, check whether the beam is safe in deflection or not. Consider effective cover to be 45 mm. Take M25 concrete and TMT bars. All the loads are at service level.

## Short answer

Mu=256.41 kNm. Doubly reinforced: Ast=1632 mm² (3-25+1-16 mm), Asc=664 mm² (4-16 mm). Deflection safe: L/d actual=9.3, allowable=18.7.

## Answer

**Given:** $L=4$ m, $b=250$ mm, $D=475$ mm, effective cover 45 mm, so $d=430$ mm and $d'=45$ mm; superimposed load 45 kN/m and point load 75 kN at mid-span (service), M25. "TMT bars" is taken as Fe500 ($f_y=500$ N/mm$^2$).

**Step 1: Loads and moment**

$$
\begin{aligned}
\text{Self weight}&=0.25\times0.475\times25=2.97\ \text{kN/m},\quad w=47.97\ \text{kN/m}\\
w_u&=1.5w=71.95\ \text{kN/m},\quad P_u=1.5\times75=112.5\ \text{kN}\\
M_u&=\frac{w_uL^2}{8}+\frac{P_uL}{4}=143.91+112.50=256.41\ \text{kN·m}
\end{aligned}
$$

**Step 2: Limiting moment** (Fe500: $x_{u,max}=0.46d$)

$$
M_{u,lim}=0.1336f_{ck}bd^2=0.1336\times25\times250\times430^2=154.40\ \text{kN·m}<M_u
$$

The section is too shallow for a singly reinforced design, so it is designed as **doubly reinforced**.

**Doubly reinforced design**

$$
M_{u2}=M_u-M_{u,lim}=256.41-154.40=102.01\ \text{kN·m}
$$

Limiting depth $x_{u,max}=0.46d=197.8$ mm. Strain in compression steel (similar triangles):

$$
\varepsilon_{sc}=0.0035\left(1-\frac{d'}{x_{u,max}}\right)=0.0035\left(1-\frac{45}{197.8}\right)=0.00270
$$

From the design stress-strain curve of Fe500 (IS 456 Fig. 23): $f_{sc}=410.2$ N/mm$^2$. Stress in the concrete displaced by the steel: $f_{cc}=0.446f_{ck}=11.15$ N/mm$^2$.

$$
A_{sc}=\frac{M_{u2}}{(f_{sc}-f_{cc})(d-d')}=\frac{102.01\times10^6}{(410.2-11.15)(430-45)}=664\ \text{mm}^2
$$

$$
A_{st1}=\frac{0.36f_{ck}\,b\,x_{u,max}}{0.87f_y}=1023\ \text{mm}^2,\qquad A_{st2}=\frac{A_{sc}(f_{sc}-f_{cc})}{0.87f_y}=609\ \text{mm}^2
$$

$$
A_{st}=A_{st1}+A_{st2}=1632\ \text{mm}^2
$$

**Provide:** tension 3-25 mm + 1-16 mm dia ($1674$ mm$^2$, one layer in the 250 mm width) and compression 4-16 mm dia ($804$ mm$^2$), with 8 mm stirrups.

Check: $0.87f_yA_{st}=435\times1674=728$ kN, which is less than the compression available at $x_{u,max}$, $0.36f_{ck}bx_{u,max}+A_{sc}(f_{sc}-f_{cc})=445+321=766$ kN, so $x_u<x_{u,max}$, OK. (Providing far more tension steel, such as 6-20 mm = 1885 mm$^2$, would make $x_u>x_{u,max}$.)

**Step 3: Deflection check** (IS 456 cl. 23.2)

$$
p_t=\frac{100A_{st}}{bd}=1.56\%,\quad f_s=0.58f_y\frac{A_{st,req}}{A_{st,prov}}=282.7\ \text{N/mm}^2,\quad k_t\approx0.80\ \text{(Fig. 4)}
$$

Compression steel $p_c=\dfrac{100A_{sc}}{bd}=0.75\%$, modification factor $k_c=\dfrac{1.6p_c}{p_c+0.275}=1.17$.

$$
\left(\frac{L}{d}\right)_{allow}=20\times k_t\times k_c=20\times0.80\times1.17=18.7,\qquad \left(\frac{L}{d}\right)_{actual}=\frac{4000}{430}=9.30
$$

Actual $<$ allowable, so the beam is **safe in deflection**.

**Answer:** $M_u=256.41$ kN·m; $A_{st}=1632$ mm$^2$ (provide 3-25 + 1-16 mm), $A_{sc}=664$ mm$^2$ (provide 4-16 mm); safe in deflection.

## Check yourself

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

1. Why is the factored moment Mu = 256.41 kN·m larger than the service moment?
   A. Only the self weight is multiplied by 1.5
   B. Service loads (including self weight) are multiplied by 1.5 and the point load moment PuL/4 is added to wuL^2/8
   C. The point load is ignored and the UDL is multiplied by 2.5
   D. Mu is found using the limiting moment of resistance instead of loads
   Answer: B. All service loads are factored by 1.5, and Mu is the sum of the UDL moment wuL^2/8 and the point-load moment PuL/4. (Key point: Factored moment)

2. Why is the beam designed as doubly reinforced?
   A. Because the span is only 4 m
   B. Because Mu,lim is greater than Mu
   C. Mu (256.41 kN·m) exceeds Mu,lim (154.40 kN·m) and the size is fixed
   D. Because TMT bars cannot be used as singly reinforced steel
   Answer: C. When Mu > Mu,lim and the section cannot be enlarged, compression steel is added to carry the excess moment Mu2. (Key point: Doubly reinforced decision)

3. How is Asc obtained for the extra moment Mu2?
   A. Asc = Mu2 / (0.87 fy d)
   B. Asc = Mu / (0.36 fck b xu,max)
   C. Asc = Mu2 / (fsc (d + d'))
   D. Asc = Mu2 / ((fsc - fcc)(d - d'))
   Answer: D. The extra moment is resisted by a couple of compression steel (net stress fsc - fcc) and equal added tension steel, with lever arm d - d'. (Key point: Compression steel area)

4. In the deflection check, what is (L/d)allow built from?
   A. 20 x kt (from pt and fs) x kc (from compression steel)
   B. 26 x kt only, ignoring compression steel
   C. 20 x pt x pc
   D. L/d is fixed at 20 regardless of steel
   Answer: A. Basic L/d of 20 for simply supported beams is multiplied by modification factors for tension steel (kt) and compression steel (kc). (Key point: Deflection check)

## Close variants set in other exams

- [Beam flexure and deflection, 4.5 m span](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Abeam-flexure-and-deflection-4-5) — Design of RCC Structure, chapter 4, asked 2×

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