---
title: "Beam flexure and deflection, 4.5 m span"
subject: "Design of RCC Structure"
chapter: 4
key: "semester_7/design_of_rcc_structure:beam-flexure-and-deflection-4-5"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/beam-flexure-and-deflection-4-5
site: BCT Engineering
---

# Beam flexure and deflection, 4.5 m span

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

- Asked in: 2082 Baisakh, 8 marks
- Video: [4 - Example 3 - Flexure Design of RC Section (with given cross section)](https://www.youtube.com/watch?v=WjDqtN6mQi8) — David Garber
- 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-flexure-and-deflection-4-5
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-4

## Question

A simply supported RCC beam of effective span 4.5 m is subjected to superimposed load of 45 kN/m excluding its self-weight with point load 75 kN at mid-span. 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 Fe 415 steel bars. All the given loads are in service level.

## Short answer

Beam size 300×600 mm, d=555 mm. Factored moment Mu=314.51 kNm; singly reinforced, required Ast=1948 mm², provide 4-25 mm bars. Deflection check: L/d actual=8.11 < allowable 19.2 (modification factor 0.96), safe.

## Answer

**Given:** $L=4.5$ m, superimposed load 45 kN/m, point load 75 kN at mid-span (service), effective cover 45 mm, M25, Fe415.

**Step 1: Section (assumed)**

Take $b=300$ mm, $D=600$ mm, so $d=600-45=555$ mm.

**Step 2: Loads**

$$
\begin{aligned}
\text{Self weight}&=0.30\times0.60\times25=4.50\ \text{kN/m}\\
w&=45+4.50=49.50\ \text{kN/m},\quad w_u=1.5w=74.25\ \text{kN/m}\\
P_u&=1.5\times75=112.5\ \text{kN}
\end{aligned}
$$

**Step 3: Factored moment**

$$
M_u=\frac{w_uL^2}{8}+\frac{P_uL}{4}=187.95+126.56=314.51\ \text{kN·m}
$$

**Step 4: Limiting moment** (IS 456 cl. 38.1; Fe415: $x_{u,max}=0.48d$)

$$
M_{u,lim}=0.138f_{ck}bd^2=0.138\times25\times300\times555^2=318.72\ \text{kN·m}\ \ge M_u
$$

A singly reinforced section is sufficient.

**Step 5: Tension steel** (IS 456 Annex G-1.1(b))

$$
M_u=0.87f_yA_{st}d\left(1-\frac{A_{st}f_y}{bdf_{ck}}\right)
$$

Solving this quadratic for $M_u=314.51$ kN·m gives $A_{st,req}=1948$ mm$^2$.

Provide 4-25 mm bars, $A_{st,prov}=1963$ mm$^2$ (fits in one layer in the 300 mm width).

- Minimum steel: $0.85bd/f_y=341$ mm$^2$, OK.
- $x_u=\dfrac{0.87f_yA_{st}}{0.36f_{ck}b}=262.6$ mm $<0.48d=266.4$ mm, OK.

**Step 6: Deflection check** (IS 456 cl. 23.2, simply supported basic ratio 20)

$$
p_t=\frac{100A_{st}}{bd}=1.18\%,\qquad f_s=0.58f_y\frac{A_{st,req}}{A_{st,prov}}=238.8\ \text{N/mm}^2
$$

Modification factor (IS 456 Fig. 4): $k_t\approx0.96$.

$$
\left(\frac{L}{d}\right)_{allow}=20\times0.96=19.2,\qquad \left(\frac{L}{d}\right)_{actual}=\frac{4500}{555}=8.11
$$

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

**Answer:** Beam 300 x 600 mm, $d=555$ mm, $M_u=314.51$ kN·m, $A_{st}=1948$ mm$^2$ (provide 4-25 mm bars); the deflection check is satisfied. Add 2-12 mm hanger bars and 8 mm stirrups (shear design to be done separately).

## Check yourself

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

1. How is the design moment obtained for UDL plus central point load?
   A. wu L^2/8 + Pu L/4 using factored loads
   B. wu L^2/8 only
   C. wu L^2/2 + Pu L
   D. Service loads without factor
   Answer: A. Factored UDL gives 187.95 kN·m and factored point load gives 126.56 kN·m, total 314.51 kN·m. (Key point: Factored moment)

2. Why is a singly reinforced section sufficient?
   A. Mu = 314.51 kN·m <= Mu,lim = 318.72 kN·m
   B. Mu is greater than Mu,lim
   C. Because Ast is under 0.85bd/fy
   D. Because d is 600 mm
   Answer: A. Mu,lim = 0.138 fck b d^2 = 318.72 kN·m exceeds Mu. (Key point: Singly reinforced check)

3. How is deflection checked?
   A. Actual L/d = 8.11 against allowable 20 x kt = 21.1
   B. By measuring cracks
   C. Comparing d to 4.5 m
   D. Using fs = fy
   Answer: A. The basic ratio 20 is modified by kt (from pt and fs); actual L/d is less, so safe. (Key point: Deflection method)

4. How is steel service stress fs estimated?
   A. 0.58 fy x Ast,req / Ast,prov
   B. 0.87 fy
   C. fy / 1.15
   D. 0.446 fck
   Answer: A. IS 456 gives fs = 0.58 fy (Ast required / Ast provided) = 238.8 N/mm^2. (Key point: Service stress)

## Close variants set in other exams

- [Beam design, 250x475 section](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Abeam-design-250x475-section) — Design of RCC Structure, chapter 4, asked 2×

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- [Limit state design of beam, 45 kN/m](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Alimit-state-design-of-beam-45) — Design of RCC Structure, chapter 4
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