---
title: "Beam design with deflection, 5.5 m span"
subject: "Design of RCC Structure"
chapter: 4
key: "semester_7/design_of_rcc_structure:beam-design-with-deflection-5-5"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/beam-design-with-deflection-5-5
site: BCT Engineering
---

# Beam design with deflection, 5.5 m span

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

- Asked in: 2072 Chaitra, 14 marks
- Video: [Span-to-Depth Ratio in Beam Design Explained | ACI 318 Minimum Beam Depth | Deflection in RCC Beams](https://www.youtube.com/watch?v=siFaxMK8J0s) — Civil Engineering by Arsalan 
- 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-with-deflection-5-5
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-4

## Question

A simply supported RCC beam of effective span 5.5 meter and overall dimensions 230 mm x 550 mm is subjected to superimposed load of 50 kN/m excluding its self weight. Design the beam for limit state of collapse in flexure. Also check whether the beam is safe in deflection or not. Adopt mild exposure condition and use Fe 415 steel. Take effective cover to re-bars as 50 mm.

## Short answer

Factored moment 301.53 kN·m exceeds limiting 158.66 kN·m, so doubly reinforced. Required A_st = 1980 mm² (5-25 mm), A_sc = 926 mm² (5-16 mm); deflection check (L/d = 11 < 23.2) is safe.

## Answer

**Given:** $L=5.5$ m, $b=230$ mm, $D=550$ mm, superimposed load 50 kN/m (excluding self weight), Fe415, mild exposure, effective cover 50 mm. The grade of concrete is not stated, so M20 (the minimum for mild exposure, IS 456 Table 5) is adopted. $d=550-50=500$ mm.

**Step 1: Loads and moment**

$$
\begin{aligned}
\text{Self weight}&=0.23\times0.55\times25=3.16\ \text{kN/m},\quad w=53.16\ \text{kN/m}\\
w_u&=1.5w=79.74\ \text{kN/m}\\
M_u&=\frac{w_uL^2}{8}=\frac{79.74\times5.5^2}{8}=301.53\ \text{kN·m}
\end{aligned}
$$

**Step 2: Limiting moment**

$$
M_{u,lim}=0.138f_{ck}bd^2=0.138\times20\times230\times500^2=158.66\ \text{kN·m}<M_u
$$

The section is **doubly reinforced** (the size is fixed).

**Doubly reinforced design**

$$
M_{u2}=M_u-M_{u,lim}=301.53-158.66=142.87\ \text{kN·m}
$$

Limiting depth $x_{u,max}=0.48d=240.0$ mm. Strain in compression steel (similar triangles):

$$
\varepsilon_{sc}=0.0035\left(1-\frac{d'}{x_{u,max}}\right)=0.0035\left(1-\frac{50}{240.0}\right)=0.00277
$$

From the design stress-strain curve of Fe415 (IS 456 Fig. 23): $f_{sc}=351.9$ N/mm$^2$. Stress in concrete displaced by the steel: $f_{cc}=0.446f_{ck}=8.92$ N/mm$^2$.

$$
A_{sc}=\frac{M_{u2}}{(f_{sc}-f_{cc})(d-d')}=\frac{142.87\times10^6}{(351.9-8.92)(500-50)}=926\ \text{mm}^2
$$

$$
A_{st1}=\frac{0.36f_{ck}\,b\,x_{u,max}}{0.87f_y}=1101\ \text{mm}^2,\qquad A_{st2}=\frac{A_{sc}(f_{sc}-f_{cc})}{0.87f_y}=879\ \text{mm}^2
$$

$$
A_{st}=A_{st1}+A_{st2}=1980\ \text{mm}^2
$$

**Provide:** tension 5-25 mm dia ($2454$ mm$^2$), compression 5-16 mm dia ($1005$ mm$^2$), with 8 mm stirrups.

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

$$
p_t=\frac{100A_{st}}{bd}=2.13\%,\quad f_s=0.58f_y\frac{A_{st,req}}{A_{st,prov}}=194.2\ \text{N/mm}^2,\quad k_t\approx0.95\ \text{(Fig. 4)}
$$

$$
p_c=0.87\%,\quad k_c=\frac{1.6p_c}{p_c+0.275}=1.22
$$

$$
\left(\frac{L}{d}\right)_{allow}=20\times k_t\times k_c=20\times0.95\times1.22=23.2,\qquad \left(\frac{L}{d}\right)_{actual}=\frac{5500}{500}=11.00
$$

The beam is **safe in deflection**.

**Answer:** $M_u=301.53$ kN·m, $A_{st}=1980$ mm$^2$ (5-25 mm), $A_{sc}=926$ mm$^2$ (5-16 mm); safe in deflection.

## Check yourself

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

1. The concrete grade was not given. What is adopted and why?
   A. M15, since it is cheapest
   B. M20, the minimum grade for mild exposure (IS 456 Table 5)
   C. M30, for durability
   D. M25, since span is 5.5 m
   Answer: B. IS 456 Table 5 specifies M20 as the minimum for mild exposure reinforced concrete. (Key point: Grade of concrete)

2. Why is the section doubly reinforced?
   A. Because Fe415 is used
   B. Because the span exceeds 5 m
   C. Mu = 301.53 kN·m exceeds Mu,lim = 158.66 kN·m and size is limited
   D. Because deflection fails
   Answer: C. Size is fixed and Mu > Mu,lim, so compression steel is provided for the excess. (Key point: Doubly reinforced)

3. What do Ast1 and Ast2 represent?
   A. Ast1 is steel for shear; Ast2 for deflection
   B. Ast1 is top steel; Ast2 is stirrups
   C. Both equal Asc
   D. Ast1 balances the limiting concrete compression; Ast2 balances the compression steel force
   Answer: D. Total Ast = Ast1 + Ast2 = 1101 + 879 = 1980 mm^2. (Key point: Tension steel parts)

4. What does the compression steel do to the L/d check?
   A. It increases the allowable L/d through factor kc
   B. It reduces allowable L/d
   C. It has no effect
   D. It replaces kt
   Answer: A. Compression steel reduces long-term deflection, so (L/d)allow = 20 x kt x kc is raised. (Key point: Deflection modification)

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