---
title: "Intermediate beam design for BM and deflection"
subject: "Design of RCC Structure"
chapter: 4
key: "semester_7/design_of_rcc_structure:intermediate-beam-design-for-bm-and"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/intermediate-beam-design-for-bm-and
site: BCT Engineering
---

# Intermediate beam design for BM and deflection

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

- Asked in: 2069 Chaitra, 10 marks
- Video: [CONCRETE BEAM DESIGN - PART 1](https://www.youtube.com/watch?v=w23K0sJY44Y) — Physics with Thapesan
- 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/intermediate-beam-design-for-bm-and
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-4

## Question

A floor consists of 125 mm thick RC slab, integrally connected with the beam as shown in figure. Design an intermediate beam for BM and deflection if the floor is subjected to live load of 4 kN/m$^2$ and floor finishes of 0.7 kN/m$^2$. [Figure: floor plan with one-brick-thick wall at the edge; slab spans between parallel beams (dashed lines) with 4 x 4 m panels; beam span 5 m]

## Short answer

T-beam with effective flange width 1883 mm. Neutral axis in flange (M_u = 159.90 kN·m). Required A_st = 1009 mm² (4-20 mm bars); deflection safe with L/d = 11.1 << 32.

## Answer

**Assumptions:** M20 and Fe415; beam simply supported with effective span $L=5$ m; slab panels 4 m x 4 m, so an intermediate beam carries slab load from 2 m on each side (width 4 m); beam web $b_w=300$ mm, overall depth $D=500$ mm, cover 50 mm, so $d=450$ mm. The slab and beam act together as a **T-beam**.

**Step 1: Loads per metre run of beam**

$$
\begin{aligned}
\text{Slab dead load}&=0.125\times25=3.125\ \text{kN/m}^2\\
\text{Floor finish}&=0.7\ \text{kN/m}^2,\quad\text{Live load}=4\ \text{kN/m}^2\\
\text{Total on slab}&=7.825\ \text{kN/m}^2\ \Rightarrow\ 7.825\times4=31.30\ \text{kN/m}\\
\text{Rib self weight}&=0.30\times(0.50-0.125)\times25=2.812\ \text{kN/m}\\
w&=34.11\ \text{kN/m},\qquad w_u=1.5w=51.17\ \text{kN/m}
\end{aligned}
$$

**Step 2: Bending moment**

$$
M_u=\frac{w_uL^2}{8}=\frac{51.17\times5^2}{8}=159.90\ \text{kN·m}
$$

**Step 3: Effective flange width** (IS 456 cl. 23.1.2, T-beam)

$$
b_f=\frac{l_0}{6}+b_w+6D_f=\frac{5000}{6}+300+6\times125=1883\ \text{mm}\ (\le4000\ \text{mm spacing})
$$

**Step 4: Position of neutral axis.** Moment capacity when $x_u=D_f$:

$$
M_{u,f}=0.36f_{ck}b_fD_f(d-0.42D_f)=673.76\ \text{kN·m}>M_u
$$

So $x_u<D_f$; design as a rectangular beam of width $b_f$.

**Step 5: Tension steel**

$$
M_u=0.87f_yA_{st}d\left(1-\frac{A_{st}f_y}{b_fdf_{ck}}\right)\ \Rightarrow\ A_{st}=1009\ \text{mm}^2
$$

Minimum steel $0.85b_wd/f_y=277$ mm$^2$ is satisfied. Provide 4-20 mm dia ($1257$ mm$^2$).

**Step 6: Deflection check** (IS 456 cl. 23.2). $b_w/b_f=0.16<0.3$, so the flange factor (Fig. 6) is 0.8. For flanged beams, the steel percentage for Fig. 4 is based on $b_fd$:

$$
p_t=\frac{100A_{st}}{b_fd}=\frac{100\times1257}{1883\times450}=0.15\%,\quad f_s=0.58f_y\frac{A_{st,req}}{A_{st,prov}}=193.2\ \text{N/mm}^2,\quad k_t=2.0\ \text{(maximum)}
$$

$$
\left(\frac{L}{d}\right)_{allow}=20\times2.0\times0.8=32.0,\qquad \left(\frac{L}{d}\right)_{actual}=\frac{5000}{450}=11.11
$$

The beam is **safe in deflection**. (Shear: $V_u=w_uL/2=127.9$ kN; stirrups are designed separately.)

**Answer:** T-beam, $b_f=1883$ mm, $b_w=300$ mm, $D=500$ mm; $A_{st}=1009$ mm$^2$ (4-20 mm bars); safe in deflection.

## Check yourself

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

1. Why is slab load multiplied by 4 m?
   A. Because the beam span is 4 m
   B. The intermediate beam carries slab load from 2 m on each side
   C. Because slab thickness is 4 cm
   D. Only half the slab load goes to it
   Answer: B. Panels are 4 m wide, so the tributary width of an intermediate beam is 2 + 2 = 4 m. (Key point: Load on beam)

2. Which formula gives the effective flange width of the T-beam?
   A. bf = l0/12 + bw
   B. bf = bw + 12 Df only
   C. bf = l0/6 + bw + 6Df, not exceeding the beam spacing
   D. bf = D + bw
   Answer: C. IS 456 cl. 23.1.2 for T-beams: l0/6 + bw + 6Df (limited by centre-to-centre spacing). (Key point: Flange width)

3. Mu,f = 673.76 kN·m (at xu = Df) exceeds Mu = 159.90 kN·m. What follows?
   A. Design as a T-beam with NA in web
   B. The beam fails
   C. Use compression steel
   D. xu < Df, so design as a rectangular beam of width bf
   Answer: D. If the moment capacity with NA at the flange bottom exceeds Mu, the NA is inside the flange. (Key point: NA test)

4. bw/bf = 0.16 < 0.3. What is the effect?
   A. A flange factor of 0.8 reduces the allowable L/d
   B. The allowable L/d is raised by 1.2
   C. No effect
   D. The beam is deep
   Answer: A. For flanged beams with bw/bf <= 0.3 the L/d is multiplied by 0.8. (Key point: Deflection factor)

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