---
title: "Design T-beam with 150 mm slab"
subject: "Design of RCC Structure"
chapter: 4
key: "semester_7/design_of_rcc_structure:design-t-beam-with-150-mm"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/design-t-beam-with-150-mm
site: BCT Engineering
---

# Design T-beam with 150 mm slab

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

- Asked in: 2065 Shrawan (old course), 13 marks
- Video: [Design of T Beams - Problem No 1](https://www.youtube.com/watch?v=kbrYjfhS4BE) — Stan Academy
- 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/design-t-beam-with-150-mm
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-4

## Question

A beam is simply supported on two walls of width 250 mm with a clear span of 6 m. If the beam has to support 150 mm deep slab and live load of intensity 3 kN/m$^2$, design a T-beam. The beam is spaced at 4 m c/c. Take M20 concrete and Fe415 steel. Design for shear is not required.

## Short answer

T-beam with effective flange width 2242 mm, web 300×500 mm, effective depth 450 mm. Neutral axis lies in flange; design as rectangular section of width bf. Provide 5-20 mm bars (1571 mm²) for factored moment 217 kNm. Deflection check satisfied with L/d actual 13.9 < allowable 32.

## Answer

**Assumptions:** beam web $b_w=300$ mm, $D=500$ mm (about $L/12$), effective cover 50 mm, $d=450$ mm; floor finish ignored (not given). M20, Fe415.

**Step 1: Effective span** (IS 456 cl. 22.2): the lesser of $c/c=6+0.25=6.25$ m and clear span $+d=6+0.45=6.45$ m, so $L_{eff}=6.25$ m.

**Step 2: Loads per metre length of beam** (each beam carries slab load from a 4 m width)

$$
\begin{aligned}
\text{Slab dead load}&=0.15\times25=3.75\ \text{kN/m}^2;\quad\text{live load}=3\ \text{kN/m}^2\\
\text{Slab load on beam}&=(3.75+3)\times4=27.00\ \text{kN/m}\\
\text{Rib self weight}&=0.30\times(0.50-0.15)\times25=2.625\ \text{kN/m}\\
w&=29.62\ \text{kN/m},\qquad w_u=1.5w=44.44\ \text{kN/m}
\end{aligned}
$$

$$
M_u=\frac{w_uL^2}{8}=\frac{44.44\times6.25^2}{8}=216.98\ \text{kN·m}
$$

**Step 3: Effective flange width** (T-beam)

$$
b_f=\frac{l_0}{6}+b_w+6D_f=\frac{6250}{6}+300+6\times150=2242\ \text{mm}\ (<4000\ \text{mm})
$$

**Step 4: Neutral axis position.** Moment capacity when $x_u=D_f$:

$$
M_{u,f}=0.36f_{ck}b_fD_f(d-0.42D_f)=936.93\ \text{kN·m}>M_u
$$

The neutral axis lies within the flange; design as a rectangular beam of width $b_f$.

**Step 5: Tension steel**

$$
M_u=0.87f_yA_{st}d\left(1-\frac{A_{st}f_y}{b_fdf_{ck}}\right)\ \Rightarrow\ A_{st}=1374\ \text{mm}^2
$$

$x_u=\dfrac{0.87f_yA_{st}}{0.36f_{ck}b_f}=30.7$ mm $<D_f$ (OK). Minimum steel $0.85b_wd/f_y=277$ mm$^2$ (OK).

Provide 5-20 mm dia bars, $A_{st,prov}=1571$ mm$^2$ ($b_w=300$ mm can hold them in one or two layers).

**Step 6: Deflection check.** $b_w/b_f=0.13<0.3$, so the flange factor is 0.8. For flanged beams, $p_t$ for Fig. 4 is based on $b_fd$: $p_t=\dfrac{100\times1571}{2242\times450}=0.16\%$, $f_s=0.58f_y\dfrac{A_{st,req}}{A_{st,prov}}=210.5$ N/mm$^2$, so $k_t=2.0$ (maximum).

$$
\left(\frac{L}{d}\right)_{allow}=20\times2.0\times0.8=32.0,\qquad\left(\frac{L}{d}\right)_{actual}=\frac{6250}{450}=13.89
$$

Safe in deflection.

**Answer:** T-beam $b_f=2242$ mm, $D_f=150$ mm, $b_w=300$ mm, $D=500$ mm, with 5-20 mm bars ($A_{st,req}=1374$ mm$^2$).

## Check yourself

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

1. How is effective span chosen?
   A. The greater of the two
   B. Clear span only
   C. c/c plus d
   D. The least of c/c distance and clear span + d
   Answer: D. IS 456 cl. 22.2: least of 6.25 m and 6.45 m = 6.25 m. (Key point: Effective span)

2. Why is slab load multiplied by 4 m?
   A. Beams are 4 m c/c, so each carries a 4 m strip of slab
   B. Span is 4 m
   C. Slab is 4 m thick
   D. Only live load
   Answer: A. Tributary width equals beam spacing. (Key point: Slab load)

3. Mu,f = 936.93 kN·m > Mu = 216.98. What is concluded?
   A. NA in web
   B. NA is within the flange; design as rectangular of width bf
   C. Beam is over-reinforced
   D. Use doubly reinforced
   Answer: B. Flange alone can resist Mu with xu < Df. (Key point: NA position)

4. bw/bf = 0.13. How is L/d affected?
   A. Multiplied by 1.2
   B. Not affected
   C. Multiplied by 0.8 for flanged beams
   D. Divided by 0.8
   Answer: C. IS 456 reduces L/d to 0.8 times for bw/bf <= 0.3. (Key point: Deflection factor)

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