---
title: "Beam supporting RCC slab, DL 25, IL 20 kN/m"
subject: "Design of Steel and Timber Structure"
chapter: 7
key: "semester_6/design_of_steel_and_timber_structure:beam-rcc-slab-6m-25dl-20il"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/beam-rcc-slab-6m-25dl-20il
site: BCT Engineering
---

# Beam supporting RCC slab, DL 25, IL 20 kN/m

Design of Steel and Timber Structure · Chapter 7: Flexure Members

- Asked in: 2073 Magh, 15 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/beam-rcc-slab-6m-25dl-20il
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-7

## Question

Design a simply supported I-section beam of span 6 m supports a RCC slab. The compression flange beam is restrained due to its connection with the slab. The beam is subjected to a dead load of 25 kN/m and an imposed load of 20 kN/m. Design the beam. Assume the beam is sufficiently stiff against bearing.

## Short answer

Provide ISMB 450. Factored UDL 68.57 kN/m gives M_u = 308.5 kN·m; plastic moment capacity M_d = 348.7 kN·m. Shear capacity 555 kN > 205.7 kN, deflection 5.55 mm < 20 mm.

## Answer

#### Data

- Steel Fe 410 (E250): $f_y = 250$ N/mm², $f_u = 410$ N/mm², $E = 2\times10^5$ N/mm², $\gamma_{m0} = 1.10$ (IS 800:2007, Table 5). Limit state design.
- Effective span $L = 6.00$ m.

#### Loads and design forces

Span $L = 6$ m. Dead load $25$ kN/m (taken to exclude the beam's own weight) and imposed load $20$ kN/m. Factored $w_u = 1.5(25 + 20) = 67.5$ kN/m plus factored self-weight. The beam is "sufficiently stiff against bearing", so no web bearing check is needed.

Moment without self-weight $M = 303.8$ kN·m, so $Z_{p,req} \approx M\gamma_{m0}/f_y = 303.8\times10^6\times1.1/250 = 1336$ cm³ (self-weight adds a little).

#### Trial section

Try **ISMB 450** (IS 808): $D = 450$ mm, $b_f = 150$ mm, $t_w = 9.4$ mm, $t_f = 17.4$ mm, $R_1 = 15$ mm; $I_{zz} = 30400$ cm⁴; $Z_e = 1350$ cm³; $Z_p \approx B t_f (D - t_f) + t_w (D - 2t_f)^2/4 = 1534$ cm³; mass $72.4$ kg/m.

Self-weight $= 72.4\times9.81/1000 = 0.710$ kN/m, factored $= 1.5\times0.710 = 1.065$ kN/m. Total factored UDL $w = 67.50 + 1.07 = 68.57$ kN/m.

$$
\begin{aligned} M_u &= \frac{wL^2}{8} = \frac{68.57\times6.00^2}{8} = 308.5\ \text{kN·m} \\ V_u &= \frac{wL}{2} = 205.7\ \text{kN} \end{aligned}
$$

#### Section classification (Table 2)

$\varepsilon = \sqrt{250/f_y} = 1.00$. Flange: $b/t_f = (150/2)/17.4 = 4.31 < 9.4\varepsilon = 9.4$. Web: $d/t_w = 41.0 < 84\varepsilon = 84$ (with $d = D - 2(t_f + R_1) = 385.2$ mm). The section is **plastic**, so $\beta_b = 1.0$.

#### Moment capacity (cl. 8.2.1.2)

The compression flange is restrained by the RCC slab along its full length, so lateral-torsional buckling is prevented and $M_d$ is the plastic moment capacity.

$$
\begin{aligned} M_d &= \frac{\beta_b Z_p f_y}{\gamma_{m0}} = \frac{1534\times10^3\times250}{1.1\times10^6} = 348.7\ \text{kN·m} \\ 1.2 Z_e f_y/\gamma_{m0} &= 1.2\times1350\times10^3\times250/(1.1\times10^6) = 368.2\ \text{kN·m} \end{aligned}
$$

The smaller value governs: $M_d = 348.7$ kN·m.

$$
M_d = 348.7\ \text{kN·m} > M_u = 308.5\ \text{kN·m}\quad \text{Safe}
$$

#### Shear capacity (cl. 8.4)

$$
\begin{aligned} A_v &= D t_w = 450\times9.4 = 4230\ \text{mm}^2 \\ V_d &= \frac{A_v f_y}{\sqrt3\,\gamma_{m0}} = \frac{4230\times250}{\sqrt3\times1.1\times10^3} = 555.0\ \text{kN} \end{aligned}
$$

$V_u = 205.7$ kN $< V_d$; also $V_u < 0.6V_d = 333.0$ kN, so the shear is low and the moment capacity is not reduced (cl. 9.2.1). Web shear buckling does not govern since $d/t_w = 41.0 < 67\varepsilon = 67$ (cl. 8.4.2.1).

#### Deflection (Table 6)

Deflection under unfactored service (imposed) loads, limit span/300 (floor and roof members, Table 6):

$$
\delta = \frac{5 w L^4}{384 E I} = \frac{5\times20.00\times(6000)^4}{384\times2\times10^5\times30400\times10^4} = 5.55\ \text{mm}
$$

$\delta = 5.55$ mm $< L/300 = 20.00$ mm. **OK.**

Web buckling and bearing are not checked because the beam is stated to be sufficiently stiff against bearing.

#### Final design

Provide **ISMB 450** (72.4 kg/m): $M_d = 348.7$ kN·m $> M_u = 308.5$ kN·m, $V_d = 555$ kN $> V_u = 205.7$ kN, $\delta = 5.6$ mm $< 20.0$ mm.

## Check yourself

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

1. Dead load 25 kN/m and imposed load 20 kN/m. The factored UDL before self-weight is:
   A. 45 kN/m
   B. 75 kN/m
   C. 67.5 kN/m
   D. 56.3 kN/m
   Answer: C. wu = 1.5(25 + 20) = 67.5 kN/m, then factored self-weight 1.07 kN/m is added to give 68.57 kN/m. (Key point: Factored load)

2. Because the slab restrains the compression flange, the capacity of ISMB 450 is:
   A. Md = chi_LT Zp fy/gamma_m0
   B. Md = Zp fy/gamma_m0 = 348.7 kN·m > Mu = 308.5 kN·m
   C. Md = Ze fy/gamma_m0 = 307 kN·m
   D. Md = 555 kN·m
   Answer: B. LTB is prevented so the plastic capacity applies, limited to 1.2 Ze fy/gamma_m0 = 368.2 kN·m. (Key point: Capacity of restrained beam)

3. Which load is used in the deflection check?
   A. Factored total 68.57 kN/m
   B. Unfactored dead load 25 kN/m
   C. Unfactored imposed load 20 kN/m, giving 5.55 mm < L/300 = 20 mm
   D. Dead plus imposed 45 kN/m factored
   Answer: C. Table 6 deflection limits apply to unfactored imposed load: 5wL^4/(384EI) = 5.55 mm. (Key point: Deflection loads)

4. Why are web bearing and buckling not checked?
   A. The slab carries all reactions
   B. The web is stocky so it never buckles
   C. Shear is zero at the support
   D. The question states the beam is sufficiently stiff against bearing
   Answer: D. The problem assumes sufficient bearing stiffness, so those checks are not required. (Key point: Bearing check omitted)

## Close variants set in other exams

- [Beam supporting RCC slab, DL 10 kN/m, IL 30 kN](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-6m-rcc-slab-10dl-30il) — Design of Steel and Timber Structure, chapter 7, asked 2×

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