---
title: "I-section beam for 9 m x 24 m hall slab"
subject: "Design of Steel and Timber Structure"
chapter: 7
key: "semester_6/design_of_steel_and_timber_structure:beam-ismb-hall-9x24"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/beam-ismb-hall-9x24
site: BCT Engineering
---

# I-section beam for 9 m x 24 m hall slab

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

- Asked in: 2072 Asoj, 12 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-ismb-hall-9x24
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-7

## Question

Design a simply supported I-section to support the slab of a hall 9m×24m with beams spaced at 3 m c/c. The thickness of the slab is 100 mm. Consider a floor finish load of 0.5 kN/m² and live load of 3 kN/m². The grade of the steel is E250. Assume that adequate lateral support is provided to the compression flange.

## Short answer

Use ISMB 450. Factored UDL 28.07 kN/m gives M_u = 284.2 kN·m; plastic moment capacity M_d = 348.7 kN·m. Shear 126.3 kN < 555 kN, deflection 12.65 mm < 30 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 $L = 9.00$ m.

#### Loads and design forces

The hall is 9 m × 24 m with beams at 3 m c/c, so the beams span the 9 m width (effective span $L = 9$ m assumed c/c of supports). RCC unit weight 25 kN/m³ (assumed).

| Item | Calculation | kN/m on beam |
|---|---|---|
| Slab 100 mm | $0.1\times25\times3$ | 7.50 |
| Floor finish | $0.5\times3$ | 1.50 |
| Live load | $3\times3$ | 9.00 |

Dead load $= 9.00$ kN/m, live load $= 9.00$ kN/m. Factored $w_u = 1.5(9.00 + 9.00) = 27.00$ kN/m plus self-weight. Lateral support is adequate (given). Support width is not given, so the bearing check is omitted.

Moment without self-weight $M = 273.4$ kN·m, so $Z_{p,req} \approx M\gamma_{m0}/f_y = 273.4\times10^6\times1.1/250 = 1203$ 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 $w = 27.00 + 1.07 = 28.07$ kN/m.

$$
\begin{aligned} M_u &= \frac{wL^2}{8} = \frac{28.07\times9.00^2}{8} = 284.2\ \text{kN·m} \\ V_u &= \frac{wL}{2} = 126.3\ \text{kN} \end{aligned}
$$

#### 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)

Adequate lateral support is given to the compression flange, so lateral-torsional buckling does not govern.

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

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

$$
M_d = 348.7\ \text{kN·m} > M_u = 284.2\ \text{kN·m}\quad \text{Safe}
$$

#### Shear capacity (cl. 8.4)

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

$V_u = 126.3$ 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):

$$
\delta = \frac{5 w L^4}{384 E I} = \frac{5\times9.00\times(9000)^4}{384\times2\times10^5\times30400\times10^4} = 12.65\ \text{mm}
$$

$\delta = 12.65$ mm $< L/300 = 30.00$ mm. **OK.**

The support width is not given, so web bearing is not checked here (provide a bearing length that satisfies $F_w \ge R$, normally 150 to 200 mm).

#### Final design

Provide **ISMB 450** (72.4 kg/m): $M_d = 348.7$ kN·m $> M_u = 284.2$ kN·m, $V_d = 555$ kN $> V_u = 126.3$ kN, $\delta = 12.6$ mm $< 30.0$ mm.

## Check yourself

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

1. Beams are 3 m c/c and the 100 mm RCC slab weighs 25 kN/m^3. What dead load per metre does the slab put on each beam?
   A. 7.50 kN/m
   B. 2.50 kN/m
   C. 0.75 kN/m
   D. 75 kN/m
   Answer: A. Slab load = thickness x unit weight x tributary width = 0.1 x 25 x 3 = 7.50 kN/m. (Key point: Load on beam from slab)

2. Dead load = 9.00 kN/m and live load = 9.00 kN/m. What is the factored UDL before adding self-weight?
   A. 18.0 kN/m
   B. 22.5 kN/m
   C. 27.0 kN/m
   D. 31.5 kN/m
   Answer: C. The limit state factor 1.5 applies to both: 1.5 x (9 + 9) = 27.0 kN/m. (Key point: Factored load)

3. Why is lateral-torsional buckling not checked, and what gives Md?
   A. Because the section is slender, Md = Ze fy / gamma_m0
   B. Because the compression flange is laterally supported, Md = beta_b Zp fy / gamma_m0 (limited to 1.2 Ze fy / gamma_m0)
   C. Because the span is short, Md = Mcr
   D. Because the beam is in shear, Md = Vd x L / 4
   Answer: B. With adequate lateral support the section reaches its plastic moment capacity, so Md = beta_b Zp fy/gamma_m0, with the smaller of this and 1.2 Ze fy/gamma_m0 governing. (Key point: Moment capacity of laterally supported beam)

4. Vu = 126.3 kN and Vd = 555 kN. What follows about the moment capacity?
   A. It must be reduced because shear is present
   B. It must be doubled
   C. It is not reduced because Vu < 0.6 Vd (low shear)
   D. It is replaced by the shear capacity
   Answer: C. When Vu is less than 0.6Vd the shear is low and the moment capacity is not reduced (cl. 9.2.1). (Key point: Shear and moment interaction)

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