---
title: "Intermediate beam for 18 m x 6 m office hall"
subject: "Design of Steel and Timber Structure"
chapter: 7
key: "semester_6/design_of_steel_and_timber_structure:office-hall-18x6-intermediate-beam"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/office-hall-18x6-intermediate-beam
site: BCT Engineering
---

# Intermediate beam for 18 m x 6 m office hall

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

- Asked in: 2070 Bhadra, 14 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/office-hall-18x6-intermediate-beam
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-7

## Question

A office hall of clear dimension 18 m × 6 m is provided with 12 cm thick RC slab over rolled steel beams 3 m c/c. A wearing coat of 2 cm thick lime concrete is provided over RC concrete slab. The compression flange would be supported throughout its length by providing grooves in slabs. Design an intermediate beam with the following data.
Live load = 5.5 kN/m²
Unit wt. of cement concrete = 25 kN/m³
Unit wt. of lime concrete = 18 kN/m³

## Short answer

Use ISMB 400 (61.5 kg/m). Factored moment 202.3 kN·m, capacity 264.0 kN·m; shear 467 kN > 128 kN; deflection 8.3 mm < L/300 (21 mm). The groove-supported RC slab provides full lateral restraint to the compression flange.

## 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 $L = 6.30$ m, bearing length of support $= 300$ mm.

#### Loads and design forces

The hall is 6 m wide (clear), so the beams span 6 m at 3 m c/c. The wall width is not given; **assume 300 mm bearing**, so effective span $L = 6 + 0.3 = 6.3$ m.

| Item | Calculation | kN/m² | kN/m (3 m width) |
|---|---|---|---|
| RC slab 120 mm | $0.12\times25$ | 3.000 | 9.00 |
| Lime concrete 20 mm | $0.02\times18$ | 0.360 | 1.08 |
| Live load | given | 5.500 | 16.50 |

Dead load $= 10.08$ kN/m, live $= 16.50$ kN/m; $w_u = 1.5(10.08 + 16.50) = 39.87$ kN/m plus self-weight. $b_1 = 300$ mm.

Moment without self-weight $M = 197.8$ kN·m, so $Z_{p,req} \approx M\gamma_{m0}/f_y = 197.8\times10^6\times1.1/250 = 870$ cm³ (self-weight adds a little).

#### Trial section

Try **ISMB 400** (IS 808): $D = 400$ mm, $b_f = 140$ mm, $t_w = 8.9$ mm, $t_f = 16.0$ mm, $R_1 = 14$ mm; $I_{zz} = 20500$ cm⁴; $Z_e = 1020$ cm³; $Z_p \approx B t_f (D - t_f) + t_w (D - 2t_f)^2/4 = 1161$ cm³; mass $61.5$ kg/m.

Self-weight $= 61.5\times9.81/1000 = 0.603$ kN/m, factored $= 1.5\times0.603 = 0.905$ kN/m. Total factored UDL $w = 39.87 + 0.90 = 40.77$ kN/m.

$$
\begin{aligned} M_u &= \frac{wL^2}{8} = \frac{40.77\times6.30^2}{8} = 202.3\ \text{kN·m} \\ V_u &= \frac{wL}{2} = 128.4\ \text{kN} \end{aligned}
$$

#### Section classification (Table 2)

$\varepsilon = \sqrt{250/f_y} = 1.00$. Flange: $b/t_f = (140/2)/16.0 = 4.38 < 9.4\varepsilon = 9.4$. Web: $d/t_w = 38.2 < 84\varepsilon = 84$ (with $d = D - 2(t_f + R_1) = 340.0$ mm). The section is **plastic**, so $\beta_b = 1.0$.

#### Moment capacity (cl. 8.2.1.2)

The compression flange is supported throughout its length by the grooves in the slab, so lateral buckling is prevented.

$$
\begin{aligned} M_d &= \frac{\beta_b Z_p f_y}{\gamma_{m0}} = \frac{1161\times10^3\times250}{1.1\times10^6} = 264.0\ \text{kN·m} \\ 1.2 Z_e f_y/\gamma_{m0} &= 1.2\times1020\times10^3\times250/(1.1\times10^6) = 278.2\ \text{kN·m} \end{aligned}
$$

The smaller value governs: $M_d = 264.0$ kN·m.

$$
M_d = 264.0\ \text{kN·m} > M_u = 202.3\ \text{kN·m}\quad \text{Safe}
$$

#### Shear capacity (cl. 8.4)

$$
\begin{aligned} A_v &= D t_w = 400\times8.9 = 3560\ \text{mm}^2 \\ V_d &= \frac{A_v f_y}{\sqrt3\,\gamma_{m0}} = \frac{3560\times250}{\sqrt3\times1.1\times10^3} = 467.1\ \text{kN} \end{aligned}
$$

$V_u = 128.4$ kN $< V_d$; also $V_u < 0.6V_d = 280.3$ kN, so the shear is low and the moment capacity is not reduced (cl. 9.2.1). Web shear buckling does not govern since $d/t_w = 38.2 < 67\varepsilon = 67$ (cl. 8.4.2.1).

#### Deflection (Table 6)

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

$$
\delta = \frac{5 w L^4}{384 E I} = \frac{5\times16.50\times(6300)^4}{384\times2\times10^5\times20500\times10^4} = 8.25\ \text{mm}
$$

$\delta = 8.25$ mm $< L/300 = 21.00$ mm. **OK.**

#### Web buckling and bearing at support (cl. 8.7.3 and 8.7.4)

Stiff bearing length $b_1 = 300$ mm. Clear web depth $d = 340.0$ mm; dispersion of load at 45° to mid-depth at an end support $n_1 = d/2 = 170.0$ mm. Effective length $0.7d$ (cl. 8.7.1.5), $\lambda = 0.7d/(t_w/\sqrt{12}) = 92.6$; curve c gives $f_{cd} = 117.2$ N/mm².

$$
\begin{aligned} F_{cdw} &= (b_1 + n_1) t_w f_{cd} = (300 + 170.0)\times8.9\times117.2/10^3 = 490.2\ \text{kN} \\ n_2 &= 2.5(t_f + R_1) = 75.0\ \text{mm} \\ F_w &= \frac{(b_1 + n_2) t_w f_y}{\gamma_{m0}} = \frac{(300 + 75.0)\times8.9\times250}{1.1\times10^3} = 758.5\ \text{kN} \end{aligned}
$$

Support reaction $R = 128.4$ kN $< F_{cdw}$ and $< F_w$. No web stiffener is needed.

#### Final design

Provide **ISMB 400** (61.5 kg/m): $M_d = 264.0$ kN·m $> M_u = 202.3$ kN·m, $V_d = 467$ kN $> V_u = 128.4$ kN, $\delta = 8.3$ mm $< 21.0$ mm.

## Check yourself

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

1. Which gives the slab and wearing coat dead load per metre of beam (3 m c/c)?
   A. 9.00 + 1.08 = 10.08 kN/m from 0.12x25x3 and 0.02x18x3
   B. 3.00 + 0.36 = 3.36 kN/m
   C. 9.00 + 10.8 = 19.8 kN/m
   D. 12.0 + 1.08 = 13.08 kN/m
   Answer: A. RC slab 0.12 x 25 x 3 = 9.00 kN/m and lime concrete 0.02 x 18 x 3 = 1.08 kN/m. (Key point: Dead load calculation)

2. With DL = 10.08 and LL = 16.50 kN/m, what is the factored UDL before self-weight?
   A. 26.58 kN/m
   B. 33.2 kN/m
   C. 39.87 kN/m
   D. 45.0 kN/m
   Answer: C. 1.5 x (10.08 + 16.50) = 39.87 kN/m. (Key point: Factored load)

3. What effect do the slab grooves supporting the compression flange have?
   A. They increase the shear capacity
   B. They prevent lateral buckling, so Md = plastic moment capacity (264.0 kN.m)
   C. They reduce the live load
   D. They require a stiffener
   Answer: B. Full lateral support means no LTB reduction and Md = beta_b Zp fy/gamma_m0. (Key point: Grooved slab restraint)

4. For ISMB 400, beta_b Zp fy/gamma_m0 = 264.0 and 1.2 Ze fy/gamma_m0 = 278.2 kN.m. What is Md and is it safe against Mu = 202.3 kN.m?
   A. 278.2 kN.m, safe
   B. 264.0 kN.m, safe
   C. 264.0 kN.m, unsafe
   D. 202.3 kN.m, safe
   Answer: B. The smaller value governs: Md = 264.0 kN.m > 202.3 kN.m. (Key point: Governing Md)

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