---
title: "Intermediate beam for office hall 16 m x 6 m"
subject: "Design of Steel and Timber Structure"
chapter: 7
key: "semester_6/design_of_steel_and_timber_structure:intermediate-beam-16x6-hall"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/intermediate-beam-16x6-hall
site: BCT Engineering
---

# Intermediate beam for office hall 16 m x 6 m

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

- Asked in: 2076 Baisakh, 14 marks
- Video: [50 फीट हाल मे बीम ? 50 feet Span Beam Design | depth and reinforcement of beam for 50 feet span](https://www.youtube.com/watch?v=Qt7T4rJd35Y) — Veer Buildhouse Engineering
- 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/intermediate-beam-16x6-hall
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-7

## Question

An office hall measuring 16m x 6m consists of beams spaced at 3m c/c. RCC slab of 12cm is cast over the beam. The imposed load is 3kN/m². The beam is supported on 300mm wall. The compression flange is supported throughout its length. Design intermediate beam and check for shear, deflection and lateral stability.

## Short answer

Provide ISMB 300. Factored moment 137.3 kN·m < moment capacity 152.4 kN·m; shear 87.2 kN < 303.1 kN; deflection 10.27 mm < L/300 = 21 mm. Lateral stability is ensured because the compression flange is continuously supported by the slab.

## 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, so the beams span 6 m and are spaced 3 m c/c. With 300 mm wall bearing, effective span $= 6.0 + 0.3 = 6.3$ m. Unit weight of RCC $= 25$ kN/m³ (assumed); no floor finish is given.

- Slab: $0.12\times25 = 3.00$ kN/m², per metre of beam $= 3.00\times3 = 9.00$ kN/m
- Imposed load: $3\times3 = 9.00$ kN/m

Factored load $w_u = 1.5(9.00 + 9.00) = 27.00$ kN/m (plus self-weight).

Moment without self-weight $M = 134.0$ kN·m, so $Z_{p,req} \approx M\gamma_{m0}/f_y = 134.0\times10^6\times1.1/250 = 589$ cm³ (self-weight adds a little).

#### Trial section

Try **ISMB 300** (IS 808): $D = 300$ mm, $b_f = 140$ mm, $t_w = 7.7$ mm, $t_f = 13.1$ mm, $R_1 = 14$ mm; $I_{zz} = 8990$ cm⁴; $Z_e = 599$ cm³; $Z_p \approx B t_f (D - t_f) + t_w (D - 2t_f)^2/4 = 670$ cm³; mass $46.0$ kg/m.

Self-weight $= 46.0\times9.81/1000 = 0.451$ kN/m, factored $= 1.5\times0.451 = 0.677$ kN/m. Total factored UDL $w = 27.00 + 0.68 = 27.68$ kN/m.

$$
\begin{aligned} M_u &= \frac{wL^2}{8} = \frac{27.68\times6.30^2}{8} = 137.3\ \text{kN·m} \\ V_u &= \frac{wL}{2} = 87.2\ \text{kN} \end{aligned}
$$

#### Section classification (Table 2)

$\varepsilon = \sqrt{250/f_y} = 1.00$. Flange: $b/t_f = (140/2)/13.1 = 5.34 < 9.4\varepsilon = 9.4$. Web: $d/t_w = 31.9 < 84\varepsilon = 84$ (with $d = D - 2(t_f + R_1) = 245.8$ 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 slab, so lateral buckling is prevented (cl. 8.2.2); this is the check for lateral stability.

$$
\begin{aligned} M_d &= \frac{\beta_b Z_p f_y}{\gamma_{m0}} = \frac{670\times10^3\times250}{1.1\times10^6} = 152.4\ \text{kN·m} \\ 1.2 Z_e f_y/\gamma_{m0} &= 1.2\times599\times10^3\times250/(1.1\times10^6) = 163.4\ \text{kN·m} \end{aligned}
$$

The smaller value governs: $M_d = 152.4$ kN·m.

$$
M_d = 152.4\ \text{kN·m} > M_u = 137.3\ \text{kN·m}\quad \text{Safe}
$$

#### Shear capacity (cl. 8.4)

$$
\begin{aligned} A_v &= D t_w = 300\times7.7 = 2310\ \text{mm}^2 \\ V_d &= \frac{A_v f_y}{\sqrt3\,\gamma_{m0}} = \frac{2310\times250}{\sqrt3\times1.1\times10^3} = 303.1\ \text{kN} \end{aligned}
$$

$V_u = 87.2$ kN $< V_d$; also $V_u < 0.6V_d = 181.9$ kN, so the shear is low and the moment capacity is not reduced (cl. 9.2.1). Web shear buckling does not govern since $d/t_w = 31.9 < 67\varepsilon = 67$ (cl. 8.4.2.1).

#### Deflection (Table 6)

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

$$
\delta = \frac{5 w L^4}{384 E I} = \frac{5\times9.00\times(6300)^4}{384\times2\times10^5\times8990\times10^4} = 10.27\ \text{mm}
$$

$\delta = 10.27$ 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 $d = 245.8$ mm; dispersion of load at 45° to mid-depth at an end support $n_1 = d/2 = 122.9$ mm. Effective length $0.7d$ (cl. 8.7.1.5), $\lambda = 0.7d/(t_w/\sqrt{12}) = 77.4$; curve c gives $f_{cd} = 140.4$ N/mm².

$$
\begin{aligned} F_{cdw} &= (b_1 + n_1) t_w f_{cd} = (300 + 122.9)\times7.7\times140.4/10^3 = 457.1\ \text{kN} \\ n_2 &= 2.5(t_f + R_1) = 67.8\ \text{mm} \\ F_w &= \frac{(b_1 + n_2) t_w f_y}{\gamma_{m0}} = \frac{(300 + 67.8)\times7.7\times250}{1.1\times10^3} = 643.6\ \text{kN} \end{aligned}
$$

Support reaction $R = 87.2$ kN $< F_{cdw}$ and $< F_w$. No web stiffener is needed.

#### Final design

Provide **ISMB 300** (46.0 kg/m): $M_d = 152.4$ kN·m $> M_u = 137.3$ kN·m, $V_d = 303$ kN $> V_u = 87.2$ kN, $\delta = 10.3$ mm $< 21.0$ mm.

## Check yourself

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

1. For the 6 m intermediate beam at 3 m c/c with a 12 cm RCC slab (25 kN/m3) and 3 kN/m2 imposed load, what is the factored UDL before adding self-weight?
   A. 18.00 kN/m
   B. 27.00 kN/m
   C. 13.50 kN/m
   D. 36.00 kN/m
   Answer: B. Slab = 0.12x25x3 = 9 kN/m and imposed = 3x3 = 9 kN/m. Factored = 1.5(9 + 9) = 27 kN/m; 18 kN/m is the unfactored value. (Key point: Load per metre of beam)

2. Why is the effective span taken as 6.3 m?
   A. Hall width 6 m plus the 300 mm wall bearing
   B. Hall length 16 m divided by the 3 m spacing
   C. Hall width 6 m plus 12 cm slab thickness
   D. Hall width 6 m plus 1.5 x 0.3 m for safety
   Answer: A. The beam spans the 6 m width and bears on 300 mm walls, so L = 6.0 + 0.3 = 6.3 m. (Key point: Effective span)

3. Why does the moment capacity use the full plastic value Md = betab Zp fy/gamma_m0 with no lateral-torsional buckling reduction?
   A. Because the section is slender so LTB does not occur
   B. Because shear is low, so LTB is ignored
   C. Because the 12 cm slab continuously supports the compression flange
   D. Because the beam is shorter than 10 m
   Answer: C. The compression flange is supported throughout its length by the slab, so it cannot buckle sideways and LTB is prevented (cl. 8.2.2). (Key point: Lateral stability check)

4. How is the deflection of the ISMB 300 checked?
   A. Factored total load against a limit of span/300
   B. Unfactored imposed load (9 kN/m) against a limit of span/300 = 21 mm
   C. Unfactored imposed load against a limit of span/150
   D. Factored load including self-weight against span/250
   Answer: B. Serviceability uses unfactored imposed loads. delta = 10.27 mm is below L/300 = 21 mm, so it is OK. (Key point: Deflection check)

## Close variants set in other exams

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