---
title: "Intermediate beam for 15 m x 6 m hall"
subject: "Design of Steel and Timber Structure"
chapter: 7
key: "semester_6/design_of_steel_and_timber_structure:intermediate-beam-15x6-110mm"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/intermediate-beam-15x6-110mm
site: BCT Engineering
---

# Intermediate beam for 15 m x 6 m hall

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

- Asked in: 2071 Bhadra, 15 marks
- Video: [24 Feet Span of Beam Reinforcement | Beam Structural Design |](https://www.youtube.com/watch?v=PnF6CeklJes) — Engr Sami 
- 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-15x6-110mm
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-7

## Question

A hall measuring 15 m × 6 m consists of beams spaced at 3 m c/c. R.C.C. slab of 110 mm is cast over the beam. The imposed load is 4 kN/m². The beam is supported on 250 mm wall. Design intermediate beam and check for shear, deflection and lateral stability.

## Short answer

Use ISMB 300 (46 kg/m). Factored moment 151.6 kN·m, capacity 152.4 kN·m; shear 303 kN > 97 kN; deflection 13.3 mm < L/300 (20.8 mm). The 110 mm 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 $L = 6.25$ m, bearing length of support $= 250$ mm.

#### Loads and design forces

The hall is 6 m wide, so the beams span 6 m at 3 m c/c. Wall bearing 250 mm: effective span $L = 6.0 + 0.25 = 6.25$ m. RCC unit weight 25 kN/m³ (assumed).

- Slab: $0.11\times25 = 2.75$ kN/m², on 3 m width $= 8.25$ kN/m
- Imposed: $4\times3 = 12.00$ kN/m

$w_u = 1.5(8.25 + 12.00) = 30.375$ kN/m plus self-weight. Stiff bearing $b_1 = 250$ mm.

Moment without self-weight $M = 148.3$ kN·m, so $Z_{p,req} \approx M\gamma_{m0}/f_y = 148.3\times10^6\times1.1/250 = 653$ 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 $w = 30.38 + 0.68 = 31.05$ kN/m.

$$
\begin{aligned} M_u &= \frac{wL^2}{8} = \frac{31.05\times6.25^2}{8} = 151.6\ \text{kN·m} \\ V_u &= \frac{wL}{2} = 97.0\ \text{kN} \end{aligned}
$$

#### 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 slab, cast over the beam, supports the compression flange throughout its length, so lateral buckling cannot occur; this satisfies the check for lateral stability (cl. 8.2.2).

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

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

$$
M_d = 152.4\ \text{kN·m} > M_u = 151.6\ \text{kN·m}\quad \text{Safe}
$$

#### Shear capacity (cl. 8.4)

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

$V_u = 97.0$ 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):

$$
\delta = \frac{5 w L^4}{384 E I} = \frac{5\times12.00\times(6250)^4}{384\times2\times10^5\times8990\times10^4} = 13.26\ \text{mm}
$$

$\delta = 13.26$ mm $< L/300 = 20.83$ mm. **OK.**

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

Stiff bearing length $b_1 = 250$ 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.

$$
\begin{aligned} F_{cdw} &= (b_1 + n_1) t_w f_{cd} = (250 + 122.9)\times7.7\times140.4/10^3 = 403.1\ \text{kN} \\ n_2 &= 2.5(t_f + R_1) = 67.8\ \text{mm} \\ F_w &= \frac{(b_1 + n_2) t_w f_y}{\gamma_{m0}} = \frac{(250 + 67.8)\times7.7\times250}{1.1\times10^3} = 556.1\ \text{kN} \end{aligned}
$$

Support reaction $R = 97.0$ 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 $> M_u = 151.6$ kN·m, $V_d = 303$ kN $> V_u = 97.0$ kN, $\delta = 13.3$ mm $< 20.8$ mm.

## Check yourself

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

1. The hall is 6 m wide and the beam bears on 250 mm walls. What effective span is used?
   A. 6.00 m
   B. 6.25 m
   C. 6.50 m
   D. 15 m
   Answer: B. Effective span = clear span + bearing = 6.0 + 0.25 = 6.25 m; the beams span the 6 m width, not the 15 m length. (Key point: Effective span)

2. Slab 8.25 kN/m and imposed 12 kN/m give 30.375 kN/m factored. How is the trial section's self-weight treated?
   A. Ignored
   B. Added unfactored: 0.451 kN/m
   C. Added after factoring by 1.5: 0.677 kN/m, giving w = 31.05 kN/m
   D. Multiplied by 3
   Answer: C. Self-weight 0.451 kN/m is factored by 1.5 to 0.677 kN/m and added to give 31.05 kN/m. (Key point: Factored load including self-weight)

3. How is the lateral stability check satisfied?
   A. By calculating Mcr and chi_LT
   B. By adding stiffeners
   C. By limiting the span to 6 m
   D. The slab cast over the beam supports the compression flange throughout, so lateral buckling cannot occur
   Answer: D. A continuously supported compression flange cannot buckle laterally, so Md is the plastic capacity 152.4 kN.m > Mu = 151.6 kN.m. (Key point: Lateral stability)

4. Which load is used in the deflection check delta = 5wL^4/(384EI)?
   A. Factored load 31.05 kN/m
   B. Unfactored imposed load 12 kN/m, limit L/300 = 20.83 mm
   C. Dead load only, 8.25 kN/m
   D. Factored imposed load 18 kN/m
   Answer: B. Deflection is checked under unfactored service (imposed) load: 13.26 mm < 20.83 mm. (Key point: Deflection load)

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