---
title: "Main beam of 12 m x 12 m floor"
subject: "Design of Steel and Timber Structure"
chapter: 7
key: "semester_6/design_of_steel_and_timber_structure:main-beam-floor-12x12"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/main-beam-floor-12x12
site: BCT Engineering
---

# Main beam of 12 m x 12 m floor

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

- Asked in: 2075 Baisakh, 10 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/main-beam-floor-12x12
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-7

## Question

A floor 12 m ×12 m in plan as shown in figure to be covered with a floor made up of secondary and main beams. The secondary beams of 5 KN/m are spaced at 3 m intervals. It supports reinforced concrete slab 125 mm thick and floor finish 0.5 KN/m². If live load on the floor is 3 KN/m², design main beam located at end of floor for bending. Compression flange of main beam is laterally restrained.
[Figure: floor plan 12 m x 12 m; main beams at 2 x 6 m spacing; secondary beams at 3 x 4 m spacing, i.e. 3 m c/c]

## Short answer

The end main beam carries three factored point loads 111.94 kN plus self-weight UDL, giving factored moment 704.2 kN·m. Provide ISMB 600 (plastic, laterally restrained); its plastic moment capacity 775.0 kN·m > 704.2 kN·m, shear also safe.

## 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 $L = 12.00$ m.

#### Loads and design forces

**Arrangement read from the figure (assumption).** The floor is 12 m × 12 m. Main beams are 6 m apart (three lines of main beams, each spanning 12 m); secondary beams are at 3 m c/c and span 6 m between main beams. The "5 kN/m" is taken as the self-weight of a secondary beam. The main beam at the **end** of the floor carries the reaction of one span of each secondary beam, at 3 m, 6 m and 9 m along its 12 m span.

Load on a secondary beam (per metre): slab $0.125\times25 = 3.125$ kN/m², finish $0.5$, live $3.0$ kN/m², so floor load $= 6.625$ kN/m². On a 3 m width: $19.88$ kN/m $+ 5.0$ (self-weight) $= 24.88$ kN/m.

Reaction of a secondary beam on the end main beam $= 24.88\times6/2 = 74.62$ kN. Factored point load $W_u = 1.5\times74.62 = 111.94$ kN at $L/4$, $L/2$ and $3L/4$ ($L = 12$ m).

Moment without self-weight $M = 671.6$ kN·m, so $Z_{p,req} \approx M\gamma_{m0}/f_y = 671.6\times10^6\times1.1/250 = 2955$ cm³ (self-weight adds a little).

#### Trial section

Try **ISMB 600** (IS 808): $D = 600$ mm, $b_f = 210$ mm, $t_w = 12.0$ mm, $t_f = 20.3$ mm, $R_1 = 20$ mm; $I_{zz} = 91800$ cm⁴; $Z_e = 3060$ cm³; $Z_p \approx B t_f (D - t_f) + t_w (D - 2t_f)^2/4 = 3410$ cm³; mass $123.0$ kg/m.

Self-weight $= 123.0\times9.81/1000 = 1.207$ kN/m, factored $= 1.5\times1.207 = 1.810$ kN/m. The main beam carries no other distributed load, so the factored UDL is its self-weight, $w = 1.81$ kN/m.

The reaction from the three point loads is $1.5W_u$. Moment at mid-span $= 1.5W_u(L/2) - W_u(L/4) = W_uL/2$; the self-weight $w$ of the main beam is added as a UDL.

$$
\begin{aligned} M_u &= \frac{W_uL}{2} + \frac{wL^2}{8} = \frac{111.94\times12}{2} + \frac{1.81\times12^2}{8} = 704.2\ \text{kN·m} \\ V_u &= 1.5W_u + \frac{wL}{2} = 178.8\ \text{kN} \end{aligned}
$$

#### Section classification (Table 2)

$\varepsilon = \sqrt{250/f_y} = 1.00$. Flange: $b/t_f = (210/2)/20.3 = 5.17 < 9.4\varepsilon = 9.4$. Web: $d/t_w = 43.3 < 84\varepsilon = 84$ (with $d = D - 2(t_f + R_1) = 519.4$ mm). The section is **plastic**, so $\beta_b = 1.0$.

#### Moment capacity (cl. 8.2.1.2)

The compression flange of the main beam is laterally restrained (given), so $M_d$ is the plastic moment capacity.

$$
\begin{aligned} M_d &= \frac{\beta_b Z_p f_y}{\gamma_{m0}} = \frac{3410\times10^3\times250}{1.1\times10^6} = 775.0\ \text{kN·m} \\ 1.2 Z_e f_y/\gamma_{m0} &= 1.2\times3060\times10^3\times250/(1.1\times10^6) = 834.5\ \text{kN·m} \end{aligned}
$$

The smaller value governs: $M_d = 775.0$ kN·m.

$$
M_d = 775.0\ \text{kN·m} > M_u = 704.2\ \text{kN·m}\quad \text{Safe}
$$

#### Shear capacity (cl. 8.4)

$$
\begin{aligned} A_v &= D t_w = 600\times12.0 = 7200\ \text{mm}^2 \\ V_d &= \frac{A_v f_y}{\sqrt3\,\gamma_{m0}} = \frac{7200\times250}{\sqrt3\times1.1\times10^3} = 944.8\ \text{kN} \end{aligned}
$$

$V_u = 178.8$ kN $< V_d$; also $V_u < 0.6V_d = 566.9$ kN, so the shear is low and the moment capacity is not reduced (cl. 9.2.1). Web shear buckling does not govern since $d/t_w = 43.3 < 67\varepsilon = 67$ (cl. 8.4.2.1).

The question asks only for the design for bending; shear is checked above for completeness, and deflection is not checked.

#### Final design

Provide **ISMB 600** (123.0 kg/m): $M_d = 775.0$ kN·m $> M_u = 704.2$ kN·m, $V_d = 945$ kN $> V_u = 178.8$ kN.

## Check yourself

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

1. How does the end main beam (12 m span) receive load from the 3 m c/c secondary beams?
   A. As a UDL of 24.88 kN/m
   B. As a single point load at mid-span
   C. As a triangular load
   D. As three equal point loads at 3 m, 6 m and 9 m
   Answer: D. The secondary beams frame into the main beam at L/4, L/2 and 3L/4, so it carries three equal point loads. (Key point: Load arrangement)

2. How is the reaction from one secondary beam found?
   A. Floor load 6.625 kN/m2 x 3 m width + 5 kN/m self-weight = 24.88 kN/m, times 6/2
   B. Floor load only x 12 m
   C. Only the 5 kN/m self-weight x 6 m
   D. Live load 3 kN/m2 x 3 m x 12 m
   Answer: A. Slab 3.125 + finish 0.5 + live 3.0 = 6.625 kN/m2, x3 = 19.88, + 5 = 24.88 kN/m; reaction = 24.88x6/2 = 74.62 kN. (Key point: Secondary beam reaction)

3. For three equal point loads Wu at L/4, L/2 and 3L/4 on a simply supported span, the mid-span moment is:
   A. Wu L/4
   B. Wu L/2
   C. 3Wu L/8
   D. Wu L
   Answer: B. Reaction = 1.5Wu, so M = 1.5Wu(L/2) - Wu(L/4) = Wu L/2 = 111.94x6 = 671.6 kN·m (704.2 with self-weight). (Key point: Maximum moment)

4. With the compression flange restrained, the ISMB 600 capacity is:
   A. Reduced by chi_LT to below Mu
   B. Md = Zp fy/gamma_m0 = 775.0 kN·m, greater than Mu = 704.2 kN·m
   C. Md = Ze fy/gamma_m0 = 695 kN·m
   D. Md = 944.8 kN·m from shear
   Answer: B. Restraint prevents LTB, so the plastic capacity 3410x250/1.1 = 775 kN·m governs, and it is safe against Mu. (Key point: Bending capacity)

## Related in Design of Steel and Timber Structure

- [I-section beam for 9 m x 24 m hall slab](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-ismb-hall-9x24) — Design of Steel and Timber Structure, chapter 7
- [Beam with slab, 20 kN/m, 350 mm wall](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-concrete-slab-20knm-350mm-wall) — Design of Steel and Timber Structure, chapter 7
- [RCC floor joist design, 15 m x 6 m hall](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Afloor-joist-15x6-rcc) — Design of Steel and Timber Structure, chapter 7
- [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×
- [Beam supporting RCC slab, DL 25, IL 20 kN/m](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-rcc-slab-6m-25dl-20il) — Design of Steel and Timber Structure, chapter 7, asked 2×
- [Beam 6 m span with 100 kN secondary beam load](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-6m-20knm-100kn-midspan) — Design of Steel and Timber Structure, chapter 7

## Related in other subjects

- [Cross wall design, 125 mm slab, 3.2 m](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Ainterior-cross-wall-125mm-3-2m) — Concrete Technology and Masonry Structure, chapter 8, asked 3×
- [Slab panel 2 for BM and SF (floor plan)](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Aslab-panel-2-for-bm-and) — Design of RCC Structure, chapter 9
- [Slab continuous over three edges, 5x4 m (225 mm beam)](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Aslab-continuous-over-three-edges-5x4-2) — Design of RCC Structure, chapter 9
- [Design T-beam with 150 mm slab](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Adesign-t-beam-with-150-mm) — Design of RCC Structure, chapter 4