---
title: "RCC floor joist design, 15 m x 6 m hall"
subject: "Design of Steel and Timber Structure"
chapter: 7
key: "semester_6/design_of_steel_and_timber_structure:floor-joist-15x6-rcc"
source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/floor-joist-15x6-rcc
site: BCT Engineering
---

# RCC floor joist design, 15 m x 6 m hall

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

- Asked in: 2081 Chaitra, 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/floor-joist-15x6-rcc
- Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-7

## Question

A hall of clear dimension 15 m × 6 m is to be covered by RCC slab flooring 12 cm thick resting over RC joist spaced at an interval of 3 m center to center. Terrazzo finishing 2 cm thick is to be provided over the RCC slab. The live load on slab is 4 kN/m². The joists are resting over 30 cm thick walls. Design the floor joist. Take unit weight of RCC and terrazzo floor finish of 24 kN/m³. Consider laterally supported beam. Check for lateral stability is required.

## Short answer

Provide ISMB 350 (Fe 410) as the floor joist. The section has moment capacity 199.3 kN·m > applied 168.1 kN·m, shear capacity 372 kN > 106.8 kN, and deflection 9.05 mm under live load < 21 mm. All checks including web bearing are satisfied.

## 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

Effective span of joist (spans the 6 m width): clear span $+$ bearing $= 6.0 + 0.3 = 6.3$ m. Joists at 3 m c/c, so each joist carries a 3 m width of floor.

| Load | Calculation | kN/m² | kN/m (3 m width) |
|---|---|---|---|
| RCC slab 120 mm | $0.12\times24$ | 2.88 | 8.64 |
| Terrazzo 20 mm | $0.02\times24$ | 0.48 | 1.44 |
| Live load | given | 4.00 | 12.00 |

Dead load $= 10.08$ kN/m (excluding joist self-weight), live load $= 12.00$ kN/m. Factored load (IS 800 Table 4, $\gamma_f = 1.5$ for DL and IL): $w_u = 1.5(10.08 + 12.00) = 33.12$ kN/m.

Moment without self-weight $M = 164.3$ kN·m, so $Z_{p,req} \approx M\gamma_{m0}/f_y = 164.3\times10^6\times1.1/250 = 723$ cm³ (self-weight adds a little).

#### Trial section

Try **ISMB 350** (IS 808): $D = 350$ mm, $b_f = 140$ mm, $t_w = 8.1$ mm, $t_f = 14.2$ mm, $R_1 = 14$ mm; $I_{zz} = 13600$ cm⁴; $Z_e = 779$ cm³; $Z_p \approx B t_f (D - t_f) + t_w (D - 2t_f)^2/4 = 877$ cm³; mass $52.4$ kg/m.

Self-weight $= 52.4\times9.81/1000 = 0.514$ kN/m, factored $= 1.5\times0.514 = 0.771$ kN/m. Total factored UDL $w = 33.12 + 0.77 = 33.89$ kN/m.

$$
\begin{aligned} M_u &= \frac{wL^2}{8} = \frac{33.89\times6.30^2}{8} = 168.1\ \text{kN·m} \\ V_u &= \frac{wL}{2} = 106.8\ \text{kN} \end{aligned}
$$

#### Section classification (Table 2)

$\varepsilon = \sqrt{250/f_y} = 1.00$. Flange: $b/t_f = (140/2)/14.2 = 4.93 < 9.4\varepsilon = 9.4$. Web: $d/t_w = 36.2 < 84\varepsilon = 84$ (with $d = D - 2(t_f + R_1) = 293.6$ mm). The section is **plastic**, so $\beta_b = 1.0$.

#### Moment capacity (cl. 8.2.1.2)

The RCC slab is cast over the joist and holds the compression flange in position along its length, so the beam is laterally supported and lateral-torsional buckling does not govern. The check for lateral stability is therefore satisfied by the full restraint of the compression flange (cl. 8.2.2: no reduction when the flange is continuously restrained).

$$
\begin{aligned} M_d &= \frac{\beta_b Z_p f_y}{\gamma_{m0}} = \frac{877\times10^3\times250}{1.1\times10^6} = 199.3\ \text{kN·m} \\ 1.2 Z_e f_y/\gamma_{m0} &= 1.2\times779\times10^3\times250/(1.1\times10^6) = 212.5\ \text{kN·m} \end{aligned}
$$

The smaller value governs: $M_d = 199.3$ kN·m.

$$
M_d = 199.3\ \text{kN·m} > M_u = 168.1\ \text{kN·m}\quad \text{Safe}
$$

#### Shear capacity (cl. 8.4)

$$
\begin{aligned} A_v &= D t_w = 350\times8.1 = 2835\ \text{mm}^2 \\ V_d &= \frac{A_v f_y}{\sqrt3\,\gamma_{m0}} = \frac{2835\times250}{\sqrt3\times1.1\times10^3} = 372.0\ \text{kN} \end{aligned}
$$

$V_u = 106.8$ kN $< V_d$; also $V_u < 0.6V_d = 223.2$ kN, so the shear is low and the moment capacity is not reduced (cl. 9.2.1). Web shear buckling does not govern since $d/t_w = 36.2 < 67\varepsilon = 67$ (cl. 8.4.2.1).

#### Deflection (Table 6)

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

$$
\delta = \frac{5 w L^4}{384 E I} = \frac{5\times12.00\times(6300)^4}{384\times2\times10^5\times13600\times10^4} = 9.05\ \text{mm}
$$

$\delta = 9.05$ 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 $d = 293.6$ mm; dispersion of load at 45° to mid-depth at an end support $n_1 = d/2 = 146.8$ mm. Effective length $0.7d$ (cl. 8.7.1.5), $\lambda = 0.7d/(t_w/\sqrt{12}) = 87.9$; curve c gives $f_{cd} = 124.1$ N/mm².

$$
\begin{aligned} F_{cdw} &= (b_1 + n_1) t_w f_{cd} = (300 + 146.8)\times8.1\times124.1/10^3 = 449.3\ \text{kN} \\ n_2 &= 2.5(t_f + R_1) = 70.5\ \text{mm} \\ F_w &= \frac{(b_1 + n_2) t_w f_y}{\gamma_{m0}} = \frac{(300 + 70.5)\times8.1\times250}{1.1\times10^3} = 682.1\ \text{kN} \end{aligned}
$$

Support reaction $R = 106.8$ kN $< F_{cdw}$ and $< F_w$. No web stiffener is needed.

#### Final design

Provide **ISMB 350** (52.4 kg/m): $M_d = 199.3$ kN·m $> M_u = 168.1$ kN·m, $V_d = 372$ kN $> V_u = 106.8$ kN, $\delta = 9.0$ mm $< 21.0$ mm.

## Check yourself

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

1. Why is the joist span taken as 6.3 m and the load width as 3 m?
   A. Joists span the 6 m width with 0.3 m bearing, and are spaced 3 m c/c so each carries 3 m of floor
   B. Joists span the 15 m length, with 3 m spacing
   C. The span is 6 m clear plus 3 m spacing
   D. The span is 6.3 m because of self-weight
   Answer: A. Effective span is clear span plus bearing (6.0 + 0.3), and a joist at 3 m c/c supports a 3 m wide strip of floor. (Key point: Effective span and load width)

2. Why is the lateral-torsional buckling check satisfied for this joist?
   A. The RCC slab continuously restrains the compression flange
   B. The joist is very short
   C. ISMB 350 is a closed section
   D. Terrazzo adds stiffness to the web
   Answer: A. The slab cast over the joist holds the compression flange in position, so no reduction for lateral buckling is needed and M_d = Z_p f_y / gamma_m0. (Key point: Lateral stability)

3. Which value governs the design moment capacity of ISMB 350 here?
   A. The smaller of Z_p f_y/gamma_m0 = 199.3 and 1.2 Z_e f_y/gamma_m0 = 212.5, so 199.3 kN·m
   B. 212.5 kN·m, the larger value
   C. Z_e f_y/gamma_m0 only
   D. The average of the two values
   Answer: A. The code limits M_d to the smaller of the plastic capacity and 1.2 Z_e f_y/gamma_m0; here 199.3 kN·m exceeds M_u = 168.1 kN·m. (Key point: Moment capacity)

4. Which load is used for the deflection check and what is the limit?
   A. Unfactored imposed load, limit span/300
   B. Factored total load, limit span/300
   C. Unfactored dead load only, limit span/100
   D. Factored imposed load, limit span/500
   Answer: A. Serviceability uses unfactored (service) imposed loads; the Table 6 limit for floor members is L/300 = 21 mm, and the joist deflects 9.05 mm. (Key point: Deflection check)

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