Main beam of 12 m x 12 m floor

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.