Intermediate beam for 18 m x 6 m office hall

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 (clear), so the beams span 6 m at 3 m c/c. The wall width is not given; assume 300 mm bearing, so effective span $L = 6 + 0.3 = 6.3$ m.

Dead load $= 10.08$ kN/m, live $= 16.50$ kN/m; $w_u = 1.5(10.08 + 16.50) = 39.87$ kN/m plus self-weight. $b_1 = 300$ mm.

Moment without self-weight $M = 197.8$ kN·m, so $Z_{p,req} \approx M\gamma_{m0}/f_y = 197.8\times10^6\times1.1/250 = 870$ cm³ (self-weight adds a little).

Trial section

Try ISMB 400 (IS 808): $D = 400$ mm, $b_f = 140$ mm, $t_w = 8.9$ mm, $t_f = 16.0$ mm, $R_1 = 14$ mm; $I_{zz} = 20500$ cm⁴; $Z_e = 1020$ cm³; $Z_p \approx B t_f (D - t_f) + t_w (D - 2t_f)^2/4 = 1161$ cm³; mass $61.5$ kg/m.

Self-weight $= 61.5\times9.81/1000 = 0.603$ kN/m, factored $= 1.5\times0.603 = 0.905$ kN/m. Total factored UDL $w = 39.87 + 0.90 = 40.77$ kN/m.

$$\begin{aligned} M_u &= \frac{wL^2}{8} = \frac{40.77\times6.30^2}{8} = 202.3\ \text{kN·m} \\ V_u &= \frac{wL}{2} = 128.4\ \text{kN} \end{aligned}$$

Section classification (Table 2)

$\varepsilon = \sqrt{250/f_y} = 1.00$. Flange: $b/t_f = (140/2)/16.0 = 4.38 < 9.4\varepsilon = 9.4$. Web: $d/t_w = 38.2 < 84\varepsilon = 84$ (with $d = D - 2(t_f + R_1) = 340.0$ 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 grooves in the slab, so lateral buckling is prevented.

$$\begin{aligned} M_d &= \frac{\beta_b Z_p f_y}{\gamma_{m0}} = \frac{1161\times10^3\times250}{1.1\times10^6} = 264.0\ \text{kN·m} \\ 1.2 Z_e f_y/\gamma_{m0} &= 1.2\times1020\times10^3\times250/(1.1\times10^6) = 278.2\ \text{kN·m} \end{aligned}$$

The smaller value governs: $M_d = 264.0$ kN·m.

$$M_d = 264.0\ \text{kN·m} > M_u = 202.3\ \text{kN·m}\quad \text{Safe}$$

Shear capacity (cl. 8.4)

$$\begin{aligned} A_v &= D t_w = 400\times8.9 = 3560\ \text{mm}^2 \\ V_d &= \frac{A_v f_y}{\sqrt3\,\gamma_{m0}} = \frac{3560\times250}{\sqrt3\times1.1\times10^3} = 467.1\ \text{kN} \end{aligned}$$

$V_u = 128.4$ kN $< V_d$; also $V_u < 0.6V_d = 280.3$ kN, so the shear is low and the moment capacity is not reduced (cl. 9.2.1). Web shear buckling does not govern since $d/t_w = 38.2 < 67\varepsilon = 67$ (cl. 8.4.2.1).

Deflection (Table 6)

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

$$\delta = \frac{5 w L^4}{384 E I} = \frac{5\times16.50\times(6300)^4}{384\times2\times10^5\times20500\times10^4} = 8.25\ \text{mm}$$

$\delta = 8.25$ 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 $d = 340.0$ mm; dispersion of load at 45° to mid-depth at an end support $n_1 = d/2 = 170.0$ mm. Effective length $0.7d$ (cl. 8.7.1.5), $\lambda = 0.7d/(t_w/\sqrt{12}) = 92.6$; curve c gives $f_{cd} = 117.2$ N/mm².

$$\begin{aligned} F_{cdw} &= (b_1 + n_1) t_w f_{cd} = (300 + 170.0)\times8.9\times117.2/10^3 = 490.2\ \text{kN} \\ n_2 &= 2.5(t_f + R_1) = 75.0\ \text{mm} \\ F_w &= \frac{(b_1 + n_2) t_w f_y}{\gamma_{m0}} = \frac{(300 + 75.0)\times8.9\times250}{1.1\times10^3} = 758.5\ \text{kN} \end{aligned}$$

Support reaction $R = 128.4$ kN $< F_{cdw}$ and $< F_w$. No web stiffener is needed.

Final design

Provide ISMB 400 (61.5 kg/m): $M_d = 264.0$ kN·m $> M_u = 202.3$ kN·m, $V_d = 467$ kN $> V_u = 128.4$ kN, $\delta = 8.3$ mm $< 21.0$ mm.