Beam 6 m span, 46 kN/m, Fe 410

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 $L = 6.00$ m, bearing length of support $= 200$ mm.

Loads and design forces

The UDL of 46 kN/m (inclusive of self-weight) is the service load; factored $w_u = 1.5\times46 = 69$ kN/m. $L = 6$ m: $M_u = 69\times36/8 = 310.5$ kN·m, $V_u = 207$ kN. The support width is not given, so a stiff bearing of 200 mm is assumed for the web checks. For deflection the whole 46 kN/m is conservatively taken as imposed.

Moment without self-weight $M = 310.5$ kN·m, so $Z_{p,req} \approx M\gamma_{m0}/f_y = 310.5\times10^6\times1.1/250 = 1366$ cm³.

Trial section

Try ISMB 450 (IS 808): $D = 450$ mm, $b_f = 150$ mm, $t_w = 9.4$ mm, $t_f = 17.4$ mm, $R_1 = 15$ mm; $I_{zz} = 30400$ cm⁴; $Z_e = 1350$ cm³; $Z_p \approx B t_f (D - t_f) + t_w (D - 2t_f)^2/4 = 1534$ cm³; mass $72.4$ kg/m.

$$\begin{aligned} M_u &= \frac{wL^2}{8} = \frac{69.00\times6.00^2}{8} = 310.5\ \text{kN·m} \\ V_u &= \frac{wL}{2} = 207.0\ \text{kN} \end{aligned}$$

Section classification (Table 2)

$\varepsilon = \sqrt{250/f_y} = 1.00$. Flange: $b/t_f = (150/2)/17.4 = 4.31 < 9.4\varepsilon = 9.4$. Web: $d/t_w = 41.0 < 84\varepsilon = 84$ (with $d = D - 2(t_f + R_1) = 385.2$ mm). The section is plastic, so $\beta_b = 1.0$.

Moment capacity (cl. 8.2.1.2)

The compression flange is restrained throughout the span (given), so lateral-torsional buckling is prevented and $M_d$ is the plastic moment capacity.

$$\begin{aligned} M_d &= \frac{\beta_b Z_p f_y}{\gamma_{m0}} = \frac{1534\times10^3\times250}{1.1\times10^6} = 348.7\ \text{kN·m} \\ 1.2 Z_e f_y/\gamma_{m0} &= 1.2\times1350\times10^3\times250/(1.1\times10^6) = 368.2\ \text{kN·m} \end{aligned}$$

The smaller value governs: $M_d = 348.7$ kN·m.

$$M_d = 348.7\ \text{kN·m} > M_u = 310.5\ \text{kN·m}\quad \text{Safe}$$

Shear capacity (cl. 8.4)

$$\begin{aligned} A_v &= D t_w = 450\times9.4 = 4230\ \text{mm}^2 \\ V_d &= \frac{A_v f_y}{\sqrt3\,\gamma_{m0}} = \frac{4230\times250}{\sqrt3\times1.1\times10^3} = 555.0\ \text{kN} \end{aligned}$$

$V_u = 207.0$ kN $< V_d$; also $V_u < 0.6V_d = 333.0$ kN, so the shear is low and the moment capacity is not reduced (cl. 9.2.1). Web shear buckling does not govern since $d/t_w = 41.0 < 67\varepsilon = 67$ (cl. 8.4.2.1).

Deflection (Table 6)

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

$$\delta = \frac{5 w L^4}{384 E I} = \frac{5\times46.00\times(6000)^4}{384\times2\times10^5\times30400\times10^4} = 12.77\ \text{mm}$$

$\delta = 12.77$ mm $< L/300 = 20.00$ mm. OK.

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

Stiff bearing length $b_1 = 200$ mm. Clear web depth $d = 385.2$ mm; dispersion of load at 45° to mid-depth at an end support $n_1 = d/2 = 192.6$ mm. Effective length $0.7d$ (cl. 8.7.1.5), $\lambda = 0.7d/(t_w/\sqrt{12}) = 99.4$; curve c gives $f_{cd} = 107.9$ N/mm².

$$\begin{aligned} F_{cdw} &= (b_1 + n_1) t_w f_{cd} = (200 + 192.6)\times9.4\times107.9/10^3 = 398.1\ \text{kN} \\ n_2 &= 2.5(t_f + R_1) = 81.0\ \text{mm} \\ F_w &= \frac{(b_1 + n_2) t_w f_y}{\gamma_{m0}} = \frac{(200 + 81.0)\times9.4\times250}{1.1\times10^3} = 600.3\ \text{kN} \end{aligned}$$

Support reaction $R = 207.0$ kN $< F_{cdw}$ and $< F_w$. No web stiffener is needed.

Final design

Provide ISMB 450 (72.4 kg/m): $M_d = 348.7$ kN·m $> M_u = 310.5$ kN·m, $V_d = 555$ kN $> V_u = 207.0$ kN, $\delta = 12.8$ mm $< 20.0$ mm.