Beam 4 m span, 45 kN/m and 100 kN midspan

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 $L = 4.00$ m, bearing length of support $= 300$ mm.

Loads and design forces

The imposed load of 45 kN/m and the concentrated load of 100 kN at mid-span are taken as the design (factored) loads; self-weight is added with $\gamma_f = 1.5$. Span $= 4.0$ m, depth limited to 350 mm, so ISMB sections up to ISMB 350 are considered. Stiff bearing at supports: 300 mm.

For the deflection check the unfactored imposed loads are $45/1.5 = 30.0$ kN/m and $100/1.5 = 66.7$ kN.

Moment without self-weight $M = 190.0$ kN·m, so $Z_{p,req} \approx M\gamma_{m0}/f_y = 190.0\times10^6\times1.1/250 = 836$ 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 $w = 45.00 + 0.77 = 45.77$ kN/m.

$$\begin{aligned} M_u &= \frac{wL^2}{8} + \frac{WL}{4} = \frac{45.77\times4.00^2}{8} + \frac{100.0\times4.00}{4} = 191.5\ \text{kN·m} \\ V_u &= \frac{wL}{2} + \frac{W}{2} = 141.5\ \text{kN} \end{aligned}$$

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 beam is laterally supported, so lateral-torsional buckling does not occur and $M_d$ is the plastic moment capacity.

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

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

$$M_d = 199.3\ \text{kN·m} > M_u = 191.5\ \text{kN·m}\quad \text{Safe}$$

Shear capacity (cl. 8.4)

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

$V_u = 141.5$ 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):

$$\delta = \frac{5 w L^4}{384 E I} + \frac{W L^3}{48 E I} = \frac{5\times30.00\times(4000)^4}{384\times2\times10^5\times13600\times10^4} + \frac{66667\times(4000)^3}{48\times2\times10^5\times13600\times10^4} = 6.94\ \text{mm}$$

$\delta = 6.94$ mm $< L/300 = 13.33$ 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 $R = 141.5$ 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 $> M_u = 191.5$ kN·m, $V_d = 372$ kN $> V_u = 141.5$ kN, $\delta = 6.9$ mm $< 13.3$ mm.