Beam 6 m span with 100 kN secondary beam load

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 $= 250$ mm.

Loads and design forces

The imposed load of $20$ kN/m (including self-weight) and the 100 kN load from the secondary beam are taken as service loads. Factored: $w_u = 1.5\times20 = 30$ kN/m, $W_u = 1.5\times100 = 150$ kN. Self-weight is already included, so it is not added again. Span $L = 6$ m, support width 250 mm.

The beam is laterally restrained at both ends and at mid-span, so the unrestrained length is $L_{LT} = 3.0$ m (simply supported segment, torsionally restrained, warping free, $K = 1.0$, Table 15). The two segments are identical, so one is checked using the maximum (mid-span) moment with no allowance for the moment gradient, which is conservative.

Design moment (self-weight included in the given load) $M = 360.0$ kN·m, so $Z_{p,req} \approx M\gamma_{m0}/f_y = 360.0\times10^6\times1.1/250 = 1584$ cm³.

Trial section

Try ISMB 500 (IS 808): $D = 500$ mm, $b_f = 180$ mm, $t_w = 10.2$ mm, $t_f = 17.2$ mm, $R_1 = 17$ mm; $I_{zz} = 45200$ cm⁴; $Z_e = 1810$ cm³; $Z_p \approx B t_f (D - t_f) + t_w (D - 2t_f)^2/4 = 2048$ cm³; mass $86.9$ kg/m.

$$\begin{aligned} M_u &= \frac{wL^2}{8} + \frac{WL}{4} = \frac{30.00\times6.00^2}{8} + \frac{150.0\times6.00}{4} = 360.0\ \text{kN·m} \\ V_u &= \frac{wL}{2} + \frac{W}{2} = 165.0\ \text{kN} \end{aligned}$$

Section classification (Table 2)

$\varepsilon = \sqrt{250/f_y} = 1.00$. Flange: $b/t_f = (180/2)/17.2 = 5.23 < 9.4\varepsilon = 9.4$. Web: $d/t_w = 42.3 < 84\varepsilon = 84$ (with $d = D - 2(t_f + R_1) = 431.6$ mm). The section is plastic, so $\beta_b = 1.0$.

Moment capacity (cl. 8.2.2)

The compression flange is not continuously restrained, so lateral-torsional buckling governs (cl. 8.2.2). Effective length for LTB $L_{LT} = 3.00$ m. $I_y = 1370$ cm⁴; $I_t = (2b_ft_f^3 + d t_w^3)/3 = 77.5$ cm⁴; $I_w = I_y h_f^2/4 = 80\times10^{10}$ mm⁶ ($h_f = D - t_f = 482.8$ mm).

$$\begin{aligned} M_{cr} &= \sqrt{\frac{\pi^2EI_y}{L_{LT}^2}\left(GI_t + \frac{\pi^2EI_w}{L_{LT}^2}\right)} = 839.8\ \text{kN·m} \\ \lambda_{LT} &= \sqrt{\frac{\beta_b Z_p f_y}{M_{cr}}} = 0.781 \\ \phi_{LT} &= 0.5[1 + 0.21(\lambda_{LT} - 0.2) + \lambda_{LT}^2] = 0.866 \\ \chi_{LT} &= \frac{1}{\phi_{LT} + \sqrt{\phi_{LT}^2 - \lambda_{LT}^2}} = 0.807 \\ f_{bd} &= \chi_{LT} f_y/\gamma_{m0} = 183.3\ \text{N/mm}^2 \\ M_d &= \beta_b Z_p f_{bd} = 375.3\ \text{kN·m} \end{aligned}$$ $$M_d = 375.3\ \text{kN·m} > M_u = 360.0\ \text{kN·m}\quad \text{Safe}$$

Shear capacity (cl. 8.4)

$$\begin{aligned} A_v &= D t_w = 500\times10.2 = 5100\ \text{mm}^2 \\ V_d &= \frac{A_v f_y}{\sqrt3\,\gamma_{m0}} = \frac{5100\times250}{\sqrt3\times1.1\times10^3} = 669.2\ \text{kN} \end{aligned}$$

$V_u = 165.0$ kN $< V_d$; also $V_u < 0.6V_d = 401.5$ kN, so the shear is low and the moment capacity is not reduced (cl. 9.2.1). Web shear buckling does not govern since $d/t_w = 42.3 < 67\varepsilon = 67$ (cl. 8.4.2.1).

Deflection (Table 6)

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

$$\delta = \frac{5 w L^4}{384 E I} + \frac{W L^3}{48 E I} = \frac{5\times20.00\times(6000)^4}{384\times2\times10^5\times45200\times10^4} + \frac{100000\times(6000)^3}{48\times2\times10^5\times45200\times10^4} = 8.71\ \text{mm}$$

$\delta = 8.71$ mm $< L/300 = 20.00$ mm. OK.

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

Stiff bearing length $b_1 = 250$ mm. Clear web depth $d = 431.6$ mm; dispersion of load at 45° to mid-depth at an end support $n_1 = d/2 = 215.8$ mm. Effective length $0.7d$ (cl. 8.7.1.5), $\lambda = 0.7d/(t_w/\sqrt{12}) = 102.6$; curve c gives $f_{cd} = 103.6$ N/mm².

$$\begin{aligned} F_{cdw} &= (b_1 + n_1) t_w f_{cd} = (250 + 215.8)\times10.2\times103.6/10^3 = 492.3\ \text{kN} \\ n_2 &= 2.5(t_f + R_1) = 85.5\ \text{mm} \\ F_w &= \frac{(b_1 + n_2) t_w f_y}{\gamma_{m0}} = \frac{(250 + 85.5)\times10.2\times250}{1.1\times10^3} = 777.7\ \text{kN} \end{aligned}$$

Support reaction $R = 165.0$ kN $< F_{cdw}$ and $< F_w$. No web stiffener is needed.

Final design

Provide ISMB 500 (86.9 kg/m): $M_d = 375.3$ kN·m $> M_u = 360.0$ kN·m, $V_d = 669$ kN $> V_u = 165.0$ kN, $\delta = 8.7$ mm $< 20.0$ mm.