Simply supported beam 6 m, 60 kN/m

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.

Loads and design forces

The UDL of 60 kN/m (inclusive of self-weight) is taken as the service load; factored load $w_u = 1.5\times60 = 90$ kN/m. Span $L = 6$ m, so $M_u = 90\times6^2/8 = 405$ kN·m and $V_u = 90\times6/2 = 270$ kN. The beam is laterally supported, so bending, shear and deflection are checked; bearing at the supports is not asked and is not checked (the support width is not given). For deflection, the whole service load is conservatively treated as imposed.

Self-weight is already included, so $M_u = 405.0$ kN·m and $Z_{p,req} \approx M_u\gamma_{m0}/f_y = 405.0\times10^6\times1.1/250 = 1782$ 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.

$$\begin{aligned} M_u &= \frac{wL^2}{8} = \frac{90.00\times6.00^2}{8} = 405.0\ \text{kN·m} \\ V_u &= \frac{wL}{2} = 270.0\ \text{kN} \end{aligned}$$

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. 8.2.1.2)

The beam is laterally supported (given), so the compression flange cannot buckle sideways and lateral stability is satisfied; $M_d$ is the plastic moment capacity.

$$\begin{aligned} M_d &= \frac{\beta_b Z_p f_y}{\gamma_{m0}} = \frac{2048\times10^3\times250}{1.1\times10^6} = 465.4\ \text{kN·m} \\ 1.2 Z_e f_y/\gamma_{m0} &= 1.2\times1810\times10^3\times250/(1.1\times10^6) = 493.6\ \text{kN·m} \end{aligned}$$

The smaller value governs: $M_d = 465.4$ kN·m.

$$M_d = 465.4\ \text{kN·m} > M_u = 405.0\ \text{kN·m}\quad \text{Safe}$$

Shear capacity (cl. 8.4)

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

$V_u = 270.0$ 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 unfactored service (imposed) loads, limit span/300 (floor and roof members, Table 6):

$$\delta = \frac{5 w L^4}{384 E I} = \frac{5\times60.00\times(6000)^4}{384\times2\times10^5\times45200\times10^4} = 11.20\ \text{mm}$$

$\delta = 11.20$ mm $< L/300 = 20.00$ mm. OK.

Bearing and web buckling at the supports are not checked because the support width is not given.

Final design

Provide ISMB 500 (86.9 kg/m): $M_d = 465.4$ kN·m $> M_u = 405.0$ kN·m, $V_d = 669$ kN $> V_u = 270.0$ kN, $\delta = 11.2$ mm $< 20.0$ mm.