Laterally supported I-section for 700 kNm

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.
• Span not given (not needed for bending design).

Loads and design forces

The factored moment is $M_u = 700$ kN·m (taken to include the beam's own weight). The beam is laterally supported, so the plastic moment capacity applies. No span, shear or load is given, so only bending is designed here, and the shear capacity is reported for reference.

Moment without self-weight $M = 700.0$ kN·m, so $Z_{p,req} \approx M\gamma_{m0}/f_y = 700.0\times10^6\times1.1/250 = 3080$ cm³.

Trial section

Try ISMB 600 (IS 808): $D = 600$ mm, $b_f = 210$ mm, $t_w = 12.0$ mm, $t_f = 20.3$ mm, $R_1 = 20$ mm; $I_{zz} = 91800$ cm⁴; $Z_e = 3060$ cm³; $Z_p \approx B t_f (D - t_f) + t_w (D - 2t_f)^2/4 = 3410$ cm³; mass $123.0$ kg/m.

Section classification (Table 2)

$\varepsilon = \sqrt{250/f_y} = 1.00$. Flange: $b/t_f = (210/2)/20.3 = 5.17 < 9.4\varepsilon = 9.4$. Web: $d/t_w = 43.3 < 84\varepsilon = 84$ (with $d = D - 2(t_f + R_1) = 519.4$ 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 plastic moment capacity governs; lateral-torsional buckling is not checked.

$$\begin{aligned} M_d &= \frac{\beta_b Z_p f_y}{\gamma_{m0}} = \frac{3410\times10^3\times250}{1.1\times10^6} = 775.0\ \text{kN·m} \\ 1.2 Z_e f_y/\gamma_{m0} &= 1.2\times3060\times10^3\times250/(1.1\times10^6) = 834.5\ \text{kN·m} \end{aligned}$$

The smaller value governs: $M_d = 775.0$ kN·m.

$$M_d = 775.0\ \text{kN·m} > M_u = 700.0\ \text{kN·m}\quad \text{Safe}$$

Shear capacity (cl. 8.4)

$$\begin{aligned} A_v &= D t_w = 600\times12.0 = 7200\ \text{mm}^2 \\ V_d &= \frac{A_v f_y}{\sqrt3\,\gamma_{m0}} = \frac{7200\times250}{\sqrt3\times1.1\times10^3} = 944.8\ \text{kN} \end{aligned}$$

The shear force is not given. The section can resist a shear of $V_d = 945$ kN; for $V \le 0.6V_d = 567$ kN the moment capacity is not reduced (cl. 9.2.1). The web does not buckle in shear since $d/t_w = 43.3 < 67\varepsilon = 67$.

Web bearing and deflection depend on the span and support, which are not given.

Final design

Provide ISMB 600 (123.0 kg/m): $M_d = 775.0$ kN·m $> M_u = 700.0$ kN·m, $V_d = 945$ kN.