Safe UDL on ISLB 600 beam, restrained vs unrestrained

Approach. IS 800:2007 cl. 8.2: design bending strength $M_d = \beta_b Z_p f_y/\gamma_{m0}$ for a laterally supported beam, and $M_d = \beta_b Z_p f_{bd}$ with $f_{bd} = \chi_{LT}f_y/\gamma_{m0}$ for an unrestrained beam. $f_y = 250$ N/mm² (Fe 410), $\gamma_{m0} = 1.10$. Safe load found from $M_d = wL^2/8$.

Section data: ISLB 600 @ 99.5 kg/m (IS 808)

$D = 600$ mm, $b_f = 210$ mm, $t_w = 10.5$ mm, $t_f = 15.5$ mm, $R_1 = 20$ mm, $I_{zz} = 72800$ cm⁴, $I_{yy} = 1820$ cm⁴, $Z_e = 2430$ cm³.

Plastic modulus (from dimensions): $Z_p = b_ft_f(D - t_f) + t_w(D - 2t_f)^2/4 = 2752$ cm³.

Classification: flange $b/t_f = 105/15.5 = 6.77 < 9.4$; web $d/t_w = 50.4 < 84$. The section is plastic ($\beta_b = 1$).

(a) Compression flange restrained against lateral buckling

$$\begin{aligned} M_d &= \frac{Z_pf_y}{\gamma_{m0}} = \frac{2752\times10^3\times250}{1.1\times10^6} = 625.6\ \text{kN·m} \\ \text{Limit } 1.2Z_ef_y/\gamma_{m0} &= 662.7\ \text{kN·m} \\ M_d &= 625.6\ \text{kN·m} \end{aligned}$$

Safe factored UDL (including self-weight): $w = 8M_d/L^2 = 8\times625.6/7^2 = 102.1$ kN/m.

Shear check: $V = wL/2 = 357.5$ kN $< V_d = D t_wf_y/(\sqrt3\gamma_{m0}) = 827$ kN, and below $0.6V_d$, so bending is not reduced.

Service load $= 102.1/1.5 = 68.1$ kN/m. Deducting the self-weight ($99.5\times9.81/1000 = 0.98$ kN/m) the superimposed service load is $\approx 67.1$ kN/m.

Answer (a): $M_d = 626$ kN·m; safe factored load $\approx 102.1$ kN/m ($\approx 68.1$ kN/m unfactored, including self-weight).

(b) Not restrained against lateral buckling (torsion restrained, flanges free to warp)

For simply supported ends with torsion restrained and warping free, $L_{LT} = 1.0L = 7000$ mm (Table 15).

$$\begin{aligned} I_t &= \frac{2b_ft_f^3 + (D-2t_f)t_w^3}{3} = 74.1\ \text{cm}^4 \\ h_f &= D - t_f = 584.5\ \text{mm},\quad I_w = \frac{I_yh_f^2}{4} = 155.4\times10^{10}\ \text{mm}^6 \\ M_{cr} &= \sqrt{\frac{\pi^2EI_y}{L_{LT}^2}\left(GI_t + \frac{\pi^2EI_w}{L_{LT}^2}\right)} = 296.1\ \text{kN·m} \end{aligned}$$

(with $G = 0.769\times10^5$ N/mm²). Rolled section, so $\alpha_{LT} = 0.21$ (Table 13):

$$\begin{aligned} \lambda_{LT} &= \sqrt{Z_pf_y/M_{cr}} = 1.524 \\ \phi_{LT} &= 0.5[1 + 0.21(\lambda_{LT} - 0.2) + \lambda_{LT}^2] = 1.801 \\ \chi_{LT} &= \frac{1}{\phi_{LT} + \sqrt{\phi_{LT}^2 - \lambda_{LT}^2}} = 0.362 \\ f_{bd} &= \chi_{LT}f_y/\gamma_{m0} = 82.3\ \text{N/mm}^2 \\ M_d &= Z_pf_{bd} = 226.7\ \text{kN·m} \end{aligned}$$

Safe factored UDL: $w = 8M_d/L^2 = 8\times226.7/49 = 37.0$ kN/m (service $\approx 24.7$ kN/m).

Answer (b): $M_d = 227$ kN·m; safe factored load $\approx 37.0$ kN/m ($\approx 24.7$ kN/m unfactored, including self-weight).

Lack of lateral restraint reduces the strength to about 36% of the restrained value.