Moment capacity of unrestrained ISMB 500

Approach. Moment capacity of a laterally unrestrained beam, IS 800:2007 cl. 8.2.2: $M_d = \beta_bZ_pf_{bd}$, $f_{bd} = \chi_{LT}f_y/\gamma_{m0}$, with $M_{cr}$ from Annex E.

Assumed data

• ISMB 500 @ 86.9 kg/m (IS 808): $D = 500$, $b_f = 180$, $t_w = 10.2$, $t_f = 17.2$ mm, $I_{yy} = 1370$ cm⁴, $Z_e = 1810$ cm³.
• Fe 410: $f_y = 250$ N/mm². Simply supported over 6 m, ends restrained against torsion with the flanges free to warp, so $L_{LT} = 1.0L = 6000$ mm (Table 15). No intermediate lateral restraint. Load applied at the shear centre level (normal loading, the usual assumption).

Step 1: Section properties

Plastic modulus $Z_p = b_ft_f(D - t_f) + t_w(D - 2t_f)^2/4 = 2048$ cm³ (plastic section, $\beta_b = 1$; $b/t_f = 5.23$, $d/t_w = 42.3$).

$$\begin{aligned} I_t &= \frac{2\times180\times17.2^3 + (500 - 34.4)\times10.2^3}{3} = 77.5\ \text{cm}^4 \\ I_w &= \frac{I_y(D - t_f)^2}{4} = \frac{1370\times10^4\times482.8^2}{4} = 79.8\times10^{10}\ \text{mm}^6 \end{aligned}$$

Step 2: Elastic critical moment

$$M_{cr} = \sqrt{\frac{\pi^2EI_y}{L_{LT}^2}\left(GI_t + \frac{\pi^2EI_w}{L_{LT}^2}\right)} = 278.7\ \text{kN·m}$$

($E = 2\times10^5$, $G = 0.769\times10^5$ N/mm²).

Step 3: Reduction factor

Rolled section: $\alpha_{LT} = 0.21$.

$$\begin{aligned} \lambda_{LT} &= \sqrt{\frac{\beta_bZ_pf_y}{M_{cr}}} = \sqrt{\frac{2047546\times250}{278.7\times10^6}} = 1.355 \\ \phi_{LT} &= 0.5[1 + 0.21(\lambda_{LT} - 0.2) + \lambda_{LT}^2] = 1.540 \\ \chi_{LT} &= \frac{1}{\phi_{LT} + \sqrt{\phi_{LT}^2 - \lambda_{LT}^2}} = 0.440 \\ f_{bd} &= \frac{0.440\times250}{1.1} = 100.1\ \text{N/mm}^2 \end{aligned}$$

Step 4: Moment capacity

$$M_d = \beta_bZ_pf_{bd} = 2047546\times100.1/10^6 = 205.0\ \text{kN·m}$$

(limit $1.2Z_ef_y/\gamma_{m0} = 494$ kN·m, not governing).

Answer: moment carrying capacity $M_d \approx 205$ kN·m, about 44% of the fully restrained value $Z_pf_y/\gamma_{m0} = 465$ kN·m. For a UDL this corresponds to a factored load $w = 8M_d/L^2 = 45.5$ kN/m including self-weight.