UDL capacity of MB 300 with flange plates

Approach. Plastic design strength of the built-up section (IS 800:2007 cl. 8.2.1), then reduction of moment capacity under high shear (cl. 9.2), and the load found from the governing check. $f_y = 250$ N/mm², $\gamma_{m0} = 1.10$. The beam is laterally restrained, so lateral-torsional buckling is ignored. Plates are assumed fully connected to the flanges (continuous welds or adequate bolts), one 200×12 plate on each flange. Effective span $L = 3.5$ m.

Section properties

ISMB 300 (IS 808): $D = 300$ mm, $b_f = 140$ mm, $t_w = 7.7$ mm, $t_f = 13.1$ mm, $I_{zz} = 8990$ cm⁴, mass 46.0 kg/m. Plastic modulus of ISMB 300 from its dimensions: $Z_{p,I} = 670$ cm³.

$$\begin{aligned} \text{Overall depth} &= 300 + 2\times12 = 324\ \text{mm} \\ y_c &= 150 + 6 = 156\ \text{mm (plate centroid from N.A.)} \\ I_{zz} &= 8990\times10^4 + 2\left[\frac{200\times12^3}{12} + 200\times12\times156^2\right] = 20677\ \text{cm}^4 \\ Z_e &= \frac{I_{zz}}{162} = 1276\ \text{cm}^3 \\ Z_p &= Z_{p,I} + 2(200\times12)(156) = 670 + 749 = 1419\ \text{cm}^3 \end{aligned}$$

Classification: plate outstand from the web face $= (200 - 7.7)/2 = 96.2$ mm, $b/t = 96.2/12 = 8.0 < 9.4\varepsilon = 9.4$ (plastic); ISMB web $d/t_w = 31.9 < 84$. So $\beta_b = 1$.

Moment capacity

$$\begin{aligned} M_d &= \frac{Z_pf_y}{\gamma_{m0}} = \frac{1419\times10^3\times250}{1.1\times10^6} = 322.6\ \text{kN·m} \\ 1.2Z_ef_y/\gamma_{m0} &= 348.1\ \text{kN·m} \end{aligned}$$

$M_d = 322.6$ kN·m. Load limited by moment: $w = 8M_d/L^2 = 8\times322.6/3.5^2 = 210.7$ kN/m.

Shear capacity

Shear is resisted by the web of the rolled section: $A_v = D t_w = 300\times7.7 = 2310$ mm² (plates ignored, conservative).

$$V_d = \frac{A_vf_y}{\sqrt3\gamma_{m0}} = \frac{2310\times250}{\sqrt3\times1.1\times10^3} = 303.1\ \text{kN}$$

Load limited by shear: $w = 2V_d/L = 2\times303.1/3.5 = 173.2$ kN/m. Shear governs (173.2 < 210.7).

Moment capacity under high shear (cl. 9.2.2)

At $w = 173.2$ kN/m: $V = V_d = 303.1$ kN $> 0.6V_d$, so the moment capacity is reduced. $M_{fd}$ is the plastic strength of the section without the shear area: $Z_{fd} = Z_p - t_wD^2/4 = 1419 - 173 = 1246$ cm³, $M_{fd} = 283.2$ kN·m.

$$\begin{aligned} \beta &= \left(\frac{2V}{V_d} - 1\right)^2 = 1.00 \\ M_{dv} &= M_d - \beta(M_d - M_{fd}) = 322.6 - 1.00(322.6 - 283.2) = 283.2\ \text{kN·m} \end{aligned}$$

Moment at this load $= wL^2/8 = 173.2\times3.5^2/8 = 265.2$ kN·m $< M_{dv} = 283.2$ kN·m. OK.

Result

Self-weight $= (46.0 + 2\times0.2\times0.012\times7850)\times9.81/1000 = 0.82$ kN/m.

Answer: the safe factored uniformly distributed load is $\approx 173$ kN/m (including self-weight; governed by shear), i.e. a service load of $\approx 115$ kN/m, or about 172 kN/m factored superimposed load after deducting the factored self-weight of 1.23 kN/m. Total factored reaction $= 303$ kN.