Beam with flange plates, 30 kN/m + 16 kN

Data

• Fe 410 (E250): $f_y = 250$ N/mm², $f_u = 410$ N/mm², $E = 2\times10^5$ N/mm², $\gamma_{m0} = 1.10$. Limit state design (IS 800:2007).
• Effective span $L = 6.00$ m, maximum overall depth 300 mm. A rolled I-section alone is not enough, so flange plates are added.

Loads and design forces

Working loads: UDL 30 kN/m and a point load of 16 kN at mid-span (taken as imposed loads). Factored: $w_u = 1.5\times30 = 45$ kN/m, $W_u = 1.5\times16 = 24$ kN. Span $L = 6$ m; the beam is laterally supported throughout. The support bearing length is not given, so no bearing check is made. M16 bolts, grade 4.6, product grade C, are used for the plate connection.

Without self-weight $M = 238.5$ kN·m; no rolled ISMB within the depth limit can supply $Z_p = M\gamma_{m0}/f_y = 1049$ cm³, so plates are added to both flanges.

Trial section

Try ISMB 250 with two plates 145 × 20 mm (one on each flange). ISMB 250: $D = 250$, $b_f = 125$, $t_w = 6.9$, $t_f = 12.5$, $R_1 = 13$ mm, $I_{zz} = 5130$ cm⁴, mass 37.3 kg/m. Overall depth $= 250 + 2\times20 = 290$ mm $\le 300$ mm.

$$\begin{aligned} y_c &= D/2 + t_p/2 = 135\ \text{mm} \\ I_{zz} &= 5130\times10^4 + 2\left[\frac{145\times20^3}{12} + 145\times20\times135^2\right] = 15720\ \text{cm}^4 \\ Z_e &= I_{zz}/(290/2) = 1084\ \text{cm}^3 \\ Z_p &= Z_{p,ISMB} + 2(b_pt_p)y_c = 458 + 783 = 1241\ \text{cm}^3 \end{aligned}$$

Mass of section $= 37.3 + 2\times0.145\times0.020\times7850 = 82.8$ kg/m, self-weight $= 0.813$ kN/m (factored 1.219 kN/m). Total factored UDL $w = 45.00 + 1.22 = 46.22$ kN/m.

$$\begin{aligned} M_u &= \frac{wL^2}{8} + \frac{WL}{4} = \frac{46.22\times6.00^2}{8} + \frac{24.0\times6.00}{4} = 244.0\ \text{kN·m} \\ V_u &= 150.7\ \text{kN} \end{aligned}$$

Section classification

$\varepsilon = 1$. Plate outstand (from the web face) $b/t = (145 - 6.9)/2/20 = 3.45 < 9.4\varepsilon$ (plastic; plates continuously connected). ISMB flange $b/t_f = 5.00$, web $d/t_w = 28.8 < 84$. Plastic section, $\beta_b = 1$.

Moment capacity

The beam is laterally supported throughout, so lateral-torsional buckling cannot occur (lateral stability is satisfied) and $M_d$ is the plastic capacity.

$$\begin{aligned} M_d &= \frac{Z_pf_y}{\gamma_{m0}} = \frac{1241\times10^3\times250}{1.1\times10^6} = 282.1\ \text{kN·m} \\ 1.2Z_ef_y/\gamma_{m0} &= 295.7\ \text{kN·m} \end{aligned}$$

Shear capacity

Shear is taken by the web of the rolled section: $V_d = D t_wf_y/(\sqrt3\gamma_{m0}) = 250\times6.9\times250/(\sqrt3\times1.1\times10^3) = 226.3$ kN; $V_u = 150.7$ kN.

$V_u > 0.6V_d = 135.8$ kN, so the moment capacity is reduced (cl. 9.2.2): $M_{fd} = 257.6$ kN·m (plastic capacity excluding the shear area $Dt_w$), $\beta = (2V/V_d - 1)^2 = 0.110$,

$$M_{dv} = M_d - \beta(M_d - M_{fd}) = 282.1 - 0.110(282.1 - 257.6) = 279.5\ \text{kN·m}$$

$M_{dv} = 279.5$ kN·m $> M_u = 244.0$ kN·m. Safe in bending and shear.

Deflection (limit span/300, Table 6)

$$\delta = \frac{5wL^4}{384EI} + \frac{WL^3}{48EI} = 18.39\ \text{mm} < \frac{L}{300} = 20.00\ \text{mm}$$

(unfactored imposed loads $w = 30.0$ kN/m, $W = 16.0$ kN, $I_{zz} = 15720$ cm⁴). OK.

Connection of flange plates (M16, grade 4.6 bolts)

Horizontal shear per unit length between plate and flange at the support: $q = \dfrac{V A_py_c}{I_{zz}} = \dfrac{150657\times2900\times135}{15720\times10^4} = 375.2$ N/mm.

M16 bolt (single shear): $V_{dsb} = 29.0$ kN; bearing on the thinner part ($t = 12.5$ mm, $k_b = 0.556$): $V_{dpb} = 91.1$ kN. Bolt value $= 29.0$ kN. With two bolts per pitch (two lines):

$$p = \frac{2\times29.0\times10^3}{375.2} = 154\ \text{mm}$$

Maximum pitch in a compression member (cl. 10.2.3.2): $12t = 12\times12.5 = 150$ mm or 200 mm, whichever is less, $= 150$ mm. Provide two lines of M16 bolts at 150 mm pitch (this pitch is based on the shear flow at the support, so it is safe along the whole span). The plates run the full length of the beam.

Final design

ISMB 250 with a 145 × 20 mm plate on each flange (overall depth 290 mm); plates connected by M16 bolts at 150 mm pitch. $M_{dv} = 279.5$ kN·m $> 244.0$ kN·m, $\delta = 18.4$ mm $< 20$ mm.