Intermediate beam for 15 m x 6 m hall

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.
• Effective span $L = 6.25$ m, bearing length of support $= 250$ mm.

Loads and design forces

The hall is 6 m wide, so the beams span 6 m at 3 m c/c. Wall bearing 250 mm: effective span $L = 6.0 + 0.25 = 6.25$ m. RCC unit weight 25 kN/m³ (assumed).

• Slab: $0.11\times25 = 2.75$ kN/m², on 3 m width $= 8.25$ kN/m
• Imposed: $4\times3 = 12.00$ kN/m

$w_u = 1.5(8.25 + 12.00) = 30.375$ kN/m plus self-weight. Stiff bearing $b_1 = 250$ mm.

Moment without self-weight $M = 148.3$ kN·m, so $Z_{p,req} \approx M\gamma_{m0}/f_y = 148.3\times10^6\times1.1/250 = 653$ cm³ (self-weight adds a little).

Trial section

Try ISMB 300 (IS 808): $D = 300$ mm, $b_f = 140$ mm, $t_w = 7.7$ mm, $t_f = 13.1$ mm, $R_1 = 14$ mm; $I_{zz} = 8990$ cm⁴; $Z_e = 599$ cm³; $Z_p \approx B t_f (D - t_f) + t_w (D - 2t_f)^2/4 = 670$ cm³; mass $46.0$ kg/m.

Self-weight $= 46.0\times9.81/1000 = 0.451$ kN/m, factored $= 1.5\times0.451 = 0.677$ kN/m. Total factored UDL $w = 30.38 + 0.68 = 31.05$ kN/m.

$$\begin{aligned} M_u &= \frac{wL^2}{8} = \frac{31.05\times6.25^2}{8} = 151.6\ \text{kN·m} \\ V_u &= \frac{wL}{2} = 97.0\ \text{kN} \end{aligned}$$

Section classification (Table 2)

$\varepsilon = \sqrt{250/f_y} = 1.00$. Flange: $b/t_f = (140/2)/13.1 = 5.34 < 9.4\varepsilon = 9.4$. Web: $d/t_w = 31.9 < 84\varepsilon = 84$ (with $d = D - 2(t_f + R_1) = 245.8$ mm). The section is plastic, so $\beta_b = 1.0$.

Moment capacity (cl. 8.2.1.2)

The slab, cast over the beam, supports the compression flange throughout its length, so lateral buckling cannot occur; this satisfies the check for lateral stability (cl. 8.2.2).

$$\begin{aligned} M_d &= \frac{\beta_b Z_p f_y}{\gamma_{m0}} = \frac{670\times10^3\times250}{1.1\times10^6} = 152.4\ \text{kN·m} \\ 1.2 Z_e f_y/\gamma_{m0} &= 1.2\times599\times10^3\times250/(1.1\times10^6) = 163.4\ \text{kN·m} \end{aligned}$$

The smaller value governs: $M_d = 152.4$ kN·m.

$$M_d = 152.4\ \text{kN·m} > M_u = 151.6\ \text{kN·m}\quad \text{Safe}$$

Shear capacity (cl. 8.4)

$$\begin{aligned} A_v &= D t_w = 300\times7.7 = 2310\ \text{mm}^2 \\ V_d &= \frac{A_v f_y}{\sqrt3\,\gamma_{m0}} = \frac{2310\times250}{\sqrt3\times1.1\times10^3} = 303.1\ \text{kN} \end{aligned}$$

$V_u = 97.0$ kN $< V_d$; also $V_u < 0.6V_d = 181.9$ kN, so the shear is low and the moment capacity is not reduced (cl. 9.2.1). Web shear buckling does not govern since $d/t_w = 31.9 < 67\varepsilon = 67$ (cl. 8.4.2.1).

Deflection (Table 6)

Deflection under unfactored service (imposed) loads, limit span/300 (floor and roof members, Table 6):

$$\delta = \frac{5 w L^4}{384 E I} = \frac{5\times12.00\times(6250)^4}{384\times2\times10^5\times8990\times10^4} = 13.26\ \text{mm}$$

$\delta = 13.26$ mm $< L/300 = 20.83$ mm. OK.

Web buckling and bearing at support (cl. 8.7.3 and 8.7.4)

Stiff bearing length $b_1 = 250$ mm. Clear web depth $d = 245.8$ mm; dispersion of load at 45° to mid-depth at an end support $n_1 = d/2 = 122.9$ mm. Effective length $0.7d$ (cl. 8.7.1.5), $\lambda = 0.7d/(t_w/\sqrt{12}) = 77.4$; curve c gives $f_{cd} = 140.4$ N/mm².

$$\begin{aligned} F_{cdw} &= (b_1 + n_1) t_w f_{cd} = (250 + 122.9)\times7.7\times140.4/10^3 = 403.1\ \text{kN} \\ n_2 &= 2.5(t_f + R_1) = 67.8\ \text{mm} \\ F_w &= \frac{(b_1 + n_2) t_w f_y}{\gamma_{m0}} = \frac{(250 + 67.8)\times7.7\times250}{1.1\times10^3} = 556.1\ \text{kN} \end{aligned}$$

Support reaction $R = 97.0$ kN $< F_{cdw}$ and $< F_w$. No web stiffener is needed.

Final design

Provide ISMB 300 (46.0 kg/m): $M_d = 152.4$ kN·m $> M_u = 151.6$ kN·m, $V_d = 303$ kN $> V_u = 97.0$ kN, $\delta = 13.3$ mm $< 20.8$ mm.