Laced two-channel column, 800 kN, 6 m

Approach. Laced column of two channels (back to back), designed to IS 800:2007 (cl. 7.1.2 for the compressive strength, cl. 7.6 for laced columns). Steel Fe 410 (E250) is assumed: $f_y = 250$ N/mm², $f_u = 410$ N/mm², $\gamma_{m0} = 1.10$. The 800 kN load is taken as the factored load.

Data

• Factored axial load $P = 800$ kN; effective length $KL = 6.0$ m (both axes; ends held in position but not restrained against rotation, $K = 1.0$, Table 11).
• Built-up member: buckling curve c, $\alpha = 0.49$ (Table 10 and Table 7). Effective slenderness of a laced column $= 1.05 \times$ actual maximum slenderness (cl. 7.6.1.5).

Step 1: Trial section

Try 2 × ISMC 225 (IS 808): $A = 33.3$ cm² each, $I_{xx} = 2710$ cm⁴, $I_{yy} = 188$ cm⁴, $C_{yy} = 2.31$ cm, $B = 80$ mm, $t_w = 6.5$ mm, $t_f = 12.4$ mm. (2 × ISMC 200 gives only about 795 kN, so it is not enough.)

Total area $A_e = 2 \times 3330 = 6660$ mm². $r_{zz} = \sqrt{I_{xx}/A} = 9.02$ cm.

$$\begin{aligned} \lambda_{zz} &= \frac{KL}{r_{zz}} = \frac{6000}{90.2} = 66.5 \\ \lambda_{eff} &= 1.05 \times 66.5 = 69.8 \end{aligned}$$

For $\lambda_{eff} = 69.8$, curve c: $f_{cc} = \pi^2E/\lambda^2 = 405$ N/mm², $\lambda_n = \sqrt{f_y/f_{cc}} = 0.786$, $\phi = 0.5[1 + 0.49(\lambda_n - 0.2) + \lambda_n^2] = 0.952$, $\chi = 0.671$.

$$\begin{aligned} f_{cd} &= \chi f_y/\gamma_{m0} = 0.671 \times 250/1.1 = 152.5\ \text{N/mm}^2 \\ P_d &= A_e f_{cd} = 6660 \times 152.5/10^3 = 1016\ \text{kN} > 800\ \text{kN} \end{aligned}$$

Step 2: Spacing of channels

To make the column equally strong about the other axis, $I_{yy} \ge I_{zz}$ (cl. 7.6.1.1). With $h$ the distance between the centroids of the two channels:

$$\begin{aligned} 2\left[I_{y} + A\left(\frac{h}{2}\right)^2\right] &\ge 2I_{x} \\ h &\ge 2\sqrt{\frac{I_x - I_y}{A}} = 2\sqrt{\frac{(2710 - 188) \times 10^4}{3330}} = 174\ \text{mm} \end{aligned}$$

Gap between webs $g = h - 2C_{yy} = 174 - 46.2 = 128$ mm. Provide g = 130 mm (centroid distance $h = 176$ mm).

$$\begin{aligned} I_{yy} &= 2[188 + 33.3(8.81)^2] = 5545\ \text{cm}^4 \\ r_{yy} &= \sqrt{I_{yy}/2A} = 9.12\ \text{cm}, \quad \lambda_{yy} = 6000/91.2 = 65.8 \end{aligned}$$

$\lambda_{yy} = 65.8 \le \lambda_{zz} = 66.5$, so the zz-axis governs and the capacity $P_d = 1016$ kN $\ge 800$ kN. Safe. Overall width of column $= 2B + g = 290$ mm.

  +-------+   gap g   +-------+
  | ISMC  |<--------->| ISMC  |
  +-------+           +-------+
  webs face each other, flanges point outward
  lacing bolted to the flanges (top and bottom)

Step 3: Lacing system (single lacing, two planes)

Transverse shear (cl. 7.6.6.1): $V_t = 2.5\% P = 0.025 \times 800 = 20.0$ kN, shared by two lacing planes: $10.00$ kN each.

Bolt lines are taken at mid-width of the flanges, so the transverse distance between lacing connections is $s = g + B = 130 + 80 = 210$ mm. Lacing angle $\theta = 45^\circ$ to the column axis (permitted 40° to 70°, cl. 7.6.4).

$$\begin{aligned} \text{Length of bar } l &= s/\sin\theta = 210/\sin 45^\circ = 297\ \text{mm} \\ \text{Force in bar } F &= \frac{V_t/2}{\sin\theta} = \frac{10.00}{\sin 45^\circ} = 14.14\ \text{kN} \end{aligned}$$

Spacing of connections along one channel $a_1 = 2s\cot\theta = 420$ mm. Check (cl. 7.6.5.1): $a_1/r_{min} = 420/23.8 = 17.7$, which is less than the smaller of 50 and $0.7\lambda_{max} = 46.6$. OK.

Flat size. Width $\ge 3d = 48$ mm for M16 bolts; take 60 mm. Thickness $\ge l/40 = 7.4$ mm (cl. 7.6.3), and for $KL/r \le 145$, $t \ge l\sqrt{12}/145 = 7.1$ mm. Provide ISF 60 × 8 mm.

$r = t/\sqrt{12} = 2.31$ mm, $l/r = 128.6 < 145$ OK. $f_{cd}$ (curve c) $= 75.5$ N/mm², $P_{d,lace} = 480 \times 75.5/10^3 = 36.2$ kN $> F = 14.14$ kN. OK.

Step 4: Connection of lacing (M16, grade 4.6 bolts)

$$\begin{aligned} V_{dsb} &= \frac{f_{ub}}{\sqrt3}\frac{0.78\,\pi d^2/4}{\gamma_{mb}} = \frac{400}{\sqrt3} \times \frac{157}{1.25 \times 10^3} = 29.0\ \text{kN} \\ V_{dpb} &= \frac{2.5\,k_b\, d\, t\, f_u}{\gamma_{mb}} = \frac{2.5 \times 0.556 \times 16 \times 8 \times 410}{1.25 \times 10^3} = 58.3\ \text{kN} \end{aligned}$$

($k_b = e/3d_0 = 30/54 = 0.556$ with edge distance 30 mm $\ge 1.5d_0$.) Bolt value $= 29.0$ kN. Bolts needed $= 14.14/29.0 = 0.49$, so provide 1 bolt M16 at each end of every bar.

Net section of flat in tension: $A_n = (60 - 18) \times 8 = 336$ mm², $T_{dn} = 0.9 f_u A_n/\gamma_{m1} = 99.2$ kN $> 14.14$ kN. OK.

Step 5: End tie plates

Tie plates are provided at the ends and designed like battens (cl. 7.7.2): effective depth (distance between outermost bolts) $\ge h = 176$ mm; thickness $\ge s/50 = 4.2$ mm. Use 3 M16 bolts on each channel at 90 mm pitch (outermost bolts 180 mm apart) with 30 mm end distance, so the plate is 290 mm wide × 240 mm deep × 6 mm thick.

With $C = 420$ mm, $N=2$: $V_b = V_tC/(NS) = 20.0$ kN and $M = V_tC/2N = 2100$ kN·mm. Bolt forces: from $M$ $= 2100 \times 90/(2 \times 90^2) = 11.7$ kN, from shear $= 20.0/3 = 6.7$ kN, resultant $= 13.4$ kN $\le 29.0$ kN. OK.

Final design

• Column: 2 × ISMC 225 back to back, gap 130 mm, $P_d = 1016$ kN $> 800$ kN.
• Lacing: single, flats 60 × 8 mm at 45°, 1 M16 bolt at each end.
• End tie plates 290 × 240 × 6 mm, 3 M16 bolts per channel.

  +--------------------+   <- end tie plate
  | \    /\    /\    /|
  |  \  /  \  /  \  / |   single lacing
  |   \/    \/    \/  |   (same direction
  |   /\    /\    /\  |    on both faces)
  +--------------------+   <- end tie plate