Column with two channels toe to toe, 1200 kN

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

Data

• Factored axial load $P = 1200$ 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 250 (IS 808): $A = 39.0$ cm² each, $I_{xx} = 3880$ cm⁴, $I_{yy} = 211$ cm⁴, $C_{yy} = 2.30$ cm, $B = 80$ mm, $t_w = 7.2$ mm, $t_f = 14.1$ mm.

Total area $A_e = 2 \times 3900 = 7800$ mm². $r_{zz} = \sqrt{I_{xx}/A} = 9.97$ cm.

$$\begin{aligned} \lambda_{zz} &= \frac{KL}{r_{zz}} = \frac{6000}{99.7} = 60.2 \\ \lambda_{eff} &= 1.05 \times 60.2 = 63.2 \end{aligned}$$

For $\lambda_{eff} = 63.2$, curve c: $f_{cc} = \pi^2E/\lambda^2 = 495$ N/mm², $\lambda_n = \sqrt{f_y/f_{cc}} = 0.711$, $\phi = 0.5[1 + 0.49(\lambda_n - 0.2) + \lambda_n^2] = 0.878$, $\chi = 0.718$.

$$\begin{aligned} f_{cd} &= \chi f_y/\gamma_{m0} = 0.718 \times 250/1.1 = 163.2\ \text{N/mm}^2 \\ P_d &= A_e f_{cd} = 7800 \times 163.2/10^3 = 1273\ \text{kN} > 1200\ \text{kN} \end{aligned}$$

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:

$$\begin{aligned} 2\left[I_{y} + A\left(\frac{h}{2}\right)^2\right] &\ge 2I_{x} \\ h &\ge 2\sqrt{\frac{I_x - I_y}{A}} = 2\sqrt{\frac{(3880 - 211) \times 10^4}{3900}} = 194\ \text{mm} \end{aligned}$$

Gap between toes $g = h - 2(B - C_{yy}) = 194 - 114.0 = 80$ mm. Provide g = 80 mm (centroid distance $h = 194$ mm).

$$\begin{aligned} I_{yy} &= 2[211 + 39.0(9.70)^2] = 7761\ \text{cm}^4 \\ r_{yy} &= \sqrt{I_{yy}/2A} = 9.97\ \text{cm}, \quad \lambda_{yy} = 6000/99.7 = 60.2 \end{aligned}$$

$\lambda_{yy} = 60.2 \le \lambda_{zz} = 60.2$, so the zz-axis governs and the capacity $P_d = 1273$ kN $\ge 1200$ kN. Safe. Overall width of column $= 2B + g = 240$ mm.

  +-------+   gap g   +-------+
  | ISMC  |<--------->| ISMC  |
  +-------+           +-------+
  flanges point towards each other (webs outside)
  lacing bolted to the flanges (top and bottom faces)

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

Transverse shear (cl. 7.6.6.1): $V_t = 2.5\% P = 0.025 \times 1200 = 30.0$ kN, shared by two lacing planes: $15.00$ kN each.

Bolt lines are taken at mid-width of the flanges, so the transverse distance between lacing connections is $s = g + B = 80 + 80 = 160$ mm. Lacing angle $\theta = 45^\circ$ to the column axis (permitted 40° to 70°, cl. 7.6.4).

$$\begin{aligned} \text{Length of bar } l &= s/\sin\theta = 160/\sin 45^\circ = 226\ \text{mm} \\ \text{Force in bar } F &= \frac{V_t/2}{\sin\theta} = \frac{15.00}{\sin 45^\circ} = 21.21\ \text{kN} \end{aligned}$$

Spacing of connections along one channel $a_1 = 2s\cot\theta = 320$ mm. Check (cl. 7.6.5.1): $a_1/r_{min} = 320/23.3 = 13.8$, which is less than the smaller of 50 and $0.7\lambda_{max} = 42.1$. OK.

Flat size. Width $\ge 3d = 48$ mm for M16 bolts; take 60 mm. Thickness $\ge l/40 = 5.7$ mm (cl. 7.6.3), and for $KL/r \le 145$, $t \ge l\sqrt{12}/145 = 5.4$ mm. Provide ISF 60 × 6 mm.

$r = t/\sqrt{12} = 1.73$ mm, $l/r = 130.6 < 145$ OK. $f_{cd}$ (curve c) $= 73.7$ N/mm², $P_{d,lace} = 360 \times 73.7/10^3 = 26.5$ kN $> F = 21.21$ kN. OK.

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

$$\begin{aligned} V_{dsb} &= \frac{f_{ub}}{\sqrt3}\frac{0.78\,\pi d^2/4}{\gamma_{mb}} = \frac{400}{\sqrt3} \times \frac{157}{1.25 \times 10^3} = 29.0\ \text{kN} \\ V_{dpb} &= \frac{2.5\,k_b\, d\, t\, f_u}{\gamma_{mb}} = \frac{2.5 \times 0.556 \times 16 \times 6 \times 410}{1.25 \times 10^3} = 43.7\ \text{kN} \end{aligned}$$

($k_b = e/3d_0 = 30/54 = 0.556$ with edge distance 30 mm $\ge 1.5d_0$.) Bolt value $= 29.0$ kN. Bolts needed $= 21.21/29.0 = 0.73$, so provide 1 bolt M16 at each end of every bar.

Net section of flat in tension: $A_n = (60 - 18) \times 6 = 252$ mm², $T_{dn} = 0.9 f_u A_n/\gamma_{m1} = 74.4$ kN $> 21.21$ 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) $\ge h = 194$ mm; thickness $\ge s/50 = 3.2$ mm. Use 3 M16 bolts on each channel at 100 mm pitch (outermost bolts 200 mm apart) with 30 mm end distance, so the plate is 240 mm wide × 260 mm deep × 6 mm thick.

With $C = 320$ mm, $N=2$: $V_b = V_tC/(NS) = 30.0$ kN and $M = V_tC/2N = 2400$ kN·mm. Bolt forces: from $M$ $= 2400 \times 100/(2 \times 100^2) = 12.0$ kN, from shear $= 30.0/3 = 10.0$ kN, resultant $= 15.6$ kN $\le 29.0$ kN. OK.

Final design

• Column: 2 × ISMC 250 toe to toe, gap 80 mm, $P_d = 1273$ kN $> 1200$ kN.
• Lacing: single, flats 60 × 6 mm at 45°, 1 M16 bolt at each end.
• End tie plates 240 × 260 × 6 mm, 3 M16 bolts per channel.

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