Bridge compression member, 1200 kN, tie bar lacing

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): $f_y = 250$ N/mm², $f_u = 410$ N/mm², $\gamma_{m0} = 1.10$.

Data

• Factored axial load $P = 1200$ kN; effective length $KL = 8.5$ m (given, same for both axes).
• 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 300 (IS 808): $A = 46.3$ cm² each, $I_{xx} = 6420$ cm⁴, $I_{yy} = 313$ cm⁴, $C_{yy} = 2.35$ cm, $B = 90$ mm, $t_w = 7.8$ mm, $t_f = 13.6$ mm.

Total area $A_e = 2 \times 4630 = 9260$ mm². $r_{zz} = \sqrt{I_{xx}/A} = 11.78$ cm.

$$\begin{aligned} \lambda_{zz} &= \frac{KL}{r_{zz}} = \frac{8500}{117.8} = 72.2 \\ \lambda_{eff} &= 1.05 \times 72.2 = 75.8 \end{aligned}$$

For $\lambda_{eff} = 75.8$, curve c: $f_{cc} = \pi^2E/\lambda^2 = 344$ N/mm², $\lambda_n = \sqrt{f_y/f_{cc}} = 0.853$, $\phi = 0.5[1 + 0.49(\lambda_n - 0.2) + \lambda_n^2] = 1.024$, $\chi = 0.629$.

$$\begin{aligned} f_{cd} &= \chi f_y/\gamma_{m0} = 0.629 \times 250/1.1 = 142.9\ \text{N/mm}^2 \\ P_d &= A_e f_{cd} = 9260 \times 142.9/10^3 = 1324\ \text{kN} > 1200\ \text{kN} \end{aligned}$$

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:

$$\begin{aligned} 2\left[I_{y} + A\left(\frac{h}{2}\right)^2\right] &\ge 2I_{x} \\ h &\ge 2\sqrt{\frac{I_x - I_y}{A}} = 2\sqrt{\frac{(6420 - 313) \times 10^4}{4630}} = 230\ \text{mm} \end{aligned}$$

Gap between webs $g = h - 2C_{yy} = 230 - 47.0 = 183$ mm. Provide g = 190 mm (centroid distance $h = 237$ mm).

$$\begin{aligned} I_{yy} &= 2[313 + 46.3(11.85)^2] = 13629\ \text{cm}^4 \\ r_{yy} &= \sqrt{I_{yy}/2A} = 12.13\ \text{cm}, \quad \lambda_{yy} = 8500/121.3 = 70.1 \end{aligned}$$

$\lambda_{yy} = 70.1 \le \lambda_{zz} = 72.2$, so the zz-axis governs and the capacity $P_d = 1324$ kN $\ge 1200$ kN. Safe. Overall width of column $= 2B + g = 370$ mm.

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

Step 3: Single lacing system using flat tie bars (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 $s = g + B = 190 + 90 = 280$ mm. Lacing angle $\theta = 45^\circ$ to the column axis (permitted 40° to 70°, cl. 7.6.4).

$$\begin{aligned} \text{Length of bar } l &= s/\sin\theta = 280/\sin 45^\circ = 396\ \text{mm} \\ \text{Force in bar } F &= \frac{V_t/2}{\sin\theta} = \frac{15.00}{\sin 45^\circ} = 21.21\ \text{kN} \end{aligned}$$

Spacing of connections along one channel $a_1 = 2s\cot\theta = 560$ mm. Check (cl. 7.6.5.1): $a_1/r_{min} = 560/26.0 = 21.5$, which is less than the smaller of 50 and $0.7\lambda_{max} = 50.5$. OK.

Tie bar (flat) size. Width $\ge 3d = 48$ mm for M16 bolts; take 60 mm. Thickness $\ge l/40 = 9.9$ mm (cl. 7.6.3), and for $KL/r \le 145$, $t \ge l\sqrt{12}/145 = 9.5$ mm. Provide ISF 60 × 10 mm.

$r = t/\sqrt{12} = 2.89$ mm, $l/r = 137.2 < 145$ OK. $f_{cd}$ (curve c) $= 68.4$ N/mm², $P_{d,lace} = 600 \times 68.4/10^3 = 41.0$ kN $> F = 21.21$ kN. OK.

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

$$\begin{aligned} V_{dsb} &= \frac{f_{ub}}{\sqrt3}\frac{0.78\,\pi d^2/4}{\gamma_{mb}} = \frac{400}{\sqrt3} \times \frac{157}{1.25 \times 10^3} = 29.0\ \text{kN} \\ V_{dpb} &= \frac{2.5\,k_b\, d\, t\, f_u}{\gamma_{mb}} = \frac{2.5 \times 0.556 \times 16 \times 10 \times 410}{1.25 \times 10^3} = 72.9\ \text{kN} \end{aligned}$$

($k_b = e/3d_0 = 30/54 = 0.556$ 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: $A_n = (60 - 18) \times 10 = 420$ mm², $T_{dn} = 0.9 f_u A_n/\gamma_{m1} = 124.0$ 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) $\ge h = 237$ mm; thickness $\ge s/50 = 5.6$ mm. Use 4 M16 bolts on each channel at 80 mm pitch (outermost bolts 240 mm apart) with 30 mm end distance, so the plate is 370 mm wide × 300 mm deep × 6 mm thick.

With $C = 560$ mm, $N=2$: $V_b = V_tC/(NS) = 30.0$ kN and $M = V_tC/2N = 4200$ kN·mm. Bolt distances from the group centre are ±40 and ±120 mm, $\Sigma r^2 = 32000$ mm². Bolt forces: from $M$ $= 4200 \times 120/32000 = 15.8$ kN, from shear $= 30.0/4 = 7.5$ kN, resultant $= 17.4$ kN $\le 29.0$ kN. OK.

Final design

• Column: 2 × ISMC 300 back to back, gap 190 mm, $P_d = 1324$ kN $> 1200$ kN.
• Lacing: single, flats 60 × 10 mm at 45°, 1 M16 bolt at each end.
• End tie plates 370 × 300 × 6 mm, 4 M16 bolts per channel.

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