Built-up column with battens, 10 m, 1080 kN

Approach. Battened column of two channels, designed to IS 800:2007 (cl. 7.1.2 for compressive strength and cl. 7.7 for battened columns). Fe 410 (E250): $f_y = 250$ N/mm², $f_u = 410$ N/mm²; bolts M16 grade 4.6 ($f_{ub} = 400$ N/mm²), $\gamma_{m0} = 1.10$, $\gamma_{mb} = 1.25$.

Data

• $P = 1080$ kN (factored). Ends restrained in position but not in direction, so $KL = 1.0 \times 10$ m $= 10000$ mm (Table 11).
• Buckling curve c, $\alpha = 0.49$ (built-up member, Table 10). Effective slenderness of a battened column $= 1.1 \times$ actual maximum slenderness (cl. 7.7.1.4).

Step 1: Trial section

Try 2 × ISMC 350: $A = 54.4$ cm² each, $I_{xx} = 10000$ cm⁴, $I_{yy} = 434$ cm⁴, $C_{yy} = 2.44$ cm, $B = 100$ mm, $t_w = 8.3$ mm, $t_f = 13.5$ mm, $r_{y} = 2.82$ cm.

$$\begin{aligned} r_{zz} &= \sqrt{I_{xx}/A} = 13.56\ \text{cm} \\ \lambda_{zz} &= 10000/135.6 = 73.8 \\ \lambda_{eff} &= 1.1 \times 73.8 = 81.1 \end{aligned}$$

Curve c at $\lambda_{eff} = 81.1$: $f_{cc} = 300$ N/mm², $\lambda_n = 0.913$, $\phi = 1.092$, $\chi = 0.592$.

$$\begin{aligned} f_{cd} &= 0.592 \times 250/1.1 = 134.5\ \text{N/mm}^2 \\ P_d &= 2A f_{cd} = 10880 \times 134.5/10^3 = 1463\ \text{kN} > 1080\ \text{kN} \end{aligned}$$

Step 2: Spacing of channels

For $I_{yy} \ge I_{zz}$: $h \ge 2\sqrt{(I_x - I_y)/A} = 2\sqrt{(10000 - 434) \times 10^4/5440} = 265$ mm.

Gap between webs $g = h - 2C_{yy} = 265 - 48.8 = 216$ mm; provide g = 220 mm, so $h = 269$ mm.

$I_{yy} = 2[434 + 54.4(13.44)^2] = 20521$ cm⁴, $r_{yy} = 13.73$ cm, $\lambda_{yy} = 72.8$. This is not more than $\lambda_{zz} = 73.8$, so the capacity $P_d = 1463$ kN is safe. Overall width $= 2B + g = 420$ mm.

Step 3: Spacing of battens (cl. 7.7.3)

The slenderness of one channel between batten connections must not exceed 50 or $0.7\lambda_{max} = 0.7 \times 73.8 = 51.6$. So $\lambda_1 \le 50.0$, and with $r_{min} = r_y = 28.2$ mm:

$$\begin{aligned} C &\le 50.0 \times 28.2 = 1410\ \text{mm} \\ \text{Bays} &\ge 10000/1410 = 7.1 \to 8\ (\ge 3) \end{aligned}$$

The main-member check of Step 7 (bending from the batten moment) needs a smaller spacing, so adopt 11 bays: $C = 10000/11 = 909$ mm, $\lambda_1 = 909/28.2 = 32.2$.

Step 4: Forces on battens (cl. 7.7.2.1)

Transverse shear $V_t = 0.025 \times 1080 = 27.0$ kN, $N = 2$ planes. Bolt lines are at mid-flange, so $S = g + B = 220 + 100 = 320$ mm.

$$\begin{aligned} V_b &= \frac{V_t C}{N S} = \frac{27.0 \times 909}{2 \times 320} = 38.4\ \text{kN} \\ M &= \frac{V_t C}{2N} = \frac{27.0 \times 909}{4} = 6136\ \text{kN·mm} \end{aligned}$$

Step 5: Size of battens (cl. 7.7.2.3)

The effective depth of a batten is the longitudinal distance between its outermost bolts. End battens: effective depth $\ge h = 269$ mm; intermediate battens $\ge 0.75h = 202$ mm; in no case $< 2B = 200$ mm. Use the same size throughout: effective depth 270 mm (outermost bolts 270 mm apart), overall depth $= 270 + 2 \times 30 = 330$ mm.

Thickness $\ge S/50 = 320/50 = 6.4$ mm; provide 8 mm. Batten plate: 420 mm × 330 mm × 8 mm.

Plate check: $\sigma_b = M/Z = 6136364/145200 = 42.3$ N/mm² $< f_y/\gamma_{m0} = 227$; $\tau = V_b/(t\,d) = 14.5$ N/mm² $< f_y/(\sqrt3\gamma_{m0}) = 131$. OK.

Step 6: Bolts (M16, grade 4.6)

$V_{dsb} = 29.0$ kN (single shear, threaded), $V_{dpb} = 58.3$ kN ($t = 8$ mm, $k_b = 0.556$). Bolt value $= 29.0$ kN.

Try 5 bolts in one line on each channel at a pitch of 67.5 mm, so the outermost bolts are $4 \times 67.5 = 270$ mm apart (= the effective depth). Bolt distances from the centre of the group: 0, ±67.5, ±135 mm, $\Sigma r^2 = 2(67.5^2 + 135^2) = 45563$ mm². Force on the farthest bolt due to $M$ $= M r_{max}/\Sigma r^2 = 6136 \times 135/45563 = 18.2$ kN; due to shear $= V_b/n = 38.4/5 = 7.7$ kN.

$$R = \sqrt{18.2^2 + 7.7^2} = 19.7\ \text{kN} \le 29.0\ \text{kN}$$

Provide 5 bolts M16 per channel at each batten. Edge distance 30 mm, pitch 67.5 mm ($\ge 2.5d = 40$ mm).

Step 7: Check of the main member (cl. 7.7.2.1)

Each channel carries $P/2 = 540$ kN with a local moment $M = 6.14$ kN·m about its own $y$-axis. $\lambda_1 = 32.2$, $f_{cd} = 208.4$ N/mm², $P_{d1} = 5440 \times 208.4/10^3 = 1134$ kN. $M_d = Z_y f_y/\gamma_{m0} = 57.3 \times 10^3 \times 250/1.1 = 13.02$ kN·m ($Z_y = 57.3$ cm³).

$$\frac{P/2}{P_{d1}} + \frac{M}{M_d} = \frac{540}{1134} + \frac{6.14}{13.02} = 0.95 < 1.0$$

The main member is safe.

Final design

• 2 × ISMC 350 back to back, gap 220 mm, $P_d = 1463$ kN $> 1080$ kN.
• 11 equal bays of 909 mm; battens 420 × 330 × 8 mm at both ends and at every bay point, each connected by 5 M16 bolts (pitch 67.5 mm) to each channel.

  |==|            |==|
  |  |------------|  |  <- end batten
  |  |            |  |
  |  |------------|  |  <- batten
  |  |            |  |  (bay C = 909 mm)
  |  |------------|  |  <- batten
  |==|            |==|
  channel     channel