Built-up column, single lacing, 8 m, 1100 kN

Reading of the doubtful value: the bolt size is printed "M10 [?]". M10 is too small for the plate and flange used here, so M20 is adopted (grade 4.6); with M10, two bolts would be needed at each end of a lacing bar.

Given/assumptions: factored load $P_u=1100$ kN; length 8 m, held in position at both ends but not restrained against rotation → $K=1.0$, $KL=8$ m. Two channels back to back, single bolted lacing; E250 ($f_y=250$, $f_u=410$).

1. Section: 2 ISMC 300 back to back (webs facing, flanges outward)

$A=2\times4564=9128$ mm², $I_{xx}=2\times6362.6\times10^4$ mm⁴, $r_{xx}=118.1$ mm, $C_{yy}=23.6$ mm, $b_f=90$ mm. Built-up members use buckling class c (IS 800:2007 Table 10, $\alpha=0.49$).

• $\lambda_{xx}=\dfrac{KL}{r_{xx}}=\dfrac{8000}{118.1}=67.8$
• Laced column: $\lambda_e=1.05\lambda=71.1$ (cl. 7.6.1.4)
• $f_{cd}=\dfrac{f_y/\gamma_{m0}}{\phi+\sqrt{\phi^2-\lambda_n^2}}$ with $\phi=0.5[1+0.49(\lambda_n-0.2)+\lambda_n^2]$, $\lambda_n=\sqrt{f_y/f_{cc}}$, $f_{cc}=\pi^2E/\lambda_e^2$ → $f_{cd}=150.4$ N/mm²
• $P_d=A f_{cd}=1373$ kN $\ge P_u=1100$ kN ✓

2. Clear gap for equal stiffness

The centroid of each section is at $x_c=23.6+s/2$ from the column axis. For $I_{yy}\ge I_{xx}$:

$$x_c\ge\sqrt{\frac{I_{xx}-I_{yy,ch}}{A_{ch}}}=115.2\ \text{mm}\ \Rightarrow\ s\ge183\ \text{mm}$$

Provide $s=190$ mm. Then $I_{yy}=2[I_{yy,ch}+A_{ch}x_c^2]=13461\times10^4$ mm⁴, $r_{yy}=121.4$ mm, $\lambda_{yy}=65.9$, $1.05\lambda_{yy}=69.2$, $P_{dy}=1402$ kN $\ge P_u$ ✓.

3. Design of lacing (single lacing, two planes)

   |\     /|
   | \   / |    theta = 45 deg to the axis
   |  \ /  |    S = distance between connection lines
   |  / \  |
   | /   \ |

• Distance between connection lines $S=s+b_f=190+90=280$ mm
• Transverse shear $V=2.5\%\,P_u=27.5$ kN; per lacing plane $V_p=13.75$ kN
• Force in lacing bar $F=\dfrac{V_p}{\sin\theta}=19.45$ kN
• Length of bar $l=\dfrac{S}{\sin\theta}=396$ mm; effective length $=l$ (between the inner end bolts)
• $t\ge l/40=10$ mm; width $\ge3d=60$ mm → flat 60 × 10 mm
• $\lambda=\dfrac{l}{t/\sqrt{12}}=137<145$ ✓ (cl. 7.6.3); $f_{cd}=68$ N/mm², $P_c=41.0$ kN $>F$ ✓
• Tension: $T_{dn}=\dfrac{0.9(w-d_0)tf_u}{\gamma_{m1}}=112.2$ kN $>F$ ✓
• Connection: M20 (4.6) $V_{dsb}=45.3$ kN, bearing on 10 mm $=99.4$ kN → 1 bolt at each end
• Lacing angle $\theta=45°$ lies between 40° and 70° ✓
• Spacing of lacing points on one section $=\dfrac{2S}{\tan\theta}=560$ mm; slenderness of the section between them $=\dfrac{560}{26.1}=21.5\le\min(50,\ 0.7\lambda_{max})$ ✓ (cl. 7.6.3)

4. Tie plates

Tie (end) plates at both ends of the lacing system: length $\ge S+$ edges $=360$ mm, depth $\ge S=280$ mm, thickness $\ge S/50=6$ mm → 360 × 280 × 6 mm at each end, connected to each section by 2 bolts.

Answer: 2 ISMC 300 back to back (webs facing, flanges outward), clear gap 190 mm, single lacing of 60 × 10 mm flats at 45°, tie plates 360 × 280 × 6 mm at the ends.