Built-up column with battens, 6 m, 1200 kN

Given/assumptions: axial load $P_u=1200$ kN (taken as factored); effective length 6 m ($KL=6$ m); two channels back to back with battens, bolted (M20, grade 4.6); Fe410 / E250 ($f_y=250$, $f_u=410$).

1. Section: 2 ISMC 250 back to back (flanges outward)

$A=7734$ mm², $I_{xx}=2\times3816.8\times10^4$ mm⁴, $r_{xx}=99.3$ mm, $C_{yy}=23.0$ mm, $b_f=80$ mm, $r_{yy,ch}=23.8$ mm. Built-up members are buckling class c.

• $\lambda_{xx}=\dfrac{KL}{r_{xx}}=\dfrac{6000}{99.3}=60.4$
• Battened column: $\lambda_e=1.1\lambda=66.4$ (cl. 7.6.1.5)
• $f_{cd}=158.0$ N/mm² (curve c, $f_y=250$, $\gamma_{m0}=1.10$) → $P_d=A f_{cd}=1222$ kN $\ge1200$ kN ✓

2. Clear gap (so that $I_{yy}\ge I_{xx}$)

The centroid of each section is at $x_c=23.0+s/2$ from the column axis:

$$x_c\ge\sqrt{\frac{I_{xx}-I_{yy,ch}}{A_{ch}}}=96.5\ \text{mm}\Rightarrow s\ge147\ \text{mm}$$

Provide $s=150$ mm. $I_{yy}=7866\times10^4$ mm⁴, $r_{yy}=100.8$ mm, $1.1\lambda_{yy}=65.4$, $P_{dy}=1234$ kN $\ge P_u$ ✓

3. Spacing of battens (cl. 7.7.1.4)

The slenderness of one section between battens must be $\le\min(50,\ 0.7\lambda_e)=46.5$:

$$C\le 46.5\times23.8=1107\ \text{mm}$$

For the length of 6000 mm provide 6 bays, i.e. 7 battens at $C=1000$ mm c/c.

4. Size of battens (cl. 7.7.2)

• Distance between the connection lines $a=S=s+b_f=230$ mm
• End battens: depth $\ge a$ → 260 mm; intermediate battens: depth $\ge\max(0.75a,\ 2b_f)$ → 260 mm
• Thickness $\ge a/50=230/50$ → 8 mm; length of batten $=a+80=310$ mm.

5. Design forces on a batten

Transverse shear $V_t=2.5\%P_u=30.0$ kN, number of batten planes $N=2$:

$$V_b=\frac{V_tC}{NS}=\frac{30.0\times1000}{2\times230}=65.22\ \text{kN},\qquad M=\frac{V_tC}{2N}=7.500\ \text{kN·m}$$

• Plate in bending: $Z=\dfrac{t d^2}{6}=90133$ mm³ (intermediate batten); $\sigma=\dfrac{M}{Z}=83.2$ N/mm² $<f_y/\gamma_{m0}=227$ N/mm² ✓ ($M_d=20.48$ kN·m)
• Shear: $\tau=\dfrac{V_b}{dt}=31.4$ N/mm² $<\dfrac{f_y}{\sqrt3\gamma_{m0}}=131$ N/mm² ✓

6. Bolts connecting battens to the sections (M20, 4.6)

4 bolts in a vertical line at each end (pitch 60 mm, end 40 mm): $V_{dsb}=45.3$ kN, bearing on 8 mm $=79.5$ kN → bolt value $45.3$ kN.

• Direct shear per bolt $=V_b/n=16.30$ kN; moment effect on the extreme bolt $=\dfrac{M y_{max}}{\Sigma y^2}=37.50$ kN
• Resultant $R=\sqrt{16.30^2+37.50^2}=40.89$ kN $<45.3$ kN ✓

Answer: 2 ISMC 250 back to back (flanges outward), clear gap 150 mm; battens 260 × 8 mm (end and intermediate), 310 mm long, at 1000 mm c/c, with 4 M20 bolts at each end.