Column with battens, 1500 kN, 8 m, fixed ends

Given/assumptions: factored load $P_u=1500$ kN; height 8 m; both ends fixed against rotation but the frame is free to sway → effective length factor $K=1.2$ (IS 800:2007 Table 11, recommended value), so $KL=1.2\times8000=9600$ mm. The column is built up of two rolled I-sections (ISMB) placed side by side and tied by battens (bolted, M20 4.6); E250 ($f_y=250$, $f_u=410$). The battens are on the flange faces; the bolt group on each I-section is centred on its web line, so $S=s+b_f$.

1. Section: 2 ISMB 350 placed side by side with a gap (webs parallel)

$A=13342$ mm², $I_{xx}=2\times13630.3\times10^4$ mm⁴, $r_{xx}=142.9$ mm, $C_{yy}=70$ mm, $b_f=140$ mm, $r_{yy,ch}=28.4$ mm. Built-up members are buckling class c.

• $\lambda_{xx}=\dfrac{KL}{r_{xx}}=\dfrac{9600}{142.9}=67.2$
• Battened column: $\lambda_e=1.1\lambda=73.9$ (cl. 7.6.1.5)
• $f_{cd}=146.0$ N/mm² (curve c, $f_y=250$, $\gamma_{m0}=1.10$) → $P_d=A f_{cd}=1948$ kN $\ge1500$ kN ✓

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

The centroid of each section is at $x_c=70.0+s/2$ from the column axis:

$$x_c\ge\sqrt{\frac{I_{xx}-I_{yy,ch}}{A_{ch}}}=140.1\ \text{mm}\Rightarrow s\ge140\ \text{mm}$$

Provide $s=150$ mm. $I_{yy}=29127\times10^4$ mm⁴, $r_{yy}=147.8$ mm, $1.1\lambda_{yy}=71.5$, $P_{dy}=2000$ kN $\ge P_u$ ✓

3. Spacing of battens (cl. 7.7.1.4)

The slenderness of one section between battens must be $\le\min(50,\ 0.7\lambda_e)=50.0$:

$$C\le 50.0\times28.4=1420\ \text{mm}$$

For the length of 8000 mm provide 6 bays, i.e. 7 battens at $C=1333$ mm c/c.

4. Size of battens (cl. 7.7.2)

• Distance between the connection lines $a=S=s+b_f=290$ mm
• End battens: depth $\ge a$ → 320 mm; intermediate battens: depth $\ge\max(0.75a,\ 2b_f)=280$ mm → 320 mm
• Thickness $\ge a/50=290/50$ → 8 mm; length of batten $=290+70+2\times35=430$ mm (to take the two bolt lines on each flange).

5. Design forces on a batten

Transverse shear $V_t=2.5\%P_u=37.5$ kN, number of batten planes $N=2$:

$$V_b=\frac{V_tC}{NS}=\frac{37.5\times1333}{2\times290}=86.21\ \text{kN},\qquad M=\frac{V_tC}{2N}=12.500\ \text{kN·m}$$

• Plate in bending: $Z=\dfrac{t d^2}{6}=136533$ mm³ (intermediate batten); $\sigma=\dfrac{M}{Z}=91.6$ N/mm² $<f_y/\gamma_{m0}=227$ N/mm² ✓ ($M_d=31.03$ kN·m)
• Shear: $\tau=\dfrac{V_b}{dt}=33.7$ N/mm² $<\dfrac{f_y}{\sqrt3\gamma_{m0}}=131$ N/mm² ✓

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

A bolt cannot be placed on the web line of an I-section, so at each end use 8 bolts in two vertical lines of 4, one line on each side of the web (35 mm from the web centre line, pitch 80 mm, end 40 mm). $V_{dsb}=45.3$ kN, bearing on 8 mm $=79.5$ kN → bolt value $45.3$ kN.

• $\Sigma r^2=8\times35^2+4\times(40^2+120^2)=73800$ mm²
• Direct shear per bolt $=V_b/8=10.78$ kN (along the column); moment on the extreme bolt (35, 120): $\dfrac{M y}{\Sigma r^2}=20.33$ kN across and $\dfrac{M x}{\Sigma r^2}=5.93$ kN along the column
• Resultant $R=\sqrt{20.33^2+(10.78+5.93)^2}=26.3$ kN $<45.3$ kN ✓

Answer: 2 ISMB 350 placed side by side with a gap (webs parallel), clear gap 150 mm; battens 320 × 8 mm (end and intermediate), 430 mm long, at 1333 mm c/c, with 8 M20 bolts (two lines of 4) at each end.