Bolted bracket connection, 200 kN at 250 mm

Assumptions: the 200 kN is the factored load, acting in the plane of the bolts at $e=250$ mm from the centroid of the bolt group (column face taken as the bolt line). The column flange is wide enough for a gauge of 100 mm. M20 bolts, grade 4.6, in two vertical lines, plate 12 mm, E250.

Bolt value

• Shear: $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{245}{1.25}\times10^{-3}=45.26$ kN
• Bearing ($e=40$, $p=70$, $t=12$ mm): $k_b=0.606$, $V_{dpb}=119.3$ kN
• Bolt value $=45.26$ kN

Trial and check

The bolts are assumed at gauge 100 mm and pitch 70 mm. For $n$ bolts the critical corner bolt receives direct shear $P/n$ plus the torsion effect of $P e$ about the centroid:

For 14 bolts: $\Sigma(x^2+y^2)=309400$ mm², $T=200\times250=50000$ kN·mm

• Direct shear $=200/14=14.29$ kN
• Corner bolt ($x=50$, $y=210$): $F_x=\dfrac{Ty}{\Sigma r^2}=33.94$ kN; vertical torsion part $=8.08$ kN
• $R=\sqrt{33.94^2+(14.29+8.08)^2}=40.64$ kN $<45.26$ kN

Answer: 14 bolts M20 (4.6) in two vertical lines of 7, gauge 100 mm, pitch 70 mm, end distance 40 mm (bracket depth about 500 mm).