Check bolted bracket on SC 250 (25 cm eccentricity)

Reading of the data: "14-16 mm diameter" is taken as 14 bolts of 16 mm diameter (M16, product grade C, class 4.6) in two vertical rows 100 mm apart, 7 per row, pitch 80 mm. The 100 kN is first treated as the design (factored) load; the working-load case is checked at the end. Plate 12 mm, Fe410.

Bolt value (IS 800:2007 cl. 10.3)

• M16: $A_{nb}=157$ mm², $d_0=18$ mm (grade C hole: $d+2$).
• Shear: $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{157}{1.25}\times10^{-3}=29.01$ kN
• Bearing ($e=40$, $p=80$): $k_b=\min(0.741,1.231,0.976,1)=0.741$; $V_{dpb}=2.5k_bdtf_u/\gamma_{mb}=116.6$ kN
• Bolt value $=29.01$ kN

Forces

• $\Sigma(x^2+y^2)=14(50^2)+2\times2(80^2+160^2+240^2)=393400$ mm²; $T=100\times250=25000$ kN·mm
• Direct shear per bolt $=100/14=7.14$ kN
• Critical bolt (corner, $x=50$, $y=240$): $F_x=\dfrac{Ty}{\Sigma r^2}=15.25$ kN; torsional vertical part $=3.18$ kN
• $R=\sqrt{15.25^2+(7.14+3.18)^2}=18.42$ kN

$R=18.42\ \text{kN}<29.01$ kN (utilisation 0.63). If 100 kN is a working load, the factored force is $1.5\times18.42=27.63$ kN $<29.01$ kN (utilisation 0.95), still safe.

Answer: the joint is safe. (End distance 40 mm ≥ $1.5d_0=27$ mm; pitch 80 mm lies between $2.5d=40$ mm and $\min(32t,300)$.)