Bracket plate on column flange, 5 M20 bolts

Assumptions (figure values are ambiguous): the bracket plate (10 mm) lies against the column flange (ISMB 300, $t_f=12.4$ mm) and the load $P$ acts in the plane of the bolts. The 5 bolts are M20, grade 4.6, in the pattern: four corner bolts at $(\pm60,\pm80)$ mm and one at the centre (gauges 60+60 and pitch 80+80; this makes the corner distance $r=\sqrt{60^2+80^2}=100$ mm). The bolt group centroid is 150 mm from the column face (90 mm edge + 60 mm), so the eccentricity of $P$ (250 mm from the face) is $e=250-150=100$ mm. Working load $P$ = factored load / 1.5.

     o       o      y = +80
         o          y =   0   centre
     o       o      y = -80
    x=-60   +60            P acts e = 100 mm from centroid

Bolt value (IS 800:2007, cl. 10.3)

• M20: $A_{nb}=245$ mm², $f_{ub}=400$, $f_u=410$, $d_0=22$ mm.
• Shear (single shear, thread in plane): $V_{dsb}=\dfrac{400}{\sqrt3}\times\dfrac{245}{1.25}\times10^{-3}=45.26$ kN
• Bearing (plate 10 mm, $e=40$, $p=80$): $k_b=\min(0.606,\,0.962,\,0.976,\,1)=0.606$; $V_{dpb}=2.5k_bdtf_u/\gamma_{mb}=99.4$ kN
• Bolt value $V_{db}=45.26$ kN (shear governs).

Force on the critical (corner) bolt, per unit factored load $P_u$

• $\Sigma r^2=4\times100^2=40000$ mm²; torsion $T=P_ue=100P_u$
• Direct shear (vertical) $F_D=P_u/5=0.2P_u$
• Torsional force $F_T=\dfrac{T r}{\Sigma r^2}=\dfrac{100P_u\times100}{40000}=0.25P_u$, with components $F_{Tx}=0.25P_u\times\tfrac{80}{100}=0.20P_u$ and $F_{Ty}=0.25P_u\times\tfrac{60}{100}=0.15P_u$
• Resultant $R=\sqrt{(0.20P_u)^2+(0.2P_u+0.15P_u)^2}=0.4031P_u$

Capacity

$$0.4031P_u\le45.26\Rightarrow P_u=112.3\ \text{kN}$$

Answer: factored load $P_u\approx112.3$ kN; safe working load $P=P_u/1.5\approx74.9$ kN.