the bolts are 20mm $\phi$ m and 4.6 grade

Step I

list bolt value=$\frac{fub\times Anb\times hn}{\sqrt{3}ymb}$

=$\frac{400\times245\times 1}{\sqrt{3}\times 1.25}$

Vdsb=45.26kN

Step II

dirct shera force per bolt=$\frac{w}{6}$ in horizontal direction

direct shear force per bolt=$\frac{2}{6}$in vertical direction

F1=$\sqrt{(\frac{w}{6})^{2}+(\frac{w}{6})^{2}}$=0.236w

Step III

force in bolt due to tortion

F2=$\frac{\sum(w.e)\gamma n}{\sum \gamma ^{2}}$

$\sum $we=w(700)+w(80)=780w

$\gamma n=\sqrt{80^{2}+80^{2}}$=113.13

$\gamma ^{2}=6(80)^{2}+4(80)^{2}$=64000

F2=$\frac{780w\times113.13}{64000}$

F2=1.37w

Step V

Resultant flow equation

R=F1+F2

45.26=0.236w+1.378w

w=28.02kN

Step VI

safe working load=$\frac{w}{1.5}$

=$\frac{28.02}{1.5}$

=18.68kN

Type II

bolt are sub to direct shera and bending tention

Step I-

Find bolt value

Step II -

No. of bolts required=$\sqrt{\frac{6m}{n\times p\times Bolt \ value}}$

M=P.e

P=eccentric load

E=eccentric distance

n=no.of rows

P=Pitch distance

Bv=list bolt value

Step III -- Total height of plate

H=2e+$\sum$P(pitch distance)

Step IV- Interaction equation

$(\frac{Vsb}{Vdb})^{2}+(\frac{Tb}{Tdb})^{2}$

Step--V direct shear=$\frac{Reaction}{No.of \ bolt}$

max Tensile force in critical bolt=$\frac{m\times yn}{\sum y^{2}}$

$m^{'}=\frac{m}{l+(\frac{2h}{2l}\times \frac{\sum y}{\sum y^{2}})}$

h=H-e