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Determine safe load p that can be carried by than as shown bolts are 20mm $\phi$ 4.6 grade.Thickness of bracket plate is 10mm
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Step I

list bolt value

shear capacity of bolt=$\frac{fub\times Anb\times nn}{\sqrt{3}Ymb}$

=$\frac{400\times 245\times 1}{\sqrt{3} \times 1.25}$

Vdsb=45.26 kN

bearing capacity of bolt=2$\times Kb\times d\times t\times $fu

find kb

$\frac{e}{3do}=\frac{40}{3\times 22}$=0.60

$\frac{P}{3do}$=0.25=$\frac{80}{3\times 22}$=0.25=0.96

$\frac{fub}{fu}=\frac{400}{410}$=0.97

Vdpb=$2\times0.6\times 20\times10\times 410$

Vdpb=98.4kN

Step II

Direct force per bolt or load in each bolt

F1=$\frac{load}{No.\ bolt \ bolt}=\frac{p}{10}$ kN=0.1P

Step III

force in bolt due to tortion

F2=$\frac{\sum(p.p)\times rn}{\sum \gamma ^{2}}$

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F2=$\frac{(p\times 200)\times 170.88}{(164000)}$

F2=0.21 P

Step IV

used resultant force equation

$R^{2}=F_{1}^{2}+F_{2}^{2}+2F_{1}F_{2}cos \theta$

45.26$^{2}=(\frac{p}{10})^{2}+(0.21p)^{2}+2(\frac{p}{10})(0.21p)$ cos69.44

P=172.48kN

Step V

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

=$\frac{172.48}{1.5}$

=114.99

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