Explanation :
P = 10 kN, e = 15 cm
r_{1} = r_{2} = r_{3} = r_{4} = 5 cm
Direct load to bolt (1) = $\frac{\mathrm P}4$
${\mathrm f}_1=\frac{10}4=2.5\;\mathrm{kN}$
Force in bolt (1) due to moment
$\begin{array}{l}\;\;\;=\frac{{\mathrm{Per}}_1}{\sum{\displaystyle{\displaystyle\mathrm r}_\mathrm i^2}}\\{\mathrm f}_2=\frac{10\times15\times5}{4\times5^2}=7.5\;\mathrm{kN}\end{array}$
Angle between force F_{1} and F_{2} = 135°
$\begin{array}{l}\mathrm R=\sqrt{\begin{array}{c}\mathrm F_1^2+\mathrm F_2^2\\+2{\mathrm F}_1{\mathrm F}_2\cos(135^\circ)\end{array}}\\\\\;\;=\sqrt{\begin{array}{c}2.5^2+7.5^2\\+2\times2.5\\\times7.5\times\cos(135^\circ)\end{array}}\\\;\;\;=5.99\;\mathrm{kN}\end{array}$