A column carries a 10 kN load through the bracket shown, at eccentricity 15 cm from the bolt-group centre. The four-bolt geometry, 10 cm dimensions and 90° angle are given in the figure. Find the resultant force in bolt 1 in kN, to one decimal place.
The symmetric geometry gives each bolt a radius of 5 cm from the group centre. Direct shear per bolt is $10/4=2.5$ kN. The eccentric moment is $10(15)=150$ kN cm, producing secondary shear $F_m=Mr/\sum r_i^2=150(5)/[4(5^2)]=7.5$ kN per bolt. At bolt 1 its components are $7.5/\sqrt2=5.303$ kN; the vertical component opposes the direct shear. The resultant is $\sqrt{5.303^2+(5.303-2.5)^2}=5.999$ kN, or 6.0 kN.
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