A rectangular steel bar of length 500 mm, width 100 mm, and thickness 15 mm is cantilevered to a 200 mm steel channel using 4 bolts, as shown.

For an external load of 10 kN applied at the tip of the steel bar, the resultant shear load on the bolt at B, is ___________ kN (round off to one decimal place).

The correct answer is 16.0.

1. Centroid of the Bolt Group:
From the given schematic, there are 4 identical bolts: A, B, C, and D arranged symmetrically. Let the centroid of this bolt group be denoted as $O$.
The vertical distance of each bolt from the horizontal centerline passing through $O$ is:
$y=50\text{mm}$
The horizontal distance of each bolt from the vertical centerline passing through $O$ is:
$x=50\text{mm}$
Thus, the radial distance ($r$) of each bolt from the centroid $O$ is equal for all four bolts:
$r=\sqrt{x^2+y^2}=\sqrt{{50}^2+{50}^2}=50\sqrt{2}\text{mm}\approx 70.71\text{mm}$

2. Primary Shear Force on Bolt B:
The external load of $F=10\text{kN}$ acts vertically downwards at the tip of the steel bar. This load is shared equally among the 4 bolts as a primary shear force ($F^{\prime }$):
$F^{\prime }=\frac{F}{4}=\frac{10\text{kN}}{4}=2.5\text{kN}$
The direction of this primary shear force $F^{\prime }$ on each bolt (including bolt B) is vertically downwards.

3. Torque (Moment) about the Centroid O:
The distance from the centroid $O$ of the bolt group to the tip where the load is applied is calculated as:
$L=100\text{mm}+300\text{mm}=400\text{mm}$
The eccentric load creates a clockwise bending moment (torque, $T$) about the centroid:
$T=F\times L=10\text{kN}\times 400\text{mm}=4000\text{kN·mm}$

4. Secondary Shear Force on Bolt B:
The secondary shear force ($F^{″}$) on each bolt due to the torque is given by:
$F^{″}=\frac{T·r}{\sum r^2}$
Since all four bolts are at the same distance $r$ from the centroid:
$\sum r^2=4r^2=4\times {\left(50\sqrt{2}\right)}^2=4\times 5000=20000{\text{mm}}^2$
Substituting these values:
$F^{″}=\frac{4000\times 50\sqrt{2}}{20000}=10\sqrt{2}\text{kN}\approx 14.14\text{kN}$
The secondary shear force is perpendicular to the radial vector of bolt B. For the top-right bolt B, the clockwise torque produces a secondary shear force directed perpendicular to the radius vector, pointing downwards and to the right at an angle of $45°$ to the vertical.

5. Resultant Shear Force on Bolt B:
The angle $\theta$ between the downward primary shear force ($F^{\prime }=2.5\text{kN}$) and the secondary shear force ($F^{″}=14.14\text{kN}$) at bolt B is $45°$.
Using the parallelogram law of vector addition:
$R_B=\sqrt{{\left(F^{\prime }\right)}^2+{\left(F^{″}\right)}^2+2F^{\prime }{F}^{″}\cos \left(\theta \right)}$
$R_B=\sqrt{{2.5}^2+{\left(10\sqrt{2}\right)}^2+2\times 2.5\times 10\sqrt{2}\times \cos \left(45°\right)}$
Since $\cos \left(45°\right)=\frac{1}{\sqrt{2}}$:
$R_B=\sqrt{6.25+200+2\times 2.5\times 10\sqrt{2}\times \frac{1}{\sqrt{2}}}$
$R_B=\sqrt{6.25+200+50}=\sqrt{256.25}\approx 16.008\text{kN}$
Rounding off to one decimal place, we obtain:
$R_B=16.0\text{kN}$