Question

A bracket connection is made with four bolts of 10 mm diameter and supports a load of 10 kN at an eccentricity of 100 mm. The maximum force to be resisted by any bolt will be ____. (all distance are in mm)

• A. 7.16 kN
• B. 6.5 kN
• C. 5 kN
• D. 6.8 kN

Answer 1

The correct option is A 7.16 kN



Diameter of bolt = 10 mm
Due to symmetry of bolts C.G of bolts will lie at O
Distance of any bolt from C.G,
$r=\sqrt{{30}^2+{40}^2}$
= 50 mm
= 0.05 m
Direct shear force on each bolt ,
$F_v=\frac{10}{4}$
= 2.5 kN
Torsional force,
$F_T$ = Torsional stress x Area

= $\left(\frac{T{r}_i}{J}\right).A_i$

$=\frac{Pe{r}_i{A}_i}{\sum A_i{r}_i^2}$

For equal area of each bolt,

$F_T=\frac{Pe{r}_i}{\sum r_i^2}=\frac{(10\times 0.1)\times 0.05}{4\times (0.05{)}^2}$

= 5 kN
Resultant force will be maximum for bolt 1 and 2 because $\theta$ is minimum for these bolt.

$F_r=\sqrt{F_v^2+F_T^2+2F_V{F}_{T}cos\theta }$

$=\sqrt{{2.5}^2+5^2+2\times 2.5\times 5\times \frac{4}{5}}$

= 7.16 kN