Question

A bracket connection is made with four bolts of 20 mm diameter and supports a load of 40 kN at an eccentricity of 60 mm. The maximum force resisted by any bolt will be _____ kN.

• A. 20.88

Answer 1

The correct option is A 20.88
Force on bolt due to direct shear

$F_1=\frac{P}{4}=\frac{40}{4}=10kN$

Force due to twisting moment

$F_2=\frac{P.e.r}{\Sigma r_i^2}$

$\tan \theta =\frac{30}{40}=\frac{3}{4}$

$\cos \theta =\frac{4}{5}$

$r_1=r_2=r_3=r_4=50mm$

$\therefore F_2=\frac{P.e.r_1}{\Sigma r_i^2}=\frac{40\times 60\times 50}{4\times {50}^2}=12kN$

$\therefore F_R=\sqrt{F_1^2+F_2^2+2F_1{F}_2\cos \theta }$

$=\sqrt{{10}^2+{12}^2+2\times 10\times 12\times \frac{4}{5}}$

$=20.88kN$