Question

For the bolted joint shown in figure. A vertical load of 10 kN is applied at the end of the beam as shown above. It has four bolts each one having an effective load bearing area of 100 mm${}^2$

What is the secondary effect in bolt '3' ?

Answer 1

Secondary effect has tensile effect on each bolt.
On bolt 3,

$\frac{P_{t3}}{L_3}(L_1^2+L_2^2+L_3^2+L_4^2)=P.e$

$\text{P = 10 kN, e = 1 m}$

$\text{L = Distance of bolt from tilting edge.}$

$L_3=L_4=0.8m,L_1=L_2=0.2m$

$P_{t3}=\frac{10\times 1\times 0.8}{(2\times (0.8{)}^2+2\times {0.2}^2)}$

$P_{t3}=5.88kN,A=100m{m}^2$

Hence,

${\sigma }_3=\frac{5.88\times {10}^3}{100}=58.8MPa={\sigma }_4$