Question

A tie member of truss consist of two ISA $90\times 90\times 8$ placed back to back connected to a gusset plate of $10mm$, has to carry a factored load of $250kN$. The grade of steel is Fe410 and one of the shear plane passes through thread.The number of bolts of diameter $18mm$ of $4.6$ class bolts required are _______.(Take $k_b=0.5$)

• A. 4

Answer 1

The correct option is A 4
Strength of bolt in double shear
$V_{dsb}=(1+0.78)\times \frac{\pi }{4}\times {18}^2\times \frac{400}{\sqrt{3}\times 1.25}=83.68kN$
Strength of bolt in bearing
$V_{dpb}=\frac{2.5k_{b}dt{f}_u}{{\gamma }_{mb}}$
$=\frac{2.5\times 0.5\times 18\times 10\times 410}{1.25}=73.8kN$
Bolt value $=min\left(V_{dsb}{V}_{dpb}\right)=73.8kN$
$\therefore$ Number of bolts $=\frac{250}{73.8}=3.38=4$