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

$V_{dpb}=\frac{2.5k_b.d.t.f_u}{{\gamma }_{mb}}$
$=\frac{2.5\times 0.5\times 18\times 10\times 410}{1.25}=73.8kN$

Bolt value,
$=min(V_{dsb},V_{dpb})=73.8kN$

$\therefore$ Number of bolts
$=\frac{250}{73.8}=3.38=4$