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Bearing Type Connections - Design Strength of Bolt

A tie member ...

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 $10 m m$ , has to carry a factored load of $250 k N$ . The grade of steel is Fe410 and one of the shear plane passes through thread.The number of bolts of diameter $18 m m$ of $4.6$ class bolts required are _______.(Take $k_{b} = 0.5$ )

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Solution

The correct option is A 4
Strength of bolt in double shear

$V_{d s b} = (1 + 0.78) \times \frac{π}{4} \times 18^{2} \times \frac{400}{\sqrt{3} \times 1.25} = 83.68 k N$

Strength of bolt in bearing

$V_{d p b} = \frac{2.5 k_{b} . d . t . f_{u}}{γ_{m b}}$
$= \frac{2.5 \times 0.5 \times 18 \times 10 \times 410}{1.25} = 73.8 k N$

Bolt value,
$= m i n (V_{d s b}, V_{d p b}) = 73.8 k N$

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

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