Question

A 16 mm thick plate is connected to two 8 mm thick plates on either side as shown in figure below through 18 mm diameter bolts. If bolte of grade 4.6 are used, then the bolt value is_______kN.
( Assume $K_b=0.5$ and both the shear planes lie in the shank of the bolt)

• A. 94.03

Answer 1

The correct option is A 94.03
Bolt is subjected to double shear

For bolt grade 4.6
$\Rightarrow f_u=4\times 100=400MPa$
and
$f_y=0.6\times 400=240MPa$

Design strength of bolt in shear

$V_{dsb}=\frac{f_{ub}}{\sqrt{3}{\gamma }_{mb}}(n_n{A}_{nb}+n_s{A}_{sb})$

$=\frac{400}{\sqrt{3}\times 1.25}(0+1\times 2\times \frac{\pi }{4}(18{)}^2)$

$=94.03\times {10}^{3}N$

$=94.03kN$

Design strength of bolt in bearing

$V_{dpb}=\frac{2.5k_{b}dt{f}_u}{{\gamma }_{mb}}$

$f_u$ = Ultimate tensile strength of plate (or) bolt which ever is lesser

$\therefore f_{ub}=400MPa\left(Bolt\right)$

$t=\text{min. of}\left[\text{thickness of main plate}\right(or)\text{Sum of thickness of cover plate}]$

$V_{dpb}=2.5\times 0.5\times 18\times 16\times \frac{400}{1.25}$

$V_{dpb}=115.2kN$

$\therefore$ Design strength of bolt is minimum of

$V_{dsb}and{V}_{dpb}$

Hence, the bolt value = design strength of the the bolt
Bolt value = 94.03 kN