A 24 mm diameter bolt of grade 4.6 is in double shear, the shearing strength of bolt will be (assume no threads in the shear planes)?

Concept:

Shearing strength of all bolts (Ps)

Where n is number of bolts, d is diameter of bolt

$({P_s} = m \times \frac{\pi }{4}{d^2}*{f_S} \times n)$

m = 1 for single shear and m = 2 for double shear

Here m = 2 because of double shear in double cover butt joint

$({f_s} = \frac{{{f_{ub}}}}{{\sqrt 3 \times 1.25}})$

(fub is ultimate stress of bolt)

Calculations:

For 4.6 grade bolts, fub = 400 MPa, fyb = 240 MPa

$(\begin{array}{l}
ShearArea = \frac{\pi }{4} \times {24^2}\\
{P_s} = 2 \times \frac{\pi }{4} \times {24^2} \times \frac{{400}}{{\sqrt 3 \times 1.25}} \times 1 \times {10^{ - 3}}kN
\end{array})$

Ps = 167.2 kN