Question

A double bolted single cover butt joint is to be designed in order to connect two plates ISF 240 x 10 and ISF 240 x 8. The plates are subjected to a factored tensile force of 340 kN. If the bolts used to make connection are M20 bolts of grade 4.6, then number of bolts required per chain line is ________. (Take Fe 410 for plates)

Answer 1

$P=340kN$

$\because$ It is a single cover butt joint, number of shear planes = 1

Design strength of M20 bolt of grade 4.6 in shear

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

$n_n=1$
$A_{nb}=0.78A_{sb}=0.78\times \frac{\pi \times 20\times 20}{4}=244.92m{m}^2$
$f_{ub}=400N/m{m}^2$
${\gamma }_{mb}=1.25$
$n_s=0$

$\Rightarrow V_{dsb}=45.25kN$

Total no. of bolts required =

$\frac{P}{V_{db}}=\frac{340}{45.25}$

$\Rightarrow n=7.51\simeq 8$

Hence, bolts required per chain line
$=\frac{8}{2}=4Nos$