A 12 mm thick plate is connected through a 20 mm diameter bolt as show...
Given
Bolt diameter = 20 mm
Bolt hole diameter, dh = 22 mm
Safe stress on connection = 120 N/mm2
Thickness of plate = 12 mm
Also, Staggered pitch, S = 50 mm
Gauge length, g = 60 mm
Net width of the plate, corresponding to ABCD
= Total width of plate – (number of bolts × dh)
= (40 + 3 × 60 + 40) – 2 × 22
= 216 mm
Net width of the plate corresponding to ABECFG
Where b = width of the plate.
n = total numbers of bolt whole in the path considered
S1, S2, and S3 are the staggered pitch between bolts BE, EC and CF respectively.
g = gauge length
= 203.25 mm
The net width of the plate corresponding ABECFG
= 204.42 mm
The net width of the plate corresponding to ABEFG
= 214.83 mm
Therefore, minimum width for net effective area = 203.25 mm
Strength of the plate = (203.25 × 12) × 120 × 10-3 kN
= 292.68 kN