Four 05 bolts are used in a bolted bracket that is loaded as

Four 0.5\" bolts are used in a bolted bracket that is loaded as shown in the figure below. Determine the force on each bolt; Determine the safety factor guarding against bearing on the plate if it is 0.25 inch thick and has yield strength of 50 kpsi and an ultimate strength of 75 kpsi.

Solution

The direct sher load on bolt is F/A = 1000/area of bolts = 1275.6 lb/in2

Thr centroid of the forces are at x=0 and y is at mid of he plate ie 2inch aboce rivet C ie 4 inch from the centre of rivet The rivets A and D are subjected to max .load since they are further away.

The perpenticular distance of force to the center of the rivet a is 6 x sin30 = 3 inch

Normal Fa load due to torqe is F e L/total length = 1000x 3x4 / 2(4+2) = 1000lb

Resultant load = sqrt ( Fdirect ² + f Torque² - 2 FdxFt cos 60 ) = 1162 lb/in2

Hence the bolt is safe This is the load coming on bolt D also.

for bolt B and C the Normal force can be calculated similarly as Fb = 1000x 1 x 2 /2(4+2) = 166.6

RESULTANT FORRCE = SQRT ( 1275.2² + 166.6 ² - 2 1275.2x 166.6 cos 60 ) = 1200.04

The bearing area is0.5 x 0.25 for each bolt = 0.125.

The max load is on the bolt A and D the bearing load allowed is 50,000x 0.125 = 6250 lb

The safety factor is 6250/ actual load coming = 6250/1162 =5.37.