A steel plate subjected to a force of 5 KN and fixed to a channel by means of three identical bolts is shown in Fig. 3.1. The bolts are made from plain carbon steel 45C8 (Syt = 380 N/mm2) and the factor of safety is 3. Specify the size of bolts.

Subject: Machine Design -I

Topic: Design against static Loads, Bolted and welded joints under eccentric loading. Power Screw

Difficulty: Medium

Step 1: Permissible shear stress

$\tau=\frac{S_sy}{fs}=\frac{0.5S_yt}{fs}=\frac{0.5*380}{3}=63.33 N/mm^2$

Step 2: Primary and secondary shear forces:

The centre of gravity of three bolts will be at the centre of bolt-2. The primary and secondary shear forces are shown in fig(b) and (c)

$P_1'=P_2'=P_3'=\frac{P}{3}=\frac{5000}{3}=1666.67 N$

$P_1''=P_3''=\frac{Pe*r_1}{r_1^2+r_3^2}=\frac{5000*305*75}{75^2+75^2}=10166.67N$

Step 3: Resultant shear force.

The resultant shear force on the bolt 3 is maximum.

$P_3=P_3'+P_3''=1666.67+10166.67=11833.34N$

Step 4: Size of bolts:

$\tau=\frac{P_3}{A} or 63.33=\frac{11833.34}{\frac{\pi}{4}d_c^2}$

$d_c=15.42 mm$

Also,

$d=\frac{d_c}{0.8}=\frac{15.42}{0.8}=19.28 or 20mm$

The standard size of the bolts is M20.