Subject: Machine Design -I

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

Difficulty: Medium

Step 1. Primary shear force:

$P_1'=P_2'=P_3'=P_4'=\frac{P}{4}=\frac{5*10^3}{4}=1250N$

Step 2 Primary shear force:

$r_1=r_2=r_3=r_4=100mm$

$C=\frac{Pe}{(r_1^2+r_2^2+r_3^2+r_4^2)}=\frac{(5*10^3)(200)}{4(100)^2}=25$

Therefore,

$P_1''=P_2''=P_3''=P_4''=C_r=25(100)=2500N$

Step 3: Resultant shear force:

The primary and secondary shear forces are shown in fig.b and fig.c. It is observed from the figure that rivet-2 is subjected to maximum resultant force. At rivet-2, the primary and secondary shear forces are additive.

Therefore,

$P_2=P_2'+P_2''=1250+2500=3750N$

Step 4: Diameter of rivets:

$P_2=\frac{\pi}{4}d^2\tau or 3750=\frac{\pi}{4}d^2(60)$

$d=8.92 or 9mm$