| written 7.9 years ago by | • modified 7.6 years ago |
Subject: Machine Design -I
Topic: Design against static Loads, Bolted and welded joints under eccentric loading. Power Screw
Difficulty: Medium

| written 7.9 years ago by | • modified 7.6 years ago |
Subject: Machine Design -I
Topic: Design against static Loads, Bolted and welded joints under eccentric loading. Power Screw
Difficulty: Medium

| written 7.6 years ago by |

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 …