For the fourbar parallelogram linkage shown in Figure 1 calc

For the fourbar parallelogram linkage shown in Figure 1, calculate the force in each of the supporting massless links AB and CD of the same length (0.5 m), at the position shown. The angular velocity omega of the link AB is 6 rad/s. The mass of connecting beam BD is 100 kg. The mass moment of inertia of BD about point G is 30 kg middot m^2.

Solution

solution:

1)here for given mechanism moving with accelaration aequation in dynamic equillibrium are

Fbx+Fdx=Max

Fby+Fdy=May

where for moment about point D

Fby*.8+May*.4-I*m2=0

m2=angular accelaration of link BD

Fdy*.8+May*.4-I*m2=0

but for given mechanism has constant angular velocity whence system has no dyanmic forces devloped

hence supporting forces on B and D are

Fbx=Fby=Fdx=Fdy=0

2)as here system in static equillibrium

Fb+Fd=100*9.81

from symmetry

Fb=Fdhence Fb=Fd=490.5 N

at point B and D for ces divide are divide along the link and perpendicular to link and they are in equalmagnitude for B and d

Fb\'=Fd\'=490.5cos45=424.78 N

perpendicular

Fb\'\'=Fd\'\'=490.5*sin45=245.24 N

perpendicular force create torque in link AB,Cd

T=Fb\'\'*.5=122.625 Nm

3)hence force in Ab and CD are

Fb\'=Fd\'=424.78 N

Tab=Tcd=122.625 Nm