In the eccentric swinging block mechanism below, the crank is in the form of an

Question

In the eccentric swinging block mechanism below, the crank is in the form of an eccentric which has a diameter of 190 mm. The friction between all contacting surfaces is 0.1. The width of link 4 is 80 mm (symmetric about B0) and the thickness of link 3 in contact is 25 mm. The radii of the revolute joints A0 and B0 is 10 mm. Determine the required input torque T12 when the crank angle ?12 is 60°. Neglect the inertial and gravitational forces. You may also neglect the friction at a revolute joint if the radius of the friction circle is less than 5 mm. A0A = 85 mm, A0B0= 280 mm, AP = 400mm, F13 = 600 N, ? = 80°.

Explanation / Answer

solution:

1)here for swinging block mechanism force is applied at P is divided into normal and parallel to link AP,

here by geometry we get that angle made force F=600 N is 80+30=110-90=20 degree to normal hence forces on link AP are

normal force Fn=Fcos20=563.81 N

parallel force=Fp=Fsin20=205.21 N

2)as in swinging block mechanism force is applied at P ,it will produced torque around point Ao and it si given by

Tn=Fn*AP=563.81*400=225526.229 N mm

Tp=Fp*AoA=205.21*85=17443.027 N mm

resultant torque is at Ao

Tr=Tn+Tp=242969.25 N mm

this torque is in clockwise diretion and it is opposed by friction circle at revolute joint at Ao and Bo

3)anticlockwise torque due to friction at revolute joint is

at Ao

Tao=normal force*radius of jounal*coefficient of friction

for Ao,normal force is F=600 N ,as it is creating torque and by equivalence condition it is passing through point Ao,normally after torque is consider and hence normal force is F=600 N

Tao=F*R*mu=600*10*.1=600 N mm(anticlockwise)

where torque at Bo is

Tbo=normal force*radius of jounal*coefficient of friction

here normal force is not F=600,as it has two component and only normal component Fn is responsible for friction at Bo and Fp will cause sliding of link and oppose sliding and create torque at Ao

hence here are two torque due to swinging block as follows

Tbon=Fn*R*mu=563.81*10*.1=563.81 N mm(anticlockwise)

friction force=Rf=mu*Fn=563.81*.1=56.381 N

this force will create torque around Ao,

Tbof=Rf*AOA=4792.385 N mm

4)hence resultant torque acting in clockwise direction is T12

T12=Tr-Tao-Tbon-Tbof=242969.25-600-563.81-4792.385=237013.055 N mm

hence resultant torque applied to crank is equal to T12 but in opposite direction

T12=237013.055 N mm

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