A 4-bar mechanism has following dimensions:

l(DA)= 300 mm l(CB) = l(AB) = 360 mm

l(DC) = 600 mm. The link ‘DC’ is fixed. The angle ADC

is 60°

The driving link ‘DA’ rotates at a speed of 100 rpm

clockwise and constant driving torque is 50 N.M. Calculate

the Velocity of point ‘B’ and angular velocity of driven link

‘CB’.

A 4-bar mechanism has following dimensions:

l(DA) = 300 mm, l(CB) = l(AB) = 360mm, l(DC) = 600 mm. the link ‘DC’ is fixed. The

angle ADC is 60o. The driving link ‘DA’ rotates at a speed of 100 rpm clockwise and

constant driving torque is 50 N-m. Calculate the velocity of point ‘B’ and angular

velocity of driven link ‘CB’.

Given Data:

N AD = 100 rpm

DA = 300 mm = 0.3 m

T A = 50 N-m

ω AD = 2 π x 100 /60 = 10.47 rad/sec.

Velocity of A w.r.t. D (V AD);

Velocity of Point B:

[1] Since the link DC is fixed, therefore points d and c are taken as one point in the

velocity diagram. Draw vector da perpendicular to DA, to some suitable scale, to

represent the velocity of A with respect to D or simply velocity of A (i.e. V AD or V A ) such

that,

Vector da =V AD = V A = 3.14 m/s

[2] Now from point a, draw vector ab perpendicular to AB represents the velocity of B

with respect to A (i.e. V BA ), and from point c draw vector cb perpendicular to CB to

represent the velocity of B with respect to C or simply velocity of B (i.e. V BC or V B ). The

Vectors ab and cb intersect at b.

[3] By measurement, we find that velocity of point B,

V B = V BC = vector cb = 2.25 m/s

[4] Angular Velocity of driven link CB

Since CB = 360 mm =0.36 m, therefore angular velocity of the driven link CB,

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