Example: EQS5
Four-bar linkage kinematic analysis


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For [inline math] and φ3(0) = 0.789 [rad], φ4(0) = 4.492 [rad] solve the algebraic equations characterizing kinematics of the four-bar mechanism:

0 = r2 cos (φ2) + r3 cos (φ3) – r4 cos (φ4) – r1
0 = r2 sin (φ2) + r3 sin (φ3) + r4 sin (φ4)

Position of the critical point T in the (x, y)-plane can be computed by evaluating the expressions:

xT = r2 cos (φ2) + r3/2 cos (φ3)
yT = r2 sin (φ2) + r3/2 sin (φ3)

System excitation

The mechanism is driven by the constant angular velocity of bar 2

φ2 = 2πt

System Parameters

r10.6r1 = 0.6[m]distance of hinges
r20.3r2 = 0.3[m]length of bar 2
r30.55r3 = 0.55[m]length of bar 3
r40.4r4 = 0.4[m]length of bar 4

Last Update

March 14, 2008