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DYNAMICS - EXAMPLE


Three Bar Linkage Problem
  Example

 

Bar AB of a classic three bar linkage system has an angular velocity of 5 rad/s. The lengths of the bars AB and DC are each 3 ft. What is the angular velocity of bar BC?

 

   
    Solution


 

This is a three step process. First find the velocity of point B. Then find the direction velocity direction of point C. Finally, the angular velocity of bar BC can be determined the information known about point B and C.

Velocity of point B,
     vB = ω × rB/A
         = 5k × 3(cos 45i + sin 45j)
         = -10.61i + 10.61j ft/s

Velocity of point C,
     vC = ωCD × rC/D
         = ωCDk × 3(-cos 60i + sin 60j)
         = -1.5ωCDj - 2.598ωCDi ft/s

To understand the motion of Bar BC, the position vector, rC/B needs to be identified.
     rC/B =(6 - 3cos45 - 3cos60)i + (3sin60 - 3cos45)j
           = 2.379i + 0.4768j

Angular velocity of bar BC is found by relating the velocity of point C to point B,
     vC = vB + vC/B = vB + ωBC × rC/B
     -1.5ωCDj - 2.598ωCDi = -10.61i + 10.61j +
                 ωBCk × (2.379i + 0.4768j)
     -2.598 ωCDi -1.5 wCDj = -10.61i + 10.61j +
                 2.379ωBCj - 0.4768ωBCi

Solving i and j terms,
     -2.598ωCD = -10.61 - 0.4768ωBC
     -1.5ωCD = 10.61 + 2.379ωBC

Solving equations for ωBC,

ωBC = -6.305 rad/s

     
   
 
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