Example - Analysis of a Frame/Machine

 

Note that the crank, AB, is a frame element since it supports bending due to the applied moment whereas the connecting rod, BC, is a two-force member (meaning a truss element).  So the resultant force in the connecting rod acts along the member BC.

 

The free body diagram (FBD #1) of the entire system (crank + connecting rod + pistion) is:

 

                              

With the Data:  d = 50 mm, e = 75 mm, f = 175 mm, and  M = 1.5 kNm.

 

From equilibrium   Σ MA  =  0      M  -  Cy (e + f)  =  0  so  Cy  = M / (e + f) 

 

                                   Cy  =  6.0 kN  (result)

 

The free body diagram (FBD #1) of the connecting rod + pistion is:

 

                         

 

Now  tan θ  =  d/f  =  0.286  so  θ  =  15.945o   and  Σ Fy  =  0  gives  - FBC sin θ + 6 = 0

 

So    FBC = 21.84 kN   and  Σ Fx  =  0  gives   21.84 cos θ – P  =  0

 

Or  P  =  21.84 cos θ      giving the result as:      P  =  21.0 kN

 

 

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