Beam Deflection (Statically Indeterminate)

 

Example:  (continued)  From geometry   y1 + y2  =  0  (See figure below.)

Note:  Mo  is the applied external moment whereas  FB  is the unknown support reaction

at B.

            

 

So         MoL2 / 2EI  +  FBL3 / 3EI  =  0    giving       FB  =  − 3Mo/2L        (result)

               

 

The final step, returning to equilibrium to determine the support reactions at A from the

FBD shown below.

 

       →  Σ Fx  =  0                                                             Ax  =  0     (result)

 

    ↑  Σ Fy  =  0    Ay  +  FB  =  0     So                             Ay  =  − FB   =   3Mo/2L      (result)                     

 

CCW  Σ MA  =  0   − MA + Mo  +  (FB)L  =  0   and        MA  =   Mo  +  (FB)L        

 

  Or    MA  =  Mo  − (3Mo/2L) L  =    − Mo/2L                  MA  =   − Mo/2     (result)

 

 

                          

 

 

                     

So the moment at A is acts counterclockwise.

 

Click here for another example.

 

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