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8                     Finite Element Modeling and Simulation with ANSYS Workbench



              Adding the two matrix equations (i.e., using superposition), we have

                                    k 1  − k 1   0  u 1     f 1 1  
                                   −    k 1 +   −        f 2 +  2 
                                                              1
                                                    
                                    k 1     k 2   k 2  u 2 =   f 1 
                                                                  1
                                                        
                                                     
                                                    
                                    0    −       k 2  u 3     2 f  2  
                                          k 2                     
            This is the same equation we derived by using the concept of equilibrium of forces.
            1.2.2.2  Assembly of Element Equations: Energy Approach
            We can also obtain the result using an energy method, for example, the principle of mini-
            mum potential energy. In fact, the energy approach is more general and considered the
            foundation of the FEM. To proceed, we consider the strain energy U stored in the spring
            system shown in Figure 1.5.

                                    1       1        1         1
                                                                   T
                                U =   1 k ∆ 1 2  +  2 k ∆ 2 2  =  ∆ 1 k ∆ 1  +  ∆ 2 k ∆ 2
                                                        T
                                                          1
                                                                    2
                                    2       2        2         2
              However,
                                             u 1                        u 2 
                         ∆ 1 = u 2 − u 1 =−[  1  1  ]   ,  ∆ 2 = u 3 − u 2 =−[  1  1  ]  
                                              u 2                        u 3 

              We have

                                      1 k
                                                                  
                     1         k 1  −  1 u    1      k 2  − 2 k u 2 
                U =   [u 1  u 2 ]         +  [u 2  u 3 ]         = (enlarging …)
                                                                    u
                     2         − 1 k  1 k   2 u    2   − 2 k  2 k   3 
                                     k 1  −k 1    0   1 u 
                     1                                
                  =   [ u 1  u 2  u 3 ]  −k 1  k 1  + k 2  −k 2   2 u                (1.8)
                     2                                
                                     0    −k 2 2  2 k    3 u  
              The potential of the external forces is

                                                                     
                                                                     F 1
                                                       [
                               Ω= −Fu   − Fu  − Fu  = − u 1  u 2  u 3 ]               (1.9)
                                                                     
                                                                     F 2
                                           22
                                     11
                                                 33
                                                                     
                                                                     F 3
                                                                     
              Thus, the total potential energy of the system is
                                            k 1   − 1 k   0   1 u             
                                                                                  F 1
                            1                                                 
                Π = U  +  Ω =  [u 1  2 u  u 3 ]  − 1 k  1 k  + k 2  − 2 k   2 u  − [u 1  u 2  u 3  ] F 2  (1.10)
                                                                                  
                            2                                                 
                                           
                                                             
                                                              
                                                                                  F 3
                                            0    − 2 k    2 k   u 3            
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