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Bars and Trusses                                                             33



                                            2  − 2  0  u 1    F 1 
                                                      
                                       EA  − 2  3  − 1  u 2 =  
                                                         
                                                             
                                                             F 2
                                        L     −   1         
                                            0  1      u 3    F 3 
                  Load and boundary conditions (BCs) are
                                           u 1 =  u 3 =  0,  F 2 =  P


                  FE equation becomes

                                            2  − 2  0   0    F 1 
                                       EA  − 2  3  −       
                                                              P
                                                      
                                        L          1  u 2 =  
                                                      
                                                             
                                                         
                                            0  − 1  1    0     F 3 
                  “Deleting” the first row and column, and the third row and column, we obtain
                                             EA  3 []{}  P
                                                   u 2 = {}
                                              L
                  Thus,

                                                    PL
                                               u 2 =
                                                    3 EA

                 and

                                              u 1     0
                                                  PL  
                                              u 2 =   1 
                                                 3 EA   
                                                        0
                                              u 3     
                  Stress in element 1 is

                                                                 1 u  
                                   σ= E  ε= EBu   = E  −  1 L  1 L    
                                                       /
                                                             /
                                                               
                                     1
                                          1
                                               11
                                                                 2 u   
                                          2 u  − u 1  E   PL    P
                                      = E      =       − 0 =
                                                          
                                                           
                                           L     L   3EA     3 A
                  Similarly, stress in element 2 is
                                                                 2 u  
                                   σ= ε= EBu     2  = E  −  1 L  1 L    
                                        E
                                                        /
                                                             /
                                                               
                                               2
                                          2
                                     2
                                                                 3 u   
                                          3 u  − u 2  E   PL    P
                                      = E       =    0  −    =−
                                           L     L    3EA   3 A
                 which indicates that bar 2 is in compression.
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