Deformation of a Circular Bar

 

Example:  (continued)  

 

Now for each segment of the bar the  axial deformation  is    δi = PiLi/AiEi  

and the sum of all deformations must add to zero.

 

                                      δ1  +  δ2  +  δ3  =  0

 

which gives                   P1L1/AE  +  P2L2/AE   +  P3L3/AE     =  0

 

or               

                      FABL1/AE  +  FBCL2/AE   +  FCDL3/AE     =  0

 

                     (24 -F1)(1.0)/AE  +  24(0.4)/AE   +  9(0.6)L3/AE     =  0

 

       F1 =  24 + 24(0.4) + 9(0.6)  =  39 kN       (result)

 

 

 

Note that the force in the section of the bar between A and B is  FAB = 24 – 39 = - 15 kN.

So that portion of the bar is in compression.

 

The other two sections (between B and C  and  between  C and D ) are in tension.

FBC=  24 kN   and  FCD = 9 kN

 

  

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