Deformation of a Circular Bar (Statically Indeterminate) – Using Superposition

 

Example:  Bar  ABCD  shown below is clamped at both ends and subjected to applied loads
P and 2P.    The bar has cross-sectional area, A and modulus of elasticity is E.  Find the
restraint loads at each end and the maximum normal stress in the bar.                                                       
                             

 

Strategy:  Use a free body diagram to identify the forces of constraint at A and at D (shown
in the top figure below.)  Next replace the original bar with three bars the first loaded by  P,

the second loaded by 2P, and the third loaded by FD (as shown in the bottom three figures.)

Then apply the load-deflection relation given by  ∆i = FiLi/AiEi  for each of the three separate

bars.  The geometrical relation is that the net deflection is zero.  Solve for the unknown reactions.

Finally pass sections to determine the axial force in each section of the bar.  Use it to calculate

the normal stress in each section as force per unit area.

 

                      

 

For equlilibrium,

 

          →ΣFx = 0    – FA – P  + 2P  +  FD  = 0   and for the data:         FA  –  FD  =  P       (1)

 

Let   ∆  denote axial deflection.  Now the total deflection sums to zero.

                                        ∆1  +   ∆2  +   ∆3  =  0

 

So                    – P(L/2)/AE  +  (2P)(3L/4)/AE  +  (FD)(L)/AE  =  0

 

which yields   FD  =  – P    (as before)    and from (1)   FA  =  0           as before.

 

The axial stresses in each section is the same as before (force in section/area).

 

Click here for another example.

 

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