Torsion of a Circular Bar (Statically Indeterminate)

 

Example:  (continued)

                                                             
                             

 

 

 

From equilibrium,

                                                 TA  –  TD  =  P       (1)

 

From equilibrium for each section,

 

For A to B:  T1  =  TA,        For B to C:  T2  =  TA + T,       For C to D:  T3  =  TA - T

 

The relative rotations for each section are given by   Ti Li / JG

 

                          ∆B/A  =  TA (L/2) / JG

 

                          ∆C/B  =  (TA + T)(L/4) /JG

 

                          ∆D/C   = (TA – T)(L/4) / JG

 

The sum of the relative rotations must add to zero.  (Geometric condition)

 

          ∆B/A  +  ∆C/B   + ∆D/C   =  0         (Cancel  L, J, and G to simplify.)

 

                  TA /2  +  TA /4  +  T/4 +   TA /4   – T/4  =  0

 

So   TA  =  0   and  by (1)       TD  =  – T     (result for support torques)

 

So the axial stresses in each section are  (for a bar with radius  c )

 

 

     τ1  =  0 ,             τ2  =  Tc/J  ,         τ3  =  – Tc/J

 

 

Thus the maximum shear stress equals  Tc/J  and occurs between B and C.   (result)

 

Click here to solve this problem again using superposition this time.

     


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