Torsion of a Circular Bar

 

Example:  The stepped shaft ABC shown below has a material with shearing modulus of

elasticity, G, of 11 x 106 psi.  Two torques,  -Ti  and -(3/2)Ti  lbin. act on the shaft.  The

following data apply:  T = 6000 lb in.,  L1 = 25 in., L2 = 18 in. d1 = 2.5 in., and d2 = 2 in. 

Find the angle of twist at C (the free end of the stepped shaft),  φC  .

                       

 

Strategy:  Use a free body diagram to determine the torque in each section of the stepped

shaft.  Then apply the relation for angle of twist    φ= TL/JG   for each section to determine

the angle of twist related to each section.  For section 1 between A and B:

                           

 

→ΣMx = 0    - T1 + (3/2)T + T = 0   and for the data:  T1 = 15000 lb in

 

Now for the section between A and B  J1 = π(d1/2)4 / 2 = 3.835 in4   

 

              φ1  =  T1 L1 / J1 G1  =  15000(25)/[(3.835)( 11 x 106)  =   0.0089 radians

                                                                                              (angle of twist at B relative to A)

                            

 

→ΣMx = 0    - T2 + T = 0   and for the data:  T1 = 6000 lb in

 

Now for the section between B and C  J2 = π(d1/2)4 / 2 = 1.571 in4  

 

           φ 2  =  T2 L2 / J2 G2  =  6000(18)/[(1.571)( 11 x 106)  =  0.00625 radians

                                                                                          (angle of twist at C relative to B)

 

The total angle of twist at C is then φ1  + φ2  =  0.015 radians      (result)

 


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