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 maximum shear stress in the stepped shaft.

 

                        torsionexamplea.jpg

 

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

shaft.  Then apply the relation for shear stress    τ = Tc/J for each section to determine

which shearing stress is the maximum.  For section 1 between A and B:

                            torxmpfbd.jpg

 

→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  

 

               τ1max   =    T1 (d1/2) / J1  =   15000(1.25)/3.835  =  4890  psi

 

                             torxmpfbd1.jpg

 

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

 

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

 

                     τ2max   =   T2 (d2/2) / J2    =   6000(1.0)/1.571  =  3820  psi

 

So              τ1max = 4890 psi  and  τ2max = 3820 psi

 

The maximum shearing stress in the stepped shaft is then  4890 psi (in section 1).   (result)

 


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