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MECHANICS - CASE STUDY SOLUTION


Drilling Pipe Angle of Twist

 

The drilling shaft is constructed from two different materials, steel and aluminum. Each material section is 0.25 inches thick with the steel on the outside. The applied torque is 15 kip-ft (equal to 15,000 lb-ft) and is constant along the 100 foot pipe length.

As the diagram at the left shows, the angle of twist, θ, will be the same for both materials at any given location along the pipe. This condition is critical in solving for the stresses.

   
  Shear Stiffness and
Polar Moment of Inertia

   

Before developing the basic equations, there are several constants that are needed. The first is the shear stiffness, G. Recall, G can be calculated from E and ν, giving

     

The polar moment of inertia call also be calculated for each material section.

     

     
    Torque Distribution in Each Section


Shaft Cross Section

 

The drilling shaft maintains a torque or moment of 15 kip-ft but the amount of the torque in each of the two sections is not known. Summing the moments around the pipe axis gives

     TA + TS = Ttotal = 15 kip-ft

Thus, there are two unknowns but only one equation from static equilibrium. A second equation can be developed from the angle of twist compatibility relationship. Each section must twist the same, or

      θsteel = θalum

Recall, the angle of twist for any circular shaft is

     

Using this in the compatibility equation gives

     

Rearranging and substituting values gives

     

     TS = 4.586 TA

This indicates the steel section takes almost 80% of the torque. Thus, the aluminum will not help much. This result can be substituted back into the moment equilibrium equation, giving

     TA + 4.586 TA = 15,000 lb-ft

     TA = 2,685 lb-ft

     TS = 12,310 lb-ft

     
    Stresses

   

The stresses in each of the two sections can now be determined using the torsion stress equation.

     

The maximum stress occurs at the outer edge of each section

     

Shear Stress in Two-Part Drilling Pipe
 

     tS-max = 14.2 (2) = 28.40 ksi

     tA-max = 4.752 (1.75) = 8.316 ksi

To better understand the stress distribution, the diagram at the left shows the shear stress as a function of the radius. Notice, there is a discontinuity at the interface between the material sections. The angle of twist and strain is equal at the interface, but the stresses are not.

     
   
 
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