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MECHANICS - EXAMPLE


Thin-walled circular tube
  Example

 

A simple 5 ft long circular steel tube is subjected to a torque (moment) of 5 kip-ft. The radius of the pipe (to the mid-point) is 1 inch and the wall thickness is 0.2 inch. What is the shear stress in the tube? Compare the shear stress based on theory developed in section "Circular Bars and Shafts" to the "Thin-walled Tube" theory developed in this section.

   
    Solution


Circular tube Cross Section
 

Shear Stress Based on Thin-walled Theory

First, it is good to check if this is a thin walled tube. The thickness is 0.2 inches and the overall dimension of the tube is 2 inches. Thus, the thickness is roughly 10% of the overall cross section. This is on the border of being thin-walled. The solution should be assumed to be approximate, and not exact.

The shear flow for a thin-walled tube is

     q = T/(2Ao)

For this tube, the area is

     Ao = π rmid2 = π 12 = π in2

Thus, the shear flow is

     q = T/(2Ao) = (5 kip-ft)(12 in/ft)/(2 π in2) =

        = 9.549 kip/in

The shear stress is

     τ = q/t = 9.549/0.2 = 47.75 ksi

     

 

 

 

 

Shear Stress Based on Circular Bar Theory

Shear stress based on circular bar theory is considered exact for circular bars and shafts. It will be interesting to see how well thin-walled theory compares.

The shear stress for a circular bar is given by

      τ = Tro/J

where J is the polar moment of inertia and ro is the outside radius. For this example, the polar moment of inertia is

     J = π ro4/2 - π ri4/2 = π [(1+0.2/2)4 - (1-0.2/2)4]/2

       = 1.269 in4

Thus, the shear stress is

     τ = (5 kip-ft)(12 in/ft)(1.1 in) / (1.269 in4)

       = 52.01 ksi

This is about 9% higher than the value calculated from thin-walled theory. Thus, for thin-walled theory to be accurate, the tube thickness should be less than 10% of the overall dimension.