Torsion of a Tube with Ring Cross-Section

Let us consider a shaft with circular cross-section of radius R. Length of shaft is L (see figure).

Select the coordinate system with the z-axis directed along the axis of the shaft, and the coordinate z=0 located at the left edge of the tube.
The tube is subjected to the externally applied torque τ. The torque is applied at the right end of the tube, the left end of the tube is rigidly clamped.

Let us use the following initial data: length L is 0.6 m, external diameter of cross-section d1 is 0.036 m, internal diameter of cross-section d2 is 0.044 m, the magnitude of the applied torque τ is 100 N-m.
Material characteristics: E = 2.1E+011 Pa, G=8.203E+010 Pa, ν = 0.28.
To find the angle of twist, let us use the following relation:
,
where ϕ0 – angle of twist of the cross-section z=0, G – shear modulus, Jp – polar moment of inertia of the ring cross-section.
Since by formulation of the task, the left end is clamped, ϕ0=0. Then, at a distance z=0.5L from the clamped edge, the angle of twist ϕ is given by the formula:
,

Thus, ϕ0.5L= 1.8009E-003 rad; Absolute value of displacement (at z=0.5L) Δumax=3.60181E-005 m.
After carrying out calculation with the help of AutoFEM, the following results are obtained:

Table 1.Parameters of finite element mesh

 

Table 2. Result "Displacement" *

Table 3. Result "Rotation, Y" *

 

 

*The results of numerical tests depend on the finite element mesh and may differ slightly from those given in the table.

 

Read more about AutoFEM Static Analysis

autofem.com

Return to contents