Torsion of a Shaft with Circular 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 shaft.
The shaft is subjected to the externally applied torque τ. The torque is applied at the right end of the shaft, the left end of the shaft is rigidly clamped.

Let us use the following initial data: length L of the shaft is 0.6 m, radius of cross-section R of the shaft is 0.02 m, the magnitude of the applied torque τ is 100 N-m.
Material characteristics: E = 2.1E+011 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=E/2(1+ν) – shear modulus, Jp=πR4 / 2 – polar moment of inertia of the circular cross-section.
Since by formulation of the task, the left end of the shaft is clamped, ϕ0=0. Then, at a distance z=0.5L from the clamped edge of the shaft, the angle of twist ϕ is given by the formula:

Thus, ϕ0.5L= 1.4551E-003 rad.
After carrying out calculation with the help of AutoFEM, the following results are obtained:

Table 1.Parameters of finite element mesh

Absolute value of displacement (at z=0.5L) Δu = 2.9488E-005 m. (mean at Sensor 1-4).

Table 2. Result "Angle of twist" *

Conclusions:

The relative error of the numerical solution compared to the analytical solution is 1.33%.

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

 

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