Cylinder Under Rotational Force

Overview

The aim of this test case is to validate the following functions:

• centrifugal force

The simulation results of SimScale were compared to the numerical results presented in [HPLA100]. The mesh used in (A) and (C) was created with the automatic-tetrahedralization-tool on the SimScale platform. The mesh used in (B) und (D) was created locally.

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Geometry

To obtain the solid body the face ABCD is rotated 45° around the z-axis. Because of the symmetry of the cylinder only one quarter was modelled.

Analysis type and Domain

Tool Type : Code_Aster

Analysis Type : Static

Mesh and Element types :

Simulation Setup

Material:

• isotropic: E = 200 GPa, ν = 0.3, ρ = 8000 kg/m³

Constraints:

• Face AA’BB’ and face CC’DD’ zero z-displacement
• Symmetry boundary condition on face ABCD and A’B’C’D’

Loads:

• Centrifugal force with a rotational speed = ω = 1 rad/s around the z-axis applied on the whole body

Reference Solution

$$u(r) = \frac{-(1+\nu)(1-2\nu)}{(1-\nu)E} \rho \omega^2 \frac{r^3}{8} + Ar + \frac{B}{r}$$

$$\sigma_{zz}(r) = \frac{-\nu}{(1-\nu)} \rho \omega^2 \frac{r^2}{2} + \frac{2 \nu E}{(1+\nu)(1-2\nu)} A$$

$$A = \frac{(3-2\nu)(1+\nu)(1-2\nu)}{4(1-\nu)E} \rho \omega^2 R^2 (1-x^2) = 7.13588 \cdot 10^{-12}$$

$$B = \frac{(3-2\nu)(1+\nu)}{8(1-\nu)E} \rho \omega^2 R^4 (1-x^2)^2 = 3.561258 \cdot 10^{-15} m^2$$

$$x = \frac {h} {2R} = \frac {0.001m} {2 \cdot 0.02 m} = 0.025$$

All stated equations used to solve the problem are derived in [HPLA100]. The parameters h and R in (5) are the thickness of the cross section and the radius of the middle surface of the cylinder respectively.

Results

Comparison of the displacement U_r and the stresses \(\sigma_{zz}\) of the inner and the outter face obtained with SimScale and the results derived from [HPLA100].

References

Last updated: January 29th, 2019

What's Next

part of: Design analysis of spherical pressure vessel