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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

CylinderUnderRotationalForce-geometry

Geometry of the cylinder

  A B C D
x [m] 0.0195 0.0205 0.0205 0.0195
y [m] 0 0 0 0
z [m] 0.01 0.01 0 0

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 :

Case Mesh type Number of nodes Element type
(A) quadratic tetrahedral 2893 3D isoparametric
(B) quadratic hexahedral 2583 3D isoparametric
(C) quadratic tetrahedral 17460 3D isoparametric
(D) quadratic hexahedral 9285 3D isoparametric
CylinderUnderRotationalForce-mesh-a

Mesh used for the SimScale case (A)

CylinderUnderRotationalForce-mesh-b

Mesh used for the SimScale case (B)

Simulation Setup

Material:

  • isotropic: E = 200 GPa, \(\nu\) = 0.3, \(\rho\) = 8000 kg/m^3

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 \(\omega\) = 1 rad/s around the z-axis applied on the whole body

Reference Solution

\[\begin{equation}\label{ref1} u(r) = \frac{-(1+\nu)(1-2\nu)}{(1-\nu)E} \rho \omega^2 \frac{r^3}{8} + Ar + \frac{B}{r} \end{equation}\]\[\begin{equation}\label{ref2} \sigma_{zz}(r) = \frac{-\nu}{(1-\nu)} \rho \omega^2 \frac{r^2}{2} + \frac{2 \nu E}{(1+\nu)(1-2\nu)} A \end{equation}\]\[\begin{equation}\label{ref3} A = \frac{(3-2\nu)(1+\nu)(1-2\nu)}{4(1-\nu)E} \rho \omega^2 R^2 (1-x^2) = 7.13588 \cdot 10^{-12} \end{equation}\]\[\begin{equation}\label{ref4} B = \frac{(3-2\nu)(1+\nu)}{8(1-\nu)E} \rho \omega^2 R^4 (1-x^2)^2 = 3.561258 \cdot 10^{-15} m^2 \end{equation}\]\[\begin{equation}\label{ref5} x = \frac {h} {2R} = \frac {0.001m} {2 \cdot 0.02 m} = 0.025 \end{equation}\]

All stated equations used to solve the problem are derived in [HPLA100]. The parameters \(h\) and \(R\) in \(\eqref{ref5}\) are the thickness of the cross section and the radius of the middle surface of the cylinder respectively.

Results

Important

The values for the comparison of the displacement \(u_{r}\) and the stresses \(\sigma_{zz}\) are averaged over on edge and an area respectively

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].

Comparison of the displacement \(u_{r}\) and the stresses \(\sigma_{zz}\)
Case Quantity [HPLA100] SimScale Error
(A) \(u_{r}(r=0.0195 m)\) [m] 2.9424E-013 2.83491E-013 3.653%
(A) \(u_{r}(r=0.0205 m)\) [m] 2.8801E-013 2.8972E-013 -0.594%
(A) \(\sigma_{zz}(r=0.0195 m)\) [N/m^2] 0.99488 0.992259 0.263%
(A) \(\sigma_{zz}(r=0.0205 m)\) [N/m^2] 0.92631 0.929711 -0.367%
(B) \(u_{r}(r=0.0195 m)\) [m] 2.9424E-013 2.94195E-013 0.015%
(B) \(u_{r}(r=0.0205 m)\) [m] 2.8801E-013 2.87966E-013 0.015%
(B) \(\sigma_{zz}(r=0.0195 m)\) [N/m^2] 0.99488 1.00893 -1.412%
(B) \(\sigma_{zz}(r=0.0205 m)\) [N/m^2] 0.92631 0.914645 1.259%
(C) \(u_{r}(r=0.0195 m)\) [m] 2.9424E-013 2.94234E-013 0.002%
(C) \(u_{r}(r=0.0205 m)\) [m] 2.8801E-013 2.88003E-013 0.002%
(C) \(\sigma_{zz}(r=0.0195 m)\) [N/m^2] 0.99488 0.995217 -0.034%
(C) \(\sigma_{zz}(r=0.0205 m)\) [N/m^2] 0.92631 0.926616 -0.033%
(D) \(u_{r}(r=0.0195 m)\) [m] 2.9424E-013 2.94237E-013 0.001%
(D) \(u_{r}(r=0.0205 m)\) [m] 2.8801E-013 2.88007E-013 0.001%
(D) \(\sigma_{zz}(r=0.0195 m)\) [N/m^2] 0.99488 0.994995 -0.012%
(D) \(\sigma_{zz}(r=0.0205 m)\) [N/m^2] 0.92631 0.926415 -0.011%