Validation Case: Cylinder Under Rotational Force

This validation case belongs to solid mechanics. The aim of this test case is to validate the following parameter:

• Centrifugal force

The simulation results of the cloud-based simulation platform SimScale were compared to the results presented in [HPLA100]${}^1$.

Geometry

The geometry used for the case is as follows:

The solid body is created by rotating face ABCD by 45° and due to the symmetry of the cylinder, only one-quarter of the cylinder is modeled.

The coordinates of points A, B, C, and D can be seen below:

Point	X	Y	Z
A	0.0195	0	0.01
B	0.0205	0	0.01
C	0.0205	0	0
D	0.0195	0	0

Analysis Type and Mesh

Tool Type: Code_Aster

Analysis Type: Static

Mesh and Element Types:

The mesh was created with the Standard meshing algorithm in SimScale.

Case	Element Type	Number of Nodes	Element Technology
A	1st Order Tetrahedral	3242	Standard
B	2nd Order Tetrahedral	20965	Reduced Integration

Simulation Setup

Material:

• Steel (linear elastic)
  • Isotropic: $E$ = 200 $GPa$
  • $\nu$ = 0.3
  • $\rho$ = 8000 $kg/m^3$

Initial and/or Boundary Conditions:

• Constraints:
  • $d_z$ = 0:
    • Face AA’BB’
    • Face CC’DD’
  • Symmetry:
    • Face ABCD
    • Face A’B’C’D’
• Load:
  • Centrifugal force with a rotational speed $\omega$ of 1 $rad/s$ around the z-axis applied to the whole body.

Reference Solution

The reference solutions for stress and displacement is calculated with the equations below:

$$\begin{array}{}\text{(1)}& u(r)=\frac{-(1+\nu )(1-2\nu )}{(1-\nu )E}\rho {\mathrm{\Omega }}^2\frac{r^3}{8}+Ar+\frac{B}{r}\end{array}$$

$$\begin{array}{}\text{(2)}& {\sigma }_{zz}(r)=\frac{-\nu }{1-\nu }\rho {\mathrm{\Omega }}^2\frac{r^2}{2}+\frac{2\nu E}{(1+\nu )(1-2\nu )}A\end{array}$$

$$\begin{array}{}\text{(3)}& A=\frac{(3-2\nu )(1+\nu )(1-2\nu )}{4(1-\nu )E}\rho {\mathrm{\Omega }}^2{R}^2(1-x^2)=7.13588\times {10}^{-12}m{m}^2\end{array}$$

$$\begin{array}{}\text{(4)}& B=\frac{(3-2\nu )(1+\nu )(1-2\nu )}{8(1-\nu )E}\rho {\mathrm{\Omega }}^2{R}^4(1-x^2{)}^2=3.561258\times {10}^{-15}m{m}^2\end{array}$$

$$\begin{array}{}\text{(5)}& x=\frac{h}{2R}=\frac{0.001m}{2\times 0.02m}=0.025\end{array}$$

Term $h$ is the thickness of the cross-section and $R$ is the radius of the middle surface of the cylinder and both are in meters ($m$).

Result Comparison

The rotational force is validated by comparing the displacement $u_r$ in meters $m$ and Cauchy stresses ${\sigma }_{zz}$ in $N/m^2$ obtained from SimScale against the reference results obtained from [HPLA100] is given below:

Case	Quantity	HPLA-100	SimScale	Error [%]
A	$u_r(r$ = 0.0195 $m)$	2.9424e-13	2.939e-13	-0.109
A	$u_r(r$ = 0.0205 $m)$	2.8801e-13	2.877e-13	-0.113
A	${\sigma }_{zz}(r$ = 0.0195 $m)$	0.99488	0.990826	-0.407
A	${\sigma }_{zz}(r$ = 0.0205 $m)$	0.92631	0.931388	+0.548
B	$u_r(r$ = 0.0195 $m)$	2.9424e-13	2.942E-13	-0.001%
B	$u_r(r$ = 0.0205 $m)$	2.8801e-13	2.880E-13	-0.001%
B	${\sigma }_{zz}(r$ = 0.0195 $m)$	0.99488	0.995056	+0.018
B	${\sigma }_{zz}(r$ = 0.0205 $m)$	0.92631	0.926469	+0.017

The stress experienced by the cylinder under the rotational force can be seen below:

Last updated: November 16th, 2022