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 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: Dynamic

Mesh and Element Types:

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

Case	Mesh Type	Number of Nodes	Element Type
A	1st order standard	3242	Standard
B	1st order hexahedrons	2583	Standard
C	2nd order standard	20965	Standard
D	2nd order hexahedrons	9285	Standard

Simulation Setup

Material:

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

Intial 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:

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

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

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

$$B = \frac{(3-2\nu)(1+\nu)(1-2\nu)}{8(1-\nu)E}\rho\Omega^2R^4(1-x^2)^2 = 3.561258\times 10^{-15} mm^2\tag{4}$$

$$x = \frac{h}{2R} =\frac{0.001m}{2\times0.02m} = 0.025\tag{5}$$

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	2.9430e-13	-0.020
A	\(u_r(r\) = 0.0205 \(m)\)	2.8801e-13 m	2.8763e-13	-0.132
A	\(\sigma_{zz}(r\) = 0.0195 \(m)\)	0.99488	0.990826	0.4078
A	\(\sigma_{zz}(r\) = 0.0205 \(m)\)	0.92631	0.931388	-0.548
B	\(u_r(r\) = 0.0195 \(m)\)	2.9424e-13	2.94195e-13	0.015
B	\(u_r(r\) = 0.0205 \(m)\)	2.8801e-13	2.87966e-13	0.015
B	\(\sigma_{zz}(r\) = 0.0195 \(m)\)	0.99488	1.00893	-1.412
B	\(\sigma_{zz}(r\) = 0.0205 \(m)\)	0.92631	0.914645	1.259
C	\(u_r(r\) = 0.0195 \(m)\)	2.9424e-13	2.94238e-13	0.001
C	\(u_r(r\) = 0.0205 \(m)\)	2.8801e-13	2.88006e-13	0.001
C	\(\sigma_{zz}(r\) = 0.0195 \(m)\)	0.99488	0.995077	-0.02
C	\(\sigma_{zz}(r\) = 0.0205 \(m)\)	0.92631	0.926488	-0.02
D	\(u_r(r\) = 0.0195 \(m)\)	2.9424e-13	2.942373-13	0.001
D	\(u_r(r\) = 0.0205 \(m)\)	2.8801e-13	2.88007e-13	0.001
D	\(\sigma_{zz}(r\) = 0.0195 \(m)\)	0.99488	0.994995	-0.012
D	\(\sigma_{zz}(r\) = 0.0205 \(m)\)	0.92631	0.926415	-0.011

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

Last updated: December 8th, 2020