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Documentation

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:

model of a quarter of a cylinder used for cylinder under rotational force validation
Figure 1: Cylinder geometry

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:

PointXYZ
A0.019500.01
B0.020500.01
C0.020500
D0.019500
Table 1: Geometry coordinates

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.

CaseMesh TypeNumber of NodesElement Type
A1st order standard3242Standard
B1st order hexahedrons2583Standard
C2nd order standard20965Standard
D2nd order hexahedrons9285Standard
Table 2: Mesh overview
generated first order standard mesh in simscale
Figure 2: Generated first order standard mesh in SimScale
uploadeded first order hexahedron mesh of cylinder which consists of hexahedrons and has 2583 nodes used for rotational force validation
Figure 3: Uploaded first order hexahedron mesh

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:

CaseQuantityHPLA-100SimScaleError [%]
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 m2.8763e-13-0.132
A\(\sigma_{zz}(r\) = 0.0195 \(m)\)0.99488 0.990826 0.4078
A\(\sigma_{zz}(r\) = 0.0205 \(m)\)0.926310.931388-0.548
B\(u_r(r\) = 0.0195 \(m)\)2.9424e-132.94195e-130.015
B\(u_r(r\) = 0.0205 \(m)\)2.8801e-132.87966e-130.015
B\(\sigma_{zz}(r\) = 0.0195 \(m)\)0.994881.00893-1.412
B\(\sigma_{zz}(r\) = 0.0205 \(m)\)0.926310.9146451.259
C\(u_r(r\) = 0.0195 \(m)\)2.9424e-132.94238e-130.001
C\(u_r(r\) = 0.0205 \(m)\)2.8801e-132.88006e-130.001
C\(\sigma_{zz}(r\) = 0.0195 \(m)\)0.994880.995077-0.02
C\(\sigma_{zz}(r\) = 0.0205 \(m)\)0.926310.926488-0.02
D\(u_r(r\) = 0.0195 \(m)\)2.9424e-132.942373-130.001
D\(u_r(r\) = 0.0205 \(m)\)2.8801e-132.88007e-130.001
D\(\sigma_{zz}(r\) = 0.0195 \(m)\)0.994880.994995-0.012
D\(\sigma_{zz}(r\) = 0.0205 \(m)\)0.926310.926415-0.011
Table 3: Stress and displacement comparison

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

stress visualization simscale post-processor
Figure 4: Stress visualization in the SimScale post-processor

Note

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