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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 the cloud-based simulation platform SimScale were compared to the results presented in [HPLA100]\(^1\).


    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:

    Table 1: Geometry coordinates

    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.

    CaseElement TypeNumber of NodesElement Technology
    A1st Order Tetrahedral3242Standard
    B2nd Order Tetrahedral20965Reduced Integration
    Table 2: Mesh overview
    generated first order standard mesh in simscale
    Figure 2: Generated first order standard mesh in SimScale

    Simulation Setup


    • 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:

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

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

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

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

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

    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 2.939e-13-0.109
    A\(u_r(r\) = 0.0205 \(m)\)2.8801e-132.877e-13-0.113
    A\(\sigma_{zz}(r\) = 0.0195 \(m)\)0.99488 0.990826 -0.407
    A\(\sigma_{zz}(r\) = 0.0205 \(m)\)0.926310.931388+0.548
    B\(u_r(r\) = 0.0195 \(m)\)2.9424e-132.942E-13-0.001%
    B\(u_r(r\) = 0.0205 \(m)\)2.8801e-132.880E-13-0.001%
    B\(\sigma_{zz}(r\) = 0.0195 \(m)\)0.994880.995056+0.018
    B\(\sigma_{zz}(r\) = 0.0205 \(m)\)0.926310.926469+0.017
    Table 3: Stress and displacement comparison

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

    stress visualization of a quarter cylinder in the simscale post-processor


    If you still encounter problems validating your simulation, then please post the issue on our forum or contact us.

    Last updated: November 16th, 2022