This validation case belongs to vibrations and the elastic support boundary condition in solid mechanics. The aim of this test case is to validate the following parameters:

- Elastic support

The simulation results of SimScale were compared to the results derived from [Schaum]\(^1\).

The geometry used for the case is as follows:

The cube has an edge length of 1 \(m\), with the upper face partitioned in half.

**Tool Type**: Code Aster

**Analysis Type**: Static Linear and Dynamic

Cases corresponding to analysis type are as follow:

Case | Analysis Type |
---|---|

A-1 | Static Linear |

A-2 | Dynamic |

B-1 | Static Linear |

B-2 | Static Linear |

B-3 | Static Linear |

**Mesh and Element Types**:

Tetrahedral meshes were computed using SimScale’s standard mesh algorithm and manual sizing.

Case | Mesh Type | Number of Nodes | Number of Elements | Element Type |
---|---|---|---|---|

A-1 | 1st Order Tetrahedral | 45 | 97 | Standard |

A-2 | 1st Order Tetrahedral | 45 | 97 | Standard |

B-1 | 1st Order Tetrahedral | 128 | 405 | Standard |

B-2 | 1st Order Tetrahedral | 128 | 405 | Standard |

B-3 | 1st Order Tetrahedral | 128 | 405 | Standard |

**Material**:

- Linear Elastic Isotropic:
- \( E = \) 205 \(GPa \)
- \( \nu = \) 0.28
- \( \rho = \) 10 \(kg/m^3 \)

**Boundary Conditions**:

- Constraints:
- Case A-1/A-2:
- Total isotropic spring stiffness \( K = \) 9810 \(N/m\) on face EFGH

- Case B-1:
- Total isotropic spring stiffness \( K = \) 4905 \(N/m\) on face EIGJ
- Total orthotropic spring stiffness \( K_x = K_y = K_z = \) 4905 \(N/m \) on face IFJH

- Case B-2:
- Distributed isotropic stiffness \( K/A = \) 9810 \(N/m^3\) on face EIGJ
- Distributed orthotropic stiffness \( K_x/A = K_y/A = K_z/A = \) 9810 \(N/m^3 \) on face IFJH

- Case B-3:
- Total isotropic spring stiffness \( K = \) 1962 \( N/m \) on face EIGJ
- Total orthotropic spring stiffness \( K_x = K_y = K_z = \) 1962 \( N/m^3 \) on face IFJH
- Distributed isotropic spring stiffness \( K/A = \) 3924 \( N/m^3 \) on face EIGJ
- Distributed orthotropic spring stiffness \( K_x/A = K_y/A = K_z/A = \) 3924 \( N/m^3 \)
- Total isotropic spring stiffness of \( K = \) 1962 \( N/m \) on face EFGH

- Case A-1/A-2:
- Loads:
- Self weight with gravitational acceleration \( g = \) 9.81 \( m/s^2 \) in the \( -Z \) direction.

The following table summarizes the elastic support boundary conditions by case:

Case | Elastic Support Type on Face EIGJ | Elastic Support Typeon Face IFJH | Elastic Support Typeon Face EFGH |
---|---|---|---|

A-1 | – | – | Isotropic Total |

A-2 | – | – | Isotropic Total |

B-1 | Isotropic Total | Orthotropic Total | – |

B-2 | Isotropic Distributed | Orthotropic Distributed | – |

B-3 | Isotropic Total+Distributed | Orthotropic Total+Distributed | Isotropic Total |

The analytical solutions for the rotation angle \(\theta_B\) and maximum shear stress \(\tau_{max}\) are given by the following equations:

Cases A-1, B-1, B-2, B-3:

\( x = \frac{mg}{k} \tag{1} \)

Case A-2:

\( x(t) = V_0 \omega Sin(\omega t) + X_0 Cos(\omega t) \tag{2} \)

\( \omega = \sqrt{ k / m} \tag{3} \)

\( V_0 = -0.01 m/s \)

\( X_0 = -0.02 m \)

\( X_eq = -0.01 m \)

\( 2 <= t <= 4 \)

The computed reference solution is:

\( x_{static} = 0.01\ m \)

\( \omega = 31.32\ Rad/s \)

\( x(t) = (-3.193*10^{-4})Sin(31.32t) – 0.01Cos(31.32t) \tag{4} \)

*\( x(t) \) corresponding to the displacement with respect to the equilibrium position.

Comparison of displacement DZ on static cases:

CASE | DZ | REF | ERROR |
---|---|---|---|

A-1 | 0.01 | 0.01 | 0 % |

B-1 | 0.01 | 0.01 | 0 % |

B-2 | 0.01 | 0.01 | 0 % |

B-3 | 0.01 | 0.01 | 0 % |

Comparison of transient displacement of face ABCD in dynamic case can be found in Figure 3. Here the vibrations from the elastic support condition can be appreciated.

References

- (2011)”McGraw-Hill Schaum’s outlines, Engineering Mechanics: Dynamics”, pg 271-273, N. W. Nelson, C. L. Best, W. J. McLean, Merle C. Potter

Note

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Last updated: October 8th, 2020

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