Documentation

This bonded contact gravitational load validation case belongs to solid mechanics. This test case aims to validate the following parameter:

- Gravitational load

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

Two beam geometries are used for this gravitational load validation. They have a cross-section of 0.05 x 0.05 \(m^2\) and 1 \(m\) length (l). The first one consists of unrotated beam geometry, shown below:

The second geometry is rotated 45º around the positive x-axis:

The coordinates for the points in the first geometry are as tabulated below:

A | B | C | D | A’ | B’ | C’ | D’ | |

x | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |

y | 0 | 0.05 | 0.05 | 0 | 0 | 0.05 | 0.05 | 0 |

z | 0.05 | 0.05 | 0 | 0 | 0.05 | 0.05 | 0 | 0 |

Similarly, for the rotated geometry, we have:

A | B | C | D | A’ | B’ | C’ | D’ | |

x | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |

y | -0.03536 | 0 | 0.03536 | 0 | -0.03536 | 0 | 0.03536 | 0 |

z | 0.03536 | 0.0707 | 0.03536 | 0 | 0.03536 | 0.0707 | 0.03536 | 0 |

**Tool Type**: Code Aster

**Analysis Type**: Linear static

**Mesh and Element Types**: The meshes for cases A and B were created in SimScale. The standard algorithm was used. The meshes from case A and B were downloaded, rotated by 45º around the positive x-axis, and imported to SimScale. They were used for cases C and D, respectively. With this method, we achieve the same meshes for the rotated and unrotated cases.

Case | Geometry | Mesh Type | Number of Nodes | Element Type |

(A) | Beam – original | Standard – tetrahedral cells | 12737 | 1st order |

(B) | Beam – original | Standard – tetrahedral cells | 91499 | 1st order |

(C) | Beam – rotated | Standard – tetrahedral cells | 12737 | 2nd order |

(D) | Beam – rotated | Standard – tetrahedral cells | 91499 | 2nd order |

Find below the mesh used for case D. It’s a standard mesh with second-order tetrahedral cells.

**Material**:

- Steel (linear elastic)
- \(E\) = 205 \(GPa\)
- \(\nu\) = 0.28
- \(\rho\) = 7870 \(kg/m³\)

**Boundary Conditions**:

- Constraints
- Fixed support on face ABCD.

- Gravity (defined under
*model*):- Cases A and B: 9.81 \(m/s²\) in the negative z-direction (0, 0, -1);
- Cases C and D: 9.81 \(m/s²\) rotated 45º around the positive x-direction (0, 1, -1)

Converting the gravitational load to a line load \((w_a)\):

$$w_{a}l = V.\rho.g \tag {1}$$

Solving \((1)\), we have:

$$w_{a}=193.01175\ N/m \tag {2}$$

The moment of inertia \(I\) is given by:

$$I = \frac {b.h^3}{12} = 5.20833⋅10^{−7}\ m^4 \tag {3}$$

The equation (4) below is derived from [Roark]\(^1\)

$$y(l) = -\frac{w_a l^4}{8 E I} = -2.2597 \cdot 10^{-4}\ m \tag {4}$$

The table below shows the SimScale results for the displacement at the free end (face A’B’C’D’) in the gravity direction. Results are compared to the analytical solution by [Roark].

Case | Quantity | [Roark] | SimScale | Error (%) |

(A) | Displacement at the free end \([m]\) | -2.2597e-4 | -2.1480e-4 | -5.20 |

(B) | Displacement at the free end \([m]\) | -2.2597e-4 | -2.2560e-4 | -0.16 |

(C) | Displacement at the free end \([m]\) | -2.2597e-4 | -2.1480e-4 | -5.20 |

(D) | Displacement at the free end \([m]\) | -2.2597e-4 | -2.2560e-4 | -0.16 |

Inspecting the displacements in the z-direction for case B:

References

- W. C. YOUNG, R. G. BUDYNAS. Roark’s Formulas for Stress and Strain. Seventh Edition. McGraw-Hill. 2002. p. 191.

Last updated: May 19th, 2021

We appreciate and value your feedback.