Documentation

This validation case belongs to solid mechanics. The aim of this test case is to validate the following parameters:

- Torque load applied through remote force boundary condition
- Shear stress distribution and maximum value
- Rotational deformation magnitude

The simulation results of SimScale were compared to the results presented in [Roark]\(^1\)

The geometry used for the case is below:

The axis of the shaft cylinder is aligned with the \(Z\) axis, with a length \(L = \) 0.5 \(m\) and a radius \(r = \) 0.1 \(m\).

**Tool Type**: Code_Aster

**Analysis Type**: Linear Static

**Mesh and Element Types**:

Tetrahedral meshes were computed using SimScale’s standard meshing algorithm and manual sizing. Table 1 shows an overview of the meshes in the validation project.

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

A | Standard | 191965 | 1st order tetrahedral |

B | Standard | 189630 | 2nd order tetrahedral |

The second order tetrahedral mesh from case B is shown in Figure 2:

**Material**:

- Linear Elastic Isotropic:
- \(E = \) 208 \(GPa\)
- \(\nu = \) 0.3
- \(G = \) 80 \(GPa \)

**Boundary Conditions**:

- Constraints:
- Face A is fixed.

- Loads:
- Torque \(T = \) 50000 \(Nm\) on face B.

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

\( \theta_B = \frac{ T L }{G J} \tag{1} \)

\( \tau_{max} = \frac{T R}{J} \tag{2} \)

\( J = \frac{\pi R^4}{2} \tag{3} \)

The computed reference solution is:

\( \theta_B = 1.9894×10^{-3}\ Rad \)

\( \tau_{max} = 31.847\ MPa \)

The plane maximum displacement \( U \) is used to compute the rotation angle through equation 4 (obtained through the cosines law):

\( \theta = ArcCos[ 1- \frac{U^2}{2R^2} ] \tag{4} \)

CASE | \(U\) | \( \theta \) | \( \theta_{ref} \) | ERROR |
---|---|---|---|---|

A | 0.000198463 | 0.00198463 | 0.0019894 | -0.24% |

B | 0.000198944 | 0.00198944 | 0.0019894 | 0.00% |

The maximum shear stress is taken from the Cauchy stress tensor, component [SIYZ]:

CASE | SIYZ | \( \tau_{ref} \) | ERROR |
---|---|---|---|

A | 31.7483 | 31.847 | – 0.26% |

B | 31.8405 | 31.847 | – 0.03% |

For both 1st order and 2nd order meshes, the results show good agreement with the analytical solution. Find below an image showing the displacement magnitude on the cylinder, for case B:

References

- (2011) “Roark’s Formulas For Stress And Strain, Eighth Edition”, W. C. Young, R. G. Budynas, A. M. Sadegh.

Note

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Last updated: July 21st, 2021

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