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:
Figure 1: Unrotated beam geometry.
The second geometry is rotated 45º around the positive x-axis:
Figure 2: Rotated beam geometry.
The coordinates for the points in the first geometry are as tabulated below:
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
Table 3: Mesh characteristics.
Find below the mesh used for case D. It’s a standard mesh with second-order tetrahedral cells.
Figure 3: Second order standard mesh used for Case D.
Simulation Setup
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)
Reference Solution
Converting the gravitational load to a line load \((w_a)\):
The equation (4) below is derived from [Roark]\(^1\)
$$y(l) = -\frac{w_a l^4}{8 E I} = -2.2597 \cdot 10^{-4}\ m \tag {4}$$
Result Comparison
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
Table 4: Comparison of SimScale’s results against an analytical solution.
Inspecting the displacements in the z-direction for case B:
Figure 4: Case B, showing the displacements in the z-direction.
References
W. C. YOUNG, R. G. BUDYNAS. Roark’s Formulas for Stress and Strain. Seventh Edition. McGraw-Hill. 2002. p. 191.
Last updated: October 8th, 2020
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