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:

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.

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:

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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