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Validation Case: Fixed Beam Under Gravitational Load

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]\(^1\).

Geometry

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:

fixed beam gravitational load validation simscale
Figure 1: Unrotated beam geometry.

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

rotated geometry fixed beam validation simscale
Figure 2: Rotated beam geometry.

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

ABCDA’B’C’D’
x00001111
y00.050.05000.050.050
z0.050.05000.050.0500
Table 1: Unrotated beam dimensions in meters.

Similarly, for the rotated geometry, we have:

ABCDA’B’C’D’
x00001111
y-0.0353600.035360-0.0353600.035360
z0.035360.07070.0353600.035360.07070.035360
Table 2: Rotated beam dimensions in meters.

Analysis Type and Mesh

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.

CaseGeometryMesh TypeNumber of NodesElement Type
(A)Beam – originalStandard – tetrahedral cells127371st order
(B)Beam – originalStandard – tetrahedral cells914991st order
(C)Beam – rotatedStandard – tetrahedral cells127372nd order
(D)Beam – rotatedStandard – tetrahedral cells914992nd order
Table 3: Mesh characteristics.

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

tetrahedral second order mesh rotated
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 \((w_a)\):

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

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]\(^1\)

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

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].

CaseQuantity[Roark]SimScaleError (%)
(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:

fixed beam under gravitational load results
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: January 29th, 2021

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