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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]\(^1\).


    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:

    Table 1: Unrotated beam dimensions in meters

    Similarly, for the rotated geometry, we have:

    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 cells914992nd order
    (C)Beam – rotatedStandard – tetrahedral cells127371st 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


    • 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


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

    Last updated: November 7th, 2023