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:

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

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

	A	B	C	D	A’	B’	C’	D’
x	0	0	0	0	1	1	1	1
y	0	0.05	0.05	0	0	0.05	0.05	0
z	0.05	0.05	0	0	0.05	0.05	0	0

Similarly, for the rotated geometry, we have:

	A	B	C	D	A’	B’	C’	D’
x	0	0	0	0	1	1	1	1
y	-0.03536	0	0.03536	0	-0.03536	0	0.03536	0
z	0.03536	0.0707	0.03536	0	0.03536	0.0707	0.03536	0

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.

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

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 \((w_a)\):

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

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].

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

Inspecting the displacements in the z-direction for case B:

Last updated: June 30th, 2020