Linear Dynamics and Non-Linear Finite Element Analysis using ANSYS Workbench

Advanced Finite Element Analysis Ravishankar Venkatasubramanian
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Project Report: Advanced Finite Element Analysis
Introduction:
In this project, we explore problems pertinent to two major systems:
1. Dynamics Analysis
2. Non Linear Analysis
Dynamic analysis are real-world problems which are analyzed by understanding either the mode shapes
and Eigen values and then using this data to find out a specific response, or by the method of time
integration, which is a more direct method of solving these problems. Although time integration is a
direct method, the mathematics involved in its use is tedious that it cannot be used for large number of
degrees of freedom (DOFs). Hence, we use modal superposition to calculate the response for large
number of DOFs. In this project, we will discuss the application of modal superposition on a cantilever
plate.
Non-Linear analysis is subdivided into two major categories:
1. Geometric Non-Linearity.
2. Material Non-Linearity
In Geometric Non-Linearity, the non-linearity exists in the way the deformation occurs. Large
deformation for example is one of the examples for Geometric Non-linearity. In this project, we discuss
and analyze one example of geometric non-linearity. On the other hand, in Material Non-Linearity, the
non-linearity exists in the properties of the material, that is, the material could be Hyperelastic, or could
be Elastoplastic. We will discuss one example on Material Non-Linearity as well.
Let us begin our discussion with an example in Dynamics Analysis:
Example 1: Dynamics Analysis (Normal Modes and Frequency Response Analysis)
References:
1. http://www.scc.kit.edu/scc/sw/msc/Nas102/prob01.pdf
2. http://web.mscsoftware.com/support/online_ex/previous_nastran/nas102/prob06.pdf
For a flat plate as shown below, perform Modal Analysis to determine the first five modes of vibration,
and its corresponding mode shapes.
Figure 1: Description of Length and Breadth Dimensions and mesh system in NASTRAN Analysis

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Figure 2: Material Properties
The system is excited by a 0.1 psi pressure load over the total surface of the plate and a 1.0 lb. force at
a corner of the tip lagging 45°. Use a modal damping of ξ = 0.03. Use a frequency step of 20 Hz between
a range of 20Hz and 1000 Hz. Perform Modal Frequency Response analysis for the mentioned loads
and boundary conditions:
Figure 3: Boundary Conditions and Load
Solution:
We use ANSYS Workbench 17.0 for the analysis and simulation. The solution was first attempted using
3D- Hexahedral elements. The initial analysis to be done was Modal Analysis. A fixed support boundary
condition is applied to the system, as shown in Figure 4.

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Figure 4: Boundary Condition for Eigen Value Analysis
After the boundary conditions are applied, the number of modes to be found are written in the Analysis
settings tab, as shown in Figure 5:
Figure 5: Analysis Settings – Modal Analysis
After this step, Eigen value analysis is run to find out the Mode shapes (Eigen vectors) and Eigen Values
(Natural Frequency), as shown in Figure 6 and Figure 7.
Figure 6: Natural Frequency for each mode

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Figure 7: Mode Shape 1 for a Natural Frequency of 131.79 Hz.
These answers of fundamental frequencies from Figure 6 match closely with the answers obtained in
the NASTRAN Example problem (from reference), as shown in Figure 8 (Natural Frequency circled in
blue):
Figure 8: Eigen Values and Natural Frequencies from NASTRAN Example Problem.
Varying the mesh density and the type of element (From Hexahedral to Tetrahedral) gives tiny change
in Modal Frequency, which is not highly significant.
Using these Eigen Values, we now move on to Frequency Response Analysis. In frequency response
analysis, we use the Eigen values to find out Frequency response as a function of these mode shapes
and modal frequencies. This method saves a lot of time compared to the direct method. Figure 9
provides the geometry, and the requisite boundary conditions applied.

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Figure 9: Boundary Conditions on the Model
Initially, hexahedral elements are used for the analysis, and the mesh density is at its coarsest. This
analysis is run with the mentioned damping ratio and the frequency range, with mentioned steps in
frequency is taken, as shown in Figure 10.
Figure 10: Analysis Settings

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With the mentioned Analysis settings, the Frequency Response Analysis (FRA) was run. The FRA
peaks are shown in Figure 11:
Figure 11: FRA peaks for Hexahedral Elements
The FRA peaks show that the displacement amplitude is highest near the resonance frequencies between
10Hz- 1000Hz. This is expected for the question as instability generally exists near the resonant
frequencies. The value of displacement can be further refined by using a finer mesh. Testing has also
been done using Tetrahedral mesh, and the displacement has been found to be close to the answer
mentioned in the NASTRAN Manual (Figure 12) for fine Hexahedral mesh. Hexahedral elements have
more nodes, and hence can handle bending better than Tetrahedral elements. A finer mesh provides
closer interpolation values, and more accurate results.
Figure 12: PATRAN Results from Reference for FRA

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FRA analysis for higher number of solution intervals, i.e., higher number of frequencies between 20Hz-
1000Hz provides a smoother curve with higher accuracy of results. But these results cannot be
compared with the results from Figure 12, which is based on a solution interval of 49.
For Tetrahedral elements, the FRA provides a higher displacement, since the tetrahedral elements
cannot handle bending and oscillatory movements as effectively as Hexahedral elements, there is a
slightly higher displacement, as shown in Figure 13.
Figure 13: FRA peaks for Tetrahedral Elements
The system is a plate system, and cannot be modeled as a 1D system, and hence, we restrict its modeling.
With increase in mesh density, there is an increase in accuracy. Hence, the most accurate mesh to use
in this case is a Fine-Hexahedral mesh.
Example 2: (Large Deformation; Geometric Non-Linearity)
Reference: http://support.midasnfx.com/files/NAFEMS-PDF/Z-shaped%20cantilever.pdf
Figure 14 shows a Z-shaped cantilever laid along the oblique line of 45˚. The total load P at all the
points on the free end D in the positive Z-direction is conservative (non-follower load). The material
properties are also given.
Figure 14: Z- Shaped Cantilever

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Solution:
This system is designed on ANSYS Workbench 17.0 using the Design Modeler. The model was initially
modeled as a 3D system, and later modeled as a 2D system and a 1D system. We will discuss about the
results of each system in detail.
Figure 15: Hexahedral Coarse Mesh
Figure 16 shows details of the boundary conditions. The fixed support boundary condition and Force is
applied in a ramped fashion, over a period of time.
Figure 16: Boundary Conditions on the Z-Shaped Cantilever

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Figure 17: Application of Ramped Force
Ramped force is essential to make sure that the system does not collapse due to sudden application of
forces up to 4000N.
Figure 18: Analysis Settings
The number of steps is set to 100 to obtain an accurate non-linear solution. The other controls are set to
program controlled as the question does not mention any other specific changes to make to the system.
The final deformation, at Load = 4000N is shown in Figure 19:

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Figure 19: Directional Deformation of the system at 4000N Load
A graph is drawn with displacement in the X-Axis and Load in the Y-axis, to get an idea about how the
system deforms with increasing force. The displacement rapidly increases with load for the first 500N.
After this force, the deformation has reached around 100mm, where the mid-section of the Z-shaped
cantilever causes ‘tension stiffening’, as shown in Figure 20. This tension stiffening continues all the
way upto the load of 4000N, and it can be see that there is very less displacement (43mm) over a large
amount of force (3500N). This can be attributed solely to the stiffening in the mid-section of the Z
shaped cantilever. Please note that the deformation shown in Figure 19 is only for the tip of the
cantilever.
Figure 20: Tip Displacement vs Load (Z Shaped Cantilever)
This graph is compared with the graph obtained from the Midas-NFX reference, shown in Figure 20.
0
500
1000
1500
2000
2500
3000
3500
4000
4500
0 20 40 60 80 100 120 140 160
Load(N)
Tip Displacement (mm)
Tip Displacement vs Load

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Figure 21: Tip Displacement vs Displacement (Midas-NFX)
Comparing both Figure 20 and Figure 21, we can conclude that the graphs are almost the same.
Increasing the mesh density in both 3D and 1D has not provided a significant increase in the accuracy
of the system. This is probably because the system is well equipped to handle bending as such, and the
slow increase in displacement due to tension stiffening has provided sufficient iterations to weed out
any numerical errors, which might creep in the analysis. The 3D Hexahedral elements can handle
bending effectively, and hence, provide accurate results. Figure 22, 23 show pictures of the final
deformation in both 2D and 1D systems.
Figure 22: Tip Deformation – Quadrilateral Shell Elements

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Figure 23: Tip Deformation – 1D Beam Elements
As seen from Figure 22, 1D beam elements can handle bending effectively, and hence, the deformation
at the tip matches with 3D Tetrahedral/Hexahedral elements. 2D triangular and Quadrilateral elements
on the other hand, cannot handle bending as effectively as beam elements, and provide slightly more
deformation. This is seen in Figure 21. There is no significant change in tip displacement with mesh
density, and hence, that has not been discussed in detail. For further information, the attached input files
provide 3 different mesh densities, along with all element types (3D, 2D and 1D).
The solution has been compared with the reference Midas-NFX, and has been found to match with the
prescribed result.
Example 3: (Material Non-Linearity: Elasto-Plastic Material)
Reference: http://www.scc.kit.edu/scc/sw/msc/Nas103/Workshop_6.pdf
Figure 24: Diagram of Problem
L=50, W=10, T=0.1.
The material used in this system has the following properties:
Young’s Modulus = 3.0E+6, Poisson’s Ratio = 0.25, Tangent Modulus = 30303, Yield Stress = 850.

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Find the Elastic and Plastic Strain at the end of the loading. The loading cycle is:
1. Load P=800
2. Load P= 1000
3. Unload P= 950
4. Unload P=0
Solution:
The modeling can be done by using a quarter of the model (symmetry), and it will yield the same result
with a lesser time duration involved. But for this problem, we use a system which has the entire bar.
Figure 25 shows the meshed bar, with a coarse hexahedral mesh.
Figure 25: Meshed Part – Hexahedral Mesh
We expect, from the loading pattern that the material yields between 1 and 2 seconds, since the Yield
stress is 850. After yielding, the material will maintain its plastic nature, even if it unloads, as done in
t=3 and t=4. The loading history is shown in Figure 26 (This is because the stress is
F/A=Load/(10*0.1)=> Stress = Load).
Figure 26: Load curve with time

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The boundary conditions applied on the system are shown in Figure 27.
Figure 27: Application of Boundary Conditions
The analysis is run, and the requested outputs are equivalent plastic strain, the total strain in the body
and the equivalent von-mises stress. The analysis is run under these conditions. Applying a fixed
boundary condition on one end provides a system which is constrained, and the simulation can proceed
as expected. The system is the same as the one where the tensile load is applied to both ends.
The equivalent plastic strain is shown in Figure 28:
Figure 28: Equivalent Plastic Strain in the system at t=4s (Final)
The system shows an equivalent plastic strain of 5.73mm, which is the same as plastic strain obtained
when the load is 1000.

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The graph between time and equivalent plastic strain is:
Figure 29: Plastic Strain vs Time
The plastic strain remains constant in unloading, as expected, as all the elastic strain is dissipated in
unloading.
The graph between time and Equivalent total strain is:
Figure 30: Total Strain vs Time
The total strain reduces over time in the unloading process as the elastic strain is dissipated over time
during the unloading procedure. Now, we can draw a graph between the equivalent stress and plastic
strain.
Figure 31: Plastic Strain and Equivalent Stress
0
200
400
600
800
1000
1200
-0.001 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007
Plastic Strain vs Equivalent Stress
Plastic Strain vs Equivalent Stress

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This graph can now be compared with Figure 32, which is the graph between Plastic Strain and
Equivalent stress in the reference:
Figure 32: Plastic Strain vs Stress (Reference Material)
As seen in this system, we can say that the solution is an almost perfect match. Increasing the mesh
density provides more accuracy in the system, but changing the element type from 3D to 2D will not
cause much improvement in results. This is because Hexahedral or any element in the 3D domain can
handle stretching efficiently. Hence, this system can be meshed with any element, 3D, 2D or 1D to get
effective results.

Linear Dynamics and Non-Linear Finite Element Analysis using ANSYS Workbench

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