linear dynamics and non-linear finite element analysis using ansys workbench

Advanced Finite Element Analysis Ravishankar Venkatasubramanian

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

http://www.scc.kit.edu/scc/sw/msc/Nas102/prob01.pdf

http://web.mscsoftware.com/support/online_ex/previous_nastran/nas102/prob06.pdf


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.


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


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.


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


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


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

http://support.midasnfx.com/files/NAFEMS-PDF/Z-shaped%20cantilever.pdf


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


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:


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


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


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.

http://www.scc.kit.edu/scc/sw/msc/Nas103/Workshop_6.pdf


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


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.


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


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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