Finite Element Analysis With Ansys BY NANGI PETROS
Transcript of Finite Element Analysis With Ansys BY NANGI PETROS
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TASK 1.1Create and present FE model of the frame (Discuss the procedure used as well)
Figure 1.1.1 Created FE model of the frame
Procedure used to create the FE model in ANSYS
Given data:
metresAB 3
metresBC 2
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PaE9
10200
29.0v
37860 mkg
12
101 4
I
The width of the frame was not given. An assumption of 0.27 metres was made for the
frames width. And then the following commands were performed from the ANSYS main
menu. PreprocessorModellingCreateAreasRectangleBy dimensions from
the dialogue box, the following dimensions was used to create BC; X1=0, X2=2, Y1=0
Y2=0.27 and AB; X1=0, X2=0.27, Y1=0, Y2=-3.
Figure 1.1.2 Creation of BC (X1=0, X2=2, Y1=0, Y2=0.27)
Figure 1.1.3 Creation of AB (X1=0, X2=0.27, Y1=0, Y2=-3)
After creating BC and AB, it was then necessary to add both areas. This process was done
with the following command, PreprocessorModellingOperateBooleansAdd
Areas from the dialogue box; Pick All.
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Figure 1.1.3 Added areas of BC and AB
TASK 1.2
Discuss real constants, material model and element chosen for FE analysis
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Element type
The element type chosen for this free vibration analysis is SHELL 93. SHELL 93 represents a
thin wall structure (like the FE model frame). SHELL 93 is particularly well suited to model
curved shells. The element has six (6) degree of freedom at each node: translation in the
nodal X, Y and Z direction and rotation about nodal X, Y and Z axes. The deformation shape
has plasticity, stress stiffening, large deflection and large strain capabilities (Adams &
Askenazi 2009).
The SHELL 93 element type, was selected with the following commands; Preprocessor
Element TypeAdd/Edit/Delete. From the popped-up box; Solid, 8 node 93 was selected
from the ANSYS library of elements (see Figure 1.2.1).
Figure 1.2.1 Shows the selected element type (SHELL 93)
Real Constant
The real constant for this free vibration analysis defines the FE model frame, with thickness
of metres0005.0 .This thickness value was assumed and used to show shell as solid by the
software tool (ANSYS). The following commands were used to input the thickness value
PreprocessorReal ConstantAdd/Edit/Delete. From the popped-up box; real constant
is defined with shell thickness at nodes I, J, K and L with thickness values of metres0005.0
(see figure 1.2.2).
Chosen Element Type;
SHELL 93
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Figure 1.2.2 Shows the assumed shell thickness input of metres0005.0 at nodes I, J, K and L
Material Model
The material model for the FE model frame is linear isotropic. This means that the extension
of the material is in direct proportion with the load applied to it (Hookes law). The material
for FE model frame is steel; since its Modulus of Elasticity is 2910200200 mNGPa . It
Poissons ratio is 0.29. Its weight density is 37860 mkg . These values were input with the
following commands PreprocessorMaterial PropsMaterial Models. From the
popped-up box, the followings were selected Structural
Linear
Elastic
Isotropic.The values for Modulus of elasticity and Poissons ratio were input. This preceded by input
the density value, which was done by clicking on Density and OK (see Figure 1.2.3 and
1.2.4).
Figure 1.2.3Elastic Modulus and Poissons ration input
Elastic Modulus
and Poissons ratio
input; 200GPa and
0.29 respectively
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Figure 1.2.4 Input for density ( 37860 mkg )
Density
Input of3
7860 mkg
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TASK 1.3
Present meshed FE model and discussed the procedure used for it
Figure 1.3.1 Mesh frame model
Mesh element type 8 node 93
Element edge length 0.05 metres
Mesh type: Manual meshing
process with H-method
H-method was selected because the variation on closeness
of natural frequency result only, was required for this
analysis (not the exact result).
Procedure used to create the mesh
Manual Meshing Process (with H-Method)
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The type of meshing used for this analysis is a manual meshing process. This means that the
area was divided by meshes of a particular edge length (0.05 metres) input manually to set up
jobs for the free vibration analysis problem. H-method was selected because the variation on
closeness of displacement only, was required for this analysis (not the exact result).
After selecting the element edge length (0.05 metres) as can be seen in Figure 1.3.2; the mesh
function was executed with the following commands PreprocessorMeshingMesh
AreasFree. From the popped-up menuPick all.
Figure 1.3.2 Selected element edge length (0.05 metres)
TASK 1.4
Elementedge length
0.05 meters
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Present your post processing results in textual and graphical form (three resonant
frequencies and mode are to be presented)
Boundary conditions used for the analysis
The analysis performed here-in is Modal analysis. The right angle frame is fixed at A.
Therefore all freedoms were constrained (DOF = 0) ad A. At C movement are arrested in
the Y direction. This constrain will help determine the natural frequencies under the action
of force inherent in the system (frame) itself.
Figure 1.4.1 Type of analysis chosen (modal)
Figure 1.4.2DOF = 0, assigned at point A of the frame
Analysis type chosen (modal)
Assigned DOF = 0
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Figure 1.4.3Movement arrested in the Y direction of point C
Figure 1.4.4 Complete constrained section of the frame
Execution calculation used includes the mode extraction method (PCG Lanczos)
For the mode extraction method; the PCG Lanczos was selected as the eigenvalue solver forthis typical un-damped modal analysis problem. This solver is useful when finding few
modes (up to about 100) of large models (ANSYS release 2008).
UY=0. Movement
arrested in the Y
direction of point C
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Figure 1.4.5 Mode of extraction method
PCG Lanczos
No. of modes
to extract 10
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Figure 1.4.6 First mode of vibration
VALUE TEXTUAL DATA
Displacement 0.742648 metres At node 271: UX = 0.742648 metres
Resonance Frequency 0.021412 Hz Load step = 1, Sub-step = 1,
Cumulative =1
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Figure 1.4.7 Second mode of vibration
VALUE TEXTUAL DATA
Displacement 1.153 metres At node 297: UX = 1.153 metres
Resonance Frequency 0.076245 Hz Load step = 1, Sub-step = 2,
Cumulative =2
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Figure 1.4.8 Third mode of vibration
VALUE TEXTUAL DATA
Displacement 0.1771596 metres At node 261: UX = 0.1771596 m
Resonance Frequency 0.216144 Hz Load step = 1, Sub-step = 3,
Cumulative =3
Conclusion
Table 1.4.1 Comparison of all three (3) modes of vibration collated
Displacement (metres) Resonance Frequency (Hz)
First mode of vibration 0.742648 0.021412
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Second mode of vibration 1.153 0.076245
Third mode of vibration 0.1771596 0.216144
Based on the vibration modes in Table 1.4.1; it is notable that large radial displacementappeared on the second mode of vibration. But failure or excessive deformation may be
caused if vibration occurs at a high resonant frequency (third mode of vibration). Resonance
is the tendency of a system, to oscillate at greateramplitude, at some frequencies than at
others.
TASK 2.1
Create a finite element (FE) model
http://en.wikipedia.org/wiki/Oscillatehttp://en.wikipedia.org/wiki/Amplitudehttp://en.wikipedia.org/wiki/Frequencyhttp://en.wikipedia.org/wiki/Frequencyhttp://en.wikipedia.org/wiki/Amplitudehttp://en.wikipedia.org/wiki/Oscillate -
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The material of the product is ASTM A-514 Steel. It is quenched and tempered steel with
high yield strength and allows thickness to be minimized. The yield strength of the ASTM A-
514 steel is 690 MPa. The load is shown as distributed load acting over the right part of the
wrench.
Table 2.1.1 Shows the available parameters to be used to create the finite element (FE)
model
X-last digit of I/C number
IC Number : P00097467
Radius of the circle mm105715
Distance between circles mm280740
Allowance from the centre of circle mm70710
Radius of the middle hexagon mm6379
Radius of the corner hexagon mm4977
Load mmN70710
Youngs Modulus (at room temperature) 2310210210 mmNGPa (efunda Inc
2011)
Poissons ratio (v) 0.27 (efunda Inc 2011)
Figure 2.1.1 Rough sketch of the FE model
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Figure 2.1.1 Created FE model
Procedure used to create the finite element (FE) model
Table 2.1.2 How the model was created in ANSYS
OPERATION PROCEDURE (all dimension in mm)
Rectangular
Dimension (in
mm)
From the ANSYS main menu, ModellingCreateAreas
RectangleBy dimension. X1=0, X2=873.049, Y1=0 and Y2=140
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Circular
Dimension (in
mm) on
different points
on the
rectangle
From the ANSYS main menu ModellingCreateAreasSolid
Circle. Left corner circular part; X=0, Y=70, Radius = 105Apply.
Middle circular part; X=436.524, Y=70 and Radius=105 Apply.
Right corner circular; X=873.049, Y=70 and Radius = 105
OK.
Adding created
areas
After creating the four different areas (three circles and one rectangle),
it is necessary to add them up as one area. This was done by performing
the following operation, Preprocessor ModellingOperate
Booleans AreasPick all
Hexagonal
dimensions (in
mm)
This step was preceded by the creation of hexagons from the centre of
the cicular part inluding middle and corners (left and right). The
following operations were used to create the hexagon, Preprocessor
ModellingCreateAreasPolygonHexagon. From the
popped-up box, dimensions were assigned. Dimension used to create
the middle hexagonal part of the wrench X=436.524, Y=70 and Radius
= 63Apply. The left corner hexagon was created with the following
dimension X=0, Y=50, Radius=49Apply. While the right corner
hexagon was with the following dimensions X=873.049, Y=70, Radius
=49OK.
Subtracting
hexagonal
areas
The hexagonal area was finally subtracted from the main area to
completely create the wrench FE model. The hexagonal areas were
subtracted by performing the following Preprocessor Modelling
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OperateBooleansSubtractAreas. The main area was
highlighted; the Apply button was clicked on, from the popped-up box.
This was followed by clicking on the hexagons and the OKbutton.
TASK 2.2
Discuss material model, element and real constant used for the analysis
Chosen type of study
The type of study chosen for the analysis is structural (stress) analysis. It is used to determine
the effects of static or dynamic loading on a component like the bicycle wrench.
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Figure 2.2.1 Study (structural) chosen from the ANSYS main menu
Chosen element for meshing (Quad 4 node 42) and why?
Quad 4 node 42 was chosen for this problem because of the need to use the PLANE42 (2D
plane stress or plane strain) element. This element is a rectangular node element which has 4
nodes each with 2 degrees of freedom (translation along the X and Y axes). It is known that,
for any element, DOF solution u is solved at nodes (usually accurate based on meshed
density). Therefore an element type with more nodes would yield a much more accurate
result. At this initial analysis, only a variation on the closeness of displacement result is
required and thus Quad 4 node 42 is chosen. A different element type will be chosen for
refine mesh later on, on subsequent task.
Structural
Analysis
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Figure 2.2.2 Element type (Plane 42) chosen from the ANSYS main menu
Assumption for Material properties
The material properties for the ASTM A-514 Steel are assumed to be linear isotopic with
Modulus of Elasticity 2310210210 mmNGPa and Poissons ratio (v) = 0.27. This
operation was performed in ANSYS with the following commands Pre-processor
Material PropertiesMaterial ModelsStructuralLinearElastic Isotropic
(see figure 1.3.3).
Figure 2.2.3 Defined material properties; Linear isotropic, Elastic modulus 210,0002
mmN .
Poisson Ratio 0.27
Plane 42 element (plane
strain/stress element)
Quad 4 node 42
(element chosen
for meshing)
Material model defined
as Linear Isotropic
Materials Elastic Modulus
210,0002
mmN . Poisson
Ratio 0.27
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Assumption for Real Constant
The geometry of the body (the wrench), has an element where one dimension is very small
compared to the other (i.e. the element is flat or thin). The stresses are negligible with respect
to the smaller dimension as they are not able to develop within the material and are small
compared to the in-plane stresses (Allan 2008). Therefore, the face of the element is not acted
by loads and so the structural element can be analysed as 2-dimensional planes stress
problem. Considering a plane stress condition for the bicycle wrench, thickness is measured
along the z-direction and it is small relative to its lateral dimensions. Since the original
thickness for the bicycle wrench is unknown; a uniform 1mm thickness is thereby assumed
for the plane stress problem
Figure 2.2.4 Selection as a plane stress problem with thickness
Figure 2.2.5 Assumed thickness selection of 1mm
This operation was performed in ANSYS with the following commands Pre-processor
Material PropertiesMaterial ModelsStructuralLinearElastic Isotropic
Selection of plane
stress with thickness
Assumed 1mm
thickness
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TASK 2.3
Present meshed model and discussed the procedure used for it
Figure 2.3.1 Shows the meshed FE model, created with two different element edge lengths
(10 and 20mm respectively)
Meshed Information
Table 2.3.1 Shows some mesh information used for the analysis
Mesh element type Quad 4 node 42
Critical region The fine meshing was increased at this region (by decreasing
element edge length to 10mm). This was done because the
region was more critical and likely to be the hub for stress
concentration
Criticalregion,Element
edgelength10mm
Non-criticalregion,
Elementedgelength
20mm
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Element edge length for
critical region
10 mm
Element edge length for
non-critical region
20 mm
Mesh type: H-refine H-refine was selected because the variation on closeness of
displacement only, was required for this analysis (not the
exact result).
Number of elements 441
Element 1has nodes: 174, 428, 525 and 525
Element 441 has nodes:273, 157, 156 and 274
Number of Nodes 526
At node 1: X = 623.61, Y = -25
At node 526: X = 665.53, Y = 22.99
The element edge length was created with the following operations MeshingSize Control
Manual SizeLinesPicked lines (e.g. the critical region was highlighted).
Figure 2.3.2 Element edge length used for the critical region (10 mm). The same procedure
was applied to the non-critical region (20mm)
TASK 2.4
Analyse the model and present the nodal displacements and Displacement plot obtained
from the analysis
Boundary condition: The side hexagonal holes in the wrench are necessary for fasteners
such as bolts, rivets, etc. It is necessary to know how stresses and deformations occur near
them. For this reason the left hexagonal hole is constrained with all DOF (degrees of
10 mm element edge
length, for the critical
region
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freedom) equal to zero (0). Pressure is expected to be applied at the right end of the wrench.
The pressure is expected to rotate the wrench in a clock-wise direction. The pressure is given
as;
mmN70
Since a thickness of mm1 is assumed;
mmmmNessure
1
170Pr
With the following commands, the all DOF constrain was applied to the left corner hexagonal
hole. SolutionDefine LoadsApplyStructuralDisplacementOn lines (left
corner hexagon was highlighted)Apply
Figure 2.4.1 All DOF = 0 constrain application, on the left corner hexagonal hole
270 mmN
All DOF
applied as 0
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Figure 2.4.2 Applied pressure value of 70 2mmN
Figure 2.4.3 Line plot based on the constrain assumption
Applied
pressure of2
70 mmN
All DOF = 0
Pressure on lines
(UDL) 70 2mmN
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Figure 2.4.3 shows the complete constrain of the 2D wrench model in a line plot. The left
corner of the wrench will be connected to a bolt, therefore all DOF (degrees of freedom) = 0.
A uniformly distributed load (pressure of 70
2
mmN ) will be required to rotate the wrench ina clock-wise direction. Please see the next page for graphical plot.
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Figure 2.4.4 Displacement plot
DISPLACEMENT VALUE POSITION/TEXTUAL DATA
Maximum displacement 62.099 mm At node 6:
UY = 31.442 mm; UX = 4.2578
Minimum displacement 0 mm At nodes 149 to 166:
UY = 0 mm; UX = 0
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Figure 2.4.5 Principal stress plot 1p
PRINCIPAL STRESS 1p VALUE POSITION/TEXTUAL DATA
Maximum 1p
23813 mmN At node 42:
23813 mmN
Minimum 1p
20 mmN At nodes 6:
20000.0 mmN
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Figure 2.4.6 Principal stress plot 2p
PRINCIPAL STRESS 2p VALUE POSITION/TEXTUAL DATA
Maximum 2p 2
182.546 mmN At node 41:2
182.546 mmN
Minimum 2p 2
612.499 mmN At nodes 81:
2612.499 mmN
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Figure 2.4.7 Shear Stress xy
SHEAR STRESSxy
VALUE POSITION/TEXTUAL DATA
Maximumxy
21163 mmN At node 88:
21163 mmN
Minimumxy
21055 mmN At nodes 130:
21055 mmN
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Figure 2.4.8 Von Mises Plot
VON MISES PLOT VALUE POSITION/TEXTUAL DATA
Maximum 23842 mmN At node 81:
23842 mmN
Minimum 2198876.0 mmN At nodes 10:
2198876.0 mmN
Procedure used to generate the graphical plot
The graphical plots were created after constrain conditions had been applied and solution
solved.
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Figure 2.4.9 Solution solved
After the solution had been solved, the following commands were used to generate the
graphical plot for displacement. General PostprocessorPlot ResultsNodal Solution
(from the popped-up box) DOF SolutionDisplacement Vector SumOK(see figure
2.4.10).
Figure 2.4.10 How the Displacement plot was generated
Similar steps were also used to generate the graphical plots for principal stresses. General
PostprocessorPlot ResultsNodal Solution (from the popped-up box) Stress 1st
Principal stress or2nd Principal Stress orShear Stress orVon Mises StressOK(e.g
see figure 2.4.11 [1st
principal stress]).
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Figure 2.4.11 How the 1st
Principal stress plot was generated
Conclusion
The given model does require changes in the dimension (especially thickness). The reason is
because, the Von Mises stress results (2
3842 mmN ) did not fulfil the maximum distortion
energy theory, which states that for a structure to be safe; the Von Mises stresses must not
exceed the yield strengh (690 2mmN ) of the material. Increasing the dimension will increase
the strength of the wrench (Allan 2008). Only then will the wrench fulfill the maximum
distortion enegy theory.
Reference:
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1. Allan Bower, F, 2008,Introduction to Finite Element Analysis in Solid Mechanics,Journal of Applied mechanics of solids, viewed 3 November 2011,
2. eFunda Inc. 2011, ASTM A514 Type A viewed 4 November 2011,
3. Engineers Edge 2011, Strength of material Mechanics of material viewed 5November 2011, < http://www.engineersedge.com/strength_of_materials.htm>