Stresses and Deformation

8/2/2019 Stresses and Deformation

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Table of Contents

Introduction 1

Steps involved with simulation 3

Description of modules 5

Element Type 6

Shear Locking 7

Hour Glassing 8

Calculations 9

Dimensions Calculation 9

Force calculations 9

Theoretical Calculations 11

Deformation of Elements 13

Computational Results 17

Discussion 20

Conclusion 21

References 22

Appendix 23


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Introduction

The main objective of this assignment is to analyse the stresses and deformationunder the given loads when the maximum elastic moment is reached in the middle

portion of the plate.

The plate section has been divided into four other classes which need to be solved

as following:

a) 3D Solid Elements

b) Plane stress Elements

c) Plane strain Elements

d) Shell Elements

In the following assignment the calculations are done in terms of X units

Where X = 1 + 0.001 * # (# = last three digits of the student number)

Therefore X in my case is 1.599 as, X = 1 + 0.001 * 599 (student no. = 3192599)

The plate shown above has a force of 5115.84N acting vertically upwards at points A

and D whereas point B is a fixed joint and point C has a roller attached onto it.


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The plate shown above has a force acting on point A of 5115.84N in downwards

direction and a roller joint at point B along with a X symmetry boundary condition on

the left plane of the plate.

Given Mechanical-Elastic Material Parameters

Youngs Modulus of Elasticity: E = 200 GPa

Poissons Ratio: v = 0.3

Plate Dimensions

Thickness = 2X

t = 2 * 1.599 = 3.198mm


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Steps involved with simulation

For this assignment we used Abaqus 6.10 Student version.

As we start the program we go on as follow:

Create a new model database

From the left side of the window the model tree appears and double clicks on

the Parts

In the appeared Create Part dialog box name the part and Select 3D Planar, Select

Deformable, Select Solid / Shell, Set approximate size = 200

Sketch

Create the geometry of the structure and do its partition by selecting a datum plane

and then clicking the create partition option.

Material

Double click on the Materials in the model tree. Name the material as Steel, Click

on the Mechanical tab and select Elasticity and then Elastic, Define Youngs

Modulus and Poissons Ratio as, E = 200 GPa, =0.3

Sections

Double click on the Sections in the model tree. Name the section, select the

material created and give the material thickness in case of shell or just click OK in

case of solid.

Section Assignments

Then expand the part in model tree and double click on Section Assignments then

Select the drew structure in the screen and Press Yes to select the section.

Assembly (Instances)

Expand the Assembly in the model tree and then double click on Instances. Then

select Dependent for the instance type on appeared dial box.

Steps


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Double click on the Steps in the model tree and name the step as step1and click

OK.

Boundary Conditions

Double click on the BCs in the model tree and then Name the boundary

conditioned Fixed/Roller and select Displacement/Rotation for this type, Select

where the boundary condition needed, Select the U1 and U2 displacements in case

of fixed and U2 in case of roller and click ok. Then double click BCs again and name

it symmetry, select Symmetry in this case and select the region on the structure

where the boundary condition is needed and then select the type of symmetry

needed and click ok.

Load

Right click on the load in model tree and select the load manager then create load

and Enter (-) or (+) 5115.84N value to the CF2 direction.

Mesh

In the model tree double click on Mesh and select the structure area and select

Yes to integrate and on Assign Element Type icon then select Standard for

element type and Quadratic for geometric order and select 3D stress / plane strain

/ plane stress for family as required.

Select Seed on toolbar and select part. Then enter the seed value as (5,4,3,2,1 as

needed).

Job

In the model tree double click on the Job and name the job 3D solid (or as the

model designed) and give description.Now from the model tree expand Job and

right click the created job and select submit.

Now we can see the status changing from submitted to running and then completed

which indicates the process was successful and there were no problems related to it.

Now right click on the created job and select Results and in the visualisation we

can see the deformation shapes, nodes, elements etc.


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Description of modules

Part

Property

Load

Mesh

Job

Part

The basic step in analysing a given frame is creating a part first. The steps are given

above of how to create a part. The frame can then be analysed in 2Dimension or

3Dimensions and as we are using only one material here we choose 3Dimensional

Solid here so the computer can analyse that the initial sketch is extruded in the

direction perpendicular to the plane.

Property

In the following step we apply properties to the material. The steps of doing so are

given above. First we put in the youngs modulus and then poisons ratio for the

material. The material is then chosen to be elastic and section is selected as truss

but the calculations are done for both beam and truss. After this step we define the

cross-sectional area. Thus in the following module we define all the properties of the

material.

Load

In the following module we choose the loads and loading conditions such as static

load, concentrated loads etc. then the boundary conditions are selected for the frameif the nodes are to be fixed, roller X,Y,Z symmetry etc.

Mesh

In the following module we select the element type for the part to be meshed. We

mesh every entity in the part together such as how much is the load applied, or in

which direction it is, boundary conditions for the frame etc. so that the frame has all

the required inputs and can be simulated upon them to get effective results. The

steps to do the following are given above.


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Job

The following module in concerned with the evaluation of the given frame. After the

job is completed in the job manager we can click on results and go to visualisation to

check the deformation, the forces applied etc. after this we can get a report on the

following plate by just going into the tool in the menu bar and can select query what

basic results are we after such as stress components, displacement etc. and then

ABAQUS saves a file in a form of report in notepad format on the disk.

Element Type

Linear or first order elements are often overly stiff in bending analysis and suffer from

numerical error called Shear Locking. Under the pure bending the edges of first

order elements do not tend to bend or curve and thus they compensate by creating

artificial shear stresses which in turn oppose the bending deformation. These

elements contain 8 nodes and it is possible to get incorrect displacements or false

stresses when the elements undergo shear locking.

Quadratic elements have 20 nodes in comparison to 8 in first order elements, which

in turn increase the accuracy of the simulation and avoid shear locking by making

edges able to bend into curves. Fully integrated quadratic elements are most

accurate while bending analysis though they require more disk space and computing

time due to the amount of data points.

It is therefore preferable to use reduced integration quadratic brick elements as

unlikely fully integrated linear brick elements they have ability to bend at the edged

while undergoing bending process. The only problem with that can be the

hourglassing which in abaqus is controlled as default for all reduced integration brick

elements.


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Shear Locking

In an ideal situation, a block of material under a pure bending experiences a curvedshape change. But the edged of the fully integrated first order elements are however

not able to bend to curves. Thus the linear elements will develop a shape shown

below under a pure bending moment.

Fig1: first order elements

To cause the angle A to change under the pure moment, an artificial shear stress

has been introduced. This means that the strain energy of the element is generating

shear deformation instead of bending deformation. So we can see that the linear fully

integrated elements becomes locked or overly stiff under the bending moment thus

causing wrong displacements, false stresses etc.

The fully integrated second order elements behave differently since they are able to

bend to curves. Thus the angle A remains to be 90 degrees after bending which in

turn does not creates any artificial shear stress and correctly simulates the behaviour

of the material block.

Fig2: Second order elements

The figure above represents fully integrated second order elements which do not

have any shear locking effect.


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Hour Glassing

To address the shear locking and to increase computational efficiency, a reduced

integration scheme is proposed and is widely implemented. For example, for the

reduced-integration first order 8 node brick element, a single integration point is used

while its fully integrated version has eight integration points.

The reduced integration first order element suffers from its own numerical difficulty

called hour glassing. The hour glassing has to be properly controlled; if not then the

results from it will be not beneficial.

We can see in the figure below that the vertical and horizontal lines and the angle A

remain unchanged which means that the normal stresses and shear stresses are

zero at the integration and that there is no strain energy generated from deformation.

Fig: first order elements with 1 integration point


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Calculations

Dimensions Calculation

Length of plate = 140 * X = 140 * 1.599 = 223.86mm

Height of plate = 20 * X = 20 * 1.599 = 31.98mm

Thickness of plate = 2 * X = 2 * 1.599 = 3.198mm

Now, Calculating the distances where the partitions are created

Dimensions of partitions on the length of the plate

First partition = 10 * 1.599 = 15.99mm

Second partition = 30 * 1.599 = 47.97mm

Third partition = 110 * 1.599 = 175.89mm

Fourth partition = 130 * 1.599 = 207.87mm

Dimension of partition on the height of the plate is = 10 * 1.599 = 15.99mm

Dimension of partition on the thickness of plate is = 1 * 1.599 = 1.599mm

Force calculations

Where,

b = 1.599mm

h = 31.98mm


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Therefore I = 4358.15mm

4

Where,

Therefore M = 81766.5

Now,

Where,

M = 81766.5

d = 31.98mm

Therefore Force = 5115.84N


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Theoretical Calculations

3D solid Element analysis:

-240.915 -284.216 -240.915 -284.216 -240.915 -284.216 -240.915

-165.915 -209.216 -165.915 -209.216 -165.915 -209.216 -165.915

-90.9155 -134.217 -90.9155 -134.217 -90.9155 -134.217 -90.9155

-15.9152 -59.2167 -15.9152 -59.2167 -15.9152 -59.2167 -15.9152

240.889 284.209 240.889 284.209 240.889 284.209 240.889

165.857 209.177 165.857 209.177 165.857 209.177 165.857

90.8238 134.144 90.8238 134.144 90.8238 134.144 90.8238

15.7907 59.1112 15.7907 59.1112 15.7907 59.1112 15.7907

Mean Stress I Y Moment E ROC Length ROC1 Deflection-262.5655 4358.15 13.986 81817.52 200000 10653.34 111.93 10652.75 0.588016

-187.5655 4358.15 9.99 81825.684 200000 10652.28 111.93 10651.69 0.588075

-112.56625 4358.15 5.994 81845.279 200000 10649.73 111.93 10649.14 0.588216

-37.56595 4358.15 1.998 81940.963 200000 10637.29 111.93 10636.7 0.588903

262.549 4358.15 13.986 81812.378 200000 10654.01 111.93 10653.42 0.587979

187.517 4358.15 9.99 81804.526 200000 10655.03 111.93 10654.45 0.587923

112.4839 4358.15 5.994 81785.404 200000 10657.53 111.93 10656.94 0.587785

37.45095 4358.15 1.998 81690.119 200000 10669.96 111.93 10669.37 0.5871


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Plane Stress analysis:

700 120.451 120.451 142.106 142.106

701 157.959 157.959 179.615 179.615

702 195.467 195.467 217.123 217.123

703 232.976 232.976 232.976 232.976

704 270.484 270.484 292.139 292.139

Mean

Stress

Y I Moment E ROC Length ROC1 Deflection

131.2785 3.9975 4358.15 143122.3 200000 6090.1061 111.93 6089.077 1.028667

168.787 6.6625 4358.15 110408.9 200000 7894.5653 111.93 7893.772 0.793518

206.295 9.3275 4358.15 96388.59 200000 9042.8755 111.93 9042.183 0.692745

232.976 11.9925 4358.15 84664.95 200000 10295.052 111.93 10294.44 0.608481

281.3115 14.6575 4358.15 83643.03 200000 10420.832 111.93 10420.23 0.601137


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Deformation of Elements

3D solid Element analysis

a) F1 = F2 = 5115.84N (full model)

b) F1 = 5115.84N, F2 = 0N ( half model)


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a) F1 = 5115.84N, F2 = 0N

Plane Strain analysis:

a) F1 = 5115.84N, F2 = 0N


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Shell Element analysis:

a) F1 = 5115.84N, F2 = 0N

Axis Symmetry analysis:


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Computational Results

3D solid Element analysis:

a) Full model

b) Half model

3D Solid H 5 C3D20R -252.788, -252.788, -287.439, -287.439, -252.788, -252.788, -287.439

3D Solid H 640 C3D20R 252.776, 252.776, 287.436, 287.436, 252.776, 252.776, 287.436,

287.436

3D Solid H 639 C3D20R 192.744, 192.744, 227.404, 227.404, 192.744, 192.744, 227.404,

227.404

3D Solid H 4 C3D20R -192.771, -192.771, -227.422, -227.422, -192.771, -192.771, -227.422, -

227.422

3D Solid H 3 C3D20R -132.754, -132.754, -167.405, -167.405, -132.754, -132.754, -167.405, -

167.405

3D Solid H 638 C3D20R 132.713, 132.713, 167.372, 167.372, 132.713, 132.713, 167.372,

167.372

3D Solid 627 C3D20R 240.889,240.889,284.209,284.209, 240.889,240.889,284.209,284.209

3D Solid 115 C3D20R -165.915,-165.915,-209.216,-209.216,-165.915,-165.915,-209.216,-209.216

3D Solid 595 C3D20R 165.857,165.857, 209.177, 209.177,165.857,165.857, 209.177, 209.177

3D Solid 147 C3D20R -90.9155,-90.9155, -134.217, -134.217,-90.9155,-90.9155, -134.217, -134.217

3D Solid 563 C3D20R 90.8238,90.8238, 134.144, 134.144,90.8238,90.8238, 134.144, 134.144

3D Solid 179 C3D20R -15.9152,-15.9152, -59.2167, -59.2167,-15.9152,-15.9152, -59.2167, -59.2167

3D Solid 531 C3D20R 15.7907,15.7907,59.1112,59.1112,15.7907,15.7907,59.1112,59.1112


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3D Solid H 637 C3D20R 72.6807, 72.6807, 107.34, 107.34, 72.6806, 72.6806, 107.34, 107.34

3D Solid H 2 C3D20R -72.7376, -72.7376, -107.388, -107.388, -72.7375, -72.7375, -107.388, -

107.388


Plane stress 250 CPS8R -254.631,-254.631, -232.976, -232.976

Plane stress 251 CPS8R -217.123,-217.123,-195.467,-196.467

Plane stress 702 CPS8R 195.467,195.467,217.123,217.123

Plane stress 703 CPS8R 232.976, 232.976, 254.631, 254.631

Plane stress 704 CPS8R 270.484,270.484,292.139,292.139

Plane stress 252 CPS8R -179.615, -179.615,-157.959,-157.959

Plane stress 701 CPS8R 157.959,157.959, 179.615, 179.615

Plane stress 700 CPS8R 120.451, 120.451,142.106,142.106

Plane Strain analysis:

PART-1-1 1009 CPE8R 1 -1.34727E-09

PART-1-1 1009 CPE8R 2 -1.298E-09

PART-1-1 2816 CPE8R 3 1.298E-09

PART-1-1 2816 CPE8R 4 1.34727E-09

PART-1-1 1010 CPE8R 1 -1.26193E-09

PART-1-1 1010 CPE8R 2 -1.21267E-09

PART-1-1 2815 CPE8R 1 1.21267E-09

PART-1-1 2815 CPE8R 2 1.26193E-09

PART-1-1 1011 CPE8R 3 -1.1766E-09

PART-1-1 1011 CPE8R 4 -1.12734E-09

PART-1-1 2813 CPE8R 3 1.04201E-09

PART-1-1 2813 CPE8R 4 1.09127E-09

PART-1-1 1013 CPE8R 3 -1.00594E-09


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PART-1-1 1013 CPE8R 4 -956.676E-12

PART-1-1 2813 CPE8R 3 1.04201E-09

PART-1-1 2813 CPE8R 4 1.09127E-09

Shell Element analysis:

SHELL-1 704 S8R -270.543, -292.203, -270.543, -292.203

SHELL-1 249 S8R 292.203, 270.543, 292.203, 270.543

SHELL-1 703 S8R -233.027, -254.687, -233.027, -254.687

SHELL-1 250 S8R 254.687, 233.027, 254.687, 233.027

SHELL-1 702 S8R -195.51, -217.17, -195.51, -217.17

SHELL-1 251 S8R 217.17, 195.51, 217.17, 195.51

Axis Symmetry analysis:

AXIS-1 385 CAX8R -0.706836, -0.637737, -0.652838, -0.585053

AXIS-1 840 CAX8R 0.433805, 0.436208, 0.386828, 0.388488

AXIS-1 386 CAX8R -0.586778, -0.519089, -0.535563, -0.470037

AXIS-1 839 CAX8R 0.422383, 0.430519, 0.37724, 0.384274

AXIS-1 387 CAX8R -0.47105, -0.408337, -0.424284, -0.365501

AXIS-1 838 CAX8R 0.395947, 0.413343, 0.354457, 0.36944


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Discussion

The main objective of the assignment is to get familiar with effects like Shear locking,

hourglassing and what type of element should be used. ABAQUS has given us the

understanding that we can repeat the analysis many times until we are satisfied with

them, as the data collected from ABAQUS is then compared to the manual

calculations for the better accuracy. Although we can repeat the calculations as

many time on ABAQUS but if the primary inputs are not fed correctly then the model

is of no use in real life even if the results match the manual calculations.

While doing the assignment there were few problems encountered such as shear

locking, hourglassing, meshing, selecting element type and boundary conditions.

When assigning the section, the section assignments would ask for a new section

every time as in shell element the thickness was to be provided while creating a new

section which was cleared after the help of the tutors.

While meshing if the seed size is small, it is difficult to find the result of the nodes

required as the calculations and time is increased hence the data also increases, for

which we should opt a smaller seed size when required.

The seed value used for all the models are different as such as for 3D solid full

model the seed value has been taken as 4 whereas for plane stress element it is

taken as 3. Due to the seed value difference the value of Y in calculations above has

changed for both the cases thus making it more accurate in case of plane stress

element.

We can see from looking at the results that the strain values for plane strain

elements are very low which is because there is no strain element in the model ie: no

strain is occurring at any point in bending process.

We can validate our results as the moment found in both the computational and

theoretical results are same thus we can say the calculations are correct.


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Conclusion

This assignment has proven very beneficial in understanding the effects of shearlocking, hourglassing, and selection of specific element type. There were few errors

found during the meshing process and sectioning which were then rectified and

corrected with the help of tutors. Abaqus has been really helpful to me in designing

of car chassis for my project as I can test the stresses on the rods used, by putting

them in abaqus, thus I can say that abaqus is really useful in real life.


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References

Lecture Notes

http://www.mscsoftware.com/events/vpd2006/na/presentations/tech_papers/27.pdf

http://www.st.bv.tum.de/content/teaching/fem1/fem1.A3.pdf

www.cmmacs.ernet.in/books/ch06.pdf

http://books.google.com/books?id=axqiAQAACAAJ&hl=en
http://www.mscsoftware.com/events/vpd2006/na/presentations/tech_papers/27.pdfhttp://www.mscsoftware.com/events/vpd2006/na/presentations/tech_papers/27.pdfhttp://www.st.bv.tum.de/content/teaching/fem1/fem1.A3.pdfhttp://www.st.bv.tum.de/content/teaching/fem1/fem1.A3.pdfhttp://www.cmmacs.ernet.in/books/ch06.pdfhttp://www.cmmacs.ernet.in/books/ch06.pdfhttp://www.cmmacs.ernet.in/books/ch06.pdfhttp://www.cmmacs.ernet.in/books/ch06.pdfhttp://books.google.com/books?id=axqiAQAACAAJ&hl=enhttp://books.google.com/books?id=axqiAQAACAAJ&hl=enhttp://books.google.com/books?id=axqiAQAACAAJ&hl=enhttp://www.cmmacs.ernet.in/books/ch06.pdfhttp://www.st.bv.tum.de/content/teaching/fem1/fem1.A3.pdfhttp://www.mscsoftware.com/events/vpd2006/na/presentations/tech_papers/27.pdf


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Appendix

Mean stress (

Calculate Moment (M):

Where, I = moment of inertia

Y = distance of mean stress from centre line.

Calculate Radius of Curvature (

Where, E= young modulus

M = moment

I = moment of inertia

Stresses and Deformation

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