Ioannis Vazaios 10123567 CIVL841 Assignment 2-Elastic Finite Element Analysis

8/12/2019 Ioannis Vazaios 10123567 CIVL841 Assignment 2-Elastic Finite Element Analysis

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Course: CIVL 841-Numerical and Analytical Methods in GeomechanicsName: Ioannis VazaiosSt.ID: 10123567

Assignment: 2-Elastic Finite Element Analysis

Exercise 2a: Derive the shape functions for a 10-Node Element

As illustrated in Figure 1, we have an element in 2 dimensions, positioned in a Cartesian,coordinate system X, Y and it is formed by 10 nodes. Its shape is arbitrary.

Figure 1: 10-Node Element in a Cartesian Coordinate System

Due to the complexity of its geometry we will have to transfer it to a local coordinate system, , as illustrated in Figure 2.

Figure 2: 10-Node Element in a local coordinate system ,

Y

X

6

9

1

3 8

5

10

2

74

6

9

1

3

8

5 10

2 74


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3/23

32

811864822

2812

6567

181622

184053

681

812

922

312

923

31

922

312

32222

323

322

812

31

32

31

0*32

31

*031

1*31

31

*32

*1* 2

4

N

22

2

5 329

29

227

272

31

32

*31

*31

32

**31

N

29

29

227

182

27

272 3

1

3

1

3

1

272

1

3

1

31

32

*032

1*32

3

1*1*

223

23222

6

N

29

29

227

182

27

31

32

*032

1*32

31

*1*223

7

N

222

8 3229

227

9

272

32

32

*31

*31

32

**32

N

322223

9 281

18648

281

6567

18162

18405

681

3

10*

3

2

3

1*

3

101*

3

131

*32

*1*

N

222210 272727

271

31

*31

*31

31

1

**1

N

Therefore, the shape functions formed for a 10-Node element are a 3 rd order polynomial,which could generally be written in the following form:

F(, )=a+b +c+d2+e+f 2+g3+h2+i 2+j3


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Figure 5: 3-Node Element in a normalized coordinate system ,

1001

11 N (2)

12 N (3)

13 N (4)

Therefore, quantities like X and Y are derived by the following relations: X=N1*X1+N2*X2+N3*X3 Y=N1*Y1+N2*Y2+N3*Y3

Secondly, we will form the Jacobian Matrix which is illustrated at Table 2.

y x

y x

J

Table 2: The Jacobian Matrix

So we derive the following relations from the shape functions for the 3-node element:

1232133

22

11 *0*1*1 x x x x x x

N x

N x

N x

1332133

22

11 *1*0*1 x x x x x x

N x

N x

N x

1

3

2

(0, 1)

(0, 0) (1, 0)


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1232133

22

11 *0*1*1 y y y y y y

N y

N y

N y

1332133

22

11 *1*0*1 y y y y y y

N y

N y

N y

And the Jacobian Matrix has the values illustrated in the following table.

1313

1212

y y x x

y y x x J

And the determinant of the Jacobian Matrix is derived by the following relation:

detJ=(x 2-x1)*(y3-y1)-(x3-x1)*(y2-y1)

Continuing, the reverse matrix of the Jacobian is :

1231

31131

det1

det1

x x x x

y y y y

J x x

y y

J y y

x x J

After we have set the Jacobian Matrix and its reverse we will continue with deriving thestrain-displacement matrix B. Assuming the strain vector [ ], we can apply the followingrelation, [ ]=[][u]T. Therefore, the usefulness of matrix B is that connects the strains withthe displacements and we get the following matrices:

33

22

11

332211

321

321

000

000

*

vu

vu

vu

x N

y N

x N

y N

x N

y N

y N

y N

y N

x N

x N

x N

xv

yu

yv xu

u T

But the shape functions N i for i=1 to 3 are functions of , thus we have to apply the chainrule and in this way we get:

J y y

x N

x N

x N

det32111

J x x

y N

y N

y N

det23111

J y y

x N

x N

x N

det13222

J x x

y N

y N

y N

det31222


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J y y

x N

x N

x N

det21333

J x x

y N

y N

y N

det12333

So the B matrix is formed and it has generally the following form:

363534333231

262524232221

161514131211

B B B B B B

B B B B B B

B B B B B B

B

And its transposed one:

636261

535251

434241

333231

232221

131211

362616

352515

342414

332313

322212

312111

X X X

X X X

X X X

X X X

X X X

X X X

B B B

B B B

B B B

B B B

B B B

B B B

B T

In this way we have derived all the necessary components to form our stiffness matrix K.Initially, we will form the product between the matrices B T and D.

X63D33X62D23X61D13X63D32X62D22X61D12X63D31X62D21X61D11




D23D33X22D23X21D13X23D32X22D22X21D12X23D31X22D21X21D11X13D32X12D23X11D13X13D32X12D22X11D12X13D31X12D21X11D11

DTB

And thus the W matrix is formed:

636261

535251

434241

333231232221

131211

W W W

W W W

W W W

W W W W W W

W W W

D BW T

And


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.. ... ... ... .326322621261316321621161

.. ... ... ... .325322521251315321521151

.. ... ... ... .324322421241314321421141

.. ... ... ... .323322321231313321321131

.. ... ... ... .322322221221312321221121

.. ... ... ... .321322121211311321121111

BW BW BW BW BW BW

BW BW BW BW BW BW

BW BW BW BW BW BW

BW BW BW BW BW BW

BW BW BW BW BW BW

BW BW BW BW BW BW

BW

In this way we have created the Matrix A, where [A]=[B] T[D][B], and matrix A has thefollowing form:

666564636261

565554535251

464544434241

363534333231

262524232221

161514131211

A A A A A A

A A A A A A

A A A A A A

A A A A A A

A A A A A A

A A A A A A

A

At this point we have to note that Matrix A is a 6x6 matrix and A ij=(number) for i, j= 1 to 6and not a function of , . Thus, by using equation (1) the product of B, B T and D matrices canget out of the integral due to the fact that it contains numerical values and not functions. So:

V

T

V

T dV B D BdV B D B K

By assuming a constant thickness t for our 3 node element the integral gets the form:

A

dA1 =X+C thus AC C AC X dA A

A 010

0

In which A is the area of our triangular element and the stiffness matrix K gets the followingform:

[K]=t*A*[B]T[D][B]

So the first element of the stiffness matrix K 11 is the product K 11 =t*A*A11 , in which A 11 isequal to:

W11=X11D11+X12D21+X13D31=)21)(1(

)1(

det00

)21)(1(

)1(

det

3232

vv

v E

J

y y

vv

v E

J

y y

W12=X11D12+X12D22+X13D32=)21)(1(det

00)21)(1(det

3232

vv Ev

J y y

vv Ev

J y y

W13=X11D13+X12D23+X13D33=)1(2det)1(2det

00 2323v

E J x x

v E

J x x

A11=W11B11+W12B21+W13B31= 2232322 ))(21())(1(2))(det21)(1(2 x xv y yv J vv E


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Thus for the plane strain case:

K11 = 2232322 ))(21())(1(2))(det21)(1(2** x xv y yv J vv E

At but detJ=2A trinagle so:

K11 = 223232 ))(2/1())(1()21)(1(4 x xv y yvvv A Et

Exercise 2c: Consider how the node numbering in Figure 2.4 (of notes) influences thebanded nature of the global equations (Consider the case where the first element on theleft has nodes 1, 2 and 5). Derive an expression for bandwidth as a function of the nodenumbers for an element with nodes i, j and k (where i


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By combining the matrices above we get the total stiffness matrix for the three elements asillustrated in figure 7.

Figure 7: Total Stiffness Matrix. The diagonal elements are illustrated with the black line and the bandwidth isillustrated with the red lines (approximately).

Continuing, we will take into consideration the second case scenario in which the firstelement is formed by nodes 1, 2 and 5. In this particular case we will have the following

stiffness matrices for each element (Figure 8).

Figure 8: 3-Stiffness Matrices for each of the three elements 1, 2, 3 for the second case scenario

By combining the matrices above we get the total stiffness matrix for the three elements asillustrated in figure 9.

1 2 3 4 5 6 7 8 9 10 Ui PiK1 K1 K1 K1 K1 K1 0 0 0 0 U1 P1K1 K1 K1 K1 K1 K1 0 0 0 0 U2 P2

K1 K1 K1+K2 K1+K2 K2 K2 K2+K3 K2+K3 K3 K3 U3 P3K1 K1 K1+K2 K1+K2 K2 K2 K2+K3 K2+K3 K3 K3 U4 P4K1 K1 K2 K2 K2 K2 K2 K2 0 0 U5 P5K1 K1 K2 K2 K2 K2 K2 K2 0 0 U6 P60 0 K2+K3 K2+K3 K2 K2 K2+K3 K2+K3 K3 K3 U7 P70 0 K2+K3 K2+K3 K2 K2 K2+K3 K2+K3 K3 K3 U8 P80 0 K3 K3 0 0 K3 K3 K3 K3 U9 P90 0 K3 K3 0 0 K3 K3 K3 K3 U10 P10

K1 1 2 3 4 5 6 Ui Pi1 u1 P12 u2 P23 u3 P34 u4 P45 u9 P96 u10 P10




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Figure 9: Total Stiffness Matrix. The diagonal elements are illustrated with the black line. The black arrows are

used as indicators of the bandwidth of the total stiffness matrix in figure7.

By comparing the total stiffness matrices in figures 7 and 9 in can be observed that thebandwidth of the matrix of the second scenario has increased due to the change of thenumbering of the nodes, a fact that is going to affect both the memory and thecomputational time required to solve the problem (Increase in the bandwidth of the matrixresults in additional memory requirements and computational time).

For the second part of this exercise we will derive an expression for bandwidth as a functionof the node numbers for an element with nodes I, j and k, in which I


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Therefore, we get the two following expressions: Half of the bandwidth=(DOF per node)*[(Maximum difference of node number

between directly connected nodes)+1] Half of the bandwidth=(Maximum difference of global DOF of an element)+1

As an example we will use the matrices in figures 7 and 9. By using the first and secondexpression respectively for the first matrix we get:

b/2 =2*[(5-2)+1]=8 (element 3)b/2 =(10-3)+1=8 (element 3)

And if we count the elements of the last column, until we get a zero value, we have 8 matrixelements. For the second matrix we get:

b/2 =2*[(5-1)+1]=10 (element 1)b/2 =(10-1)+1=10 (element 1)

And by counting the elements of the last column, we get 10 matrix elements. In this way wealso have a quantitative expression to support our argument above, which due to thechange of the numbering of the nodes at element 1 we get an increased bandwidth asillustrated in figures 7 and 9.

Exercise 2d: Determine the equivalent nodal forces for a uniform body force over a 6-nodetriangle.

In this particular exercise first we will illustrate how we can derive the nodal forces for auniform body force over a 3-node element. For the simplicity of the calculations we will

assume only one component for the body force and this is going to be the one along the xaxis as illustrated in figure 10.

Figure 10: Triangular 3-node and 6-node elements with a uniform body force along the X axis

For the case of the 3 node triangle we know that the shape functions are the following: N1=1-- N2= N3=

x

x


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In order to perform the integrations that follow we will consider the simple case of a singleintegration point with coordinates (1/3, 1/3) in the normalized coordinate system , .Therefore, we get the following for each respective node:

xtriangle xtrianglei xtriangle x A x

A AW N Ad d J N dA N F 3

15.0

3

1

3

1122

1

1 11111

xtriangle xtriangle

i xtriangle x

A

x A AW N Ad d J N dA N F 31

5.031

221

112222

xtriangle xtrianglei

xtriangle x A

x A AW N Ad d J N dA N F 31

5.031

221

113333

The same could be done for the body force component along the Y axis y.Continuing, we will follow the same procedure for the case of a 6-node triangle, which hasthe following shape functions:

N1=(1-- )(1-2 -2) N2=(2-1) N3= (2-1) N4=4(1--) N5=4 N6=4 (1--)

Again, we will a perform a single point integration at (1/3, 1/3) of the normalized coordinatesystem , . Therefore, we get the following for each respective node:

xtriangle xtriangle

xtrianglei xtriangle x A x

A A

AW N Ad d J N dA N F

91

31

31

5.03

12

3

121

3

1

3

1122

1

1 11111

xtriangle xtriangle

xtrianglei

xtriangle x

A

x

A A


91

31

31

5.0131

231

221

112222

xtriangle xtriangle

xtrianglei

xtriangle x

A

x

A A


91

31

31

5.0131

231

221

113333

xtriangle xtriangle

xtrianglei

xtriangle x A

x

A A


94

31

34

5.031

31

131

4221

114444


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xtriangle

xtrianglei

xtriangle x

A

x

A


94

5.031

31

4221

115555

xtriangle xtriangle

xtrianglei

xtriangle x

A

x

A A


94

31

34

5.031

31

131

4221

116666

Furthermore, the weight of the element is calculated by the following expression:

Wx=Aelement x (Intentionally we have ignored the thickness of the element t element )

Thus by adding the nodal forces we should get the weight of the element. Therefore:

F1+F2+F3+ F4+F5+F6=Atriangle x(-1/9-1/9-1/9+4/9+4/9+4/9)= A triangle x=Wx. The same applies ifwe want to calculate the equivalent nodal forces for a uniform body along the y axis y.

Exercise 2e: Derive the equivalent nodal forces for a 6-noded and 10-noded triangle for thefollowing cases.

For the first case of a 6-node element we can observe that the three nodes that are going tobe loaded due to the linear distributed load P y are 1, 2 and 4. Thus the shape functions thatwe are going to use are the following: N1=(1-- )(1-2 -2 ) N2=(2-1) 4(1--)

But all the nodes are located at the lowest boundary of the element where =0, thus we getthe following relations:

021

121

21

223211

1

0

432

31

0

1

0

1

0

11

p

pd pd pd N p F y

Py=P P

y


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61

31

21

31

2212

1

0

341

0

1

0

1

0

23222 p p pd pd pd N p F y

31

41

31

4434414*

1

0

431

0

1

0

1

0

322

44 p p pd pd pd N p F y

Therefore, the equivalent nodal forces are: Node 1: F 1=0 Node 4: F 4=p/3 Node 2: F 2=p/6

So by performing a check we get that 0+p/3+p/6= p/2 =1/2*1*p which is the resultant forceof the linearly distributed load.

For the second case we will examine the 10-node triangle. The uniform load p x is applied atthe lowest boundary thus the nodes which are going to be examined are 1, 4, 7 and 2.Therefore, the shape functions that have to be deployed are the following:

2233221 227

29

1892

111 N

232 29

29

N

6

276

5432

2726

1354 N

29

29

227

182

27 2237 N

But as it has already been mentioned the nodes are located at the lowest boundary where=0 and the components of each relation above that include are reduced to 0 and we getthe following relations:

889

824

822

88

89

34

111

89

34

1129

1892

111

1

0

4321

0

1

0

3211

y y y

y y y

p p p

pd pd N p F

884

812

89

2

1

2

3

8

9

22

3

8

9

2

9

2

91

0

234

1

0

1

0

23

22

y y

y y y y

p p

p pd pd N p F

8

3

18*2162

18*3405

18*4243

18162

21

18405

31

18243

41

181622

184053

681

1

0

2341

0

1

0

44

y y

y y y

p p

pd pd N p F


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Figure 12: Stress and Strain Results for a 100 element mesh



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By comparing the analytical with the numerical analyses results we get the following table 3.

Table 3: Analytical vs Numerical Solution

FEM1 FEM2 FEM3 FEM4100 25 4 1

p (kPa) -100 -100 -100 -100 -100E (kPa) 1000 1000 1000 1000 1000

v 0.1 0.1 0.1 0.1 0.1

syy (kPa) -100 -100 -100 -100 -100sxx (kPa) -11.11 -11.11 -11.11 -11.11 -11.11

eyy -0.0978 -9.778E-02 -9.778E-02 -9.778E-02 -9.778E-02exx 0.0 1E-17 1E-17 1E-17 1E-17

Analytical


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From the table above it can be observed that the simplicity of the problem makes it easy forthe numerical solutions to easily converge with the analytical one, despite the fact that wehave limited the mesh from 100 elements to 1. Therefore, the mesh in this particular casedoes not affect the stress and strain results and in all four cases we have achieved to get thesame vertical stress, strain and horizontal stress result as with the analytical solution.

However, this is mainly because we have a uniformly distributed load and the fact that wehave used solid, quadrilateral, full integration elements to simulate the block. In thehypothetical case of a concentrated load applied at the middle of the block the results areexpected to vary a lot depending on the density of the mesh. Choosing another type ofelements may also make the results vary depending on the density of the mesh even in thecase of the uniform load.

Exercise 2g: Stresses around a circular opening in elastic solid (Kirsch Problem)

For this particular exercise we have assumed the model with the geometrical featuresillustrated at the following figure.

Figure 16: Elastic medium with an opening of a radius= 1m

Because we have assumed a whole model without taking advantage of the vertical andhorizontal axis of symmetry, the boundaries have been set far from the hole in order tominimize the boundary effects. The boundaries are not fixed in any direction and for thisparticular case we have assumed a uniformly distributed load of the same magnitude p=100kPa applied on all four sides of the model, thus we have an isotropic case (and the stresses

are independent to the angle ) to simplify things. Our medium has a linear elastic behaviorwith the following properties: Youngs Elasticity M odulus E=100,000 kPa Poissons ratio v=0.1

In addition, the following analyses were performed: A refined mesh around the hole was generated (Number of elements in total: 42,214

and 80 elements along the periphery of the hole) A less refined mesh around the hole was generated (Number of elements in total:

25,762 and 40 elements along the periphery of the hole) A sparse mesh around the hole was generated (Number of elements in total: 18,462

and 20 elements along the periphery of the hole)

60 m

60 m


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Therefore, different meshes were generated in order to check the effect of the density ofthe mesh on the results of the radial and tangential stress/strain when compared to theanalytical solution. In all cases, the type of elements which was used was plane strain,quadrilateral, full integration elements. All these three analyses were performed following a 2-step sequence. In the first step we

apply the load all over the elastic medium and then we remove the material within theboundaries of the hole. Moreover, an additional analysis was performed using the 25,762-element model in which we reversed the sequence. Thus, in the first step, the material fromthe hole was removed and in the second step we were applying the load. The results of allthe analyses performed are illustrated in the following figures.

Figure 17: Radial stress vs. radial distance

Figure 18: Tangential stress vs. radial distance


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Figure 19: Final stresses occurring in the normal and reverse case

Figure 20: Final strains occurring in the normal and reverse case

From figures 17 and 18 we can see that the more refined the mesh becomes the closes weget to the analytical solution for both the radial and tangential stress at the hole boundary.More specifically, for the least dense mesh the radial stress that we get at the boundary ofthe hole is approximately S rr10 kPa and the tangential stress S 190 kPa , for the morerefined case S rr5 kPa and S 195 kPa and for the most refined mesh we get S rr2.5 kPa andS 198 kPa while the analytical solution is S rr=0 kPa and S =200 kPa.From figures 19 and 20 we can observe that for the same model if the sequence of stepsperformed in the analysis change that does not affect the final results of either the stressesor strains which are the same in both cases. Of course in the geotechnical problems this nottrue. The medium is already in an initial stress condition (geostatic stresses), which vary withthe depth, and when an analysis is performed must be taken into consideration.


22/23

So the correct sequence of events must be simulated in order to get the correct deflectionsof the cavity formed as the excavation process progress, as it is a time dependentphenomenon (ways to simulate it in 2 dimensions is either decrease the elasticity modulusof the material inside the hole or decrease the inner hole pressure to get the pre-convergence effect).

Exercise 2h: How does the size of a 2D element change the stiffness if the aspect ratioremains the same?

In order to examine the effect of the element size to the stiffness of the particular element,we will assume that we have a 3-node, triangular element. This kind of element has a 6X6stiffness matrix (as we have shown in exercise 2b earlier) and it is the elements of this matrixthat form the stiffness of the element. Thus, in order to had a better idea we should find allthese elements that form the aforementioned matrix. That would be very complicatedthough and, because of this fact, we will only examine how the first element of the stiffnessmatrix (we have already found it in exercise 2b) is affected by the change of the size of the 3-

node element.Continuing, for this particular exercise we will assume four elements. The way they areformed is presented at the following table.

Table 4: Four 3-node element cases in which elements 2, 3 and 4 have a 10 times larger area than element 1

It can be observed at the table above that elements 2, 3 and 4 have an area 10 times largerthan this of the element 1 but they differ in shape. Continuing, we examine the ratiosK11i/K11 1 for i=2-4 in order to see the effect of the changing size and shape of the elementon the first element of the stiffness matrix. The results are presented in table 5.

Table 5: K11 i/K11 1 for i=2-4 for five different values of the Poisson s ratio

The reason why the results are presented in this way is because we get to eliminate theeffect of the elasticity modulus (assuming that we have the same material) and the thicknessof the element (which can be assumed that remains constant).

Element Node X Y A1 0 02 0 1 0.53 1 01 0 02 3.16 0 5.03 0 3.161 0 02 4.48 0 5.03 0 2.231 0 02 2.23 0 5.03 0 4.48

1

2

3

4

Poisson's ratio 0.1 0.2 0.3 0.4 0.5

k11 2/k11 1 0.99856 0.99856 0.99856 0.99856 0.99856

k11 3/k11 1 0.961828 0.90904 0.83279 0.712969 0.49729

k11 4/k11 1 1.542502 1.59529 1.67154 1.791361 2.00704


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Thus, the only quantities we have to take into consideration are the Poisson s ratio and thecoordinates of the nodes in the Cartesian system. If we assume that Poisson s ratio is v=0.1then we can see that for element 2, which has the same form as element 1 and it has justbeen magnified by 10 times, the first element of the stiffness matrix in both cases 1and 2 isapproximately the same as the ratio is approximately 1. However, if the shape of the

element changes, the value of K11 can vary greatly as we can observe at table5. Additionally,for elements 1 and 2, regardless the value of the Poisson, ratio K11 2/K11 1 remains the sameand does not change, while in all other cases this ratio varies greatly.Therefore, it can be assumed that between two elements that have the same shape andform changing the size does not really affect the first element of the stiffness matrix. On theother hand if both the shape form and size change then the first element of the stiffnessmatrix can vary greatly.

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Ioannis Vazaios 10123567 CIVL841 Assignment 2-Elastic Finite Element Analysis