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Section 2: FINITE ELEMENT ANALYSIS –FUNDAMENTALS

Displacement Based Finite Element Analysis

Fi it l t l i i l d li t t l t i ll di t (iFinite element analysis involves modeling structural components using small, discrete (i.e., ”finite”) interconnected elements. A displacement interpolation function is assigned to each finite element and every element is directly or indirectly connected to all other elements through common interfaces that are typically nodes, and/or common boundaries (lines or g yp y (surfaces).

The total set of equations obtained by describing the physical behavior of these nodes represents a system of equations that is best described in a matrix format The solution stepsrepresents a system of equations that is best described in a matrix format. The solution steps are as follows:

1. Select the element type which infers the displacement function is specified 2. Discretize the component3. Define a constitutive relationship (stress – strain relationship)4. Assemble the element stiffness matrix5. Assemble the component, or global stiffness matrix6 Solve the system of equations for unknonwn nodal displacements6. Solve the system of equations for unknonwn nodal displacements7. Solve for Element Strains and Stresses

Before we start we review a few fundamentals of matrix algebra and elasticity.


Matrix NotationGiven

nn

nn

xaxaxabxaxaxab

+++=+++=

KKKKK

K

K

22221212

12121111

in short hand the above equation appears as

nnnnnn xaxaxab +++= K2211

where

{ } [ ]{ }xab =

{ }

=bb

b 2

1

{ }

= n

n

aaaaaa

aK

K

22221

11211

{ }

=xx

x 2

1

nbK

nnnn aaa K

KKKK

21

nxK


Review – Matrix Algebra

We will see that the finite element method requires solving large systems of linear equations using matrix algebra tools.

First, review matrix multiplication

×+××+××+×

92421102719261825176521

=

×+××+××+×

=

2554471047294628452109843

2x2 2x3 2x3

The above can be represented in a general format:

[A] [B] = [C]n x pm x n m x p


To review some basic matrix algebra, consider the following system of equations 8

9

10

equations

17462

=+=+

yxyx

5

6

7

(x,y)= (1,4)

or

17162 +−= xy

1

2

3

4

44+−= xy

0 1 2 3 4 5 60

In a matrix format the system of equations are represented as

[ ]{ } { }=

=

176

4112

bxaoryx

[ ] { }

=

=

=

176

,,4112

byx

xa


MATLAB code MATLAB t t

We can make use of MATLAB to solve the system of equations. Given below are the input commands and the resulting output.

MATLAB code

>> a = [2 1;1 4]

MATLAB output

a =2 1

>> b = [6;17]

2 11 4

b =>> b = [6;17] b 6

17

>> determinant = det(a) determinant =7

Note |a| ≠ 0

>> solution = a\b solution =1.00004.0000


Now consider a different system of equations

1624 =+ yx 8

9

10

621624

=+=+

yxyx

4

5

6

7

or

6282

+−=+−=

xyxy

0 1 2 3 4 5 60

1

2

3

or

0 1 2 3 4 5 6

Here we can see clearly that the lines being represented by the equations are parallel. In a matrix format:

1624 [ ]{ } { }bxaoryx

=

=

6

161224

wherewhere[ ] { }

=

=

=

616

,,1224

byx

xa


MATLAB code MATLAB output

Again using of MATLAB to solve the system of equations leads to

a =4 22 1

>> a = [4 2;2 1]MATLAB code MATLAB output

2 1b =

166

>> b = [16;6]

6determinant =

0Warning: Matrix is singular to working

>> determinant = det(a)

>> solution = a\b g g gprecision.

solution =Note |a| = 0Inf

-InfNote |a| = 0


Consider the solution of a system with three equations and three variables represented graphically in the following figure

2.5

3

MATLAB Code:

>> a=[-1 1 2;3 -1 1;3 3 1]

MATLAB Output:

a =-1 1 2

1.5

2

>> b=[2;6;2]

-1 1 23 -1 13 3 1

b =

11.05

1.11.15

1 05

-1

-0.95

-0.91

[ ] b 262

0.90.95

1

-1.1

-1.05

>> solution=a\b solution =1-12

− 2211 x

2

=

−26

133113

zy


Matrix Algebra Applied to FEA

For displacement based stress analysis

DKKKF 1112111 ...

=

n

DD

KKKKK

FF

MM

M

M2

1

2212

11211

2

1 ...

nnnnn DKKF

MMM

1 ...

Stiffness matrix -Symmetric since Kij = Kji

Nodal DisplacementsNodal Forces


Matrix Transpose

– The transpose of a matrix is found by interchanging rows and columns, e.g.

[ ] [ ]

=

=

642531

654321

TAA

– Transpose of a product

65

[ ] [ ]( ) [ ] [ ]TTT ABBA =


Matrix Differentiation

One can differentiate a matrix by differentiating every element of the matrix in theOne can differentiate a matrix by differentiating every element of the matrix in the conventional manner. Consider

[ ]

42

23 32 xxx

The derivative d[a]/dx of this matrix is

[ ]

=5

42

32

xxxxxxa

The derivative d[a]/dx of this matrix is

[ ]

3

2 343 xxad[ ]

=4

3

513144

xxx

dxad


Similarly, one can take the partial derivative of a matrix as follows

[ ]

32 223 xyzxyx

[ ]

∂∂

=∂∂

32

52

42

zyxyyxzzxyy

xxa

=

00304

343

2

4

22

yzxz

yxzyx

In structural analysis we differentiate strain energy potential functions that have the form

y

xaa1

via matrix multiplication

[ ]

=

yx

aaaa

yxU2221

1211

21

( )22212

211 2

21 yaxyaxaU ++=


Partial differentiation leads to

yaxaU1211 +=

∂ yx 1211∂

yaxayU

2212 +=∂∂

Or in matrix format

∂∂

xaaxU

1211

y∂

With U defined in t o dimensions from the pre io s page

=

∂∂∂

yx

aaaa

yUx

2221

1211

With U defined in two dimensions from the previous page

[ ]yx

aaaa

yxU21 1211

=

{ } [ ]{ }XaX

yaa

T

212 2221

=


then

[ ]{ }XU∂

Here xi represents x and y using index notation or

[ ]{ }Xaxi

=∂

{ }yx

xi

=

The above holds only if [a] is a symmetric matrix. We can do the same in three dimensions or in n-dimensions.

{ }X=


Matrix Integration

One can differentiate a matrix by differentiating every element of the matrix in the y g yconventional manner. Consider

[ ]

= 3

2

144343

xxxx

a

The integration of this matrix is 23

[ ]

4513

144x

xxa

[ ]

=∫5

42

23

32

32

xxxxxxxxx

dxa

We often integrate the expression

[ ] [ ][ ] dddXAX T∫∫∫This triple product will be symmetric if [A] is symmetric

[ ] [ ][ ] dzdydxXAX∫∫∫


Other Matrix Terminology

• Banded matrix– If all non-zero terms are contained within a band along the diagonal,

the matrix is said to be banded

• Sparse matrix– If a matrix has relatively few non-zero terms (as is common in FEA), y ( ),

the matrix is said to be sparse

• Singular matrixg– If the determinant of the matrix equals zero, the matrix is said to be

singular. As we saw, if [A] is singular, then the system of equations [A]{x}={b} has no unique solution.


Stress at a PointOne can use the state of stress at aOne can use the state of stress at a point depicted by the cube at the right to demonstrate that equilibrium is defined by the indicial expression:

0=+∂

∂j

i

ij Bxσ

013

31

2

21

1

11 =+∂∂

+∂∂

+∂∂ B

xxxσσσ

In expanded notation:

0

0

332313

23

32

2

22

1

12

=+∂

+∂

+∂

=+∂∂

+∂∂

+∂∂

B

Bxxxσσσ

σσσ

03321

=+∂

+∂

+∂

Bxxx


In terms of a coordinate system labeled (x, y, z) then:

0=+∂∂

+∂

∂+

∂∂

xz

zx

y

yx

x

xx Bxxxσσσ

0

0

+∂

+∂

+∂

=+∂

∂+

∂

∂+

∂

∂

zzyzxz

yz

zy

y

yy

x

xy

B

Bxxx

σσσ

σσσ

0=+∂

+∂

+∂ z

z

zz

y

y

x

xz Bxxx


Strain Displacement Relationships

F El i i h f ll i li l i hi b i d di l

∂ ∂∂∂1

From Elasticity the following nonlinear relationship between strain and displacement was derived

∂∂

∂∂

+∂

∂+

∂∂

=i

l

j

l

i

j

j

iij x

uxu

xu

xu

21ε

The nonlinear expression for the strain displacement relationship is utilized in the advanced course in Finite Element Analysis. The expression can be linearized as follows:

∂

∂+

∂∂

=i

j

j

iij x

uxu

21ε

We will utilize the linear form in this class.


Define a Stress-Strain Relationship

Stress-strain relationships are key to producing results that are meaningful to the engineerStress strain relationships are key to producing results that are meaningful to the engineer. If we identify stress using indicial notation as follows:

σij – stress on the i’th face of the stress cube in the j’th direction

And strain as

∂

∂+

∂∂

= jiij x

uxu

21ε

Then in Elasticity or Continuum Mechanics we learned that stress is related to strain in the following manner.

∂∂ ij xx2

Here γijkl is a fourth order tensor of material constants. Refer to CVE 604 Elasticity notes posted to my web page. In a uniaxial format

klijklij εγσ =

p y p g

xx EorE εσεσ == 1111


In a matrix format stress is related to strain in the following manner:

However due to symmetry of the material, this 6 x 6 matrix is symmetric and populated by a number of zeroe entries.a u be o e oe e t es.


Element Types

Initially we will categorize elements by their geometric nature, i.e.,:

1. Simple line elements (springs)

2 T di i l l t d t t l t i d l t2. Two dimensional elements used to represent plane strain and plane stress

3. Three dimensional elements (see figure)

4. Axisymmetric elementsy

The component being analyzed is discretized and a choice is made as to which type of element is used based onwhich type of element is used based on boundary conditions (loads and supports) as well other issues that will become evident as the course proceeds.


Types of Elements - Examples

One-Dimensional ElementsLine, Rods, Beams, Trusses

Two-Dimensional ElementsTriangular, Quadrilateral

Plates, Shells, 2-D Continua

Three-Dimensional ElementsTetrahedral, Rectangular Prism (Brick)

3-D Continua


Three DimensionalOne-Dimensional Frame Elements

Two-Dimensional Triangular Elements

Three-Dimensional Brick Elements


Common Sources of Error in FEA

1. Domain Approximation

2 Element Interpolation/Approximation2. Element Interpolation/Approximation

3. Numerical Integration Errors(Including Spatial and Time Integration)

4. Computer Errors (Round-Off, Etc., )


Measures of Accuracy and Convergence

Because FEA produces an approximate solution, we must establish metrics for accuracy. Consider the following generalized defintion for error (should we use precision?):

Error = |(Exact Solution) - (FEM Solution)||( ) ( )|

Thus we need methods that will speed convergence to the exact solution. We will limit error in one of two ways:

Increase the Number of Elements (h-convergence) (brute force approach)

or

Increase the Approximation Order (p convergence) (elegant approach)Increase the Approximation Order (p-convergence) (elegant approach)

Error will tend to zero as the number of elements, or the approximation order goes to infinity.


Two-Dimensional Discretization Refinement

(Node)•

(Discretization with 228 Elements)

(Triangular Element)•

•

(Discretization with 912 Elements)


Displacement Interpolation FunctionsDisplacements throughout an element are defined by an interpolation function(s) that isDisplacements throughout an element are defined by an interpolation function(s) that is dependent on displacement of the nodes. These functions can be linear, quadratic or cubic polynomials. The linear displacement function for a spring element will have the form

For a two dimensional element the the displacement function is a function of the coordinates in is plane (either the x-y plane or the x1-x2 plane) The displacement

xccu 21 +=

coordinates in is plane (either the x y plane or the x1 x2 plane). The displacement functions are expressed in terms of the unknown nodal displacements, i.e., if nen is the number of element nodes, then the displacement at the ith node is defined as

{ } enxiyixii niqqqq ,,2,1,, L==

The displacement at any point within the element is then

{ } enxiyixii qqqq ,,,,,

{ } { }[ ]{ }

wvuu ,,=

Where [ f ] is a matrix of shape functions, which we will talk about in detail.

[ ]{ } eni niqf ,,2,1 L==


Element Stiffness Matrix

Three methods to determine element stiffness matrix:Three methods to determine element stiffness matrix:

1. Direct Equilibrium Method2. Work or Energy Method3. Methods of Weighted Residuals

Using any of the methods will produce the element stiffness matrix given in their matrix format as:format as:

n

n

dd

kkkkkk

ff

K

K

2

1

22221

11211

2

1

=

nnnnn

n

n dkkkf

fK

K

KKKKK2

21

222212


The individual stiffness matrix for each element in the discretized model must be

Global Stiffness Matrix

combined in some fashion in order to conduct the structural analysis of a component.

We will utilize superposition to accomplish this (often referred to as the direct stiffness method). Force equilibrium is imposed at each node and compatibility (continuity) is ) q p p y ( y)required at the nodes and along the boundaries of two and three dimensional elements. We will not allow the creation of new surfaces in this class (see Fracture Mechanics). This process produces a global force-displacement relationship for the component that appears as follows:appears as follows:

=

n

n

DD

KKKKKK

FF

K

K

2

1

22221

11211

2

1

=

nnnnnn DKKKFK

K

KKKKK

21

The matris [K], the global stiffnes matrix, is singular (its determinant is zero). To remove this singularity boundary conditions must be imposed. This will remove rigid body motion from the analysis.

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