CHAPTER02-Review Topics [Compatibility Mode].pdf

2/1/2011

1

Tuesday, February 01, 2011

Review Topics Dr.Eng.Mohammad R. KHEDMATI 1

Review Topics



Contents

Matrix AlgebraMatrix AlgebraMatrix OperationsMatrix OperationsBasic ElasticityBasic Elasticity

2/1/2011

2



Matrix Algebra

A matrix is an m x n array of numbers arranged in m rows and n columns.

m = n A square matrix. m = 1 A row matrix. n = 1 A column matrix. aij Element of matrix a row i, column j



Matrix Operations

Multiplication of a matrix by a scalar. [a] = k [c] aij = kcijAddition of matrices.Matrices must be of same order (m x n)Add them term by term[c] = [a] +[b] cij = aij + bij

2/1/2011

3



Matrix Operations

Multiplication of two matricesIf [a] is m x n then [b] must have n rows[c] = [a] [b]

n

ij ie eje 1

c a b=

=



Transpose of a matrix:Interchange of rows and columns

If [a] is m x n then [a]T is n x m

If [a] = [a]T then [a] is symmetric. [a] must be a square matrix

Matrix Operations

T

ij jia a =

2/1/2011

4



Matrix Operations

The identity matrix (or unit matrix) is denoted by the symbol [I]:

[a][I] = [I][a] = [a]

[ ]1 0 0

I 0 1 00 0 1

=



Matrix Operations

The inverse of a matrix is such that:

[ ][ ] 1a a [I ]- =

2/1/2011

5



Matrix Operations

Differentiating a matrix:

[ ] ijdad a

dx dx

=



Matrix Operations


[ ]

=

yx

aaaa

xU2221

1211

y 21

=

yx

aaaa

yUxU

2221

1211

We usually have an expression of the following form in structural analysis theory:

For which we have:

2/1/2011

6



Matrix Operations


{ } [ ]{ }XaXU T21

=

[ ]{ }XaxU

i=

In a general form, we have:

For which we have:



Matrix Operations

Integrating a matrix.

ij[a]dx a dx =

2/1/2011

7



Finding the Inverse of a Matrix

Need to find the determinant

Need to find the co-factors of [a]

determinant of matrixa [a]=



Cofactors

Cofactors of [aij] are given by:

Then :

[ ]ijwhere matrix d is the first minor

of a and is matrix a

with row i and column j deleted.

i jijC ( 1) d

+= -

[ ]T1ij

C[a ]

a- =

2/1/2011

8



Sets of Linear Algebraic Eqs.

Cramers RuleInverse MethodGaussian EliminationGauss-Seidel Iteration



Cramers Rule[ ]

o r i n i n d e x n o t a t i o n :

L e t m a t r i x b e m a t r i xw i t h c o l u m n i r e p a c e d b y .

T h e n :

n

i j j ij 1

( i )

( i )

i

a { x } { c }

a x c

[ d ] [ a ]{ c }

dx

a

=

=

=

=

2/1/2011

9



Example:

Consider the following equations:

1 2 3

1 2 3

2 3

x 3x 2x 22x 4x 2x 1

4x x 3

- + - =

- + =

+ =



Example:

=

-

--

3

1

2

x

x

x

140

242

231

3

2

1

:formmatrixIn

2/1/2011

10



1.41041

140

242

231143

241

232

)1(

1 =--

=

-

--

-

-

==a

dx

Solution:



1.1

140

242

231130

212

221

a

dx

)2(

2 =

-

--

--

==

2/1/2011

11



4.1

140

242

231340

142

231

a

dx

)3(

3 -=

-

--

-

-

==



Inverse Method

[ ]{ } { }[ ] [ ]{ } [ ] { }[ ]{ } [ ] { }{ } [ ] { }

1 1

1

1

a x c

a a x a c

I x a c

x a c

- -

-

-

=

=

=

=

2/1/2011

12



Example

=

-

--

312

140242231

3

2

1

xxx



Example

-=

--=

-

--=

-

4.11.11.4

312

2.04.08.02.01.02.02.01.12.1

312

140242231

3

2

1

3

2

1

1

3

2

1

xxx

xxx

xxx

2/1/2011

13



Gaussian Elimination

=

n

2

1

n

2

1

nn2n1n

n22221

n11211

c

c

c

x

x

x

aaa

aaa

aaa

MM

K

MMM

K

K

General System of n equations with n unknowns:



Steps in Gaussian Elimination

Eliminate the coefficient of x1 in every equation except the first one. Select a11 as the pivot element. Add the multiple -a21/ a11 of the first row

to the second row. Add the multiple -a31/ a11 of the first row

to the third row. Continue this procedure through the nth

row

2/1/2011

14



After this Step:

=

n

2

1

n

2

1

nn2n

n222

n11211

c

c

c

x

x

x

aa0

aa0

aaa

MM

K

MMM

K

K




Eliminate the coefficient of x2 in every equation below the second one. Select a22 as the pivot element. Add the multiple -a 32/ a 22 of the second

row to the third row. Add the multiple -a 42/ a 22 of the second

row to the fourth row. Continue this procedure through the nth

row

2/1/2011

15



After This Step:

=

n

2

1

n

2

1

nn3n

n333

n22322

n1131211

c

c

c

x

x

x

aa00

aa00

aaa0

aaaa

MM

K

MMMM

K

L

K




Repeat the process for the remaining rows until we have a triangularized system of equation.

=

-- 1nn

4

3

2

1

n

4

3

2

1

1nnn

n444

n33433

n2242322

n114131211

c

cccc

x

xxxx

a0000

aa000aaa00aaaa0aaaaa

MML

MMMMMLLLL

2/1/2011

16



Solve Using Back-substitution

-=

=

+=+

-

-

n

1irrir1n,1

iii

1nnn

1nn

n

xaaa1x

acx



Example

=

649

xxx

111012122

3

2

1

2/1/2011

17



Eliminate the coefficient of x1 in every equation except the first one. Select a11 =2 as the pivot element.Add the multiple -a21/ a11 = -2/2 = -1 of the

first row to the second row.Add the multiple -a31/ a11 = -1/2=-0.5 of

the first row to the third row.



Step 1

-=

--

5.15

9

xxx

5.000110

122

3

2

1

2/1/2011

18




Eliminate the coefficient of x2 in every equation below the second one. Select a22 as the pivot element. (Already done in this example.)



Step 2

-=

--

5.15

9

xxx

5.000110

122

3

2

1

2/1/2011

19



Solve Using Back-substitution( )( )

( )

( ) 12

3)2(29x

21

35x

32

12

3

acx

2

2

33

33

=--

=

=-

+-=

==

=



Gauss-Seidel Iteration

( )

( )

( )1n1n.n22n11nnnn

n

nn2323121222

2

nn1313212111

1

xaxaxaca1x

xaxaxaca1x

xaxaxaca1x

:forminequationsWrite

------=

----=

----=

L

M

L

L

2/1/2011

20



Gauss-Seidel IterationAssume a set of initial values for unknowns. Substitute into RHS of first equation. Solve for new value of x1Use new value of x1and assumed values of other xs to solve for x2 in second equation.Continue till new values of all variables are obtained.Iterate until convergence.

RHS=Right Hand Side



Example

1x1x1x21x

6x2x6xx4x5xx4x2xx4

4321

43

432

321

21

-====

=+-=-+-

=-+-

=-

2/1/2011

21



Example

( ) ( )

( ) ( )( ) ( )( )

( ) ( ) 16.067.1221x221x

672.1168.1641xx64

1x

68.114354

1xx541x

43124

1x241x

34

423

312

21

-=+-=+-=

=-++=++=

=++=++=

=+=+=



Example

( ) ( )

( ) ( )

( ) ( )( )

( ) ( ) 28.0.0944.1221x221x

944.116.0899.1641xx64

1x

899.1672.1922.0541xx54

1x

922.068.1241x24

1x

34

423

312

21

-=+-=+-=

=-++=++=

=++=++=

=+=+=

2/1/2011

22



Iteration x1 x2 x3 x4 0 0.5 1.0 1.0 -1.0 1 0.75 1.68 1.672 -0.16 2 0.922 1.899 1.944 -0.028 3 0.975 1.979 1.988 -0.006 4 0.988 1.9945 1.9983 -0.0008 Exact 1.0 2.0 2.0 0.0 4



Basic Elasticity

For Linear, homogeneous, isotropic material behavior.

2/1/2011

23



Differential Equations of Equilibrium

xyx xzb

xy y yzb

yzxz zb

X 0x y z

Y 0x y z

Z 0x y z

ts t+ + + =

t s t

+ + + =

tt s+ + + =

Body ForcesUnit: unit of force/unit of volume



Strain-Displacement Relationships

(u,v,w) are the x, y and z components of displacement

x xy

y xz

z yz

u u v x y xv u w y z xw w v z y z

= = +

= = +

= = +

2/1/2011

24



Stress-Strain Relationships

( ) ( )

x x

y y

z z

xy xy

yz yz

zx zx

1 0 0 01 0 0 0

1 0 0 01 2E 0 0 0 0 0

21 1 21 20 0 0 0 0

21 20 0 0 0 0

2

-n n n n -n ns e n n -ns e - ns e = t g+n - n - n t g

t g - n



3D Stress-Strain Matrix

[ ] ( ) ( )

( )Note :

1 0 0 01 0 0 0

1 0 0 01 2E 0 0 0 0 0D 21 1 2

1 20 0 0 0 02

1 20 0 0 0 02

EG2 1

-n n n n -n n n n -n - n

= +n - n - n - n

=+n

2/1/2011

25



Plane StressPlane Stress Matrix

[ ] ( )

1 0ED 1 0

110 0

2

n

= n - n - n

0=== yzxzz tts



Plane StrainPlane Strain Matrix

[ ] ( ) ( )

1 0ED 1 0

1 1 21 20 0

2

-n n

= n -n +n - n - n

0=== yzxzz gge

Documents

CHAPTER02-Review Topics [Compatibility Mode].pdf