CHAPTER02-Review Topics [Compatibility Mode].pdf
Transcript of CHAPTER02-Review Topics [Compatibility Mode].pdf
-
2/1/2011
1
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 1
Review Topics
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 2
Contents
Matrix AlgebraMatrix AlgebraMatrix OperationsMatrix OperationsBasic ElasticityBasic Elasticity
-
2/1/2011
2
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 3
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
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 4
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
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 5
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=
=
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 6
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
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 7
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
=
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 8
Matrix Operations
The inverse of a matrix is such that:
[ ][ ] 1a a [I ]- =
-
2/1/2011
5
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 9
Matrix Operations
Differentiating a matrix:
[ ] ijdad a
dx dx
=
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 10
Matrix Operations
Differentiating a matrix:
[ ]
=
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
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 11
Matrix Operations
Differentiating a matrix:
{ } [ ]{ }XaXU T21
=
[ ]{ }XaxU
i=
In a general form, we have:
For which we have:
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 12
Matrix Operations
Integrating a matrix.
ij[a]dx a dx =
-
2/1/2011
7
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 13
Finding the Inverse of a Matrix
Need to find the determinant
Need to find the co-factors of [a]
determinant of matrixa [a]=
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 14
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
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 15
Sets of Linear Algebraic Eqs.
Cramers RuleInverse MethodGaussian EliminationGauss-Seidel Iteration
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 16
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
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 17
Example:
Consider the following equations:
1 2 3
1 2 3
2 3
x 3x 2x 22x 4x 2x 1
4x x 3
- + - =
- + =
+ =
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 18
Example:
=
-
--
3
1
2
x
x
x
140
242
231
3
2
1
:formmatrixIn
-
2/1/2011
10
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 19
1.41041
140
242
231143
241
232
)1(
1 =--
=
-
--
-
-
==a
dx
Solution:
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 20
1.1
140
242
231130
212
221
a
dx
)2(
2 =
-
--
--
==
-
2/1/2011
11
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 21
4.1
140
242
231340
142
231
a
dx
)3(
3 -=
-
--
-
-
==
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 22
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
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 23
Example
=
-
--
312
140242231
3
2
1
xxx
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 24
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
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 25
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:
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 26
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
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 27
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
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 28
Steps in Gaussian Elimination
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
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 29
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
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 30
Steps in Gaussian Elimination
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
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 31
Solve Using Back-substitution
-=
=
+=+
-
-
n
1irrir1n,1
iii
1nnn
1nn
n
xaaa1x
acx
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 32
Example
=
649
xxx
111012122
3
2
1
-
2/1/2011
17
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 33
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.
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 34
Step 1
-=
--
5.15
9
xxx
5.000110
122
3
2
1
-
2/1/2011
18
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 35
Steps in Gaussian Elimination
Eliminate the coefficient of x2 in every equation below the second one. Select a22 as the pivot element. (Already done in this example.)
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 36
Step 2
-=
--
5.15
9
xxx
5.000110
122
3
2
1
-
2/1/2011
19
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 37
Solve Using Back-substitution( )( )
( )
( ) 12
3)2(29x
21
35x
32
12
3
acx
2
2
33
33
=--
=
=-
+-=
==
=
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 38
Gauss-Seidel Iteration
( )
( )
( )1n1n.n22n11nnnn
n
nn2323121222
2
nn1313212111
1
xaxaxaca1x
xaxaxaca1x
xaxaxaca1x
:forminequationsWrite
------=
----=
----=
L
M
L
L
-
2/1/2011
20
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 39
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
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 40
Example
1x1x1x21x
6x2x6xx4x5xx4x2xx4
4321
43
432
321
21
-====
=+-=-+-
=-+-
=-
-
2/1/2011
21
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 41
Example
( ) ( )
( ) ( )( ) ( )( )
( ) ( ) 16.067.1221x221x
672.1168.1641xx64
1x
68.114354
1xx541x
43124
1x241x
34
423
312
21
-=+-=+-=
=-++=++=
=++=++=
=+=+=
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 42
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
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 43
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
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 44
Basic Elasticity
For Linear, homogeneous, isotropic material behavior.
-
2/1/2011
23
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 45
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
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 46
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
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 47
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
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 48
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
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 49
Plane StressPlane Stress Matrix
[ ] ( )
1 0ED 1 0
110 0
2
n
= n - n - n
0=== yzxzz tts
Tuesday, February 01, 2011
Review Topics Dr.Eng.Mohammad R. KHEDMATI 50
Plane StrainPlane Strain Matrix
[ ] ( ) ( )
1 0ED 1 0
1 1 21 20 0
2
-n n
= n -n +n - n - n
0=== yzxzz gge