matrix multiplication : when the number of columns of the...
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Matrices Review
Matrix Multiplication : When the number of columns of the first matrix is the same as the number of rows in the second matrix then matrix multiplication can be performed.Here is an example of matrix multiplication for two 2x2 matrices
Here is an example of matrices multiplication for a 3x3 matrix
When A has dimensions mxn, B has dimensions nxp. Then the product of A and B is the matrix C, which has dimensions mxp.
Transpose of Matrices : The transpose of a matrix is found by exchanging rows for columns i.e. Matrix A = (aij) and the transpose of A is: AT=(aij) Where i is the row number and j is the column number.For example, The transpose of a matrix would be:
In the case of a square matrix (m=n), the transpose can be used to check if a matrix is symmetric. For a symmetric matrix A = AT
The Determinant of a Matrix : Determinants play an important role in finding the inverse of a matrix and also in solving systems of linear equations. Determinant of a 2x2 matrix Assuming A is an arbitrary 2x2 matrix A, where the elements are given by:
Determinant of a 3x3 matrix The determinant of a 3x3 matrix is more difficult
For a 2x2 matrix the matrix inverse isInverse Matrix
Example:
TACosSin
SinCosA
SinCosACosSinSinCos
A
=⎥⎦
⎤⎢⎣
⎡ −=
=+=⇒⎥⎦
⎤⎢⎣
⎡−
=
−
θθθθ
θθθθθθ
11
1
1
22
For a 3x3 matrix
⎥⎦
⎤⎢⎣
⎡−
=θθθθ
CosSinSinCos
A
θθτθσθσσ cossin2sincos 221 XYYXX ++=
( )θθτθθσθθστ 2211 sincoscossincossin −+⋅⋅+⋅⋅−= xyyxyx
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−−=
θθθθθθθθθθθθθθ
22
22
22
sincoscossincossincossin2cossin
cossin2sincos][T
Coordinate Transformations
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
=⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
xy
y
x
yx
y
x
Tτσσ
τσσ
][
11
1
1
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
=⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡−
11
1
11][
yx
y
x
xy
y
x
Tτσσ
τσσ
θθτθσθσσ cossin2cossin 221 XYYXY −+=
⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡−
=⎥⎦
⎤⎢⎣
⎡yx
yx
θθθθ
cossinsincos
''
⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡yx
Tyx''
Theory of Matrix Method for Stress Calculations in 2-D
From equations of rotational transformation of axis, we obtain the following:
inversely ⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡ −=⎥
⎦
⎤⎢⎣
⎡''
cossinsincos
yx
yx
θθθθ
T
TT ⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡ −=⎥
⎦
⎤⎢⎣
⎡−
θθθθ
cossinsincos1Hence
τxy
σx
τyx
σx
σy
τyx
τxy
σy
⎥⎦
⎤⎢⎣
⎡
yyxy
yxxx
σττσ
X
θ
YY'
X'
AC = Area AAB = A cos θBC = A sin θ
A
B C
θ
θ 'xσxσ
yσxyτ
{ } { }
{ }
{ }
{ } ⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡ −+⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡−=
⎥⎦
⎤⎢⎣
⎡−+⎥
⎦
⎤⎢⎣
⎡+⎥
⎦
⎤⎢⎣
⎡−⎥
⎦
⎤⎢⎣
⎡−=
⎥⎦
⎤⎢⎣
⎡+⎥
⎦
⎤⎢⎣
⎡+⎥
⎦
⎤⎢⎣
⎡=
Σ=
''
'
'''
cossinsincos
sincos
0
cossin
sincos
sincos0
0
0
yx
x
yxy
xyx
yxxy
xy
xy
x
ACy
x
BCy
x
ABy
x
AA
AAAA
FF
FF
FFF
τσ
θθθθ
θθ
σττσ
θθ
τθθ
σθστ
θτσ
Canceling area A out and pre-multiplying by transformation T (where
the identity matrix. The order of the matrix multiplication doesmatter in the final outcome., we have
ITT T =⊗
Using and force equilibrium equation, we obtain expressions for stress transformations as follows: θ 'xσ
xσ
yσxyτ
⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡−
=⎥⎦
⎤⎢⎣
⎡θθ
σττσ
θθθθ
τσ
sincos
cossinsincos
''
'
yxy
xyx
yx
x
I=⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡−
×⎥⎦
⎤⎢⎣
⎡ −1001
cossinsincos
cossinsincos
θθθθ
θθθθ
{ } ⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡ −+⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡−=
''
'
cossinsincos
sincos
0yx
x
yxy
xyx AAτσ
θθθθ
θθ
σττσ
For the forces in the X axis we will use the same procedure. YY'
X
X'θ
θB C
D
BD = Area ABC = A cos θCD = A sin θ
θ
'yσ ''yxτ
yσ
xσxyτ
{ } { }
{ }
{ }
{ } ⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡ −+⎥
⎦
⎤⎢⎣
⎡−⎥
⎦
⎤⎢⎣
⎡=
⎥⎦
⎤⎢⎣
⎡+⎥
⎦
⎤⎢⎣
⎡−+⎥
⎦
⎤⎢⎣
⎡−⎥
⎦
⎤⎢⎣
⎡=
⎥⎦
⎤⎢⎣
⎡+⎥
⎦
⎤⎢⎣
⎡+⎥
⎦
⎤⎢⎣
⎡=
Σ=
'
''
'''
cossinsincos
cossin
0
sincos
cossin
cossin0
0
0
y
yx
yxy
xyx
yxyy
xy
xy
x
BDy
x
BCy
x
CDy
x
AA
AAAA
FF
FF
FFF
στ
θθθθ
θθ
σττσ
θθ
τθθ
σθστ
θτσ
⎥⎦
⎤⎢⎣
⎡−⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡−
=⎥⎦
⎤⎢⎣
⎡θθ
σττσ
θθθθ
στ
cossin
cossinsincos
'
''
yxy
yxx
y
xy
⎥⎦
⎤⎢⎣
⎡ −⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡−
=⎥⎦
⎤⎢⎣
⎡θθθθ
σττσ
θθθθ
σττσ
cossinsincos
cossinsincos
'''
'''
yxy
xyx
yyx
yxx
Combining the above expressions
⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡TT
yxy
yxxT
yyx
xyx
σττσ
σττσ
'''
'''
or
⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡−
=⎥⎦
⎤⎢⎣
⎡θθ
σττσ
θθθθ
τσ
sincos
cossinsincos
''
'
yxy
xyx
yx
x
⎥⎦
⎤⎢⎣
⎡−⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡−
=⎥⎦
⎤⎢⎣
⎡θθ
σττσ
θθθθ
στ
cossin
cossinsincos
'
''
yxy
xyx
y
yx
-x
-y
z
σz
σx
σy
τxy
τxzτyx
τyz
τzxτzy
σ
The general three dimensional state of stress consists of three unequal principal stresses acting at a point (triaxial state of stresses).
J
L
K
The plane JKL is assumed to be a principal plane and σ is the principal stress acting normal to the plane.
Letα, β and γ are the angles between the vector σand the x, y and z axis respectively and
State of Stresses in Three Dimensions
γβα cos cos cos === mlk
Under equilibrium conditions
mlkm
mlkl
mlkk
zzyzxz
zyyyxy
zxyxxx
⋅+⋅+⋅=⋅
⋅+⋅+⋅=⋅
⋅+⋅+⋅=⋅
σττσ
τστσ
ττσσ
( )( )
( ) 0
0
0
=⋅−+⋅−⋅−
=⋅−⋅−+⋅−
=⋅−⋅−⋅−
mlk
mlk
mlk
zyzxz
zyyyxy
zxyxxx
σσττ
τσστ
ττσσAs k, l and m are different than zero (non-trivial solution)
0=⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−−−−−−−−
mlk
zzyzxz
zyyyxy
zxyxxx
σστττσστττσσ
0=−−−−−−−−−
zzyzxz
zyyyxy
zxyxxx
σστττσστττσσ
The determinant must be equal to zero
Solution of the determinant results in a cubic equation in σ
1222 =++ mlk
The eigenvalues of the stress matrix are the principal stresses.The eigenvectors of the stress matrix are the principal directions.
1222 =++ mlk 321 σσσ >>
2223
2222
1
322
13
2
0
xyzxzyyzxyzxzxyzyx
yzxzxyzxzyyx
zyx
I
I
IIII
τστστστττσσσ
τττσσσσσσ
σσσσσσ
⋅−⋅−⋅−⋅⋅⋅+⋅⋅=
−−−⋅+⋅+⋅=
++==−⋅+⋅−
0=⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−−−−−−−−
mlk
zzyzxz
zyyyyx
zxyxxx
σστττσστττσσ
The three roots are the three principal stresses σ1 , σ2 , σ3.
I1, I2, and I3 are known as stress invariants as they do not change in value when the axes are rotated to new positions.
I1 has been seen before for the two dimensional state of stress. It states the useful relationship that the sum of the normal stresses for any orientation in the coordinate system is equal to the sum of the normal stresses for any other orientation
321111 σσσσσσσσσ ++=++=++ zyxzyx
zyzzx
yzyyx
xzxyx
zzx
xzx
zzzy
yzy
yxy
xyx
zyx
I
I
I
στττστττσ
σττσ
σττσ
σττσ
σσσ
=
++=
++=
3
2
1
0322
13 =−⋅+⋅− III σσσ
( )3
60cos2
3cos2
103,2
11
IA
IA
+±⋅⋅−=
+⋅⋅=
ασ
ασ
( ) ( )3
321
31
22
1
233
23
33
A
IIII
Cos
IIA
⋅⎥⎥⎦
⎤
⎢⎢⎣
⎡+⋅⎟
⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛⋅
=
⎟⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛=
α
Stress Invariants for Principal Stresses
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
3
2
1
σσ
σσ
3213
1332212
3211
σσσσσσσσσ
σσσ
⋅⋅=⋅+⋅+⋅=
++=
III
⎟⎠⎞
⎜⎝⎛ −−−
=2
,2
,2
max 133221max
σσσσσστThe solution are the eigenvaluesof the stress tensor
Example: determine the principal stresses for the state of stress (in MPa).
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−
−
28000020024002400
0=−−−−−−−−−
zzyzx
yzyyx
xzxyx
σστττσστττσσThe solution are the eigenvalues of the
stress tensor;Substituting:
Solution:
0)280(00
0200)240(0)240(
=−−
−−−−−
σσ
σ
( ) ( )( ) 0))240()240(()200()280( =−⋅−−−⋅⋅−− σσσOne solution σ3=-280MPa is a principal stress because τxz and τyz are zero, then the other two principal stresses are easy to find by solving the quadratic equation inside the square brackets for σ
160360 2601002
)240(4)200()200( 0)240(200
21
2222
MPa- σMPa σ ==±=
⋅+−±−−==−−
σ
σσσ
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−
−
28000020024002400
000,128,16 600,113 80 321 =−=−= III
( )
( )
( ) 9.2793806015.104.1962
0.1603806015.104.1962
99.35938015.104.1962
15.10 8620.0)3( 4.196
3
2
1
−=−
+−⋅⋅−=
−=−
++⋅⋅−=
=−
+⋅⋅=
===
Cos
Cos
Cos
CosA
σ
σ
σ
αα
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−
280000160000360
Example 2: Determine the maximum principal stresses and the maximum shear stress for the following triaxialstress state.
MPa⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−
−=
102530253040304020
σ
[ ]⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−
−=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
==102530
253040304020
_
zzyzxz
zyyyxy
zxyxxx
TensorStressστττστττσ
σ
Solution
3 21 2 3 0I I Iσ σ σ− + − =
1
2 2 22
2 2 23 2
x y z
x y x z y z xy xz yz
x y z xy xz yz x yz y xz z xy
I
I
I
σ σ σ
σ σ σ σ σ σ τ τ τ
σ σ σ τ τ τ σ τ σ τ σ τ
= + +
= + + − − −
= + − − −
= 20 + 30 –10 = 40 MPa
= -3025 MPa
= -89500 MPa
MPaMPaMPaMPa
5.58)8.513.65(2/18.51 5.26 3.65
max
321
=+=−===
τσσσ
Solution to Example
-800000
-600000
-400000
-200000
0
200000
400000
600000
-100 -80 -60 -40 -20 0 20 40 60 80 100
Stress (MPa)
Sigm
a (M
Pa)
-51.8 MPa
65.3 MPa
26.5 MPa
Mohr’s Circles for 3-D AnalysisMohr’s circles can make visualization of the stress condition clearer to the designer. Note that the principal stress values are alwaysordered by convention so the σ1 is the largest value in the tensile direction and σ3 is the largest value in the compressive direction. Note also that there is one dominant peak shear stress in this diagram.
Be forewarned the principal stresses and this peak shear stress are going to play a strong role in determining the factor of safety in mechanical design.
A Mohr’s circle can be generated for triaxial stress states, but it is often unnecessary, as it is sufficient to know the values of theprincipal stresses.
The principal stresses must be ordered from larger to smaller.
τ
2
2
2
3113
3223
2112
σστ
σστ
σστ
−=
−=
−=
Compare 2-D and 3-D Mohr’s Circle. If σz is zero, does it have an effect in 3D?
σ2σ3 σ1
Consider σ3=0 then the plane will be an angle α from σ1, in the direction of σ2 (clockwise). Point P
Consider σ2=0 then the plane will be an angle α from σ1, in the direction of σ3 (clockwise). Point Q
The required system of stresses, fall within P and Q. Loci determined by the center in
221 σσ +
231 σσ +
232 σσ +
2α2α
P
Q
232 σσ +
σ2σ3 σ1
Consider σ1=0 then the plane will be an angle β from σ2, in the direction of σ3 (anticlockwise). Point R
Consider σ3=0 then the plane will be an angle β from σ2, in the direction of σ1 (clockwise). Point S
The required system of stresses, fall within R and S. Loci determined by the center in
2β2β
231 σσ +
Example:
Use Mohr’s Circle to obtain the principal stresses and maximum shear of a component subjected to the following stresses:
ccwncompressio
tensiontension
xy
z
y
x
4025
3090
=−=
==
τσ
σσ
ccw counterclockwise
Stress on ANY Inclined Plane (3-D)
The stress on a plane (S) can be decomposed into its normal component (Sn) and its shear component (Ss).
22222 τσ +=+= sn SSS
zyx sssS ++=nIf α, β and γ are the angles between the vector Snand the x, y and z axis respectively and
γβα cos cos cos === mlk then
[ ] [ ]T
zzyzxz
zyyyxy
zxyxxx
zzzyzx
yzyyyx
xzxyxx
TT ×⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
×=⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
στττστττσ
στττστττσ
111111
111111
111111
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
=
zzyzxz
zyyyxy
zxyxxx
στττστττσ
σ
It can be proved that
There is no easy Mohr’s circle graphical solution for problems of triaxial stress state. Solution for maximum principal stresses and maximum shear stress is analytical.
Mohr’s Circles for 3-D Analysis
Consider the x, y and z axis to coincide with the axis of the principal stresses σ1, σ2 and σ3.
If α, β and γ are the angles between the normal to the plane and the x, y and zaxis respectively and
γβα cos cos cos === mlk1222 =++ mlk
We would like to find graphically the normal stress and shear stress on the plane.
x
y
z
σ3
σ1
σ2
σ
Octahedral Plane and StressesAn octahedral plane is a plane that makes three identical angles with the principal planes.
( ) ( ) ( ) ( )
22
12
213
232
221
223
22
21
2
1321233
222
211
321
62931
33
31
II
n
Innn
nnnn
op
opop
op
−=
−+−+−=−++⋅=
=++
=⋅+⋅+⋅=
====
τ
σσσσσσσσσστ
σσσσσσσ
mlkm
mlkl
mlkk
zzyzxz
zyyyxy
zxyxxx
⋅+⋅+⋅=⋅
⋅+⋅+⋅=⋅
⋅+⋅+⋅=⋅
σττσ
τστσ
ττσσ
Mean and Deviatoric Stresses
When describing the materials behavior of metals, one concludes that in certain cases some stress components play a more important role than other components. Plastic behavior of metals, is reported to be independent of the average (mean) normal stress.
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
==++
=++
=
M
M
M
opzyx
M
M
I
σσ
σ
σσσσσσσσ
000000
3331321
Mean stress matrix
Deviatoric stress MzzMyyMxx σσσσσσσσσ −=−=−= ′′′
The shear components do not change
⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢
⎣
⎡
−−
−−
−−
=
yyxxzzyzxz
zyzzxxyyxy
zxyxzzyyxx
D
σσσττ
τσσστ
ττσσσ
31
31
32
31
31
32
31
31
32
Deviatoric Stress Matrix
Deviatoric stresses play an important role in the theory of plasticity. They influence the yielding of ductile materials.
The principal stresses obtained only from the deviatoric matrix is
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−
−=
M
M
M
DP
σσσσ
σσ
3
2
1
,
000000
σ
Example
For a given stress matrix representing the state of stress at a certain point
[ ] MPa⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
203022321
σFind the stress invariant, the principal stresses, the principal directions, the octahedral stress and the shear stress associated with the octahedral stress.
Solution:
223022223201221
5302122221
5221
2223
2222
1
−=×××+×−×−×−××=
−=−−−×+×+×=
=++=
I
I
I
02255 23 =+⋅−+⋅− σσσ14.2
00.214.5
3
2
1
−===
σσσ
( )( )
( ) 0214.50300214.52
032114.5
111
111
111
=⋅−+⋅−⋅−=⋅−⋅−+⋅−
=⋅−⋅−⋅−
mlkmlk
mlk
14.200.214.5
3
2
1
−===
σσσ
[ ] MPa⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
203022321
σ
( )( )
( ) 0
0
0
=⋅−+⋅−⋅−
=⋅−⋅−+⋅−
=⋅−⋅−⋅−
mlk
mlk
mlk
zyzxz
zyyyxy
zxyxxx
σσττ
τσστ
ττσσ
627.0418.0657.0
1
1
1
===
mlk
121
21
21 =++ mlk
555.0832.0
0
2
2
2
−===
mlk
546.0364.0
714.0
3
3
3
==−=
mlk
35
31 ==
IMeanσ
980
80)5(6529
62922
22
12
=
=−×−×=
−=
op
op
op II
τ
τ
τ