Download CAUCHY`S FORMULA AND EIGENVAULES (PRINCIPAL

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CAUCHY'S FORMULA AND EIGENVAULES (PRINCIPAL STRESSES) (05)
I
II
Main Topics
A
Cauchy’s formula
B
Principal stresses (eigenvectors and eigenvalues)
Cauchy's formula
A
Relates traction vector components to stress tensor components (see
Figures 5.1, 5.2, 5.3 for derivation)
B
Ti = σji nj
(5.1)
1
Meaning of terms




Ti=traction vector component: T = T1i + T2 j + T3 k
a
b
σij = stress component
c
n =unit normal vector. The components nj of the unit normal
d
2
3
are the direction cosines between n and the coordinate axes.
Fi Fi A1 Fi A2 Fi A3
=
+
+
A A1 A A2 A A3 A
This represents the physics directly
The traction component that acts in the i-direction reflects the
contribution of the stresses that act in that direction.
4 Note that the j's "cancel out"
C
5 Note that the subscripts on the T and the n differ
6
σ is symmetric (σij =σji), so …
Ti = σij nj
Standard form of Cauchy’s formula
1
2
3
Stephen Martel
The subscript j's still "cancel out"
The subscripts on the T and the n still differ
Easier(?) to remember than “B”
5-1
University of Hawaii
GG711c
D
E
1/22/10
Full expansion
Ti = σji nj
T1 = σ11 n1 + σ21 n2 + σ31 n3
T2 = σ12 n1 + σ22 n2 + σ32 n3
T3 = σ13 n1 + σ23 n2 + σ33 n3
Matrix form
T1  σ11 σ 21 σ 31 n1 
  
 
T2  = σ12 σ 22 σ 32 n 2 
T3  σ13 σ 23 σ 33 n 3 
2
Ti = σij nj
= σ11 n1 + σ12 n2 + σ13 n3
= σ21 n1 + σ22 n2 + σ23 n3
= σ31 n1 + σ32 n2 + σ33 n3
 T1   σ 11 σ 12
  
T2  = σ 21 σ 22
T  σ
 3   31 σ 32
σ 13   n1 
 
σ 23  n2 
σ 33  n3 
€
Stephen Martel
5-2
University of Hawaii
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1/22/10
3
Derivation of Cauchy’s Equation
5.1
C
C
Area A3
x2
#32
C
Area A1
"2
#12
#31
"
O
#11
O
#13
"3
#33
B
"1
O
B
x1
D
O
D
Area A
#22
#21
#23
x3
B
Area A2
D
!F1 = 0, so (!F1)/A = 0.
"1(A/A) = (#11)(A1/A) + (#21)(A2/A) + (#31)(A3/A).
Similarly, !F2 = 0 and !F3 = 0, so
"2(A/A) = (#12)(A1/A) + (#22)(A2/A) + (#32)(A3/A).
"3(A/A) = (#13)(A1/A) + (#23)(A2/A) + (#33)(A3/A).
Stephen Martel
5-3
University of Hawaii
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4
C
Derivation of Cauchy’s Equation
5.2
C
Area A3
x2
C
Area A1
P’’’
P’’’
P’
O
B
O
"2
P
P’
O
OP is normal to BCD
"1
B
x1
"3
D
O
P’’
D
x3
Area A
B
P’’
Area A2
D
Note that ! DCB of area A projects onto the x1-x2 plane as ! OCB,
onto the x2-x3 plane as ! OCD, and onto the x3-x1 plane as ! OBD.
BOP’ is perpendicular to CD, and because CD is a line in BCD,
BOP’ is perpendicular to BCD. Similarly, COP’’ is perpendicular to BD, so
COP’’ is perpendicular to BCD. The intersection of BOP’ and COP’’ is
perpendicular to BCD, and that intersection is OP.
"1, "2, "3, are angles between OP and x1, x2, and x3, respectively.
A1 = 1/2 (base OCD)(height OCD) = (CD)(OP’) = OP’
A
1/2 (base DCB)(height CBD)
Stephen Martel
(CD)(BP’)
5-4
BP’
University of Hawaii
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Derivation of Cauchy’s Equation
P’
5
5.3
P’
!1
P
P
!1
O
"
x1
!1
B
O
Triangles BOP and BP’O
are similar right triangles;
they both have angle OBP
(i.e., ") in common.
"
x1
B
Therefore, angle BP’O = !1.
A1 = OP’ = cos !1 = n1
A
BP’
Similarly, A2 = OP’’ = cos !2 = n2 and A3 = OP’’’ = cos !3 = n3 so
A
CP’’
A
DP’’’
#1(A/A) = ($11)(A1/A) + ($21)(A2/A) + ($31)(A3/A)
becomes
#1 = ($11)(n1) + ($21)( n2) + ($31)(n3).
Similarly,
#2(A/A) = ($12)(A1/A) + ($22)(A2/A) + ($32)(A3/A)
becomes
#2 = ($12)(n1) + ($22)( n2) + ($32)(n3),
and
#3(A/A) = ($13)(A1/A) + ($23)(A2/A) + ($33)(A3/A)
becomes
#3 = ($13)(n1) + ($23)( n2) + ($33)(n3).
So #i = $ji nj, but $ij= $ji, so #i = $ij nj
.
.
Stephen Martel
5-5
University of Hawaii
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1/22/10
6
III Principal stresses from tensor and matrix perspectives
Consider a plane with a normal vector n defined by direction cosines n1, n2,
and n3. The components of traction T on the plane, by Cauchy’s formula, are
Ti = σij nj. They also are simply the components of T: T1=Tn1, T2=Tn2, and
T3=Tn3. The components can be equated:
σ 11 σ 12

σ 21 σ 22
σ
 31 σ 32
 n1 
σ 13   n1 
 
 
σ 23  n2  = T n 2  .
n 
σ 33  n3 
 3
(5.2)
The right side of (5.2) can be subtracted from the left side to yield:
σ 11 − T
σ 12
σ 13   n1 

 
σ 22 − T
σ 23  n2  = 0 .
 σ 21
 σ
σ 32
σ 33 − T  n 3 
 31
(5.3)
Equation (5.3) can be rewritten
[σ -IT] [n] =0,
where I is the identity matrix
1 0
I=0 1
0 0
0
0.
1
For any square matrix [A], [A][I] = [A].
(5.4)
(5.5)
According to theorems of linear algebra, equation (5.3) can be solved only if
the determinant |σ -IT| equals zero:
σ11 − T
σ12
σ 13
σ 21
σ 22 − T
σ 23 = 0
σ 31
σ 32
(5.6)
σ 33 − T
In many cases the components of σ are known but T is must be solved for.
Problems of the form of equation (5.4) are common in many branches of
mathematics, engineering, and physics, and they have a special name:
eigenvalue problems. The values of T (i.e., |T|, the principal values) that solve
the equation are called eigenvalues, and the vectors n (the principal directions)
that give the directions of T are called eigenvectors. Because these problems
are so common, many mathematics packages, including Matlab, have special
routines to solve for eigenvalues and eigenvectors.
Solving (5.6) by hand requires finding the roots of a cubic equation (not
easy), so we consider the easier 2-D case, which yields a quadratic equation.
Stephen Martel
5-6
University of Hawaii
GG711c
σ 11 − T
σ 12
=0
σ 21
σ 22 − T
1/22/10
Note :
7
a b
= ad− bc
c d
(5.7)
(σ 11 − T)(σ 22 − T) − (σ 12 )(σ 21 ) = 0
(5.8)
T 2 − T( σ 11 + σ 22 ) + (σ 11 )(σ 22 ) − (σ 12 )(σ 21 ) = 0
(5.9)
T 2 − T( σ 11 + σ 22 ) + [(σ 11)(σ 22 ) − (σ 12 )2 ] = 0
(5.10a)
T 2 − T(I1 ) + [I2 ] = 0
or
(5.10b)
The term T in equations (5.10) is solved using the quadratic formula:
€
T=
T=
T=
(σ 11 + σ 22 ) ± ( σ 11 + σ 22 ) 2 − 4[(σ 11 )(σ 22 ) − (σ 12 2 )] I1 ± I1 2 − 4 I 2
=
2
2
2 + 2σ
(σ 11 + σ 22 ) ± ( σ 11
2 − 2σ
(σ 11 + σ 22 ) ± ( σ 11
11σ 22
11σ 22
+ σ 22
2 ) + 4[ σ
11 )(σ 22 ) − (σ 12
(5.12)
2]
12
(5.13)
(σ 11 + σ 22 ) ± ( σ 11 − σ 22 ) + 4[σ 12
2
Stephen Martel
(5.11)
2 )]
2
2
T=
+ σ 22
2
2 ) − 4[(σ
2]
=
2−
I1 ± I1
2
5-7
4I 2
(5.14)
University of Hawaii
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8




 σ11 − σ 22  2
 I1 2
 σ11 + σ 22 
 I1 
2
 
 [ ] [ ]
  

 +σ
T =
± 
12 = c ± r =  2  ±  2  − I2 = σ1,σ 2
2
2


 

  





(5.15)
An inspection of the diagram below shows that the first term in brackets in
equation (5.15) is the mean normal stress (i.e., the center of the Mohr circle)
and the second term in brackets is the maximum possible shear stress (i.e., the
radius of the Mohr circle). So the principal stresses lie at the end of a horizontal
diameter through the Mohr circle. The terms c, r, I1, and I2 are called invariants
and are independent of the frame of reference.
Stephen Martel
5-8
University of Hawaii
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9
Example
Suppose the stress state at a point is given by
10 3 
σ ij = 
,
 3 2
where dimensions are in MPa.
Solving for the principal values using eq. (14) yields
T=
(10 + 2)
 10 − 2  2
±
+ 32 = 6 ± 25 = 11 and 1


2
2
Now we substitute these back into (5.3)
10 −11
3   n1  0 

  =  
2 −11 n2  0 
 3
10 −1
3   n1  0 

  =  
2 −1 n2  0 
 3
−1

3
9

3
⇒
⇒
3   n1   0
  =  
−9  n2   0
3   n1  0 
  =  
1  n2  0 
for T = σ1 = 11MPa.
for T = σ2 = 1MPa.
These relations yield
(a)
-n1 + 3n2 = 0 (σ1 = 11MPa) (b)
3n1 + n2 = 0
(σ2 = 1MPa).
From (a), for an eigenvalue (principal value) of 11 MPa, n1 = 3n2.
From (b), for an eigenvalue (principal value) of 1 MPa, n2 = -3 n1.
x2 2 MPa
1 MPa
x2’
3 MPa
n1 = cos !1n
n2 = cos !2n
x2
11 MPa
3 MPa
x1’
10 MPa
x1
1
1
3
x1
= sin !1n
!2n
!1n
So n2/n1 = tan !1n
-1
x1 !1n = tan (n2/n1)
= atan2(n2,n1)
-3
For σ1 = 11MPa*
x n
n 
n 
1
= tan−1 2  = tan−1 2  = tan−1 2  = tan−1  = 18.5
x n
n 
 3n 
x1,normal
x1, x1'
 3
 1 
 1
 2
For σ2 = 1Mpa*
x n
n 
 −3n 
1  = tan−1(−3) = −71.5
θ
=θ
= tan−1 2  = tan−1 2  = tan−1
x n
n 
 n 
x1,normal
x1, x 2'
 1 
 1
 1 
θ
€
€
=θ
The two eigenvectors are perpendicular, as they are supposed to be.
* In the first expression for θ, the normal direction is the x1’ direction, and n1
and n2 are the direction cosines for a unit vector along x1’. In the second
expression for θ, the normal direction is the x2’ direction, and n1 and n2 are the
direction cosines for a unit vector along x2’.
Stephen Martel
5-9
University of Hawaii
GG711c
V
1/22/10
10
Matrix treatments of stress transformation
In matrix form, σ i′j ′ = ai ′k a j ′lσ kl becomes (Mal & Singh, 1991, p. 37)
σ' = [a] [σ] [aT],
(5.16)
where
a ′
1 1 a1′ 2

a = a2′1 a2′2
a
 3′1 a3′2
a ′
1 1 a2′1
T
a = a1′2 a2′2
a
 1′ 3 a2′3
a1′ 3 
a2′3 
a3′3 
(5.17)
a3′1 
a3′2 
a3′3 
(5.18)
The proper order of matrix multiplication is essential in order to reproduce
the expansions of lecture 17: [a] [σ] [aT]≠ [aT] [σ][a]!
In MATLAB, equation (5.16) would be written:
sigmaprime = a * sigma * a'
The term a' signifies [aT]. Matlab also has a function “eig” to find eigenvectors
(given in terms of the direction cosines) and eigenvalues.
[V,D] = eig (sigma)
Stephen Martel
5-10
University of Hawaii
GG711c
1/22/10
11
Example
»sigmaxy = [10 3;3 2]
sigmaxy =
10
3
3
2
»a = [3/sqrt(10) 1/sqrt(10);-1/sqrt(10) 3/sqrt(10)]
a=
0.9487
-0.3162
0.3162
0.9487
»sigmaprime = a*sigmaxy*a'
The first row of matrix “a” is the negative
of the first column of matrix V below.
The second row of matrix “a” is the negative
of the second column of matrix V
below.
sigmaprime =
11.0000 -0.0000
-0.0000 1.0000
»[V,D] = eig(sigmaxy)
V=
-0.9487 0.3162
-0.3162 -0.9487
D=
11
0
Column 1 in V relates to column 1 in D
Column 2 in V relates to column 2 in D
0
1
The direction cosines (eigenvectors) in the first column of V correspond to the
eigenvalue in the first column of D.
The direction cosines (eigenvectors) in the second column of V correspond to
the eigenvalue in the second column of D.
Stephen Martel
5-11
University of Hawaii