Cauchy's Formula and Eighenvalues

8/12/2019 Cauchy's Formula and Eighenvalues

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GG711c 1/22/10 1

Stephen Martel 5-1 University of Hawaii

CAUCHY'S FORMULA AND EIGENVAULES (PRINCIPAL STRESSES) (05)

I Main Topics

A Cauchy’s formulaB Principal stresses (eigenvectors and eigenvalues)

II Cauchy's formula

A Relates traction vector components to stress tensor components (see

Figures 5.1, 5.2, 5.3 for derivation)

B Ti = ! ji n j (5.1)

1 Meaning of terms

a Ti=traction vector component:! T = T 1

! i + T 2

! j + T 3

! k

b !ij = stress component c n =unit normal vector. The components n j of the unit normal

are the direction cosines between n and the coordinate axes.

dF

i

A=

F i

A1

A1

A+

F i

A2

A2

A+

F i

A3

A3

A

2 This represents the physics directly

3 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

5 Note that the subscripts on the T and the n differ

6 ! is symmetric (!ij =! ji), so …

C Ti = !ij n j Standard form of Cauchy’s formula

1 The subscript j's stil l cancel out

2 The subscripts on the T and the n still differ

3 Easier(?) to remember than “B”



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D Full expansion

Ti = ! ji n j Ti = !ij n j

T1 = !11 n1 + !21 n2 + !31 n3 = !11 n1 + !12 n2 + !13 n3

T2 = !12 n1 + !22 n2 + !32 n3 = !21 n1 + !22 n2 + !23 n3

T3 = !13 n1 + !23 n2 + !33 n3 = !31 n1 + !32 n2 + !33 n3

E Matrix formT 1

T 2

T 3

"

#

$ $ $

%

&

' ' '

=

( 11

( 21

( 31

( 12

( 22

( 32

( 13

( 23

( 33

"

#

$ $ $

%

&

' ' '

n1

n2

n3

"

#

$ $ $

%

&

' ' '

T 1

T 2

T 3

!

"

# # #

$

%

& & & =

' 11

' 12

' 13

' 21 ' 22 ' 23

' 31

' 32

' 33

!

"

# # #

$

%

& & &

n1

n2

n3

!

"

# # #

$

%

& & &



GG711c 1/22/10 3


!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).

B

C

D

O

x1

"1

"2

"3

x

2

x3

"

#32

B

C

O#33

#31

C

D

O

#12

#13

#11

B

D

O#22

#23

#21

Derivation of Cauchy’s Equation 5.1

Area A1

Area AArea A2

Area A3



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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, soCOP’’ is perpendicular to BCD. The intersection of BOP’ and COP’’ isperpendicular 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) (CD)(BP’) BP’

Area A1

Area A2

Area A3

Derivation of Cauchy’s Equation

C

D

OP’

B

C

O

P’’’

B

D

O

P’’

B

C

D

O x1

x2

x3

PP’ "1

"2

"3

P’’’

P’’

OP is normal to BCD

5.2

Area A



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Derivation of Cauchy’s Equation

O B O B

P’

x1

P P

!1 "

Triangles BOP and BP’Oare similar right triangles;

they both have angle OBP(i.e., ") in common.

P’

x1!1

!1

"

Therefore, angle BP’O = !1.

A1 = OP’ = cos !1 = n1A BP’

Similarly,

5.3

1(A/A) = ($11)(A1/A) + ($21)(A2/A) + ($31)(A3/A) becomes

1 = ($11)(n1) + ($21)( n2) + ($31)(n3). Similarly,


2 = ($12)(n1) + ($22)( n2) + ($32)(n3), and


3 = ($13)(n1) + ($23)( n2) + ($33)(n3).

So #i = $ ji n j, but $ij= $ ji, so #i = $ij n j

.

.

A2 = OP’’ = cos !2 = n2A CP’’

and soA3 = OP’’’ = cos !3 = n3A DP’’’



GG711c 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 = !ij n j. They also are simply the components of T: T1=Tn1, T2=Tn2, andT3=Tn3. The components can be equated:

! 11

! 12

! 13

! 21 ! 22 ! 23

! 31

! 32

! 33

"

#

$ $ $

%

&

' ' '

n1

n2

n3

"

#

$ $ $

%

&

' ' ' = T

n1

n2

n3

"

#

$ $ $

%

&

' ' ' .

(5.2)

The right side of (5.2) can be subtracted from the left side to yield:! 11 " T ! 12 ! 13

! 21 ! 22 " T ! 23

! 31 ! 32 ! 33 " T

#

$

% % %

&

'

( ( (

n1

n2

n3

#

$

% % %

&

'

( ( ( = 0 .

(5.3)

Equation (5.3) can be rewritten

[! -IT] [n] =0, where I is the identity matrix(5.4)

I =

1 0 0

0 1 0

0 0 1

. For any square matrix [A], [A][I] = [A].(5.5)

According to theorems of linear algebra, equation (5.3) can be solved only if

the determinant |! -IT| equals zero:" 11 #T " 12 " 13

" 21 " 22 # T " 23

" 31 " 32 " 33# T

= 0 (5.6)

In many cases the components of ! 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.



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! 11 " T ! 12

! 21

! 22 " T

= 0

Note :

a b

c d =ad "bc

(5.7)

(! 11 " T )(! 22 " T ) " (! 12)(! 21) =0 (5.8)

T 2! T (" 11 +" 22 )+ (" 11)(" 22) ! (" 12)(" 21) = 0 (5.9)

T 2! T (" 11 +" 22 )+ [(" 11)(" 22) ! (" 12 )

2] = 0 (5.10a) or T

2"T ( I

1)+ [ I

2] = 0 (5.10b)

The term T in equations (5.10) is solved using the quadratic formula:

T =(! 11 +! 22 )± (! 11 +! 22)

2" 4[(! 11)(! 22 )" (! 12

2 )]

2=

I 1 ± I 12" 4 I 2

2 (5.11)

T =(! 11 +! 22 )± (! 11

2+ 2! 11! 22 +! 22

2) " 4[(! 11)(! 22) " (! 122 )]

2 (5.12)

T =(! 11 +! 22 )± (! 11

2" 2! 11! 22 +! 22

2) + 4[! 122 ]

2 (5.13)

T =(! 11 +! 22 )± (! 11 "! 22)

2+ 4[! 12

2]

2=

I 1 ± I 12" 4 I 2

2 (5.14)



GG711c 1/22/10 8


T =

" 11

+" 22

2

#

$ %

&

' ( ±

" 11

)" 22

2

*

+ ,

-

. /

2

+" 12

2

#

$

%

%

&

'

(

( = c[ ]± r[ ] =

I 1

2

#

$ %

&

' ( ±

I 1

2

*

+ ,

-

. /

2

) I 2

#

$

%

%

&

'

(

( ="

1,"

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, I 1, and I

2 are called invariants

and are independent of the frame of reference.



GG711c 1/22/10 9


Example

Suppose the stress state at a point is given by

! ij =10 3

3 2

"

# $

%

& ' , where dimensions are in MPa.

Solving for the principal values using eq. (14) yields

T =(10 + 2)

2±

10 "2

2

# $

% &

2

+ 32= 6 ± 25 =11 and 1

Now we substitute these back into (5.3)10 !11 3

3 2 !11

"

# $

%

& ' n1

n2

"

# $

%

& ' =

0

0

"

# $

%

& ' " !1 3

3 !9

"

# $

%

& ' n1

n2

"

# $

%

& ' =

0

0

"

# $

%

& ' for T = !1 = 11MPa.

10 !1 3

3 2 !1

"

# $

%

& ' n1

n2

"

# $

%

& ' =

0

0

"

# $

%

& ' " 9 3

3 1

!

" #

$

% & n1

n2

!

" #

$

% & =

0

0

!

" #

$

% & for T = !2 = 1MPa.

These relations yield

(a) -n1 + 3n2 = 0 (!1 = 11MPa) (b) 3n1 + n2 = 0 (!2 = 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

x1

x2

x1x1

x2’

x1’

2 MPa 1 MPa

11 MPa3 MPa

3 MPa

10 MPa1

1

-3

3

n1 = cos !1nn2 = cos !2n = sin !1n

So n2/n1 = tan !1n!1n = tan-1 (n2/n1)

= atan2(n2,n1)

!2n

!1n

For !1 = 11MPa*

" x

1,normal

=" x

1, x

1'

= tan#1 x

2 n

x1

n

$

%

& &

'

(

) ) = tan

#1 n2

n1

$

%

& &

'

(

) ) = tan

#1 n2

3n2

$

%

& &

'

(

) ) = tan

#1 1

3

$

% &

'

( ) =18.5

!

For !2 = 1Mpa*

" x

1,normal

=" x

1, x

2'

= tan#1 x

2 n

x1

n

$

%

& &

'

(

) ) = tan

#1 n2

n1

$

%

& &

'

(

) ) = tan

#1 #3n1

n1

$

%

& &

'

(

) ) = tan

#1 #3( ) = #71.5!

The two eigenvectors are perpendicular, as they are supposed to be.* In the first expression for #, 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 #, the normal direction is the x2’ direction, and n1 and n2 are the

direction cosines for a unit vector along x2’.



GG711c 1/22/10 10


V Matrix treatments of stress transformation

In matrix form, " # i # j = a # i k a # j l" kl becomes (Mal & Singh, 1991, p. 37)

!' = [a] [!] [aT], (5.16)

where

a=

a " 1 1 a " 1 2

a " 1 3

a " 2 1 a " 2 2

a " 2 3

a " 3 1 a " 3 2

a " 3 3

#

$

%

%

&

'

(

( (5.17)

aT

=

a " 1 1 a " 2 1

a " 3 1

a " 1 2 a " 2 2

a " 3 2

a " 1 3 a " 2 3

a " 3 3

#

$

%

%

&

'

(

( (5.18)

The proper order of matrix multiplication isss nt i l

in order to reproducethe expansions of lecture 17: [a] [!] [aT]! [aT] [!][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)



GG711c 1/22/10 11


Example

»sigmaxy = [10 3;3 2]

sigmaxy =

10 33 2

»a = [3/sqrt(10) 1/sqrt(10);-1/sqrt(10) 3/sqrt(10)]

a =0.9487 0.3162 The first row of matrix “a” is the negative

-0.3162 0.9487 of the first column of matrix V below.

The second row of matrix “a” is the negative

»sigmaprime = a*sigmaxy*a' 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 Column 1 in V relates to column 1 in D

-0.3162 -0.9487 Column 2 in V relates to column 2 in D

D =

11 0

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.

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Cauchy's Formula and Eighenvalues

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