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10/25/11
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19. Principal Stresses
I Main Topics A Cauchy’s formula B Principal stresses (eigenvectors and eigenvalues) C Example
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19. Principal Stresses
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hKp://hvo.wr.usgs.gov/kilauea/update/images.html
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19. Principal Stresses
II Cauchy’s formula A Relates tracPon
(stress vector) components to stress tensor components in the same reference frame
B 2D and 3D treatments analogous
C τi = σij nj = njσij
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Note: all stress components shown are posiPve
19. Principal Stresses II Cauchy’s formula (cont.)
C τ i = njσji 1 Meaning of terms
a τ i = tracPon component
b n j = direcPon cosine of angle between n- ‐direcPon and j- ‐direcPon
c σ ji = tracPon component
d τ i and σji act in the same direcPon
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nj = cosθnj = anj
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19. Principal Stresses
II Cauchy’s formula (cont.) D Expansion (2D) of τi = nj σji
1 τx = nx σxx + ny σyx
2 τy = nx σxy + ny σyy
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nj = cosθnj = anj
19. Principal Stresses
II Cauchy’s formula (cont.) E DerivaPon:
ContribuPons to τx
1
2
3
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τx =w 1( )σxx +w 2( )σyx
τx =nxσxx + nyσyx
Fx
An
=Ax
An
⎛⎝⎜
⎞⎠⎟
Fx1( )
Ax
+Ay
An
⎛⎝⎜
⎞⎠⎟
Fx2( )
Ay
Note that all contribuPons must act in x- direcPon ‐
nx = cosθnx = anx ny = cosθny = any
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19. Principal Stresses
II Cauchy’s formula (cont.) E DerivaPon:
ContribuPons to τy
1
2
3
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nx = cosθnx = anx ny = cosθny = any
τy =w 3( )σxy +w 4( )σyy
τy =nx σxy + nyσyy
Fy
An
=Ax
An
⎛⎝⎜
⎞⎠⎟
Fy3( )
Ax
+Ay
An
⎛⎝⎜
⎞⎠⎟
Fy4( )
Ay
Note that all contribuPons must act in y- direcPon ‐
19. Principal Stresses II Cauchy’s formula (cont.)
F AlternaPve forms 1 τi = njσji 2 τi = σjinj 3 τi = σijnj
4
5 Matlab a t = s’*n b t = s*n
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nj = cosθnj = anj
τx
τy
τz
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥=
σxx σyx σzx
σxy σyy σxy
σxz σyz σzz
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
nx
ny
nz
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
τx =nxσxx + nyσyx
τy =nxσxy + nyσyy
3D
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19. Principal Stresses
III Principal stresses (eigenvectors and eigenvalues)
A
B
C
The form of (C ) is [A][X=λ[X], and [σ] is symmetric
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τx
τy
⎡
⎣⎢⎢
⎤
⎦⎥⎥=
σxx σyx
σxy σyy
⎡
⎣⎢⎢
⎤
⎦⎥⎥
nx
ny
⎡
⎣⎢⎢
⎤
⎦⎥⎥
τx
τy
⎡
⎣⎢⎢
⎤
⎦⎥⎥= τ
→ nx
ny
⎡
⎣⎢⎢
⎤
⎦⎥⎥
σxx σyx
σxy σyy
⎡
⎣⎢⎢
⎤
⎦⎥⎥
nx
ny
⎡
⎣⎢⎢
⎤
⎦⎥⎥=λ
nx
ny
⎡
⎣⎢⎢
⎤
⎦⎥⎥
Let λ= τ→
Cauchy’s Formula
Vector components
9. EIGENVECTORS, EIGENVALUES, AND FINITE STRAIN
III Eigenvalue problems, eigenvectors and eigenvalues (cont.) E AlternaPve form of an eigenvalue equaPon
1 [A][X]=λ[X] SubtracPng λ[IX] = λ[X] from both sides yields: 2 [A- Iλ][X]=0 (same form as [‐ A][X]=0)
F SoluPon condiPons and connecPons with determinants 1 Unique trivial soluPon of [X] = 0 if and only if |A- Iλ|≠0 ‐2 Eigenvector soluPons ([X] ≠ 0) if and only if |A- Iλ|=0 ‐
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From previous notes
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9. EIGENVECTORS, EIGENVALUES, AND FINITE STRAIN III Determinant (cont.)
D Geometric meanings of the real matrix equaPon AX = B = 0 1 |A| ≠ 0 ;
a [A]- 1 ‐ exists b Describes two lines (or 3
planes) that intersect at the origin
c X has a unique soluPon 2 |A| = 0 ;
a [A]- 1‐ does not exist b Describes two co- linear ‐
lines that that pass through the origin (or three planes that intersect a line or plane through the origin)
c X has no unique soluPon
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Parallel lines have parallel normals
nx(1) ny(1) x d1=0nx(2) ny(2) y d2=0
AX = B = 0
=
|A| = nx(1) * ny(2) - ny(1) * nx(2) = 0 n1 x n2 = 0
Intersecting lines have non-parallel normals
nx(1) ny(1) x d1=0nx(2) ny(2) y d2=0
AX = B = 0
=
|A| = nx(1) * ny(2) - ny(1) * nx(2) ≠ 0 n1 x n2 ≠ 0
From previous notes
9. EIGENVECTORS, EIGENVALUES, AND FINITE STRAIN
III Eigenvalue problems, eigenvectors and eigenvalues (cont.) J CharacterisPc equaPon: |A- Iλ|=0 ‐
3 Eigenvalues of a symmetric 2x2 matrix
a
b
c
d
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λ1, λ2 =a + d( )± a + d( )2 −4 ad −b2( )
2
Radical term cannot be negaPve. Eigenvalues are real.
A = a bb d
⎡⎣⎢
⎤⎦⎥
λ1, λ2 =a + d( )± a + 2ad + d( )2 −4ad + 4b2
2
λ1, λ2 =a + d( )± a −2ad + d( )2 + 4b2
2
λ1, λ2 =a + d( )± a −d( )2 + 4b2
2
From previous notes
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9. EIGENVECTORS, EIGENVALUES, AND FINITE STRAIN
L DisPnct eigenvectors (X1, X2) of a symmetric 2x2 matrix are perpendicular Since the leo sides of (2a) and (2b) are equal, the right sides must be equal too. Hence, 4 λ1 (X2•X1) =λ2 (X1•X2) Now subtract the right side of (4) from the leo 5 (λ1 – λ2)(X2•X1) =0 • The eigenvalues generally are different, so λ1 – λ2 ≠ 0. • This means for (5) to hold that X2•X1 =0. • Therefore, the eigenvectors (X1, X2) of a symmetric 2x2
matrix are perpendicular
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From previous notes
19. Principal Stresses III Principal stresses (eigenvectors and eigenvalues)
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σxx σyx
σxy σyy
⎡
⎣⎢⎢
⎤
⎦⎥⎥
nx
ny
⎡
⎣⎢⎢
⎤
⎦⎥⎥=λ
nx
ny
⎡
⎣⎢⎢
⎤
⎦⎥⎥
D Meaning 1 Since the stress tensor is symmetric, a
reference frame with perpendicular axes defined by nx and ny pairs can be found such that the shear stresses are zero
2 This is the only way to saPsfy the equaPon above; otherwise σxy ny ≠ 0, and σxy nx ≠0
3 For different (principal) values of λ, the orientaPon of the corresponding principal axis is expected to differ
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19. Principal Stresses
V Example Find the principal stresses
given
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σij =σxx = −4MPa σxy = −4MPa
σyx = −4MPa σyy = −4MPa
⎡
⎣⎢⎢
⎤
⎦⎥⎥
19. Principal Stresses
V Example
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σij =σxx = −4MPa σxy = −4MPa
σyx = −4MPa σyy = −4MPa
⎡
⎣⎢⎢
⎤
⎦⎥⎥
λ1, λ2 =a + d( )± a −d( )2 + 4b2
2
λ1, λ2 = −4 ±642
= −4 ±4 =0,−8
First find eigenvalues (in MPa)
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19. Principal Stresses IV Example
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σij =σxx = −4MPa σxy = −4MPa
σyx = −4MPa σyy = −4MPa
⎡
⎣⎢⎢
⎤
⎦⎥⎥
λ1, λ2 = −4 ±642
= −4 ±4 =0,−8
Then solve for eigenvectors (X) using [A- Iλ][X]=0 ‐
For λ1 =0 : −4 −0 −4−4 −4 −0
⎡⎣⎢
⎤⎦⎥
nx
ny
⎡
⎣⎢⎢
⎤
⎦⎥⎥= 0
0⎡⎣⎢
⎤⎦⎥⇒ −4nx −4ny =0 ⇒ nx = −ny
For λ2 = −8 :−4 − −8( ) −4
−4 σyy − −8( )⎡
⎣⎢⎢
⎤
⎦⎥⎥
nx
ny
⎡
⎣⎢⎢
⎤
⎦⎥⎥= 0
0⎡⎣⎢
⎤⎦⎥⇒ 4nx −4ny =0 ⇒ nx =ny
Eigenvalues (MPa)
19. Principal Stresses IV Example
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λ1 =0MPaλ2 = −8MPa
Eigenvectors nx = −ny
nx =ny
Eigenvalues
σij =σxx = −4MPa σxy = −4MPa
σyx = −4MPa σyy = −4MPa
⎡
⎣⎢⎢
⎤
⎦⎥⎥
nx2 + ny
2 =1
2nx2 =1
nx = 2 2
ny = 2 2
Note that X1•X2 = 0 Principal direcPons are perpendicular
nx2 + ny
2 =1
2nx2 =1
nx = 2 2
ny = −2 2 x x
y y
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19. Principal Stresses V Example (values in MPa)
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σxx = - 4‐ τxn = - 4 ‐ σx’x’ = - 8‐ τx’n = - 8 ‐σxy = - 4‐ τxs = - 4 ‐ σx’y’ = 0 τx’s = 0 σyx = - 4‐ τyn = + 4 σy’x’ = - 0‐ τy’s = +0 σyy = - 4‐ τys = - 4 ‐ σy’y’ = 0 τy’n = 0
n s
n
s
σ1 σ2
19. Principal Stresses V Example
Matrix form/Matlab
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>> sij = [- 4 - 4;- 4 - 4] ‐ ‐ ‐ ‐sij = - 4 - 4 ‐ ‐ - 4 - 4 ‐ ‐>> [v,d]=eig(sij) v = 0.7071 - 0.7071 ‐ 0.7071 0.7071 d = - 8 0 ‐ 0 0
Eigenvectors (in columns)
Corresponding eigenvalues (in columns)
σ1 σ2
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