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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 τi = njσji
1 Meaning of terms a τi = tracPon
component b nj = direcPon cosine
of angle between n-‐direcPon and j-‐direcPon
c σji = tracPon component
d τi 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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τ x = w1( )σ xx + w
2( )σ yx
τ x = nxσ xx + nyσ yx
FxAn
=Ax
An
⎛⎝⎜
⎞⎠⎟Fx
1( )
Ax
+Ay
An
⎛⎝⎜
⎞⎠⎟Fx
2( )
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
τ y = w3( )σ xy + w
4( )σ yy
τ y = nxσ xy + nyσ yy
FyAn
=Ax
An
⎛⎝⎜
⎞⎠⎟Fy
3( )
Ax
+Ay
An
⎛⎝⎜
⎞⎠⎟Fy
4( )
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
τ xτ yτ z
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
=
σ xx σ yx σ zx
σ xy σ yy σ xy
σ xz σ yz σ zz
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
nxnynz
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
τ x = nxσ xx + nyσ yx
τ y = nxσ xy + nyσ yy
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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τ xτ y
⎡
⎣⎢⎢
⎤
⎦⎥⎥=
σ xx σ yx
σ xy σ yy
⎡
⎣⎢⎢
⎤
⎦⎥⎥
nxny
⎡
⎣⎢⎢
⎤
⎦⎥⎥
τ xτ y
⎡
⎣⎢⎢
⎤
⎦⎥⎥= τ
→ nxny
⎡
⎣⎢⎢
⎤
⎦⎥⎥
σ xx σ yx
σ xy σ yy
⎡
⎣⎢⎢
⎤
⎦⎥⎥
nxny
⎡
⎣⎢⎢
⎤
⎦⎥⎥= λ
nxny
⎡
⎣⎢⎢
⎤
⎦⎥⎥
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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σ xx σ yx
σ xy σ yy
⎡
⎣⎢⎢
⎤
⎦⎥⎥
nxny
⎡
⎣⎢⎢
⎤
⎦⎥⎥= λ
nxny
⎡
⎣⎢⎢
⎤
⎦⎥⎥
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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σ ij =σ xx = −4MPa σ xy = −4MPaσ yx = −4MPa σ yy = −4MPa
⎡
⎣⎢⎢
⎤
⎦⎥⎥
19. Principal Stresses
V Example
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σ ij =σ xx = −4MPa σ xy = −4MPaσ yx = −4MPa σ yy = −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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σ ij =σ xx = −4MPa σ xy = −4MPaσ yx = −4MPa σ yy = −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
⎡
⎣⎢
⎤
⎦⎥
nxny
⎡
⎣⎢⎢
⎤
⎦⎥⎥= 0
0⎡
⎣⎢
⎤
⎦⎥ ⇒ −4nx − 4ny = 0⇒ nx = −ny
For λ2 = −8 :−4 − −8( ) −4
−4 σ yy − −8( )⎡
⎣⎢⎢
⎤
⎦⎥⎥
nxny
⎡
⎣⎢⎢
⎤
⎦⎥⎥= 0
0⎡
⎣⎢
⎤
⎦⎥ ⇒ 4nx − 4ny = 0⇒ nx = ny
Eigenvalues (MPa)
19. Principal Stresses IV Example
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λ1 = 0MPaλ2 = −8MPa
Eigenvectors nx = −nynx = ny
Eigenvalues
σ ij =σ xx = −4MPa σ xy = −4MPaσ yx = −4MPa σ yy = −4MPa
⎡
⎣⎢⎢
⎤
⎦⎥⎥
nx2 + ny
2 = 1
2nx2 = 1
nx = 2 2
ny = 2 2Note 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