19.&principal&stresses - soest ·...

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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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hKp://hvo.wr.usgs.gov/kilauea/update/images.html

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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


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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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



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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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


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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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


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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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


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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V Example Find the principal stresses

given

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σ ij =σ xx = −4MPa σ xy = −4MPaσ yx = −4MPa σ yy = −4MPa

⎡

⎣⎢⎢

⎤

⎦⎥⎥


V Example

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⎡

⎣⎢⎢

⎤

⎦⎥⎥

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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⎡

⎣⎢⎢

⎤

⎦⎥⎥

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


⎡

⎣⎢⎢

⎤

⎦⎥⎥

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

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