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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