19. Principal Stresses - SOEST 2 19. Principal Stresses II Cauchy’s formula A Relates tracQon...

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19.PrincipalStresses

I MainTopics

A Cauchy’sformulaB Principalstresses(eigenvectorsandeigenvalues)C Example

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

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II Cauchy’sformulaA RelatestracQon(stressvector)componentstostresstensorcomponentsinthesamereferenceframe

B 2Dand3Dtreatmentsanalogous

C τi=σijnj=njσij=njσji

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Note:allstresscomponentsshownareposiQve


II Cauchy’sformula(cont.)C τi=njσji

1  Meaningoftermsa τi=tracQon

componentb nj=direcQoncosine

ofanglebetweenn-direcQonandj-direcQon

c σji=stresscomponent

d τiandσjiactinthesamedirecQon

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nj=cosθnj=anj

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II Cauchy’sformula(cont.)

D Expansion(2D)ofτi=njσji1  τx=nxσxx+nyσyx2  τy=nxσxy+nyσyy

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nj=cosθnj=anj



E DerivaQon:ContribuQonstoτx

1

2

3

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nx=cosθnx=anxny=cosθny=any

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

NotethatallcontribuQonsmustactinx-direcQon

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E DerivaQon:ContribuQonstoτy

1

2

3

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nx=cosθnx=anxny=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

NotethatallcontribuQonsmustactiny-direcQon

19.PrincipalStressesII Cauchy’sformula(cont.)

F AlternaQveforms1  τi=njσji2  τi=σjinj3  τi=σijnj

4

5  Matlaba t=s’*nb t=s*n

6Notethatthestressmatrix(tensor)transformsthenormalvectortotheplane(n)tothetracQonvectoracQngontheplane(τ)

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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

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

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A Nowweseek(a)theorientaQonoftheunitnormal(givenbynxandny)toanyspecialplanewheretheassociatedtracQonvectorisperpendicular(normal)tothatplane,and(b)themagnitude(λ)ofthattracQonvector.

B ThesetracQonvectorshavenoshearcomponentandhencecorrespondtotheprincipalstresses.

C TheorientaQonsofthespecialtracQonvectorsarecalledeigenvectors,andthemagnitudesofthesespecialtracQonvectorsarecalledeigenvalues.

D AneigenvectorpointsinthesamedirecQonasthenormaltotheplane,sothetransformaQonofthenormalvectortothetracQonvectorbyCauchy’sformuladoesnotinvolvearotaQon.

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nj=cosθnj=anj

III Principalstresses(eigenvectorsandeigenvalues)

NotethatthetracQonvectorbelowparallelsthenormalvectortotheplane


E Thex-andy-componentsofsuchaprincipaltracQonvectorareobtainedbyprojecQngthevectorontothex-andy-axes:

Sincethemagnitudeoftheeigenvectorisascalar,boththenormaltotheplaneandtheeigenvectorpointinthesamedirecQon.

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nj=cosθnj=anj

τ x

τ y

⎡

⎣⎢⎢

⎤

⎦⎥⎥= τ

→ nxny

⎡

⎣⎢⎢

⎤

⎦⎥⎥

III Principalstresses(eigenvectorsandeigenvalues)

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IIIPrincipalstresses(eigenvectorsandeigenvalues)

F

G

H

Theformof(H)is[A][X]=λ[X],and[σ]issymmetric

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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→


IIIPrincipalstresses(eigenvectorsandeigenvalues)

Frompreviousnotes

Subtracttherightsidefrombothsides

I

orJ,where

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σ xx − T σ yx − 0σ xy − 0 σ yy − T

⎡

⎣⎢⎢

⎤

⎦⎥⎥

nxny

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 0

0⎡

⎣⎢

⎤

⎦⎥

σ − IT[ ] n[ ] = 0[ ] I[ ] = 1 00 1

⎡

⎣⎢

⎤

⎦⎥

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Now,abriefinterludetoshowhowtosolveanalyQcallyfortheeigenvaluesin2D

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9.EIGENVECTORS,EIGENVALUES,ANDFINITESTRAINIIIDeterminant(cont.)

D GeometricmeaningsoftherealmatrixequaQonAX=B=01 |A|≠0;

a [A]-1existsb Describestwolines(or3

planes)thatintersectattheorigin

c XhasauniquesoluQon2  |A|=0;

a [A]-1doesnotexistb Describestwoco-linear

linesthatthatpassthroughtheorigin(orthreeplanesthatintersectalineorplanethroughtheorigin)

c XhasnouniquesoluQon

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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

Frompreviousnotes

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9.EIGENVECTORS,EIGENVALUES,ANDFINITESTRAIN

IIIEigenvalueproblems,eigenvectorsandeigenvalues(cont.)EAlternaQveformofaneigenvalueequaQon

1[A][X]=λ[X]SubtracQngλ[IX]=λ[X]frombothsidesyields:

2[A-Iλ][X]=0(sameformas[A][X]=0)F SoluQoncondiQonsandconnecQonswithdeterminants

1UniquetrivialsoluQonof[X]=0ifandonlyif|A-Iλ|≠0

2EigenvectorsoluQons([X]≠0)ifandonlyif|A-Iλ|=0

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Frompreviousnotes


IIIEigenvalueproblems,eigenvectorsandeigenvalues(cont.)

GCharacterisQcequaQon:|A-Iλ|=0

1Eigenvaluesofasymmetric2x2matrix

a

b

c

d

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λ1,λ2 =a + d( ) ± a + d( )2 − 4 ad − b2( )

2

RadicaltermcannotbenegaQve.Eigenvaluesarereal.

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

Frompreviousnotes

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VISoluQonsforsymmetricmatrices(cont.)BAnydisQncteigenvectors(X1,X2)ofasymmetricnxnmatrixareperpendicular(X1•X2=0)1a AX1=λ1X1 1b AX2=λ2X2AX1parallelsX1,AX2parallelsX2(propertyofeigenvectors)

DopngAX1byX2andAX2byX1cantestwhetherX1andX2areorthogonal.

2a X2•AX1=X2•λ1X1=λ1(X2•X1)2b X1•AX2=X1•λ2X2=λ2(X1•X2)

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Frompreviousnotes


LDisQncteigenvectors(X1,X2)ofasymmetric2x2matrixareperpendicularSincethelersidesof(2a)and(2b)areequal,therightsidesmustbeequaltoo.Hence,4  λ1(X2•X1)=λ2(X1•X2)Nowsubtracttherightsideof(4)fromtheler5  (λ1–λ2)(X2•X1)=0• Theeigenvaluesgenerallyaredifferent,soλ1–λ2≠0.• Thismeansfor(5)toholdthatX2•X1=0.• Therefore,theeigenvectors(X1,X2)ofasymmetric2x2

matrixareperpendicular

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Frompreviousnotes

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Endofbriefinterlude

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IV ExampleFindtheprincipalstresses

given

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

⎡

⎣⎢⎢

⎤

⎦⎥⎥

θ ′x x = −45!,θ ′x y = 45!,θ ′y x = −135!,θ ′y y = −45!

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IV Example

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⎡

⎣⎢⎢

⎤

⎦⎥⎥

λ1,λ2 =σ xx +σ yy( ) ± σ xx −σ yy( )2 + 4σ xy

2

2

λ1,λ2 = −4MPa ± 642

MPa = 0MPa,−8MPa

Firstfindeigenvalues

19.PrincipalStressesIV Example

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⎡

⎣⎢⎢

⎤

⎦⎥⎥

λ1,λ2 = −4MPa ± 642

MPa = 0MPa,−8MPa

Thensolveforeigenvectors(thedimensionsofstressareunnecessarybelowandaredropped)

For λ1 = 0 :−4 − 0 −4−4 −4 − 0

⎡

⎣⎢

⎤

⎦⎥

nxny

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 0

0⎡

⎣⎢

⎤

⎦⎥⇒

−4nx − 4ny = 0⇒ nx = −ny−4nx − 4ny = 0⇒ nx = −ny

For λ2 = −8 :−4 − −8( ) −4

−4 −4 − −8( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

nxny

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 0

0⎡

⎣⎢

⎤

⎦⎥⇒

4nx − 4ny = 0⇒ nx = ny−4nx + 4ny = 0⇒ nx = ny

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19.PrincipalStressesIV Example

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λ1 = 0MPAλ2 = −8MPA

Eigenvectorsnx = −nynx = ny

Eigenvalues


⎡

⎣⎢⎢

⎤

⎦⎥⎥

19.PrincipalStressesIV Example(valuesinMPa)

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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 τys=+4 σy’x’=-0 τy’s=+0

σyy=-4 τyn=-4 σy’y’=0 τy’n=0

ns

n

s

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IV ExampleMatrixform

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σ ′x ′x σ ′x ′y

σ ′y ′x σ ′y ′y

⎡

⎣⎢⎢

⎤

⎦⎥⎥=

a ′x ′x a ′x ′y

a ′y ′x a ′y ′y

⎡

⎣⎢⎢

⎤

⎦⎥⎥

σ xx σ xy

σ yx σ yy

⎡

⎣⎢⎢

⎤

⎦⎥⎥

a ′x ′x a ′x ′y

a ′y ′x a ′y ′y

⎡

⎣⎢⎢

⎤

⎦⎥⎥

T

σ ′i ′j⎡⎣ ⎤⎦ = a[ ] σ ′i ′j⎡⎣ ⎤⎦ a[ ]T

Thisexpressionisvalidin2Dand3D!

19.PrincipalStressesIV ExampleMatrixform/Matlab

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>>sij=[-4-4;-4-4]sij=-4-4-4-4>>[vec,val]=eig(sij)vec=0.7071-0.70710.70710.7071val=-8000

Eigenvectors(incolumns)

Correspondingeigenvalues(incolumns)

19. Principal Stresses - SOEST 2 19. Principal Stresses II Cauchy’s formula A Relates tracQon...

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