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19.PrincipalStresses
I MainTopics
A Cauchy’sformulaB Principalstresses(eigenvectorsandeigenvalues)C Example
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19.PrincipalStresses
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hKp://hvo.wr.usgs.gov/kilauea/update/images.html
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19.PrincipalStresses
II Cauchy’sformulaA RelatestracQon(stressvector)componentstostresstensorcomponentsinthesamereferenceframe
B 2Dand3Dtreatmentsanalogous
C τi=σijnj=njσij=njσji
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Note:allstresscomponentsshownareposiQve
19.PrincipalStresses
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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19.PrincipalStresses
II Cauchy’sformula(cont.)
D Expansion(2D)ofτi=njσji1 τx=nxσxx+nyσyx2 τy=nxσxy+nyσyy
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nj=cosθnj=anj
19.PrincipalStresses
II Cauchy’sformula(cont.)
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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19.PrincipalStresses
II Cauchy’sformula(cont.)
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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19.PrincipalStresses
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
19.PrincipalStresses
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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19.PrincipalStresses
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→
19.PrincipalStresses
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
9.EIGENVECTORS,EIGENVALUES,ANDFINITESTRAIN
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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9.EIGENVECTORS,EIGENVALUES,ANDFINITESTRAIN
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
9.EIGENVECTORS,EIGENVALUES,ANDFINITESTRAIN
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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19.PrincipalStresses
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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19.PrincipalStresses
IV Example
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σ ij =σ xx = −4MPa σ xy = −4MPaσ yx = −4MPa σ yy = −4MPa
⎡
⎣⎢⎢
⎤
⎦⎥⎥
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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σ ij =σ xx = −4MPa σ xy = −4MPaσ yx = −4MPa σ yy = −4MPa
⎡
⎣⎢⎢
⎤
⎦⎥⎥
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
σ ij =σ xx = −4MPa σ xy = −4MPaσ yx = −4MPa σ yy = −4MPa
⎡
⎣⎢⎢
⎤
⎦⎥⎥
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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19.PrincipalStresses
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)
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