Lec.19.pptx - 19. Principal Stresses I Main Topics...

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Unformatted text preview: 10/25/11 19. Principal Stresses I Main Topics A Cauchy’s formula B Principal stresses (eigenvectors and eigenvalues) C Example 10/25/11 GG303 1 19. Principal Stresses hKp://hvo.wr.usgs.gov/kilauea/update/images.html 10/25/11 GG303 2 1 10/25/11 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 10/25/11 Note: all stress components shown are posiPve GG303 3 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 10/25/11 GG303 nj = cosθnj = anj 4 2 10/25/11 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 nj = cosθnj = anj 10/25/11 GG303 5 19. Principal Stresses II Cauchy’s formula (cont.) Note that all contribuPons must act in x ­direcPon E DerivaPon: ContribuPons to τx 1 τ x = w(1)σ xx + w(2)σ yx 2 Fx ⎛ Ax ⎞ Fx(1) ⎛ Ay ⎞ Fx( 2 ) = + An ⎜ An ⎟ Ax ⎜ An ⎟ Ay ⎝⎠ ⎝⎠ 3 τ x = nxσ xx + nyσ yx 10/25/11 GG303 nx = cosθnx = anx ny = cosθny = any 6 3 10/25/11 19. Principal Stresses II Cauchy’s formula (cont.) Note that all contribuPons must act in y ­direcPon E DerivaPon: ContribuPons to τy 1 τ y = w( 3)σ xy + w( 4 )σ yy 2 3 4 ⎛ A ⎞ Fy( ) ⎛ Ay ⎞ Fy( ) =⎜ x⎟ +⎜ ⎟ An ⎝ An ⎠ Ax ⎝ An ⎠ Ay Fy nx = cosθnx = anx ny = cosθny = any 3 τ y = nxσ xy + nyσ yy 10/25/11 GG303 7 19. Principal Stresses τ x = nxσ xx + nyσ yx τ y = nxσ xy + nyσ yy II Cauchy’s formula (cont.) F AlternaPve forms 1  τi = njσji 2  τi = σjinj 3  τi = σijnj ⎡ τ x ⎤ ⎡ σ xx ⎥⎢ τ y ⎥ = ⎢ σ xy ⎢ ⎥⎢ ⎢τ ⎥ ⎣σ ⎣ z ⎦ ⎢ xz 4 ⎢ ⎢ σ yx σ yy σ yz σ zx ⎤ ⎡ nx ⎤ ⎥⎢ ⎥ σ xy ⎥ ⎢ ny ⎥ ⎥⎢ ⎥ σ zz ⎥ ⎣ nz ⎦ ⎥ ⎦⎢ 5  Matlab a t = s’*n b t = s*n 10/25/11 3D nj = cosθnj = anj GG303 8 4 10/25/11 19. Principal Stresses III Principal stresses (eigenvectors and eigenvalues) A B C ⎡ τx ⎢ ⎢ τy ⎣ ⎤ ⎡ σ xx ⎥=⎢ ⎥ ⎢ σ xy ⎦⎣ ⎡ τx ⎢ ⎢ τy ⎣ ⎤ → ⎡ nx ⎤ Vector components ⎥= τ ⎢ ⎥ → ⎥ ⎢ ny ⎥ Let λ = τ ⎦ ⎣ ⎦ ⎡ σ xx ⎢ ⎢ σ xy ⎣ σ yx ⎤ ⎡ nx ⎤ ⎥⎢ ⎥ σ yy ⎥ ⎢ ny ⎥ ⎣ ⎦ ⎦ Cauchy’s Formula ⎡ nx ⎤ σ yx ⎤ ⎡ nx ⎤ ⎥⎢ ⎥ = λ⎢ ⎥ σ yy ⎥ ⎢ ny ⎥ ⎢ ny ⎥ ⎦ ⎣ ⎦ ⎦⎣ The form of (C ) is [A][X=λ[X], and [σ] is symmetric 10/25/11 GG303 9 9. EIGENVECTORS, EIGENVALUES, AND From previous notes 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 10/25/11 GG303 10 5 10/25/11 9. EIGENVECTORS, EIGENVALUES, AND FINITE STRAIN From previous notes 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 10/25/11 Intersecting lines have non-parallel normals AX = B = 0 nx(1) nx(2) ny(1) ny(2) x y = |A| = nx(1) * ny(2) - ny(1) * nx(2) ≠ 0 n1 x n2 ≠ 0 Parallel lines have parallel normals AX = B = 0 nx(1) nx(2) ny(1) ny(2) x y = d1=0 d2=0 |A| = nx(1) * ny(2) - ny(1) * nx(2) = 0 n1 x n2 = 0 GG303 d1=0 d2=0 11 9. EIGENVECTORS, EIGENVALUES, AND FINITE STRAIN From previous notes III Eigenvalue problems, eigenvectors and eigenvalues (cont.) J CharacterisPc equaPon: |A ­Iλ|=0 3 Eigenvalues of a symmetric 2x2 matrix a λ1 , λ2 = b λ1 , λ2 = c λ1 , λ2 = d 10/25/11 ( a + d ) ± ( a + d )2 − 4 ( ad − b 2 ) ⎡a b⎤ A=⎢ ⎥ ⎣b d⎦ 2 ( a + d ) ± ( a + 2ad + d )2 − 4 ad + 4b 2 2 ( a + d ) ± ( a − 2ad + d )2 + 4b 2 λ1 , λ2 = 2 ( a + d ) ± ( a − d )2 + 4 b 2 2 GG303 Radical term cannot be negaPve. Eigenvalues are real. 12 6 10/25/11 9. EIGENVECTORS, EIGENVALUES, AND From previous notes 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 10/25/11 GG303 13 19. Principal Stresses III Principal stresses (eigenvectors and eigenvalues) ⎡ σ xx ⎢ ⎢ σ xy ⎣ ⎡ nx ⎤ σ yx ⎤ ⎡ nx ⎤ ⎥⎢ ⎥ = λ⎢ ⎥ σ yy ⎥ ⎢ ny ⎥ ⎢ 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 10/25/11 GG303 14 7 10/25/11 19. Principal Stresses V Example Find the principal stresses given σ ij ⎡ σ xx = −4 MPa σ xy = −4 MPa =⎢ ⎢ σ yx = −4 MPa σ yy = −4 MPa ⎣ 10/25/11 ⎤ ⎥ ⎥ ⎦ GG303 15 19. Principal Stresses V Example ⎡ σ xx = −4 MPa σ xy = −4 MPa σ ij = ⎢ ⎢ σ yx = −4 MPa σ yy = −4 MPa ⎣ ⎤ ⎥ ⎥ ⎦ First find eigenvalues (in MPa) λ1 , λ2 = ( a + d ) ± ( a − d )2 + 4 b 2 λ1 , λ2 = −4 ± 10/25/11 2 64 = −4 ± 4 = 0, −8 2 GG303 16 8 10/25/11 19. Principal Stresses IV Example ⎡ σ xx = −4 MPa σ xy = −4 MPa σ ij = ⎢ ⎢ σ yx = −4 MPa σ yy = −4 MPa ⎣ ⎤ ⎥ ⎥ ⎦ 64 Eigenvalues (MPa) = −4 ± 4 = 0, −8 2 Then solve for eigenvectors (X) using [A ­Iλ][X]=0 λ1 , λ2 = −4 ± ⎡ ⎤ ⎡ −4 − 0 −4 ⎤ ⎢ nx ⎥ ⎡ 0 ⎤ For λ1 = 0 : ⎢ =⎢ ⎥n ⎥ ⇒ −4 nx − 4 ny = 0 ⇒ nx = − ny −4 − 0 ⎦ ⎢ y ⎥ ⎣ 0 ⎦ ⎣ −4 ⎣ ⎦ ⎡ −4 − ( −8 ) ⎤⎡ n ⎤ ⎡ −4 ⎤ ⎥ ⎢ x ⎥ = ⎢ 0 ⎥ ⇒ 4 nx − 4 ny = 0 ⇒ nx = ny For λ2 = −8 : ⎢ ⎢ −4 σ yy − ( −8 ) ⎥ ⎢ ny ⎥ ⎣ 0 ⎦ ⎦ ⎣ ⎦⎣ 10/25/11 GG303 17 19. Principal Stresses IV Example ⎡ σ xx = −4 MPa σ xy = −4 MPa σ ij = ⎢ ⎢ σ yx = −4 MPa σ yy = −4 MPa ⎣ 2 2 nx + ny = 1 2 2 nx = 1 Eigenvalues λ1 = 0 MPa ⎤ ⎥ ⎥ ⎦ λ2 = −8 MPa y y nx = 2 2 ny = − 2 2 nx = ny 2 2 nx + ny = 1 x x Eigenvectors nx = − ny 2 2 nx = 1 nx = 2 2 ny = 2 2 Note that X1•X2 = 0 Principal direcPons are perpendicular 10/25/11 GG303 18 9 10/25/11 19. Principal Stresses V Example (values in MPa) 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 σyy =  ­ 4 τys =  ­4 σy’y’ = 0 n τy’s = +0 τy’n = 0 s σ1 σ2 n s 10/25/11 GG303 19 19. Principal Stresses V Example Matrix form/Matlab σ1 10/25/11 σ2 >> 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 GG303 Eigenvectors (in columns) Corresponding eigenvalues (in columns) 20 10 ...
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This note was uploaded on 12/05/2011 for the course GEOLOGY 300 taught by Professor Stephenmartel during the Fall '11 term at University of Hawaii, Manoa.

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