GG612 Lecture 3. Outline
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1 GG61 Lecture 3 Strain and Stress Should complete infinitesimal strain by adding rota>on. Outline Matrix Opera+ons Strain 1 General concepts Homogeneous strain 3 Matrix representa>ons 4 Squares of line lengths 5 E (strain matrix) 6 ε (infinitesimal strain) 7 Coaxial finite strain 8 Non- coaxial finite strain Stress 1 Stress vector Stress at a point 3 Principal stresses 1
2 Main Theme Representa>on of complicated quan>>es describing strain and stress at a point in a clear manner Vector Conven>ons X = ini>al posi>on X = final posi>on U = displacement X U X
3 Matrix Inverses AA - 1 = A - 1 A = [I] [AB] - 1 = [B - 1 ][A - 1 ] ABB - 1 A - 1 =A[I]A - 1 =[I] [AB][AB] - 1 = [I] [B - 1 A - 1 ]= [AB] - 1 Matrix Inverses and Transposes a b = [a T ][b] a 1 a a n [AB] T = [B T ][A T ] b 1 b b n = a 1 b 1 + a b + + a n b n A nxq = AB = a 1 a ; B = qxm a n a 1 b 1 a1 b a 1 b m a b 1 a b a b m a n b 1 an b 1 a n b m [ AB] T = b 1 B T A T b = b m [ AB] T = B T A T b 1 b b m a 1 b 1 a b 1 a n b 1 a 1 b a b a n b 1 a 1 b m a b m a n b m a 1 a a n = b 1 a 1 b1 a b 1 a n b a 1 b a b a n b m a 1 bm a 1 b m a 1 3
4 Rota>on Matrix [R] Rota>ons change the orienta>ons of vectors but not their lengths X X = X X cosθ XX X X = X X X = RX X X = [RX] [RX] X X = [RX] T [RX] X X = [X T R T ] [RX] [X T ] [X]= [X T R T ] [RX] [X T ][I][X]= [X T ][R T ] [R][X] [I] = [R T ] [R] But [I] = [R - 1 ] [R], so [R T ] = [R - 1 ] X X R = cosθ sinθ Rota>on Matrix [R] D Example sinθ ; X cosθ [ ] = [ R] [ X] X θ z X y x x y = cosθ sinθ sinθ cosθ x y x = cosθx + sinθy y = sinθx + cosθy x + y = x + y R T = cosθ sinθ sinθ cosθ RR T = cosθ sinθ sinθ cosθ cosθ sinθ sinθ = cosθ R T = R 1 4
5 General Concepts Deforma>on = Rigid body mo>on + Strain Rigid body mo>on Rigid body transla>on Treated by matrix addi>on [X ] = [X] + [U] Rigid body rota>on Changes orienta>on of lines, but not their length Axis of rota>on does not rotate; it is an eigenvector Treated by matrix mul>plica>on [X ] = [R] [X] Transla>on Transla>on + Rota>on Transla>on + Rota>on + Strain Normal strains General Concepts Change in line length Extension (elonga>on) = Δs/s 0 Stretch = S = s /s 0 Quadra>c elonga>on = Q = (s /s 0 ) Shear strains Change in right angles Dimensions: Dimensionless 5
6 Homogeneous strain Parallel lines to parallel lines (D and 3D) Circle to ellipse (D) Sphere to ellipsoid (3D) Homogeneous strain Matrix Representa>on (D) [ X ] = [ F] [ X] x y = a b c d x y [ X] = [ F] 1 [ X ] x y = a b c d 1 x y y x y x 6
7 d x = x x dx + x y dy d y = y x dx + y y dy d x d y = Matrix Representa>ons: Posi>ons (D) x x y x x y y y [ d X ] = [ F] [ dx] dx dy d x d y = Matrix Representa>ons: Posi>ons (D) x x y x x y y y dx dy If deriva>ves are constant (e.g., at a point) x y = a b c d [ X ] = [ F] [ X] x y 7
8 du = u u dx + x y dy dv = v v dx + x y dy Matrix Representa>ons Displacements (D) du dv = u x v x u y v y dx dy [ du ] = [ J u ][ dx] u = u x x + u y y v = v x x + v y y Matrix Representa>ons Displacements (D) u v = u x v x u y v y x y [ U ] = [ J u ][ X] If deriva>ves are constant (e.g., at a point) 8
9 Matrix Representa>ons Posi>ons and Displacements (D) U = X X U = FX X = FX IX U = [ F I ] X [ F I ] J u [ J u ] = a 1 b c d 1 Matrix Representa>ons Posi>ons and Displacements Lagrangian: f(x) [ X ] = [ F] [ X] U = X X = FX X [ ] X U = FX IX = F I Eulerian: g(x ) [ X] = F 1 [ ] X U = X X = X F 1 X U = I F 1 X 9
10 s = X Squares of Line Lengths X cos( θ X X ) s = X X = X T X s = X T X s = X X s = FX [ ] T [ FX] s = X T F T FX E (strain matrix) s s s s E = dx T F T F I dx = dx T [ E]dX FT F I 10
11 ε (Infinitesimal Strain Matrix, D) E F T F I = 1 E = 1 u x + 1 u y [ J u + I ] T J u + I v x v y + 1 [ ] I u x + 1 v x u y v y If par>al deriva>ves << 1, their squares can be dropped to obtain the infinitesimal strain matrix ε u ε = 1 x + u u y + v u y + v v y + v y ε (Infinitesimal Strain Matrix, D) J u = u x v x u y v y u ε = 1 x + u v x + u y u y + v v y + v y = 1 J + J T u u u x v x u u y = 1 x + u u y + v u v v y x + u v y y + v + 1 x u u y v v y x u v y y v y J u = ε + ω ε is symmetric ω is an>- symmetric Linear superposi>on 11
12 ε = ε (Infinitesimal Strain Matrix, D) Meaning of components u 1 u y + v 1 u y + v v y y Pure strain without rota>on v x = u y du ds dv ds = ε xx ε yx ε xy ε yy 1 0 = ε xx ε yx ds x First column in ε: rela>ve displacement vector for unit element in x- direc>on ε yx is displacement in the y- direc>on of right end of unit element in x- direc>on ε = ε (Infinitesimal Strain Matrix, D) Meaning of components u 1 u y + v 1 u y + v v y ds y Pure strain without rota>on v x = u y du ds dv ds = ε xx ε yx ε xy ε yy 0 1 = ε xy ε yy x Second column in ε: rela>ve displacement vector for unit element in y- direc>on ε xy is displacement in the x- direc>on of upper end of unit element in y- direc>on 1
13 ε = ε (Infinitesimal Strain Matrix, D) Meaning of components u 1 u y + v 1 u y + v v y Pure strain without rota>on ε 11 = ε xx = elonga>on of line parallel to x- axis ε 1 = ε xy (Δθ)/ ε 1 = ε yx (Δθ)/ ε = ε yy = elonga>on of line parallel to y- axis Δθ = (ψ ψ 1 ) = 1 v x + u y Shear strain > 0 if angle between +x and +y axes decreases ω = ω (Infinitesimal Strain Matrix, D) Meaning of components 0 1 u y v 1 v x u y 0 y Pure rota>on without strain v x = u y du ds dv ds 0 ω z ω z = 0 ω z ω z x ds ω z << 1 radian First column in ω: rela>ve displacement vector for unit element in x- direc>on ω xy is displacement in the y- direc>on of right end of unit element in x- direc>on 13
14 ω = ω (Infinitesimal Strain Matrix, D) Meaning of components 0 1 u y v 1 v x u y 0 y ds Pure rota>on without strain v x = u y du ds dv ds 0 ω z ω z = 0 ω z ω z ω z << 1 radian x Second column in ω: rela>ve displacement vector for unit element in y- direc>on ω yx is displacement in the nega%ve x- direc>on of upper end of unit element in y- direc>on ε = ε (Infinitesimal Strain Matrix, D) Meaning of components u 1 u y + v 1 u y + v v y Pure rota>on without strain ε 11 = ε xx = elonga>on of line parallel to x- axis ε 1 = ε xy (Δθ)/ ε 1 = ε yx (Δθ)/ ε = ε yy = elonga>on of line parallel to y- axis Δθ = (ψ ψ 1 ) = 1 v x + u y Shear strain > 0 if angle between +x and +y axes decreases 14
15 Coaxial Finite Strain F = a b b d ; F [ ][ X] = λ[ X] 1 1 F = 1 F = F T All values of X X are posi>ve if X 0 F is posi>ve definite F has an inverse Eigenvalues > 0 F has a square root Coaxial Finite Strain F = a b b d ; F [ ][ X] = λ[ X] 1 1 F = 1 Strain ellipse 1 Eigenvectors (X) of F are perpendicular because F is symmetric (X 1 X = 0) X 1, X solve d(x X )/dθ = 0 3 X 1, X along major axes of strain ellipse 4 X 1 =X 1 ; X =X 5 Principal strain axes do not rotate Reciprocal Strain ellipse 15
16 Non- coaxial Finite Strain The vectors that transform from the axes of the reciprocal strain ellipse to the principal axes of the strain ellipse rotate The rota>on is given by the matrix that rotates the principal axes of the reciprocal strain ellipse to those of the strain ellipse Non- coaxial Finite Strain 1 [X ]=[F][X] X X =[X][F T F][X] 3 [F T F] is symmetric 4 Eigenvectors of [F T F] give principal strain direc>ons 5 Square roots of eigenvalues of [F T F] give principal stretches 6 [X]=[F- 1][X ] 7 X X=[X ][F - 1 ] T [F - 1 ][X ] 8 [F - 1 ] T [F - 1 ] is symmetric 9 Eigenvectors of [F - 1 ] T [F - 1 ]give principal strain direc>ons 10 Square roots of eigenvalues of [F - 1 ] T [F - 1 ] give (reciprocal) principal stretches 16
17 Non- coaxial Finite Strain 1 The strain ellipse and the reciprocal strain ellipse have the same eigenvalues but different eigenvectors. [F T F]=[[F - 1 ] T [F - 1 ]] [[F - 1 ] T [F - 1 ]] - 1 = [[F - 1 ] - 1 [F - 1 ] T ] - 1 ]=FF T. Non- coaxial Finite Strain 1 Strain ellipse and reciprocal strain ellipse have equal eigenvalues, different eigenvectors. [F T F]=[[F - 1 ] T [F - 1 ]] [[F - 1 ] T [F - 1 ]] - 1 = [[F - 1 ] - 1 [F - 1 ] T ] - 1 ] = FF T. F = Strain ellipse Reciprocal Strain ellipse F T F = ;FF T = 4 17
18 F T F = Coaxial vs. Non- coaxial Strain F = 1 F T F = F = FF T = FF T = 4 >> [X,L] = eig(f) X = L = >> [X1,L1] = eig(f*f) X1 = L1 = >> [X,L0]=eig(F) X = L0 = >> [X,L1]=eig(F'*F) X = L1 = >> [X1,L]=eig(F*F ) X1 = L = Coaxial vs. Non- coaxial Strain Coaxial F = F T (F is symmetric) FF T = F T F = F (F is symmetric) FX = λx [FF T ]X = λ X [F T F]X = λ X F = U = V F=[1 1; 1 ]; F =[ 3; 3 5]; >> [X,L] = eig(f) X = L = >> [X1,L1] = eig(f*f) X1 = L1 = Non- coaxial F F T (F is not symmetric) F T F F T F (but both symmetric) FX = λx [FF T ]X 1 = λ 1 X 1 ; λ 1 = λ λ [FF T ]X = λ X ; X X 1 X F = RU = R[F T F] 1/ = VR = [F T F] 1/ R F=[1 1; 0 ]; F T F=[1 1; 1 5]; FF T =[ ; 4]; >> [X,L0]=eig(F) X = L0 = >> [X,L]=eig(F'*F) X = L = >> [X1,L1]=eig(F*F ) X1 = L1 =
19 Polar Decomposi>on Theorem Suppose (1) [F]= [R][U], where R is a rota>on matrix and U is a symmetric stretch matrix. Then () F T F = [RU] T [RU] = U T R T RU =U T R - 1 RU=U T U However, U is postulated to be posi>ve definite, so (3) U T U = U = F T F Since F T F gives squares of line lengths, if U gives strains without rota>ons, it too should give the same squares of line lengths. Hence (4) U = [F T F ] 1/ From equa>on (1): (5) R = FU - 1 Polar Decomposi>on Theorem Suppose (1) [F]= [V] [R * ], where R * is a rota>on matrix and V is a symmetric stretch matrix. Then () FF T = [VR * ] [VR * ] T = VR * R *T V T = VR * R *- 1 V T = VV T However, V is postulated to be posi>ve definite, so (3) VV T = V = FF T Since FF T gives squares of line lengths, if V gives strains without rota>ons, it too should give the same squares of line lengths. Hence (4) V = [FF T ] 1/ From equa>on (1): (5) R * = V - 1 F 19
20 Polar Decomposi>on Theorem Proof that the polar decomposi>ons are unique. Suppose different decomposi>ons exist F = R 1 U 1 = R U X X = [ FX] FX F T F = R 1 U 1 [ ] = [ FX] T [ FX] = X T F T FX [ ] T [ R 1 U 1 ] = U T 1 R T 1 R 1 U 1 = U T 1 R 1 1 R 1 U 1 = U 1 T IU 1 = U 1 T U 1 = U 1 U 1 = U 1 F T F = [ R U ] T [ R U ] = U T R T R U = U T R 1 R U U 1 = U = U T IU = U T U = U U = U U 1 = U = U F = R 1 U 1 = R U 1 R 1 = R = R Polar Decomposi>on Theorem The same procedure can be followed to show that the decomposi>on F=VR* is unique. These results are very important: F can be decomposed into only one symmetric matrix that is pre- mul>plied by a unique rota>on matrix, and F can be decomposed into only one symmetric matrix that is post- mul>plied by a unique rota>on matrix. 0
21 Polar Decomposi>on Theorem Proof that F = RU = VR Polar Decomposi>on Theorem Comparison of eigenvectors and eigenvalues 1
22 1 Stress vector Stress state at a point 3 Stress transforma>ons 4 Principal stresses Stress 16. STRESS AT A POINT hwp://hvo.wr.usgs.gov/kilauea/update/images.html /14/1 GG303 44
23 16. STRESS AT A POINT I Stress vector (trac>on) on a plane A τ = lim F / A A 0 B Trac>on vectors can be added as vectors C A trac>on vector can be resolved into normal (τ n ) and shear (τ s ) components 1 A normal trac>on (τ n ) acts perpendicular to a plane A shear trac>on (τ s ) acts parallel to a plane D Local reference frame 1 The n- axis is normal to the plane The s- axis is parallel to the plane /14/1 GG STRESS AT A POINT III Stress at a point (cont.) A Stresses refer to balanced internal "forces (per unit area)". They differ from force vectors, which, if unbalanced, cause accelera>ons B "On - in conven>on": The stress component σ ij acts on the plane normal to the i- direc>on and acts in the j- direc>on 1 Normal stresses: i=j Shear stresses: i j /14/1 GG
24 16. STRESS AT A POINT III Stress at a point C Dimensions of stress: force/unit area D Conven>on for stresses 1 Tension is posi>ve Compression is nega>ve 3 Follows from on- in conven>on 4 Consistent with most mechanics books 5 Counter to most geology books /14/1 GG STRESS AT A POINT III Stress at a point C D σ ij = σ xx σ yx σ xy σ yy σ xx σ xy σ xz σ ij = σ yx σ yy σ yz σ zx σ zy σ zz - D 4 components 3- D 9 components E In nature, the state of stress can (and usually does) vary from point to point F For rota>onal equilibrium, σ xy = σ yx, σ xz = σ zx, σ yz = σ zy /14/1 GG
25 16. STRESS AT A POINT IV Principal Stresses (these have magnitudes and orienta>ons) A Principal stresses act on planes which feel no shear stress B The principal stresses are normal stresses. C Principal stresses act on perpendicular planes D The maximum, intermediate, and minimum principal stresses are usually designated σ 1, σ, and σ 3, respec>vely. E Principal stresses have a single subscript. /14/1 GG STRESS AT A POINT IV Principal Stresses (cont.) F Principal stresses represent the stress state most simply G H σ 1 0 σ ij = 0 σ σ xx σ xy σ xz σ ij = σ yx σ yy σ yz σ zx σ zy σ zz - D components 3- D 3 components /14/1 GG
26 19. Principal Stresses hwp://hvo.wr.usgs.gov/kilauea/update/images.html /14/1 GG Mohr Circle for Trac>ons From King et al., 1994 (Fig. 11) Coulomb stress change caused by the Landers rupture. The le}- lateral ML=6.5 Big Bear rupture occurred along dowed line 3 hr 6 min a}er the Landers main shock. The Coulomb stress increase at the future Big Bear epicenter is.-.9 bars. hwp://earthquake.usgs.gov/research/modeling/papers/landers.php /14/1 GG
27 19. Principal Stresses II Cauchy s formula A Relates trac>on (stress vector) components to stress tensor components in the same reference frame B D and 3D treatments analogous C τ i = σ ij n j = n j σ ij Note: all stress components shown are posi>ve /14/1 GG Principal Stresses II Cauchy s formula (cont.) C τ i = n j σ ji 1 Meaning of terms a τ i = trac>on component b n j = direc>on cosine of angle between n- direc>on and j- direc>on c σ ji = trac>on component d τ i and σ ji act in the same direc>on n j = cosθ nj = a nj /14/1 GG
28 II Cauchy s formula (cont.) 19. Principal Stresses D Expansion (D) of τ i = n j σ ji 1 τ x = n x σ xx + n y σ yx τ y = n x σ xy + n y σ yy n j = cosθ nj = a nj /14/1 GG II Cauchy s formula (cont.) E Deriva>on: Contribu>ons to τ x 19. Principal Stresses Note that all contribu>ons must act in x- direc>on 1 3 τ x = w ( 1) σ xx + w ( ) σ yx F x A n = A x A n ( 1) F x + A y A x τ x = n x σ xx + n y σ yx A n ( ) F x A y n x = cosθ nx = a nx n y = cosθ ny = a ny /14/1 GG
29 II Cauchy s formula (cont.) E Deriva>on: Contribu>ons to τ y 19. Principal Stresses Note that all contribu>ons must act in y- direc>on 1 3 τ y = w ( 3) σ xy + w ( 4) σ yy F y A n = A x A n ( 3) F y + A y A x τ y = n x σ xy + n y σ yy A n ( 4) F y A y n x = cosθ nx = a nx n y = cosθ ny = a ny /14/1 GG Principal Stresses II Cauchy s formula (cont.) F Alterna>ve forms 1 τ i = n j σ ji τ i = σ ji n j 3 τ i = σ ij n j τ x = n x σ xx + n y σ yx τ y = n x σ xy + n y σ yy 4 τ x τ y τ z σ xx σ yx σ zx = σ xy σ yy σ xy σ xz σ yz σ zz 5 Matlab a t = s *n b t = s*n n x n y n z 3D n j = cosθ nj = a nj /14/1 GG
30 19. Principal Stresses III Principal stresses (eigenvectors and eigenvalues) A τ x τ y = σ xx σ yx σ xy σ yy n x n y Cauchy s Formula B C τ x τ y σ xx σ xy = τ σ yx σ yy n x n y n x n y Vector components Let λ = τ = λ n x n y The form of (C ) is [A][X=λ[X], and [σ] is symmetric /14/1 GG EIGENVECTORS, EIGENVALUES, AND FINITE STRAIN From previous notes III Eigenvalue problems, eigenvectors and eigenvalues (cont.) J Characteris>c equa>on: A- Iλ =0 a b 3 Eigenvalues of a symmetric x matrix A = b d ( a λ 1,λ = a + d) ± ( a + d) 4( ad b ) b ( λ 1,λ = a + d) ± ( a + ad + d) 4ad + 4b c ( λ 1,λ = a + d) ± ( a ad + d) + 4b d ( λ 1,λ = a + d) ± ( a d) + 4b Radical term cannot be nega>ve. Eigenvalues are real. /14/1 GG
31 9. EIGENVECTORS, EIGENVALUES, AND From previous notes FINITE STRAIN L Dis>nct eigenvectors (X 1, X ) of a symmetric x matrix are perpendicular Since the le} sides of (a) and (b) are equal, the right sides must be equal too. Hence, 4 λ 1 (X X 1 ) =λ (X 1 X ) Now subtract the right side of (4) from the le} 5 (λ 1 λ )(X X 1 ) =0 The eigenvalues generally are different, so λ 1 λ 0. This means for (5) to hold that X X 1 =0. Therefore, the eigenvectors (X 1, X ) of a symmetric x matrix are perpendicular /14/1 GG Principal Stresses III Principal stresses (eigenvectors and eigenvalues) σ xx σ xy σ yx σ yy n x n y = λ D Meaning 1 Since the stress tensor is symmetric, a reference frame with perpendicular axes defined by n x and n y pairs can be found such that the shear stresses are zero This is the only way to sa>sfy the equa>on above; otherwise σ xy n y 0, and σ xy n x 0 3 For different (principal) values of λ, the orienta>on of the corresponding principal axis is expected to differ n x n y /14/1 GG
32 19. Principal Stresses V Example Find the principal stresses given σ ij = σ xx = 4MPa σ yx = 4MPa σ xy = 4MPa σ yy = 4MPa /14/1 GG Principal Stresses V Example σ xx = 4MPa σ xy = 4MPa σ ij = σ yx = 4MPa σ yy = 4MPa First find eigenvalues (in MPa) λ 1,λ = a + d ( ) ± a d λ 1,λ = 4 ± 64 ( ) + 4b = 4 ± 4 = 0, 8 /14/1 GG
33 19. Principal Stresses IV Example σ xx = 4MPa σ xy = 4MPa σ ij = σ yx = 4MPa σ yy = 4MPa λ 1,λ = 4 ± 64 = 4 ± 4 = 0, 8 Eigenvalues (MPa) Then solve for eigenvectors (X) using [A- Iλ][X]= For λ 1 = 0 : ( 8) 4 For λ = 8 : 4 σ yy 8 n x n y ( ) = 0 4n x 4n y = 0 n x = n y 0 n x n y = 0 4n x 4n y = 0 n x = n y 0 /14/1 GG Principal Stresses IV Example σ xx = 4MPa σ ij = σ yx = 4MPa σ xy = 4MPa σ yy = 4MPa Eigenvalues λ 1 = 0MPa λ = 8MPa Eigenvectors n x = n y n x = n y n x + n y = 1 n x = 1 n x = n y = y x y x n x + n y = 1 n x = 1 n x = n y = Note that X 1 X = 0 Principal direc>ons are perpendicular /14/1 GG
34 V Example Matrix form/matlab σ 1 σ 19. Principal Stresses >> sij = [- 4-4;- 4-4] sij = >> [v,d]=eig(sij) v = d = Eigenvectors (in columns) Corresponding eigenvalues (in columns) /14/1 GG Summary of Strain and Stress Different quan>>es with different dimensions (dimensionless vs. force/unit area) Both can be represented by the orienta>on and magnitude of their principal values Strain describes changes in distance between points and changes in right angles Matrices of co- axial strain and stress are symmetric: eigenvalues are orthogonal and do not rotate Asymmetric strain matrices involve rota>on Infinitesimal strains can be superposed linearly Finite strains involve matrix mul>plica>on 34
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