19. Principal Stresses

Transcription

1 19. Principal Stresses I Main Topics A Cauchy s formula B Principal stresses (eigenvectors and eigenvalues) C Example 10/24/18 GG303 1

2 19. Principal Stresses hkp://hvo.wr.usgs.gov/kilauea/update/images.html 10/24/18 GG303 2

3 19. Principal Stresses II Cauchy s formula A Relates trac<on (stress vector) components to stress tensor components in the same reference frame B 2D and 3D treatments analogous C τ i = σ ij n j = n j σ ij = n j σ ji Note: all stress components shown are positive 10/24/18 GG303 3

4 19. Principal Stresses II Cauchy s formula (cont.) C τ i = n j σ ji 1 Meaning of terms a τ i = traction component b n j = direction cosine of angle between n- direction and j- direction c σ ji = stress component d τ i and σ ji act in the same direction n j = cosθ nj = a nj 10/24/18 GG303 4

5 19. Principal Stresses II Cauchy s formula (cont.) D Expansion (2D) of τ i = n j σ ji 1 τ x = n x σ xx + n y σ yx 2 τ y = n x σ xy + n y σ yy n j = cosθ nj = a nj 10/24/18 GG303 5

6 II Cauchy s formula (cont.) E Derivation: Contributions to τ x 19. Principal Stresses Note that all contribunons must act in x-direcnon 1 τ x = w ( 1) σ xx + w ( 2) σ yx 2 F x A n = A x A n ( 1) F x + A y A x A n ( 2) F x A y 3 τ x = n x σ xx + n y σ yx n x = cosθ nx = a nx n y = cosθ ny = a ny 10/24/18 GG303 6

7 II Cauchy s formula (cont.) E Deriva?on: Contribu?ons to τ y 19. Principal Stresses Note that all contributions must act in y-direction 1 τ y = w ( 3) σ xy + w ( 4) σ yy 2 F y A n = A x A n ( 3) F y + A y A x A n ( 4) F y A y 3 τ y = n x σ xy + n y σ yy n x = cosθ nx = a nx n y = cosθ ny = a ny 10/24/18 GG303 7

8 19. Principal Stresses II Cauchy s formula (cont.) F Alterna>ve forms 1 τ i = n j σ ji 2 τ i = σ ji n j 3 τ i = σ ij n j 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 6 Note that the stress matrix (tensor) transforms the normal vector to the plane (n) to the trac>on vector ac>ng on the plane (τ) n x n y n z n j = cosθ nj = a nj 10/24/18 GG303 8

9 A B 19. Principal Stresses III Principal stresses (eigenvectors and eigenvalues) C Now we seek (a) the orientation of the unit normal (given by n x and n y ) to any special plane where the associated traction vector is perpendicular (normal) to that plane, and (b) the magnitude (λ) of that traction vector. These traction vectors have no shear component and hence correspond to the principal stresses. The orientations of the special traction vectors are called eigenvectors, and the magnitudes of these special traction vectors are called eigenvalues. D An eigenvector points in the same direction as the normal to the plane, so the transformation of the normal vector to the traction vector by Cauchy s formula does not involve a rotation. Note that the traction vector below parallels the normal vector to the plane n j = cosθ nj = a nj 10/24/18 GG303 9

10 19. Principal Stresses III Principal stresses (eigenvectors and eigenvalues) E The x- and y- components of such a principal traction vector are obtained by projecting the vector onto the x- and y- axes: τ x τ y = n x τ n y Since the magnitude of the eigenvector is a scalar, both the normal to the plane and the eigenvector point in the same direction. n j = cosθ nj = a nj 10/24/18 GG303 10

11 19. Principal Stresses III Principal stresses (eigenvectors and eigenvalues) F τ x τ y = τ n x n y Let λ = τ G H τ x τ y σ xx σ xy = σ xx σ yx σ xy σ yy σ yx σ yy n x n y = λ n x n y n x n y The form of (H ) is [A][X]=λ[X], and [σ] is symmetric 10/24/18 GG303 11

12 19. Principal Stresses III Principal stresses (eigenvectors and eigenvalues) From previous notes Subtract the right side from both sides I or J σ xx T σ yx 0 σ xy 0 σ yy T n x n y = 0 0 [ σ IT ][ n] = [ 0], where [ I ] = /24/18 GG303 12

13 Now, a brief interlude to show how to solve analytically for the eigenvalues in 2D 10/24/18 GG303 13

14 9. EIGENVECTORS, EIGENVALUES, AND FINITE STRAIN From previous notes III Determinant (cont.) D Geometric meanings of the real matrix equation 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 solution 2 A = 0 ; a b c [A] -1 does not exist Describes two co-linear lines that that pass through the origin (or three planes that intersect a line or plane through the origin) X has no unique solution 10/24/18 GG303 14

15 9. EIGENVECTORS, EIGENVALUES, AND From previous notes FINITE STRAIN III Eigenvalue problems, eigenvectors and eigenvalues (cont.) E AlternaGve form of an eigenvalue equagon 1 [A][X]=λ[X] SubtracGng λ[ix] = λ[x] from both sides yields: 2 [A-Iλ][X]=0 (same form as [A][X]=0) F SoluGon condigons and connecgons with determinants 1 Unique trivial solugon of [X] = 0 if and only if A-Iλ = 0 2 Eigenvector solugons ([X] 0) if and only if A-Iλ =0 10/24/18 GG303 15

16 9. EIGENVECTORS, EIGENVALUES, AND FINITE STRAIN From previous notes III Eigenvalue problems, eigenvectors and eigenvalues (cont.) G CharacterisHc equahon: A-Iλ =0 1 Eigenvalues of a symmetric 2x2 matrix a b c d ( λ 1,λ 2 = a + d ) ± ( a + d) 2 4 ad b 2 λ 1,λ 2 = a + d λ 1,λ 2 = a + d ( ) 10/24/18 GG ( ) ± ( a + 2ad + d) 2 4ad + 4b 2 ( ) ± ( a 2ad + d) 2 + 4b 2 λ 1,λ 2 = a + d 2 2 ( ) ± ( a d) 2 + 4b 2 2 A = a b b d Radical term cannot be negative. Eigenvalues are real.

17 9. EIGENVECTORS, EIGENVALUES, AND FINITE STRAIN From previous notes VI Solu7ons for symmetric matrices (cont.) B Any dis7nct eigenvectors (X 1, X 2 ) of a symmetric nxn matrix are perpendicular (X 1 X 2 = 0) 1a AX 1 =λ 1 X 1 1b AX 2 =λ 2 X 2 AX 1 parallels X 1, AX 2 parallels X 2 (property of eigenvectors) DoRng AX 1 by X 2 and AX 2 by X 1 can test whether X 1 and X 2 are orthogonal. 2a X 2 AX 1 = X 2 λ 1 X 1 = λ 1 (X 2 X 1 ) 2b X 1 AX 2 = X 1 λ 2 X 2 = λ 2 (X 1 X 2 ) 10/24/18 GG303 17

18 9. EIGENVECTORS, EIGENVALUES, AND From previous notes FINITE STRAIN L Distinct eigenvectors (X 1, X 2 ) of a symmetric 2x2 matrix are perpendicular Since the left sides of (2a) and (2b) are equal, the right sides must be equal too. Hence, 4 λ 1 (X 2 X 1 ) =λ 2 (X 1 X 2 ) Now subtract the right side of (4) from the left 5 (λ 1 λ 2 )(X 2 X 1 ) =0 The eigenvalues generally are different, so λ 1 λ 2 0. This means for (5) to hold that X 2 X 1 =0. Therefore, the eigenvectors (X 1, X 2 ) of a symmetric 2x2 matrix are perpendicular 10/24/18 GG303 18

19 End of brief interlude 10/24/18 GG303 19

20 19. Principal Stresses IV Example Find the principal stresses given σ ij = σ xx σ yx σ xy σ yy θ x x = 45!,θ x y = 45!,θ y x = 135!,θ y y = 45! 10/24/18 GG303 20

21 19. Principal Stresses IV Example σ xx σ xy σ ij = σ yx σ yy First find eigenvalues ( ) ± ( σ xx σ ) 2 2 yy + 4σ xy λ 1,λ 2 = σ xx + σ yy λ 1,λ 2 ± MPa = 0MPa, 8MPa 10/24/18 GG303 21

22 IV σ ij = 19. Principal Stresses Example σ xx σ yx σ xy σ yy λ 1,λ 2 ± 64 MPa = 0MPa, 8MPa 2 Then solve for eigenvectors (the dimensions of stress are unnecessary below and are dropped) For λ 1 = 0 : For λ 2 = 8 : ( ) 4 4 ( 8) 4 n x n y 10/24/18 GG = 0 0 n x n y = 0 0 4n 4n = 0 n = n x y x y 4n x 4n y = 0 n x = n y 4n x 4n y = 0 n x = n y 4n x + 4n y = 0 n x = n y

23 19. Principal Stresses IV Example σ xx σ ij = σ yx σ xy σ yy Eigenvalues λ 1 = 0MPA λ 2 = 8MPA Eigenvectors n x = n y n x = n y 10/24/18 GG303 23

24 IV Example (values in MPa) 19. Principal Stresses σ xx = - 4 τ xn = - 4 σ x x = - 8 τ x n = - 8 σ xy = - 4 τ xs = - 4 σ x y = 0 τ x s = 0 σ yx = - 4 τ ys = + 4 σ y x = -0 τ y s = +0 σ yy = - 4 τ yn = -4 σ y y = 0 τ y n = 0 s n n s 10/24/18 GG303 24

25 IV Example Matrix form 19. Principal Stresses σ x x σ x y σ y x σ y y = a a x x a x y y x a y y σ xx σ yx σ xy σ yy a x x a x y a y x a y y T σ i j = a [ ] σ ij [ ] T a This expression is valid in 2D and 3D! 10/24/18 GG303 25

26 IV Example Matrix form/matlab 19. Principal Stresses >> sij = [-4-4;-4-4] sij = >> [vec,val]=eig(sij) vec = val = Eigenvectors (in columns) Corresponding eigenvalues (in columns) 10/24/18 GG303 26