Strain analysis.

Transcription

1 Strain analysis

2 Plates vs. continuum Gordon and Stein, 1991 Most plates are rigid at the <1 mm/yr level => until know we have studied a purely discontinuous approach where plates are individual rigid entities. But: Some plates are affected by GIA deformation Some plates are affected by broad tectonic deformation Some plate boundaries are broad Need for a continuous approach = strain

3 Strain - 1 D case Solid rod in a 1-dimensional coordinate system (0,x). Let us take two points A (x) and B (x +dx), very close to each other. Apply a traction P in the x- direction at the end of the rod. The rod will elongate, points A and B will be displaced. Assume infinitesimal displacements. The amount of displacement depends on the position of the point along the rod: the closer to the place where the force is applied, the larger the displacement will be. Solid rod in a 1-dimensional coordinate system (0,x). Let us consider two points A (x) and B (x+dx), very close to each other. Therefore, B will be displaced slightly more than A. The displacement at A is noted u(x), the displacement at B is noted u(x+δx)

4 Strain - 1D case Because the displacement δx is infinitesimal, we can write: and: u(x + dx) = u(x) + u x dx A " B " = dx + u(x) + u x dx ) u(x) = dx + u ( x dx Strain is defined as [change of length] / [original length]. Therefore: e x = ΔL L = AB AB = AB dx + u x dx ( * dx ) dx = u x Hence, infinitesimal strain is the spatial gradient of the displacement

5 D case: velocity gradient tensor Assuming: Local -dimensional Cartesian frame Infinitesimal displacements Velocity (as a function of position X) can be expanded in a series: v(x + δx) = v(x) + v X δx With the two components: v x (X + δx) = v x (X) + v x x δx + v x y δy v y (X + δx) = v y (X) + v y x δx + v y y δy Therefore, one can write: v x x v(x + δx) = v(x) + V.δX with V = v y x v x y ) ) v y ) y () = velocity gradient tensor

6 Velocity gradient tensor Given sites (i and j) separated by (small) distance δx, equation: v(x + δx) = v(x) V.δX v j v i Becomes: v i v j = v ij = V.δX We can write it for 3 sites as: V ij = A.F with: " V ij = # v x1 v y1 v x13 v y13 " x 1 y x 1 y 1, A = x 13 y # 0 0 x 13 y 13 " v x v x F = v y # v y x y x y (assuming point 1 is reference) Observables = Velocity differences Model matrix = Position differences Unknown = Velocity gradients Which can be solved using least-squares: ( ) 1 F = A T C 1 V A AT C V C F 1 1 V ij

7 Velocity gradient tensor: covariance Observables are velocity differences.the covariance associated with v ij can then be found using: " # v xij v yij " = # A " # v xi v yi v xj v yi One usually knows the covariance matrix C ind of the individual site velocities -- the variance propagation law then gives: C Vij = A.C ind.a T C Vij is then propagated in the least squares estimation (see previous slide) to get the covariance matrix of the unknowns C F.

8 Velocity gradient tensor: covariance of unknowns Given that: C Vij = A.C ind.a T One can write the variance on the east (for instance) component of the velocity difference as: σ eij = σ ei + σ e j σ ei e j Therefore, if the covariance terms are large enough, velocities can be non-significant while relative velocities (and velocity gradient tensor) are

9 Strain and rotation rates Displacements actually combine strain and a rigid-body rotation. Tensor theory states that any second-rank tensor can be decomposed into a symmetric and an antisymmetric tensor. One can therefore write: V = 1,. V =.. 1. ( - T [ V + V ] + 1 T [ V V ] v x x v x y + v y x ) + * 1 ( v x y + v y x v y y )/, + 1. * ( 0-0 v y x v x y 1 ( ) + 0 * v x y v y x )/ + 1 * Symmetric tensor Antisymmetric tensor

10 Strain and rotation rates In other words: # V = e xx e xy e xy e yy # ( + 0 ω ( ω 0 With: E = 1 W = 1 T [ V + V ] T [ V V ] e xx = v x x,e = v y yy y,e = e = 1 xy yx ω = 1 v x y v ( y * x ) E strain rate tensor v x y + v y x ( * ) W rigid rotation

11 Estimating strain rate Given sites (i and j), one can write the velocity difference (w.r.t. site i) as a function of the site position (w.r.t. site I) as: v ij = V.δX v ij = (E + W ).δx v x ij = e xx δx + e xy δy + ωδy ( ) v yij = e xy δx + e yy δy ωδx For 3 sites as, this can be written as: V ij = A.S with: " V ij = # v x1 v y1 v x13 v y13 Observables = velocity differences " x 1 y 1 0 y 1 0 x 1 y 1 x 1, A = x 13 y 13 0 y 13 # 0 y 13 y 13 x 13 Model matrix = Position differences Which can be solved using least-squares: " e xx e xy S = e yy # ω ( ) 1 S = A T C 1 V A AT C V C S 1 1 V ij

12 Estimating strain: A better approach First, estimate velocity gradient components (and covariance) using least squares (see previous slides). Then, recall that: e xx = v x x,e yy = v y y,e xy = e yx = 1 # v x y + v y x (,ω = 1 # v x y v y x ( Therefore, strain and rotation rate components can be found using the following transformation matrix: # v x x ( # e xx * v ( (, / x ( e xy ( =, / y ( e yy (, / v y ( (, / ω x ( v y ( A ( y Propagate velocity gradient covariance to strain and rotation rate covariance using: C E,W = AC V A T

13 Tensor = independent of the coordinate system = retains its properties independently from the ref. system. Find reference system where: Shear strain (e xy ) is zero Two other components are maximal Equivalent to diagonalize E: Eigenvectors = principal axes of strain rate tensor Eigenvalues = principal strain rates Representing strain

14 Principal strains Principal values of E: (= principal strains) det(e λi) = 0 det e λ xx e xy ( e xx λ) e yy λ λ = e xx + e yy e ( xy * = 0 e yy λ) ( ) e xy e xy = 0 ± + -, e xx e yy. 0 / + e xy Principal directions: Use rotation matrix A: A = cosθ sinθ sinθ ) cosθ ( Expand AEA T and write that shear component = 0 to find: e 1 = (e xx e yy )sinθ cosθ + e xy (cos θ sin θ) = 0 tan(θ) = e xy e xx e yy sinθ cosθ = 1 sin θ ( ) cos θ sin θ = cos θ ( )

15 Principal strains Principal strains: One maximal, one minimal By convention, extension is taken positive e 1,e = e xx + e yy ± # e xx e yy ( + e xy Principal angle (direction of e 1 ): e xy tan(θ) = e xx e yy

16 Back transformation Rotate principal strain rate tensor by angle -θ using rotation matrix A: cosθ sinθ A = ) sinθ cosθ( Recall that tensor rotation is given by: e AE p A T = 1 cos θ + e sin θ e 1 sinθ cosθ + e sinθ cosθ e 1 sinθ cosθ + e sinθ cosθ ) e 1 sin θ + e cos θ ( Which gives: e xx = e 1 cos θ + e sin θ = e + e 1 + e e 1 e yy = e 1 sin θ + e cos θ = e + e 1 e e 1 e xy = e 1 sinθ cosθ + e sinθ cosθ = e e 1 cos( θ) cos( θ) sin( θ) sinθ cosθ = 1 sin( θ) cos θ sin θ = cos( θ)

17 Shear strain Shear strain (from previous transformation) is maximal when sin(θ)=1, giving: sin( θ) e xy = e 1 e e xy,max = e 1 e = e xx e yy Recall the expression for e 1 (which we set to zero to find the principal angle θ): ( * ) + e xy e 1 = (e xx e yy )sinθ cosθ + e xy (cos θ sin θ) = 1 (e xx e yy )sin( θ) + e xy cos( θ) Angle at which shear strain is maximal is obtained by differentiating w.r.t. θ: e 1 = max de 1 dθ = 0 1 (e yy e xx )cos( θ S ) e xy sin( θ S ) = 0 tan θ S ( ) = e yy e xx e xy

18 Shear strain Maximum shear strain: e xy,max = # e xx e yy ( + e xy Maximum shear angle: tan( θ S ) = e e yy xx e xy θ S = θ ± 45 o

19 Example 1: pure extension Strain rate tensor: e xx = 0.00 ppb/yr e xy = ppb/yr e yy = ppb/yr Principal strains: e 1 = ppb/yr (most extensional) e = 0.00 ppb/yr (most compressional) theta = (e 1, CW from north) Rotation: omega = 0.00 deg/my

20 Example : simple shear Strain rate tensor: e xx = 0.00 ppb/yr e xy = ppb/yr e yy = ppb/yr Principal strains: e 1 = ppb/yr (most extensional) e = ppb/yr (most compressional) theta = (e 1, CW from north) Rotation: omega = 0.09 deg/my

21 Strain invariants As any second rank tensor, strain rate tensor has 3 invariants = quantities that remain unchanged regardless of the reference system First invariant = tensor trace I E = tr( E) = e xx + e yy = e 1 + e Second invariant = sometimes used to represent strain magnitude II E = 1 ( tr ( E ) tr(ee) ) = e xx e yy e xy = e 1 e Third invariant = determinant, same as second invariant in case of x symetric tensor III E = det( E) = e xx e yy e xy = e 1 e

22 Example: current strain rates in Asia The 3x10-9 yr -1 line coincides with the 95 significance level A large part of Asia shows strain rates that are not significant at the 95 confidence level and are lower than 3x10-9 yr -1

23 In practice Discretize space Using polygons with vertices corresponding to data points For instance: Delaunay triangulation: Circumcircle of every triangle does not contain any other point of the triangulation Minimize sliver triangles Using arbitrary polygons: requires interpolation Calculate strain within each polygon WARNING: assumes homogeneous and continuous strain within each polygon

24 For instance GPS velocities and principal strain rates in Central Asia

25 WARNING! Result depends on the size of the elements used to discretize the domain Example: Generate a random D velocity field (e.g. case of non significant residual velocities showing only noise) Triangulate with smaller elements in the middle of the domain Calculate strain Result will be apparently larger strain rates where elements are smaller