Lecture 13: Strain part 2 GEOS 655 Tectonic Geodesy Jeff Freymueller Strain and Rotaon Tensors • We described the deformaon as the sum of two tensors, a strain tensor and a rotaon tensor. ⎛ ⎞ ⎛ ⎞ 1 ∂ui ∂u j 1 ∂ui ∂u j ui (x0 + dx) = ui (x0 ) + ⎜ + ⎟ dx j + ⎜ − ⎟ dx j 2⎝ ∂x j ∂xi ⎠ 2⎝ ∂x j ∂xi ⎠ ui (x0 + dx) = ui (x0 ) + εij dx j + ωij dx j ⎡ ⎛ ⎞ ⎛ ⎞⎤ ⎡ ⎛ ⎞ ⎛ ⎞⎤ ⎡ ⎤ ∂u1 1 ∂u1 ∂u2 1 ∂u1 ∂u3 1 ∂u1 ∂u2 1 ∂u1 ∂u3 ∂u1 ∂u1 ∂u1 0 ⎢ ⎥ ⎢ ⎜ + ⎟ ⎜ + ⎟⎥ ⎢ ⎜ − ⎟ ⎜ − ⎟⎥ x x x ∂x1 2⎝ ∂x2 ∂x1 ⎠ 2⎝ ∂x3 ∂x1 ⎠ 2⎝ ∂x2 ∂x1 ⎠ 2⎝ ∂x3 ∂x1 ⎠ ⎢ ∂ 1 ∂ 2 ∂ 3 ⎥ ⎢ ⎥ ⎢ ⎥ € ⎢ ⎛ ⎞ ⎛ ⎞⎥ ⎢ ⎛ ⎞ ⎛ ⎞⎥ ⎢∂ u2 ∂u2 ∂u2 ⎥ 1 ∂u1 ∂u2 ∂u2 1 ∂u2 ∂u3 1 ∂u2 ∂u1 1 ∂u2 ∂u3 = ⎢ ⎜ + ⎟ ⎜ + ⎟⎥ + ⎢ ⎜ − ⎟ 0 ⎜ − ⎟⎥ ⎢ ∂x1 ∂x2 ∂x3 ⎥ 2⎝ ∂x2 ∂x1 ⎠ ∂x2 2⎝ ∂x3 ∂x2 ⎠ 2⎝ ∂x1 ∂x2 ⎠ 2⎝ ∂x3 ∂x2 ⎠ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ∂u3 ∂u3 ∂u3 ⎢ 1⎛ ∂u ∂u ⎞ 1⎛ ∂u ∂u ⎞ ∂u ⎥ ⎢ 1⎛ ∂u ∂u ⎞ 1⎛ ∂u ∂u ⎞ ⎥ ⎢ ⎥ ⎜ 1 + 3 ⎟ ⎜ 2 + 3 ⎟ 3 ⎜ 3 − 1 ⎟ ⎜ 3 − 2 ⎟ 0 ⎣ ∂x1 ∂x2 ∂x3 ⎦ ⎢ ⎥ ⎢ ⎥ ⎣ 2⎝ ∂x3 ∂x1 ⎠ 2⎝ ∂x3 ∂x2 ⎠ ∂x3 ⎦ ⎣ 2⎝ ∂x1 ∂x3 ⎠ 2⎝ ∂x2 ∂x3 ⎠ ⎦ symmetric, strain anti-symmetric, rotation € OK, So What is a Tensor, Anyway • Tensor, not tonsure! è • Examples of tensors of various ranks: – Rank 0: scalar – Rank 1: vector – Rank 2: matrix – A tensor of rank N+1 is like a set of tensors of rank N, like you can think of a matrix as a set of column vectors • The stress tensor was actually the first tensor -- the mathemacs was developed to deal with stress. • The mathemacal definiAon is based on transformaon properAes. And WriAng Them in Index Notaon • We can write the strain tensor ⎡ε ε ε ⎤ ⎛ ⎞ 11 12 13 1 ∂ui ∂u j ⎢ ⎥ εij = ⎜ + ⎟ = ⎢ε 21 ε22 ε23⎥ 2⎝ ∂x j ∂xi ⎠ ⎣⎢ε 31 ε32 ε33⎦⎥ • The strain tensor is symmetric, so only 6 components are independent. Symmetry requires that: € ε21 = ε12 ε31 = ε13 ε23 = ε32 • The rotaon tensor is anA-symmetric and has only 3 ⎡ 0 ω ω ⎤ independent components: ⎛ ⎞ 12 13 1 ∂u ∂u j ⎢ ⎥ € i 0 ωij = ⎜ − ⎟ = ⎢− ω12 ω23⎥ 2⎝ ∂x j ∂xi ⎠ ⎣⎢− ω13 −ω23 0 ⎦⎥ € EsAmang Strain and Rotaon from GPS Data • This is preYy easy. We can esAmate all of the components of the strain and rotaon tensors directly from the GPS data. – We can write equaons in terms of the 6 independent strain tensor components and 3 independent rotaon tensor components – Or we can write equaons in terms of the 9 components of the displacement gradient tensor • Write the moAons (relave to a reference site or reference locaon) in terms of distance from reference site: ui (x0 + dx) − ui (x0 ) = εij dx j + ωij dx j • Here x0 is the posiAon of the reference locaon, and dx is the vector from reference locaon to where we have data € The Equaons in 2D • The equaons below are wriYen out for the 2D case (velocity and strain/rotaon rate): vx = Vx + ε˙ xxΔx + ε˙ xyΔy + ω˙ Δy vy = Vy + ε˙ xyΔx + ε˙ yyΔy −ω˙ Δx – For 2D strain, we have 4 parameters (3 strain rates, 1 rotaon rate) and we need ≥2 sites with €horizontal data. – For 3D, we would have 9 parameters and would require ≥3 sites with 3D data. VelociAes Relave to Eurasia Chen et al. (2004), JGR Displacement and Strain • Displacements (or rates) are a combinaon of rigid body translaon, rotaon and internal deformaon ⎡ ⎤ ⎡ 1 ⎤ ⎡ veast ⎤ ve,body ε˙11 2 (ε˙12 + ω˙ ) ⎡ x⎤ ⎢ ⎥ = ⎢ ⎥ + ⎢ 1 ⎥⎢ ⎥ ⎣v north ⎦ ⎣v n,body⎦ ⎣ 2 (ε˙12 −ω˙ ) ε˙ 22 ⎦⎣ y⎦ ε = strain tensor components (x, y) = position ω = rotation v = velocity € Uniform Strain Rate 0 1 0 1 0 ε!1 − − = ε!2 + ε!2 0 0 1 0 0 ε!2 Deforming Block Model • Based on GPS data from 44 sites • Four blocks moving relave to each other on major faults, plus uniform strain – Block moAons are predominantly strike-slip • Models with spaal variaons in strain do not fit significantly beYer than uniform strain • Models with all slip concentrated on a few faults fit worse than the deforming block model. 4.4+/-1.1 7.4+/-0.7 5.9+/-0.7 Chen et al. Deforming Block Model Rotaon Tensor ⎡ 1⎛ ∂u ∂u ⎞ 1⎛ ∂u ∂u ⎞⎤ ⎢ 0 ⎜ 1 − 2 ⎟ ⎜ 1 − 3 ⎟⎥ ⎢ 2⎝ ∂x2 ∂x1 ⎠ 2⎝ ∂x3 ∂x1 ⎠⎥ ⎢ 1⎛ ∂u ∂u ⎞ 1⎛ ∂u ∂u ⎞⎥ ⎜ 2 − 1 ⎟ 0 ⎜ 2 − 3 ⎟ ⎢ 2 x x 2 x x ⎥ ⎢ ⎝ ∂ 1 ∂ 2 ⎠ ⎝ ∂ 3 ∂ 2 ⎠⎥ ⎛ ⎞ ⎛ ⎞ ⎢ 1 ∂u3 ∂u1 1 ∂u3 ∂u2 ⎥ ⎢ ⎜ − ⎟ ⎜ − ⎟ 0 ⎥ ⎣ 2⎝ ∂x1 ∂x3 ⎠ 2⎝ ∂x2 ∂x3 ⎠ ⎦ ⎡ 1⎛ ∂u ∂u ⎞ 1⎛ ∂u ∂u ⎞⎤ ⎢ 0 ⎜ 1 − 2 ⎟ ⎜ 1 − 3 ⎟⎥ ⎢ 2⎝ ∂x2 ∂x1 ⎠ 2⎝ ∂x3 ∂x1 ⎠⎥ ⎢ ⎛ ⎞ ⎛ ⎞⎥ In terms of our earlier 1 ∂u2 ∂u1 1 ∂u2 ∂u3 ⎜ − ⎟ 0 ⎜ − ⎟ = βij x − x € ⎢ 2 x x 2 x x ⎥ ( ) coordinate rotation matrix: ⎢ ⎝ ∂ 1 ∂ 2 ⎠ ⎝ ∂ 3 ∂ 2 ⎠⎥ ⎛ ⎞ ⎛ ⎞ ⎢ 1 ∂u3 ∂u1 1 ∂u3 ∂u2 ⎥ ⎢ ⎜ − ⎟ ⎜ − ⎟ 0 ⎥ ⎣ 2⎝ ∂x1 ∂x3 ⎠ 2⎝ ∂x2 ∂x3 ⎠ ⎦ € Rotaon as a Vector • We can represent the rotaon tensor as a vector. Why? – Rotaon can be described by a vector; think angular velocity vector – With only three independent components, the number of terms adds up • We can write a cross product operator using the “permutaon tensor” (actually a tensor of rank 3, 1 like a matrix with 3 dimensions): Ωk = − 2 eijkωij ⎧ ⎫ e123 = e231 = e312 =1 ⎪ ⎪ eijk = ⎨ e213 = e321 = e132 = −1 ⎬ ⎪ otherwise : 0 ⎪ ⎩⎪ = ⎭⎪ More on the Permutaon Tensor • We can write the vector cross product in terms of this permutaon tensor (a × b)k = eijkaib j • Remember this rule? – The permutaon tensor is like a short-hand for that rule € • The permutaon tensor can be wriYen in terms of the Kronecker delta (this is someAmes useful): eijkeist = δ jsδkt −δ jtδks € Ω and ωij • We can write Ω in terms of u: 1 1 Ωk = − 2 eijkωij = − 2 ekijωij ⎛ ⎞ 1 1 ∂ui 1 ∂u j Ωk = − 2 ⎜ 2 ekij − 2 ekij ⎟ ⎝ ∂x j ∂xi ⎠ ⎛ ⎞ 1 1 ∂ui 1 ∂u j Ωk = 2 ⎜ 2 ekji + 2 ekij ⎟ Notice the permutation ⎝ ∂x j ∂xi ⎠ ⎛ ∂u ⎞ 1 i These two terms are equal. Why? Ωk = 2 ⎜ ekji ⎟ ⎝ ∂x j ⎠ 1 Ω = 2 (∇ × u) ∂ ∂ ∂ It is the curl of u ∇ = iˆ + ˆj + kˆ ∂x ∂y ∂z € € ωij and Ω • We can turn this around and write ωij in terms of Ω. Start with: 1 e 1 Ωk = − 2 ijkωij emnkΩk = − 2 emnkeijkωij • MulAply both sides by e 1 mnk emnkΩk = − 2 (δmiδnj −δmjδni )ωij 1 emnkΩk = − 2 (δmiδnjωij −δmjδniωij ) 1 emnkΩk = − 2 (ωmn −ωnm ) € 1 emnkΩk = − 2 (ωmn + ωmn ) emnkΩk = −ωmn ωij = −eijkΩk € One Final Manipulaon • Let’s go back to the original term in the Taylor Series expansion: (rot ) ui = ωij dx j = −eijkΩkdx j = eikjΩkdx j u(rot ) = Ω × dx – This makes it a bit more clear that this is a rotaon. € Strain Tensor Axial Strains Shear Strains ⎡ ∂u 1⎛ ∂u ∂u ⎞ 1⎛ ∂u ∂u ⎞⎤ ⎢ 1 ⎜ 1 + 2 ⎟ ⎜ 1 + 3 ⎟⎥ ⎢ ∂x1 2⎝ ∂x2 ∂x1 ⎠ 2⎝ ∂x3 ∂x1 ⎠⎥ ⎢ 1⎛ ∂u ∂u ⎞ ∂u 1⎛ ∂u ∂u ⎞⎥ ⎜ 1 + 2 ⎟ 2 ⎜ 2 + 3 ⎟ ⎢ 2 x x x 2 x x ⎥ ⎢ ⎝ ∂ 2 ∂ 1 ⎠ ∂ 2 ⎝ ∂ 3 ∂ 2 ⎠⎥ ⎛ ⎞ ⎛ ⎞ ⎢ 1 ∂u1 ∂u3 1 ∂u2 ∂u3 ∂u3 ⎥ ⎢ ⎜ + ⎟ ⎜ + ⎟ ⎥ ⎣ 2⎝ ∂x3 ∂x1 ⎠ 2⎝ ∂x3 ∂x2 ⎠ ∂x3 ⎦ € Examples of Strains uniaxial strains An example Including shear strain Principal Axes of Strain • For any strain, there is a ε coordinate system where ε2 1 the strains are uniaxial – no shear strain. • Called the principal axes of the strain tensor. € – Strains in those direcAons are € the principal strains. • Principal axes are the eigenvectors of the strain tensor and principal strains are the eigenvalues. Strain in a ParAcular DirecAon • Let’s look at how we can use the strain tensor to get a line length change in an arbitrary direcAon. Start with a simple example in 2D: ⎡ε 11 0⎤ ⎢ ⎥ ε11 > 0 ⎣ 0 0⎦ • If we start with a circle, it will be deformed into an ellipse. € Strain in a ParAcular DirecAon ΔL ⎡ε 11 0⎤ ⎢ ⎥ ε11 > 0 u 0 0 x ⎣ ⎦ r u x θ € x r θ ux = xε11 = (r ⋅ cosθ)ε11 uy = 0 2 ΔL = ux cosθ = rε11 cos θ 2 ΔL /L = ε11 cos θ € Strain in a ParAcular DirecAon • For a strain only in the x direcAon, ⎡ε 11 0⎤ 2 ⎢ ⎥ ε11 > 0 ΔL /L = ε11 cos θ ⎣ 0 0⎦ • If we do the same for a strain in only in the y direcAon, we get something similar, € ⎡0 0 ⎤ 2 ⎢ ⎥ ε22 > 0 ΔL /L = ε22 sin θ ⎣0 ε22⎦ • In general, we get an equaon for an ellipse: 2 2 € err = ε11 cos θ + ε22 sin θ + ε12 sin2θ € Suppose we have mulAple lines • If we measure line length changes for several lines of different orientaons, θi θi • If we have lines of three or more different orientaons, we can esAmate the three horizontal strains. 2 2 (err )i = ε11 cos θi + ε22 sin θi + ε12 sin2θi € Suppose we have mulAple lines • This is the basis for esAmang strain from EDM networks • The lines don’t have to have the same origin. I just drew it that way for convenience. • The more lines of different orientaons, the beYer the esAmate becomes 2 2 (err )i = ε11 cos θi + ε22 sin θi + ε12 sin2θi € Chen and Freymueller (2002) • Evaluated strain rates for four different near- fault EDM networks along the San Andreas fault • All sites were so close to the fault we assume uniform strain within the network EsAmated Strain Rate Tensors EsAmang Strain from Line Lengths • Obviously, what we looked at is a 2D problem. We assumed the state of plane strain, in which all deformaon is horizontal. • Line lengths or line length changes are invariant under rotaons, so we can learn nothing about the rotaon tensor from them – But we also can esAmate strain without worrying about rotaon • We have to assume uniform strain, which might be the wrong assumpAon.