Lecture 13: Strain part 2

GEOS 655 Tectonic Geodesy Jeff Freymueller Strain and Rotaon

• We described the deformaon as the sum of two tensors, a strain 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: – 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 definion is based on transformaon properes. And Wring 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 an-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 ⎦⎥

€ Esmang Strain and Rotaon from GPS Data • This is prey easy. We can esmate 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 tensor • Write the moons (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 posion 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 wrien 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. Velocies 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 moons are predominantly strike-slip • Models with spaal variaons in strain do not fit significantly beer 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 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 wrien in terms of the Kronecker delta (this is somemes 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 • Mulply 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 direcons are € the principal strains. • Principal axes are the eigenvectors of the strain tensor and principal strains are the eigenvalues. Strain in a Parcular Direcon • Let’s look at how we can use the strain tensor to get a line length change in an arbitrary direcon. 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 Parcular Direcon 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 Parcular Direcon

• For a strain only in the x direcon,

⎡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 direcon, 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 mulple 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 esmate the three horizontal strains.

2 2 (err )i = ε11 cos θi + ε22 sin θi + ε12 sin2θi

€ Suppose we have mulple lines

• This is the basis for esmang 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 beer the esmate 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 Esmated Strain Rate Tensors Esmang 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 esmate strain without worrying about rotaon • We have to assume uniform strain, which might be the wrong assumpon. • An alternave might be to assume simple shear (no axial strains), such as for glacial flow or some strike-slip fault moon problems. Strain and Angle Changes

• If we strain an object, lines within the object will, in general, rotate:

• But a uniform strain (uniform scaling up or down in size) does not rotate any lines. So that means we can’t esmate all three components of the strain tensor from angle changes. Esmang Strain from Angle Changes • We can’t esmate the dilataon (Δ). Rewrite our three horizontal strains this way:

Δ = ε11 + ε22 Dilatation

γ = ε −ε Pure shear with N-S contraction, E-W extension 1 11 22 Pure shear with NW-SE contraction, NE-SW extension γ 2 = 2ε12 (or simple shear with displacement along x-axis) • The gammas are called the engineering shear strains (the epsilons are the tensor strains). The factor of 2 is from historical pracce. € – Older papers oen use the gammas. Newer work usually uses tensor strains. Be careful of the factor of 2! Illustraons of Engineering Shears

γ1 = ε11 −ε22 > 0 γ 2 = 2ε12 > 0

€ € What Can Be Determined From the Data? • There is a general lesson here: – You might want to know all components of the strain and rotaon tensors. – When your data only determine some parts of your parameters, it makes sense to re-parameterize the problem so that you directly esmate what your data can determine • The alternave is to put possibly arbitrary constraints on your parameters – The discussion of what you can learn about a plate rotaon from a single site was another example. – What we are doing is breaking up our parameters into two sets (determined and impossible to determine) that are (ideally) orthogonal in some way. More examples of this

• If you have GPS displacement/velocity data, you can determine all components of the displacement gradient tensor (or strain + rotaon) • If you have line length changes, you can determine all components of strain, but nothing about rotaon – So don’t try to esmate the displacement gradient tensor; esmate the strain tensor instead • If you have angle changes, you can’t determine rotaon or dilataon – So don’t try to esmate the strain tensor; esmate the gammas instead. Axis of Maximum Shear

• One other thing you can get out of the gammas is the axis of maximum shear

γ1 tan2Θs = γ 2 1 −1 Θs = 2 tan (γ1 γ 2) 2 2 Magnitude of maximum shear γ = γ1 + γ 2

– If γ1 = 0, the maximum shear is along the x-axis.

–€ If γ2 = 0, the maximum shear is 45° from the x-axis. (Quantave) Rotaon of a Line Segment • Define the rotaon vector of the x+dx ξ+dξ line by the equaon at le. In index notaon it is: θ eijkdx j dξk Θi = dxn dxn • Because ξi = xi + ui, ⎛ ⎞ ∂ξk ∂uk x ξ dξk = dxm = ⎜δ km + ⎟ dxm ∂xm ⎝ ∂xm ⎠ € • Note: curly d and straight d are Define the rotation vector Θ not the same (paral vs. total derivaves). dx × dξ Θ = € • Also the magnitudes of dx and dξ dx ⋅ dξ are equal for a pure rotaon. Θ = sinθ ≈ θ

€ Total vs. Partial

• There is a subtlety to partial that you might or might not know. • Assume a function of space and time, f(x,y,t), where the variables x and y also depend on time. – Example: A function that depends on the angle of the sun, which depends on your position (as a function of time if you are moving), and with time directly – The function may have a direct dependence on time, and an indirect dependence that comes from x(t) and y(t). • The total derivative is

df ∂f ∂f dx ∂f dy = + + dt ∂t ∂x dt ∂y dt

€ (Quantave) Rotaon of a Line Segment • Define the rotaon vector of the x+dx ξ+dξ line by the equaon at le. In index notaon it is: θ eijkdx j dξk Θi = dxn dxn • Because ξi = xi + ui, ⎛ ⎞ ∂ξk ∂uk x ξ dξk = dxm = ⎜δ km + ⎟ dxm ∂xm ⎝ ∂xm ⎠ € • Note: curly d and straight d are Define the rotation vector Θ not the same (paral vs. total derivaves). dx × dξ Θ = € • Also the magnitudes of dx and dξ dx ⋅ dξ are equal for a pure rotaon. Θ = sinθ ≈ θ

€ Rotaon of Line Segment • Substung this in gives: ⎡ ⎛ ⎞⎤ 1 ∂uk Θi = ⎢e ijkdx j ⎜δ km + ⎟⎥ dx m dxn dxn ⎣ ⎝ ∂xm ⎠⎦ ⎡ ⎛ ⎞⎤ 1 ∂uk Θi = ⎢e ijkdx j ⎜δ km dxm + dxm ⎟⎥ dxn dxn ⎣ ⎝ ∂xm ⎠⎦ ⎡ ⎤ 1 ∂uk Θi = ⎢e ijkdx j dxk + eijkdx j dxm ⎥ dxn dxn ⎣ ∂xm ⎦ This term is the velocity This term is just x x x = 0 gradient tensor, or strain + rotation dx j dxm Θi = [0 + eijk (εkm + ωkm )] dxn dxn These are just unit vectors Θi = eijk (εkm + ωkm )dxˆ j dxˆ m

€ What Does This Mean? • Here’s the equaon:

Θi = eijk (εkm + ωkm )dxˆ j dxˆ m • This equaon tells us that a line segment can rotate because of either a strain or a rotaon. – The rotaon part is obvious – The strain part we saw earlier (graphically). Depending on the values € of the strain components, the rotaon might be zero or non-zero. – Also, the two terms might cancel out!