Lecture 13: Strain part 2
GEOS 655 Tectonic Geodesy Jeff Freymueller Strain and Rota on Tensors
• We described the deforma on as the sum of two tensors, a strain tensor and a rota on 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 mathema cs was developed to deal with stress. • The mathema cal defini on is based on transforma on proper es. And Wri ng Them in Index Nota on
• 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 rota on 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 ⎦⎥
€ Es ma ng Strain and Rota on from GPS Data • This is pre y easy. We can es mate all of the components of the strain and rota on tensors directly from the GPS data. – We can write equa ons in terms of the 6 independent strain tensor components and 3 independent rota on tensor components – Or we can write equa ons in terms of the 9 components of the displacement gradient tensor • Write the mo ons (rela ve to a reference site or reference loca on) in terms of distance from reference site:
ui (x0 + dx) − ui (x0 ) = εij dx j + ωij dx j • Here x0 is the posi on of the reference loca on, and dx is the vector from reference loca on to where we have data
€ The Equa ons in 2D
• The equa ons below are wri en out for the 2D case (velocity and strain/rota on 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 rota on rate) and we need ≥2 sites with €horizontal data. – For 3D, we would have 9 parameters and would require ≥3 sites with 3D data. Veloci es Rela ve to Eurasia
Chen et al. (2004), JGR Displacement and Strain
• Displacements (or rates) are a combina on of rigid body transla on, rota on and internal deforma on
⎡ ⎤ ⎡ 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 rela ve to each other on major faults, plus uniform strain – Block mo ons are predominantly strike-slip • Models with spa al varia ons in strain do not fit significantly be er 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 Rota on 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 ⎠ ⎦
€ Rota on as a Vector • We can represent the rota on tensor as a vector. Why? – Rota on 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 “permuta on 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 Permuta on Tensor • We can write the vector cross product in terms of this permuta on tensor
(a × b)k = eijkaib j • Remember this rule?
– The permuta on tensor is like a short-hand for that rule € • The permuta on tensor can be wri en in terms of the Kronecker delta (this is some mes 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 • Mul ply 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 Manipula on
• 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 rota on.
€ 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 direc ons are € the principal strains. • Principal axes are the eigenvectors of the strain tensor and principal strains are the eigenvalues. Strain in a Par cular Direc on • Let’s look at how we can use the strain tensor to get a line length change in an arbitrary direc on. 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 Par cular Direc on 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 Par cular Direc on
• For a strain only in the x direc on,
⎡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 direc on, we get something similar, € ⎡0 0 ⎤ 2 ⎢ ⎥ ε22 > 0 ΔL /L = ε22 sin θ ⎣0 ε22⎦ • In general, we get an equa on for an ellipse: 2 2 € err = ε11 cos θ + ε22 sin θ + ε12 sin2θ
€ Suppose we have mul ple lines
• If we measure line length changes for several lines of different
orienta ons, θi θi • If we have lines of three or more different orienta ons, we can es mate the three horizontal strains.
2 2 (err )i = ε11 cos θi + ε22 sin θi + ε12 sin2θi
€ Suppose we have mul ple lines
• This is the basis for es ma ng 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 orienta ons, the be er the es mate 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 Es mated Strain Rate Tensors Es ma ng Strain from Line Lengths
• Obviously, what we looked at is a 2D problem. We assumed the state of plane strain, in which all deforma on is horizontal. • Line lengths or line length changes are invariant under rota ons, so we can learn nothing about the rota on tensor from them – But we also can es mate strain without worrying about rota on • We have to assume uniform strain, which might be the wrong assump on. • An alterna ve might be to assume simple shear (no axial strains), such as for glacial flow or some strike-slip fault mo on 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 es mate all three components of the strain tensor from angle changes. Es ma ng Strain from Angle Changes • We can’t es mate the dilata on (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 prac ce. € – Older papers o en use the gammas. Newer work usually uses tensor strains. Be careful of the factor of 2! Illustra ons 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 rota on tensors. – When your data only determine some parts of your parameters, it makes sense to re-parameterize the problem so that you directly es mate what your data can determine • The alterna ve is to put possibly arbitrary constraints on your parameters – The discussion of what you can learn about a plate rota on 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 + rota on) • If you have line length changes, you can determine all components of strain, but nothing about rota on – So don’t try to es mate the displacement gradient tensor; es mate the strain tensor instead • If you have angle changes, you can’t determine rota on or dilata on – So don’t try to es mate the strain tensor; es mate 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. (Quan ta ve) Rota on of a Line Segment • Define the rota on vector of the x+dx ξ+dξ line by the equa on at le . In index nota on 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 (par al vs. total deriva ves). dx × dξ Θ = € • Also the magnitudes of dx and dξ dx ⋅ dξ are equal for a pure rota on. Θ = sinθ ≈ θ
€ Total vs. Partial Derivative
• There is a subtlety to partial derivatives 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
€ (Quan ta ve) Rota on of a Line Segment • Define the rota on vector of the x+dx ξ+dξ line by the equa on at le . In index nota on 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 (par al vs. total deriva ves). dx × dξ Θ = € • Also the magnitudes of dx and dξ dx ⋅ dξ are equal for a pure rota on. Θ = sinθ ≈ θ
€ Rota on of Line Segment • Subs tu ng 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 equa on:
Θi = eijk (εkm + ωkm )dxˆ j dxˆ m • This equa on tells us that a line segment can rotate because of either a strain or a rota on. – The rota on part is obvious – The strain part we saw earlier (graphically). Depending on the values € of the strain components, the rota on might be zero or non-zero. – Also, the two terms might cancel out!