Earth Deformation Homework 1

Earth Deformation Homework 1 Michal Dichter October 7, 14 Problem 1 (T+S Problem -5) We assume the setup of Figure -4 from Turcotte and Schubert: We are given the following values: hcc = 5 km hsb = 7 km ρm = kg/m ρcc = 7 kg/m ρs = 45 kg/m We need to solve equation (-1): hsb = hcc ρm ρcc ρm ρs 1 1 α Plugging in the values above we get: α wb 1.4 w

Problem (Rotational symmetry of a crystal) Let us take some matrix A: A = a b c d e f g h i Matrix of a 9-degree counter-clockwise rotation about x: R x (9) = 1 1 1 Matrix of a 9-degree counter-clockwise rotation about y: 1 R y (9) = 1 1 Matrix of a 9-degree counter-clockwise rotation about z: 1 R z (9) = 1 1 The matrix A rotated 9 degrees about x: A rot.x = R x AR 1 x = The matrix A rotated 9 degrees about y: A rot.y = R y AR 1 y = g i h a c b d f e i h g f e d c b a The matrix A rotated 9 degrees about z: A rot.z = R z AR 1 z = e d f b a c h g i The problem states that we must have: A rot.x = A rot.y = A rot.z = A Thus we can conclude that all of the off-diagonal components must be zero and that a = e = i such that the matrix A must be of the form: a A = a a We therefore conclude that the matrix A has only one independent component!

Problem (T+S Problem -1) We can construct the following diagram: σ yy x O σ yx A x' σ y ' x ' σ xx σ xy y ' σ y ' y ' B y Force balance in the ˆx direction tells us that: σ yx OA σ xx OB σ y x cos(θ)ab + σ y y sin(θ)ab = We can divide both sides by AB and rearrange: σ y x cos θ σ y y sin θ = σ yx cos θ σ xx sin θ Now multiply both sides by sin θ: ( ) σ y x sin θ cos θ σ y y sin θ = σ yx sin θ cos θ σ xx sin θ Force balance in the ŷ direction tells us that: σ yy OA σ xy OB σ y x sin(θ)ab σ y y cos(θ)ab = We can divide both sides by AB and rearrange: σ y x sin θ + σ y y cos θ = σ yy cos θ σ xy sin θ Now multiply both sides by cos θ: ( ) σ y x sin θ cos θ + σ y y cos θ = σ yy cos θ σ xy sin θ cos θ Now use the symmetry of the stress tensor to say that σ xy = σ yx and subtract ( ) from ( ): σ y y (sin θ + cos θ) = σ xx sin θ + σ yy cos θ σ xy sin θ cos θ Now recall that sin θ + cos θ = 1 and sin θ cos θ = sin θ. So we see that: σ y y = σ xx sin θ + σ yy cos θ σ xy sin θ

Problem 4 (Stress tensor) We are given the following stress tensor: We can calculate the pressure p: σ = p σ xx + σ yy + σ zz 1 1 1 1 = 1 + 1 + So we know the isotropic part of σ: 4 σ iso = 4 4 And the deviatoric part of σ: The eigenvalues of σ are: σ dev = σ σ iso = And the corresponding eigenvectors are: σ I = σ II = σ III = I = 1 II = 1 1 1 III = 1 1 1 1 1 1 1 It is easy to see that these three eigenvectors are all mutually orthogonal (the dot product of any two equals zero) they form an orthonormal basis. = 4

Problem 5 (Mohr circle) We can construct the following schematic diagram: τ n point B τ n =C +μ σ n point A a b R B R A C σ A α M σ B σ 1 A α N We can calculate the slope µ from point A to point B as a function of the angle α: ( ) τn (R B R A ) sin α ( ) µ = = σ n (σ B σa ) + (R B R A )(1 + cos α) A B σ 1 B σ n We also know that the line perpendicular to τ n = C + µσ n drawn from M to A (or from N to B) has a slope of 1/µ: ( ) 1 ( ) µ = τn R A sin α = σ n (σ A + R A + R A cos α) (σ A + R = tan α A) M A Recall that we know the following values: σ A 1 σ A = 11 MPa = MPa σ B 1 σ B = 17 MPa = 4 MPa R A = σa 1 σ A R B = σb 1 σ B = 54 MPa = 8 MPa We can now solve ( ) and ( ) for µ and α: µ.65 α 1.9 radians 79.65 degrees

We can use some geometry to solve for C (see the labels on the diagram): We know a so we can further say: cos(α π/) = a h h = σa + R A + R A cos α cos(α π/) 155 MPa Now: So: b = h cos α 145 MPa C 45.64 MPa Now we suspect that a rd sample had a pre-existing crack. Then C = and we can calculate the two points where the failure criterion curve intersects the Mohr circle. For the 1 st test: 7.6 < α < 86.7. For the nd test: 74. < α < 85.. N.B. You can solve the whole problem by plotting Test 1 and Test on graph paper! Problem 6 (Strain accumulation at the San Andreas Fault) We are given the deformation gradient tensor: ( ).15.4 D =..15 We can decompose the deformation gradient tensor into a symmetric tensor and an antisymmetric tensor: ( ) ( ).15.1..1 D = +.1.15.1. We know that the San Andreas Fault trends N65 W so let us rotate the strain tensor counterclockwise into that coordinate system: ɛ = ( cos 65 sin 65 sin 65 cos 65 ) (.15.1.1.15 = (.4.19.19.4 ) ( cos 65 sin 65 sin 65 cos 65 ) 1 We see that the fault-shear strains (off-diagonal components) are rather large compared to the fault-normal strains (diagonal components). This is exactly what we expect for a strikeslip fault such as the San Andreas! The dilatation equals the trace (sum of the diagonal elements) of ɛ. Thus we can see =. )

Problem 7 (T+S Problem -19) Equation (-1) tells us the functional form w(x) of the plate deflection: ( w(x) = w e x/α cos x α + sin x ) α To get to the bending moment M we need to calculate the second derivative of w(x): d w dx = w e x/α (sin x α α cos x ) α The bending moment M is proportional to the second derivative of w(x): M = D d w dx We want to calculate the maximum value of M so we must take a derivative and set it equal to zero: dm dx = 4D α e x/α cos x α = The above is true for: x max = ± πα Now we can plug back in for M m and get equation (-18): M max.416 Dw α We can use the values given to estimate the maximum bending moment in the lithosphere: M max 1.6 1 17 N The maximum bending (fiber) stress σ max xx is then given by equation (-86): σ max xx = ± 6M h 8.1 18 N/m The + corresponds to the tensile stress at the top of the plate and the corresponds to the compressive stress at the bottom of the plate.

Problem 8 (T+S Problem -) We are told that the Amazon River Basin has a width w = 4 km. We are to model the basin as an elastic plate subject to a central line load (see Figures -9 and -). We can use equation (-15) to solve for the flexural parameter α: x b = πα 4 km = πα α 64 km We know that (ρ m ρ s ) = 7 kg/m so we can solve for the flexural rigidity D: [ 4D α = (ρ m ρ s )g ] 1/4 D.9 1 Nm Now we can solve for the thickness T e of the elastic lithosphere: D = ET e 1(1 ν ) Note the rather small value of T e here. T e 17 km Problem 9 (Dabbahu laccolith) Start from the flexure equation: D d4 w dx 4 dw = q(x) P dx Note that P = here. We can separate variables and integrate the equation four times to get: w(x) = qx4 4D + αx + βx + γx + δ The Greek letters are constants to be determined by four boundary conditions: w(±l/) = dw = dx x=± L The first two BCs tell us that α = γ =. The second two BCs tell us: β = ql 48D So we get: If we define: δ = ql4 84D w(x) = qx4 4D + ql x 48D + ql4 84D w ql4 84D

Then we can rewrite w(x) as: ( w(x) = w 1 8 L x + 16 ) L 4 x4 If we write w(x) as a bx + cx 4 then we can do a polynomial fit for the Dabbahu data. The parameter a tells us a value for w I got w 148 mm. We can use b and c to calculate L as follows: b c = L L 45 km Now we can use equation (-17) to calculate the flexural rigidity D: D = gl4 (ρ c ρ mag ) 56 D 6.5 1 19 Nm We can use our value of D to solve for the plate thickness h: D = Eh 1(1 ν ) h. km Finally we can solve for the magma pressure p: w = q D = ρ cgh p D p 57 MPa 5 Dabbahu laccolith B B' Annual uplift mm 15 1 5 1 1 Perpendicular distance from fault km

Two kilometers or so to the roof of the laccolith seems reasonable given that Dabbahu sits on a plate triple junction (Somalian-Nubian-Arabian). We expect dykes to nucleate at areas of great tensional stress. We know how the normal strain ɛ xx (x, y) relates to the normal stress σ xx (x, y): σ xx (x, y) = E 1 ν ɛ xx(x, y) And we know how ɛ xx (x, y) relates to the deflection w(x): ɛ xx (x, y) = y d w dx Now we can take the second derivative of the deflection w(x) at evaluate that at y = +h/ (the bottom of the plate): ɛ xx (x) = qhl 48D ( 1 1 ) L x So the normal stresses σ xx (x) at the bottom of the plate can be found recall that D = Eh /1(1 ν ): σ xx (x) = E ( 1 ν ɛ xx(x) = ql 4h 1 1 ) L x We want the value of x [ L/, L/] that produces the greatest negative (tensional) value of stress. We can see that σ xx (x) plots as a concave-downward parabola. Thus the greatest negative stresses must occur at the endpoints of the model domain at x = L/ and x = +L/. Problem 1 (Broken plate flexure) Blue: V = 1.5 1 1 N/m and M = 5. 1 17 N. Purple: V = N/m and M = 5. 1 17 N. Yellow: V = 1.5 1 1 N/m and M = N. 5 Flexure of plates for different end loads V and M 5 w x km 1 15 5 5 1 15 5 5 4 Distance from application of V and M km The flexural response depends upon the flexural rigidity D of the plate which scales as the cube(!) of the elastic plate thickness. In other words: A modest change to T e greatly changes the flexural response of the plate.

Problem 11 (Adding topography) Flexure of a plate w only topo V M red 5 5 w x km 1 15 5 5 5 1 15 5 5 4 Distance from LHS of rectangular topo km 5 Flexure of a plate w only topo V M red 5 1 w x km 15 5 5 4 45 1 4 5 6 7 8 9 1 Distance from LHS of rectangular topo km We cannot use the shape of the Ganges Foredeep Basin (circled in green) to study slab-pull on the subducting Indian Plate. This is because the flexural response of the plate around the Foredeep Basin looks the same with or without loading at x =. Thus topography alone controls the shape of the Basin at least for the case at hand.