Hand-in problems in the course Geodynamics at Uppsala university VT 2010 Peter Schmidt February 1, 2010
This document gives the problems from the text book: Geodynamics, second edition, by Turcotte & Schubert, used for the hand-in assignment in the course Geodynamics at Uppsala university VT 2010. However, to understand the context of many of the problems, as well as to be able to solve the the student needs to read the text preceeding the problem in the text book, especially for the later chapters. As the course proceeds and the deadline for a given hand-in is passed this document will be updated with the solution to the problems of the hand-in.
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Contents Hand-in I, Chapter 2 1.1. Problem 2-4 of Turcotte and Schubert . 1.2. Problem 2-7 of Turcotte and Schubert . 1.3. Problem 2-9 of Turcotte and Schubert . 1.4. Problem 2-11 of Turcotte and Schubert 1.5. Problem 2-14 of Turcotte and Schubert 1.6. Problem 2-15 of Turcotte and Schubert 1.7. Problem 2-20 of Turcotte and Schubert 1.8. Problem 2-21 of Turcotte and Schubert 1.9. Inner matrix products . . . . . . . . .
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1 1 2 3 4 5 7 8 8 9
Hand-in II, Chapter 3: Elasticity and flexure 2.1. Problem 3-1 of Turcotte and Schubert . 2.2. Problem 3-2 of Turcotte and Schubert . 2.3. Problem 3-7 of Turcotte and Schubert . 2.4. Problem 3-16 of Turcotte and Schubert 2.5. Problem 3-19 of Turcotte and Schubert 2.6. Problem 3-21 of Turcotte and Schubert 2.7. Problem 3-22 of Turcotte and Schubert
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10 10 10 10 10 11 11 11
Hand-in III, Chapter 4: Heat transfer. 3.1. Problem 4-8 of Turcotte and Schubert . 3.2. Problem 4-31 of Turcotte and Schubert 3.3. Problem 4-33 of Turcotte and Schubert 3.4. Problem 4-39 of Turcotte and Schubert 3.5. Problem 4-57 of Turcotte and Schubert
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12 12 12 12 12 12
Hand-in IV, Chapter 6, Part I: Fluid mechanics 4.1. Problem 6-1 of Turcotte and Schubert . . 4.2. Problem 6-2 of Turcotte and Schubert . . 4.3. Problem 6-3 of Turcotte and Schubert . . 4.4. Problem 6-4 of Turcotte and Schubert . . 4.5. Problem 6-5 of Turcotte and Schubert . . 4.6. Problem 6-6 of Turcotte and Schubert . . 4.7. Problem 6-9 of Turcotte and Schubert . . 4.8. Problem 6-11 of Turcotte and Schubert . 4.9. Problem 6-12 of Turcotte and Schubert .
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13 13 13 13 13 13 13 13 13 14
Hand-in V, Chapter 6, PartII: Fluid mechanics 5.1. Problem 6-22 of Turcotte and Schubert . 5.2. Problem 6-23 of Turcotte and Schubert . 5.3. Problem 6-24 of Turcotte and Schubert . 5.4. Problem 6-26 of Turcotte and Schubert . 5.5. Problem 6-33 of Turcotte and Schubert . 5.6. Problem 6-36 of Turcotte and Schubert .
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15 15 15 15 15 16 16
Hand-in VI, Chapter 7: Rock Rheology 6.1. Problem 7-10 of Turcotte and Schubert 6.2. Problem 7-12 of Turcotte and Schubert 6.3. Problem 7-17 of Turcotte and Schubert 6.4. Problem 7-24 of Turcotte and Schubert
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17 17 17 17 18
Hand-in VII, Chapter 8: Faulting 7.1. Problem 8-3 of Turcotte and Schubert . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2. Problem 8-6 of Turcotte and Schubert . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3. Problem 8-8 of Turcotte and Schubert . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19 19 19 19
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Hand-in I, Chapter 2: Stress and strain in solids; isostacy; vectors and tensors. 2-4. A sedimentary basin has a thickness of 4 km. Assuming that the crustal stretching model is applicable and that hcc = 35 km, ρm = 3300 kg m−3 , ρcc = 2750 kg m−3 , and ρs = 2550 kg m−3 , determine the stretching factor.
The crustal stretching model considers a section of width wo of the continental crust of original thickness hcc under tension. The tension stretches the segment to a new width wb with thickness hcb , thus creating a basin of depth hsb , which is assumed to subsequently be filled by a sediment of density ρs (see figure 1).
Figure 1: The crustal stretching model.
The stretching factor α for the crustal stretching model is defined as α=
wb wo
(2-6)
If we assume that the segment is stretched at constant volume it must hold that wb hcb = wo hcc hcc ⇒ α= hcb
(2-7) (2-8)
Assuming that the region after the stretching and the fill-in of sediments is at isostatic equilibrium, we find that the equilibrium equation becomes ρc hcc = ρs hsb + ρc hcb + ρm (hcc − hsb − hcb ) Using equation (2-8) we then find that ρc hcc ⇒ ⇒ ⇒
hcc hcc + ρm hcc − hsb − = ρs hsb + ρc α α
hcc (ρc − ρm ) = hcc (ρc − ρm ) + hsb (ρm − ρs ) α hcc (ρc − ρm ) α= hcc (ρc − ρm ) + hsb (ρm − ρs ) 35000 · (2750 − 3300) α= = 1.18 35000 · (2750 − 3300) + 4000 · (3300 − 2550)
1
(2-9)
2-7. Consider a continental block to have a thickness of 70 km corresponding to a major mountain range. If the continent has a density of 2800 kg m−3 , determine the tensional stress in the continental block.
Consider a continental block ”floating” in the mantle as displayed in figure 2.
Figure 2
The horizontal force, per unit width, acting on the edge of the block Fm is then given by the lithostatic pressure pL (or normal stress, σxx ) in the mantle, integrated over the edge of the block (note that since we are considering the force per unit width we only need to integrate over depth) Z b Z b 1 (2-14) Fm = pL dy = ρm g ydy = ρm gb2 2 0 0 The horizontal force Fc , per unit width, on a vertical cross-section inside the block can then be found by integrating the horizontal normal stress σxx over the cross-section. Let us assume that σxx is made up of two parts: 1. The lithostatic contribution: ρc gy 2. A constant tectonic contribution: ∆σxx (i.e. the deviatoric stress we wish to find) I.e. we have Z Fc =
h
h
Z σxx dy =
0
(ρc gy + ∆σxx ) dy = 0
1 ρc gh2 + ∆σxx h 2
From isostatic considerations it then holds that ρc ρc h = ρm b ⇒ b=h ρm
(*)
Equating equations (2-14), (2-16), and using (*)we than find that
⇒
1 1 1 ρ2c 2 ρc gh2 + ∆σxx h = ρm gb2 = gh 2 2 2 ρm 1 ρc ∆σxx = ρc gh −1 2 ρm
And putting in the numerics we find ∆σxx
1 = · 2800 · 10 · 70000 · 2
(2-16)
2800 − 1 ≈ −1.48 × 108 Pa = −148 MPa 3300
2
N.B. The answer given in the book has an erroneous unit (N instead of N/m) 2-9. Assume that the friction law given in Equation (2-23) is applicable to the strike-slip fault illustrated in Figure 2-10 with f = 0.3. Also assume that the normal stress σxx is lithostatic with ρc = 2750 kg m−3 . If the fault is 10 km deep, what is the force (per unit length of the fault) resisting motion on the fault? What is the mean tectonic shear stress over this depth σ ¯zx required to overcome this frictional resistance?
Let the length of the fault be Lz , and consider a strip along the fault plane of height δy, i.e. of area A = Lz δy. Let the shear stress on the fault plane be σxz , the force resisting motion on the strip δF is then simply δF = Aσxz = Lz δyσxz Or expressed per unit length of the fault δF = δyσxz Lz which is equal to the situation displayed in figure 2-10 (assuming δA = δy · 1)
Figure 3: Figure 2-10
Assuming that the shear stress on the fault plane is given by σxz = f σxx
(2-23)
and that the normal stress on the fault plane is lithostatic (i.e. due to the overburden only) σxx = ρgy
(2-24)
we have that the force per unit length resisting motion on the strip is δF = δyf ρgy Lz If we then let the width of the strip become infinitesimal we find that dF = f ρgydy Lz Which can be integrated over the depth of the fault to yield Z y F 1 = f ρgy 0 dy 0 = f ρgy 2 Lz 2 0 Putting in the numerics we find F 1 = · 0.3 · 2750 · 10 · 100002 = 4.152 × 1011 N m−1 = 412.5 GN m−1 Lz 2 3
The mean tectonic shear stress σ ¯xz over the depth of the fault, required to overcome this frictional resistance then needs to be of at least the same magnitude as this force divided by the depth of the fault (to get from force to stress), hence 4.152 × 1011 F/Lz = = 4.152 × 107 Pa = 41.52 MPa σ ¯xz = y 10000 2-11. Consider a rock mass resting on an inclined bedding plane as shown in Figure 2-12. By balancing the forces acting on the block parallel to the inclined plane, show that the tangential force per unit area σx0 y0 on the plane supporting the block is ρgh sin θ (ρ is the density and h is the thickness of the block). Show that the sliding condition is θ = tan−1 f
(2-26)
The only force acting on the rock mass is the gravitational force, i.e. Fy = −mg where the minus sign comes from the gravitational force acting in the opposite direction to the y-direction. Consider then a vertical column through the rock mass of width δx0 against the inclined plane. The width of this column in the x-direction is then (see figure 4) δx = δx0 cos θ and the height is δy =
h cos θ
finally for the width in the z-direction we find δz = δz 0 hence the volume of the column is V = δxδyδz = ·δx0 cos θ ·
h δz 0 = δx0 δz 0 h cos θ
Figure 4: Modified from figure 2-12 of Turcotte and Schubert 4
Let the density of the rock mass be ρ, then the mass of the column is simply given by m = ρV = ρδx0 δz 0 h And hence the gravitational force on the column Fy = −ρδx0 δz 0 hg or alternatively expressed per unit area of the inclined plane Fy = −ρhg = σy0 y δx0 δz 0 If we then choose a coordinate system, x0 y 0 rotated clockwise by an angle θ, i.e. such that the x0 -axis is parallel to inclined plane, we find that the force Fy in this coordinate system has the components Fx0 Fy = − 0 0 sin θ = ρhg sin θ = σx0 y0 0 0 δx δz δx δz Fy Fy 0 = 0 0 cos θ = ρhg cos θ = σy0 y0 δx0 δz 0 δx δz where we have recognized that the force components per unit area is nothing else than the normal and shear stress on the inclined plane in the x0 y 0 -coordinate system. Assuming a friction law of the form (f )
−σx0 y0 = f σy0 y0
(2-23)
where the minus sign comes from the fact that the frictional stress is positive in the negative x0 -direction, and using the condition of sliding is (f ) σx0 y0 ≥ σx0 y0 we find that at the onset of sliding f= ⇒
σx0 y0 −hρg sin θ = = tan θ 0 0 −σy y −hρg cos θ
θ = tan−1 f
(2-26)
2-14. The state of stress at a point on a fault plane is σyy = 150 MPa, σxx = 200 MPa, σxy = 0 (y is depth and x points westward). What are the normal stress and tangential stress on the fault plane if the fault strikes N-S and dips 35o to the west?
The geometry of the problem can be seen in figure 5
Figure 5 5
We are given σyy and σxx and want to find the normal and tangential stresses to the plane. As the strike of the fault is aligned to the z-axis we can consider the problem in the xy-plane only, i.e. as a 2D-problem instead. Thus by rotating the coordinate system by an angle θ to the x0 y 0 -system (see figure 6), we can find the sought for stresses as the normal stress σy0 y0 and the tangential stress σx0 y0
Figure 6
Using tensor notation the rotation is given by σ 0 = RσRT where
R=
cos θ − sin θ
sin θ cos θ
is the rotation matrix for a clockwise rotation where as the rotation matrix for an anti-clockwise rotation has opposite signs on the sines, and σ σxy σ = xx σyx σyy is the 2D-stress tensor, hence σx0 x0 σx0 y 0 cos θ sin θ σxx σxy cos θ − sin θ 0 σ = = σy0 x0 σy0 y0 − sin θ cos θ σyx σyy sin θ cos θ cos θ sin θ σxx cos θ + σxy sin θ −σxx sin θ + σxy cos θ = − sin θ cos θ σyx cos θ + σyy sin θ −σyx sin θ + σyy cos θ σxx cos2 θ + σyy sin2 θ + 2σxy sin θ cos θ (σyy − σxx ) cos θ sin θ + σxy (cos2 θ − sin2 θ) = (σyy − σxx ) cos θ sin θ + σxy (cos2 θ − sin2 θ) σxx sin2 θ + σyy cos2 θ − 2σxy sin θ cos θ Hence the sought for stresses are σy0 y0 = σxx sin2 θ + σyy cos2 θ − 2σxy sin θ cos θ = σxx sin2 θ + σyy cos2 θ − σxy sin 2θ 1 σx0 y0 = (σyy − σxx ) cos θ sin θ + σxy (cos2 θ − sin2 θ) = (σyy − σxx ) sin 2θ + σxy cos 2θ 2 Putting in the numerics we find that σy0 y0 = 200 · sin2 (35o ) + 150 · cos2 (35o ) = 166 MPa σx0 y0 =
(150 − 200) · sin(2 ∗ 35o ) = −23.5 MPa 2
6
(2-41) (2-40)
2-15. Show that the sum of the normal stresses on any two orthogonal planes is a constant. Evaluate the constant.
Actually this is generally not true. For the statement to hold true we need to demand that the intersection of the planes is parallel for all of the possible configurations (a simple example of a case in which the statement does not hold can be found below). I.e. the orientation of the planes are only allowed to change in 2D, due to a rotation around the intersection axis, being fixed in the third dimension. We can relate the normal stresses on the planes via a finite rotation around the intersection axis. Choose a coordinate system such that the intersection is parallel to the z-axis, we can then have that σ 0 = RσRT where
R=
cos θ − sin θ
sin θ cos θ
is the rotation matrix,and σ=
σxx σyx
σxy σyy
is the 2D-stress tensor, hence σx0 x0 σx0 y 0 cos θ sin θ σxx σxy cos θ − sin θ 0 σ = = σy0 x0 σy0 y0 − sin θ cos θ σyx σyy sin θ cos θ cos θ sin θ σxx cos θ + σxy sin θ −σxx sin θ + σxy cos θ = − sin θ cos θ σyx cos θ + σyy sin θ −σyx sin θ + σyy cos θ σxx cos2 θ + σyy sin2 θ + 2σxy sin θ cos θ (σyy − σxx ) cos θ sin θ + σxy (cos2 θ − sin2 θ) = (σyy − σxx ) cos θ sin θ + σxy (cos2 θ − sin2 θ) σxx sin2 θ + σyy cos2 θ − 2σxy sin θ cos θ Thus σx0 x0 + σy0 y0 = σxx cos2 θ + σyy sin2 θ + 2σxy sin θ cos θ + σxx sin2 θ + σyy cos2 θ − 2σxy sin θ cos θ = σxx (cos2 θ + sin2 θ) + σyy (sin2 θ + cos2 θ) = σxx + σyy Hence the sum of the normal stresses for a particular choice of orientation of the two orthogonal planes is simply the sum of the normal stresses for any orientation of the planes such that the direction of the intersection between the planes is preserved. Note that since we are only rotating in the xy-plane we could as well have written σx0 x0 + σy0 y0 + σz0 z0 = ... + σzz = ... = σxx + σyy + σzz as σz0 z0 = σzz for this choice of coordinate system. Hence we find that the sum of the three normal stresses are constant. This however is true in general (i.e. for all rotations) and the sum of the normal components of the stress tensor is generally referred to as the first invariant of the stress tensor. As a simple example of a case in which the statement does not hold true consider the stress state: σxx and σyy 6= 0, σzz = σxy = σxz = σyz = 0. Consider two orthogonal planes such that their intersection is parallel to the z-axis. According to our derivation above the sum of the normal stresses on the planes are then simply σxx + σyy . Next consider two orthogonal planes such that the their intersection is parallel to the x-axis, and the normals to the planes points in the y- and z-direction, the sum of the the normal stresses on these planes are then σyy .
7
2-20. The measured horizontal principal stresses at a depth of 200 m are given in Table 2-1 as a function of distance from the San Andreas fault. What are the values of maximum shear stress at each distance? Distance from fault [km] 2 4 22 34
Maximum principal stress [MPa] 9 14 18 22
Minimum principal stress [MPa] 8 8 8 11
Table 1: Table 2-1 of Turcotte and Schubert: Stress measurements at 200 m depth vs. distance from the San Andreas fault.
The maximum shear stress in 2D as a function of the principal stresses is given by σxy |max =
1 (σ1 − σ2 ) 2
(2-60)
Hence Distance from fault [km] 2 4 22 34
σxy |max 1 2 (9-8) = 0.5 MPa 1 2 (14-8) = 3.0 MPa 1 2 (18-8) = 5.0 MPa 1 2 (22-11) = 5.5 MPa
2-21. Uplift and subsidence of large areas are also accompanied by horizontal or lateral strain because of the curvature of the Earth’s surface. Show that the lateral strain accompanied by an uplift ∆y is given by =
∆y R
(2-74)
where R is the radius of the Earth
Given that the radius of the Earth is much larger than any possible uplift or subsidence of any surface area we can consider the lateral strain as a normal strain. I.e. as the ratio of lateral change divided by the original size.
Figure 7 8
Let the horizontal distance for an area at a radius R be L(R) and that of the same area uplifted a distance ∆y be L(R + ∆y) (see Figure 7). We have that L(R) = Rθ L(R + ∆y) = (R + ∆y)θ Hence =
L(R + ∆y) − L(R) (R + ∆y)θ − Rθ ∆y = = L(R) Rθ R
(2-74)
whereas for subsidence we have that =
L(R − ∆y) − L(R) (R − ∆y)θ − Rθ ∆y ∆y = =− =− L(R − ∆y) (R − ∆y)θ R − ∆y R
where we have used that ∆y R An alternative derivation is to express L(R+∆y) as L(R)(1+) = Rθ(1+) = (R+∆y)θ ⇒ 1+ = 1+∆y/R ⇒ = ∆y/R Given 1 A= 2
3 1
1 , −1
and
1 B = 3 2
2 0 −1
what is the inner matrix product C = A · B? Is the inner product D = B · A defined?
1 C=A·B= 2
3 1
1 1 3 −1 2
2 12 0 = 3 −1
Since the size of AT is equal to the size of B both C and D are defined. 1 2 5 5 1 3 1 D = B · A = 3 0 = 3 9 2 1 −1 2 −1 0 5
9
1 5
−1 3 3
Hand-in II, Chapter 3: Elasticity and flexure 3-1. Determine the surface stress after the erosion of 10 km of granite. Assume that the initial state of stress is lithostatic and that ρ = 2700 kg m−3 and ν = 0.25
The wording was a bit confusing to the class last year. Perhaps this helps: A surface is initially unstressed. It is then covered by sediments of thickness 5 km. What is the stress at the original surface (which is now at 5 km depth)? 3-2. An unstressed surface is covered with sediments with a density of 2500 kg m−3 to a depth of 5 km. If the surface is laterally constrained and has a Poisson’s ratio of 0.25, what are the three components of stress at the original surface?
3-7. What is the displacement of a plate pinned at both ends (w = 0 at x = 0, L) with equal and opposite bending moments applied at the ends? The problem is illustrated in Figure 3-16.
3-16. Show that the cross-sectional area of a two-dimensional laccolith is given by (p − ρgh)L5 /720D
10
Additional task: Evaluate the bending moment directly underneath the line load, i.e. at x = 0. How does this compare to the maximum value you calculated above? Can you explain this? 3-19. (a) Consider a lithospheric plate under a line load. Show that the absolute value of the bending moment is a maximum at xm = α cos−1 0 =
π α 2
(3-137)
and that its value is Mm = −
2Dw0 −π/2 Dw0 e = −0.416 2 α2 α
(3-138)
(b) Refraction studies show that the Moho is depressed about 10 km beneath the center of the Hawaiian Islands. Assuming that this is the value of w0 and that h = 34 km, E = 70 Gpa, ν = 0.25, ρm − ρw = 2300 kg m−3 , and g = 10 m s−2 , determine the maximum bending stress in the lithosphere.
3-21. An ocean basin has a depth of 5.5 km. If it is filled to sea level with sediments of density 2600 kg m−3 , what is the maximum depth of the resulting sedimentary basin? assume ρm = 3300 kg m−3 .
3-22. The Amazon River in Brazil has a width of 400 km. Assuming that the basin is caused by a line load at its center and that the elastic lithosphere is not broken, determine the thickness of the elastic lithosphere. Assume E = 70 GPa, ν = 0.25, and ρm − ρs = 700 kg m−3 .
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Hand-in III, Chapter 4: Heat transfer. 4-8 Consider one-dimensional steady-state heat conduction in a half-space with heat production that decreases exponentially with depth. The surface heat flow-heat production relation is qo = qm + ρHo hr . What is the heat flow - heat production relation at depth y = h∗ ? Let q ∗ and H ∗ be the upward heat flux and heat production at y = h∗ .
4-31 Derive an expression for the thickness of the thermal boundary layer if we define it to be the distance where θ = 0.01
Hint: The governing equation is linear; hence, you can add any two solution together and the sum will also satisfy the equation. You just have to make sure that the boundary conditions are satisfied. 4-33 One way of determining the effects of erosion on subsurface temperatures is to consider the instantaneous removal of a thickness l of ground. Prior to the removal T = To + βy where y is the depth, β is the geothermal gradient, and To is the surface temperature. After removal, the new surface is maintained at temperature To . Show that the subsurface temperature after the removal of the surface layer is given by y √ T = To + βy + βlerf 2 κt How is the surface heat flow affected by the removal of surface material.
4-39 One of the estimates for the age of the Earth given by Lord Kelvin in the 1860’s assumed that Earth was initially molten at a constant temperature Tm and that it subsequently cooled by conduction with a constant surface temperature To . The age of the Earth could then be determined from the present surface thermal gradient (dT /dy)o . Reproduce Kelvin’s results assuming Tm − To = 1700 K, C = 1 kJ kg−1 K−1 , L = 400 kJ kg−1 , κ = 1 mm2 s−1 , and (dT /dy)o = 25 K km−1 . In addition, determine the thickness of the solidified lithosphere.
4-57 If petroleum formation requires temperatures between 380 and 430 K, how deep would you drill in a sedimentary basin 20 Ma old? Assume To = 285 K, T1 = 1600 K, κm = 1 mm2 s−1 , ks = 2 W m−1 K−1 , and km = 3.3 W m−1 K−1
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Hand-in IV, Chapter 6, Part I: Fluid mechanics 6-1 Show that the mean velocity in the channel is given by u ¯=−
h2 dp uo + 12µ dx 2
(6-17)
6-2 Derive a general expression for the shear stress τ at any location y in the channel. What are the simplified forms of τ for Couette flow and for the case uo = 0?
6-3 Find the point in the channel at which the velocity is a maximum.
6-4 Consider the steady, unidirectional flow of a viscous fluid down the upper face of an inclined plane. Assume that the flow occurs in a layer of constant thickness h, as shown in Figure 6-3. Show that the velocity profile is given by u=
ρg sin α 2 h − y2 2µ
(6-18)
where y is the coordinate measured perpendicular to the inclined plane (y=h is the surface of the plane), α is the inclination of the plane to the horizontal, and g is the acceleration of gravity. First show that dτ = −ρg sin α dy
(6-19)
and then apply the no-slip condition at y = h and the free-surface condition, τ = 0, at y = 0. What is the mean velocity in the layer? What is the thickness of a layer whose rate of flow down the incline (per unit width in the direction perpendicular to the plane in figure 6-3) is Q?
6-5 For an asthenosphere with a viscosity µ = 4 × 1019 Pa s and a thickness h = 200 km, what is the shear stress on the base of the lithosphere if there is no counterflow (∂p/∂x = 0)? Assume uo = 50 mm yr−1 and that the base of the asthenosphere has zero velocity.
6-6 Assume that the base stress obtained in probelm 6-5 is acting on 6000 km of lithosphere with a thickness of 100 km. What tensional stress in the lithosphere (hL = 100 km) must be applied at a trench to overcome this basal drag?
6-9 Determine the rate at which magma flows up a two-dimensional channel of width d under the buoyant pressure gradient −(ρs − ρl )g. Assume laminar flow.
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6-11 Show that the constant of integration A in the above post glacial rebound solution is given by A=−
λ 2π
2
ρgwmo −t/τr e 2µ
(6-106)
6-12 The ice sheet over Hudson Bay, Canada, had an estimated thickness of 2 km. At the present time there is a negative free-air gravity anomaly in this region of 0.3 mm s−2 a) Assuming that the ice (density of 1000 kg m−3 ) was in isostatic equilibrium and displaced mantle rock with a density of 3300 kg m−3 , determine the depression of the land surface wmo b) Assuming that the negative free-air gravity anomaly is due to incomplete rebound, determine w at the present time c) Applying the periodic analysis given above, determine the mantle viscosity. Assume that the ice-sheet melted 10000 years ago and that the appropriate wave length for the Hudson Bay ice sheet was 500 km d) Dicuss the difference between the viscosity obtained in c) and that obtained for Scandinavia
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Hand-in V, Chapter 6, PartII: Fluid mechanics 6-22 The Stokes drag D on a sphere can only depend on the velocity of the sphere u, its radius a, and the viscosity µ and the density of the fluid. Show by dimensional analysis that ρua D = f (6-232) ρu2 a2 µ Where f is an arbitrary function. Because the equations of slow viscous flow are linear, D can only be directly proportional to u. Use this fact together with equation (6-232) to conclude that D ∝ µua
(6-233)
6-23 Consider a spherical bubble of a low-viscosity fluid with density ρb rising or falling through a much more viscous fluid with density ρf and viscosity µf because of buoyancy force. For this problem the appropriate boundary conditions at the surface of the sphere, r = a, are ur = 0 and τrθ = 0. Using Equations (6-210), (6-211), and (6-220), show that a ur = U −1 + cos θ (6-234) r a uθ = U 1 − sin θ (6-235) 2r By integrating Equation (6-196), show that on r = a p=
µf U cos θ a
(6-236)
The drag force is obtained by carring out the integral Z π ∂ur 2 cos θ sin θdθ D = 2πa p − 2µf ∂r r=a 0
(6-237)
Show that D = 4πµf aU
(6-238)
and demonstrate that the terminal velocity of the bubble in the fluid is U=
a2 g(ρf − ρb ) 3µf
(6-239)
(Problem and information not in older versions of Turcotte and Schubert.) 6-24 Determine the radius of the plume conduit, the volume flux, the heat flux, the mean ascent velocity, and the plume head volume for the Azores plume. Assume that Tp − T1 = 200 K, αv = 3 x 10−5 K−1 , µp = 1019 Pa s, ρm = 3300 kg m−3 , µm = 1021 Pa s, and cp = 1.25 kJ kg−1 K−1
(Problem 6-24 in older versions of Turcotte and Schubert.) 6-26 Consider the unidirectional flow driven by a constant horizontal pressure gradient through a channel with stationary plane parallel walls, as discussed in section 6-2. Determine the temperature distribution 15
in the channel, the wall heat flux, the heat transfer coefficient, and the Nusselt number by assuming, as in the pipe flow problem above, that the temperature of both walls and the fluid varies linearly with distance x along the channel. You will need to form the temperature equation in two dimensions that balances horizontal heat advection against vertical heat conduction, as given in Equation (4-156)
(Problem 6-31 in older versions of Turcotte and Schubert.) 6-33 Consider convection in a fluid layer heated from below. The mean surface heat flux q¯ is transferred through the cold thermal boundary layer by conduction. Therefore we can write q¯ =
k(Tc − To ) δ
(6-386)
where δ is a characteristic thermal boundary layer thickness. Show that δ = 1.7 Ra−1/3 b
(6-387)
Calculate δ for an upper mantle convection cell given the parameter values used in this section.
(Problem 6-34 in older versions of Turcotte and Schubert.) N.B. To solve this problem the viscosity of the mantle is needed, assume a value of 1021 Pa s. 6-36 Apply the two-dimensional boundary-layer model for heated-from-below convection to the entire mantle. Calculate the mean surface heat flux, the mean horizontal velocity, and the mean surface thermal boundary layer thickness. Assume T1 − To = 3000 K, b = 2880 km, k = 4 W m−1 K−1 , κ = 1 mm2 s−1 , αv = 3 x 10−5 K−1 , g = 10 m s−2 , and ρo = 4000 kg m−3
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Hand-in VI, Chapter 7: Rock Rheology 7-10 Show that the effective viscosity µef f for the channel flow of a power-law fluid is given by µef f =
τ = du/dy
or µef f µef f,wall
=
2y h
p1 − po L
h2 4(n + 2)¯ u
2y h
1−n (7-127)
1−n (7-128)
where µef f,wall is the value of µef f at y = ±h/2. Plot µef f /µef f,wall as a function of y/h for n = 1, 3, and 5.
7-12 Consider an ice sheet of thickness h lying on bedrock with a slope α, as shown in figure 7-16. The ice will creep slowly down hill under the force of its own weight. Determine the velocity profile u(y) of the ice. The viscosity of ice has the temperature dependence given in Equation (7-130). Assume that the temperature profile in the ice is linear with the surface temperature To (at y = 0) and the bedrock-ice interface temperature T (at y = h). Assume that there is no melting at the base of the ice sheet so that the no-slip condition applies, that is u = 0 at y = h, and utilize the approximation given in Equation (7-133)
(N.B. Look out for errors/missprints in equations: 7-213 and 7-216 of Turcotte and Schubert) 7-17 The way which subsolidus convection with temperature-dependent viscosity regulates the Earth’s thermal history can be quantatively assessed using the following simple model. Assume that the Earth can be characterized by the mean temperature T¯ and that Equation (7-200) gives the rate of cooling. Let the model Earth begin its thermal evolution at time t = 0 with a high temperature T¯(0) and cool there after. Disregard the heating due to the decay of radioactive isotopes and assume that the Earth cools by vigorous convection. Show that the mean surface heat flux q¯ can be related to the mean temperature by q¯ = 0.74k
ρgαv µκ
1/3
T¯ − To
4/3
(7-213)
Use Equations (6-316) and (6-337) and assume that the total temperature drop driving convection is twice the difference between the mean temperature T¯ and the surface temperature To . Following Equation (7100), assume that the viscosity is given by Ea µ = C T¯ exp (7-214) RT¯ and write the cooling formula as dT¯ 2.2κ ρgαv 1/3 ¯ Ea =− T exp − dt a Cκ 3RT¯
(7-215)
Equation (7-215) was obtained assuming (T¯ − To )4/3 ' T¯4/3 , a valid simplification since To T¯. Integrate the cooling formula and show that Ea Ea 2.2κ ρgαv 1/3 Ei − Ei = t (7-216) ¯ ¯ a Cκ 3RT 3RT (0) Where Ei is the exponential integral. Calculate and plot T¯/T¯(0) versus t for representative values of the parameters in Equation (7-216). Discuss the role of the temperature dependence in the cooling history 17
7-24 Consider a long circular cylinder of elastic-perfectly plastic material that is subjected to a torque at its outer surface r = a. The state of stress in the cylinder can be characterized by an azimuthal shear stress τ . Determine the torque for which an elastic core of radius c remains. Assume that the yield stress in shear is σo . In the elastic region the shear stress is proportional to the distance from the axis of the cylinder r. What is the torque for the onset of plastic yielding? What is the maximum torque that can be sustained by the cylinder?
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Hand-in VII, Chapter 8: Faulting 8-3 A number of criteria have been proposed to relate the brittle fracture of rock to the state of stress. One of these is the Coulomb-Navier criterion, which states that failure occurs on a plane when the shear stress τ attains the value |τ | = S + µσn
(8-37)
where S is the inherent shear strength of the rock and µ is the coefficient of internal friction. Consider a two-dimensional state of stress with principal stresses σ1 and σ2 and show that |τ | − µσn has a maximum value for a plane whose normal makes an angle θ to the larger principal stress given by tan 2θ =
−1 µ
(8-38)
Show also that the quantity |τ | − µσn for this plane is |τ | − µσn =
1 1 (σ1 − σ2 )(1 + µ2 )1/2 − (σ1 + σ2 )µ 2 2
(8-39)
According to the Coulomb-Navier criterion, failure will occur if this quantity equals S, that is the failure criterion takes the form σ1 [1 + µ2 ] − µ − σ2 [1 + µ2 ] + µ = 2S (8-40) What is the compressive strength of the rock in terms of µ and S? From Equation (8-38) it is seen that θ must exceed 45o , so that the direction of shear fracture makes an acute angle with σ1 . The Coulomb-Navier criterion is found to be reasonably valid for igneous rocks under compression.
8-6 The spring force on the slider in Figure 8-18 at the time of slip initiation is fs Fn . What is the spring force on the slider block at the end of slip?
8-8 Compute the wave energy released in a magnitude 8.5 earthquake and compare it with the amount of heat lost through the surface of the Earth in an entire year.
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