November 16, 2011
Outline 1 Ocean Floor Topography Reality and model 2 3 Assumptions Results Conclusion
Ocean Floor Topography Ocean Floor Topography Reality and model Figure: From Geodynamics Turcotte & Schubert Depth = 2αρ m(t m T s ) (ρ m ρ w ) ( κx πu 0 ) 1/2
Mendocino Fracture Zone Ocean Floor Topography Reality and model Figure: Mendocino Fracture Zone
Relevent problem in reality Ocean Floor Topography Reality and model Sandwell and Schubert, 1982 Figure: What brings up the problem?
Depth-Age Relation Depth-Age Relation d(t) = d ref + 2αρ m(t m T s ) (ρ m ρ w ) for ages less than about 70 Myr. ( κt ) 1/2 π Parker and Oldenburg, 1973 Davis and Lister, 1974 Oxburgh and Turcotte, 1978 Parsons and Sclater, 1977
Depth Difference d(t) = d ref + 2αρm(Tm Ts) (ρ m ρ w ) h A = d(t A ) d(0) = 2αρm(Tm Ts) (ρ m ρ w ) h B = d(t B ) d(t B ) = 2αρm(Tm Ts) (ρ m ρ w ) ( κta π δ B = h A h B = 2αρm(Tm Ts) ( κ ) 1/2 (ρ m ρ w ) π ( κt ) 1/2 π ) 1/2 ( κ π ) 1/2 (t 1/2 B t 1/2 B ) ( t 1/2 + t 1/2 B t1/2 B ) t = t A 0 = t B t B Sandwell and Schubert, 1982
Model Flexure Equation D d 4 w dx 4 + P d 2 w dx 2 + g(ρ m ρ w )w = q a (x)
Model Flexure Equation D d4 w dx 4 P d2 w dx 2 D d 4 w dx 4 + P d 2 w dx 2 + g(ρ m ρ w )w = q a (x) Strength of plate End load g(ρ m ρ w )w Restricting force q a (x) Applied load
Model Flexure Equation D d4 w dx 4 P d2 w dx 2 D d 4 w dx 4 + P d 2 w dx 2 + g(ρ m ρ w )w = q a (x) Strength of plate End load g(ρ m ρ w )w Restricting force q a (x) Applied load Flexure Equation D d 4 w dx 4 + g(ρ m ρ w )w = 0
Flexure Model Figure: Model used to calculate the flexural topography Governing Equation d D 4 w 1 1 dx 4 + g(ρ m ρ w )w 1 = 0 x < x 1 d D 4 w i i dx 4 + g(ρ m ρ w )w i = 0 x i 1 < x < x i d D 4 w N+1 N+1 + g(ρ dx 4 m ρ w )w N+1 = 0 x N < x where D i is the flexural rigidity of ith block.
Physical Constrains(Boundary Conditions) w i w i+1 = δ i
Physical Constrains(Boundary Conditions) w i w i+1 = δ i Finite at infinity far away lim x w 1,N < Const.
Physical Constrains(Boundary Conditions) w i w i+1 = δ i Finite at infinity far away lim x w 1,N < Const. Continuous Slope
Physical Constrains(Boundary Conditions) w i w i+1 = δ i Finite at infinity far away Continuous Slope lim x w 1,N < Const. dw i dx = dw i+1 dx
Physical Constrains(Boundary Conditions) w i w i+1 = δ i Finite at infinity far away Continuous Slope Continuous Moment lim x w 1,N < Const. dw i dx = dw i+1 dx
Physical Constrains(Boundary Conditions) w i w i+1 = δ i Finite at infinity far away Continuous Slope Continuous Moment lim x w 1,N < Const. dw i dx = dw i+1 dx d D 2 w i d i = D 2 w i+1 dx 2 i+1 dx 2
Physical Constrains(Boundary Conditions) w i w i+1 = δ i Finite at infinity far away Continuous Slope Continuous Moment Continuous Shear force lim x w 1,N < Const. dw i dx = dw i+1 dx d D 2 w i d i = D 2 w i+1 dx 2 i+1 dx 2
Physical Constrains(Boundary Conditions) w i w i+1 = δ i Finite at infinity far away Continuous Slope Continuous Moment Continuous Shear force lim x w 1,N < Const. dw i dx = dw i+1 dx d D 2 w i d i = D 2 w i+1 dx 2 i+1 dx 2 d D 3 w i d i = D 3 w i+1 dx 3 i+1 dx 3
Differential Equations Consider the nth order linear homogeneous differential equation: L[y] = a n y (n) + a n 1 y (n 1) + + a 1 y = 0 where y = y(x). Many years ago, one genius found that y = e rx is a solution of the equation. Then we have: (a n r n + a n 1 r n 1 + + a 1 r)e rx = 0 A polynomial of degree n has n zeros, say we have n roots r 1,, r n. If e r 1x,, e rnx are linearly independent, the general solution of L[y] = 0 is
Differential Equations Consider the nth order linear homogeneous differential equation: L[y] = a n y (n) + a n 1 y (n 1) + + a 1 y = 0 where y = y(x). Many years ago, one genius found that y = e rx is a solution of the equation. Then we have: (a n r n + a n 1 r n 1 + + a 1 r)e rx = 0 A polynomial of degree n has n zeros, say we have n roots r 1,, r n. If e r 1x,, e rnx are linearly independent, the general solution of L[y] = 0 is y = c 1 e r 1x + c 2 e r 2x + + c n e rnx
Analytical Solution 1 Governing Equation D i d 4 w i dx 4 + g(ρ m ρ w )w i = 0
Analytical Solution 1 Governing Equation 4th polynomial D i d 4 w i dx 4 + g(ρ m ρ w )w i = 0 D i r 4 i + g(ρ m ρ w ) = 0
Analytical Solution 1 Governing Equation 4th polynomial Roots D i d 4 w i dx 4 + g(ρ m ρ w )w i = 0 D i r 4 i + g(ρ m ρ w ) = 0 r i1 = r i3 = ( g(ρm ρ w ) ) 1/4 D i e iπ/4 r i2 = ( ) 1/4 g(ρm ρw ) D i e i5π/4 r i4 = ( g(ρm ρ w ) D ( i g(ρm ρw ) D i ) 1/4 e i3π/4 ) 1/4 e i7π/4
Analytical Solution 2 General Solution w i (x) = c i1 e r 1x + c i2 e r 2x + c i3 e r 3x + c i4 e r 4x Wavelength of the flexure λ and flexture rigidity are given as: { λ = 2π( 4D D = g(ρ m ρ w ) )1/4 Eh 3 e 12(1 ν 2 ) where h e is the effective elastic thickness as: h e = 2(κt) 1/2 erfc 1 ( Tm T e T m T s )
Analytical Solution 3 Roots r i1 = 2π λ i (1 + i) r i2 = 2π λ i ( 1 + i) r i3 = 2π λ i ( 1 i) r i4 = 2π λ i (1 i)
Analytical Solution 3 Roots r i1 = 2π λ i (1 + i) r i2 = 2π λ i ( 1 + i) r i3 = 2π λ i ( 1 i) r i4 = 2π λ i (1 i) Flexure Topography w i (x) = c i1 e r1x + c i2 e r2x + c i3 + c i4 e r 4x ( ) = e 2πx/λ i A i sin 2πx λ i + B i cos 2πx ( λ i ) +e 2πx/λ i C i sin 2πx λ i + D i cos 2πx λ i with help of e iθ = cos θ + i sin θ. A i, B i, C i, D i are determined by physical constrains.
2 Blocks situation Flexure Topography ( w 1 = e 2πx/λ 1 C 1 sin 2πx w 2 = e 2πx/λ 2 λ 1 ) + D 1 cos 2πx λ 1 ( ) A 2 sin 2πx λ 2 + B 2 cos 2πx λ 2 x 1 < 0 x 2 > 0 Based on the four physical constrains, we could solve the four unknown parameter.
Stress and Moment Moment M i (x) = D i d 2 w i (x) dx 2 Bending stress σ xxi (x) = 6M i(x) h 2 e = Eh e d 2 w i (x) 2(1 ν 2 ) dx 2 Shear stress(average) σ xzi (x) = D i h e d 3 w i (x) dx 3
Time to Deal with Real World Assumptions Results Conclusion Model Assumptions FZ scarp fossilized (no slip on plane) No lateral heat conduction
Model Results Assumptions Results Conclusion 400 w(x) Deflection Due to Flexure 300 Flexural Topography (m) 200 100 0 100 200 300 400 2 1.5 1 0.5 0 0.5 1 1.5 2 Distance across FZ (m) x 10 5
Model Results Assumptions Results Conclusion 400 w(x) Deflection Due to Flexure 300 Flexural Topography (m) 200 100 0 100 200 300 400 2 1.5 1 0.5 0 0.5 1 1.5 2 Distance across FZ (m) x 10 5 4100 Depth Profile No Slip 4200 4300 Actual Topography (m) 4400 4500 4600 4700 4800 4900 5000 2 1.5 1 0.5 0 0.5 1 1.5 2 Distance across FZ (m) x 10 5
Model Results Assumptions Results Conclusion 400 w(x) Deflection Due to Flexure 6 x 107 Bending Stress σ xx 300 4 Flexural Topography (m) 200 100 0 100 200 Bending Stress (Pa) 2 0 2 4 300 400 2 1.5 1 0.5 0 0.5 1 1.5 2 Distance across FZ (m) x 10 5 6 2 1.5 1 0.5 0 0.5 1 1.5 2 Distance across FZ (m) x 10 5 4100 Depth Profile No Slip 4200 4300 Actual Topography (m) 4400 4500 4600 4700 4800 4900 5000 2 1.5 1 0.5 0 0.5 1 1.5 2 Distance across FZ (m) x 10 5
Model Results Assumptions Results Conclusion 400 w(x) Deflection Due to Flexure 6 x 107 Bending Stress σ xx 300 4 Flexural Topography (m) 200 100 0 100 200 Bending Stress (Pa) 2 0 2 4 300 400 2 1.5 1 0.5 0 0.5 1 1.5 2 Distance across FZ (m) x 10 5 6 2 1.5 1 0.5 0 0.5 1 1.5 2 Distance across FZ (m) x 10 5 4100 Depth Profile No Slip 12 x 106 Shear Stress σ xz 4200 10 Actual Topography (m) 4300 4400 4500 4600 4700 4800 Shear Stress (Pa) 8 6 4 2 0 4900 2 5000 2 1.5 1 0.5 0 0.5 1 1.5 2 Distance across FZ (m) x 10 5 4 2 1.5 1 0.5 0 0.5 1 1.5 2 Distance across FZ (m) x 10 5
Topography Assumptions Results Conclusion Crustal ages: 46 Myr x < 0 and 54 Myr x > 0 4100 Depth Profile No Slip 4200 4300 Actual Topography (m) 4400 4500 4600 4700 4800 4900 5000 2 1.5 1 0.5 0 0.5 1 1.5 2 Distance across FZ (m) x 10 5
Deflection Due to Flexture Assumptions Results Conclusion Flexural wavelengths: 342km(x < 0) and 358km(x > 0) 400 w(x) Deflection Due to Flexure 300 Flexural Topography (m) 200 100 0 100 200 300 400 2 1.5 1 0.5 0 0.5 1 1.5 2 Distance across FZ (m) x 10 5
Deflection Due to Flexture Assumptions Results Conclusion Flexural rigidities: 4.99 10 10 Pa km 4 and 6.00 10 10 Pa km 4 Effective elastic thicknesses: 20.5 km and 21.8 km 6 x 107 Bending Stress σ xx 12 x 106 Shear Stress σ xz 4 10 8 Bending Stress (Pa) 2 0 2 Shear Stress (Pa) 6 4 2 0 4 2 6 2 1.5 1 0.5 0 0.5 1 1.5 2 Distance across FZ (m) x 10 5 4 2 1.5 1 0.5 0 0.5 1 1.5 2 Distance across FZ (m) x 10 5
Model VS. Reality Assumptions Results Conclusion Age of block Age of Left Block 42.3 Myr & Age of Right Block 23.3 Myr 4000 Mendocino Fracture Zone Model Prediction 4 Actual Topography (m) 4500 5000 23.3 Myr depth to seafloor (km) 4.5 5 42.3 Myr 5500 8 6 4 2 0 2 4 6 8 Distance across FZ (m) x 10 4 5.5 39.6 39.8 40 40.2 40.4 40.6 latitude (degrees N)
4000 4500 5000 5500 8 6 4 2 0 2 4 6 8 4 4.5 5 5.5 42.3 Myr 23.3 Myr 39.6 39.8 40 40.2 40.4 40.6 latitude (degrees N) Model VS Reality 2 Assumptions Results Conclusion Mendocino Fracture Zone Model Prediction Actual Topography (m) Distance across FZ (m) x 10 4 depth to seafloor (km) Comparisons between theoretical and actual bathymetric profiles for Mendocino-Pioneer FZ Pair (Sandwell and Schubert): Model fits multiple FZs fairly well Asymmetric flexure Elastic coupling (apparent tilting)
Conclusion Assumptions Results Conclusion 1 No significant slip on the fossil fault planes of the Mendocino and Pioneer FZs.
Conclusion Assumptions Results Conclusion 1 No significant slip on the fossil fault planes of the Mendocino and Pioneer FZs. 2 The lithosphere bends in the vicinity of a FZ as a consequence of the frozen-in scarp and the differential subsidence of the lithosphere far from the FZ.
Conclusion Assumptions Results Conclusion 1 No significant slip on the fossil fault planes of the Mendocino and Pioneer FZs. 2 The lithosphere bends in the vicinity of a FZ as a consequence of the frozen-in scarp and the differential subsidence of the lithosphere far from the FZ. 3 Good fits can be obtained by modeling this flexure as a thin elastic plate with age dependent effective elastic thickness.
Conclusion Assumptions Results Conclusion 1 No significant slip on the fossil fault planes of the Mendocino and Pioneer FZs. 2 The lithosphere bends in the vicinity of a FZ as a consequence of the frozen-in scarp and the differential subsidence of the lithosphere far from the FZ. 3 Good fits can be obtained by modeling this flexure as a thin elastic plate with age dependent effective elastic thickness. 4 We can use this model to calculate T e, the isotherm which defines the base of the elastic lithosphere. ( T = 450C)
Conclusion Assumptions Results Conclusion 1 No significant slip on the fossil fault planes of the Mendocino and Pioneer FZs. 2 The lithosphere bends in the vicinity of a FZ as a consequence of the frozen-in scarp and the differential subsidence of the lithosphere far from the FZ. 3 Good fits can be obtained by modeling this flexure as a thin elastic plate with age dependent effective elastic thickness. 4 We can use this model to calculate T e, the isotherm which defines the base of the elastic lithosphere. ( T = 450C) 5 Maximum bending stress ( 100 MPa) much less than stresses encountered in subduction zones.
Conclusion Assumptions Results Conclusion 1 No significant slip on the fossil fault planes of the Mendocino and Pioneer FZs. 2 The lithosphere bends in the vicinity of a FZ as a consequence of the frozen-in scarp and the differential subsidence of the lithosphere far from the FZ. 3 Good fits can be obtained by modeling this flexure as a thin elastic plate with age dependent effective elastic thickness. 4 We can use this model to calculate T e, the isotherm which defines the base of the elastic lithosphere. ( T = 450C) 5 Maximum bending stress ( 100 MPa) much less than stresses encountered in subduction zones. 6 In the case of multiple fracture zones, there is elastic coupling when two adjacent FZs are spaced less than a flexural wavelength apart. This is true for the Mendocino-Pioneer FZ pair.
Questions? Assumptions Results Conclusion