fracture zone flexure - university of california, san diego

OutlineBackgroundDerivations

Calculations and PlotsConclusions

Fracture Zone Flexure

By: Xiaohua Xu& Zhao Chen

December 12, 2012

By: Xiaohua Xu & Zhao Chen Fracture Zone Flexure



Outline

1 Background

2 DerivationsThe Dominating EquationsParameters in this ProblemSolution to the Equations

3 Calculations and PlotsPlots to actually dataBending Stress and Shear Stress

4 Conclusions




Background

Observation

1. Constant scarp heights.2. A characteristic ridge-trough topographic FZ signature.3. Flexural amplitude increasing with age.

Model

1. Differential subsidence of lithosphere far from the FZ.2. Flexure of a thin elastic plate.

Age-dependent effective elastic thickness.

Frozen-in scarp.




Sketches of the Model




The Dominating EquationsParameters in this ProblemSolution to the Equations

Outline

1 Background



4 Conclusions





The Dominating Equations

D1d4w1

dx4+ g (ρm − ρw )w1 = 0 x < 0

D2d4w2

dx4+ g (ρm − ρw )w2 = 0 x > 0





Boundary Conditions

The boundary condition requires that w vanishes as x → ±∞.Other boundary conditions are derived out by the displacement being δand the continuity of the slope, moment and shear force:

w1 − w2 = δ

dw1

dx− dw2

dx= 0

−D1d2w1

dx2+ D2

d2w2

dx2= 0

−D1d3w1

dx3+ D2

d3w2

dx3= 0





Outline

1 Background



4 Conclusions





Parameters in this Problem

The flexural rigidity is:

D =Ehe

3

12 (1− ν2)

Where he is the effective elastic thickness related to age t by:

he = 2(κt)1/2erfc−1

(Tm − Te

Tm − Ts

)





Parameters in this Problem

δ = hA−hB =2αρm (Tm − Ts)

(ρm − ρw )

(κπ

)1/2·{

(tB′ − tB)1/2 + tB1/2 − tB′

1/2}





Outline

1 Background



4 Conclusions





Methods for Solving these Equations

Let’s assume that the solution is wi = er , thus we have

Di r4 + g (ρm − ρw ) = 0

It is not hard to find that:

rij =

(g (ρm − ρw )

Di

)1/4

e(2j−1)π

4 i =2π√

2

λie

(2j−1)π4 i , j = 1, 2, 3, 4

Then, by superposition:

wi (x) =4∑

j=1

erijx





Form of the solution to the Equations

The form of the solutions to these ODEs is:

wi (x) = e−2πx/λi

{Ai1 sin

2πx

λi+ Ai2 cos

2πx

λi

}

+e2πx/λi

{Ai3 sin

2πx

λi+ Ai4 cos

2πx

λi

}where the flexure wavelength λ is given by:

λi = 2π

(4Di

g (ρm − ρw )

)1/4





Determine the Coefficients

From the BC that w vanishes at infinity, it is easy to know thatA11 = A12 = A23 = A24 = 0.For BCs at x = 0, we first need to derive out that:

dw1

dx=

2π

λ1

e2πx/λ1

{A13 sin

2πx

λ1

+ A14 cos2πx

λ1

}+

2π

λ1

e2πx/λ1

{A13 cos

2πx

λ1

− A14 sin2πx

λ1

}

dw2

dx= −

2π

λ2

e−2πx/λ2

{A21 sin

2πx

λ2

+ A22 cos2πx

λ2

}+

2π

λ2

e−2πx/λ2

{A21 cos

2πx

λ2

− A22 sin2πx

λ2

}

d2w1

dx2= 2

(2π

λ1

)2e2πx

/λ1

{A13 cos

2πx

λ1

− A14 sin2πx

λ1

}

d2w2

dx2= −2

(2π

λ2

)2e−2πx

/λ2

{A21 cos

2πx

λ2

− A22 sin2πx

λ2

}

d3w1

dx3= 2

(2π

λ1

)3e2πx

/λ1

{A13 cos

2πx

λ1

− A14 sin2πx

λ1

}− 2

(2π

λ1

)3e2πx

/λ1

{A13 sin

2πx

λ1

+ A14 cos2πx

λ1

}

d3w2

dx3= 2

(2π

λ2

)3e−2πx

/λ2

{A21 cos

2πx

λ2

− A22 sin2πx

λ2

}+ 2

(2π

λ2

)3e−2πx

/λ2

{A21 sin

2πx

λ2

+ A22 cos2πx

λ2

}






Take them into the BCs, we can get the equations:

A14 − A22 = δ1

1

λ1(A14 + A13) +

1

λ2(−A22 + A21) = 0

−D11

λ12A13 − D2

1

λ22A21 = 0

−D11

λ13 (A13 − A14) + D2

1

λ23 (A21 + A22) = 0






Put these equations into Matlab and use the ”SOLVE” function and thenuse ”LATEX” function to get its LATEX form, we get:

A13 = −D1 δ λ2

2(−D2 λ1

2 + D1 λ22)

D12λ2

4 + 2D1 D2 λ13λ2 + 2D1 D2 λ1

2λ22 + 2D1 D2 λ1 λ2

3 + D22λ1

4

A14 = −δ(D1

2λ24 + D1 D2 λ1

2λ22 + 2D1 D2 λ1 λ2

3)

D1 2λ24 + 2D1 D2 λ1

3λ2 + 2D1 D2 λ12λ2

2 + 2D1 D2 λ1 λ23 + D2

2λ14

A21 =D2 δ λ1

2(−D2 λ1

2 + D1 λ22)

D12λ2

4 + 2D1 D2 λ13λ2 + 2D1 D2 λ1

2λ22 + 2D1 D2 λ1 λ2

3 + D22λ1

4

A22 =D2 δ λ1

2(D2 λ1

2 + 2D1 λ1 λ2 + D1 λ22)

D12λ2

4 + 2D1 D2 λ13λ2 + 2D1 D2 λ1

2λ22 + 2D1 D2 λ1 λ2

3 + D22λ1

4




Plots to actually dataBending Stress and Shear Stress

Outline

1 Background



4 Conclusions





The parameters occurred in the derivation





Some derivations

From the equation

d(t) = dref +2αρm(Tm − Ts)

(ρm − ρw )(κt

π)1/2 + W

we plot as following:





Reality Data





t = 5Myr ,∆t = 28Myr , Model V.S. Reality





t = 0− 100Myr ,∆t = 28Myr , Model





Outline

1 Background



4 Conclusions





Derivation of the Stress

The bending stress is defined as:

σxx(x) =−Ehe

2(1− µ2)

d2w(x)

dx2=

6M(x)

h2e

where M(x) is the bending moment:

M(x) = −D d2w(x)

dx2

Also, the average shear stress is the shear force divided by he :

σxz(x) =−Dhe

d3w(x)

dx3

Note that both D and he is different on both sides.





t = 5Myr ,∆t = 28Myr , Bending Stress & Shear Stress





t = 0− 100Myr ,∆t = 28Myr , Bending Stress





t = 0− 100Myr ,∆t = 28Myr , Shear Stress




Conclusions

Conclusions

This model of FZ flexure due to frozen-in scarp and differentialsubsidence of lithosphere explains the ridge-trough topo signature inthe vicinity of the FZ. (Ex. Mendocino and Pioneer FZs)

Heat conduction has a small but not negligible effect on the flexure.∆T → Pressure → Flexure

Fracture zones are seismically inactive and does not slip significantlyover a long period of time(130Myr).

Flexural amplitude increases with age, but gains most of itsamplitude at younger age.

Results of the modeling found that the base of the elasticlithosphere is approximation defined by 450 degrees.




About Temperature-Flexure effect




References

Sandwell D. and Schubert G., Lithosphere Flexure at FractureZones, J. Geophys. Res., 87, 4657-4667, 1982.

Turcotte D. and Schubert G., Geodynamics 2nd Edition, CambridgeUniversity Press 2002

Sandwell D., Notes of SIO234, ????, 2012




Thanks




Questions


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