8. Plate Flexure

Ge 163 4/30/14-

Outline

• Motivation • 2-D flexural equation for a thin elastic plate • Flexure of a plate subject to an end load • “Isostatic” restoring force • Loading by a ‘line source’ (i.e. seamount chain) • Detail observations of Hawaiian chain • Flexure and compensation in the frequency domain • Elastic thickness versus plate age • Bending of a plate with plastic failure • Details of oceanic trenches

• Appendix A: Derivation of the 2-D flexural equation for a thin elastic plate • Appendix B: More details on a bending plate with plastic failure

Free-Air Gravity

Two-dimensional flexure of an elastic plate (section 3-9 in Turcotte & Schubert [2002]) See Appendix in this slide pack for details

Plates are subject to variety of loads -- such as volcanoes (seamounts) that force the lithosphere to bend under their own weight. By relating observed flexure with known loads

elastic properties thickness of plates

Will require that the plate be in equilibrium to all forces and torques

Loaded by Va. Static force balance requires that the ends have an upward for -Va/2.

Assume thin plate h LAssume deflections are small w L (necessary when we use linear elasticity)

Look at Forces and Torques exerted on a small segment from x to x + dxDownward load, per unit length in the z-dir, between x and x + dxq(x)dxV is the net shear force (per unit z-dir)P is the net horiontal force (per unit z-dir)

Summary for deriving equation governing flexure of a 2-D plate

1. Vertical force balance (i.e. dVdx

= −q )

2. Torque balance (i.e. dM − Pdw = Vdx )3. ⇒ 2nd diff. eq. for Moment, M

d2Mdx2 = −q + P

d 2wdx2

4. M ← εxx ← d2wdx2

but M = σ xxydy−h /2

h /2

∫5. 4th order diff.eq. for w , displacement

D d 4wdx4 = q(x) − P d

2wdx2

�

Dd4wdx 4 = q(x) − P d

2wdx 2

D ≡Eh3

12(1−ν 2)D is the flexural rigidity with units N - mE is Young's modulus andν is Poisson's ratio

Bending of a plate under a load

Concentrated load, Va, at x=L

Plate embedded on left → w = 0 x = 0

dwdx

= 0 x = 0

no Torque at x = L d2wdx2 = 0 x = L

Weightless plate q = 0, except at x = L

Using the equation dVdx

= −q = 0 →V = constant = Va

Dd 4wdx4 = q(x) − P d

2wdx2

with ( q = P = 0) it canbe shown that:

w =Vax

2

2DL −

x3

⎛⎝⎜

⎞⎠⎟

measure w(x), if we know Vainfer D

Modification needed for hydrostatic restoring force

The net force acting on the lithospheric plate isq = qa − (ρm − ρw )gwρw would alternatively be ρc is the material fillingthe gap was crustal density∴

Dd 4wdx4 + P

d 2wdx2 + (ρm − ρ)gw = qa (x)

Some dimensional values for the rigidity

D =Eh3

12(1−ν 2 )E = 70 GPah = 100 kmν = 0.25D = 6 ×1024 N-m

Bending Under a Line Load

Figure 3-29

Assume no horizontal force, P = 0Since load only exists at a point (line), q = 0

⇒ Dd 4wdx4 + (ρm − ρw )gw = 0

general solution

w = ex /α c1 cos xα+ c2 sin x

α⎛⎝⎜

⎞⎠⎟+ e− x /α c3 cos x

α+ c4 sin x

α⎛⎝⎜

⎞⎠⎟

c1,c2 ,c3,c4 are determined from BC's

α =4D

(ρm − ρw )g⎡

⎣⎢

⎤

⎦⎥

1/4

α = "flexural parameter", dimensions of lengthw→ 0 as x→∞ ⇒ c1 = c2 = 0

symmetry at x = 0 ∴ dwdx

= 0 ⇒ c3 = c4

w = c3e− x /α cos x

α+ sin x

α⎛⎝⎜

⎞⎠⎟

x ≥ 0

It can be shown for a line load

Dd 3wdx3 = −

Vo2

(the 1/2 because half the load is supported

by half the plate)

→Vo =8Dc3

α 3

w =Voα

3

8De− x /α cos x

α+ sin x

α⎛⎝⎜

⎞⎠⎟

Alternatively, the maximum deflection is (at x = 0 )

wo =Voα

3

8D

w = woe− x /α cos x

α+ sin x

α⎛⎝⎜

⎞⎠⎟

Distance of the forebulge, xb , is determined where the slopeis zero xb = πα

For Hawiian chainxb ≅ 250 km→α = 80 kmρm − ρw = 2300 kg m-3

g = 10 m s-2

α =4D

(ρm − ρw )g⎡

⎣⎢

⎤

⎦⎥

1/4

→ D = 2.4 ×1023 N-m

D =Eh3

12(1−ν 2 )E = 70 GPaν = 0.25⇒ h = 34 km(Assuming a fractured plate, one gets a thicker plate, ≈ 49 km)

Watts [2001]

Watts [2001] Sediments ~2 km/s

Watts [2001]

Watts [2001]

Watts [2001]

An alternative approach to understanding flexure and compensation. We just looked at problems in the spatial domain, now let’s now look effective compensation in the spectral domain. (Turcotte & Schubert, section 3-16)

Consider an elastic plate (2-D) which is periodically loaded with crustal thickness variations

h = ho sin2π xλ= ho sin kx (k is the wavenumber, k = 2π

λ )

Assume h is smaller than elastic plate thickness.

This topography gives rise to a load (i.e. "ρgh"). Assumingthat the load was created by a layer of crustal density, ρc )qa (x) = ρcgho sin kx

Using the 2-D flexure equation

Dd 4wdx4 + P

d 2wdx2 + (ρm − ρc )gw = qa (x)

with no horizontal forces:

Dd 4wdx4 + (ρm − ρc )gw = ρcgho sin kx

Clearly, the deflection caused by the topographic loadwill be a sine function in phase with the original load.Assumew = wo sin kxSubstituting into the governing equation and solving for wo

wo =ho

ρm

ρc−1+ D

ρcgk 4

Notice that Dρcg

⎛⎝⎜

⎞⎠⎟

1/4

has the dimensions of length

If λ 2π Dρcg

⎛⎝⎜

⎞⎠⎟

1/4

wo hoLarge k (short wavelength) causes virtually no deformationof the lithosphere

If λ 2π Dρcg

⎛⎝⎜

⎞⎠⎟

1/4

w =ρcho

(ρm − ρc )= wo∞

This is the isostatic limit.One can define a degree of compensation

C =wwo∞

=(ρm − ρc )

ρm − ρc +Dgk 4

An effective way of looking at these wavelength dependent compensation problems is through the gravity field.

There are two effects on the gravity:1. Through topography → Δgt2. Through deflection of an internal interface ("Moho") at a mean depth, bm → Δgm

For the first contribution to gravity (from topography)use the Bouguer gravity relation Δg = 2πρcGh

∴Δgt = 2πρcGho sin kx

The deflection of the Moho gives rise to a mass anomalyat depth

σ=(ρc − ρm )w =−(ρm − ρc )hoρm

ρc−1+ D

ρcgk 4⎡

⎣⎢

⎤

⎦⎥

Aside: It can be shown (recall our solution ofLaplace's equation in a spherical geomtry), that the gravity anomaly (2-D) at a depth y is:g = 2πGσ oe

−ky sin kx

With this we get:

Δgm =2πG(ρm − ρc )hoe

−kbm

ρm

ρc−1+ D

ρcgk 4⎡

⎣⎢

⎤

⎦⎥

sin kx

The free-air gravity anomaly is (summation of the twosources of gravity):Δgfa = Δgt + Δgm

Δgfa = 2πρcG 1− e−kbm

1+ D(ρm − ρc )

k 4

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

ho sin kx

If

λ 2π D(ρm − ρc )g⎡

⎣⎢

⎤

⎦⎥

1/4

The free-air gravity becomes:Δgfa = 2πGρcho sin kx Notice that this is just the Bouguercorrection, so that the Bouguer anomly vanishes:ΔgB = 0[The mass of the local topography is uncompensated and sothe Bouguer gravity anonaly is zero].

However, for long wavelength topography:

λ 2π D(ρm − ρc )g⎡

⎣⎢

⎤

⎦⎥

1/4

and λ bm

The free-air gravity anomaly becomes:Δgfa = 0and the Bouguer gravity anomly becomes:ΔgB = −2πρcGho sin kx[The surface topgraphy is totally compensated].

One can look at the 'transfer function', Z, or 'admittance' betweentopography gtravity:

Z(k) = Δg(k)h(k)

Z(k) = 2πρcG 1− e−kbm

1+ D(ρm − ρc )

k 4

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

Watts [2001]

Watts [2001]

Major Conclusion, elastic thickness (h, Te) increases with age

Watts [2001]

Oceanic Trenches

Bending of the Elastic Lithosphere at an Ocean Trench

Again we start with the 2-D flexural equation

Dd 4wdx4 + (ρm − ρw )gw = 0

Use general solution, but consider thatw→ 0 as x→∞:

w = e− x /α c3 cos xα+ c4 sin x

α⎛⎝⎜

⎞⎠⎟

at x = 0, bending moment is − Mo

since M = −Dd 2wdx2 ⇒ c4 =

−Moα2

2Dat x = 0 shear force is −Vo

since dMdx

= V & M = −Dd 2wdx2 ⇒ c3 = (Voα + Mo ) α

2

2D\

∴w =α 2e− x /α

2D−Mo sin x

α+ (Voα + Mo )cos x

α⎧⎨⎩

⎫⎬⎭

If Mo = 0, then reduces to problem of a fractured platewith end load

�

α ∝ D1/ 4

Not very useful since neither Vo and Mo can bedetermined directly.

Quantities which can be measured are the half-widthof the forebulge xb − xo and the height of the forebulge, wb

It can be shown that xb − xo =π4α

For the Mariana:xb − xo = 55 kmα = 70 kmD = 1.4 ×1023 N-m ⇒ h = 28 km

Watts [2001]

Watts [2001]

Mofjeld et al., 2004

-When abyssal hills oriented >30° from trench axis new large amplitude bending induced faults generated - When < 30° fabric re-activiated

Tensional outer rise events: Open Symbols If follows (<30 year) large under-thrusting event: Solid square Compressional: Solid

Christensen & Ruff [1988]

A model for the bending of the lithosphere at oceanic trenches in which the elastic plate undergoes plastic failure.

At low confining pressures rocks are brittle (i.e. they fracture under large stress). When confining pressure approaches a rock’s brittle strength, a transition to brittle to plastic behavior occurs. Also, yield stress decreases with increasing temperature.

σo is the yield stress

An idealization of this of this behavior is the elastic-perfectly plastic rheology. σ=Eε along AB

Elastic ?

Partly plastic?

Billen & Gurnis [2005]

Billen & Gurnis [2005]

Billen & Gurnis [2005]

Billen & Gurnis [2005]

A dynamic Viscous model of a subduction zone

Zhong & Gurnis [1994]

Zhong & Gurnis [1994]

Zhong & Gurnis [1994]

Appendix A

•  Derivation of the 2-D flexural equation for a thin elastic plate

Summary for deriving equation governing flexure of a 2-D plate

1. Vertical force balance (i.e. dVdx

= −q )

2. Torque balance (i.e. dM − Pdw = Vdx )3. ⇒ 2nd diff. eq. for Moment, M

d2Mdx2 = −q + P

d 2wdx2

4. M ← εxx ← d2wdx2

but M = σ xxydy−h /2

h /2

∫5. 4th order diff.eq. for w , displacement

D d 4wdx4 = q(x) − P d

2wdx2

Assume thin plate h LAssume deflections are small w L (necessary when we use linear elasticity)

Look at Forces and Torques exerted on a small segment from x to x + dxDownward load, per unit length in the z-dir, between x and x + dxq(x)dxV is the net shear force (per unit z-dir)P is the net horiontal force (per unit z-dir)

Bending moment M acting on the cross-section of the plate is the integrated effect of the moments exerted by thenormal stress σ xx (σ xx also known as fiber stresses)

A force balance in the vertical direction between x and x + dxq(x)dx + dV = 0

→ dVdx

= −q

M and M + dM combine to give net counter-clockwise Torque dMV and V+dV are separated by an infiniesimal moment arm, dx → net Torque VdxP (constant with x ) (moment arm is − dw in going from x to dx) → − Pdw

Balance all torques:

dM − Pdw = Vdx (2 terms on left are counter-clockwise while right side is clockwise)

dMdx

= V + Pdwdx

eliminate shear force by ddx

( )d 2Mdx2 =

dVdx

+ Pd 2wdx2

→

d 2Mdx2 = −q + P

d 2wdx2

If we can relate M to w, then we can put the previousdiff. eq. in terms of w

The force on an element of tickness dy is σ xxdy;this force exerts a torque σ xxydy about the mid-plane:

M = σ xxydy−h /2

h /2

∫

Strains in plate σ xx accompanied by ε xx bending is 2-D , ε zz = 0 Zero stress normal to the plate, σ yy = 0 at surface then plate →σ yy throughout

⇒ Example of plane stress (i.e. only on zero component of principal stress)

εxx =1E

σ xx −νσ zz( )

εzz =1E

σ zz −νσ xx( ) (E is Young's modulus, ν is Poisson's ratio)

if ε zz = 0

→σ xx =E

(1−ν 2 )ε xx

∴M = σ xxydy−h /2

h /2

∫ =E

(1−ν 2 )ε xxy-h/2

h/2

∫ dy

ε xx depends on -distance from the mid-plane, y -local radius of curvature, R

assume (infinitesimal)and mid-plane remains at , then

Δ = −yφ = −yR

minus sign ⇒ contraction when y is positive

ε xx = −Δ

=yR

it can be shown that1R≈ −

d 2wdx2

∴ε xx = −yd 2wdx2

M =E

(1−ν 2 )ε xxydy−h /2

h /2

∫

=−E

(1−ν 2 )d 2wdx2 y2dy

−h /2

h /2

∫

=−E

(1−ν 2 )d 2wdx2

y3

3⎡

⎣⎢

⎤

⎦⎥−h /2

h /2

=−Eh3

12(1−ν 2 )d 2wdx2

D ≡Eh3

12(1−ν 2 ) "Flexural Rigidity" of the plate

∴M = −Dd 2wdx2

Recall from our vertical force balance combined with the Torque balance:d 2Mdx2 = −q + P

d 2wdx2

with M = −Dd 2wdw2

⇒

Dd 4wdx4 = q(x) − P d

2wdx2

Appendix B

•  More details on a bending plate with plastic failure

A model for the bending of the lithosphere at oceanic trenches in which the elastic plate undergoes plastic failure.

At low confining pressures rocks are brittle (i.e. they fracture under large stress). When confining pressure approaches a rock’s brittle strength, a transition to brittle to plastic behavior occurs. Also, yield stress decreases with increasing temperature.

σo is the yield stress

An idealization of this of this behavior is the elastic-perfectly plastic rheology. σ=Eε along AB

For uniaxial loading, yield condition is σ = σ o

For 3-D, more complicated. The Tresca (maximum shear stress criterion)states that the solid yields when the shear stress reaches a critical value, σ *

In 3-D, maximum shear stress is 12

σ1 − σ 3( ),where σ1 and σ 3 are the maximum and minimum principal stresses.Thus, the Tresca yield condition is:

σ * =12

σ1 − σ 3( )But this must reduce to the yield condition for uniaxial stress when

σ 2 = σ 3 = 0 so σ * =12σ1 =

12σ o

∴ σ o = σ1 − σ 3

Now let us return to bending of the purely elastic plate, inwhich we showed

σ xx = −Ey

1−ν 2

d 2wdx2

The principle stress in the plate are σ xx , σ yy , and σ zz . For our 2-D bending of a plate, σ yy = 0 and ε zz = 0, giving σ zz = νσ xx . Thusσ1 = σ xx σ 2 = σ zz = νσ xx σ1 = σ yy = 0

Consequently for the Tresca criterion we getσ xx,o = σ o

Now, the plate curvature corresponding to the onset of plasticity is:

d 2wdx2 = −

2σ xx,o(1−ν2 )

Eh

If (already derived): M = −Eh3

12(1−ν 2 )d 2wdx2 then

Mo =σ xx,oh

2

6 . If bending moment exceeds this, plastic failure occurs

When the plate is entirely plastic, there is a maximum (orcritical bending moment), Mc . Turcotte & Schubertshow that this is

Mc =σ xx,oh

2

4and that at the onset of plastic failure, that

Mo =23Mc

If the plate curvature at the onset of plastic failure

is d 2wdx2

⎛⎝⎜

⎞⎠⎟ o

, then it can be shown that:

d 2wdx2

d 2wdx2

⎛⎝⎜

⎞⎠⎟ o

= 3− 2MMo

⎛⎝⎜

⎞⎠⎟

−1/2

The curvature approaches ∞ as M → Mc = 1.5Mo

Elastic ?

Partly plastic?