Download CHAPTER 3 ELASTICITY AND FLEXURE

Elastic Bending of Tectonic Plates
Two-Dimensional Bending of Flexure of Plates
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Some loads cause the lithosphere to bend.
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Flexure can be used to deduce elastic properties and thicknesses of plates.
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A simple example of plate bending is cylindrical bending from heavy loads in which
the width of the plate >> length.
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The deflection can be determined by requiring it to be in equilibrium under the
action of all the forces and torques exerted on it.
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Forces:
Q (downward force per unit area)
V (net shear force per unit length)
P (horizontal force per unit length)
Because the net acceleration is zero, the net vertical force is zero. Hence:
q(x)dx + dV =0 or q = -dV/dx
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The net bending moment, M per unit length, is the integrated effect of moments
exerted by normal forces on the cross-section (see figure next slide).
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The moments M and M+dM give a net counterclockwise torque dM while V (shear)
forces exert a torque Vdx in a clockwise sense. The P forces exert a counter
clockwise torque (-Pdw).
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The net torque is zero so dM –Pdw = Vdx (w is deflection of the plate)
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Differentiating twice with respect to x, one obtains the following relationship:
d2M/dx2 = -q + P* d2w/dx2
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M is inversely proportional to the local radius of curvature R (1/R = -d2w/dx2)
-As the plate is deflected downward, the top part is contracted (σxx is positive) and the
lower part is extended (σxx is negative). σxx is zero on the midplane at y=0 (neutral
unstrained surface)
-The bending moment is M= ∫(-h/2 to h/2)σxxy dy where h is the thickness of the plate.
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The change in length (Δ l) = -y(l/R) where y is the distance from the midplane and R s
the radius of curvature.
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R Φ = dx, hence the subtending angle Φ = -d2w/dx2 * dx
1/R = Φ/l = Φ/dx = - d2w/dx2 = D
M= D/R (flexural rigidity/ radius of curvature)
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So… the general equation for the deflection of the plate is
D d4w/dx4 = q(x) –P d2w/dx2
Bending of Plates under Applied Moments and Vertical
Loads
•Let us consider a plate embedded on only one end.
•If we assume the plate is weightless, q=0, V=0 (no applied force), P=0.
•Therefore, M is constant and equal to the applied torque.
•If w=0 at x=0 and dw/dx =0 at x=0 we find that
W=-Mqx2/2D (the plate has the shape of a parabola)
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The bending of a plate embedded at one end with uniform loading:
Q= constant, P=0
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D d4w/dx4 = q/D is the governing equation.
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With the four boundary conditions, w=dw/dx =0 at x=0 and d2w/dx2 =0 at x=L, one
obtains the following equation:
W=qx2/D (x2/24 –Lx/6 + L2/4)
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For this particular example, the maximum bending stress at x=0 is 3qL2/h2. The ratio
of shear stress to maximum bending stress at x=0 is h/3L, which tells us that shear
stresses in plates are small when compared to bending stresses.
Buckling of a Plate under a Horizontal Load
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Let us consider a plate pinned at both ends with an applied horizontal force, P.
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The governing equation is
D d4w/dx4 + P d2w/dx2 =0
Integrate twice:
D d2w/dx2 + Pw = c1x + c2
Buckling of a Plate under a Horizontal Load, con’t
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Solution of the differential equation gives the general solution of w=c1 sin (P/D)1/2 x
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If c is not zero, sin (P/D)1/2L =0 so (P/D)1/2L = nπ, n= 1, 2, 3…
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This tells us that P=(n2π2D)/L2, a series of values of which the smallest is Pc =
(π2)D/L2
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This makes a half sine curve where w=c1 sin πx/L. c1, the amplitude, can not be
determined by linear analysis.
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Plate flexure theory in mountain belts is much more complicated due to layering.
Deformation of Strata Overlying an Igneous Intrusion
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A laccolith is a sill-like igneous intrusion with a round lense-like shape that forms
when magma intrudes in layered rocks and magma raises and deforms the
overburden.
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The loading of the plate is the part of the upward pressure force in excess of the
lithostatic pressure.
Q= -p + ρgh
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If we consider x=0 to be the center of the laccolith, x= +/- L/2
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With some algebra we obtain…
W=- (p-ρgh)/24D (x4– L2x2/2 + L4/16)
where the maximum deflection at x=0 is
Wo = -(p – ρgh)L4/384D.
Application to the Earth’s Lithosphere
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We must remember to include the hydrostatic restoring force. For example, when
oceanic lithosphere is depressed, water fills in space vacated by the mantle’s moved
rocks. Continental crust is thickened by the amount that the Moho is depressed.
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If water fills, the weight per unit area of a vertical column is ρwg(hw + w) + ρmgh
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Deeper, where there is no plate deflection in the mantle, the weight per unit area is
ρwghw + ρmg(h+w)
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so there is an upward hydrostatic force of (ρm –ρw)gw which tries to restore the
lithosphere to its original configuration
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When the net force per unit area is q=qa- (ρm–ρw)gw, where qa is the applied load, the
governing equation becomes:
Dd4w/dx4 +Pd2w/dx2 + (ρm-ρw)gw = qa(x).
Application to the Earth’s Lithosphere
Periodic Loading (to answer the question of how
mountains and valleys affect the lithosphere)
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The elevation of the Earth’s topography can be approximated by using the following formula :
h=ho sin 2 π x/λ
where h is the topographic height and λ is the wavelength.
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The load is qa(x) = ρcgho sin 2π x/λ, so the governing equation is:
Dd4w/dx4 + (ρm – ρc)gw = ρcgho sin 2π x/λ
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Because the loading is periodic, the deflection is also:
Wo= ho/ ((ρm/ρc) -1 + D/ρg (2π/λ)4)
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If the wavelength is short (λ<< 2π (D/ρcg) then
(1/4wo << ho )
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Thus, short wavelength topography causes virtually no deformation of the lithosphere.
Application to the Earth’s Lithosphere
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If the wavelength is long (λ >> 2 π (D/ρcg)),
1/4w = ρcho/(ρm-ρc)
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For long wavelength topography, the
lithosphere has no rigidity, and the
topography is in hydrostatic equilibrium.
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The degree of compensation, C, of the
topographic load is the ratio of the deflection
of the lithosphere to its maximum or
hydrostatic deflection.
Stability of the Earth’s Lithosphere under an End Load
(Buckling)
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When the lithosphere is subjected to a horizontal force, P, the governing equation is:
Dd4w/dx4 + Pd2w/dx2 + (ρm– ρw) gw = 0.
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This equation can be satisfied by a sinusoidal deflection, giving:
D(2π/λ)4 –P(2π/λ)2+ (ρm-ρw)g =0
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The solution provides, Pc, the minimum value of P for which the initially horizontal
lithosphere will become unstable and accept the sinusoidal shape.
Pc = 4Dg(ρm – ρw) ½
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The critical wavelength is
λc = 2π (2D/ρc)1/2= 2π (Eh3/(12(1-v2)( ρm-ρw)g))1/4
Stability of the Earth’s Lithosphere under an End Load (Buckling)
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Now, to determine whether buckling of the lithosphere can lead to the formation of a
series of synclines and anticlines:
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Let us consider an elastic lithosphere with a thickness of 50 km. Taking E=100 GPa,
v= .25, ρm = 3300 kg/m3, and ρw = 1000 kg/m3, we find that the critical stress is 6.4
GPa. Therefore the lithosphere can support a horizontal compression of 6.4 GPa
without buckling. Because the stress is very large, one can conclude that such
buckling does not occur.
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However, the buckling of thin elastic layers may contribute to the formation of folded
structures.
Bending of the Elastic Lithosphere under the Loads of
Island Chains
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Volcanic islands provide loads that cause the lithosphere to bend. An example
is Hawaii, which has very deep seas surrounding the island (Hawaiian Deep).
Bending of the Elastic Lithosphere under the Loads of Island Chains,con’t
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The governing equation is:
Dd4w/dx4 + (ρm –ρw)gw = 0
whose solution is:
W= e(x/α) (c1 cos (x/α) + c2 sin (x/α)) + e(-x/α)(c3 cos (x/α) + c4 sin (x/α))
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α is the flexural parameter and is equal to (4D/ ((ρm –ρw)g)¼
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The above equation becomes
w=c3 e(-x/α)(cos (x/α) + sin (w/α)),
where x is positive and c3 is Voα3/8D, the maximum amplitude of deflection.
Bending of the Elastic Lithosphere under the Loads of Island Chains,con’t
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In the case of Hawaii, where volcanism may have weakened the lithosphere, we
might also consider a case where the lithosphere is fractured along the line of the
ridge.
For the same load as the first case, the fractured model has twice the deflection.
Bending of the Elastic Lithosphere at an Ocean Trench
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Prior to subduction, considerable bending of the lithosphere occurs. To model this, we
consider an elastic plate acted upon by an end load Vo, and a bending moment Mo.
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Solution of the governing equation gives:
w= e(-x/α) [c3 cos (x/α) + c4 sin (x/α)]
where
c4 = -Mo α2/2D and c3 = (Voα + Mo) α2/2D
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Vo and Mo can not be determined directly, but the half-width of the forebulge is π/4 α, from
which we can derive very complicated formulas to use.
Bending of the Elastic Lithosphere at an Ocean Trench, con’t
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The plot of (w/wb) vs. (x-xo)/(xb/xo) defines a universal flexure profile which is valid
for any two-dimensional elastic flexure of the lithosphere under end loading.
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A comparison of the universal flexure profile and subduction at the Marianas is
excellent.
Flexure and the Structure of Sedimentary Basins
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Lithospheric flexure is also associated with the
structure of many sedimentary basins. Sedimentary
basins are areas of the Earth’s surface that have
been depressed and the depressions filled with
sediments. They are often important reservoirs of
petroleum.
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The loading of mountain belts can be used to
explain some foreland basins, as in the
Appalachian and Andes Mountains.