Elasticity and Flexure
Linear Elasticity - stresses are linearly proportional to strains. Strain is recoverable.
Lame parameters - material properties (proportionality constants) which relate stress to strain.
Young's modulus and Poisson's ratio - related to Lame parameters. Young's modulus determines strain in the same direction as stress. Poisson's ratio determines strain in orthogonal directions to stress. Young's modulus for rocks is about 10 to 100 GPa. Poisson's ratio is 0.1 to 0.4. Poisson's ratio for an incompressible liquid is 0.5.
Uniaxial Stress - only one of the principal stresses is nonzero.
An example is Hooke's law where stress is equal to Young's modulus times strain. Orthogonal strains are equal to minus Poisson's ratio times the strain parallel to the nonzero stress (maximum principal strain).
If the maximum principal strain is a decrease in length than the other principal strains are increases in length.
Uniaxial Strain - only one of the principal strains is nonzero (works for erosion or deposition where adjacent material prevents strains in the horizontal directions).
For overburden, the vertical stress is sediment density times thickness of sediment times gravitation acceleration.
Horizontal normal stresses are equal to the weight of overburden times Poisson's ratio divided by 1 minus Poisson's ratio.
vertical stress = (1-v)E/((1+v)(1-2v)) vertical strain
Plane Stress - one of the principal stresses is zero. Usual used to study horizontal tectonic stresses assuming that changes in vertical stress are essentially zero.
Strain parallel to the zero stress direction is minus Poisson's ratio divided by Young's modulus times the sum of the nonzero principal stresses.
Strains parallel to the nonzero principal stress directions are equal to the parallel principal stress minus Poisson's ratio times the orthogonal nonzero principal stress all divided by Young's modulus.
Plane Strain - one of the principal strains is zero. Stress parallel to the zero principal strain is equal to Poisson' ratio times the sum of the other two principal stresses.
Pure Shear and Simple Shear
Shear stress = E/(1+v) shear strain or shear stress = 2G shear strain.
Pure shear and simple shear differ by a solid body rotation that does not effect the state of stress.
Isotropic Stress - all 3 principal stresses are equal
Pressure is equal to bulk modulus times dilatation. Bulk modulus (K) is the inverse of compressibility.
Change in density is equal to density times pressure times compressibility.
K = E/(3(1-2v))
Two-Dimensional Bending or Flexure of Plates
The deflection of a plate can be determined by requiring it to be in equilibrium under the action of all the forces and torques exerted on it.
Force Balance - dV/dx = -q
Where V is shear force and q is applied vertical force
Torque Balance - d2M/dx2 = -q + Pd2w/dx2
Where M is bending moment related to fiber stresses, P is horizontal force, and w is deflection
Bending moment is equal to the integral of fiber stress times distance from the neutral plane from the bottom to the top of the plate.
In 2D bending of a thin plate, the vertical stress is zero and the strain in the neglected dimension is also zero. Thus, fiber stress is equal to E/(1-v2) times the fiber strain.
Fiber strain is related to curvature of the plate times distance from the neutral plane. Plate curvature is equal to d2w/dx2 or
Fiber Stress = E/(1-v2) Fiber Strain = - E/(1-v2) y d2w/dx2
Thus, the bending moment, M, is
M = -Eh3/(12(1-v2) d2w/dx2 or M = -D d2w/dx2
Where D is flexural rigidity.
The torque balance now becomes
D d4w/dx4 = q - Pd2w/dx2
Buckling of a Plate under a Horizontal Load
Assume that plate is pinned at both ends and no vertical forces
D d4w/dx4 + Pd2w/dx2 = 0
Solution by integrating and using pinned plate boundary conditions is
W = c1 sin(3.1459x/L)
where L is the length of the plate.
The smallest horizontal load to buckle the plate is
Pc = D(3.14159)2/L2
Bucking of the plate also depends on the layers above and below (even if they are fluid). The flexural rigidity (D) for the lithosphere is large, thus horizontal forces (10-100 MPa) are not large enough to buckle it.
Application to the Earthís Lithosphere
As the lithosphere bends, it pushes asthenosphere out of the way. Thus, there is an upward hydrostatic restoring force, which depends on the density difference between the asthenosphere and the material filling the depression (water, sediments or rocks) and the amount of deflection (w).
D d4w/dx4 + Pd2w/dx2 + (rm-rw)gw = qa(x)
Periodic Loading
Deflection of the lithosphere to a periodic load (e.g., topography) will vary spatially with the same wavelength as the topography.
wo = ho/(rm/rc ó 1 + (D/rcg)(2*3.14159/L)4
where wo is the maximum deflection, ho is the maximum topography, and L is the wavelength of the topography.
Wohydrostatic = rcho/(rm ó rc)
Degree of compensation, C, is the ratio of deflection to hydrostatic deflection. Degree of compensation depends on the wavelength of the load.
Bending of the Lithosphere under the Loads of Island Chains
D d4w/dx4 + (rm-rw)gw = 0
where w goes to zero at infinity and dw/dx = 0 at x = 0
w = wo exp(-x/a) (cos(x/a) + sin(x/a))
Where wo = Voa3/(8D) and Vo is the island load
a is the flexural parameter
a4 = 4D/(rm-rw)g
half-width of depression, xo, is a*(3*(3.14159)/4)
The peripheral bulge is
xb = 3.14159 a
wb = -0.0432wo
For a broken plate, d2w/dx2 = 0 at x = 0. For a given D and applied load the deflection is deeper and narrower.
w = wo exp(-x/a) cos(x/a)
wo = Voa3/(4D)
xo, is a*(3.14159)/2
xb = a*(3*(3.14159)/4)
wb = -0.0670wo