Geodynamics - Plates 3

Lecture Notes 20 Plates (continued) Response to a line load (e.g. volcanic chain under water) Va �w �c h �m { Figure 20.1 D ⎧0 at x ≠ 0 d 4w + (ρ m − ρ w )gw = ⎨ 4 dx ⎩V0 at x = 0 Solution to homogeneous equation w = e x /α (C1 cos x α x x x + C2 sin ) + e−x /α (C3 cos + C4 sin ) α ⎤ ⎡ 4D with α = ⎢ ⎥ ⎣ (ρ m − ρ w )g ⎦ α α 1/ 4 α is flexural parameter. Invoke symmetry, boundedness. Determine solution only for x ≥ 0 . dw = 0 at x = 0 dx C3 = C4 w→0 C1 = C2 = 0 x →∞ x w = C3e−x /α (cos x α x + sin ) α Now to evaluate C3 , go back to the end load problem (or original definition) dM dw =V +P dx dx C3 depends on V0 . d 3w 1V =− 3 2D dx 1 d 3w V0 = D 3 2 dx = x =0 4DC3 α3 V0 is negative load, half supported by each side. w= V0α 3 − x /α x x e (cos + sin ) 8D α α V0α 3 w0 = 8D w = w0 e− x /α (cos x α x + sin ) α wb 0 x 0/ � w0 w Figure 20.2 x b/ � x/� x0 = tan−1 (−1) = xb = sin −1 (0) = π α α 3π 4 wb = −w0 e− π = −0.0432w0 x ; 3o ; 300km α= 314 π km ; 100km Suppose plate is fractured. V0/2 x=0 Figure 20.3 Same equation. Different boundary conditions ⇒ different profile. M = 0 at x = 0 d 2w =0 dx 2 w= V0α 3 − x /α x e cos α 4D w0 = V0α 3 4D wb 0 x 0/ � x b/ � x/� w0 w Figure 20.4 x0 = π 2 α, xb = 3π α, 4 wb = −0.067w0 h ; 35km unbroken h ; 50km broken Thickness of broken lithosphere is about 1.5 times thickness of the unbroken lithosphere. Bending at a trench. wb x0 -V0 xb -M0 Figure 20.5 w = e− x /α (C3 cos x α x + C4 sin ) α M 0α 2 C4 = − 2D C3 = (V0α + M 0 ) w= α2 2D (entireV0 supported) α 2 e− x / α ⎧ x x⎫ ⎨−M 0 sin + (V0α + M 0 )cos ⎬ 2D ⎩ α α⎭ Observables -- x0 , xb , wb tan x0 = 1 + αV0 / M 0 tan xb = −1 − 2 M 0 / αV0 α α xb − x0 = π 4 α x wb = α 2 e− x wb = − b /α 2D α2M0 2D ⎧ x x ⎫ ⎨−M 0 sin b + (V0α + M 0 )cos b ⎬ ⎩ α α⎭ e−[( xb − x0 )/α ]e− x0 /α sin (( xb − x0 ) / α ) cos (x0 / α ) Plates Summary: Successes – “moats” and “rises” ubiquitous. Elastic flexure seems to apply. Problems: • • • Geophysical measurements of w are difficult. Reference level? Other sources of topography. Alternative approach – use gravity Consider same observed topography Load Crust Stiff "Airy" root Floppy Figure 20.6 Fourier Series t(x) → t(k) Apply load t c = t 0 cos kx Get displacement D → w = w0 cos kx w0 = t0 D 4 ρm −1+ k ρc ρc g δ g = g0 cos kx g0 : ρ c (t c − e− kH w0 ) where H is crustal thickness and w0 depends on k and D. δ g(k) → δ g(x) Compare δ g(x) with data.