Page 319 - Physical Principles of Sedimentary Basin Analysis

9.6 Damped flexure of a plate above a viscous mantle 301

Exercise 9.8 Show that the compressive force F needed to increase the deflection by a
factor f from an uncompressed state is F = ( f − 1)F c /f .Howlargemust F be for the
deflection to be increased by 1% when k = k min ?

9.6 Damped flexure of a plate above a viscous mantle
An elastic plate that is rapidly loaded does not instantly deflect under the load because
of the viscous damping of the ductile mantle. For example if the lithosphere is rapidly
loaded by a volcanic island then some time will pass before the full elastic deflection is
reached. A similar situation is the rapid deglaciation of a continent where the postglacial
rebound will continue for some time after the deglaciation. We would like to know how
long the lithosphere needs to adjust to changes in the surface load. A simple approach to
this problem is to represent the viscous damping from the mantle by a dashpot. The dashpot
is similar to the shock absorber in a car and it produces a resisting stress that is proportional
to the velocity.
dw M
dw
σ = η = (9.80)
dt l dt
where the parameter η is the viscosity μ divided by the mantle length scale l.The last
equality (9.80) is the viscosity μ multiplied with the strain rate approximated as (1/l)
dw/dt. The stress σ acts on the base of the lithosphere and it is in addition to the restoring
pressure from buoyancy. The equation for the deflection of the lithosphere is therefore
4
∂ w ∂w
+ gw + η = q(x). (9.81)
∂x 4 ∂t
We can use the equation to study rapid loading and unloading. Let us first study rapid
unloading when the plate has been deflected for a very long time by a periodic load q(x) =
q 0 cos(kx), where k = 2π/λ is the wave number. We then have that ∂w/∂t = 0 before the
plate is unloaded and that the initial deflection is
q 0 cos(kx)
w e (x) = (9.82)
4
Dk + g
as shown in Section 9.4. When the load is suddenly removed we have to solve equa-
tion (9.81)for q(x) = 0 with w e (x) as the initial condition. The deflection as a function of
time is then
w(x, t) = w e (x)e −t/t 0 (k) , (9.83)

where the characteristic time t 0 is dependent on the wave number,
η
t 0 (k) = . (9.84)
4
Dk + g
It is seen that the deflections of the plate decay to zero as time goes to infinity. The
time ln2 t 0 (k) is the half-life for the rebound at the given wave number, and most of the