Page 327 - Physical Principles of Sedimentary Basin Analysis

9.9 Flexure of a viscoelastic plate 309

observations cannot be easily explained by the viscoelastic model. One possible solution
to the problem is that the load itself should be included in the plate, which would make
the plate thicker and more rigid. This problem also suggests that simple elastic and
viscoelastic models may be too simple to account for certain observations of flexure.
The viscoelastic model shows that elastic support of surface loads decays to zero with
increasing time. It is therefore possible that old loads are supported mainly by buoyancy,
which may explain why simple isostatic equilibrium calculations often are so successful.
Note 9.8 The deflection (9.122) is obtained by inserting the guess for a solution (9.121)
into equation (9.120) for the deflection. We then get the following equation for the
unknown function Y(t):
1 1
4
˙
˙
Dk Y + Y + Y = , (9.127)
t e t e
which can be rewritten as
˙
t 0 Y + Y = 1. (9.128)
This equation has the solution

Y(t) = 1 + Ce −t/t 0 , (9.129)

where C is an integration constant that is obtained from the initial condition, which is
the elastic deflection at t = 0. We have already found the elastic deflection for the same
periodic load, equation (9.58), which can be rewritten as

q 0 t e
w(x) = cos(kx). (9.130)
g t 0
The viscoelastic deflection at t = 0is
q 0
w(x) = 1 + C cos(kx) (9.131)
g
and we therefore get that C = (t e /t 0 ) − 1.
Exercise 9.11 Show that the deflection (9.122) can be approximated by

4
λ c
w(x, t) ≈ w iso 1 − e −t/t e ≈ w iso (9.132)
λ
in the regime λ
λ c . Hint: use that 1/(1 + x) ≈ 1 − x for small x.
Exercise 9.12 Show that the deflection can be approximated by

w(x, t) = w iso 1 − e −t/t 0 (9.133)

in the regime λ
λ c .