Page 324 - Physical Principles of Sedimentary Basin Analysis

306 Flexure of the lithosphere

Solution: (a) A starting point for obtaining the deflection is equation (9.98),
2
∂ ˙w 1
D =˙ + τ (9.110)
τ
∂x 2 t e
where the torque from force F is
τ(x, t) = (l − x)F(t). (9.111)
An integration with respect to time gives
2
∂ w 1 t
1

= F(t) + F(t )dt (l − x). (9.112)
∂x 2 t e 0 D
x
This equation can be rewritten in the dimensionless (unit) coordinate ˆ = x/l as
2
∂ w
x
= 6w 2 (t)(1 −ˆ) (9.113)
∂ ˆ x 2
where
1 l
t
3

w 2 (t) = F(t) + F(t )dt . (9.114)
t e 0 6D
The integration with respect to ˆ then leads to
x
3
x
w(ˆ, t) = w 2 (t)(1 −ˆx) + c 1 ˆx + c 2 (9.115)
where c 1 and c 2 are two time-dependent integration constants. The clamped end (ˆx = 0)
has the two boundary conditions w = 0 and ∂w/∂ ˆ = 0. From the first boundary condition
x
(w = 0) we get that c 2 =−w 2 , and from the second boundary condition (∂w/∂ ˆx = 0) we
get that c 1 = 3w 2 . The deflection then becomes

3
w(ˆ, t) = w 2 (t) (1 −ˆx) + 3ˆx − 1 , (9.116)
x
which is the same solution as in Exercise 9.4, except for the time-dependence in w 2 .
(b) The solution can be decomposed into an elastic deflection w e (x) and a viscous
deflection w v (x, t) where
l
3
3
x
w e (ˆx, t) = F(t) (1 −ˆ) + 3ˆx − 1 (9.117)
6D
and
1 t l 3

3

x
w v (ˆx, t) = F(t )dt (1 −ˆ) + 3ˆx − 1 . (9.118)
t e 0 6D
The elastic deflection is proportional to the force F(t) (see Exercise 9.4), and the viscous
deflection is zero at time t = 0.
(c) When F(t) = F 0 (constant) we get that w e (x, t) = w v (x, t) after the time span t = t e .
This justifies the interpretation that t e is the time span needed for the viscous deflection to
be of similar size to the elastic deflections.