Page 92 - Physical Principles of Sedimentary Basin Analysis

74 Linear elasticity and continuum mechanics

along the z-axis in a 1D model, when the overpressure p e is used instead of the fluid
pressure.
(b) Let the overpressure p e (z) be known as a function of depth z. Show that the
displacement along the z-axis becomes
( − α f )g 2 α z

u(z) =− z + p e (z )dz (3.191)
2(2G + λ) (2G + λ) 0
when the bulk density is constant.
Exercise 3.33 Show that the displacement u(z) along the z-axis is

( s − w ) 1 2

u(z) =− z + φ 0 z 0 (z 0 − z − z 0 e −z/z 0 ) (3.192)
(2G + λ) 2
for a hydrostatic basin, when the bulk density is = φ w + (1 − φ) s and the porosity
as a function of depth is φ = φ 0 exp(−z/z 0 ). The displacement field is zero (u = 0) at
the surface (z = 0) and there is zero stress at the surface, σ zz =−(2G + λ)du/dz = 0
for z = 0.

3.21 Particle paths and streamlines

The particle paths are the traces made by flowing particles. They are always tangential to
the flow field, and are therefore the solution of
dr
= v(r, t) (3.193)
dt
when the vector field is known. The vector equation (3.193) can also be written as
dx dy dz
= = = dt (3.194)
v x v y v z
which later will be a useful representation of the particle paths. The same equation also
follows from the vector product
v × dr = 0 (3.195)

where dr = (dx, dy, dz) is a small line segment along the path.
A streamline is the path given by the flow field at a specific point in time. The streamlines
therefore coincide with the particle paths when the flow field is independent of time. The
following flow field illustrates the difference between particle paths and streamlines:
dx x dy y
= v x = and = v y = (3.196)
dt t 0 + t dt t 0
where t 0 is a parameter. An integration of equations (3.196) gives the particle path
t

x(t) = x 0 · 1 + and y(t) = y 0 e t/t 0 (3.197)
t 0