Geoscience Reference
In-Depth Information
central Australia comprises a series of east-west-trending intra-continental basins
and arches, with the gravity highs corresponding to the arches and the lows to
the sedimentary basins. The crust beneath this region apparently has an average
thickness of about 40 km, but the Moho is depressed by up to 10 km beneath
the basins and is similarly elevated beneath the arches. However, there is no
indication of faulting at the bases of the basins. There has been no plate-boundary
activity in the region since the late Proterozoic. Subsidence of the basins started
about 1000 Ma ago and continued for some 700 Ma, and the subsidence rate
has increased with time (Fig. 10.49(c)). These facts, taken together, indicate that
neither an extensional nor a thermal model is appropriate for these basins and
arches. It has been proposed that they formed instead as a result of compression
of the lithosphere.
Let us initially consider a simple problem: an elastic plate, of flexural rigidity
D
, subjected only to a constant horizontal force
H
per unit width. The deformation
of this plate
w
satisfies Eq. (5.56):
D
d
4
w
d
x
4
+
H
d
2
w
d
x
2
=
0
(10.13)
The solution to this equation is obtained by integrating twice, giving
D
d
2
w
d
x
2
+
Hw
=
c
1
x
+
c
2
(10.14)
where
c
1
and
c
2
are constants of integration.
If we assume the plate to be of a finite length
l
with d
2
w
d
x
2
/
=
0 and
w
=
0
both at
x
l
(i.e., the plate is fixed at 0 and
l
), then both
c
1
and
c
2
must be zero. The solution to Eq. (10.14)isthen sinusoidal:
w
=
c
3
sin
H
=
0 and at
x
=
D
x
+
c
4
cos
H
D
x
(10.15)
where
c
3
and
c
4
are constants. Because the plate is fixed at
x
=
0,
c
4
must equal
zero. The condition that
w
must also equal zero at
x
l
is then possible only
when
c
3
is zero (in which case there is no deformation at all) or when
H
D
l
=
n
π
=
for
n
=
1
,
2
,...
(10.16)
The smallest value of
H
for which deformation occurs is therefore given by
n
l
2
.For horizontal forces
less than this value, there is no deformation. At this critical value, the plate
deforms into a sine curve given by
w
=
c
3
sin
π
x
l
2
D
=
1. This critical value of the horizontal force is
π
/
(10.17)
However, this simple calculation is not directly applicable to the lithospheric
plates (or to layers of rock) because the lithosphere is hydrostatically supported
by the underlying mantle. A hydrostatic restoring force (see Eqs. (5.58) and
(5.59)) must be included in Eq. (10.13)inorder for us to be able to apply it to the
