Civil Engineering Reference
In-Depth Information
Poisson's Ratio
µ
, and any temperature increase
that occurs. In ideal conditions
σ
3
is related to
σ
1
as follows:
exceeds the stress. For conditions where
σ
3
is low
or tensile (negative), failure occurs more readily
compared to conditions at greater depth where
both
σ
1
and
σ
3
are compressive (positive).
The Mohr diagram also shows the orientation
of the fracture with respect to the stress direc-
tion (Figure 2.2(c)). Since principal stresses are
oriented mutually at right angles, sets of joints
tend to form in orthogonal directions.
µ
1
µ
σ
1
E
1
µ
εT
σ
3
=
+
(2.2)
−
−
where
E
is the modulus of deformation of the
rock,
ε
is the coefficient of thermal expansion and
T
is the temperature rise. The first component of
equation (2.2) shows the value of the horizontal
stress due to gravitational loading; if the Poisson's
ratio is 0.25, for example, then
σ
3
2.3 Effects of discontinuities on slope
stability
While the orientation of discontinuities is the
prime geological factor influencing stability, and
is the subject of this chapter, other properties
such as persistence and spacing are significant
in design. For example, Figure 2.3 shows three
slopes excavated in a rock mass containing two
joint sets: set J1 dips at 45
◦
out of the face, and
set J2 dips at 60
◦
into the face. The stability of
these slopes differs as follows. In Figure 2.3(a), set
J1, which is widely spaced and has a persistence
greater than the slope height, forms a potentially
unstable plane failure over the full height of the
cut. In Figure 2.3(b) both sets J1 and J2 have
low persistence and are closely spaced so that
while small blocks ravel from the face, there is
no overall slope failure. In Figure 2.3(c), set J2 is
persistent and closely spaced, and forms a series
of thin slabs dipping into the face that create a
toppling failure.
The significance of Figure 2.3 is that, while an
analysis of the orientation of joint sets J1 and J2
would show identical conditions on a stereonet,
there are other characteristics of these discon-
tinuities that must also be considered in design.
These characteristics, which are discussed further
in Chapter 3, should be described in detail as part
of the geological data collection program for rock
slope design.
0.33
σ
1
.If
the rock were not free to expand, and a temperat-
ure rise of 100
◦
C occurs in a rock with a modulus
value of 50 GPa and an
ε
value of 15
=
10
−
6
/
◦
C,
then a thermal stress of 100 GPa will be gener-
ated. In reality, the values of the principal stresses
will be modified by tectonic action resulting in
deformation such as folding and faulting.
On Figure 2.2(a), the value of
σ
1
is defined by
equation (2.1), and the value of
σ
3
varies with
depth as follows. The value of
σ
3
is tensile at
depths less than 1.5 km where the sediments have
not been consolidated into rock, and below this
depth,
σ
3
increases as defined by equation (2.2),
assuming no temperature change.
The formation of joints in rock during the
burial-uplift process shown in Figure 2.2(a) will
depend on the rock strength in comparison to the
applied stresses. A method of identifying condi-
tions that will cause the rock to fracture is to use
a Mohr diagram (Figure 2.2(b)). In Figure 2.2(b)
the rock strength is shown as a straight line under
compressive stress, and curved under tensile stress
because micro-fractures within the rock act as
stress concentrators that diminish the strength
when a tensile stress is applied. On the Mohr
diagram, circles represent the
σ
1
and
σ
3
stresses
at different depths, and where the circle inter-
sects the strength line, failure will occur. The
stress conditions show that fracture will occur at
a depth of 2 km (
σ
1
×
0 MPa)
because there is no horizontal confining stress act-
ing. However, at a depth of 5 km where the rock
is highly confined (
σ
1
=
=
52 MPa;
σ
3
=
2.4 Orientation of discontinuities
The first step in the investigation of discontinuities
in a slope is to analyze their orientation and
25 MPa)
the rock will not fracture because the strength
130 MPa;
σ
3
=
