Unit 14: Structural Mechanics in Construction and Civil Engineering
Chapter 16
Gravity Retaining Walls
Chapter 14 has been concerned mainly with the addition of direct and
bending stress when these two types of stress occur within a material,
e.g. the variation of stress across the face of a column section which
arises due to eccentricity of loading.
The principles involved in those cases are used in much the same way on
occasions where, for example, a wall, resting on soil or on a concrete
footing and acted upon by horizontal forces, is transmitting to the soil or
footing stresses which consist of
1. Direct stress from the wall’s weight.
2. Stress due to the overturning moment.
Before these resultant stresses are considered, it will be necessary to
study the effect of the combined action of the vertical and horizontal (or
inclined) forces on the overall behaviour of the wall.
As a result of that action the wall fail in three ways:
i)
ii)
iii)
Sliding
Overturning
Overstressing.
Since, however, the main purpose of a retaining wall is to provide
resistance to the horizontal (or inclined) forces caused by the retaining
material, the nature of the pressures these forces exert on the wall will be
investigated first.
Chapter 16
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Unit 14: Structural Mechanics in Construction and Civil Engineering
16.1 Horizontal Forces
16.1.1
Wind Pressure
This is the simplest case, because wind pressure is assumed to be
uniform. Therefore, the total resultant force acts at the centre of the area
over which the pressure is applied and this force, P = p x area newtons,
or since, in the case of a wall, 1m length of wall is generally considered,
P = p x 1 x H = p x H newtons
Where
p = unit wind pressure in N/m2
H = the height of the part of the wall subject to that wind
pressure (see Fig. 1) in metres.
Figure 1
Wind loads. It is based on several factors, including not only the basic
speed of the wind and the type of topography in the locality of the wall,
but also a statistical factor which takes into account the probability of the
basic wind speed being exceeded within the projected life span of the
wall.
Chapter 16
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Unit 14: Structural Mechanics in Construction and Civil Engineering
16.1.2
Liquid Pressure
Consider the vertical surface AB, shown in Fig. 2(a), to be the face of a
wall which is retaining a liquid. It can be shown that a cubic metre of
liquid, situated at a depth h metres below the surface, exerts a pressure
of w x h kN outwards on all its six side surfaces. w in this case is the
equivalent density or the unit weight of the liquid in kN/m3. (i.e. density
× gravity)
Thus the intensity of outward pressure varies directly with the depth and
will have a maximum value of w x H kN/m2 at H m, the maximum depth
as indicated in Fig. 2 (b).
(a)
(b)
Figure 2
At the surface of the liquid (where h= 0), the pressure will be zero. So, as
the maximum is wH kN/m2, the average pressure between A and B is
kN/m2.
In dealing with retaining walls generally, it is convenient, as we said
earlier, to consider the forces acting on one metre length of wall, that is,
an area of wall H m high and 1 m measured perpendicular to the plane of
the diagrams (Fig. 2(a) and (b)). Thus, as the “wetted area” concerned is
H m2, the total force caused by water pressure on a one metre strip of
wall is
“Wetted area” × Average rate of pressure
Chapter 16
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Unit 14: Structural Mechanics in Construction and Civil Engineering
This total resultant force on the wall’s vertical surface from the liquid is
(as will be seen from Fig. 2(b)) the resultant of a large number of forces,
which range from zero at the top to wH at the base. The resultant will
therefore act at a point rd of H from the base, as shown in Fig. 3.
Figure 3
Note: If the liquid does not reach the top of the wall, as, for example, in
Fig. 4, then the resultant force is calculated with H as the depth of the
liquid and not as the height of the wall. The force is again
acts at a point
and it
(one-third the depth of the liquid) from the wall’s base.
Figure 4
Chapter 16
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Unit 14: Structural Mechanics in Construction and Civil Engineering
Example 1
A masonry dam retains water on its vertical face. The wall is, as shown in
Fig. 5, 3.7 m, but the water level reaches only 0.7 m from the top of the
wall. What is the resultant water pressure per metre run of wall?
Answer
Figure 5
The equivalent density w of water = 10 kN/m3
Therefore
acting at 1 m above base
In cases where the wall in contact with the water is not vertical, as is the
case shown in Fig. 6, the wetted area will be larger than in the case of a
vertical back, and the resultant pressure will thus be increased to
(i.e. the wetted area will be L m2 instead of H m2 considering one metre
run of wall).
Chapter 16
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Unit 14: Structural Mechanics in Construction and Civil Engineering
Figure 6
16.1.3
Soil Pressure
It is obvious that pressures on walls from retained soil or other granular
materials cannot be determined with quite the same accuracy as with
water. Soils vary in weight and character; they behave quite differently
under varying conditions of moisture, etc., and, in general, the resultant
pressures on vertical and non-vertical surfaces from soils are obtained
from various soil pressure theories. Numerous theories exist for the
calculation of soil pressures, and these theories vary in the assumptions
which they make and the estimated pressure which they determine. A
great deal of research is still being directed upon this subject, but most of
this is beyond the scope of a volume of this type, and therefore only the
well tried Rankine’s theory will be dealt with in detail.
RANKINE’S THEORY OF SOIL PRESSURE
It has been seen that a cubic metre of liquid at a depth h below the
surface presses outwards horizontally by an amount wh kN/m2 (w being
the equivalent density of liquid). In the case of soil weighing w kN/m3, the
Chapter 16
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Unit 14: Structural Mechanics in Construction and Civil Engineering
outward pressure at a depth of h m below the surface will be less than wh
kN/m2, since some of the soil is “self-supporting”.
Consider, for example, the soil retaining by the vertical face AB in Fig. 7.
Figure 7
If the retaining face AB was removed, then some of the soil would
probably collapse at once, and in the course of time the soil would
assume a line BC, as shown. The angle Ф made between the horizontal
and the line BC varies with different types of soil, and is called the angle
of repose or the angle of internal friction of the soil. It can be said,
therefore, that only part of the soil was in fact being retained by the wall
and was exerting pressure on the wall. Thus, it follows that the amount of
pressure on the wall from the soil depends upon the angle of repose for
the type of soil concerned, and Rankine’s theory states in general terms
that the outward pressure per square metre at a depth of h m due to a
level fill of soil is
[
]
as compared with (wh) kN/m2 in the case of liquids. Thus, by similar
reasoning as used in the case of liquid pressure, the maximum pressure
at the bottom of the wall is given by
Maximum pressure =
[
]
Average pressure =
[
]
Chapter 16
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Unit 14: Structural Mechanics in Construction and Civil Engineering
The soil act at this average rate on an area of H m2 of wall, so the total
resultant force per metre run of wall is
[
and this acts, as shown in Fig. 8, at
]
above the base of the wall.
Figure 8
Example 2
Soil weighing 15 kN/m3 and having an angle of repose Ф of 30o, exerts
pressure on a 4.5 m high vertical face. What is the resultant horizontal
force per metre run of wall?
Answer
sin Ф = sin 30o = 0.5 and
[
Chapter 16
]
[
]
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Unit 14: Structural Mechanics in Construction and Civil Engineering
Exercise 16.1
1: Analysis of a trapezoidal gravity dam structure
Figure 9 shows a section through a mass concrete gravity dam which is to
be used to form one end of a freshwater reservoir. Assuming that the
water level in the reservoir is at the top of the dam, determine the
maximum height H if the dam is not to overturn.
Assume:
Density of concrete
= 2400 kg/m3
Density of water
= 1000 kg/m3
Figure 9
Answer
In our solution we will consider a 1 m length of the dam. We also note
that the dam is formed from a trapezoidal section. In order to calculate
the restoring moment for the structure, we will need to determine the
position of its centroid. This can be difficult for complex shapes, however
in order to simplify the problem we can consider the section as being
made up of a triangular and a rectangular part. Finally we will need to
determine the forces and moments exerted by the water pressure on the
structure. The load from the water will be a triangular distribution which
will be zero at the water surface and a maximum at the base. The
maximum force at the base will be given by:
Maximum force of water
Where
Chapter 16
F = mg = gh
m = mass (kg)
g = gravity (9.81 ms-2)
 = density of water = 1000 kg/m3
h = height (m)
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Unit 14: Structural Mechanics in Construction and Civil Engineering
So
F = 1000 × 9.81 × H = 9.81 H kN/m
Noting that rotation will take place about point A (at the toe of the dam),
the forces and lever arms required for our calculations are shown in figure
10.
Figure 10
From figure 10 we note:
W1 = Force from section 1 (triangular part of structure)
W2 = Force from section 2 (rectangular part of structure)
W3 = Force from water
To solve we must calculate the overturning and restoring moments about
point A, per metre length, of the dam.
Overturning moment:
(
)
(Note: W = mass × gravity = density × volume × gravity and the volume
is taken over 1 m length of dam)
Restoring moments:
(
Chapter 16
(
))
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Unit 14: Structural Mechanics in Construction and Civil Engineering
(
(
))
For equilibrium: Overturning moment = Restoring moment
From which
Chapter 16
H = 25.17m
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