﻿ Craig's Soil Mechanics 7th Edition - Mecânica dos Solos 2 - 42
458 pág.

# Craig's Soil Mechanics 7th Edition

Disciplina:Mecânica dos Solos 2867 materiais5.534 seguidores
Pré-visualização50 páginas
```surcharge. The pressures in soil 2 are calculated using Ka ¼ 0:36,Kp ¼ 2:78
and �0 ¼ (20� 9:8) ¼ 10:2 kN/m3 (see Table 6.1). The active and passive pressure
distributions are shown in Figure 6.7. In addition, there is equal hydrostatic pressure
on each side of the wall below the water table.

Sloping soil surface

The Rankine theory will now be applied to cases in which the soil surface slopes at a
constant angle � to the horizontal. It is assumed that the active and passive pressures act
in a direction parallel to the sloping surface. Consider a rhombic element of soil, with
sides vertical and at angle � to the horizontal, at depth z in a semi-infinite mass. The
vertical stress and the active or passive pressure are each inclined at � to the appropriate
sides of the element, as shown in Figure 6.8(a). Since these stresses are not normal to
their respective planes (i.e. there are shear components), they are not principal stresses.
In the active case, the vertical stress at depth z on a plane inclined at angle � to the

horizontal is given by

�z ¼ �z cos�

and is represented by the distance OA on the stress diagram (Figure 6.8(b)). If lateral
expansion of the soil is sufficient to induce the state of plastic equilibrium, the Mohr
circle representing the state of stress in the element must pass through point A (such
that the greater part of the circle lies on the side of A towards the origin) and touch the

Table 6.1

Soil Depth
(m)

Pressure (kN/m2)

Active pressure
1 0 0:24� 50 ¼ 12:0
1 6 (0:24� 50)þ (0:24� 18� 6) ¼ 12:0þ 25:9 ¼ 37:9
2 6 0:36[50þ (18� 6)]� (2� 10� ﬃﬃﬃﬃﬃﬃﬃﬃﬃ0:36p ) ¼ 56:9� 12:0 ¼ 44:9
2 9 0:36[50þ (18� 6)]� (2� 10� ﬃﬃﬃﬃﬃﬃﬃﬃﬃ0:36p )þ (0:36� 10:2� 3)

¼ 56:9� 12:0þ 11:0 ¼ 55:9
Passive pressure
1 0 0
1 1.5 4:17� 18� 1:5 ¼ 112:6
2 1.5 (2:78� 18� 1:5)þ (2� 10� ﬃﬃﬃﬃﬃﬃﬃﬃﬃ2:78p ) ¼ 75:1þ 33:3 ¼ 108:4
2 4.5 (2:78� 18� 1:5)þ (2� 10� ﬃﬃﬃﬃﬃﬃﬃﬃﬃ2:78p )

þ (2:78� 10:2� 3) ¼ 75:1þ 33:3þ 85:1 ¼ 193:5

170 Lateral earth pressure

failure envelope for the soil. The active pressure pa is then represented by OB (numer-
ically equal to OB0) on the diagram. When c ¼ 0 the relationship between pa and �z,
giving the active pressure coefficient, can be derived from the diagram:

Ka ¼ pa
�z
¼ OB
OA

¼ OB
0

OA

Now

OD ¼ OCcos�

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ
ðOC2 sin2 ��OC2 sin2 �Þ

q

Figure 6.8 Active and passive states for sloping surface.

Rankine’s theory of earth pressure 171

therefore

Ka ¼ cos� �
ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ
ðcos2 � � cos2 �Þ

p
cos� þ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ
ðcos2 � � cos2 �Þ

p ð6:8Þ
Thus the active pressure, acting parallel to the slope, is given by

pa ¼ Ka�z cos � ð6:9Þ

and the total active thrust on a vertical wall surface of height H is

Pa ¼ 1
2
Ka�H

2 cos� ð6:10Þ

In the passive case, the vertical stress �z is represented by the distance OB
0 in

Figure 6.8(b). The Mohr circle representing the state of stress in the element, after a
state of plastic equilibrium has been induced by lateral compression of the soil, must
pass through B0 (such that the greater part of the circle lies on the side of B0 away from
the origin) and touch the failure envelope. The passive pressure pp is then represented
by OA0 (numerically equal to OA) and when c ¼ 0 the passive pressure coefficient
(equal to pp/�z) is given by

Kp ¼ cos� þ
ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ
ðcos2 � � cos2 �Þ

p
cos� �

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ
ðcos2 � � cos2 �Þ

p ð6:11Þ
Then the passive pressure, acting parallel to the slope, is given by

pp ¼ Kp�z cos� ð6:12Þ

and the total passive resistance on a vertical wall surface of height H is

Pp ¼ 1
2
Kp�H

2 cos� ð6:13Þ

The active and passive pressures can, of course, be obtained graphically from
Figure 6.8(b). The above formulae apply only when the shear strength parameter c
is zero; when c is greater than zero the graphical procedure should be used.
The directions of the two sets of failure planes can be obtained from Figure 6.8(b).

In the active case, the coordinates of point A represent the state of stress on a plane
inclined at angle � to the horizontal, therefore point B0 is the origin of planes, also
known as the pole. (A line drawn from the origin of planes intersects the circumference
of the circle at a point whose coordinates represent the state of stress on a plane
parallel to that line.) The state of stress on a vertical plane is represented by the
coordinates of point B. Then the failure planes, which are shown in Figure 6.8(a),
are parallel to B0F and B0G (F and G lying on the failure envelope). In the passive case,
the coordinates of point B0 represent the state of stress on a plane inclined at angle �

172 Lateral earth pressure

to the horizontal, and therefore point A is the origin of planes: the state of stress on
a vertical plane is represented by the coordinates of point A0. Then the failure planes
in the passive case are parallel to AF and AG.
Referring to Equations 6.8 and 6.11, it is clear that both Ka and Kp become equal to

unity when � ¼ �; this is incompatible with real soil behaviour. Use of the theory is
inappropriate, therefore, in such circumstances.

Example 6.3

A vertical wall 6m high, above the water table, retains a 20� soil slope, the retained
soil having a unit weight of 18 kN/m3; the appropriate shear strength parameters are
c0 ¼ 0 and �0 ¼ 40�. Determine the total active thrust on the wall and the directions of
the two sets of failure planes relative to the horizontal.
In this case the total active thrust can be obtained by calculation. Using Equation 6.8,

Ka ¼ cos 20
� �

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ
ðcos2 20� � cos2 40�Þ

p
cos 20� þ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃðcos2 20� � cos2 40�Þp ¼ 0:265

Then

Pa ¼ 1
2
Ka�H

2 cos�

¼ 1
2
� 0:265� 18� 62 � 0:940 ¼ 81 kN=m

The result can also be determined using a stress diagram (Figure 6.9). Draw the
failure envelope on the �/� plot and a straight line through the origin at 20� to the
horizontal. At a depth of 6m,

�z ¼ �z cos� ¼ 18� 6� 0:940 ¼ 102 kN=m2

and this stress is set off to scale (distance OA) along the 20� line. The Mohr circle is
then drawn as in Figure 6.9 and the active pressure (distance OB or OB0) is scaled from
the diagram, i.e.

pa ¼ 27 kN=m2

Then

Pa ¼ 1
2
paH ¼ 1

2
� 27� 6 ¼ 81 kN=m

The failure planes are parallel to B0F and B0G in Figure 6.9. The directions of these
lines are measured as 59� and 71�, respectively, to the horizontal (adding up to
90� þ �).

Rankine’s theory of earth pressure 173

Earth pressure at-rest

It has been shown that active pressure is associated with lateral expansion of the soil
and is a minimum value; passive pressure is associated with lateral compression of the
soil and is a maximum value. The active and passive values may thus be referred to as
limit pressures. If the lateral strain in the soil is zero, the corresponding lateral pressure
is called the earth pressure at-rest and is usually expressed in terms of effective stress by
the equation

p0 ¼ K0�0z ð6:14Þ

where K0 is defined as the coefficient of earth pressure at-rest, in terms of effective stress.
Since the at-rest condition does not involve failure of the soil, the Mohr circle

representing the vertical and horizontal stresses does not touch the failure envelope
and the horizontal stress cannot be evaluated. The value of K0, however, can be deter-
mined experimentally by means of a triaxial test in which the axial stress and the all-
round pressure are increased simultaneously such that the lateral strain in the specimen is
maintained at zero: the hydraulic triaxial apparatus is most suitable for this purpose. For
soft clays, methods of measuring lateral pressure in situ have been developed by Bjerrum
and Andersen [3] and,```