7

CHAPTER 1. LIMITING EQUILIBRIUM IN A SOIL MASS

1.1 Basic problems of limiting equilibrium

Limiting (or plastic) equilibrium occurs in a soil mass where the failure condition ( = (f is reached either on a continuous surface or on the whole mass.

There are three basic problems of limiting equilibrium in a soil mass:

a. earth pressure

Fig. 1.1

In fig. 1.1 is represented a retaining wall, which has the purpose of supporting the soil behind it. The wall is free to have displacements and to tilt under the pressure exerted by the soil. For a certain displacement (a, a continuous failure surface BC develops, linking the bottom of the wall with the surface. The wall is moving away from the soil behind, which tends to expand or dilate. The action exerted by the soil on the wall is called active earth pressure. The resultant of the active earth pressure on the face AB of the wall is the active thrust Pa.

Fig. 1.2

In fig. 1.2 is represented the foundation of an arch. Under the force R exerted by the arch the foundation is moving against the soil behind, which is compressed. For a certain displacement (p of the foundation, a continuous failure surface develops, linking the base of the foundation with the ground surface. The resistance opposed by the soil to the foundation is called passive earth pressure or passive resistance. The resultant of the passive earth pressure on the face AB of the foundation is the passive resistance Pp.

The inter-action between the two elements (the wall or the foundation and the soil) is summarized in the table 1.1.

Table 1.1Role of the elementEarth pressure case

Soil is pushingWall is resistingActive case

Foundation is pushingSoil is resistingPassive case

There is also a significant difference in the displacements required to generate the two limit cases.

Fig. 1.3

Fig. 1.3 shows an experimental set-up consisting of a large box equipped with a wall hinged at the bottom of the box and with soil behind the wall. With a system of screws and gauges placed in front of the wall, it is possible to induce the movement of the wall in both directions: away from the soil (thus generating the active earth pressure) or against the soil (thus generating the passive earth pressure).

Fig. 1.4

Fig. 1.4 represents the results of such an experiment, under the form of a force-displacement diagram.

On the basis of experiments of such kind, the displacements required to reach the active or the passive case, in function of the height H of the wall, for various types of soils have been established.

Table 1.2 gives some usual values.

Table 1.2

Type of soilActive case

(aPassive case

(p

Dense, cohesionless 0,0005 H0,005 H

Loose, cohesionless 0,002 H 0,01 H

Stiff, cohesive 0,01 H 0,02 H

Soft, cohesive 0,02 H 0,04 H

As one can observe from the diagram in the fig. 1, for no movement of the wall ((=0), the total force exerted by the soil, Po, is situated between the total active thrust Pa and the total passive resistance Pp.

Pa < Po < Pp

(1.1)

Po corresponds to the earth pressure at rest. A common example for the at rest case is represented by an underground structure as shown in the fig. 7.5. The earth pressures at rest on both sides of the structure are identical, hence there is no movement and no possibility of developing neither the active nor the passive case.

Fig. 1.5

The at rest condition corresponds to the elastic equilibrium in the soil mass.

b. slope stability

By performing an excavation between two levels, the slope AB resulted (fig.1.6).

Fig. 1.6

For the design of such a work, is compulsory to cheek the slope stability. Indeed, under certain circumstances, such as: a too steep slope, a too large surcharge on the ground surface, a decrease of the shear strength of the soil produced by an increase of the water content etc, a continuous failure surface BC could develop, causing the loss of the stability of the slope.Slope stability represents another basic problem of the limiting equilibrium in a soil mass.

c. ultimate bearing capacity

The ultimate bearing capacity is defined as the least pressure which would cause shear failure of the supporting soil immediately below and adjacent to a foundation.

Fig. 1.7 represents a strip footing under a wall subjected to a centric load N. The base pressure is . If p increases to an ultimate value pf, a failure surface AC develops between the edges of the footing and the ground surface.

Fig.1.7When the base pressure reaches the value pcr, the ultimate bearing capacity of the foundation is reached, the foundation either tilts (fig.1.8) or moves downward (fig.1.9), in both cases the loss of the bearing capacity being due to the slipping toward the ground surface of a part of the soil mass under the foundation, relative to the rest of the mass.

Fig.1.8Fig.1.9

The three problems previously described have in common the occurrence of a failure surface along which a part of the soil mass slips relative to the rest of the mass.

Another common feature of the three examples is the condition of plane strain, meaning that the strains in the longitudinal direction of the structure are assumed to be zero.

To establish the state of stresses in a soil mass in a limit equilibrium, is necessary to solve a system of equations formed by the equilibrium condition and by the failure condition. For a plane problem, the system becomes:

(1.2)

(

Rigorous solutions of the system (1.2) were obtained in the Theory of Plasticity for a relatively small number of problems. Well known are the solutions of the Russian V.V. Sokolovski.

In what follows, a particular problem of limit equilibrium will be considered, whose solutions were formulated by the Scottish engineer J. Rankine (1820-1872), namely the limit equilibrium in a semi-infinite soil mass limited by an horizontal surface or by a sloping surface. Rankines solutions can be used, under certain conditions, in all of the three basic problems of limit equilibrium.

1.2 Limiting equilibrium state in a soil mass with horizontal surface (Rankines Theory)

A non-cohesive soil mass is considered, with horizontal surface (fig. 1.10). In a point M, at depth Z, stresses and , induced by the own weight of the soil, are:

where KO is the coefficient of earth pressure at rest.

Fig.1.10

When the surface is horizontal and the soil mass is semi-infinite, any vertical plane, including the one passing through the point M is a plane of symmetry. No shear stresses exist on horizontal and vertical planes, and and are principal stresses:

The relation corresponds to the liniar behavior of soil, to the elastic equilibrium.

There are two ways by which the limit equilibrium state can be reached in the considered point M (fig. 1.11).

a. by a gradual decrease of the principal stress in horizontal direction, . For a certain ration between and , the failure condition is met (the Mohr circle touches the Coulombs line). The state of stresses in this case, when < , is called active limit state.

b. by a gradual increase of the principal stress in horizontal direction, . At first, the diameter of the Mohr circle is smaller than the one corresponding to the elastic equilibrium. When becomes equal to , the Mohr circle reduces to a point. After wards, the diameter of the circle increases gradually and for a certain ratio between and the failure condition is met. The state of stresses in this case, when > , is called passive limit state.

Fig. 1.11

In order to make the transition from the at rest condition to the limit equilibrium condition, Rankine introduced in the soil mass a fictitious, thin wall, infinitely long, perfectly smooth. This last condition was needed in order not to have friction between the thin wall and the soil and, hence, to keep the vertical plane as a principal plane.

Once the wall is installed, half of the semi-infinite soil mass can be removed and, in order not to change the state of stresses, replaced by a horizontal pressure increasing linearly with depth: (fig. 1.12).

To generate one of the two limit states, it would be sufficient to move the wall away from the soil behind (active limit state) or against the soil (passive limit state).

Fig. 1.12

1.2.1 Active limit state

To make the wall move away from the soil behind, the horizontal pressures on the wall should decrease from the initial values corresponding to the at-rest condition. The soil dilates or expands outwards.

For a certain value of the displacement of the walls the principal stress reaches the value for which failure condition is fulfilled (fig. 1.13).

Fig. 1.13

The relation between and corresponding to the active limit state can be obtained from the condition of tangency of the Mohr circle to Colombs line.

For non-cohesive soils (fig. 1.14)

(1.4)

For cohesive soils (fig. 1.15)

(1.5)

In order to find the directions of the failure planes, is necessary to establish first the pole. To get the pole, one has to know a point in the Mohr circle expressing the total stress on a plane of known direction. Then, a parallel drawn from that point to the direction of the plane will intersect the circle in the pole. Lets take the point B, of coordinates (). It defines the total stress on the horizontal plane. The parallel drawn from B to the horizontal line confounds itself with the O axis and intersects the circle in the point A which is the pole.

Fig. 1.14

The directions of the failure planes are obtained by linking the pole will the points of tangency T and T which belong to the failure line. The failure planes make an angle of (45o + ) with the horizontal.

Fig. 1.16

The depth z was arbitrarily chosen. For another depth, another Mohr circle tangent to the failure line will correspond, and two failure planes making the angle (45o + ) with the horizontal. The active limit state in the whole soil mass, is characterized by the presence of two sets of failure planes each inclined at (45o + ) to the horizontal plane (which is the plane of maximum principal stress) and having between them the angle (90o - ) (fig. 1.16).

The cohesion modifies, as it was shown, the ratio between and , but the directions of the failure planes remain unchanged.

1.2.2 Passive limit state

The movement of the fictitious wall against the soil mass behind, made possible by an increase of the horizontal pressures acting on the wall, produces a compression of the soil. As it is known, soil works much better in compression than in tension. Therefore, the displacement of the wall required to induce the passive limit state is significantly larger than . For a certain value of the maximum principal stress reaches the value corresponding to the fulfillment of the failure condition (fig. 1.17).

Fig. 1.17

The relation between and corresponding to the passive limit state can be obtained from the condition of tangency of the Mohr circle to the Coulombs line.

For non-cohesive soils (fig. 1.18)

(1.6)

For cohesive soils

(1.7)

To find the pole, a parallel is drawn from the point A of the circle, expressing the total stress on the horizontal plane, to the direction of that plane. The parallel intersects the circle in the point B which is the pole. The directions of the failure planes are obtained by linking the pole with the points of tangency of the Mohr circle to the Coulombs line, T and T. The failure planes makes the angle (45o - ) with the horizontal.

Fig. 1.18

Assuming that the passive limit state is occurring in the whole soil mass, two sets of failure planes, each inclined at (45o - ) to the horizontal, will form (fig. 1.19).

Fig.1.191.2.3 The use of Rankines theory to compute the earth pressure on a vertical wall

Active pressure

A vertical retaining wall of height H, limited by a horizontal ground surface is considered (fig.1.20).

Fig.1.20

In the case of non-cohesive soil, the relation (1.4) can be used directly to compute the active earth pressure by making a change of notations. At the base of the wall:

(1.8)

The resultant of the earth pressure, the total active thrust is:

(1.9)

In the case of cohesive soil:

(1.10)

The active pressure diagram is obtained by the superposition of two diagrams (fig. 1.21). The pressure pa is zero at a particular depth zo, which is obtained from the relation:

(1.11)

Fig. 1.21

For a depth Hcr = 2zo the resultant of the active pressure is zero (the triangle of negative values abc annulates the triangle of positive values cde). Hcr, called critical height, represents the maximum the critical height on which a vertical cut in a cohesive soil mass could stand unsupported.

The total active thrust Pa is obtained by summing up the pressure on the height of the wall. There are two variants (fig. 1.22): when considering the ability of the cohesive soil to work in tension, the total active thrust Pa is the area of the trapez of pressures defg:

(1.12)

when the ability of the cohesive soil to work in tension is not considered, the diagram of tensions abc is disregarded and the total active thrust is the area of the diagram of compressions cfg:

(1.13)

Fig. 1.22

Passive pressure

In the case of non-cohesive soil the relation (1.6) can be used directly to compute the passive earth pressure by making a change of notations.

At the base of the wall (fig. 1.23):

(1.14)

Fig. 1.23

The total passive resistance Pp is the area of the passive pressures diagram:

(1.15)

Fig. 1.24

In the case of cohesive soil (fig. 1.24):

(1.16)

The total passive resistance Pp:

(1.17)

The influence of the water table

If the soil below the water table is in the fully drained condition, the active and passive pressures must be expressed in terms of the effective weight of the soil and the effective stress parameters c and .

For example, if the water table is at the surface and if no seepage is taking place:

(1.18)

(1.19)

The hydrostatic pressure due to the water in the soil press must be considered in addition to the active or passive pressure.

For the undrained condition in a fully saturated soil, the active and passive pressures are calculated using the parameters cu and with the total unit weight, which means that the water in the soil pores is not considered separately.

1.3 Limiting equilibrium state in a soil mass with sloping surface (Rankines Theory)

A semi-infinite soil mass with sloping surface of constant angle to the horizontal is considered. Since the slop is infinitely long, stresses acting on a vertical plane passing through the soil mass will be identical with the ones passing through any other vertical plane. Also, the total stress in a point of a plane parallel to the sloping surface will be the same as for any other point of the respective plane.

A prism of soil in considered with unit width, unit length (plane strain problem), and height z (fig.1.25). The prism is acted upon by the following forces:

the own weight G;

the reactions R on the base cd parallel to the sloping surface;

the lateral forces F on the two vertical planes ac and bd Equilibrium conditions are expressed and lead to the following results:

from the sum of forces on a direction parallel to the sloping surface S = 0;

from the sum of forces on vertical direction: V = G;

from the moment in respect to the middle of the base cd: T = 0.

Fig. 1.25

It follows that the force E acting on the vertical plane is parallel to the sloping surface and the force V acting on a plane parallel to the sloping surface is vertical. Therefore, in any point of the soil mass, the vertical plane and the plane parallel to the sloping surface are conjugated planes and forces acting on these planes are conjugated forces.

This is the basic assumption of Rankines theory.

The pressure on the base cd of the prism, at the depth is:

(1.20)

The components of the stress p are:

(1.21)

The angle of deviation between the normal component and the vertical is defined by:

(1.22)

It results that on a plane parallel to the surface, the angle of deviation of the total stress in a point (the inclination of the total stress in respect to the normal on the plane) is equal to , the inclination of the soil surface.

The lateral pressures on the vertical faces ac and bd of the prism are in direct proportion to the vertical pressures p, varying between two limit values, corresponding to the active pressures on to the passive pressures (fig. 1.26). This can be put into evidence by means of the Mohr circles.

Fig. 1.26

In a system of coordinates (), the total vertical stress in a point at depth z is represented by the vector, which makes the angle with the horizontal (because ).

The Coulombs line is known. The problem is to build Mohr circles corresponding to the limit equilibrium, which should fulfill three conditions:

to have the center on the axis

to pass through the point N, at the extremity of the vector ;

to be tangent to the Coulombs line.

There are two circles which fulfill these conditions, corresponding to the two limit states: active and passive (fig. 1.27).

Fig. 1.27

a. Active stateThe Mohr circle corresponding to the active case is represented in the fig. 1.28.

Fig. 1.28

The point N in the circle represents the total stress p on a plane of known direction. A parallel to the plane confounds with the vector (since both the vector and the sloping surface have the same inclination to the horizontal). The parallel intersects the circle in a point P which is the pole.

Once the pole is defined, the total stress on the vertical plane is found by drawing a vertical line from P, which meets the circle in N.

O N = paBecause of the symmetrical position in respect to the axis, ON = OPThe ratio between pa and p should be found. For this purpose, the Mohr circle is used (fig. 1.29).

Fig. 1.29

And, since

(1.23)

Check

For

In order to define the directions of the failure planes, the pole is linked to the points of tangency of the Mohr circle to the Coulombs line. In order to define the directions of the principal planes, the pole is linked to the points A and, respectively B, on the axis O, representing the principal stresses (fig. 1.30).

Fig. 1.30

For a considered point M in the soil mass, at the depth z, MK and ML are the failure planes (fig.1.31). The reactions on the failure planes are R1 and R2. Since MK and ML are failure planes, along which the failure condition is fulfilled in every point, the reaction should be inclined in respect to the normal with an angle . From the Mohr circle resulted that the angle KML is equal to (90o - ). It follows that the direction of R1 is parallel to ML and that the direction of R2 is parallel to MK. Hence, the failure planes are conjugated planes.

Fig. 1.31

All above findings are valid for any point in the soil mass. To the active limit state in the whole soil mass correspond two sets of failure planes (fig. 1.32).

Fig. 1.32

The results can be used for the computation of the earth pressure on a retaining wall, under two conditions (fig. 1.33):

the wall should be vertical

the inclination of the active earth pressure on the wall should make an angle with the normal to the wall (the condition that the vertical plane and the sloping surface are conjugated planes).

Fig. 1.33

The active earth pressure at the base of the wall is:

(1.24)

The angle made by the earth pressure with the normal is usually called and represents the friction angle between the wall and the soil.

Obviously, the friction between the wall and the soil has nothing to do with the inclination of the soil surface.

The condition is an artificial one and puts a severe limitation on the use of the Rankines theory for the computation of earth pressure in the case of a sloping surface.

b. Passive state

The same steps as in the previous case are followed.

The Mohr circle corresponding to the passive state is drawn (fig. 1.34). The total stress p on the plane cd is represented by the vector ON.

Fig. 1.34

The pole P is obtained by drawing from N a parallel with the sloping surface. The total stress on the vertical plane is obtained by drawing from the pole P a vertical.

ON = p

ON = pp = OP

OB = OC1 cos

r = OC1 sin

(1.25)

Check

For = 0

PT2, P are the directions of the failure planes, PB and PD are the directions of the principal planes. (fig. 1.35).

Fig. 1.35

The failure planes as conjugated planes for a point M of the soil mass, at depth z, are shown in fig. 1.36.

Fig. 1.36

The two sets of failure plane, for the case when the whole mass of soil is reaching the passive state are shown in the fig. 1.37.

Fig. 1.37

The use of the Rankines theory for the computation of the passive earth pressure on a vertical wall, requires, as in the active case, to make.

(1.26)

PAGE 30

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