8

CHAPTER 2. LATERAL EARTH PRESSURERankines theory previously presented considers the state of stress in a soil mass when the condition of plastic equilibrium has been reached, meaning that the shear failure is on the point of occurring throughout the entire mass which is subjected to lateral expansion or compression. The theory was applied for the computation of lateral earth pressure on vertical retaining walls of finite depth. Some limitations of the use of the Rankines theory were shown, the main important one being the need to take the angle of friction between the soil and the wall equal to the angle of inclination of the soil surface.

A more general theory of earth pressure was developed by Coulomb (1776) and it will be presented in what follows.

2.1 Coulombs theory of earth pressure

2.1.1 Active case

Coulombs theory involves consideration of the stability, as a whole, of the wedge of soil, between a retaining wall and a trial failure plane (fig. 2.1a). The face AB of the wall makes an angle with the horizontal and the soil surface is inclined at angle to the horizontal. The soil is considered cohesionless.

As a result of a movement of the wall away from the soil, a failure surface will develop. The main assumption of the Coulombs theory is that the failure surface is plane. The shear strength of the soil, defined by , is considered to be fully mobilized along the failure surface BC. From all possible plane failure surfaces originated in the point B, the one leading to the maximum total thrust should be found.

Fig. 2.1 a

Fig. 2.1 b

A trial failure plane BC is chosen, at angle to the horizontal. The equilibrium of the soil wedge ABC is considered, under the following forces:

the own weight (G);

the reaction to the force (P) between the soil and the wall; because at failure the soil wedge tends to move down the plane BC, the reaction P acts at angle below the normal to the wall, where is the angle of friction between the wall and the soil;

the reaction (R) on the failure plane; at failure, when the shear strength of the soil has been fully mobilized, the direction of R is at angle below the normal the failure plane, R being the resultant of the normal and shear forces on the failure plane.

The directions of all three forces and the magnitude of G are known, therefore the problem is statically determined. The triangle of forces can be drawn (fig. 2.1b) and the value of P determined for the trial in question.

Using the sine rule:

where

(2.1)

In relation (2.1) the only variable is . In order to find the maximum value of P, corresponding to a particular value of , the first derivate should be computed and annulated.

(2.2)

From the relation (2.2), a particular value is found and, then:

P() = Pmax = Pa

(2.3)

Pa is usually expressed as:

(2.4)

where Ka is the coefficient of active earth pressure.

Ka is a function of angles and is given in tables in various handbooks.

For the particular case: , value which was previously obtained by applying Rankines theory.

In this case, the total active thrust is:

(2.5)

Culmanns graphical solution

A graphical transposition of the Coulombs method was developed by Culmann.

A line BD, inclined at angle to the horizontal is drawn (fig. 2.2). Since the line BD defines the stable slope of the non-cohesive soil with an angle of internal friction , the failure surface should be located above the line BD, called reference line.

Fig. 2.2

Another line BE, called orientation line, at an angle to the BD line is drawn.

Trial failure surface BC1, BC2, BC3... are considered.

The weight G1 of the wedge ABC1 is computed and represented at a chosen scale on the reference line BD. From the extremity of the vector G1, a parallel to the orientation line BE is drawn. The parallel intersects the failure plane BC1 in the point P1. The vector OP1 represents the reaction P1 corresponding to the wedge of soil ABC1.

One should check that the triangle BG1P1 in the Culmanns construction is equal to the triangle of forces G1, P1, R1, but rotated at an angle (90o + ) to the vertical. The side BG1 is equal to the weight G1. The two angles adjacent to the side BG1 are: BG1 P1 = and P1BG1 = (. The two triangles are equal.

The procedure is repeated for the wedges ABC2, ABC3, ABC4, giving the corresponding forces P2, P3, P4. The extremities of the vectors P1, P2, P3 are linked with a continuous curve. A tangency to the curve parallel to the reference line defines the maximum value Pmax = Pa.

The direction of the failure plane is obtained by linking the point of tangency with the point B.

Active earth pressure distribution

The analytical approach by Coulomb or its graphical transposition by Culmann lead to the determination of the total active thrust Pa.

In practice, however, is necessary to know also the distribution of the active earth pressure on the wall.

In the case of a plane soil surface, it is commonly accepted a linear distribution with depth of the earth pressure (fig. 2.3).

Fig. 2.3

The area of the triangle of pressures should be equal to the force Pa which was established analytically or graphically. From that condition, the ordinate of the pressure diagram at the base, paH, is found.

AE = AB

(2.6)

For the particular case

(2.7)

2.1.2 Passive case

The basic assumption of a plane failure surface is preserved. As a result of a move of the wall against the soil, part of the soil mass behind the wall slips relative to the rest of the mass along the plane surface BC (fig. 2.4a).

From all possible failure planes passing through the point B, should be found the one corresponding to the minimum resistance of the soil behind the wall, which is the passive resistance.

The computation is following the same lines as in the active case.

To the arbitrarily chosen angle of the failure plane to the horizontal, the wedge ABC corresponds.

The reaction P on the wall acts at angle above the normal to the wall.

At failure, when the shear strength of the soil has been fully mobilized, the direction R is at angle above the normal to the failure plane, R being the resultant of the normal and shear forces on the failure plane.

Fig. 2.4 a & b

The triangle of forces can be drawn (fig. 2.4b) and the value of P determined for the trial in question.

Using the sine rule:

where:

(2.8)

From the relation (2.7), a particular value is found.

P () = Pmin = Pp

(2.9)

Pp is usually expressed as:

(2.10)

where: Kp is the coefficient of passive resistance

Kp is a function of angles and is given in tables in various handbooks.

For the particular case: value which was previously obtained by applying Rankines theory.

In this case, the total passive resistance is:

(2.11)

Culmanns graphical solution

A line BD, inclined at angle, below the horizontal, called reference line, is drawn (fig. 2.5).A line BE, called orientation line, at an angle to the BD line, is drawn.

Trial failure surfaces BC1, BC2, BC3 ., are considered.

The weight G1 of the wedge ABC1 is computed and represented at a chosen scale on the reference line BD. From the extremity of the vector G1, a parallel to the orientation line BE is drawn. The parallel intersects the failure plane BC1 in the point P1. The vector OP1 represents the reaction P1 corresponding to the wedge of soil ABC1. The triangle BG1P1 in the Culmanns construction is equal to the triangle of forces G1 P1 R1, but rotated at an angle (900 - ) to the vertical.

The procedure is repeated for the wedges ABC2, ABC3, ABC4, giving the corresponding forces P2, P3, P4 .

Fig. 2.5

The extremities of the vectors P1, P2, P3 are linked with a continuous curve. A tangency to the curve parallel to the reference line defines the minimum value Pmin = Pp.

The direction of the failure plane is obtained by linking the point of tangency with the point B.

Passive earth resistance distribution

As for the active case, it is assumed a linear distribution with depth of the passive earth resistance (fig. 2.6).

Fig. 2.6

The area of the triangle of passive earth resistances should be equal to the force Pp which was established analytically or graphically. From that condition the ordinate of the passive earth resistances diagram at the base, ppH, is found.

(2.12)

For the particular case

(2.13)

2.1.3 The influence of the cohesion on the lateral earth pressure

The influence of the cohesion on the lateral earth pressure was already outlined in the p. 1.2.3, when Rankines theory was used for the case of vertical wall limited by horizontal soil surface.

In the general case of an inclined wall limited by a sloping soil surface, Coulombs theory can be extended to account for the presence of cohesion (fig. 2.7a).

The shear strength fully mobilized on the failure surface is:

(2.14)

As a result, on the failure surface, besides the reaction R at an angle below the normal (resultant of the force N and of the force N tan) will occur a force C due to the cohesion:

(2.15)

Fig. 2.7 a &b

The total reaction Q on the failure surface is the sum of R and C (fig. 2.7b).

For a given failure plan, the cohesion force C is known both in magnitude and direction. The equilibrium of the soil wedge will imply to decompose the resultant of forces G and C on the directions of P and R (fig. 2.8). A number of trial failure planes would have to be selected and polygons of forces similar to the one in fig. 2.8 to be constructed. In this way, the maximum active thrust Pa is obtained (fig. 2.9).

Fig. 2.8Fig. 2.9

A more refined computation will consider also the influence of the adhesion force Ca developed along the wall and the fact that cracks might develop within the tension zone at a depth . Hence, not only that there is no shear strength to be mobilized above the zo line, but a hydrostatic pressure Ew might be involved in the equilibrium.

A trial failure plane BC1 is considered.

There are 4 forces known both in magnitude and direction: G1 = weight of the prism of soil ABC1C11

, where ca can be taken 0,5 c

(2.16)

(2.17)

There are two forces for which only the direction is known: P and R. They are found from the polygon of forces expressing the equilibrium condition (fig. 2.10).

The construction is repeated for other trial failure planes BC2, BC3etc, to find the failure surface for which Pmax = Pa (fig. 2.11).

Fig. 2.10Fig. 2.11

2.1.4 Earth pressure in the case of a uniformly distributed surcharge pressure over the entire surface of the soil mass

If a uniformly distributed surcharge pressure of q per unit area acts over the entire surface of the soil mass (fig. 8.12) an increase of the active earth pressure is expected. In order to determine the additional pressure, the Coulombs approach is used.

To the weight of the wedge of soil ABC the surcharge q is added:

(2.18)

where

(2.19)

The physical meaning of He is the following one: the surcharge pressure q is fictitiously replaced by a layer of soil of unit weight and height h; as a result, the wall will be extended fictitiously with a height He.

To draw He, the following procedure is used (fig. 2.13).

compute ;

draw a parallel to the soil surface at the distance h;

the parallel intersects the extension of the face AB of the wall in the point ;

draw from a horizontal;

the distance between the horizontal line drawn from and the top of the wall is He, the equivalent height.

To check the outlined procedure, one should refer to the fig. 2.13.

Fig. 2.13

In the triangle

EMBED Equation.3 A: He =

EMBED Equation.3 = A sin

The relation (2.19) was found.

Particular case:

He = h =

(2.20)

In the relation (8.18) the multiplier () does not depend on . The triangle of forces will include Gtotal Ptotal R, where Paq is the total active thrust including the effect of the surcharge q. The sine rule is applied, leading to:

(2.21)

By derivation,, the term () which is constant in respect to is preserved. Therefore, Paq and P will be in the same relation as Gtotal and G:

(2.22)

To determine the modified diagram of active earth pressures, taking into account the surcharge pressure q, the following procedure is used (fig. 2.14).

The pressure diagram corresponding only to the weight of the wedge ABC is drawn. The ordinate at the base is paH = . The equivalent height is determined.

The pressure p at the base of the fictitious wall of height He is computed:

The additional pressure p is constant along the height H of the wall.

The total pressure diagram, including the effect of the surcharge pressure has two parts:

the triangle of pressures representing Pa the rectangle of pressures representing

Fig. 2.14

Particular case: (fig. 2.15)

(2.23)

Fig. 2.15

A similar procedure is used in the case of the passive resistance. For instance, when (fig. 2.16)

(2.24)

Fig. 2.16

The case of stratified soil mass

The concept of equivalent height is used also to compute the earth pressure in the case of a stratified soil mass.

In fig. 2.17 is shown a vertical wall limited by a horizontal surface on which is acting a uniformly distributed surcharge pressure q. There are two layers of soil behind the wall, with characteristics and, respectively .

Fig. 2.17

The following approximate method is applied in order to compute the active earth pressure diagram.

Computation starts with the top layer 1. The surcharge pressure q is replaced by layer of soil with unit weight and height . The ordinates of the earth pressure diagram at the top and at the base of the layer 1 are then computed:

(2.25)

(2.26)

Passing to the layer 2, the vertical pressure of the layer 1, including the fictitious layer of height he1 replacing the surcharge pressure q, represents a surcharge pressure for the layer 2. According to the known rule, this must be replaced by a layer of soil with unit weight and equivalent height

(2.27)

The ordinates of the earth pressure diagram at the top and at the base of the layer 2 are computed.

(2.28)

(2.29)

If the soil would have been formed by n layers, the relations for the n layer are:

(2.30)

(2.31)

As one can realize, the earth pressures diagram has jumps at the lines of separation between layers. The jumps occur when the angles of internal friction of various layers are different. If the angle would have been constant and only the unit weights different, the diagram would present only changes in slope from one layer to the next one.

In the general case of a wall with inclined face and sloping surface of soil, the Coulombs method is used, starting with the upper layer, transformed then in a surcharge pressure for the next layer a.s.o.

The earth pressures diagrams with jumps at the change of the layer are considered as conventional ones because admit than in the same points of the wall are acting simultaneously two pressures ( and a.s.o.). In reality, the earth pressures diagram should be a continuous one, without jumps. Nevertheless, for practical purpose the procedure described is accepted.

2.1.5 Earth pressure in the case of a concentrated, linearly distributed, surcharge on the surface of the soil mass

The surcharge Q is distributed per unit length of the wall and is located at a certain distance from the crest of the wall (fig. 2.18).

Fig. 2.18

To establish the magnitude of the total active thrust PaQ, including the effect of the surcharge Q, the graphical method of Culmann is used.

Trial failure planes AC1, AC2...etc are considered in turn and the weights of the soil wedges ABC1, ABC2, including, when occurs, the surcharge Q, are plotted on the reference line. Culmanns curve has a jump when the failure plane passes through the point of application of the surcharge Q. In most cases, this gives the maximum value of the thrust.

PaQ = Pa + Pa

(2.32)

There are, however, situations when the failure plane does not pass through the point of application of the surcharge Q or when the surcharge is so far in respect to the crest A of the wall that it does not influence the magnitude of the active thrust.

Influence of the position of the surcharge Q on the magnitude of the active thrust

Two Culmanns curves are drawn (fig. 2.19).

The curve 1 corresponds to the case when the surcharge Q is considered to act from the very first wedge.

The curve 2 corresponds to the case where there is no surcharge.

Except the initial part, the two curves are parallel.

Tangents are drawn to each of the two curves, by means of which three zones are defined on the soil surface.

Fig. 2.19

Zone I

AC1The point C1 is obtained by linking B with the point of tangency to the curve 1. PaQ is the maximum total active thrust. No matter where Q is located inside this zone, the additional thrust is constant.

Zone II

C1C2The tangency to the curve 2 intersects the curve 1 in the point M. Linking the point B with the point M and extending the line BM, the point C2 is defined on the surface. When the surcharge Q acts in the zone II, the failure plane is passing through the point of application of the surcharge. The additional active thrust Pa varies between a maximum, corresponding to the point C1 and zero, corresponding to the point C2.

Zone III

To the right of C2. For a given failure plane BC; one can realize that the thrust P, given by the intersection with the curve 1 is smaller than the thrust Pa corresponding to the curve 2, when there is no surcharge. In conclusion, for any point to the right of C2, the surcharge does not influence the magnitude of the active thrust.

The variation of the additional thrust Pa in the zones I and II is shown in the fig. 2.20.

Fig. 2.20

Changes of the active earth pressures diagram

a. The surcharge Q acts in zone I (fig. 2.21)

Fig. 2.21

From the point of application of the surcharge Q a parallel is drawn to the line of natural slope, defining the point M and another parallel is drawn to the failure line BC1, defining the point N. Changes in the pressures diagram occur between the ordinates corresponding to the points M and N. The area of the triangle of additional pressures is equal to Pa.

b. The surcharge acts in zone II (fig. 2.22)

The failure plane passes through the point of application of the surcharge Q. A parallel is drawn from this point to the line of natural slope, defining the point M. Changes in the pressures diagram occur below the ordinate corresponding to the point M. The area of the triangle of additional pressures is Pa.

Fig. 2.22

Both diagrams shown in the fig. 2.21 and 2.22 are conventional ones, since they admit jumps in the point M but, nevertheless, they are accepted in the practice.

2.1.6 Earth pressure in cases with particular geometry

a. Face of the wall with two slopes (fig. 2.23a)

The Culmanns procedure is used. Computation starts with the face AB1, considered as a distinct wall. The total active thrust is obtained by drawing the tangency to the Culmanns curve parallel to the reference line. The corresponding pressure diagram on the face AB1 is then computed (fig. 2.23b).

Fig. 2.23 a&b

The face B1B2 of the wall is considered as part of a fictitious wall with the face B1B2. The Culmanns construction is used for this wall, the total active thrust on the face B1B2 and the corresponding pressure diagram are obtained. From the total pressure diagram on the fictitious wall B1B2 is then considered only the part corresponding to the real face B1B2 in order to define Pa2.

b. Ground surface with two slopes (fig. 2.24)

The magnitude of the total active thrust can be computed by replacing the wall AB with a fictitious wall B chosen in such a way that the weight of the failure wedge remains unchanged. For that purpose, the point B is linked to the point C, where the change in the slope of the ground surface occurs, and then a parallel to BC is drawn from A, until it reaches in the extension of the slope CD.

Fig. 2.24

B fulfills the required condition:

because the two triangles have the same base (BC) and the same height h (the distance between the parallels A and BC).

The total active thrust Pa on the fictitious wall AB and one slope ground surface CD can be found using Culmanns method, with the observation that the orientation line should consider the inclination of the face AB.

To determine the earth pressure diagram, the one-slope ground surface ACM is considered and the Culmanns method is used, thus obtaining the failure surface BC1 and the pressure diagram abc. The total active thrust corresponding to the wall AB limited by the one-slope ground surface ACM is larger than Pa previously obtained, because it involves the additional soil mass between the surface CD and the line CM, which in reality does not exist.

A parallel CN to the failure plane C1B is drawn. A correction should be made in the pressure diagram below the ordinate corresponding to the point N. From the triangle of pressures abc should be subtracted a triangle of pressures npc having the area equal to .

The base of the triangle npc is:

(2.33)

The total active thrust on the wall AB, limited by the two-slopes ground surface is the area of pressures abpn and is acting in the centre of gravity of that area.

2.2 The influence of the wall-soil friction and of the deformation condition of the wall on the lateral earth pressure

2.2.1 Soil-wall friction

The friction between the soil and the wall is considered by means of the friction angle, which is the angle to the normal of the total active thrust Pa. Depending on the relative movement between the soil and the wall, the force Pa will be below or above the normal.

When the wall moves away from the soil and tilts (fig. 2.25), Pa is above the normal, with a component along the wall Pav which is a friction force and a component normal to the wall Pah. In this case, which is the most common one, can be taken as positive.

Fig. 2.25

There is also a case when the wall is founded on a very compressible soil (fig. 2.26). As a result, the wall will settle. The friction force Pav is opposing to the downward movement of the wall. As a consequence, the force Pa will be below the normal.

Fig. 2.26

Computations performed with different values of have shown that has a small influence on the magnitude of Pa. Instead, the variation of the horizontal component of Pa which is Pah = Pa cos can have a significant influence on the stability conditions of the wall.

In general:

(2.34)

In practice, it is recommended:

< <

(2.35)

2.2.2 Deformation condition of the wall

Fig. 2.27

A vertical wall limited by horizontal soil surface is considered (fig. 2.27). As shown in the p. 1.2.1, the active state can be developed only within a wedge of soil between the wall and a failure plane passing through the lower end of the wall and at an angle of () to the horizontal. The remainder of the soil mass is in a state of elastic equilibrium. A certain minimum value of lateral strain is necessary in order to develop the active state within the wedge. A uniform strain within the wedge can be produced by a rotational movement (B) of the wall, away from the soil, about its lower end and a deformation of this type, of sufficient magnitude constitutes the minimum deformation condition for the development of the active state and for an active earth pressure varying linearly with depth (fig. 2.28).

Fig. 2.28

Fig. 2.29

If the wall deforms by rotation about its upper end (fig. 2.29) the condition for the development of the active state is not satisfied since adequate strain in the soil near the surface cannot take place. The soil adjacent to the wall in its upper part cannot reach the limiting equilibrium and remains in elastic state. The prism of soil limited by the failure surface moves downward. To render possible such a movement, the failure surface should intercept the horizontal ground surface under a right angle. The failure surface cannot be taken any longer as a plane one, like in Coulombs theory, but with sufficient approximation can be taken as a logarithmic spiral. The soil mass at the lower end of the wall is obstructed to slip by the friction forces developed at the contact with the part of the soil mass remained in elastic state. On that basis, the soil mass which tends to slip transfers the pressure to the upper part of the wall. This is an expression of the arching effect (the part of the soil mass which can move is discharged by transferring the pressure to the upper part which is blocked). Due to the arching effect, the distribution of the earth pressure is no longer linear, but parabolic. The point of application of the total active thrust is situated between 0,45 H and 0,55 H from the bottom of the wall (fig. 2.30).

Fig. 2.30

2.3 Lateral earth pressure in the hypothesis of curved failure surfaces

2.3.1 Active case

A vertical wall limited by a horizontal ground surface is considered (fig. 2.31). The crest of the wall can move upward and away from the soil, thus creating the condition for the development of the active state. Due to the friction between the soil and the wall, shear stresses occur along the wall. Consequently, the vertical face of the wall cannot be a principal direction, as was the case in the Rankines theory. The last failure plane with an angle () to the horizontal, corresponding to the Rankine active zone is the one passing through the crest of the wall. Between this plane and the wall the failure surfaces are curved, as a result of the influence of the shear stresses along the wall. The failure surface is a mixed one, composed by a curve and by a plane making the angle () to the horizontal.

Fig. 2.31

In the active case, the curvature is slight and the error involved in assuming a plane surface is relatively small, less than 5%, which is fully allowable.

In conclusion, in the active case it is not justified to compute the lateral earth pressure considering curved failure surfaces.

2.3.2 Passive case

A vertical wall limited by a horizontal ground surface is considered (fig. 2.32). The crest of the wall can move downward and against the soil, thus creating the condition for the development of the passive state.

The last plane corresponding to the Rankine passive zone passes through the crest of wall. The wedge of soil influenced by the shear stresses on the wall is much larger than in the active case. Between the results of the computations performed using Coulombs theory (plane failure surface) and those obtained in the Theory of Plasticity with curved failure surface, differences are great not only concerning the shape of the failure surface but, more important, concerning the magnitude of the total passive resistance (40% or even more).

Fig. 2.32

In what follows, an approximate method is presented, in which is considered that the failure surface is assumed to consist of a circular arch BC (centre 0, radius R) and a straight line DC, which is a failure line belonging to the Rankine passive zone, at an angle () to the horizontal (fig.2.33).

Fig. 2.33

If the soil behind the wall has both internal friction and cohesion, the computation is made in two steps.

a. soil with internal friction and weight (

The failure plane drawn from the crest of the wall A, at an angle () to the horizontal separates the Rankine passive zone from the zone of curved failure surface. An arbitrary point D is selected on that plane (fig. 2.34).

Fig. 2.34

In order to find the centre 0 of the circular arc BD, a normal in the point D to the line DC is drawn as well as a normal in the middle of the chord BD. At the intersection of the two normals, the centre 0 is found. When the wall displacement is such that the total passive resistance is fully mobilized, the soil within the triangle ADC is in the passive Rankine state, both angles EAD and ECD being (). The horizontal force (Ep) on the vertical plane DE, therefore, is the Rankine passive value given by equation (2.36), acting horizontally at a distance DE/3 above the point D.

Ep =

(2.36)

It is necessary, then, to analyse only the stability of the soil mass ABDE. The forces to be considered are:

the weight (G) of ABDE, acting through the centroid of the section;

the force Ep on DE;

the reaction (P) to the face between the soil and the wall, acting at angle above the normal and at a distance of AB/3 above B;

the reaction (Q) on the failure surface BD when the shear strength along the circular arc BD is fully mobilized, the reaction Q being assumed to act at an angle to the normal.

In order to find the direction and the point of application of the reaction Q, the arc BD is divided in a serie of elements of length ds, with the elementary reaction qds applied making the angle with the normal (the radius) in the point of application.

From the centre 0 of the circular arc of radius R, a normal OT is drawn on the line AD, which is the line of action of the elementary reaction qds in the point D.

OT = OD sin = R sin

A circle of radius r = R sin is drawn. This circle is referred to as the friction circle or -circle.

Since all elementary reactions make the angle with the radius R, it is obvious that they are tangent to the - circle, which can be defined as the locus of the points of tangency of the lines of action of the elementary reactions qds.

In order to find the direction of the force Q, which is the resultant of the elementary reactions qds, the moment in respect to the centre 0 is taken:

(2.37)

, since a polygonal contour is larger than the straight line between the extremities of the contour (fig. 2.35).

Fig. 2.35

It results than d > R sin

By approximation, and d = R sin (the error involved is small and the assumption made is on the safe side).

The values of forces G and Ep are known and their resultant (S) is determined graphically (fig. 2.36). For equilibrium the lines of action of forces S, P and R must intersect. Therefore, the line of action of Q must pass through the intersection of S and P and be tangential to the - circle. In this way, the direction of Q is fixed. The polygon of forces can now be completed and the value of P determined.

Fig. 2.36

The point D on the failure line passing through A was chosen arbitrarily. The analysis must be repeated for a number of failure surfaces corresponding to other points D1, D2etc. By plotting the values of P against the ground surface taken as a reference line (fig. 2.37) a curve is obtained. The tangential to the curves gives Pmin = Pp.

Fig. 2.37

b. soil having only cohesionThe superposition of effects will be used:

(2.38)

where is the passive resistance due to the internal friction and to the weight of the soil (determined as it was shown)

is the additional passive resistance due to cohesion.

In order to find for the same failure surface BDC, the following procedure is used (fig. 2.38).

Fig. 2.38

The passive resistance corresponding to the wedge EDC is:

(2.39)

The forces to be considered to analyse the stability of the soil mass ABDE are:

Ca - the resultant of the adhesion between the soil and the wall; Ca = ca

C - the resultant of the cohesion forces on the circular arc BD;

- the passive earth resistance due to the cohesion on the vertical plane DE, corresponding to the failure surface DC;

- the reaction on the failure surface, tangential to the - circle

- the reaction to the face between the soil and the wall, due to cohesion, acting at angle above the normal and at a distance of AB/2 above B.

In order to find the magnitude, the point of application and the direction of the force C, the circular arc BD is divided in elements ds on which are acting the elementary forces cds, tangential to the elements and directed contrary to the slide.

Two elements disposed symmetrically in relation to the middle of the chord are considered (fig. 2.39). Components of the two elementary forces cds, normal to the chord and parallel to the chord, are drawn. As one can realize, the components normal to the chord are equal but opposite in sign, while the components parallel to the chord are equal and of the same sign.

Fig. 2.39

Therefore, the force C = c

To find the line of action of the force C, the moment in respect to the centre of the circular arc is considered.

(2.40)

Where: a is the distance from the line of action of the force C to the centre 0.

The force P is obtained as follows:

Forces Ca and C are composed giving a resultant Ct Ct is composed with giving a resultant S

The forces S, P and Q must intersect, therefore the line of action of Q must pass through the intersection of S and P and be tangential to the -circle. The polygon of forces is completed and P is determined.

The analysis is repeated for the same points D1, D2, for which P was previously determined (for the case .

By summing P and P for each trial and representing the forces P+ P against the reference line (fig. 2.40) a curve of variation is drawn. The tangential to the curve parallel to the reference line defines Ptot min = Ptot p, which is the total passive resistance including the effect of both internal friction and cohesion.

Fig. 2.40

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