EARTH PRESSURES ON RETAINING STRUCTURES

Transcription

1 EARTH PRESSURES ON RETAINING STRUCTURES 12.1 Active Pressure and Passive Pressure When a sudden change in level of the ground surface is to be provided for some purpose a retaining structure is commonly employed to enable this to be done without risk of the higher portion of the ground collapsing on to the lower portion. The higher portion of the ground is commonly referred to as the backfill since this soil is usually backfilled after construction of the retaining wall. Some cross-sections of typical retaining walls are shown in Figure The forces produced by the lateral earth pressures acting on the walls are indicated by the arrows. This chapter is mainly concerned with techniques for the calculation of the magnitude and distribution of lateral earth pressures acting on such walls under conditions likely to be encountered in practice. Such information is essential for the safe design of a retaining wall. In the early part of this chapter the case depicted in Figure 12.1(a) (i.e. vertical back of wall, horizontal backfill and earth pressure acting horizontally)will be examined. In order to illustrate the way in which the magnitude of the earth pressure acting on a retaining wall may vary, a somewhat idealised example is provided in Figure Part (a) of the figure depicts an element of soil at a depth z below the horizontal ground surface in a uniform deposit of dry sand. The vertical and horizontal stresses on this element are given by σ v and σ h respectively where: and where σ v = ρ d gz σ h = Koσ h ρ is the dry density of the sand K0 d is the coefficient of earth pressure at rest (See section 3.1 Geomechanics 1) Both of these stresses are clearly principal stresses and they display a linear increase with depth. No distinction here is made between total and effective stresses since there are no pore pressures.

2 12-2 Figure Typical Retaining Walls Figure Effect of Lateral Wall Movement

3 12-3 In Figure 12.2(b) a portion of the soil has been replaced by a retaining wall and it is assumed that this has been done without disturbing the soil and without changing any of the stresses acting on a vertical plane AB. This means that the distribution of horizontal stress still remains triangular as shown. If this stress is integrated over the depth z the horizontal force P o acting on the wall may be found: P o = = 1 2 σ z per unit length of wall h 1 2 K o ρ d gz2 (12.1) The other forces acting on the retaining wall will be the weight of the wall W w and resisting force R acting on the base BC of the wall. The force R acts vertically since BC is a principal plane. Clearly these three forces are not in equilibrium. The wall will move towards the left in the diagram by translation or anticlockwise rotation under the action of the force P o. As the wall moves a horizontal shear force will be mobilised along the base BC and force P o will decrease until equilibrium is reached. Figure 12.2(c) illustrates the way in which the force on the wall decreases as the movement occurs (Terzaghi, 1934). The force on the wall reaches a minimum value P A after a certain amount of movement has taken place. This minimum force corresponds to a state of failure being developed in the backfill as a result of lateral yielding of the soil. Associated with the wall movement there would generally be a tendency for a shear stress to develop on the plane AB along the back of the retaining wall. For the time being it will be assumed that the back of the wall is perfectly smooth so that the horizontal force will continue to act horizontally at right angles to the back of the wall. This minimum force P A is known as the active force or active thrust and is derived from the active earth pressure acting on the back of the retaining wall. It is widely assumed that retaining walls should be designed to resist this active thrust. In the preceding discussion the wall yielded towards the left under the effects of the earth pressure from the backfill. If, on the other hand the wall was pushed horizontally towards the right the earth pressure on the back of the wall would increase instead of decreasing. The horizontal force resulting from this earth pressure increases to a maximum value P P as shown in Figure 12.2(c). This maximum would be reached when the backfill soil is in a state of failure as a result of lateral compression. The force P p is known as the passive force or passive thrust and it is derived from the passive earth pressure. For design purposes P P is the maximum

4 12-4 possible thrust that the wall can mobilise The Coulomb Theory A number of theories have been proposed for the calculation of the active thrust P A against a retaining wall. One of the simplest and most widely accepted is the Coulomb theory which was published in 1773 (see Terzaghi (1943), and Taylor (1948)). In this theory the equilibrium of a sliding wedge of soil behind the retaining wall is examined. The wedge is shown by the triangle ABD in Figure 12.3(a). As the wall yields towards the left it is assumed that a wedge of soil ABD will slide down a plane BD inclined at some angle θ to the horizontal. In sliding down the plane a shear force acting up the plane is mobilised to resist the motion. When this shear force T(= N tan φ ) is fully mobilised the dry sandy soil will be in a state of failure. When this condition is reached the reaction P of the wall on the soil (equal and opposite to the force of the soil on the wall) will be the active thrust P A. Two major assumptions are made with this theory - the surface of failure BD of the soil wedge is plane and the force P acts in a known direction. The forces acting on the wedge are the wall reaction P acting horizontally (since the back of the wall is smooth), the force W due to the mass of the soil wedge ABD and the force F on the failure plane BD. The force F has two components N and T as shown in Figure 12.3(a). Summing forces in the vertical direction yields: W - T sin θ - N cos θ = 0 that is 1 2 ρ d g H2 cotθ - N (cos θ + sin θ tan φ) = 0 where ρ d is the density of the dry sand H is the height of the retaining wall N = 1 2 ρ d gh2 cot θ (cos θ + sin θ tan φ) (12.2)

5 12-5 Figure Forces acting on a wedge in dry sand - active case

6 12-6 Summing forces in the horizontal direction yields: P - N sin θ + T cos θ = 0 P = N (sin θ - cos θ tan φ) and if equation (12.2) is substituted into the above equation P = 1 2 ρ d gh2 cot θ (sin θ - cos θ tan φ) (cos θ + sin θ tan φ) (12.3) Equation (12.3) shows the way in which P varies with the wedge angle θ and this relationship is illustrated in Figure 12.3(b). In order to find the particular value of P corresponding to P A it is necessary to find the critical wedge angle (θ cr ) at which the maximum value of P is required for equilibrium of the wedge. This may be found by maximising P in equation (12.3). For the condition that: dp dθ = 0 it is found that: θ = θ cr = 45 + φ /2 (12.4) If the equation (12.4) is substituted into equation (12.3) the value of the active thrust P A per unit length of wall may be found: P A = 1 2 ρ d g H2 ( 1 - sin φ 1 + sin φ ) (12.5) P A = 1 2 ρ d g H2 K A (12.6) where K A is known as the coefficient of active earth pressure for the conditions assumed in the analysis.

7 12-7 Figure Rankine approach for Active Earth Pressure

8 The Rankine Theory With the Rankine theory which was presented in 1860 an expression for the lateral earth pressure is developed from a consideration of the stresses existing in a soil when a state of failure throughout the soil has been produced as a result of lateral yielding. The development may be illustrated by means of the Mohr circle. In Figure 12.4(a) the lines OF represent the failure envelopes for the dry sand which has a friction angle of φ. The initial stresses at a point a distance z below the horizontal ground surface are represented by the Mohr circle M R Q where: OQ = vertical stress = σ 1 = ρ d gz OM = horizontal stress = σ 3 = K0ρ d gz As lateral yielding takes place the value of σ 3 will decrease whereas σ 1 will remain constant. The stress path representing this lateral yielding is RS. Had the initial stress conditions been isotropic (i.e. K0 = 1) the stress path would be QS. Point S corresponds to the failure Mohr circle with points of tangency with the failure envelope at points F. The inclinations of the failure planes are given by the lines drawn through the origin of planes (N) to the failure points (F). From the geometry of Figure 12.4(a) it is seen that the failure lines, for any depth below the ground surface are inclined to the horizontal by an angle of (45 + φ /2), in agreement with the angle found by means of the Coulomb theory. With the origin of planes located at point N it is clear that the minor principal stress σ 3 (= ON) represents the stress on the vertical plane. From the geometry of Figure 12.4(a) the major and minor principal stresses for the failure circle are related as follows: or 1 sinφ ON = OQ 1 + sinφ 1 sinφ σ 3 = OQ 1 + sinφ

9 sinφ or active pressure p A = ρ gz 1 + sinφ d (12.7) = ρ d gzk A Since the active pressure P A is directly proportional to depth z below the ground surface the distribution of active pressure is linear as shown in Figure 12.4(b). For a retaining wall of height H the active thrust P A may be found by integration of the active pressure P A over the height H to yield: P A = 1 2 ρ d g H2 K A (12.6) in agreement with the answer obtained form the Coulomb theory. Since the active pressure displays a triangular shaped distribution in Figure 12.4(b) the active thrust acts through a point equal to one third of the wall height above the base of the retaining wall Effect of Surcharge Pressure and Submergence on Active Pressure When the groundwater table is coincident with the top of the backfill as illustrated in Figure 12.5(b) a distinction needs to be made between effective stresses and total stresses in the calculation of earth pressure. With a submerged sand backfill the calculation is invariably made in terms of effective stresses and a hydrostatic pore pressure distribution is taken from the level of the water table. This means, in effect, that the long term earth pressure is calculated. The short term earth pressure is of little practical interest in cases where the sand is sufficiently permeable that any pore pressures developed as a result of the stress changes (caused by lateral yielding of the wall - see Chapter 7 Geomechanics 1) dissipate immediately. The vertical stress σ v at any depth z below the surface of the backfill is: σ v = σ 1 = ρ sat gz + q where ρ sat q is the saturated density of the sand is the surcharge pressure applied to the surface of the backfill.

10 12-10 (a) Positions of Mohr circles for any depth Z Figure Effect of surcharge and submergence on active pressure in sand

11 12-11 The effective vertical stress σ v ' is: σ v' = σ 1 ' = ρ sat gz + q - ρ w gz = ρ b gz + q The effective vertical stress is plotted at point A in Figure 12.5(a). The failure Mohr circle corresponding to this stress has been drawn tangent to the failure line which is inclined at an angle of φ' to the horizontal. The origin of planes is located at point B so the effective stress acting on a vertical plane is given by OB, the minor principal effective stress at failure. That is: σ 3f ' = p' A = σ 1 ' K A = ρ b gz K A + q K A where K A = 1 - sin φ' 1 + sin φ' The total active pressure at depth z acting on the back of the retaining wall is: p A = p A ' + u = ρ b gz K A + q K A + ρ w gz (12.8) This total active pressure is represented by the distance OC in Figure 12.5(a). Equation (12.8) shows that two terms of the expression show a direct proportionality with z whereas the surcharge term is constant with increasing depth. The distribution of active pressure is illustrated in Figure 12.5(b). The total active thrust P A may be found by integration of the pressure over the height H of the retaining wall. It should be noted that the active pressure coefficient K A does not appear in the pore water pressure term. P A = 1 2 ρ b gh2 K A + qhk A ρ w gh2 (12.9)

12 12-12 Example A retaining wall 3m high retains a backfill of saturated sand as shown in Figure The water table is coincident with the top of the wall. A surcharge consisting of a 0.5m thickness of dry sand exists above the saturated sand. Assuming that the back of the wall in perfectly smooth determine the magnitude and location of the active thrust. Dry density of sand ρ d = 1.8 Mg/m 3 Saturated density of sand ρ sat = 2.1 Mg/m 3 Drained friction angle φ d = 30 The magnitude of the active thrust may be calculated by means of equation (12.9) using the following terms: K A = 1 - sin φ d 1 + sin φ d = = 1 3 Since the active thrust is being determined for drained conditions in the backfill the value of K A is defined in terms of the drained friction angle φ d. Surcharge pressure q = 1.8 x 9.81 x 0.5 = 8.83 kn/m 2 Substituting into equation (3.9) P A = 1 2 x 1.1 x 9.81 x 9 x x 3 x x 1.0 x 9.81 x 9 = = kn/m of wall The largest component of the total active thrust in the calculation above is the force produced by the water pressure. In order to reduce the total force that a retaining wall must withstand designers try to ensure that water will be kept drained from the backfill by provision, for example, of drainage holes through the retaining wall.

13 12-13 Figure 12.6 Figure Active pressure with a cohesive soil

14 12-14 The three components of P A calculated above act at distances above the base of the wall of 1m, 1.5m and 1m respectively. If the resultant thrust P A acts at a distance of Bm above the base of the wall, then: P A x B = x x x 1.0 = kn B = = 1.06 m above base of wall 12.5 Active Pressure in Cohesive Soils For a soil for which the shear strength is expressed in terms of both c and φ components (either total or effective stresses) the active pressure at any depth below the top of the backfill may be found from the failure Mohr circle shown in Figure The failure circle has been drawn for a particular value of the vertical stress σ v. The origin of planes is located at point A. The failure plane is inclined in the same direction as in the case of cohesionless soil namely at an angle of (45 + φ /2) to the horizontal. The stress acting on the vertical plane is given by the distance OA, which is the active pressure p A. From the geometry of the figure, p and σ A v may be related as follows: p A 1 sinφ 1 sinφ = σ v 2c 1+ sinφ 1+ sinφ 0.5 p = σ K 2c K (12.10) A v A A Using this basic equation the expressions for active pressure at any depth z below the top of the backfill may be written for various conditions.

15 12-15 For the long term active pressure from a saturated clay when the water table is coincident with the top of the backfill and a surcharge pressure q is applied over the surface of the backfill: p A ( ρ gz + q) K c K + ρ gz = 2 (12.11) b A d A w In equation (12.11) the cd is the drained cohesion and K A is calculated from the drained friction angle φ d. The active thrust may be calculated by integration of the active pressure over the height of the wall. For the calculation of the short term active pressure the undrained strength parameters cu and φu should be used. With a saturated soil the value of φu will be zero (as discussed in section 8.7.3, Geomechanics 1) which means that K A will be equal to unity. If there is no surcharge the active pressure will be: p A = ρ sat gz - 2c u (12.12) This pressure distribution which is illustrated in Figure 12.8, indicates that at some depth z o the active pressure is zero. z o = 2c u ρ sat g (12.13) The negative portion of the active pressure diagram between the backfill surface and this depth of z o is normally ignored in the calculation of the active thrust. It is assumed that over this depth a tension crack will form so that no earth pressure will be exerted on the wall Calculation of Passive Pressure By using techniques similar to those discussed above a series of expressions may be developed for the passive case. This will be demonstrated firstly by using a Coulomb approach in which the forces acting on a wedge of soil are put into equilibrium. The same assumptions as used in the active case will apply, namely a horizontal backfill surface and a retaining wall with a smooth vertical back.

16 12-16 Figure Short term active pressure distribution for a saturated clay Figure Forces acting on a wedge of soil - passive case

17 12-17 Consider the most general case in which a surcharge pressure q is applied to the surface of the backfill and the backfill soil possesses both cohesive and frictional strength components. In the passive case where the wall is pushed into the backfill, the soil wedge ABD will tend to move up the plane BD so the shear force T will act down the failure plane BD as shown in Figure W + Q + T sin θ - N cos θ = 0 where Q is the force per unit length of wall due to the surcharge acting over the top of the wedge. That is: N cos θ = W + Q + T sin θ = 1 2 ρ gh2 cotθ + qhcotθ + sinθ (N tan φ + chcosecθ) where ρ is the relevant soil density N = 1 2 ρ gh2 cotθ + qhcotθ + ch (cos θ - sin θ tan φ) (12.14) For horizontal force equilibrium P - T cos θ - N sin θ = 0 P = N sin θ + cos θ (N tan φ + ch cosec θ) (12.15) If the N in equation (12.14) is substituted into equation (12.15) a somewhat lengthy expression is obtained for P, which varies with the wedge angle θ shown in Figure The value of θ at which the minimum value of P occurs may be found by minimising the expression to give: θ crit = 45 - φ /2 and if this angle is substituted into the original expression for P, the passive thrust P P is found: p P = 1 2 ρgh2 1 + sin φ 1 - sin φ + q H 1 + sin φ sin φ + 2cH 1+ sinφ 1 sinφ

18 12-18 Figure Determination of passive thrust Figure Determination of passive pressure using the Mohr Circle

19 12-19 Figure p p = 1 2 ρ gh2 K p + q H K p + 2c K p (12.16) where K p is the coefficient of passive earth pressure. If the Rankine approach is used, an expression for the passive pressure p p at a depth of z below the top of the backfill may be developed. In the passive pressure situation the horizontal stress is greater than the vertical stress consequently at failure as shown in Figure σ 3 = σ v = ρ gz and σ 1 = p p The origin of planes is now located at point A on the Mohr circle so the inclination of the failure plane is (45 - φ /2) to the horizontal in agreement with the critical angle found in the Coulomb approach. From the geometry of the diagram in Figure the major principal stresses are related as follows:

20 12-20 σ 1 = p p = σ v K p + 2c K p (12.17) This general form of expression for passive pressure may be used to write a number of expressions for various conditions. For example for a submerged sand backfill for which the drained cohesion c d = 0, the total long term passive pressure at depth z is: p p = (ρ b gz + q) Kp + ρ b gz (12.18) where Kp = 1 + sin φ d 1 - sin φ d For short term conditions with a saturated clay backfill with no surcharge pressure the value of the passive coefficient Kp = 1 since the undrained friction angle φ u = 0 and the short term passive pressure is: p p = ρ sat gz + 2c u (12.19) As in the active case, integration of all of these passive pressure distributions over the height of the retaining wall gives values of passive thrust in agreement with the answers obtained with the Coulomb theory. Example Draw the distribution of passive pressure against the smooth wall when the backfill consists of the soils shown in Figure Since the undrained strength parameters only have been provided for the clay the short term passive pressure only can be calculated. Regarding the saturated sand it will be assumed that even for the short term case no excess hydrostatic pore pressures are present. For both the dry and saturated sands: Kp = 1 + sin φ d 1 - sin φ d = = 3

21 12-21 The passive pressure in the dry sand is: p p = ρ d gz Kp = 1.8 x 9.81 x z x 3 = 53.0 z kn/m 2 = 42.4 kn/m 2 at z = 0.8m For the saturated sand the calculation will be carried out in terms of effective stresses. The dry sand will be treated as a surcharge and the distance z will be measured from the top of the saturated sand stratum. p p = (ρ b gz + q) Kp + ρ w gz = (1.2 x 9.81 x z x 9.81 x 0.8) x x 9.81 x z = z kn/m 2 = 42.4 kn/m 2 at z = 0 = 96.5 kn/m 2 at z = 1.2m For the saturated clay the calculation must be carried out in terms of total stresses. The dry saturated sands above will be treated as a surcharge and the distance z will be measured from the top of the clay stratum. p p = (ρ sat gz + q) + 2c u = (2.1 x 9.81 x z x 9.81 x x 9.81 x 1.2) + 2 x 80 = 20.6 z kn/m 2 = kn/m 2 at z = 0 = kn/m 2 at z = 2.0m With these figures the figures the distribution of passive pressure may be drawn as shown in Figure

22 Effect of Seepage on Earth Pressure With a submerged backfill in the active pressure case the force derived from the water pressure is often the major component of the total thrust acting on the wall. This point was illustrated in the example in the section This component of the total thrust may be significantly reduced if the steps are taken to drain the water from backfill. This reduction in the thrust occurs even when steady state seepage is taking place. In this situation the expressions which have been developed in the early part of this chapter cannot be used for the determination of the thrust on the wall. The calculation can most conveniently be carried out by means of a graphical method using Coulomb wedge approach. This is illustrated in Fig in which the forces acting on a trail wedge ABD of submerged soil are shown. The forces U1 and U2 which are derived from the pore water pressure must be determined from the appropriate flow net. Force W is the saturated weight of the wedge ABD. Force F is the resultant of the shear force T and the effective normal force N. The effective force acting on the back of the retaining wall AB is indicated by P. The magnitude of P required to satisfy force equilibrium is found from the force polygon in the lower part of Fig The process is repeated for other values of the wedge angle to facilitate the plotting of a diagram of P against θ similar to Fig 12.13(b). From such a diagram the active thrust on the wall may be determined.

23 12-23 Fig Forces acting on soils wedge when seepage occurs Fig Example The soil behind and below a retaining wall 3 m high( Fig 12.14) consists of a saturated sand through which steady state seepage is taking place. The perched water table at the top of the backfill is being maintained by means of rainfall. If the back of the wall is smooth, determine the total thrust acting on the wall. Also calculate the thrust a long time after the rainfall stops, when all of the water in the backfill has been drained away. Dry density of sand ρ d = 1.8 Mg/m 3 Saturated density of sand ρ sat = 1.8 Mg/m 3 30 = φ Effective friction angle, The flow net for this problem has been sketched in fig The total head values have been written on each equipotential line for an assumed datum at the level of the downstream water table at the base of the wall. From the total head values the pressure heads acting along AB may be calculated and the pressure head distribution has been

24 12-24 sketched on the left side of Fig Integration of this pressure yields a force U1 of 25.4 kn per m of wall. Four trial wedges have been chosen for analysis- AB1, AB2, AB3 and AB4 which will be referred to as wedges 1, 2, 3 and 4 respectively. For each wedge the total weight W is calculated and the force U2 derived from pore water pressure acting on the sloping plane of the wedge is determined from the flow net. The results of these calculations are summarized below: Wedge Wedge Angle θ W-kN U 2 - kn The force polygons for the four wedges may now be drawn as shown in the upper part of Fig The total thrust on the wall which is equal and opposite to force P is plotted as a function of the wedge angle θ in the lower portion of Fig From this plot the total active thrust may be determined. P A = 48.4 kn/m of wall Since total thrust P= effective thrust P + pore pressure force U 1 P = P A - U 1 = 23.0 kn/m of wall While the water force is still the dominant component of the total thrust its relative magnitude has been decreased by the occurrence of steady state seepage. This may be illustrated by a comparison with the problem in section 12.4 in which the drainage of the water from the backfill is prevented by an impervious layer at the base of the retaining wall. In this latter problem (from page neglecting the surcharge) P = kn/m U 1 = kn/m When all of the water has been drained from the backfill a dry sand will be left. P A 1 = ρ d gh 2 2 K = (1/2) x 1.8 x 9.8 x 9 x (1/3) A

25 12-25 = 26.5 kn/m This influence of water in the backfill upon the total thrust on the wall may be illustrated by the following comparison: Saturated backfill - no seepage, P A = 60.3 kn/m Saturated backfill - seepage, P A = 48.4 kn/m Dry backfill P A = 26.5 kn/m

26 12-26 Fig Forces acting on Wedges (Fig.12.14) 12.8 Effect of Wall Friction In the treatment presented so far in this chapter the backs of the retaining walls have been assumed to be smooth. In other words no shear has been acting along the back of the wall and the earth pressure acts at right angles to the wall surface. In reality however, the backs of the retaining walls are rough. This means that a shear stress will act along the back of the wall as the wall moves. This is illustrated for the active case in the Fig 12.16(a). The wall yields under the effect of the backfill pressure. As the yielding continues the backfill soil tends to follow the wall until finally a failure state for the soil is reached. The movement of the soil wedge is downwards along the failure plane as illustrated in the figure. The movement develops shear stresses acting downwards on the wall are illustrated in the right hand sketch. The force PAN which is the normal component of the active thrust per unit length of wall acts at right angles to the back of the wall. The shear force generated on the back of the wall as a result of wall friction is the force PAN tan δ where δ is the angle of wall friction. If the soil is cohesive in character, adhesion may also act on the back of the wall. The adhesion force Cw per unit length of wall, which is also indicated in the sketch is determined as the wall adhesion( stress cw multiplied by the height of wall). For the passive case the situation is illustrated in Fig 12.16(b). In this case the movement of the soil wedge is upwards along the failure plane as the wall is pushed into the backfill. This movement generates shear stresses which act upwards on the back of the wall. The resulting forces acting on the wall are illustrated in the right hand portion of the diagram. The force PPN is the normal component of the passive thrust which acts right angles to the back of the wall. The shear force generated by the wall friction is equal to PPN tan δ. Again if the backfill soil is cohesive in character, an adhesion force Cw will also be present. The shear stress τ acting on the back of a retaining wall possesses two components- the adhesion and the wall friction. τw = cw + σn tan δ in which the σn = pan, where pan is the normal stress acting on the back of the wall. For the active case σn = pan where pan is the normal component of the active pressure at a particular depth. For the passive case σn = ppn When wall friction or wall adhesion is introduced into the earth pressure calculation then the simple expressions for the active or passive pressure coefficients which were presented earlier in this chapter can no longer be used. Other expressions for the earth pressure coefficients such as the one given in equation(12.20) must therefore be developed.

27 12-27 Fig Effect of Wall Friction on Earth Pressure Fig Active Thrust- Gravity Retaining Wall

28 12-28 Fig Forces acting on Soil Wedge General case This is an expression developed from the Coulomb theory for the active thrust produced by a dry cohesionless soil on the back of a retaining wall for the situation shown in Fig P A 1 = ρ d gh 2 2 ( sin( β + δ )) cosecβ sin 1/ 2 sin + ( β φ ) ( φ + δ ) sin( φ i) sin( β i) Where ρd is the mass density of the cohesionless backfill and the other symbols are given in Fig For the more general active case in which the backfill soil is cohesive in character and surface of the backfill is uneven and subjected to surcharge loads as illustrated in Fig the problem may be solved graphically. Using a Coulomb approach a trial wedge ABD as shown in Fig is selected. A force polygon is drawn for the forces acting on this soil wedge thus enabling determination of the force P required to maintain equilibrium. The process is repeated for a number of wedge angles θ until the maximum value of force P is found. This maximum value then corresponds to the active thrust. This procedure is very similar to that used in Fig Alternatively one of the many graphical constructions which have been proposed for the solution of this general problem may be used for determination of the active thrust These methods, such as those Rebhann, Poncelet or Culmann are described in details in other soil mechanics texts( Jumikis(1962)). The coulomb wedge technique which works so successfully in the calculation of an active thrust is not entirely satisfactory for the calculation of passive thrust when wall friction is taken into account. For low angles of wall friction(less than about 10 of φ /3)

29 12-29 the Coulomb assumption of a plane surface of sliding is acceptable. As the angle of wall friction(δ) increases relative to the angle of shearing resistance (φ) of the soil the surface of sliding becomes increasingly curved as illustrated in Fig In this case the Coulomb assumption of a plane surface of sliding is unacceptable. The correct passive thrust may be calculated only when allowance is made for the curvature in the actual surface of sliding. Two commonly used methods for this calculation are the logarithmic spiral method and the friction circle method. In the former, portion of the surface of sliding is assumed to follow the shape of a logarithmic spiral. In the latter, portion of the surface of sliding it is assumed to follow the arc of a circle. These two methods have been described in detail by Tarzaghi (1943). The magnitude of the error introduced by the Coulomb assumption of a plane surface of sliding is illustrated in Fig This figure indicates that for a cohesionless soil the Coulomb theory produces too high a value of the passive pressure coefficient for relatively large values of the angle of wall friction. Fig Surface of Sliding - Passive Case (Rough Wall)

30 12-30 Fig Influence of Wall Friction K p for Sand Example For the retaining structure illustrated in Fig determine the magnitude, location and direction of the active thrust acting on the wall. The properties of the dry cohesionless backfill are as follows: Dry density, ρ d = 1800 Kg/m 3 Angle of shearing resistance, φ = 30 In this case the Coulomb expression given in equation (12.20) may be used. The only term which needs to be evaluated for substitution in this equation is the angle of wall friction δ. In the absence of other data the angle of wall friction is commonly taken to be approximately 2/3 of the angle of shearing resistance. In this example the angle of wall friction be assumed to be 24.

31 12-31 P A = 1 2 ρ gh d 2 ( sin ( β + δ )) cos ecβ sin 1 / 2 sin + ( β φ ) ( φ + δ ) sin ( φ i) sin ( β i) = 1 2 x 1.8x9.81x9 o ( sin ( 134 ) cos ec110 1 / 2 sin 74 o sin 60 + sin o o sin( 26 o ( 100 ) o ) x x (0.719) = 79.5 x = 38.9 kn/m Once the angle of wall friction has been chosen the direction of the active thrust is determined. In this case the active thrust acts at an angle of 44 to the horizontal as shown in Fig Since the active pressure distribution on the back of this wall is triangular the resultant will act through a point equal to one-third of the wall height above the base. This means that the active thrust will act on the back of wall at a distance of one meter above the base as shown in Fig Rankin Approach of Earth Pressure for an Inclined Backfill Fig 12.21(a) represents the section of an infinitely long slope inclined at an angle i to the horizontal. AB represents a plane parallel to the ground surface and at any depth z below the ground surface. The vertical stress σv acting on the plane AB may be determined from equilibrium considerations. ρgzb Vertical stress σ v = = ρ gz cosi bseci

32 12-32 When ρ equals mass density if the soil forming the slope, the other symbols are as shown in the diagram. If this stress is resolved normal and parallel to plane AB the normal stress σn and shear stress τ may be determined Normal stress Shear stress σ = n σ v cosi = ρgz cos 2 i τ = σ v sin i = ρgz cos i sin i These stresses may be represented on a shear stress-normal stress plot as shown in Fig 12.21(b). Point S on this diagram represents the stress on the plane AB. The horizontal co-ordinate of point S represents the normal stress σn, the vertical co-ordinate of point S represents the shear stress τ, and the inclined distance OS represents the vertical stress σv. If failure of the soil forming the slope is produced by means of lateral yielding then the new stress state may be represented by means of a failure Mohr circle drawn on the shear stress-normal stress diagram This circle SRU must be tangent to the failure line. Since failure was produced by means of lateral yielding this circle represents the active case. On this circle point R represents the origin of planes. As discussed in Geomechanics 1, the stress acting on a vertical plane may be found by drawing a line parallel to that plane through the origin of planes. This means that points U on the circle represents the stress acting on a vertical plane at a depth of z below the ground surface. This stress which is active pressure pa acts parallel to the ground surface as illustrated in Fig 12.21(c). If the vertical plane is assumed to represent the vertical back of a retaining wall, then the Rankine approach discussed above for the determination of active pressure will yield the correct answer only for the case in which the angle of wall friction δ is equal to the angle of the backfill slope i. For this special case the active thrust acting on the back of the wall will equal to active thrust calculated by means of a Coulomb wedge analysis.

33 12-33 (a) Fig For cases in which the angle of wall friction is not equal to the angle of the backfill slope the Rankine approach cannot be used for the determination of active pressure. For the passive case the soil forming the slope may be brought into the state of failure by means of lateral compression. The corresponding failure circle representing the passive state shown by circle STV in Fig 12.21(b). For this circle the point T represents the origin of planes. Point V on this circle represents the stress acting on a vertical plane at depth z below the ground surface. This stress which is the passive pressure pp acts in a direction parallel to the ground surface as illustrated in Fig 12.21(c). If the vertical plane is looked

34 12-34 upon as the vertical back of a retaining wall then as in the active case, the passive pressure acts in a direction parallel to the ground surface. It is clear from the discussion in section 12.8 relating to angles of wall friction for the passive case that the wall friction in this case is acting in the wrong direction. The Rankine approach to the determination of passive pressure in cases where wall friction is present is therefore quite inappropriate Design Criteria for Retaining Walls In addition to the structural design of the wall, a number of points relating to the backfill and foundation solid must be checked to ensure that the wall is stable (see Huntington (1961)). (i) Bearing Pressures The maximum bearing pressure, which usually occurs beneath the toe of the retaining wall, should not exceed the allowable bearing capacity for the foundation soil. One of the methods of determining this maximum bearing pressure f1 is illustrated in Fig The resultant R of the active thrust PA and the weight of the retaining wall W is found as illustrated in the figure. This resultant force R is resolved into its components in the vertical and horizontal directions as indicated by the forces RV and RH respectively. In general the vertical force RV will apply an eccentric load of the base of the retaining wall. The eccentricity in Fig is indicated by e. When e is less than B/6 the bearing pressure f1 and f2 may be determined by means of the following expressions: R = V 6e f (12.21) B B f R = V 6e 1 B B 2 (12.22) These expressions are based upon the assumption that the bearing pressure distribution is trapezoidal as illustrated in Fig with the force RV passing through the centroid of the trapezoidal shape. The ultimate bearing capacity of the foundation soil must exceed the maximum bearing pressure f1 with a factor of safety not less than 2.0.

35 12-35 Fig Assumed Contact Stress Distribution (ii) Resistance to overturning It is sometimes stated that the factor of safety against overturning should be at least equal to 2.0. Because there are various ways of calculating this factor of safety as discussed by Huntington(1961) it is often preferable to satisfy this criterion by ensuring that the resultant force R( see Fig 12.22) falls within the middle third of the base of the retaining wall.

36 12-36 (iii) Lateral Sliding The horizontal force RH in Fig will tend to produce horizontal sliding of the retaining wall on the foundation soil. This force is registered by the shear force which will be generated at the contact between the retaining wall and the soil. It is usually recommended that a factor of safety of about 2.0 against horizontal sliding be applied in the design of the wall. The factor of safety F may be determined from the following expression, F = R V tan φ + R H cb Where c and φare the relevant strength parameters for the wall base-foundation soil contact. (iv) Deep Seated Failure It is essential to provide an adequate factor of safety against a deep seated failure in the soil which could produce complete collapse of the retaining wall regardless of the magnitude of the earth pressure acting on the retaining wall. The techniques for carrying out this calculation are discussed in Chapter 11. REFERENCES Huntington,W.C.(1962), Earth Pressures and Retaining Walls, John Wiley, N.Y. Junikis, A.R.(1962), Soil Mechanics, Can Nostrand, N.J. Taylor, D.W.(1948), Fundamentals of Soil Mechanics, John Wiley, N.Y.

37 12-37 Terzaghi,K.(1934), Large Retaining Wall Test, I-Pressure of Dry Sand, Engineering News- Record, vol 112, pp Terzaghi,K.(1943), Theoretical Soil Mechanics, John Wiley, N.Y.