Module 6 (Lecture 23) LATERAL EARTH PRESSURE

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1 Module 6 (Lecture 23) LATERAL EARTH PRESSURE Topics 1.1 PASSIVE PRESSURE 1.2 RANKINE PASSIVE EARTH PRESSURE 1.3 RANKINE PASSIVE EARTH PRESSURE-INCLINED BACKFILL 1.4 COULOMB S PASSIVE EARTH PRESSURE 1.5 COMMENTS ON THE FAILURE SURFACE ASSUMPTION FOR COULOMB S PRESSURE CALCULATIONS PASSIVE PRESSURE RANKINE PASSIVE EARTH PRESSURE Figure 6.25a shows a vertical frictionless retaining wall with a horizontal backfill. At depth z, the vertical pressure on a soil element is σσ vv = γγγγ. Initially, if the wall does not yield at all, the lateral stress at that depth will be σσ h = KK oo σσ vv. This state of stress is illustrated by the Mohr s circle aa in figure 6.25b. Now, if the wall is pushed into the soil mass by an amount Δxx, as shown in figure 6.25a, the vertical stress at depth z will stay the same; however, the horizontal stress will increase. Thus σσ h will be greater than KK oo σσ vv. The state of stress can now be represented by the Mohr s circle b in figure 6.25b. If the wall moves farther inward (that is, Δxx is increased still more), the stresses at depth z will ultimately reach the state represented by Mohr s circle c (figure 6.25b). Note that this Mohr s circle touches the Mohr-Coulomb failure envelope, which implies that the soil behind the wall will fail by being pushed upward. The horizontal stress, σσ h, at this point is referred to as the Rankine passive pressure, or σσ h = σσ pp.

2 Figure 6.25 Rankine passive pressure For Mohr s circle c in figure 6.25b, the major principal stress is σσ pp, and the minor principal stress is σσ vv. Substituting them into equation (84 from chapter 1) yields σσ pp = σσ vv tan φφ + 2cc tan 45 + φφ [6.55] 2 2 Now, let KK pp = Rankine passive earth pressure coefficient tan φφ [6.56] 2 (See table 9). Hence, from equation (55), σσ pp = σσ vv KK pp + 2cc KK pp [6.57]

3 Table 9 Variation of Rankine KK pp Soil friction angle, φφ (deg) KK pp = tan 2 (45 + φφ/2)

4 Equation (57) produces figure 6.25c, the passive pressure diagram for the wall shown in figure 6. 25a. Note that at zz = 0, σσ vv = 0 and σσ pp = 2cc KK pp and at zz = HH, σσ vv = γγγγ and σσ pp = γγγγkk pp + 2cc KK pp The passive force per unit length of the wall can be determined from the area of the pressure diagram, or PP pp = 1 2 γγhh2 + 2cccc KK pp [6.58] The approximate magnitudes of the wall movements, Δxx, required to develop failure under passive conditions are Soil type Wall movement for passive condition, Δxx Dense sand 0.005H Loose sand 0.01H Stiff clay 0.01H Soft clay 0.05H Example 10 A 3-m high wall is shown in figure 6.26a. Determine the Rankine passive force per unit length of the wall.

5 Figure 6.26 Solution For the top layer KK pp(1) = tan φφ 1 2 = tan2 ( ) = 3 From the bottom soil layer KK pp(2) = tan φφ 2 2 = tan2 ( ) = 2.56 σσ pp = σσ vv KK pp + 2cc KK pp Where σσ vv = effective vertical stress at zz = 0, σσ vv = 0, cc 1 = 0, σσ pp = 0 at zz = 2 m, σσ vv = (15.72)(2) = kn/m 2, c 1 = 0 So, for the top soil layer σσ pp = 31.44KK pp(1) + 2(0) KK pp(1) = 31.44(3) = kn/m 2 At this depth, that is, zz = 2m, for the bottom soil layer σσ pp = σσ vv KK pp(2) + 2cc KK pp(2) = 31.44(2.56) + 2(10) 2.56 = = kn/m 2 Again, at zz = 3 m, σσ vv = (15.72)(2) + (γγ sat γγ ww )(1) = ( )(1) = kn/m 2

6 Hence σσ pp = σσ vv KK pp(2) + 2cc KK pp(2) = 40.49(2.56) + (2)(10)(1.6) = kn/m 2 Note that, because a water table is present, the hydrostatic stress, u, also has to be taken into consideration. Forzz = 0, to 2 m, uu = 0; zz = 3 m, uu = (1)(γγ ww ) = 9.81 kn/m 2. The passive pressure diagram is plotted in figure 6.26b. The passive force per unit length of the wall can be determined from the area of the pressure diagram as follows: Area no. Area (2)(94.32) = (112.49)(1) = (1)( ) = (9.81)(1) = kn/m PP pp = RANKINE PASSIVE EARTH PRESSURE-INCLINED BACKFILL For a frictionless vertical retaining wall (figure 6.10) with a granular backfill (cc = 0), the Rankine passive pressure at any depth can be determined in a manner similar to that done in the case of active pressure, or σσ pp = γγγγkk pp [6.59] and the passive force PP pp = 1 2 γγhh2 KK pp [6.60] Where KK pp = cos αα cos αα+ cos 2 αα cos 2 φφ cos αα cos 2 αα cos 2 φφ [6.61] As in the case of the active force, the resultant force, PP pp, is inclined at an angle αα with the horizontal and intersects the wall at a distance of HH/3 from the bottom of the wall. The values of KK pp (passive earth pressure coefficient) for various values of αα and φφ are given in table 10. If the backfill of the frictionless vertical retaining wall is a cc φφ soil (figure 6.10), then

7 σσ aa = γγγγkk pp = γγγγkk pp cos αα [6.62] Where KK pp = 1 cos 2 φφ 2 cos2 cc αα + 2 cos φφ sin φφ γγγγ + 4 cos 2 αα(cos 2 αα cos 2 φφ) + 4 cc γγγγ 2 cos 2 φφ + 8 cc γγγγ cos2 αα sin φφ cos φφ Table 10 Passive Earth Pressure Coefficient, KK pp [equation (61) φφ(deg) 1 [6.63] αα(deg) Table 11 Values of KK pp cc/γγγγ φφ(deg) αα(deg)

8 The variation of KK pp with φφ, αα, cc/γγγγ is given in table 11 (Mazindrani and Ganjali, 1997). COULOMB S PASSIVE EARTH PRESSURE Coulomb (1776) also represent an analysis for determining the passive earth pressure (that is, when the wall moves into the soil mass) for walls possessing friction (δδ = angle of wall friction) and retaining a granular backfill material similar. To understand the determination of Coulomb s passive force, PP pp, consider the wall shown in figure 6.27a. As in the case of active pressure, Coulomb assumed that the potential failure surface in soil is a plane. For a trial failure wedge of soil, such as AAAAAA 1, the forces per unit length of the wall acting on the wedge Figure 6.27 Coulomb s passive pressure

9 1. The weight of the wedge, W 2. The resultant, R, of the normal and shear forces on the plane BBBB 1 3. The passive force, PP pp Figure 6.27 shows the force triangle at equilibrium for the trial wedge AAAAAA 1. From this force triangle, the value of PP pp can be determined because the direction of all three forces and the magnitude of one force are known. Similar force triangles for several trial wedges, such as AAAAAA 1, AAAAAA 2, AAAAAA 3, can be constructed, and the corresponding values of PP pp can be determined. The top part of figure 6.27a shows the nature of the variation of the PP pp values for different wedges. The minimum value of PP pp in this diagram is Coulomb s passive force. Mathematically, this can be expressed as PP pp = 1 2 γγhh2 KK pp [6.64] Table 12 Values of KK pp [equation (65)] for ββ = 9999 aaaaaa αα = 00 Where δδ(deg) φφ(deg) KK pp = Coulomb spassive pressure coefficient = sin 2 (ββ φφ) 2 [6.65] sin 2 sin (φφ +δδ)sin (φφ αα ) ββ sin (ββ+δδ) 1 sin (ββ +δδ)sin (ββ +αα ) The values of the passive pressure coefficient, KK pp, for various values of φφ and δδ are given in table 12 (ββ = 90 and αα = 0 ).

10 Note that the resultant passive force, PP pp, will act at a distance of HH/3 from the bottom of the wall and will be inclined at an angle δδ to the normal drawn to the back face of the wall. COMMENTS ON THE FAILURE SURFACE ASSUMPTION FOR COULOMB S PRESSURE CALCULATIONS Coulomb s pressure calculation methods for active and passive pressure. The fundamental assumption for these analyses is the acceptance of plane failure surfaces. However, for walls with friction, this assumption does not hold in practice. The nature of actual failure surfaces in the soil mass for active and passive pressure is shown in figure 6.28a and b, respectively (for a vertical wall with a horizontal backfill). Note that the failure surfaces BC are curved and that the failure surfaces CD are planes. Figure Nature of failure surface in soil with wall friction for (a0 active pressure case and (b) passive pressure case Although the actual failure surface in soil for the case of active pressure is somewhat different from that assumed in the calculation of the Coulomb pressure, the results are not greatly different. However, in the cases of passive pressure, as the value of δδ increases, Coulomb s method of calculation gives increasingly erroneous values of PP pp. This factor of error could lead to an unsafe condition because the values of PP pp would become higher than the soil resistance. Caquot and Kerisel (1948) developed a chart (figure 6.29) for estimating the value of the passive pressure coefficient (KK pp ) with curved failure surface in granular soil (cc = 0), such as that shown in figure In their solution, the portion BC of the failure surface was assumed to be an arc of a logarithmic spiral. While using figure 6.29, the following points should be kept in mind:

11 Figure 6.29 Caquot and Kerisel s passive pressure coefficient, KK pp, for granular soil 1. The curves are for φφ = δδ. 2. If δδ/φφ is less than one, then KK pp(δδ) = RR KK pp(δδ=φφ) Where RR = reduction factor The reduction factor R is given in table The passive pressure is PP pp = 1 2 γγhh2 KK pp(δδ)

12 Table 13 Reduction factor, R, for use in conjunction with figure δδ/φφ φφ(deg)