Chapter 12: Lateral Earth Pressure

Transcription

1 Part 4: Lateral Earth Pressure and Earth-Retaining Structures Chapter 12: Lateral Earth Pressure Introduction Vertical or near-vertical slopes of soil are supported by retaining walls, cantilever sheetpile walls, sheet-pile bulkheads, braced cuts, and other, similar structures. The proper design of those structures requires an estimation of lateral earth pressure, which is a function of several factors, such as: (a) the type and amount of wall movement, (b) the shear strength parameters of the soil, (c) the unit weight of the soil, and (d) the drainage conditions in the backfill. The following Figure shows a retaining wall of height H. For similar types of backfill: a. The wall may be restrained from moving (Figure a). The lateral earth pressure on the wall at any depth is called the atrest earth pressure. b. The wall may tilt away from the soil that is retained (Figure b). With sufficient wall tilt, a triangular soil wedge behind the wall will fail. The lateral pressure for this condition is referred to as active earth pressure. c. The wall may be pushed into the soil that is retained (Figure c). With sufficient wall movement, a soil wedge will fail. The lateral pressure for this condition is referred to as passive earth pressure. Engr. Yasser M. Almadhoun Page 1

2 Lateral Earth Pressure at Rest Consider a vertical wall of height H, as shown in Figure 12.3, retaining a soil having a unit weight of g. A uniformly distributed load, q/unit area, is also applied at the ground surface. At any depth z below the ground surface, the vertical subsurface stress is: If the wall is at rest and is not allowed to move at all, either away from the soil mass or into the soil mass (i.e., there is zero horizontal strain), the lateral pressure at a depth z is: Engr. Yasser M. Almadhoun Page 2

3 For normally consolidated soil, the relation for K o (Jaky, 1944) is: For overconsolidated soil, the at-rest earth pressure coefficient may be expressed as: The total force, P o, per unit length of the wall given in Figure 12.3a can now be obtained from the area of the pressure diagram given in Figure 12.3b and is: The location of the line of action of the resultant force, P o, can be obtained by taking the moment about the bottom of the wall. Thus: Engr. Yasser M. Almadhoun Page 3

4 If the water table is located at a depth z, H, the at-rest pressure diagram shown in Figure 12.3b will have to be somewhat modified, as shown in Figure If the effective unit weight of soil below the water table equals (i.e., sat w ), then: Hence, the total force per unit length of the wall can be determined from the area of the pressure diagram. Specifically: So, Engr. Yasser M. Almadhoun Page 4

5 Active Pressure Rankine Active Earth Pressure The Rankine active earth pressure calculations are based on the assumption that the wall is frictionless. The lateral earth pressure involves walls that do not yield at all. However, if a wall tends to move away from the soil a distance Δx, as shown in the following Figure, the soil pressure on the wall at any depth will decrease. For a wall that is frictionless, the horizontal stress, σ h, at depth z will equal K o σ o (=K o z) when Δx is zero. However, with Δx > 0, σ h will be less than K o σ o. Engr. Yasser M. Almadhoun Page 5

6 Rankine active-pressure coefficient K a = tan 2 (45 1 sin ) = sin The pressure distribution shows that at z = 0 the active pressure equals 2c k a, indicating a tensile stress that decreases with depth and becomes zero at a depth z = zc, or: The depth zc is usually referred to as the depth of tensile crack, because the tensile stress in the soil will eventually cause a crack along the soilwall interface. Thus, the total Rankine active force per unit length of the wall before the tensile crack occurs is: After the tensile crack appears, the force per unit length on the wall will be caused only by the pressure distribution between depths z = zc and z = H, as shown by the hatched area in the previous Figure. This force may be expressed as: Or: Engr. Yasser M. Almadhoun Page 6

7 However, it is important to realize that the active earth pressure condition will be reached only if the wall is allowed to yield sufficiently. The necessary amount of outward displacement of the wall is as given as under: Soil type Wall movement for passive condition, Δx granular 0.001H 0.004H cohesive.01h 0.04H If there exists a surcharge load acting downward on the top surface of the backfill: The Rankine active stress at depth z can be calculated as follows: σ h(active) = (q + γh)k a 2c K a The Rankine active force per unit length of the wall at depth z can be calculated as follows: P a = (qh γh2 ) K a 2c K a H Example 12.2 See example 2.2 in textbook, page 602. Example 12.3 See example 12.3 in textbook, page 604. Engr. Yasser M. Almadhoun Page 7

8 A Generalized Case for Rankine Active Pressure Granular Backfill In the previous section, the relationship was developed for Rankine active pressure for a retaining wall with a vertical back and a horizontal backfill. That can be extended to general cases of frictionless walls with inclined backs and inclined backfills. The previous Figure shows a retaining wall whose back is inclined at an angle with the vertical. The granular backfill is inclined at an angle α with the horizontal. The active force Pa for unit length of the wall then can be calculated as: Where Ka can be found from the Table 12.1 or using this equation: And the horizontal and vertical Rankine active forces (Pa(h) and Pa(v) respectively) for unit length of the wall is: P a(h) = P a cos(β a + θ) P a(v) = P a sin(β a + θ) Engr. Yasser M. Almadhoun Page 8

9 Example 12.4 See example 12.4 in textbook, page 612. Granular Backfill with Vertical Back Face of Wall If the backfill of a frictionless retaining wall is a granular soil (c = 0) and rises at an angle α with respect to the horizontal, the active earth-pressure coefficient may be expressed in the form (or can be found from Table 12.3): At any depth z, the Rankine active pressure may be expressed as: Also, the total force per unit length of the wall is: Note that, in this case, the direction of the resultant force Pa is inclined at an angle α with the horizontal and intersects the wall at a distance H/3 from the base of the wall. Engr. Yasser M. Almadhoun Page 9

10 Rankine Active Pressure with Vertical Wall Backface and Inclined c - ϕ Soil Backfill For a frictionless retaining wall with a vertical back face ( = 0) and inclined backfill of c ϕ soil (see the following Figure) at an angle α with the horizontal, the active pressure at any depth z can be given as: where: Some values of K a are given in Table Engr. Yasser M. Almadhoun Page 10

11 For a problem of this type, the depth of tensile crack is given as: Example 12.5 See example 12.5 in textbook, page 613. Coulomb s Active Earth Pressure To apply Coulomb s active earth pressure theory, let us consider a retaining wall with its back face inclined at an angle β with the horizontal, as shown in Figure 12.12a. The backfill is a granular soil that slopes at an angle α with the horizontal. Also, let δ be the angle of friction between the soil and the wall (i.e., the angle of wall friction). To find the active force, consider a possible soil failure wedge ABC1. The forces acting on this wedge (per unit length at right angles to the cross section shown) are as follows: 1. The weight of the wedge, W. Engr. Yasser M. Almadhoun Page 11

12 2. The resultant, R, of the normal and resisting shear forces along the surface, BC1. The force R will be inclined at an angle ϕ to the normal drawn to BC1. 3. The active force per unit length of the wall, Pa, which will be inclined at an angle δ to the normal drawn to the back face of the wall. The maximum value of Pa thus determined is Coulomb s active force (see top part of Figure 12.12a), which may be expressed as: where, Engr. Yasser M. Almadhoun Page 12

13 The values of the active earth pressure coefficient, Ka, for a vertical retaining wall (β = 90 ) with horizontal backfill (α = 0 ) are given in Table Note that the line of action of the resultant force (Pa) will act at a distance H/3 above the base of the wall and will be inclined at an angle δ to the normal drawn to the back of the wall. In the actual design of retaining walls, the value of the wall friction angle δ is assumed to be between ϕ /2 and 2/3ϕ. The active earth pressure coefficients for various values of ϕ, α, and β with δ = 1 2 and 2 3 are respectively given in Tables 12.6 and Engr. Yasser M. Almadhoun Page 13

14 If a uniform surcharge of intensity q is located above the backfill, as shown in Figure 12.13, the active force, Pa, can be calculated as: where: Engr. Yasser M. Almadhoun Page 14

15 Example 12.6 See example 12.6 in textbook, page 620. Example 12.7 See example 12.7 in textbook, page 621. Passive Pressure Rankine Passive Earth Pressure The horizontal stress, σ h, at this point is referred to as the Rankine passive pressure, or σ h = σ p. For Mohr s circle, the major principal stress is σ p, and the minor principal stress is σ o : Rankine active-pressure coefficient K a = tan 2 ( sin ) = 2 1 sin Then, we have: Engr. Yasser M. Almadhoun Page 15

16 The passive pressure diagram for the wall shown in the following Figure. Note that: The passive force per unit length of the wall can be determined from the area of the pressure diagram, or: The approximate magnitudes of the wall movements, Δx, required to develop failure under passive conditions are as follows: If the backfill behind the wall is a granular soil (i.e., c = 0), then, the passive force per unit length of the wall will be: Engr. Yasser M. Almadhoun Page 16

17 Example See example in textbook, page 636. Rankine Passive Earth Pressure Vertical Backface and Inclined Backfill Granular Soil For a frictionless vertical retaining wall (as the following Figure) with a granular backfill (c = 0), the Rankine passive pressure at any depth can be determined in a manner similar to that done in the case of active pressure in a preceeding section. The pressure is: and the passive force is: where: Engr. Yasser M. Almadhoun Page 17

18 As in the case of the active force, the resultant force, P p, is inclined at an angle α with the horizontal and intersects the wall at a distance H/3 from the bottom of the wall. The values of K p (the passive earth pressure coefficient) for various values of α and ϕ are given in Table C ϕ Soil If the backfill of the frictionless vertical retaining wall is a c ϕ soil, then: where: The variation of K p with ϕ, α, and c / z is given in Table Engr. Yasser M. Almadhoun Page 18

19 Coulomb s Passive Earth Pressure To understand the determination of Coulomb s passive force, Pp, consider the wall shown in Figure 12.21a. 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 ABC1, the forces per unit length of the wall acting on the wedge are 1. The weight of the wedge, W, 2. The resultant, R, of the normal and shear forces on the plane BC1, and 3. The passive force, Pp. The minimum value of Pp in this diagram is Coulomb s passive force, mathematically expressed as: where, Engr. Yasser M. Almadhoun Page 19

20 The values of the passive pressure coefficient, Kp, for various values of ϕ and δ are given in Table (β = 90, α = 0 ). Note that the resultant passive force, Pp, will act at a distance H/3 from the bottom of the wall and will be inclined at an angle δ to the normal drawn to the back face of the wall. Engr. Yasser M. Almadhoun Page 20

21 Problems Problem (1) Consider the retaining wall shown in the Figure below. Calculate the Coulomb s active force per unit length of the wall and the location of the line of action of that resultant at which it acts at the retaining wall. Solution: Z Engr. Yasser M. Almadhoun Page 21

22 P a = 1 2 γ eqk a H 2 sin β if γ eq = [ sin(β + α) ] (2q H ), then: P a = 1 2 (γ + [ sin β sin(β + α) ] (2q H )) K ah 2 eqn (1) P a = 1 2 K aγh 2 sin β + K a Hq [ ] eqn (2) sin(β + α) P a = P a(1) + P a(2) eqn (3) K a can be found using the following equation (equation in textbook): K a = K a = sin 2 (β + ) 2 sin 2 β sin 2 (β δ ) [1 + sin( + δ ) sin( α) sin( δ ) sin( + α) ] sin 2 ( ) sin 2 85 sin 2 (85 20 ) [1 + sin( ) 2 sin( 5 ) sin(30 20 ) sin( ) ] K a = Or even from Table 12.6 in textbook: Engr. Yasser M. Almadhoun Page 22

23 Now, substitute into equation (1): P a = 1 2 (γ + [ sin β sin(β + α) ] (2q H )) K ah 2 eqn (1) P a = 1 2 (18 + [ sin 85 sin( ) ] (2 96 )) (0.3578)(6.1)2 6.1 = KN/m To determine the location of the line of action of the resultant at which it acts at the retaining wall: Recall equations (2) & (3): P a = 1 2 K aγh 2 sin β + K a Hq [ ] eqn (2) sin(β + α) P a = P a(1) + P a(2) eqn (3) z P a = H 3 P a(1) + H 2 P a(2) z P a = H 3 (1 2 K aγh 2 ) + H 2 (K sin β ahq [ sin(β + α) ]) z (328.5) = ( ) z = 2.68 m ( [ sin 85 sin( ) ]) Engr. Yasser M. Almadhoun Page 23

24 Problem (2) Consider the retaining wall shown in the Figure below. Calculate the Rankine passive force per unit length of the wall and the location of the line of action of that resultant at which it acts on the retaining wall. Solution: Rankine passive lateral earth pressure coefficients: K p(1) = tan 2 ( K p(2) = tan 2 ( ) = tan2 ( ) = ) = tan2 ( ) = 2.56 Rankine passive lateral earth pressures: *at h = 0.00 m σ v = γh = 0.00 KN/m 2 *at h = 2.00 m (just before the GWT): σ v = γh = = KN/m 2 Engr. Yasser M. Almadhoun Page 24

25 σ a = σ v K p(1) = = KN/m 2 *at h = 2.00 m (just after the GWT): σ v = γh = = KN/m 2 σ a = σ v K p(2) c K p(2) = (10.00) 2.56 *at h = 3.00 m: = KN/m 2 σ v = γh = ( ) = KN/m 2 σ a = σ v K p(2) c K p(2) = (10.00) 2.56 = KN/m 2 u = γ w H = = 9.81 KN/m 2 Rankine passive forces: P p(1) = 1 2 Hσ a = 1 (2.00)(94.32) = KN 2 P p(2) = Hσ a = (1.00)(112.4) = KN P p(3) = 1 2 Hσ a = 1 (1.00)( ) = KN 2 P p(3) = 1 2 Hσ a = 1 (1.00)(9.81) = 4.91 KN 2 Rankine passive resultant: P p = P p(1) + P p(2) + P p(3) + P p(4) P p = = c. 26 KN Location of the Rankine passive resultant: z P p = z 1 3 P p(1) + z 2 2 P p(2) + z 3 3 P p(3) + z 4 3 P p(4) z (223.26) = ( ) (94.32) + ( ) (112.40) + ( 1.00 ) (11.63) + ( ) (4.91) z = m Engr. Yasser M. Almadhoun Page 25