Module 6 (Lecture 22) LATERAL EARTH PRESSURE

Transcription

1 Module 6 (Lecture ) LATERAL EARTH PRESSURE 1.1 LATERAL EARTH PRESSURE DUE TO SURCHARGE 1. ACTIVE PRESSURE FOR WALL ROTATION ABOUT TOP- BRACED CUT 1.3 ACTIVE EARTH PRESSURE FOR TRANSLATION OF RETAINING WALL-GRANULAR BACKFILL

2 LATERAL EARTH PRESSURE DUE TO SURCHARGE In several instances, the theory of elasticity is used to determine the lateral earth pressure on retaining structures caused by various types of surcharge loading, such as line loading (figure 6.18a) and strip loading (figure 6.18b). Figure 6.18 Lateral earth pressure caused by (a) line load and (b) strip load According to the theory of elasticity, the stress at any depth, z, on a retaining structure caused by a line load of intensity q/ unit length (figure 6.18a) may be given as σσ = qq aa bb [6.36] ππππ (aa +bb ) Where σσ = horizontal stress at depth zz = bbbb (See figure 6.18a for explanations of the terms a and b). However, because soil is not a perfectly elastic medium, some deviations from equation (36) may be expected. The modified forms of this equation generally accepted for use with soils are as follows:

3 σσ = 4qq ππππ And σσ = qq HH aa bb (aa +bb ) 0.03bb (0.16+bb ) for aa > 0.4 [6.37] for aa 0.4 [6.38] Figure 6.18b shows a strip load with an intensity of qq/unit area located at a distance bb from a wall of height H. based on the theory of elasticity, the horizontal stress, σσ, at any depth z on a retaining structure is σσ = qq (ββ sin ββ cos αα) [6.39] ππ (The angles αα and ββ are defied in figure 6.18b). However, in the case of soils, the right-hand side of equation (39) is doubled to account for the yielding soil continuum, or σσ = qq ππ (ββ sin ββ cos αα) [6.40] The total force per unit length (P) due to the strip loading only (Jarquio, 1981) may be expressed as PP = qq 90 [HH(θθ θθ 1 )] [6.41] Where θθ 1 = tan 1 bb (deg) [6.4] HH θθ = tan 1 aa +bb (deg) [6.43] HH Example 8 Refer to figure 6.18b. Here, aa = m, bb = 1 m, qq = 40 kn/m, and HH = 6 m. Determine the total pressure on the wall caused by the strip loading only. Solution From equations (4 and 43), θθ 1 = tan 1 1 = θθ = tan = 6.7 6

4 From equation (41), PP = qq 90 [HH(θθ θθ 1 )] = 40 [6( )] = 4.63 kn/m 90 ACTIVE PRESSURE FOR WALL ROTATION ABOUT TOP-BRACED CUT In the preceding sections, we have seen that a retaining wall rotates about its bottom (figure 6.19a). With sufficient yielding of the wall, the lateral earth pressure is approximately equal to the obtained by Rankine theory of Coulomb s theory. In contrast to retaining walls, braced cuts show a different type of wall yielding (see figure 6.19b). in this case, deformation of the wall gradually increases with the depth of excavation. The variation of the amount of deformation depends on several factors, such as the type of soil, the depth of excavation, and the workmanship. However, with very little wall yielding at the top of the cut, the lateral earth pressure will be lose to the at-rest pressure. At the bottom of the wall, with a much larger degree of yielding, the lateral earth pressure will be substantially lower than the Rankine active earth pressure. As a result, the distribution of lateral earth pressure will vary substantially in comparison to the linear distribution assumed in the case of retaining walls. Figure 6.19 Nature of yielding of walls: (a) retaining wall; (b) braced cut The total lateral force per unit length of the wall, PP aa, imposed on a wall may be evaluated theoretically by using Terzaghi (1943) general wedge theory (figure 6.0). The failure surface is assumed to be the arc of a logarithmic spiral, defined as

5 Figure 6.0 Braced cut analysis by general wedge theory-wall rotation about top rr = rr oo ee θθ tan φφ [6.44] Where φφ = angle of friction of soil In figure 6.0, H is the height of the cut. The unit weight, angle of friction, and cohesion of the soil are equal to γγ, φφ, and cc, respectively. Following are the forces per unit length of the cut acting on the trial failure wedge: 1. weight of the wedge, W. resultant of the normal and shear forces along aaaa, RR 3. cohesive force along aaaa, CC 4. adhesive force along aaaa, CC aa. PP aa,which is the active force acting a distance nn aa HH from the bottom of the wall and is inclined at an angle δδ to the horizontal The adhesive force is CC aa = cc aa HH [6.4] Where cc aa = unit adhesion A detailed outline for the evaluation of PP aa is beyond the scope of this text; those interested should check a soil mechanics text for more information (for example, Das, 1998), Kim and Preber (1969) provided tabulated values of PP aa / 1 γγhh determined by using the principles of general wedge theory, and these values are given in table 8. In developing the theoretical values in table 8, it was assumed that

6 cc aa cc = tan δδ tan φφ [6.46] ACTIVE EARTH PRESSURE FOR TRANSLATION OF RETAINING WALL- GRANULAR BACKFILL Under certain circumstances, retaining walls may undergo lateral translation, as shown in figure 6.1. A solution to the distribution of active pressure for this case was provided by Dubrova (1963) and was also described by Harr (1966). The solution of Dubrova assumes the validity of Coulomb s solution [equations and 6)]. In order to understand this procedure, let us consider a vertical wall with a horizontal granular backfill (figure 6.). For rotation about the top of the wall, the resultant R of the normal and shear forces along the rupture line AAAA is inclined at an angle φφ to the normal drawn to AAAA. According to Dubrova there exists infinite number of quasi-rupture lines such as AA CC, AAC, for which the resultant force R is inclined an angle ψψ, where ψψ = φφφφ HH [6.47 Figure 6.1 Lateral translation of retaining wall

7 Figure 6. Quasi-rupture lines behind retaining wall Table 8 PP aa / 11 γγhh ] vvvvvvvvvvvv φφ, δδ, nn aa, aaaaaa cc/γγhh nn aa = 0.3 cc/γγγγ nn aa = 0.4 cc/γγγγ nn aa = 0. cc/γγγγ nn aa = 0.6 cc/γγγγ φφ, in degrees (1) δδ, in degrees () 0 (3) 0.1 (4) 0. () 0 (6) 0.1 (7) 0. (8) 0 (9) 0.1 () 0. (11) 0 (1) 0.1 (13) 0. (14)

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9 After Kim and Preber (1969) Now, refer to equation ( and 6) for Coulomb s active pressure. For ββ = 90 and αα = 0, the relationship for Coulomb s active force can also be rewritten as PP aa = γγ HH 1 cos δδ cos ψψ +(tan ψ+tan ψψ tan δδ) 0. [6.48] The force against the wall at any z is then given as PP aa = γγ zz 1 cos δδ cos ψψ +(tan ψψ+tan ψψ tan δδ) 0. [6.49] The active pressure at any depth z for wall rotation about the top is

10 σσ aa (zz) = ddpp aa dddd γγ cos δδ zzcos ψψ 1+mm sin ψψ) zz φφ cos ψψ HH(1 mm sin ψψ) (sinψψ + mm) [6.0] Where mm = 1 + tan δδ tan ψψ 0. [6.1] For frictionless walls, δδ = 0 and equation (0) simplifies to σσ aa (zz) = γγ tan 4 ψψ zz φφzz HH cos ψψ [] For wall rotation about the bottom, a similar expression can be found in the form σσ aa (zz) = γγγγ cos φφ cos δδ 1+mm sin φφ [6.3] For translation of the wall, the active pressure can then be taken as σσ aa (zz) translation = 1 [σσ aa(zz) motion about top + σσ aa (zz) motion about bottom ] [6.4] An experimental verification of this procedure was provided by Matsuzawa and Hazarika (1996). The results were obtained from large-scale model tests and are shown in figure 6.3. The theory and experimental results show go agreement. Figure 6.3 Experimental verification of Dubrova s theory for lateral wall translation by large-scale model tests Example 9 Consider a frictionless wall 16 ft high. For the granular backfill, γγ = 1 lb/ft 3 and φφ = 36. Calculate the plot the variation of σσ aa (zz) for a translation mode of the wall movement.

11 Solution For a frictionless wall, δδ = 0. Hence, m is equal to one [equation (1)]. So for rotation about the top, from equation (), σσ aa (zz) = σσ aa(zz) = γγ tan 4 φφφφ HH zz φφzz For rotation about the bottom, from equation (3), HH cos φφφφ HH σσ aa (zz) = σσ cos φφ aa(zz) = γγγγ 1 + sin φφ σσ aa (zz) translation = σσ aa(zz) + σσ aa(zz) The following table can now be prepared with γγ = 1 lb/ft 3, φφ = 36, and HH = 16 ft. zz(ft) σσ aa(zz)(lb/ft ) σσ aa (zz)(lb/ ft ) σσ aa (zz) translation (lb /ft The plot of σσ aa (zz) versus z is shown in figure 6.4.

12 Figure.6.4