(Lecture 21) LATERAL EARTH PRESSURE

Transcription

1 Module 6 (Lecture 21) LATERAL EARTH PRESSURE Topics 1.1 COULOMB S ACTIVE EARTH PRESSURE 1.2 ACTIVE EARTH PRESSURE FOR EARTHQUAKE CONDITIONS COULOMB S ACTIVE EARTH PRESSURE The Rankine active earth pressure calculations discussed in the preceding sections were based on the assumption that the wall is frictionless. In 1776, Coulomb proposed a theory to calculate the lateral earth pressure on a retaining wall with granular soil backfill. This theory takes wall friction into consideration. To apply Coulomb s active earth pressure theory, let us consider a retaining wall with its back face inclined at an angle ββ with the horizontal, as shown in figure The backfill is a granular soil that sloes at an angle αα with the horizontal. Also, let δδ be the angle of friction between the soil and the wall (that is, angle of wall friction).

2 Figure 6.12 Coulomb s active presure Under active pressure the wall will move away from the soil mass (to the left in figure 6.12a). Coulomb assumed that, in such a case, the failure surface in the soil mass would be a plane (e.g., BBBB 1, BBBB 2, ) to find the active force in our example, consider a possible soil failure wedge AAAAAA 1. The forces acting on this wedge, AAAAAA 1 (per unit length at right angles to the cross section shown), are as follows: PP aa = 1 2 KK aaγγhh 2 1. Weight of the wedge, W. 2. The resultant, R, of the normal and resisting shear forces along the surface, BBBB 1. The force R will be inclined at an angle φφ to the normal drawn to the surface BBBB The active force per unit length of the wall, PP aa. The force PP aa will be inclined at an angle δδ to the normal drawn to the back face of the wall. For equilibrium purposes, a force triangle can be drawn, as shown in figure 6.12b. Note that θθ 1 is the angle that BBBB 1 makes with the horizontal. Because the magnitude of W as well as the directions of all three forces are known, the value of PP aa can now be determined. Similarly, the active forces of other trial wedges, such as AAAAAA 2, AAAAAA 3, ) can be determined. The maximum value of PP aa thus determined is Coulomb s active force (see top part of figure 6.12), which may be expressed as PP aa = 1 2 KK aaγγhh 2 [6.25] Where

3 KK aa = Coulomb s active earth pressure coefficient = sin 2 (ββ+φφ) [6.26] sin 2 sin (φφ +δδ)sin (φφ αα ) ββ sin (ββ δδ) 1+ sin (ββ +δδ)sin (αα ββ ) And HH = height of the wall The values of the active earth pressure coefficient, KK aa, for a vertical retaining wall (ββ = 90 ) with horizontal backfill (αα = 0) are given in table 4. Note that the line of action of the resultant (PP aa ) will act at a distance of HH/3 above the base of the wall and will be inclined a angle δδ to the normal drawn to the back of the wall. In the actual design of retaining walls, the value of the wall friction angle, δδ, is assumed to be between φφ/2 and 2 ϕ. The active earth pressure coefficients for various values of 3 φφ, αα, and ββ with δδ = 1 φφ and 2 φφ are given in table 5 and 6. These coefficients are very 2 3 useful design considerations. If a uniform surcharge of intensity q is located above the backfill, as shown in figure 6.13, the active force, PP aa, can be calculated as Figure 6.13 Coulomb s active pressure with a surcharge on the backfill

4 PP aa = 1 KK 2 aaγγ eeee HH 2 equation (26) [6.27] Table 4 Value of KK aa [equation (26)] for 9999, αα = 00 δδ(deg) φφ(deg) Table 5 Values of KK aa [equation (26)]. Note: δδ = 22 φφ 33 ββ(deg) αα(deg) φφ(deg)

5

6

7 Where γγ eeee = γγ + sin ββ 2qq sin (ββ+αα) HH cos αα [6.28] The derivation of equation (27 and 28) are contained in other soil mechanics texts (e.g., Das, 1987).

8 Table 6 Values of KK aa.note: δδ = φφ/22 ββ(deg) αα(deg) φφ(deg)

9

10

11 Example 6 Consider the retaining wall shown in figure 6.12a. given: HH = 4.6 m; unit weight of soil = 16.5 kn/m 3 ; angle of friction of soil= 30 ; wall friction-angle, δδ = 2 3 φφ; soil cohesion, cc = 0; αα = 0, and ββ = 90. Calculate the Coulomb s active force per unit length of the wall. Solution From equation (25) PP aa = 1 2 γγhh2 KK aa From table 5, for αα = 0, ββ = 90, φφ = 30, and δδ = 2 3 φφ = 30, KK aa = Hence PP aa = 1 2 (16.5)(4.6)2 (0.297) = kn/m ACTIVE EARTH PRESSURE FOR EARTHQUAKE CONDITIONS Coulomb s active earth pressure theory can be extended to take into account the forces caused by an earthquake. Figure 6.14 shows a condition of active earthquake with a granular backfill (cc = 0). Note that the forces acting on the soil failure wedge in figure 6.14 re essentially the same as those shown in figure 6.12a, with the addition of kk h WW and kk vv WW in the horizontal and vertical directions, respectively, kk h and kk vv may be defined as

12 Figure 6.14 Derivation of equation (31) kk h = kk vv = horizontal earthquake acceleration component acceleration due to gravity,gg vertical earthquake acceleration component acceleration due to gravity,gg [6.29] [6.30] The relation for the active force per unit length of the wall (PP aaaa ) can be determined as PP aaaa = 1 2 γγhh2 (1 kk vv )KK aaaa [6.31] Where KK aaaa = active earth pressure coefficient = sin 2 (φφ+ββ θθ ) cos θθ sin 2 β sin (ββ θθ δδ) 1+ sin (φφ +δδ)sin (φφ θθ αα ) sin ββ δδ θθ sin (αα+ββ ) 2 [6.32] θθ = tan 1 kk h 1 kk v [6.33] Note that for no earthquake condition kk h = 0, kk vv = 0, and θθ = 0

13 Hence KK aaaa = KK aa [as given by equation (26)]. The variation of KK aaaa cos δδ with kk h for the case of kk vv = 0, ββ = 90, αα = 0, and δδ = φφ/2 is shown in figure Some values of KK aaaa for ββ = 90 and kk vv = 0 are given in table 7. Figure 6.15 Figure 6.15 Variation of KK aaaa cos δδ with kk h (note: kk vv = 0, ββ = 90, αα = 0, and δδ = φφ/ 2). (Note: KK aaaa cos δδ is the component of earth pressure coefficient at right angles to the back face of the wall Table 7 Values of KK aaaa [equation (32)] for ββ = 90 = kk vv = 0 φφ(deg) kk h δδ(deg) αα(deg)

14 φφ/ φφ/

15 φφ/ φφ φφ Equation (31) is usually referred to as the Mononobe-Okabe solution. Unlike the case shown in figure 6.12a, the resultant earth pressure in this situation, as calculated by equation (31) does not act at a distance of HH/3 from the bottom of the wall. The following procedure may be used to obtain the location of the resultant force PP aaaa : 1. Calculate PP aaaa by using equation (31) 2. Calculate PP aa by using equation (25) 3. Calculate

16 Δ PP aaaa = PP aaaa PP aa [6.34] 4. Assume that PP aa acts at a distance of HH/3 from the bottom of the wall (figure 6.16). 5. Assume that ΔPP aaaa acts at a distance of 0.6H from the bottom of the wall (figure 6.16). 6. Calculate the location of the resultant as zz = (0.6HH)(Δ PP aaaa )+ HH 3 (PP aa ) PP aaaa [6.35] Figure 6.16 Determining the line of action of PP aaaa Example 7 Refer to figure For kk vv = 0 and kk h = 0.3, determine:

17 Figure a. PP aaaa b. The location of the resultant, zz, from the bottom of the wall Solution Part a From equation (31) PP aaaa = 1 2 γγhh2 (1 kk vv )KK aaaa Here, γγ = 105 lb/ft 3, HH = 10 ft, and kk vv = 0. As δδ = φφ/2, we can use figure 6.15 to determine KK aaaa. For kk h = 0.3, KK aaaa 0.472, so PP aaaa = 1 2 (105)(10)2 (1 0)(0.472) = 2478 lb/ft Part b From equation (25), PP aa == 1 2 γγhh2 KK aa From equation (26) with δδ = 17.5, ββ = 90, and αα = 0, KK aa (table 6), so PP aa = 1 2 (105)(10)2 (0.246) = 1292 lb/ft PP aaaa = PP aaaa PP aa = = 1186 lb/ft

18 From equation (35), zz = (0.6HH)( PP aaaa )+(HH/3)(PP aa ) PP aaaa = [(0.6)(10)](1292)+(10/3)(11186 ) 2478 = 4.72 ft