Module 7 (Lecture 27) RETAINING WALLS

Module 7 (Lecture 27) RETAINING WALLS Topics 1.1 RETAINING WALLS WITH METALLIC STRIP REINFORCEMENT Calculation of Active Horizontal and vertical Pressure Tie Force Factor of Safety Against Tie Failure Total Length of Tie 1.2 STEP-BY-STEP DESIGN PROCEDURE (METALLIC STRIP REINFORCEMENT Internal Stability External Stability Internal Stability Check Tie thickness Tie length External Stability Check Check for overturning Check for sliding Check for bearing capacity

RETAINING WALLS WITH METALLIC STRIP REINFORCEMENT Reinforced earth walls are flexible walls. Their main components are 1. Backfill, which is granular soil 2. Reinforcing strips, which are thin, wide strips placed at regular intervals 3. A cover on the front face, which is referred to as the skin Figure 25 is a diagram of a reinforced earth wall. Note that, at any depth, the reinforcing stripes or ties are placed with a horizontal spacing of SS HH center-to-center; the vertical spacing of the strips or ties is SS VV center-to-center. The skin can be constructed with sections of relatively flexible thin material. Lee et al. (1973) showed that, with a conservative design, a 0.2-in.thick ( 5 mm) galvanized steel skin would be enough to hold a wall about 45-50 ft (14-15 m) high. In most cases, precast concrete slabs can be used as skin. The slabs are grooved to fit into each other so that soil cannot flow out between the joints. When metal skins are used, they are bolted together, and reinforcing strips are placed between the skins. Figure 7.25 Reinforced earth retaining wall Calculation of Active Horizontal and vertical Pressure Figure 26a shows a retaining wall with a granular backfill having a unit weight of γγ 1 and a friction angle of φφ 1. Below the base of the retaining wall, the in situ soil has been excavated and reccopacted, with granular soil used as backfill. Below the backfill, the in situ soil has a unit weight of γγ 2 and a friction angle of φφ 2, and cohesion of cc 2. A surcharge having an intensity of q per unit area lies atop the retaining wall. The wall has reinforcement ties at depths zz = 0, SS VV,.., NNNN VV. The height of the wall is NNNN VV = HH.

Figure 7.26 Analysis of a reinforced earth retaining wall According to the Rankine active pressure theory σσ aa = σσ vv KK aa 2cc KK aa σσ aa = Rankine active pressure at any depth zz For dry granular soils with no surcharge at the top, cc = 0, σσ vv = γγ 1 zz, and KK aa = tan 2 (45 φφ 1 /2). Thus σσ aa(1) = γγ 1 zzkk aa [7.30] When a surcharge is added at the top, as shown in figure 29, σσ vv = σσ vv(1) = γγ 1 zz Due to soil only +σσ vv(2) Due to the surcharge [7.31] The magnitude of σσ vv(2) can be calculated by using the 2:1 method of stress distribution described in equation (14 chapter 4) and figure 6 (from chapter 4). It is shown in figure 27a. According to Laba and Kennedy (1986),

Figure 7.27 (a) notation for the relationship of σσ vv(2) - equations (32 and 33); (b) notation for the relationship of σσ aa(2) - equations (35 and 36); σσ vv(2) = qqqq aa +zz (forzz 2bb ) [7.32] And σσ vv(2) = qqqq aa + zz 2 +bb (forzz > 2bb ) [7.33] Also, when a surcharge is added at the top, the lateral pressure at any depth is σσ aa = σσ aa(1) = KK aa γγ 1 zz Due to soil only +σσ aa(2) Due to the surcharge [7.34] According to Laba and Kennedy (1986), σσ aa(2) may be expressed (figure 30b) as σσ aa(2) = MM 2qq (ββ sinββcos2αα) ππ (in radius) [7.35] MM = 1.4 0.4bb 0.14HH 1 [7.36] The net active (lateral) pressure distribution on the retaining wall calculated by using equations. (34, 35 and 36) is shown in figure 29b. Tie Force Refer again to figure 29. The tie force per unit length of the wall developed at any depth z is

TT = active earth pressure at depth zz area of the wall to be supported by the tie = (σσ aa )(SS VV SS HH ) [7.37] Factor of Safety Against Tie Failure The reinforcement ties at each level and thus the walls cold fail by either (a0 tie breaking or (b) tie pullout. The factor of safety against tie breaking may be determined as FFFF (BB) = yield or breaking strength of each tie maximum tie force in any tie = wwww ff yy σσ aa SS VV SS HH [7.38] ww = width of each tie tt = thickness of each tie ff yy = yield or breaking strength of the tie material A factor of safety of about 2.5-3 is generally recommended for ties at all levels. Reinforcing ties at any depth, z, will fail by pullout if the frictional resistance developed along their surfaces is less than the force to which the ties are being subjected. The effective length of the ties along which the frictional resistance is developed may be conservatively taken as the length that extends beyond the limits of the Rankine active failure zone, which is the zone AAAAAA in figure 29. Line BBBB in figure 29 makes an angle of 45 + φφ 1 /2 with the horizontal. Now, the maximum friction force FF RR that can be realized for a tie at depth z is FF RR = 2ll ee wwσσ vv tan φφ μμ [7.39] ll ee = effective length σσ vv = effective vertical pressure at a depth zz φφ μμ = soil tie friction angle Thus the factor of safety against tie pullout at any depth z is FFFF (PP) = FF RR TT [7.40] FFFF (PP) = factor of safety against tie pullout

Substituting equations (37 and 39) into equation (40) yields FFFF (PP) = 2ll eewwσσ vv tan φφ μμ σσ aa SS VV SS HH [7.41] Total Length of Tie The total length of ties at any depth is LL = ll rr + ll ee [7.42] ll rr = length with the Rankine failure zone ll ee = effective length For a given FFFF (PP) from equation (41), ll ee = FFFF (PP)σσ aa SS VV SS HH 2wwσσ vv tan φφ μμ [7.43] Again, at any depth z, ll rr = (HH zz) tan 45+ φφ 1 2 [7.44] So, combining equations (42, 43, and 44) gives LL = (HH zz) + FFFF(PP)σσaa SSVV SSHH tan 45+ φφ 1 2 2wwσσ vv tan φφ μμ [7.45] STEP-BY-STEP DESIGN PROCEDURE (METALLIC STRIP REINFORCEMENT) Following is a step-by-step procedure for the design of reinforced earth retaining walls. General: 1. Determine the height of the wall, H, and the properties of the granular backfill material, such as unit weight (γγ 2 ) and angle of friction φφ 1 ). 2. Obtain the soil-tie friction angle, φφ μμ and the required values of FFFF (BB) and FFFF (PP). Internal Stability: 3. Assume values for horizontal and vertical tie spacing. Also assume the width of reinforcing strip, w, to be used. 4. Calculate σσ aa from equations (34, 35, and 36). 5. Calculate the tie forces at various levels from equation (37). 6. For known values of FFFF (BB), calculate the thickness of ties, t, to resist the tie breakout:

TT = σσ aa SS VV SS HH = wwww ff yy FFFF (BB) Or tt = (σσ aa SS VV SS HH )[FFFF (BB) ] [7.46] wwww yy The convention is to keep the magnitude of t the same at all levels, so σσ aa, in equation (46) should equal σσ aa(max ). 7. For the known values of φφ μμ and FFFF (PP), determine the length, L, of the ties at various levels from equation (45). 8. The magnitudes of SS VV, SS HH, tt, ww, and LL may be changed to obtain the most economical design. External Stability: 9. A check for overturning can be done as follows with reference to figure 28. Taking the moment about B yields the overturning moment for the unit length of the wall. MM OO = PP aa zz [7.47]

Figure 7.28 Stability check for the retaining wall HH PP aa = active force = σσ aa dddd 0 The resisting moment per unit length of the wall is MM RR = WW 1 xx 1 + WW 2 xx 2 +. +qqqq bb + aa [7.48] WW 1 = (area AAAAAAAAAA)(1)(γγ 1 ) WW 2 = (area FFFFFFFF)(1)(γγ 1 ) So 2 FFFF (overturning ) = MM RR MM OO WW 1 xx 1 +WW 2 xx 2 +.+qqqq bb + aa 2 HH 0 σσ aa dddd zz [7.49] 10. The check for sliding can be done by using equation (11), or

FFFF (sliding ) = WW 1xx 1 +WW 2 +qqaa [tan (kkφφ 1 ] PP aa [7.50] kk 2 3. 11. Check for ultimate bearing capacity failure. The ultimate bearing capacity can be given as qq uu = cc 2 NN cc + 1 2 γγ 2LL 2 NN γγ [7.51a] The bearing capacity factors NN cc and NN γγ correspond to the soil friction angle, φφ 2 (table 4 chapter 3). In equation (51a) LL 2 is the effective length, or LL 2 = LL 2 2ee [7.51b] ee = eccentricity ee = LL 2 MM RR MM OO 2 Σ VV [7.51c] Example 6 Σ VV = WW 1 + WW 2 + qqqq The vertical stress at zz = HH, from equation (31), is σσ vv(hh) = γγ 1 HH + σσ vv(2) [7.52] So the factor of safety against bearing capacity failure is FFFF (bearing capacity ) = qq ult σσ vv(hh) [7.53] Generally, minimum values of FFFF (overturning ) = 3, FFFF (sliding ) = 3, and FFFF (bearing capacity failure ) = 3 to 5 are recommended. A 30-ft-high retaining wall galvanized steel-strip reinforcement in a granular backfill has to be constructed. Referring to figure 29, given: Granular backfill: φφ 1 = 36

γγ 1 = 105 lb/ft 3 Foundation soil: φφ 2 = 28 γγ 2 = 110 lb/ft 3 cc 2 = 1000 lb/ft 2 Galvanized steel reinforcement: Width of strip, ww = 3 in. SS VV = 2 ft center to center SS HH = 3 ft center to center ff yy = 35,000 lb/in 2 φφ μμ = 20 Required FFFF (BB) = 3 Required FFFF (PP) = 3 Check for the external and internal stability. Assume the corrosion rate of the galvanized steel to be 0.001 in./year ad the life span of the structure to be 50 years. Solution Internal Stability Check a. Tie thickness: Maximum tie force, TT max = σσ aa(max ) SS VV SS HH σσ aa(max ) = γγγγkk aa = γγγγtan 2 45 + φφ 1 2 So TT (max ) = γγγγ tan 2 45 + φφ 1 2 SS VVSS HH From equation (46), for tie break, tt = (σσ aa SS VV SS HH )[FFFF (BB) ] wwff yy = γγγγ tan 2 45+ φφ 1 2 SS VVSS HH wwff yy

Or tt = (105)(30)tan 2 45 36 2 (2)(3) (3) = 0.0117 ft = 0.14 in. 3 12 ft (35,000 144 lb /ft2 ) If the rate of corrosion is 0.001 in./yr and the life span of the structure is 50 yr, then the actual thickness, t, o the ties will be tt = 0.14 + (0.001)(50) = 0.19 in. So a tie thickness of 0.2 in. would be enough. b. Tie length: Refer to equation (45). For this case, σσ aa = γγ 1 zzkk aa and σσ vv = γγ 1 zz, so LL = (h zz) + FFFF(PP)γγ1zzKKaa SSVVSSHH tan 45+ φφ 1 2 2wwγγ 1 zz tan φφ μμ Not the following table can be prepared. (Note: FFFF (PP) = 3, HH = 30 ft, ww = 3 in., and φφ μμ = 20 ). zz(ft) 5 10 15 20 25 30 Tie length, L (ft) [equation (45)] 38.45 35.89 33.34 30.79 28.25 25.7 So use a tie length of LL = 4444 ffff External Stability Check a. Check for overturning: Refer to figure 29. For this case, using equation (49), FFFF (overturning ) = WW 1xx 1 HH σσ aa dddd zz 0 WW 1 = γγ 1 HHHH = (105)(30)(40) = 126,000 lb xx 1 = 20 ft

HH PP aa = σσ aa dddd = 1 γγ 2 1KK aa HH 2 = 1 2 (105)(0.26)(30)2 = 12,285 lb/ft 0 zz = 30 3 = 10 ft FFFF (overturning ) = (126,000)(20) (12,285)(10) = 20.5 > 3 OK Figure 7.29 b. Check for sliding: From equation (50) FFFF (sliding ) = WW 1 tan (kkφφ 1 ) PP aa = 2 126,000tan 3 (36) = 4.57 > 3OK 12,285 c. Check for bearing capacity: For φφ 2 = 28, NN cc = 25.8, NN γγ = 16.78 (Table 4 from chapter 3). From equation (51a), qq ult = cc 2 NN cc + 1 2 γγ 2LL NN γγ

ee = LL 2 MM RR MM OO Σ VV = 40 2 (126,000 20) (12,285 10) = 0.975 ft 126,000 From equation (52), σσ VV(HH) = γγ 1 HH = (105)(30) = 3150 lb/ft 2 FFFF (bearing capacity ) = qq ult = 60,791 = 19.3 > 5 OK qq vv(hh ) 3150