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115 copper chromate or approved equivalent preservative. The Back wall height, H2 2.2 m bottom row of timber should be treated for direct burial. Traffic surcharge, q 9.4 kN/m2 The color may be green or brown, but not mixed. Bridge vertical dead load, DL 45 kN/m Forming elements in the back of the timber face may con- Bridge vertical live load, LL 50 kN/m sist of wood (minimum 250 mm nominal thickness treated Bridge horizontal load, F2 2.25 kN/m to an acceptable level with copper chromate or approved Span 24 m (simple span) equivalent), fiberglass, plastic, or other approved material. Length of concrete approach slab 4.25 m The typical reinforcement used is a nonwoven geotex- tile, although other geosynthetics that satisfy the design Trial design parameters: criteria can also be used. Nails should be 16d galvanized ring shank nails and Sill width, B 1.5 m should be placed at the top and bottom of the timbers at Clear distance, d 0.3 m 0.3 m (1 ft) intervals. Sill type integrated sill Compaction should be consistent with project embank- Facing modular concrete blocks ment specifications, except that no compaction is Facing block size 200 mm x 200 mm x 400 mm allowed within 0.3 to 0.6 m (1 to 2 ft) of the wall face. Batter of facing 1/35 (6 mm setback for each Shimming of timber to maintain verticality is permissible. block) All reinforcement overlaps should be at least 0.3 m (1 ft) Reinforcement spacing 0.2 m wide and should be perpendicular to the wall face. All exposed fabric should be painted with a latex paint Note: As the batter of 1/35 corresponds to an angle of 1.6, matching the color of the timbers. less than 8, the abutment wall is to be designed as a ver- To improve connection strength on the top lifts, the geo- tical wall, and the coefficient of earth pressure is to follow textile can be wrapped around the facing timbers and then the general Rankine case, per Section 4.2d, NHI manual. covered or protected with wooden panels. This technique has been described by Keller and Devin (2003). The configuration of trial design for the GRS abutment is shown in Figure 3-4. Natural Rock Facing: Step 2: Establish soil properties Do not exceed the height and slope angles delineated in the design without evidence that higher or steeper Reinforced fill: features will be stable. The selected fill satisfies the following criteria: Rocks should be placed by skilled operators and should 100 percent passing 100 mm (4 in.) sieve, 0-60 percent be placed in fairly uniform lifts. passing No. 40 (0.425 mm) sieve, and 0-15 percent Care should be exercised in placing the infill. The infill- passing No. 200 (0.075 mm) sieve; PI 6. ing should be as complete as possible. The friction angle of the fill = 35, as determined by one set of the standard direct shear test on the portion DESIGN EXAMPLES finer than 2 mm (No. 10) sieve, using a sample com- pacted to 95 percent of AASHTO T-99, Methods C Two design examples are given here to illustrate the or D, at optimum moisture content. design computation procedure of the recommended design method. Design Example 1 has an integrated sill with a tall test = 35, rf = 18.8 kN/m3, Ka(rf) = tan2 (45 - rf/2) upper wall, whereas Design Example 2 has an isolated sill = 0.28 with a short upper wall. Note: The "design friction angle" is taken as one degree lower than test, i.e., design = rf = 34 (see The Recom- Design Example 1: GRS Abutment with an mend Design Method Step 2). Integrated Sill and a Tall Upper Wall Retained earth: Step 1: Establish abutment geometry, external loads and trial design parameters re = 30, re = 18.8 kN/m3, Ka(re) = tan2 (45 - re/2) = 0.33 Wall heights and external loads: Foundation soil: Total abutment height, H 9.7 m Load-bearing wall height, H1 7.5 m fs = 30, fs = 20.0 kN/m3, qaf = 300 kN/m2

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116 Step 3: Establish design requirements With unit weight of concrete, concrete = 23.6 kN/m3, the fol- lowing forces acting on the sill are determined: External stability design requirements: Sliding 1.5 V1 = (B t) concrete Eccentricity l/6 V1= (1.5 m 0.65 m) 23.6 kN/m3 = 23.01 kN/m Sill pressure allowable bearing of the reinforced fill qallow = 180 kPa (as determined in Step 4 below) V2 = [(fw + b) fh] concrete Average contact pressure at the foundation level allow- V2 = [(0.8 m + 0.4 m) 0.1 m] 23.6 kN/m3 able bearing pressure of the foundation soil, qaf = 300 kPa = 2.83 kN/m Internal stability design requirements: V3 = [b (H2 - fh - t)] concrete Factor of safety against pullout FSpullout 1.5 V3 = [0.4 m (2.2 m - 0.1 m - 0.65 m)] 23.6 kN/m3 Facing connection strength is OK with reinforcement = 13.69 kN/m spacing = 0.2 m (see The Recommended Design Method, Step 3). DL = 45 kN/m (from Step 1) LL = 50 kN/m (from Step 1) Step 4: Determine allowable bearing pressure of reinforced fill Fq = Ka(rf) q H2 Fq = 0.28 9.4 kN/m2 2.2 m = 5.79 kN/m Determine the allowable bearing pressure of the rein- forced fill, qallow, with the following conditions: F1 = 1/2 Ka(rf) rf H22 design = rf = 34 F1 = 1/2 (0.28) 18.8 kN/m3 (2.2 m)2 = 12.74 kN/m Reinforcement spacing = 0.2 m (uniform spacing with F2 = 2.25 kN/m (from Step 1) no truncation) Integrated sill, sill width = 1.5 m Check factor of safety against sliding: (1) From Table 3-1, for = 34 and reinforcement spac- ing = 0.2 m, allowable bearing pressure = 180 kPa. Va = sum of vertical forces acting on the sill (2) From Figure 3-1, the correction factor for a sill Va = V1 + V2 + V3 + DL + LL width of 1.5 m is 1.0; thus the corrected allowable Va = 23.01 kN/m + 2.83 kN/m + 13.69 kN/m bearing pressure = 180 kPa 1.0 = 180 kPa. + 45 kN/m + 50 kN/m = 134.53 kN/m (3) No reduction for an integrated sill. Thus, qallow = 180 kPa. Fa = sum of horizontal forces acting on the sill Fa = Fq + F1 + F2 Fa = 5.79 kN/m + 12.74 kN/m + 2.25 kN/m Step 5: Establish trial reinforcement length = 20.78 kN/m Select a preliminary reinforcement length = 0.7 * total ( Va - LL ) tan rf FSsliding = abutment height Fa FSsliding = (134.53 kN/m - 50 kN/m) tan 34 L = 0.7 H = 0.7 9.7 m = 6.8 m (use 7.0 m) / 20.78 kN/m = 2.74 > 1.5 (OK) Check eccentricity requirement: Step 6: Evaluate stability of footing/sill The preliminary sill configuration and forces acting on the MOA = sum of overturning moments about point A sill are shown in Figure 3-5. The dimensions of the sill are MOA = Fq (H2/2) + F1 (H2/3) + F2 (t + fh) MOA = 5.79 kN/m (2.2 m/2) + 12.74 kN/m B 1.5 m (2.2 m/3) + 2.25 kN/m (0.65 m + 0.1 m) d 0.3 m = 17.40 kN/m m H2 2.2 m t 0.65 m MRA = sum of resisting moments about point A b 0.4 m MRA = V1 (B/2) + V2 [(fw + b)/2 + (B - b - fw)] fw 0.8 m + V3 [(b/2) + (B - b)] + (DL + LL) fh 0.1 m [(fw/2) + (B - b - fw)]

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117 I1 is the influence depth caused by the horizontal forces MRA = 23.01 kN/m (1.5 m/2) + 2.83 kN/m [(0.8 m in the back wall (see Figure 3-6): + 0.4 m)/2 + (1.5 m - 0.4 m - 0.8 m)] I1 = (d + B - 2 e) tan (45 + rf/2) + 13.69 kN/m [(0.4 m/2) + (1.5 m - 0.4 m)] I1 = [0.3 m + 1.5 m - 2 (0.11 m)] tan (45 + 34/2) + (45 kN/m + 50 kN/m) [(0.8 m/2) + (1.5 m = 2.97 m - 0.4 m - 0.8 m)] = 104.10 kN/m m e = eccentricity at the base of the sill Check factor of safety against sliding for the reinforced B M RA - MOA volume: e = - 2 Va V = sum of vertical forces acting on the foundation soil e = (1.5 m/2) - (104.10 kN/m m - 17.40 kN/m V = V4 + V5 + Vq + Va m)/134.53 kN/m = 0.11 m V = 987 kN/m + 215.07 kN/m + 48.88 kN/m + 134.53 kN/m = 1385.48 kN/m B/6 = 1.5 m/6 = 0.25 m F = sum of horizontal forces acting on the foundation e < B/6 (OK) soil F = F3 + F4 + Fa Check allowable bearing pressure of the reinforced fill: F = 125.63 kN/m + 174.49 kN/m + 20.78 kN/m = 320.90 kN/m psill = applied pressure from the sill Va p sill = ( V - LL - Vq ) tan fs B - 2e ' FSsliding = F psill = 134.53 kN/m/[1.5 m - (2 0.11 m)] = 105.1 kN/m2 psill = 105.1 kPa < qallow = 180 kPa (OK) FSsliding = (1385.48 kN/m - 50 kN/m - 48.88 kN/m) tan (30)/320.90 kN/m = 2.31 > 1.5 (OK) Step 7: Check external stability of reinforced fill Check eccentricity requirement for the reinforced volume: with the preliminary reinforcement length established in Step 5 MO = sum of overturning moments about point C The forces needed to evaluate the external stability of the MO = F3 (H1/2) + F4 (H1/3) + Fa (H1 - I1/3) abutment are shown in Figure 3-4. These forces are MO = 125.63 kN/m (7.5 m/2) + 174.49 kN/m calculated as follows. (7.5 m/3) + 20.78 kN/m [7.5 m - (2.97 m/3)] = 1042.62 kN/m m V4 = (L H1) rf MR = sum of resisting moments about point C V4 = (7 m 7.5 m) 18.8 kN/m3 = 987 kN/m MR = V4 (L/2) + (V5 + Vq) [(L - d - B)/2 + (d + B)] + (MRA + Va d) V5 = [(L - d - B) H2] rf MR = 987 kN/m (7 m/2) + (215.07 kN/m V5 = [(7 m - 0.3 m - 1.5 m) 2.2 m] 18.8 kN/m3 = + 48.88 kN/m) [(7 m - 0.3 m - 1.5 m)/2 215.07 kN/m + (0.3 m + 1.5 m)] + [104.10 kN/m m + 134.53 kN/m (0.3 m)] = 4760.34 kN/m m Vq = (L - d - B) q Vq = (7 m - 0.3 m - 1.5 m) 9.4 kN/m2 = 48.88 kN/m MS = moment about point C caused by traffic surcharge MS = Vq [(L - d - B)/2 + (d + B)] F3 = [Ka(re) (q + re H2)] H1 MS = 48.88 kN/m [(7 m - 0.3 m - 1.5 m)/2 + (0.3 m F3 = [0.33 (9.4 kN/m2 + 18.8 kN/m3 2.2 m)] 7.5 m + 1.5 m)] = 215.07 kN/m m = 125.63 kN/m e = eccentricity at the base of the reinforced volume F4 = 1/2 Ka(re) re H12 L ( MR - MS ) - MO F4 = 1/2 (0.33) 18.8 kN/m3 (7.5 m)2 e= - 2 V - Vq = 174.49 kN/m e = (7 m/2) - [(4760.34 kN/m m - 215.07 kN/m m) Va = 134.53 kN/m (from Step 6) - 1042.62 kN/m m]/(1385.48 kN/m - 48.88 kN/m) Fa = 20.78 kN/m (from Step 6) = 0.88 m

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118 L/6 = 7 m/6 = 1.17 m Le = length of embedment in the resistant zone behind the failure surface at depth z e < L/6 (OK) Le = L - La Check allowable bearing pressure of the foundation soil: La = length of embedment in the active zone at depth z La = (H1 - z) tan (45 - rf/2) Calculate the "influence length" D1 at the foundation level and compare with the effective reinforcement length, L. Li = length of embedment within the influence area inside the resistant zone; this length can be measured D1 = d + B + H1/2 = d + (B - 2e) + H1/2 directly from the design drawing. D1 = 0.3 m + [1.5 m - 2 (0.11 m)] + 7.5 m/2 = 5.33 m C = reinforcement effective unit perimeter; C = 2 for L = L - 2e strips, grids, and sheets. L = 7.0 m - 2 (0.88 m) = 5.24 m Rc = coverage ratio; Rc = 1.0 for 100 percent coverage Because D1 at the foundation level is greater than L of reinforcement. (L = L-2e), thus the contact pressure on the founda- tion level, pcontact, is calculated as follows. h = horizontal pressure at depth z h = Ka(rf) (vs + v + q) + h V p contact = L - 2e vs = vertical soil pressure at depth z pcontact = 1385.48 kN/m/[7.0 m - 2 (0.88 m)] vs = (rf H2) + (rf z) = 264.40 kN/m2 qaf = 300 kN/m2 (from Step 2) v = distributed vertical pressure from sill pcontact = 264.40 kN/m2 < qaf = 300 kN/m2 (OK) v = Va / D D = effective width of applied load at depth z Step 8: Evaluate internal stability at each For z z2: D = (B - 2e) + z reinforcement level For z > z2: D = d + (B - 2e) + z/2 z2 = 2 d With geosynthetic reinforcement, the coefficient of lateral earth pressure is constant throughout the entire wall h = supplement horizontal pressure at depth z height, per Section 4.3b, NHI manual. For z I1: h = 2 Fa (I1 - z)/(I12) For z > I1: h = 0 The internal stability is evaluated by checking the rein- forcement pullout failure. Tmax = maximum tensile force in the reinforcement at depth z Check reinforcement pullout failure: Tmax = h s Pr = pullout resistance s = vertical reinforcement spacing Pr = F* (v Le) C Rc F* = pullout resistance factor FSpullout = factor of safety against reinforcement pullout F* = 2/3 tan rf FSpullout = Pr / Tmax F* = 2/3 tan(34) = 0.45 Let depth z be measured from the top of the load-bearing = a scale effect correction factor ranging from 0.6 to wall. Reinforcement no. 25 at z = 2.5 m (see Figure 3-6) 1.0 for geosynthetic reinforcement; for geotextile, is would serve as an example for determining the FSpullout. defaulted to 0.6, per Section 3.3b, NHI manual. vs = (rf H2) + (rf z) (v Le) = normal force at the soil-reinforcement inter- vs = (18.8 kN/m3 2.2 m) + (18.8 kN/m3 2.5 m) face at depth z (excluding traffic surcharge) = 88.36 kN/m2 (v Le) = (vs Le) + (v Li) z2 = 2 d = 2 (0.3 m) = 0.6 m

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119 The recommended combined safety factor is Fs = 5.5 for z = 2.5 m > z2 = 0.6 m, D = d + (B - 2e) + z/2 reinforcement 0.2 m and Fs = 3.5 for reinforcement D = 0.3 m + [1.5 m - 2 (0.11 m)] + 2.5 m/2 spacing of 0.4 m. = 2.83 m From Table 3-2, h(max) is 59.84 kN/m2, which occurs at v = Va / D reinforcement No. 1 with depth z = 7.3 m and s = 0.2 m. v = 134.53 kN/m/ 2.83 m = 47.54 kN/m2 T@=1.0 percent = h(max) s z = 2.5 m < I1 = 2.97 m, h = 2 Fa (I1 - z)/(I12) T@=1.0 percent = 59.84 kN/m2 (0.2 m) = 11.97 kN/m h = 2 (20.78 kN/m) (2.97 m - 2.5 m)/(2.97 m)2 = 2.21 kN/m2 The uniform reinforcement spacing is 0.2 m, hence Fs = 5.5. h = Ka(rf) (vs + v + q) + h Tult = Fs T@=1.0 percent h = 0.28 (88.36 kN/m2 + 47.54 kN/m2 + 9.4 kN/m2) Tult = 5.5 (11.97 kN/m) = 65.84 kN/m + 2.21 kN/m2 = 42.89 kN/m2 A reinforcement with minimum "working" stiffness, Tmax = h s T@=1.0 percent = 12.0 kN/m and minimum ultimate strength Tmax = 42.89 kN/m2 (0.2 m) = 8.58 kN/m (per ASTM D4595), Tult = 65.8 kN/m is required. La = (H1 - z) tan (45 - rf/2) La = (7.5 m - 2.5 m) tan (45 - 34/2) = 2.66 m Step 10: Design of back/upper wall Le = L - La Reinforced fill: Same as that of the load- Le = 7 m - 2.66 m = 4.34 m bearing wall Reinforcement: Same as that of the load- Calculate the normal force at z = 2.5 m: bearing wall Reinforcement length: 4.25 m (length of approach (v Le) = (vs Le) + (v Li) slab) + 1.5 m = 5.75 m (see (v Le) = (88.36 kN/m2 4.34 m) + (47.54 kN/m2 The Recommended Design 0.17 m) = 391.56 kN/m Method Step 10) Pr = F* (v Le) C Rc Reinforcement layout: Vertical spacing = 0.3 m Pr = (0.45) (0.6) 391.56 kN/m (2) (1) Wrapped-face with wrapped = 211.44 kN/m return at least 0.5 m (18 in.) in the horizontal direction and FSpullout = Pr / Tmax anchored in at least 100 mm of FSpullout = 211.44 kN/m/ 8.58 kN/m = 24.64 fill material. FSpullout = 24.64 > 1.5 (OK) A compressible layer of about 50 mm thick should be installed between the wrapped face and the rigid back wall. The values of FSpullout for all the reinforcements in the load-bearing wall are summarized in Table 3-2. Step 11: Check angular distortion between abutments Step 9: Determine the required reinforcement stiffness and strength abutment = abutment settlement abutment = 1.5 percent H1 The minimum "working" reinforcement stiffness is deter- abutment = 0.015 (7.5 m) = 0.1125 m mined as foundation = foundation settlement (as determined by con- T@=1.0 percent h(max) s ventional settlement computation methods) Egfoundation = 0.01 m The minimum ultimate reinforcement strength is deter- mined as total = total settlement total = abutment + foundation Tult Fs T@=1.0 percent total = 0.1125 m + 0.01 m = 0.1225 m

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120 TABLE 3-2 Design Example 1--tabulated results of FSpullout at each reinforcement level No. Depth s vs D v h h Tmax La Le Li ( v Le) Pr FSpullout z 2 2 2 2 (m) (m) (kN/m ) (m) (kN/m ) (kN/m ) (kN/m ) (kN/m) (m) (m) (m) (kN/m) (kN/m) 1 7.3 0.2 178.6 5.23 25.72 0.00 59.84 11.97 0.11 6.89 5.12 1363.00 735.49 61.45 2 7.1 0.2 174.84 5.13 26.22 0.00 58.93 11.79 0.21 6.79 4.92 1315.65 709.93 60.24 3 6.9 0.2 171.08 5.03 26.75 0.00 58.02 11.60 0.32 6.68 4.71 1268.98 684.75 59.01 4 6.7 0.2 167.32 4.93 27.29 0.00 57.12 11.42 0.43 6.57 4.50 1222.99 659.93 57.76 5 6.5 0.2 163.56 4.83 27.85 0.00 56.23 11.25 0.53 6.47 4.30 1177.67 635.48 56.51 6 6.3 0.2 159.8 4.73 28.44 0.00 55.34 11.07 0.64 6.36 4.09 1133.02 611.39 55.24 7 6.1 0.2 156.04 4.63 29.06 0.00 54.46 10.89 0.74 6.26 3.89 1089.03 587.65 53.95 8 5.9 0.2 152.28 4.53 29.70 0.00 53.59 10.72 0.85 6.15 3.68 1045.68 564.25 52.65 9 5.7 0.2 148.52 4.43 30.37 0.00 52.72 10.54 0.96 6.04 3.47 1002.96 541.20 51.33 10 5.5 0.2 144.76 4.33 31.07 0.00 51.86 10.37 1.06 5.94 3.27 960.87 518.49 49.99 11 5.3 0.2 141 4.23 31.80 0.00 51.02 10.20 1.17 5.83 3.06 919.39 496.11 48.62 12 5.1 0.2 137.24 4.13 32.57 0.00 50.18 10.04 1.28 5.72 2.85 878.51 474.05 47.24 13 4.9 0.2 133.48 4.03 33.38 0.00 49.35 9.87 1.38 5.62 2.65 838.21 452.31 45.82 14 4.7 0.2 129.72 3.93 34.23 0.00 48.54 9.71 1.49 5.51 2.44 798.48 430.87 44.38 15 4.5 0.2 125.96 3.83 35.13 0.00 47.74 9.55 1.60 5.40 2.23 759.30 409.72 42.92 16 4.3 0.2 122.2 3.73 36.07 0.00 46.95 9.39 1.70 5.30 2.03 720.64 388.86 41.42 17 4.1 0.2 118.44 3.63 37.06 0.00 46.17 9.23 1.81 5.19 1.82 682.49 368.28 39.88 18 3.9 0.2 114.68 3.53 38.11 0.00 45.41 9.08 1.91 5.09 1.62 644.82 347.95 38.31 19 3.7 0.2 110.92 3.43 39.22 0.00 44.67 8.93 2.02 4.98 1.41 607.61 327.87 36.70 20 3.5 0.2 107.16 3.33 40.40 0.00 43.95 8.79 2.13 4.87 1.20 570.82 308.02 35.04 21 3.3 0.2 103.4 3.23 41.65 0.00 43.25 8.65 2.23 4.77 1.00 534.41 288.37 33.34 22 3.1 0.2 99.64 3.13 42.98 0.00 42.57 8.51 2.34 4.66 0.79 498.35 268.91 31.59 23 2.9 0.2 95.88 3.03 44.40 0.34 42.25 8.45 2.45 4.55 0.58 462.58 249.61 29.54 24 2.7 0.2 92.12 2.93 45.91 1.28 42.56 8.51 2.55 4.45 0.38 427.08 230.45 27.07 25 2.5 0.2 88.36 2.83 47.54 2.22 42.90 8.58 2.66 4.34 0.17 391.76 211.40 24.64 26 2.3 0.2 84.6 2.73 49.28 3.16 43.28 8.66 2.76 4.24 0.00 358.29 193.34 22.34 27 2.1 0.2 80.84 2.63 51.15 4.10 43.69 8.74 2.87 4.13 0.00 333.77 180.10 20.61 28 1.9 0.2 77.08 2.53 53.17 5.04 44.15 8.83 2.98 4.02 0.00 310.05 167.30 18.95 29 1.7 0.2 73.32 2.43 55.36 5.98 44.65 8.93 3.08 3.92 0.00 287.13 154.94 17.35 30 1.5 0.2 69.56 2.33 57.74 6.93 45.20 9.04 3.19 3.81 0.00 265.01 143.00 15.82 31 1.3 0.2 65.8 2.23 60.33 7.87 45.81 9.16 3.30 3.70 0.00 243.68 131.49 14.35 32 1.1 0.2 62.04 2.13 63.16 8.81 46.50 9.30 3.40 3.60 0.00 223.16 120.42 12.95 33 0.9 0.2 58.28 2.03 66.27 9.75 47.26 9.45 3.51 3.49 0.00 203.44 109.78 11.62 34 0.7 0.2 54.52 1.93 69.70 10.69 48.11 9.62 3.62 3.38 0.00 184.52 99.57 10.35 35 0.5 0.2 50.76 1.78 75.58 11.63 49.64 9.93 3.72 3.28 0.00 166.39 89.79 9.04 36 0.3 0.2 47 1.58 85.15 12.57 52.21 10.44 3.83 3.17 0.00 149.07 80.44 7.70 37 0.1 0.2 43.24 1.38 97.49 13.51 55.55 11.11 3.93 3.07 0.00 132.55 71.52 6.44 Angular distortion = total / span length = 0.1225 m / Back wall height, H2 2.2 m 24 m = 0.0051 Facing modular concrete blocks (200 mm x 200 mm x Tolerable angular distortion for simple span = 0.005 400 mm) Front batter 1/35 Angular distortion = 0.0051 0.005 (OK) Sill type integrated sill Sill width 1.5 m Design summary: Sill clear distance 0.3 m Embedment 200 mm (one facing block height) The configuration for the trial design is shown in Figure 3-7. Reinforcement: Abutment configuration: Minimum stiffness at =1.0 percent, T@=1.0 percent = 12.0 kN/m Load-bearing wall Minimum ultimate strength, Tult = 65.8 kN/m height, H1 7.5 m Length in load-bearing wall = 7.0 m

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121 Compressible Layer 1.5 m 4.25 m 1.5 m 0.3 m Approach Slab (Back wall) 0.3 m 2.2 m Sill 1 Front batter 35 0.2 m (Load-bearing wall) Reinforcement: 7.5 m T @ =1% = 12 kN/m T ult = 65.8 kN/m 7m 0.2 m (Embedment) Figure 3-7. Design Example 1--configuration of the trial design. Length of top three layers in load-bearing wall = 7.5 m Span length 10 m (simple span) Vertical spacing in load-bearing wall = 0.2 m Approach slab is not used Length in back wall = 5.75 m Vertical spacing in back wall = 0.3 m Trial design parameters: Sill width, B 0.6 m Design Example 2: A GRS Abutment with Clear distance, d 0.3 m an Isolated Sill and a Short Lower Wall Sill type isolated sill Facing modular concrete blocks Step 1: Establish abutment geometry, external Facing block size 200 mm x 200 mm x 400 mm loads and trial design parameters Batter of facing 1/35 (6 mm setback for each block) Wall heights and external loads: Reinforcement spacing 0.2 m Total abutment height, H 3.0 m Note: As the batter of 1/35 corresponds to an angle of Load-bearing wall height, H1 2.4 m 1.6, less than 8, the abutment wall is to be designed as a Back wall height, H2 0.6 m vertical wall, and the coefficient of earth pressure is to fol- Traffic surcharge, q 9.4 kN/m2 low the general Rankine case. Bridge vertical dead load, DL 35 kN/m Bridge vertical live load, LL 40 kN/m The configuration of the initial trial design for the GRS Bridge horizontal load, F2 1.75 kN/m abutment is shown in Figure 3-8.

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122 Traffic Surcharge: q = 9.4 kN/m 2 LL Vq d = 0.3 m (Back Wall) DL V5 H2 = 0.6 m Fq F1 t = 0.3 m V1 A H' = Total Abutment Ht. = 3.0 m Va Retained Earth 2e' Soil Unit Wt. = 18 kN/m3 B' = B - 2e' re = 30 B = 0.6 m K a(re) = 0.33 (Load-Bearing Wall) H1 = 2.4 m F3 V4 Reinforced Fill F4 Soil Unit Wt. = 20 kN/m3 rf = 36 K a(rf) = 0.26 C L - 2e 2e Foundation Soil L = 2.10 m (Preliminary Reinforcement Length) Figure 3-8. Design Example 2--configuration of the abutment. Step 2: Establish soil properties Retained earth: Reinforced fill: re = 30, re = 18 kN/m3, Ka(re) = tan2 (45 - re/2) = 0.33 The selected fill satisfies the following criteria: 100 percent passing 100 mm (4 in.) sieve, 0-60 percent Foundation soil: passing No. 40 (0.425 mm) sieve, and 0-15 percent passing No. 200 (0.075 mm) sieve; PI 6. fs = 30, fs = 20.0 kN/m3, qaf = 300 kN/m2 The friction angle of the fill = 37, as determined by one set of the standard direct shear test on the portion Step 3: Establish design requirements finer than 2 mm (No. 10) sieve, using a sample com- pacted to 95 percent of AASHTO T-99, Methods C or External stability design requirements: D, at optimum moisture content. Sliding 1.5 test = 37, rf = 20 kN/m3, Ka(rf) = tan2 (45 - rf/2) Eccentricity l/6 = 0.26 Sill pressure allowable bearing of the reinforced fill qallow = 345 kPa (as determined in Step 4 below) Note: The "design friction angle" is taken as one Average contact pressure at the foundation level degree lower than test, i.e., design = rf = 36 (see The allowable bearing pressure of the foundation soil, Recommended Design Method Step 2). qaf = 300 kPa

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123 Internal stability design requirements: Fq and F1 are included in this design example conserva- tively. Factor of safety against pullout FSpullout 1.5 Facing connection strength is OK with reinforce- Fq = Ka(rf) q H2 ment spacing = 0.2 m (see The Recommended Fq = 0.26 9.4 kN/m2 0.6 m = 1.47 kN/m Design Method Step 3) F1 = 1/2 Ka(rf) rf H22 F1 = 1/2 (0.26) 20 kN/m3 (0.6 m)2 = 0.94 kN/m Step 4: Determine allowable bearing pressure of reinforced fill F2 = 1.75 kN/m (from Step 1) Determine the allowable bearing pressure of the rein- Check factor of safety against sliding: forced fill qallow with the following conditions: Va = sum of vertical forces acting on the sill design = rf = 36 Va = V1 + DL + LL Reinforcement spacing = 0.2 m (uniform spacing with Va = 4.25 kN/m + 35 kN/m + 40 kN/m = 79.25 kN/m no truncation) Isolated sill, sill width = 0.6 m Fa = sum of horizontal forces acting on the sill Fa = Fq + F1 + F2 (1) From Table 3-1, for = 36 and reinforcement Fa = 1.47 kN/m + 0.94 kN/m + 1.75 kN/m = 4.16 kN/m spacing = 0.2 m, allowable bearing pressure = 200 kPa. ( Va - LL ) tan rf (2) From Figure 3-1, the correction factor for a sill FSsliding = Fa width of 0.6 m is 2.3; thus the corrected allowable bearing pressure = 200 kPa 2.3 = 460 kPa. FSsliding = (79.25 kN/m - 40 kN/m) tan 36/ 4.16 kN/m (3) Reduction factor of 0.75 applies for an isolated sill. = 6.85 > 1.5 (OK) Thus, qallow = 0.75 460 kPa = 345 kPa. Check eccentricity requirement: Step 5: Establish trial reinforcement length MOA = sum of overturning moments about point A Select a preliminary reinforcement length = 0.7 * total MOA = Fq (H2/2) + F1 (H2/3) + F2 (t) abutment height MOA = 1.47 kN/m (0.6 m/2) + 0.94 kN/m (0.6 m/3) + 1.75 kN/m (0.3 m) = 1.15 kN/m m L = 0.7 H = 0.7 3.0 m = 2.1 m MRA = sum of resisting moments about point A MRA = V1 (B/2) + (DL + LL) (B/2) MRA = 4.25 kN/m (0.6 m/2) + (35 kN/m + 40 kN/m) Step 6: Evaluate stability of footing/sill (0.6 m/2) = 23.78 kN/m m The preliminary sill configuration and forces acting on the sill are shown in Figure 3-8. The dimensions of the sill are e = eccentricity at the base of the sill B M RA - MOA e = - B 0.6 m 2 Va d 0.3 m e = (0.6 m/2) - (23.78 kN/m m - 1.15 kN/m H2 0.6 m m)/79.25 kN/m = 0.01 m t 0.3 m B/6 = 0.6 m/6 = 0.10 m With unit weight of concrete, concrete = 23.6 kN/m , the fol- 3 lowing forces acting on the sill are determined: e < B/6 (OK) V1 = (B t) concrete Check allowable bearing pressure of the reinforced fill: V1= (0.6 m 0.3 m) 23.6 kN/m3 = 4.25 kN/m psill = applied pressure from the sill DL = 35 kN/m (from Step 1) Va p sill = LL = 40 kN/m (from Step 1) B - 2e '

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124 psill = 79.25 kN/m/[0.6 m - (2 0.01 m)] = 136.64 kN/m2 F3 = [Ka(re) (q + re H2)] H1 F3 = [0.33 (9.4 kN/m2 + 18 kN/m3 0.6 m)] 2.4 m psill = 136.64 kPa < qallow = 345 kPa (OK) = 16.0 kN/m F4 = 1/2 Ka(re) re H12 Step 7: Check external stability of reinforced fill F4 = 1/2 (0.33) 18 kN/m3 (2.4 m)2 = 17.11 kN/m with the preliminary reinforcement length established in Step 5 Va = 79.25 kN/m (from Step 6) The forces needed to evaluate the external stability of the Fa = 4.16 kN/m (from Step 6) abutment are shown in Figure 3-8. These forces are cal- culated as follows: I1 is the influence depth caused by the horizontal forces in the back wall (see Figure 3-9): V4 = (L H1) rf I1 = (d + B - 2 e) tan (45 + rf/2) V4 = (2.1 m 2.4 m) 20 kN/m3 = 100.8 kN/m I1 = [0.3 m + 0.6 m - 2 (0.01 m)] tan (45 + 36/2) = 1.73 m V5 = [(L - d - B) H2] rf V5 = [(2.1 m - 0.3 m - 0.6 m) 0.6 m] 20 kN/m3 Check factor of safety against sliding for the reinforced = 14.4 kN/m volume: Vq = (L - d - B) q V = sum of vertical forces acting on the foundation Vq = (2.1 m - 0.3 m - 0.6 m) 9.4 kN/m2 = 11.28 kN/m soil B' = 0.58 m d = 0.3 m 2e' = 0.02 m Failure surface z2 = 0.6 m 11 z 10 Influence line (2V:1H slope) 9 l1 = 1.73 m (at no. 7) (at no. 7) 8 La = 0.71 m Le = 1.69 m 7 Li = 0.66 m 6 (at no. 7) 5 4 3 2 1 45 + rf /2 D 1 = 2.08 m L = 2.40 m (Second Trial) Figure 3-9. Design Example 2 - Notations of the quantities for internal stability evaluation.

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125 V = V4 + V5 + Vq + Va volume is checked again with the new reinforcement V = 100.8 kN/m + 14.4 kN/m + 11.28 kN/m length of 2.4 m as follows: + 79.25 kN/m = 205.73 kN/m V4 = (L H1) rf F = sum of horizontal forces acting on the foundation V4 = (2.4 m 2.4 m) 20 kN/m3 = 115.2 kN/m soil F = F3 + F4 + Fa V5 = [(L - d - B) H2] rf F = 16.0 kN/m + 17.11 kN/m + 4.16 kN/m V5 = [(2.4 m - 0.3 m - 0.6 m) 0.6 m] 20 kN/m3 = 37.27 kN/m = 18.0 kN/m ( V - LL - Vq ) tan fs Vq = (L - d - B) q FSsliding = F Vq = (2.4 m - 0.3 m - 0.6 m) 9.4 kN/m2 = 14.10 kN/m FSsliding = (205.73 kN/m - 40 kN/m - 11.28 kN/m) V = sum of vertical forces acting on the foundation tan (30)/37.27 kN/m = 2.39 > 1.5 (OK) soil V = V4 + V5 + Vq + Va Check eccentricity requirement for the reinforced volume: V = 115.2 kN/m + 18.0 kN/m + 14.10 kN/m + 79.25 kN/m = 226.55 kN/m MO = sum of overturning moments about point C MO = F3 (H1/2) + F4 (H1/3) + Fa (H1 - I1/3) MR = sum of resisting moments about point C MO = 16.0 kN/m (2.4 m/2) + 17.11 kN/m (2.4 m/3) MR = V4 (L/2) + (V5 + Vq) [(L - d - B)/2 + 4.16 kN/m [2.4 m - (1.73 m/3)] + (d + B)] + (MRA + Va d) = 40.47 kN/m m MR = 115.2 kN/m (2.4 m/2) + (18.0 kN/m + 14.10 kN/m) [(2.4 m - 0.3 m - 0.6 m)/2 MR = sum of resisting moments about point C + (0.3 m + 0.6 m)] + [23.78 kN/m m MR = V4 (L/2) + (V5 + Vq) [(L - d - B)/2 + 79.25 kN/m (0.3 m)] = 238.76 kN/m m + (d + B)] + (MRA + Va d) MR = 100.8 kN/m (2.1 m/2) + (14.4 kN/m MS = moment about point C caused by traffic sur- + 11.28 kN/m) [(2.1 m - 0.3 m - 0.6 m)/2 charge + (0.3 m + 0.6 m)] + [23.78 kN/m m MS = Vq [(L - d - B)/2 + (d + B)] + 79.25 kN/m (0.3 m)] = 191.92 kN/m m MS = 14.10 kN/m [(2.4 m - 0.3 m - 0.6 m)/2 + (0.3 m + 0.6 m)] = 23.27 kN/m m MS = moment about point C caused by traffic surcharge MS = Vq [(L - d - B)/2 + (d + B)] e = eccentricity at the base of the reinforced volume MS = 11.28 kN/m [(2.1 m - 0.3 m - 0.6 m)/2 L ( MR - MS ) - MO + (0.3 m + 0.6 m)] = 16.92 kN/m m e= - 2 V - Vq e = (2.4 m/2) - [(238.76 kN/m m - 23.27 kN/m m) e = eccentricity at the base of the reinforced volume - 40.47 kN/m m]/(226.55 kN/m - 14.10 kN/m) L ( MR - MS ) - MO = 0.38 m e= - 2 V - Vq L/6 = 2.4 m/6 = 0.4 m e = (2.1 m/2) - [(191.92 kN/m m - 16.92 kN/m m) - 40.47 kN/m m]/(205.73 kN/m - 11.28 kN/m) e < L/6 (OK) = 0.36 m Check allowable bearing pressure of the foundation soil: L/6 = 2.1 m/6 = 0.35 m Calculate the "influence length" D1 at the foundation e > L/6 (NG) level and compare with the effective reinforcement length, L. As a second trial, the length of the reinforcement was increased to 2.4 m to ensure the eccentricity require- D1 = d + B + H1/2 = d + (B - 2e) + H1/2 ment. The eccentricity requirement of the reinforced D1 = 0.3 m + [0.6 m - 2 (0.01 m)] + 2.4 m/2 = 2.08 m

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126 L = L - 2e C = reinforcement effective unit perimeter; C = 2 for L = 2.4 m - 2 (0.38 m) = 1.64 m strips, grids, and sheets Because D1 at the foundation level is greater than L Rc = coverage ratio; Rc = 1.0 for 100 percent coverage (L = L-2e), the contact pressure on the foundation of reinforcement level, pcontact, is calculated as follows: h = horizontal pressure at depth z V h = Ka(rf) (vs + v + q) + h p contact = L - 2e vs = vertical soil pressure at depth z pcontact = 226.55 kN/m/[2.4 m - 2 (0.38 m)] vs = (rf H2) + (rf z) = 138.14 kN/m2 qaf = 300 kN/m2 (from Step 2) v = distributed vertical pressure from sill pcontact = 138.14 kN/m2 qaf = 300 kN/m2 (OK) v = Va / D D = effective width of applied load at depth z Step 8: Evaluate internal stability at each For z z2: D = (B - 2e) + z reinforcement level For z > z2: D = d + (B - 2e) + z/2 z2 = 2 d With geosynthetic reinforcement, the coefficient of lateral earth pressure is constant throughout the entire wall h = supplement horizontal pressure at depth z height, per Section 4.3b, NHI manual. For z I1: h = 2 Fa (I1 - z)/(I12) For z > I1: h = 0 The internal stability is evaluated by checking the rein- forcement pullout failure. Tmax = maximum tensile force in the reinforcement at depth z Tmax = h s Check reinforcement pullout failure: s = vertical reinforcement spacing Pr = pullout resistance Pr = F* (v Le) C Rc FSpullout = factor of safety against reinforcement pullout FSpullout = Pr / Tmax F* = pullout resistance factor F* = 2/3 tan rf Let depth z be measured from the top of the load-bearing F* = 2/3 tan(36) = 0.48 wall. Reinforcement no. 7 at z = 1.0 m (see Figure 3-9) would serve as an example for determining the = a scale effect correction factor ranging from 0.6 FSpullout. to 1.0 for geosynthetic reinforcement; for geotextile, is defaulted to 0.6, per Section 3.3b, NHI manual. vs = (rf H2) + (rf z) vs = (20 kN/m3 0.6 m) + (20 kN/m3 1.0 m) (v Le) = normal force at the soil-reinforcement inter- = 32.0 kN/m2 face at depth z (excluding traffic surcharge) z2 = 2 d = 2 (0.3 m) = 0.6 m (v Le) = (vs Le) + (v Li) z = 1.0 m > z2 = 0.6 m, D = d + (B - 2e) + z/2 D = 0.3 m + [0.6 m - 2 (0.01 m)] + 1.0 m/2 = 1.38 m Le = length of embedment in the resistant zone behind the failure surface at depth z v = Va / D Le = L - La v = 79.25 kN/m/ 1.38 m = 57.43 kN/m2 La = length of embedment in the active zone at depth z La = (H1 - z) tan (45 - rf/2) z = 1.0 m I1 = 1.73 m, h = 2 Fa (I1 - z)/(I12) h = 2 (4.16 kN/m) (1.73 m - 1.0 m)/(1.73 m)2 Li = length of embedment within the influence area = 2.03 kN/m2 inside the resistant zone; this length can be measured directly from the design drawing. h = Ka(rf) (vs + v + q) + h

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127 The minimum ultimate reinforcement strength is deter- h = 0.26 (32.0 kN/m2 + 57.43 kN/m2 + 9.4 kN/m2) mined as + 2.03 kN/m2 = 27.73 kN/m2 Tult Fs T@=1.0 percent Tmax = h s Tmax = 27.73 kN/m2 (0.2 m) = 5.55 kN/m The recommended combined safety factor is Fs = 5.5 for reinforcement 0.2 m, and Fs = 3.5 for reinforcement La = (H1 - z) tan (45 - rf/2) spacing of 0.4 m. La = (2.4 m - 1.0 m) tan (45 - 36/2) = 0.71 m From Table 3-3, h(max) is 37.57 kN/m2, which occurs Le = L - La at reinforcement no. 11 with depth z = 0.2 m and Le = 2.4 m - 0.71 m = 1.69 m s = 0.2 m. Calculate the normal force at z = 1.0 m: T@=1.0 percent = h(max) s T@=1.0 percent = 37.57 kN/m2 (0.2 m) = 7.51 kN/m (v Le) = (vs Le) + (v Li) (v Le) = (32.0 kN/m2 1.69 m) + (57.43 kN/m2 The uniform reinforcement spacing is 0.2 m, hence 0.66 m) = 91.98 kN/m Fs = 5.5. Pr = F* (v Le) C Rc Tult = Fs T@=1.0 percent Pr = (0.48) (0.6) 91.98 kN/m (2) (1) = 52.98 kN/m Tult = 5.5 (7.51 kN/m) = 41.31 kN/m FSpullout = Pr / Tmax A reinforcement with minimum "working" stiffness, FSpullout = 52.98 kN/m/ 5.55 kN/m = 9.55 T@=1.0 percent = 7.5 kN/m and minimum ultimate strength (per ASTM D4595), Tult = 41.3 kN/m is required. FSpullout = 9.55 > 1.5 (OK) Step 10: Design of back/upper wall The values of FSpullout for all the reinforcements in the load-bearing wall are summarized in Table 3-3. Reinforced fill: Same as that of the load-bearing wall Reinforcement: Same as that of the load-bearing Step 9: Determine the required reinforcement wall stiffness and strength Reinforcement length: 1.3 m (without an approach slab, the back wall reinforcement The minimum "working" reinforcement stiffness is deter- length is to be flush with the rein- mined as forcement in the load-bearing wall) T@=1.0 percent h(max) s Reinforcement layout: Vertical spacing = 0.2 m TABLE 3-3 Design Example 2--tabulated results of FSpullout at each reinforcement level No. Depth s vs D v h h Tmax La Le Li ( v Le) Pr FSpullout z (m) (m) (kN/m2) (m) (kN/m2) (kN/m2) (kN/m2) (kN/m) (m) (m) (m) (kN/m) (kN/m) 1 2.2 0.2 56.0 1.97 40.21 0.00 27.42 5.48 0.10 2.30 1.87 203.84 118.48 21.61 2 2.0 0.2 52.0 1.87 42.36 0.00 26.94 5.39 0.20 2.20 1.67 184.82 107.42 19.94 3 1.8 0.2 48.0 1.77 44.75 0.00 26.52 5.30 0.31 2.09 1.47 166.09 96.54 18.20 4 1.6 0.2 44.0 1.67 47.43 0.31 26.49 5.30 0.41 1.99 1.26 147.58 85.78 16.19 5 1.4 0.2 40.0 1.57 50.45 0.88 26.80 5.36 0.51 1.89 1.06 129.16 75.07 14.01 6 1.2 0.2 36.0 1.47 53.88 1.45 27.22 5.44 0.61 1.79 0.86 110.70 64.34 11.82 7 1.0 0.2 32.0 1.37 57.81 2.01 27.77 5.55 0.71 1.69 0.66 91.99 53.47 9.63 8 0.8 0.2 28.0 1.27 62.35 2.58 28.48 5.70 0.82 1.58 0.46 72.79 42.31 7.43 9 0.6 0.2 24.0 1.17 67.68 3.15 29.39 5.88 0.92 1.48 0.25 52.77 30.67 5.22 10 0.4 0.2 20.0 0.97 81.62 3.72 32.54 6.51 1.02 1.38 0.05 31.85 18.52 2.84 11 0.2 0.2 16.0 0.77 102.79 4.29 37.57 7.51 1.12 1.28 0.00 20.46 11.89 1.58

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128 Step 11: Check angular distortion between Abutment configuration: abutments Load-bearing wall abutment = abutment settlement height, H1 2.4 m abutment = 1.5 percent H1 Back wall abutment = 0.015 (2.4 m) = 0.036 m height, H2 0.6 m Facing modular concrete blocks foundation = foundation settlement (as determined by con- (200 mm x 200 mm x 400 mm) ventional settlement computation methods) Egfoundation = 0.01 m Front batter 1/35 Sill type isolated sill total = total settlement Sill width 0.6 m total = abutment + foundation Sill clear distance 0.3 m total = 0.036 m + 0.01 m = 0.046 m Embedment 200 mm (one facing block height) Angular distortion = total / span length = 0.046 m / 10 m = 0.0046 Reinforcement: Tolerable angular distortion for simple span = 0.005 Minimum stiffness at =1.0 percent, T@=1.0 percent Angular distortion = 0.0046 < 0.005 (OK) = 7.5 kN/m Minimum ultimate strength, Tult = 41.3 kN/m Design summary: Length in load-bearing wall = 2.4 m Vertical spacing in load-bearing wall = 0.2 m The configuration for the trial design is shown in Length in back wall = 1.3 m Figure 3-10. Vertical spacing in back wall = 0.2 m 0.3 m 0.6 m 1.3 m 0.6 m 0.2 m 0.3 m Sill 1 Front batter 35 0.2 m (Load-bearing wall) 2.4 m Reinforcement: T@ = 1% = 7.5 kN/m T ult = 41.3 kN/m 0.2 m 2.40 m (Embedment) Figure 3-10. Design Example 2--configuration of the trial design.