Seismic Design of Geosynthetic-Reinforced Soil Bridge Abutments with Modular Block Facing (2012)

Below is the uncorrected machine-read text of this chapter, intended to provide our own search engines and external engines with highly rich, chapter-representative searchable text of each book. Because it is UNCORRECTED material, please consider the following text as a useful but insufficient proxy for the authoritative book pages.

CHAPTER 2 ASD SEISMIC DESIGN OF GEOSYNTHETIC-REINFORCED SOIL (GRS) BRIDGE ABUTMENTS INTRODUCTION Allowable Stress Design (ASD) is a method which ensures stresses developed in a structure due to service loads do not exceed the elastic limit. Factors of safety are then used to ensure the stresses remain within allowable limits. The following sections describe a step-by-step ASD design method via an example GRS bridge abutment that has the same configuration as the abutment tested on the shake table. The design method presented in the following sections for GRS bridge abutments has been developed based on NCHRP Report 556, Technical Bulletin MSE-9 produced by The Reinforced Earth Company and AASHTO LRFD Bridge Design Specifications (2007). ESTABLISH DESIGN PARAMETERS The configuration of the GRS abutment tested is shown in Figure 2.1. The width of the GRS abutment is 3.25 m, and its length (normal to the figure) is 3.25 m. Seismic Considerations The pseudo-static forces presented in this example are functions of Am, the average maximum horizontal acceleration occurring in the reinforced soil structure and the soil behind the retained soil. The acceleration Am is a function of the free-field maximum horizontal acceleration, A. The value of the horizontal seismic coefficient, kh, presented in this example is equal to the average maximum acceleration, Am. The vertical seismic coefficient, kv, is assumed to be zero in this example for simplicity. Bathurst and Cai (1995) have indicated that over a wide range of horizontal seismic coefficient values the assumption that kv = 0 is reasonably accurate and, in

43 fact, results in a slightly more conservative value of PAE than values calculated assuming that the vertical component of seismic earth force acts upward (kv < 0). Free field acceleration, 20.0=A Average maximum acceleration, ( ) 25.045.1 =−= AAAm Note: ( )AAAm −= 45.1 for 0.05 < A < 0.45 otherwise use Am = A (Ref. MSE-9) Horizontal seismic coefficient, 250.Ak mh == Vertical seismic coefficient, 0=vk Figure 2.1: Abutment Configuration Wall Heights and External Loads (See Figure 2.1)

44 Total abutment height, H = 3.6 m Load bearing wall height, H1 = 3.2 m Back wall height, H2 = 0.4 m The bridge vertical dead load, Qd, is taken as one-half of the weight of the simply supported bridge. The live load, Ql, and the traffic surcharge load, q, are taken as zero since there will be no live load and traffic load applied to the bridge during the shake table test. Bridge vertical dead load, Qd = 82.92 kN/m Bridge vertical live load, Ql = 0.0 kN/m Traffic surcharge load, q = 0.0 kPa Trial Design Parameters (See Figure 2.1) Sill width, B = 0.75 m Clear distance, d = 0.3 m Sill type: isolated sill Facing: modular concrete blocks Facing block size: 200 mm x 200 mm x 400 mm Wall thickness, D = 0.2 m Batter of Facing, ψ = 0° Reinforcement Length, L = 2.8 m Reinforcement spacing, s = 0.2 m ESTABLISH SOIL PROPERTIES Reinforced Fill Based on NCHRP Report 556 requirements, the fill must satisfy the following criteria: 100 percent passing 100 mm sieve 0-60 percent passing No. 40 (0.425 mm) sieve 0-15 percent passing No. 200 (0.075 mm) sieve PI ≤ 6

45 0 10 20 30 40 50 60 70 80 90 100 0.01 0.1 1 10 100 Grain Size, D (mm) Pe rc en t Fi ne r Sample 1 Sample 2 Sieve No. 200 Sieve No. 4 Figure 2.2: Grain Size Distribution of Backfill Soil The fill, CA-6, satisfies the grain size distribution as shown in Figure 2.2 The friction angle of the fill, °= 44testφ , is determined by the standard direct shear test on the portion finer than 2 mm (No. 10) sieve, using a sample compacted to 95 percent of AASHTO T-99, Method C or D, at optimum moisture content. Friction angle of the reinforced fill: °= 44rfφ Unit weight of the reinforced fill: 52.21=rfγ kN/m 3 Active earth pressure coefficient of reinforced fill: 1800245tan 2 .)/(K rfa(rf) =−°= φ Retained Earth Friction angle of the retained earth: °= 44reφ Unit weight of the retained earth: 52.21=reγ kN/m 3 Active earth pressure coefficient of retained earth: ( ) 18002/45tan 2 .K rea(re) =−°= φ Angle of inclination above horizontal of retained earth: β = 0°

46 Foundation Soil Friction angle of the foundation soil: °= 44fsφ Unit weight of the foundation soil: 52.21=fsγ kN/m 3 Allowable bearing capacity of the foundation soil: qaf = 300 kPa ESTABLISH DESIGN REQUIREMENTS External Stability Design Requirements Factor of safety against sliding: FSsliding ≥ 1.1 Factor of safety against overturning: FSoverturning ≥ 1.5 Eccentricity of GRS abutment: e ≤ L/6 Average sill pressure, psill ≤ allowable bearing pressure of the reinforced fill, qallow, as determined in Section 2.5 Average contact pressure at the foundation level, pcontact ≤ allowable bearing pressure of the foundation soil, qaf Internal Stability Requirements Factor of safety against geosynthetic pullout: FSpullout ≥ 1.1 Factor of safety against geosynthetic breakage: FSbreakage ≥ 1.1 DETERMINE ALLOWABLE BEARING PRESSURE OF REINFORCED FILL Determine the allowable bearing pressure of the reinforced fill below the sill, qallow, with the following conditions: -Design friction angle of the reinforced fill, °= 44rfφ -Reinforcement spacing, 2.0=s m (uniform spacing with no truncation) -Isolated sill -Sill width, 75.0=B m

47 (1) From Table 3-1 NCHRP 556 (See Appendix A), for °= 44rfφ and reinforcement spacing, s = 0.2 m, allowable bearing pressure for sill = 380 kPa. (using linear interpolation in Table 3-1) (2) From Figure 3-1 NCHRP 556 (See Appendix A), the correction factor for a sill width of 0.75 m is 1.75; thus the corrected allowable bearing pressure kPa66575.1kPa380 =×= (3) Reduction factor for an isolated sill, 0.75. Thus, 49966575.0 =×=allowq kPa Note that qallow is the allowable bearing capacity for static loading conditions. The dynamic allowable bearing capacity may be different from the static one. Vesic (1973) suggested a reduction of 2º in the soil’s friction angle in Terzaghi’s bearing capacity equation to account for bearing capacity reduction due to dynamic loads applied to a shallow foundation underlain by unreinforced soil. Due to lack of dynamic tests on GRS bridge abutments, it is assumed that the above experimental observation by Vesic applies to a dynamically loaded shallow foundation (sill) situated on the top surface of a GRS wall (i.e, bridge abutment). Subsequent dynamic testing to verify this assumption is needed. Use °=°−°= 42244dyφ with B = 0.75 m and isolated sill 433=→ allowq kPa using Table 3-1 and Figure 3-1 from Appendix A EVALUATE STABILITY OF SILL For the stability of the sill alone, the sill should be treated as a gravity wall, being assigned seismic coefficients, kh and kv. However, the actual accelerations applied to the sill at the top of the reinforced soil structure are unknown, therefore its stability will be evaluated using the “free field” acceleration, A. The preliminary sill configuration and forces acting on the sill are shown in Figure 2.3. The dimensions of the sill are: B = 0.75 m

48 H2 = 0.4 m t = 0.2 m b = 0.2 m Center of gravity of sill (with reference to pt. A): y = 0.142 m x = 0.433 m Figure 2.3: Static and Dynamic Forces Acting on Sill With a unit weight of concrete, 5623.γconcrete = kN/m 3, the following forces acting on the sill can be determined: Weight of sill, Ws ( )( ) ( ) concretev2s γktbBHbW ×±××−+×= 1 ( )( ) ( ) 48.456.23012.02.075.04.02.0 =×±××−+×=sW kN/m Inertial force of sill, 1isP AWP sis ×=1 90.020048.41 =×= .Pis kN/m

49 Inertial force of dead load, Fd AQF dd ×= ( ) 17.3320084.165 =×= .Fd kN/m Qd = 82.92 kN/m is the dead load reaction supported by the abutment, and is equal to one-half of the bridge weight. The bridge constructed for the shake table test has elastomeric bearing pads on the abutment side and slide bearings on the opposite end. The slide bearings do not resist horizontal motion, therefore the inertial force, Fd, assumes that the full bridge inertial force is applied to the GRS abutment, as a result: 84.16592.822 =× kN/m is substituted here for Qd in the calculation of Fd. Inertial force of live load, Fl AQF ll ×= 0.020.00.0 =×=lF kN/m Static traffic surcharge pressure, P2q 22 HqKP aq ××= 0.06.00.0180.02 =××=qP kN/m Static soil pressure, P2 a(rf)rf KHγ.P ×××= 2 22 50 31018004052.2150 22 ....P =×××= kN/m Pseudo-static pressure, Paes ( ) ( ) ( )( )rfarfaerfvaes KKHγk.P −×××±×= 22150 ( ) 18.0)180.0286.0(4.052.21015.0 2 =−×××±×=aesP kN/m Where:

50 ( ) ( ) ( ) ( ) ( ) ( )θψδψθ βψθψδ βθφδφ θψφ +−××         +×+− −−×+ +−+ = − coscoscos coscos sinsin 1)(cos 2 2 2 rfrf rf rfaeK ( ) ( ) ( ) ( )( ) ( ) ( )°+°−°×°×°       °+°×°+°−° °−°−°×°+° +°−°+° = − 3.1103.29cos0cos3.11cos 00cos3.1103.29cos 03.1144sin3.2944sin 13.11044cos 2 2 2 rfaeK ( ) 286.0=rfaeK and: °=      ± =      ± = −− 311 01 20.0 tan 1 tan 11 . k A θ v also: Angle of friction between soil and concrete: ( ) ( ) °=°×=×= 3.29443/23/2 rfφδ It is noteworthy that Paes is the pressure against the sill. The fill behind the sill is only 0.4 m high in this example. Without traffic load, the lateral pressure should be very small. The traffic surcharge load must also be included in the total dynamic earth pressure. The total dynamic earth pressure acting at 0.6H2 above the base of the sill is: 18.0 31.0 0.0 118.01 2 2 =    +=      + P P P qaes kN/m (Ref. MSE-9) Check Factor of Safety Against Sill Sliding       +++++ ××+ = 2 2 221 sliding 1 )3/2tan()( FS P P PPPPF WQ q aesqisd rfsd φ 1.142.1 18.031.00.090.017.33 )443/2tan()48.492.82( FSsliding ≥=++++ °××+ = →OK

51 The bridge live load, Ql, and its inertial component, Fl, are not included in sliding analysis as their inclusion would tend to increase the factor of safety against sliding. Check Factor of Safety Against Sill Overturning Sum of resisting moments about point A: (See Figure 2.3) xWfQM sdRA ×+×=∑ 74.24433.048.4275.092.82 =×+×=∑ ARM kN-m/m Sum of overturning moments about point A: (See Figure 2.3) yPH P P PHPtFM is q aesdOA ×+××      ++×+×=∑ 12 2 2 22 )6.0(1)3/( 85.6142.090.0)4.06.0(18.0)3/4.0(31.02.017.33 =×+××+×+×=∑ AOM kN-m/m →≥=== ∑ ∑ 5.161.3 85.6 74.24 FS goverturnin A A O R M M OK The bridge live load, Ql, and its inertial component, Fl, are usually not included in overturning analysis as their inclusion would have little or no effect on the factor of safety against overturning. (In the current analysis Ql = Fl = 0 kN/m) Check Eccentricity and Bearing at Base of Sill For the eccentricity and bearing stability calculations at the base of the sill, 50% of the bridge live load is included. Although AASHTO LRFD Bridge Design Specifications (2007) allows omission of live loads for seismic stability analysis, it is likely that traffic loads may exist during a seismic event. Therefore, 50% of the maximum live load applied for seismic analysis should conservatively represent the conditions associated with rush hour automobile traffic. Nonetheless, Ql in the current analysis is 0 kN/m. As indicated earlier, the live load, Ql, and the traffic surcharge load, q, have been taken as zero since there were no live load or traffic load applied to the bridge during the shake table test.

52 Eccentricity at base of sill, e' sld OR WQQ MMB e AA ++ − −=′ ∑ ∑ 5.02 Sum of resisting moments about point A: (See Figure 2.3) ( ) xWfQQM sldRA ×+×+=∑ 5.0 ( ) 74.24433.048.4275.00.05.092.82 =×+××+=∑ ARM kN-m/m Sum of overturning moments about point A: (See Figure 2.3) ( ) yPH P P PHPtFFM is q aesldOA ×+××      ++×+×+=∑ 12 2 2 22 )6.0(1)3/(5.0 ( ) 142.090.0)4.06.0(18.0)3/4.0(31.02.00.05.017.33 ×+××+×+××+=∑ AOM 17.0 48.40.092.82 85.674.24 2 75.0 = ++ − −=′e m Applied pressure from sill, psill eB WQQ p sldsill ′− ++ = 2 5.0 213 17.0275.0 48.40.05.092.82 = ×− +×+ =sillp kPa ≤ qallow = 433 kPa → OK EVALUATE EXTERNAL STABILITY OF GRS ABUTMENT The evaluation of external stability of the GRS abutment considers the sill to be an integral part of the reinforced fill and is analyzed using the same acceleration, Am, that is applied to the reinforced fill volume. The static and dynamic forces used in external stability calculations of the GRS abutment are shown in Figure 2.4.

53 Figure 2.4: Static and Dynamic Forces Acting on Soil Mass From before: Qd = 82.92 kN/m, Fd = 33.17 kN/m, Ws = 4.48 kN/m With reference to Figure 2.4, the inertial force of sill, Pis2, is: msis AWP ×=2 12.125.048.42 =×=isP kN/m Weight of overlying fill, W2 ( ) rfvkHBdLW γ×±××−−= 1)( 22 ( ) 06.1552.21014.0)75.03.08.2(2 =×±××−−=W kN/m The effective zone (Figure 2.5) is assumed to be H1 by H/2 based on the Technical Bulletin MSE-9 produced by The Reinforced Earth Company and AASHTO LRFD Bridge Design

54 Specifications (2007). With reference to Figure 2.5, the inertial force of the overlying fill is calculated using the effective weight of the overlying fill, W2eff. ( ) rfveff kHBdHW γ×±××−−= 1)2/( 22 ( ) ( ) 46.652.21014.075.03.02/6.32 =×±××−−=effW kN/m Inertial force of overlying fill, Pi2 meffi AWP ×= 22 62.125.046.62 =×=iP kN/m Figure 2.5: Effective Weight of Soil Mass Weight of reinforced fill, W ( ) rfvkHDLW γ×±××+= 1)( 1 ( ) 59.20652.21012.3)2.08.2( =×±××+=W kN/m

55 The calculated weight of the reinforced fill, W, includes the weight of the facing blocks which are assumed to here have the same unit weight as the reinforced fill. With reference to Figure 2.5, the effective weight of reinforced soil, Weff, is: ( ) rfveff kHHW γ×±××= 12/ 1 ( ) 96.12352.21012.32/6.3 =×±××=effW kN/m Inertial force of reinforced soil, Pir meffir AWP ×= 99.3025.096.123 =×=irP kN/m Static soil pressure, P 2 )(5.0 HKP reare ′×××= γ 10.256.3180.052.215.0 2 =×××=P kN/m Dynamic horizontal thrust, Pae ( ) ( ) ( )( )reareaerevae KKHγk.P −×××±×= 2150 ( ) 19.27)180.0375.0(6.352.21015.0 2 =−×××±×=aeP kN/m Where: ( ) ( ) ( ) ( ) ( )θδθ βθδ βθφδφ θφ +×         ×+ −−×+ +− = − coscos coscos sinsin 1)(cos 2 2 rere re reaeK ( ) ( ) ( ) ( ) ( ) 375.01444cos14cos 0cos1444cos 01444sin4444sin 1)1444(cos 2 2 = °+°×°       °×°+° °−°−°×°+° +°−° = − reaeK and:

56 °=      ± =      ± = −− 14 01 25.0 tan 1 tan 11 v h k k θ °== 44reφδ (soil-to-soil) Check Factor of Safety Against Abutment Sliding aeiirisd fssd PPPPPF WWWQ ×+++++ ×+++ = 5.0 )tan()( FS 22 2 sliding φ →≥= ×+++++ °×+++ = 1.183.2 19.275.010.2562.199.3012.117.33 )44tan()59.20606.1548.492.82( FSsliding OK The bridge live load, Ql, and its inertial component, Fl, are not included in sliding analysis as their inclusion would tend to increase the factor of safety against sliding. Check Factor of Safety Against Abutment Overturning Sum of resisting moments about point C: (See Figure 2.4) D)/2)((LWB)dDd)/2B((LW )xd(DWD)d(fQM 2 sdRC +×++++−−× +++×+++×=∑ +++×+++×=∑ )433.03.02.0(48.4)2.03.0275.0(92.82CRM )2/)2.08.2((59.206)75.03.02.02/)3.075.08.2((06.15 +×++++−−× = 410.33 kN-m/m Sum of overturning moments about point C: (See Figure 2.4) ( ) ( ) +×+×××+×=∑ )/(HPH.P.H/PM iraeOC 260503 1 ( ) ( ) ( )yHPtHFH/HP isdi +×++×++× 121122 2 ( ) ( ) ( )+×+×××+×=∑ 2/2.399.306.36.019.275.03/6.310.25COM ( ) ( ) ( )142.02.312.12.02.317.332.32/4.062.1 +×++×++× = 231.10 kN-m/m ∑= CC OR MM /FS goverturnin

57 ∑ →≥== OK5.178.110.231/33.410FS goverturnin The bridge live load, Ql, and its inertial component, Fl, are usually not included in overturning analysis as their inclusion would have little or no effect on the factor of safety against overturning. (In the current analysis Ql = Fl = 0 kN/m) Check Eccentricity and Bearing at Base of Abutment The eccentricity and bearing requirements under the reinforced soil mass are calculated using static conditions only. A seismic event is considered temporary and transient, therefore, bearing pressures at the foundation level are assumed not to increase significantly during a seismic event. Sum of resisting moments about point C: (See Figure 2.6) +++×+++×+=∑ )()()( xdDWDdfQQM sldRC )2/)(()2/)((2 DLWBdDdBLW +×++++−−× +++×+++×+=∑ )433.03.02.0(48.4)2.03.0275.0()0.092.82(CRM )2/)2.08.2((59.206)75.03.02.02/)3.075.08.2((06.15 +×++++−−× = 410.33 kN-m/m Sum of overturning moments about point C: (See Figure 2.6) )3/(HPM CO ×=∑ 12.30)3/6.3(10.25 =×=∑ COM kN-m/m Eccentricity at base of abutment, e WWWQQ MML e sld OR CC ++++ − −= ∑ ∑ 22 17.0 59.20606.1548.40.092.82 12.3033.410 2 8.2 = ++++ − −=e m 47.06/8.26/ ==L m

58 e = 0.17 m ≤ L / 6 = 0.47 m →OK The influence length, D1 at foundation level: (See Figure 2.7) 2/)2( 11 HeBdD +′−+= 31.22/2.3)17.0275.0(3.01 =+×−+=D m Effective reinforcement length, L′ (See Figure 2.7) eLL 2−=′ 46.217.028.2 =×−=′L m The contact pressure on the foundation level, pcontact, is calculated by dividing the total vertical load in the reinforced volume by D1 or L′ , whichever is smaller. (Ref. NCHRP Report 556) 1 2 D WWWQQ p sldcontact ++++ = 79.133 31.2 59.20606.1548.40.092.82 = ++++ =contactp kPa pcontact = 133.79 kPa ≤ qaf = 300 kPa → OK Figure 2.6: Static Forces Acting on Soil Mass

59 STATIC INTERNAL STABILITY AT EACH REINFORCEMENT LEVEL The first phase in evaluating internal stability of the GRS abutment is the calculation of tensile forces resulting from static forces alone. The second phase, (Section 2.9 below), consists of calculating the overall dynamic force, Pi, which includes forces from the reinforced mass as well as the forces transmitted from the sill. The dynamic force, Pi, is then distributed among the reinforcement layers proportional to their resistant area. The effects of both static and dynamic loading are then combined to evaluate the overall internal stability of the GRS abutment. See Figures 2.6 and 2.7 for notations of the quantities used in the evaluation of static internal stability. Figure 2.7: Calculating Vertical Stresses in the Reinforced Soil Zone

60 Pullout resistance, Pr RcCLFPr ev ×××××= )( * σα Where: ∗F = Pullout resistance factor rfF φtan67.0 * = 64.044tan67.0* =°=F α = Scale effect correction factor 6.0=α for geotextile reinforcement )( ev L×σ = Normal force at the soil-reinforcement interface at depth z (excluding traffic surcharge) )(Δ)()( ivevsev LLL ×+×=× σσσ Le = Length of embedment in resistant zone behind the failure surface at depth z Le = L – La La = Length of embedment in the active zone at depth z La = )2/45tan()( 1 rfzH φ−°×− Li = Length of embedment within the influence area inside the resistant zone. (See Figure 2.7) C = Reinforcement effective unit perimeter C = 2 for strips, grids and sheets Rc = Coverage ratio Rc = 1.0 for 100% coverage of reinforcement =hσ Horizontal pressure at depth z

61 ( ) ( ) hvvsrfah qK σσσσ ΔΔ +++×= =vsσ Vertical soil pressure at depth z )()( 2 zH rfrfvs ×+×= γγσ =vσΔ Distributed vertical pressure from sill 2/)(Δ DQQW ldsv ++=σ D2 = Effective width of applied load at depth z For zeBDzz +′−=≤ )2(: 22 For 2/)2(: 22 zeBdDzz +′−+=> 6.03.0222 =×=×= dz m =3z Influence depth of horizontal forces transferred from sill )2/45tan()2(3 rfeBdz φ+°×′−+= 67.1)2/4445tan()17.0275.03.0(3 =+°××−+=z m ∆σh = Supplement horizontal pressure at depth z For )/()(2Δ: 23323 zzzPzz h −××=≤ σ For 0Δ:3 => hzz σ =maxT Maximum tensile force in the reinforcement at depth z sT hmax ×=σ kN/m Tmax must be calculated for each reinforcement layer as shown in Table 2.1 s = Vertical spacing of reinforcement s = 0.2 m

62 Table 2.1: Static Internal Stability No. z (m) L (m) s (m) σvs (kN/m2) D2 (m) Δσv (kN/m2) Δσh (kN/m2) σh (kN/m2) Tmax (kN/m) La (m) Le (m) Li (m) (σv*Le) (kN/m) Pr (kN/m) FSpullout 16 0.20 2.80 0.20 12.91 0.77 114.21 0.27 23.18 4.64 1.27 1.53 0.00 19.71 15.23 3.28 15 0.40 2.80 0.20 17.22 0.97 90.54 0.24 19.66 3.93 1.19 1.61 0.00 27.74 21.43 5.45 14 0.60 2.80 0.20 21.52 1.17 75.00 0.21 17.61 3.52 1.10 1.70 0.06 41.13 31.77 9.02 13 0.80 2.80 0.20 25.82 1.27 69.08 0.18 17.28 3.46 1.02 1.78 0.25 63.03 48.69 14.09 12 1.00 2.80 0.20 30.13 1.37 64.02 0.16 17.12 3.42 0.93 1.87 0.43 83.84 64.77 18.92 11 1.20 2.80 0.20 34.43 1.47 59.65 0.13 17.08 3.42 0.85 1.95 0.62 103.94 80.30 23.51 10 1.40 2.80 0.20 38.74 1.57 55.84 0.10 17.14 3.43 0.76 2.04 0.80 123.60 95.49 27.86 9 1.60 2.80 0.20 43.04 1.67 52.48 0.07 17.28 3.46 0.68 2.12 0.99 143.03 110.50 31.98 8 1.80 2.80 0.20 47.34 1.77 49.51 0.04 17.49 3.50 0.59 2.21 1.17 162.40 125.46 35.87 7 2.00 2.80 0.20 51.65 1.87 46.86 0.01 17.75 3.55 0.51 2.29 1.36 181.84 140.48 39.56 6 2.20 2.80 0.20 55.95 1.97 44.47 0.00 18.09 3.62 0.42 2.38 1.54 201.43 155.62 43.00 5 2.40 2.80 0.20 60.26 2.07 42.32 0.00 18.48 3.70 0.34 2.46 1.73 221.28 170.95 46.25 4 2.60 2.80 0.20 64.56 2.17 40.36 0.00 18.91 3.78 0.25 2.55 1.91 241.44 186.53 49.33 3 2.80 2.80 0.20 68.86 2.27 38.58 0.00 19.36 3.87 0.17 2.63 2.10 261.97 202.39 52.27 2 3.00 2.80 0.20 73.17 2.37 36.95 0.00 19.84 3.97 0.08 2.72 2.28 282.92 218.57 55.08 1 3.20 2.80 0.20 77.47 2.47 35.45 0.00 20.35 4.07 0.00 2.80 2.47 304.32 235.10 57.77 DYNAMIC INTERNAL STABILITY AT EACH REINFORCEMENT LEVEL Active zone weight, Wa (See Figure 2.8) Wa = Area of active zone envelope rfγ× ( )( ) ( ) ( )( )[ ] rfa HHHHW γ×××−×= 5.03.05.03.01 (Ref. MSE-9) ( )( ) ( ) ( )( )[ ] 52.216.35.06.33.05.06.33.02.3 ×××××−××=aW = 53.46 kN/m Dynamic force, Pi msldai AWQQWP ×+++= )5.067.0( 80.3025.0)48.40.05.092.8246.5367.0( =×+×++×=iP kN/m The 0.67 multiplier in front of the calculation for the active zone weight, Wa, is a correction factor to adjust the idealized (bilinear) active zone weight to the actual active zone weight (The Reinforced Earth Company, Technical Bulletin MSE-9, 1995).

63 The ultimate tensile strength of the geotextile used in the GRS abutment tested is: Tult = 70 kN/m (GEOTEX 4x4 fabric) The reduction factor for tensile strength of fabric: RF = 1.331 The allowable tensile strength of the geotextile is calculated as: 59.52331.1/70/ === RFTT ultal kN/m The maximum tensile force in the reinforcement at depth z is calculated as: Figure 2.8: Assumed Active Zone for Calculating Dynamic Forces in the Reinforcement Layers

64 ∑ = = ×= ni i e e imd i i L L PT 1 Where: = ie L Length of embedment in resistant zone behind the dynamic failure surface at depth z as shown in Figure 2.8 TTotal = Static + Dynamic tensile forces in the reinforcement at depth z Ttotal = Tmax + Tmd Define the factor of safety against geosynthetic breakage as: FSbreakage = Tal / Ttotal Define the factor of safety against geosynthetic pullout as: FSpullout = Pr / Ttotal FSbreakage and FSpullout are calculated for all geosynthetic layers as shown in Table 2.2 The factors of safety obtained in Table 2.2 are above the 1.1 limit at every reinforcement level, thus no further reinforcement is required.

65 Table 2.2: Overall (Static + Dynamic) Internal Stability No. z (m) L (m) s (m) Lei (m) Tmd (kN/m) Tmax (kN/m) Ttotal (kN/m) FSbreakage Pr (kN/m) FSpullout 16 0.20 2.80 0.20 1.72 1.61 4.64 6.25 8.42 15.23 2.44 15 0.40 2.80 0.20 1.72 1.61 3.93 5.54 9.49 21.43 3.87 14 0.60 2.80 0.20 1.72 1.61 3.52 5.13 10.25 31.77 6.19 13 0.80 2.80 0.20 1.72 1.61 3.46 5.07 10.38 48.69 9.61 12 1.00 2.80 0.20 1.72 1.61 3.42 5.03 10.45 64.77 12.87 11 1.20 2.80 0.20 1.72 1.61 3.42 5.02 10.47 80.30 15.98 10 1.40 2.80 0.20 1.72 1.61 3.43 5.04 10.44 95.49 18.96 9 1.60 2.80 0.20 1.84 1.72 3.46 5.18 10.16 110.50 21.34 8 1.80 2.80 0.20 1.96 1.83 3.50 5.33 9.86 125.46 23.53 7 2.00 2.80 0.20 2.08 1.95 3.55 5.50 9.57 140.48 25.56 6 2.20 2.80 0.20 2.20 2.06 3.62 5.68 9.26 155.62 27.41 5 2.40 2.80 0.20 2.32 2.17 3.70 5.87 8.96 170.95 29.14 4 2.60 2.80 0.20 2.44 2.28 3.78 6.06 8.67 186.53 30.76 3 2.80 2.80 0.20 2.56 2.40 3.87 6.27 8.39 202.39 32.29 2 3.00 2.80 0.20 2.68 2.51 3.97 6.48 8.12 218.57 33.75 1 3.20 2.80 0.20 2.80 2.62 4.07 6.69 7.86 235.10 35.15