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

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CHAPTER 3 LRFD SEISMIC DESIGN OF GEOSYNTHETIC-REINFORCED SOIL (GRS) BRIDGE ABUTMENTS INTRODUCTION Load and Resistance Factor Design (LRFD) is a method which takes variability in the behavior of structural elements and loads into account in an explicit manner. While relying on an extensive use of statistical methods, LRFD sets forth results in a usable manner by comparing factored loads to design strengths. The design method presented in the following sections for GRS bridge abutments has been developed based on the AASHTO LRFD Bridge Design Specifications (2007), NCHRP Report 556 and Technical Bulletin MSE-9 produced by The Reinforced Earth Company. Accelerations to be Considered in Design For both external and internal stability, the dynamic forces related to the reinforced soil mass, sill, and bridge superstructure must be accounted for separately. The dynamic loads from the bridge deck are calculated by the bridge designer, along with the static bridge loadings. The loads are expressed in terms of the maximum free field acceleration, A, expected at the site for the earthquake and class of risk under consideration. The GRS abutment wall (backfill soil, geosynthetic reinforcement, and facing units) forms a single monolithic structure. All of these components along with the sill and the bridge shall be assigned the same class of risk, and the same acceleration. The average maximum acceleration, Am, assigned to the reinforced soil mass supporting the sill is a function of the free-field acceleration: AAAm )45.1( −= (Eqn. 11.10.7.1-1) 1 1 AASHTO LRFD Bridge Design Specifications (2007)

67 The external stability of the sill is checked twice: (1) assuming that the sill is a separate entity, and (2) the sill is included in the overall stability of the GRS abutment. With respect to its own stability, the sill should be treated as a gravity wall, being assigned seismic coefficients kh and kv. However, since the actual accelerations reaching the sill at the top of the GRS abutment are unknown, its stability shall be confirmed using the free field acceleration, A. With respect to overall stability of the GRS abutment, the sill is considered an integral part of the reinforced soil mass and will be analyzed using the same assumptions as the reinforced soil mass. Bridge Superstructure Loads The dynamic bridge loads from the superstructure must be divided into vertical and horizontal loads, due to dead loads and traffic loads. Although past editions of AASHTO Standard Specifications omit live loads in the analysis of seismic stability, it is possible that there will be live load on the bridge during an earthquake. Though it is unlikely that the maximum live load condition (fully loaded trucks) will coincide with the earthquake, it is acceptable to assume that 50% of the maximum live load is applied during an earthquake. LRFD SEISMIC DESIGN EXAMPLE OF A GRS BRIDGE ABUTMENT The following section describes a step-by-step LRFD design method via an example GRS bridge abutment that has the same configuration as the abutment tested on the shake table. Figure 3.1 shows the configuration of the GRS bridge abutment used.

68 Figure 3.1: LRFD Example Problem Configuration External Stability Verifying external stability is done in two steps. In the first step, the stability of the sill is examined with respect to sliding, overturning, and bearing capacity. In the second step, the stability of the reinforced soil mass is verified with respect to sliding, overturning, and bearing capacity. The two calculation procedures are presented separately. External Stability of the Sill Loads Transmitted From the Bridge Deck. For sill stability calculations with respect to sliding and overturning, only the dead load, Qd, of the bridge and the horizontal inertia of the dead load, Fd, shall be considered. The inclusion of the bridge live load, Ql, and its inertial

69 component, Fl, would have tendency to increase the factor of safety for sliding and have little or no effect on overturning and are therefore omitted. The force Fd is calculated as follows and is applied at the location of bearing as shown in Figure 3.2. AQF dd = For bearing pressure calculation and surcharge effect for internal stability calculations, the dead load, Qd, plus 50% of the live load, 0.5Ql, are applied vertically. Simultaneously, the inertia of the dead load and live load, Fd+l, is applied horizontally: AQQF ldld )5.0( +=+ Figure 3.2: Static and Dynamic Forces Acting on Sill Sill Inertia Force. The weight of the sill, Ws, (including its backwall) generates the inertia force, Pis, given by: AWP sis = Forces from the Backfill. For the external stability of the sill, the static and dynamic forces exerted on the backwall of the sill by the backfill overlying the reinforced soil mass shall be considered. The dynamic force is calculated using the free-field acceleration, A. The static earth

70 pressure, FT, (Figure 3.2) is calculated using Rankine analysis; and the dynamic (pseudo-static) force, ΔPAE, is calculated using the Mononobe-Okabe formula: EQaaeAE KKHP γγ )(2 1 2 1 −=∆ (Eqn. 11.10.7.1-2) In which ( ) ( ) ( ) ( ) 2 2 2 coscos sinsin 1 )cos(coscos )(cos −       −++ −−+ + ++ −− = βθβδ θφδφ θβδβθ βθφ i i K ae (Eqn.A11.1.1.1-2) φ φ sin1 sin1 + − =aK vv h k A k k − = − = −− 1 tan 1 tan 11θ =φ Friction angle of soil =β Slope of the GRS wall facing to the vertical (negative for inclination towards the reinforced soil) =i Backfill slope angle (typically 0º for GRS bridge abutments) =δ Angle of friction between soil and abutment =EQγ Load factor for earthquake loads from Table 3.4.1-1 The dynamic (pseudo-static) force, ΔPAE, is applied at 0.6H2 above the base of the sill as shown in Figure 3.2. Note that the traffic surcharge must also be incorporated into the total dynamic earth pressure (as illustrated in Section 2.6 based on Equation 18, Chapter 1). Traffic surcharge was omitted in the current analysis for simplicity. For the example problem shown in Figure 3.1, assume Ql = 0.0 kN/m and Qd = 82.92 kN/m as given by the bridge engineer. Assume A=0.2 for the example GRS abutment. Acceleration coefficients are given in Figures 3.10.2-1 thru 3.10.2-3 AASHTO LRFD Bridge Design Specifications (2007).

71 Apply a seismic horizontal load Fd (Figure 3.2): 17.33)20.0(84.165 === AQF dd 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 (rollers) that do not resist horizontal motion on the other end. Due to this configuration, the inertial force, Fd, assumes that the full bridge inertial force is applied to the GRS abutment and therefore 84.16592.822 =× kN/m is substituted here for Qd in the calculation of Fd. Use Eqn. 11.10.7.1-2 to calculate ΔPAE EQaaeAE KKHP γγ )(2 1 2 21 −=∆ Use Eqn. A11.1.1.1–2 to calculate Kae ( ) ( ) ( ) ( ) ( ) )3.1103.29cos()0(cos3.11cos 00cos3.1103.29cos 03.1144sin3.2944sin 1)03.1144(cos 2 2 2 °+°+°°°       °−°°+°+° °−°−°°+° +°−°−° = − aeK = 0.286 Where: °= − = − = −− 3.11 01 20.0 tan 1 tan 11 vk Aθ Vertical acceleration coefficient, kv = 0: Angle of friction between soil and concrete: °=°== 3.2944 3 2 3 2φδ Friction angle of soil, °= 44φ Slope of wall to the vertical, °= 0β Backfill slope angle, °= 0i 180.0 44sin1 44sin1 = °+ °− =aK )180.0286.0( 2 1 2 21 −=∆→ EQAE HP γγ EQAE HP γγ 2 21053.0=∆ For extreme event I 1=→ EQγ (Table 3.4.1-1)

72 18.0)1()4.0)(52.21(053.0 2 ==∆ AEP kN/m Use 09.0)18.0(5.05.0 ==∆ AEP kN/m (Article 11.10.7.1) aT KHF 2 212 1 γ= (Eqn. 3.11.5.8.1-1) 31.0)180.0()4.0)(52.21( 2 1 2 ==TF kN/m 90.0)20.0(48.4 === AWP sis kN/m Sill Sliding. (Ignore Ql) (Article 10.6.3.4) =RR Factored resistance against failure by sliding epepnR RRRR φφφ ττ +== (Eqn. 10.6.3.4–1) Rn = Nominal sliding resistance against failure by sliding Ignore passive resistance: epep Rφ 80.0=τφ (Cast–in–place concrete on sand) (Table 10.5.5.2.2–1) δτ tanVR = (Eqn. 10.6.3.4-2) (Article 11.10.5.3) → Use fφδ tantan = for concrete cast against soil fφ is the internal friction angle of drained soil V is the total vertical force (kN/m) ds QWV += (Ignore Ql for sliding and overturning) ( ) fR φδτ tan92.8248.4tan)92.8248.4( +=+= 40.8444tan)92.8248.4( =°+=τR kN/m 52.67)40.84)(80.0( === ττφ RRR kN/m 52.67=RR kN/m (factored resistance against failure by sliding) Factored driving forces (horizontal) AETisd PFPF Δ5.0+++= Factored driving forces (horizontal) 47.3409.031.090.017.33 =+++= kN/m →Factored driving forces 47.34= kN/m < Factored resistance 52.67= kN/m Okay (No Sliding)

73 Sill Overturning. (Ignore Ql) Moments are taken about point A in Figure 3.2: Factored driving moments )24.0(09.0)133(.31.0)142.0(90.0)2.0(17.33 +++= = 6.82 kN-m/m Resisting moment 74.24)433.0(48.4)275.0(92.82 =+= kN-m/m → Factored driving moments 82.6= kN-m/m < Resisting moment 74.24= kN-m/m Okay (No Overturning) Bearing Capacity of Sill. (Consider 0.5 Ql) Determine the allowable dynamic bearing pressure of the reinforced fill from Table 3-1 NCHRP Report 556 (See Appendix A) For an isolated sill with 75.0=B m and °= 44φ → qallow-static = 499 kPa From Das’ book “Principles of Soil Dynamics”, “…the minimum value of the ultimate dynamic bearing capacity of shallow foundations on dense sands obtained between static to impact loading range can be estimated by using a friction angle dyφ , such that °−= 2φφdy ” (Vesic, 1973). 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): Use °=°−°= 42244dyφ and 75.0=B m and a 0.75 reduction factor for isolated sill 433=→ −dynamicallowq kPa (Used linear interpolation in Table 3-1) Factored resistance nbR qq φ== (Eqn. 10.6.3.1.1–1) Assume: dynamicallown qq −= 55.0=bφ (Table 10.5.5.2.2–1) (Plate Load - The findings reported in NCHRP Report 556 are based on experimental procedures resembling the plate load test) 238)433(55.0 ==Rq kPa

74 For the eccentricity and bearing stability calculations at the base of the sill, 50% of the bridge live load, Ql is included while the inertia of the dead load and reduced live load, Fd+l, is applied horizontally. From Figure 3.2: ( )∑ =+++= 74.24)433.0(48.4)275.0(0)275.0(092.82ARM kN-m/m ( ) ( ) ( ) ( )∑ =+++= 24.009.0133.031.0142.09.02.017.33AOM 6.82 kN-m/m ∑ =++= 4.87048.492.82V kN/m 17.0 4.87 82.674.24 2 75.0 2 = − −= − −= ∑ ∑ ∑ V MMB e AA OR m ( ) 41.017.0275.02' =−=−= eBB m Applied stress 213 41.0 4.87 ' === ∑ B V kPa Applied stress = 213 kPa < Factored resistance = 238 kPa. Okay (No Bearing Capacity Failure) External Stability of Reinforced Mass Forces Transmitted From the Bridge Deck. Only dead load, Qd, and the inertia of the dead load, Fd, are considered in the external stability calculation. If included in the calculation, live loads would have a tendency to increase the safety factor with respect to sliding of the reinforced soil mass and would have little or no effect on overturning. (The Reinforced Earth Company Technical Bulletin MSE-9) Sill Inertia Force. For overall stability of the GRS bridge abutment, the sill, including its backwall, is considered an integral part of the GRS abutment. Therefore, as for the reinforced soil mass, the inertia of the sill is calculated using the acceleration Am as follows: hsmsis kWAWP == Inertia Forces of the Reinforced Soil Mass. Let Weff denote the effective weight of the reinforced soil mass and W2eff the effective weight of the overlying fill, then assume an inertia force at the center of gravity of each weight equal to:

75 heffEQmeffEQir kWAWP γγ == (Reinforced Soil Mass) heffEQmeffEQI kWAWP 222 γγ == (Overlying Fill) See Figure 3.3 for the area included in the calculation of effective weights. Figure 3.3: Effective Weight of Soil Mass Forces Transmitted From the Retained Soil. The static earth pressure, P, exerted by the retained soil is applied at H/3 above the base as shown in Figure 3.4. One-half of the horizontal dynamic force, ΔPAE, exerted by the retained soil is applied at 0.6H above the base as shown in Figure 3.4. The dynamic force, ΔPAE, is calculated using the acceleration Am. Calculations. (For sliding and overturning ignore Ql) From Figure 3.4: ( )( )( ) 59.20652.212.30.3 ==W kN/m

76 The calculated weight of the reinforced fill, W, includes the weight of the facing blocks which are assumed to have the same unit weight as the reinforced fill. ( ) mEQmeffEQir AHHAWP 115.0 γγγ == ( )( )( )( )( ) 99.3025.052.212.36.35.01 ==irP kN/m ( )( ) 06.1552.214.075.12 ==W kN/m ( )( ) 62.125.046.6122 === meffEQi AWP γ kN/m Also from Figure 3.4: aKHP 2 25.0 γ= (Eqn. 3.11.5.8.1–1) 180.0 44sin1 44sin1 = °+ °− =aK 10.25)180.0()6.3)(52.21)(5.0( 2 ==P kN/m Factored 19.27)1)(180.0375.0()6.3)(52.21(5.0)( 2 1 22 2 =−=−=∆ EQaaeAE KKHP γγ kN/m For Kae use °==°=°=°=°= 44,0,0,14,44 22 φδβθφ i (soil-to-soil) 375.0 )0cos()1444cos( )1444sin()4444sin( 1 )14044cos()0cos(14cos )01444(cos 2 2 =      °°+° °−°°+° + °+°+°°° °−°−° = − aeK Use 60.13)19.27(5.05.0 ==∆ aeP kN 12.1)25.0(48.4 === msis AWP kN

77 Figure 3.4: Static and Dynamic Forces Acting on Soil Mass Sliding of Reinforced Mass. (Article 10.6.3.4) =RR Factored resistance against failure by sliding epepR RRR φφ ττ += (Ignore passive resistance: epep Rφ ) (Eqn. 10.6.3.4–1) 9.0=τφ (for soil-on-soil) (Table 10.5.5.2.2–1) δτ tanVR = (Eqn. 10.6.3.4–2) (Article 11.10.5.3) → Use 966.044tantantan 3 =°== φδ V is the vertical force (kN/m) 05.30906.1548.492.8259.206 =+++=V kN/m 54.298)966.0(05.309 ==τR kN/m 69.268)54.298)(9.0( === ττφ RRR kN/m 69.268=RR kN/m (factored resistance) Factored driving forces 60.10560.1310.2562.112.117.3399.30 =+++++= kN/m Factored driving forces 60.105= kN/m < Factored resistance 69.268= kN/m Okay (No Sliding)

78 Overturning of Reinforced Mass. Moments are taken about point C in Figure 3.4: Factored driving moments ( ) ( ) ( ) ( )4.362.1342.312.14.317.336.199.30 +++= ( ) ( ) 11.23116.260.132.110.25 =++ kN-m/m Resisting moment 33.410)125.2(06.15)933.0(48.4)775.0(92.82)5.1(59.206 =+++= kN-m/m → Factored driving moments 11.231= kN-m/m < Resisting moment 33.410= kN-m/m Okay (No Overturning) Bearing Capacity of Reinforced Mass. The eccentricity and bearing requirements under the reinforced soil mass are calculated using static conditions only as shown in Figure 3.5. 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. Figure 3.5: Static Forces Acting on Soil Mass

79 Factored resistance nbR qq φ= (Eqn. 10.6.3.1.1 – 1) 5.0=bφ (Table 10.5.5.2.2 – 1) γγγγ wmwqqmfcmn cNBcNDcNq '5.0 3++= (Eqn. 10.6.3.1.2a – 1) In this example c = 0 and Df = 0. γγγγ iSNN m = 6.224443 =→°= γφ N (Table 10.6.3.1.2a – 1) 1== γwwq CC Assuming deep GWT (Table 10.6.3.1.2a – 2) 6.0 3 3 4.014.01 =−=−= L B Sγ (Table 10.6.3.1.2a – 3) 1=qd for 0=fD (Table 10.6.3.1.2a – 4) )1( cot 1 +         + −= n fcBLV H i φγ (Eqn. 10.6.3.1.2a – 8) From AASHTO Figure C10.6.3.1.2a – 1: use °= 90θ 290sin2 2 =°=→ n (Eqn. 10.6.3.1.2a – 9) H = Unfactored horizontal load (static) = 10.25 kN/m V = Unfactored vertical load (static) lds QQWWW ++++ 2 05.309092.8248.406.1559.206 =++++= kN/m 776.0 05.309 10.25 1 )12( =    −= + γi 57.104)776.0)(6.0)(6.224( ==mNγ Consider static eccentricity only: [ ] 05.309 )2.1)(10.25(33.410 2 8.2 2 − −= − −= ∑V MML e CC oR 17.0=e m < 47.0 6 8.2 6 == L m

80 46.2)17.0(28.22' =−=−= eLB m γγγ wmn CNBq '5.0 3= 93.2767)1)(57.104)(46.2)(52.21(5.0 ==nq kPa Factored resistance: 97.1383)93.2767(5.0 ==Rq kPa Applied stress 63.125 46.2 05.309 2 == − = eL V kPa Applied stress = 125.63 kPa< Factored resistance = 1383.97 kPa Okay (No Bearing Capacity Failure) Internal Stability Internal stability calculations are done in three steps: (1) calculate the tensile forces in the reinforcement layers due to the application of static loads using the usual static analysis, (2) calculate the internal dynamic load, Pi (function of the reinforced soil mass and the concentrated load transmitted by the sill) and then distribute Pi among the reinforcement layers in proportion to their resistant lengths, and (3) add the tensile loads calculated in steps 1 and 2. The dynamic force, Pi, is proportional to the "active zone" of the reinforced soil mass, through its own weight and the load it carries. This active zone is confined within an idealized bilinear envelope shown in Figure 3.6. To calculate the weight of the actual active zone, the weight of the idealized (bilinear) active zone envelope is multiplied by the coefficient 0.67. The applied loads from the sill are directly added to obtain the total vertical load. The total vertical load is then multiplied by the acceleration Am to obtain the dynamic force, Pi, to be distributed among the reinforcing layers. The weight of the active zone envelope is a function of the geometry of the structure and the sill. Further confirmation of the assumed shape of the active zone is needed, however, similar active zone shapes are assumed in the design of reinforced earth abutments with inextensible reinforcement. The load sustained by the active zone is a combination of the vertical bridge loads, consisting of the dead load, Qd, and 50% of the live load, 0.5Ql, and the weight of the sill, Ws, which includes the backwall. Thus:

81 [ ] msldai AWQQWP +++= 5.067.0 Figure 3.6: Assumed Active Zone for Calculating Dynamic Forces in the Reinforcement Layers Refer to Figures 3.6 and 3.7: Maximum reinforcement load VHmax ST σ= Factored horizontal stress at each reinforcement level is: )( HrVPH K σσγσ ∆+= (Eqn. 11.10.6.2.1 – 1) Pγ is a load factor: Pγ = 1.35 (max) to 1.0 (minimum) (Table 3.4.1-2) Use Pγ = 1.35

82 vv z σγσ ∆+= 11 1D Pv V =∆σ Consider 100% of Ql for reinforcement force calculations: 40.8792.82048.4 =++=++= dlsv QQWP kN/m For 1121 ' zBDzz +=→≤ (See Figure 3.7) For d zB Dzz + + =→> 2 ' 1 121 57.0)09.0(275.02' =−=−= eBB m From Figure 3.7: 58.0 48.492.82 )733.0)(48.4()575.0)(92.82( = + + =d m Figure 3.7: Calculating Vertical Stresses in the Reinforced Soil Zone

83 1 40.87 Dv =∆σ Take 1= a r K K (Figure 11.10.6.2.1-3) 180.0 44sin1 44sin1 = °+ °− ==→ ar KK →= vapH K σγσ )( 11 vaPH zK σγγσ ∆+= ( 0=∆ Hσ kPa in this example)       += 1 1 40.87 52.21)180.0)(35.1( D zHσ 1 1 24.21 23.5 D zH +=σ 1121 ' zBDzz +=→≤ d zB Dzz + + =→> 2 ' 1 121 ,max VH ST σ= 20.0=vS m Table 3.1 shows sample calculations of Tmax, the static factored force applied to the geosynthetic fabric, for selected layers Table 3.1: Tmax Calculated for Select Reinforcement Layers Layer z1 (m) D1 (m) σH (kPa) Tmax (kN/m) 1 3.20 2.47 25.34 5.07 8 1.80 1.77 21.41 4.28 12 1.00 1.37 20.73 1.28 16 0.20 0.77 28.63 5.73 Check Static Pullout vσ at any depth is vv z σγσ ∆+= 11 1 11 40.87 D zv += γσ The geosynthetic layer effective length (see Figure 3.6) must satisfy the following equation:

84 cv e CRF T L ασφ * max≥ (Eqn. 11.10.6.3.2 – 1] For static load use 9.0=φ Also use: 647.044tan67.0tan67.0 1 * =°== φF (Fig 11.10.6.3.2 – 1) 6.0=α for geotextile (Table 11.10.6.3.2 – 1) C = 2 1=cR for geotextile (Article 11.10.6.4 – 1) For layer 16, 20.01 =z m, 77.01 =D m, 81.11777.0 40.87 )20.0(52.21 =+=vσ kPa 72.1=eL m Check Le using Eqn. 11.10.6.3.2-1 as follows: m07.0 )1)(2)(81.117)(6.0)(647.0)(9.0( 73.5 m72.1 =>=eL (Okay) Layer 12, 00.11 =z m, 37.11 =D m, 32.8537.1 40.87 )00.1(52.21 =+=vσ kPa m02.0 )1)(2)(32.85)(6.0)(647.0)(9.0( 28.1 m72.1 =>=eL (Okay) Layer 8, 80.11 =z m, 77.11 =D m, 11.8877.1 40.87 )80.1(52.21 =+=vσ kPa m07.0 )1)(2)(11.88)(6.0)(647.0)(9.0( 28.4 m96.1 =>=eL (Okay) Layer 1, 2.31 =z m, 77.11 =D m, 24.11877.1 40.87 )2.3(52.21 =+=vσ kPa m06.0 )1)(2)(24.118)(6.0)(647.0)(9.0( 07.5 m80.2 =>=eL (Okay) Check Reinforcement Strength (Static) Tal = Nominal long-term reinforcement design strength DCRID ultult al RFRFRF T RF T T ×× == (Eqn. 11.10.6.4.3b – 1) calmax RTT φ≤ (Eqn. 11.10.6.4.1 – 1)

85 For this example use a reinforcement with ultT = 70 kN/m (GEOTEX 4x4 fabric) DCRID RFRFRFRF ××= Use 1.1=== DCRID RFRFRF 331.1=→ RF 59.52 331.1 70 === RF T T ultal kN/m Use 9.0=φ and 1=cR ( )( )( ) 33.47159.529.0 ==cal RTφ kN/m Layer 16, Tmax = 5.73 kN/m < 47.33 kN/m (Okay) Layer 12, Tmax = 1.28 kN/m < 47.33 kN/m (Okay) Layer 8, Tmax = 4.28 kN/m < 47.33 kN/m (Okay) Layer 1, Tmax = 5.07 kN/m < 47.33 kN/m (Okay) Dynamic Reinforcement Forces Wa = Weight of the assumed active zone (see Figure 3.6) ( )( ) ( ) ( )( )[ ] 11 5.03.05.03.0 γ×××−×= HHHHWa ( )( )( ) ( )( )( )( )( )[ ] 46.5352.216.35.06.33.05.06.33.02.3 =×−=aW kN/m Internal Dynamic Force Pi: [ ] msldai AWQQWP +++= 5.067.0 Factored [ ] msldaEQi AWQQWP +++= 5.067.0γ Take 1=EQγ , ( ) ( )[ ]( ) 80.3025.048.405.092.8246.5367.0 =+++=iP kN mdmaxtotal TTT += , where Tmax is static and Tmd is dynamic ∑ = = ej m j eji md L LP T 1 γ (Eqn. 11.10.7.2 – 1) Use 1=γ for extreme event (Table3.4.1 – 1)

86 ∑ = = ej m j ej md L L T 1 )80.30)(1( (See Figure 3.6) Layer 16 61.1 92.32 72.1 )80.30( ==→ mdT kN/m Layer 12 61.1 92.32 72.1 )80.30( ==→ mdT kN/m Layer 8 83.1 92.32 96.1 )80.30( ==→ mdT kN/m Layer 1 62.2 92.32 80.2 )80.30( ==→ mdT kN/m Required ultimate resistance to static load: c rs R RFT S φ max≥ (Eqn. 11.10.7.2 – 3) Required ultimate resistance to dynamic load: c DIDmd rt R RFRFT S φ ≥ (Eqn. 11.10.7.2 – 3) Required ultimate resistance: rtrsult SST += (Eqn. 11.10.7.2 – 5) Use 2.1=φ (combined static/dynamic) 1=cR 331.1=FR 1.1=IDRF 1.1=DRF For layer 16 36.6 )1)(2.1( )331.1)(73.5( =≥rsS kN/m 62.1 )1)(2.1( )1.1)(1.1)(61.1( =≥rtS kN/m Required ultimate resistance 98.762.136.6 =+=ultT kN/m 98.7=ultT kN/m < ultimate strength of selected geotextile = 70 kN/m (Okay) Similarly, Layer 12, 42.1 )1)(2.1( )331.1)(28.1( =≥rsS kN/m

87 62.1 )1)(2.1( )1.1)(1.1)(61.1( =≥rtS kN/m 70kN/m04.362.142.1 <=+=ultT kN/m (Okay) Layer 8, 75.4 )1)(2.1( )331.1)(28.4( =≥rsS kN/m 85.1 )1)(2.1( )1.1)(1.1)(83.1( =≥rtS kN/m 70kN/m6.685.175.4 <=+=ultT kN/m (Okay) Layer 1, 62.5 )1)(2.1( )331.1)(07.5( =≥rsS kN/m 64.2 )1)(2.1( )1.1)(1.1)(62.2( =≥rtS kN/m 70kN/m26.864.262.5 <=+=ultT kN/m (Okay) Finally, check the effective length, Le, for pullout )8.0( cv Total e CRF T L ασφ ∗ ≥ Use 2.1=φ (combined static/dynamic pullout resistance) (Table 11.5.6 – 1) 452.0=∗F , 6.0=α , 2=C , 1=cR Layer 16 m08.0 )1)(2)(81.117)(6.0)(647.0)(8.0)(2.1( 61.173.5 m72.1 = + >=→ eL (Okay) Layer 12 m05.0 )1)(2)(32.85)(6.0)(647.0)(8.0)(2.1( 61.128.1 m72.1 = + >=→ eL (Okay) Layer 8 m09.0 )1)(2)(11.88)(6.0)(647.0)(8.0)(2.1( 83.128.4 m96.1 = + >=→ eL (Okay) Layer 1 m09.0 )1)(2)(24.118)(6.0)(647.0)(8.0)(2.1( 62.207.5 m8.2 = + >=→ eL (Okay)