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70 APPENDIX F Example Introduction No seismic considerations. Precast panel units: 5-ft wide 5-ft tall 0.5-ft thick. This example problem demonstrates analysis and design of a Type of reinforcement: Grade 65 (Fy = 65 ksi) with zinc MSE wall using LRFD and the corresponding metal loss mod- coating of 86 m for galvanized reinforcements. Nine cases els and resistance factors based on recommendations described are considered in this example with reinforcement types in this report. The example is adapted from Berg et al. (2009), and sizes summarized as follows: Appendix E.3. Various designs are executed employing ribbed- steel-strip- or steel-grid-type reinforcements, that may be plain Fill steel or galvanized, and construction that may incorporate high Case Quality Galvanized Type of Reinforcement quality, good quality, or marginal quality fills. Both the simpli- 1 High Yes 50-mm wide 4-mm fied and the coherent gravity methods will be used to compute thick ribbed strips reinforcement tension. Results from this example illustrate the 2 High Yes W11 W11 (longitudinal effects that these parameters have on the amount of reinforce- transverse, welded ment needed to meet the demand (i.e., the applicable load case). wire fabric) For the purpose of this illustration it is assumed that fill quality 3 Good Yes 50-mm wide 4-mm refers to electrochemical properties, and the mechanical prop- thick ribbed strips erties of the fill (e.g., unit weight, shear strength) are the same 4 Good Yes W11 W11 (longitudinal for all of the fill qualities considered. transverse, welded The MSE wall has a sloping backfill surcharge and includes a wire fabric) segmental precast concrete panel face as shown in Figure F-1. 5 Marginal Yes W20 W11 (longitudinal The analysis is based on principles of MSE design described transverse, welded by Berg et al. (2009). Table F-1 presents a summary of steps wire fabric) involved in the analysis. This appendix describes details 6 High No 50-mm wide 6-mm thick related to evaluation of the internal stability of the wall as this ribbed strips bears on the calculation of tension (yield) resistance and the 7 High No W20 W11 (longitudinal corresponding reinforcement cross-section. Berg et al. (2009), transverse, welded Appendix E-3, also describes details of the external stability wire fabric) analysis, design of facing elements, overall and compound sta- 8 Good No 50-mm wide 8-mm bility analysis at the service limit state, and design of the wall thick ribbed strips drainage system. 9 Good No W20 W11 (longitudinal transverse, welded Step 1. Establish Project Requirements wire fabric) Exposed wall height, He = 28 ft. Step 2. Evaluate Project Parameters Length of wall = 850 ft. Design life = 75 years or 50 years as appropriate in consid- Reinforced backfill, r = 34, r = 125 pcf, coefficient of eration of fill quality and whether or not reinforcements uniformity, Cu = 7.0. are galvanized. Retained backfill, f = 30, f = 125 pcf.

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71 Due to the 2H:1V backslope, the initial length of reinforce- ment is assumed to be 0.8 H or 24 ft. The length of the rein- 2 forcement is assumed to be constant throughout the height to limit differential settlements across the reinforced zone because differential settlements could overstress the reinforcements. Step 4. Estimate Unfactored Loads To compute the numerical values of various forces and Retained backfill moments, the parameters provided in Step 2 are used. Earth (above or behind pressures transferred from the retained fill are not considered H He Reinforced backfill the reinforced ' r, r backfill) for internal stability analysis with the simplified method, but ' f, f are related to reinforcement tension with the Coherent Grav- ity method. Using the values of the various friction angles, the coefficients of lateral earth pressure for the retained fill are L computed as follows: d Leveling Foundation Soil Coefficient of active earth pressure per Eq. 3.11.5.3-1 of Pad ' fd, fd AASHTO (2009) is Figure F-1. Configuration showing various sin 2 ( + f ) parameters for analysis of an MSE wall with Ka = sloping backfill (not-to-scale). sin sin ( - ) 2 Where, per Eq. 3.11.5.3-2 of AASHTO (2009), the various parameters in above equation are as follows: Step 3. Estimate Depth of Embedment and Length of Reinforcement sin ( f + ) sin ( f - ) 2 = 1 + Based on Table C.11.10.2.2.-1 of AASHTO (2009), the sin ( - ) sin ( + ) minimum embedment depth = H/20 for walls with hori- zontal ground in front of wall, in other words, 1.4 ft for = friction angle between fill and wall taken as specified, exposed wall height of 28 ft. For this design, assume embed- = angle (nominal) of fill to horizontal, ment, d = 2.0 ft. Thus, design height of the wall, H = He + = angle of back face of wall to horizontal, and d = 28 ft + 2.0 ft = 30 ft. f = effective angle of internal friction of retained backfill. Table F-1. Summary of steps in analysis of MSE wall with sloping backfill. Step Item 1 Establish project requirements 2 Establish project parameters 3 Estimate wall embedment depth and length of reinforcement 4 Estimate unfactored loads 5 Summarize applicable load and resistance factors 6 Evaluate external stability of MSE wall--not discussed, see Berg et al. (2009) 6.1 Evaluation of sliding resistance 6.2 Evaluation of limiting eccentricity 6.3 Evaluation of bearing resistance 6.4 Settlement analysis 7 Evaluate internal stability of MSE wall 7.1 Estimate critical failure surface, variation of Kr and F* for internal stability 7.2 Establish vertical layout of soil reinforcements 7.3 Calculate horizontal stress and maximum tension at each reinforcement level 7.4 Establish nominal and factored long-term tensile resistance of soil reinforcement 7.5 Establish nominal and factored pullout resistance of soil reinforcement 7.6 Establish number of soil reinforcing strips at each level of reinforcement 8 Design of facing elements--not discussed, see Berg et al. (2009) 9 Check overall and compound stability at the service limit state--not discussed, see Berg et al. (2009) 10 Design wall drainage system--not discussed, see Berg et al. (2009)

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72 Table F-5.1. Summary of applicable load factors. Load Factors Load Combination (after AASHTO, 2009, Tables 3.4.1-1 and 3.4.1-2) EV EH Strength I (maximum) 1.35 1.5 Strength I (minimum) 1.00 0.9 Service I 1.00 1.0 Note: EV and EH = vertical earth load and horizontal earth load, respectively. For this example problem, compute the coefficient of limits states. Based on Table 11.5.5-1 from AASHTO (2009) active earth pressure for the retained fill, Kaf, using = 26.56 a resistance factor, p = 0.9 is applied to the nominal pullout (for the 2:1 backslope), vertical backface, = 90, and = resistance. Table F-5.2 summarizes the applicable resistance as follows: factors for tensile resistance, t, for galvanized, or plain steel, strip and grid reinforcements with different fill conditions as 2 sin ( 30 + 26.56 ) sin ( 30 - 26.56 ) recommended in this report. These resistance factors apply to = 1 + sin ( 90 - 26.56 ) sin ( 90 + 26.56 ) design lives (tdesign) up to 100 years unless otherwise noted. 2 (0.834 )(0.060) Step 6. Evaluate External Stability = 1 + = 1.563 (0.894 )(0.894 ) of MSE Wall Not included because metal loss is not relevant to these sin ( + f ) 2 sin ( 90 +30 ) 2 calculations. See Berg et al. (2009), Appendix E.3. K af = = sin 2 sin ( - ) 1.563( sin 90 )2 [ sin ( 90 - 26.56 )] Step 7. Evaluate Internal Stability Analysis 0.750 of MSE Wall = = 0.537 (1.563)(1.0)(0.894 ) 7.1 Estimate critical failure surface, variation of Kr , and F* for internal stability Step 5. Summarize Applicable Load and For the simplified method the variation of Kr depends on Resistance Factors the stiffness of the reinforcements and is different for strip- Table F-5.1 summarizes the load factors applied to cal- or grid-type reinforcements. For the case of inextensible steel culations of reinforcement load using the Simplified Method. ribbed strips, the profile of the critical failure surface, the vari- For the internal stability analysis using the Simplified Method ation of internal lateral horizontal stress coefficient, Kr, and only the maximum values of the load factors for the Strength I the variation of the pullout resistance factor, F, are as shown load case apply. However the coherent gravity method requires in Figure F-2 wherein other definitions, such as measurement consideration as to whether maximum or minimum values of depths Z and Zp as well as heights H and H1 are also shown. render the most critical loading conditions. In most cases, the The variation of Kr and F are with respect to depth Z that is proper choice can be readily identified by inspection at the measured from the top of the reinforced soil zone. For the onset. computation of Kr, the value of Ka is based on the angle of Appropriate resistance factors have to be used for compu- internal friction of the reinforced backfill, r, and the assump- tation of factored resistances during evaluation of strength tion that the backslope angle = 0; thus, Ka = tan2(45 - 34/2) Table F-5.2. Summary of applicable resistance factors for evaluation of tensile resistances. Reinforcement Fill Quality High Good Marginal Galvanized Strips 0.80 0.65 NA Galvanized Grids 0.70 0.55 0.301 Plain Steel Strips 0.452 0.451 NA Plain Steel Grids 0.352 0.351 NA 1t = 50 yrs design 2t = 75 yrs design

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73 0.3H1 H 1.7 Ka Kr 2.000 F* Z=0 Z=0 Z Z=20 ft 1.2 Ka Z=20 ft tan (1) = 0.675 Zp Zp (tan )(0.3H ) H = H1= H + H 1 - 0.3 tan Zp at start of resistant zone, Zp-s= Z + Latan Zp at end of resistant zone, Zp-e= Z + Ltan Use average Zp over the resistance zone, Zp-ave, for computing pullout resistance Zp-ave = Z + 0.5(Latan + Ltan) = Z + 0.5 tan (La + L) Ka is computed assuming that the backslope angle is zero, i.e., = 0 per Article C11.10.6.2.1 of AASHTO (2009) Figure F-2. Geometry definition, location of critical failure surface, and variation of Kr and F* parameters for steel ribbed strips. = 0.283. Hence, the value of Kr varies from 1.7(0.283) = 0.481 the walls is locally adjusted as necessary to fit the height of at Z = 0 ft to 1.2(0.283) = 0.340 at Z = 20 ft. For steel strips, the wall. F = 1.2+log10Cu. Using Cu = 7.0 as given in Step 2, For internal stability computations, each layer of reinforce- F = 1.2 + log10(7.0) = 2.045 > 2.000. Therefore, use F = 2.000. ment is assigned a tributary area, Atrib as follows: For the case of inextensible grids (i.e. welded wire fabric) the value of Kr varies from 2.5Ka at Z = 0 (2.5(0.283) = 0.707) to A trib =( w p )( S vt ) 1.2Ka at Z = 20 ft. (1.2(0.283)=0.34). For grid-type reinforce- ments, the value of F varies from 20(t/St) at Z = 0 ft to 10(t/St) where wp is the panel width of the precast facing element, and at Z 20 ft, where t is the diameter of the transverse wires and Svt is the vertical tributary spacing of the reinforcements based St is the transverse wire spacing. on the location of the reinforcements above and below the level The coherent gravity method uses Kr = K0 for the top of the reinforcement under consideration. The computation of 20 feet (Z 20 ft), where K0 is the coefficient of lateral earth Svt is summarized in Table F-7.1 wherein Svt = Z+ - Z-. Note that pressure at-rest, approximated as 1- sin r = 1 - sin 34 = wp = 5.00 ft per Step 1. 0.441 in this example. At depths Z > 20 ft Kr = Ka = 0.283 (in this example). 7.3 Calculate horizontal stress and maximum tension at each reinforcement level 7.2 Establish vertical layout of soil reinforcements The horizontal spacing of the reinforcements is based on Using the definition of depth Z as shown in Figure F-2 the maximum tension (Tmax) at each level of reinforcements, the following vertical layout of the soil reinforcements is which requires computation of the horizontal stress, H, at chosen: each reinforcement level. The reinforcement tensile and pull- out resistances are then compared with Tmax and an appropri- Z = 1.25 ft, 3.75 ft, 6.25 ft, 8.75 ft, 11.25 ft, 13.75 ft, 16.25 ft, ate reinforcement pattern is adopted. This section demonstrates 18.75 ft, 21.25 ft t, 23.75 ft, 26.25 ft, and 28.75 ft. the calculation of horizontal stress, H, and maximum ten- sion, Tmax. The above layout leads to 12 levels of reinforcements. For the Simplified Method the horizontal stress, H, at any The vertical spacing was chosen based on a typical vertical depth within the MSE wall is based on only the soil load as spacing, Sv, of approximately 2.5 ft that is commonly used summarized in Table F-7.2. in the industry for steel ribbed strip- or grid-type reinforce- ment. The vertical spacing near the top and bottom of H = H-soil + H-surcharge

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74 Table F-7.1. Summary of computations for Svt. Level Z (ft) Z - (ft) Z + (ft) Svt (ft) 1 1.25 0 1.25+0.5(3.751.25)=2.50 2.50 2 3.75 3.75-0.5(3.751.25)=2.50 3.75+0.5(6.25-3.75)=5.00 2.50 3 6.25 6.25-0.5(6.25-3.75)=5.00 6.25+0.5(8.75-6.25)=7.50 2.50 4 8.75 8.75-0.5(8.75-6.25)=7.50 8.75+0.5(11.25-8.75)=10.00 2.50 5 11.25 11.25-0.5(11.25-8.75)=10.00 11.25+0.5(13.75-11.25)=12.50 2.50 6 13.75 13.75-0.5(13.75-11.25)=12.50 13.75+0.5(16.25-13.75)=15.00 2.50 7 16.25 16.25-0.5(16.25-13.75)=15.00 16.25+0.5(18.75-16.25)=17.50 2.50 8 18.75 18.76-0.5(18.75-16.25)=17.50 18.75+0.5(21.25-18.75)=20.00 2.50 9 21.25 21.25-0.5(21.25-18.75)=20.00 21.25+0.5(23.75-21.25)=22.50 2.50 10 23.75 23.75-0.5(23.75-21.25)=22.50 23.75+0.5(26.25-23.75)=25.00 2.50 11 26.25 26.25-0.5(26.25-23.75)=25.00 26.25+0.5(28.75-26.25)=27.50 2.50 12 28.75 28.75-0.5(28.75-26.25)=27.50 30.00 2.50 Using the unit weight of the reinforced soil mass and At Z = 8.75 ft, the following depths are computed: heights Z and S as shown in Figure F-3(b), the equation for Z- = 7.50 ft (from Table F-7.1) horizontal stress at any depth Z within the MSE wall can be Z+ = 10.00 ft (from Table F-7.1) written as follows: Obtain Kr by linear interpolation between 1.7Ka = 0.481 at Z = 0.00 ft and 1.2Ka = 0.340 at Z = 20.00 ft as follows: H = K r ( r Z ) P-EV + K r ( r S ) P-EV = K r [ r ( Z + S ) P-EV ] At Z- = 7.50 ft, Kr(Z-) = 0.340 + (20.00 ft - 7.50 ft)(0.481- Once the horizontal stress is computed at any given level of 0.340)/20.00 ft = 0.428 reinforcement, the maximum tension, Tmax, is computed as At Z+ = 10.00 ft, Kr(Z+) = 0.340 + (20.00 ft - 10.00 ft)(0.481- follows: 0.340)/20.00 ft = 0.411 Tmax = ( H )( A trib ) Compute H-soil = [Kr v-soil]P-EV as follows: where Atrib is the tributary area for the soil reinforcement at a P-EV = 1.35 from Table F-5.1 given level. At Z- = 7.50 ft, For the coherent gravity method the factored horizontal v-soil(Z-) = (0.125 kcf)(7.50 ft) = 0.94 ksf stress at each reinforcement level is computed as H-soil(Z-) = [Kr(Zp-)v-soil(z-)]P-EV = (0.428)(0.94 ksf)(1.35) = H = Krv 0.54 ksf where v is the pressure due to resultant vertical forces at the At Z+ = 10.00 ft, reinforcement level being evaluated, determined using a uni- v-soil(Z+) = (0.125 kcf)(10.00 ft) = 1.25 ksf form pressure distribution over an effective width (L-2e) as H-soil(Z+) = [Kr(Zp+)v-soil(z+)]P-EV = (0.411)(1.25 ksf)(1.35) = specified in AASHTO (2009), Article 10.6.1.3, where e is the 0.69 ksf load eccentricity. The vertical effective stress at each level of H-soil = 0.5(0.54 ksf + 0.69 ksf) = 0.62 ksf reinforcement shall consider the local equilibrium of all Compute H-surcharge = [Kr 2]P-EV as follows: forces at that level only. Forces used to compute v (EV and 2 = (1/2)(0.7Htan)(f) (from Figure F-3(b)) EH) are factored as described in Table F-5.1. 2 = (1/2)(0.730 ft)[tan (26.56)](0.125 kcf) = 0.656 ksf The computations for Tmax using the simplified method for P-EV = 1.35 from Table F-5.1 Case 1 are illustrated at z = 8.75 ft, which is Level 4 in the assumed vertical layout of reinforcement. Assume Strength I At Z- = 7.50 ft, (max) load combination for illustration purposes and use H-surcharge = [Kr(Z-) 2]P-EV = (0.428)(0.656 ksf)(1.35) = appropriate load factors from Table F-5.1. 0.38 ksf Table F-7.2. Summary of load components leading to horizontal stress. Load Component Load Type Horizontal Stress Soil load from reinforced mass, v-soil EV H-soil = [Kr v-soil] P-EV Surcharge load due to backslope, 2 EV H-surcharge = [Kr 2] P-EV

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75 V2=L(h-z)r/2 (2/3)L h = z + Ltan() V1=rzL L/2 FTV z FTH A (a) (b) Figure F-3. Legend for computation of forces and moments for (a) internal stability analysis with the coherent gravity method, and (b) internal stability analysis with the simplified method (not-to-scale). At Z+ = 10.00 ft, The maximum tension at Level 4 is computed as follows: H-surcharge = [Kr(Z+) 2]P-ES = (0.411)(0.656 ksf)(1.35) Tmax = (H)(Atrib) = (0.99 ksf)(12.50 ft2) = 12.37 k for panel = 0.36 ksf of 5-ft width H-surcharge = 0.5(0.38 ksf + 0.36 ksf) = 0.37 ksf Compute H = H-soil + H-surcharge as follows: Using similar computations, the various quantities can H = 0.62 ksf + 0.37 ksf = 0.99 ksf be developed at other levels of reinforcements and load Based on Table F-7.1, the vertical tributary spacing at Level 4 combinations. is Svt = 2.50 ft The computations for Tmax using the coherent method for The panel width, wp, is 5.00 ft (given in Step 1) Case 1 are similar, however v is computed based on the equi- The tributary area, Atrib, is computed as follows: librium of forces and moments using a force diagram similar Atrib = (2.50 ft)(5.00 ft) = 12.50 ft2 to Figure F-3(a) for each level of reinforcement. Equations for

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76 unfactored vertical forces and moments for coherent gravity Location of the resultant from 9.10 ft method are as follows: Point A = (MRA - MOA)/VA Eccentricity of Vertical load @ 2.90 ft Moment arm Z = 30 ft = 0.5L - a Force LRFD (Length units) Effective width @ Z = 30 ft = L-2eL 18.20 ft (Force/length units) Load Type @ Point A v @t Z = 30 ft =V/(L-2eL) = v 10.19 ksf V1 = (r)(z)(L) EV L/2 Horizontal stress, H, and tensile force, Tmax, are computed 1 using Kr appropriate to the coherent gravity method, and V2 = (L)(L tan )(f) EV (2/3)L 2 tributary area as demonstrated for the simplified method. FTV = (1/2)(r)(h2)(Kaf)(sin) EH L FTH = (1/2)(r)(h2)(Kaf)(cos) EH h/3 7.4 Establish nominal and factored long-term Note: h = z + Ltan tensile resistance of soil reinforcement Compute unfactored vertical forces and moments at Z = 30 ft The nominal tensile resistance of soil reinforcements is (about Point A in Figure F-3(a)) based on the design life and estimated loss of steel over the design life during corrosion. Table F-7.3 is a summary of the V1 = 90.00 k/ft MV1 = 90 12 = 1080.00 k-ft/ft metal loss models recommended in this report and the esti- V2 = 18.00 k/ft MV2 = 18 16 = 288.00 k-ft/ft mated metal loss per side (i.e., sacrificial steel requirements) for FTV = 26.48 k/ft MFTV = 26.48 24 = 635.44 k-ft/ft each case considered in this example. For galvanized reinforce- FTH = 52.95 k/ft MFTH = 52.95 14 = 741.35 k-ft/ft ments it is assumed that steel corrosion is initiated subsequent Compute factored moments and forces at Z = 30 ft. (Checks to depletion of zinc. For fill materials that meet AASHTO requirements and an initial zinc thickness, zi = 86 m, zinc life with Strength I maximum and minimum load factors are is computed as 16 years using the AASHTO metal loss model necessary. Strength I Max was determined to govern for this described in Table 3 of this report and per Article 11.10.6.4.2a case and only these calculations are shown here.) of AASHTO (2009). Vertical load @Z = 30 ft, VAb1 = V1+V2 145.80 k/ft Considering Case 1 (described in Step 1) and a design life of Vertical load @Z = 30 ft, VAb2 = FTV 39.72 k/ft 75 years, the anticipated thickness loss is calculated as follows: Total vertical @Z = 30 ft, 185.52 k/ft E R = 708 m ( 2 sides ) = 1416 m ( 0.056 in.) , and V = R = VAb1+VAb2 E C = 4 mm - 1.416 mm = 2.58 mm ( 0.102 in.) Resisting moments about Point A, 1846.80 k-ft/ft MRA1 = MV1+MV2 Based on a 50 mm wide strip, the cross-sectional area at Resisting moments about Point A, 953.17 k-ft/ft the end of 75 years will be equal to (50 mm) (2.58 mm) = MRA2 = MFTV 129 mm2 (0.2 in.2) Total resisting moment @ Point A, 2799.97 k-ft/ft For Grade 65 steel with Fy = 65 ksi, the nominal tensile MRA = MRA1+MRA2 resistance at the end of a 75 year design life will be Tn = 65 ksi Overturning moments @ Point A, 1112.03 k-ft/ft (0.200 in.2) = 13.00 k/strip. Using the resistance factor, t = 0.75 MOA = MFTH as listed in Table F-5.2 for galvanized strip-type reinforce- Net moment at Point A, 1687.94 k-ft/ft ments in high quality fill, the factored tensile resistance, MA = MRA - MOA Tr = 13.00 k/strip (0.75) = 9.75 k/strip. Table F-7.3. Basis for computing sacrificial steel requirements. Fill tdesign X (tdesign) Reinforcement Recommended Metal Loss Model Quality (years) (m) Galvanized, zi = 86 m High X (m) = (tdesign 16 years) x 12 75 708 m/yr Galvanized, zi = 86 m Good X (m) = (tdesign 16 years) x 12 75 708 m/yr Galvanized, zi = 86 m Marginal X (m) = (tdesign 10 years) x 28 50 1120 m/yr Plain Steel High X (m) = tdesign x 13 m/yr 75 975 Plain Steel Good X (m) = 80 x tdesign0.8 50 1829

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77 Considering Case 2, the cross-sectional area at the end of Compute nominal pullout resistance as follows: 75 years for a Wll cold drawn wire will be equal to (0.374 Pr = (F)(2)(b)(Le)[(v-soil)(P-EV)] in (initial diameter of W11) - 0.056 in (loss of diameter))2/ Pr = (1.0)(1.420)(2)(0.164 ft)(13.41 ft)(2.175 ksf) 4 = 0.079 in2. Tn = 65 ksi (0.079 in2) = 5.14 k/wire. Using = 13.58 k/strip t = 0.70 as listed in Table F-5.2 for galvanized grids in high quality fill, the factored tensile resistance, Tf = 5.14 k/wire Compute factored pullout resistance as follows: (0.70) = 3.61 k/wire. Prr = Pr = (0.90)(13.58 k/strip) = 12.23 k/strip Using similar computations, the various quantities can be 7.5 Establish nominal and factored pullout developed at other levels of reinforcements and load combi- resistance of soil reinforcement nations. These calculations are similar for grid-type reinforce- The nominal pullout resistance, Pr, of galvanized steel, ments but with the appropriate factor for F. Calculations of ribbed, and strip-type soil reinforcements is computed with pullout resistance are the same using either the simplified or the following equation: coherent gravity methods. Pr = ( F )( 2b )( L e )[( v )( P-EV )] 7.6 Establish number of soil reinforcements at For Case 1, the following parameters are constant at all lev- each level of reinforcement els of reinforcements: Based on Tmax, Tr, and Prr, the number of strip reinforce- b = 1.969 in. = 0.164 ft ments at any given level of reinforcements can be computed = 1.0 for inextensible reinforcement per Table 11.10.6.3.2-1 as follows: of AA ASHTO (2009) Based on tensile resistance considerations, the number of The computations for Pr are illustrated for Case 1 at strip reinforcements, Nt, is computed as follows: z = 8.75 ft which is Level 4 as measured from the top of the wall. N t = Tmax Tr Assume Strength I (max) load combination for illustration pur- poses and use appropriate load factors from Table F-5.1. Based on pullout resistance considerations, the number of strip reinforcements, Np, is computed as follows: Compute effective (resisting) length, Le, as follows: N p = Tmax Prr Since Z < H1/2, active length La = 0.3(H1) and Le = L - La = L - 0.3(H1) Based on Tmax, Tr and Prr, the number of longitudinal wires H1= H + H for grid-type reinforcements at any given level of reinforce- H = ( tan )(0.3 H ) = (0.5)(0.3 30 ft ) = 5.29 ft ments can be computed as follows: 1 - 0.3 tan 1 - 0.3( 0.5 ) Assume spacing of the longitudinal wires, Sl = 6 in. = 0.5 ft H1 = H + H = 30.00 ft + 5.29 ft = 35.29 ft Based on tensile resistance considerations, the number of Active length, La = 0.3(35.29 ft) = 10.59 ft longitudinal wires, Nt, is computed as follows: Effective (resisting) length, Le = 24.00 ft - 10.59 ft = 13.41 ft N t = Tmax Tr Compute (v)(P-EV) As per Figure F-3(b), v = r(Zp-ave) Based on pullout resistance considerations, the number of Zp-ave = Z + 0.5 tan (La + L) = 8.75 ft + 0.5[tan(26.56)] longitudinal wires, Np, is computed as follows: (10.59 ft + 24.00 ft) = 17.40 ft N p = 1 + ( Tmax Prr ) ( S1 ) Per Article 11.10.6.3.2 of AASHTO (2009), use unfactored vertical stress for pullout resistance. Thus, Considering Case 1 and the Level 4 reinforcement at P-EV = 1.00 Z = 8.75 ft, the number of strip reinforcements can be com- v(P-EV) = (0.125 kcf)(17.40 ft) (1.00) = 2.175 ksf puted as follows: Obtain F at Z = 8.75 ft Tmax = 12.36 k for panel of 5-ft width, Tr = 10.41 k/strip, Obtain F by linear interpolation between 2.000 at Z = 0 Prr = 12.23 k/strip and 0.675 at Z = 20.00 ft as follows: Nt = Tmax/Tr = (12.36 k for panel of 5-ft width)/(10.41 k/strip) F = 0.675 + (20.00 ft - 8.75 ft)(2.000 - 0.675)/20 ft = 1.420 = 1.19 strips for panel of 5-ft width

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78 Table F-7.4 (a). Simplified method--computed reinforcement requirements. Fill Reinforcement tdesign As per 5-ft. wide panel Case Galvanized Quality Type (years) (in.2) 1 High Yes Strip 75 8.1 2 High Yes Grid 75 7.1 3 Good Yes Strip 75 9.0 4 Good Yes Grid 75 8.9 5 Marginal Yes Grid 50 17.0/16.41 6 High No Strip 75 13.0 7 High No Grid 75 13.8 8 Good No Strip 50 16.7 9 Good No Grid 50 19.2 Table F-7.4 (b). Coherent gravity method--computed reinforcement requirements. Fill Reinforcement tdesign As per 5 ft. wide panel Case Galvanized Quality Type (years) (in2) 1 High Yes Strip 75 9.0 2 High Yes Grid 75 7.3 3 Good Yes Strip 75 10.2 4 Good Yes Grid 75 8.9 5 Marginal Yes Grid 50 17.6/17.01 6 High No Strip 75 14.4 7 High No Grid 75 14.0 8 Good No Strip 50 18.0 9 Good No Grid 50 19.6 1 Model I/Model II-- demonstrates that resistance factors are calibrated with respect to different models to render similar designs. Np = Tmax/Prr = (12.36 k for panel of 5-ft width)/(12.23 k/ same probability that reinforcement resistance may fall below strip) = 1.01 strips for panel of 5-ft width acceptable levels before the end of the design life (pf = 0.01). Since Nt > Np, tension breakage is the governing criteria This example demonstrates that the designs executed with and therefore the governing value, Ng, is 1.19. Round up to Models I or II and corresponding resistance factors are select two strips at Level 4 for each panel of 5-ft width. indeed similar. These results demonstrate the advantages of using galvanized The computations in Sections 7.4 to 7.6 are repeated at steel to reduce the sacrificial steel requirements. Reinforcement each level of reinforcement. Tables of results from the com- requirements for plain steel are between 1.5 and 2.0 times putations at all levels of reinforcement for Strength I (max) higher in terms of cross-sectional area (As) compared to when load combination and Cases 19 are included at the end of galvanized steel reinforcements are used in similar fill condi- this appendix. The last column of the tables for Cases 1, 3, 6, tions (e.g., Case 1 compared to Case 6, Case 2 compared to and 8 provides horizontal spacing of the reinforcing strips, Case 7, Case 3 compared to Case 8, and Case 4 compared to which is obtained by dividing the panel width, wp, by the gov- Case 9). Designs achieved using the simplified method of analy- erning number of strips, Ng. sis are close to those rendered with the coherent gravity method Tables F-7.4(a) and (b) summarize the steel requirements when the same resistance factors are applied. This is expected computed using the Simplified and Coherent Gravity Meth- because the simplified method was calibrated to render results ods, respectively, for Cases 19 in terms of the steel area (As) similar to the coherent gravity method. However, when com- required for each 5-ft width of the wall (corresponding to the paring details of the designs achieved with the coherent gravity width of the precast concrete facing panel). compared to the simplified methods, the distributions of Both Models I and II are used to compute nominal steel the reinforcements are different. Use of the coherent gravity requirements for Case 5. As described in the report, Model method results in fewer reinforcements placed near the top of II renders twice the nominal sacrificial steel compared to the wall and more reinforcements placed near the bottom com- Model I, but resistance factors are calibrated to render the pared to designs achieved with the simplified method.

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CASE 1 H= 30 ft Ka= 0.283 p= 0.9 X= 708 m = 0.027874 in L= 24 ft P-EV = 1.35 t= 0.8 t= 75 yrs tan 0.5 2= 0.65625 ksf b= 0.164 ft zi= 86 m H= 5.294118 ft F*min=tan( r)= 0.674502089 CRz0-z2= 15 m/yr H1= 35.29412 ft F*max= 2 CRz2+= 4 m/yr La= 10.58824 ft CRsteel= 12 m/yr r= 125 pcf s= 4 mm = 0.15748 in Cu= 7 Fy= 65 ksi GALVANIZED STEEL STRIPS AND HIGH QUALITY FILL ( min >10,000 cm) SIMPLIFIED METHOD Level Z Zp-ave H Tmax F* Le pP r tTn Np Nt Ng Sh (ft) (ft) ksf k/ 5 ft wide panel dim (ft) k/strip k/strip - - - (ft) 1 1.25 9.90 0.52 6.45 1.917 13.41 9.39 10.41 0.7 0.6 2 2.50 2 3.75 12.40 0.69 8.61 1.751 13.41 10.75 10.41 0.8 0.8 2 2.50 3 6.25 14.90 0.85 10.57 1.586 13.41 11.69 10.41 0.9 1.0 2 2.50 4 8.75 17.40 0.99 12.36 1.420 13.41 12.23 10.41 1.0 1.2 2 2.50 5 11.25 19.90 1.12 13.95 1.254 13.41 12.35 10.41 1.1 1.3 2 2.50 6 13.75 22.19 1.23 15.36 1.089 14.25 12.70 10.41 1.2 1.5 2 2.50 7 16.25 24.31 1.33 16.58 0.923 15.75 13.04 10.41 1.3 1.6 2 2.50 8 18.75 26.44 1.41 17.62 0.757 17.25 12.74 10.41 1.4 1.7 2 2.50 9 21.25 28.56 1.52 18.98 0.675 18.75 13.33 10.41 1.4 1.8 2 2.50 10 23.75 30.69 1.66 20.77 0.675 20.25 15.47 10.41 1.3 2.0 2 2.50 11 26.25 32.81 1.81 22.56 0.675 21.75 17.76 10.41 1.3 2.2 3 1.67 12 28.75 34.94 1.95 24.36 0.675 23.25 20.22 10.41 1.2 2.3 3 1.67 Asum = 8.06

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CASE 2 H= 30 ft Ka= 0.283 p= 0.9 X= 708 m = 0.027874016 in L= 24 ft P-EV = 1.35 t= 0.7 t= 75 yrs tan 0.5 2= 0.65625 ksf St= 1 ft zi= 86 m * H= 5.294118 ft F min= 0.311666667 Sl= 0.5 ft CRz0-z2= 15 m/yr H1= 35.29412 ft F*max= 0.623333333 Tranverse W11 0.374 in diameter CRz2+= 4 m/yr La= 10.58824 ft CRsteel= 12 m/yr r= 125 pcf Longitudinal W11 0.374 in diameter Fy= 65 ksi GALVANIZED GRIDS AND HIGH QUALITY FILL ( min > 10,000 -cm) SIMPLIFIED METHOD Level Z Zp-ave H Tmax F* Le pPr tTn Np Nt Ng Bar Mat (ft) (ft) ksf k/ 5 ft wide panel dim (ft) k/ft k/wire - - - - 1 1.25 9.90 0.75 9.32 0.604 13.41 18.03 3.62 2.0 2.6 3 3W11 + W11 x 1.0' 2 3.75 12.40 0.96 12.06 0.565 13.41 21.13 3.62 2.1 3.3 4 4W11 + W11 x 1.0' 3 6.25 14.90 1.15 14.31 0.526 13.41 23.64 3.62 2.2 4.0 4 4W11 + W11 x 1.0' 4 8.75 17.40 1.29 16.08 0.487 13.41 25.57 3.62 2.3 4.4 5 5W11 + W11 x 1.0' 5 11.25 19.90 1.39 17.36 0.448 13.41 26.90 3.62 2.3 4.8 5 5W11 + W11 x 1.0' 6 13.75 22.19 1.45 18.16 0.409 14.25 29.10 3.62 2.2 5.0 6 6W11 + W11 x 1.0' 7 16.25 24.31 1.48 18.47 0.370 15.75 31.89 3.62 2.2 5.1 6 6W11 + W11 x 1.0' 8 18.75 26.44 1.46 18.30 0.331 17.25 33.98 3.62 2.1 5.1 6 6W11 + W11 x 1.0' 9 21.25 28.56 1.52 18.98 0.312 18.75 37.56 3.62 2.0 5.2 6 6W11 + W11 x 1.0' 10 23.75 30.69 1.66 20.77 0.312 20.25 43.58 3.62 2.0 5.7 6 6W11 + W11 x 1.0' 11 26.25 32.81 1.81 22.56 0.312 21.75 50.05 3.62 1.9 6.2 7 7W11 + W11 x 1.0' 12 28.75 34.94 1.95 24.36 0.312 23.25 56.96 3.62 1.9 6.7 7 7W11 + W11 x 1.0' Asum = 7.14

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CASE 9 H= 30 ft Ka= 0.283 p= 0.9 X= 1829.22 m = 0.072016544 in L= 24 ft P-EV = 1.35 t= 0.35 t= 50 yrs tan 0.5 2= 0.65625 ksf St= 1.0 ft H= 5.294118 ft F*min= 0.311666667 Sl= 0.5 ft H1= 35.29412 ft F*max= 0.623333333 Tranverse W11 0.374 in diameter La= 10.58824 ft r= 125 pcf Longitudinal W20 0.505 in diameter Fy= 65 ksi PLAIN STEEL GRIDS AND GOOD QUALITY FILL (3000 -cm< min < 10,000 -cm) SIMPLIFIED METHOD Level Z Zp-ave H Tmax F* Le pPr tTn Np Nt Ng Bar Mat (ft) (ft) ksf k/ 5 ft wide panel dim (ft) k/ft k/wire - - - - 1 1.25 9.90 0.75 9.32 0.604 13.41 18.03 2.33 2.0 4.0 5 5W20 + W11 x 1.0' 2 3.75 12.40 0.96 12.06 0.565 13.41 21.13 2.33 2.1 5.2 6 6W20 + W11 x 1.0' 3 6.25 14.90 1.15 14.31 0.526 13.41 23.64 2.33 2.2 6.1 7 7W20 + W11 x 1.0' 4 8.75 17.40 1.29 16.08 0.487 13.41 25.57 2.33 2.3 6.9 7 7W20 + W11 x 1.0' 5 11.25 19.90 1.39 17.36 0.448 13.41 26.90 2.33 2.3 7.5 8 8W20 + W11 x 1.0' 6 13.75 22.19 1.45 18.16 0.409 14.25 29.10 2.33 2.2 7.8 8 8W20 + W11 x 1.0' 7 16.25 24.31 1.48 18.47 0.370 15.75 31.89 2.33 2.2 7.9 8 8W20 + W11 x 1.0' 8 18.75 26.44 1.46 18.30 0.331 17.25 33.98 2.33 2.1 7.9 8 8W20 + W11 x 1.0' 9 21.25 28.56 1.52 18.98 0.312 18.75 37.56 2.33 2.0 8.2 9 9W20 + W11 x 1.0' 10 23.75 30.69 1.66 20.77 0.312 20.25 43.58 2.33 2.0 8.9 9 9W20 + W11 x 1.0' 11 26.25 32.81 1.81 22.56 0.312 21.75 50.05 2.33 1.9 9.7 10 10W20 + W11 x 1.0' 12 28.75 34.94 1.95 24.36 0.312 23.25 56.96 2.33 1.9 10.5 11 11W20 + W11 x 1.0' Asum = 19.23

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CASE 1 H= 30 ft Kaf= 0.283 p= 0.9 X= 708 m = 0.027874 in L= 24 ft k0f= 0.440807097 t= 0.8 t= 75 yrs tan 0.5 P-EV = 1.35 b= 0.164 ft zi= 86 m H= 5.294118 ft V2= 18 k/ft CRz0-z2= 15 m/yr * H1= 35.29412 ft F min=tan( r)= 0.674502089 CRz2+= 4 m/yr La= 10.58824 ft F*max= 2 CRsteel= 12 m/yr r= 125 pcf Kab = 0.537 s= 4 mm = 0.15748 in Cu= 7 b= 125 pcf Fy= 65 ksi P-EH = 1.5 GALVANIZED STRIPS AND HIGH QUALITY FILL ( min > 10,000 -cm) COHERENT GRAVITY METHOD Level Z Zp-ave H Tmax F* Le pPr tTn Np Nt Ng Sh (ft) (ft) ksf k/ 5 ft wide panel dim (ft) k/strip k/strip - - - (ft) 1 1.25 9.90 0.47 5.86 1.917 13.41 9.39 10.41 0.6 0.6 2 2.50 2 3.75 12.40 0.65 8.07 1.751 13.41 10.75 10.41 0.8 0.8 2 2.50 3 6.25 14.90 0.82 10.21 1.586 13.41 11.69 10.41 0.9 1.0 2 2.50 4 8.75 17.40 0.98 12.26 1.420 13.41 12.23 10.41 1.0 1.2 2 2.50 5 11.25 19.90 1.14 14.24 1.254 13.41 12.35 10.41 1.2 1.4 2 2.50 6 13.75 22.19 1.29 16.13 1.089 14.25 12.70 10.41 1.3 1.5 2 2.50 7 16.25 24.31 1.44 17.94 0.923 15.75 13.04 10.41 1.4 1.7 2 2.50 8 18.75 26.44 1.57 19.65 0.757 17.25 12.74 10.41 1.5 1.9 2 2.50 9 21.25 28.56 1.77 22.10 0.675 18.75 13.33 10.41 1.7 2.1 3 1.67 10 23.75 30.69 2.04 25.51 0.675 20.25 15.47 10.41 1.6 2.5 3 1.67 11 26.25 32.81 2.35 29.36 0.675 21.75 17.76 10.41 1.7 2.8 3 1.67 12 28.75 34.94 2.70 33.73 0.675 23.25 20.22 10.41 1.7 3.2 4 1.25 Asum = 8.99

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CASE 2 H= 30 ft Kaf= 0.283 p= 0.9 X= 708 m = 0.027874016 in L= 24 ft k0f= 0.440807097 t= 0.7 t= 75 yrs tan 0.5 P-EV = 1.35 St= 1.0 ft zi= 86 m H= 5.294118 ft V2= 18 k/ft Sl= 0.5 ft CRz0-z2= 15 m/yr H1= 35.29412 ft F*min= 0.311666667 Tranverse W11 0.374 in diameter CRz2+= 4 m/yr La= 10.58824 ft F*max= 0.623333333 CRsteel= 12 m/yr r= 125 pcf Kab = 0.537 Longitudinal W11 0.374 in diameter b= 125 pcf Fy= 65 ksi P-EH = 1.5 GALVANIZED GRIDS AND HIGH QUALITY FILL ( min > 10,000 -cm) COHERENT GRAVITY METHOD Level Z Zp-ave H Tmax F* Le pPr tTn Np Nt Ng Sh (ft) (ft) ksf k/ 5 ft wide panel dim (ft) k/ft k/wire - - - - 1 1.25 9.90 0.47 5.86 0.604 13.41 18.03 3.62 1.7 1.6 2 2W11 + W11 x 1.0' 2 3.75 12.40 0.65 8.07 0.565 13.41 21.13 3.62 1.8 2.2 3 3W11 + W11 x 1.0' 3 6.25 14.90 0.82 10.21 0.526 13.41 23.64 3.62 1.9 2.8 3 3W11 + W11 x 1.0' 4 8.75 17.40 0.98 12.26 0.487 13.41 25.57 3.62 2.0 3.4 4 4W11 + W11 x 1.0' 5 11.25 19.90 1.14 14.24 0.448 13.41 26.90 3.62 2.1 3.9 4 4W11 + W11 x 1.0' 6 13.75 22.19 1.29 16.13 0.409 14.25 29.10 3.62 2.1 4.5 5 5W11 + W11 x 1.0' 7 16.25 24.31 1.44 17.94 0.370 15.75 31.89 3.62 2.1 5.0 5 5W11 + W11 x 1.0' 8 18.75 26.44 1.57 19.65 0.331 17.25 33.98 3.62 2.2 5.4 6 6W11 + W11 x 1.0' 9 21.25 28.56 1.77 22.10 0.312 18.75 37.56 3.62 2.2 6.1 7 7W11 + W11 x 1.0' 10 23.75 30.69 2.04 25.51 0.312 20.25 43.58 3.62 2.2 7.0 8 8W11 + W11 x 1.0' 11 26.25 32.81 2.35 29.36 0.312 21.75 50.05 3.62 2.2 8.1 9 9W11 + W11 x 1.0' 12 28.75 34.94 2.70 33.73 0.312 23.25 56.96 3.62 2.2 9.3 10 10W11 + W11 x 1.0' Asum = 7.25

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CASE 3 H= 30 ft Kaf= 0.283 p= 0.9 X= 708 m = 0.027874 in L= 24 ft k0f= 0.440807097 t= 0.65 t= 75 yrs tan 0.5 P-EV = 1.35 b= 0.164 ft zi= 86 m H= 5.294118 ft V2= 18 k/ft CRz0-z2= 15 m/yr H1= 35.29412 ft F*min=tan( r)= 0.674502089 CRz2+= 4 m/yr La= 10.58824 ft F*max= 2 CRsteel= 12 m/yr r= 125 pcf Kab = 0.537 s= 4 mm = 0.15748 in Cu= 7 b= 125 pcf Fy= 65 ksi P-EH = 1.5 GALVANIZED STRIPS AND GOOD QUALITY FILL (3000 -cm< min < 10,000 -cm) COHERENT GRAVITY METHOD Level Z Zp-ave H Tmax F* Le pP r tTn Np Nt Ng Sh (ft) (ft) ksf k/ 5 ft wide panel dim (ft) k/strip k/strip - - - (ft) 1 1.25 9.90 0.47 5.86 1.917 13.41 9.39 8.46 0.6 0.7 2 2.50 2 3.75 12.40 0.65 8.07 1.751 13.41 10.75 8.46 0.8 1.0 2 2.50 3 6.25 14.90 0.82 10.21 1.586 13.41 11.69 8.46 0.9 1.2 2 2.50 4 8.75 17.40 0.98 12.26 1.420 13.41 12.23 8.46 1.0 1.4 2 2.50 5 11.25 19.90 1.14 14.24 1.254 13.41 12.35 8.46 1.2 1.7 2 2.50 6 13.75 22.19 1.29 16.13 1.089 14.25 12.70 8.46 1.3 1.9 2 2.50 7 16.25 24.31 1.44 17.94 0.923 15.75 13.04 8.46 1.4 2.1 3 1.67 8 18.75 26.44 1.57 19.65 0.757 17.25 12.74 8.46 1.5 2.3 3 1.67 9 21.25 28.56 1.77 22.10 0.675 18.75 13.33 8.46 1.7 2.6 3 1.67 10 23.75 30.69 2.04 25.51 0.675 20.25 15.47 8.46 1.6 3.0 4 1.25 11 26.25 32.81 2.35 29.36 0.675 21.75 17.76 8.46 1.7 3.5 4 1.25 12 28.75 34.94 2.70 33.73 0.675 23.25 20.22 8.46 1.7 4.0 4 1.25 Asum = 10.23

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CASE 4 H= 30 ft Kaf= 0.283 p= 0.9 X= 708 m = 0.027874016 in L= 24 ft k0f= 0.440807097 t= 0.55 t= 75 yrs tan 0.5 P-EV = 1.35 St= 1.0 ft zi= 86 m H= 5.294118 ft V2= 18 k/ft Sl= 0.5 ft CRz0-z2= 15 m/yr H1= 35.29412 ft F*min= 0.311666667 Tranverse W11 0.374 in diameter CRz2+= 4 m/yr La= 10.58824 ft F*max= 0.623333333 CRsteel= 12 m/yr r= 125 pcf Kab = 0.537 Longitudinal W11 0.374 in diameter b= 125 pcf Fy= 65 ksi P-EH = 1.5 GALVANIZED GRIDS AND GOOD QUALITY FILL (3000 -cm< min < 10,000 -cm) COHERENT GRAVITY METHOD Level Z Zp-ave H Tmax F* Le pPr tTn Np Nt Ng Sh (ft) (ft) ksf k/ 5 ft wide panel dim (ft) k/ft k/wire - - - - 1 1.25 9.90 0.47 5.86 0.604 13.41 18.03 2.84 1.7 2.1 3 3W11 + W11 x 1.0' 2 3.75 12.40 0.65 8.07 0.565 13.41 21.13 2.84 1.8 2.8 3 3W11 + W11 x 1.0' 3 6.25 14.90 0.82 10.21 0.526 13.41 23.64 2.84 1.9 3.6 4 4W11 + W11 x 1.0' 4 8.75 17.40 0.98 12.26 0.487 13.41 25.57 2.84 2.0 4.3 5 5W11 + W11 x 1.0' 5 11.25 19.90 1.14 14.24 0.448 13.41 26.90 2.84 2.1 5.0 6 6W11 + W11 x 1.0' 6 13.75 22.19 1.29 16.13 0.409 14.25 29.10 2.84 2.1 5.7 6 6W11 + W11 x 1.0' 7 16.25 24.31 1.44 17.94 0.370 15.75 31.89 2.84 2.1 6.3 7 7W11 + W11 x 1.0' 8 18.75 26.44 1.57 19.65 0.331 17.25 33.98 2.84 2.2 6.9 7 7W11 + W11 x 1.0' 9 21.25 28.56 1.77 22.10 0.312 18.75 37.56 2.84 2.2 7.8 8 8W11 + W11 x 1.0' 10 23.75 30.69 2.04 25.51 0.312 20.25 43.58 2.84 2.2 9.0 9 9W11 + W11 x 1.0' 11 26.25 32.81 2.35 29.36 0.312 21.75 50.05 2.84 2.2 10.3 11 11W11 + W11 x 1.0' 12 28.75 34.94 2.70 33.73 0.312 23.25 56.96 2.84 2.2 11.9 12 12W11 + W11 x 1.0' Asum = 8.90

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CASE 5(a). Model I H= 30 ft Kaf= 0.283 p= 0.9 X= 1120 m = 0.044094488 in L= 24 ft k0f= 0.440807097 t= 0.3 t= 50 yrs tan 0.5 P-EV = 1.35 St= 1.0 ft zi= 86 m H= 5.294118 ft V2= 18 k/ft Sl= 0.5 ft CRz0-z2= 8.6 m/yr * H1= 35.29412 ft F min = 0.311666667 Tranverse W11 0.374 in diameter CRz2+= 8.6 m/yr La= 10.58824 ft F*max= 0.623333333 CRsteel= 28 m/yr r= 125 pcf Kab = 0.537 Longitudinal W20 0.505 in diameter b= 125 pcf Fy= 65 ksi P-EH = 1.5 GALVANIZED GRIDS AND MARGINAL QUALITY FILL (1,000 -cm< min < 3,000 -cm) COHERENT GRAVITY METHOD Level Z Zp-ave H Tmax F* Le pPr tTn Np Nt Ng Sh (ft) (ft) ksf k/ 5 ft wide panel dim (ft) k/ft k/wire - - - - 1 1.25 9.90 0.47 5.86 0.604 13.41 18.03 2.66 1.7 2.2 3 3W20 + W11 x 1.0' 2 3.75 12.40 0.65 8.07 0.565 13.41 21.13 2.66 1.8 3.0 4 4W20 + W11 x 1.0' 3 6.25 14.90 0.82 10.21 0.526 13.41 23.64 2.66 1.9 3.8 4 4W20 + W11 x 1.0' 4 8.75 17.40 0.98 12.26 0.487 13.41 25.57 2.66 2.0 4.6 5 5W20 + W11 x 1.0' 5 11.25 19.90 1.14 14.24 0.448 13.41 26.90 2.66 2.1 5.4 6 6W20 + W11 x 1.0' 6 13.75 22.19 1.29 16.13 0.409 14.25 29.10 2.66 2.1 6.1 7 7W20 + W11 x 1.0' 7 16.25 24.31 1.44 17.94 0.370 15.75 31.89 2.66 2.1 6.7 7 7W20 + W11 x 1.0' 8 18.75 26.44 1.57 19.65 0.331 17.25 33.98 2.66 2.2 7.4 8 8W20 + W11 x 1.0' 9 21.25 28.56 1.77 22.10 0.312 18.75 37.56 2.66 2.2 8.3 9 9W20 + W11 x 1.0' 10 23.75 30.69 2.04 25.51 0.312 20.25 43.58 2.66 2.2 9.6 10 10W20 + W11 x 1.0' 11 26.25 32.81 2.35 29.36 0.312 21.75 50.05 2.66 2.2 11.0 12 12W20 + W11 x 1.0' 12 28.75 34.94 2.70 33.73 0.312 23.25 56.96 2.66 2.2 12.7 13 13W20 + W11 x 1.0' Asum = 17.63

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CASE 5(b). Model II H= 30 ft Kaf= 0.283 p= 0.9 X= 2240 m = 0.088188976 in L= 24 ft k0f= 0.440807097 t= 0.5 t= 50 yrs tan 0.5 P-EV = 1.35 St= 1.0 ft zi= 86 m H= 5.294118 ft V2= 18 k/ft Sl= 0.5 ft CRz0-z2= 8.6 m/yr * H1= 35.29412 ft F min = 0.311666667 Tranverse W11 0.374 in diameter CRz2+= 8.6 m/yr La= 10.58824 ft F*max= 0.623333333 CRsteel= 56 m/yr r= 125 pcf Kab = 0.537 Longitudinal W20 0.505 in diameter b= 125 pcf Fy= 65 ksi P-EH = 1.5 GALVANIZED GRIDS AND MARGINAL QUALITY FILL (1,000 -cm< min < 3,000 -cm) COHERENT GRAVITY METHOD Level Z Zp-ave H Tmax F* Le pPr tTn Np Nt Ng Sh (ft) (ft) ksf k/ 5 ft wide panel dim (ft) k/ft k/wire - - - - 1 1.25 9.90 0.47 5.86 0.604 13.41 18.03 2.76 1.7 2.1 3 3W20 + W11 x 1.0' 2 3.75 12.40 0.65 8.07 0.565 13.41 21.13 2.76 1.8 2.9 3 3W20 + W11 x 1.0' 3 6.25 14.90 0.82 10.21 0.526 13.41 23.64 2.76 1.9 3.7 4 4W20 + W11 x 1.0' 4 8.75 17.40 0.98 12.26 0.487 13.41 25.57 2.76 2.0 4.4 5 5W20 + W11 x 1.0' 5 11.25 19.90 1.14 14.24 0.448 13.41 26.90 2.76 2.1 5.2 6 6W20 + W11 x 1.0' 6 13.75 22.19 1.29 16.13 0.409 14.25 29.10 2.76 2.1 5.9 6 6W20 + W11 x 1.0' 7 16.25 24.31 1.44 17.94 0.370 15.75 31.89 2.76 2.1 6.5 7 7W20 + W11 x 1.0' 8 18.75 26.44 1.57 19.65 0.331 17.25 33.98 2.76 2.2 7.1 8 8W20 + W11 x 1.0' 9 21.25 28.56 1.77 22.10 0.312 18.75 37.56 2.76 2.2 8.0 9 9W20 + W11 x 1.0' 10 23.75 30.69 2.04 25.51 0.312 20.25 43.58 2.76 2.2 9.3 10 10W20 + W11 x 1.0' 11 26.25 32.81 2.35 29.36 0.312 21.75 50.05 2.76 2.2 10.7 11 11W20 + W11 x 1.0' 12 28.75 34.94 2.70 33.73 0.312 23.25 56.96 2.76 2.2 12.2 13 13W20 + W11 x 1.0' Asum = 17.03

OCR for page 70
CASE 6 H= 30 ft Kaf= 0.283 p= 0.9 X= 975 m = 0.038386 in L= 24 ft k0f= 0.440807097 t= 0.45 t= 75 yrs tan 0.5 P-EV = 1.35 b= 0.164 ft H= 5.294118 ft V2= 18 k/ft H1= 35.29412 ft min=tan( r)= 0.674502089 La= 10.58824 ft F*max= 2 r= 125 pcf Kab = 0.537 s= 6 mm = 0.23622 in Cu= 7 b= 125 pcf Fy= 65 ksi P-EH = 1.5 PLAIN STEEL STRIPS AND HIGH QUALITY FILL ( min > 10,000 -cm) COHERENT GRAVITY METHOD Level Z Zp-ave H Tmax F* Le pPr tTn Np Nt Ng Sh (ft) (ft) ksf k/ 5 ft wide panel dim (ft) k/strip k/strip - - - (ft) 1 1.25 9.90 0.47 5.86 1.917 13.41 9.39 9.18 0.6 0.6 2 2.50 2 3.75 12.40 0.65 8.07 1.751 13.41 10.75 9.18 0.8 0.9 2 2.50 3 6.25 14.90 0.82 10.21 1.586 13.41 11.69 9.18 0.9 1.1 2 2.50 4 8.75 17.40 0.98 12.26 1.420 13.41 12.23 9.18 1.0 1.3 2 2.50 5 11.25 19.90 1.14 14.24 1.254 13.41 12.35 9.18 1.2 1.6 2 2.50 6 13.75 22.19 1.29 16.13 1.089 14.25 12.70 9.18 1.3 1.8 2 2.50 7 16.25 24.31 1.44 17.94 0.923 15.75 13.04 9.18 1.4 2.0 2 2.50 8 18.75 26.44 1.57 19.65 0.757 17.25 12.74 9.18 1.5 2.1 3 1.67 9 21.25 28.56 1.77 22.10 0.675 18.75 13.33 9.18 1.7 2.4 3 1.67 10 23.75 30.69 2.04 25.51 0.675 20.25 15.47 9.18 1.6 2.8 3 1.67 11 26.25 32.81 2.35 29.36 0.675 21.75 17.76 9.18 1.7 3.2 4 1.25 12 28.75 34.94 2.70 33.73 0.675 23.25 20.22 9.18 1.7 3.7 4 1.25 Asum = 14.41

OCR for page 70
CASE 7 H= 30 ft Kaf= 0.283 p= 0.9 X= 975 m = 0.038385827 in L= 24 ft k0f= 0.440807097 t= 0.35 t= 75 yrs tan 0.5 P-EV = 1.35 St= 1.0 ft H= 5.294118 ft V2= 18 k/ft S l= 0.5 ft H1= 35.29412 ft F*min= 0.311666667 Tranverse W11 0.374 in diameter La= 10.58824 ft F*max= 0.623333333 r= 125 pcf Kab = 0.537 Longitudinal W20 0.505 in diameter b= 125 pcf Fy= 65 ksi P-EH = 1.5 PLAIN STEEL GRIDS AND HIGH QUALITY FILL ( min > 10,000 -cm) COHERENT GRAVITY METHOD Level Z Zp-ave H Tmax F* Le p Pr tTn Np Nt Ng Sh (ft) (ft) ksf k/ 5 ft wide panel dim (ft) k/ft k/wire - - - - 1 1.25 9.90 0.47 5.86 0.604 13.41 18.03 3.28 1.7 1.8 2 2W20 + W11 x 1.0' 2 3.75 12.40 0.65 8.07 0.565 13.41 21.13 3.28 1.8 2.5 3 3W20 + W11 x 1.0' 3 6.25 14.90 0.82 10.21 0.526 13.41 23.64 3.28 1.9 3.1 4 4W20 + W11 x 1.0' 4 8.75 17.40 0.98 12.26 0.487 13.41 25.57 3.28 2.0 3.7 4 4W20 + W11 x 1.0' 5 11.25 19.90 1.14 14.24 0.448 13.41 26.90 3.28 2.1 4.3 5 5W20 + W11 x 1.0' 6 13.75 22.19 1.29 16.13 0.409 14.25 29.10 3.28 2.1 4.9 5 5W20 + W11 x 1.0' 7 16.25 24.31 1.44 17.94 0.370 15.75 31.89 3.28 2.1 5.5 6 6W20 + W11 x 1.0' 8 18.75 26.44 1.57 19.65 0.331 17.25 33.98 3.28 2.2 6.0 6 6W20 + W11 x 1.0' 9 21.25 28.56 1.77 22.10 0.312 18.75 37.56 3.28 2.2 6.7 7 7W20 + W11 x 1.0' 10 23.75 30.69 2.04 25.51 0.312 20.25 43.58 3.28 2.2 7.8 8 8W20 + W11 x 1.0' 11 26.25 32.81 2.35 29.36 0.312 21.75 50.05 3.28 2.2 9.0 9 9W20 + W11 x 1.0' 12 28.75 34.94 2.70 33.73 0.312 23.25 56.96 3.28 2.2 10.3 11 11W20 + W11 x 1.0' Asum = 14.02

OCR for page 70
CASE 8 H= 30 ft Kaf= 0.283 p= 0.9 X= 1829.22 m = 0.072017 in L= 24 ft k0f= 0.440807097 t= 0.45 t= 50 yrs tan 0.5 P-EV = 1.35 b= 0.164 ft H= 5.294118 ft V2= 18 k/ft H1= 35.29412 ft min=tan( r)= 0.674502089 La= 10.58824 ft F*max= 2 r= 125 pcf Kab = 0.537 s= 8 mm = 0.314961 in Cu= 7 b= 125 pcf Fy= 65 ksi P-EH = 1.5 PLAIN STEEL STRIPS AND GOOD QUALITY FILL (3000 -cm< min < 10,000 -cm) COHERENT GRAVITY METHOD Level Z Zp-ave H Tmax F* Le pPr tTn Np Nt Ng Sh (ft) (ft) ksf k/ 5 ft wide panel dim (ft) k/strip k/strip - - - (ft) 1 1.25 9.90 0.47 5.86 1.917 13.41 9.39 9.84 0.6 0.6 2 2.50 2 3.75 12.40 0.65 8.07 1.751 13.41 10.75 9.84 0.8 0.8 2 2.50 3 6.25 14.90 0.82 10.21 1.586 13.41 11.69 9.84 0.9 1.0 2 2.50 4 8.75 17.40 0.98 12.26 1.420 13.41 12.23 9.84 1.0 1.2 2 2.50 5 11.25 19.90 1.14 14.24 1.254 13.41 12.35 9.84 1.2 1.4 2 2.50 6 13.75 22.19 1.29 16.13 1.089 14.25 12.70 9.84 1.3 1.6 2 2.50 7 16.25 24.31 1.44 17.94 0.923 15.75 13.04 9.84 1.4 1.8 2 2.50 8 18.75 26.44 1.57 19.65 0.757 17.25 12.74 9.84 1.5 2.0 2 2.50 9 21.25 28.56 1.77 22.10 0.675 18.75 13.33 9.84 1.7 2.2 3 1.67 10 23.75 30.69 2.04 25.51 0.675 20.25 15.47 9.84 1.6 2.6 3 1.67 11 26.25 32.81 2.35 29.36 0.675 21.75 17.76 9.84 1.7 3.0 3 1.67 12 28.75 34.94 2.70 33.73 0.675 23.25 20.22 9.84 1.7 3.4 4 1.25 Asum = 17.98

OCR for page 70
CASE 9 H= 30 ft Kaf= 0.283 p= 0.9 X= 1829.22 m = 0.072016544 in L= 24 ft k0f= 0.440807097 t= 0.35 t= 50 yrs tan 0.5 P-EV = 1.35 St= 1.0 ft H= 5.294118 ft V2= 18 k/ft Sl= 0.5 ft * H1= 35.29412 ft F min= 0.311666667 Tranverse W11 0.374 in diameter La= 10.58824 ft F*max= 0.623333333 r= 125 pcf Kab = 0.537 Longitudinal W20 0.505 in diameter b= 125 pcf Fy= 65 ksi P-EH = 1.5 PLAIN STEEL GRIDS AND GOOD QUALITY FILL (3000 -cm< min < 10,000 -cm) COHERENT GRAVITY METHOD Level Z Zp-ave H Tmax F* Le pPr tTn Np Nt Ng Sh (ft) (ft) ksf k/ 5 ft wide panel dim (ft) k/ft k/wire - - - - 1 1.25 9.90 0.47 5.86 0.604 13.41 18.03 2.33 1.7 2.5 3 3W20 + W11 x 1.0' 2 3.75 12.40 0.65 8.07 0.565 13.41 21.13 2.33 1.8 3.5 4 4W20 + W11 x 1.0' 3 6.25 14.90 0.82 10.21 0.526 13.41 23.64 2.33 1.9 4.4 5 5W20 + W11 x 1.0' 4 8.75 17.40 0.98 12.26 0.487 13.41 25.57 2.33 2.0 5.3 6 6W20 + W11 x 1.0' 5 11.25 19.90 1.14 14.24 0.448 13.41 26.90 2.33 2.1 6.1 7 7W20 + W11 x 1.0' 6 13.75 22.19 1.29 16.13 0.409 14.25 29.10 2.33 2.1 6.9 7 7W20 + W11 x 1.0' 7 16.25 24.31 1.44 17.94 0.370 15.75 31.89 2.33 2.1 7.7 8 8W20 + W11 x 1.0' 8 18.75 26.44 1.57 19.65 0.331 17.25 33.98 2.33 2.2 8.4 9 9W20 + W11 x 1.0' 9 21.25 28.56 1.77 22.10 0.312 18.75 37.56 2.33 2.2 9.5 10 10W20 + W11 x 1.0' 10 23.75 30.69 2.04 25.51 0.312 20.25 43.58 2.33 2.2 11.0 11 11W20 + W11 x 1.0' 11 26.25 32.81 2.35 29.36 0.312 21.75 50.05 2.33 2.2 12.6 13 13W20 + W11 x 1.0' 12 28.75 34.94 2.70 33.73 0.312 23.25 56.96 2.33 2.2 14.5 15 15W20 + W11 x 1.0' Asum = 19.63