Geofoam Applications in the Design and Construction of Highway Embankments (2004)

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7-1 CHAPTER 7 DESIGN EXAMPLES Contents Introduction...................................................................................................................................7-2 Design Example 1 – Trapezoidal Geofoam Embankment............................................................7-2 Step 1 – Background Investigation ...........................................................................................7-2 Step 2 – Select Preliminary Eps Type and Assume a Preliminary Pavement System Design..7-3 Step 3 – Determine a Preliminary Fill Mass Arrangement .......................................................7-4 Step 4 – Foundation Soil Settlement Analysis..........................................................................7-4 Step 5 – Bearing Capacity.......................................................................................................7-11 Step 6 – External Slope Stability ............................................................................................7-12 Step 7 – External Seismic Stability.........................................................................................7-12 Step 8 – Hydrostatic Uplift (Flotation) ...................................................................................7-14 Step 9 – Translation Due to Water (External).........................................................................7-15 Step 10 – Translation Due to Wind (External)........................................................................7-17 Step 11 – Translation Due to Water (Internal)........................................................................7-19 Step 12 – Translation Due to Wind (Internal).........................................................................7-21 Step 13 – Seismic Stability (Internal) .....................................................................................7-23 Step 14 – Load Bearing...........................................................................................................7-26 Step 15 – Pavement System Design........................................................................................7-35 Step 16 – Determine if the Final Pavement System Design Results in a Significant Change in Overburden Compared to the Preliminary Pavement System Design Developed in Step 2………………………………………………………………………………7-36 Step 17 –Final Embankment Design.......................................................................................7-36 Design Example 2 – Lateral Pressures on an Abutment .............................................................7-36

7-2 References...................................................................................................................................7-39 Figures ........................................................................................................................................7-40 Tables..........................................................................................................................................7-49 ___________________________________________________________________________ INTRODUCTION This chapter presents two design examples that illustrate the design of an EPS-block geofoam embankment. The design principles and methods are discussed in detail in Chapters 3 through 6 and summarized in the flow chart illustrated in Figure 3.3. Example 1 presents the complete design for a trapezoidal geofoam embankment. Example 2 shows how to calculate earth pressures generated by an EPS-block geofoam bridge approach fill on an abutment. In each example, detailed calculations are shown with the appropriate equation and design chart numbers that appear in Chapters 3 through 6. Additionally, tables have been extensively used to summarize design calculation input values and results. These tables can serve as the basis for developing computer design spreadsheets. Only SI units have been used in the design examples. Appendix F includes conversion factors that can be used to convert between SI units and I-P units. DESIGN EXAMPLE 1 – TRAPEZOIDAL GEOFOAM EMBANKMENT STEP 1 – BACKGROUND INVESTIGATION • Geometric Requirements:  Two lane roadway with each lane 3.7 m wide and two 1.8 m wide shoulders. Therefore, the required total width at the top of the embankment is 11 m (TW=11 m).  Side slope of 3 (H) : 1 (V).  Height of embankment, H, of 5 m. • Site Conditions:

7-3  Site is in Urbana, Illinois.  Subsurface conditions shown in Figure 7.1.  The 100-year flood water level is expected to be 1.12 m. It is anticipated that water will only accumulate on one side of the embankment. • Design Requirements  Maximum allowable settlement is 400 mm during 20 years service life.  Use peak horizontal bedrock acceleration for 10 percent probability of exceedance in 50 years to evaluate seismic stability.  Use AASHTO H 20-44 standard loading to estimate live loads from traffic for geofoam load bearing calculations.  Local transportation agency prefers a flexible pavement system design based on a 75 percent level of reliability.  The proposed roadway embankment will be located along a low-volume road with an estimated traffic level of 300,000 equivalent single axle loads (ESAL). STEP 2 – SELECT PRELIMINARY EPS TYPE AND ASSUME A PRELIMINARY PAVEMENT SYSTEM DESIGN • Preliminary EPS Type:  Start with an EPS50. From Table 6.2, EPS50 has a dry density of 20 kg/m3, which is equivalent to a dry unit weight of 3EPS Dry 0.2 kN/mγ = .  From the Design Loads section of Chapter 3, assume a saturated unit weight of 3 EPS Abs 1 kN/mγ = to account for potential water absorption. • Preliminary Pavement System Design:  From the Design Loads section of Chapter 3, assume a thickness of pavement of pavementT 610 mm= .

7-4  From the Design Loads section of Chapter 3, assume an overall unit weight for the pavement system of 3pavementγ 20 kN/m= . STEP 3 – DETERMINE A PRELIMINARY FILL MASS ARRANGEMENT • Start the design process with fill mass (see Figure 3.2) consisting of only EPS blocks and soil cover on the sides of the embankment. This will provide an indication of the suitability of using an EPS-block geofoam embankment. • Thickness of EPS, TEPS, equals pavementH T 5 m 0.6 m 4.4 m− = − = . • From the Embankment Cover section of Chapter 5, assume a total (moist) unit weight for the soil cover, γcover, of 18.8 kN/m³ and a thickness, Tcover, of 400 mm. • Figure 7.1 shows the preliminary fill mass arrangement. Figure 7.1 Embankment geometry, subsurface conditions, and preliminary pavement system and fill mass arrangement. STEP 4 – FOUNDATION SOIL SETTLEMENT ANALYSIS • From Figure 5.2, the proposed embankment will have the following geometric dimensions:  slope change in the horizontal direction 3a H * 5 m 15 m slope change in the vertical direction 1 = = ∗ =  11 mb 5.5 m 2 = = • Divide the soft clay into 10 sublayers.  Each layer will be 15 m 1.5 m thick 10 layers = • Determine the magnitude of settlement at the center and edge of the embankment as shown in Figure 5.2. Therefore, determine the change in effective vertical stress, Z′∆σ , of each soil layer at the center and edge of the embankment as shown in Figure 7.2. To

7-5 demonstrate the use of the procedure to estimate ∆σ′z and the overall settlement described in Chapter 5, a detailed calculation will be performed for Layer 5. Figure 7.2 Subdivision of soft clay layer for settlement analysis. • Determine the total increase in vertical stress at the center of the embankment, ∆σZ@center, at mid-height of Layer 5.  Determine ∆σZ caused by zone I as shown in Figure 5.2.  From Equation (5.13), 3 fill EPS EPSq T 1 kN/m 4.4 m 4.4 kPaγ= ∗ = ∗ = Note that γEPS Abs is used for γEPS in settlement calculations.  From Equation (5.14), 3 pavement pavement pavementq γ T 20 kN/m 0.61 m 12.2 kPa= ∗ = ∗ =  From Equation (5.12), I fill pavementq q q 4.4 kPa + 12.2 kPa = 16.6 kPa= + =  From Fig. 5.3 and Equation (5.11), b 5.5 m2 arctan 2 arctan 1.3674 radians z 6.75 m α    = ∗ = ∗ =        From Equation (5.10), ( ) ( ) I I Z q sin 16.6 kPa = 1.3674 radians + sin (1.3674 radians) = 12.40 kPa ∆σ = α + απ π  Determine ∆σZ caused by zone II as shown in Figure 5.2.  From Equation (5.19),

7-6 3cover cover cover T 0.4 mq γ 18.8 kN/m 7.9 kPa cos cos 18.43 = ∗ = ∗ =θ o fillq 4.4 kPa as previously determined=  From Equation (5.18), II fill cover q q q 4.4 kPa + 7.9 kPa = 12.3 kPa= + =  From Fig. 5.4 and Equation (5.16), 5.5 marctan 0.6836 radians 6.75 m δ  = =    From Equation (5.17), a + b 15 m + 5.5 marctan arctan 0.6836 0.5691 radians z 6.75 m α δ   = − = − =        From Equation (5.15), ( ) II II Z q x sin 2 2 0.5 a 12.3 20.5 m = 0.5691 radians sin 2 0.6836 radians 2 0.5 15 m = 1.13 kPa  σ = α − δ π ∗   − ∗ π ∗   Determine the total increase in vertical stress at the center of the embankment, ∆σZ@center.  From Equation (5.20), ( ) ( ) I IIZ@center Z Z 2 12.40 kPa 2 1.13 kPa 14.66 kPa ∆σ = ∆σ + ∗∆σ = + ∗ = • Determine the total increase in vertical stress at the edge of the embankment, ∆σZ@edge, at the mid-height of Layer 5.  Determine ∆σZ caused by zone II as shown in Figure 5.2  From Equation (5.23) and Fig. 5.6,

7-7 a 15 marctan arctan 1.1479 radians z 6.75 m δ    = = =        From Equation (5.22), ( ) ( )( ) II II Z q 12.3 kPasin 2δ sin 2 1.1479 radians 2 2 1.47 kPa ∆σ = = ∗π π =  Determine ∆σZ caused by zone III as shown in Figure 5.2.  From Equation (5.25) and Fig. 5.7, ( )15 m 2 5.5 ma + 2barctan arctan 1.3168 radians z 6.75 m δ  + ∗ = = =         From Equation (5.26), ( ) ( )2 15 m 2 5.5 m2a+2barctan arctan 75.45 0.0908 radians z 6.75 m α δ  ∗ + ∗ = − = − =        o  From Equation (5.24), ( ) III III Z q x sin 2 2 0.5 a 12.3 41 m 0.0908 radians sin 2 1.3168 radians 2 0.5 15 m 0.02 kPa  ∆σ = α − δ π ∗   = − ∗ π ∗  = Note that qIII = qII  Determine ∆σZ caused by zone I as shown in Figure 5.2.  From Equation (5.28) and Fig. 5.8, a 15 marctan arctan 1.1479 radians z 6.75 m δ    = = =        From Equation (5.29), a+2b 15 m (2*5.5 m)arctan arctan 65.77 0.1689 radians z 6.75 m α δ +   = − = − =       o

7-8  From Equation (5.27), ( ) I I Z q sin cos 2 16.6 = [0.1689 radians sin 0.1689 radians cos(0.1689 radians + (2*1.1479 radians))] 0.20 kPa  ∆σ = α + α α + δ π +π =  Determine the total increase in vertical stress at the edge of the embankment, ∆σZ@edge.  From Equation (5.30), I II IIIZ@edge Z Z Z 0.20 kPa 1.47 kPa 0.02 kPa 1.69 kPa ∆σ = ∆σ + ∆σ + ∆σ = + + = • Determine geostatic or insitu effective vertical stress at mid-height of each sublayer. This is the preconstruction effective vertical stress, σ′vo . For Layer 5, ( ) ( )3 3vo sat wγ γ z = 16 kN/m 9.81 kN/m 6.75 m 41.78 kPa′σ = − ∗ − ∗ = • Determine the postconstruction effective vertical stress, at mid-height of each sublayer, σ′vf . For Layer 5,  At the center of the embankment, from Equation (5.32), vf vo z vo Z@center 41.78 kPa 14.66 kPa 56.44 kPa ′ ′ ′ ′σ = σ + ∆σ = ∆σ + ∆σ = + =  At edge of embankment, from Equation (5.32), vf vo Z vo Z@edge 41.78 kPa 1.69 kPa 43.47 kPa ′ ′ ′ ′σ = σ + σ = σ + ∆σ = + = Note the use of geofoam results in a small increase in effective vertical stress after construction. • Determine settlement due to the consolidation of the soft clay, Sp. For Layer 5,

7-9  From Figure 7.1, OCR = p vo 1 ′σ =′σ . Therefore, p vo1 1 41.78 kPa 41.78 kPa′ ′σ = ∗σ = ∗ =  At the center of the embankment OCR = 1, therefore, use Equation (5.3) to determine Sp. For Layer 5, c vf 5 o o p CSp L log 1+e 0.35 56.44 kPa 1.5 m log 0.0254 m 1 1.7 41.78 kPa = 25.4 mm ′σ= ′σ = ∗ ∗ =+  At the edge of the embankment, OCR = 1, therefore, use Equation (5.3) to determine Sp. For Layer 5, 5 0.35 43.47 kPaSp 1.5 m log 0.0033 m 1 1.7 41.78 kPa 3.3 mm = ∗ ∗ =+ = • Tables 7.1 and 7.2 provide a summary of Sp values for each layer and total Sp at the center and edge of the embankment, respectively. • Determine settlement due to secondary compression. From Figure 7.1, c o p o c C 0.04 C 0.35 e 1.7 t =15 years L 15 m C α = = = =  For roadway embankments, the critical time for obtaining settlement estimates is typically the life of the pavement system. A maximum settlement of 400 mm during a 20 year duration is allowed as part of the design requirements for this embankment. Therefore, use t = 20 years.  From Equation (5.7),

7-10 ( ) ( )c c s o o p C / C C 0.04 0.35t 20 yrsS L log = 15 m log 1 e t 1 1.7 15 yrs 0.0097 m = 9.7 mm α ∗ ∗  = ∗ ∗  + +   = Since secondary compression is a function of time and not on effective stress, the magnitude of secondary consolidation will be the same at both the center and edge of the embankment. Table 7.1 Summary of settlement due to consolidation of the soft clay at the center of the trapezoidal embankment. Table 7.2 Summary of settlement due to consolidation of the soft clay at the edge of the trapezoidal embankment. • Determine the total settlement due to consolidation and secondary compression of the soft clay foundation using Equation (5.8),  At the center of the embankment, Total p sS = S + S = 369.4 mm + 9.7 mm = 379.1 mm  At the edge of the embankment, Total p sS = S + S = 32.5 mm + 9.7 mm = 42.2 mm For an embankment consisting of soil fill instead of EPS blocks with a total (moist) unit weight of 19 kN/m3, the total settlement is expected to be 1,113 and 214 mm at the center and edge of the embankment, respectively. Therefore, in order to meet the maximum allowed settlement requirement for this embankment of 400 mm, the fill mass can either consist entirely of EPS blocks or a combination of EPS blocks and soil fill. For this example, the fill mass is assumed to only consist of EPS blocks. • Settlement due to long-term vertical deformation (creep) of the fill mass will be negligible if the applied vertical stress is such that it produces an immediate geofoam

7-11 strain of less than 1 percent. The immediate or elastic vertical strain can be estimated from Equation (5.9). However, load bearing design, which is performed as part of internal stability, is based on selecting an EPS block that will provide an immediate vertical strain of less than 1 percent. Therefore, it can be assumed that long-term vertical deformation of the EPS blocks will be negligible. STEP 5 – BEARING CAPACITY • Determine the normal stress applied by the pavement system at the top of the embankment, σn,pavement. σn,pavement = qpavement determined in Step 4 = 12.2 kPa • Determine the normal stress applied by the traffic surcharge at the top of the embankment. σn,traffic = γsoil fill * 0.61 m per (1) If γsoil fill =18.9 kN/m3 as an estimate, σn,traffic = 18.9 kN/m3 *0.61 m = 11.5 kN/m2 • From Equation (5.44), Tw =11 m, and γEPS = 1 kN/m3, the minimum required undrained shear strength to satisfy a factor of safety of 3 is: ( ) ( ) ( ) ( ) n, pavement n, traffic W EPS EPS u W EPS 3 *T γ *T3s * 5 T +T 2 1 kN/m *4.4 m12.2 kPa 11.5 kPa *11 m3 * 5 11 m 4.4 m 2 11.48 kPa   σ +σ = +        +  = +  +    = From Figure 7.1, su foundation soil = 15 kPa su foundation soil > su REQ so the factor of safety against bearing capacity failure of 3 is exceeded.

7-12 Alternatively, Figure 5.10 can be used. For TEPS = 4.4 m and an 11 m roadway width, Figure 5.10 indicates that s u REQ = 15.3 kPa. Thus, Figure 5.10 indicates that the su foundation soil of 15 kPa does not meet the factor of safety against bearing capacity failure of 3. However, Figure 5.10 is based on a Tpavement of 1,000 mm and γpavement of 20 kN/m3 or σn,pavement of 21.5 kPa. In this example problem, the preliminary pavement system also has a γpavement of 20 kN/m3 but Tpavement is 610 mm or σn,pavement of 12.2 kPa. Therefore, Figure 5.10 yields a more conservative su REQ and Equation 5.44 provides a better estimate su REQ. STEP 6 – EXTERNAL SLOPE STABILITY • Determine the static external slope stability factor of safety. For the roadway width of 11 m, Fig. 5.14 can be used. For su = 15 kPa, Fig. 5.14 indicates that the factor of safety exceeds 1.5 for both TEPS = 3.1 m and TEPS = 6.1 m. Therefore for TEPS = 4.4 m, the factor of safety will exceed the required 1.5. STEP 7 – EXTERNAL SEISMIC STABILITY • As shown in the Design Requirements of Step 1, a 10 percent probability of exceedance in 50 years is required for this example. • Select a peak horizontal bedrock acceleration, arock, with a 10 percent probability of exceedance in 50 years. From the USGS zip code earthquake ground motion hazard on the USGS website (www.USGS.gov),  arock = 0.04 g for Urbana, Illinois. • Estimate the ground surface acceleration. This is the acceleration at the base of the embankment, abase. Because the foundation soil consists of soft clay, Figure 5.18 can be used.  abase = 0.13 g

7-13 • Estimate the acceleration at the top of the embankment, aemb. As presented in the External Seismic Stability of Trapezoidal Embankments section of Chapter 5, the EPS- block geofoam can be assumed to behave as a deep cohesionless soil. Fig. 5.17 can be used to estimate aemb from abase.  aemb = 0.09 g • Estimate the acceleration at the center of gravity of the slide mass as determined from the critical static failure surface. As noted in Chapter 5, if a circular failure surface is used for the external static stability analysis, the center of gravity of the sliding mass is usually located near the center or mid-height of the slide mass. For preliminary analysis, the acceleration at the base of the embankment can be used for external seismic stability analysis if the site has soft soil and the bedrock acceleration has been corrected for amplification through the soft soil. • Estimate the horizontal seismic coefficient, kh, at the center of gravity of the slide mass. As indicated previously, the acceleration at the base of the embankment can be used to provide a conservative estimate of the horizontal seismic coefficient, kh, for preliminary design.  baseh a 0.13 gk = 0.13g g= = • Determine the pseudo-static factor of safety, FS′, for the critical static failure surface and ensure that it exceeds 1.2. For an embankment with a top width of 11m and a kh = 0.13, Figures 5.22 and 5.23 can be used to obtain an estimate of FS′. For a 3H:1V embankment at su = 15 kPa and TEPS = 4.4 m, Figure 5.22 provides a FS′ = 1.5 for a kh = 0.1 and Figure 5.23 provides a FS′ = 1.1 for a kh=0.2. Therefore, by linear interpolation, at kh = 0.13, the FS′ will be approximately 1.2, which meets the required FS′.

7-14 STEP 8 – HYDROSTATIC UPLIFT (FLOTATION) • Determine the weight of the EPS-block geofoam, WEPS. For simplicity, assume the EPS blocks extend for the full height of the embankment, i.e., TEPS = H. The 100-year flood water level is expected to be 1.12 m. It is anticipated that water will only accumulate on one side of the embankment,  ( ) ( )w w 3 EPS EPS Dry H T B 5 m 11 m + 41 m W 0.2 kN/m 2 2 = 26 kN/m of roadway γ+= ∗ = ∗ • Determine the vertical weight component of water on the embankment face above the base of the embankment on the accumulated water side, Ww. w total total w 1 shW (h+S ) (h+S ) 2 sv γ = ∗ ∗ ∗ ∗   where sh:sv is the horizontal to vertical slope change of the embankment. 3 w 1 3W (1.12 m + 0.38 m) (1.12 m + 0.38 m) 9.81 kN/m 33.1 kN/m of roadway. 2 1  = ∗ ∗ ∗ ∗ =   • Determine the additional overburden force required above the EPS blocks to obtain a factor of safety of 1.2, OREQ. From Equation (5.74), ( ) ( ) REQ w total w EPS w 3 1O 1.2 γ (h+S ) B W W 2 1 = 1.2 9.81 kN/m (1.12 m+0.38 m) 41 m 26 kN/m + 33.1 kN/m 2 = 302.9 kN/m of roadway     = ∗ ∗ ∗ ∗ − +           ∗ ∗ ∗ ∗ −       Alternatively, for a 3H:1V embankment slope with no tailwater, Figure 5.49 can be used to estimate OREQ. For a road width = 11 m, H = 5 m, totalh+SAccumulated water level 1.12 m + 0.38 m 1.5 0.3 Embankment height H 5 m 5 = = = =

7-15 Figure 5.49 indicates OREQ = 310 kN/m of roadway • Determine if the pavement system and soil cover provide an adequate overburden force to resist hydrostatic uplift.  Determine the weight of the soil cover, Wcover. From Equation (5.64), EPS cover cover cover 3 T TW 2 γ sin cos 4.4 m 0.4 m = 2 18.8 kN/m sin 18.4 cos 18.4 = 220.1 kN/m of roadway  = ∗ ∗ ∗ θ θ   ∗ ∗ ∗  o o  Because the calculation of WEPS is based on the assumption that the EPS blocks extends the full height of the embankment, the weight of the EPS equivalent to the height of the pavement system must be subtracted from the total overburden weight. Therefore, use Equation (5.67). ( ) ( ) ( ) ( ) REQ pavement pavement w EPS pavement w cover 3 3 O T T T T W 302.9 kN/m 20 kN/m 0.6 m 11 m 0.2 kN/m 0.6 m 11 m 220.1 kN/m γ γ< ∗ ∗ − ∗ ∗ + < ∗ ∗ − ∗ ∗ + 302.9 kN/m < 350.8 kN/m of roadway. Therefore, the pavement system and soil cover provide sufficient overburden force to resist hydrostatic uplift. STEP 9 – TRANSLATION DUE TO WATER (EXTERNAL) • Determine the lowest interface friction angle, δ, between the EPS/foundation soil or, if a separation material is placed between the EPS fill and foundation soil, the lowest interface friction between the EPS/separation material and separation material/foundation soil. The exact type of separation material, if one is required, will typically not be known until construction begins. Four possible interface cases between the EPS blocks and the foundation soil, which are summarized in Table 7.3, will be considered herein. As shown in Table 7.3, the lowest and most critical interface δ is 20 degrees. Therefore, δ = 20 degrees will be used in the analysis of translation due to water and translation due to wind in Steps 9 and 10, respectively.

7-16 Table 7.3 Summary of interface friction angles, δ, considered between the EPS blocks and the foundation soil. • Determine the additional overburden force required above the EPS blocks to obtain a factor of safety against translation due to water of 1.2, OREQ. From Equation (5.77) and the values of WEPS = 26 kN/m of roadway and WW = 33.11 kN/m of roadway determined in Step 8, ( ) ( ) ( ) ( ) ( ) 2 w total REQ total w w EPS w 2 3 11.2 γ (h+S ) 12O (h+S ) γ B W W tan 2 11.2 9.81 kN/m (1.12 m+0.38 m) 12 = (1.12 m+0.38 m) 9.81 kN/m 41 m 2tan 20 26 kN/m 33.11 kN/ m δ   ∗    = + ∗ ∗ − −     ∗     + ∗ ∗   − − o 278.9 kN/m of roadway= Alternatively, for a 3H:1V embankment slope with no tailwater, Fig. 5.54 can be used to estimate OREQ. For a road width = 11 m, H = 5 m, totalh + SAccumulated water level 1.12 m + 0.38 m 1.5 0.3, 20 Embankment height H 5 m 5 δ= = = = = o OREQ = 300 kN/m of roadway. • Determine if the pavement system and soil cover will provide an adequate overburden force to resist translation due to water.  Determine the weight of the soil cover, Wcover. Wcover was determined to be 220.14 kN/m of roadway in Step 8.  Because the calculation of WEPS is based on the assumption that the EPS blocks extend the full height of the embankment, the weight of the EPS equivalent to the height of the pavement system must be subtracted from the total overburden

7-17 weight. Therefore, use Equation (5.67). Note that the right side of the equation was determined in Step 8 to be 350.82 kN/m of roadway. Therefore, ( ) ( )REQ pavement pavement w EPS pavement w coverO T T T T Wγ γ< ∗ ∗ − ∗ ∗ + ( ) ( )3 3278.9 kN/m < 20 kN/m 0.6 m 11 m 0.2 kN/m 0.6 m 11 m 220.1 kN/m ∗ ∗ − ∗ ∗ + 278.9 kN/m < 350.8 kN/m of roadway. Therefore, the pavement system and soil cover will provide sufficient overburden force. STEP 10 – TRANSLATION DUE TO WIND (EXTERNAL) • As presented in Chapter 5, it is recommended that the translation due to wind failure mechanism not be considered until further research is performed on the applicability of Equations (3.4) and (3.5) to EPS-block geofoam embankments. However, the wind loading failure mechanism will be evaluated herein to demonstrate the use of the applicable equations and Figure 5.58. • Determine the upwind and downwind pressures, pu and pD, respectively, on the sides of the embankment.  Obtain the design wind speed, V. From ANSI/ASCE 7-95 (8), V = 40 m/s  From Equation (3.4), pU = 0.75 V2 sin θu = 0.75 (40 m/s)2 sin 18.4˚ = 378.8 kPa  From Equation (3.5), pD = 0.75 V2 sinθD = 0.75 (40 m/s)2 sin 18.4˚ = 378.8 kPa • Determine the upwind and downwind force, RU and RD, respectively, on the sides of the embankment.  RU = pU * H = 378.8 kPa * 5 m = 1,894.0 kN/m of roadway  RD = pD * H = 378.8 kPa * 5 m = 1,894.0 kN/m of roadway

7-18 • Determine the additional overburden force required above the EPS blocks to obtain a factor of safety against translation due to wind of 1.2, OREQ. From Equation (5.81) and the value of WEPS = 26 kN/m of roadway determined in Step 8, ( ) ( )U D REQ EPS 1.2 R R 1.2 1,894.0 kN/m + 1,894.0 kN/m O W 26 kN/m tan tan 20δ ∗ + ∗= − = −o 12,462.9 kN/m of roadway= Alternatively, Fig. 5.58 can be used to estimate OREQ. For a 3H:1V, 5 m high embankment with V=40 m/s and δ = 20 degrees, OREQ = 12,500 kN/m of roadway. • Determine if the pavement system and soil cover provide adequate overburden force.  Determine the weight of the soil cover, Wcover. Wcover was determined to be 220.1 kN/m of roadway in Step 8.  Because the calculation of WEPS is based on the assumption that the EPS blocks extend the full height of the embankment, the weight of the EPS equivalent to the height of the pavement system must be subtracted from the total overburden weight. Therefore, use Equation (5.67). Note that the right side of the equation was determined in Step 8 to be 350.8 kN/m of roadway. Therefore, ( ) ( )REQ pavement pavement w EPS pavement w coverO T T T T Wγ γ< ∗ ∗ − ∗ ∗ + ( ) ( )3 312,462.9 kN/m < 20 kN/m 0.6 m 11 m 0.2 kN/m 0.6 m 11 m 220.1 kN/m ∗ ∗ − ∗ ∗ + 12,462.9 kN/m is not < 350.8 kN/m of roadway. Therefore, the pavement system and soil cover will not provide sufficient force. However, as indicated previously, it is recommended that the translation due to wind failure mechanism not be considered until further research is performed on the applicability of Equations (3.4) and (3.5). The wind failure mechanism was

7-19 evaluated herein to demonstrate the use of these equations. Thus, the wind failure mechanism will not be considered in the design of this embankment. STEP 11 – TRANSLATION DUE TO WATER (INTERNAL) • Determine the levels within the embankment that will be used to analyze the potential for translation due to water. The 100-year flood water level is 1.12 m. The thickness of EPS blocks typically range between 610 mm to 1,000 mm. Because the 100-year flood water level is greater than the typical thickness range of EPS blocks, the translation due to water failure mechanism should be evaluated. Check the potential for sliding at h=0.37 m above the embankment and foundation soil interface.  Accumulated water level = h + Stotal = 0.37 m + 0.38 m = 0.75 m  From Fig. 5.47, determine the new geometric parameters. Tw = 11m remains the same H = 5 m – 0.75 m = 4.25 m w w sh 3 B = T + 2 H 11 m + 2 4.25 m sv 1 = 36.5 m       ∗ ∗ = ∗ ∗             • Determine the weight of EPS-block geofoam, WEPS, for the new embankment height to be analyzed. For simplicity, assume the EPS blocks extend the full height of the new embankment height, i.e., TEPS = H. ( ) ( )w w 3 EPS EPS Dry H T B 4.25 m 11 m + 36.5 m W 0.2 kN/m 2 2 γ+= ∗ = ∗ = 20.2 kN/m of roadway. • Determine the vertical component of water weight on the embankment face above the base of the embankment on the accumulated water side, Ww. 3 w total w 1 sh 1 3W h (h+S ) γ 0.75 m 0.75 m 9.81 kN/m 2 sv 2 1    = ∗ ∗ ∗ ∗ = ∗ ∗ ∗ ∗       = 8.3 kN/m of roadway.

7-20 • Assume the interface friction angle, δ, between the EPS blocks is 30 degrees (from Chapter 2). • Determine the additional overburden force required above the EPS blocks to obtain a factor of safety against translation due to water of 1.2, OREQ. From Equation (5.77), ( ) ( ) ( ) 2 w total REQ total w w EPS w 11.2 γ (h+S ) 12O (h+S ) γ B W W tan 2δ   ∗    = + ∗ ∗ − −   ( ) ( ) ( ) 3 2 3 11.2 9.81 kN/m (0.75 m) 12 0.75 m 9.81 kN/m 36.5 m 20.2 kN/m 2tan 30 8.3 kN/m   ∗    = + ∗ ∗ −   − o = 111.5 kN/m of roadway. Alternatively, for a 3H:1V embankment slope with no tailwater, Fig. 5.54 can be used to estimate OREQ. For a road width = 11m, H = 4.25, totalh+SAccumulated water level 0.75 0.18, 30 Embankment height H 4.25 δ= = = = o OREQ = 120 kN/m of roadway. • Determine if the pavement system and soil cover provide adequate overburden force.  Determine the weight of the soil cover, Wcover. Thickness of EPS, TEPS = H – Tpavement = 4.25 m – 0.6 m = 3.65 m.  From Equation (5.64), 3EPS cover cover cover T T 3.65 m 0.4 mW 2 γ 2 18.8 kN/m sinθ cosθ sin18.4 cos18.4    = ∗ ∗ ∗ = ∗ ∗ ∗      o o = 182.6 kN/m of roadway.  Because the calculation of WEPS is based on the assumption that the EPS blocks extend the full height of the embankment, the weight of the EPS equivalent to the

7-21 height of the pavement system must be subtracted from the total overburden weight. Therefore, use Equation (5.67). ( ) ( )REQ pavement pavement w EPS pavement w coverO T T T T Wγ γ= ∗ ∗ − ∗ ∗ + 111.5 kN/m < (20 kN/m³ * 0.6 m * 11 m) – (0.2 kN/m³ * 0.6 m *11 m) + 182.6 kN/m 111.5 kN/m < 313.2 kN/m of roadway. Therefore, the pavement system and soil cover will provide sufficient overburden force. STEP 12 – TRANSLATION DUE TO WIND (INTERNAL) • As presented in Chapter 5, it is recommended that the translation due to wind failure mechanism not be considered until further research is performed on the applicability of Equations (3.4) and (3.5) to EPS-block geofoam embankments. However, the wind loading failure mechanism will be evaluated herein to demonstrate the use of the applicable equations and Figure 5.58. • Determine the levels within the embankment that will be used to analyze the potential for translation due to wind. Determine the potential for sliding at mid-height of the embankment.  From Figure 5.57, determine new geometric parameters. w w w 1H = 5 m = 2.5 m 2 T 11 m (remains the same) sh 3B T 2 H 11 m + 2 2.5 m sv 1 = 26 m ∗ =       = + ∗ ∗ = ∗ ∗             • Determine the upwind and downwind pressures, pU and pD, respectively, on the sides of the embankment.  Obtain a design wind speed, V. From Step 10, V = 40 m/s

7-22  From Equation (3.4), pU = 0.75 V² sin θu = 0.75 (40 m/s) ² sin 18.4˚ = 378.8 kPa  From Equation (3.5), pD = 0.75 V² sin θD = 0.75 (40 m/s) ² sin 18.4˚ = 378.8 kPa • Determine the upwind and downwind force, RU and RD, respectively, on the sides of the embankment.  RU = pU * H = 378.8 kPa * 2.5 m = 947.0 kN/m  RD = pD * H = 378.8 kPa * 2.5 m = 947.0 kN/m • Determine weight of EPS-block geofoam, WEPS, for the new embankment height to be analyzed. For simplicity, assume the EPS blocks extend the full height of the new embankment height, i.e., TEPS = H. ( ) ( )w w 3 EPS EPS Dry H T B 2.5 m 11 m + 26 m W 0.2 kN/m 2 2 γ+= ∗ = ∗ = 9.2 kN/m of roadway. • From Chapter 2, assume the interface friction angle, δ, between the EPS blocks is 30 degrees. • Determine the additional overburden force required above the EPS blocks to obtain a factor of safety of 1.2, OREQ. From Equation (5.81), ( ) ( )U D REQ EPS 1.2 R R 1.2 947.0 kN/m + 947.0 kN/m O W 9.2 kN/m tan tan 30δ ∗ + ∗= − = −o = 3,927.4 kN/m of roadway. Alternatively, Fig. 5.58 can be used to estimate OREQ. For a 3H:1V, 2.5 m high embankment with V = 40 m/s and δ = 30 degrees, OREQ = 3,900 kN/m of roadway. • Determine if the pavement system and soil cover provide an adequate overburden force.

7-23  Determine the weight of the soil cover, Wcover. Thickness of EPS, TEPS = H – Tpavement = 2.5 m – 0.6 m = 1.9 m  From Equation (5.64), 3EPS cover cover cover T T 1.9 m 0.4 mW 2 γ 2 18.8 kN/m sin cos sin18.4 cos18.4    = ∗ ∗ ∗ = ∗ ∗ ∗  θ θ    o o = 94.7 kN/m of roadway. • Because the calculation of WEPS is based on the assumption that the EPS blocks extend the full height of the embankment, the weight of the EPS equivalent to the height of the pavement system must be subtracted from the total overburden weight. Therefore, use Equation (5.67). ( ) ( )REQ pavement pavement w EPS pavement w coverO T T T T Wγ γ= ∗ ∗ − ∗ ∗ + 3,927.4 kN/m < (20 kN/m3 * 0.6 m * 11 m) – (0.2 kN/m3 * 0.6 m * 11 m) + 94.7 kN/m 3,927.4 kN/m is not < 225.4 kN/m. Therefore, the pavement system and soil cover will not provide sufficient force. However, as presented in Chapter 5, it is recommended that the translation due to wind failure mechanism not be considered until further research is performed on the applicability of Equations (3.4) and (3.5). The wind failure mechanism was evaluated herein to demonstrate the use of these equations. Thus, the wind failure mechanism will not be considered in the design of this embankment. STEP 13 – SEISMIC STABILITY (INTERNAL) • Identify the critical interface friction angle, δ, for each of the three failure modes shown in Fig. 6.2 and briefly described below.  Mode I: Determine the lowest interface friction angle between the pavement system/EPS or, if a separation material is placed between the pavement system and EPS blocks, the lowest interface friction between the pavement

7-24 system/separation material and separation material/EPS. The type of separation material, if one is required, will typically not be initially known. Four possible interface cases between the pavement system and EPS blocks, which are summarized in Table 7.4 will be considered herein. As shown in Table 7.4, the critical δ is 25 degrees.  Mode II: From testing described in Chapter 2, δ = 30 degrees for sliding along an EPS/EPS interface.  Mode III: A δ = 20 degrees was determined to be the critical interface friction angle between the EPS blocks and the foundation soil in Step 9. Table 7.4 Summary of geosynthetic interface friction angles, δ, considered between the pavement system and the EPS blocks. • Estimate the horizontal seismic coefficient, kh, at the center of gravity of the slide mass of each failure mode.  Mode I: The center of gravity along the height of the slide mass corresponds to approximately the mid-height of the pavement system, Zcenter = ½ * Tpavement = ½ * (0.61 m) = 0.305 m By linear interpolation, the acceleration at the center of gravity of the pavement system is close to the acceleration at the top of the embankment, aemb, of 0.09 g, which was determined in Step 7. Therefore, the horizontal seismic coefficient for Mode I is emb h a 0.09 gk 0.09 g g = = =  Mode II: Based on the assumption that the bottom of the failure surface is located near the base of the embankment, the center of gravity along the height of the slide mass, Zcenter, is approximately,

7-25 ( ) ( )pavement pavement EPS EPS center pavement EPS Z Z γ Z γ γ γ ∗ + ∗= + where pavementZ and EPSZ is the vertical distance from the top of the embankment to the center of gravity of the pavement system and EPS fill, respectively. ( ) ( )3 3 center 3 3 0.305 m 20 kN/m 2.81 m 1 kN/m Z 0.424 m 20 kN/m 1 kN/m ∗ + ∗= =+ By linear interpolation, the acceleration at the center of gravity is close to the acceleration at the top of the embankment, aemb, of 0.09 g, which was determined in Step 7. Therefore, the horizontal seismic coefficient for Mode II is approximately the same as for Mode I, i.e., kh = 0.09.  Mode III: The horizontal seismic coefficient at the center of gravity will be similar to that of Mode II, i.e., kh = 0.09, because the failure surface for Mode III is near the base of the embankment as in Mode II.  A summary of the critical interface friction angles and horizontal seismic coefficients for the three failure modes is presented in Table 7.5. Table 7.5 Summary of critical interface friction angles and horizontal seismic coefficients for the three internal seismic stability failure modes. • Determine the pseudo-static factor of safety against internal seismic stability, FS′, and ensure that it exceeds 1.2. An estimate of FS′ for preliminary design can be obtained from Fig. 6.6. The FS′ relationship for kh = 0.10 can be used to obtain an estimate of FS′ at kh = 0.09. A summary of FS′ values is shown in Table 7.5. Note from Fig. 6.6 that it is only necessary to determine FS′ for Mode I and the mode with the critical δ between Mode II and III. As shown in Table 7.5, FS′ exceeds 1.2 for all three potential failure modes.

7-26 STEP 14 – LOAD BEARING • Sub-Step 1: Estimate traffic loads. For this section of roadway, use AASHTO H 20-44 standard loading. Rear axle load = 106.8 kN Load per dual set, LLD = 106.8 kN 53.4 kN 2 = • Sub-Step 2: Add impact allowance to traffic loads. Use impact coefficient, I, of 0.3 and Equation (6.3), QD = LLD * (1 + I) = 53.4 kN * (1+0.3) = 69.4 kN • Sub-Step 3: Estimate traffic load stresses on top of the EPS blocks.  Consider an asphalt concrete pavement with and without a concrete separation layer. Also consider an asphalt thickness of 76 and 178 mm for preliminary design. The vertical stress due to traffic loads, σLL, at the top of the EPS blocks can be obtained from Fig. 6.15 for an asphalt concrete pavement system and from Fig. 6.17 for an asphalt concrete pavement system with a 102 mm concrete separation layer. A summary of σLL values per dual tire set, is presented in Table 7.6.  Determine if the applied vertical stresses overlap between the two interior dual tire sets shown as No. 2 and 3 in Figure 7.3. These two dual tire sets are the closest and thus the stresses exerted by these two dual tires will overlap first. The composite pavement system with the 76 mm asphalt layer will be used to demonstrate the procedure. QD = 69.4 kN from Sub-step 2 and σLL = 19 kPa as determined in Sub- step 3 and shown in Table 7.6. Determine circular contact area from Equation (6.6),

7-27 2D D CD LL Q Q 69.4 kNA 3.65 m q 19 kPaσ= = = = Figure 7.3 Cross-section of rear axle of two standard H 20-44 trucks on the proposed 11 m wide roadway embankment. Determine an equivalent rectangular loaded area using Fig. 6.18, 2 CDAArea 3.65 mL = 2.64 m 0.5227 0.5227 0.5227 ′ = = = L = 0.8712 L = 0.8712 2.64 m = 2.3 m B = 0.6 L = 0.6 2.64 1.58 m ′∗ ∗ ′∗ ∗ = From Fig. 7.4, stress overlap occurs if center-to-center wheel spacing ≤ B where B is 0.6*L′ or 1.58 m for this example. The actual center-to-center spacing between the two dual tire sets (Set No. 2 and 3) is 0.61 m * 2 = 1.22 m. Because the actual center-to-center spacing of the two dual tire sets is less than B, i.e., 1.22 < 1.58 m, the stresses overlap (see Fig. 7.4). Overlap = 1.58 m – 1.22 m = 0.36 m. The combined rectangular width = (2 * B) – 0.36 m = (2 * 1.58 m) – 0.36 m = 2.80 m The combined rectangular area = 2.8 m * 2.3 m = 6.44 m² The combined load of the two dual tire sets = 2 * 69.4 kN = 138.8 kN The combined vertical stress = 2 138.8 kN 21.55 kPa 6.44 m = Figure 7.4. Determination of stress overlap between two sets of dual tires.  Determine if stresses overlap between an interior dual tire set and an exterior dual tire set, e.g., Set No. 1 and 2 (see Figure 7.3), or, if stresses overlap between the two interior dual tires, and an exterior dual tire set, e.g., Set No. 1 and

7-28 Combined set No. 2 and 3. The composite pavement system with the 76 mm asphalt layer and the flexible pavement system with the 178 mm asphalt layer will be used to demonstrate the procedure. For the composite pavement system with the 76 mm asphalt layer, the center-to-center spacing between the combined interior dual tires (Combined Set No. 2 and 3) and the exterior dual tire set (Set No. 1) is 2.44 m. The spacing between the two loaded area is ( ) ( )2.44 m 0.5 2.8 m 0.5 1.58 m 0.25 m. − ∗ + ∗ =  Because there is spacing between the two loaded areas, no overlap occurs. For the flexible pavement system with the 178 mm asphalt layer, the stresses from the two interior dual tires (Set No. 2 and 3) do not overlap. Therefore, since the actual center-to-center spacing between one interior dual tire set and an exterior dual tire set (between Set No. 1 and 2) is greater than B, i.e., 1.83 m > 1.58 m, stress overlap does not occur.  A summary of combined stresses, if applicable, is presented in Table 7.6. • Sub-Step 4: Estimate gravity stresses at top of EPS blocks.  Tpavement = 0.61 m, γpavement = 20 kN/m³  From Equation (6.8), σDL = Tpavement * γpavement = 0.61 m * 20 kN/m³ = 12.2 kPa • Sub-Step 5: Calculate the total vertical stresses at top of EPS blocks.  From Equation (6.9), σtotal = σLL + σDL

7-29 A summary of σtotal for the various pavement systems is presented in Table 7.6. If applied stresses were found to overlap between adjacent tire sets in Sub-step 3, the largest σLL is used. • Sub-Step 6: Determine minimum required elastic limit stress for EPS under pavement system.  Use factor of safety, FS = 1.2.  From Equation (6.10), σe ≥ σtotal * FS = σtotal * 1.2. A summary of the required σe for the various pavement systems is presented in Table 7.6. • Sub-Step 7: Select appropriate EPS block to satisfy the required EPS elastic limit stress for placement underneath the pavement system.  Use Table 6.2 to select an EPS block that exhibits an elastic limit stress greater than or equal to the required σe determined in Sub-step 6. A summary of the required EPS block types is presented in Table 7.6. • Sub-Step 8: Select preliminary pavement system type and determine if a separation layer is required.  A summary of a cost analysis for the four pavement systems analyzed is presented in Table 7.7. The unit costs for asphalt and granular base were obtained from (10). The unit cost for the concrete separation layer was obtained from the Indiana State Route 109 and the Utah Interstate-15 case histories presented in Chapter 11. The cost of the EPS-block geofoam was obtained from Chapter 12. The cost analysis shows the following results: (1) an asphalt concrete pavement system without a concrete separation layer is more cost effective than a composite pavement system that includes a concrete separation layer, (2) a

7-30 flexible pavement system consisting of 178 mm thick asphalt concrete and 432 mm granular base is the most cost effective of the four pavement systems considered, and (3) EPS70 is required for the most cost effective flexible pavement system. • Sub-Step 9: Estimate traffic stresses at various depths within the EPS blocks. This sub-step starts the second phase of the load bearing analysis because it focuses on the load bearing capacity of the EPS below a depth of 610 mm in the EPS whereas the first phase focused on the load bearing capacity of the EPS within the upper 610 mm of the EPS mass.  Summarize the equivalent rectangular loaded areas on top of the EPS blocks from each dual tire set or from combined dual tire sets if the applied stresses were found to overlap in Sub-step 3. Because the pavement system with an asphalt thickness of 178 mm and a 432 mm thick granular base was determined to be the most economical, only this pavement system is analyzed. From Sub- step 3 and from Table 7.6, no stress overlap occurs within the pavement system for this pavement system. The equivalent rectangular loaded areas on top of the EPS blocks are shown in Figure 7.5. Figure 7.5 Plan view of equivalent rectangular loaded area from live load stresses on top of the EPS blocks. Table 7.6. Summary of applied vertical stresses on top of the EPS blocks, minimum required elastic limit stress, and required EPS type for the pavement systems analyzed. Table 7.7. Cost comparison between the pavement systems analyzed.

7-31  Determine the depths within the EPS where stress overlap occurs. Using the 1H:2V assumed stress distribution method, stress overlap between two adjacent loaded areas occurs at a depth Z equal to the spacing, S, between the two loaded areas. Therefore, stress overlap between the adjacent dual tire sets shown in Fig. 7.5 will occur at S = Z0 = 0.12 m and 0.73 m. However, stresses will overlap first at the smaller Z0 value. At Z0 = 0.12 m stresses between the two sets of interior tires (Set No. 2 and 3) overlap.  Estimate traffic stresses at any depth where stress overlap occurs. If stress overlap occurs at Z0 < 0.61 m, the EPS selected in Sub-step 7 for the preliminary pavement system selected in Sub-step 8 must be checked to verify that the EPS can support the combined stresses.  From Figures 7.5 and 6.19, the combined rectangular loaded area dimensions at Z0 = 0.12 m are combined 2 3 2 0B B B S Z = 1.1 m + 1.1 m + 0.12 m + 0.12 m = 2.44 m = + + + combined 2 0 1 2 3 4L L Z (Note that L L L L 1.61 m) = 1.61 m + 0.12 m = 1.73 m = + = = = = combined 2 3Q Q Q 69.4 kN + 69.4 kN = 138.8 kN= + = From Equation (6.11), the vertical stress induced by traffic loading is ( )( ) ( )( )combinedZ,LL combined combined Q 138.8 kN B L 2.44 m 1.73 m = 32.9 kPa σ = =  From Equation (6.11), the vertical stress caused by traffic loading by the exterior dual tire sets (Set No. 1 and 4) is

7-32 ( )( ) ( )( )Z,LL Q 69.4 kN B+z L+Z 1.1 m + 0.12 m 1.61 m+0.12 m = 32.9 kPa σ = =  Check to ensure stress overlap does not occur between the new combined rectangular loaded area and the adjacent exterior dual tire sets. At Z0 = 0.12 m the width of the loaded area from the two exterior dual tire sets is B1 = B4 = 1.1 m + Z0 = 1.1 m + 0.12 m = 1.22 m Therefore, the spacing between the center combined loaded area and the adjacent exterior dual tire sets is 0.61 m as shown in Fig. 7.6. Therefore, stress overlap will occur at 0.12 m + 0.61 m = 0.73 m below the top of the EPS blocks. Figure 7.6. Equivalent rectangular loaded areas from live load stresses at 0.12 m below the top of the EPS blocks.  Estimate traffic stresses at Z = 0.61 m. From Equation (6.11), the vertical stress caused by the interior combined loaded area from dual tire sets 2 and 3 is ( )( ) ( )( )2,3Z,LL 2,3 2,3 Q B 0.61 m 0.12 m L 0.61 m 0.12 m σ = + − + − ( )( ) 138.8 kN 2.44 m+0.49 m 1.73 0.49 m = + = 21.3 kPa From Equation (6.11), the vertical stress caused by the exterior dual tires, Set 1 and 4, is ( )( ) ( )( )Z,LL Q 69.4 kN 18.28 kPa B+z L+Z 1.1 m +0.61 m 1.61 m+0.61 m σ = = =  Estimate traffic stresses at Z0 = 0.73 m. At this depth, the vertical stresses imposed by the exterior dual tires, Set No. 1 and 4, will overlap with the combined interior loaded area of dual tire sets No. 2 and 3.

7-33 ( )combined 1 2,3 4 1 3B B B B S S 0.73 m 0.12 m= + + + + + − 1.22 m + 2.44 m + 1.22 m + 0.61 m + 0.61 m + (0.61 m) = 6.71 m = combined 1 0L L Z 1.61 m +0.73 m = 2.34 m= + = combined 1 2,3 4Q Q Q Q 69.4 kN + 138.8 kN + 69.4 kN = 277.6 kN= + + = ( )( ) ( )( )combinedZ,LL combined combined Q 277.6 kN 17.68 kPa B L 6.71 m 2.34 m σ = = =  Estimate traffic stresses at 1 m intervals after 0.73 m. Table 7.8 provides a summary of traffic stresses, σLL, within the EPS. Table 7.8. Summary of applied vertical stresses, minimum required elastic limit stress, and required EPS types at various depths within the EPS blocks. • Sub-Step 10: Estimate the gravity stresses at various depths within the EPS blocks. The procedure for estimating the gravity stresses will be illustrated for Z0 = 0.73 m.  Determine the surcharge at the center of the embankment from the pavement system and any road hardware placed on top of the roadway, qt. In this example, no excessive surcharge loads from road hardware are anticipated. Therefore, use Equation (6.19), 3 t pavement pavement pavementq q T 20 kN/m 0.61 m = 12.2 kPaγ= = ∗ = ∗  From Fig. 5.3 and Equation (6.18) determine α, b 5.5 m2 arctan 2 arctan 2.8777 radians Z 0.73 m α    = ∗ = ∗ =        From Equation (6.17), determine the increase in vertical stress due to the pavement system gravity load, σZ,DL.

7-34 ( ) ( )tZ,DL q 12.2 kPasin 2.8777 radians+sin (2.8777 radians)∆σ = α + α =π π = 12.2 kPa  From Equation (6.20), determine the total gravity stress from the pavement system and the EPS blocks, ( ) ( ) ( ) ( )3Z,DL Z,DL EPSZ γ 12.2 kPa 0.73 m 1 kN/mσ = ∆σ + ∗ = + ∗ = 12.93 kPa  Table 7.8 provides a summary of the total gravity stresses within the EPS. • Sub-Step 11: Calculate total stresses at various depths within the EPS blocks. The procedure for determining the total stresses will be shown for Z0 = 0.73 m.  From Equation (6.21), total Z,LL Z,DL 17.68 kPa + 12.93 kPa = 30.61 kPa σ = σ + σ = Table 7.8 provides a summary of the total stresses, σtotal, within the EPS. • Sub-Step 12: Determine the minimum required elastic limit stress for EPS at various depths within the EPS blocks.  The procedure for determining the required elastic limit stress will be shown for Z0 = 0.73 m. From Equation (6.22), e total 1.2 30.61 kPa 1.2 = 36.73 kPaσ ≥ σ ∗ = ∗ • Sub-Step 13: Select appropriate EPS block to satisfy the required EPS elastic limit stress at various depths within the EPS blocks. Table 7.8 provides a summary of EPS block types obtained from Table 6.2 that meet or exceed the minimum required elastic limit stress. Since stress overlap occur at Z0 = 0.12 m, which is less than 0.61 m, the EPS selected in Sub-step 7 for the preliminary pavement system selected in Sub-step 8 must be checked to verify that the EPS can

7-35 support the combined stresses. As shown in Table 7.8, both an EPS70 and EPS50 was selected at Z = 0.61 m for the two dual tire load combinations analyzed. Therefore, the EPS with the larger elastic limit stress, EPS70, is selected to ensure adequate load bearing. This is the same EPS type selected in Sub-step 7 for the preliminary pavement system of 178 mm of asphalt concrete and 432 mm of granular base selected in Sub- step 8. Therefore, an EPS70 can be used directly below the pavement system for a depth of 0.61 m. As shown in Table 7.8, an EPS40 can be used below the EPS70 for depths greater than 0.61 m. STEP 15 – PAVEMENT SYSTEM DESIGN • The local transportation agency prefers a flexible pavement system. The proposed roadway will be located along a low-volume road with an estimated traffic level of 300,000 equivalent single axle loads, i.e., ESAL = 300,000. The local transportation agency would like the pavement designed based on a 75 percent level of reliability. The pavement design is to be based on the AASHTO 1993 design procedure.  Determine the design structural number, SN. From Table 4.2 and based on an EPS70, which will be used for the initial 0.61 m of the embankment below the pavement system, SNREQ = 5.  Verify that the preliminary pavement system will meet the required structural number. It is assumed that the following material layer coefficients have been obtained from the local transportation agency’s design manual. Asphalt concrete a1 = 0.44 Crushed stone aggregate base a2 = 0.14 The preliminary pavement system consists of 178 mm (7 in.) of asphalt concrete and 432 mm (17 in.) of crushed stone base for a total thickness of 610 mm. From Equation (4.1),

7-36 ( ) ( )1 1 2 2SN = a D a D 0.44 7 in. 0.14 17 in. = 5.46 + = ∗ + ∗ SN ≥ SNREQ 5.46 ≥ 5 Therefore, the pavement system consisting of 178 mm of asphalt concrete and 432 mm of granular base meets the required SN and can be used for the roadway. STEP 16 – DETERMINE IF THE FINAL PAVEMENT SYSTEM DESIGN RESULTS IN A SIGNIFICANT CHANGE IN OVERBURDEN COMPARED TO THE PRELIMINARY PAVEMENT SYSTEM DESIGN DEVELOPED IN STEP 2.  The final pavement system determined in Step 15 has the same thickness as the preliminary pavement system of 0.61 m. However, as will be shown in Figure 7.7, if a crown of 2 percent is used for the top of the pavement, the pavement system thickness at the center of the roadway will need to be increased by 110 mm. Therefore, the total pavement system thickness at the center of the roadway will be 720 mm. Consideration should be given to re-checking the design procedure because of this increase in pavement thickness at the centerline of the roadway. However, this re-checking will not be shown here. For this example, the increase in stress due to the crown may not be significant because of stress distribution with depth. STEP 17 –FINAL EMBANKMENT DESIGN. Figure 7.7 provides a cross-section of the proposed design. Figure 7.7 Cross-section of the proposed EPS-block geofoam roadway embankment. DESIGN EXAMPLE 2 – LATERAL PRESSURES ON AN ABUTMENT The design requirements for abutments as well as design examples can be found in (11). The steps of the abutment design procedure are summarized in Chapter 6. The purpose of this design example is to demonstrate how to calculate earth pressures generated by an EPS-block

7-37 geofoam bridge approach fill on an abutment. Determination of earth pressures is required in Step 2 of the abutment design procedure. • Estimate the lateral earth pressures generated by an EPS-block geofoam bridge approach fill on the abutment shown in Figure 7.8. The bridge approach detail is the one used as part of the bridge approach rehabilitation project for the bridge over the N.F. Shoshone river in Wyoming, which was presented in Chapter 11. Figure 7.8 Bridge approach configuration. The passive pressure of the soil in front of the abutment is ignored because of the large displacement required to mobilize the passive resistance. A live load surcharge equal to 0.61 m of earth acts on the surface of the backfill. The weight of the approach slab and sand base is considered as a dead load surcharge. Figure 7.9 provides a summary of loadings applied to the abutment. Figure 7.9 Summary of loadings applied to the abutment. • Determine the horizontal forces generated by the live and dead load surcharges.  Horizontal pressure and force due to live load surcharge. ωL = 0.61 m * γt = 0.61 m * 18.8 kN/m3 = 11.47 kN/m2 2L L 1 1H H = 11.47 kN/m 2.795 m = 3.21 kN/m of wall 10 10 ω ′= ∗ ∗ ∗ ∗ Note that it is assumed that the lateral pressure imposed by the live load surcharge is equal to 1/10 times the vertical stress (12).  Horizontal pressure and force due to the concrete approach slab surcharge. ωD,Conc = 0.305 m * γConc = 0.305 m * 23.6 kN/m3 = 7.2 kN/m2 2 D,Conc D,Conc 1 1H H = 7.2 kN/m 2.795 m=2.01 kN/m of wall 10 10 ω ′= ∗ ∗ ∗ ∗  Horizontal pressure and force due to sand base surcharge. ωD,Sand = 0.205 m * γt = 0.205 m * 18.8 kN/m3 = 3.85 kN/m2

7-38 2 D,Sand D,Sand 1 1H H = 3.85 kN/m 2.795 m=1.08 kN/m of wall 10 10 ω ′= ∗ ∗ ∗ ∗ • Determine the horizontal force generated by the EPS-block geofoam fill. As indicated in Chapter 6, the horizontal force from the EPS blocks is neglected because it is negligible. • Determine the horizontal force generated by the soil backfill behind the EPS/soil interface.  As indicated in Chapter 6, the lateral earth pressure force, PA, generated by the soil behind the EPS/soil interface is conservatively assumed to be transmitted without dissipation through the geofoam to the back of the abutment.  Determine the coefficient of active earth pressure, KA. From Equation (6.23) and based on the friction angle of the EPS/soil interface, δ, equal to the friction angle of the soil φ, the following is obtained: ( ) ( ) ( ) ( )( ) 2 A 1sin -φ sinK sin φ+δ sin φ sin +δ sin    θ   θ =   θ + θ  ( ) ( ) ( ) ( )( ) 2 1sin 45 35 sin 45 0.0173 0.02 sin 35 35 sin 35 sin 45 35 sin 45    −    = = ≈ + + +    o o o o o o o o o  Determine the lateral earth pressure force, PA, from Equation (6.24) ( )22 3A t a1 1P H K 18.8 kN/m 2.795 m 0.022 2γ ′= ∗ ∗ ∗ = ∗ ∗ ∗ = 1.47 kN/m of wall

7-39  It can be seen from Figure 7.9 that the largest horizontal force is applied by the live load surcharge. REFERENCES 1. American Association of State Highway and Transportation Officials, Standard Specifications for Highway Bridges, 16th, American Association of State Highway and Transportation Officials, Washington, D.C. (1996). 2. Refsdal, G., “Frost Protection of Road Pavements.” Frost Action in Soils - No. 26, Committee on Permafrost, ed., Oslo, Norway (1987) pp. 3-19. 3. Jutkofsky, W. S., Sung, J. T., and Negussey, D., “Stabilization of an Embankment Slope with Geofoam.” Transportation Research Record 1736, Transportation Research Board, Washington, D.C. (2000) pp. 94-102. 4. Bartlett, S., Negussey, D., Kimble, M., and Sheeley, M., “Use of Geofoam as Super- Lightweight Fill for I-15 Reconstruction (Paper Pre-Print).” Transportation Research Record 1736, Transportation Research Board, Washington, D.C. (2000). 5. Terzaghi, K., Peck, R. B., and Mesri, G., Soil Mechanics in Engineering Practice, 3rd, John Wiley & Sons, Inc., New York (1996). 6. Koerner, R. M., Designing with Geosynthetics, 4th, Prentice Hall, Upper Saddle River, N.J. (1998). 7. Scarborough, J. A., Filz, G. M., Mitchell, J. K., Brandon, T. L., Hoppe, E. J., and Hite, S. L., “Design of High Reinforced Embankments Constructed with Poor Quality Soil and Degradable Shale.” Geosynthetics '99 (1999:Boston, MA) Specifying Geosynthetics and Developing Design Details,1999, Vol. 1 pp. 491-504. 8. Minimum Design Loads for Buildings and Other Structures, ANSI/ASCE 7-95, Approved June 6, 1996, American Society of Civil Engineers, New York (1996). 9. “Foundations & Earth Structures, Design Manual 7.02 Revalidated by Change 1 September 1986.” Naval Facilities Command, Alexandria, VA (1986) 253 pp. 10. RSMeans Company Inc., RSMeans Site Work & Landscape Cost Data, 20th Annual Edition, 2001, RSMeans Company Inc., Kingston, MA (2000) 638 pp. 11. Barker, R. M., Duncan, J. M., Rojiani, K. B., Ooi, P. S. K., Tan, C. K., and Kim, S. G., “Manuals for the Design of Bridge Foundations, NCHRP Report 343.” Transportation Research Board, National Research Council, Washington, D.C. (1991) 308 pp. 12. Horvath, J. S., “Designing with Geofoam Geosynthetic, Seminar Notes.” (1999).

FIGURE 7.1 Proj 24-11.doc S = 15 kPa = 16 kN/m e = 1.7 C = 0.35 C = 0.04 = 1 t = 15 years C /C =0.04 sat 3 0 c r p c Sand Soft Clay 15m Soi l C ov er Pavement System Soil Cover T = 11 m T =0.61 m T =4.4 m H = 5 m EPS blocks pavement EPS p vo u OCR = W 7-40

FIGURE 7.2 PROJ 24-11.doc 7-41

FIGURE 7.3 PROJ 24-11.doc Center of Embankment Dual Tires 1.83 m1.83 m 0.61 m 1 2 3 4 T = 11 mw Tire Set No. 7-42

FIGURE 7.4 PROJ 24-11.doc 0.5 B 0.5 B 0.5 B BB Dual Tires Pavement System Center-to Center Spacing B = Width of equivalent rectangular loaded area for one set of dual tires. Note: If center-to-center spacing < B, stresses imposed by the two dual tire sets overlap. 0.5 B 7-43

FIGURE 7.5 PROJ 24-11.doc Note: L1 = L2 = L3 = L4 = 1.61 m. These loaded area lengths are not shown in this figure. z B = 1.1 m S = 0.73 m B = 1.1 m Dual Tires 1.22 m Q = 69.4 kN B = 1.1 m B = 1.1 m S = 0.12 m S = 0.73 m1 1 2 2 3 3 4 Q = 69.4 kNQ = 69.4 kNQ = 69.4 kN1 2 3 4 1.83 m 1.83 m 7-44

FIGURE 7.6 PROJ 24-11.doc Q = 69.4 kNQ = 138.8 kNQ = 69.4 kN 42,31 2.44 m 2.44 m z = 0.12 m S = 0.61 m B = 1.22 mB = 2.44 m S = 0.61 m B = 1.22 m1 1 2,3 3 4 7-45

FIGURE 7.7 PROJ 24-11.doc 100 mm Sand Bed and/or geotextile (if necessary) EPS 40 400 mm Soil Cover 610 mm (min) Gasoline and Diesel Resistant Geomembrane 100 mm Sand 5 m 3 1 2% 2% 720 mm Aggregate Base 178 mm Hot Mix Asphalt Pavement Traffic Lane 3.7 m Traffic Lane 3.7 m Shoulder 1.8 m 1.8 m Shoulder 11 m EPS 70 7-46

FIGURE 7.8 Proj 24-11.doc 3030 mm2460 mm Leveling Sand = 45θ o Base Course Concrete Approach Slab EPS 70 Sand Base Soil Backfill =18.8 kN/m =35 6240 mm Bituminous Pavement 50 mm φ γ ot 3 33 05 m m 27 45 m m 20 5 m m 30 5 m m 7-47

FIGURE 7.9 Proj 24-11.doc HL D, ConcH D, SandH A, SoilP = 1.47 kN/m H = 27 95 m m 93 1. 7 m m 13 97 .5 m m 0.61 m Live Load Surcharge, ( = 18.8 kN/m )3L γt Concrete Approach Slab, ( = 23.6 kN/m )D, Conc γConc 3 Sand Base, ( = 18.8 kN/m )D, Sand γt 3 ω ω ω = 3.21 kN/m = 2.01 kN/m = 1.08 kN/m 7-48

TABLE 7.1 PROJ 24-11.doc Layer No. Layer Thickness (m) z (m) IZ∆σ (kPa) IIZ∆σ (kPa) IIIZ∆σ (kPa) ∆σZ@center (kPa) σ′vo (kPa) σ′vf (kPa) σ′p (kPa) Sp (m) 1 1.5 0.75 16.58 0.01 0.01 16.59 4.64 21.24 4.64 0.1284 2 1.5 2.25 16.20 0.12 0.12 16.44 13.93 30.37 13.93 0.0658 3 1.5 3.75 15.19 0.41 0.41 16.02 23.21 39.23 23.21 0.0443 4 1.5 5.25 13.82 0.78 0.78 15.39 32.50 47.88 32.50 0.0327 5 1.5 6.75 12.40 1.13 1.13 14.66 41.78 56.44 41.78 0.0254 6 1.5 8.25 11.09 1.40 1.40 13.90 51.07 64.97 51.07 0.0203 7 1.5 9.75 9.95 1.61 1.61 13.16 60.35 73.51 60.35 0.0167 8 1.5 11.25 8.98 1.74 1.74 12.46 69.64 82.10 69.64 0.0139 9 1.5 12.75 8.15 1.82 1.82 11.8 78.92 90.72 78.92 0.0118 10 1.5 14.25 7.44 1.87 1.87 11.18 88.21 99.39 88.21 0.0101 Total Sp = 0.3694 7-49

TABLE 7.2 PROJ 24-11.doc Layer No. Layer Thickness (m) z (m) IZ∆σ (kPa) IIZ∆σ (kPa) IIIZ∆σ (kPa) ∆σZ@edge (kPa) σ′vo (kPa) σ′vf (kPa) σ′p (kPa) Sp (m) 1 1.5 0.75 0 0.20 0 0.20 4.64 4.84 4.64 0.0035 2 1.5 2.25 0.01 0.57 0 0.58 13.93 14.51 13.93 0.0035 3 1.5 3.75 0.04 0.92 0 0.97 23.21 24.18 23.21 0.0034 4 1.5 5.25 0.10 1.22 0.01 1.33 32.50 33.83 32.50 0.0034 5 1.5 6.75 0.20 1.47 0.02 1.69 41.78 43.47 41.78 0.0033 6 1.5 8.25 0.33 1.65 0.03 2.01 51.07 53.08 51.07 0.0033 7 1.5 9.75 0.47 1.79 0.05 2.32 60.35 62.67 60.35 0.0032 8 1.5 11.25 0.63 1.88 0.08 2.59 69.64 72.23 69.64 0.0031 9 1.5 12.75 0.79 1.93 0.10 2.83 78.92 81.75 78.92 0.0030 10 1.5 14.25 0.95 1.96 0.14 3.04 88.21 91.25 88.21 0.0029 Total Sp = 0.0325 7-50

TABLE 7.3 PROJ 24-11.doc Case Number Description of Interface Potential Type of Interface Materials Estimated δ (degrees) Source of δ Notes 1 EPS-block geofoam placed directly on the soil foundation. EPS/clay 27 (2) (1) EPS/sand 30 (3) (4) 2 Sand placed between the EPS blocks and soil foundation to serve as both a stable construction platform and a leveling material. sand/clay 20 (5) EPS/sand 30 (3) (4) sand/ geotextile 26 (6) (2) 3 Sand over a geotextile placed between the EPS blocks and soil foundation to serve as both a stable construction platform and leveling material. geotextile/clay 26 (7) (3) EPS/geotextile 25 Ch. 2 of this study 4 Geotextile placed between the EPS blocks and soil foundation. geotextile/clay 26 (7) (3) Notes: (1) A δ = 27˚ was provided for EPS and general soil interfaces. The type of soil was not provided. (2) A δ = 26˚ was provided for a concrete sand with φ = 30˚ and a nonwoven, heat bonded geotextile. (3) δ based on test results between a Trevira 1155 nonwoven geotextile and a red, sandy silt with 50 to 60 percent passing the U.S. No. 200 sieve and a liquid limit and plasticity index of 50 and 10, respectively. 7-51

TABLE 7.4 PROJ 24-11.doc Case Number Description of Interface Potential Type of Interface Materials Estimated δ (degrees) Source of δ Notes 1 Pavement system placed directly on the EPS blocks Crushed stone or sand/EPS 30 (3) (4) (1) crushed stone or sand/geotextile 26 (6) (2) 2 Geotextile placed between the pavement system and the EPS blocks geotextile/EPS 25 Ch. 2 of this study crushed stone or sand/concrete 29 (9) (3) 3 Concrete separation layer placed between the pavement system and the EPS blocks concrete/EPS 66 (4) crushed stone or sand/geomembrane 25 (6) (4) 4 Geomembrane placed between the pavement system and EPS blocks geomembrane/EPS 52 Ch. 2 of this study Notes: (1) δ based on sand/EPS interface. See Chapter 2. (2) A δ = 26˚ was provided for a concrete sand with φ = 30˚ and a nonwoven, heat bonded geotextile. (3) A range of δ = 29˚ to 31˚ was provided for mass concrete on clean gravel, gravel-sand mixtures, and coarse sand. (4) δ based on a concrete sand with φ = 30˚ and a smooth PVC geomembrane. 7-52

TABLE 7.5 PROJ 24-11.doc Failure Mode δ (degrees) kh FS′ I 25 0.09 4.7 II 30 0.09 Not critical III 20 0.09 3.6 7-53

TABLE 7.6 PROJ 24-11.doc Pavement System and Thickness of Asphalt Concrete σLL (kPa) σLL if stress overlap occurs between interior dual tire sets (kPa) σLL if stress overlap occurs between interior and exterior dual tire sets (kPa) Largest σLL (kPa) σDL* (kPa) σtotal (kPa) Required Elastic Limit Stress, σe (kPa) EPS TYPE Needed** Flexible, 76 mm 64 No Overlap No Overlap 64 12.2 76.2 91.44 EPS100 Flexible, 178 mm 39 No Overlap No Overlap 39 12.2 51.2 61.44 EPS70 Composite, 76 mm 19 21.55 No Overlap 21.55 12.2 33.75 40.5 EPS50 Composite, 178 mm 16 18.76 No Overlap 18.76 12.2 30.96 37.15 EPS50 *Based on a 0.610 m pavement system at 20 kN/m3 **EPS40 not recommended directly beneath paved areas. 7-54

TABLE 7.7 PROJ 24-11.doc EPS Type Needed Cost $/m2 per 610 mm Thickness Pavement System and Thickness of Asphalt Concrete Asphalt Thickness (mm) Cost $/m2 per mm Thicknes s Cost $/m2 Concrete Separation Layer Thickness (mm) Cost $/m2 per mm Thickness Cost $/m2 Granular Base Thickness (mm)*** Cost $/m3 Cost $/m2 Total Cost $/m2 EPS100 $39.65 Flexible, 76 mm 76 $0.10 $7.60 0 $0.36 $0.00 534 $26.00 $13.88 $61.13 EPS70 $30.50 Flexible, 178 mm 178 $0.10 $17.80 0 $0.36 $0.00 432 $26.00 $11.23 $59.53 EPS50 $26.23 Composite, 76 mm 76 $0.10 $7.60 102 $0.36 $36.72 432 $26.00 $11.23 $81.78 EPS50 $26.23 Composite, 178 mm 178 $0.10 $17.80 102 $0.36 $36.72 330 $26.00 $8.58 $89.33 ***based on 610 mm pavement system 7-55

TABLE 7.8 PROJ 24-11.doc Dual Tire Load Combination Q (kN) σLL (kPa) qt (kPa)* ∆σZ,DL Pavement (kPa) Ζ(m) (kPa) σZ,DL Total (kPa) σtotal (kPa) Required Elastic Limit Stress, σe (kPa). EPS Type Needed 2 and 3 combined 138.8 32.88 12.20 12.20 0.12 12.32 45.20 54.24 EPS70 1 or 4 single 69.4 32.88 12.20 12.20 0.12 12.32 45.20 54.24 EPS70 2 and 3 combined 138.8 21.34 12.20 12.19 0.61 12.80 34.14 40.97 EPS50** 1 or 4 single 69.4 18.28 12.20 12.19 0.61 12.80 31.08 37.30 EPS40 All combined 277.6 17.68 12.20 12.20 0.73 12.93 30.61 36.73 EPS40 All combined 277.6 10.78 12.20 12.06 1.73 13.79 24.57 29.48 EPS40 All combined 277.6 7.34 12.20 11.71 3.73 14.44 21.79 26.15 EPS40 *Based on a 0.610 m pavement system at 20 kN/m3 **Indicates that this is the critical EPS type for this depth. 7-56