VIII. Sample Design Calculation... Page 2. Copyright 2011 Blaise J. Fitzpatrick, P.E. Page 1 of 24

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1 Mechanically Stabilized Earth Structures Part 3 By Blaise J. Fitzpatrick, P.E. Fitzpatrick Engineering Associates, P.C. VIII. Sample Design Calculation... Page Copyright 0 Blaise J. Fitzpatrick, P.E. Page of 4

2 Sample Design Calculation A SunCam online continuing education course Sample Design Calculation based on AASHTO/FHWA Allowable Stress Design as noted in the NHI , Mechanically Stabilized Earth Walls and Reinforced Soil Slopes - Design and Construction Guidelines, March 00. Design a 0-ft tall MSE wall with the following geometry and loading conditions o Level toe with -ft embedment depth below final grade o Level backfill, =0-degrees o Live load traffic surcharge of q=50-lb/ft Segmental retaining wall block for this sample calculation has a mechanical connection and high strength geotextile reinforcement will be used to reinforce the soil. Figure General cross section of MSE wall. Traffic Surcharge, q=50-psf 8' Reinforced Soil Retained Soil ' Soil Design Parameters 3 Foundation Soil. Reinforced Soil =35-degrees c =0-lb/ft =5-lb/ft 3. Retained Soil r =8-degrees c r =0-lb/ft r =0-lb/ft 3 3. Foundation Soil f =8-degrees c f =0-lb/ft f =0-lb/ft 3 Copyright 0 Blaise J. Fitzpatrick, P.E. Page of 4

3 External Stability Calculations Figure Free Body Diagram of MSE wall. A SunCam online continuing education course Assumed for maximum horizontal stress componets and bearing capacity V Assumed loading for sliding, overtuning and pullout components 0' V F O H/ H/3 F L Determine Active Earth Pressure Coefficient (K a (ext) ) based on Rankine Earth Pressure Theory and Horizontal and Vertical Forces acting on the MSE wall. K a ( ext) cos cos cos r cos Eq. cos cos cos r cos(0) cos 0 cos 8 K a(ext) cos(0) cos(0) cos 0 cos 8 = 0.36 F = ½ r H K a(ext) = ½ (0 lb/ft 3 ) (0-ft) (0.36) = 7,943 lb/ft Eq. F = q H K a(ext) = (50 lb/ft ) (0-ft) (0.36) =,805 lb/ft Eq. 3 L = 0.7 H = 0.7 (0-ft) = 4-ft Eq. 4 V = H L = (5 lb/ft 3 ) (0-ft) (4-ft) = 35,000 lb/ft Eq. 5 V = q L = (50 lb/ft ) (4-ft) = 3,500 lb/ft Eq. 6 Copyright 0 Blaise J. Fitzpatrick, P.E. Page 3 of 4

4 ~ External Factors of Safety for Overturning, Sliding and Bearing Capacity Factor of Safety for Overturning - FS OT Moments are taken about Point O (see Figure ) Resisting moment does not include traffic surcharge, q L V MomentsResisting M R FS OT =. 0 Eq. 7 Moments Overeturning M H H O F F 3 M R = V L = (35,000 lb/ft) (4-ft) = 45,000 lb-ft M O = 3 F H + F H = (7,943 lb/ft) (0-ft) + (,805 lb/ft) (0-ft) = 7,003 lb-ft 3 FS OT = M M R O 45,000 lb - ft 7,003 lb - ft = 3.45 >.0 therefore meets requirement. Factor of Safety for Sliding FS SL FS SL = Resisting force does not include traffic surcharge, q Horizontal Resisting Force F Horizontal Driving Force F o FR 35,000 lb/ft FS SL tan 8 = F D 7,943 lb/ft,805 lb/ft R D V tan f.5 F F =.9 >.5 therefore meets requirement. Eq. 8 Factor of Safety for Bearing Capacity FS BC R BC = Resultant of Vertical Forces Resisting moment for bearing capacity (M R(BC) ) includes traffic surcharge, q e = eccentricity (feet) B = effective foundation width v = vertical overburden stress (lb/ft ) q ult = ultimate bearing capacity (lb/ft ) Copyright 0 Blaise J. Fitzpatrick, P.E. Page 4 of 4

5 R BC = Resultant of Vertical Forces = V + V = 35,000 lb/ft + 3,500 lb/ft = 38,500 lb/ft Eq. 9 M R (BC) = (RBC ) (L) = (38,500 lb/ft) (4-ft) = 69,500 lb/ft Eq. 0 L M R M O L e R 6 BC 4 ft 69,500lb ft 7,003lb ft e =.84-ft Eq. 38,500 lb / ft L 4 ft L =.33-ft e.84-ft <.33-ft (therefore eccentricity is okay) B = L - e = 4-ft (.84-ft) = 0.3-ft Eq. v = o V V F sin 35,000 lb/ft 3,500 lb/ft 7,943 lb/ft (sin 0 ) = L e 4 - ft - (.84 - ft) = 3,734 lb/ft Eq. 3 MSEW Screen Shot of Meyerhof stress distribution, vertical overburden stress and eccentricity. Bearing capacity factors can be found in most Foundation Engineering text books or by applying the following formulas, in this example for f =8-degrees: N q = tan (45 + ) e tan = 4.7 Eq. 4 N c = (N q ) cot = 5.80 Eq. 5 N = (N q + ) tan = 6.7 Eq. 6 q ult = c f N c + f B N (this is the ultimate bearing capacity) Eq. 7 q ult = (0-psf) (5.80) + (0-lb/ft 3 ) (0.3-ft) (6.7) = 9,48 lb/ft FS BC = q ult ' v = 9,48lb/ft 3,734 lb/ft =.54 >.0 therefore meets requirement. Eq. 8 e L. 84 ft 4 ft = 0.3 Eq. 9 Copyright 0 Blaise J. Fitzpatrick, P.E. Page 5 of 4

6 ~ Internal Factors of Safety for Sliding, Overstress, Pullout and Connection Determine the reinforcement Long Term Design Strength of a High Strength Geotextile T ult = Ultimate Geosynthetic Strength (from manufacturer) = 7,00 lb/ft RF CR = Creep Reduction Factor (from manufacturer) =.68 RF ID = Installation Damage Reduction Factor (from manufacturer) =.0 RF D = Durability Reduction Factor (from manufacturer) =.0 LTDS = Tult RFcr RFid RFd = 7,00 lb/ft = 3,74 lb/ft Eq. 0 Determine Internal Active Earth Pressure Coefficient K a(int) = sin sin = sin 35 sin 35 = 0.7 Eq. Determine Reinforcement Pullout Properties F* = tan = tan 35 o = Pullout Resistant Factor Eq. R c =.0 00% Geosynthetic Coverage C i = 0.9 Coefficient of Interaction (from manufacturer) C ds = 0.8 Coefficient of Direct Sliding (from manufacturer) C =.0 (for geotextile or geogrid) Reinforcement effective unit parameter =.0 Scale correction factor (from manufacturer) Calculate maximum tensile force in each reinforcement layer. We need to determine horizontal stress ( H ) along the potential failure line from the weight of the reinforced fill (Z) plus, if present uniform surcharge loads (q) and concentrated surcharge loads v and H. Z = distance from top of wall to reinforcement layer = surcharge load due to sloping backfill H = horizontal surcharge load due to footing Layer 9 z= 3.5 z= 9.5 v = vertical surcharge load due to footing q = traffic surcharge = 50-lb/ft 0' Layer 6 z= 5.5 H = horizontal stress = K a(int) v + H Layer 3 v = vertical stress = Z + + q + v 4' Copyright 0 Blaise J. Fitzpatrick, P.E. Page 6 of 4

7 Given no slopes or footing loads:, H and v = 0-psf For this sample problem we will calculate the vertical and horozontal stress at three geosynthetic layers located at a depth of 3.5-ft, 9.5-ft and 5.5-ft from the top of wall. v(n) = Z (n) + + q + v Eq. 3 v(3) = (5 lb/ft 3 ) (5.5-ft) + 0-lb/ft + 50-lb/ft + 0-lb/ft =,88 lb/ft v(6) = (5 lb/ft 3 ) (9.5-ft) + 0-lb/ft + 50-lb/ft + 0-lb/ft =,438 lb/ft v(9) = (5 lb/ft 3 ) (3.5-ft) + 0-lb/ft + 50-lb/ft + 0-lb/ft = 688 lb/ft h(n) = K a(int) v(n) + H Eq. 4 h(3) = (0.7) (,88 lb/ft ) + 0-lb/ft = 593 lb/ft h(6) = (0.7) (,438 lb/ft ) + 0-lb/ft = 390 lb/ft h(9) = (0.7) (688 lb/ft ) + 0-lb/ft = 86 lb/ft Calculate the maximum tension (T max ) in each reinforcement layer. The maximum vertical spacing (S v ) between reinforcement layers is limited to.0-ft. T max(n) = ( h(n) ) (S v ) Eq. 5 T max(3) = (593 lb/ft ) (.0-ft) =,86 lb/ft T max(6) = (390 lb/ft ) (.0-ft) = 779 lb/ft T max(9) = (86 lb/ft ) (.0-ft) = 373 lb/ft Calculate the Factor of Safety for Reinforcement Overstress in each reinforcement layer. FS OS(n) = FS OS(3) = FS OS(6) = FS OS(9) = LTDS T ( n) max( n) 3,74 lb/ft,86 lb/ft 3,74 lb/ft 779 lb/ft 3,74 lb/ft 373 lb/ft Eq. 6 =.76 > >.50 ok = 4.0 > >.50 ok = > >.50 ok Copyright 0 Blaise J. Fitzpatrick, P.E. Page 7 of 4

8 Observation: The factor of safety for reinforcement overstress increases in the upper portion of the wall due to the lower stress level. In this example we are using one type of reinforcement for design, however in a real application two or three types of reinforcement types may be used with highest strength reinforcement in the lower portion of the wall where the stress is high and lower strength reinforcement in the upper portion of the wall where horizontal stresses are lower. Stability with respect to reinforcements pullout requires the following criteria be satisfied T max = F* Z p L e C R c Eq. 7 FS po where: FS PO = Safety factor against pullout >.5 Rc = Coverage ratio T max = Maximum reinforcement tension = Scale correction factor C = for geogrid or geotextile F* = Pullout resistance factor Z p = Overburden pressure, including distributed dead load surcharges, neglecting traffic loads (see Figure 30 in NHI ). Le = Length of embedment in the resisting zone. Note that the boundary between the resisting and active zones may be modified by concentrated loadings. Required embedment length in reinforced zone beyond the potential failure surface is determined from: L e(n) > C F * c i.5 T max( n) Z R ( n) Determine the active zone (L a ) for each reinforcement layer. c Eq. 8 L a(n) = (H z (n) ) tan(45-/) Eq. 9 L a(3) = (0-ft 5.5-ft) tan(45-35 o /) =.34-ft L a(6) = (0-ft 9.5-ft) tan(45-35 o La Le /) = 5.47-ft L a(9) = (0-ft 3.5-ft) tan(45-35 o /) = 8.59-ft Determine the actual design embedment length for (L e ) for each reinforcement layer. Active Zone Layer 9 z 9 =3.5 z 6 =9.5 L e(n) = L L a(n) Eq. 30 L e(3) = 4-ft.34-ft =.66-ft 0' Layer 6 Resistant Zone z 3 =5.5 L e(6) = 4-ft 5.47-ft = 8.53-ft L e(9) = 4-ft 8.59-ft = 5.4-ft Layer 3 Theoretical failure plane at 45+/ 4' Copyright 0 Blaise J. Fitzpatrick, P.E. Page 8 of 4

9 Determine the Factor of Safety for Pullout for each reinforcement layer. C F * ci Z FS po(n) = T ( n) max( n) L e( n) R c Eq. 3 FS po(3) = 3 (.0)(0.7)(0.9)(5 lb / ft )(5.5 ft)(.66,86 lb / ft ft)(.0)(.0) = >.50 ok FS po(6) = 3 (.0)(0.7)(0.9)(5 lb / ft )(9.5 ft)( lb / ft ft)(.0)(.0) = >.50 ok FS po(9) = 3 (.0)(0.7)(0.9)(5 lb / ft )(3.5 ft)( lb / ft ft)(.0)(.0) = 8.00 >.50 ok Stability analysis with respect to Internal Sliding along individual reinforcement layers. FS SL(n) = FR tan = F D V ( n ) tan F F (3) (3) Eq. 3 = tan - (C ds *tan ) = tan - (0.8 tan 35 o ) = 9.56 o Eq. 33 Previously defined or calculated terms/values to be used in internal sliding calculations: K a(ext) = 0.36 = 5-lb/ft 3 z (3) = 5.5-ft z (9) = 3.5-ft = 9.56 o q traffic = 50-lb/ft r = 0-lb/ft 3 z (6) = 9.5-ft L = 4-ft Internal Factor of Safety for Sliding on Layer 3 q=50-psf F (3) = ½ r z (3) K a(ext) F (3) = ½ (0 lb/ft 3 ) (5.5-ft) (0.36) = 4,77 lb/ft F (3) = q z (3) K a(ext) F (3) = (50 lb/ft ) (5.5-ft) (0.36) =,399 lb/ft V (3) = z (3) L V (3) = (5 lb/ft 3 ) (5.5-ft) (4-ft) = 7,5 lb/ft z V(3) Layer 3 z 3 / F(3) F(3) z 3 /3 FS SL(3) = V ( n ) tan F (3) F (3) 4' FS SL(3) = 7,5 lb/ft(tan9.56) 4,77lb/ft,399 lb/ft =.463 >.50 ok Copyright 0 Blaise J. Fitzpatrick, P.E. Page 9 of 4

10 Internal Factor of Safety for Sliding on Layer 6 A SunCam online continuing education course q=50-psf F (6) = ½ r z (6) K a(ext) F (6) = ½ (0 lb/ft 3 ) (9.5-ft) (0.36) =,79 lb/ft F (6) = q z (6) K a(ext) F (6) = (50 lb/ft ) (9.5-ft) (0.36) = 857 lb/ft z V(6) Layer 6 z 6 / F(6) F(6) z 6 /3 V (6) = z (6) L V (6) = (5 lb/ft 3 ) (9.5-ft) (4-ft) = 6,65 lb/ft FS SL(6) = V ( n ) tan F (6) F (6) 4' FS SL(6) = 6,65 lb/ft(tan9.56),79 lb/ft 857 lb/ft = 3.55 >.50 ok Internal Factor of Safety for Sliding on Layer 9 q=50-psf F (9) = ½ r z (9) K a(ext) F (6) = ½ (0 lb/ft 3 ) (3.5-ft) (0.36) = 43 lb/ft F (9) = q z (9) K a(ext) F (9) = (50 lb/ft ) (3.5-ft) (0.36) = 36 lb/ft z V(9) Layer 9 z 9 / F(9) F(9) z 9 /3 V (9) = z (9) L V (9) = (5 lb/ft 3 ) (3.5-ft) (4-ft) = 6,5 lb/ft FS SL(9) = V ( n ) tan F (9) F (9) 4' FS SL(9) = 6,5 lb/ft(tan9.56) 43 lb/ft 36 lb/ft = 6.36 >.50 ok Observation: The factor of safety for internal direct sliding increases as you move from the bottom of wall to the top of wall. This is due to the fact that the reinforcement length remains constant at L-4-ft throughout the wall height while the horizontal force on a given reinforcement layer decreases as reinforcement elevation increases. In short internal direct sliding controls the design lengths at the bottom of the wall, whereas pullout failure controls the design lengths at the top of wall as previously calculated. Copyright 0 Blaise J. Fitzpatrick, P.E. Page 0 of 4

11 Screen shot from program MSEW 3.0 with respect to internal sliding and external sliding. Screen shot from program MSEW 3.0 with respect to bearing capacity. Copyright 0 Blaise J. Fitzpatrick, P.E. Page of 4

12 Internal eccentricity calculations for each reinforcement layer. Note ASSHTO/FHWA does not require an evaluation of e/l, the calculation provide here is for information only. Reinforcement Layer 3 M R(3) = V(3) L = (7,5 lb/ft) (4-ft) = 89,875 lb-ft M O(3) = 3 F(3) z (3) + F(3) z (3) = 3 (4,47 lb/ft)(5.5-ft) + (,399 lb/ft)(5.5-ft) = 35,49 lb-ft FS OT(3) = e M M M R(3) O(3) M L R( 3) O(3) L RBC(3) 6 89,875 lb - ft = 3.45 >.0 35,49 lb - ft 4 - ft 89,875 lb - ft 35,49 lb - ft e =.308-ft 7,5 lb/ft e L ft = ft z Layer 3 V(3) 7,5 lb/ft Layer 3 q=50-psf F(3)=,3 99 lb /ft F(3) =4,7 7 lb /ft z 3 /3 z 3 / Reinforcement Layer 6 4' M R(6) = V(6) L = (6,65 lb/ft) (4-ft) = 6,375 lb-ft M O(6) = 3 F(3) z (6) + F(3) z (6) = 3 (,79 lb/ft)(5.5-ft) + (857 lb/ft)(5.5-ft) = 9,745 lb-ft FS OT(3) = M M R(6) O(6) 6,375 lb - ft =.94 >.0 9,745 lb - ft Layer 6 q=50-psf e M M L R( 3) O(6) L RBC(6) ft 6,375 lb - ft 9,745 lb - ft e = ft 6,65 lb/ft z V(6) 6,65 lb/ft Layer 6 F(6)=857 lb/ft z 6 / z 6 /3 F(6)=,7 9 lb /ft e ft L 4 - ft = ' Copyright 0 Blaise J. Fitzpatrick, P.E. Page of 4

13 Reinforcement Layer 9 A SunCam online continuing education course M R(9) = V(9) L = (6,5 lb/ft) (4-ft) = 4,875 lb-ft M O(9) = 3 F(3) z (9) + F(3) z (9) = 3 (34 lb/ft)(5.5-ft) + (36 lb/ft)(5.5-ft) = 836 lb-ft FS OT(3) = M M R(6) O(6) 4,875 lb - ft = 5.5 > lb - ft Layer 9 q=50-psf L e M M R( 9) O(9) L RBC(9) ft 4,875 lb - ft 836 lb - ft e = 0.37-ft 6,5 lb/ft z V(9) Layer 9 6,5 lb/ft F(9)= 34 lb /ft z 9 /3 z 9 / F(9)=3 6 lb /ft e ft L 4 - ft = ' Observation: The factor of safety for overturning increases dramatically as you move from the bottom of wall to the top of wall. Resisting and overturning moments both decrease as you move from the bottom of wall to the top wall with overturning moments decreasing at a much faster rate than the resisting moments; this is due to the fact that the reinforcement length remains constant at L-4-ft throughout the wall height. In actual design overturning for external or internal stability is performed to determine the eccentricity needed for bearing capacity analysis. Furthermore overturning never controls the design reinforcement length that is typically controlled by either sliding at the bottom of the wall pullout at the top of wall or overall global stability. Copyright 0 Blaise J. Fitzpatrick, P.E. Page 3 of 4

14 Stability analysis with respect to Connection Capacity Connection capacity results for the high strength geotextile indicate the connection capacity is T conn(n) =,585-lb/ft + N tan 3 o Eq. 34 N (n) = z (n) block (this is the normal load on the connection) Eq. 35 block RF D = -lb/ft 3 (provided by block manufacturer based on connection test data) =.0 (reduction factor for durability at the connection) RF CR =.45 (reduction factor for creep at the connection) The mode of failure should still be considered to be pullout if longitudinal ribs in geogrids do not rupture, with longitudinal being defined as the direction of the applied load, or for geotextiles if significant ripping of the geotextile perpendicular to the direction of loading does not occur. T sc is defined as the peak load per unit reinforcement width obtained in the connection strength test, where pullout is known to be the mode of failure, or the load at which the end of the reinforcement between the facing blocks deflects 5 mm. T ultconn is defined as the peak load per unit reinforcement width where rupture is the mode of failure in the connection strength test. T Lot is the ultimate wide width tensile strength (ASTM D-4595) for the reinforcement material lot used for the connection strength testing. Copyright 0 Blaise J. Fitzpatrick, P.E. Page 4 of 4

15 CR u(n) = CR s(n) = T T T ultconn (n) T Lot (n) sc(n) Lot (n) A SunCam online continuing education course Eq. 36 Eq. 37 T ac(rup) = T Lot RF D CR RF u CR (connection capacity based of reinforcement rupture or break) Eq. 38 T ac (po) = (T lot ) (CR S ) (connection capacity based of reinforcement pullout) Eq. 39 FS (rup) = T ac(rup) T max (connection factor of safety based of reinforcement rupture or break) Eq. 40 FS (po) = T ac(po) T max (connection factor of safety based of reinforcement pullout) Eq. 4 Reinforcement Layer 3 T max(3) =,86-lb/ft (previously calculated when determining FS OS ) z (3) = 5.5-ft N (3) = (5.5-ft) (-lb/ft 3 ) =,736-lb/ft T conn(3) =,585-lb/ft +,736-lb/ft (tan 3 o ) CR u(3) = 3,68- lb/ft 7,00 - lb/ft = 3,68-lb/ft = CR s(3) = T ac(rup) = 3,68- lb/ft 7,00 - lb/ft (7,00 - lb/ft) (0.5039) (.0) (.45) = =,75-lb/ft T ac(po) = ( 7,00 - lb/ft ) (0.5039) = 3,68-lb/ft FS (rup) = FS (po) =,75- lb/ft,86 - lb/ft 3,68 - lb/ft,86 - lb/ft =.9 >.50 ok = 3.06 >.50 ok Copyright 0 Blaise J. Fitzpatrick, P.E. Page 5 of 4

16 Reinforcement Layer 6 T max(6) = 779-lb/ft (previously calculated when determining FS OS ) z (6) = 9.5-ft N (6) = (9.5-ft) (-lb/ft 3 ) =,064-lb/ft T conn(6) =,585-lb/ft +,064-lb/ft (tan 3 o ) = 3,4-lb/ft CR u(6) = 3,4 - lb/ft 7,00 - lb/ft = CR s(6) = T ac(rup) = 3,4 - lb/ft 7,00 - lb/ft (7,00 - lb/ft) (0.4478) (.0) (.45) = =,0-lb/ft T ac(po) = ( 7,00 - lb/ft ) ( ) = 3,4-lb/ft FS (rup) =,0 - lb/ft lb/ft =.59 >.50 ok CR s(9) = T ac(rup) = (7,00 - lb/ft) (0.397) (.0) (.45) = =,768-lb/ft T ac(po) = ( 7,00 - lb/ft ) (0.397) =,8-lb/ft FS (rup) = FS (po) = FS (po) = 3,4 - lb/ft lb/ft = 4.4 >.50 ok Reinforcement Layer 9 T max(9) = 373-lb/ft (previously calculated when determining FS OS ) z (9) = 3.5-ft N (9) = (3.5-ft) (-lb/ft 3 ) = 39-lb/ft T conn(9) =,585-lb/ft + 39-lb/ft (tan 3 o ) =,8-lb/ft CR u(9) =,8- lb/ft 7,00 - lb/ft = 0.397,8- lb/ft 7,00 - lb/ft,768 - lb/ft lb/ft,8- lb/ft lb/ft = 4.75 >.50 ok = 7.57 >.50 ok Copyright 0 Blaise J. Fitzpatrick, P.E. Page 6 of 4

17 Global Stability Analysis A SunCam online continuing education course A global stability analyses is presented for the wall by modeling the same wall height (H=0-ft), live load traffic loading conditions of q=50-psf and soil strength/unit weight properties. The geosynthetic-reinforcement length and vertical spacing modeled in the global stability analyses was determined by hand calculation and verified with program MSEW (3.0) using the FHWA NHI methodology as shown in Part of this course. Global stability analyses are sensitive to soil strength parameters therefore changes, differences or variances between assumed strength parameters that may have been made in stability analyses and actual site specific strength parameters can significantly affect the type and length of the geosynthetic-reinforcement required. Commentary on Cohesion and Factor of Safety in Global Stability Analyses Cohesion has major effects on stability and the long-term effective strength value of cohesion is not always certain. Furthermore, cohesive backfill in made-made embankments (if at all used) is likely to be normally consolidated. Consequently, it is often assumed in practice that for design purposes the apparent cohesion is zero (c=0-psf), especially under drained loading conditions. If cohesive fill is used, extreme care should be used when specifying the cohesion value. Cohesion has significant effects on stability and thus the required reinforcement strength. In fact, a small value of cohesion will indicate that no reinforcement at all is needed at the upper portion of a MSE wall. However, over the long-run cohesion of manmade structures tends to drop and nearly diminish. Since long-term stability of MSE walls is of major concern, it is perhaps wise to ignore the cohesion altogether. It is therefore recommended to limit the design value of cohesion to 00-psf but only in residual soil and no cohesion in fill soil. Global stability for this example is analyzed using the commercially available two-dimensional slope stability program ReSSA (v3.0), screen shots provided from program ReSSA by permission of the software developer Adama Engineering ( Global stability analyses were performed using Bishop's method of slices (Bishop, 955), which accounts for moment equilibrium used for circular searches; and Spencer's method (Spencer, 973), which accounts for force and moment equilibrium used for circular and non-circular searches. Spencer s factor of safety best represents the most precise factor of safety against global stability for non-circular slip surfaces, although factor of safety results for circular and non-circular slip surfaces are often similar using Bishop, Janbu, or Spencer's method if the slip surface is uniform as in the case of circular surfaces. It is generally accepted to use Bishop's method of slices for circular searches. Copyright 0 Blaise J. Fitzpatrick, P.E. Page 7 of 4

18 Once the geometry, soil properties and loading are input the engineer needs to set upper and lower limits to find the minimum factor of safety. The ReSSA program will display all internal, compound internal and deep seated circles to be analyzed. Copyright 0 Blaise J. Fitzpatrick, P.E. Page 8 of 4

19 The critical slip surface or slip surface with the lowest safety factor is shown; this is the factor of safety. For this section Fs=.49. A safety map can be generated that shows all slip surfaces and associated safety factors along with the section factor of safety, Fs= Copyright 0 Blaise J. Fitzpatrick, P.E. Page 9 of 4

20 Translational analysis looks at internal sliding planes based on a -part wedge. Reinforcement strength and length is key to this analysis. A safety map can be generated that shows all slip surfaces and associated safety factors along with the section factor of safety, Fs= Copyright 0 Blaise J. Fitzpatrick, P.E. Page 0 of 4

21 3-part wedge analysis set up with passive/active search box, entrance/exit angles and incremental sensitivity. Safety map generated showing all slip surfaces and associated safety factors along with the section factor of safety, Fs= Copyright 0 Blaise J. Fitzpatrick, P.E. Page of 4

22 Compilation of all three analyses: Fs=.49 Bishop Circular Fs=.63 Spencer -Part Wedge Fs=.33 Spencer 3-Part Wedge Global stability results indicate the most critical slip surface passes below and behind the reinforced zone. Safety maps for all three analyses (circular, -part wedge and 3-part wedge) indicate safety factors within the reinforced zone are greater than.50. The lowest factors of safety is Fs=.33, which would meet the minimum requirement for most jurisdictions unless otherwise specified to be greater than Fs= Copyright 0 Blaise J. Fitzpatrick, P.E. Page of 4

23 REFERENCES ADAMA Engineering, Inc., ) Mechanically Stabilized Earth Walls MSEW v3.0 and ) Reinforced Slope Stability Analysis ReSSA v3.0, Newark, DL, Bishop, A.W., "The Use of the Slip Circle in the Stability Analysis of Slopes", Geotechnique, 955, Vol. No., pp Bernardi, M. and Fitzpatrick, B.J., Connection Capacity of Precast Concrete Segmental Units and Geogrids: Testing and Design Considerations. Geosynthetics Conference 99, Boston, Massachusetts, April 8 to 9, 999. Bowles, J.E., "Foundation Analysis and Design", 4th Ed, McGraw Hill, NY, New York, 988, pp Collin, J.G., "Design Manual for Segmented Retaining Walls Second Edition Second Printing ", National Concrete Masonry Association, 30 Horse Pen Road Herndon, VA USA, 997, 89 pp. Das, B. M., "Principles of Foundation Engineer", PWS Kent Boston, Massachusetts, 984, pp. 595 pp. Design Manual - Soil Mechanics, Foundations, and Earth Structures, NAVFAC DM-7, Department of the Navy, Naval Facilities Engineering Command, March 97. Elias, V., Christopher, B.R., Berg, R.R., "Mechanically Stabilized Earth Walls and Reinforced Soil Slopes Design and Construction Guidelines", Publication No. FHWA NHI , March 00, 394 pp. Fitzpatrick, B.J., Letter to the Editor. Geosynthetics Magazine, October Issue 006. Fitzpatrick, B.J., How Global Stability Analysis Can Make a World of Difference on Your Next Project, Retaining Ideas Volume 4 Issue, Anchor Retaining Walls 999. Holtz R.D. and Kovacs, W.D., "An Introduction to Geotechnical Engineering", Prentice Hall, Englewood Cliffs, New Jersey, 98, 733 pp. Janbu, N., "Slope Stability Computations", The Embankment Dam Engineering, Casagrande Volume, John Wiley and Sons, Inc., New York, NY, 973, pp Simac, M.R., Fitzpatrick, B., (00) Design and Procurement Challenges for MSE Structures: Options going forward Earth Retention 00, Seattle, Washington, August 00. Simac, M.R., Fitzpatrick, B., (008) Part 3B - Three Challenges in Building SRWs and other Reinforced-Soil Structures Construction Observation Aspects Geosynthetics Magazine, Vol. 6, No. 5, August-September Copyright 0 Blaise J. Fitzpatrick, P.E. Page 3 of 4

24 Simac, M.R., Fitzpatrick, B., (008) Part 3A - Three Challenges in Building SRWs and other Reinforced-Soil Structures Construction Aspects Geosynthetics Magazine, Vol. 6, No. 4, June-July 008. Simac, M.R., Fitzpatrick, B., (008) Part B Three Challenges in Building SRWs and other Reinforced-Soil Structures MSE Designer Aspects Geosynthetics Magazine, Vol. 6, No. 3, April-May 008. Simac, M.R., Fitzpatrick, B., (008) Part A Three Challenges in Building SRWs and other Reinforced-Soil Structures - Site Design and Geotechnical Aspects Geosynthetics Magazine, Vol. 6, No., February-March 008. Simac, M.R., Fitzpatrick, B., (007) Part Three Challenges in Building SRWs and other Reinforced-Soil Structures - Options for buying the SRW Geosynthetics Magazine, Vol 5, No 5, October-November 007. Spencer, E., "A Method of Analysis of the Stability of Embankments Assuming Parallel Inter- Slice Forces", Geotechnique, 967, XVII, No., pp Copyright 0 Blaise J. Fitzpatrick, P.E. Page 4 of 4