Calculations for Reinforced Soil Wall (StaTitle Hand Calculations for modular block wall of height 10.75 m Reference BS 8006-1:2010 Date : Designed by: Checked by : Approved by : Design Input Parameters Reinforced Soil Data Angle of Internal friction ᶲ1 Unit wt Y1 Retained Backfill Soil Data Angle of Internal friction ᶲ2 Unit wt Foundation Soil Data Cohesion Angle of Internal Friction ᶲ3 Unit wt Crash Barrier Data Strip load due to crash barrier Q Live Load Ql 32 ° 18.5 kN/cu.m 30 ° 18.5 kN/cu.m 0 kPa 30° 18.5 kN/cu.m 15.45 kPa 23 kPa Live Load should be considered as per provisions of IRC:78-2014 Water table is considered below the influence zone. General Shear Failure is considered. COMPUTATION OF EXTERNA 10.75 m HIGH WALL BY STATIC ANALYSIS Coefficient of active earth pressure (k a): For reinforced soil: ka = (l-sinᶲ)/(l+sinᶲ) kal = 0.307 Wall batter ᶿ= For retained backfill soil: ᶲ1 = 32 4.23 ka2 0.18 kN/m Strip load due to Crash Barrier V2 Arm = Q x b*fts XI = 3.333 = Mechanical wall height H = 10.60 m Minimum embedment depth 1 m Unit weight of foundation soil Vf 18 Foundation properties : kN/m3 Summary of partial factors to be used Partial factors to be applied tan0' Soil material factors : to be applied C to be applied Cu Sliding across surface of reinforcement (fs) Soil/reinforcement interaction factors Pullout resistance of reinforcement (fp) Foundation bearing capacity : to be applied Partial factors of safety Sliding along base of the structure or any horizontal surface where there is soil-soil contact Partial load factors for load combinations associated with walls combinations A Effects Mass of the reinforced soil body (ffs) Mass of the backfill on top of the reinforced soil wall (ffs) Earth pressure behind the structu(ffs) Traffic load: On reinforced soil b(fq) Behind reinforced soil block (fq) 1.5*10.75*7.5 kN/m = 2267.5 51 IRC:SP:102-2014 CASE :A LOADS Self weight of Reinforced Soil Wall Lever Vj= Yi*H*L*ff5 = 18.75 m Length of reinforcement L= 7.8 .6*1. 75*10.34 < = 657.CU fs is the partial factor against base sliding L is the effective base width for sliding Sliding force (Rh) = Resisting force = (Rv * tanO'p)/fms = 886.5 = 534.91 Hence structure is safe in sliding stability 52 IRC.5 - 123.6 X2 = 0. C'.B For long term stability where there is soil to soil contact at the base of the structure fsRh< Rv(tan0'p/fms)+(C*L/fms) Rh is the horizontal factored disturbing force Rv is the vertical factored resultant force 0'p is the peak angle of shearing resistance under effective stress conditions fms is the partial materials factor applied to tan0'p.08 kN/m b= 1.5*10.56 kN/m Check for Sliding along the base CASE.6*1.8 X3 = 3.SP.6*1.75*1.5*0.78 kN/m 886.2 kN/m Resultant Vertical Load Rv = V1+V2+V3 = 2566.5 kN/m = 37.91 kN/m (Pi+P2) 789.5*1.1/2 * ka2 *v2 * H2*ffs = 0.22 kN/m Earth Pressure due to Live Load : P2 = ka2 * = 0.75*1.333*23*10.333*18. 102-2014 Check for Bearing Failure Overturning Moment Mo = (PI * H/3 + P2 * H/2) CASE :A .5 kN/m - 262.45 kN/m Horizontal Forces Earth pressure behind reinforced soil block Pi .8 Ql * H * ffs Vertical load due to Live Load V3 = Ql x L*ffs = 23*7.width of the Strip Load V2 = 15. 8 2566.t Ultimate bearing capacity of foundation soil quit = {cf for cp = 30° Nc = 30.5 (L-2e) yf Ny = 1440.4 Ny = 22.14 Nq = 18.46 = 3.40 L-2e = 5.14 Hence foundation is safe against bearing capacity failure 53 IRCiSP.4 kN-m/m Resisting Moment Mr = (VI * L/2 +Lc/2 x V2+V3*L/2) = 9641.29 < 1047.80 kN/m2 qr< 1047.46 5.14 kN/m2 456.50 quit = qNq+ 0.752 < < Bearing pressure qr due to Meyerhof distribution qr = Rv L-2e L is the reinforcement length at the base of the wall Rv is the resultant of all factored vertical loads = 2566.75 m HIGH WALL BY STATIC ANALYSIS Check for Rupture For reinforced soil: kal = 0.29 kN/m2 + Y* Dm fms is partial material factor applied to qu.307 . 102-2014 COMPUTATIOM OF INTERNAL STABILITY FOR 10.048 m 2.504 = qr < quit/ fms 466.= 2578.8 7062.3 KN-m/m Eccentricity (e) of resultant load Rv about the centre line of the base of width L (Mr-M0) e= L/2 I(V1+V2+V3) = 3.80- = 1. *L7fms) Rh is the horizontal factored disturbing force Rv is the vertical factored resultant force 0'p is the peak angle of shearing resistance under effectiv fms is the partial materials facto 0'p.55 m L 7.57 . C'.3 ( Px+Pz)* ffs 635.0 r1 32 ° For retained backfill soil : ka2 = 0.53 < 54 848.80 kN/m Rv = (vx + v2) 1508. * H Check for internal Sliding Calculation for bottom layer of Geogrid For long term stability where thercontact at the base of the structure fsRh< Rv(a1*tan0'p/fm$)+(C.333 H 10.CU fs is the partial factor against base sliding L' is the effective base width for sliding Rh = = 1.8 kN/m Horizontal Forces Pa = 1/2 * ka2*v2* H2 = 343.60 m Foundation Properties Dm = 1m r 18 kN/m3 LOADS Self weight of Reinforced soil wall Vj = Vi*H*L = 1483.05 kN/m 826.72 kN/m Vertical load due to live load Vj = Q| x L 174.P due to LL P2 = ka2 * Q.02 kN/m 4) E.33 kN/m Vertical load due to strip load Qx b V2 = 24. 07 KN/m .836 KN/m Tj “ TW + Tsi Tj = 70.307 Svj 0.5*(E2-E1)+E1 0.1 kN-m/m ffs Resisting moment Mr = (Vx * L/2 +Lc/2 x V2+V3*L/2)*ffs Mr = 9481.505 m ovj = Rvj / (L-2e) = Tpj 446.102-2014 1.875 m Tsj = 0.81 m SvJ = 0.73 KN/m Dj = 6.SP.23 kN/m Considering the crash barrier as a strip load and where Tsj = (ka * Svj * ff * SJ / Dj Dj = ((hj +b)/2) + d calculating Tsj for the bottom most Grid layer Ka 0.5 SL = 24. e = Elevation of Geogrid Layer El = 0.0 KN-m/m Eccentricity.IRC:SP:102-2014 MOMENTS Overturning Moment Mo = Mo = (Pi * H/3*ffs + P2 * H/2 *ffs) 2449.2 m E2 = 0.42 kN/m2 = kai * ovj * Svj = 69.505 m ff = 55 IRC. 8006- 1.5 hj = 10. 8006-1 : 20 fn 1.53 kN/m 31.94 TD= Tult/(RFd*RFid*RFcr) = 78.1 from Table 9 of BS: 8006 Check for Pullout Inclination of failure surface w.0 Strength Reduction factors Durability (RFd) Installation damage(RFjd) (based on type of soil) Creep (RFcr) 1.t horizontal Elevation from bottom wall batter (w) 6 = 45 + 0/: = 45 + 32/2 = El = 0. 8006-1 2 1 O.203 m 4.51 20.5 from Table 11 of BS.23° E.r.0 60.K 56 1RC:SP: 102-2014 from Table 11 of BS: 8006-1 : 20 p = 0.Geogrid D Geogrid Type #2 Geogrid Type #1 Tult(kPa) 40.3 from Table 11 of BS. Le = L h E x Tan(co) Tan{i//) Effective Length Le = 7.4 .1 ( Varies based on temp of soil) T design (kPa) =TU|t/(RFdxRFidxRFc For Type #6 : 1.1 >1.41 fn=TD/Tj 1.55 m ff = fp < fp < 1.5 from Table 11 of BS: 8006 Pj * p* Le*(ffs*Vl*hj)/Tj*fn 28.1 from Table 9 of BS.47 m Tt Perimeter of the jth layer: J\J n C cohesion of the soil =0 KN/m2 ffs 1. 0m length is provided at an elevation of 0.07 KN/m As Tj is greater than TU|tconn(n) a secondary reinforcement of 1.3 < KN/m ^ Tu|tconn (n) 61. 70.77 KN/m T.81 KN/m acs 19.15 m Density of plain concrete = W„ = Block unit width front to back = w{n) 72.8 .Case A Check for Connection Strength The ultimate connection strength Tu|tconn (n) at each geosynthetic reinforcement shall be calculated as Tultconn (n) — acs ^A/w(n)tanAcs where TU|tconn(n) = ultimate connection strength acs = apparent minimum connect Acs = apparent angle of friction f acs . Acs Ww(n) shall be considered from Type #6 = (H-Ej)*Yu*Wu Connection calculations for bottom most Grid : H= yu wall height = 10.406/0.609m = Therefore 5^for the bottom most layer for connection Tj = 70.505 56.71 KN/m Acs 30 1 ultconn (n) From Case A 61.3 Tj 56.07*0. Reinforced Soil Wall (Static) ° ° . 15 m +0.e 10.5 uuuuuuu .i.6 m road crust Ultimate limit state Serveciability limit state combinations B C 1.5 m 1. K .m m a) External anti infernal stability m (Critical case for sliding) effective stress conditions O. base of width L I(V1+V2+V3) L/6 1.K 466.29 kN/m2 Nc +q Nq+ 0.27 O.K .5 (L-2e) yf Ny} O. 8 m b = 1.8 m 80.= 18.8 m X3 = 3.6 m X2 = 0.5 KN/m3 Lever Arm where XI = 3.K .84 kN/m of shearing resistance under effective stress conditions O. 6 m d = 0.98m2 Density of Conc where hj = SL = A*yc (10.014 m < L/6 < 1.28 3.9 2524.27 O.786 1.75-0.5 from Table L/2- 12 of BS 8006-1 : 2010 (Mr-M0) I(V1+V2+V3)*1.5 3.8 m where Crash barrier area (A) = 0.8 7031.6m b = 1.2) for bottom grid 25 KN/m3 .80- 2.= 1.K First layer from bottom Second layer from bottom (for bottom grid) calculating the tension due to the st 1. 0 1.0 1.88 52.15 Geogrid Type #6 150.K jlli’ C t r-.82 78. 8006-1 /ms from Table 11 of BS: 8006-1 : 2010 where a1 = 0.51 41.1 1.15 1. from Table 11 of BS.0 100. 8006.8 from Table 11 of BS: 8006 Pj * p* Le*(ffs*Vl*hj)/Tj*fn Where fp = 1.15 1.35 from Table 9 of BS: 8006 62. 8006-1 : 2010 from Table 11 of BS.3 from Table 11 of BS: 8006-1: 2010 .a' tan from Table 9 of BS.-J |jfr.53 O.Geogrid Data Geogrid Type #3 Geogrid Type Geogrid Type #4 #5 80.0 120. 305 m Ei = 0.203J/2) 0.203+((0.K .203 m 0.609-0.406 m O.n) = ultimate connection strength acs = apparent minimum connection strength between geogrid reinforcement and block unit Acs = apparent angle of friction for connection of geogrid reinforcement and block unit shall be considered from Type #6 grid to block connection report 24 KN/m 0.