Abutment Design Example to Eurocodes and UK National Annexes

Scroll to top Index1.Actions from Bearings2.Actions on Back of Abutment3.Stability Check4.Wall Design5.Base Slab Design6.Curtain Wall(Upstand Wall) Design

Design the fixed and free end cantilever abutments to the 20m span deck shown to carry Load Model 1 and vehicles SV80, SV100 and SV196 for Load Model 3. Analyse the abutments using a unit strip method. The bridge site is located south east of Oxford (to establish the range of shade air temperatures). Vehicle collision on large abutments need not be considered as they are assumed to have sufficient mass to withstand the collision loads for global purposes.

The ground investigation report shows suitable founding strata about 9.5m below the proposed road level and 1.5m below existing ground level. Test results show the founding strata to be a well drained, cohesionless soil having an angle of shearing resistance (') = 34, a critical state angle of shearing resistance ('c) = 30 and a weight density = 19kN/m3.Backfill material will be Class 6N with an angle of shearing resistance ('bf;k) = 35 and weight density (bf;k) = 19kN/m3.

The proposed deck consists of 11No. Y4 prestressed concrete beams at 1m centres and concrete deck slab as shown.

EN 1997-1:2004 Clause 2.4.7.3.4.1(1)P - Use Design Approach 1 only for verification of resistance for structural and ground limit states in persistent and transient situations (STR and GEO).Consider Combination 1: A1 + M1 + R1 and Combination 2: A2 + M2 + R1A grillage analysis gave the following characteristic reactions for the various load cases:Critical Vertical Reaction Under One BeamCharacteristic Reaction(kN)ULS Reaction(kN)Concrete Deck180240Surfacing4560gr1a290430gr2220310gr5270400gr6210300Total Vertical Reaction on Each AbutmentCharacteristic Reaction(kN)A1(G;sup / G;inf)A2(G;sup / G;inf)Concrete Deck19001.35 / 0.951.0 / 1.0Surfacing3201.2 / 0.951.0 / 1.0gr1a14901.35 / 01.15 / 0gr211201.35 / 01.15 / 0gr519301.35 / 01.15 / 0gr614701.35 / 01.15 / 0Characteristic loading on 1m length of abutment:Deck Dead Load = 1900 / 11.6 = 164kN/mMaximum Surfacing = 1.55 320 / 11.6 = 43kN/mMinimum Surfacing = 0.6 320 / 11.6 = 17kN/mgr1a on Deck = 1490 / 11.6 = 128kN/mgr2 on Deck = 1030 / 11.6 = 89kN/mgr5 on Deck = 1930 / 11.6 = 166kN/mgr6 on Deck = 1470 / 11.6 = 127kN/mFrom UK NA to BS EN 1991-1-5:2003 Figures NA.1 and NA.2 the minimum and maximum shade air temperatures are -17 and +34C respectively. For bridge deck type 3 the corresponding minimum (Te,min) and maximum (Te,max) effective bridge temperatures are -11 and +36C from BS EN 1991-1-5:2003 Figure 6.1.Hence the temperature range = 11 + 36 = 47C.Form EN 1991-1-5 Table C.1 - Coefficient of thermal expansion for a concrete deck = 10 10-6 per C.However CIRIA Report C660 ("Early-age thermal crack control in concrete") suggests that a value of 10 10-6 per C is unsuitable for some of the concrete aggregates used in the UK and suggest a value of 12 10-6 per C should be used if the type of aggregate has not been specified.Hence the range of movement at the free end of the 20m span deck = 47 12 10-6 20 103 = 11.3mm.The design thermal movement in the deck will be [(11.3 / 2) F] = [11.3 1.35 /2] = 8mm.

Option 1 - Elastomeric Bearing:With a maximum ultimate reaction = 240 + 60 + 430 = 730kN then a suitable elastomeric bearing would be Ekspan's Elastomeric Pad :Bearing EKR35: Maximum Load = 1053kN Shear Deflection = 13.3mm Shear Stiffness = 12.14kN/mm Bearing Thickness = 19mmNote: the required shear deflection (8mm) should be limited to between 30% to 50% of the thickness of the bearing. The figure quoted in the catalogue for the maximum shear deflection is 70% of the thickness. A tolerance is also required for setting the bearing if the ambient temperature is not at the mid range temperature. The design shade air temperature range will be -17 to +34C which would require the bearings to be installed at a shade air temperature of [(34+17)/2-17] = 9C to achieve the 8mm movement. If the bearings are set at a maximum shade air temperature (T0) of 16C then, by proportion the deck will expand 8(34-16)/[(34+17)/2] = 6mm and contract 8(16+17)/[(34+17)/2] = 10mm. Let us assume that this maximum shade air temperature of 16C for fixing the bearings is specified for T0 in the Contract and design the abutments accordingly.Horizontal load at bearing for 10mm contraction = 12.14 10 = 121kN.This is an ultimate load hence the characteristic horizontal load = 121 / 1.35 = 90kN.If a fixed abutment is used then the movement will take place at one end so:Total horizontal load on each abutment = 11 90 = 990 kN 990 / 11.6 = 85kN/m.If no fixed abutment is used then the movement will take place at both ends so:Total horizontal load on each abutment = 85/2 = 43kN/m. Option 2 - Sliding Bearing:With a maximum ultimate reaction of 730kN and longitudinal movement of 8mm then a suitable bearing from the Ekspan EA Series would be /80/210/25/25: Maximum Load = 800kN Base Plate A dimension = 210mm Base Plate B dimension = 365mm Movement X = 12.5mmAverage characteristic permanent load reaction = (1900 + 320) / 11 = 2220 / 11 = 200kNContact pressure under base plate = 200000 / (210 365) = 3N/mm2As the mating surface between the stainless steel and PTFE is smaller than the base plate then the pressure between the sliding faces will be in the order of 5N/mm2.Ekspan recommend a coefficient of friction = 0.05, however use a coefficient of friction = 0.08 for long term exposure conditions.Hence total horizontal load on each abutment when the deck expands or contracts = 2220 0.08 = 180kN 180 / 11.6 = 16kN/m.Braking and Acceleration Force - BS EN 1991-2:2003 Clause 4.4.1:(2) Characteristic Force for LM1 = 0.6Q1(2Q1k)+0.1q1qq1w1L = 0.6 1 (2 300) + 0.1 1 9 3 20 = 414kNFor global effects, braking force on 1m width of abutment = 414 / 11.6 = 36kN/m.(NA. 2.18.1) Characteristic Force for LM3 (SV196) = Qlk,s = w = 0.25 (165kN 9axles + 180kN 2axles + 100kN 1axle) = 486kNFor global effects, braking force on 1m width of abutment = 486 / 11.6 = 42kN/m.When this load is applied on the deck it will act at bearing shelf level, and will not affect the free abutment if sliding bearings are used.Note: Braking forces should not be taken into account at the surfacing level of the carriageway over the backfill (See BS EN 1991-2:2003 Cl. 4.9.2)

Loading at Rear of AbutmentBackfill

For Stability calculations use active earth pressures = Ka bf;k hFor Design of Structural Members use at-rest earth pressures = K0 bf;k hSLSCombination 1Combination 2Partial factors for soil parameters M1.01.01.25'bf;d = tan-1[tan('bf;k)/M]35.035.029.3Ka = (1-Sin'bf;d) / (1+Sin'bf;d)0.2710.2710.343K0 = 1-Sin'bf;d0.4260.4260.511Partial factors for soil weight G (sup/inf)1.0/1.01.35/0.951.0/1.0Backfill density (bf;d) = bf;k G;sup19.025.6519.0Backfill density (bf;d) = bf;k G;inf19.018.119.0Model factors Sd;K 1.01.21.2Hap;d = bf;dKaSd;KZ2/22.575Z2kN/m4.171Z2kN/m3.91Z2kN/mSurcharge - Use Horizontal Surcharge Model in PD 6694-1:2011 Figure 2:Carriageway width = 7.3m there are 2 notional lanes of effective width Weff of 3m with 1.3m wide remaining area (see Table 4.1 of BS EN 1991-2:2003).The vehicle model for loads on backfill behind abutments is positioned in each notional lane (see Clause NA.2.34.2 UK NA to BS EN 1991-2) the effective number of lanes (Nlane) in the surcharge model will be 2.

From PD 6694-1:2011 Table 7 :Normal TrafficLine Load kN/m = Hsc;F = F.Kd.Nlane/Wabut = 2330Kd2/11.6 = 113.79KdNote: Df is used for determining the distibution of the Line Load in the wall for a metre strip analysis, but is not included in the calculation when considering the overall stability of the wall.UDL kN/m2 = h;ave = h.Wlane.Nlane/Wabut = 20Kd32/11.6 = 10.34KdSV/196 Traffic (SV/196 lane 1 + Frequent value of Normal Traffic in Lane 2)Line Load kN/m (at ground level) = Hsc;F = F.Kd.(1 + 1)/Wabut = 2330Kd1.75/11.6 = 99.57KdUDL kN/m2 = h;ave = (h1+1.h2).Wlane/Wabut = (30+0.7520)Kd3/11.6 = 11.64KdSLSCombination 1Combination 2Partial Factor on Surcharge Q1.01.351.15Assume Abutments are to be backfilled in accordance with the Highways Agency Manual of Contract Documents for Highway Works (MCHW), then compaction pressures due to construction vehicles are deemed to be coverered if the surcharge model in Figure 2 and Table 7 of PD 6694-1:2011 (as shown above) is employed (see PD 6694-1:2011 Clause 7.3.3).

1) Stability Check

Initial Sizing for Base DimensionsThere are a number of publications that will give guidance on base sizes for free standing cantilever walls, Reynolds's Reinforced Concrete Designer's Handbook being one such book.Alternatively a simple spreadsheet will achieve a result by trial and error.Load Combinations

Backfill + Construction surcharge(Not used - backfilled to MCHW - see PD 6694-1:2011 Clause 7.3.3)

Backfill + Normal Traffic Surcharge + Deck Permanent load + Deck contraction/shrinkage

Backfill + Normal Traffic Surcharge + Deck Permanent load + gr1a on deck

Backfill + SV/100 and SV/196 Surcharge + Deck Permanent load + gr1a (frequent value) on deck

Backfill + Normal Traffic Surcharge (frequent value) + Deck Permanent load + gr5 on deck

Backfill + Normal Traffic Surcharge (frequent value) + Deck Permanent load + gr2 (1LM1 with braking on deck)(Braking not applied to free abutment if sliding bearings are provided)

Backfill + Deck Permanent load + gr6 (LM3 with braking on deck)(Braking not applied to free abutment if sliding bearings are provided)Example of Stability Calculations:CASE 6 - Fixed Abutment

Density of reinforced concrete = 25kN/m3.SLS (G = Q = Sd;K = 1.0)Weight of wall stem = G twall Zwall conc = 1.0 1.0 6.5 25 = 162.5kN/mWeight of base = G Wbase Zbase conc = 1.0 6.4 1.0 25 = 160kN/mWeight of backfill = G Wheel Zheel bf;d = 1.0 4.3 8.5 19 = 694.5kN/mBackfill Force Hap;d = G Sd;K Ka bf;d Z2/2 = 1.0 1.0 0.271 19 9.52 / 2 = 232kN/mFrequent value of Surcharge UDL Force Hsc;udl = 1 Q h;ave Z = 0.75 1.0 (10.34 0.271) 9.5 = 20 kN/mFrequent value of Surcharge Line Load Force Hsc;F = 1 Q Hsc;F = 0.75 1.0 (113.79 0.271) = 23 kN/mDeck Maximum Permanent load (concrete + surfacingmax) = G VDL = 1.0 (164 + 43) = 207kN/mDeck Minimum Permanent load (concrete + surfacingmin) = G VDL = 1.0 (164 + 17) = 181kN/mDeck Vertical Traffic load (gr2) = Q Vtraffic = 1.0 89 = 89kN/mDeck Horizontal Traffic load (gr2) = Q Hbraking = 1.0 36 = 36kN/mRestoring Effects:MinimumVLever ArmMoment About AStem162.51.6260Base1603.2512Backfill694.54.252952Deck (VDLmin)1811.55281 = 1198 = 4005MaximumVLever ArmMoment About AStem162.51.6260Base1603.2512Backfill694.54.252952Deck (VDLmax + Vtraffic)2961.55459 = 1313 = 4183Overturning Effects:HLever ArmMoment About AHap;d2323.167735Hsc;udl204.7595Hsc;F239.5219Hbraking367.5270 = 311 = 1319For sliding effects:'c = 30Partial factor on M on tan('c;k) = 1.0Coefficient of friction = d = tan('c;k) / M = tan(30)/1.0 = 0.58Sliding resistance = dVmin = Rx;d = 0.58 1198 = 695kN/mActive Force = H = 311kN/m < 695 OKBearing Pressure:PD 6694-1 Cl. 5.2.2 requires no uplift at SLSCheck bearing pressure at toe and heel of base slab = (V / A) (V e y / I) where V e is the moment about the centre of the base.V = 1313kN/mA = 6.4m2/mI / y = 6.42 / 6 = 6.827m3/mNett moment = 4183 - 1319 = 2864kNm/mEccentricity (e) of V about centre-line of base = 3.2 - (2864 / 1313) = 1.019mPressure under base = (1313 / 6.4) (1313 1.019 / 6.827)Pressure under toe = 205 + 196 = 401kN/m2 Pressure under heel = 205 - 196 = 9kN/m2 > 0 OK (no uplift)AlsoBS EN 1997-1:2004, 2.4.8(4), allows the serviceability limit state for settlement to be verified by ensuring that a sufficiently low fraction of the ground strength is mobilized. This requirement can be deemed to be satisfied if the maximum pressure under a foundation at SLS does not exceed one third of the design resistance R/A' calculated in accordance with BS EN 1997-1:2004, Annex D, using characteristic values of ', cu and ' and representative values of horizontal and vertical actions.From Annex D.4 for Drained Conditions:R/A' = c'Ncbcscic + q'Nqbqsqiq + 0.5'B'Nbsic' = 0' = G;inf = 19 1.0 = 19kN/m3q' = 1.5 19 = 28.5 kN/m2 (Foundation 1.5m below existing ground level)Nq = etan'dtan2(45 + 'd / 2)'d = ' = 34 (M = 1.0)Nq = etan34tan2(45 + 34 / 2) = 29.4N = 2(Nq - 1)tan'd = 2 (29.4 - 1) tan34 = 38.3bq = b = 1.0 ( = 0)B' = B - 2e = 6.4 - 2 1.019 = 4.362sq = 1 + (B' / L')sin'd = 1 + (4.362 / 11.6) sin34 = 1.21s = 1 - 0.3(B' / L') = 1 - 0.3(4.362 / 11.6) = 0.89m = (2 + B' / L') / (1 + B' / L') = (2 + 4.362 / 11.6) / (1 + 4.362 / 11.6) = 1.73iq = [1 - H / (V + A'c'dcot'd)]m = [1 - 311 / 1313]1.73 = 0.63i = [1 - H / (V + A'c'dcot'd)]m+1 = [1 - 311 / 1313]2.73 = 0.48R/A' = 0 + (28.5 29.4 1.0 1.21 0.63) + (0.5 19 4.362 38.3 1.0 0.89 0.48 = 639 + 678 = 1317 kN/m21/3(R/A') = 1317 / 3 = 439 kN/m2 > 401 kN/m2 settlement check OK. ULS Check Combination 1 and Combination 2. G;supG;infQSd;KCombination 1Combination 1 Surfacing1.351.20.950.951.351.2Combination 21.001.001.151.2Comb.1Comb.2Min. weight of wall stem = G;inf twall Zwall conc154162.5Max. weight of wall stem = G;sup twall Zwall conc219162.5Min. weight of base = G;inf Wbase Zbase conc152160Max. weight of base = G;sup Wbase Zbase conc216160Min. weight of backfill = G;inf Wheel Zheel bf;d660694.5Max. weight of backfill = G;sup Wheel Zheel bf;d937.5694.5Ka0.2710.343Backfill Force Hap;d = G;sup Sd;K Ka bf;d Z2/2376353Frequent value of Surcharge UDL Force Hsc;udl = 1 Q h;ave Z2729Frequent value of Surcharge Line Load Force Hsc;F = 1 Q Hsc;F3133.7Deck Maximum Permanent load (concrete + surfacingmax) = G;sup VDL273207Deck Minimum Permanent load (concrete + surfacingmin) = G;inf VDL172181Deck Vertical Traffic load (gr2) = Q;sup Vtraffic120102Deck Horizontal Traffic load (gr2) = Q Hbraking4941

Combination 1Restoring Effects :MinimumVLever ArmMoment About AStem1541.6246Base1523.2486Backfill6604.252805Deck (VDLmin)1721.55267 = 1138 = 3804MaximumVLever ArmMoment About AStem2191.6350Base2163.2691Backfill937.54.253984Deck (VDLmax + Vtraffic)3931.55609 = 1765.5 = 5634Overturning Effects:HLever ArmMoment About AHap;d3763.1671191Hsc;udl274.75128Hsc;F319.5295Hbraking497.5368 = 483 = 1982For sliding effects:'c = 30Partial factor on M on tan('c;k) = 1.0Coefficient of friction = d = tan('c;k) / M = tan(30)/1.0 = 0.58Sliding resistance = dVmin = Rx;d = 0.58 1138 = 660kN/mActive Force = H = 483kN/m < 660 OKBearing Pressure:EN 1997-1:2004 Cl. 6.5.4 restrict eccentricity of loading to 1/3 of the width of the footing at ULSV = 1765.5kN/mNett moment = 5634 - 1982 = 3652kNm/mEccentricity (e) of V about centre-line of base = 3.2 - (3652 / 1765.5) = 1.131mBase width / 3 = 6.4 / 3 = 2.131 > 1.131 OKAssume rectangular pressure distribution under the base as described in EN 1997-1:2004 Annex DEffective base width B' = B - 2e = 6.4 - 2 1.131 = 4.138mPressure under base = (1765.5 / 4.138) = 427kN/m2R/A' = c'Ncbcscic + q'Nqbqsqiq + 0.5'B'Nbsic' = 0' = G;inf = 19 0.95 = 18.1kN/m3q' = 1.5 18.1 = 27.2kN/m2 (Foundation 1.5m below existing ground level)Nq = etan'dtan2(45 + 'd / 2)'d = ' = 34 (M = 1.0)Nq = etan34tan2(45 + 34 / 2) = 29.4N = 2(Nq - 1)tan'd = 2 (29.4 - 1) tan34 = 38.3bq = b = 1.0 ( = 0)B' = 4.138sq = 1 + (B' / L')sin'd = 1 + (4.138 / 11.6) sin34 = 1.20s = 1 - 0.3(B' / L') = 1 - 0.3(4.138 / 11.6) = 0.89m = (2 + B' / L') / (1 + B' / L') = (2 + 4.138 / 11.6) / (1 + 4.138 / 11.6) = 1.74iq = [1 - H / (V + A'c'dcot'd)]m = [1 - 483 / 1765.5]1.74 = 0.57i = [1 - H / (V + A'c'dcot'd)]m+1 = [1 - 483 / 1765.5]2.74 = 0.42R/A' = 0 + (27.2 29.4 1.0 1.20 0.57) + (0.5 18.1 4.138 38.3 1.0 0.89 0.42 = 547 + 536 = 1083 kN/m2 > 427 kN/m2 OK.

Combination 2Restoring Effects :MinimumVLever ArmMoment About AStem162.51.6260Base1603.2512Backfill694.54.252952Deck (VDLmin)1811.55281 = 1198 = 4005MaximumVLever ArmMoment About AStem162.51.6260Base1603.2512Backfill694.54.252952Deck (VDLmax + Vtraffic)3091.55479 = 1326 = 4203Overturning Effects:HLever ArmMoment About AHap;d3533.1671118Hsc;udl294.75138Hsc;F33.79.5320Hbraking417.5308 = 457 = 1884For sliding effects:'c = 30Partial factor on M on tan('c;k) = 1.25Coefficient of friction = d = tan('c;k) / M = tan(30)/1.25 = 0.46Sliding resistance = dVmin = Rx;d = 0.46 1198 = 551kN/mActive Force = H = 457kN/m < 551 OKBearing Pressure:EN 1997-1:2004 Cl. 6.5.4 restrict eccentricity of loading to 1/3 of the width of the footing at ULSV = 1326kN/mNett moment = 4203 - 1884 = 2319kNm/mEccentricity (e) of V about centre-line of base = 3.2 - (2319 / 1326) = 1.451mBase width / 3 = 6.4 / 3 = 2.131 > 1.451 OKAssume rectangular pressure distribution under the base as described in EN 1997-1:2004 Annex DEffective base width B' = B - 2e = 6.4 - 2 1.451 = 3.498mPressure under base = (1326 / 3.498) = 379kN/m2R/A' = c'Ncbcscic + q'Nqbqsqiq + 0.5'B'Nbsic' = 0' = G;inf = 19 1.0 = 19kN/m3q' = 1.5 19 = 28.5kN/m2 (Foundation 1.5m below existing ground level)Nq = etan'dtan2(45 + 'd / 2)'d = tan-1[tan('k)/M] = tan-1[tan34/1.25] = 28.4Nq = etan28.4tan2(45 + 28.4 / 2) = 15.4N = 2(Nq - 1)tan'd = 2 (15.4 - 1) tan28.4 = 15.6bq = b = 1.0 ( = 0)B' = 3.498sq = 1 + (B' / L')sin'd = 1 + (3.498 / 11.6) sin28.4 = 1.14s = 1 - 0.3(B' / L') = 1 - 0.3(3.498 / 11.6) = 0.91m = (2 + B' / L') / (1 + B' / L') = (2 + 3.498 / 11.6) / (1 + 3.498 / 11.6) = 1.768iq = [1 - H / (V + A'c'dcot'd)]m = [1 - 457 / 1326]1.768 = 0.47i = [1 - H / (V + A'c'dcot'd)]m+1 = [1 - 457 / 1326]2.768 = 0.31R/A' = 0 + (28.5 15.4 1.0 1.14 0.47) + (0.5 19 3.498 15.6 1.0 0.91 0.31) = 235 + 146 = 381 kN/m2 > 379 kN/m2 OK.Analysing Load Cases 2 to 7 for the fixed abutment and the free abutment using a simple spreadsheet the following results were obtained:Notation:Case 2, 6 and 7 - results of fixed abutment with dowels and free abutment with sliding bearings.Case 2a, 6a and 7a - results of fixed abutment with dowels and free abutment with elastomeric bearings.Case 2b, 6b and 7b - results of both abutments with elastomeric bearings.All other cases are not affected by the bearing arrangement.Fixed Abutment:SlidingSLSComb.1Comb.2Resistance692657553Case 2Case 2a306375476569453522Case 3290454437Case 4289453436Case 5275435416Case 6 & 6a311483458Case 7& 7a274433402BearingPressureSLSToeSLSHeelComb.1Vd/A' / R/A'Comb.2Vd/A' / R/A'Case 2Case 2a35943523-52384 / 1054478 / 773341 / 354447 / 257Case 339330418 / 1199370 / 426Case 437736402 / 1188354 / 419Case 539242418 / 1281362 / 473Case 6 & 6a4019427 / 1082380 / 380Case 7 & 7a37745403 / 1265336 / 485Free Abutment:SlidingSLSComb.1Comb.2Resistance711675568Case 2Case 2aCase 2b312382339486580522463532490Case 3297465447Case 4269464447Case 5282445426Case 6 & 6aCase 6b282307445479426455Case 7 & 7aCase 7b239260387415363387BearingPressureSLSToeSLSHeelComb.1Vd/A' / R/A'Comb.2Vd/A' / R/A'Case 2Case 2aCase 2b36143639025-49-3386 / 1070478 / 793417 / 957342 / 360444 / 262375 / 320Case 339531420 / 1212371 / 431Case 437937404 / 1202355 / 424Case 539443421 / 1294364 / 477Case 6 & 6aCase 6b3643915023390 / 1257416 / 1145337 / 453366 / 405Case 7 & 7aCase 7b3343659267368 / 1480395 / 1387300 / 578326 / 539Note:1) Numbers in bold indicate failed results.2) Slight differences in results between the example and the spreadsheet for Case 6 are due to rounding off errors in the example.It can be seen that the use of elastomeric bearings (Case 2) will govern the critical design load cases on the abutments. We shall assume that there are no specific requirements for using elastomeric bearings and design the abutments for the lesser load effects by using sliding bearings.

2) Wall and Base DesignLoads on the back of the wall are calculated using 'at rest' earth pressures. Serviceability and Ultimate load effects need to be calculated for the load cases 2 to 7 shown above. Again, these are best carried out using a simple spreadsheet.Using the Fixed Abutment Load Case 6 again as an example of the calculations:Wall DesignSLSCombination 1Combination 2Partial factors for soil parameters M1.01.01.25'bf;d = tan-1[tan('bf;k)/M]35.035.029.3K0 = 1-Sin'bf;d0.4260.4260.511Partial factors for soil weight G;sup 1.01.351.0Backfill density (bf;d) = bf;k G;sup19.025.719.0Model factors Sd;K 1.01.21.2Hap;d = bf;dK0Sd;KZ2/24.047Z2kN/m6.569Z2kN/m5.825Z2kN/mConsider a section at the base of the wall (Z = 8.5m)Backfill:Hap;d(kN) =292475421Moment (kNm) (lever arm = 8.5/3) =82712951193Frequent value of Normal Surcharge:1Q;sup =0.751.0130.8631Q;supHsc;F = 1Q;sup113.79KdDf =243334Moment (kNm) (lever arm = 8.5) =2042812891Q;suph;aveZ = 1Q;sup10.34KdZ =283839Moment (kNm) (lever arm = 4.25) =119162166Deck Permanent Load Reaction:G;sup for concrete =1.01.351.0Deck concrete = 164221164G;sup for surfacing =1.01.21.0Deck surfacing = 435243Moment = V e (e = 0.5 - 0.45 = 0.05) = 101410Deck Variable Reaction (gr2):Q;sup =1.01.351.15Variable Vertical Reaction = 89120102Moment = V e (e = 0.5 - 0.45) = 465Variable Horizontal Reaction (Braking) = 364941Moment = H 6.5 = 234319267Comb.1 shear at base of wall = H = 475 + 33 + 38 + 49 = 595kNComb.2 shear at base of wall = H = 421 + 33 + 38 + 41 = 533kNSLS moment at base of wall = M = 827 + 204 + 119 + 10 + 4 + 234 = 1398kNm (837 permanent + 561 variable)ULS Comb.1 moment at base of wall = M = 1295 + 281 + 162 + 14 + 6 + 319 = 2077kNmULS Comb.2 moment at base of wall = M = 1193 + 289 + 166 + 10 + 5 + 267 = 1930kNmAnalysing the fixed abutment and free abutment with Load Cases 2 to 7 using a simple spreadsheet the following results were obtained for the design moments and shear at the base of the wall:(Note - Slight differences in results between the example and the spreadsheet for Case 6 are due to rounding off errors in the example.)Fixed Abutment:SLS Moment(Permanent)SLS Moment(Variable)SLS Moment(Total)Case 28405391379Case 38404421282Case 48404271267Case 58403351175Case 68405651405Case 78402791119MomentULS Comb.1ShearULS Comb.1MomentULS Comb.2ShearULS Comb.2Case 220865901908534Case 319545691811518Case 419345701792519Case 518095451664494Case 621205941928535Case 717345311525469

Free Abutment:SLS Moment(Permanent)SLS Moment(Variable)SLS Moment(Total)Case 28785511429Case 38784511329Case 48784371315Case 58783421220Case 68783381216Case 78787885MomentULS Comb.1ShearULS Comb.1MomentULS Comb.2ShearULS Comb.2Case 221636061979547Case 320295841880531Case 420095851860532Case 518815601729507Case 618765601724507Case 714284891266434

Concrete to BS 8500:2006Use strength class C32/40 with water-cement ratio 0.5 and minimum cement content of 340kg/m3 for exposure condition XD2.Nominal cover to reinforcement = 60mm (45mm minimum cover plus a tolerance c of 15mm).Reinforcement to BS 4449:2005 Grade B500B: fy = 500N/mm2Design for critical moments and shear in Free Abutment:

Check slenderness of abutment wall to see if second order effects need to be considered:EN 1992-1-1 clause 5.8.3.1 = 0/i lim = 20A.B.C/nUse suggested values when ef not known:A = 0.7, B = 1.1, C = 0.7n = NEd / (Acfcd)NEd = 164 + 43 + 166 = 373kNfcd = ccfck / c = 0.85 32 / 1.5 = 18.1N/mm2Ac = 106mm2 (per metre width)n = 373 103 / (106 18.1) = 0.021lim = 20 0.7 1.1 0.7 / 0.021 = 74.40 = 2 = 2 6.63 = 13.26m (cantilever with sliding bearings to deck)i = (1/12) = 0.289m = 13.26 / 0.289 = 45.9 < 74.4 OK, second order effects need not be considered.EN 1992-1-1 & EN 1992-2It is usual to design reinforced concrete for the ultimate limit state and check for serviceability conditions.MULS = 2163kNm/m, VULS = 606kN/m, MSLS = 1429kNm/m [878(permanent)+551(variable)]cl. 3.1.6(101)PDesign compressive strength = fcd = ccfck / ccl. 3.1.7cc = 0.85cl. 2.4.2.4Table 2.1N: c = 1.5, s = 1.15fcd = 0.85 32 / 1.5 = 18.1 N/mm2Table 3.1c2 = 0.002, cu2 = 0.0035, n = 2.0Try 40mm dia. reinforcement at 150mm centres (8378mm2/m):Nominal cover to reinforcement in rear face of wall = 60mmd = 1000 - 60 - 20 = 920mmFig. 3.3Using parabolic-rectangular diagram:Average stress fav = fcd[1-c2 / {cu2(n+1)}] = 18.1 [1 - 0.002 / {0.0035 (2 + 1)}] = 14.7 N/mm2 Assuming steel yields then:M = fsz = fykAsz / s = Fcz = favbXzDepth to neutral axis X = fykAs / (favbs)X = 500 8378 / (14.7 1000 1.15) = 247.8mmCheck that steel will yield:Cl. 3.2.7(4)Modulus of Elasticity Es = 200 kN/mm2Steel strain at yield = s,yield = fyk / s / Es = 500 / 1.15 / 200000 = 0.00217from linear strain relationship:s = cu2(d/X - 1) = 0.0035 ( 920 / 247.8 - 1) = 0.009 > 0.00217 steel will yield.Hence Mult = favbXz = favbX(d - X)Where = 1 - [0.5cu22 - c22 / {(n+1)(n+2)}] / [cu22 - cu2c2 / (n+1)] = 1 - [0.5 0.00352 - 0.0022 / {(2 + 1) (2 + 2)}] / 0.00352 - 0.0035 0.002 / (2 + 1)] = 0.416Mult = 14.7 1000 247.8 (920 - 0.416 247.8) 10-6 = 2976 kNm > 2163 OK

Check Serviceability Limit StateCharacteristic Combination SLS Design Moment = 1429kNm/m (878 +551)Check stresses in the concrete and reinforcement at:i) Early Age (before creep has occurred)ii) Long term after all the creep has taken place.i) Before creep has occurred the cracked section properties will be based on the short-term modulus for all actions.EN 1992-1-1 Table 3.1Ecm = 22[(fck + 8) / 10]0.3 = 22[(32 + 8) / 10]0.3 = 33.4 kN/mm2Ec,eff = Ecm = 33.4 kN/mm2Modular Ratio m = Es / Ecm = 200 / 33.4 = 6.0Let dc = depth to neutral axis then equating strains for cracked section:s = c(d - dc) / dcEquating forces:AsEss = 0.5bdccEc,effHence dc = [-AsEs + {(AsEs)2 + 2bAsEsEc,effd}0.5] / bEc,effdc = [-8378 200000 + {(8378 200000)2 + 2 1000 8378 200000 33400 920}0.5] / (1000 33400) = 258mmCracked second moment of area = As(d-dc)2 + Ec,effbdc3 / 3EsINA = 8378 (920 - 258)2 + 33.4 1000 2583 / (3 200) = 4.63 109 mm4 (steel units)Approximate concrete stress c = M / zc + N / AcN (Case 2) = 164 + 43 = 207 kNc {1429 106 258 / (4.63 109 6.0)} + {207 103 / (258 103)} = 13.3 + 0.8 = 14.1 N/mm2cl. 7.2(102)Limiting concrete stress = k1fckk1 = 0.6Limiting concrete stress = 0.6 32 = 19.2 N/mm2 > 14.1 OKEN 1992-1-1ii) After all creep has taken place the cracked section properties will be based on the long-term and short-term modulus for the various actions.Short-term modulus = EcmLong-term modulus = Ecm / (1+)Effective modulus Ec,eff = (Mqp + Mst)Ecm / {Mst + (1 + )Mqp} Table 3.1fcm = fck + 8 = 32 + 8 = 40 N/mm2Cl. 3.1.4Relative humidity of the ambient environment = 80% (outside conditions)Age of concrete at initial loading t0 = say 7 days (after formwork has been released and waterproofing system applied to rear face of wall)Annex B (B.6)&(B.8c)h0 = 2Ac / U = 2 (11600 1000) / (11600 + 2 1000) = 17061 = [35 / fcm]0.7 = [35 / 40]0.7 = 0.91 2 = [35 / fcm]0.2 = [35 / 40]0.2 = 0.97 (B.3b)RH = [1 + 1 {(1 - RH / 100) / (0.1 h01/3)}] 2RH = [1 + 0.91 {(1 - 80 / 100) / ( 0.1 17061/3)}] 0.97 = 1.118(B.4)(fcm) = 16.8 / fcm0.5 = 16.8 / 400.5 = 2.656(B.5)(t0) = 1 / (0.1 + t00.2) = 1 / ( 0.1 + 70.2) = 0.635(B.2)0 = RH (fcm) (t0) = 1.118 2.656 0.635 = 1.886Moment due to long-term actions = Mqp = 878 kNmMoment due to short-term actions = Mst = 551 kNmHence Effective Modulus Ec,eff = {(878 + 551) 33.4} / {551 + 878 ( 1 + 1.886)} = 15.5 kN/mm2Modular Ratio m = Es / Ec,eff = 200 / 15.5 = 12.9Let dc = depth to neutral axis then equating strains for cracked section:s = c(d - dc) / dcEquating forces:AsEss = 0.5bdccEc,effHence dc = [-AsEs + {(AsEs)2 + 2bAsEsEc,effd}0.5] / bEc,effdc = [-8378 200000 + {(8378 200000)2 + 2 1000 8378 200000 15500 920}0.5] / (1000 15500) = 351mmCracked second moment of area = As(d-dc)2 + Ec,effbdc3 / 3EsINA = 8378 (920 - 351)2 + 15.5 1000 3513 / (3 200) = 3.83 109 mm4 (steel units)Concrete stress c M / zc + N / Acc = {1429 106 351 / (3.83 109 12.9)} + (207 103 / (351 103) = 10.2 + 0.6 = 10.8 N/mm2cl. 7.2(102)Limiting concrete stress = k1fckk1 = 0.6Limiting concrete stress = 0.6 32 = 19.2 N/mm2 > 10.8 OKcl. 7.2(5)Limiting steel stress = k3fykk3 = 0.8Limiting steel stress = 0.8 500 = 400 N/mm2Steel stress s = M / zss = 1429 106 (920 - 351) / (3.83 109) = 212 N/mm2 < 400 >OKCrack Control:Consider worst condition before creep has occurred and Quasi-Permanent Combination Moment + 2 temperature effects = 878 + 0.5(16 6.63) = 931 kNm Cl. 7.3.4(1)Crack width wk = sr,max(sm - cm)Cl. 7.3.4(3)Spacing Limit = 5(c+/2) = 5(60 + 40/2) = 400mm > 150mm OKsr,max = k3c + k1k2k4 / p,effk1 = 0.8 (high bond bars)k2 = 0.5 (for bending)k3 = 3.4 (recommended value)k4 = 0.425 (recommended value)Cl. 7.3.2(3)hc,eff is the lesser of:i) 2.5(h-d) = 2.5(1000 - 920) = 200ii) (h-x)/3 = (1000 - 258) / 3 = 247iii) h/2 = 1000 / 2 = 500 hc,eff = 200 mmand Ac,eff = 200 1000 = 200000 mm2Cl. 7.3.4(2)p,eff = As / Ac,eff = 8378 / 200000 = 0.0419sr,max = k3c + k1k2k4 / p,effsr,max = (3.4 60) + (0.8 0.5 0.425 40) / 0.0419 = 204 + 162 = 366Cl. 7.3.4(2)(sm - cm) = [s - {ktfct,eff(1 + ep,eff) /p,eff}] / Es 0.6s / Eskt = 0.4 for permanent loading e = Es / Ecm = 200 / 33.4 = 6.0s = 931 106 (920 - 258) / (4.63 109) = 133 N/mm2Table 3.1fct,eff = fctm = 0.3 fck(2/3) = 0.3 32(2/3) = 3.02 N/mm2(sm - cm) = [133 - {0.4 3.02 (1 + 6.0 0.0419) / 0.0419}] / 200000 = 0.485 10-30.6s / Es = 0.6 133 / 200000 = 0.399 10-3 < 0.485 10-3 OKCrack width wk = sr,max(sm - cm) = 366 0.485 10-3 = 0.18 mm NA EN 1992-2 Table NA.2Recommended value of wmax = 0.3 mm > 0.18 mm OKHence B40 bars at 150 centres are adequate for the rear face at the base of the wall.Shear CapacityCl. 6.2.2(101)Shear Capacity of Wall with B40 dia. reinforcement @ 150c/c VRd,c = [CRd,ck(1001fck)1/3]bwdCRd,c = 0.18 / c = 0.18 / 1.5 = 0.12k = 1 + (200 / d)0.5 2.0k = 1 + (200 / 920)0.5 = 1.47 < 2.01 = Asl / bwd 0.021 = 8378 / (1000 920) = 0.009 < 0.02Cl. 3.1.2(102)Pfck = 32 ( < Cmax = C50/60)VRd,c = [0.12 1.47 (100 0.009 32)1/3] 1000 920 10-3 = 497 kN ( < 606 kN Fail : see below)Minimum VRd,c = (vmin)bwd = 0.035k3/2fck1/2bwd = 0.035 1.473/2 321/2 1000 920 10-3 = 325 kNcl 6.2.2(6)Check that the maximum allowable shear force is not exceeded:Maximum allowable shear force = 0.5bwdfcd = 0.6[1 - fck / 250] = 0.6 [1 - 32 / 250] = 0.523fcd = ccfck/ccc = 1.0 [see NA to Cl. 3.1.6(101)P] fcd = 1.0 32 / 1.5 = 21.3 N/mm2Maximu VEd = 0.5 1000 920 0.523 21.3 10-3 = 5124 kN >> 606 kNVRd,c = 497 kN < VEd = 606 kN Fail. It would be necessary to increase the longitudinal reinforcement to B40 at 125 c/c however the UK National Annex allows an alternative approach.NA to 1992-2 Cl. 6.2.2(101)Alternative Solution:If the reduction factor is not used to reduce the applied shear force actions then the allowable shear force VRd,c may be enhanced if the section being considered is within 2d of the support.i) Consider a section at (a = 0.829m) from the bottom of wall :Maximum ULS shear force from spreadsheet for Case 2 = 511 kNShear enhancement factor = (2d/a) = 2 0.92 / 0.829 = 2.22VRd,c = 2.22 497 = 1103 kN ( > 511 kN OK)ii) Consider a section at (a = 1.657m) from the bottom of wall :Maximum ULS shear force from spreadsheet for Case 2 = 426 kNShear enhancement factor = (2d/a) = 2 0.92 / 1.657 = 1.11VRd,c with no enhancement = 497 kN > 426 kN , by inspection, all sections will be suitable to resist shear using B40 bars at 150 centres.Early Thermal CrackingConsidering the effects of casting the wall stem onto the base slab by complying with the early thermal cracking of concrete to C660 then B32 horizontal lacer bars @ 100 c/c will be required in both faces in the bottom 0.5m of the wall, reducing to B25 bars @ 200 above 1.3m from the bottom of the wall. Minimum Wall ReinforcementEN 1992-1-1 Clause 9.6.2 - Vertical reinforcement:As,vmin = 0.002Ac = 0.002 106 = 2000 mm2/m (1000 mm2/m in each face). Use B16 @ 150 c/c (As = 1340mm2/m).EN 1992-1-1 Clause 9.6.3 - Horizontal reinforcement:As,vmin = 0.001Ac or 25% of vertical reinforcement = 0.001 106 = 1000 mm2/m (in each face) or 25% 8378 = 2095mm2/m. B20 @ 150 c/c = 2094mm2/m, but B32 @ 100 c/c reducing to B25 bars @ 150 are required to resist early thermal cracking.Hence early thermal cracking and long-term creep and shrinkage crack control require greater areas of reinforcement than the minimum wall reinforcement. Base DesignMaximum bending and shear effects in the base slab will occur at sections near the front and back of the wall. Calculations need to be carried out for serviceability and ultimate limit states using 'at rest pressures'Using the Fixed Abutment Load Case 6 again as an example of the calculations:CASE 6 - Fixed Abutment Serviceability Limit State

Weight of wall stem = 162.5kN/mWeight of base = 160kN/mWeight of backfill = 694.5kN/mB/fill Force Hap;d = G Sd;K K0 bf;d Z2 / 2 = 1.0 1.0 0.426 19 9.52 / 2 = 365kN/mFrequent value of Surcharge UDL Force Hsc;udl = 1 Q h;ave Z = 0.75 1.0 (10.34 0.426) 9.5 = 31kN/mDispersion Factor Df for line load = (1 + Z / 2) / (1 + Z) = (1 + 9.5 / 2) / (1 + 9.5) = 0.55 < 0.67 Df = 0.67Frequent value of Surcharge Line Load Force Hsc;F = 1 Q Hsc;F = 0.75 1.0 (113.79 0.426 0.67) = 24kN/mDeck Maximum Permanent load (concrete + surfacingmax) = 207kN/mDeck Vertical Traffic load (gr2) = 89kN/mDeck Horizontal Traffic load (gr2) = 36kN/mRestoring Effects:VLever ArmMoment About AStem162.51.6260Base1603.2512Backfill694.54.252952Deck Vertical Reaction2961.55459 = 1313 = 4183Overturning Effects:

HLever ArmMoment About ABackfill3653.1671156Surcharge UDL314.75147Surcharge Line Load249.5228Deck Horizontal Reaction367.5270 = 456 = 1801Bearing Pressure at toe and heel of base slab = (V / A) (V e y / I)V = 1313kN/mA = 6.4m2/mI / y = 6.42 / 6 = 6.827m3/mNett moment = 4183 - 1801 = 2382kNm/mEccentricity (e) of V about centre-line of base = 3.2 - (2382 / 1313) = 1.386mPressure under base = (1313 / 6.4) (1313 1.386 / 6.827)Pressure under toe = 205 + 267 = 472kN/m2Pressure under heel = 205 - 267 = -62kN/m2 (uplift)Adjust for uplift:Reduced length of pressure under base = 3(B/2 - e) = 3 (6.4 / 2 - 1.386) = 5.442 mPressure under toe = 2 1313 / 5.442 = 483 kN/m2Pressure at front face of wall = 483 4.342 / 5.442} = 385kN/m2Pressure at rear face of wall = 483 3.342 / 5.442} = 297kN/m2

SLS Moment at a-a = (385 1.12 / 2) + ([483 - 385] 1.12 / 3) - (25 1.0 1.12 / 2) = 257kNm/m (tension in bottom face). SLS Moment at b-b = (297 3.3422 / 6) - (695 4.3 / 2) - (25 1.0 4.32 / 2) = -1173kNm/m (tension in top face). CASE 6 - Fixed Abutment Ultimate Limit State

Weight of wall stem Comb.1 = G;sup 162.5 = 1.35 162.5 = 219 kN/mWeight of wall stem Comb.2 = G;sup 162.5 = 1.0 162.5 = 163 kN/mWeight of base Comb.1 = G;sup 160 = 1.35 160 = 216 kN/mWeight of base Comb.2 = G;sup 160 = 1.0 160 = 160 kN/mWeight of backfill Comb.1 = G;sup 694.5 = 1.35 694.5 = 938 kN/mWeight of backfill Comb.2 = G;sup 694.5 = 1.0 694.5 = 695 kN/mB/fill Force Hap;d Comb.1 = G;sup Sd;K K0 bf;d Z2 / 2 = 1.35 1.2 0.426 19 9.52 / 2 = 592kN/mB/fill Force Hap;d Comb.2 = G;sup Sd;K K0 bf;d Z2 / 2 = 1.0 1.2 0.511 19 9.52 / 2 = 526kN/mFrequent value of Surcharge UDL Force Hsc;udl Comb.1 = 1 Q;sup h;ave Z = 0.75 1.35 (10.34 0.426) 9.5 = 42kN/mFrequent value of Surcharge UDL Force Hsc;udl Comb.2 = 1 Q;sup h;ave Z = 0.75 1.15 (10.34 0.511) 9.5 = 43kN/mDispersion Factor Df for line load = (1 + Z / 2) / (1 + Z) = (1 + 9.5 / 2) / (1 + 9.5) = 0.55 < 0.67 Df = 0.67Frequent value of Surcharge Line Load Force Hsc;F Comb.1 = 1 Q;sup Hsc;F = 0.75 1.35 (113.79 0.426 0.67) = 33kN/mFrequent value of Surcharge Line Load Force Hsc;F Comb.2 = 1 Q;sup Hsc;F = 0.75 1.15 (113.79 0.511 0.67) = 34kN/mDeck Maximum Permanent load (concrete + surfacingmax) Comb.1 = G;sup 164 + G;sup 43 = 1.35 164 + 1.2 43 = 273kN/mDeck Maximum Permanent load (concrete + surfacingmax) Comb.2 = G;sup 164 + G;sup 43 = 1.0 164 + 1.0 43 = 207kN/mDeck Vertical Traffic load (gr2) Comb.1 = Q;sup 89 = 1.35 89 = 120 kN/mDeck Vertical Traffic load (gr2) Comb.2 = Q;sup 89 = 1.15 89 = 102 kN/mDeck Horizontal Traffic load (gr2) Comb.1 = Q;sup 36 = 1.35 36 = 49 kN/mDeck Horizontal Traffic load (gr2) Comb.2 = Q;sup 36 = 1.15 36 = 41 kN/mRestoring Effects:VComb.1/Comb.2Lever ArmMoment About AComb.1/Comb.2Stem219/1631.6350/261Base216/1603.2691/512Backfill938/6954.253987/2954Deck Vertical Reaction393/3091.55609/479 = 1766/1327 = 5637/4206Overturning Effects:

HLever ArmMoment About ABackfill592/5263.1671875/1666Surcharge UDL42/434.75200/204Surcharge Line Load33/349.5314/323Deck Horizontal Load49/417.5368/308 = 716/644 = 2757/2501Assume rectangular pressure distribution under the base as described in EN 1997-1:2004 Annex DCombination 1:V = 1766 kN/m Nett moment = 5637 - 2757 = 2880 kNm/mEccentricity (e) of V about centre-line of base = 3.2 - (2880 / 1766) = 1.569mEffective base width B' = B - 2e = 6.4 - 2 1.569 = 3.262mPressure under base = (1766 / 3.262) = 541 kN/m2Combination 2:V = 1327 kN/m Nett moment = 4206 - 2501 = 1705 kNm/mEccentricity (e) of V about centre-line of base = 3.2 - (1705 / 1327) = 1.915mEffective base width B' = B - 2e = 6.4 - 2 1.915 = 2.57mPressure under base = (1327 / 2.57) = 516 kN/m2

Combination 1:ULS Shear at a-a = (541 1.1) - (1.35 1.0 1.1 25) = 558 kN/mULS Shear at b-b = 541 (3.262 - 2.1) - (1.35 1.0 4.3 25) - 938} = -454 kN/m ULS Moment at a-a = (541 1.12 / 2) - (1.35 25 1.0 1.12 / 2) = 307 kNm/m (tension in bottom face). ULS Moment at b-b = (541 (3.262 - 2.1)2 / 2) - (1.35 25 1.0 4.32 / 2) - (938 4.3 / 2) = -1963 kNm/m (tension in top face). Combination 2:ULS Shear at a-a = (516 1.1) - (1.0 1.0 1.1 25) = 540 kN/mULS Shear at b-b = 516 (2.57 - 2.1) - (1.0 1.0 4.3 25) - 695} = 560 kN/m ULS Moment at a-a = (516 1.12 / 2) - (1.0 25 1.0 1.12 / 2) = 297 kNm/m (tension in bottom face). ULS Moment at b-b = (516 (2.57 - 2.1)2 / 2) - (1.0 25 1.0 4.32 / 2) - (695 4.3 / 2) = -1668 kNm/m (tension in top face). Analysing the fixed abutment and the free abutment with Load Cases 2 to 7 using a simple spreadsheet the following results were obtained:

Fixed Abutment Base:Section a-aULS ShearComb.1/Comb.2SLSMomentULS MomentComb.1/Comb.2Case 2509/501235280/276Case 3539/515253297/283Case 4521/497244286/273Case 5527485250290/267Case 6558/541258307/298Case 7445/433235272/238Section b-bULS ShearComb.1/Comb.2SLSMomentULS MomentComb.1/Comb.2Case 2480/58811841962/1676Case 3364/46610711835/1610Case 4372/46810691830/1607Case 5290/3659731716/1519Case 6453/56111781961/1668Case 7281/3119341663/1436

Free Abutment Base:Section a-aULS ShearComb.1/Comb.2SLSMomentULS MomentComb.1/Comb.2Case 2511/500236281/275Case 3542/516254298/284Case 4523/498246288/274Case 5530/487251292/268Case 6494/455234272/250Case 7440/372212242/205Section b-bULS ShearComb.1/Comb.2SLSMomentULS MomentComb.1/Comb.2Case 2479/58212392054/1765Case 3365/46411241924/1693Case 4373/46711221918/1690Case 5291/36510231800/1597Case 6327/39410401815/1611Case 7172/1927411428/1266Early Thermal CrackingConsidering the effects of casting the base slab onto the blinding concrete by complying with the early thermal cracking of concrete to C660 then a minimum steel area of B25 distribution bars @ 200 c/c will be required to comply with clause 7.3.2(2) of BS EN 1992-1-1. Design for shear and bending effects at section a-a for the Fixed Abutment and b-b for the Free Abutment using a simple spreadsheet for slab member capacities:Section a-a: Muls = 307 kNm/m, Vuls = 558 kN/m, Msls = 258 kNm/m (Mperm = 181 kNm/m + Mvar = 77 kNm/m)B25's @ 150 c/c give Muls = 1262 kNm/m > 307 OK, Vuls = 731 kN/m (at d from support) > 558 OK, Msls = 1103 kNm/m > 258 OKSection b-b: Muls = 2054 kNm/m, Vuls = 582 kN/m, Msls = 1239 kNm/m (Mperm = 860 kNm/m + Mvar = 380 kNm/m)B40's @ 150 c/c give Muls = 2975 kNm/m > 2054 OK, Vuls = 996 kN/m (at d from support) > 582 OK, Msls = 2065 kNm/m > 1239 OKLocal EffectsCurtain Wall (Abutment Upstand Wall)This wall is designed to be cast onto the top of the abutment after the deck has been built. Loading will be applied from the backfill, surcharge and braking loads on top of the wall.EN 1991-2 Clause 4.9.2(2):Braking Force = 0.6Q1Q1k from LM1 axle = 0.6 300 = 180kNTo allow for load distribution effects assume a 45 dispersal down the wall, with maximum dispersal of the width of the abutment (11.6m). Positioning the axle in the centre of the 3.0m notional lane gives a distribution width of 7.65m at the base of the wall.

Shear at the base of the wall:due to braking = 180 / 7.65 = 23.5kN/mdue to backfill = K0 bf;d Z2 / 2 = 0.426 19.0 3.02 / 2 = 36kN/mTotal ULS shear = 1.35 (23.5 + 1.2 36) = 90kN/m Bending moment at the base of the wall:due to braking = 180 3.0 / 7.65 = 70.6kNm/mdue to backfill = K0 bf;d Z3 / 6 = 0.426 19.0 3.03 / 6 = 36kNm/mTotal SLS moment = 70.6 + 36 = 107kNm/mTotal ULS moment (combination 1) = (1.35 70.6) + (1.35 1.2 36) = 154kNm/m Check effects of surcharge + backfill at base of curtain wall:Normal traffic surcharge:Df = (1 + Z/2) / (1 + Z) = (1 + 3/2) / (1 + 3) = 0.625 < 0.67 Df = 0.67Line Load = 113.79KdDf = 113.79 0.426 0.67 = 32.5kN/mUDL = 10.34Kd = 10.34 0.426 = 4.4kN/m2Total ULS shear = 1.35 (32.5 + 4.4 3.0 + 1.2 36) = 120kN/mTotal SLS moment = 32.5 3.0 + 4.4 3.02 / 2 + 36 = 97.5 + 19.8 + 36 = 153kNm/mTotal ULS moment = 1.35 (97.5 + 19.8) + (1.35 1.2 36) = 158 + 58 = 216kNm/mSV/196 traffic surcharge:Line Load = 99.57KdDf = 99.57 0.426 0.67 = 28.4kN/mUDL = 11.64Kd = 11.64 0.426 = 5.0kN/m2Total ULS shear = 1.35 (28.4 + 5.0 3.0 + 1.2 36) = 117kN/mTotal SLS moment = 28.4 3.0 + 5.0 3.02 / 2 + 36 = 85.2 + 22.5 + 36 = 144kNm/mTotal ULS moment = 1.35 (85.2 + 22.5) + (1.35 1.2 36) = 158 + 58 = 204kNm/mHence Normal Traffic Surcharge + Backfill has worst effect on curtain wall.400 thick curtain wall with B32 @ 150 c/c :Mult = 601 kNm/m > 216 kNm/m OKMsls = 317 kNm/m > 153 kNm/m OKVult = 261 kN/m > 120 kN/m Shear OK