Abutment Design Example

March 25, 2018 | Author: zrilek1 | Category: Lane, Creep (Deformation), Bearing (Mechanical), Concrete, Friction
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Abutment Design Example to Eurocodes and UK National Annexes⇑ Scroll to top Index 1.Actions from Bearings 2.Actions on Back of Abutment 3.Stability Check 4.Wall Design 5.Base Slab Design 6.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 (φ') = 34°, a critical state angle of shearing resistance (φ'cν) = 30° and a weight density = 19kN/m3. Backfill material will be Class 6N with an angle of shearing resistance (φ'bf;k) = 35° and weight density (γbf;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 “+” R1 A grillage analysis gave the following characteristic reactions for the various load cases: Critical Vertical Reaction Under One Beam Characteristic Reaction (kN) ULS Reaction (kN) Concrete Deck 180 240 Surfacing 45 60 gr1a 290 430 gr2 220 310 gr5 270 400 gr6 210 300 Total Vertical Reaction on Each Abutment Characteristic Reaction (kN) A1 (γG;sup / γG;inf) A2 (γG;sup / γG;inf) Concrete Deck 1900 1.35 / 0.95 1.0 / 1.0 Surfacing 320 1.2 / 0.95 1.0 / 1.0 gr1a 1490 1.35 / 0 1.15 / 0 gr2 1120 1.35 / 0 1.15 / 0 gr5 1930 1.35 / 0 1.15 / 0 gr6 1470 1.35 / 0 1.15 / 0 Characteristic loading on 1m length of abutment: Deck Dead Load = 1900 / 11.6 = 164kN/m Maximum Surfacing = 1.55 × 320 / 11.6 = 43kN/m Minimum Surfacing = 0.6 × 320 / 11.6 = 17kN/m gr1a on Deck = 1490 / 11.6 = 128kN/m gr2 on Deck = 1030 / 11.6 = 89kN/m gr5 on Deck = 1930 / 11.6 = 166kN/m gr6 on Deck = 1470 / 11.6 = 127kN/m From 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 +34°C respectively. For bridge deck type 3 the corresponding minimum (Te,min) and maximum (Te,max) effective bridge temperatures are -11 and +36°C from BS EN 1991-1-5:2003 Figure 6.1. Hence the temperature range = 11 + 36 = 47°C. 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) γF] = ±[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 = 19mm Note: 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 +34°C which would require the bearings to be installed at a shade air temperature of [(34+17)/2-17] = 9°C to achieve the ± 8mm movement. If the bearings are set at a maximum shade air temperature (T0) of 16°C 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 16°C 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 = 12.5mm Average characteristic permanent load reaction = (1900 + 320) / 11 = 2220 / 11 = 200kN Contact pressure under base plate = 200000 / (210 × 365) = 3N/mm2 As 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 = 0.6αQ1(2Q1k)+0.1αq1qq1w1L = 0.6 × 1 (2 × 300) + 0.1 × 1 × 9 × 3 × 20 = 414kN For 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 = δw = 0.25 × (165kN × 9axles + 180kN × 2axles + 100kN × 1axle) = 486kN For 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 Abutment Backfill For Stability calculations use active earth pressures = Ka γbf;k h For Design of Structural Members use at-rest earth pressures = K0 γbf;k h SLS Combination 1 Combination 2 Partial factors for soil parameters γM 1.0 1.0 k γG.0° 29.426 0.d = tan-1[tan(φ'bf.K 1.d) = γbf.0/1.k)/γM] 35.sup 19.65 19.0 Model factors γSd.25 φ'bf.k γG.95 1.35/0.0/1.d) = γbf.271 0.426 0.0 Backfill density (γbf.3° Ka = (1-Sinφ'bf.0 18.343 K0 = 1-Sinφ'bf.0° 35.d 0.inf 19.d) 0.d) / (1+Sinφ'bf.0 25.1 19.511 Partial factors for soil weight γG (sup/inf) 1.0 1.0 Backfill density (γbf.0 1.1.2 .271 0. F = F.2 Hap.2 UK NA to BS EN 1991-2) ∴ the effective number of lanes (Nlane) in the surcharge model will be 2.1 of BS EN 1991-2:2003).57Kd UDL kN/m2 = σh. but is not included in the calculation when considering the overall stability of the wall.34.2.79Kd Note: Df is used for determining the distibution of the Line Load in the wall for a metre strip analysis.6 = 11. UDL kN/m2 = σh.91Z2kN/m Surcharge .Use Horizontal Surcharge Model in PD 6694-1:2011 Figure 2: Carriageway width = 7.KZ2/2 2.64Kd .575Z2kN/m 4.Kd.ave = (σh1+ψ1. The vehicle model for loads on backfill behind abutments is positioned in each notional lane (see Clause NA.(1 + ψ1)/Wabut = 2×330Kd×1.Wlane/Wabut = (30+0.75×20)×Kd×3/11.Kd.Nlane/Wabut = 20×Kd×3×2/11.6 = 113.Nlane/Wabut = 2×330Kd×2/11.Wlane.F = F.3m wide remaining area (see Table 4.34Kd SV/196 Traffic (SV/196 lane 1 + Frequent value of Normal Traffic in Lane 2) Line Load kN/m (at ground level) = Hsc.1.σh2).d = γbf.3m ∴ there are 2 notional lanes of effective width Weff of 3m with 1.dKaγSd. From PD 6694-1:2011 Table 7 : Normal Traffic Line Load kN/m = Hsc.ave = σh.171Z2kN/m 3.6 = 99.6 = 10.75/11. see PD 6694-1:2011 Clause 7. Reynolds's Reinforced Concrete Designer's Handbook being one such book.3. 1) Stability Check Initial Sizing for Base Dimensions There are a number of publications that will give guidance on base sizes for free standing cantilever walls.SLS Combination 1 Combination 2 Partial Factor on Surcharge γQ 1.3). Alternatively a simple spreadsheet will achieve a result by trial and error.0 1. Load Combinations Backfill + Construction surcharge (Not used .backfilled to 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) Backfill + Normal Traffic Surcharge + Deck Permanent load + Deck contraction/shrinkage .15 Assume Abutments are to be backfilled in accordance with the Highways Agency Manual of Contract Documents for Highway Works (MCHW).3.35 1. 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 (ψ1LM1 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 .3 × 8.0 × (164 + 43) = 207kN/m Deck Minimum Permanent load (concrete + surfacingmin) = γG × VDL = 1.79 × 0.0) Weight of wall stem = γG × twall × Zwall × γconc = 1.34 × 0.0 × 1.271 × 19 × 9.d = γG × γSd.271) × 9.0 × 6.0 × 6. SLS (γG = γQ = γSd.F = 0.4 × 1.F = ψ1 × γQ × Hsc.0 × 89 = 89kN/m Deck Horizontal Traffic load (gr2) = γQ × Hbraking = 1.0 × 25 = 160kN/m Weight of backfill = γG × Wheel × Zheel × γbf.5kN/m Weight of base = γG × Wbase × Zbase × γconc = 1.6 260 Base 160 3.52 / 2 = 232kN/m Frequent value of Surcharge UDL Force Hsc.0 × 1.5 × 19 = 694.0 × 4.0 × (113.K = 1.5kN/m Backfill Force Hap.Fixed Abutment Density of reinforced concrete = 25kN/m3.75 × 1.ave × Z = 0.0 × (164 + 17) = 181kN/m Deck Vertical Traffic load (gr2) = γQ × Vtraffic = 1.0 × 36 = 36kN/m Restoring Effects: Minimum V Lever Arm Moment About A Stem 162.K × Ka × γbf.5 × 25 = 162.d × Z2/2 = 1.d = 1.5 = 20 kN/m Frequent value of Surcharge Line Load Force Hsc.5 1.75 × 1.271) = 23 kN/m Deck Maximum Permanent load (concrete + surfacingmax) = γG × VDL = 1.0 × (10.2 .0 × 0.udl = ψ1 × γQ × σh. 25 2952 Deck (VDLmin) 181 1.25 .5 4.6 260 Base 160 3.55 281 ∑= 1198 ∑= 4005 Maximum V Lever Arm Moment About A Stem 162.2 512 Backfill 694.512 Backfill 694.5 1.5 4. udl 20 4.55 459 ∑= 1313 ∑= 4183 Overturning Effects: H Lever Arm Moment About A Hap.d 232 3.75 95 Hsc.5 219 Hbraking 36 .F 23 9.2952 Deck (VDLmax + Vtraffic) 296 1.167 735 Hsc. 73 iq = [1 .42 / 6 = 6.63 iγ = [1 .0) Nq = eπtan34tan2(45 + 34 / 2) = 29.3 × 1.362 / 11.(2864 / 1313) = 1.1)tanφ'd = 2 × (29.7.6) × sin34 = 1.3 bq = bγ = 1.827m3/m Nett moment = 4183 .89 m = (2 + B' / L') / (1 + B' / L') = (2 + 4.48 R/A' = 0 + (28.311 / 1313]1.0 Coefficient of friction = μd = tan(φ'cν.8(4).58 × 1198 = 695kN/m Active Force = ∑H = 311kN/m < 695 ∴ OK Bearing Pressure: PD 6694-1 Cl.63) + (0.1) × tan34 = 38.019m Pressure under base = (1313 / 6.0 × 0.0 × 1.827) Pressure under toe = 205 + 196 = 401kN/m2 Pressure under heel = 205 .6) = 0.5γ'B'Nγbγsγiγ c' = 0 γ' = γ × γG.0.196 = 9kN/m2 > 0 ∴ OK (no uplift) Also BS EN 1997-1:2004.58 Sliding resistance = μd∑Vmin = Rνx.5 270 ∑= 311 ∑= 1319 For sliding effects: φ'cν = 30° Partial factor on γM on tan(φ'cν.4 .4.21 × 0.362 / 11.H / (V + A'c'dcotφ'd)]m+1 = [1 .4 for Drained Conditions: R/A' = c'Ncbcscic + q'Nqbqsqiq + 0.2 requires no uplift at SLS Check 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. From Annex D. 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 1997-1:2004.1319 = 2864kNm/m Eccentricity (e) of V about centre-line of base = 3.21 sγ = 1 .89 × 0. V = 1313kN/m A = 6.5 kN/m2 (Foundation 1.0 (α = 0) B' = B .311 / 1313]2.4m2/m I / y = 6.362 × 38.2 × 1.d = 0. using characteristic values of φ'.6) / (1 + 4.5 × 19 = 28.4) ± (1313 × 1.0 = 0.0.73 = 0. cu and γ' and representative values of horizontal and vertical actions.362 sq = 1 + (B' / L')sinφ'd = 1 + (4.H / (V + A'c'dcotφ'd)]m = [1 .2e = 6.362 / 11.4 . Annex D.5 × 19 × 4.48 = 639 + 678 = 1317 kN/m2 .5m below existing ground level) Nq = eπtanφ'dtan2(45 + φ'd / 2) φ'd = φ' = 34° (γM = 1.k) / γM = tan(30°)/1.3(4.inf = 19 × 1.2 . 2.019 / 6.5 × 29.0 = 19kN/m3 q' = 1.362 / 11.4 Nγ = 2(Nq .6) = 1.3(B' / L') = 1 .019 = 4.2. allows the serviceability limit state for settlement to be verified by ensuring that a “sufficiently low fraction of the ground strength is mobilized”.73 = 0.4 × 1.k) = 1. 5. 35 1.15 1.inf × twall × Zwall × γconc 154 162.1/3(R/A') = 1317 / 3 = 439 kN/m2 > 401 kN/m2 ∴ settlement check OK.5 Max.K Combination 1 Combination 1 Surfacing 1.2 0.inf γQ γSd.sup × twall × Zwall × γconc 219 . weight of wall stem = γG.95 0.sup γG.00 1.2 Combination 2 1.1 Comb. ULS Check Combination 1 and Combination 2.2 Comb. weight of wall stem = γG. γG.95 1.35 1.00 1.2 Min. d 937.inf × Wheel × Zheel × γbf.d × Z2/2 376 353 Frequent value of Surcharge UDL Force Hsc.271 0.343 Backfill Force Hap.sup × Wheel × Zheel × γbf.5 Max. weight of base = γG.F = ψ1 × γQ × Hsc.5 Ka 0.K × Ka × γbf.udl = ψ1 × γQ × σh.sup × Wbase × Zbase × γconc 216 160 Min.d 660 694. weight of backfill = γG.ave × Z 27 29 Frequent value of Surcharge Line Load Force Hsc.inf × Wbase × Zbase × γconc 152 160 Max.d = γG.162.5 694.5 Min. weight of backfill = γG.sup × γSd.7 Deck Maximum Permanent load (concrete + surfacingmax) = γG. weight of base = γG.sup × VDL 273 207 .F 31 33. sup × Vtraffic 120 102 Deck Horizontal Traffic load (gr2) = γQ × Hbraking 49 41 Combination 1 Restoring Effects : Minimum V Lever Arm Moment About A Stem 154 1.25 2805 .inf × VDL 172 181 Deck Vertical Traffic load (gr2) = γQ.6 246 Base 152 3.2 486 Backfill 660 4.Deck Minimum Permanent load (concrete + surfacingmin) = γG. 6 350 Base 216 3.Deck (VDLmin) 172 1.2 691 Backfill 937.25 3984 Deck (VDLmax + Vtraffic) 393 1.55 267 ∑= 1138 ∑= 3804 Maximum V Lever Arm Moment About A Stem 219 1.55 609 .5 4. 5 ∑= 5634 Overturning Effects: H Lever Arm Moment About A Hap.75 128 Hsc.∑= 1765.5 295 Hbraking 49 7.d 376 3.F 31 9.udl 27 4.5 368 ∑= 483 ∑= .167 1191 Hsc. 131m Base width / 3 = 6.(3652 / 1765.3 × 1.5]2.89 m = (2 + B' / L') / (1 + B' / L') = (2 + 4. 6.2 × 29.138 sq = 1 + (B' / L')sinφ'd = 1 + (4.138 / 11.5 × 18.4 Nγ = 2(Nq .5.42 = 547 + 536 = 1083 kN/m2 > 427 kN/m2 ∴ OK.1) × tan34 = 38.74 = 0.2kN/m2 (Foundation 1.57 iγ = [1 . Combination 2 Restoring Effects : Minimum V Lever Arm Moment About A Stem 162.20 sγ = 1 .4 restrict eccentricity of loading to 1/3 of the width of the footing at ULS V = 1765.58 × 1138 = 660kN/m Active Force = ∑H = 483kN/m < 660 ∴ OK Bearing Pressure: EN 1997-1:2004 Cl.138) = 427kN/m2 R/A' = c'Ncbcscic + q'Nqbqsqiq + 0.74 iq = [1 .0) Nq = eπtan34tan2(45 + 34 / 2) = 29.5]1.5 / 4.1982 = 3652kNm/m Eccentricity (e) of V about centre-line of base = 3.58 Sliding resistance = μd∑Vmin = Rνx.138 / 11.inf = 19 × 0.138 / 11.4 × 1.k) = 1.483 / 1765.95 = 18.5m below existing ground level) Nq = eπtanφ'dtan2(45 + φ'd / 2) φ'd = φ' = 34° (γM = 1.1kN/m3 q' = 1.1 = 27.138 / 11.4 .5γ'B'Nγbγsγiγ c' = 0 γ' = γ × γG.3(B' / L') = 1 .5 .89 × 0.1)tanφ'd = 2 × (29.5 × 18.483 / 1765.0.2 × 1.131 > 1.4 / 3 = 2.d = 0.0.138 × 38.2 .6) = 1.131 = 4.0 (α = 0) B' = 4.5kN/m Nett moment = 5634 .57) + (0.74 = 0.5) = 1.3 bq = bγ = 1.6) = 0.0 = 0.6) × sin34 = 1.H / (V + A'c'dcotφ'd)]m+1 = [1 .H / (V + A'c'dcotφ'd)]m = [1 .0 × 0.4 .131 ∴ OK Assume rectangular pressure distribution under the base as described in EN 1997-1:2004 Annex D Effective base width B' = B .k) / γM = tan(30°)/1.1982 For sliding effects: φ'cν = 30° Partial factor on γM on tan(φ'cν.0 Coefficient of friction = μd = tan(φ'cν.1 × 4.0 × 1.20 × 0.6) / (1 + 4.2e = 6.138m Pressure under base = (1765.42 R/A' = 0 + (27.3(4. 1.6 260 Base 160 .25 2952 Deck (VDLmin) 181 1.6 260 Base 160 3.5 1.5 4.2 512 Backfill 694.55 281 ∑= 1198 ∑= 4005 Maximum V Lever Arm Moment About A Stem 162. 167 1118 Hsc.2 512 Backfill 694.3.75 138 Hsc.25 2952 Deck (VDLmax + Vtraffic) 309 1.udl 29 4.5 4.55 479 ∑= 1326 ∑= 4203 Overturning Effects: H Lever Arm Moment About A Hap.d 353 3.F . 451m Base width / 3 = 6.5 320 Hbraking 41 7.25 = 0.451 = 3.768 iq = [1 .1884 = 2319kNm/m Eccentricity (e) of V about centre-line of base = 3.25] = 28.6) = 0.6 bq = bγ = 1.4 .d = 0.4 / 3 = 2.4tan2(45 + 28.5 × 19 = 28.5m below existing ground level) Nq = eπtanφ'dtan2(45 + φ'd / 2) φ'd = tan-1[tan(φ'k)/γM] = tan-1[tan34/1.4 Nγ = 2(Nq .0.6) / (1 + 3.4 = 15.5kN/m2 (Foundation 1.768 = 0.451 ∴ OK Assume rectangular pressure distribution under the base as described in EN 1997-1:2004 Annex D Effective base width B' = B .1) × tan28. 6.46 Sliding resistance = μd∑Vmin = Rνx.457 / 1326]1.4° Nq = eπtan28.498 / 11.5.498 / 11.4 .0 = 19kN/m3 q' = 1.5γ'B'Nγbγsγiγ c' = 0 γ' = γ × γG.k) / γM = tan(30°)/1.3(B' / L') = 1 .6) = 1.4 restrict eccentricity of loading to 1/3 of the width of the footing at ULS V = 1326kN/m Nett moment = 4203 .47 .498 / 11.33.2 × 1.498 / 11.0 (α = 0) B' = 3.46 × 1198 = 551kN/m Active Force = ∑H = 457kN/m < 551 ∴ OK Bearing Pressure: EN 1997-1:2004 Cl.1)tanφ'd = 2 × (15.131 > 1.91 m = (2 + B' / L') / (1 + B' / L') = (2 + 3.0.4 / 2) = 15.(2319 / 1326) = 1.3(3.4 = 1.5 308 ∑= 457 ∑= 1884 For sliding effects: φ'cν = 30° Partial factor on γM on tan(φ'cν.H / (V + A'c'dcotφ'd)]m = [1 .6) × sin28.498 sq = 1 + (B' / L')sinφ'd = 1 + (3.7 9.14 sγ = 1 .k) = 1.inf = 19 × 1.2 .2e = 6.498) = 379kN/m2 R/A' = c'Ncbcscic + q'Nqbqsqiq + 0.25 Coefficient of friction = μd = tan(φ'cν.498m Pressure under base = (1326 / 3. 14 × 0.4 × 1.results of both abutments with elastomeric bearings. Case 2a.91 × 0.31) = 235 + 146 = 381 kN/m2 > 379 kN/m2 ∴ OK. 6b and 7b .H / (V + A'c'dcotφ'd)]m+1 = [1 . 6a and 7a .47) + (0.1 Comb.5 × 19 × 3.768 = 0.results of fixed abutment with dowels and free abutment with elastomeric bearings.0 × 0.iγ = [1 .results of fixed abutment with dowels and free abutment with sliding bearings.498 × 15. Case 2b.2 Resistance 692 657 553 Case 2 Case 2a 306 375 476 569 453 522 Case 3 290 454 437 Case 4 289 453 436 .0 × 1. 6 and 7 . Fixed Abutment: Sliding SLS Comb.6 × 1. All other cases are not affected by the bearing arrangement.457 / 1326]2. 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 2.5 × 15.31 R/A' = 0 + (28. 2 Vd/A' / R/A' Case 2 Case 2a 359 435 23 -52 384 / 1054 478 / 773 341 / 354 447 / 257 .1 Vd/A' / R/A' Comb.Case 5 275 435 416 Case 6 & 6a 311 483 458 Case 7& 7a 274 433 402 Bearing Pressure SLS Toe SLS Heel Comb. Case 3 393 30 418 / 1199 370 / 426 Case 4 377 36 402 / 1188 354 / 419 Case 5 392 42 418 / 1281 362 / 473 Case 6 & 6a 401 9 427 / 1082 380 / 380 Case 7 & 7a 377 45 403 / 1265 336 / 485 Free Abutment: Sliding SLS . Comb.2 Resistance 711 675 568 Case 2 Case 2a Case 2b 312 382 339 486 580 522 463 532 490 Case 3 297 465 447 Case 4 269 464 447 Case 5 282 445 426 Case 6 & 6a Case 6b .1 Comb. 2 Vd/A' / R/A' Case 2 Case 2a Case 2b 361 436 390 25 -49 -3 386 / 1070 478 / 793 417 / 957 .282 307 445 479 426 455 Case 7 & 7a Case 7b 239 260 387 415 363 387 Bearing Pressure SLS Toe SLS Heel Comb.1 Vd/A' / R/A' Comb. 342 / 360 444 / 262 375 / 320 Case 3 395 31 420 / 1212 371 / 431 Case 4 379 37 404 / 1202 355 / 424 Case 5 394 43 421 / 1294 364 / 477 Case 6 & 6a Case 6b 364 391 50 23 390 / 1257 416 / 1145 337 / 453 366 / 405 Case 7 & 7a Case 7b 334 365 . Using the Fixed Abutment Load Case 6 again as an example of the calculations: Wall Design SLS Combination 1 Combination 2 Partial factors for soil parameters γM 1. Serviceability and Ultimate load effects need to be calculated for the load cases 2 to 7 shown above. Again.0 1.3° K0 = 1-Sinφ'bf. 2) Slight differences in results between the example and the spreadsheet for Case 6 are due to rounding off errors in the example.d = tan-1[tan(φ'bf.511 .426 0. 2) Wall and Base Design Loads on the back of the wall are calculated using 'at rest' earth pressures. 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.0° 29.k)/γM] 35.426 0. these are best carried out using a simple spreadsheet.0° 35.0 1.92 67 368 / 1480 395 / 1387 300 / 578 326 / 539 Note: 1) Numbers in bold indicate failed results. It can be seen that the use of elastomeric bearings (Case 2) will govern the critical design load cases on the abutments.25 φ'bf.d 0. KZ2/2 4.d) = γbf.0 Model factors γSd.2 1.sup 19.d(kN) = 292 475 421 Moment (kNm) (lever arm = 8.825Z2kN/m Consider a section at the base of the wall (Z = 8.047Z2kN/m 6.2 Hap.0 1.0 25.7 19.0 1.5m) Backfill: Hap.569Z2kN/m 5.dK0γSd.35 1.k γG.K 1.Partial factors for soil weight γG.d = γbf.sup 1.5/3) = 827 1295 1193 Frequent value of Normal Surcharge: ψ1γQ.0 Backfill density (γbf.sup = . 79KdDf = 24 33 34 Moment (kNm) (lever arm = 8.sup113.0 1.25) = 119 162 166 Deck Permanent Load Reaction: γG.F = ψ1γQ.sup10.aveZ = ψ1γQ.supHsc.0 Deck concrete = 164 221 164 .013 0.supσh.5) = 204 281 289 ψ1γQ.0.sup for concrete = 1.35 1.34KdZ = 28 38 39 Moment (kNm) (lever arm = 4.863 ψ1γQ.75 1. sup for surfacing = 1.35 1.5 .0 Deck surfacing = 43 52 43 Moment = Σ V × e (e = 0.5 .45 = 0.2 1.0 1.0.γG.0 1.45) = 4 6 5 Variable Horizontal Reaction (Braking) = 36 49 .sup = 1.15 Variable Vertical Reaction = 89 120 102 Moment = V × e (e = 0.05) = 10 14 10 Deck Variable Reaction (gr2): γQ.0. ) Fixed Abutment: SLS Moment (Permanent) SLS Moment (Variable) SLS Moment (Total) Case 2 840 539 1379 Case 3 840 442 1282 Case 4 840 427 .5 = 234 319 267 Comb.1 moment at base of wall = Σ M = 1295 + 281 + 162 + 14 + 6 + 319 = 2077kNm ULS Comb.2 shear at base of wall = Σ H = 421 + 33 + 38 + 41 = 533kN SLS moment at base of wall = Σ M = 827 + 204 + 119 + 10 + 4 + 234 = 1398kNm (837 permanent + 561 variable) ULS Comb.1 shear at base of wall = Σ H = 475 + 33 + 38 + 49 = 595kN Comb.Slight differences in results between the example and the spreadsheet for Case 6 are due to rounding off errors in the example.41 Moment = H × 6.2 moment at base of wall = Σ M = 1193 + 289 + 166 + 10 + 5 + 267 = 1930kNm Analysing 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 . 2 Case 2 2086 590 1908 534 Case 3 1954 569 .1 Moment ULS Comb.1 Shear ULS Comb.2 Shear ULS Comb.1267 Case 5 840 335 1175 Case 6 840 565 1405 Case 7 840 279 1119 Moment ULS Comb. 1811 518 Case 4 1934 570 1792 519 Case 5 1809 545 1664 494 Case 6 2120 594 1928 535 Case 7 1734 531 1525 469 Free Abutment: SLS Moment (Permanent) SLS Moment (Variable) . 1 .SLS Moment (Total) Case 2 878 551 1429 Case 3 878 451 1329 Case 4 878 437 1315 Case 5 878 342 1220 Case 6 878 338 1216 Case 7 878 7 885 Moment ULS Comb. 2 Shear ULS Comb.Shear ULS Comb.1 Moment ULS Comb.2 Case 2 2163 606 1979 547 Case 3 2029 584 1880 531 Case 4 2009 585 1860 532 Case 5 1881 560 1729 507 Case 6 1876 560 1724 . 85 × 32 / 1. EN 1992-1-1 & EN 1992-2 .8.B.021 = 74.1 × 0. Reinforcement to BS 4449:2005 Grade B500B: fy = 500N/mm2 Design 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 5.26 / 0. C = 0.289m λ = 13.4 0 = 2 × = 2 × 6. second order effects need not be considered.7 n = NEd / (Acfcd) NEd = 164 + 43 + 166 = 373kN fcd = αccfck / γc = 0.021 λlim = 20 × 0.5 and minimum cement content of 340kg/m3 for exposure condition XD2.7.289 = 45.C/√n Use suggested values when φef not known: A = 0.4 ∴ OK.1 λ = 0/i ≤ λlim = 20A. Nominal cover to reinforcement = 60mm (45mm minimum cover plus a tolerance Δc of 15mm).1) = 0.5 = 18.3. B = 1.507 Case 7 1428 489 1266 434 Concrete to BS 8500:2006 Use strength class C32/40 with water-cement ratio 0.1N/mm2 Ac = 106mm2 (per metre width) n = 373 × 103 / (106 × 18.1.26m (cantilever with sliding bearings to deck) i = √(1/12) = 0.7 / √0.7 × 1.9 < 74.63 = 13. 5 = 18.00352 .8 × (920 .002 / (2 + 1)] = 0.βX) Where β = 1 . 3.7(4) Modulus of Elasticity Es = 200 kN/mm2 Steel strain at yield = εs.0 Try 40mm dia.416 × 247.5 × 0.It is usual to design reinforced concrete for the ultimate limit state and check for serviceability conditions.00352 .85 cl.0.8mm Check that steel will yield: Cl.20 = 920mm Fig.5.15) = 247.4 Table 2.0. MSLS = 1429kNm/m [878(permanent)+551(variable)] cl.416 Mult = 14.εcu2εc2 / (n+1)] β = 1 . n = 2.0035 × (2 + 1)}] = 14.[0.1 × [1 .60 .7 N/mm2 Assuming steel yields then: M = fsz = fykAsz / γs = Fcz = favbXz Depth to neutral axis X = fykAs / (favbγs) X = 500 × 8378 / (14.002 / {0.1) = 0. γs = 1.0.yield = fyk / γs / Es = 500 / 1.1) = 0.00217 from linear strain relationship: εs = εcu2(d/X . 3. 3.εc22 / {(n+1)(n+2)}] / [εcu22 . Hence Mult = favbXz = favbX(d .7 × 1000 × 1. MULS = 2163kNm/m.4.1 εc2 = 0.15 / 200000 = 0.0035.1N: γc = 1.00217 ∴ steel will yield. VULS = 606kN/m.1.1. reinforcement at 150mm centres (8378mm2/m): Nominal cover to reinforcement in rear face of wall = 60mm d = 1000 .3 Using parabolic-rectangular diagram: Average stress fav = fcd[1-εc2 / {εcu2(n+1)}] = 18.8) × 10-6 = 2976 kNm > 2163 ∴ OK Check Serviceability Limit State Characteristic Combination SLS Design Moment = 1429kNm/m (878 +551) Check stresses in the concrete and reinforcement at: i) Early Age (before creep has occurred) .[0.1 N/mm2 Table 3.7 αcc = 0.6(101)P Design compressive strength = fcd = αccfck / γc cl. 3.0022 / {(2 + 1) × (2 + 2)}] / 0.002.0035 ( 920 / 247. εcu2 = 0.2.0035 × 0.5εcu22 . 2.7 × 1000 × 247.85 × 32 / 1.15 fcd = 0.8 .009 > 0.0.2. 7 = 0.effd}0.1 Ecm = 22[(fck + 8) / 10]0.7 = [35 / 40]0.2(102) Limiting concrete stress = k1fck k1 = 0.97 (B.1 ∴ OK EN 1992-1-1 ii) 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 = Ecm Long-term modulus = Ecm / (1+φ) Effective modulus Ec.1 × 17061/3)}] × 0. EN 1992-1-1 Table 3.eff = Ecm = 33.4 = 6.63 × 109 × 6.8 = 14.8c) h0 = 2Ac / U = 2 × (11600 × 1000) / (11600 + 2 × 1000) = 1706 α1 = [35 / fcm]0.2 N/mm2 > 14.1 N/mm2 cl.4 Relative 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.4 kN/mm2 Modular Ratio m = Es / Ecm = 200 / 33.6)&(B.eff Hence dc = [-AsEs + {(AsEs)2 + 2bAsEsEc.97 = 1.3 = 22[(32 + 8) / 10]0.4 kN/mm2 Ec.6 Limiting concrete stress = 0.dc) / dc Equating forces: AsEsεs = 0. 3.63 × 109 mm4 (steel units) Approximate concrete stress σc = M / zc + N / Ac N (Case 2) = 164 + 43 = 207 kN σc ≅ {1429 × 106 × 258 / (4.5] / bEc.ii) Long term after all the creep has taken place.6 × 32 = 19. i) Before creep has occurred the cracked section properties will be based on the short-term modulus for all actions.1 × h01/3)}] × α2 φRH = [1 + 0.4 × 1000 × 2583 / (3 × 200) = 4.3 = 33. 7.258)2 + 33.91 α2 = [35 / fcm]0.eff = (Mqp + Mst)Ecm / {Mst + (1 + φ)Mqp} Table 3.1.5bdcεcEc.effbdc3 / 3Es INA = 8378 × (920 .91 × {(1 .0 Let dc = depth to neutral axis then equating strains for cracked section: εs = εc(d .eff dc = [-8378 × 200000 + {(8378 × 200000)2 + 2 × 1000 × 8378 × 200000 × 33400 × 920}0.2 = 0.RH / 100) / (0.80 / 100) / ( 0.0)} + {207 × 103 / (258 × 103)} = 13.5] / (1000 × 33400) = 258mm Cracked second moment of area = As(d-dc)2 + Ec.3b) φRH = [1 + α1 × {(1 .2 = [35 / 40]0.1 fcm = fck + 8 = 32 + 8 = 40 N/mm2 Cl.3 + 0.118 . 351)2 + 15.(B.351) / (3.5bdcεcEc.83 × 109 mm4 (steel units) Concrete stress σc ≅ M / zc + N / Ac σc = {1429 × 106 × 351 / (3.886)} = 15.max(εsm .4} / {551 + 878 × ( 1 + 1.5 × 1000 × 3513 / (3 × 200) = 3.5) β(t0) = 1 / (0.effbdc3 / 3Es INA = 8378 × (920 .2) φ0 = φRH × β(fcm) × β(t0) = 1.5 = 16.5 kN/mm2 Modular Ratio m = Es / Ec.8 N/mm2 cl.3.2(102) Limiting concrete stress = k1fck k1 = 0.656 (B.2 N/mm2 > 10.63) = 931 kNm Cl.8 × 500 = 400 N/mm2 Steel stress σs = M / zs σs = 1429 × 106 × (920 .2) = 1 / ( 0.1 + 70.83 × 109 × 12.8 / 400.9 Let dc = depth to neutral axis then equating strains for cracked section: εs = εc(d . 7.4(1) Crack width wk = sr.6 × 32 = 19.effd}0.5] / (1000 × 15500) = 351mm Cracked second moment of area = As(d-dc)2 + Ec.5(16 × 6.eff Hence dc = [-AsEs + {(AsEs)2 + 2bAsEsEc.5] / bEc.83 × 109) = 212 N/mm2 < 400 ∴ >OK Crack Control: Consider worst condition before creep has occurred and Quasi-Permanent Combination Moment + ψ2 × temperature effects = 878 + 0.5 = 2.8 Limiting steel stress = 0.635 = 1.6 = 10.2) = 0.9)} + (207 × 103 / (351 × 103) = 10.656 × 0. 7.dc) / dc Equating forces: AsEsεs = 0.1 + t00.4) β(fcm) = 16.118 × 2.eff = 200 / 15.εcm) .6 Limiting concrete stress = 0.5 = 12. 7.2 + 0.eff = {(878 + 551) × 33.8 ∴ OK cl.8 / fcm0.635 (B.eff dc = [-8378 × 200000 + {(8378 × 200000)2 + 2 × 1000 × 8378 × 200000 × 15500 × 920}0.2(5) Limiting steel stress = k3fyk k3 = 0.886 Moment due to long-term actions = Mqp = 878 kNm Moment due to short-term actions = Mst = 551 kNm Hence Effective Modulus Ec. 18 / 1.485 × 10-3 ∴ OK Crack width wk = sr.5(1000 .4 × 60) + (0.5 (for bending) k3 = 3.0419}] / 200000 = 0.Cl.0419 sr.6σs / Es = 0.eff = fctm = 0.3.max(εsm .2 Recommended value of wmax = 0.425 × 40) / 0.02 × (1 + 6.18 mm NA EN 1992-2 Table NA.5 = 1.485 × 10-3 = 0.2(3) hc.0 × 0.eff = As / Ac.c = 0. 7.63 × 109) = 133 N/mm2 Table 3.εcm) = [133 .0419 = 204 + 162 = 366 Cl.3 × 32(2/3) = 3.2(101) Shear Capacity of Wall with B40 dia. reinforcement @ 150c/c VRd. 6.eff = 200 × 1000 = 200000 mm2 Cl. 7.12 k = 1 + (200 / d)0. 7.258) / 3 = 247 iii) h/2 = 1000 / 2 = 500 ∴ hc.3.4(2) (εsm .4(3) Spacing Limit = 5(c+φ/2) = 5(60 + 40/2) = 400mm > 150mm ∴ OK sr.02 N/mm2 (εsm .258) / (4.4 (recommended value) k4 = 0.eff}] / Es ≥ 0.3 × fck(2/3) = 0.εcm) = [σs .max = k3c + k1k2k4φ / ρp.0 k = 1 + (200 / 920)0.eff is the lesser of: i) 2.eff = 200 mm and Ac. Shear Capacity Cl.8 (high bond bars) k2 = 0.0 σs = 931 × 106 × (920 .3 mm > 0.eff sr.max = (3.920) = 200 ii) (h-x)/3 = (1000 .eff k1 = 0.4 × 3.8 × 0.425 (recommended value) Cl.ck(100ρ1fck)1/3]bwd CRd.5 = 0.6σs / Es kt = 0.4 for permanent loading αe = Es / Ecm = 200 / 33.18 / γc = 0.{0.1 fct.485 × 10-3 0.eff = 8378 / 200000 = 0.2.c = [CRd.5 ≤2.4 = 6.399 × 10-3 < 0.0419) / 0.5 × 0.{ktfct. 7.eff(1 + αeρp.0 .4(2) ρp.εcm) = 366 × 0.5(h-d) = 2.max = k3c + k1k2k4φ / ρp.3.47 < 2.eff) /ρp.6 × 133 / 200000 = 0.18 mm ∴ OK Hence B40 bars at 150 centres are adequate for the rear face at the base of the wall.3. Use B16 @ 150 c/c (As = 1340mm2/m).6 × [1 .523 × 21.035k3/2fck1/2bwd = 0.12 × 1.002 × 106 = 2000 mm2/m (1000 mm2/m in each face).3 × 10-3 = 5124 kN >> 606 kN VRd.002Ac = 0.c may be enhanced if the section being considered is within 2d of the support.c = (vmin)bwd = 0.2(102)P fck = 32 ( < Cmax = C50/60) VRd.0 [see NA to Cl.6.009 × 32)1/3] × 1000 × 920 × 10-3 = 497 kN ( < 606 kN Fail : see below) Minimum VRd. It would be necessary to increase the longitudinal reinforcement to B40 at 125 c/c however the UK National Annex allows an alternative approach.32 / 250] = 0. 3.11 VRd.001Ac or 25% of vertical reinforcement = 0.02 Cl.5 = 21.2 .657m) from the bottom of wall : Maximum ULS shear force from spreadsheet for Case 2 = 426 kN Shear enhancement factor = (2d/a) = 2 × 0.c with no enhancement = 497 kN > 426 kN ∴.5 × 1000 × 920 × 0.1.829m) from the bottom of wall : Maximum ULS shear force from spreadsheet for Case 2 = 511 kN Shear enhancement factor = (2d/a) = 2 × 0.vmin = 0. by inspection.c = [0.009 < 0.c = 2. EN 1992-1-1 Clause 9.Horizontal reinforcement: As.6(101)P] fcd = 1.1. but B32 @ 100 c/c reducing to B25 bars @ 150 are required to resist early thermal cracking.657 = 1.fck / 250] = 0.5m of the wall.vmin = 0.829 = 2.02 ρ1 = 8378 / (1000 × 920) = 0.2.c = 497 kN < VEd = 606 kN ∴ Fail.0 × 32 / 1. Hence early thermal cracking and long-term creep and shrinkage crack control require greater areas of reinforcement than the minimum wall reinforcement. all sections will be suitable to resist shear using B40 bars at 150 centres.523 fcd = αccfck/γc αcc = 1. 3.2(6) Check that the maximum allowable shear force is not exceeded: Maximum allowable shear force = 0.473/2 × 321/2 × 1000 × 920 × 10-3 = 325 kN cl 6.5bwdνfcd ν = 0. Minimum Wall Reinforcement EN 1992-1-1 Clause 9.92 / 1.001 × 106 = 1000 mm2/m (in each face) or 25% × 8378 = 2095mm2/m.2. i) Consider a section at (a = 0.3m from the bottom of the wall.2(101) Alternative Solution: If the reduction factor β is not used to reduce the applied shear force actions then the allowable shear force VRd.035 × 1. . 6. reducing to B25 bars @ 200 above 1.3 N/mm2 Maximu VEd = 0.3 .92 / 0.ρ1 = Asl / bwd ≤0.22 VRd.6.47 × (100 × 0.22 × 497 = 1103 kN ( > 511 kN ∴ OK) ii) Consider a section at (a = 1. Early Thermal Cracking Considering 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. B20 @ 150 c/c = 2094mm2/m.6[1 .Vertical reinforcement: As. NA to 1992-2 Cl. F = 0.52 / 2 = 365kN/m Frequent value of Surcharge UDL Force Hsc.426 × 0.75 × 1.67 Frequent value of Surcharge Line Load Force Hsc.K × K0 × γbf.426 × 19 × 9.67) = 24kN/m Deck Maximum Permanent load (concrete + surfacingmax) = 207kN/m Deck Vertical Traffic load (gr2) = 89kN/m Deck Horizontal Traffic load (gr2) = 36kN/m Restoring Effects: V Lever Arm Moment About A Stem 162.5 .5 / 2) / (1 + 9. 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 .55 < 0.0 × 0.79 × 0.6 260 Base 160 3.426) × 9.F = ψ1 × γQ × Hsc.d = γG × γSd.Fixed Abutment Serviceability Limit State Weight of wall stem = 162.ave × Z = 0.67 ∴ Df = 0.5kN/m Weight of base = 160kN/m Weight of backfill = 694.udl = ψ1 × γQ × σh.5 = 31kN/m Dispersion Factor Df for line load = (1 + Z / 2) / (1 + Z) = (1 + 9.34 × 0.5 1.Base Design Maximum bending and shear effects in the base slab will occur at sections near the front and back of the wall.2 512 Backfill 694.0 × 1.d × Z2 / 2 = 1.5) = 0.5kN/m B/fill Force Hap.0 × (113.0 × (10.75 × 1. 25 2952 Deck Vertical Reaction 296 1.75 147 Surcharge Line Load 24 9.167 1156 Surcharge UDL 31 4.5 228 .55 459 ∑= 1313 ∑= 4183 Overturning Effects: H Lever Arm Moment About A Backfill 365 3.4. (25 × 1.442} = 385kN/m2 Pressure at rear face of wall = 483 × 3.4) ± (1313 × 1.442 m Pressure under toe = 2 × 1313 / 5.442 = 483 kN/m2 Pressure at front face of wall = 483 × 4.267 = -62kN/m2 (uplift) Adjust for uplift: Reduced length of pressure under base = 3(B/2 .4m2/m I / y = 6. CASE 6 .386m Pressure under base = (1313 / 6.386 / 6.Deck Horizontal Reaction 36 7.342 / 5.(25 × 1.32 / 2) = -1173kNm/m (tension in top face).Fixed Abutment Ultimate Limit State .5 270 ∑= 456 ∑= 1801 Bearing Pressure at toe and heel of base slab = (V / A) ± (V × e × y / I) V = 1313kN/m A = 6.342 / 5.(695 × 4.4 / 2 .827m3/m Nett moment = 4183 .3422 / 6) .386) = 5.0 × 1.12 / 2) = 257kNm/m (tension in bottom face).(2382 / 1313) = 1.3 / 2) .827) Pressure under toe = 205 + 267 = 472kN/m2 Pressure under heel = 205 . SLS Moment at b-b = (297 × 3.e) = 3 × (6.12 / 3) .12 / 2) + ([483 .1801 = 2382kNm/m Eccentricity (e) of V about centre-line of base = 3.0 × 4.385] × 1.42 / 6 = 6.442} = 297kN/m2 SLS Moment at a-a = (385 × 1.1.2 . 35 × 162.35 × 160 = 216 kN/m Weight of base Comb.1/Comb.79 × 0.sup × Hsc.15 × (113.15 × (10.2 = γG.F = 0.5 = 938 kN/m Weight of backfill Comb.sup × σh.ave × Z = 0.sup × 164 + γG.0 × 43 = 207kN/m Deck Vertical Traffic load (gr2) Comb.sup × 36 = 1.1 = ψ1 × γQ.79 × 0.1 = γG.d × Z2 / 2 = 1.0 × 162.426 × 0.35 × 694.75 × 1.5 / 2) / (1 + 9.sup × 160 = 1.sup × 162.67) = 33kN/m Frequent value of Surcharge Line Load Force Hsc.1 = γQ.511 × 19 × 9.35 × 1.35 × 164 + 1.F Comb.2 691/512 Backfill 938/695 4.5 = 1.52 / 2 = 526kN/m Frequent value of Surcharge UDL Force Hsc.34 × 0.1 = γQ.2 = γG.5 = 1.0 × 1.0 × 164 + 1.2 = ψ1 × γQ.5 = 219 kN/m Weight of wall stem Comb.1 = γG.K × K0 × γbf.1 = γG.6 350/261 Base 216/160 3.ave × Z = 0.75 × 1.sup × 36 = 1.sup × 694.5) = 0.55 < 0.sup × 89 = 1.511) × 9.d Comb.sup × 43 = 1.15 × 89 = 102 kN/m Deck Horizontal Traffic load (gr2) Comb.F = 0.5 = 695 kN/m B/fill Force Hap.25 3987/2954 .udl Comb.2 Stem 219/163 1.35 × 89 = 120 kN/m Deck Vertical Traffic load (gr2) Comb.5 = 163 kN/m Weight of base Comb.5 = 42kN/m Frequent value of Surcharge UDL Force Hsc.sup × 164 + γG.Weight of wall stem Comb.2 = γG.35 × (113.sup × 43 = 1.426) × 9.35 × (10.2 Lever Arm Moment About A Comb.sup × γSd.2 × 0.2 = γG.35 × 36 = 49 kN/m Deck Horizontal Traffic load (gr2) Comb.67) = 34kN/m Deck Maximum Permanent load (concrete + surfacingmax) Comb.2 = ψ1 × γQ.0 × 694.1 = γG.F Comb.K × K0 × γbf.sup × 89 = 1.15 × 36 = 41 kN/m Restoring Effects: V Comb.sup × 694.sup × 162.52 / 2 = 592kN/m B/fill Force Hap.udl Comb.1 = γG.sup × 160 = 1.1/Comb.5 = 1.34 × 0.511 × 0.sup × γSd.75 × 1.75 × 1.2 = γG.sup × σh.426 × 19 × 9.67 ∴ Df = 0.67 Frequent value of Surcharge Line Load Force Hsc.0 × 160 = 160 kN/m Weight of backfill Comb.2 × 43 = 273kN/m Deck Maximum Permanent load (concrete + surfacingmax) Comb.2 = γQ.sup × Hsc.2 × 0.2 = γQ.1 = ψ1 × γQ.5 = 1.d Comb.d × Z2 / 2 = 1.5 = 43kN/m Dispersion Factor Df for line load = (1 + Z / 2) / (1 + Z) = (1 + 9. 75 200/204 Surcharge Line Load 33/34 9.167 1875/1666 Surcharge UDL 42/43 4.Deck Vertical Reaction 393/309 1.55 609/479 ∑= 1766/1327 ∑= 5637/4206 Overturning Effects: H Lever Arm Moment About A Backfill 592/526 3.5 314/323 Deck Horizontal Load 49/41 . 12 / 2) = 297 kNm/m (tension in bottom face).(1.12 / 2) .(1.3 / 2) = -1668 kNm/m (tension in top face).57 .1)2 / 2) .0 × 1.7.2 × 1.569m Effective base width B' = B .57 . ULS Moment at b-b = (541 × (3.0 × 1.262 .4 .0 × 1.(1.12 / 2) .2.1 × 25) = 540 kN/m ULS Shear at b-b = 516 × (2.5 368/308 ∑= 716/644 ∑= 2757/2501 Assume rectangular pressure distribution under the base as described in EN 1997-1:2004 Annex D Combination 1: V = 1766 kN/m Nett moment = 5637 .569 = 3.0 × 4. Combination 2: ULS Shear at a-a = (516 × 1. .0 × 4.938} = -454 kN/m ULS Moment at a-a = (541 × 1.262m Pressure under base = (1766 / 3.1) .695} = 560 kN/m ULS Moment at a-a = (516 × 1.32 / 2) .(1.2757 = 2880 kNm/m Eccentricity (e) of V about centre-line of base = 3.3 / 2) = -1963 kNm/m (tension in top face).(938 × 4.262) = 541 kN/m2 Combination 2: V = 1327 kN/m Nett moment = 4206 .1)2 / 2) .1) .1) .0 × 4.(1.(695 × 4.35 × 25 × 1.915 = 2.(1.12 / 2) = 307 kNm/m (tension in bottom face).262 .2 .35 × 25 × 1.2 × 1.1 × 25) = 558 kN/m ULS Shear at b-b = 541 × (3.2.57m Pressure under base = (1327 / 2.3 × 25) .0 × 25 × 1.32 / 2) .0 × 1.3 × 25) .57) = 516 kN/m2 Combination 1: ULS Shear at a-a = (541 × 1.2 .35 × 1.0 × 25 × 1.2e = 6.4 .1) .2501 = 1705 kNm/m Eccentricity (e) of V about centre-line of base = 3.35 × 1.2.(2880 / 1766) = 1.2.0 × 1.(1.0 × 1.(1705 / 1327) = 1. ULS Moment at b-b = (516 × (2.2e = 6.(1.915m Effective base width B' = B .0 × 4. 2 SLS Moment ULS Moment Comb.1/Comb.1/Comb.2 Case 2 509/501 235 280/276 Case 3 539/515 253 297/283 Case 4 521/497 244 286/273 Case 5 527485 250 290/267 Case 6 558/541 258 307/298 .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-a ULS Shear Comb. 1/Comb.2 Case 2 480/588 1184 1962/1676 Case 3 364/466 1071 1835/1610 Case 4 372/468 1069 1830/1607 Case 5 290/365 973 1716/1519 Case 6 .Case 7 445/433 235 272/238 Section b-b ULS Shear Comb.2 SLS Moment ULS Moment Comb.1/Comb. 1/Comb.2 Case 2 511/500 236 281/275 Case 3 542/516 254 298/284 Case 4 523/498 246 288/274 .453/561 1178 1961/1668 Case 7 281/311 934 1663/1436 Free Abutment Base: Section a-a ULS Shear Comb.1/Comb.2 SLS Moment ULS Moment Comb. 2 Case 2 479/582 1239 2054/1765 Case 3 365/464 1124 1924/1693 Case 4 .1/Comb.2 SLS Moment ULS Moment Comb.Case 5 530/487 251 292/268 Case 6 494/455 234 272/250 Case 7 440/372 212 242/205 Section b-b ULS Shear Comb.1/Comb. 6 × 300 = 180kN To allow for load distribution effects assume a 45° dispersal down the wall.0m notional lane gives a distribution width of 7. 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 = 996 kN/m (at d from support) > 582 ∴ OK.373/467 1122 1918/1690 Case 5 291/365 1023 1800/1597 Case 6 327/394 1040 1815/1611 Case 7 172/192 741 1428/1266 Early Thermal Cracking Considering 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. Loading will be applied from the backfill. Msls = 1239 kNm/m (Mperm = 860 kNm/m + Mvar = 380 kNm/m) B40's @ 150 c/c give Muls = 2975 kNm/m > 2054 ∴ OK.9. surcharge and braking loads on top of the wall. Msls = 1103 kNm/m > 258 ∴ OK Section b-b: Muls = 2054 kNm/m. Msls = 2065 kNm/m > 1239 ∴ OK Local Effects Curtain Wall (Abutment Upstand Wall) This wall is designed to be cast onto the top of the abutment after the deck has been built.3. . with maximum dispersal of the width of the abutment (11.2(2) of BS EN 1992-1-1. 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 = 582 kN/m.6αQ1Q1k from LM1 axle = 0.2(2): Braking Force = 0.6m). EN 1991-2 Clause 4. Positioning the axle in the centre of the 3. Vuls = 731 kN/m (at d from support) > 558 ∴ OK. Vuls = 558 kN/m.65m at the base of the wall. 5 × 3.35 × 1.65 = 70.0 + 1.34Kd = 10.35 × 70.6 + 36 = 107kNm/m Total ULS moment (combination 1) = (1.426 × 19.5 + 19.2 × 36) = 158 + 58 = 204kNm/m Hence Normal Traffic Surcharge + Backfill has worst effect on curtain wall.02 / 2 = 36kN/m Total ULS shear = 1.5 + 4.35 × (23.426 × 0.625 < 0.5kN/m due to backfill = K0 γbf.5) + (1.35 × (97.2 + 22.426 = 4.0 × 3.5kN/m UDL = 10.Shear at the base of the wall: due to braking = 180 / 7.0 × 3.8 + 36 = 153kNm/m Total ULS moment = 1.02 / 2 + 36 = 85.4kN/m2 Total ULS shear = 1.2 × 36) = 90kN/m Bending moment at the base of the wall: due to braking = 180 × 3.0 × 3.5 + 19.4 × 3.67 ∴ Df = 0.0kN/m2 Total ULS shear = 1.0 + 1.35 × (32.64 × 0.65 = 23.426 × 0.67 = 32.426 × 19.2 × 36) = 120kN/m Total SLS moment = 32.35 × 1.4 × 3.5 + 1.79KdDf = 113.79 × 0.57 × 0.8) + (1.34 × 0.4 + 5.03 / 6 = 36kNm/m Total SLS moment = 70.64Kd = 11.0 / 7.35 × (28.2 × 36) = 117kN/m Total SLS moment = 28.d Z3 / 6 = 0.6kNm/m due to backfill = K0 γbf.57KdDf = 99.35 × 1.d Z2 / 2 = 0.426 = 5.2 + 22.4 × 3.4kN/m UDL = 11. 400 thick curtain wall with B32 @ 150 c/c : Mult = 601 kNm/m > 216 kNm/m ∴ OK Msls = 317 kNm/m > 153 kNm/m ∴ OK Vult = 261 kN/m > 120 kN/m ∴ Shear OK .0 + 5.6) + (1.5 + 36 = 144kNm/m Total ULS moment = 1.67 Line Load = 113.67 = 28.02 / 2 + 36 = 97.35 × (85.2 × 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) = 0.0 + 4.0 × 3.2 × 36) = 158 + 58 = 216kNm/m SV/196 traffic surcharge: Line Load = 99.
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