Prestressed Concrete
Since concrete is weak in tension in normal reinforced concrete construction cracks develop in the tension z Prestressing involves inducing compressive stresses in the zone which will tend to become tensile under ext The prestressing force also reduces the magnitude of the principal tensile stress in the web so that thin-web The prestressing force has to be produced by a high tensile steel, and it is necessary to use high quality con There are two methods of prestressing concrete : 1) Pre-cast Pre-tensioned 2) Pre-cast Post-tensioned Both methods involve tensioning cables inside a concrete beam and then anchoring the stressed cables to t 1) Pre-tensioned Beams
Stage 1 Tendons and reinforcement are positioned in the beam mould. Stage 2 Tendons are stressed to about 70% of their ultimate strength. Stage 3 Concrete is cast into the beam mould and allowed to cure to the required initial strength. Stage 4 When the concrete has cured the stressing force is released and the tendons anchor themselves in the conc 2) Post-tensioned Beams
Stage 1 Cable ducts and reinforcement are positioned in the beam mould. The ducts are usually raised towards the Stage 2 Concrete is cast into the beam mould and allowed to cure to the required initial strength. Stage 3 Tendons are threaded through the cable ducts and tensioned to about 70% of their ultimate strength. Stage 4 Wedges are inserted into the end anchorages and the tensioning force on the tendons is released. Grout is t
Loss of Prestress When the tensioning force is released and the tendons are anchored to the concrete a series of effects resu a. relaxation of the steel tendons b. elastic deformation of the concrete c. shrinkage and creep of the concrete d. slip or movement of the tendons at the anchorages during anchoring e. other causes in special circumstances , such as when steam curing is used with pre-tensioning. Total losses in prestress can amount to about 30% of the initial tensioning stress.
Prestressed Concrete Beam Design to BS 5400 Part 4
Problem:
Design a simply supported prestressed concrete Y beam which carries a 150mm thick concrete slab and 100mm of s
γconc. = 24kN/mm3
25 units of HB to be considered at SLS for load combination 1 only (BS 5400 Pt4 Cl. 4.2.2) Loading per beam (at 1.0m c/c) Nominal Dead Loads :
slab = 24 x 0.15 x 1.0
= 3.6 kN/m
beam = say Y5 beam
= 10.78 kN/m
surfacing = 24 x 0.1 x= 2.4 kN/m Nominal Live Load :
HA = 10 x 1.0 + 33.0
= 10 kN/m + 33kN
25 units HB = 25 x 10= 62.5 kN per wheel Load factors for serviceability and ultimate limit state from BS 5400 Part 2 (or BD 37/01) Table 1: SLS Comb.1 γfL concrete1.0 1.0
Dead Load Superimposed Dead Load
γfL surfacing1.2 1.2 γfL HA
Live Load
Temperature Difference
γfL
-
Comb.3 1.15
1.75
1.75
1.2
1.0
γfL HB
1.1
-
-
1.0
0.8
1.15
1.5
Concrete Grades Beam fcu = 50 N/mm2, fci = 40 N/mm2 Slab fcu = 40 N/mm2 BS 5400 Pt. 4 Section Properties cl.7.4.1 Property
Modular ratio effect for different concrete strengths between beam and slab may be ig Beam Section
Composite Section
Area(mm2)
4.49E+05
5.99E+05
Centroid(mm)
456
623
2nd Moment of Area(mm4)
5.29E+10
1.04E+11
Modulus @ Level 1(mm3)
1.16E+08
1.66E+08
Modulus @ Level 2(mm3)
8.91E+07
2.42E+08
Modulus @ Level 3(mm3)
-
1.79E+08
Temperature Difference Effects Apply temperature differences given in BS 5400 Pt2 Fig.9 (Group 4)to a simplified beam section. Cl. 5.4.6 - Coefficient of thermal expansion = 12 x 10-6 per ºC. From BS 5400 Pt4 Table 3 : Ec = 34 kN/mm2 for fcu = 50N/mm2 Hence restrained temperature stresses per °C = 34 x 10 3 x 12 x 10-6 = 0.408 N/mm2
a) Positive temperature difference Force F to restrain temperature strain : 0.408 x 1000 x [ 150 x ( 3.0 + 5.25 ) ] x 10-3 + 0.408 x ( 300 x 250 x 1.5 + 750 x 200 x 1.25 ) x 10-3 = 504.9 + 122.4 = 627.3 kN Moment M about centroid of section to restrain curvature due to temperature strain : 0.408 x 1000 x [ 150 x ( 3.0 x 502 + 5.25 x 527 ) ] x 10-6 + 0.408 x ( 300 x 250 x 1.5 x 344 - 750 x 200 x 1.25 x 556 ) x 10-6 = 261.5 - 26.7 = 234.8 kNm
b) Reverse temperature difference Force F to restrain temperature strain : - 0.408 x [ 1000 x 150 x ( 3.6 + 2.3 ) + 300 x 90 x ( 0.9 + 1.35 ) ] x 10 -3 - 0.408 x 300 x ( 200 x 0.45 + 150 x 0.45 ) x 10-3 - 0.408 x 750 x [ 50 x ( 0.9 + 0.15 ) + 240 x ( 1.2 + 2.6 ) ] x 10 -3 = - 385.9 - 19.3 - 295.1 = - 700.3 kN Moment M about centroid of section to restrain curvature due to temperature strain : - 0.408 x [ 150000 x ( 3.6 x 502 + 2.3 x 527 ) + 27000 x ( 0.9 x 382 + 1.35 x 397 ) ] x 10 -6 - 0.408 x 300 x ( 200 x 0.45 x 270 - 150 x 0.45 x 283 ) x 10-6 + 0.408 x 750 x [ 50 x ( 0.9 x 358 + 0.15 x 366 ) + 240 x ( 1.2 x 503 + 2.6 x 543 ) ] x 10 -6 = - 194.5 - 0.6 + 153.8 = - 41.3 kNm
Differential Shrinkage Effects BS 5400 Pt.4
Use cl.6.7.2.4 Table 29 :
cl.7.4.3.4
Total shrinkage of insitu concrete = 300 x 10-6
Assume that 2/3 of the total shrinkage of the precast concrete takes place b hence the differential shrinkage is 200 x 10-6 BS 5400 Pt.4
Force to restrain differential shrinkage : F = - ε diff x Ecf x Acf x φ
cl.7.4.3.5
F = -200 x 10-6 x 34 x 1000 x 150 x 0.43 = -439 kN Eccentricity acent = 502mm Restraint moment Mcs = -439 x 0.502 = -220.4 kNm
Self weight of beam and weight of deck slab is supported by the beam. When the deck slab concrete has cured then
Dead Loading (beam and slab) Total load for serviceability limit state = (1.0 x 3.6)+(1.0 x 10.78) = 14.4kN/m Design serviceability moment = 14.4 x 242 / 8 = 1037 kNm
Combination 1 Loading Super. & HA live load for SL= [(1.2 x 2.4)+(1.2 x 10)]udl & [(1.2 x 33)]kel = (2.88 + 12.0)udl & 39.6kel = 14.9 kN/m & 39.6kN Super. & HB live load for SL = 2.88 & 4 wheels @ 1.1 x 62.5 = 2.9 kN/m & 4 wheels @ 68.75 kN Total load for ultimate limit = [(1.15 x 3.6)+(1.15 x 10.78)+(1.75 x 2.4)+(1.5 x 10)]udl & [(1.5 x 33)]kel = (4.14 + 12.40 + 4.20 + 15.0)udl & 49.5kel = 35.7 kN/m & 49.5kN
HA Design serviceability mo = 14.9 x 24.02 / 8 + 39.6 x 24 / 4 =1310 kNm 25 units HB Design SLS mom= 2.9 x 24.02 / 8 + 982.3(from grillage analysis) = 1191.1 kNm Design ultimate moment
= 35.7 x 24.02 / 8 + 49.5 x 24 / 4 = 2867 kNm
Combination 3 Loading Super. & HA live load for SL= [(1.2 x 2.4)+(1.0 x 10)]udl & [(1.0 x 33)]kel = (2.88 + 10.0)udl & 33kel = 12.9 kN/m & 33kN Total load for ultimate limit = [(1.15 x 3.6)+(1.15 x 10.78)+(1.75 x 2.4)+(1.25 x 10)]udl & [(1.25 x 33)]kel = (4.14 + 12.40 + 4.20 + 12.5)udl & 41.3kel = 33.2 kN/m & 41.3kN Design serviceability moment= 12.9 x 24.02 / 8 + 33 x 24 / 4 = 1127 kNm Allowable stresses in precast concrete At transfer : Compression ( Table 23 )
cl.
0.5fci (= - 1.0 (eqn. 1) x Zlevel 1 + (eqn. 2) x Zlevel 2 gives : P >= A x (20 x Zlevel 1 - 1.0 x Zlevel 2) / (Zlevel 1 + Zlevel 2) P = 449.22 x 103 x ( 20 x 116.02 - 89.066) / ( 116.02 + 89.066) x 10-3 = 4888 kN Allow 10% for loss of force before and during transfer, then the initial force P o = 4888 / 0.9 = 5431kN Using 15.2mm class 2 relaxation standard strand at maximum initial force of 174kN (0.75 x P u) Area of tendon = 139mm2 Nominal tensile strength = fpu =1670 N/mm2 Hence 32 tendons required. Initial force Po = 32 x 174 = 5568 kN P = 0.9 x 5568 = 5011 kN
Substituting P = 5011 kN in (eqn. 2) e 0 hence O.K.) Level 1, combination 3 : f = 17.08 - 17.67 = - 0.59 N/mm2 (> - 3.2 hence O.K.)
Level 2, combination 1 : f = - 0.98 + 1037 / 89.066 + 1310 / 242.424 + 1.64 = 17.71 (< 25 O.K Level 3. combination 3 : f = (1127 / 179.402) + (0.8 x 3.15) = 8.8 N/mm2 (< 25 O.K.) Ultimate Capacity of Beam and Deck Slab (Composite Section)
Ultimate Design Moment = γf3 x M = 1.1 x 2867 = 3154 kNm cl. 6.3.3
Only steel in the tension zone is to be considered :
Centroid of tendons in tension zone = (6x60 + 10x110 + 8x160 + 4x260) / 28 = 135mm Effective depth from Level 3 = 1200 - 135 = 1065mm Assume that the maximum design stress is developed in the tendons, then : Tensile force in tendons Fp = 0.87 x 28 x 139 x 1670 x 10-3 = 5655 kN Compressive force in concrete flange : Ff = 0.4 x 40 x 1000 x 150 x 10-3 = 2400 kN Let X = depth to neutral axis. Compressive force in concrete web : Fw = 0.4 x 50 x [393 - (393 - 200) x (X - 150) / (671 x 2)] x (X - 150) x 10-3 Fw = ( -2.876X2 + 8722.84X - 1243717) x 10-3 Equating forces to obtain X : 5655 = 2400 + ( -2.876X2 + 8722.84X - 1243717) x 10-3 X = 659 mm Stress in tendon after losses = fpe = 4144 x 103 / (32 x 139) = 932 N/mm2 Prestrain εpe = fpe / Es = 932 / 200 x 103 = 0.0047
Determine depth to neutral axis by an iterative strain compatibility analysis Try X = 659 mm as an initial estimate Width of web at this depth = 247mm εpb6 = ε6 + εpe = -459 x 0.0035 / 659 + 0.0047 = 0.0022 εpb5 = ε5 + εpe = -359 * 0.0035 / 659 + 0.0047 = 0.0028 εpb4 = ε4 + εpe = 281 * 0.0035 / 659 + 0.0047 = 0.0062 εpb3 = ε3 + εpe = 381 * 0.0035 / 659 + 0.0047 = 0.0067 εpb2 = ε2 + εpe = 431 * 0.0035 / 659 + 0.0047 = 0.0069 εpb1 = ε1 + εpe = 481 * 0.0035 / 659 + 0.0047 = 0.0072 fpb6 = 0.0022 x 200 x 103 = 444 N/mm2 fpb5 = 0.0028 x 200 x 103 = 551 N/mm2 fpb4 = 1162 + 290 x (0.0062 - 0.0058) / 0.0065 = 1178 N/mm2 fpb3 = 1162 + 290 x (0.0067 - 0.0058) / 0.0065 = 1201 N/mm2 fpb2 = 1162 + 290 x (0.0069 - 0.0058) / 0.0065 = 1213 N/mm2 fpb1 = 1162 + 290 x (0.0072 - 0.0058) / 0.0065 = 1225 N/mm2
Tensile force in tendons : Fp6 = 2 x 139 x 444 x 10-3 = 124 Fp5 = 2 x 139 x 551 x 10-3 = 153 Fp4 = 4 x 139 x 1178 x 10-3 = 655 Fp3 = 8 x 139 x 1201 x 10-3 = 1336 Fp2 = 10 x 139 x 1213 x 10-3= 1686 Fp1 = 6 x 139 x 1225 x 10-3 = 1022 4976 kN Compressive force in concrete : Ff = 0.4 x 40 x 1000 x 150 x 10-3= 2400 Fw = 0.4 x 50 x 0.5 x (393 + 247) x (659 - 150) x 10-3= 3258 5658 kN Fc > Ft therefore reduce depth to neutral axis and repeat the calculations. Using a depth of 565mm will achieve equilibrium. The following forces are obtained : Fp6 = 134
Ff = 2400
Fp5 = 168
Fw = 2765
Fp4 = 675
Fc = 5165
Fp3 = 1382 Fp2 = 1746 Fp1 = 1060 Ft = 5165 Taking Moments about the neutral axis : Fp6 = 134 x -0.365 =-49 Fp5 = 168 x -0.265 = -45
Fp4 = 675 x 0.375 = 253 Fp3 = 1382 x 0.475 = 656 Fp2 = 1746 x 0.525 =917 Fp1 = 1060 x 0.575 = 610 Ff = 2400 x 0.49 = 1176 Fw = 3258 x 0.207 = 674 Mu = 4192 kNm > 3154 kNm hence O.K. cl. 6.3.3.1
Mu / M = 4192 / 3154 = 1.33 ( > 1.15 ) hence strain in outermost tendon O.K.
Abutment Design Example to BD 30
Design the fixed and free end cantilever abutments to the 20m span deck shown to carry HA and 45 units o
The ground investigation report shows suitable founding strata about 9.5m below the proposed road level. T Backfill material will be Class 6N with an effective angle of internal friction (ϕ') = 35o and density (γ) = 19k
The proposed deck consists of 11No. Y4 prestressed concrete beams and concrete deck slab as shown. Loading From the Deck A grillage analysis gave the following reactions for the various load cases: Critical Rea Nominal Reaction( Concrete Deck Surfacing HA udl+kel 45 units HB
Total Reaction on Each Abutment
180 30 160 350
Ultimate Reaction(kN) 230 1900 60 265 500
320 1140 1940
Nominal loading on 1m length of abutment: Deck Dead Load = (1900 + 320) / 11.6 = 191kN/m HA live Load on Deck = 1140 / 11.6 = 98kN/m HB live Load on Deck = 1940 / 11.6 = 167kN/m
From BS 5400 Part 2 Figures 7 and 8 the minimum and maximum shade air temperatures are -19 and +37 For a Group 4 type strucutre (see fig. 9) the corresponding minimum and maximum effective bridge temper Hence the temperature range = 11 + 36 = 47oC. From Clause 5.4.6 the range of movement at the free end of the 20m span deck = 47 x 12 x 10 -6 x 20 x 10 The ultimate thermal movement in the deck will be ± [(11.3 / 2) γf3 γfL] = ±[11.3 x 1.1 x 1.3 /2] = ± 8mm
Option 1 - Elastomeric Bearing: With a maximum ultimate reaction = 230 + 60 + 500 = 790kN then a suitable elastomeric bearing would b
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 th A tolerance is also required for setting the bearing if the ambient temperature is not at the mid range temp Horizontal load at bearing for 10mm contraction = 12.14 x 10 = 121kN. This is an ultimate load hence the nominal horizontal load = 121 / 1.1 / 1.3 = 85kN at each bearing. Total horizontal load on each abutment = 11 x 85 = 935 kN ≡ 935 / 11.6 = 81kN/m. Alternatively using BS 5400 Part 9.1 Clause 5.14.2.6: H = AGδr/tq Using the Ekspan bearing EKR35 Maximum Load = 1053kN
Area = 610 x 420 = 256200mm2 Nominl hardness = 60 IRHD Bearing Thickness = 19mm
Shear modulus G from Table 8 = 0.9N/mm2 H = 256200 x 0.9 x 10-3 x 10 / 19 = 121kN This correllates with the value obtained above using the shear stiffness from the manufacturer's data sheet.
Option 2 - Sliding Bearing: With a maximum ultimate reaction of 790kN and longitudinal movement of ± 8mm then a suitable bearing from the Ekspan Maximum Load = 800kN Base Plate A dimension = 210mm Base Plate B dimension = 365mm Movement ± X = 12.5mm BS 5400 Part 2 - Clause 5.4.7.3: Average nominal dead load reaction = (1900 + 320) / 11 = 2220 / 11 = 200kN Contact pressure under base plate = 200000 / (210 x 365) = 3N/mm2 As the mating surface between the stainless steel and PTFE is smaller than the base plate then the pressure From Table3 of BS 5400 Part 9.1 the Coefficient of friction = 0.08 for a bearing stress of 5N/mm 2 Hence total horizontal load on each abutment when the deck expands or contracts = 2220 x 0.08 = 180kN Traction and Braking Load - BS 5400 Part 2 Clause 6.10: Nominal Load for HA = 8kN/m x 20m + 250kN = 410kN Nominal Load for HB = 25% of 45units x 10kN x 4axles = 450kN 450 > 410kN hence HB braking is critical. Braking load on 1m width of abutment = 450 / 11.6 = 39kN/m. When this load is applied on the deck it will act on the fixed abutment only. Skidding Load - BS 5400 Part 2 Clause 6.11: Nominal Load = 300kN 300 < 450kN hence braking load is critical in the longitudinal direction. When this load is applied on the deck it will act on the fixed abutment only. Loading at Rear of Abutment Backfill For Stability calculations use active earth pressures = Ka γ h Ka for Class 6N material = (1-Sin35) / (1+Sin35) = 0.27 Density of Class 6N material = 19kN/m3
Active Pressure at depth h = 0.27 x 19 x h = 5.13h kN/m2 Hence Fb = 5.13h2/2 = 2.57h2kN/m Surcharge - BS 5400 Part 2 Clause 5.8.2: For HA loading surcharge = 10 kN/m2 For HB loading surcharge = 20 kN/m2 Assume a surchage loading for the compaction plant to be equivalent to 30 units of HB Hence Compaction Plant surcharge = 12 kN/m2. For surcharge of w kN/m2 : Fs = Ka w h = 0.27wh kN/m 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, Reynolds's Rein Alternatively a simple spreadsheet will achieve a result by trial and error. Load Combinations
Backfill + Construction surcharge
Backfill + Construc tion surcharg e
Sackfill Backfill + HA + HA surcharg surcharg e + Deck e + dead Braking load + behind Deck abutmen contracti t + Deck on dead load
Backfill Backfill + HB + HA surcharg surcharg e + Deck e + Deck dead dead load load + HB on deck
Fixed Abutment Only Backfill + HA surcharge + Deck dead load + HA on deck + Braking on deck CASE 1 - Fixed Abutment Density of reinforced concrete = 25kN/m3. Weight of wall stem = 1.0 x 6.5 x 25 = 163kN/m Weight of base = 6.4 x 1.0 x 25 = 160kN/m Weight of backfill = 4.3 x 6.5 x 19 = 531kN/m Weight of surcharge = 4.3 x 12 = 52kN/m Backfill Force Fb = 0.27 x 19 x 7.52 / 2 = 144kN/m Surcharge Force Fs = 0.27 x 12 x 7.5 = 24 kN/m Restoring Effects: Weight Stem Base Backfill Surcharge ∑= Overturning Effects: F Backfill Surcharge ∑ =168
Lever A 163 160 531 52
1.6 3.2 4.25 4.25
Moment About A 261 512 2257 221 ∑ =3251
2.5 3.75
Moment About A 361 91 ∑ =452
906 Lever A 144 24
Factor of Safety Against Overturning = 3251 / 452 = 7.2 > 2.0 ∴ OK. For sliding effects: Active Force = Fb + Fs = 168kN/m Frictional force on underside of base resisting movement = W tan(φ) = 906 x tan(30o) = 523kN/m Factor of Safety Against Sliding = 523 / 168 = 3.1 > 2.0 ∴ OK.
Bearing Pressure: Check bearing pressure at toe and heel of base slab = (P / A) ± (P x e / Z) where P x e is the moment abou P = 906kN/m A = 6.4m2/m Z = 6.42 / 6 = 6.827m3/m Nett moment = 3251 - 452 = 2799kNm/m Eccentricity (e) of P about centre-line of base = 3.2 - (2799 / 906) = 0.111m Pressure under base = (906 / 6.4) ± (906 x 0.111 / 6.827) Pressure under toe = 142 + 15 = 157kN/m2 < 400kN/m2 ∴ OK. Pressure under heel = 142 - 15 = 127kN/m2 Hence the abutment will be stable for Case 1. Analysing the fixed abutment with Load Cases 1 to 6 and the free abutment with Load Cases 1 to 5 using a Fixed Abutment: F of S Overtur Case Case Case Case Case Case Case
1 2 2a 3 4 5 6
7.16 2.87 4.31 3.43 4.48 5.22 3.8
F of S Sliding Bearing Pressure Bearing at Pressure Toe at Heel 3.09 156 2.13 386 2.64 315 2.43 351 2.63 322 3.17 362 2.62 378
Free Abutment:
F of S Overturning F of S Sli Bearing PrBearing Pressure at Heel Case 1 7.15 3.09 168 Case 2 2.91 2.14 388 Case 2a 4.33 2.64 318 Case 3 3.46 2.44 354 Case 4 4.5 2.64 325 Case 5 5.22 3.16 365 It can be seen that the use of elastomeric bearings (Case 2) will govern the critical design load cases on the 2) Wall and Base Design
Loads on the back of the wall are calculated using 'at rest' earth pressures. Serviceability and Ultimate load Using the Fixed Abutment Load Case 1 again as an example of the calculations: Wall Design Ko = 1 - Sin(ϕ') = 1 - Sin(35o) = 0.426 γfL for horizontal loads due to surcharge and backfill from BS 5400 Part 2 Clause 5.8.1.2:
Serviceability = 1.0 Ultimate = 1.5 γf3 = 1.0 for serviceability and 1.1 for ultimate (from BS 5400 Part 4 Clauses 4.2.2 and 4.2.3) Backfill Force Fb on the rear of the wall = 0.426 x 19 x 6.52 / 2 = 171kN/m Surcharge Force Fs on the rear of the wall = 0.426 x 12 x 6.5 = 33kN/m At the base of the Wall: Serviceability moment = (171 x 6.5 / 3) + (33 x 6.5 / 2) = 371 + 107 = 478kNm/m Ultimate moment = 1.1 x 1.5 x 478 = 789kNm/m Ultimate shear = 1.1 x 1.5 x (171 + 33) = 337kN/m Analysing the fixed abutment with Load Cases 1 to 6 and the free abutment with Load Cases 1 to 5 using a Fixed Abutment: Moment SLS Dead Case Case Case Case Case Case
1 2a 3 4 5 6
Free Abutment: Moment SLS Dead Case 1 Case 2a Case 3 Case 4 Case 5
371 829 829 829 829 829
Moment SLS Liv 108 258 486 308 154 408
Moment ULS 790 1771 2097 1877 1622 1985
Shear ULS 337 566 596 602 543 599
394 868 868 868 868
Moment SLS Liv 112 265 495 318 159
Moment ULS 835 1846 2175 1956 1694
Shear ULS 350 581 612 619 559
Concrete to BS 8500:2006 Use strength class C32/40 with water-cement ratio 0.5 and minimum cement content of 340kg/m3 for expo Nominal cover to reinforcement = 60mm (45mm minimum cover plus a tolerance Δc of 15mm).Reinforceme
Design for critical moments and shear in Free Abutment: Reinforced concrete walls are designed to BS 5400 Part 4 Clause 5.6. Check classification to clause 5.6.1.1: Ultimate axial load in wall from deck reactions = 2400 + 600 + 2770 = 5770 kN 0.1fcuAc = 0.1 x 40 x 103 x 11.6 x 1 = 46400 kN > 5770 ∴ design as a slab in accordance with clause 5.4
Bending
BS 5400 Part 4 Clause 5.4.2 → for reisitance moments in slabs design to clause 5.3.2.3: z = {1 - [ 1.1fyAs) / (fcubd) ]} d Use B40 @ 150 c/c: As = 8378mm2/m, d = 1000 - 60 - 20 = 920mm z = {1 - [ 1.1 x 500 x 8378) / (40 x 1000 x 920) ]} d = 0.875d < 0.95d ∴ OK Mu = (0.87fy)Asz = 0.87 x 500 x 8378 x 0.875 x 920 x 10-6 = 2934kNm/m > 2175kNn/m ∴ Carrying out the crack control calculation to Clause 5.8.8.2 gives a crack width of 0.2mm < 0.25mm. Also the steel reinforcement and concrete stresses meet the limitations required in clause 4.1.1.3 Shear Shear requirements are designed to BS 5400 clause 5.4.4: v = V / (bd) = 619 x 103 / (1000 x 920) = 0.673 N/mm2 No shear reinforcement is required when v < ξsvc ξs = (500/d)1/4 = (500 / 920)1/4 = 0.86 vc = (0.27/γm)(100As/bwd)1/3(fcu)1/3 = (0.27 / 1.25) x ({100 x 8378} / {1000 x 920})1/3 x (40)
ξsvc = 0.86 x 0.72 = 0.62 N/mms < 0.673 hence shear reinforcement should be provided, however check sh ULS shear at Section 7H/8 for load case 4 = 487 kN v = V / (bd) = 487 x 103 / (1000 x 920) = 0.53 N/mm2 < 0.62 Hence height requiring strengthening = 1.073 x (0.673 - 0.62) / (0.673 - 0.53) = 0.4m < d. Provide a 500 x 500 splay at the base of the wall with B32 @ 150c/c bars in sloping face. Early Thermal Cracking
Considering the effects of casting the wall stem onto the base slab by complying with the early thermal cracking of concrete Minimum area of secondary reinforcement to Clause 5.8.4.2 = 0.12% of bad = 0.0012 x 1000 x 920 = 110 Base Design
Maximum bending and shear effects in the base slab will occur at sections near the front and back of the wa Using the Fixed Abutment Load Case 1 again as an example of the calculations: CASE 1 - Fixed Abutment Serviceability Limit State γfL = 1.0 γf3 = 1.0 Weight of wall stem = 1.0 x 6.5 x 25 x 1.0 = 163kN/m Weight of base = 6.4 x 1.0 x 25 x 1.0 = 160kN/m Weight of backfill = 4.3 x 6.5 x 19 x 1.0 = 531kN/m Weight of surcharge = 4.3 x 12 x 1.0 = 52kN/m B/fill Force Fb = 0.426 x 19 x 7.52 x 1.0 / 2 = 228kN/m Surcharge Force Fs = 0.426 x 12 x 7.5 x 1.0 = 38 kN/m
Restoring Effects: Weight Stem Base Backfill Surcharge
Lever ArmMoment About A 163 1.6 160 3.2 531 4.25 52 4.25
261 512 2257 221
∑= Overturning Effects: F Backfill Surcharge ∑=
906
266
∑=
Lever ArmMoment About A 228 2.5 38 3.75 ∑= 713
3251
570 143
Bearing Pressure at toe and heel of base slab = (P / A) ± (P x e / Z) P = 906kN/m A = 6.4m2/m Z = 6.42 / 6 = 6.827m3/m Nett moment = 3251 - 713 = 2538kNm/m Eccentricity (e) of P about centre-line of base = 3.2 - (2538 / 906) = 0.399m Pressure under base = (906 / 6.4) ± (906 x 0.399 / 6.827) Pressure under toe = 142 + 53 = 195kN/m2 Pressure under heel = 142 - 53 = 89kN/m2 Pressure at front face of wall = 89 + {(195 - 89) x 5.3 / 6.4} = 177kN/m2 Pressure at rear face of wall = 89 + {(195 - 89) x 4.3 / 6.4} = 160kN/m2
SLS Moment at a-a = (177 x 1.12 / 2) + ([195 - 177] x 1.12 / 3) - (25 x 1.0 x 1.12 / 2) = 99kNm/m (tensio
SLS Moment at b-b = (89 x 4.32 / 2) + ([160 - 89] x 4.32 / 6) - (25 x 1.0 x 4.32 / 2) - (531 x 4.3 / 2) - (52
CASE 1 - Fixed Abutment Ultimate Limit State γfL for concrete = 1.15 γfL for fill and surcharge(vetical) = 1.2 γfL for fill and surcharge(horizontal) = 1.5 Weight of wall stem = 1.0 x 6.5 x 25 x 1.15 = 187kN/m Weight of base = 6.4 x 1.0 x 25 x 1.15 = 184kN/m Weight of backfill = 4.3 x 6.5 x 19 x 1.2 = 637kN/m Weight of surcharge = 4.3 x 12 x 1.2 = 62kN/m Backfill Force Fb = 0.426 x 19 x 7.52 x 1.5 / 2 = 341kN/m Surcharge Force Fs = 0.426 x 12 x 7.5 x 1.5 = 58 kN/m Restoring Effects:
Weight
Lever Arm
Moment About A
Stem Base Backfill Surcharge ∑=
Overturning Effects: F Backfill Surcharge ∑=
187 184 637 62 1070
1.6 3.2 4.25 4.25 ∑=
Lever Arm 341 58 399
299 589 2707 264 3859
Moment About A 2.5 3.75 ∑=
853 218 1071
Bearing Pressure at toe and heel of base slab = (P / A) ± (P x e / Z) P = 1070kN/m A = 6.4m2/m Z = 6.42 / 6 = 6.827m3/m Nett moment = 3859 - 1071 = 2788kNm/m Eccentricity (e) of P about centre-line of base = 3.2 - (2788 / 1070) = 0.594m Pressure under base = (1070 / 6.4) ± (1070 x 0.594 / 6.827) Pressure under toe = 167 + 93 = 260kN/m2 Pressure under heel = 167 - 93 = 74kN/m2 Pressure at front face of wall = 74 + {(260 - 74) x 5.3 / 6.4} = 228kN/m2 Pressure at rear face of wall = 74 + {(260 - 74) x 4.3 / 6.4} = 199kN/m2 γf3 = 1.1 ULS Shear at a-a = 1.1 x {[(260 + 228) x 1.1 / 2] - (1.15 x 1.1 x 25)} = 260kN/m ULS Shear at b-b = 1.1 x {[(199 + 74) x 4.3 / 2] - (1.15 x 4.3 x 25) - 637 - 62} = 259kN/m
ULS Moment at a-a = 1.1 x {(228 x 1.12 / 2) + ([260 - 228] x 1.12 / 3) - (1.15 x 25 x 1.0 x 1.12 / 2)} = 14
SLS Moment at b-b = 1.1 x {(74 x 4.32 / 2) + ([199 74] x 4.32 / 6) - (1.15 x 25 x 1.0 x 4.32 / 2) - (637 x 4.3 / 2) - (62 x 4.3 / 2)} = -769kNm/m (tension in top face).
Analysing the fixed abutment with Load Cases 1 to 6 and the free abutment with Load Cases 1 to 5 using a Fixed Abutment Base:
Section a-a ULSShear Case 1 Case 2a Case 3 Case 4 Case 5 Case 6
Free Abutment Base: Section a-a ULSShear Case 1 Case 2a Case 3 Case 4 Case 5
Section b-b SLSMomen ULSMoment 261 99 528 205 593 235 550 208 610 241 637 255
ULS SheaSLS MomeULS Moment 147 259 447 302 458 980 340 553 1178 314 495 1003 348 327 853 365 470 1098
SLSMomen ULSMoment 267 101 534 207 598 236 557 211 616 243
Section b-b ULSShearSLSMoment ULSMoment 151 266 475 305 466 1029 342 559 1233 317 504 1055 351 335 901
Design for shear and bending effects at sections a-a and b-b for the Free Abutment: Bending BS 5400 Part 4 Clause 5.7.3 → design as a slab for reisitance moments to clause 5.3.2.3: z = {1 - [ 1.1fyAs) / (fcubd) ]} d Use B32 @ 150 c/c: As = 5362mm2/m, d = 1000 - 60 - 16 = 924mm z = {1 - [ 1.1 x 500 x 5362) / (40 x 1000 x 924) ]} d = 0.92d < 0.95d ∴ OK Mu = (0.87fy)Asz = 0.87 x 500 x 5362 x 0.92 x 924 x 10-6 = 1983kNm/m > 1922kNm/m ∴ (1983kNm/m also > 1834kNm/m ∴ B32 @ 150 c/c suitable for fixed abutment.
For the Serviceability check for Case 3 an approximation of the dead load moment can be obtained by remo Carrying out the crack control calculation to Clause 5.8.8.2 gives a crack width of 0.27mm > 0.25mm This could be corrected by reducing the bar spacing, but increase the bar size to B40@150 c/c as this is req Using B40@150c/c the crack control calculation gives a crack width of 0.17mm < 0.25mm ∴ OK. Also the steel reinforcement and concrete stresses meet the limitations required in clause 4.1.1.3 Shear Shear on Toe - Use Fixed Abutment Load Case 6: By inspection B32@150c/c will be adequate for the bending effects in the toe (Muls = 365kNm < 1983kNm) Shear requirements are designed to BS 5400 clause 5.7.3.2(a) checking shear at d away from the front face of the wall to clause 5.4.4.1: ULS Shear on toe = 1.1 x {(620 + 599) x 0.5 x 0.176 1.15 x 1 x 0.176 x 25} = 112kN
v = V / (bd) = 112 x 103 / (1000 x 924) = 0.121 N/mm2 No shear reinforcement is required when v < ξsvc Reinforcement in tension = B32 @ 150 c/c ξs = (500/d)1/4 = (500 / 924)1/4 = 0.86 vc = (0.27/γm)(100As/bwd)1/3(fcu)1/3 = (0.27 / 1.25) x ({100 x 5362} / {1000 x 924})1/3 x (40) ξsvc = 0.86 x 0.62 = 0.53 N/mms > 0.121N/mms ∴ OK
Shear on Heel - Use Free Abutment Load Case 3: Shear requirements are designed at the back face of the wall to clause 5.4.4.1: Length of heel = (6.5 - 1.1 - 1.0) = 4.4m ULS Shear on heel = 1.1 x {348 x 0.5 x (5.185 - 2.1) - 1.15 x 1 x 4.4 x 25 - 1.2 x 4.4 x (8.63 x 19 + 10)}
Using B32@150 c/c then: v = V / (bd) = 559 x 103 / (1000 x 924) = 0.605 N/mm2 No shear reinforcement is required when v < ξsvc ξs = (500/d)1/4 = (500 / 924)1/4 = 0.86 vc = (0.27/γm)(100As/bwd)1/3(fcu)1/3 = (0.27 / 1.25) x ({100 x 5362} / {1000 x 924})1/3 x (40) ξsvc = 0.86 x 0.62 = 0.53 N/mms < 0.605N/mms ∴ Fail
Rather than provide shear reinforcement try increasing bars to B40 @ 150 c/c (also required for crack con vc = (0.27/γm)(100As/bwd)1/3(fcu)1/3 = (0.27 / 1.25) x ({100 x 8378} / {1000 x 920})1/3 x (40) ξsvc = 0.86 x 0.716 = 0.616 N/mms > 0.605N/mms ∴ OK Early Thermal Cracking
Considering the effects of casting the base slab onto the blinding concrete by complying with the early thermal cracking of Minimum area of main reinforcement to Clause 5.8.4.1 = 0.15% of bad = 0.0015 x 1000 x 924 = 1386 mm Local Effects
Curtain Wall This wall is designed to be cast onto the top of the abutment after the deck has been built. Loading will be a HB braking load to BS 5400 clause 6.10 = 25% x 45units x 10kN on each axle = 112.5kN per axle. Assume a 45o dispersal to the curtain wall and a maximum dispersal of the width of the abutment (11.6m) 1st axle load on back of abutment = 112.5 / 3.0 = 37.5kN/m 2nd axle load on back of abutment = 112.5 / 6.6 = 17.0kN/m 3rd & 4th axle loads on back of abutment = 2 x 112.5 / 11.6 = 19.4kN/m Maximum load on back of abutment = 37.5 + 17.0 + 19.4 = = 73.9kN/m Bending and Shear at Base of 3m High Curtain Wall Horizontal load due to HB surcharge = 0.426 x 20 x 3.0 = 25.6 kN/m
Horizontal load due to backfill = 0.426 x 19 x 3.02 / 2 = 36.4 kN/m SLS Moment = (73.9 x 3.0) + (25.6 x 1.5) + (36.4 x 1.0) = 297 kNm/m (36 dead + 261 live) ULS Moment = 1.1 x {(1.1 x 73.9 x 3.0) + (1.5 x 25.6 x 1.5) + (1.5 x 36.4 x 1.0)} = 392 kNm/m ULS Shear = 1.1 x {(1.1 x 73.9) + (1.5 x 25.6) + (1.5 x 36.4)} = 192kN/m
400 thick curtain wall with B32 @ 150 c/c : Mult = 584 kNm/m > 392 kNm/m ∴ OK SLS Moment produces crack width of 0.21mm < 0.25 ∴ OK ξsvc = 0.97 N/mm2 > v = 0.59 N/mm2 ∴ Shear OK
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s develop in the tension zone at working loads and therefore all concrete in tension is ignored in design. become tensile under external loads. This compressive stress neutralizes the tensile stress so that no resultant tension the web so that thin-webbed I - sections may be used without the risk of diagonal tension failures and with further sav ary to use high quality concrete to resist the higher compressive stresses that are developed.
g the stressed cables to the concrete.
or themselves in the concrete.
sually raised towards the neutral axis at the ends to reduce the eccentricity of the stressing force.
r ultimate strength.
ons is released. Grout is then pumped into the ducts to protect the tendons.
te a series of effects result in a loss of stress in the tendons. The effects are:
ith pre-tensioning.
ncrete slab and 100mm of surfacing, together with a nominal live load udl of 10.0 kN/m 2 and kel of 33kN/m . The span of the bea
= 10.78 kN/m
ULS Comb.1
Comb.3
1.25 -
-
een beam and slab may be ignored.
cast concrete takes place before the deck slab is cast and that the residual shrinkage is 100 x 10 -6 ,
x Ecf x Acf x φ
f
ab concrete has cured then any further loading (superimposed and live loads) is supported by the composite section of the beam
1.25 x 33)]kel
Comb.3 (HB)
-0.60 -17.67*
due to the self weight at this section is near zero and initial stress conditions are: .....................(eqn. 1) .....................(eqn. 2)
/ 0.9 = 5431kN
ete at transfer :
1 + 449220 x 2192 / 52.905 x 109) / 31 ]
rete shrinkage εcs = 300 x 10-6
6 x 196 x 32 x 139 = 262
8 x 10-6 x 12.76 x 196 x 32 x 139 = 550
- 923 = 4144 kN (Pe/P = 0.82)
estress at :
ects of design loads, differential shrinkage and temperature difference :
2 hence O.K.)
424 + 1.64 = 17.71 (< 25 O.K.)
ost tendon O.K.
o carry HA and 45 units of HB loading. Analyse the abutments using a unit strip method. The bridge site is located sout
the proposed road level. Test results show the founding strata to be a cohesionless soil having an angle of shearing res 35o and density (γ) = 19kN/m3.
deck slab as shown.
2400 600 1880 2770
eratures are -19 and +37 oC respectively. m effective bridge temperatures are -11 and +36 oC from tables 10 and 11.
= 47 x 12 x 10 -6 x 20 x 103 = 11.3mm. x 1.1 x 1.3 /2] = ± 8mm.
stomeric bearing would be Ekspan's Elastomeric Pad Bearing EKR35:
0% of the thickness of the bearing. The figure quoted in the catalogue for the maximum shear deflection is 70% of the ot at the mid range temperature. The design shade air temperature range will be -19 to +37 oC which would require the
N at each bearing.
manufacturer's data sheet.
able bearing from the Ekspan EA Series would be /80/210/25/25:
se plate then the pressure between the sliding faces will be in the order of 5N/mm 2. ess of 5N/mm 2 = 2220 x 0.08 = 180kN ≡ 180 / 11.6 = 16kN/m.
tilever walls, Reynolds's Reinforced Concrete Designer's Handbook being one such book.
30o) = 523kN/m
P x e is the moment about the centre of the base.
oad Cases 1 to 5 using a simple spreadsheet the following results were obtained:
127 5 76 39 83 81 43
120 7 78 42 84 82 l design load cases on the abutments. We shall assume that there are no specific requirements for using elastomeric be
eability and Ultimate load effects need to be calculated for the load cases 1 to 6 shown above. Again, these are best ca
2 and 4.2.3)
oad Cases 1 to 5 using a simple spreadsheet the following results were obtained for the design moments and shear at
ent of 340kg/m3 for exposure condition XD2. Δc of 15mm).Reinforcement to BS 4449:2005 Grade B500B:
fy = 500N/mm2
75kNn/m ∴ OK
n clause 4.1.1.3 ∴ serviceability requirements are satisfied.
})1/3 x (40)1/3 = 0.72
ovided, however check shear at distance H/8 (8.63 / 8 = 1.079m) up the wall.
y thermal cracking of concrete to BD 28 then B16 horizontal lacer bars @ 150 c/c will be required in both faces in the bottom half of the 0012 x 1000 x 920 = 1104 mm2/m (use B16 @ 150c/c - As = 1340mm2/m)
e front and back of the wall. Different load factors are used for serviceability and ultimate limit states so the calculation
/ 2) = 99kNm/m (tension in bottom face).
2) - (531 x 4.3 / 2) - (52 x 4.3 / 2) = -443kNm/m (tension in top face).
= 259kN/m
25 x 1.0 x 1.12 / 2)} = 148kNm/m (tension in bottom face).
oad Cases 1 to 5 using a simple spreadsheet the following results were obtained:
768 1596 1834 1700 1402 1717
ULSMoment 816 1678 1922 1786 1480
2kNm/m ∴ OK
can be obtained by removing the surcharge and braking loads. The spreadsheet result gives the dead load SLS mome 0.27mm > 0.25mm ∴ Fail. 40@150 c/c as this is required to avoid the use of links (see below).
n clause 4.1.1.3 ∴ serviceability requirements are satisfied.
})1/3 x (40)1/3 = 0.62
x 4.4 x (8.63 x 19 + 10)} = 559kN
})1/3 x (40)1/3 = 0.62
lso required for crack control as shown above). })1/3 x (40)1/3 = 0.716
he early thermal cracking of concrete to BD 28 then B16 distribution bars @ 250 c/c will be required. x 1000 x 924 = 1386 mm2/m (use B20 @ 200c/c - As = 1570mm2/m).
een built. Loading will be applied from the backfill, surcharge and braking loads on top of the wall. 112.5kN per axle. of the abutment (11.6m) then:
+ 261 live) } = 392 kNm/m
116.020 89.066
166.156 242.424 179.402
at no resultant tension exists, (or only very small values, within the tensile strength of the concrete). Cracking is there es and with further savings in self-weight.
N/m . The span of the beam is 24.0m centre to centre of bearings and the beams are spaced at 1.0m intervals.
site section of the beam and slab.
dge site is located south east of Oxford (to establish the range of shade air temperatures).
n angle of shearing resistance (φ) = 30 o and a safe bearing capacity of 400kN/m2.
eflection is 70% of the thickness. which would require the bearings to be installed at a shade air temperature of 9oC to achieve the ± 8mm movement. If
or using elastomeric bearings and design the abutments for the lesser load effects by using sliding bearings.
gain, these are best carried out using a simple spreadsheet.
moments and shear at the base of the wall:
s in the bottom half of the wall.
tates so the calculations need to be carried out for each limit state using 'at rest pressures'
e dead load SLS moment for Case 3 as 723kNm, thus the live load moment = 1233 - 723 = 510kNm.
ete). Cracking is therefore eliminated under working load and all of the concrete may be assumed effective in carrying
± 8mm movement. If the bearings are set at a maximum shade air temperature of 16oC then, by proportion the deck
ng bearings.
ed effective in carrying load. Therefore lighter sections may be used to carry a given bending moment, and prestressed
by proportion the deck will expand 6mm and contract 10mm. Let us assume that this maximum shade air temperature
ment, and prestressed concrete may be used over much longer spans than reinforced concrete.
shade air temperature of 16 oC for fixing the bearings is specified in the Contract and design the abutments accordingly.
abutments accordingly.
Prestressed Concrete Since concrete is weak in tension in normal reinforced concrete construction cracks develop in the tension zone
Prestressing involves inducing compressive stresses in the zone which will tend to become tensile under externa
The prestressing force also reduces the magnitude of the principal tensile stress in the web so that thin-webbed
The prestressing force has to be produced by a high tensile steel, and it is necessary to use high quality concret There are two methods of prestressing concrete : 1) Pre-cast Pre-tensioned 2) Pre-cast Post-tensioned
Both methods involve tensioning cables inside a concrete beam and then anchoring the stressed cables to the c 1) Pre-tensioned Beams Stage 1 Tendons and reinforcement are positioned in the beam mould. Stage 2 Tendons are stressed to about 70% of their ultimate strength. Stage 3 Concrete is cast into the beam mould and allowed to cure to the required initial strength. Stage 4
When the concrete has cured the stressing force is released and the tendons anchor themselves in the concrete 2) Post-tensioned Beams
Stage 1
Cable ducts and reinforcement are positioned in the beam mould. The ducts are usually raised towards the neut Stage 2 Concrete is cast into the beam mould and allowed to cure to the required initial strength. Stage 3 Tendons are threaded through the cable ducts and tensioned to about 70% of their ultimate strength. Stage 4
Wedges are inserted into the end anchorages and the tensioning force on the tendons is released. Grout is then
Loss of Prestress When the tensioning force is released and the tendons are anchored to the concrete a series of effects result in a. relaxation of the steel tendons b. elastic deformation of the concrete c. shrinkage and creep of the concrete d. slip or movement of the tendons at the anchorages during anchoring e. other causes in special circumstances , such as when steam curing is used with pre-tensioning. Total losses in prestress can amount to about 30% of the initial tensioning stress.
Prestressed Concrete Beam Design to BS 5400 Part 4
Problem:
Design a simply supported prestressed concrete Y beam which carries a 150mm thick concrete slab and 100mm of surfa
γconc. = 24kN/mm3 25 units of HB to be considered at SLS for load combination 1 only (BS 5400 Pt4 Cl. 4.2.2) Loading per beam (at 1.0m c/c) Nominal Dead Loads :
Nominal Live Load :
slab = 24 x 0.15 x 1.0
= 3.6 kN/m
beam = say Y5 beam
= 10.78 kN/m
surfacing = 24 x 0.1 x 1.0
= 2.4 kN/m
HA = 10 x 1.0 + 33.0
= 10 kN/m + 33kN
25 units HB = 25 x 10 / 4 per wheel
= 62.5 kN per wheel
Load factors for serviceability and ultimate limit state from BS 5400 Part 2 (or BD 37/01) Table 1:
Comb.1 Dead Load Superimposed Dead Load
γfL concrete1.0 γfL surfacing1.2
Live Load
γfL HA
1.2 γfL HB
Temperature Difference
γfL
1.1 -
Concrete Grades Beam fcu = 50 N/mm2, fci = 40 N/mm2 Slab fcu = 40 N/mm2 BS 5400 Pt. 4 Section Properties cl.7.4.1 Property Area(mm2) Centroid(mm)
Modular ratio effect for different concrete strengths between beam and slab may be ignore Beam Section 4.49E+05
456
2nd Moment of Area(mm4)
5.29E+10
Modulus @ Level 1(mm3)
1.16E+08
Modulus @ Level 2(mm3)
8.91E+07
Composite Section
Modulus @ Level 3(mm3)
-
Temperature Difference Effects Apply temperature differences given in BS 5400 Pt2 Fig.9 (Group 4)to a simplified beam section. Cl. 5.4.6 - Coefficient of thermal expansion = 12 x 10-6 per ºC. From BS 5400 Pt4 Table 3 : Ec = 34 kN/mm2 for fcu = 50N/mm2 Hence restrained temperature stresses per °C = 34 x 10 3 x 12 x 10-6 = 0.408 N/mm2
a) Positive temperature difference Force F to restrain temperature strain : 0.408 x 1000 x [ 150 x ( 3.0 + 5.25 ) ] x 10-3 + 0.408 x ( 300 x 250 x 1.5 + 750 x 200 x 1.25 ) x 10-3 = 504.9 + 122.4 = 627.3 kN Moment M about centroid of section to restrain curvature due to temperature strain : 0.408 x 1000 x [ 150 x ( 3.0 x 502 + 5.25 x 527 ) ] x 10-6 + 0.408 x ( 300 x 250 x 1.5 x 344 - 750 x 200 x 1.25 x 556 ) x 10-6 = 261.5 - 26.7 = 234.8 kNm
b) Reverse temperature difference Force F to restrain temperature strain : - 0.408 x [ 1000 x 150 x ( 3.6 + 2.3 ) + 300 x 90 x ( 0.9 + 1.35 ) ] x 10 -3 - 0.408 x 300 x ( 200 x 0.45 + 150 x 0.45 ) x 10-3 - 0.408 x 750 x [ 50 x ( 0.9 + 0.15 ) + 240 x ( 1.2 + 2.6 ) ] x 10 -3 = - 385.9 - 19.3 - 295.1 = - 700.3 kN Moment M about centroid of section to restrain curvature due to temperature strain : - 0.408 x [ 150000 x ( 3.6 x 502 + 2.3 x 527 ) + 27000 x ( 0.9 x 382 + 1.35 x 397 ) ] x 10 -6 - 0.408 x 300 x ( 200 x 0.45 x 270 - 150 x 0.45 x 283 ) x 10-6 + 0.408 x 750 x [ 50 x ( 0.9 x 358 + 0.15 x 366 ) + 240 x ( 1.2 x 503 + 2.6 x 543 ) ] x 10 -6 = - 194.5 - 0.6 + 153.8 = - 41.3 kNm
Differential Shrinkage Effects BS 5400 Pt.4
Use cl.6.7.2.4 Table 29 :
cl.7.4.3.4
Total shrinkage of insitu concrete = 300 x 10-6
Assume that 2/3 of the total shrinkage of the precast concrete takes place before the deck slab is cast and that the re hence the differential shrinkage is 200 x 10-6 BS 5400 Pt.4
Force to restrain differential shrinkage : F = - ε diff x Ecf x Acf x φ
cl.7.4.3.5
F = -200 x 10-6 x 34 x 1000 x 150 x 0.43 = -439 kN
Eccentricity acent = 502mm Restraint moment Mcs = -439 x 0.502 = -220.4 kNm
Self weight of beam and weight of deck slab is supported by the beam. When the deck slab concrete has cured then any
Dead Loading (beam and slab) Total load for serviceability limit state = (1.0 x 3.6)+(1.0 x 10.78) = 14.4kN/m Design serviceability moment = 14.4 x 242 / 8 = 1037 kNm
Combination 1 Loading Super. & HA live load for SL= [(1.2 x 2.4)+(1.2 x 10)]udl & [(1.2 x 33)]kel = (2.88 + 12.0)udl & 39.6kel = 14.9 kN/m & 39.6kN Super. & HB live load for SL = 2.88 & 4 wheels @ 1.1 x 62.5 = 2.9 kN/m & 4 wheels @ 68.75 kN Total load for ultimate limit = [(1.15 x 3.6)+(1.15 x 10.78)+(1.75 x 2.4)+(1.5 x 10)]udl & [(1.5 x 33)]kel = (4.14 + 12.40 + 4.20 + 15.0)udl & 49.5kel = 35.7 kN/m & 49.5kN HA Design serviceability mo = 14.9 x 24.02 / 8 + 39.6 x 24 / 4
=1310 kNm 25 units HB Design SLS mom= 2.9 x 24.02 / 8 + 982.3(from grillage analysis) = 1191.1 kNm Design ultimate moment
= 35.7 x 24.02 / 8 + 49.5 x 24 / 4
= 2867 kNm
Combination 3 Loading Super. & HA live load for SL= [(1.2 x 2.4)+(1.0 x 10)]udl & [(1.0 x 33)]kel = (2.88 + 10.0)udl & 33kel = 12.9 kN/m & 33kN Total load for ultimate limit = [(1.15 x 3.6)+(1.15 x 10.78)+(1.75 x 2.4)+(1.25 x 10)]udl & [(1.25 x 33)]kel = (4.14 + 12.40 + 4.20 + 12.5)udl & 41.3kel = 33.2 kN/m & 41.3kN Design serviceability moment= 12.9 x 24.02 / 8 + 33 x 24 / 4 = 1127 kNm Allowable stresses in precast concrete At transfer : cl.6.3.2.2 b)
Compression ( Table 23 )
0.5fci (= - 1.0
(eqn. 1) x Zlevel 1 + (eqn. 2) x Zlevel 2 gives : P >= A x (20 x Zlevel 1 - 1.0 x Zlevel 2) / (Zlevel 1 + Zlevel 2) P = 449.22 x 103 x ( 20 x 116.02 - 89.066) / ( 116.02 + 89.066) x 10-3 = 4888 kN Allow 10% for loss of force before and during transfer, then the initial force P o = 4888 / 0.9 = 5431kN Using 15.2mm class 2 relaxation standard strand at maximum initial force of 174kN (0.75 x P u) Area of tendon = 139mm2 Nominal tensile strength = fpu =1670 N/mm2 Hence 32 tendons required. Initial force Po = 32 x 174 = 5568 kN P = 0.9 x 5568 = 5011 kN Substituting P = 5011 kN in (eqn. 2) e 0 hence O.K.) Level 1, combination 3 : f = 17.08 - 17.67 = - 0.59 N/mm2 (> - 3.2 hence O.K.) Level 2, combination 1 : f = - 0.98 + 1037 / 89.066 + 1310 / 242.424 + 1.64 = 17.71 (< 25 O.K.) Level 3. combination 3 : f = (1127 / 179.402) + (0.8 x 3.15) = 8.8 N/mm2 (< 25 O.K.) Ultimate Capacity of Beam and Deck Slab (Composite Section)
Ultimate Design Moment = γf3 x M = 1.1 x 2867 = 3154 kNm cl. 6.3.3
Only steel in the tension zone is to be considered :
Centroid of tendons in tension zone = (6x60 + 10x110 + 8x160 + 4x260) / 28 = 135mm Effective depth from Level 3 = 1200 - 135 = 1065mm
Assume that the maximum design stress is developed in the tendons, then : Tensile force in tendons Fp = 0.87 x 28 x 139 x 1670 x 10-3 = 5655 kN Compressive force in concrete flange : Ff = 0.4 x 40 x 1000 x 150 x 10-3 = 2400 kN Let X = depth to neutral axis. Compressive force in concrete web : Fw = 0.4 x 50 x [393 - (393 - 200) x (X - 150) / (671 x 2)] x (X - 150) x 10-3 Fw = ( -2.876X2 + 8722.84X - 1243717) x 10-3 Equating forces to obtain X : 5655 = 2400 + ( -2.876X2 + 8722.84X - 1243717) x 10-3 X = 659 mm Stress in tendon after losses = fpe = 4144 x 103 / (32 x 139) = 932 N/mm2 Prestrain εpe = fpe / Es = 932 / 200 x 103 = 0.0047
Determine depth to neutral axis by an iterative strain compatibility analysis Try X = 659 mm as an initial estimate Width of web at this depth = 247mm εpb6 = ε6 + εpe = -459 x 0.0035 / 659 + 0.0047 = 0.0022 εpb5 = ε5 + εpe = -359 * 0.0035 / 659 + 0.0047 = 0.0028 εpb4 = ε4 + εpe = 281 * 0.0035 / 659 + 0.0047 = 0.0062 εpb3 = ε3 + εpe = 381 * 0.0035 / 659 + 0.0047 = 0.0067 εpb2 = ε2 + εpe = 431 * 0.0035 / 659 + 0.0047 = 0.0069 εpb1 = ε1 + εpe = 481 * 0.0035 / 659 + 0.0047 = 0.0072 fpb6 = 0.0022 x 200 x 103 = 444 N/mm2 fpb5 = 0.0028 x 200 x 103 = 551 N/mm2
fpb4 = 1162 + 290 x (0.0062 - 0.0058) / 0.0065 = 1178 N/mm2 fpb3 = 1162 + 290 x (0.0067 - 0.0058) / 0.0065 = 1201 N/mm2 fpb2 = 1162 + 290 x (0.0069 - 0.0058) / 0.0065 = 1213 N/mm2 fpb1 = 1162 + 290 x (0.0072 - 0.0058) / 0.0065 = 1225 N/mm2 Tensile force in tendons : Fp6 = 2 x 139 x 444 x 10-3 = 124 Fp5 = 2 x 139 x 551 x 10-3 = 153 Fp4 = 4 x 139 x 1178 x 10-3 = 655 Fp3 = 8 x 139 x 1201 x 10-3 = 1336 Fp2 = 10 x 139 x 1213 x 10-3= 1686 Fp1 = 6 x 139 x 1225 x 10-3 = 1022 4976 kN Compressive force in concrete : Ff = 0.4 x 40 x 1000 x 150 x 10-3= 2400 Fw = 0.4 x 50 x 0.5 x (393 + 247) x (659 - 150) x 10-3= 3258 5658 kN Fc > Ft therefore reduce depth to neutral axis and repeat the calculations. Using a depth of 565mm will achieve equilibrium. The following forces are obtained : Fp6 = 134
Ff = 2400
Fp5 = 168
Fw = 2765
Fp4 = 675
Fc = 5165
Fp3 = 1382 Fp2 = 1746 Fp1 = 1060 Ft = 5165 Taking Moments about the neutral axis : Fp6 = 134 x -0.365 =-49 Fp5 = 168 x -0.265 = -45 Fp4 = 675 x 0.375 = 253 Fp3 = 1382 x 0.475 = 656 Fp2 = 1746 x 0.525 =917 Fp1 = 1060 x 0.575 = 610 Ff = 2400 x 0.49 = 1176 Fw = 3258 x 0.207 = 674 Mu = 4192 kNm > 3154 kNm hence O.K. cl. 6.3.3.1
Mu / M = 4192 / 3154 = 1.33 ( > 1.15 ) hence strain in outermost tendon O.K.
Abutment Design Example to BD 30
Design the fixed and free end cantilever abutments to the 20m span deck shown to carry HA and 45 units of HB
The ground investigation report shows suitable founding strata about 9.5m below the proposed road level. Test
Backfill material will be Class 6N with an effective angle of internal friction (ϕ') = 35o and density (γ) = 19kN/m
Nominal Reaction(kN) Concrete Deck
Ultimate Reaction(kN) Nominal Reaction(kN)
Ultimate Reaction(kN)
180
230
1900
2400
30
60
320
600
HA udl+kel
160
265
1140
1880
45 units HB
350
500
1940
2770
Surfacing
Nominal loading on 1m length of abutment: Deck Dead Load = (1900 + 320) / 11.6 = 191kN/m HA live Load on Deck = 1140 / 11.6 = 98kN/m HB live Load on Deck = 1940 / 11.6 = 167kN/m
From BS 5400 Part 2 Figures 7 and 8 the minimum and maximum shade air temperatures are -19 and +37 oC re
For a Group 4 type strucutre (see fig. 9) the corresponding minimum and maximum effective bridge temperatur Hence the temperature range = 11 + 36 = 47oC. From Clause 5.4.6 the range of movement at the free end of the 20m span deck = 47 x 12 x 10 -6 x 20 x 103 = The ultimate thermal movement in the deck will be ± [(11.3 / 2) γf3 γfL] = ±[11.3 x 1.1 x 1.3 /2] = ± 8mm.
Option 1 - Elastomeric Bearing:
With a maximum ultimate reaction = 230 + 60 + 500 = 790kN then a suitable elastomeric bearing would be Ek
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 be
A tolerance is also required for setting the bearing if the ambient temperature is not at the mid range temperatu Horizontal load at bearing for 10mm contraction = 12.14 x 10 = 121kN. This is an ultimate load hence the nominal horizontal load = 121 / 1.1 / 1.3 = 85kN at each bearing. Total horizontal load on each abutment = 11 x 85 = 935 kN ≡ 935 / 11.6 = 81kN/m. Alternatively using BS 5400 Part 9.1 Clause 5.14.2.6: H = AGδr/tq Using the Ekspan bearing EKR35
Maximum Load = 1053kN
Area = 610 x 420 = 256200mm2
Nominl hardness = 60 IRHD
Bearing Thickness = 19mm
Shear modulus G from Table 8 = 0.9N/mm2 H = 256200 x 0.9 x 10-3 x 10 / 19 = 121kN This correllates with the value obtained above using the shear stiffness from the manufacturer's data sheet. Option 2 - Sliding Bearing:
With a maximum ultimate reaction of 790kN and longitudinal movement of ± 8mm then a suitable bearing from the Ekspan EA S
Maximum Load = 800kN
Base Plate A dimension = 210mm
Base Plate B dimension = 365mm
Movement ± X = 12.5mm
BS 5400 Part 2 - Clause 5.4.7.3: Average nominal dead load reaction = (1900 + 320) / 11 = 2220 / 11 = 200kN Contact pressure under base plate = 200000 / (210 x 365) = 3N/mm2
As the mating surface between the stainless steel and PTFE is smaller than the base plate then the pressure bet From Table3 of BS 5400 Part 9.1 the Coefficient of friction = 0.08 for a bearing stress of 5N/mm 2
Hence total horizontal load on each abutment when the deck expands or contracts = 2220 x 0.08 = 180kN ≡ 18 Traction and Braking Load - BS 5400 Part 2 Clause 6.10: Nominal Load for HA = 8kN/m x 20m + 250kN = 410kN
Nominal Load for HB = 25% of 45units x 10kN x 4axles = 450kN 450 > 410kN hence HB braking is critical. Braking load on 1m width of abutment = 450 / 11.6 = 39kN/m. When this load is applied on the deck it will act on the fixed abutment only. Skidding Load - BS 5400 Part 2 Clause 6.11: Nominal Load = 300kN 300 < 450kN hence braking load is critical in the longitudinal direction. When this load is applied on the deck it will act on the fixed abutment only. Loading at Rear of Abutment Backfill For Stability calculations use active earth pressures = Ka γ h Ka for Class 6N material = (1-Sin35) / (1+Sin35) = 0.27 Density of Class 6N material = 19kN/m3 Active Pressure at depth h = 0.27 x 19 x h = 5.13h kN/m2 Hence Fb = 5.13h2/2 = 2.57h2kN/m Surcharge - BS 5400 Part 2 Clause 5.8.2: For HA loading surcharge = 10 kN/m2 For HB loading surcharge = 20 kN/m2 Assume a surchage loading for the compaction plant to be equivalent to 30 units of HB Hence Compaction Plant surcharge = 12 kN/m2. For surcharge of w kN/m2 : Fs = Ka w h = 0.27wh kN/m 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, Reynolds's Reinforce Alternatively a simple spreadsheet will achieve a result by trial and error. Load Combinations
Backfill + Construction surcharge
Backfill + Sackfill + HA surcharge Backfill + Backfill + Construct + Deck dead load + HA HB ion Deck contraction surcharg surcharg Fixed Abutment Only surcharg e+ e + Deck e Braking dead Backfill + HA surcharge + Deck dead load + HA on deck + Braking on deck behind load CASE 1 - Fixed Abutment abutment + Deck Density of reinforced concrete = 25kN/m3. dead Weight of wall stem = 1.0 x 6.5 x 25 = 163kN/m load
Backfill + HA surcharg e + Deck dead load + HB on deck
Weight of base = 6.4 x 1.0 x 25 = 160kN/m Weight of backfill = 4.3 x 6.5 x 19 = 531kN/m Weight of surcharge = 4.3 x 12 = 52kN/m Backfill Force Fb = 0.27 x 19 x 7.52 / 2 = 144kN/m Surcharge Force Fs = 0.27 x 12 x 7.5 = 24 kN/m Restoring Effects: Weight
Lever Arm
Moment About A
Stem
163
1.6
Base
160
3.2
Backfill
531
4.25
52
4.25
Surcharge ∑=
906
∑ =3251
Overturning Effects: F Backfill Surcharge ∑ =168
Lever Arm
Moment About A 144
2.5
24
3.75 ∑ =452
Factor of Safety Against Overturning = 3251 / 452 = 7.2 > 2.0 ∴ OK. For sliding effects: Active Force = Fb + Fs = 168kN/m Frictional force on underside of base resisting movement = W tan(φ) = 906 x tan(30o) = 523kN/m Factor of Safety Against Sliding = 523 / 168 = 3.1 > 2.0 ∴ OK. Bearing Pressure:
Check bearing pressure at toe and heel of base slab = (P / A) ± (P x e / Z) where P x e is the moment about th P = 906kN/m A = 6.4m2/m Z = 6.42 / 6 = 6.827m3/m Nett moment = 3251 - 452 = 2799kNm/m Eccentricity (e) of P about centre-line of base = 3.2 - (2799 / 906) = 0.111m Pressure under base = (906 / 6.4) ± (906 x 0.111 / 6.827) Pressure under toe = 142 + 15 = 157kN/m2 < 400kN/m2 ∴ OK. Pressure under heel = 142 - 15 = 127kN/m2 Hence the abutment will be stable for Case 1.
Analysing the fixed abutment with Load Cases 1 to 6 and the free abutment with Load Cases 1 to 5 using a sim Fixed Abutment: F of S Overturning
F of S Sliding
Case 1
7.16
3.09
Case 2
2.87
2.13
Case 2a
4.31
2.64
Case 3
3.43
2.43
Case 4
4.48
2.63
Case 5
5.22
3.17
Case 6
3.8
2.62
Free Abutment: F of S Overturning
F of S Sliding
Case 1
7.15
3.09
Case 2
2.91
2.14
Bearing Pressure at Toe
Case 2a
4.33
2.64
Case 3
3.46
2.44
Case 4
4.5
2.64
Case 5
5.22
3.16
It can be seen that the use of elastomeric bearings (Case 2) will govern the critical design load cases on the abu 2) Wall and Base Design
Loads on the back of the wall are calculated using 'at rest' earth pressures. Serviceability and Ultimate load effe Using the Fixed Abutment Load Case 1 again as an example of the calculations: Wall Design Ko = 1 - Sin(ϕ') = 1 - Sin(35o) = 0.426 γfL for horizontal loads due to surcharge and backfill from BS 5400 Part 2 Clause 5.8.1.2: Serviceability = 1.0 Ultimate = 1.5 γf3 = 1.0 for serviceability and 1.1 for ultimate (from BS 5400 Part 4 Clauses 4.2.2 and 4.2.3) Backfill Force Fb on the rear of the wall = 0.426 x 19 x 6.52 / 2 = 171kN/m Surcharge Force Fs on the rear of the wall = 0.426 x 12 x 6.5 = 33kN/m At the base of the Wall: Serviceability moment = (171 x 6.5 / 3) + (33 x 6.5 / 2) = 371 + 107 = 478kNm/m Ultimate moment = 1.1 x 1.5 x 478 = 789kNm/m Ultimate shear = 1.1 x 1.5 x (171 + 33) = 337kN/m
Analysing the fixed abutment with Load Cases 1 to 6 and the free abutment with Load Cases 1 to 5 using a sim Fixed Abutment: Moment
Moment
Moment
SLS Dead
SLS Live
ULS
Case 1
371
108
790
Case 2a
829
258
1771
Case 3
829
486
2097
Case 4
829
308
1877
Case 5
829
154
1622
Case 6
829
408
1985
Free Abutment: Moment
Moment
Moment
SLS Dead
SLS Live
ULS
Case 1
394
112
835
Case 2a
868
265
1846
Case 3
868
495
2175
Case 4
868
318
1956
Case 5
868
159
1694
Concrete to BS 8500:2006
Use strength class C32/40 with water-cement ratio 0.5 and minimum cement content of 340kg/m3 for exposure
Nominal cover to reinforcement = 60mm (45mm minimum cover plus a tolerance Δc of 15mm).Reinforcement t
Design for critical moments and shear in Free Abutment: Reinforced concrete walls are designed to BS 5400 Part 4 Clause 5.6. Check classification to clause 5.6.1.1: Ultimate axial load in wall from deck reactions = 2400 + 600 + 2770 = 5770 kN 0.1fcuAc = 0.1 x 40 x 103 x 11.6 x 1 = 46400 kN > 5770 ∴ design as a slab in accordance with clause 5.4
Bending BS 5400 Part 4 Clause 5.4.2 → for reisitance moments in slabs design to clause 5.3.2.3: z = {1 - [ 1.1fyAs) / (fcubd) ]} d Use B40 @ 150 c/c: As = 8378mm2/m,
d = 1000 - 60 - 20 = 920mm
z = {1 - [ 1.1 x 500 x 8378) / (40 x 1000 x 920) ]} d = 0.875d < 0.95d ∴ OK Mu = (0.87fy)Asz = 0.87 x 500 x 8378 x 0.875 x 920 x 10-6 = 2934kNm/m > 2175kNn/m ∴ OK Carrying out the crack control calculation to Clause 5.8.8.2 gives a crack width of 0.2mm < 0.25mm.
Also the steel reinforcement and concrete stresses meet the limitations required in clause 4.1.1.3 ∴ serviceabil
Shear Shear requirements are designed to BS 5400 clause 5.4.4: v = V / (bd) = 619 x 103 / (1000 x 920) = 0.673 N/mm2 No shear reinforcement is required when v < ξsvc ξs = (500/d)1/4 = (500 / 920)1/4 = 0.86 vc = (0.27/γm)(100As/bwd)1/3(fcu)1/3 = (0.27 / 1.25) x ({100 x 8378} / {1000 x 920})1/3 x (40)1/3 = 0.72
ξsvc = 0.86 x 0.72 = 0.62 N/mms < 0.673 hence shear reinforcement should be provided, however check shear ULS shear at Section 7H/8 for load case 4 = 487 kN v = V / (bd) = 487 x 103 / (1000 x 920) = 0.53 N/mm2 < 0.62 Hence height requiring strengthening = 1.073 x (0.673 - 0.62) / (0.673 - 0.53) = 0.4m < d. Provide a 500 x 500 splay at the base of the wall with B32 @ 150c/c bars in sloping face. 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 B
Minimum area of secondary reinforcement to Clause 5.8.4.2 = 0.12% of bad = 0.0012 x 1000 x 920 = 1104 mm Base Design
Maximum bending and shear effects in the base slab will occur at sections near the front and back of the wall. D Using the Fixed Abutment Load Case 1 again as an example of the calculations: CASE 1 - Fixed Abutment Serviceability Limit State γfL = 1.0
γf3 = 1.0
Weight of wall stem = 1.0 x 6.5 x 25 x 1.0 = 163kN/m Weight of base = 6.4 x 1.0 x 25 x 1.0 = 160kN/m Weight of backfill = 4.3 x 6.5 x 19 x 1.0 = 531kN/m Weight of surcharge = 4.3 x 12 x 1.0 = 52kN/m B/fill Force Fb = 0.426 x 19 x 7.52 x 1.0 / 2 = 228kN/m Surcharge Force Fs = 0.426 x 12 x 7.5 x 1.0 = 38 kN/m
Restoring Effects: Weight
Lever ArmMoment About A
Stem
163
1.6
261
Base
160
3.2
512
Backfill
531
4.25
2257
52
4.25
221
Surcharge ∑=
906
∑=
3251
Overturning Effects: F
Lever ArmMoment About A
Backfill Surcharge ∑=
228
2.5
570
38
3.75
143
266
∑=
713
Bearing Pressure at toe and heel of base slab = (P / A) ± (P x e / Z) P = 906kN/m A = 6.4m2/m Z = 6.42 / 6 = 6.827m3/m Nett moment = 3251 - 713 = 2538kNm/m Eccentricity (e) of P about centre-line of base = 3.2 - (2538 / 906) = 0.399m Pressure under base = (906 / 6.4) ± (906 x 0.399 / 6.827) Pressure under toe = 142 + 53 = 195kN/m2 Pressure under heel = 142 - 53 = 89kN/m2 Pressure at front face of wall = 89 + {(195 - 89) x 5.3 / 6.4} = 177kN/m2 Pressure at rear face of wall = 89 + {(195 - 89) x 4.3 / 6.4} = 160kN/m2
SLS Moment at a-a = (177 x 1.12 / 2) + ([195 - 177] x 1.12 / 3) - (25 x 1.0 x 1.12 / 2) = 99kNm/m (tension in
SLS Moment at b-b = (89 x 4.32 / 2) + ([160 - 89] x 4.32 / 6) - (25 x 1.0 x 4.32 / 2) - (531 x 4.3 / 2) - (52 x 4
CASE 1 - Fixed Abutment Ultimate Limit State γfL for concrete = 1.15 γfL for fill and surcharge(vetical) = 1.2 γfL for fill and surcharge(horizontal) = 1.5 Weight of wall stem = 1.0 x 6.5 x 25 x 1.15 = 187kN/m Weight of base = 6.4 x 1.0 x 25 x 1.15 = 184kN/m Weight of backfill = 4.3 x 6.5 x 19 x 1.2 = 637kN/m Weight of surcharge = 4.3 x 12 x 1.2 = 62kN/m Backfill Force Fb = 0.426 x 19 x 7.52 x 1.5 / 2 = 341kN/m Surcharge Force Fs = 0.426 x 12 x 7.5 x 1.5 = 58 kN/m Restoring Effects:
Weight
Lever Arm
Moment About A
Stem
187
1.6
Base
184
3.2
Backfill
637
4.25
62
4.25
Surcharge ∑=
1070
∑=
Overturning Effects: F Backfill Surcharge ∑=
Lever Arm
Moment About A
341
2.5
58
3.75
399
∑=
Bearing Pressure at toe and heel of base slab = (P / A) ± (P x e / Z) P = 1070kN/m A = 6.4m2/m Z = 6.42 / 6 = 6.827m3/m Nett moment = 3859 - 1071 = 2788kNm/m Eccentricity (e) of P about centre-line of base = 3.2 - (2788 / 1070) = 0.594m Pressure under base = (1070 / 6.4) ± (1070 x 0.594 / 6.827) Pressure under toe = 167 + 93 = 260kN/m2 Pressure under heel = 167 - 93 = 74kN/m2 Pressure at front face of wall = 74 + {(260 - 74) x 5.3 / 6.4} = 228kN/m2 Pressure at rear face of wall = 74 + {(260 - 74) x 4.3 / 6.4} = 199kN/m2 γf3 = 1.1 ULS Shear at a-a = 1.1 x {[(260 + 228) x 1.1 / 2] - (1.15 x 1.1 x 25)} = 260kN/m ULS Shear at b-b = 1.1 x {[(199 + 74) x 4.3 / 2] - (1.15 x 4.3 x 25) - 637 - 62} = 259kN/m
ULS Moment at a-a = 1.1 x {(228 x 1.12 / 2) + ([260 - 228] x 1.12 / 3) - (1.15 x 25 x 1.0 x 1.12 / 2)} = 148kN
SLS Moment at b-b = 1.1 x {(74 x 4.32 / 2) + ([199 74] x 4.32 / 6) - (1.15 x 25 x 1.0 x 4.32 / 2) - (637 x 4.3 / 2) - (62 x 4.3 / 2)} = -769kNm/m (tension in top face).
Analysing the fixed abutment with Load Cases 1 to 6 and the free abutment with Load Cases 1 to 5 using a sim Fixed Abutment Base:
Section a-a
Section b-b ULSShear SLSMoment
ULSMomen ULS Shear
Case 1
261
99
147
259
Case 2a
528
205
302
458
Case 3
593
235
340
553
Case 4
550
208
314
495
Case 5
610
241
348
327
Case 6
637
255
365
470
Free Abutment Base: Section a-a
Section b-b ULSShear SLSMoment
ULSMoment
ULSShear
Case 1
267
101
151
266
Case 2a
534
207
305
466
Case 3
598
236
342
559
Case 4
557
211
317
504
Case 5
616
243
351
335
Design for shear and bending effects at sections a-a and b-b for the Free Abutment: Bending BS 5400 Part 4 Clause 5.7.3 → design as a slab for reisitance moments to clause 5.3.2.3: z = {1 - [ 1.1fyAs) / (fcubd) ]} d Use B32 @ 150 c/c: As = 5362mm2/m,
d = 1000 - 60 - 16 = 924mm
z = {1 - [ 1.1 x 500 x 5362) / (40 x 1000 x 924) ]} d = 0.92d < 0.95d ∴ OK Mu = (0.87fy)Asz = 0.87 x 500 x 5362 x 0.92 x 924 x 10-6 = 1983kNm/m > 1922kNm/m ∴ OK (1983kNm/m also > 1834kNm/m ∴ B32 @ 150 c/c suitable for fixed abutment.
For the Serviceability check for Case 3 an approximation of the dead load moment can be obtained by removing Carrying out the crack control calculation to Clause 5.8.8.2 gives a crack width of 0.27mm > 0.25mm ∴ Fail.
This could be corrected by reducing the bar spacing, but increase the bar size to B40@150 c/c as this is require Using B40@150c/c the crack control calculation gives a crack width of 0.17mm < 0.25mm ∴ OK.
Also the steel reinforcement and concrete stresses meet the limitations required in clause 4.1.1.3 ∴ serviceabil Shear Shear on Toe - Use Fixed Abutment Load Case 6: By inspection B32@150c/c will be adequate for the bending effects in the toe (Muls = 365kNm < 1983kNm) Shear requirements are designed to BS 5400 clause 5.7.3.2(a) checking shear at d away from the front face of the wall to clause 5.4.4.1: ULS Shear on toe = 1.1 x {(620 + 599) x 0.5 x 0.176 1.15 x 1 x 0.176 x 25} = 112kN
v = V / (bd) = 112 x 103 / (1000 x 924) = 0.121 N/mm2 No shear reinforcement is required when v < ξsvc Reinforcement in tension = B32 @ 150 c/c ξs = (500/d)1/4 = (500 / 924)1/4 = 0.86 vc = (0.27/γm)(100As/bwd)1/3(fcu)1/3 = (0.27 / 1.25) x ({100 x 5362} / {1000 x 924})1/3 x (40)1/3 = 0.62 ξsvc = 0.86 x 0.62 = 0.53 N/mms > 0.121N/mms ∴ OK Shear on Heel - Use Free Abutment Load Case 3: Shear requirements are designed at the back face of the wall to clause 5.4.4.1: Length of heel = (6.5 - 1.1 - 1.0) = 4.4m
ULS Shear on heel = 1.1 x {348 x 0.5 x (5.185 - 2.1) - 1.15 x 1 x 4.4 x 25 - 1.2 x 4.4 x (8.63 x 19 + 10)} = 5
Using B32@150 c/c then: v = V / (bd) = 559 x 103 / (1000 x 924) = 0.605 N/mm2 No shear reinforcement is required when v < ξsvc ξs = (500/d)1/4 = (500 / 924)1/4 = 0.86 vc = (0.27/γm)(100As/bwd)1/3(fcu)1/3 = (0.27 / 1.25) x ({100 x 5362} / {1000 x 924})1/3 x (40)1/3 = 0.62 ξsvc = 0.86 x 0.62 = 0.53 N/mms < 0.605N/mms ∴ Fail Rather than provide shear reinforcement try increasing bars to B40 @ 150 c/c (also required for crack control vc = (0.27/γm)(100As/bwd)1/3(fcu)1/3 = (0.27 / 1.25) x ({100 x 8378} / {1000 x 920})1/3 x (40)1/3 = 0.716 ξsvc = 0.86 x 0.716 = 0.616 N/mms > 0.605N/mms ∴ OK Early Thermal Cracking
Considering the effects of casting the base slab onto the blinding concrete by complying with the early thermal cracking of conc
Minimum area of main reinforcement to Clause 5.8.4.1 = 0.15% of bad = 0.0015 x 1000 x 924 = 1386 mm2/m Local Effects Curtain Wall
This wall is designed to be cast onto the top of the abutment after the deck has been built. Loading will be appli HB braking load to BS 5400 clause 6.10 = 25% x 45units x 10kN on each axle = 112.5kN per axle.
Assume a 45o dispersal to the curtain wall and a maximum dispersal of the width of the abutment (11.6m) then 1st axle load on back of abutment = 112.5 / 3.0 = 37.5kN/m 2nd axle load on back of abutment = 112.5 / 6.6 = 17.0kN/m 3rd & 4th axle loads on back of abutment = 2 x 112.5 / 11.6 = 19.4kN/m
Maximum load on back of abutment = 37.5 + 17.0 + 19.4 = = 73.9kN/m Bending and Shear at Base of 3m High Curtain Wall Horizontal load due to HB surcharge = 0.426 x 20 x 3.0 = 25.6 kN/m Horizontal load due to backfill = 0.426 x 19 x 3.02 / 2 = 36.4 kN/m
SLS Moment = (73.9 x 3.0) + (25.6 x 1.5) + (36.4 x 1.0) = 297 kNm/m (36 dead + 261 live) ULS Moment = 1.1 x {(1.1 x 73.9 x 3.0) + (1.5 x 25.6 x 1.5) + (1.5 x 36.4 x 1.0)} = 392 kNm/m ULS Shear = 1.1 x {(1.1 x 73.9) + (1.5 x 25.6) + (1.5 x 36.4)} = 192kN/m
400 thick curtain wall with B32 @ 150 c/c : Mult = 584 kNm/m > 392 kNm/m ∴ OK SLS Moment produces crack width of 0.21mm < 0.25 ∴ OK ξsvc = 0.97 N/mm2 > v = 0.59 N/mm2 ∴ Shear OK
s develop in the tension zone at working loads and therefore all concrete in tension is ignored in design.
become tensile under external loads. This compressive stress neutralizes the tensile stress so that no resultant tension ex
n the web so that thin-webbed I - sections may be used without the risk of diagonal tension failures and with further saving
ary to use high quality concrete to resist the higher compressive stresses that are developed.
g the stressed cables to the concrete.
or themselves in the concrete.
sually raised towards the neutral axis at the ends to reduce the eccentricity of the stressing force.
ir ultimate strength.
dons is released. Grout is then pumped into the ducts to protect the tendons.
ete a series of effects result in a loss of stress in the tendons. The effects are:
pre-tensioning.
ncrete slab and 100mm of surfacing, together with a nominal live load udl of 10.0 kN/m 2 and kel of 33kN/m . The span of the beam
= 3.6 kN/m = 10.78 kN/m = 2.4 kN/m = 10 kN/m + 33kN = 62.5 kN per wheel
01) Table 1: SLS
ULS
Comb.3
Comb.1
Comb.3
1.0
1.15
1.15
1.2
1.75
1.75
1.0
1.5
1.25
-
-
-
0.8
-
1.0
ween beam and slab may be ignored. Composite Section 5.99E+05
623 1.04E+11 1.66E+08 2.42E+08
1.79E+08
- 700.3 kN
deck slab is cast and that the residual shrinkage is 100 x 10 -6 ,
x Ecf x Acf x φ
ff
lab concrete has cured then any further loading (superimposed and live loads) is supported by the composite section of the beam an
.5 x 33)]kel
(1.25 x 33)]kel
(HA)
(HB)
-0.60 -16.71
-0.60 -17.67*
due to the self weight at this section is near zero and initial stress conditions are: .....................(eqn. 1) .....................(eqn. 2)
/ 0.9 = 5431kN
rete at transfer :
568 = 5067 kN
rete shrinkage εcs = 300 x 10-6
-6 x 196 x 32 x 139 = 262
48 x 10-6 x 12.76 x 196 x 32 x 139 = 550
7 - 923 = 4144 kN (Pe/P = 0.82)
restress at :
nkage and temperature difference :
ost tendon O.K.
to carry HA and 45 units of HB loading. Analyse the abutments using a unit strip method. The bridge site is located south e
the proposed road level. Test results show the founding strata to be a cohesionless soil having an angle of shearing resista
35o and density (γ) = 19kN/m3.
eratures are -19 and +37 oC respectively.
m effective bridge temperatures are -11 and +36 oC from tables 10 and 11.
= 47 x 12 x 10 -6 x 20 x 103 = 11.3mm.
3 x 1.1 x 1.3 /2] = ± 8mm.
astomeric bearing would be Ekspan's Elastomeric Pad Bearing EKR35:
50% of the thickness of the bearing. The figure quoted in the catalogue for the maximum shear deflection is 70% of the th
ot at the mid range temperature. The design shade air temperature range will be -19 to +37 oC which would require the be
kN at each bearing.
manufacturer's data sheet.
able bearing from the Ekspan EA Series would be /80/210/25/25:
se plate then the pressure between the sliding faces will be in the order of 5N/mm 2.
ress of 5N/mm 2
s = 2220 x 0.08 = 180kN ≡ 180 / 11.6 = 16kN/m.
ntilever walls, Reynolds's Reinforced Concrete Designer's Handbook being one such book.
261 512 2257 221
361 91
(30o) = 523kN/m
P x e is the moment about the centre of the base.
Load Cases 1 to 5 using a simple spreadsheet the following results were obtained:
Bearing Pressure at Toe
Bearing Pressure at Heel
156
127
386
5
315
76
351
39
322
83
362
81
378
43
Bearing Pressure at Bearing Pressure at Heel 168
120
388
7
318
78
354
42
325
84
365
82
al design load cases on the abutments. We shall assume that there are no specific requirements for using elastomeric bear
eability and Ultimate load effects need to be calculated for the load cases 1 to 6 shown above. Again, these are best carrie
2 and 4.2.3)
Load Cases 1 to 5 using a simple spreadsheet the following results were obtained for the design moments and shear at the
Shear ULS 337 566 596 602 543 599
Shear
ULS 350 581 612 619 559
tent of 340kg/m3 for exposure condition XD2. Δc of 15mm).Reinforcement to BS 4449:2005 Grade B500B:
75kNn/m ∴ OK
n clause 4.1.1.3 ∴ serviceability requirements are satisfied.
f y = 500N/mm2
0})1/3 x (40)1/3 = 0.72
rovided, however check shear at distance H/8 (8.63 / 8 = 1.079m) up the wall.
0.4m < d.
y thermal cracking of concrete to BD 28 then B16 horizontal lacer bars @ 150 c/c will be required in both faces in the bottom half of the wal
0012 x 1000 x 920 = 1104 mm2/m (use B16 @ 150c/c - As = 1340mm2/m)
e front and back of the wall. Different load factors are used for serviceability and ultimate limit states so the calculations n
2
/ 2) = 99kNm/m (tension in bottom face).
2) - (531 x 4.3 / 2) - (52 x 4.3 / 2) = -443kNm/m (tension in top face).
299 589 2707 264 3859
853 218 1071
= 259kN/m
25 x 1.0 x 1.12 / 2)} = 148kNm/m (tension in bottom face).
Load Cases 1 to 5 using a simple spreadsheet the following results were obtained:
SLS Momen ULS Moment 447
768
980
1596
1178
1834
1003
1700
853
1402
1098
1717
SLSMoment
ULSMoment
475
816
1029
1678
1233
1922
1055
1786
901
1480
2kNm/m ∴ OK
t can be obtained by removing the surcharge and braking loads. The spreadsheet result gives the dead load SLS moment f
0.27mm > 0.25mm ∴ Fail.
B40@150 c/c as this is required to avoid the use of links (see below).
n clause 4.1.1.3 ∴ serviceability requirements are satisfied.
4})1/3 x (40)1/3 = 0.62
x 4.4 x (8.63 x 19 + 10)} = 559kN
4})1/3 x (40)1/3 = 0.62
also required for crack control as shown above).
0})1/3 x (40)1/3 = 0.716
the early thermal cracking of concrete to BD 28 then B16 distribution bars @ 250 c/c will be required.
x 1000 x 924 = 1386 mm2/m (use B20 @ 200c/c - As = 1570mm2/m).
een built. Loading will be applied from the backfill, surcharge and braking loads on top of the wall.
112.5kN per axle.
of the abutment (11.6m) then:
d + 261 live)
)} = 392 kNm/m
116.020
166.156
89.066
242.424 179.402
o resultant tension exists, (or only very small values, within the tensile strength of the concrete). Cracking is therefore elim
nd with further savings in self-weight.
The span of the beam is 24.0m centre to centre of bearings and the beams are spaced at 1.0m intervals.
section of the beam and slab.
site is located south east of Oxford (to establish the range of shade air temperatures).
gle of shearing resistance (φ) = 30 o and a safe bearing capacity of 400kN/m2.
ction is 70% of the thickness.
h would require the bearings to be installed at a shade air temperature of 9oC to achieve the ± 8mm movement. If the bea
sing elastomeric bearings and design the abutments for the lesser load effects by using sliding bearings.
, these are best carried out using a simple spreadsheet.
ments and shear at the base of the wall:
he bottom half of the wall.
s so the calculations need to be carried out for each limit state using 'at rest pressures'
ad load SLS moment for Case 3 as 723kNm, thus the live load moment = 1233 - 723 = 510kNm.
cking is therefore eliminated under working load and all of the concrete may be assumed effective in carrying load. Theref
movement. If the bearings are set at a maximum shade air temperature of 16oC then, by proportion the deck will expand
carrying load. Therefore lighter sections may be used to carry a given bending moment, and prestressed concrete may be
the deck will expand 6mm and contract 10mm. Let us assume that this maximum shade air temperature of 16 oC for fixing
ssed concrete may be used over much longer spans than reinforced concrete.
ture of 16 oC for fixing the bearings is specified in the Contract and design the abutments accordingly.
