Prestressed Concrete Since concrete is weak in tension in normal reinforced concrete construction c Prestressing involves inducing compressive stresses in the zone which will ten The prestressing force has to be produced by a high tensile steel, and it is n There are two methods of prestressing concrete : 1) Pre-cast Pre-tensioned 2) Pre-cast Post-tensioned 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 initi Stage 4 When the concrete has cured the stressing force is released and the tendons an Stage 1 Cable ducts and reinforcement are positioned in the beam mould. The ducts are Stage 2 Concrete is cast into the beam mould and allowed to cure to the required initi Stage 3 Tendons are threaded through the cable ducts and tensioned to about 70% of the Stage 4 Wedges are inserted into the end anchorages and the tensioning force on the te Loss of Prestress When the tensioning force is released and the tendons are anchored to the conc 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 Total losses in prestress can amount to about 30% of the initial tensioning st Prestressed Concrete Beam Design to BS 5400 Part 4 Problem: The prestressing force also reduces the magnitude of the principal tensile str Both methods involve tensioning cables inside a concrete beam and then anchori 1) Pre-tensioned Beams 2) Post-tensioned Beams Design a simply supported prestressed concrete Y beam which carries a 150mm t γconc. = 24kN/mm 3
Transcript
Prestressed Concrete
Since concrete is weak in tension in normal reinforced concrete construction cracks develop in the tension zone at working loads and therefore all concrete in tension is ignored in design.Prestressing involves inducing compressive stresses in the zone which will tend to become tensile under external loads. This compressive stress neutralizes the tensile stress so that no resultant tension exists, (or only very small values, within the tensile strength of the concrete). Cracking is therefore eliminated under working load and all of the concrete may be assumed effective in carrying load. Therefore lighter sections may be used to carry a given bending moment, and prestressed concrete may be used over much longer spans than reinforced concrete.
The prestressing force has to be produced by a high tensile steel, and it is necessary to use high quality concrete to resist the higher compressive stresses that are developed.There are two methods of prestressing concrete :1) Pre-cast Pre-tensioned2) Pre-cast Post-tensioned
Stage 1Tendons and reinforcement are positioned in the beam mould.Stage 2Tendons are stressed to about 70% of their ultimate strength.Stage 3Concrete is cast into the beam mould and allowed to cure to the required initial strength.Stage 4When the concrete has cured the stressing force is released and the tendons anchor themselves in the concrete.
Stage 1Cable ducts and reinforcement are positioned in the beam mould. The ducts are usually raised towards the neutral axis at the ends to reduce the eccentricity of the stressing force.Stage 2Concrete is cast into the beam mould and allowed to cure to the required initial strength.Stage 3Tendons are threaded through the cable ducts and tensioned to about 70% of their ultimate strength.Stage 4Wedges are inserted into the end anchorages and the tensioning force on the tendons is released. Grout is then pumped into the ducts to protect the tendons.
Loss of PrestressWhen the tensioning force is released and the tendons are anchored to the concrete a series of effects result in a loss of stress in the tendons. The effects are:
a. relaxation of the steel tendonsb. elastic deformation of the concretec. shrinkage and creep of the concreted. slip or movement of the tendons at the anchorages during anchoringe. 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:
The prestressing force also reduces the magnitude of the principal tensile stress in the web so that thin-webbed
Both methods involve tensioning cables inside a concrete beam and then anchoring the stressed cables to the concrete.
1) Pre-tensioned Beams
2) Post-tensioned Beams
Design a simply supported prestressed concrete Y beam which carries a 150mm thick concrete slab and 100mm of surfacing, together with a nominal live load udl of 10.0 kN/m2 and kel of 33kN/m . The span of the beam is 24.0m centre to centre of bearings and the beams are spaced at 1.0m intervals.
γconc. = 24kN/mm3
25 units of HB to be considered at SLS for load combination 1 only (BS 5400 Pt4 Cl. 4.2.2)
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= 2.4 kN/m
Nominal Live Load : HA = 10 x 1.0 + 33.0 = 10 kN/m + 33kN
25 units HB = 25 x = 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 Comb.3
Dead Load 1.0 1.15 1.15
Superimposed Dead Load 1.2 1.75 1.75
Live Load 1.2 1.0 1.5
1.1 -
Temperature Difference - 0.8 - 1.0
BS 5400 Pt. 4
Section Properties
cl.7.4.1 Modular ratio effect for different concrete strengths between beam and slab may be ignored.
Property Beam Section Composite Section
Loading per beam (at 1.0m c/c)
γfL concrete1.0
γfL surfacing1.2
γfL HA
γfL HB
γfL
Concrete Grades
Beam fcu = 50 N/mm2,
fci = 40 N/mm2
Slab fcu = 40 N/mm2
### ###
Centroid(mm) 456 623
### ###
### ###
### ###
Modulus @ Level 3(mm3) - ###
Temperature Difference Effects
Apply temperature differences given in BS 5400 Pt2 Fig.9 (Group 4)to a simplified beam section.
Force F to restrain temperature strain :
Moment M about centroid of section to restrain curvature due to temperature strain :
Force F to restrain temperature strain :
Moment M about centroid of section to restrain curvature due to temperature strain :
= - 194.5 - 0.6 + 153.8 = - 41.3 kNm
Area(mm2)
2nd Moment of Area(mm4)
Modulus @ Level 1(mm3)
Modulus @ Level 2(mm3)
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 103 x 12 x 10-6 = 0.408 N/mm2
a) Positive temperature difference
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
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
- 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
- 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
Differential Shrinkage Effects
BS 5400 Pt.4 Use cl.6.7.2.4 Table 29 :
cl.7.4.3.4
BS 5400 Pt.4
cl.7.4.3.5
Self weight of beam and weight of deck slab is supported by the beam. When the deck slab concrete has cured then any further loading (superimposed and live loads) is supported by the composite section of the beam and slab.
Total load for serviceability limit state = (1.0 x 3.6)+(1.0 x 10.78) = 14.4kN/m
Combination 1 Loading
Super. & HA live load for
Super. & HB live load for
Total load for ultimate limi
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 residual shrinkage is 100 x 10-6 ,
hence the differential shrinkage is 200 x 10-6
Force to restrain differential shrinkage : F = - εdiff x Ecf x Acf x φ
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
Dead Loading (beam and slab)
Design serviceability moment = 14.4 x 242 / 8 = 1037 kNm
= [(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
= 2.88 & 4 wheels @ 1.1 x 62.5
= 2.9 kN/m & 4 wheels @ 68.75 kN
= [(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
25 units HB Design SLS m
Design ultimate moment
Combination 3 Loading
Super. & HA live load for
Total load for ultimate limi
Design serviceability mom
Allowable stresses in precast concrete
cl.6.3.2.2 b)
cl.6.3.2.4 b)
At serviceability limit state :
cl.7.4.3.2 Compression (1.25 x Table 22)
= 14.9 x 24.02 / 8 + 39.6 x 24 / 4
=1310 kNm
= 2.9 x 24.02 / 8 + 982.3(from grillage analysis)
= 1191.1 kNm
= 35.7 x 24.02 / 8 + 49.5 x 24 / 4
= 2867 kNm
= [(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
= [(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
Using straight, fully bonded tendons (constant force and eccentricity).
Allow for 20% loss of prestress after transfer.
Initial prestress at Level 1 to satisfy class 2 requirement for SLS (Comb. 3).
The critical section at transfer occurs at the end of the transmission zone. The moment due to the self weight at this section is near zero and initial stress conditions are:
Hence 32 tendons required.
Stresses at Level 1 due to SLS loads (N/mm2) :
Dead Load M / Z = (1037 x 106) / (116.020 x 106)
Super. & Live Load M / Z = M / (166.156 x 106)
Reverse Temperature = γ
Stress at transfer = ( 17.67 - 3.2 ) / 0.8 = 18.1 N/mm2 (use allowable stress of 20 N/mm2)
P/A + Pe/Zlevel 1 = 20
P/A - Pe/Zlevel 2 >= - 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 Po = 4888 / 0.9 = 5431kN
Using 15.2mm class 2 relaxation standard strand at maximum initial force of 174kN (0.75 x Pu)
Area of tendon = 139mm2
Nominal tensile strength = fpu =1670 N/mm2
Initial force Po = 32 x 174 = 5568 kN
P = 0.9 x 5568 = 5011 kN
Substituting P = 5011 kN in (eqn. 2)
Arrange 32 tendons symmetrically about the Y-Y axis to achieve an eccentricity of about 216mm.
Taking moments about bottom of beam :
1000 = 2000
900 = 1800
260 = 1040
160 = 1280
10 @ 110 = 1100
60 = 360
7580
Allowing for 1% relaxation loss in steel before transfer and elastic deformation of concrete at transfer :
cl. 6.7.2.3
Initial stresses due to prestress at end of transmission zone :
Stress due to self weight of beam at mid span :
e <= Zlevel 2 / A + Zlevel 2 / P = (89.066 x 106 / 449.22 x 103) + (89.066 x 106 / 5011 x 103)
e = 198 + 18 = 216 mm
2 @
2 @
4 @
8 @
6 @
32
e = 456 - 7580 / 32 = 456 - 237 =219mm
P = 0.99 Po / [ 1 + Es x (Aps / A) x (1 + A x e2 / I) / Eci ]
P = 0.99 x Po / [ 1 + 196 x ( 32 x 139 / 449220) x (1 + 449220 x 2192 / 52.905 x 109) / 31 ]
P = 0.91 Po = 0.91 x 5568 = 5067 kN
Level 1 : P / A x ( 1 + A x e / Zlevel 1 ) = 11.3 x ( 1 + 219 / 258 ) = 20.89 N/mm2
Level 2 : P / A x ( 1 - A x e / Zlevel 2 ) = 11.3 x ( 1 - 219 / 198 ) = - 1.20 N/mm2
Moment due to self weight of beam at mid span = 10.78 x 242 / 8 = 776.2 kNm
@ Level 1 = - 776.2 / 116.02 = - 6.69 N/mm2
Initial stresses at mid span :
cl. 6.7.2.5
Loss of force after transfer due to :
cl. 6.7.2.2 Steel relaxation = 0.02 x 5568 = 111
cl. 6.7.2.4
cl. 6.7.2.5
Total Loss = 111 + 262 + 550 = 923 kN
Final stresses due to prestress after all loss of prestress at :
Level 1
Level 2
Combined stresses in final condition for worst effects of design loads, differential shrinkage and temperature difference :
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 :
Taking Moments about the neutral axis :
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
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
Fp6 = 134 Ff = 2400
Fp5 = 168 Fw = 2765
Fp4 = 675 Fc = 5165
Fp3 = 1382
Fp2 = 1746
Fp1 = 1060
Ft = 5165
Fp6 = 134 x -0.365 =-49
Fp5 = 168 x -0.265 = -45
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 loading. 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).
The proposed deck consists of 11No. Y4 prestressed concrete beams and concrete deck slab as shown.
Loading From the DeckA grillage analysis gave the following reactions for the various load cases:
Critical Reaction U Total Reaction on Each Abutment
The ground investigation report shows suitable founding strata about 9.5m below the proposed road level. Test results show the founding strata to be a cohesionless soil having an angle of shearing resistance (φ) = 30
Backfill material will be Class 6N with an effective angle of internal friction (ϕ') = 35o and density (γ) = 19kN/m
Deck Dead Load = (1900 + 320) / 11.6 = 191kN/mHA live Load on Deck = 1140 / 11.6 = 98kN/mHB 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 temperatures are -11 and +36
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
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 Ekspan's Elastomeric Pad Bearing EKR35:
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.
Alternatively using BS 5400 Part 9.1 Clause 5.14.2.6:
Using the Ekspan bearing EKR35
This correllates with the value obtained above using the shear stiffness from the manufacturer's data sheet.
Nominal Load for HA = 8kN/m x 20m + 250kN = 410kNNominal Load for HB = 25% of 45units x 10kN x 4axles = 450kN450 > 410kN hence HB braking is critical.
When this load is applied on the deck it will act on the fixed abutment only.
Nominal Load = 300kN300 < 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.
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 -19 to +37
Total horizontal load on each abutment = 11 x 85 = 935 kN ≡ 935 / 11.6 = 81kN/m.
Option 2 - Sliding Bearing:With a maximum ultimate reaction of 790kN 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
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 between the sliding faces will be in the order of 5N/mm
From Table3 of BS 5400 Part 9.1 the Coefficient of friction = 0.08 for a bearing stress of 5N/mmHence 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:
Braking load on 1m width of abutment = 450 / 11.6 = 39kN/m.
Skidding Load - BS 5400 Part 2 Clause 6.11:
For Stability calculations use active earth pressures = Ka γ hKa for Class 6N material = (1-Sin35) / (1+Sin35) = 0.27
Assume a surchage loading for the compaction plant to be equivalent to 30 units of HB
1) Stability Check
Initial Sizing for Base Dimensions
Alternatively a simple spreadsheet will achieve a result by trial and error.
Load Combinations
Fixed Abutment OnlyBackfill + HA surcharge + Deck dead load + HA on deck + Braking on deck
Weight of wall stem = 1.0 x 6.5 x 25 = 163kN/mWeight of base = 6.4 x 1.0 x 25 = 160kN/mWeight of backfill = 4.3 x 6.5 x 19 = 531kN/mWeight of surcharge = 4.3 x 12 = 52kN/m
F Lever Arm Moment About ABackfill 144 2.5 361Surcharge 24 3.75 91
∑ =168 ∑ =452
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
Hence Compaction Plant surcharge = 12 kN/m2.
For surcharge of w kN/m2 :Fs = Ka w h = 0.27wh kN/m
There are a number of publications that will give guidance on base sizes for free standing cantilever walls, Reynolds's Reinforced Concrete Designer's Handbook being one such book.
Backfill + Construction surcharge
Backfill + Construction surcharge
Sackfill + HA surcharge + Deck dead load + Deck contraction
Backfill + HA surcharge + Braking behind abutment + Deck dead load
Backfill + HB surcharge + Deck dead load
Backfill + HA surcharge + Deck dead load + HB on deck
CASE 1 - Fixed Abutment
Density of reinforced concrete = 25kN/m3.
Backfill Force Fb = 0.27 x 19 x 7.52 / 2 = 144kN/m
For sliding effects:Active Force = Fb + Fs = 168kN/m
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 the centre of the base.P = 906kN/m
Nett moment = 3251 - 452 = 2799kNm/mEccentricity (e) of P about centre-line of base = 3.2 - (2799 / 906) = 0.111mPressure under base = (906 / 6.4) ± (906 x 0.111 / 6.827)
Hence the abutment will be stable for Case 1.
Fixed Abutment:
F of S Overturning F Be Bearing Pressure at HeelCase 1 7.16 3.09 156Case 2 2.87 2.13 386Case 2a 4.31 2.64 315Case 3 3.43 2.43 351Case 4 4.48 2.63 322Case 5 5.22 3.17 362Case 6 3.8 2.62 378
Free Abutment:
F of S Overturning F of S Sli Bearing PrBearing Pressure at HeelCase 1 7.15 3.09 168Case 2 2.91 2.14 388Case 2a 4.33 2.64 318Case 3 3.46 2.44 354Case 4 4.5 2.64 325Case 5 5.22 3.16 365It can be seen that the use of elastomeric bearings (Case 2) will govern the critical design load cases on the abutments. We shall assume that there are no specific requirements for using elastomeric bearings and design the abutments for the lesser load effects by using sliding bearings.
2) Wall and Base Design
Loads on the back of the wall are calculated using 'at rest' earth pressures. Serviceability and Ultimate load effects need to be calculated for the load cases 1 to 6 shown above. Again, these are best carried out using a simple spreadsheet.Using the Fixed Abutment Load Case 1 again as an example of the calculations:Wall Design
Factor of Safety Against Overturning = 3251 / 452 = 7.2 > 2.0 ∴ OK.
Frictional force on underside of base resisting movement = W tan(φ) = 906 x tan(30o) = 523kN/mFactor of Safety Against Sliding = 523 / 168 = 3.1 > 2.0 ∴ OK.
A = 6.4m2/m
Z = 6.42 / 6 = 6.827m3/m
Pressure under toe = 142 + 15 = 157kN/m2 < 400kN/m2 ∴ OK.
Pressure under heel = 142 - 15 = 127kN/m2
Analysing the fixed abutment with Load Cases 1 to 6 and the free abutment with Load Cases 1 to 5 using a simple spreadsheet the following results were obtained:
Ko = 1 - Sin(ϕ') = 1 - Sin(35o) = 0.426γfL for horizontal loads due to surcharge and backfill from BS 5400 Part 2 Clause 5.8.1.2:
Serviceability = 1.0Ultimate = 1.5
Surcharge Force Fs on the rear of the wall = 0.426 x 12 x 6.5 = 33kN/mAt the base of the Wall:
Fixed Abutment:
Moment Moment Moment ShearSLS Dead SLS Live ULS ULS
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
Bending
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
Serviceability moment = (171 x 6.5 / 3) + (33 x 6.5 / 2) = 371 + 107 = 478kNm/mUltimate moment = 1.1 x 1.5 x 478 = 789kNm/mUltimate 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 simple spreadsheet the following results were obtained for the design moments and shear at the base of the wall:
Use strength class C32/40 with water-cement ratio 0.5 and minimum cement content of 340kg/m
Nominal cover to reinforcement = 60mm (45mm minimum cover plus a tolerance Δc of 15mm).Reinforcement to BS 4449:2005 Grade B500B: fy = 500N/mm2
Design for critical moments and shear in Free Abutment:
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
Shear
Shear requirements are designed to BS 5400 clause 5.4.4:
ULS shear at Section 7H/8 for load case 4 = 487 kN
Hence height requiring strengthening = 1.073 x (0.673 - 0.62) / (0.673 - 0.53) = 0.4m < d.
Early Thermal Cracking
Base Design
Maximum bending and shear effects in the base slab will occur at sections near the front and back of the wall. Different load factors are used for serviceability and ultimate limit states so the calculations need to be carried out for each limit state using 'at rest pressures'Using the Fixed Abutment Load Case 1 again as an example of the calculations:
Weight of wall stem = 1.0 x 6.5 x 25 x 1.0 = 163kN/mWeight of base = 6.4 x 1.0 x 25 x 1.0 = 160kN/mWeight of backfill = 4.3 x 6.5 x 19 x 1.0 = 531kN/mWeight of surcharge = 4.3 x 12 x 1.0 = 52kN/m
Surcharge Force Fs = 0.426 x 12 x 7.5 x 1.0 = 38 kN/m
Restoring Effects:
Weight Le Moment About AStem 163 1.6 261Base 160 3.2 512Backfill 531 4.25 2257Surcharge 52 4.25 221
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) ]} dUse B40 @ 150 c/c:
As = 8378mm2/m, d = 1000 - 60 - 20 = 920mmz = {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
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 at distance H/8 (8.63 / 8 = 1.079m) up the wall.
v = V / (bd) = 487 x 103 / (1000 x 920) = 0.53 N/mm2 < 0.62
Provide a 500 x 500 splay at the base of the wall with B32 @ 150c/c bars in sloping face.
Considering the effects of casting the wall stem onto the base slab by complying with the early 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 wall.
Minimum area of secondary reinforcement to Clause 5.8.4.2 = 0.12% of bad = 0.0012 x 1000 x 920 = 1104 mm2/m (use B16 @ 150c/c - As = 1340mm2/m)
F Le Moment About ABackfill 228 2.5 570Surcharge 38 3.75 143∑ = 266 ∑ = 713
Bearing Pressure at toe and heel of base slab = (P / A) ± (P x e / Z)P = 906kN/m
Nett moment = 3251 - 713 = 2538kNm/mEccentricity (e) of P about centre-line of base = 3.2 - (2538 / 906) = 0.399mPressure under base = (906 / 6.4) ± (906 x 0.399 / 6.827)
Weight of wall stem = 1.0 x 6.5 x 25 x 1.15 = 187kN/mWeight of base = 6.4 x 1.0 x 25 x 1.15 = 184kN/mWeight of backfill = 4.3 x 6.5 x 19 x 1.2 = 637kN/mWeight of surcharge = 4.3 x 12 x 1.2 = 62kN/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
A = 6.4m2/m
Z = 6.42 / 6 = 6.827m3/m
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 bottom face).
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 4.3 / 2) = -443kNm/m (tension in top face).
CASE 1 - Fixed Abutment Ultimate Limit StateγfL for concrete = 1.15γfL for fill and surcharge(vetical) = 1.2γfL for fill and surcharge(horizontal) = 1.5
Backfill Force Fb = 0.426 x 19 x 7.52 x 1.5 / 2 = 341kN/m
Overturning Effects:F Lever Arm Moment About ABackfill 341 2.5 853Surcharge 58 3.75 218∑ = 399 ∑ = 1071
Bearing Pressure at toe and heel of base slab = (P / A) ± (P x e / Z)P = 1070kN/m
Nett moment = 3859 - 1071 = 2788kNm/mEccentricity (e) of P about centre-line of base = 3.2 - (2788 / 1070) = 0.594mPressure under base = (1070 / 6.4) ± (1070 x 0.594 / 6.827)
ULS Shear at a-a = 1.1 x {[(260 + 228) x 1.1 / 2] - (1.15 x 1.1 x 25)} = 260kN/m
/ 2) - (62 x 4.3 / 2)} = -769kNm/m (tension in top face).
A = 6.4m2/m
Z = 6.42 / 6 = 6.827m3/m
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 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)} = 148kNm/m (tension in bottom face).
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
Design for shear and bending effects at sections a-a and b-b for the Free Abutment:
Bending
Use B32 @ 150 c/c:
For the Serviceability check for Case 3 an approximation of the dead load moment can be obtained by removing the surcharge and braking loads. The spreadsheet result gives the dead load SLS moment for Case 3 as 723kNm, thus the live load moment = 1233 - 723 = 510kNm.
This could be corrected by reducing the bar spacing, but increase the bar size to B40@150 c/c as this is required to avoid the use of links (see below).
ShearShear on Toe - Use Fixed Abutment Load Case 6:By inspection B32@150c/c will be adequate for the
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 -
Analysing the fixed abutment with Load Cases 1 to 6 and the free abutment with Load Cases 1 to 5 using a simple spreadsheet the following results were obtained:
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
As = 5362mm2/m, d = 1000 - 60 - 16 = 924mmz = {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.
Carrying out the crack control calculation to Clause 5.8.8.2 gives a crack width of 0.27mm > 0.25mm
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
bending effects in the toe (Muls = 365kNm < 1983kNm)
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) = 4.4mULS 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)} = 559kN
Using B32@150 c/c then:
Early Thermal Cracking
Local Effects
Curtain WallThis wall is designed to be cast onto the top of the abutment after the deck has been built. Loading will be applied from the backfill, surcharge and braking loads on top of the wall.HB braking load to BS 5400 clause 6.10 = 25% x 45units x 10kN on each axle = 112.5kN per axle.
1st axle load on back of abutment = 112.5 / 3.0 = 37.5kN/m2nd axle load on back of abutment = 112.5 / 6.6 = 17.0kN/m3rd & 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/mBending and Shear at Base of 3m High Curtain Wall
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
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 ∴ FailRather than provide shear reinforcement try increasing bars to B40 @ 150 c/c (also required for crack control as shown above).
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
Considering the effects of casting the base slab onto the blinding concrete by complying with the early thermal cracking of concrete to BD 28 then B16 distribution bars @ 250 c/c will be required.
Minimum area of main reinforcement to Clause 5.8.4.1 = 0.15% of bad = 0.0015 x 1000 x 924 = 1386 mm2/m (use B20 @ 200c/c - As = 1570mm2/m).
Assume a 45o dispersal to the curtain wall and a maximum dispersal of the width of the abutment (11.6m) then:
Horizontal load due to HB surcharge = 0.426 x 20 x 3.0 = 25.6 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/mULS Shear = 1.1 x {(1.1 x 73.9) + (1.5 x 25.6) + (1.5 x 36.4)} = 192kN/m
Last Updated : 05/09/09For more information :
Horizontal load due to backfill = 0.426 x 19 x 3.02 / 2 = 36.4 kN/m
400 thick curtain wall with B32 @ 150 c/c :Mult = 584 kNm/m > 392 kNm/m ∴ OKSLS 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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Since concrete is weak in tension in normal reinforced concrete construction cracks develop in the tension zone at working loads and therefore all concrete in tension is ignored in design.Prestressing involves inducing compressive stresses in the zone which will tend to become tensile under external loads. This compressive stress neutralizes the tensile stress so that no resultant tension exists, (or only very small values, within the tensile strength of the concrete). Cracking is therefore eliminated under working load and all of the concrete may be assumed effective in carrying load. Therefore lighter sections may be used to carry a given bending moment, and prestressed concrete may be used over much longer spans than reinforced concrete.
The prestressing force has to be produced by a high tensile steel, and it is necessary to use high quality concrete to resist the higher compressive stresses that are developed.
When the concrete has cured the stressing force is released and the tendons anchor themselves in the concrete.
Cable ducts and reinforcement are positioned in the beam mould. The ducts are usually raised towards the neutral axis at the ends to reduce the eccentricity of the stressing force.
Tendons are threaded through the cable ducts and tensioned to about 70% of their ultimate strength.
Wedges are inserted into the end anchorages and the tensioning force on the tendons is released. Grout is then pumped into the ducts to protect the tendons.
When the tensioning force is released and the tendons are anchored to the concrete a series of effects result in a loss of stress in the tendons. The effects are:
e. other causes in special circumstances , such as when steam curing is used with pre-tensioning.
The prestressing force also reduces the magnitude of the principal tensile stress in the web so that thin-webbed I - sections may be used without the risk of diagonal tension failures and with further savings in self-weight.
Both methods involve tensioning cables inside a concrete beam and then anchoring the stressed cables to the concrete.
Design a simply supported prestressed concrete Y beam which carries a 150mm thick concrete slab and 100mm of surfacing, together with a nominal live load udl of 10.0 kN/m2 and kel of 33kN/m . The span of the beam is 24.0m centre to centre of bearings and the beams are spaced at 1.0m intervals.
= 10.78 kN/m
Load factors for serviceability and ultimate limit state from BS 5400 Part 2 (or BD 37/01) Table 1:
ULS
Comb.1 Comb.3
1.25
- -
Modular ratio effect for different concrete strengths between beam and slab may be ignored.
Apply temperature differences given in BS 5400 Pt2 Fig.9 (Group 4)to a simplified beam section.
= - 385.9 - 19.3 - 295.1 = - 700.3 kN
Self weight of beam and weight of deck slab is supported by the beam. When the deck slab concrete has cured then any further loading (superimposed and live loads) is supported by the composite section of the beam and slab.
Assume that 2/3 of the total shrinkage of the precast 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
= [(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
= [(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
(HB)
-0.60
-17.67*
The critical section at transfer occurs at the end of the transmission zone. The moment due to the self weight at this section is near zero and initial stress conditions are:
.....................(eqn. 1)
.....................(eqn. 2)
= 4888 / 0.9 = 5431kN
Arrange 32 tendons symmetrically about the Y-Y axis to achieve an eccentricity of about 216mm.
Allowing for 1% relaxation loss in steel before transfer and elastic deformation of concrete at transfer :
P = 0.99 x Po / [ 1 + 196 x ( 32 x 139 / 449220) x (1 + 449220 x 2192 / 52.905 x 109) / 31 ]
Final stresses due to prestress after all loss of prestress at :
Combined stresses in final condition for worst effects of design loads, differential shrinkage and temperature difference :
Allowing for 2% relaxation loss in steel after transfer, concrete shrinkage εcs = 300 x 10-6
) = 300 x 10-6 x 196 x 32 x 139 = 262
) = 1.03 x 48 x 10-6 x 12.76 x 196 x 32 x 139 = 550
= 5067 - 923 = 4144 kN (Pe/P = 0.82)
Centroid of tendons in tension zone = (6x60 + 10x110 + 8x160 + 4x260) / 28 = 135mm
Mu / M = 4192 / 3154 = 1.33 ( > 1.15 ) hence strain in outermost tendon O.K.
Design the fixed and free end cantilever abutments to the 20m span deck shown to carry HA and 45 units of HB loading. 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).
The proposed deck consists of 11No. Y4 prestressed concrete beams and concrete deck slab as shown.
2400600
18802770
The ground investigation report shows suitable founding strata about 9.5m below the proposed road level. Test results show the founding strata to be a cohesionless soil having an angle of shearing resistance (φ) = 30
and density (γ) = 19kN/m3.
From BS 5400 Part 2 Figures 7 and 8 the minimum and maximum shade air temperatures are -19 and +37oC respectively.
For a Group 4 type strucutre (see fig. 9) the corresponding minimum and maximum effective bridge temperatures are -11 and +36oC from tables 10 and 11.
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 = 11.3mm.] = ±[11.3 x 1.1 x 1.3 /2] = ± 8mm.
With a maximum ultimate reaction = 230 + 60 + 500 = 790kN then a suitable elastomeric bearing would be Ekspan's Elastomeric Pad Bearing EKR35:
This is an ultimate load hence the nominal horizontal load = 121 / 1.1 / 1.3 = 85kN at each bearing.
This correllates with the value obtained above using the shear stiffness from the manufacturer's data sheet.
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 -19 to +37oC which would require the bearings to be installed at a shade air temperature of 9
With a maximum ultimate reaction of 790kN and longitudinal movement of ± 8mm then a suitable bearing from the Ekspan EA Series would be /80/210/25/25:
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.
From Table3 of BS 5400 Part 9.1 the Coefficient of friction = 0.08 for a bearing stress of 5N/mm2
Hence total horizontal load on each abutment when the deck expands or contracts = 2220 x 0.08 = 180kN ≡ 180 / 11.6 = 16kN/m.
There are a number of publications that will give guidance on base sizes for free standing cantilever walls, Reynolds's Reinforced Concrete Designer's Handbook being one such book.
Check bearing pressure at toe and heel of base slab = (P / A) ± (P x e / Z) where P x e is the moment about the centre of the base.
1275
7639838143
1207
78428482
It can be seen that the use of elastomeric bearings (Case 2) will govern the critical design load cases on the abutments. We shall assume that there are no specific requirements for using elastomeric bearings and design the abutments for the lesser load effects by using sliding bearings.
Loads on the back of the wall are calculated using 'at rest' earth pressures. Serviceability and Ultimate load effects need to be calculated for the load cases 1 to 6 shown above. Again, these are best carried out using a simple spreadsheet.
o) = 523kN/m
Analysing the fixed abutment with Load Cases 1 to 6 and the free abutment with Load Cases 1 to 5 using a simple spreadsheet the following results were obtained:
Analysing the fixed abutment with Load Cases 1 to 6 and the free abutment with Load Cases 1 to 5 using a simple spreadsheet the following results were obtained for the design moments and shear at the base of the wall:
Use strength class C32/40 with water-cement ratio 0.5 and minimum cement content of 340kg/m3 for exposure condition XD2.
of 15mm).Reinforcement to BS 4449:2005 Grade B500B: fy = 500N/mm2
Maximum bending and shear effects in the base slab will occur at sections near the front and back of the wall. Different load factors are used for serviceability and ultimate limit states so the calculations need to be carried out for each limit state using 'at rest pressures'
Also the steel reinforcement and concrete stresses meet the limitations required in clause 4.1.1.3 ∴ serviceability requirements are satisfied.
x (40)1/3 = 0.72
< 0.673 hence shear reinforcement should be provided, however check shear at distance H/8 (8.63 / 8 = 1.079m) up the wall.
Considering the effects of casting the wall stem onto the base slab by complying with the early 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 wall.
d = 0.0012 x 1000 x 920 = 1104 mm2/m (use B16 @ 150c/c - As = 1340mm2/m)
/ 2) - (531 x 4.3 / 2) - (52 x 4.3 / 2) = -443kNm/m (tension in top face).
/ 3) - (1.15 x 25 x 1.0 x 1.12 / 2)} = 148kNm/m (tension in bottom face).
76815961834170014021717
ULSMoment816
1678192217861480
For the Serviceability check for Case 3 an approximation of the dead load moment can be obtained by removing the surcharge and braking loads. The spreadsheet result gives the dead load SLS moment for Case 3 as 723kNm, thus the live load moment = 1233 - 723 = 510kNm.
This could be corrected by reducing the bar spacing, but increase the bar size to B40@150 c/c as this is required to avoid the use of links (see below).
Analysing the fixed abutment with Load Cases 1 to 6 and the free abutment with Load Cases 1 to 5 using a simple spreadsheet the following results were obtained:
Carrying out the crack control calculation to Clause 5.8.8.2 gives a crack width of 0.27mm > 0.25mm ∴ Fail.
Also the steel reinforcement and concrete stresses meet the limitations required in clause 4.1.1.3 ∴ serviceability requirements are satisfied.
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)} = 559kN
This wall is designed to be cast onto the top of the abutment after the deck has been built. Loading will be applied from the backfill, surcharge and braking loads on top of the wall.HB braking load to BS 5400 clause 6.10 = 25% x 45units x 10kN on each axle = 112.5kN per axle.
x (40)1/3 = 0.62
x (40)1/3 = 0.62
(also required for crack control as shown above).
x (40)1/3 = 0.716
Considering the effects of casting the base slab onto the blinding concrete by complying with the early thermal cracking of concrete to BD 28 then B16 distribution bars @ 250 c/c will be required.
d = 0.0015 x 1000 x 924 = 1386 mm2/m (use B20 @ 200c/c - As = 1570mm2/m).
dispersal to the curtain wall and a maximum dispersal of the width of the abutment (11.6m) then:
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
116.020 166.15689.066 242.424
179.402
Since concrete is weak in tension in normal reinforced concrete construction cracks develop in the tension zone at working loads and therefore all concrete in tension is ignored in design.Prestressing involves inducing compressive stresses in the zone which will tend to become tensile under external loads. This compressive stress neutralizes the tensile stress so that no resultant tension exists, (or only very small values, within the tensile strength of the concrete). Cracking is therefore eliminated under working load and all of the concrete may be assumed effective in carrying load. Therefore lighter sections may be used to carry a given bending moment, and prestressed concrete may be used over much longer spans than reinforced concrete.
- sections may be used without the risk of diagonal tension failures and with further savings in self-weight.
Design a simply supported prestressed concrete Y beam which carries a 150mm thick concrete slab and 100mm of surfacing, together with a nominal live load udl of 10.0 kN/m2 and kel of 33kN/m . The span of the beam is 24.0m centre to centre of bearings and the beams are spaced at 1.0m intervals.
Self weight of beam and weight of deck slab is supported by the beam. When the deck slab concrete has cured then any further loading (superimposed and live loads) is supported by the composite section of the beam and slab.
Design the fixed and free end cantilever abutments to the 20m span deck shown to carry HA and 45 units of HB loading. 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).
The ground investigation report shows suitable founding strata about 9.5m below the proposed road level. Test results show the founding strata to be a cohesionless soil having an angle of shearing resistance (φ) = 30o and a safe bearing capacity of 400kN/m2.
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. oC which would require the bearings to be installed at a shade air temperature of 9oC to achieve the ± 8mm movement. If the bearings are set at a maximum shade air temperature of 16
It can be seen that the use of elastomeric bearings (Case 2) will govern the critical design load cases on the abutments. We shall assume that there are no specific requirements for using elastomeric bearings and design the abutments for the lesser load effects by using sliding bearings.
Loads on the back of the wall are calculated using 'at rest' earth pressures. Serviceability and Ultimate load effects need to be calculated for the load cases 1 to 6 shown above. Again, these are best carried out using a simple spreadsheet.
Analysing the fixed abutment with Load Cases 1 to 6 and the free abutment with Load Cases 1 to 5 using a simple spreadsheet the following results were obtained for the design moments and shear at the base of the wall:
Maximum bending and shear effects in the base slab will occur at sections near the front and back of the wall. Different load factors are used for serviceability and ultimate limit states so the calculations need to be carried out for each limit state using 'at rest pressures'
Considering the effects of casting the wall stem onto the base slab by complying with the early 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 wall.
For the Serviceability check for Case 3 an approximation of the dead load moment can be obtained by removing the surcharge and braking loads. The spreadsheet result gives the dead load SLS moment for Case 3 as 723kNm, thus the live load moment = 1233 - 723 = 510kNm.
Prestressing involves inducing compressive stresses in the zone which will tend to become tensile under external loads. This compressive stress neutralizes the tensile stress so that no resultant tension exists, (or only very small values, within the tensile strength of the concrete). Cracking is therefore eliminated under working load and all of the concrete may be assumed effective in carrying load. Therefore lighter sections may be used to carry a given bending moment, and prestressed concrete may be used over much longer spans than reinforced concrete.
and kel of 33kN/m . The span of the beam is 24.0m centre to centre of bearings and the beams are spaced at 1.0m intervals.
C to achieve the ± 8mm movement. If the bearings are set at a maximum shade air temperature of 16oC then, by proportion the deck will expand 6mm and contract 10mm. Let us assume that this maximum shade air temperature of 16
It can be seen that the use of elastomeric bearings (Case 2) will govern the critical design load cases on the abutments. We shall assume that there are no specific requirements for using elastomeric bearings and design the abutments for the lesser load effects by using sliding bearings.
For the Serviceability check for Case 3 an approximation of the dead load moment can be obtained by removing the surcharge and braking loads. The spreadsheet result gives the dead load SLS moment for Case 3 as 723kNm, thus the live load moment = 1233 - 723 = 510kNm.
Prestressing involves inducing compressive stresses in the zone which will tend to become tensile under external loads. This compressive stress neutralizes the tensile stress so that no resultant tension exists, (or only very small values, within the tensile strength of the concrete). Cracking is therefore eliminated under working load and all of the concrete may be assumed effective in carrying load. Therefore lighter sections may be used to carry a given bending moment, and prestressed concrete may be used over much longer spans than reinforced concrete.
C to achieve the ± 8mm movement. If the bearings are set at a maximum shade air temperature of 16oC then, by proportion the deck will expand 6mm and contract 10mm. Let us assume that this maximum shade air temperature of 16
Prestressing involves inducing compressive stresses in the zone which will tend to become tensile under external loads. This compressive stress neutralizes the tensile stress so that no resultant tension exists, (or only very small values, within the tensile strength of the concrete). Cracking is therefore eliminated under working load and all of the concrete may be assumed effective in carrying load. Therefore lighter sections may be used to carry a given bending moment, and prestressed concrete may be used over much longer spans than reinforced concrete.
C then, by proportion the deck will expand 6mm and contract 10mm. Let us assume that this maximum shade air temperature of 16oC for fixing the bearings is specified in the Contract and design the abutments accordingly.
Prestressing involves inducing compressive stresses in the zone which will tend to become tensile under external loads. This compressive stress neutralizes the tensile stress so that no resultant tension exists, (or only very small values, within the tensile strength of the concrete). Cracking is therefore eliminated under working load and all of the concrete may be assumed effective in carrying load. Therefore lighter sections may be used to carry a given bending moment, and prestressed concrete may be used over much longer spans than reinforced concrete.
C for fixing the bearings is specified in the Contract and design the abutments accordingly.
Prestressed Concrete
Since concrete is weak in tension in normal reinforced concrete construction cracks develop in the tension zone at working loads and therefore all concrete in tension is ignored in design.
Prestressing involves inducing compressive stresses in the zone which will tend to become tensile under external loads. This compressive stress neutralizes the tensile stress so that no resultant tension exists, (or only very small values, within the tensile strength of the concrete). Cracking is therefore eliminated under working load and all of the concrete may be assumed effective in carrying load. Therefore lighter sections may be used to carry a given bending moment, and prestressed concrete may be used over much longer spans than reinforced concrete.
The prestressing force has to be produced by a high tensile steel, and it is necessary to use high quality concrete to resist the higher compressive stresses that are developed.
There are two methods of prestressing concrete :
1) Pre-cast Pre-tensioned
2) Pre-cast Post-tensioned
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.
Stage 1
Cable ducts and reinforcement are positioned in the beam mould. The ducts are usually raised towards the neutral axis at the ends to reduce the eccentricity of the stressing force.
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 pumped into the ducts to protect the tendons.
Loss of Prestress
When the tensioning force is released and the tendons are anchored to the concrete a series of effects result in a loss of stress in the tendons. The effects are:
The prestressing force also reduces the magnitude of the principal tensile stress in the web so that thin-webbed I - sections may be used without the risk of diagonal tension failures and with further savings in self-weight.
Both methods involve tensioning cables inside a concrete beam and then anchoring the stressed cables to the concrete.
1) Pre-tensioned Beams
2) Post-tensioned Beams
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:
25 units of HB to be considered at SLS for load combination 1 only (BS 5400 Pt4 Cl. 4.2.2)
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 1.0 = 2.4 kN/m
Nominal Live Load : 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
Design a simply supported prestressed concrete Y beam which carries a 150mm thick concrete slab and 100mm of surfacing, together with a nominal live load udl of 10.0 kN/m2 and kel of 33kN/m . The span of the beam is 24.0m centre to centre of bearings and the beams are spaced at 1.0m intervals.
γconc. = 24kN/mm3
Loading per beam (at 1.0m c/c)
γfL concrete1.0
Superimposed Dead Load
Live Load 1.2
1.1
Temperature Difference -
BS 5400 Pt. 4
Section Properties
cl.7.4.1 Modular ratio effect for different concrete strengths between beam and slab may be ignored.
Property Beam Section Composite Section
4.49E+05
Centroid(mm) 456
###
###
###
Modulus @ Level 3(mm3) -
Temperature Difference Effects
γfL surfacing1.2
γfL HA
γfL HB
γfL
Concrete Grades
Beam fcu = 50 N/mm2,
fci = 40 N/mm2
Slab fcu = 40 N/mm2
Area(mm2)
2nd Moment of Area(mm4)
Modulus @ Level 1(mm3)
Modulus @ Level 2(mm3)
Apply temperature differences given in BS 5400 Pt2 Fig.9 (Group 4)to a simplified beam section.
Force F to restrain temperature strain :
Moment M about centroid of section to restrain curvature due to temperature strain :
Force F to restrain temperature strain :
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 103 x 12 x 10-6 = 0.408 N/mm2
a) Positive temperature difference
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
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
- 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
Moment M about centroid of section to restrain curvature due to temperature strain :
= - 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
BS 5400 Pt.4
cl.7.4.3.5
- 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
- 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
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 residual shrinkage is 100 x 10-6 ,
hence the differential shrinkage is 200 x 10-6
Force to restrain differential shrinkage : F = - εdiff x Ecf x Acf x φ
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 further loading (superimposed and live loads) is supported by the composite section of the beam and slab.
Total load for serviceability limit state = (1.0 x 3.6)+(1.0 x 10.78) = 14.4kN/m
Combination 1 Loading
Super. & HA live load for
Super. & HB live load for
Total load for ultimate limi
HA Design serviceability
25 units HB Design SLS m
Design ultimate moment
Combination 3 Loading
Dead Loading (beam and slab)
Design serviceability moment = 14.4 x 242 / 8 = 1037 kNm
= [(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
= 2.88 & 4 wheels @ 1.1 x 62.5
= 2.9 kN/m & 4 wheels @ 68.75 kN
= [(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
= 14.9 x 24.02 / 8 + 39.6 x 24 / 4
=1310 kNm
= 2.9 x 24.02 / 8 + 982.3(from grillage analysis)
= 1191.1 kNm
= 35.7 x 24.02 / 8 + 49.5 x 24 / 4
= 2867 kNm
Super. & HA live load for
Total load for ultimate limi
Design serviceability mom
Allowable stresses in precast concrete
cl.6.3.2.2 b)
cl.6.3.2.4 b)
At serviceability limit state :
cl.7.4.3.2 Compression (1.25 x Table 22)
Comb.1 Comb.1 Comb.3
- 8.94 - 8.94 - 8.94
= [(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
= [(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
Using straight, fully bonded tendons (constant force and eccentricity).
Allow for 20% loss of prestress after transfer.
Initial prestress at Level 1 to satisfy class 2 requirement for SLS (Comb. 3).
The critical section at transfer occurs at the end of the transmission zone. The moment due to the self weight at this section is near zero and initial stress conditions are:
Hence 32 tendons required.
Super. & Live Load M / Z = M / (166.156 x 106)
Reverse Temperature = γfL x -1.69 = 0.8 x -1.69
Stress at transfer = ( 17.67 - 3.2 ) / 0.8 = 18.1 N/mm2 (use allowable stress of 20 N/mm2)
P/A + Pe/Zlevel 1 = 20
P/A - Pe/Zlevel 2 >= - 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 Po = 4888 / 0.9 = 5431kN
Using 15.2mm class 2 relaxation standard strand at maximum initial force of 174kN (0.75 x Pu)
Area of tendon = 139mm2
Nominal tensile strength = fpu =1670 N/mm2
Initial force Po = 32 x 174 = 5568 kN
P = 0.9 x 5568 = 5011 kN
Substituting P = 5011 kN in (eqn. 2)
Arrange 32 tendons symmetrically about the Y-Y axis to achieve an eccentricity of about 216mm.
Taking moments about bottom of beam :
1000 = 2000
900 = 1800
260 = 1040
160 = 1280
10 @ 110 = 1100
60 = 360
7580
Allowing for 1% relaxation loss in steel before transfer and elastic deformation of concrete at transfer :
cl. 6.7.2.3
Initial stresses due to prestress at end of transmission zone :
e <= Zlevel 2 / A + Zlevel 2 / P = (89.066 x 106 / 449.22 x 103) + (89.066 x 106 / 5011 x 103)
e = 198 + 18 = 216 mm
2 @
2 @
4 @
8 @
6 @
32
e = 456 - 7580 / 32 = 456 - 237 =219mm
P = 0.99 Po / [ 1 + Es x (Aps / A) x (1 + A x e2 / I) / Eci ]
P = 0.99 x Po / [ 1 + 196 x ( 32 x 139 / 449220) x (1 + 449220 x 2192 / 52.905 x 109) / 31 ]
P = 0.91 Po = 0.91 x 5568 = 5067 kN
Level 1 : P / A x ( 1 + A x e / Zlevel 1 ) = 11.3 x ( 1 + 219 / 258 ) = 20.89 N/mm2
Stress due to self weight of beam at mid span :
Initial stresses at mid span :
cl. 6.7.2.5
Loss of force after transfer due to :
cl. 6.7.2.2 Steel relaxation = 0.02 x 5568 = 111
cl. 6.7.2.4
cl. 6.7.2.5
Total Loss = 111 + 262 + 550 = 923 kN
Final stresses due to prestress after all loss of prestress at :
Level 1
Level 2
Combined stresses in final condition for worst effects of design loads, differential shrinkage and temperature difference :
Level 2 : P / A x ( 1 - A x e / Zlevel 2 ) = 11.3 x ( 1 - 219 / 198 ) = - 1.20 N/mm2
Moment due to self weight of beam at mid span = 10.78 x 242 / 8 = 776.2 kNm
@ Level 1 = - 776.2 / 116.02 = - 6.69 N/mm2
@ Level 2 = 776.2 / 89.066 = 8.71 N/mm2
Allowing for 2% relaxation loss in steel after transfer, concrete shrinkage εcs = 300 x 10-6
and concrete specific creep ct = 1.03 x 48 x 10-6 per N/mm2
Concrete shrinkage = (εcs x Es x Aps ) = 300 x 10-6 x 196 x 32 x 139 = 262
Concrete creep = ( ct x fco x Es x Aps ) = 1.03 x 48 x 10-6 x 12.76 x 196 x 32 x 139 = 550
Final force after all loss of prestress = Pe = 5067 - 923 = 4144 kN (Pe/P = 0.82)
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 :
Taking Moments about the neutral axis :
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
Fp6 = 134 Ff = 2400
Fp5 = 168 Fw = 2765
Fp4 = 675 Fc = 5165
Fp3 = 1382
Fp2 = 1746
Fp1 = 1060
Ft = 5165
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
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 loading. 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).
Concrete Deck 180 230 1900 2400
Surfacing 30 60 320 600
HA udl+kel 160 265 1140 1880
45 units HB 350 500 1940 2770
Nominal loading on 1m length of abutment:
Ff = 2400 x 0.49 = 1176
Fw = 3258 x 0.207 = 674
Mu = 4192 kNm > 3154 kNm hence O.K.
The ground investigation report shows suitable founding strata about 9.5m below the proposed road level. Test results show the founding strata to be a cohesionless soil having an angle of shearing resistance (φ) = 30o and a safe bearing capacity of 400kN/m2.
Backfill material will be Class 6N with an effective angle of internal friction (ϕ') = 35o and density (γ) = 19kN/m3.
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.
Alternatively using BS 5400 Part 9.1 Clause 5.14.2.6:
Using the Ekspan bearing EKR35
This correllates with the value obtained above using the shear stiffness from the manufacturer's data sheet.
BS 5400 Part 2 - Clause 5.4.7.3:
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 +37oC respectively.
For a Group 4 type strucutre (see fig. 9) the corresponding minimum and maximum effective bridge temperatures are -11 and +36oC from tables 10 and 11.
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 = 11.3mm.
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 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 -19 to +37oC which would require the bearings to be installed at a shade air temperature of 9oC to achieve the ± 8mm movement. If the bearings are set at a maximum shade air temperature of 16oC then, by proportion the deck will expand 6mm and contract 10mm. Let us assume that this maximum shade air temperature of 16oC for fixing the bearings is specified in the Contract and design the abutments accordingly.
Total horizontal load on each abutment = 11 x 85 = 935 kN ≡ 935 / 11.6 = 81kN/m.
H = AGδr/tq
· 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
Option 2 - Sliding Bearing:
With a maximum ultimate reaction of 790kN and longitudinal movement of ± 8mm then a suitable bearing from the Ekspan EA Series would be /80/210/25/25:
Average nominal dead load reaction = (1900 + 320) / 11 = 2220 / 11 = 200kN
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.
When this load is applied on the deck it will act on the fixed abutment only.
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
Assume a surchage loading for the compaction plant to be equivalent to 30 units of HB
1) Stability Check
Initial Sizing for Base Dimensions
Alternatively a simple spreadsheet will achieve a result by trial and error.
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 between the sliding faces will be in the order of 5N/mm2.
From Table3 of BS 5400 Part 9.1 the Coefficient of friction = 0.08 for a bearing stress of 5N/mm2
Hence total horizontal load on each abutment when the deck expands or contracts = 2220 x 0.08 = 180kN ≡ 180 / 11.6 = 16kN/m.
Traction and Braking Load - BS 5400 Part 2 Clause 6.10:
Braking load on 1m width of abutment = 450 / 11.6 = 39kN/m.
Skidding Load - BS 5400 Part 2 Clause 6.11:
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
Hence Compaction Plant surcharge = 12 kN/m2.
For surcharge of w kN/m2 :
Fs = Ka w h = 0.27wh kN/m
There are a number of publications that will give guidance on base sizes for free standing cantilever walls, Reynolds's Reinforced Concrete Designer's Handbook being one such book.
Backfill + HA surcharge + Deck dead load + HA on deck + Braking on deck
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
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
Surcharge 52 4.25
∑ = 906 ∑ =3251
Overturning Effects:
F Lever Arm Moment About A
Backfill 144 2.5
Surcharge 24 3.75
∑ =168 ∑ =452
Backfill + Construction surcharge
Backfill + Construction
Sackfill + HA surcharge + Deck dead load + Deck contraction
Backfill + HA surcharge +
Backfill + HB surcharge + Deck
Backfill + HA surcharge + Deck
CASE 1 - Fixed Abutment
Density of reinforced concrete = 25kN/m3.
Backfill Force Fb = 0.27 x 19 x 7.52 / 2 = 144kN/m
For sliding effects:
Active Force = Fb + Fs = 168kN/m
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 the centre of the base.
P = 906kN/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)
Hence the abutment will be stable for Case 1.
Fixed Abutment:
F of S Overturning F of S Sliding Bearing Pressure at Toe
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
Factor of Safety Against Overturning = 3251 / 452 = 7.2 > 2.0 ∴ OK.
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.
A = 6.4m2/m
Z = 6.42 / 6 = 6.827m3/m
Pressure under toe = 142 + 15 = 157kN/m2 < 400kN/m2 ∴ OK.
Pressure under heel = 142 - 15 = 127kN/m2
Analysing the fixed abutment with Load Cases 1 to 6 and the free abutment with Load Cases 1 to 5 using a simple spreadsheet the following results were obtained:
Case 1 7.15 3.09
Case 2 2.91 2.14
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 abutments. We shall assume that there are no specific requirements for using elastomeric bearings and design the abutments for the lesser load effects by using sliding bearings.
2) Wall and Base Design
Loads on the back of the wall are calculated using 'at rest' earth pressures. Serviceability and Ultimate load effects need to be calculated for the load cases 1 to 6 shown above. Again, these are best carried out using a simple spreadsheet.
Using the Fixed Abutment Load Case 1 again as an example of the calculations:
Wall Design
Serviceability = 1.0
Ultimate = 1.5
Surcharge Force Fs on the rear of the wall = 0.426 x 12 x 6.5 = 33kN/m
At the base of the Wall:
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
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:
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
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 simple spreadsheet the following results were obtained for the design moments and shear at the base of the wall:
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
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
Bending
Shear
Use strength class C32/40 with water-cement ratio 0.5 and minimum cement content of 340kg/m3 for exposure condition XD2.
Nominal cover to reinforcement = 60mm (45mm minimum cover plus a tolerance Δc of 15mm).Reinforcement to BS 4449:2005 Grade B500B: fy = 500N/mm2
Design for critical moments and shear in Free Abutment:
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
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 ∴ serviceability requirements are satisfied.
Shear requirements are designed to BS 5400 clause 5.4.4:
ULS shear at Section 7H/8 for load case 4 = 487 kN
Hence height requiring strengthening = 1.073 x (0.673 - 0.62) / (0.673 - 0.53) = 0.4m < d.
Early Thermal Cracking
Base Design
Maximum bending and shear effects in the base slab will occur at sections near the front and back of the wall. Different load factors are used for serviceability and ultimate limit states so the calculations need to be carried out for each limit state using 'at rest pressures'
Using the Fixed Abutment Load Case 1 again as an example of the calculations:
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
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
Surcharge 52 4.25 221
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 at distance H/8 (8.63 / 8 = 1.079m) up the wall.
v = V / (bd) = 487 x 103 / (1000 x 920) = 0.53 N/mm2 < 0.62
Provide a 500 x 500 splay at the base of the wall with B32 @ 150c/c bars in sloping face.
Considering the effects of casting the wall stem onto the base slab by complying with the early 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 wall.
Minimum area of secondary reinforcement to Clause 5.8.4.2 = 0.12% of bad = 0.0012 x 1000 x 920 = 1104 mm2/m (use B16 @ 150c/c - As = 1340mm2/m)
CASE 1 - Fixed Abutment Serviceability Limit State
γfL = 1.0 γf3 = 1.0
B/fill Force Fb = 0.426 x 19 x 7.52 x 1.0 / 2 = 228kN/m
Bearing Pressure at toe and heel of base slab = (P / A) ± (P x e / Z)
P = 906kN/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)
A = 6.4m2/m
Z = 6.42 / 6 = 6.827m3/m
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 bottom face).
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 4.3 / 2) = -443kNm/m (tension in top face).
CASE 1 - Fixed Abutment Ultimate Limit State
γfL for concrete = 1.15
γfL for fill and surcharge(vetical) = 1.2
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
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
Surcharge 62 4.25
∑ = 1070 ∑ =
Overturning Effects:
F Lever Arm Moment About A
Backfill 341 2.5
Surcharge 58 3.75
∑ = 399 ∑ =
γfL for fill and surcharge(horizontal) = 1.5
Backfill Force Fb = 0.426 x 19 x 7.52 x 1.5 / 2 = 341kN/m
Bearing Pressure at toe and heel of base slab = (P / A) ± (P x e / Z)
P = 1070kN/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)
ULS Shear at a-a = 1.1 x {[(260 + 228) x 1.1 / 2] - (1.15 x 1.1 x 25)} = 260kN/m
/ 2) - (62 x 4.3 / 2)} = -769kNm/m (tension in top face).
Fixed Abutment Base:
Section a-a Section b-b
ULSShear SLSMoment ULSMomenULS 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
A = 6.4m2/m
Z = 6.42 / 6 = 6.827m3/m
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 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)} = 148kNm/m (tension in bottom face).
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
Analysing the fixed abutment with Load Cases 1 to 6 and the free abutment with Load Cases 1 to 5 using a simple spreadsheet the following results were obtained:
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
Use B32 @ 150 c/c:
For the Serviceability check for Case 3 an approximation of the dead load moment can be obtained by removing the surcharge and braking loads. The spreadsheet result gives the dead load SLS moment for Case 3 as 723kNm, thus the live load moment = 1233 - 723 = 510kNm.
This could be corrected by reducing the bar spacing, but increase the bar size to B40@150 c/c as this is required to avoid the use of links (see below).
Shear
Shear on Toe - Use Fixed Abutment Load Case 6:
By inspection B32@150c/c will be adequate for the
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
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
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.
Carrying out the crack control calculation to Clause 5.8.8.2 gives a crack width of 0.27mm > 0.25mm ∴ Fail.
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 ∴ serviceability requirements are satisfied.
bending effects in the toe (Muls = 365kNm < 1983kNm)
Rather than provide shear reinforcement try increasing bars to B40 @ 150 c/c (also required for crack control as shown above).
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
Considering the effects of casting the base slab onto the blinding concrete by complying with the early thermal cracking of concrete to BD 28 then B16 distribution bars @ 250 c/c will be required.
This wall is designed to be cast onto the top of the abutment after the deck has been built. Loading will be applied from the backfill, surcharge and braking loads on top of the wall.
HB braking load to BS 5400 clause 6.10 = 25% x 45units x 10kN on each axle = 112.5kN per axle.
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
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
Minimum area of main reinforcement to Clause 5.8.4.1 = 0.15% of bad = 0.0015 x 1000 x 924 = 1386 mm2/m (use B20 @ 200c/c - As = 1570mm2/m).
Assume a 45o dispersal to the curtain wall and a maximum dispersal of the width of the abutment (11.6m) then:
Horizontal load due to backfill = 0.426 x 19 x 3.02 / 2 = 36.4 kN/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
Since concrete is weak in tension in normal reinforced concrete construction cracks develop in the tension zone at working loads and therefore all concrete in tension is ignored in design.
Prestressing involves inducing compressive stresses in the zone which will tend to become tensile under external loads. This compressive stress neutralizes the tensile stress so that no resultant tension exists, (or only very small values, within the tensile strength of the concrete). Cracking is therefore eliminated under working load and all of the concrete may be assumed effective in carrying load. Therefore lighter sections may be used to carry a given bending moment, and prestressed concrete may be used over much longer spans than reinforced concrete.
The prestressing force has to be produced by a high tensile steel, and it is necessary to use high quality concrete to resist the higher compressive stresses that are developed.
When the concrete has cured the stressing force is released and the tendons anchor themselves in the concrete.
Cable ducts and reinforcement are positioned in the beam mould. The ducts are usually raised towards the neutral axis at the ends to reduce the eccentricity of the stressing force.
Tendons are threaded through the cable ducts and tensioned to about 70% of their ultimate strength.
Wedges are inserted into the end anchorages and the tensioning force on the tendons is released. Grout is then pumped into the ducts to protect the tendons.
When the tensioning force is released and the tendons are anchored to the concrete a series of effects result in a loss of stress in the tendons. The effects are:
The prestressing force also reduces the magnitude of the principal tensile stress in the web so that thin-webbed I - sections may be used without the risk of diagonal tension failures and with further savings in self-weight.
Both methods involve tensioning cables inside a concrete beam and then anchoring the stressed cables to the concrete.
e. other causes in special circumstances , such as when steam curing is used with pre-tensioning.
= 3.6 kN/m
= 10.78 kN/m
= 2.4 kN/m
= 10 kN/m + 33kN
= 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 ULS
Comb.3 Comb.1 Comb.3
1.0 1.15 1.15
Design a simply supported prestressed concrete Y beam which carries a 150mm thick concrete slab and 100mm of surfacing, together with a nominal live load udl of 10.0 kN/m2 and kel of 33kN/m . The span of the beam is 24.0m centre to centre of bearings and the beams are spaced at 1.0m intervals.
1.2 1.75 1.75
1.0 1.5 1.25
- - -
0.8 - 1.0
Modular ratio effect for different concrete strengths between beam and slab may be ignored.
Composite Section
5.99E+05
623
###
###
2.42E+08
###
Apply temperature differences given in BS 5400 Pt2 Fig.9 (Group 4)to a simplified beam section.
= 261.5 - 26.7 = 234.8 kNm
= - 385.9 - 19.3 - 295.1 = - 700.3 kN
- 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
Assume that 2/3 of the total shrinkage of the precast concrete takes place before the deck slab is cast and that the residual shrinkage is 100 x 10-6 ,
Force to restrain differential shrinkage : F = - εdiff x Ecf x Acf x φ
Self weight of beam and weight of deck slab is supported by the beam. When the deck slab concrete has cured then any further loading (superimposed and live loads) is supported by the composite section of the beam and slab.
= [(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
(HA) (HB)
= [(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
-0.60 -0.60
-16.71 -17.67*
The critical section at transfer occurs at the end of the transmission zone. The moment due to the self weight at this section is near zero and initial stress conditions are:
.....................(eqn. 1)
.....................(eqn. 2)
= 4888 / 0.9 = 5431kN
Using 15.2mm class 2 relaxation standard strand at maximum initial force of 174kN (0.75 x Pu)
Arrange 32 tendons symmetrically about the Y-Y axis to achieve an eccentricity of about 216mm.
Allowing for 1% relaxation loss in steel before transfer and elastic deformation of concrete at transfer :
/ 52.905 x 109) / 31 ]
P = 0.91 Po = 0.91 x 5568 = 5067 kN
Final stresses due to prestress after all loss of prestress at :
Combined stresses in final condition for worst effects of design loads, differential shrinkage and temperature difference :
Allowing for 2% relaxation loss in steel after transfer, concrete shrinkage εcs = 300 x 10-6
) = 300 x 10-6 x 196 x 32 x 139 = 262
) = 1.03 x 48 x 10-6 x 12.76 x 196 x 32 x 139 = 550
Mu / M = 4192 / 3154 = 1.33 ( > 1.15 ) hence strain in outermost tendon O.K.
Design the fixed and free end cantilever abutments to the 20m span deck shown to carry HA and 45 units of HB loading. 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).
The ground investigation report shows suitable founding strata about 9.5m below the proposed road level. Test results show the founding strata to be a cohesionless soil having an angle of shearing resistance (φ) = 30o and a safe bearing capacity of 400kN/m2.
') = 35o and density (γ) = 19kN/m3.
This is an ultimate load hence the nominal horizontal load = 121 / 1.1 / 1.3 = 85kN at each bearing.
This correllates with the value obtained above using the shear stiffness from the manufacturer's data sheet.
From BS 5400 Part 2 Figures 7 and 8 the minimum and maximum shade air temperatures are -19 and +37oC respectively.
For a Group 4 type strucutre (see fig. 9) the corresponding minimum and maximum effective bridge temperatures are -11 and +36oC from tables 10 and 11.
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 = 11.3mm.
] = ±[11.3 x 1.1 x 1.3 /2] = ± 8mm.
With a maximum ultimate reaction = 230 + 60 + 500 = 790kN then a suitable elastomeric bearing would be Ekspan's Elastomeric Pad Bearing EKR35:
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 -19 to +37oC which would require the bearings to be installed at a shade air temperature of 9oC to achieve the ± 8mm movement. If the bearings are set at a maximum shade air temperature of 16oC then, by proportion the deck will expand 6mm and contract 10mm. Let us assume that this maximum shade air temperature of 16oC for fixing the bearings is specified in the Contract and design the abutments accordingly.
With a maximum ultimate reaction of 790kN and longitudinal movement of ± 8mm then a suitable bearing from the Ekspan EA Series would be /80/210/25/25:
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.
From Table3 of BS 5400 Part 9.1 the Coefficient of friction = 0.08 for a bearing stress of 5N/mm2
Hence total horizontal load on each abutment when the deck expands or contracts = 2220 x 0.08 = 180kN ≡ 180 / 11.6 = 16kN/m.
There are a number of publications that will give guidance on base sizes for free standing cantilever walls, Reynolds's Reinforced Concrete Designer's Handbook being one such book.
Check bearing pressure at toe and heel of base slab = (P / A) ± (P x e / Z) where P x e is the moment about the centre of the base.
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
Frictional force on underside of base resisting movement = W tan(φ) = 906 x tan(30o) = 523kN/m
Analysing the fixed abutment with Load Cases 1 to 6 and the free abutment with Load Cases 1 to 5 using a simple spreadsheet the following results were obtained:
168 120
388 7
318 78
354 42
325 84
365 82
It can be seen that the use of elastomeric bearings (Case 2) will govern the critical design load cases on the abutments. We shall assume that there are no specific requirements for using elastomeric bearings and design the abutments for the lesser load effects by using sliding bearings.
Loads on the back of the wall are calculated using 'at rest' earth pressures. Serviceability and Ultimate load effects need to be calculated for the load cases 1 to 6 shown above. Again, these are best carried out using a simple spreadsheet.
Shear
ULS
337
566
596
602
543
599
Shear
= 1.0 for serviceability and 1.1 for ultimate (from BS 5400 Part 4 Clauses 4.2.2 and 4.2.3)
Analysing the fixed abutment with Load Cases 1 to 6 and the free abutment with Load Cases 1 to 5 using a simple spreadsheet the following results were obtained for the design moments and shear at the base of the wall:
ULS
350
581
612
619
559
Use strength class C32/40 with water-cement ratio 0.5 and minimum cement content of 340kg/m3 for exposure condition XD2.
Nominal cover to reinforcement = 60mm (45mm minimum cover plus a tolerance Δc of 15mm).Reinforcement to BS 4449:2005 Grade B500B: fy = 500N/mm2
> 2175kNn/m ∴ OK
Also the steel reinforcement and concrete stresses meet the limitations required in clause 4.1.1.3 ∴ serviceability requirements are satisfied.
Maximum bending and shear effects in the base slab will occur at sections near the front and back of the wall. Different load factors are used for serviceability and ultimate limit states so the calculations need to be carried out for each limit state using 'at rest pressures'
1/3 x (40)1/3 = 0.72
< 0.673 hence shear reinforcement should be provided, however check shear at distance H/8 (8.63 / 8 = 1.079m) up the wall.
Considering the effects of casting the wall stem onto the base slab by complying with the early 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 wall.
d = 0.0012 x 1000 x 920 = 1104 mm2/m (use B16 @ 150c/c - As = 1340mm2/m)
/ 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
SLS MomeULS Moment
447 768
980 1596
1178 1834
1003 1700
853 1402
1098 1717
/ 3) - (1.15 x 25 x 1.0 x 1.12 / 2)} = 148kNm/m (tension in bottom face).
Analysing the fixed abutment with Load Cases 1 to 6 and the free abutment with Load Cases 1 to 5 using a simple spreadsheet the following results were obtained:
SLSMoment ULSMoment
475 816
1029 1678
1233 1922
1055 1786
901 1480
For the Serviceability check for Case 3 an approximation of the dead load moment can be obtained by removing the surcharge and braking loads. The spreadsheet result gives the dead load SLS moment for Case 3 as 723kNm, thus the live load moment = 1233 - 723 = 510kNm.
This could be corrected by reducing the bar spacing, but increase the bar size to B40@150 c/c as this is required to avoid the use of links (see below).
> 1922kNm/m ∴ OK
Carrying out the crack control calculation to Clause 5.8.8.2 gives a crack width of 0.27mm > 0.25mm ∴ Fail.
Also the steel reinforcement and concrete stresses meet the limitations required in clause 4.1.1.3 ∴ serviceability requirements are satisfied.
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)} = 559kN
1/3 x (40)1/3 = 0.62
1/3 x (40)1/3 = 0.62
(also required for crack control as shown above).
1/3 x (40)1/3 = 0.716
Considering the effects of casting the base slab onto the blinding concrete by complying with the early thermal cracking of concrete to BD 28 then B16 distribution bars @ 250 c/c will be required.
This wall is designed to be cast onto the top of the abutment after the deck has been built. Loading will be applied from the backfill, surcharge and braking loads on top of the wall.
HB braking load to BS 5400 clause 6.10 = 25% x 45units x 10kN on each axle = 112.5kN per axle.
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
116.020 166.156
d = 0.0015 x 1000 x 924 = 1386 mm2/m (use B20 @ 200c/c - As = 1570mm2/m).
dispersal to the curtain wall and a maximum dispersal of the width of the abutment (11.6m) then:
89.066 242.424
179.402
Since concrete is weak in tension in normal reinforced concrete construction cracks develop in the tension zone at working loads and therefore all concrete in tension is ignored in design.
Prestressing involves inducing compressive stresses in the zone which will tend to become tensile under external loads. This compressive stress neutralizes the tensile stress so that no resultant tension exists, (or only very small values, within the tensile strength of the concrete). Cracking is therefore eliminated under working load and all of the concrete may be assumed effective in carrying load. Therefore lighter sections may be used to carry a given bending moment, and prestressed concrete may be used over much longer spans than reinforced concrete.
- sections may be used without the risk of diagonal tension failures and with further savings in self-weight.
Design a simply supported prestressed concrete Y beam which carries a 150mm thick concrete slab and 100mm of surfacing, together with a nominal live load udl of 10.0 kN/m2 and kel of 33kN/m . The span of the beam is 24.0m centre to centre of bearings and the beams are spaced at 1.0m intervals.
Self weight of beam and weight of deck slab is supported by the beam. When the deck slab concrete has cured then any further loading (superimposed and live loads) is supported by the composite section of the beam and slab.
Design the fixed and free end cantilever abutments to the 20m span deck shown to carry HA and 45 units of HB loading. 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).
The ground investigation report shows suitable founding strata about 9.5m below the proposed road level. Test results show the founding strata to be a cohesionless soil having an angle of shearing resistance (φ) = 30o and a safe bearing capacity of 400kN/m2.
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 -19 to +37oC which would require the bearings to be installed at a shade air temperature of 9oC to achieve the ± 8mm movement. If the bearings are set at a maximum shade air temperature of 16oC then, by proportion the deck will expand 6mm and contract 10mm. Let us assume that this maximum shade air temperature of 16oC for fixing the bearings is specified in the Contract and design the abutments accordingly.
It can be seen that the use of elastomeric bearings (Case 2) will govern the critical design load cases on the abutments. We shall assume that there are no specific requirements for using elastomeric bearings and design the abutments for the lesser load effects by using sliding bearings.
Loads on the back of the wall are calculated using 'at rest' earth pressures. Serviceability and Ultimate load effects need to be calculated for the load cases 1 to 6 shown above. Again, these are best carried out using a simple spreadsheet.
Analysing the fixed abutment with Load Cases 1 to 6 and the free abutment with Load Cases 1 to 5 using a simple spreadsheet the following results were obtained for the design moments and shear at the base of the wall:
Maximum bending and shear effects in the base slab will occur at sections near the front and back of the wall. Different load factors are used for serviceability and ultimate limit states so the calculations need to be carried out for each limit state using 'at rest pressures'
Considering the effects of casting the wall stem onto the base slab by complying with the early 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 wall.
For the Serviceability check for Case 3 an approximation of the dead load moment can be obtained by removing the surcharge and braking loads. The spreadsheet result gives the dead load SLS moment for Case 3 as 723kNm, thus the live load moment = 1233 - 723 = 510kNm.
Prestressing involves inducing compressive stresses in the zone which will tend to become tensile under external loads. This compressive stress neutralizes the tensile stress so that no resultant tension exists, (or only very small values, within the tensile strength of the concrete). Cracking is therefore eliminated under working load and all of the concrete may be assumed effective in carrying load. Therefore lighter sections may be used to carry a given bending moment, and prestressed concrete may be used over much longer spans than reinforced concrete.
and kel of 33kN/m . The span of the beam is 24.0m centre to centre of bearings and the beams are spaced at 1.0m intervals.
C to achieve the ± 8mm movement. If the bearings are set at a maximum shade air temperature of 16oC then, by proportion the deck will expand 6mm and contract 10mm. Let us assume that this maximum shade air temperature of 16oC for fixing the bearings is specified in the Contract and design the abutments accordingly.
It can be seen that the use of elastomeric bearings (Case 2) will govern the critical design load cases on the abutments. We shall assume that there are no specific requirements for using elastomeric bearings and design the abutments for the lesser load effects by using sliding bearings.
For the Serviceability check for Case 3 an approximation of the dead load moment can be obtained by removing the surcharge and braking loads. The spreadsheet result gives the dead load SLS moment for Case 3 as 723kNm, thus the live load moment = 1233 - 723 = 510kNm.
Prestressing involves inducing compressive stresses in the zone which will tend to become tensile under external loads. This compressive stress neutralizes the tensile stress so that no resultant tension exists, (or only very small values, within the tensile strength of the concrete). Cracking is therefore eliminated under working load and all of the concrete may be assumed effective in carrying load. Therefore lighter sections may be used to carry a given bending moment, and prestressed concrete may be used over much longer spans than reinforced concrete.
C to achieve the ± 8mm movement. If the bearings are set at a maximum shade air temperature of 16oC then, by proportion the deck will expand 6mm and contract 10mm. Let us assume that this maximum shade air temperature of 16oC for fixing the bearings is specified in the Contract and design the abutments accordingly.
Prestressing involves inducing compressive stresses in the zone which will tend to become tensile under external loads. This compressive stress neutralizes the tensile stress so that no resultant tension exists, (or only very small values, within the tensile strength of the concrete). Cracking is therefore eliminated under working load and all of the concrete may be assumed effective in carrying load. Therefore lighter sections may be used to carry a given bending moment, and prestressed concrete may be used over much longer spans than reinforced concrete.
C then, by proportion the deck will expand 6mm and contract 10mm. Let us assume that this maximum shade air temperature of 16oC for fixing the bearings is specified in the Contract and design the abutments accordingly.
Prestressing involves inducing compressive stresses in the zone which will tend to become tensile under external loads. This compressive stress neutralizes the tensile stress so that no resultant tension exists, (or only very small values, within the tensile strength of the concrete). Cracking is therefore eliminated under working load and all of the concrete may be assumed effective in carrying load. Therefore lighter sections may be used to carry a given bending moment, and prestressed concrete may be used over much longer spans than reinforced concrete.
C for fixing the bearings is specified in the Contract and design the abutments accordingly.
