1. CONTEXTO TEÓRICO-CONCEPTUAL DEL PROBLEMA DE INVESTIGACIÓN
1.3 CATEGORÍAS DE ANÁLISIS
1.3.5.3 Finalidades y funciones de la educación inicial
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:
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:
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)
Comb.1
cl.7.4.1 Modular ratio effect for different concrete strengths between beam and slab may be ignored.
Property Beam Section Composite Section
Modulus @ Level 3(mm3)
-Temperature Difference Effects
Apply temperature differences given in BS 5400 Pt2 Fig.9 (Group 4)to a simplified beam section.
Force F to restrain temperature strain :
Moment M about centroid of section to restrain curvature due to 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
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
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
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 SL
Super. & HB live load for SL
Total load for ultimate limit
HA Design serviceability mo 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
25 units HB Design SLS mom
Design ultimate moment
Combination 3 Loading Super. & HA live load for SL
Total load for ultimate limit
cl.7.4.3.2 Compression (1.25 x Table 22)
Total Stress at Level 1 = -17.42
Hence Combination 3 is critical
Prestressing Force and Eccentricity
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:
1.25 x 0.4fcu = 25 N/mm2
Hence 32 tendons required.
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
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
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 :
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
6 @
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
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, combination 1 : f = - 0.98 + 1037 / 89.066 + 1310 / 242.424 + 1.64 = 17.71 (< 25 O.K.)
Ultimate Capacity of Beam and Deck Slab (Composite Section)
cl. 6.3.3 Only steel in the tension zone is to be considered :
Effective depth from Level 3 = 1200 - 135 = 1065mm
Concrete shrinkage = (εcs x Es x Aps ) = 300 x 10-6 x 196 x 32 x 139 = 262
Centroid of tendons in tension zone = (6x60 + 10x110 + 8x160 + 4x260) / 28 = 135mm
Assume that the maximum design stress is developed in the tendons, then : Tensile force in tendons Fp = 0.87 x 28 x 139 x 1670 x 10-3 = 5655 kN
Compressive force in concrete flange :
Let X = depth to neutral axis.
Compressive force in concrete web :
Equating forces to obtain X :
X = 659 mm
Determine depth to neutral axis by an iterative strain compatibility analysis Try X = 659 mm as an initial estimate
Width of web at this depth = 247mm
Ff = 0.4 x 40 x 1000 x 150 x 10-3 = 2400 kN
Tensile force in tendons :
4976 kN
Compressive force in concrete :
5658 kN
Fc > Ft therefore reduce depth to neutral axis and repeat the calculations.
Using a depth of 565mm will achieve equilibrium.
The following forces are obtained :
Taking Moments about the neutral axis :
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).
Fp5 = 168 Fw = 2765
Mu = 4192 kNm > 3154 kNm hence O.K.
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:
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.
Nominal Reaction(kN) Ultimate Reaction(kN)Nominal Reaction(kN) Ultimate Reaction(kN)
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.
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:
Average nominal dead load reaction = (1900 + 320) / 11 = 2220 / 11 = 200kN
Nominal Load for HA = 8kN/m x 20m + 250kN = 410kN 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.
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:
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/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:
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