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Nominal Reaction(kN)
2770 Nominal loading on 1m length of abutment:
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.
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 [(37+19)/2 -19] = 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 8x(37-16)/[(37+19)/2] = 6mm and contract 8x(16+19)/[(37+19)/2] = 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.
Horizontal load at bearing for 10mm contraction = 12.14 x 10 = 121kN.
Total horizontal load on each abutment = 11 x 85 = 935 kN ≡ 935 / 11.6 = 81kN/m.
H = 256200 x 0.9 x 10-3 x 10 / 19 = 121kN
Contact pressure under base plate = 200000 / (210 x 365) = 3N/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 HA = 8kN/m x 20m + 250kN = 410kN
Nominal Load for HB = 25% of 45units x 10kN x 4axles = 450kN
When this load is applied on the deck it will act on the fixed abutment only. Skidding Load - BS 5400 Part 2 Clause 6.11:
Active Pressure at depth h = 0.27 x 19 x h = 5.13h kN/m2
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/m
Weight of base = 6.4 x 1.0 x 25 = 160kN/m
Weight of backfill = 4.3 x 6.5 x 19 = 531kN/m
Weight of surcharge = 4.3 x 12 = 52kN/m
Backfill Force Fb = 0.27 x 19 x 7.52 / 2 = 144kN/m
Surcharge Force Fs = 0.27 x 12 x 7.5 = 24 kN/m Restoring Effects:
∑ = 906 ∑ = 3251
∑ = 452
Frictional force on underside of base resisting movement = W tan(φ) = 906 x tan(30o) = 523kN/m
Factor of Safety Against Sliding = 523 / 168 = 3.1 > 2.0 ∴ OK. Bearing Pressure:
Check bearing pressure at toe and heel of base slab = (P / A) ± (P x e / Z) where P x e is the moment about the centre of the base.
Pressure under base = (906 / 6.4) ± (906 x 0.111 / 6.827)
Pressure under heel = 142 - 15 = 127kN/m2 Hence the abutment will be stable for Case 1. Analysing the fixed abutment with Load Cases 1 to 6 and the free abutment with Load Cases 1 to 5 using a simple spreadsheet the following results were obtained:
F of SOverturning
F of SSliding
BearingPressure at Toe
BearingPressure at Heel
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.
γf3 = 1.0 for serviceability and 1.1 for ultimate (from BS 5400 Part 4 Clauses 4.2.2 and 4.2.3) Backfill Force Fb on the rear of the wall = 0.426 x 19 x 6.52 / 2 = 171kN/m
Surcharge Force Fs on the rear of the wall = 0.426 x 12 x 6.5 = 33kN/m
At the base of the Wall:
Serviceability moment = (171 x 6.5 / 3) + (33 x 6.5 / 2) = 371 + 107 = 478kNm/m
Ultimate moment = 1.1 x 1.5 x 478 = 789kNm/m
Ultimate shear = 1.1 x 1.5 x (171 + 33) = 337kN/m
MomentSLS Dead
MomentSLS Live
MomentULS
ShearULS
559 Concrete to BS 8500:2006
Reinforcement to BS 4449:2005 Grade B500B: fy = 500N/mm2 Design for critical moments and shear in Free Abutment: Reinforced concrete walls are designed to BS 5400 Part 4 Clause 5.6.
0.1fcuAc = 0.1 x 40 x 103 x 11.6 x 1 = 46400 kN > 5770 ∴ design as a slab in accordance with clause 5.4 Bending BS 5400 Part 4 Clause 5.4.2 → for reisitance moments in slabs design to clause 5.3.2.3:
Use B40 @ 150 c/c:
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 Shear requirements are designed to BS 5400 clause 5.4.4:
v = V / (bd) = 619 x 103 / (1000 x 920) = 0.673 N/mm2
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
Hence height requiring strengthening = 1.073 x (0.673 - 0.62) / (0.673 - 0.53) = 0.4m < d.
Provide a 500 x 500 splay at the base of the wall with B32 @ 150c/c bars in sloping face. Early Thermal Cracking Considering the effects of casting the wall stem onto the base slab by complying with the early thermal cracking of concrete to 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) 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: CASE 1 - Fixed Abutment Serviceability Limit State
Weight of wall stem = 1.0 x 6.5 x 25 x 1.0 = 163kN/m
Weight of base = 6.4 x 1.0 x 25 x 1.0 = 160kN/m
Weight of backfill = 4.3 x 6.5 x 19 x 1.0 = 531kN/m
Weight of surcharge = 4.3 x 12 x 1.0 = 52kN/m
B/fill Force Fb = 0.426 x 19 x 7.52 x 1.0 / 2 = 228kN/m
Surcharge Force Fs = 0.426 x 12 x 7.5 x 1.0 = 38 kN/m Restoring Effects:
∑ = 266
∑ = 713
Pressure under base = (906 / 6.4) ± (906 x 0.399 / 6.827)
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).
Weight of wall stem = 1.0 x 6.5 x 25 x 1.15 = 187kN/m
Weight of base = 6.4 x 1.0 x 25 x 1.15 = 184kN/m
Weight of backfill = 4.3 x 6.5 x 19 x 1.2 = 637kN/m
Weight of surcharge = 4.3 x 12 x 1.2 = 62kN/m
Backfill Force Fb = 0.426 x 19 x 7.52 x 1.5 / 2 = 341kN/m
Surcharge Force Fs = 0.426 x 12 x 7.5 x 1.5 = 58 kN/m Restoring Effects:
∑ = 1070 ∑ = 3859
∑ = 399
∑ = 1071
Pressure under base = (1070 / 6.4) ± (1070 x 0.594 / 6.827)
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
ULS Shear at a-a = 1.1 x {[(260 + 228) x 1.1 / 2] - (1.15 x 1.1 x 25)} = 260kN/m
ULS Shear at b-b = 1.1 x {[(199 + 74) x 4.3 / 2] - (1.15 x 4.3 x 25) - 637 - 62} = 259kN/m
ULS Moment at a-a = 1.1 x {(228 x 1.12 / 2) + ([260 - 228] x 1.12 / 3) - (1.15 x 25 x 1.0 x 1.12 / 2)} = 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 / 2) - (62 x 4.3 / 2)} = -769kNm/m (tension in top face).
1480 Design for shear and bending effects at sections a-a and b-b for the Free Abutment: Bending BS 5400 Part 4 Clause 5.7.3 → design as a slab for reisitance moments to clause 5.3.2.3:
z = {1 - [ 1.1 x 500 x 5362) / (40 x 1000 x 924) ]} d = 0.92d < 0.95d ∴ OK
Mu = (0.87fy)Asz = 0.87 x 500 x 5362 x 0.92 x 924 x 10-6 = 1983kNm/m > 1922kNm/m ∴ OK
(1983kNm/m also > 1834kNm/m ∴ B32 @ 150 c/c suitable for fixed abutment. For the Serviceability check for Case 3 an approximation of the dead load moment can be obtained by removing 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.
Also the steel reinforcement and concrete stresses meet the limitations required in clause 4.1.1.3 ∴ serviceability requirements are satisfied. Shear
v = V / (bd) = 112 x 103 / (1000 x 924) = 0.121 N/mm2
vc = (0.27/γm)(100As/bwd)1/3(fcu)1/3 = (0.27 / 1.25) x ({100 x 5362} / {1000 x 924})1/3 x (40)1/3 = 0.62
ξsvc = 0.86 x 0.62 = 0.53 N/mms > 0.121N/mms ∴ OK Shear on Heel - Use Free Abutment Load Case 3:
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
v = V / (bd) = 559 x 103 / (1000 x 924) = 0.605 N/mm2
ξsvc = 0.86 x 0.62 = 0.53 N/mms < 0.605N/mms ∴ Fail
vc = (0.27/γm)(100As/bwd)1/3(fcu)1/3 = (0.27 / 1.25) x ({100 x 8378} / {1000 x 920})1/3 x (40)1/3 = 0.716
ξsvc = 0.86 x 0.716 = 0.616 N/mms > 0.605N/mms ∴ OK
Early Thermal Cracking Considering the effects of casting the base slab onto the blinding concrete by complying with the early thermal cracking of 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). Local Effects Curtain Wall
HB braking load to BS 5400 clause 6.10 = 25% x 45units x 10kN on each axle = 112.5kN per axle.
Assume a 45o dispersal to the curtain wall and a maximum dispersal of the width of the abutment (11.6m) then:
1st axle load on back of abutment = 112.5 / 3.0 = 37.5kN/m
2nd axle load on back of abutment = 112.5 / 6.6 = 17.0kN/m
3rd & 4th axle loads on back of abutment = 2 x 112.5 / 11.6 = 19.4kN/m
Horizontal load due to HB surcharge = 0.426 x 20 x 3.0 = 25.6 kN/m
Horizontal load due to backfill = 0.426 x 19 x 3.02 / 2 = 36.4 kN/m
SLS Moment = (73.9 x 3.0) + (25.6 x 1.5) + (36.4 x 1.0) = 297 kNm/m (36 dead + 261 live)
ULS Moment = 1.1 x {(1.1 x 73.9 x 3.0) + (1.5 x 25.6 x 1.5) + (1.5 x 36.4 x 1.0)} = 392 kNm/m
ULS Shear = 1.1 x {(1.1 x 73.9) + (1.5 x 25.6) + (1.5 x 36.4)} = 192kN/m
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
Last Updated : 28/02/11For more information :