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).
Vehicle collision on the abutments need not be considered as they are assumed to have sufficient mass to withstand the collision loads for global purposes (See BD 60/04 Clause 2.2).
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.
The proposed deck consists of 11No. Y4 prestressed concrete beams and concrete deck slab as shown.
Loading From the Deck
A grillage analysis gave the following reactions for the various load cases:
Critical Reaction Under One Beam
Nominal Reaction
(kN)
Ultimate Reaction
(kN)
Concrete Deck
180
230
Surfacing
30
60
HA udl+kel
160
265
45 units HB
350
500
Total Reaction on Each Abutment
Nominal Reaction
(kN)
Ultimate Reaction
(kN)
Concrete Deck
1900
2400
Surfacing
320
600
HA udl+kel
1140
1880
45 units HB
1940
2770
Nominal loading on 1m length of abutment:
Deck Dead Load = (1900 + 320) / 11.6 = 191kN/m
HA live Load on Deck = 1140 / 11.6 = 98kN/m
HB live Load on Deck = 1940 / 11.6 = 167kN/m
From BS 5400 Part 2 Figures 7 and 8 the minimum and maximum shade air temperatures are -19 and +37oC respectively.
For a Group 4 type structure (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 × 12 × 10-6 × 20 × 103 = 11.3mm.
The ultimate thermal movement in the deck will be ± [(11.3 / 2) γf3 γfL] = ±[11.3 × 1.1 × 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 :
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 8×(37-16)/[(37+19)/2] = 6mm and contract 8×(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 × 10 = 121kN.
This is an ultimate load hence the nominal horizontal load = 121 / 1.1 / 1.3 = 85kN at each bearing.
Total horizontal load on each abutment = 11 × 85 = 935 kN ≡ 935 / 11.6 = 81kN/m.
Alternatively using BS 5400 Part 9.1 Clause 5.14.2.6:
H = AGδr/tq
Using the Ekspan bearing EKR35
Shear modulus G from Table 8 = 0.9N/mm2
H = 256200 × 0.9 × 10-3 × 10 / 19 = 121kN
This correllates with the value obtained above using the shear stiffness from the manufacturer's data sheet.
Option 2 - Sliding Bearing:
With a maximum ultimate reaction of 790kN and longitudinal movement of ± 8mm then a suitable bearing from the Ekspan EA Series would be /80/210/25/25:
BS 5400 Part 2 - Clause 5.4.7.3:
Average nominal dead load reaction = (1900 + 320) / 11 = 2220 / 11 = 200kN
Contact pressure under base plate = 200000 / (210 × 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 × 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 × 20m + 250kN = 410kN
Nominal Load for HB = 25% of 45units × 10kN × 4axles = 450kN
450 > 410kN hence HB braking is critical.
Braking load on 1m width of abutment = 450 / 11.6 = 39kN/m.
When this load is applied on the deck it will act on the fixed abutment only.
Skidding Load - BS 5400 Part 2 Clause 6.11:
Nominal Load = 300kN
300 < 450kN hence braking load is critical in the longitudinal direction.
When this load is applied on the deck it will act at bearing shelf level. If sliding bearings are used then friction forces at the free end should be considered as a relieving effect on longitudinal and skidding loads and therefore ignored when designing the fixed bearing.
Loading at Rear of Abutment
Backfill
For Stability calculations use active earth pressures = Ka γ h
Ka for Class 6N material = (1-Sin35) / (1+Sin35) = 0.27
Density of Class 6N material = 19kN/m3
Active Pressure at depth h = 0.27 × 19 × h = 5.13h kN/m2
Hence Fb = 5.13h2/2 = 2.57h2kN/m
Surcharge - BS 5400 Part 2 Clause 5.8.2:
For HA loading surcharge = 10 kN/m2
For HB loading surcharge = 20 kN/m2
Assume a surchage loading for the compaction plant to be equivalent to 30 units of HB
Hence Compaction Plant surcharge = 12 kN/m2.
For surcharge of w kN/m2 :
Fs = Ka w h = 0.27wh kN/m
1) Stability Check
Initial Sizing for Base Dimensions
There are a number of publications that will give guidance on base sizes for free standing cantilever walls, Reynolds's Reinforced Concrete Designer's Handbook being one such book.
Alternatively a simple spreadsheet will achieve a result by trial and error.
Load Combinations
Backfill + Construction surcharge
Wall backfilled up to bearing shelf level only.
Backfill + 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
Backfill + HA surcharge + Deck dead load + HA on deck + Braking on deck
(Not applied to free abutment if sliding bearings are provided)
CASE 1 - Fixed Abutment
Density of reinforced concrete = 25kN/m3.
Weight of wall stem = 1.0 × 6.5 × 25 = 163kN/m
Weight of base = 6.4 × 1.0 × 25 = 160kN/m
Weight of backfill = 4.3 × 6.5 × 19 = 531kN/m
Weight of surcharge = 4.3 × 12 = 52kN/m
Backfill Force Fb = 0.27 × 19 × 7.52 / 2 = 144kN/m
Surcharge Force Fs = 0.27 × 12 × 7.5 = 24 kN/m
Restoring Effects:
Weight
Lever Arm
Moment About A
Stem
163
1.6
261
Base
160
3.2
512
Backfill
531
4.25
2257
Surcharge
52
4.25
221
∑ =
906
∑ =
3251
Overturning Effects:
F
Lever Arm
Moment About A
Backfill
144
2.5
361
Surcharge
24
3.75
91
∑ =
168
∑ =
452
BD 30 Clause 5.2.4.2 refers to CP 2: 1951 Earth retaining structures for Safety Factors.
Factor of Safety Against Overturning = 3251 / 452 = 7.2 > 2.0 ∴ OK.
For sliding effects:
Active Force = Fb + Fs = 168kN/m
Frictional force on underside of base resisting movement = W tan(φ) = 906 × 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 × e / Z) where P × e is the moment about the centre of the base.
P = 906kN/m
A = 6.4m2/m
Z = 6.42 / 6 = 6.827m3/m
Nett moment = 3251 - 452 = 2799kNm/m
Eccentricity (e) of P about centre-line of base = 3.2 - (2799 / 906) = 0.111m
Pressure under base = (906 / 6.4) ± (906 × 0.111 / 6.827)
Pressure under toe = 142 + 15 = 157kN/m2 < 400kN/m2 ∴ OK.
Pressure under heel = 142 - 15 = 127kN/m2
Hence the abutment will be stable for Case 1.
Analysing the fixed abutment with Load Cases 1 to 6 and the free abutment with Load Cases 1 to 5 using a simple spreadsheet the following results were obtained:
Fixed Abutment:
F of S
Overturning
F of S
Sliding
Bearing
Pressure at Toe
Bearing
Pressure at Heel
Case 1
7.16
3.09
156
127
Case 2
2.87
2.13
386
5
Case 2a
4.31
2.64
315
76
Case 3
3.43
2.43
351
39
Case 4
4.48
2.63
322
83
Case 5
5.22
3.17
362
81
Case 6
3.80
2.62
378
43
F of S
Overturning
F of S
Sliding
Case 1
7.16
3.09
Case 2
2.87
2.13
Case 2a
4.31
2.64
Case 3
3.43
2.43
Case 4
4.48
2.63
Case 5
5.22
3.17
Case 6
3.80
2.62
Bearing
Pressure at Toe
Bearing
Pressure at Heel
Case 1
156
127
Case 2
386
5
Case 2a
315
76
Case 3
351
39
Case 4
322
83
Case 5
362
81
Case 6
378
43
Free Abutment:
F of S
Overturning
F of S
Sliding
Bearing
Pressure at Toe
Bearing
Pressure at Heel
Case 1
7.15
3.09
168
120
Case 2
2.91
2.14
388
7
Case 2a
4.33
2.64
318
78
Case 3
3.46
2.44
354
42
Case 4
4.50
2.64
325
84
Case 5
5.22
3.16
365
82
F of S
Overturning
F of S
Sliding
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
4.50
2.64
Case 5
5.22
3.16
Bearing
Pressure at Toe
Bearing
Pressure at Heel
Case 1
168
120
Case 2
388
7
Case 2a
318
78
Case 3
354
42
Case 4
325
84
Case 5
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.
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
Ko = 1 - Sin(ϕ') = 1 - Sin(35o) = 0.426
γfL for horizontal loads due to surcharge and backfill from BS 5400 Part 2 Clause 5.8.1.2:
Serviceability = 1.0
Ultimate = 1.5
γf3 = 1.0 for serviceability and 1.1 for ultimate (from BS 5400 Part 4 Clauses 4.2.2 and 4.2.3)
Backfill Force Fb on the rear of the wall = 0.426 × 19 × 6.52 / 2 = 171kN/m
Surcharge Force Fs on the rear of the wall = 0.426 × 12 × 6.5 = 33kN/m
At the base of the Wall (tension in the rear face):
Serviceability moment = (171 × 6.5 / 3) + (33 × 6.5 / 2) = 371 + 107 = 478kNm/m
Ultimate moment = 1.1 × 1.5 × 478 = 789kNm/m
Ultimate shear = 1.1 × 1.5 × (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:
Fixed Abutment:
Moment
SLS Dead
Moment
SLS Live
Moment
ULS
Shear
ULS
Case 1
371
108
790
337
Case 2a
829
258
1771
566
Case 3
829
486
2097
596
Case 4
829
308
1877
602
Case 5
829
154
1622
543
Case 6
829
408
1985
599
Free Abutment:
Moment
SLS Dead
Moment
SLS Live
Moment
ULS
Shear
ULS
Case 1
394
112
835
350
Case 2a
868
265
1846
581
Case 3
868
495
2175
612
Case 4
868
318
1956
619
Case 5
868
159
1694
559
Concrete to BS 8500:2006
Use strength class C32/40 with water-cement ratio 0.5 and minimum cement content of 340kg/m3 for 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:
Reinforced concrete walls are designed to BS 5400 Part 4 Clause 5.6.
Check classification to clause 5.6.1.1:
Ultimate axial load in wall from deck reactions = 2400 + 600 + 2770 = 5770 kN
0.1fcuAc = 0.1 × 40 × 103 × 11.6 × 1 = 46400 kN > 5770 ∴ design as a slab in accordance with clause 5.4
Bending
BS 5400 Part 4 Clause 5.4.2 → for reisitance moments in slabs design to clause 5.3.2.3:
z = {1 - [ 1.1fyAs) / (fcubd) ]} d
Use B40 @ 150 c/c in rear face at base of wall:
As = 8378mm2/m, d = 1000 - 60 - 20 = 920mm
z = {1 - [ 1.1 × 500 × 8378) / (40 × 1000 × 920) ]} d = 0.875d < 0.95d ∴ OK
Mu = (0.87fy)Asz = 0.87 × 500 × 8378 × 0.875 × 920 × 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 × 103 / (1000 × 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) × ({100 × 8378} / {1000 × 920})1/3 × (40)1/3 = 0.72
ξsvc = 0.86 × 0.72 = 0.62 N/mm2 < 0.673 hence shear reinforcement should be provided, however check shear at distance H/8 (8.63 / 8 = 1.079m) up the wall.
ULS shear at Section 7H/8 for load case 4 = 487 kN
v = V / (bd) = 487 × 103 / (1000 × 920) = 0.53 N/mm2 < 0.62
Hence height requiring strengthening = 1.073 × (0.673 - 0.62) / (0.673 - 0.53) = 0.4m < d.
Provide a 500 × 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 × 1000 × 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
γfL = 1.0 γf3 = 1.0
Weight of wall stem = 1.0 × 6.5 × 25 × 1.0 = 163kN/m
Weight of base = 6.4 × 1.0 × 25 × 1.0 = 160kN/m
Weight of backfill = 4.3 × 6.5 × 19 × 1.0 = 531kN/m
Weight of surcharge = 4.3 × 12 × 1.0 = 52kN/m
B/fill Force Fb = 0.426 × 19 × 7.52 × 1.0 / 2 = 228kN/m
Surcharge Force Fs = 0.426 × 12 × 7.5 × 1.0 = 38 kN/m
Restoring Effects:
Weight
Lever Arm
Moment About A
Stem
163
1.6
261
Base
160
3.2
512
Backfill
531
4.25
2257
Surcharge
52
4.25
221
∑ =
906
∑ =
3251
Overturning Effects:
F
Lever Arm
Moment About A
Backfill
288
2.5
570
Surcharge
38
3.75
143
∑ =
266
∑ =
713
Bearing Pressure at toe and heel of base slab = (P / A) ± (P × e / Z)
P = 906kN/m
A = 6.4m2/m
Z = 6.42 / 6 = 6.827m3/m
Nett moment = 3251 - 713 = 2538kNm/m
Eccentricity (e) of P about centre-line of base = 3.2 - (2538 / 906) = 0.399m
Pressure under base = (906 / 6.4) ± (906 × 0.399 / 6.827)
Pressure under toe = 142 + 53 = 195kN/m2
Pressure under heel = 142 - 53 = 89kN/m2
Pressure at front face of wall = 89 + {(195 - 89) × 5.3 / 6.4} = 177kN/m2
Pressure at rear face of wall = 89 + {(195 - 89) × 4.3 / 6.4} = 160kN/m2
SLS Moment at a-a = (177 × 1.12 / 2) + ([195 - 177] × 1.12 / 3) - (25 × 1.0 × 1.12 / 2) = 99kNm/m (tension in bottom face).
SLS Moment at b-b = (89 × 4.32 / 2) + ([160 - 89] × 4.32 / 6) - (25 × 1.0 × 4.32 / 2) - (531 × 4.3 / 2) - (52 × 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
γfL for fill and surcharge(horizontal) = 1.5
Weight of wall stem = 1.0 × 6.5 × 25 × 1.15 = 187kN/m
Weight of base = 6.4 × 1.0 × 25 × 1.15 = 184kN/m
Weight of backfill = 4.3 × 6.5 × 19 × 1.2 = 637kN/m
Weight of surcharge = 4.3 × 12 × 1.2 = 62kN/m
Backfill Force Fb = 0.426 × 19 × 7.52 × 1.5 / 2 = 341kN/m
Surcharge Force Fs = 0.426 × 12 × 7.5 × 1.5 = 58 kN/m
Restoring Effects:
Weight
Lever Arm
Moment About A
Stem
187
1.6
299
Base
184
3.2
589
Backfill
637
4.25
2707
Surcharge
62
4.25
264
∑ =
1070
∑ =
3859
Overturning Effects:
F
Lever Arm
Moment About A
Backfill
341
2.5
853
Surcharge
58
3.75
218
∑ =
399
∑ =
1071
Bearing Pressure at toe and heel of base slab = (P / A) ± (P x e / Z)
P = 1070kN/m
A = 6.4m2/m
Z = 6.42 / 6 = 6.827m3/m
Nett moment = 3859 - 1071 = 2788kNm/m
Eccentricity (e) of P about centre-line of base = 3.2 - (2788 / 1070) = 0.594m
Pressure under base = (1070 / 6.4) ± (1070 × 0.594 / 6.827)
Pressure under toe = 167 + 93 = 260kN/m2
Pressure under heel = 167 - 93 = 74kN/m2
Pressure at front face of wall = 74 + {(260 - 74) × 5.3 / 6.4} = 228kN/m2
Pressure at rear face of wall = 74 + {(260 - 74) × 4.3 / 6.4} = 199kN/m2
γf3 = 1.1
ULS Shear at a-a = 1.1 × {[(260 + 228) × 1.1 / 2] - (1.15 × 1.1 × 25)} = 260kN/m
ULS Shear at b-b = 1.1 × {[(199 + 74) × 4.3 / 2] - (1.15 × 4.3 × 25) - 637 - 62} = 259kN/m
ULS Moment at a-a = 1.1 × {(228 × 1.12 / 2) + ([260 - 228] × 1.12 / 3) - (1.15 × 25 × 1.0 × 1.12 / 2)} = 148kNm/m (tension in bottom face).
ULS Moment at b-b = 1.1 × {(74 × 4.32 / 2) + ([199 - 74] × 4.32 / 6) - (1.15 × 25 × 1.0 × 4.32 / 2) - (637 × 4.3 / 2) - (62 × 4.3 / 2)} = -769kNm/m (tension in top face).
Analysing the fixed abutment with Load Cases 1 to 6 and the free abutment with Load Cases 1 to 5 using a simple spreadsheet the following results were obtained:
Fixed Abutment Base:
Section a-a
ULS Shear
SLS Moment
ULS Moment
Case 1
261
99
147
Case 2a
528
205
302
Case 3
593
235
340
Case 4
550
208
314
Case 5
610
241
348
Case 6
637
255
365
Section b-b
ULS
Shear
SLS
Moment
ULS
Moment
Case 1
259
447
768
Case 2a
458
980
1596
Case 3
553
1178
1834
Case 4
495
1003
1700
Case 5
327
853
1402
Case 6
470
1098
1717
Free Abutment Base:
Section a-a
ULS
Shear
SLS
Moment
ULS
Moment
Case 1
267
101
151
Case 2a
534
207
305
Case 3
598
236
342
Case 4
557
211
317
Case 5
616
243
351
Section b-b
ULS
Shear
SLS
Moment
ULS
Moment
Case 1
266
475
816
Case 2a
466
1029
1678
Case 3
559
1233
1922
Case 4
504
1055
1786
Case 5
335
901
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 - [ 1.1fyAs) / (fcubd) ]} d
Use B32 @ 150 c/c:
As = 5362mm2/m, d = 1000 - 60 - 16 = 924mm
z = {1 - [ 1.1 × 500 × 5362) / (40 × 1000 × 924) ]} d = 0.92d < 0.95d ∴ OK
Mu = (0.87fy)Asz = 0.87 × 500 × 5362 × 0.92 × 924 × 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.
Carrying out the crack control calculation to Clause 5.8.8.2 gives a crack width of 0.27mm > 0.25mm ∴ Fail.
This could be corrected by reducing the bar spacing, but increase the bar size to B40@150 c/c as this is required to avoid the use of links (see below).
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.
Shear
Shear on Toe - Use Fixed Abutment Load Case 6:
By inspection B32@150c/c will be adequate for the bending effects in the toe (Muls = 365kNm < 1983kNm)
Shear requirements are designed to BS 5400 clause 5.7.3.2(a) checking shear at d away from the front face of the wall to clause 5.4.4.1:
ULS Shear on toe = 1.1 × {(620 + 599) × 0.5 × 0.176 - 1.15 × 1 × 0.176 × 25} = 112kN
v = V / (bd) = 112 × 103 / (1000 × 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) × ({100 × 5362} / {1000 × 924})1/3 × (40)1/3 = 0.62
ξsvc = 0.86 × 0.62 = 0.53 N/mm2 > 0.121N/mm2 ∴ OK
Shear on Heel - Use Free Abutment Load Case 3:
Shear requirements are designed at the back face of the wall to clause 5.4.4.1:
Length of heel = (6.5 - 1.1 - 1.0) = 4.4m
ULS Shear on heel = 1.1 × {348 × 0.5 × (5.185 - 2.1) - 1.15 × 1 × 4.4 × 25 - 1.2 × 4.4 × (8.63 × 19 + 10)} = 559kN
Using B32@150 c/c then:
v = V / (bd) = 559 × 103 / (1000 × 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) × ({100 × 5362} / {1000 × 924})1/3 × (40)1/3 = 0.62
ξsvc = 0.86 × 0.62 = 0.53 N/mm2 < 0.605N/mm2 ∴ Fail
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) × ({100 × 8378} / {1000 × 920})1/3 × (40)1/3 = 0.716
ξsvc = 0.86 × 0.716 = 0.616 N/mm2 > 0.605N/mm2 ∴ 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 × 1000 × 924 = 1386 mm2/m (use B20 @ 200c/c - As = 1570mm2/m).
Local Effects
Curtain Wall
This wall is designed to be cast onto the top of the abutment after the deck has been built. Loading will be applied from the backfill, surcharge and braking loads on top of the wall.
HB braking load to BS 5400 clause 6.10 = 25% × 45units × 4 × 10kN on 2 axles = 225kN per axle.
To allow for load distribution effects assume a 45o dispersal to the curtain wall and a 45o dispersal down the wall, with maximum dispersal of the width of the abutment (11.6m).
This crude analysis will slightly underestimate the peak values in the wall below the load, but allowance can be made when designing the reinforcement to ensure there is spare capacity. Then:
1st axle load on back of abutment = 225 / 3.0 = 75kN/m
Dispersed to the base of the curtain wall = 225 / 9.0 = 25 kN/m
2nd axle load on back of abutment = 225 / 6.6 = 34.1kN/m
Dispersed to the base of the curtain wall = 225 / 11.6 = 19.4 kN/m
For load effects at the top of the curtain wall:
Maximum load on back of abutment = 75 + 34.1 = 109.1kN/m
For load effects at the base of the curtain wall:
Maximum load on back of abutment = 25 + 19.4 = 44.4kN/m
Bending and Shear at Base of 3m High Curtain Wall
Horizontal load due to HB surcharge = 0.426 × 20 × 3.0 = 25.6 kN/m
Horizontal load due to backfill = 0.426 × 19 × 3.02 / 2 = 36.4 kN/m
SLS Moment = (44.4 × 3.0) + (25.6 × 1.5) + (36.4 × 1.0) = 208 kNm/m (36 dead + 172 live)
ULS Moment = 1.1 × {(1.1 × 44.4 × 3.0) + (1.5 × 25.6 × 1.5) + (1.5 × 36.4 × 1.0)} = 285 kNm/m
ULS Shear = 1.1 × {(1.1 × 44.4) + (1.5 × 25.6) + (1.5 × 36.4)} = 156kN/m
400 thick curtain wall with B32 @ 150 c/c :
Mult = 584 kNm/m > 285 kNm/m ∴ OK
SLS Moment produces crack width of 0.14mm < 0.25 ∴ OK
ξsvc = 0.97 N/mm2 > v = 0.48 N/mm2 ∴ Shear OK
Contact David Childs