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Abutment Design Example to Eurocodes and UK National Annexes

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Design the fixed and free end cantilever abutments to the 20m span deck shown to carry Load Model 1 and vehicles SV80, SV100 and SV196 for Load Model 3. 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 large abutments need not be considered as they are assumed to have sufficient mass to withstand the collision loads for global purposes.

The ground investigation report shows suitable founding strata about 9.5m below the proposed road level and 1.5m below existing ground level. Test results show the founding strata to be a well drained, cohesionless soil having an angle of shearing resistance (φ') = 34°, a critical state angle of shearing resistance (φ') = 30° and a weight density = 19kN/m3.

Backfill material will be Class 6N with an angle of shearing resistance (φ'bf;k) = 35° and weight density (γbf;k) = 19kN/m3.

The proposed deck consists of 11No. Y4 prestressed concrete beams at 1m centres and concrete deck slab as shown.

EN 1997-1:2004 Clause 2.4.7.3.4.1(1)P - Use Design Approach 1 only for verification of resistance for structural and ground limit states in persistent and transient situations (STR and GEO).
Consider Combination 1: A1 “+” M1 “+” R1 and Combination 2: A2 “+” M2 “+” R1

A grillage analysis gave the following characteristic reactions for the various load cases:

Critical Vertical Reaction Under One Beam

Characteristic Reaction
(kN)

ULS Reaction
(kN)

Concrete Deck

180

240

Surfacing

45

60

gr1a

290

430

gr2

220

310

gr5

270

400

gr6

210

300

Total Vertical Reaction on Each Abutment

Characteristic Reaction
(kN)

A1
G;sup / γG;inf)

A2
G;sup / γG;inf)

Concrete Deck

1900

1.35 / 0.95

1.0 / 1.0

Surfacing

320

1.2 / 0.95

1.0 / 1.0

gr1a

1490

1.35 / 0

1.15 / 0

gr2

1120

1.35 / 0

1.15 / 0

gr5

1930

1.35 / 0

1.15 / 0

gr6

1470

1.35 / 0

1.15 / 0

Maximum Surfacing = 1.55 × 320 / 11.6 = 43kN/m
Minimum Surfacing = 0.6 × 320 / 11.6 = 17kN/m
gr1a on Deck = 1490 / 11.6 = 128kN/m
gr2 on Deck = 1030 / 11.6 = 89kN/m
gr5 on Deck = 1930 / 11.6 = 166kN/m
gr6 on Deck = 1470 / 11.6 = 127kN/m

From UK NA to BS EN 1991-1-5:2003 Figures NA.1 and NA.2 the minimum and maximum shade air temperatures are -17 and +34°C respectively.
For bridge deck type 3 the corresponding minimum (Te,min) and maximum (Te,max) effective bridge temperatures are -11 and +36°C from BS EN 1991-1-5:2003 Figure 6.1.
Hence the temperature range = 11 + 36 = 47°C.
Form EN 1991-1-5 Table C.1 - Coefficient of thermal expansion for a concrete deck = 10 × 10-6 per °C.
However CIRIA Report C660 ("Early-age thermal crack control in concrete") suggests that a value of 10 × 10-6 per °C is unsuitable for some of the concrete aggregates used in the UK and suggest a value of 12 × 10-6 per °C should be used if the type of aggregate has not been specified.
Hence the range of movement at the free end of the 20m span deck = 47 × 12 × 10-6 × 20 × 103 = 11.3mm.
The design thermal movement in the deck will be ± [(11.3 / 2) γF] = ±[11.3 × 1.35 /2] = ± 8mm.

Option 1 - Elastomeric Bearing:
With a maximum ultimate reaction = 240 + 60 + 430 = 730kN then a suitable elastomeric bearing would be Ekspan's Elastomeric Pad :Bearing EKR35:

• 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 -17 to +34°C which would require the bearings to be installed at a shade air temperature of [(34+17)/2-17] = 9°C to achieve the ± 8mm movement.
If the bearings are set at a maximum shade air temperature (T0) of 16°C then, by proportion the deck will expand 8×(34-16)/[(34+17)/2] = 6mm and contract 8×(16+17)/[(34+17)/2] = 10mm.
Let us assume that this maximum shade air temperature of 16°C for fixing the bearings is specified for T0 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 characteristic horizontal load = 121 / 1.35 = 90kN.
If a fixed abutment is used then the movement will take place at one end so:
Total horizontal load on each abutment = 11 × 90 = 990 kN ≡ 990 / 11.6 = 85kN/m.
If no fixed abutment is used then the movement will take place at both ends so:
Total horizontal load on each abutment = 85/2 = 43kN/m.

Option 2 - Sliding Bearing:
With a maximum ultimate reaction of 730kN and longitudinal movement of ± 8mm then a suitable bearing from the Ekspan EA Series would be /80/210/25/25:

• Base Plate A dimension = 210mm
• Base Plate B dimension = 365mm
• Movement ± X = 12.5mm

Average characteristic permanent 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.
Ekspan recommend a coefficient of friction = 0.05, however use a coefficient of friction = 0.08 for long term exposure conditions.
Hence total horizontal load on each abutment when the deck expands or contracts = 2220 × 0.08 = 180kN ≡ 180 / 11.6 = 16kN/m.

Braking and Acceleration Force - BS EN 1991-2:2003 Clause 4.4.1:
(2) Characteristic Force for LM1 = 0.6αQ1(2Q1k)+0.1αq1qq1w1L = 0.6 × 1 (2 × 300) + 0.1 × 1 × 9 × 3 × 20 = 414kN
For global effects, braking force on 1m width of abutment = 414 / 11.6 = 36kN/m.
(NA. 2.18.1) Characteristic Force for LM3 (SV196) = Qlk,s = δw = 0.25 × (165kN × 9axles + 180kN × 2axles + 100kN × 1axle) = 486kN
For global effects, braking force on 1m width of abutment = 486 / 11.6 = 42kN/m.
When this load is applied on the deck it will act at bearing shelf level, and will not affect the free abutment if sliding bearings are used.

Note: Braking forces should not be taken into account at the surfacing level of the carriageway over the backfill (See BS EN 1991-2:2003 Cl. 4.9.2)

Backfill

For Stability calculations use active earth pressures = Ka γbf;k h

For Design of Structural Members use at-rest earth pressures = K0 γbf;k h

SLS

Combination 1

Combination 2

Partial factors for soil parameters γM

1.0

1.0

1.25

φ'bf;d = tan-1[tan(φ'bf;k)/γM]

35.0°

35.0°

29.3°

Ka = (1-Sinφ'bf;d) / (1+Sinφ'bf;d)

0.271

0.271

0.343

K0 = 1-Sinφ'bf;d

0.426

0.426

0.511

Partial factors for soil weight γG (sup/inf)

1.0/1.0

1.35/0.95

1.0/1.0

Backfill density (γbf;d) = γbf;k γG;sup

19.0

25.65

19.0

Backfill density (γbf;d) = γbf;k γG;inf

19.0

18.1

19.0

Model factors γSd;K

1.0

1.2

1.2

Hap;d = γbf;dKaγSd;KZ2/2

2.575Z2kN/m

4.171Z2kN/m

3.91Z2kN/m

Surcharge - Use Horizontal Surcharge Model in PD 6694-1:2011 Figure 2:
Carriageway width = 7.3m ∴ there are 2 notional lanes of effective width Weff of 3m with 1.3m wide remaining area (see Table 4.1 of BS EN 1991-2:2003).
The vehicle model for loads on backfill behind abutments is positioned in each notional lane (see Clause NA.2.34.2 UK NA to BS EN 1991-2) ∴ the effective number of lanes (Nlane) in the surcharge model will be 2.

From PD 6694-1:2011 Table 7 :

Normal Traffic

Line Load kN/m = Hsc;F = F.Kd.Nlane/Wabut = 2×330Kd×2/11.6 = 113.79Kd

Note:   Df is used for determining the distibution of the Line Load in the wall for a metre strip analysis, but is not included in the calculation when considering the overall stability of the wall.

UDL kN/m2 = σh;ave = σh.Wlane.Nlane/Wabut = 20×Kd×3×2/11.6 = 10.34Kd

SV/196 Traffic (SV/196 lane 1 + Frequent value of Normal Traffic in Lane 2)

Line Load kN/m (at ground level) = Hsc;F = F.Kd.(1 + ψ1)/Wabut = 2×330Kd×1.75/11.6 = 99.57Kd

UDL kN/m2 = σh;ave = (σh11h2).Wlane/Wabut = (30+0.75×20)×Kd×3/11.6 = 11.64Kd

SLS

Combination 1

Combination 2

Partial Factor on Surcharge γQ

1.0

1.35

1.15

Assume Abutments are to be backfilled in accordance with the Highways Agency Manual of Contract Documents for Highway Works (MCHW), then compaction pressures due to construction vehicles are deemed to be coverered if the surcharge model in Figure 2 and Table 7 of PD 6694-1:2011 (as shown above) is employed (see PD 6694-1:2011 Clause 7.3.3).

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.

Backfill + Construction surcharge
(Not used - backfilled to MCHW - see PD 6694-1:2011 Clause 7.3.3)

Backfill + Normal Traffic Surcharge + Deck Permanent load + Deck contraction/shrinkage

Backfill + Normal Traffic Surcharge + Deck Permanent load + gr1a on deck

Backfill + SV/100 and SV/196 Surcharge + Deck Permanent load + gr1a (frequent value) on deck

Backfill + Normal Traffic Surcharge (frequent value) + Deck Permanent load + gr5 on deck

Backfill + Normal Traffic Surcharge (frequent value) + Deck Permanent load + gr2 (ψ1LM1 with braking on deck)
(Braking not applied to free abutment if sliding bearings are provided)

Backfill + Deck Permanent load + gr6 (LM3 with braking on deck)
(Braking not applied to free abutment if sliding bearings are provided)

Example of Stability Calculations:
CASE 6 - Fixed Abutment

Density of reinforced concrete = 25kN/m3.

SLSG = γQ = γSd;K = 1.0)

Weight of wall stem = γG × twall × Zwall × γconc = 1.0 × 1.0 × 6.5 × 25 = 162.5kN/m
Weight of base = γG × Wbase × Zbase × γconc = 1.0 × 6.4 × 1.0 × 25 = 160kN/m
Weight of backfill = γG × Wheel × Zheel × γbf;d = 1.0 × 4.3 × 8.5 × 19 = 694.5kN/m
Backfill Force Hap;d = γG × γSd;K × Ka × γbf;d × Z2/2 = 1.0 × 1.0 × 0.271 × 19 × 9.52 / 2 = 232kN/m
Frequent value of Surcharge UDL Force Hsc;udl = ψ1 × γQ × σh;ave × Z = 0.75 × 1.0 × (10.34 × 0.271) × 9.5 = 20 kN/m
Frequent value of Surcharge Line Load Force Hsc;F = ψ1 × γQ × Hsc;F = 0.75 × 1.0 × (113.79 × 0.271) = 23 kN/m
Deck Maximum Permanent load (concrete + surfacingmax) = γG × VDL = 1.0 × (164 + 43) = 207kN/m
Deck Minimum Permanent load (concrete + surfacingmin) = γG × VDL = 1.0 × (164 + 17) = 181kN/m
Deck Vertical Traffic load (gr2) = γQ × Vtraffic = 1.0 × 89 = 89kN/m
Deck Horizontal Traffic load (gr2) = γQ × Hbraking = 1.0 × 36 = 36kN/m

Restoring Effects:

Minimum

V

Lever Arm

Stem

162.5

1.6

260

Base

160

3.2

512

Backfill

694.5

4.25

2952

Deck (VDLmin)

181

1.55

281

∑ =

1198

∑ =

4005

Maximum

V

Lever Arm

Stem

162.5

1.6

260

Base

160

3.2

512

Backfill

694.5

4.25

2952

Deck (VDLmax + Vtraffic)

296

1.55

459

∑ =

1313

∑ =

4183

Overturning Effects:

H

Lever Arm

Hap;d

232

3.167

735

Hsc;udl

20

4.75

95

Hsc;F

23

9.5

219

Hbraking

36

7.5

270

∑ =

311

∑ =

1319

For sliding effects:
φ' = 30°
Partial factor on γM on tan(φ'cν;k) = 1.0
Coefficient of friction = μd = tan(φ'cν;k) / γM = tan(30°)/1.0 = 0.58
Sliding resistance = μd∑Vmin = Rνx;d = 0.58 × 1198 = 695kN/m
Active Force = ∑H = 311kN/m < 695 ∴ OK

Bearing Pressure:
PD 6694-1 Cl. 5.2.2 requires no uplift at SLS
Check bearing pressure at toe and heel of base slab = (V / A) ± (V × e × y / I) where V × e is the moment about the centre of the base.
V = 1313kN/m
A = 6.4m2/m
I / y = 6.42 / 6 = 6.827m3/m
Nett moment = 4183 - 1319 = 2864kNm/m
Eccentricity (e) of V about centre-line of base = 3.2 - (2864 / 1313) = 1.019m
Pressure under base = (1313 / 6.4) ± (1313 × 1.019 / 6.827)
Pressure under toe = 205 + 196 = 401kN/m2
Pressure under heel = 205 - 196 = 9kN/m2 > 0 ∴ OK (no uplift)

Also

BS EN 1997-1:2004, 2.4.8(4), allows the serviceability limit state for settlement to be verified by ensuring that a “sufficiently low fraction of the ground strength is mobilized”. This requirement can be deemed to be satisfied if the maximum pressure under a foundation at SLS does not exceed one third of the design resistance R/A' calculated in accordance with BS EN 1997-1:2004, Annex D, using characteristic values of φ', cu and γ' and representative values of horizontal and vertical actions.
From Annex D.4 for Drained Conditions:
R/A' = c'Ncbcscic + q'Nqbqsqiq + 0.5γ'B'Nγbγsγiγ
c' = 0
γ' = γ × γG;inf = 19 × 1.0 = 19kN/m3
q' = 1.5 × 19 = 28.5 kN/m2 (Foundation 1.5m below existing ground level)
Nq = eπtanφ'dtan2(45 + φ'd / 2)
φ'd = φ' = 34° (γM = 1.0)
Nq = eπtan34tan2(45 + 34 / 2) = 29.4
Nγ = 2(Nq - 1)tanφ'd = 2 × (29.4 - 1) × tan34 = 38.3
bq = bγ = 1.0 (α = 0)
B' = B - 2e = 6.4 - 2 × 1.019 = 4.362
sq = 1 + (B' / L')sinφ'd = 1 + (4.362 / 11.6) × sin34 = 1.21
sγ = 1 - 0.3(B' / L') = 1 - 0.3(4.362 / 11.6) = 0.89
m = (2 + B' / L') / (1 + B' / L') = (2 + 4.362 / 11.6) / (1 + 4.362 / 11.6) = 1.73
iq = [1 - H / (V + A'c'dcotφ'd)]m = [1 - 311 / 1313]1.73 = 0.63
iγ = [1 - H / (V + A'c'dcotφ'd)]m+1 = [1 - 311 / 1313]2.73 = 0.48
R/A' = 0 + (28.5 × 29.4 × 1.0 × 1.21 × 0.63) + (0.5 × 19 × 4.362 × 38.3 × 1.0 × 0.89 × 0.48 = 639 + 678 = 1317 kN/m2

1/3(R/A') = 1317 / 3 = 439 kN/m2 > 401 kN/m2 ∴ settlement check OK.

ULS Check Combination 1 and Combination 2.

γG;sup

γG;inf

γQ

γSd;K

Combination 1
Combination 1 Surfacing

1.35
1.2

0.95
0.95

1.35

1.2

Combination 2

1.00

1.00

1.15

1.2

Comb.1

Comb.2

Min. weight of wall stem = γG;inf × twall × Zwall × γconc

154

162.5

Max. weight of wall stem = γG;sup × twall × Zwall × γconc

219

162.5

Min. weight of base = γG;inf × Wbase × Zbase × γconc

152

160

Max. weight of base = γG;sup × Wbase × Zbase × γconc

216

160

Min. weight of backfill = γG;inf × Wheel × Zheel × γbf;d

660

694.5

Max. weight of backfill = γG;sup × Wheel × Zheel × γbf;d

937.5

694.5

Ka

0.271

0.343

Backfill Force Hap;d = γG;sup × γSd;K × Ka × γbf;d × Z2/2

376

353

Frequent value of Surcharge UDL Force Hsc;udl = ψ1 × γQ × σh;ave × Z

27

29

Frequent value of Surcharge Line Load Force Hsc;F = ψ1 × γQ × Hsc;F

31

33.7

Deck Maximum Permanent load (concrete + surfacingmax) = γG;sup × VDL

273

207

Deck Minimum Permanent load (concrete + surfacingmin) = γG;inf × VDL

172

181

Deck Vertical Traffic load (gr2) = γQ;sup × Vtraffic

120

102

Deck Horizontal Traffic load (gr2) = γQ × Hbraking

49

41

Combination 1

Restoring Effects :

Minimum

V

Lever Arm

Stem

154

1.6

246

Base

152

3.2

486

Backfill

660

4.25

2805

Deck (VDLmin)

172

1.55

267

∑ =

1138

∑ =

3804

Maximum

V

Lever Arm

Stem

219

1.6

350

Base

216

3.2

691

Backfill

937.5

4.25

3984

Deck (VDLmax + Vtraffic)

393

1.55

609

∑ =

1765.5

∑ =

5634

Overturning Effects:

H

Lever Arm

Hap;d

376

3.167

1191

Hsc;udl

27

4.75

128

Hsc;F

31

9.5

295

Hbraking

49

7.5

368

∑ =

483

∑ =

1982

For sliding effects:
φ' = 30°
Partial factor on γM on tan(φ'cν;k) = 1.0
Coefficient of friction = μd = tan(φ'cν;k) / γM = tan(30°)/1.0 = 0.58
Sliding resistance = μd∑Vmin = Rνx;d = 0.58 × 1138 = 660kN/m
Active Force = ∑H = 483kN/m < 660 ∴ OK

Bearing Pressure:
EN 1997-1:2004 Cl. 6.5.4 restrict eccentricity of loading to 1/3 of the width of the footing at ULS
V = 1765.5kN/m
Nett moment = 5634 - 1982 = 3652kNm/m
Eccentricity (e) of V about centre-line of base = 3.2 - (3652 / 1765.5) = 1.131m
Base width / 3 = 6.4 / 3 = 2.131 > 1.131 ∴ OK
Assume rectangular pressure distribution under the base as described in EN 1997-1:2004 Annex D
Effective base width B' = B - 2e = 6.4 - 2 × 1.131 = 4.138m
Pressure under base = (1765.5 / 4.138) = 427kN/m2

R/A' = c'Ncbcscic + q'Nqbqsqiq + 0.5γ'B'Nγbγsγiγ
c' = 0
γ' = γ × γG;inf = 19 × 0.95 = 18.1kN/m3
q' = 1.5 × 18.1 = 27.2kN/m2 (Foundation 1.5m below existing ground level)
Nq = eπtanφ'dtan2(45 + φ'd / 2)
φ'd = φ' = 34° (γM = 1.0)
Nq = eπtan34tan2(45 + 34 / 2) = 29.4
Nγ = 2(Nq - 1)tanφ'd = 2 × (29.4 - 1) × tan34 = 38.3
bq = bγ = 1.0 (α = 0)
B' = 4.138
sq = 1 + (B' / L')sinφ'd = 1 + (4.138 / 11.6) × sin34 = 1.20
sγ = 1 - 0.3(B' / L') = 1 - 0.3(4.138 / 11.6) = 0.89
m = (2 + B' / L') / (1 + B' / L') = (2 + 4.138 / 11.6) / (1 + 4.138 / 11.6) = 1.74
iq = [1 - H / (V + A'c'dcotφ'd)]m = [1 - 483 / 1765.5]1.74 = 0.57
iγ = [1 - H / (V + A'c'dcotφ'd)]m+1 = [1 - 483 / 1765.5]2.74 = 0.42
R/A' = 0 + (27.2 × 29.4 × 1.0 × 1.20 × 0.57) + (0.5 × 18.1 × 4.138 × 38.3 × 1.0 × 0.89 × 0.42 = 547 + 536 = 1083 kN/m2 > 427 kN/m2 ∴ OK.

Combination 2

Restoring Effects :

Minimum

V

Lever Arm

Stem

162.5

1.6

260

Base

160

3.2

512

Backfill

694.5

4.25

2952

Deck (VDLmin)

181

1.55

281

∑ =

1198

∑ =

4005

Maximum

V

Lever Arm

Stem

162.5

1.6

260

Base

160

3.2

512

Backfill

694.5

4.25

2952

Deck (VDLmax + Vtraffic)

309

1.55

479

∑ =

1326

∑ =

4203

Overturning Effects:

H

Lever Arm

Hap;d

353

3.167

1118

Hsc;udl

29

4.75

138

Hsc;F

33.7

9.5

320

Hbraking

41

7.5

308

∑ =

457

∑ =

1884

For sliding effects:
φ' = 30°
Partial factor on γM on tan(φ'cν;k) = 1.25
Coefficient of friction = μd = tan(φ'cν;k) / γM = tan(30°)/1.25 = 0.46
Sliding resistance = μd∑Vmin = Rνx;d = 0.46 × 1198 = 551kN/m
Active Force = ∑H = 457kN/m < 551 ∴ OK

Bearing Pressure:
EN 1997-1:2004 Cl. 6.5.4 restrict eccentricity of loading to 1/3 of the width of the footing at ULS
V = 1326kN/m
Nett moment = 4203 - 1884 = 2319kNm/m
Eccentricity (e) of V about centre-line of base = 3.2 - (2319 / 1326) = 1.451m
Base width / 3 = 6.4 / 3 = 2.131 > 1.451 ∴ OK
Assume rectangular pressure distribution under the base as described in EN 1997-1:2004 Annex D
Effective base width B' = B - 2e = 6.4 - 2 × 1.451 = 3.498m
Pressure under base = (1326 / 3.498) = 379kN/m2

R/A' = c'Ncbcscic + q'Nqbqsqiq + 0.5γ'B'Nγbγsγiγ
c' = 0
γ' = γ × γG;inf = 19 × 1.0 = 19kN/m3
q' = 1.5 × 19 = 28.5kN/m2 (Foundation 1.5m below existing ground level)
Nq = eπtanφ'dtan2(45 + φ'd / 2)
φ'd = tan-1[tan(φ'k)/γM] = tan-1[tan34/1.25] = 28.4°
Nq = eπtan28.4tan2(45 + 28.4 / 2) = 15.4
Nγ = 2(Nq - 1)tanφ'd = 2 × (15.4 - 1) × tan28.4 = 15.6
bq = bγ = 1.0 (α = 0)
B' = 3.498
sq = 1 + (B' / L')sinφ'd = 1 + (3.498 / 11.6) × sin28.4 = 1.14
sγ = 1 - 0.3(B' / L') = 1 - 0.3(3.498 / 11.6) = 0.91
m = (2 + B' / L') / (1 + B' / L') = (2 + 3.498 / 11.6) / (1 + 3.498 / 11.6) = 1.768
iq = [1 - H / (V + A'c'dcotφ'd)]m = [1 - 457 / 1326]1.768 = 0.47
iγ = [1 - H / (V + A'c'dcotφ'd)]m+1 = [1 - 457 / 1326]2.768 = 0.31
R/A' = 0 + (28.5 × 15.4 × 1.0 × 1.14 × 0.47) + (0.5 × 19 × 3.498 × 15.6 × 1.0 × 0.91 × 0.31) = 235 + 146 = 381 kN/m2 > 379 kN/m2 ∴ OK.

Analysing Load Cases 2 to 7 for the fixed abutment and the free abutment using a simple spreadsheet the following results were obtained:
Notation:
Case 2, 6 and 7 - results of fixed abutment with dowels and free abutment with sliding bearings.
Case 2a, 6a and 7a - results of fixed abutment with dowels and free abutment with elastomeric bearings.
Case 2b, 6b and 7b - results of both abutments with elastomeric bearings.
All other cases are not affected by the bearing arrangement.

Fixed Abutment:

Sliding

SLS

Comb.1

Comb.2

Resistance

692

657

553

Case 2
Case 2a

306
375

476
569

453
522

Case 3

290

454

437

Case 4

289

453

436

Case 5

275

435

416

Case 6 & 6a

311

483

458

Case 7& 7a

274

433

402

Bearing
Pressure

SLS
Toe

SLS
Heel

Comb.1
Vd/A' / R/A'

Comb.2
Vd/A' / R/A'

Case 2
Case 2a

359
435

23
-52

384 / 1054
478 / 773

341 / 354
447 / 257

Case 3

393

30

418 / 1199

370 / 426

Case 4

377

36

402 / 1188

354 / 419

Case 5

392

42

418 / 1281

362 / 473

Case 6 & 6a

401

9

427 / 1082

380 / 380

Case 7 & 7a

377

45

403 / 1265

336 / 485

Sliding

SLS

Comb.1

Comb.2

Resistance

692

657

553

Case 2
Case 2a

306
375

476
569

453
522

Case 3

290

454

437

Case 4

289

453

436

Case 5

275

435

416

Case 6 & 6a

311

483

458

Case 7 & 7a

274

433

402

Bearing
Pressure

SLS
Toe

SLS
Heel

Case 2
Case 2a

359
435

23
-52

Case 3

393

30

Case 4

377

36

Case 5

392

42

Case 6 & 6a

401

9

Case 7 & 6a

377

45

Bearing
Pressure

Comb.1
Vd/A' / R/A'

Comb.2
Vd/A' / R/A'

Case 2
Case 2a

384 / 1054
478 / 773

341 / 354
447 / 257

Case 3

418 / 1199

370 / 426

Case 4

402 / 1188

354 / 419

Case 5

418 / 1281

362 / 473

Case 6 & 6a

427 / 1082

380 / 380

Case 7 & 6a

403 / 1265

336 / 485

Free Abutment:

Sliding

SLS

Comb.1

Comb.2

Resistance

711

675

568

Case 2
Case 2a
Case 2b

312
382
339

486
580
522

463
532
490

Case 3

297

465

447

Case 4

269

464

447

Case 5

282

445

426

Case 6 & 6a
Case 6b

282
307

445
479

426
455

Case 7 & 7a
Case 7b

239
260

387
415

363
387

Bearing
Pressure

SLS
Toe

SLS
Heel

Comb.1
Vd/A' / R/A'

Comb.2
Vd/A' / R/A'

Case 2
Case 2a
Case 2b

361
436
390

25
-49
-3

386 / 1070
478 / 793
417 / 957

342 / 360
444 / 262
375 / 320

Case 3

395

31

420 / 1212

371 / 431

Case 4

379

37

404 / 1202

355 / 424

Case 5

394

43

421 / 1294

364 / 477

Case 6 & 6a
Case 6b

364
391

50
23

390 / 1257
416 / 1145

337 / 453
366 / 405

Case 7 & 7a
Case 7b

334
365

92
67

368 / 1480
395 / 1387

300 / 578
326 / 539

Sliding

SLS

Comb.1

Comb.2

Resistance

711

675

568

Case 2
Case 2a
Case 2b

312
382
339

486
580
522

463
532
490

Case 3

297

465

447

Case 4

269

464

447

Case 5

282

445

426

Case 6 & 6a
Case 6b

282
307

445
479

426
455

Case 7 & 7a
Case 7b

239
260

387
415

363
387

Bearing
Pressure

SLS
Toe

SLS
Heel

Case 2
Case 2a
Case 2b

361
436
390

25
-49
-3

Case 3

395

31

Case 4

379

37

Case 5

394

43

Case 6 & 6a
Case 6b

364
391

50
23

Case 7 & 7a
Case 7b

343
365

90
67

Bearing
Pressure

Comb.1
Vd/A' / R/A'

Comb.2
Vd/A' / R/A'

Case 2
Case 2a
Case 2b

386 / 1070
478 / 793
417 / 957

342 / 360
444 / 262
375 / 320

Case 3

420 / 1212

371 / 431

Case 4

404 / 1202

355 / 424

Case 5

421 / 1294

364 / 477

Case 6 & 6a
Case 6b

390 / 1257
416 / 1145

337 / 453
366 / 405

Case 7 & 7a
Case 7b

377 / 1489
395 / 1387

308 / 584
326 / 539

Note:
1) Numbers in bold indicate failed results.
2) Slight differences in results between the example and the spreadsheet for Case 6 are due to rounding off errors in the example.

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 2 to 7 shown above. Again, these are best carried out using a simple spreadsheet.

Using the Fixed Abutment Load Case 6 again as an example of the calculations:
Wall Design

SLS

Combination 1

Combination 2

Partial factors for soil parameters γM

1.0

1.0

1.25

φ'bf;d = tan-1[tan(φ'bf;k)/γM]

35.0°

35.0°

29.3°

K0 = 1-Sinφ'bf;d

0.426

0.426

0.511

Partial factors for soil weight γG;sup

1.0

1.35

1.0

Backfill density (γbf;d) = γbf;k γG;sup

19.0

25.7

19.0

Model factors γSd;K

1.0

1.2

1.2

Hap;d = γbf;dK0γSd;KZ2/2

4.047Z2kN/m

6.569Z2kN/m

5.825Z2kN/m

Consider a section at the base of the wall (Z = 8.5m)
Backfill:

Hap;d(kN) =

292

475

421

Moment (kNm) (lever arm = 8.5/3) =

827

1295

1193

Frequent value of Normal Surcharge:

ψ1γQ;sup =

0.75

1.013

0.863

ψ1γQ;supHsc;F = ψ1γQ;sup113.79KdDf =

24

33

34

Moment (kNm) (lever arm = 8.5) =

204

281

289

ψ1γQ;supσh;aveZ = ψ1γQ;sup10.34KdZ =

28

38

39

Moment (kNm) (lever arm = 4.25) =

119

162

166

γG;sup for concrete =

1.0

1.35

1.0

Deck concrete =

164

221

164

γG;sup for surfacing =

1.0

1.2

1.0

Deck surfacing =

43

52

43

Moment = Σ V × e (e = 0.5 - 0.45 = 0.05) =

10

14

10

Deck Variable Reaction (gr2):

γQ;sup =

1.0

1.35

1.15

Variable Vertical Reaction =

89

120

102

Moment = V × e (e = 0.5 - 0.45) =

4

6

5

Variable Horizontal Reaction (Braking) =

36

49

41

Moment = H × 6.5 =

234

319

267

Comb.1 shear at base of wall = Σ H = 475 + 33 + 38 + 49 = 595kN
Comb.2 shear at base of wall = Σ H = 421 + 33 + 38 + 41 = 533kN
SLS moment at base of wall = Σ M = 827 + 204 + 119 + 10 + 4 + 234 = 1398kNm (837 permanent + 561 variable)
ULS Comb.1 moment at base of wall = Σ M = 1295 + 281 + 162 + 14 + 6 + 319 = 2077kNm
ULS Comb.2 moment at base of wall = Σ M = 1193 + 289 + 166 + 10 + 5 + 267 = 1930kNm

Analysing the fixed abutment and free abutment with Load Cases 2 to 7 using a simple spreadsheet the following results were obtained for the design moments and shear at the base of the wall:
(Note - Slight differences in results between the example and the spreadsheet for Case 6 are due to rounding off errors in the example.)

Fixed Abutment:

SLS Moment
(Permanent)

SLS Moment
(Variable)

SLS Moment
(Total)

Case 2

840

539

1379

Case 3

840

442

1282

Case 4

840

427

1267

Case 5

840

335

1175

Case 6

840

565

1405

Case 7

840

279

1119

Moment
ULS Comb.1

Shear
ULS Comb.1

Moment
ULS Comb.2

Shear
ULS Comb.2

Case 2

2086

590

1908

534

Case 3

1954

569

1811

518

Case 4

1934

570

1792

519

Case 5

1809

545

1664

494

Case 6

2120

594

1928

535

Case 7

1734

531

1525

469

Free Abutment:

SLS Moment
(Permanent)

SLS Moment
(Variable)

SLS Moment
(Total)

Case 2

878

551

1429

Case 3

878

451

1329

Case 4

878

437

1315

Case 5

878

342

1220

Case 6

878

338

1216

Case 7

878

7

885

Moment
ULS Comb.1

Shear
ULS Comb.1

Moment
ULS Comb.2

Shear
ULS Comb.2

Case 2

2163

606

1979

547

Case 3

2029

584

1880

531

Case 4

2009

585

1860

532

Case 5

1881

560

1729

507

Case 6

1876

560

1724

507

Case 7

1428

489

1266

434

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:

Check slenderness of abutment wall to see if second order effects need to be considered:
EN 1992-1-1 clause 5.8.3.1
λ = 0/i ≤ λlim = 20A.B.C/√n
Use suggested values when φef not known:
A = 0.7, B = 1.1, C = 0.7
n = NEd / (Acfcd)
NEd = 164 + 43 + 166 = 373kN
fcd = αccfck / γc = 0.85 × 32 / 1.5 = 18.1N/mm2
Ac = 106mm2 (per metre width)
n = 373 × 103 / (106 × 18.1) = 0.021
λlim = 20 × 0.7 × 1.1 × 0.7 / √0.021 = 74.4
0 = 2 × = 2 × 6.63 = 13.26m (cantilever with sliding bearings to deck)
i = √(1/12) = 0.289m
λ = 13.26 / 0.289 = 45.9 < 74.4 ∴ OK, second order effects need not be considered.

EN 1992-1-1 & EN 1992-2

It is usual to design reinforced concrete for the ultimate limit state and check for serviceability conditions.
MULS = 2163kNm/m, VULS = 606kN/m, MSLS = 1429kNm/m [878(permanent)+551(variable)]

cl. 3.1.6(101)P

Design compressive strength = fcd = αccfck / γc

cl. 3.1.7

αcc = 0.85

cl. 2.4.2.4

Table 2.1N:   γc = 1.5,   γs = 1.15

fcd = 0.85 × 32 / 1.5 = 18.1 N/mm2

Table 3.1

εc2 = 0.002,   εcu2 = 0.0035,   n = 2.0

Try B40's @ 150mm c/c (8378mm2/m) in rear face at base of wall:
Nominal cover to reinforcement in rear face of wall = 60mm
d = 1000 - 60 - 20 = 920mm

Fig. 3.3

Using parabolic-rectangular diagram:
Average stress fav = fcd[1-εc2 / {εcu2(n+1)}] = 18.1 × [1 - 0.002 / {0.0035 × (2 + 1)}] = 14.7 N/mm2

Assuming steel yields then:
M = fsz = fykAsz / γs = Fcz = favbXz
Depth to neutral axis X = fykAs / (favs)
X = 500 × 8378 / (14.7 × 1000 × 1.15) = 247.8mm

Check that steel will yield:

Cl. 3.2.7(4)

Modulus of Elasticity Es = 200 kN/mm2
Steel strain at yield = εs,yield = fyk / γs / Es = 500 / 1.15 / 200000 = 0.00217
from linear strain relationship:
εs = εcu2(d/X - 1) = 0.0035 ( 920 / 247.8 - 1) = 0.009 > 0.00217 ∴ steel will yield.

Hence Mult = favbXz = favbX(d - βX)
Where β = 1 - [0.5εcu22 - εc22 / {(n+1)(n+2)}] / [εcu22 - εcu2εc2 / (n+1)]
β = 1 - [0.5 × 0.00352 - 0.0022 / {(2 + 1) × (2 + 2)}] / 0.00352 - 0.0035 × 0.002 / (2 + 1)] = 0.416
Mult = 14.7 × 1000 × 247.8 × (920 - 0.416 × 247.8) × 10-6 = 2976 kNm > 2163 ∴ OK

Check Serviceability Limit State

Characteristic Combination SLS Design Moment = 1429kNm/m (878 +551)

Check stresses in the concrete and reinforcement at:
i) Early Age (before creep has occurred)
ii) Long term after all the creep has taken place.

i) Before creep has occurred the cracked section properties will be based on the short-term modulus for all actions.

EN 1992-1-1 Table 3.1

Ecm = 22[(fck + 8) / 10]0.3 = 22[(32 + 8) / 10]0.3 = 33.4 kN/mm2
Ec,eff = Ecm = 33.4 kN/mm2
Modular Ratio m = Es / Ecm = 200 / 33.4 = 6.0

Let dc = depth to neutral axis then equating strains for cracked section:
εs = εc(d - dc) / dc
Equating forces:
AsEsεs = 0.5bdcεcEc,eff
Hence dc = [-AsEs + {(AsEs)2 + 2bAsEsEc,effd}0.5] / bEc,eff
dc = [-8378 × 200000 + {(8378 × 200000)2 + 2 × 1000 × 8378 × 200000 × 33400 × 920}0.5] / (1000 × 33400) = 258mm
Cracked second moment of area = As(d-dc)2 + Ec,effbdc3 / 3Es
INA = 8378 × (920 - 258)2 + 33.4 × 1000 × 2583 / (3 × 200) = 4.63 × 109 mm4 (steel units)
Approximate concrete stress σc = M / zc + N / Ac
N (Case 2) = 164 + 43 = 207 kN
σc ≅ {1429 × 106 × 258 / (4.63 × 109 × 6.0)} + {207 × 103 / (258 × 103)} = 13.3 + 0.8 = 14.1 N/mm2

cl. 7.2(102)

Limiting concrete stress = k1fck
k1 = 0.6
Limiting concrete stress = 0.6 × 32 = 19.2 N/mm2 > 14.1 ∴ OK

EN 1992-1-1

ii) After all creep has taken place the cracked section properties will be based on the long-term and short-term modulus for the various actions.
Short-term modulus = Ecm
Long-term modulus = Ecm / (1+φ)
Effective modulus Ec,eff = (Mqp + Mst)Ecm / {Mst + (1 + φ)Mqp}

Table 3.1

fcm = fck + 8 = 32 + 8 = 40 N/mm2

Cl. 3.1.4

Relative humidity of the ambient environment = 80% (outside conditions)
Age of concrete at initial loading t0 = say 7 days (after formwork has been released and waterproofing system applied to rear face of wall)

Annex B (B.6)&(B.8c)

h0 = 2Ac / U = 2 × (11600 × 1000) / (11600 + 2 × 1000) = 1706
α1 = [35 / fcm]0.7 = [35 / 40]0.7 = 0.91
α2 = [35 / fcm]0.2 = [35 / 40]0.2 = 0.97

(B.3b)

φRH = [1 + α1 × {(1 - RH / 100) / (0.1 × h01/3)}] × α2
φRH = [1 + 0.91 × {(1 - 80 / 100) / ( 0.1 × 17061/3)}] × 0.97 = 1.118

(B.4)

β(fcm) = 16.8 / fcm0.5 = 16.8 / 400.5 = 2.656

(B.5)

β(t0) = 1 / (0.1 + t00.2) = 1 / ( 0.1 + 70.2) = 0.635

(B.2)

φ0 = φRH × β(fcm) × β(t0) = 1.118 × 2.656 × 0.635 = 1.886
Moment due to long-term actions = Mqp = 878 kNm
Moment due to short-term actions = Mst = 551 kNm
Hence Effective Modulus Ec,eff = {(878 + 551) × 33.4} / {551 + 878 × ( 1 + 1.886)} = 15.5 kN/mm2

Modular Ratio m = Es / Ec,eff = 200 / 15.5 = 12.9
Let dc = depth to neutral axis then equating strains for cracked section:
εs = εc(d - dc) / dc
Equating forces:
AsEsεs = 0.5bdcεcEc,eff
Hence dc = [-AsEs + {(AsEs)2 + 2bAsEsEc,effd}0.5] / bEc,eff
dc = [-8378 × 200000 + {(8378 × 200000)2 + 2 × 1000 × 8378 × 200000 × 15500 × 920}0.5] / (1000 × 15500) = 351mm
Cracked second moment of area = As(d-dc)2 + Ec,effbdc3 / 3Es
INA = 8378 × (920 - 351)2 + 15.5 × 1000 × 3513 / (3 × 200) = 3.83 × 109 mm4 (steel units)
Concrete stress σc ≅ M / zc + N / Ac
σc = {1429 × 106 × 351 / (3.83 × 109 × 12.9)} + (207 × 103 / (351 × 103) = 10.2 + 0.6 = 10.8 N/mm2

cl. 7.2(102)

Limiting concrete stress = k1fck
k1 = 0.6
Limiting concrete stress = 0.6 × 32 = 19.2 N/mm2 > 10.8 ∴ OK

cl. 7.2(5)

Limiting steel stress = k3fyk
k3 = 0.8
Limiting steel stress = 0.8 × 500 = 400 N/mm2
Steel stress σs = M / zs
σs = 1429 × 106 × (920 - 351) / (3.83 × 109) = 212 N/mm2 < 400 ∴ >OK

Crack Control:

Consider worst condition before creep has occurred and
Quasi-Permanent Combination Moment + ψ2 × temperature effects = 878 + 0.5(16 × 6.63) = 931 kNm

Cl. 7.3.4(1)

Crack width wk = sr,maxsm - εcm)

Cl. 7.3.4(3)

Spacing Limit = 5(c+φ/2) = 5(60 + 40/2) = 400mm > 150mm ∴ OK
sr,max = k3c + k1k2k4φ / ρp,eff
k1 = 0.8 (high bond bars)
k2 = 0.5 (for bending)
k3 = 3.4 (recommended value)
k4 = 0.425 (recommended value)

Cl. 7.3.2(3)

hc,eff is the lesser of:
i) 2.5(h-d) = 2.5(1000 - 920) = 200
ii) (h-x)/3 = (1000 - 258) / 3 = 247
iii) h/2 = 1000 / 2 = 500
∴ hc,eff = 200 mm
and Ac,eff = 200 × 1000 = 200000 mm2

Cl. 7.3.4(2)

ρp,eff = As / Ac,eff = 8378 / 200000 = 0.0419

sr,max = k3c + k1k2k4φ / ρp,eff
sr,max = (3.4 × 60) + (0.8 × 0.5 × 0.425 × 40) / 0.0419 = 204 + 162 = 366

Cl. 7.3.4(2)

sm - εcm) = [σs - {ktfct,eff(1 + αeρp,eff) /ρp,eff}] / Es ≥ 0.6σs / Es
αe = Es / Ecm = 200 / 33.4 = 6.0
σs = 931 × 106 × (920 - 258) / (4.63 × 109) = 133 N/mm2

Table 3.1

fct,eff = fctm = 0.3 × fck(2/3) = 0.3 × 32(2/3) = 3.02 N/mm2

sm - εcm) = [133 - {0.4 × 3.02 × (1 + 6.0 × 0.0419) / 0.0419}] / 200000 = 0.485 × 10-3
0.6σs / Es = 0.6 × 133 / 200000 = 0.399 × 10-3 < 0.485 × 10-3 ∴ OK
Crack width wk = sr,maxsm - εcm) = 366 × 0.485 × 10-3 = 0.18 mm

NA EN 1992-2 Table NA.2

Recommended value of wmax = 0.3 mm > 0.18 mm ∴ OK

Hence B40 bars at 150 centres are adequate for the rear face at the base of the wall.

Shear Capacity

Cl. 6.2.2(101)

Shear Capacity of Wall with B40 dia. reinforcement @ 150c/c
VRd,c = [CRd,ck(100ρ1fck)1/3]bwd
CRd,c = 0.18 / γc = 0.18 / 1.5 = 0.12
k = 1 + (200 / d)0.5 ≤2.0
k = 1 + (200 / 920)0.5 = 1.47 < 2.0
ρ1 = Asl / bwd ≤0.02
ρ1 = 8378 / (1000 × 920) = 0.009 < 0.02

Cl. 3.1.2(102)P

fck = 32 ( < Cmax = C50/60)
VRd,c = [0.12 × 1.47 × (100 × 0.009 × 32)1/3] × 1000 × 920 × 10-3 = 497 kN ( < 606 kN Fail : see below)
Minimum VRd,c = (vmin)bwd = 0.035k3/2fck1/2bwd = 0.035 × 1.473/2 × 321/2 × 1000 × 920 × 10-3 = 325 kN

cl 6.2.2(6)

Check that the maximum allowable shear force is not exceeded:
Maximum allowable shear force = 0.5bwdνfcd
ν = 0.6[1 - fck / 250] = 0.6 × [1 - 32 / 250] = 0.523
fcd = αccfckc
αcc = 1.0 [see NA to Cl. 3.1.6(101)P]
fcd = 1.0 × 32 / 1.5 = 21.3 N/mm2
Maximu VEd = 0.5 × 1000 × 920 × 0.523 × 21.3 × 10-3 = 5124 kN >> 606 kN
VRd,c = 497 kN < VEd = 606 kN ∴ Fail. It would be necessary to increase the longitudinal reinforcement to B40 at 125 c/c however the UK National Annex allows an alternative approach.

NA to 1992-2 Cl. 6.2.2(101)

Alternative Solution:
If the reduction factor β is not used to reduce the applied shear force actions then the allowable shear force VRd,c may be enhanced if the section being considered is within 2d of the support.

i) Consider a section at (a = 0.829m) from the bottom of wall :
Maximum ULS shear force from spreadsheet for Case 2 = 511 kN

Shear enhancement factor = (2d/a) = 2 × 0.92 / 0.829 = 2.22
VRd,c = 2.22 × 497 = 1103 kN ( > 511 kN ∴ OK)

ii) Consider a section at (a = 1.657m) from the bottom of wall :
Maximum ULS shear force from spreadsheet for Case 2 = 426 kN

Shear enhancement factor = (2d/a) = 2 × 0.92 / 1.657 = 1.11
VRd,c with no enhancement = 497 kN > 426 kN ∴, by inspection, all sections will be suitable to resist shear using B40 bars at 150 centres.

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 C660 then B32 horizontal lacer bars @ 100 c/c will be required in both faces in the bottom 0.5m of the wall, reducing to B25 bars @ 200 above 1.3m from the bottom of the wall.

Minimum Wall Reinforcement

EN 1992-1-1 Clause 9.6.2 - Vertical reinforcement:
As,vmin = 0.002Ac = 0.002 × 106 = 2000 mm2/m (1000 mm2/m in each face). Use B16 @ 150 c/c (As = 1340mm2/m).
EN 1992-1-1 Clause 9.6.3 - Horizontal reinforcement:
As,vmin = 0.001Ac or 25% of vertical reinforcement = 0.001 × 106 = 1000 mm2/m (in each face) or
25% × 8378 = 2095mm2/m. B20 @ 150 c/c = 2094mm2/m, but B32 @ 100 c/c reducing to B25 bars @ 150 are required to resist early thermal cracking.
Hence early thermal cracking and long-term creep and shrinkage crack control require greater areas of reinforcement than the minimum wall reinforcement.

Base Design

Maximum bending and shear effects in the base slab will occur at sections near the front and back of the wall. Calculations need to be carried out for serviceability and ultimate limit states using 'at rest pressures'
Using the Fixed Abutment Load Case 6 again as an example of the calculations:

CASE 6 - Fixed Abutment Serviceability Limit State

Weight of wall stem = 162.5kN/m
Weight of base = 160kN/m
Weight of backfill = 694.5kN/m
B/fill Force Hap;d = γG × γSd;K × K0 × γbf;d × Z2 / 2 = 1.0 × 1.0 × 0.426 × 19 × 9.52 / 2 = 365kN/m
Frequent value of Surcharge UDL Force Hsc;udl = ψ1 × γQ × σh;ave × Z = 0.75 × 1.0 × (10.34 × 0.426) × 9.5 = 31kN/m
Dispersion Factor Df for line load = (1 + Z / 2) / (1 + Z) = (1 + 9.5 / 2) / (1 + 9.5) = 0.55 < 0.67 ∴ Df = 0.67
Frequent value of Surcharge Line Load Force Hsc;F = ψ1 × γQ × Hsc;F = 0.75 × 1.0 × (113.79 × 0.426 × 0.67) = 24kN/m
Deck Maximum Permanent load (concrete + surfacingmax) = 207kN/m
Deck Vertical Traffic load (gr2) = 89kN/m
Deck Horizontal Traffic load (gr2) = 36kN/m

Restoring Effects:

V

Lever Arm

Stem

162.5

1.6

260

Base

160

3.2

512

Backfill

694.5

4.25

2952

Deck Vertical Reaction

296

1.55

459

∑ =

1313

∑ =

4183

Overturning Effects:

H

Lever Arm

Backfill

365

3.167

1156

Surcharge UDL

31

4.75

147

24

9.5

228

Deck Horizontal Reaction

36

7.5

270

∑ =

456

∑ =

1801

Bearing Pressure at toe and heel of base slab = (V / A) ± (V × e × y / I)
V = 1313kN/m
A = 6.4m2/m
I / y = 6.42 / 6 = 6.827m3/m
Nett moment = 4183 - 1801 = 2382kNm/m
Eccentricity (e) of V about centre-line of base = 3.2 - (2382 / 1313) = 1.386m
Pressure under base = (1313 / 6.4) ± (1313 × 1.386 / 6.827)
Pressure under toe = 205 + 267 = 472kN/m2
Pressure under heel = 205 - 267 = -62kN/m2 (uplift)
Reduced length of pressure under base = 3(B/2 - e) = 3 × (6.4 / 2 - 1.386) = 5.442 m
Pressure under toe = 2 × 1313 / 5.442 = 483 kN/m2
Pressure at front face of wall = 483 × 4.342 / 5.442} = 385kN/m2
Pressure at rear face of wall = 483 × 3.342 / 5.442} = 297kN/m2

SLS Moment at a-a = (385 × 1.12 / 2) + ([483 - 385] × 1.12 / 3) - (25 × 1.0 × 1.12 / 2) = 257kNm/m (tension in bottom face).
SLS Moment at b-b = (297 × 3.3422 / 6) - (695 × 4.3 / 2) - (25 × 1.0 × 4.32 / 2) = -1173kNm/m (tension in top face).

CASE 6 - Fixed Abutment Ultimate Limit State

Weight of wall stem Comb.1 = γG;sup × 162.5 = 1.35 × 162.5 = 219 kN/m
Weight of wall stem Comb.2 = γG;sup × 162.5 = 1.0 × 162.5 = 163 kN/m
Weight of base Comb.1 = γG;sup × 160 = 1.35 × 160 = 216 kN/m
Weight of base Comb.2 = γG;sup × 160 = 1.0 × 160 = 160 kN/m
Weight of backfill Comb.1 = γG;sup × 694.5 = 1.35 × 694.5 = 938 kN/m
Weight of backfill Comb.2 = γG;sup × 694.5 = 1.0 × 694.5 = 695 kN/m
B/fill Force Hap;d Comb.1 = γG;sup × γSd;K × K0 × γbf;d × Z2 / 2 = 1.35 × 1.2 × 0.426 × 19 × 9.52 / 2 = 592kN/m
B/fill Force Hap;d Comb.2 = γG;sup × γSd;K × K0 × γbf;d × Z2 / 2 = 1.0 × 1.2 × 0.511 × 19 × 9.52 / 2 = 526kN/m
Frequent value of Surcharge UDL Force Hsc;udl Comb.1 = ψ1 × γQ;sup × σh;ave × Z = 0.75 × 1.35 × (10.34 × 0.426) × 9.5 = 42kN/m
Frequent value of Surcharge UDL Force Hsc;udl Comb.2 = ψ1 × γQ;sup × σh;ave × Z = 0.75 × 1.15 × (10.34 × 0.511) × 9.5 = 43kN/m
Dispersion Factor Df for line load = (1 + Z / 2) / (1 + Z) = (1 + 9.5 / 2) / (1 + 9.5) = 0.55 < 0.67 ∴ Df = 0.67
Frequent value of Surcharge Line Load Force Hsc;F Comb.1 = ψ1 × γQ;sup × Hsc;F = 0.75 × 1.35 × (113.79 × 0.426 × 0.67) = 33kN/m
Frequent value of Surcharge Line Load Force Hsc;F Comb.2 = ψ1 × γQ;sup × Hsc;F = 0.75 × 1.15 × (113.79 × 0.511 × 0.67) = 34kN/m
Deck Maximum Permanent load (concrete + surfacingmax) Comb.1 = γG;sup × 164 + γG;sup × 43 = 1.35 × 164 + 1.2 × 43 = 273kN/m
Deck Maximum Permanent load (concrete + surfacingmax) Comb.2 = γG;sup × 164 + γG;sup × 43 = 1.0 × 164 + 1.0 × 43 = 207kN/m
Deck Vertical Traffic load (gr2) Comb.1 = γQ;sup × 89 = 1.35 × 89 = 120 kN/m
Deck Vertical Traffic load (gr2) Comb.2 = γQ;sup × 89 = 1.15 × 89 = 102 kN/m
Deck Horizontal Traffic load (gr2) Comb.1 = γQ;sup × 36 = 1.35 × 36 = 49 kN/m
Deck Horizontal Traffic load (gr2) Comb.2 = γQ;sup × 36 = 1.15 × 36 = 41 kN/m

Restoring Effects:

V
Comb.1/Comb.2

Lever Arm

Comb.1/Comb.2

Stem

219/163

1.6

350/261

Base

216/160

3.2

691/512

Backfill

938/695

4.25

3987/2954

Deck Vertical Reaction

393/309

1.55

609/479

∑ =

1766/1327

∑ =

5637/4206

Overturning Effects:

H

Lever Arm

Backfill

592/526

3.167

1875/1666

Surcharge UDL

42/43

4.75

200/204

33/34

9.5

314/323

49/41

7.5

368/308

∑ =

716/644

∑ =

2757/2501

Assume rectangular pressure distribution under the base as described in EN 1997-1:2004 Annex D
Combination 1:
V = 1766 kN/m
Nett moment = 5637 - 2757 = 2880 kNm/m
Eccentricity (e) of V about centre-line of base = 3.2 - (2880 / 1766) = 1.569m
Effective base width B' = B - 2e = 6.4 - 2 × 1.569 = 3.262m
Pressure under base = (1766 / 3.262) = 541 kN/m2

Combination 2:
V = 1327 kN/m
Nett moment = 4206 - 2501 = 1705 kNm/m
Eccentricity (e) of V about centre-line of base = 3.2 - (1705 / 1327) = 1.915m
Effective base width B' = B - 2e = 6.4 - 2 × 1.915 = 2.57m
Pressure under base = (1327 / 2.57) = 516 kN/m2

Combination 1:
ULS Shear at a-a = (541 × 1.1) - (1.35 × 1.0 × 1.1 × 25) = 558 kN/m
ULS Shear at b-b = 541 × (3.262 - 2.1) - (1.35 × 1.0 × 4.3 × 25) - 938} = -454 kN/m
ULS Shear at c-c = 1.35 × (6.4 - 3.262) × (8.5 × 19 + 1.0 × 25) = 790 kN/m
ULS Moment at a-a = (541 × 1.12 / 2) - (1.35 × 25 × 1.0 × 1.12 / 2) = 307 kNm/m (tension in bottom face).
ULS Moment at b-b = (541 × (3.262 - 2.1)2 / 2) - (1.35 × 25 × 1.0 × 4.32 / 2) - (938 × 4.3 / 2) = -1963 kNm/m (tension in top face).

Combination 2:
ULS Shear at a-a = (516 × 1.1) - (1.0 × 1.0 × 1.1 × 25) = 540 kN/m
ULS Shear at b-b = 516 × (2.57 - 2.1) - (1.0 × 1.0 × 4.3 × 25) - 695} = 560 kN/m
ULS Shear at c-c = 1.0 × (6.4 - 2.57) × (8.5 × 19 + 1.0 × 25) = 714 kN/m
ULS Moment at a-a = (516 × 1.12 / 2) - (1.0 × 25 × 1.0 × 1.12 / 2) = 297 kNm/m (tension in bottom face).
ULS Moment at b-b = (516 × (2.57 - 2.1)2 / 2) - (1.0 × 25 × 1.0 × 4.32 / 2) - (695 × 4.3 / 2) = -1668 kNm/m (tension in top face).

Analysing the fixed abutment and the free abutment with Load Cases 2 to 7 using a simple spreadsheet the following results were obtained:

Fixed Abutment Base:

Section a-a

ULS Shear
Comb.1/Comb.2

SLS
Moment

ULS Moment
Comb.1/Comb.2

Case 2

509/501

235

280/276

Case 3

539/515

253

297/283

Case 4

521/497

244

286/273

Case 5

527485

250

290/267

Case 6

558/541

258

307/298

Case 7

445/433

235

272/238

Section b-b

ULS Shear
Comb.1/Comb.2

SLS
Moment

ULS Moment
Comb.1/Comb.2

Case 2

480/588

1184

1962/1676

Case 3

364/466

1071

1835/1610

Case 4

372/468

1069

1830/1607

Case 5

290/365

973

1716/1519

Case 6

453/561

1178

1961/1668

Case 7

281/311

934

1663/1436

Section c-c

ULS Shear
Comb.1/Comb.2

Case 2

777/718

Case 3

737/675

Case 4

730/671

Case 5

694/627

Case 6

790/715

Case 7

665/583

Free Abutment Base:

Section a-a

ULS Shear
Comb.1/Comb.2

SLS
Moment

ULS Moment
Comb.1/Comb.2

Case 2

511/500

236

281/275

Case 3

542/516

254

298/284

Case 4

523/498

246

288/274

Case 5

530/487

251

292/268

Case 6

494/455

234

272/250

Case 7

440/372

212

242/205

Section b-b

ULS Shear
Comb.1/Comb.2

SLS
Moment

ULS Moment
Comb.1/Comb.2

Case 2

479/582

1239

2054/1765

Case 3

365/464

1124

1924/1693

Case 4

373/467

1122

1918/1690

Case 5

291/365

1023

1800/1597

Case 6

327/394

1040

1815/1611

Case 7

172/192

741

1428/1266

Section c-c

ULS Shear
Comb.1/Comb.2

Case 2

793/733

Case 3

755/691

Case 4

747/687

Case 5

711/643

Case 6

702/643

Case 7

563/499

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 C660 then a minimum steel area of B25 distribution bars @ 200 c/c will be required to comply with clause 7.3.2(2) of BS EN 1992-1-1.

Design for bending effects at section a-a for the Fixed Abutment and b-b for the Free Abutment and for shear at section c-c for the Free Abutment using a simple spreadsheet for slab member capacities:

Section a-a: Muls = 307 kNm/m, Vuls = 558 kN/m, Msls = 258 kNm/m (Mperm = 181 kNm/m + Mvar = 77 kNm/m)
B25's @ 150 c/c give Muls = 1262 kNm/m > 307 ∴ OK, Vuls = 731 kN/m (at d from support) > 558 ∴ OK, Msls = 1103 kNm/m > 258 ∴ OK

Section b-b: Muls = 2054 kNm/m, Vuls = 582 kN/m, Msls = 1239 kNm/m (Mperm = 860 kNm/m + Mvar = 380 kNm/m)
B40's @ 150 c/c give Muls = 2975 kNm/m > 2054 ∴ OK, Vuls = 996 kN/m (at d from support) > 582 ∴ OK, Msls = 2065 kNm/m > 1239 ∴ OK

Section c-c: Vuls = 793 kN/m < Vuls = 996 kN/m (at d from support using B40's @ 150 c/c) ∴ OK

Local Effects

Curtain Wall (Abutment Upstand 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.

EN 1991-2 Clause 4.9.2(2):
Braking Force = 0.6αQ1Q1k from LM1 axle = 0.6 × 300 = 180kN
To allow for load distribution effects assume a 45° dispersal down the wall, with maximum dispersal of the width of the abutment (11.6m). Positioning the axle in the centre of the 3.0m notional lane gives a distribution width of 7.65m at the base of the wall.

Shear at the base of the wall:
due to braking = 180 / 7.65 = 23.5kN/m
due to backfill = K0 γbf;d Z2 / 2 = 0.426 × 19.0 × 3.02 / 2 = 36kN/m
Total ULS shear = 1.35 × (23.5 + 1.2 × 36) = 90kN/m

Bending moment at the base of the wall:
due to braking = 180 × 3.0 / 7.65 = 70.6kNm/m
due to backfill = K0 γbf;d Z3 / 6 = 0.426 × 19.0 × 3.03 / 6 = 36kNm/m
Total SLS moment = 70.6 + 36 = 107kNm/m
Total ULS moment (combination 1) = (1.35 × 70.6) + (1.35 × 1.2 × 36) = 154kNm/m

Check effects of surcharge + backfill at base of curtain wall:

Normal traffic surcharge:
Df = (1 + Z/2) / (1 + Z) = (1 + 3/2) / (1 + 3) = 0.625 < 0.67 ∴ Df = 0.67
Line Load = 113.79KdDf = 113.79 × 0.426 × 0.67 = 32.5kN/m
UDL = 10.34Kd = 10.34 × 0.426 = 4.4kN/m2
Total ULS shear = 1.35 × (32.5 + 4.4 × 3.0 + 1.2 × 36) = 120kN/m
Total SLS moment = 32.5 × 3.0 + 4.4 × 3.02 / 2 + 36 = 97.5 + 19.8 + 36 = 153kNm/m
Total ULS moment = 1.35 × (97.5 + 19.8) + (1.35 × 1.2 × 36) = 158 + 58 = 216kNm/m

SV/196 traffic surcharge:
Line Load = 99.57KdDf = 99.57 × 0.426 × 0.67 = 28.4kN/m
UDL = 11.64Kd = 11.64 × 0.426 = 5.0kN/m2
Total ULS shear = 1.35 × (28.4 + 5.0 × 3.0 + 1.2 × 36) = 117kN/m
Total SLS moment = 28.4 × 3.0 + 5.0 × 3.02 / 2 + 36 = 85.2 + 22.5 + 36 = 144kNm/m
Total ULS moment = 1.35 × (85.2 + 22.5) + (1.35 × 1.2 × 36) = 158 + 58 = 204kNm/m

Hence Normal Traffic Surcharge + Backfill has worst effect on curtain wall.

400 thick curtain wall with B32 @ 150 c/c :
Mult = 601 kNm/m > 216 kNm/m ∴ OK
Msls = 317 kNm/m > 153 kNm/m ∴ OK
Vult = 261 kN/m > 120 kN/m ∴ Shear OK

Last Updated : 22/02/18

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