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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) = 30^{o} and a safe bearing capacity of 400kN/m^{2}.

Backfill material will be Class 6N with an effective angle of internal friction (ϕ') = 35^{o} and density (γ) = 19kN/m^{3}.

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 +37^{o}C respectively.

For a Group 4 type structure (see fig. 9) the corresponding minimum and maximum effective bridge temperatures are -11 and +36^{o}C from tables 10 and 11.

Hence the temperature range = 11 + 36 = 47^{o}C.

From Clause 5.4.6 the range of movement at the free end of the 20m span deck = 47 × 12 × 10^{-6} × 20 × 10^{3} = 11.3mm.

The ultimate thermal movement in the deck will be ± [(11.3 / 2) γf_{3} γf_{L}] = ±[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 :

- 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 +37^{o}C which would require the bearings to be installed at a shade air temperature of [(37+19)/2 -19] = 9^{o}C to achieve the ± 8mm movement.

If the bearings are set at a maximum shade air temperature of 16^{o}C 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 16^{o}C 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 = AGδ

Using the Ekspan bearing EKR35

- Maximum Load = 1053kN
- Area = 610 × 420 = 256200mm
^{2} - Nominl hardness = 60 IRHD
- Bearing Thickness = 19mm

Shear modulus G from Table 8 = 0.9N/mm^{2}

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:

- Maximum Load = 800kN
- Base Plate A dimension = 210mm
- Base Plate B dimension = 365mm
- Movement ± X = 12.5mm

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/mm^{2}

As the mating surface between the stainless steel and PTFE is smaller than the base plate then the pressure between the sliding faces will be in the order of 5N/mm^{2}.

From Table3 of BS 5400 Part 9.1 the Coefficient of friction = 0.08 for a bearing stress of 5N/mm^{2}

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 = K_{a} γ h

K_{a} for Class 6N material = (1-Sin35) / (1+Sin35) = 0.27

Density of Class 6N material = 19kN/m^{3}

Active Pressure at depth h = 0.27 × 19 × h = 5.13h kN/m^{2}

Hence Fb = 5.13h^{2}/2 = 2.57h^{2}kN/m

__Surcharge__ - BS 5400 Part 2 Clause 5.8.2:

For HA loading surcharge = **10 kN/m ^{2}**

For HB loading surcharge =

Assume a surchage loading for the compaction plant to be equivalent to 30 units of HB

Hence Compaction Plant surcharge =

For surcharge of w kN/m

Fs = K

**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/m^{3}.

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 × 7.5^{2} / 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(30^{o}) = 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 = 6.4m^{2}/m

Z = 6.4^{2} / 6 = 6.827m^{3}/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/m^{2} < 400kN/m^{2} ∴ OK.

Pressure under heel = 142 - 15 = 127kN/m^{2}

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**
K

γ

Serviceability = 1.0

Ultimate = 1.5

γ

Backfill Force Fb on the rear of the wall = 0.426 × 19 × 6.5

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 =

Ultimate moment = 1.1 × 1.5 × 478 =

Ultimate shear = 1.1 × 1.5 × (171 + 33) =

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/m^{3} for exposure condition XD2.

Nominal cover to reinforcement = 60mm (45mm minimum cover plus a tolerance Δ_{c} of 15mm).

Reinforcement to BS 4449:2005 Grade B500B: f_{y} = 500N/mm^{2}

**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

0.1f_{cu}A_{c} = 0.1 × 40 × 10^{3} × 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 - [ 1.1f_{y}A_{s}) / (f_{cu}bd) ]} d

Use **B40 @ 150 c/c** in rear face at base of wall:

A_{s} = 8378mm^{2}/m, d = 1000 - 60 - 20 = 920mm

z = {1 - [ 1.1 × 500 × 8378) / (40 × 1000 × 920) ]} d = 0.875d < 0.95d ∴ OK

Mu = (0.87f_{y})A_{s}z = 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 × 10^{3} / (1000 × 920) = 0.673 N/mm^{2}

No shear reinforcement is required when v < ξ_{s}v_{c}

ξ_{s} = (500/d)^{1/4} = (500 / 920)^{1/4} = 0.86

v_{c} = (0.27/γ_{m})(100A_{s}/b_{w}d)^{1/3}(f_{cu})^{1/3} = (0.27 / 1.25) × ({100 × 8378} / {1000 × 920})^{1/3} × (40)^{1/3} = 0.72

ξ_{s}v_{c} = 0.86 × 0.72 = 0.62 N/mm^{2} < 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 × 10^{3} / (1000 × 920) = 0.53 N/mm^{2} < 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 b_{a}d = 0.0012 × 1000 × 920 = 1104 mm^{2}/m (use **B16 @ 150c/c** - A_{s} = 1340mm^{2}/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**

γ_{fL} = 1.0 γ_{f3} = 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 × 7.5^{2} × 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 = 6.4m^{2}/m

Z = 6.4^{2} / 6 = 6.827m^{3}/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/m^{2}

Pressure under heel = 142 - 53 = 89kN/m^{2}

Pressure at front face of wall = 89 + {(195 - 89) × 5.3 / 6.4} = 177kN/m^{2}

Pressure at rear face of wall = 89 + {(195 - 89) × 4.3 / 6.4} = 160kN/m^{2}

SLS Moment at a-a = (177 × 1.1^{2} / 2) + ([195 - 177] × 1.1^{2} / 3) - (25 × 1.0 × 1.1^{2} / 2) = 99kNm/m (tension in bottom face).

SLS Moment at b-b = (89 × 4.3^{2} / 2) + ([160 - 89] × 4.3^{2} / 6) - (25 × 1.0 × 4.3^{2} / 2) - (531 × 4.3 / 2) - (52 × 4.3 / 2) = -443kNm/m (tension in top face).

__CASE 1__ - Fixed Abutment **Ultimate Limit State**

γ_{fL} for concrete = 1.15

γ_{fL} for fill and surcharge(vetical) = 1.2

γ_{fL} 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 × 7.5^{2} × 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 = 6.4m^{2}/m

Z = 6.4^{2} / 6 = 6.827m^{3}/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/m^{2}

Pressure under heel = 167 - 93 = 74kN/m^{2}

Pressure at front face of wall = 74 + {(260 - 74) × 5.3 / 6.4} = 228kN/m^{2}

Pressure at rear face of wall = 74 + {(260 - 74) × 4.3 / 6.4} = 199kN/m^{2}

γ_{f3} = 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 × 1.1^{2} / 2) + ([260 - 228] × 1.1^{2} / 3) - (1.15 × 25 × 1.0 × 1.1^{2} / 2)} = 148kNm/m (tension in bottom face).

ULS Moment at b-b = 1.1 × {(74 × 4.3^{2} / 2) + ([199 - 74] × 4.3^{2} / 6) - (1.15 × 25 × 1.0 × 4.3^{2} / 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 - [ 1.1f_{y}A_{s}) / (f_{cu}bd) ]} d

Use B32 @ 150 c/c:

A_{s} = 5362mm^{2}/m, d = 1000 - 60 - 16 = 924mm

z = {1 - [ 1.1 × 500 × 5362) / (40 × 1000 × 924) ]} d = 0.92d < 0.95d ∴ OK

Mu = (0.87f_{y})A_{s}z = 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 (M_{uls} = 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 × 10^{3} / (1000 × 924) = 0.121 N/mm^{2}

No shear reinforcement is required when v < ξ_{s}v_{c}

Reinforcement in tension = **B32 @ 150 c/c**

ξ_{s} = (500/d)^{1/4} = (500 / 924)^{1/4} = 0.86

v_{c} = (0.27/γ_{m})(100A_{s}/b_{w}d)^{1/3}(f_{cu})^{1/3} = (0.27 / 1.25) × ({100 × 5362} / {1000 × 924})^{1/3} × (40)^{1/3} = 0.62

ξ_{s}v_{c} = 0.86 × 0.62 = 0.53 N/mm^{2} > 0.121N/mm^{2} ∴ 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 × 10^{3} / (1000 × 924) = 0.605 N/mm^{2}

No shear reinforcement is required when v < ξ_{s}v_{c}

ξ_{s} = (500/d)^{1/4} = (500 / 924)^{1/4} = 0.86

v_{c} = (0.27/γ_{m})(100A_{s}/b_{w}d)^{1/3}(f_{cu})^{1/3} = (0.27 / 1.25) × ({100 × 5362} / {1000 × 924})^{1/3} × (40)^{1/3} = 0.62

ξ_{s}v_{c} = 0.86 × 0.62 = 0.53 N/mm^{2} < 0.605N/mm^{2} ∴ Fail

Rather than provide shear reinforcement try increasing bars to **B40 @ 150 c/c** (also required for crack control as shown above).

v_{c} = (0.27/γ_{m})(100A_{s}/b_{w}d)^{1/3}(f_{cu})^{1/3} = (0.27 / 1.25) × ({100 × 8378} / {1000 × 920})^{1/3} × (40)^{1/3} = 0.716

ξ_{s}v_{c} = 0.86 × 0.716 = 0.616 N/mm^{2} > 0.605N/mm^{2} ∴ 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 b_{a}d = 0.0015 × 1000 × 924 = 1386 mm^{2}/m (use **B20 @ 200c/c** - A_{s} = 1570mm^{2}/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 45^{o} dispersal to the curtain wall and a 45^{o} 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 × 3.0^{2} / 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** :

M_{ult} = 584 kNm/m > 285 kNm/m ∴ OK

SLS Moment produces crack width of 0.14mm < 0.25 ∴ OK

ξ_{s}v_{c} = 0.97 N/mm^{2} > v = 0.48 N/mm^{2} ∴ Shear OK

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