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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'_{cν}) = 30° and a weight density = 19kN/m^{3}.

Backfill material will be Class 6N with an angle of shearing resistance (φ'_{bf;k}) = 35° and weight density (γ_{bf;k}) = 19kN/m^{3}.

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__

(γ_{G;sup} / γ_{G;inf})

__A2__

(γ_{G;sup} / γ_{G;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

Characteristic loading on 1m length of abutment:

Deck Dead Load = 1900 / 11.6 = **164kN/m**

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 (T_{e,min}) and maximum (T_{e,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 × 10^{3} = 11.3mm.

The design thermal movement in the deck will be ± [(11.3 / 2) γ_{F}] = ±[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:

- 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 -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 (T_{0}) 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 T_{0} 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:

- Maximum Load = 800kN
- 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/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}.

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 = 0.6α_{Q1}(2Q_{1k})+0.1α_{q1}q_{q1}w_{1}L = 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) = Q

_{lk,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 = K_{a} γ_{bf;k} h

For Design of Structural Members use at-rest earth pressures = K_{0} γ_{bf;k} h

__SLS__

__Combination 1__

__Combination 2__

Partial factors for soil parameters γ_{M}

1.0

1.0

1.25

φ'_{bf;d} = tan^{-1}[tan(φ'_{bf;k})/γ_{M}]

35.0°

35.0°

29.3°

K_{a} = (1-Sinφ'_{bf;d}) / (1+Sinφ'_{bf;d})

0.271

0.271

0.343

K_{0} = 1-Sinφ'_{bf;d}

0.426

0.426

0.511

Partial factors for soil weight γ_{G} (_{sup}/_{inf})

1.0/1.0

1.35/0.95

1.0/1.0

Backfill density (γ_{bf;d}) = γ_{bf;k} γ_{G;sup}

19.0

25.65

19.0

Backfill density (γ_{bf;d}) = γ_{bf;k} γ_{G;inf}

19.0

18.1

19.0

Model factors γ_{Sd;K}

1.0

1.2

1.2

H_{ap;d} = γ_{bf;d}K_{a}γ_{Sd;K}Z^{2}/2

2.575Z^{2}kN/m

4.171Z^{2}kN/m

3.91Z^{2}kN/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 W_{eff} 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 (N_{lane}) in the surcharge model will be 2.

From PD 6694-1:2011 Table 7 :

__Normal Traffic__

Line Load kN/m = H_{sc;F} = F.K_{d}.N_{lane}/W_{abut} = 2×330K_{d}×2/11.6 = 113.79K_{d}

*Note: D _{f} 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/m^{2} = σ_{h;ave} = σ_{h}.W_{lane}.N_{lane}/W_{abut} = 20×K_{d}×3×2/11.6 = 10.34K_{d}

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

Line Load kN/m (at ground level) = H_{sc;F} = F.K_{d}.(1 + ψ_{1})/W_{abut} = 2×330K_{d}×1.75/11.6 = 99.57K_{d}

UDL kN/m^{2} = σ_{h;ave} = (σ_{h1}+ψ_{1}.σ_{h2}).W_{lane}/W_{abut} = (30+0.75×20)×K_{d}×3/11.6 = 11.64K_{d}

SLS

Combination 1

Combination 2

Partial Factor on Surcharge γ_{Q}

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.

__Load Combinations__

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 (ψ_{1}LM1 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/m^{3}.

__ SLS__ (γ

_{G}= γ

_{Q}= γ

_{Sd;K}= 1.0)

Weight of wall stem = γ_{G} × t_{wall} × Z_{wall} × γ_{conc} = 1.0 × 1.0 × 6.5 × 25 = 162.5kN/m

Weight of base = γ_{G} × W_{base} × Z_{base} × γ_{conc} = 1.0 × 6.4 × 1.0 × 25 = 160kN/m

Weight of backfill = γ_{G} × W_{heel} × Z_{heel} × γ_{bf;d} = 1.0 × 4.3 × 8.5 × 19 = 694.5kN/m

Backfill Force H_{ap;d} = γ_{G} × γ_{Sd;K} × K_{a} × γ_{bf;d} × Z^{2}/2 = 1.0 × 1.0 × 0.271 × 19 × 9.5^{2} / 2 = 232kN/m

Frequent value of Surcharge UDL Force H_{sc;udl} = ψ_{1} × γ_{Q} × σ_{h;ave} × Z = 0.75 × 1.0 × (10.34 × 0.271) × 9.5 = 20 kN/m

Frequent value of Surcharge Line Load Force H_{sc;F} = ψ_{1} × γ_{Q} × H_{sc;F} = 0.75 × 1.0 × (113.79 × 0.271) = 23 kN/m

Deck Maximum Permanent load (concrete + surfacing_{max}) = γ_{G} × V_{DL} = 1.0 × (164 + 43) = 207kN/m

Deck Minimum Permanent load (concrete + surfacing_{min}) = γ_{G} × V_{DL} = 1.0 × (164 + 17) = 181kN/m

Deck Vertical Traffic load (gr2) = γ_{Q} × V_{traffic} = 1.0 × 89 = 89kN/m

Deck Horizontal Traffic load (gr2) = γ_{Q} × H_{braking} = 1.0 × 36 = 36kN/m

Restoring Effects:

__Minimum__

V

Lever Arm

Moment About A

Stem

162.5

1.6

260

Base

160

3.2

512

Backfill

694.5

4.25

2952

Deck (V_{DLmin})

181

1.55

281

∑ =

1198

∑ =

4005

__Maximum__

V

Lever Arm

Moment About A

Stem

162.5

1.6

260

Base

160

3.2

512

Backfill

694.5

4.25

2952

Deck (V_{DLmax} + V_{traffic})

296

1.55

459

∑ =

1313

∑ =

4183

Overturning Effects:

H

Lever Arm

Moment About A

H_{ap;d}

232

3.167

735

H_{sc;udl}

20

4.75

95

H_{sc;F}

23

9.5

219

H_{braking}

36

7.5

270

∑ =

311

∑ =

1319

For sliding effects:

φ'_{cν} = 30°

Partial factor on γ_{M} on tan(φ'_{cν;k}) = 1.0

Coefficient of friction = μ_{d} = tan(φ'_{cν;k}) / γ_{M} = tan(30°)/1.0 = 0.58

Sliding resistance = μ_{d}∑V_{min} = R_{νx;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 = 6.4m^{2}/m

I / y = 6.4^{2} / 6 = 6.827m^{3}/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/m^{2}

Pressure under heel = 205 - 196 = 9kN/m^{2} > 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', c_{u} and γ' and representative values of horizontal and vertical actions.

From Annex D.4 for Drained Conditions:

R/A' = c'N_{c}b_{c}s_{c}i_{c} + q'N_{q}b_{q}s_{q}i_{q} + 0.5γ'B'N_{γ}b_{γ}s_{γ}i_{γ}

c' = 0

γ' = γ × γ_{G;inf} = 19 × 1.0 = 19kN/m^{3}

q' = 1.5 × 19 = 28.5 kN/m^{2} (Foundation 1.5m below existing ground level)

N_{q} = e^{πtanφ'd}tan^{2}(45 + φ'_{d} / 2)

φ'_{d} = φ' = 34° (γ_{M} = 1.0)

N_{q} = e^{πtan34}tan^{2}(45 + 34 / 2) = 29.4

N_{γ} = 2(N_{q} - 1)tanφ'_{d} = 2 × (29.4 - 1) × tan34 = 38.3

b_{q} = b_{γ} = 1.0 (α = 0)

B' = B - 2e = 6.4 - 2 × 1.019 = 4.362

s_{q} = 1 + (B' / L')sinφ'_{d} = 1 + (4.362 / 11.6) × sin34 = 1.21

s_{γ} = 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

i_{q} = [1 - H / (V + A'c'_{d}cotφ'_{d})]^{m} = [1 - 311 / 1313]^{1.73} = 0.63

i_{γ} = [1 - H / (V + A'c'_{d}cotφ'_{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/m^{2}

1/3(R/A') = 1317 / 3 = 439 kN/m^{2} > 401 kN/m^{2} ∴ settlement check OK.

__ ULS__ Check Combination 1 and Combination 2.

γ_{G;sup}

γ_{G;inf}

γ_{Q}

γ_{Sd;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 = γ_{G;inf} × t_{wall} × Z_{wall} × γ_{conc}

154

162.5

Max. weight of wall stem = γ_{G;sup} × t_{wall} × Z_{wall} × γ_{conc}

219

162.5

Min. weight of base = γ_{G;inf} × W_{base} × Z_{base} × γ_{conc}

152

160

Max. weight of base = γ_{G;sup} × W_{base} × Z_{base} × γ_{conc}

216

160

Min. weight of backfill = γ_{G;inf} × W_{heel} × Z_{heel} × γ_{bf;d}

660

694.5

Max. weight of backfill = γ_{G;sup} × W_{heel} × Z_{heel} × γ_{bf;d}

937.5

694.5

K_{a}

0.271

0.343

Backfill Force H_{ap;d} = γ_{G;sup} × γ_{Sd;K} × K_{a} × γ_{bf;d} × Z^{2}/2

376

353

Frequent value of Surcharge UDL Force H_{sc;udl} = ψ_{1} × γ_{Q} × σ_{h;ave} × Z

27

29

Frequent value of Surcharge Line Load Force H_{sc;F} = ψ_{1} × γ_{Q} × H_{sc;F}

31

33.7

Deck Maximum Permanent load (concrete + surfacing_{max}) = γ_{G;sup} × V_{DL}

273

207

Deck Minimum Permanent load (concrete + surfacing_{min}) = γ_{G;inf} × V_{DL}

172

181

Deck Vertical Traffic load (gr2) = γ_{Q;sup} × V_{traffic}

120

102

Deck Horizontal Traffic load (gr2) = γ_{Q} × H_{braking}

49

41

__Combination 1__

Restoring Effects :

__Minimum__

V

Lever Arm

Moment About A

Stem

154

1.6

246

Base

152

3.2

486

Backfill

660

4.25

2805

Deck (V_{DLmin})

172

1.55

267

∑ =

1138

∑ =

3804

__Maximum__

V

Lever Arm

Moment About A

Stem

219

1.6

350

Base

216

3.2

691

Backfill

937.5

4.25

3984

Deck (V_{DLmax} + V_{traffic})

393

1.55

609

∑ =

1765.5

∑ =

5634

Overturning Effects:

H

Lever Arm

Moment About A

H_{ap;d}

376

3.167

1191

H_{sc;udl}

27

4.75

128

H_{sc;F}

31

9.5

295

H_{braking}

49

7.5

368

∑ =

483

∑ =

1982

For sliding effects:

φ'_{cν} = 30°

Partial factor on γ_{M} on tan(φ'_{cν;k}) = 1.0

Coefficient of friction = μ_{d} = tan(φ'_{cν;k}) / γ_{M} = tan(30°)/1.0 = 0.58

Sliding resistance = μ_{d}∑V_{min} = R_{νx;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/m^{2}

R/A' = c'N_{c}b_{c}s_{c}i_{c} + q'N_{q}b_{q}s_{q}i_{q} + 0.5γ'B'N_{γ}b_{γ}s_{γ}i_{γ}

c' = 0

γ' = γ × γ_{G;inf} = 19 × 0.95 = 18.1kN/m^{3}

q' = 1.5 × 18.1 = 27.2kN/m^{2} (Foundation 1.5m below existing ground level)

N_{q} = e^{πtanφ'd}tan^{2}(45 + φ'_{d} / 2)

φ'_{d} = φ' = 34° (γ_{M} = 1.0)

N_{q} = e^{πtan34}tan^{2}(45 + 34 / 2) = 29.4

N_{γ} = 2(N_{q} - 1)tanφ'_{d} = 2 × (29.4 - 1) × tan34 = 38.3

b_{q} = b_{γ} = 1.0 (α = 0)

B' = 4.138

s_{q} = 1 + (B' / L')sinφ'_{d} = 1 + (4.138 / 11.6) × sin34 = 1.20

s_{γ} = 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

i_{q} = [1 - H / (V + A'c'_{d}cotφ'_{d})]^{m} = [1 - 483 / 1765.5]^{1.74} = 0.57

i_{γ} = [1 - H / (V + A'c'_{d}cotφ'_{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/m^{2} > 427 kN/m^{2} ∴ OK.

__Combination 2__

Restoring Effects :

__Minimum__

V

Lever Arm

Moment About A

Stem

162.5

1.6

260

Base

160

3.2

512

Backfill

694.5

4.25

2952

Deck (V_{DLmin})

181

1.55

281

∑ =

1198

∑ =

4005

__Maximum__

V

Lever Arm

Moment About A

Stem

162.5

1.6

260

Base

160

3.2

512

Backfill

694.5

4.25

2952

Deck (V_{DLmax} + V_{traffic})

309

1.55

479

∑ =

1326

∑ =

4203

Overturning Effects:

H

Lever Arm

Moment About A

H_{ap;d}

353

3.167

1118

H_{sc;udl}

29

4.75

138

H_{sc;F}

33.7

9.5

320

H_{braking}

41

7.5

308

∑ =

457

∑ =

1884

For sliding effects:

φ'_{cν} = 30°

Partial factor on γ_{M} on tan(φ'_{cν;k}) = 1.25

Coefficient of friction = μ_{d} = tan(φ'_{cν;k}) / γ_{M} = tan(30°)/1.25 = 0.46

Sliding resistance = μ_{d}∑V_{min} = R_{νx;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/m^{2}

R/A' = c'N_{c}b_{c}s_{c}i_{c} + q'N_{q}b_{q}s_{q}i_{q} + 0.5γ'B'N_{γ}b_{γ}s_{γ}i_{γ}

c' = 0

γ' = γ × γ_{G;inf} = 19 × 1.0 = 19kN/m^{3}

q' = 1.5 × 19 = 28.5kN/m^{2} (Foundation 1.5m below existing ground level)

N_{q} = e^{πtanφ'd}tan^{2}(45 + φ'_{d} / 2)

φ'_{d} = tan^{-1}[tan(φ'_{k})/γ_{M}] = tan^{-1}[tan34/1.25] = 28.4°

N_{q} = e^{πtan28.4}tan^{2}(45 + 28.4 / 2) = 15.4

N_{γ} = 2(N_{q} - 1)tanφ'_{d} = 2 × (15.4 - 1) × tan28.4 = 15.6

b_{q} = b_{γ} = 1.0 (α = 0)

B' = 3.498

s_{q} = 1 + (B' / L')sinφ'_{d} = 1 + (3.498 / 11.6) × sin28.4 = 1.14

s_{γ} = 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

i_{q} = [1 - H / (V + A'c'_{d}cotφ'_{d})]^{m} = [1 - 457 / 1326]^{1.768} = 0.47

i_{γ} = [1 - H / (V + A'c'_{d}cotφ'_{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/m^{2} > 379 kN/m^{2} ∴ 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

__BearingPressure__

SLS

Toe

SLS

Heel

Comb.1

V_{d}/A' / R/A'

Comb.2

V_{d}/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

__BearingPressure__

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

__BearingPressure__

Comb.1

V_{d}/A' / R/A'

Comb.2

V_{d}/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

__BearingPressure__

SLS

Toe

SLS

Heel

Comb.1

V_{d}/A' / R/A'

Comb.2

V_{d}/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

__BearingPressure__

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

__BearingPressure__

Comb.1

V_{d}/A' / R/A'

Comb.2

V_{d}/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 γ_{M}

1.0

1.0

1.25

φ'_{bf;d} = tan^{-1}[tan(φ'_{bf;k})/γ_{M}]

35.0°

35.0°

29.3°

K_{0} = 1-Sinφ'_{bf;d}

0.426

0.426

0.511

Partial factors for soil weight γ_{G;sup}

1.0

1.35

1.0

Backfill density (γ_{bf;d}) = γ_{bf;k} γ_{G;sup}

19.0

25.7

19.0

Model factors γ_{Sd;K}

1.0

1.2

1.2

H_{ap;d} = γ_{bf;d}K_{0}γ_{Sd;K}Z^{2}/2

4.047Z^{2}kN/m

6.569Z^{2}kN/m

5.825Z^{2}kN/m

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

Backfill:

H_{ap;d}(kN) =

292

475

421

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

827

1295

1193

Frequent value of Normal Surcharge:

ψ_{1}γ_{Q;sup} =

0.75

1.013

0.863

ψ_{1}γ_{Q;sup}H_{sc;F} = ψ_{1}γ_{Q;sup}113.79K_{d}D_{f} =

24

33

34

Moment (kNm) (lever arm = 8.5) =

204

281

289

ψ_{1}γ_{Q;sup}σ_{h;ave}Z = ψ_{1}γ_{Q;sup}10.34K_{d}Z =

28

38

39

Moment (kNm) (lever arm = 4.25) =

119

162

166

Deck Permanent Load Reaction:

γ_{G;sup} for concrete =

1.0

1.35

1.0

Deck concrete =

164

221

164

γ_{G;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):

γ_{Q;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/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:**

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 ≤ λ_{lim} = 20A.B.C/√n

Use suggested values when φ_{ef} not known:

A = 0.7, B = 1.1, C = 0.7

n = N_{Ed} / (A_{c}f_{cd})

N_{Ed} = 164 + 43 + 166 = 373kN

f_{cd} = α_{cc}f_{ck} / γ_{c} = 0.85 × 32 / 1.5 = 18.1N/mm^{2}

A_{c} = 10^{6}mm^{2} (per metre width)

n = 373 × 10^{3} / (10^{6} × 18.1) = 0.021

λ_{lim} = 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.

M_{ULS} = 2163kNm/m, V_{ULS} = 606kN/m, M_{SLS} = 1429kNm/m [878(permanent)+551(variable)]

cl. 3.1.6(101)P

Design compressive strength = f_{cd} = α_{cc}f_{ck} / γ_{c}

cl. 3.1.7

α_{cc} = 0.85

cl. 2.4.2.4

Table 2.1N: γ_{c} = 1.5, γ_{s} = 1.15

f_{cd} = 0.85 × 32 / 1.5 = 18.1 N/mm^{2}

Table 3.1

ε_{c2} = 0.002, ε_{cu2} = 0.0035, n = 2.0

**Try B40's @ 150mm c/c** (8378mm^{2}/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 f_{av} = f_{cd}[1-ε_{c2} / {ε_{cu2}(n+1)}] = 18.1 × [1 - 0.002 / {0.0035 × (2 + 1)}] = 14.7 N/mm^{2}

Assuming steel yields then:

M = f_{s}z = f_{yk}A_{s}z / γ_{s} = F_{c}z = f_{av}b*X*z

Depth to neutral axis *X* = f_{yk}A_{s} / (f_{av}bγ_{s})

*X* = 500 × 8378 / (14.7 × 1000 × 1.15) = 247.8mm

Check that steel will yield:

Cl. 3.2.7(4)

Modulus of Elasticity E_{s} = 200 kN/mm^{2}

Steel strain at yield = ε_{s,yield} = f_{yk} / γ_{s} / E_{s} = 500 / 1.15 / 200000 = 0.00217

from linear strain relationship:

ε_{s} = ε_{cu2}(d/*X* - 1) = 0.0035 ( 920 / 247.8 - 1) = 0.009 > 0.00217 ∴ steel will yield.

Hence M_{ult} = f_{av}b*X*z = f_{av}b*X*(d - β*X*)

Where β = 1 - [0.5ε_{cu2}^{2} - ε_{c2}^{2} / {(n+1)(n+2)}] / [ε_{cu2}^{2} - ε_{cu2}ε_{c2} / (n+1)]

β = 1 - [0.5 × 0.0035^{2} - 0.002^{2} / {(2 + 1) × (2 + 2)}] / 0.0035^{2} - 0.0035 × 0.002 / (2 + 1)] = 0.416

M_{ult} = 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

E_{cm} = 22[(f_{ck} + 8) / 10]^{0.3} = 22[(32 + 8) / 10]^{0.3} = 33.4 kN/mm^{2}

E_{c,eff} = E_{cm} = 33.4 kN/mm^{2}

Modular Ratio m = E_{s} / E_{cm} = 200 / 33.4 = 6.0

Let d_{c} = depth to neutral axis then equating strains for cracked section:

ε_{s} = ε_{c}(d - d_{c}) / d_{c}

Equating forces:

A_{s}E_{s}ε_{s} = 0.5bd_{c}ε_{c}E_{c,eff}

Hence d_{c} = [-A_{s}E_{s} + {(A_{s}E_{s})^{2} + 2bA_{s}E_{s}E_{c,eff}d}^{0.5}] / bE_{c,eff}

d_{c} = [-8378 × 200000 + {(8378 × 200000)^{2} + 2 × 1000 × 8378 × 200000 × 33400 × 920}^{0.5}] / (1000 × 33400) = 258mm

Cracked second moment of area = A_{s}(d-d_{c})^{2} + E_{c,eff}bd_{c}^{3} / 3E_{s}

I_{NA} = 8378 × (920 - 258)^{2} + 33.4 × 1000 × 258^{3} / (3 × 200) = 4.63 × 10^{9} mm^{4} (steel units)

Approximate concrete stress σ_{c} = M / z_{c} + N / A_{c}

N (Case 2) = 164 + 43 = 207 kN

σ_{c} ≅ {1429 × 10^{6} × 258 / (4.63 × 10^{9} × 6.0)} + {207 × 10^{3} / (258 × 10^{3})} = 13.3 + 0.8 = 14.1 N/mm^{2}

cl. 7.2(102)

Limiting concrete stress = k_{1}f_{ck}

k_{1} = 0.6

Limiting concrete stress = 0.6 × 32 = 19.2 N/mm^{2} > 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 = E_{cm}

Long-term modulus = E_{cm} / (1+φ)

Effective modulus E_{c,eff} = (M_{qp} + M_{st})E_{cm} / {M_{st} + (1 + φ)M_{qp}}

Table 3.1

f_{cm} = f_{ck} + 8 = 32 + 8 = 40 N/mm^{2}

Cl. 3.1.4

Relative humidity of the ambient environment = 80% (outside conditions)

Age of concrete at initial loading t_{0} = say 7 days (after formwork has been released and waterproofing system applied to rear face of wall)

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

h_{0} = 2A_{c} / U = 2 × (11600 × 1000) / (11600 + 2 × 1000) = 1706

α_{1} = [35 / f_{cm}]^{0.7} = [35 / 40]^{0.7} = 0.91

α_{2} = [35 / f_{cm}]^{0.2} = [35 / 40]^{0.2} = 0.97

(B.3b)

φ_{RH} = [1 + α_{1} × {(1 - RH / 100) / (0.1 × h_{0}^{1/3})}] × α_{2}

φ_{RH} = [1 + 0.91 × {(1 - 80 / 100) / ( 0.1 × 1706^{1/3})}] × 0.97 = 1.118

(B.4)

β(f_{cm}) = 16.8 / f_{cm}^{0.5} = 16.8 / 40^{0.5} = 2.656

(B.5)

β(t_{0}) = 1 / (0.1 + t_{0}^{0.2}) = 1 / ( 0.1 + 7^{0.2}) = 0.635

(B.2)

φ_{0} = φ_{RH} × β(f_{cm}) × β(t_{0}) = 1.118 × 2.656 × 0.635 = 1.886

Moment due to long-term actions = M_{qp} = 878 kNm

Moment due to short-term actions = M_{st} = 551 kNm

Hence Effective Modulus E_{c,eff} = {(878 + 551) × 33.4} / {551 + 878 × ( 1 + 1.886)} = 15.5 kN/mm^{2}

Modular Ratio m = E_{s} / E_{c,eff} = 200 / 15.5 = 12.9

Let d_{c} = depth to neutral axis then equating strains for cracked section:

ε_{s} = ε_{c}(d - d_{c}) / d_{c}

Equating forces:

A_{s}E_{s}ε_{s} = 0.5bd_{c}ε_{c}E_{c,eff}

Hence d_{c} = [-A_{s}E_{s} + {(A_{s}E_{s})^{2} + 2bA_{s}E_{s}E_{c,eff}d}^{0.5}] / bE_{c,eff}

d_{c} = [-8378 × 200000 + {(8378 × 200000)^{2} + 2 × 1000 × 8378 × 200000 × 15500 × 920}^{0.5}] / (1000 × 15500) = 351mm

Cracked second moment of area = A_{s}(d-d_{c})^{2} + E_{c,eff}bd_{c}^{3} / 3E_{s}

I_{NA} = 8378 × (920 - 351)^{2} + 15.5 × 1000 × 351^{3} / (3 × 200) = 3.83 × 10^{9} mm^{4} (steel units)

Concrete stress σ_{c} ≅ M / z_{c} + N / A_{c}

σ_{c} = {1429 × 10^{6} × 351 / (3.83 × 10^{9} × 12.9)} + (207 × 10^{3} / (351 × 10^{3}) = 10.2 + 0.6 = 10.8 N/mm^{2}

cl. 7.2(102)

Limiting concrete stress = k_{1}f_{ck}

k_{1} = 0.6

Limiting concrete stress = 0.6 × 32 = 19.2 N/mm^{2} > 10.8 ∴ OK

cl. 7.2(5)

Limiting steel stress = k_{3}f_{yk}

k_{3} = 0.8

Limiting steel stress = 0.8 × 500 = 400 N/mm^{2}

Steel stress σ_{s} = M / z_{s}

σ_{s} = 1429 × 10^{6} × (920 - 351) / (3.83 × 10^{9}) = 212 N/mm^{2} < 400 ∴ >OK

**Crack Control**:

Consider worst condition before creep has occurred and

Quasi-Permanent Combination Moment + ψ_{2} × temperature effects = 878 + 0.5(16 × 6.63) = 931 kNm

Cl. 7.3.4(1)

Crack width w_{k} = s_{r,max}(ε_{sm} - ε_{cm})

Cl. 7.3.4(3)

Spacing Limit = 5(c+φ/2) = 5(60 + 40/2) = 400mm > 150mm ∴ OK

s_{r,max} = k_{3}c + k_{1}k_{2}k_{4}φ / ρ_{p,eff}

k_{1} = 0.8 (high bond bars)

k_{2} = 0.5 (for bending)

k_{3} = 3.4 (recommended value)

k_{4} = 0.425 (recommended value)

Cl. 7.3.2(3)

h_{c,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

∴ h_{c,eff} = 200 mm

and A_{c,eff} = 200 × 1000 = 200000 mm^{2}

Cl. 7.3.4(2)

ρ_{p,eff} = A_{s} / A_{c,eff} = 8378 / 200000 = 0.0419

s_{r,max} = k_{3}c + k_{1}k_{2}k_{4}φ / ρ_{p,eff}

s_{r,max} = (3.4 × 60) + (0.8 × 0.5 × 0.425 × 40) / 0.0419 = 204 + 162 = 366

Cl. 7.3.4(2)

(ε_{sm} - ε_{cm}) = [σ_{s} - {k_{t}f_{ct,eff}(1 + α_{e}ρ_{p,eff}) /ρ_{p,eff}}] / E_{s} ≥ 0.6σ_{s} / E_{s}

k_{t} = 0.4 for permanent loading

α_{e} = E_{s} / E_{cm} = 200 / 33.4 = 6.0

σ_{s} = 931 × 10^{6} × (920 - 258) / (4.63 × 10^{9}) = 133 N/mm^{2}

Table 3.1

f_{ct,eff} = f_{ctm} = 0.3 × f_{ck}^{(2/3)} = 0.3 × 32^{(2/3)} = 3.02 N/mm^{2}

(ε_{sm} - ε_{cm}) = [133 - {0.4 × 3.02 × (1 + 6.0 × 0.0419) / 0.0419}] / 200000 = 0.485 × 10^{-3}

0.6σ_{s} / E_{s} = 0.6 × 133 / 200000 = 0.399 × 10^{-3} < 0.485 × 10^{-3} ∴ OK

Crack width w_{k} = s_{r,max}(ε_{sm} - ε_{cm}) = 366 × 0.485 × 10^{-3} = 0.18 mm

NA EN 1992-2 Table NA.2

Recommended value of w_{max} = 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

V_{Rd,c} = [C_{Rd,c}k(100ρ_{1}f_{ck})^{1/3}]b_{w}d

C_{Rd,c} = 0.18 / γ_{c} = 0.18 / 1.5 = 0.12

k = 1 + (200 / d)^{0.5} ≤2.0

k = 1 + (200 / 920)^{0.5} = 1.47 < 2.0

ρ_{1} = A_{sl} / b_{w}d ≤0.02

ρ_{1} = 8378 / (1000 × 920) = 0.009 < 0.02

Cl. 3.1.2(102)P

f_{ck} = 32 ( < C_{max} = C50/60)

V_{Rd,c} = [0.12 × 1.47 × (100 × 0.009 × 32)^{1/3}] × 1000 × 920 × 10^{-3} = 497 kN ( < 606 kN Fail : see below)

Minimum V_{Rd,c} = (v_{min})b_{w}d = 0.035k^{3/2}f_{ck}^{1/2}b_{w}d = 0.035 × 1.47^{3/2} × 32^{1/2} × 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 = 0.5b_{w}dνf_{cd}

ν = 0.6[1 - f_{ck} / 250] = 0.6 × [1 - 32 / 250] = 0.523

f_{cd} = α_{cc}f_{ck}/γ_{c}

α_{cc} = 1.0 [see NA to Cl. 3.1.6(101)P]

f_{cd} = 1.0 × 32 / 1.5 = 21.3 N/mm^{2}

Maximu V_{Ed} = 0.5 × 1000 × 920 × 0.523 × 21.3 × 10^{-3} = 5124 kN >> 606 kN

V_{Rd,c} = 497 kN < V_{Ed} = 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 V_{Rd,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

V_{Rd,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

V_{Rd,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:

A_{s,vmin} = 0.002A_{c} = 0.002 × 10^{6} = 2000 mm^{2}/m (1000 mm^{2}/m in each face). Use B16 @ 150 c/c (A_{s} = 1340mm^{2}/m).

EN 1992-1-1 Clause 9.6.3 - Horizontal reinforcement:

A_{s,vmin} = 0.001A_{c} or 25% of vertical reinforcement = 0.001 × 10^{6} = 1000 mm^{2}/m (in each face) or

25% × 8378 = 2095mm^{2}/m. B20 @ 150 c/c = 2094mm^{2}/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 H_{ap;d} = γ_{G} × γ_{Sd;K} × K_{0} × γ_{bf;d} × Z^{2} / 2 = 1.0 × 1.0 × 0.426 × 19 × 9.5^{2} / 2 = 365kN/m

Frequent value of Surcharge UDL Force H_{sc;udl} = ψ_{1} × γ_{Q} × σ_{h;ave} × Z = 0.75 × 1.0 × (10.34 × 0.426) × 9.5 = 31kN/m

Dispersion Factor D_{f} for line load = (1 + Z / 2) / (1 + Z) = (1 + 9.5 / 2) / (1 + 9.5) = 0.55 < 0.67 ∴ D_{f} = 0.67

Frequent value of Surcharge Line Load Force H_{sc;F} = ψ_{1} × γ_{Q} × H_{sc;F} = 0.75 × 1.0 × (113.79 × 0.426 × 0.67) = 24kN/m

Deck Maximum Permanent load (concrete + surfacing_{max}) = 207kN/m

Deck Vertical Traffic load (gr2) = 89kN/m

Deck Horizontal Traffic load (gr2) = 36kN/m

Restoring Effects:

V

Lever Arm

Moment About A

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

Moment About A

Backfill

365

3.167

1156

Surcharge UDL

31

4.75

147

Surcharge Line Load

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 = 6.4m^{2}/m

I / y = 6.4^{2} / 6 = 6.827m^{3}/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/m^{2}

Pressure under heel = 205 - 267 = -62kN/m^{2} (uplift)

Adjust for 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/m^{2}

Pressure at front face of wall = 483 × 4.342 / 5.442} = 385kN/m^{2}

Pressure at rear face of wall = 483 × 3.342 / 5.442} = 297kN/m^{2}

SLS Moment at a-a = (385 × 1.1^{2} / 2) + ([483 - 385] × 1.1^{2} / 3) - (25 × 1.0 × 1.1^{2} / 2) = 257kNm/m (tension in bottom face).

SLS Moment at b-b = (297 × 3.342^{2} / 6) - (695 × 4.3 / 2) - (25 × 1.0 × 4.3^{2} / 2) = -1173kNm/m (tension in top face).

__CASE 6__ - Fixed Abutment **Ultimate Limit State**

Weight of wall stem Comb.1 = γ_{G;sup} × 162.5 = 1.35 × 162.5 = 219 kN/m

Weight of wall stem Comb.2 = γ_{G;sup} × 162.5 = 1.0 × 162.5 = 163 kN/m

Weight of base Comb.1 = γ_{G;sup} × 160 = 1.35 × 160 = 216 kN/m

Weight of base Comb.2 = γ_{G;sup} × 160 = 1.0 × 160 = 160 kN/m

Weight of backfill Comb.1 = γ_{G;sup} × 694.5 = 1.35 × 694.5 = 938 kN/m

Weight of backfill Comb.2 = γ_{G;sup} × 694.5 = 1.0 × 694.5 = 695 kN/m

B/fill Force H_{ap;d} Comb.1 = γ_{G;sup} × γ_{Sd;K} × K_{0} × γ_{bf;d} × Z^{2} / 2 = 1.35 × 1.2 × 0.426 × 19 × 9.5^{2} / 2 = 592kN/m

B/fill Force H_{ap;d} Comb.2 = γ_{G;sup} × γ_{Sd;K} × K_{0} × γ_{bf;d} × Z^{2} / 2 = 1.0 × 1.2 × 0.511 × 19 × 9.5^{2} / 2 = 526kN/m

Frequent value of Surcharge UDL Force H_{sc;udl} Comb.1 = ψ_{1} × γ_{Q;sup} × σ_{h;ave} × Z = 0.75 × 1.35 × (10.34 × 0.426) × 9.5 = 42kN/m

Frequent value of Surcharge UDL Force H_{sc;udl} Comb.2 = ψ_{1} × γ_{Q;sup} × σ_{h;ave} × Z = 0.75 × 1.15 × (10.34 × 0.511) × 9.5 = 43kN/m

Dispersion Factor D_{f} for line load = (1 + Z / 2) / (1 + Z) = (1 + 9.5 / 2) / (1 + 9.5) = 0.55 < 0.67 ∴ D_{f} = 0.67

Frequent value of Surcharge Line Load Force H_{sc;F} Comb.1 = ψ_{1} × γ_{Q;sup} × H_{sc;F} = 0.75 × 1.35 × (113.79 × 0.426 × 0.67) = 33kN/m

Frequent value of Surcharge Line Load Force H_{sc;F} Comb.2 = ψ_{1} × γ_{Q;sup} × H_{sc;F} = 0.75 × 1.15 × (113.79 × 0.511 × 0.67) = 34kN/m

Deck Maximum Permanent load (concrete + surfacing_{max}) Comb.1 = γ_{G;sup} × 164 + γ_{G;sup} × 43 = 1.35 × 164 + 1.2 × 43 = 273kN/m

Deck Maximum Permanent load (concrete + surfacing_{max}) Comb.2 = γ_{G;sup} × 164 + γ_{G;sup} × 43 = 1.0 × 164 + 1.0 × 43 = 207kN/m

Deck Vertical Traffic load (gr2) Comb.1 = γ_{Q;sup} × 89 = 1.35 × 89 = 120 kN/m

Deck Vertical Traffic load (gr2) Comb.2 = γ_{Q;sup} × 89 = 1.15 × 89 = 102 kN/m

Deck Horizontal Traffic load (gr2) Comb.1 = γ_{Q;sup} × 36 = 1.35 × 36 = 49 kN/m

Deck Horizontal Traffic load (gr2) Comb.2 = γ_{Q;sup} × 36 = 1.15 × 36 = 41 kN/m

Restoring Effects:

V

Comb.1/Comb.2

Lever Arm

Moment About A

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

Moment About A

Backfill

592/526

3.167

1875/1666

Surcharge UDL

42/43

4.75

200/204

Surcharge Line Load

33/34

9.5

314/323

Deck Horizontal Load

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/m^{2}

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/m^{2}

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 Moment at a-a = (541 × 1.1^{2} / 2) - (1.35 × 25 × 1.0 × 1.1^{2} / 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 × 4.3^{2} / 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 Moment at a-a = (516 × 1.1^{2} / 2) - (1.0 × 25 × 1.0 × 1.1^{2} / 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 × 4.3^{2} / 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

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

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

Section a-a: M_{uls} = 307 kNm/m, V_{uls} = 558 kN/m, M_{sls} = 258 kNm/m (M_{perm} = 181 kNm/m + M_{var} = 77 kNm/m)

B25's @ 150 c/c give M_{uls} = 1262 kNm/m > 307 ∴ OK, V_{uls} = 731 kN/m (at d from support) > 558 ∴ OK, M_{sls} = 1103 kNm/m > 258 ∴ OK

Section b-b: M_{uls} = 2054 kNm/m, V_{uls} = 582 kN/m, M_{sls} = 1239 kNm/m (M_{perm} = 860 kNm/m + M_{var} = 380 kNm/m)

B40's @ 150 c/c give M_{uls} = 2975 kNm/m > 2054 ∴ OK, V_{uls} = 996 kN/m (at d from support) > 582 ∴ OK, M_{sls} = 2065 kNm/m > 1239 ∴ 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 = 0.6α_{Q1}Q_{1k} 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 = K_{0} γ_{bf;d} Z^{2} / 2 = 0.426 × 19.0 × 3.0^{2} / 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 = K_{0} γ_{bf;d} Z^{3} / 6 = 0.426 × 19.0 × 3.0^{3} / 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:

D_{f} = (1 + Z/2) / (1 + Z) = (1 + 3/2) / (1 + 3) = 0.625 < 0.67 ∴ D_{f} = 0.67

Line Load = 113.79K_{d}D_{f} = 113.79 × 0.426 × 0.67 = 32.5kN/m

UDL = 10.34K_{d} = 10.34 × 0.426 = 4.4kN/m^{2}

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 × 3.0^{2} / 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 = 99.57K_{d}D_{f} = 99.57 × 0.426 × 0.67 = 28.4kN/m

UDL = 11.64K_{d} = 11.64 × 0.426 = 5.0kN/m^{2}

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 × 3.0^{2} / 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** :

M_{ult} = 601 kNm/m > 216 kNm/m ∴ OK

M_{sls} = 317 kNm/m > 153 kNm/m ∴ OK

V_{ult} = 261 kN/m > 120 kN/m ∴ Shear OK

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Last Updated : 14/12/16

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