F Bridge Design| Reinforced Concrete Bridge Deck Design to Eurocodes

## Reinforced Concrete Deck Example to Eurocodes

Reinforced Concrete Deck Design to EN 1992-2 & UK National Annex

Problem:

Design a simply supported reinforced concrete deck slab using a unit strip method. The deck carries a 100mm depth of surfacing, together with a traffic load (LM1) udl of 5.5 kN/m2 and tandem axle load of 100kN (300kN/3m lane width). The deck should also be designed to carry the SV80 model vehicle. The span of the deck is 12.0m centre to centre of bearings. γconc. = 25kN/m3

SV80 model vehicle also to be considered.
Use C32/40 concrete to BS 8500.
Use Grade B500B reinforcement to BS 4449.

A Solution:

EN 1992-1-1 Table 4.1 gives the Exposure Class XD1 as suitable for deck soffits.

NA to BS EN 1992-1-1 clause 4.4.1.2(5) says values of cmin,dur shall be taken from BS 8500-1

BS 8500-1 Table A.5

Nominal cover for C32/40 concrete = cnom = cmin,dur + Δcdev = 45 + Δcdev with
maximum water-cement ratio = 0.55 and minimum cement content of 320 kg/m3
Fixing tolerence for reinforcement Δcdev = 15mm for insitu concrete
∴ Nominal cover cnom = 45 + 15 = 60mm

Design for a 1 metre width of deck (unit strip)
Note: The loading has been simplified to demonstrate the method of designing the slab
(See BS EN 1991-1-1 to 1991-1-7, 1991-2 and National Annex for full design loading.)

EN 1991-1-1 Annex A

Permanent Actions (Gk):

deck slab = 25 × 0.65 × 1.0 = 16.3 kN/m

NA Table NA.1 cl. 5.2.3(3)

It will be assumed that the depth of surfacing could vary considerably as a result of future resurfacing. Clause 5.2.3(3) makes an allowance of up to 55% additional weight.

surfacing = 1.55 × 24 × 0.1 × 1.0 = 3.7 kN/m

EN 1991-2
Table 4.2

Variable Actions (Qk):

Traffic Load Model 1 (Q1k = 1 & q1k = 0.61)

= 5.5 kN/m(udl) + 2 × (300kN/3m) axles @ 1.2m centres.

NA Fig NA.1

= 6 × (130kN/3m) axles @ 1.2m c/c

NA Cl. 2.16.3

DAF for 130kN axle = 1.16 ∴ axle load = 1.16×130/3 = 50.3kN

Combination of Actions

a) Ultimate Limit State : EN 1990 clause 6.4.3.2 requires one combination of actions to be considered for the "STR" limit state:

Eqn 6.10

Ed = E(ΣγGjGkj + γpP + γQ,1Qk,1 + ΣγQ,iΨ0,iQk,i)

Table NA.A.2.4

γsup

γinf

Ψ0

1.35

0.95

-

1.20

0.95

-

Temperature Difference

1.55

0

0

Traffic Groups

1.35

0

N/A

Key:

EN 1990 Table A.2.1 Note 3 says the value of Ψ0 of 0.6 may be reduced to 0 when considering ULS.

b) Serviceability Limit State : EN 1990 clause 6.5.3 requires three combinations of actions to be considered for the serviceability limit state:

1. Characteristic combination  Ed = E(ΣGkj + P + Qk,1 + ΣΨ0,iQk,i)   for limiting stresses
2. Frequent combination  Ed = E(ΣGkj + P + Ψ1,1Qk,1 + ΣΨ2,iQk,i)
3. Quasi-permanent combination  Ed = E(ΣGkj + P + ΣΨ2,iQk,i)   for crack control

Table NA.A.2.1

Action

Ψ0

Ψ1

Ψ2

0.75

0.75

0

0

0

0

Thermal

0.6

0.6

0.5

Temperature Difference Effects

Apply temperature differences given in EN 1991-1-5 Figure 6.2c(Type 3a) to a 1m wide deck section.
EN 1991-1-5 Table C.1 - Coefficient of thermal expansion = 10 × 10-6 per °C.
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.
∴ use 12 × 10-6 per °C
From EN 1992-1-1 Table 3.1 :
fck = 32  →  fcm = fck +8 = 40  →  Ecm = 22[(fcm) / 10]0.3 = 22 × 40.3 = 33.35 kN/mm2
Hence restrained temperature stresses per °C = 33.35 × 103 × 12 × 10-6 = 0.4 N/mm2
Interpolating values of ΔT from EN 1991-1-5 Table B.3 for a 0.65m depth of slab with 100mm surfacing we get: Section Properties
Area = 1000 × 650 = 0.65 × 106 mm2
Second Moment of Area = 1000 × 6503 / 12 = 22.9 × 109 mm4

a) Heating temperature difference
Force F to restrain temperature strain :
0.4 × 103 × [ 150 × ( 3.0 + 5.05 ) + (195 × 1.5) + (195 × 1.05)] × 10-3 = 683 kN
Moment M about centroid of section to restrain curvature due to temperature strain :
0.4 × 103 × [150 × (3.0 × 250 + 5.05 × 275) + 175 × (0.3 × 87.5 + 1.35 × 116.7) - (20 × 0.15 × 6.7) - (195 × 1.05 × 260)] × 10-6 = 119.9kNm b) Cooling temperature difference
Force F to restrain temperature strain :
- 0.4 × 103 × [ 130 × ( 1.8 + 2.5 + 1.5 + 1.9 ) + 163 × ( 0.9 + 0.75 )] × 10-3 = - 508 kN
Moment M about centroid of section to restrain curvature due to temperature strain :
- 0.4 × 103 × [130 × ( 1.8 × 260 + 2.5 × 282 - 1.5 × 260 - 1.9 × 282 ) +
163 × ( 0.9 × 141 - 0.75 × 141 )] × 10-6 = -14.3kNm Note:
1) Sign convention is compressive stresses are positive.
2) The deck is simply supported and allowed to expand and contract freely. Therefore there will be no secondary stresses due to the curvature and axial strain in the deck.

SLS = Serviceability Limit State
ULS = Ultimate Limit State

Design SLS moment = ∑Gkj = (16.3 + 3.7) × 122 / 8 = 360 kNm

Design ULS moment = ∑(γGjGkj) = [(1.35 × 16.3) + (1.2 × 3.7)] × 122 / 8] = 476 kNm

Variable Actions (per metre width of deck)

Traffic Group gr1a Moment at leading axle = 5.5 × (6 × 6.3 - 6.32/2) + 100 × 5.7 × (6.3 + 5.1) / 12 = 640 kNm

Design SLS moment characteristic combination = Qk1 = 640 kNm

Design ULS moment = γQ1Qk1 = 1.35 × 640 = 864 kNm

Traffic Group gr5 Model LM1 is positioned 5m clear of LM3 and will be off the deck.

Moment at X = 143.4 × 5.7 - 50.3 × (2.4 + 1.2) = 636kNm

Design SLS moment characteristic combination = Qk1 = 636 < 640 kNm
gr1a governs

Design ULS moment = γQ1Qk1 = 1.35 × 636 = 859 < 864 kNm
gr1a governs

Combinations of Actions

a) Ultimate Limit State

Ed = E(ΣγGjGkj + γpP + γQ,1Qk,1 + ΣγQ,iΨ0,iQk,i)

Design ULS mid span moment = 476 + 0 + 864 + 0 = 1340 kNm

b) Serviceability Limit State

Characteristic Combination

Ed = E(ΣGkj + P + Qk,1 + ΣΨ0,iQk,i)

Design SLS mid span moment:

= 360 + 0 + 640 + 0.6 × (differential temperature effects)

= 1000 kNm + 0.6 × (differential temperature effects)

Quasi-permanent Combination

Ed = E(ΣGkj + P + ΣΨ2,iQk,i)

Design SLS mid span moment = 360 + 0 + 0.5 × 0 = 360 kNm

Ultimate Capacity of Deck Slab

Ultimate Design Moment = 1340 kNm EN 1992-1-1 & EN 1992-2

It is usual to design reinforced concrete for the ultimate limit state and check for serviceability conditions.

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 32mm dia. reinforcement at 125mm centres:
Nominal cover to reinforcement in deck soffit = 60mm

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 × 6434 / (14.7 × 1000 × 1.15) = 190.6mm

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 ( 574 / 190.6 - 1) = 0.007 > 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 × 190.6 × (574 - 0.416 × 190.6) × 10-6 = 1386 kNm > 1340 ∴ OK

Check Serviceability Limit State

Characteristic Combination SLS Design Moment = 1000 kNm (360 + 640) + 0.6 × (differential temperature effects)

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 = [-6434 × 200000 + {(6434 × 200000)2 + 2 × 1000 × 6434 × 200000 × 33400 × 574}0.5] / (1000 × 33400) = 175mm
Cracked second moment of area = As(d-dc)2 + Ec,effbdc3 / 3Es
INA = 6434 × (574 - 175)2 + 33.4 × 1000 × 1753 / (3 × 200) = 1.32 × 109 mm4 (steel units)
Concrete stress σc = M / zc + 0.6 × (differential temperature effects)
σc = {1000 × 106 × 175 / (1.32 × 109 × 6.0)} + (0.6 × 2.49) = 22.1 + 1.5 = 23.6 N/mm2 (Heating temperature difference)

cl. 7.2(102)

Limiting concrete stress = k1fck
k1 = 0.6
Limiting concrete stress = 0.6 × 32 = 19.2 N/mm2 < 23.6 ∴ Fail

The concrete strength class will need to be increased to class C40/50
Re-working the example gives:
Ecm = 35.2 kN/mm2    m = 5.7    dc = 171 mm    INA = 1.34 × 109 mm4
σc = 22.4 + 1.5 = 23.9 N/mm2
Limiting stress = 0.6 × 40 = 24 N/mm2 > 23.9 ∴ 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 = 40 + 8 = 48 N/mm2

Cl. 3.1.4

Relative humidity of the ambient environment = 80% (outside conditions)
Age of concrete at initial loading t0 = 6 days (when soffit formwork is released)

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

h0 = 650
α1 = [35 / fcm]0.7 = [35 / 48]0.7 = 0.80
α2 = [35 / fcm]0.2 = [35 / 48]0.2 = 0.94

(B.3b)

φRH = [1 + α1 × {(1 - RH / 100) / (0.1 × h01/3)}] × α2
φRH = [1 + 0.8 × {(1 - 80 / 100) / ( 0.1 × 6501/3)}] × 0.94 = 1.113

(B.4)

β(fcm) = 16.8 / fcm0.5 = 16.8 / 480.5 = 2.425

(B.5)

β(t0) = 1 / (0.1 + t00.2) = 1 / ( 0.1 + 60.2) = 0.653

(B.2)

φ0 = φRH × β(fcm) × β(t0) = 1.113 × 2.425 × 0.653 = 1.762
Moment due to long-term actions = Mqp = 360 kNm
Moment due to short-term actions = Mst = 640 kNm
Hence Effective Modulus Ec,eff = {(360 + 640) × 35.2} / {640 + 360 × ( 1 + 1.762)} = 21.5 kN/mm2

Modular Ratio m = Es / Ec,eff = 200 / 21.5 = 9.3
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 = [-6434 × 200000 + {(6434 × 200000)2 + 2 × 1000 × 6434 × 200000 × 21500 × 574}0.5] / (1000 × 21500) = 209mm
Cracked second moment of area = As(d-dc)2 + Ec,effbdc3 / 3Es
INA = 6434 × (574 - 209)2 + 21.5 × 1000 × 2093 / (3 × 200) = 1.18 × 109 mm4 (steel units)
Concrete stress σc = M / zc + 0.6 × (differential temperature effects)
σc = {1000 × 106 × 209 / (1.18 × 109 × 9.3)} + (0.6 × 2.49) = 19.1 + 1.5 = 20.6 N/mm2 (Heating temperature difference)

cl. 7.2(102)

Limiting concrete stress = k1fck
k1 = 0.6
Limiting concrete stress = 0.6 × 40 = 24.0 N/mm2 > 20.6 ∴ OK

cl. 7.2(5)

Limiting steel stress = k3fyk
k3 = 0.8
Limiting steel stress = 0.8 × 500 = 400 N/mm2
Tensile stress due to cooling temperature difference =
9.3 × [{(1.52 + 0.06) × (130 - 60 - 16) / 130} - 0.06] = 5.5 N/mm2
Steel stress σs = M / zs + 0.6 × (differential temperature effects)
σs = {1000 × 106 × (574 - 209) / (1.18 × 109)} + (0.6 × 5.5) = 309 + 3
σs = 312 N/mm2 < 400 ∴ OK

Crack Control:
Consider worst condition before creep has occurred and
Quasi-Permanent Combination Moment = 360 kNm (no secondary effects from temperature difference as deck is simply supported single span)

Cl. 7.3.4(1)

Crack width wk = sr,maxsm - εcm)

Cl. 7.3.4(3)

Spacing Limit = 5(c+φ/2) = 5(60 + 32/2) = 380mm > 125mm ∴ 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(650 - 574) = 190
ii) (h-x)/3 = (650 - 175) / 3 = 158
iii) h/2 = 650 / 2 = 325
∴ hc,eff = 158 mm
and Ac,eff = 158 × 1000 = 158000 mm2

Cl. 7.3.4(2)

ρp,eff = As / Ac,eff = 6434 / 158000 = 0.0407

sr,max = k3c + k1k2k4φ / ρp,eff
sr,max = (3.4 × 60) + (0.8 × 0.5 × 0.425 × 32) / 0.0407 = 204 + 134 = 338

Cl. 7.3.4(2)

sm - εcm) = [σs - {ktfct,eff(1 + αeρp,eff) /ρp,eff}] / Es ≥ 0.6σs / Es
αe = Es / Ecm = 200 / 35.2 = 5.7
σs = 360 × 106 × (574 - 171) / (1.34 × 109) = 108 N/mm2

Table 3.1

fct,eff = fctm = 3.5 N/mm2

sm - εcm) = [108 - {0.4 × 3.5 × (1 + 5.7 × 0.0407) / 0.0407}] / 200000 = 0.328 × 10-3
0.6σs / Es = 0.6 × 108 / 200000 = 0.324 × 10-3 < 0.328 × 10-3 ∴ OK
Crack width wk = sr,maxsm - εcm) = 338 × 0.328 × 10-3 = 0.11 mm

NA EN 1992-2 Table NA.2

Recommended value of wmax = 0.3 mm > 0.11 mm ∴ OK

Hence B32 bars at 125 centres are adequate for the mid span. The concrete strength class will need to be C40/50

Shear Design

Shear is designed for ultimate limit state.

cl. 5.4.4

VEd = shear force due to ultimate actions.
Maximum VEd due to permanent actions = Σ(γGjGkj)
VEd = [(1.35 × 16.3) + (1.2 × 3.7)] × 12 / 2 = 159 kN

Traffic Group gr1a Cl. 6.2.2(6)

Reduction factor β for tandem axle at support = av / 2d = 0.5d / 2d = 0.25
Minimum reduction is applied at 2d = 2 × 0.574 = 1.148m from support
Only one axle should be considered for reduced load effect (see PD 6687-2:2008 Cl. 7.2.2)
Maximum VEd due to gr1a = Σ(γQ1Qk1) = 1.35 × [100 × (0.25 + 10.8/ 12) + (5.5 × 6.0)] = 188 kN

Traffic Group gr5 TS axles of LM1 are not fully on the deck ∴ ignore the load effects from these axles [see EN 1991-2 Clause 4.3.2(1a)]

Cl. 6.2.2(6)

Reduction factor β for axle at support = av / 2d = 0.5d / 2d = 0.25
Minimum reduction is applied at 2d = 2 × 0.574 = 1.148m from support
Only one axle should be considered for reduced load effect (see PD 6687-2:2008 Cl. 7.2.2)
Maximum VEd due to gr5 = Σ(γQ1Qk1) = 1.35 × {50.3 × [0.25 + (10.8 + 9.6 + 8.4 + 7.2 + 6) / 12] + 5.5 × 1.0 × 0.5 / 12} = 255 kN

Maximum VEd from variable actions is from traffic group gr5 = 255 kN
Maximum Combination VEd = Σ(γGjGkj + γQ1Qk1) = 159 + 255 = 414 kN

Cl. 6.2.2(101)

Shear Capacity of Slab: Try B32 dia. reinforcement @ 125c/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 / 574)0.5 = 1.59 < 2.0
ρ1 = Asl / bwd ≤0.02
ρ1 = 6434 / (1000 × 574) = 0.011 < 0.02

Cl. 3.1.2(102)P

fck = 40 ( < Cmax = C50/60)
VRd,c = [0.12 × 1.59 × (100 × 0.011 × 40)1/3] × 1000 × 574 × 10-3 = 387 kN ( < 414 kN Fail : see below)
Minimum VRd,c = (vmin)bwd = 0.035k3/2fck1/2bwd = 0.035 × 1.593/2 × 401/2 × 1000 × 574 × 10-3 = 255 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 - 40 / 250] = 0.504
fcd = αccfckc
αcc = 1.0 [see NA to Cl. 3.1.6(101)P]
fcd = 1.0 × 40 / 1.5 = 26.7 N/mm2
Maximu VEd = 0.5 × 1000 × 574 × 0.504 × 26.7 × 10-3 = 3857 kN >> 414 kN
VRd,c = 387 kN < VEd = 414 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.

Traffic Group gr5 Ignore effects from LM1

i) Consider a section at d (a = 0.574m) from the support :
Maximum ULS shear force due to gr5 = 1.35 × [50.3 × (11.426 + 10.226 + 9.026 + 7.826 + 6.626 + 5.426)/12] = 286 kN

Maximum VEd from permanent actions = {[(1.35 × 16.3) + (1.2 × 3.7)] × 12/2} - {0.574 × [(1.35 × 16.3) + (1.2 × 3.7)]} = 143 kN
Maximum Combination VEd = Σ(γGjGkj + γQ1Qk1) = 143 + 286 = 429 kN

Enhanced CRd,c = (0.18/γc) × (2d/a) = (0.18 / 1.5) × 2 = 0.24
VRd,c = [CRd,ck(100ρ1fck)1/3]bwd
VRd,c = [0.24 × 1.59 × (100 × 0.011 × 40)1/3] × 1000 × 574 × 10-3 = 773 kN ( > 429 kN ∴ OK)

ii) Consider a section at 2d (a = 1.148m say 1.15m) from the support (no enhancement):
Maximum ULS shear force due to gr5 = 1.35 × [50.3 × (10.85 + 9.65 + 8.45 + 7.25 + 6.05 + 4.85) / 12] = 267 kN

Maximum VEd from permanent actions = {[(1.35 × 16.3) + (1.2 × 3.7)] × 12/2} - {1.15 × [(1.35 × 16.3) + (1.2 × 3.7)]} = 128 kN
Maximum Combination VEd = Σ(γGjGkj + γQ1Qk1) = 128 + 267 = 395 kN
VRd,c = 387 kN ≈ 395 kN (2% error say OK for conservative nature of unit strip analysis)

Hence B32 bars at 125 centres are adequate for shear at the ends of the deck.

Note: Intermediate sections between mid span and the ends of the deck will have a smaller moment than at mid span and a small shear than at the ends of the deck. These sections need to be checked to determine where the reinforcement may be reduced to B25 at 125c/c.

cl. 7.3.2(2)

Minimum area of reinforcement As,min = kckfct,effAct / σct
kc = 0.4
k = 1.0 - {[(h - 300) / (800 - 300)] × (1.0 - 0.65)} = 0.755

Table 3.1

fct,eff = fctm = 3.5 N/mm2
Act = 1000 × 650 / 2 = 325000 mm2
σs = fyk = 500 N/mm2
As,min = 0.4 × 0.755 × 3.5 × 325000 / 500 = 687 mm2

cl. 9.2.1.1(1)

Minimum area of longitudinal reinforcement As,min = 0.26btdfctm / fyk > 0.0013btd
As,min = 0.26 × 1000 × 574 × 3.5 / 500 = 1045 mm2
0.0013btd = 0.0013 × 1000 × 574 = 746 > 1045 ∴ OK

Hence minimum longitudinal steel (bottom of slab) = B16 @ 125 c/c (As = 1608 mm2 > 1045)
Minimum distribution steel = B12 @ 125 c/c longitudinally (As = 905 mm2 > 687) & B12 @ 150 c/c transversely (As = 754 mm2 > 687)

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