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Reinforced Concrete Deck Example to Eurocodes
Reinforced Concrete Deck Design to EN 19922 & 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/m^{2} 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.
γ_{conc.} = 25kN/m^{3}
SV80 model vehicle also to be considered.
Use C32/40 concrete to BS 8500.
Use Grade B500B reinforcement to BS 4449.
A Solution:
EN 199211 Table 4.1 gives the Exposure Class XD1 as suitable for deck soffits.
NA to BS EN 199211 clause 4.4.1.2(5) says values of c_{min,dur} shall be taken from BS 85001
BS 85001 Table A.5
Nominal cover for C32/40 concrete = c_{nom} = c_{min,dur} + Δc_{dev} = 45 + Δc_{dev} with
maximum watercement ratio = 0.55 and minimum cement content of 320 kg/m^{3}
Fixing tolerence for reinforcement Δc_{dev} = 15mm for insitu concrete
∴ Nominal cover c_{nom} = 45 + 15 = 60mm
Loading
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 199111 to 199117, 19912 and National Annex for full design loading.)
EN 199111 Annex A
Permanent Actions (G_{k}):
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 19912
Table 4.2
Variable Actions (Q_{k}):
Traffic Load Model 1 (Q_{1k} = 1 & q_{1k} = 0.61)
= 5.5 kN/m(udl) + 2 × (300kN/3m) axles @ 1.2m centres.
NA Fig NA.1
Traffic Load Model 3 (SV80)
= 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
E_{d} = E(Σγ_{Gj}G_{kj} + γ_{p}P + γ_{Q,1}Q_{k,1} + Σγ_{Q,i}Ψ_{0,i}Q_{k,i})
Table NA.A.2.4
Group of Loads
γ_{sup}
γ_{inf}
Ψ_{0}
Dead Load
1.35
0.95

Superimposed Dead Load
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 Ψ_{0} of 0.6 may be reduced to 0 when considering ULS.
^{‡} Traffic loads are the leading action
b) Serviceability Limit State : EN 1990 clause 6.5.3 requires three combinations of actions to be considered for the serviceability limit state:
 Characteristic combination E_{d} = E(ΣG_{kj} + P + Q_{k,1} + ΣΨ_{0,i}Q_{k,i}) for limiting stresses
 Frequent combination E_{d} = E(ΣG_{kj} + P + Ψ_{1,1}Q_{k,1} + ΣΨ_{2,i}Q_{k,i})
 Quasipermanent combination E_{d} = E(ΣG_{kj} + P + ΣΨ_{2,i}Q_{k,i}) for crack control
Table NA.A.2.1
Action
Ψ_{0}
Ψ_{1}
Ψ_{2}
Traffic load group gr1a
0.75
0.75
0
Traffic load group gr5
0
0
0
Thermal
0.6
0.6
0.5
Temperature Difference Effects
Apply temperature differences given in EN 199115 Figure 6.2c(Type 3a) to a 1m wide deck section.
EN 199115 Table C.1  Coefficient of thermal expansion = 10 × 10^{6} per °C.
CIRIA Report C660 ("Earlyage 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 199211 Table 3.1 :
f_{ck} = 32 → f_{cm} = f_{ck} +8 = 40 → E_{cm} = 22[(f_{cm}) / 10]^{0.3} = 22 × 4^{0.3} = 33.35 kN/mm^{2}
Hence restrained temperature stresses per °C = 33.35 × 10^{3} × 12 × 10^{6} = 0.4 N/mm^{2}
Interpolating values of ΔT from EN 199115 Table B.3 for a 0.65m depth of slab with 100mm surfacing we get:
Section Properties
Area = 1000 × 650 = 0.65 × 10^{6} mm^{2}
Second Moment of Area = 1000 × 650^{3} / 12 = 22.9 × 10^{9} mm^{4}
a) Heating temperature difference
Force F to restrain temperature strain :
0.4 × 10^{3} × [ 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 × 10^{3} × [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 × 10^{3} × [ 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 × 10^{3} × [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.
Dead + Superimposed Dead Loading (per metre width of deck)
SLS = Serviceability Limit State
ULS = Ultimate Limit State
Design SLS moment = ∑G_{kj} = (16.3 + 3.7) × 12^{2} / 8 = 360 kNm
Design ULS moment = ∑(γ_{Gj}G_{kj}) = [(1.35 × 16.3) + (1.2 × 3.7)] × 12^{2} / 8] = 476 kNm
Variable Actions (per metre width of deck)
Traffic Group gr1a
Footway loading will not affect the unit strip loading.
Moment at leading axle = 5.5 × (6 × 6.3  6.3^{2}/2) + 100 × 5.7 × (6.3 + 5.1) / 12 = 640 kNm
Design SLS moment characteristic combination = Q_{k1} = 640 kNm
Design ULS moment = γ_{Q1}Q_{k1} = 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 = Q_{k1} = 636 < 640 kNm
∴ gr1a governs
Design ULS moment = γ_{Q1}Q_{k1} = 1.35 × 636 = 859 < 864 kNm
∴ gr1a governs
Combinations of Actions
a) Ultimate Limit State
E_{d} = E(Σγ_{Gj}G_{kj} + γ_{p}P + γ_{Q,1}Q_{k,1} + Σγ_{Q,i}Ψ_{0,i}Q_{k,i})
Design ULS mid span moment = 476 + 0 + 864 + 0 = 1340 kNm
b) Serviceability Limit State
Characteristic Combination
E_{d} = E(ΣG_{kj} + P + Q_{k,1} + ΣΨ_{0,i}Q_{k,i})
Design SLS mid span moment:
= 360 + 0 + 640 + 0.6 × (differential temperature effects)
= 1000 kNm + 0.6 × (differential temperature effects)
Quasipermanent Combination
E_{d} = E(ΣG_{kj} + P + ΣΨ_{2,i}Q_{k,i})
Design SLS mid span moment = 360 + 0 + 0 = 360 kNm
Ultimate Capacity of Deck Slab
Ultimate Design Moment = 1340 kNm
EN 199211 & EN 19922
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 = 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 32mm dia. reinforcement at 125mm centres:
Nominal cover to reinforcement in deck soffit = 60mm
Fig. 3.3
Using parabolicrectangular diagram:
Average stress f_{av} = f_{cd}[1ε_{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}bXz
Depth to neutral axis X = f_{yk}A_{s} / (f_{av}bγ_{s})
X = 500 × 6434 / (14.7 × 1000 × 1.15) = 190.6mm
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 ( 574 / 190.6  1) = 0.007 > 0.00217 ∴ steel will yield.
Hence M_{ult} = f_{av}bXz = f_{av}bX(d  βX)
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 × 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 shortterm modulus for all actions.
EN 199211 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} = [6434 × 200000 + {(6434 × 200000)^{2} + 2 × 1000 × 6434 × 200000 × 33400 × 574}^{0.5}] / (1000 × 33400) = 175mm
Cracked second moment of area = A_{s}(dd_{c})^{2} + E_{c,eff}bd_{c}^{3} / 3E_{s}
I_{NA} = 6434 × (574  175)^{2} + 33.4 × 1000 × 175^{3} / (3 × 200) = 1.32 × 10^{9} mm^{4} (steel units)
Concrete stress σ_{c} = M / z_{c} + 0.6 × (differential temperature effects)
σ_{c} = {1000 × 10^{6} × 175 / (1.32 × 10^{9} × 6.0)} + (0.6 × 2.49) = 22.1 + 1.5 = 23.6 N/mm^{2} (Heating temperature difference)
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} < 23.6 ∴ Fail
The concrete strength class will need to be increased to class C40/50
Reworking the example gives:
E_{cm} = 35.2 kN/mm^{2} m = 5.7 d_{c} = 171 mm I_{NA} = 1.34 × 10^{9} mm^{4}
σ_{c} = 22.4 + 1.5 = 23.9 N/mm^{2}
Limiting stress = 0.6 × 40 = 24 N/mm^{2} > 23.9 ∴ OK
EN 199211
ii) After all creep has taken place the cracked section properties will be based on the longterm and shortterm modulus for the various actions.
Shortterm modulus = E_{cm}
Longterm 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 = 40 + 8 = 48 N/mm^{2}
Cl. 3.1.4
Relative humidity of the ambient environment = 80% (outside conditions)
Age of concrete at initial loading t_{0} = 6 days (when soffit formwork is released)
Annex B (B.6)&(B.8c)
h_{0} = 650
α_{1} = [35 / f_{cm}]^{0.7} = [35 / 48]^{0.7} = 0.80
α_{2} = [35 / f_{cm}]^{0.2} = [35 / 48]^{0.2} = 0.94
(B.3b)
φ_{RH} = [1 + α_{1} × {(1  RH / 100) / (0.1 × h_{0}^{1/3})}] × α_{2}
φ_{RH} = [1 + 0.8 × {(1  80 / 100) / ( 0.1 × 650^{1/3})}] × 0.94 = 1.113
(B.4)
β(f_{cm}) = 16.8 / f_{cm}^{0.5} = 16.8 / 48^{0.5} = 2.425
(B.5)
β(t_{0}) = 1 / (0.1 + t_{0}^{0.2}) = 1 / ( 0.1 + 6^{0.2}) = 0.653
(B.2)
φ_{0} = φ_{RH} × β(f_{cm}) × β(t_{0}) = 1.113 × 2.425 × 0.653 = 1.762
Moment due to longterm actions = M_{qp} = 360 kNm
Moment due to shortterm actions = M_{st} = 640 kNm
Hence Effective Modulus E_{c,eff} = {(360 + 640) × 35.2} / {640 + 360 × ( 1 + 1.762)} = 21.5 kN/mm^{2}
Modular Ratio m = E_{s} / E_{c,eff} = 200 / 21.5 = 9.3
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} = [6434 × 200000 + {(6434 × 200000)^{2} + 2 × 1000 × 6434 × 200000 × 21500 × 574}^{0.5}] / (1000 × 21500) = 209mm
Cracked second moment of area = A_{s}(dd_{c})^{2} + E_{c,eff}bd_{c}^{3} / 3E_{s}
I_{NA} = 6434 × (574  209)^{2} + 21.5 × 1000 × 209^{3} / (3 × 200) = 1.18 × 10^{9} mm^{4} (steel units)
Concrete stress σ_{c} = M / z_{c} + 0.6 × (differential temperature effects)
σ_{c} = {1000 × 10^{6} × 209 / (1.18 × 10^{9} × 9.3)} + (0.6 × 2.49) = 19.1 + 1.5 = 20.6 N/mm^{2} (Heating temperature difference)
cl. 7.2(102)
Limiting concrete stress = k_{1}f_{ck}
k_{1} = 0.6
Limiting concrete stress = 0.6 × 40 = 24.0 N/mm^{2} > 20.6 ∴ 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}
Tensile stress due to cooling temperature difference =
9.3 × [{(1.52 + 0.06) × (130  60  16) / 130}  0.06] = 5.5 N/mm^{2}
Steel stress σ_{s} = M / z_{s} + 0.6 × (differential temperature effects)
σ_{s} = {1000 × 10^{6} × (574  209) / (1.18 × 10^{9})} + (0.6 × 5.5) = 309 + 3
σ_{s} = 312 N/mm^{2} < 400 ∴ OK
Crack Control:
Consider worst condition before creep has occurred and
QuasiPermanent Combination Moment = 360 kNm (no secondary effects from temperature difference as deck is simply supported single span)
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 + 32/2) = 380mm > 125mm ∴ 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(hd) = 2.5(650  574) = 190
ii) (hx)/3 = (650  175) / 3 = 158
iii) h/2 = 650 / 2 = 325
∴ h_{c,eff} = 158 mm
and A_{c,eff} = 158 × 1000 = 158000 mm^{2}
Cl. 7.3.4(2)
ρ_{p,eff} = A_{s} / A_{c,eff} = 6434 / 158000 = 0.0407
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 × 32) / 0.0407 = 204 + 134 = 338
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 (no shortterm loading included in design moment of 360 kNm)
α_{e} = E_{s} / E_{cm} = 200 / 35.2 = 5.7
σ_{s} = 360 × 10^{6} × (574  171) / (1.34 × 10^{9}) = 108 N/mm^{2}
Table 3.1
f_{ct,eff} = f_{ctm} = 3.5 N/mm^{2}
(ε_{sm}  ε_{cm}) = [108  {0.4 × 3.5 × (1 + 5.7 × 0.0407) / 0.0407}] / 200000 = 0.328 × 10^{3}
0.6σ_{s} / E_{s} = 0.6 × 108 / 200000 = 0.324 × 10^{3} < 0.328 × 10^{3} ∴ OK
Crack width w_{k} = s_{r,max}(ε_{sm}  ε_{cm}) = 338 × 0.328 × 10^{3} = 0.11 mm
NA EN 19922 Table NA.2
Recommended value of w_{max} = 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
V_{Ed} = shear force due to ultimate actions.
Maximum V_{Ed} due to permanent actions = Σ(γ_{Gj}G_{kj})
V_{Ed} = [(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 = a_{v} / 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 66872:2008 Cl. 7.2.2)
Maximum V_{Ed} due to gr1a = Σ(γ_{Q1}Q_{k1}) = 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 19912 Clause 4.3.2(1a)]
Cl. 6.2.2(6)
Reduction factor β for axle at support = a_{v} / 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 66872:2008 Cl. 7.2.2)
Maximum V_{Ed} due to gr5 = Σ(γ_{Q1}Q_{k1}) = 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 V_{Ed} from variable actions is from traffic group gr5 = 255 kN
Maximum Combination V_{Ed} = Σ(γ_{Gj}G_{kj} + γ_{Q1}Q_{k1}) = 159 + 255 = 414 kN
Cl. 6.2.2(101)
Shear Capacity of Slab: Try B32 dia. reinforcement @ 125c/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 / 574)^{0.5} = 1.59 < 2.0
ρ_{1} = A_{sl} / b_{w}d ≤0.02
ρ_{1} = 6434 / (1000 × 574) = 0.011 < 0.02
Cl. 3.1.2(102)P
f_{ck} = 40 ( < C_{max} = C50/60)
V_{Rd,c} = [0.12 × 1.59 × (100 × 0.011 × 40)^{1/3}] × 1000 × 574 × 10^{3} = 387 kN ( < 414 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 × 1.59^{3/2} × 40^{1/2} × 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 = 0.5b_{w}dνf_{cd}
ν = 0.6[1  f_{ck} / 250] = 0.6 × [1  40 / 250] = 0.504
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 × 40 / 1.5 = 26.7 N/mm^{2}
Maximu V_{Ed} = 0.5 × 1000 × 574 × 0.504 × 26.7 × 10^{3} = 3857 kN >> 414 kN
V_{Rd,c} = 387 kN < V_{Ed} = 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 19922 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.
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 V_{Ed} 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 V_{Ed} = Σ(γ_{Gj}G_{kj} + γ_{Q1}Q_{k1}) = 143 + 286 = 429 kN
Enhanced C_{Rd,c} = (0.18/γ_{c}) × (2d/a) = (0.18 / 1.5) × 2 = 0.24
V_{Rd,c} = [C_{Rd,c}k(100ρ_{1}f_{ck})^{1/3}]b_{w}d
V_{Rd,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 V_{Ed} 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 V_{Ed} = Σ(γ_{Gj}G_{kj} + γ_{Q1}Q_{k1}) = 128 + 267 = 395 kN
V_{Rd,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 A_{s,min} = k_{c}kf_{ct,eff}A_{ct} / σ_{ct}
k_{c} = 0.4
k = 1.0  {[(h  300) / (800  300)] × (1.0  0.65)} = 0.755
Table 3.1
f_{ct,eff} = f_{ctm} = 3.5 N/mm^{2}
A_{ct} = 1000 × 650 / 2 = 325000 mm^{2}
σ_{s} = f_{yk} = 500 N/mm^{2}
A_{s,min} = 0.4 × 0.755 × 3.5 × 325000 / 500 = 687 mm^{2}
cl. 9.2.1.1(1)
Minimum area of longitudinal reinforcement A_{s,min} = 0.26b_{t}df_{ctm} / f_{yk} > 0.0013b_{t}d
A_{s,min} = 0.26 × 1000 × 574 × 3.5 / 500 = 1045 mm^{2}
0.0013b_{t}d = 0.0013 × 1000 × 574 = 746 > 1045 ∴ OK
Hence minimum longitudinal steel (bottom of slab) = B16 @ 125 c/c (A_{s} = 1608 mm^{2} > 1045)
Minimum distribution steel = B12 @ 125 c/c longitudinally (A_{s} = 905 mm^{2} > 687) & B12 @ 150 c/c transversely (A_{s} = 754 mm^{2} > 687)
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