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Reinforced Concrete Deck Design to BS 5400 Part 4

*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 nominal HA live load udl of 17.5 kN/m^{2} and knife edge load of 33kN/m .

The deck should also be designed to
carry 30 units of HB load. The span of the deck is 12.0m centre to centre of bearings.

γ_{conc.}
= 25kN/m^{3}

30 units of HB also to be considered.

Use C32/40 concrete to BS 8500.

Use Grade B500B reinforcement to BS 4449.

BS 8500-1 cl. A.2.1 Table A.1

Exposure Class XD1 used for bridge deck soffits.

cl. A.3

Fixing tolerence for reinforcement Δ_{c} = 15mm for insitu concrete.

Table A.5

Nominal cover for C32/40 concrete = 45 + Δ_{c} = **60mm** with

maximum water-cement ratio = 0.55 and minimum cement content of 320 kg/m^{3}

__ Loading__ per metre width of deck (unit strip)

(Note: The loading has been simplified to demonstrate the method of designing the slab (See BS 5400 Pt2, or BD 37/01 for full design loading).

Nominal Dead Loads :

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

surfacing = 24 × 0.1 × 1.0 = 2.4 kN/m

Nominal Live Load :

HA = 17.5 × 1.0 + 33.0 = 17.5 kN/m(udl) + 33kN(kel)

30 units HB = 30 × 10 / 4 per wheel = 75 kN per wheel

Load factors for serviceability and ultimate limit state from BS 5400 Part 2 Table 1:

SLS

ULS

Comb 1

Comb 3

Comb 1

Comb 3

Dead Load

γ_{fL} concrete

1.0

1.0

1.15

1.15

Superimposed Dead Load

γ_{fL} surfacing

1.2

1.2

1.75

1.75

Live Load

γ_{fL} HA

1.2

1.0

1.5

1.25

γ_{fL} HB

1.1

1.0

1.3

1.1

Temperature Difference

γ_{fL}

—

0.8

—

1.0^{#}

Key: ^{#} It is usually assumed that there is local plasticity at the critical sections at Ultimate Limit State and the self equilibrating stresses due to non linear temperature distribution can be ignored in combination loadings. (Ref:"Concrete bridge engineering:performance and advances" by R.J.Cope).

Temperature Difference Effects

Apply temperature differences given in BS 5400 Pt2 Fig.9 (Group 4) to a 1m wide deck section.

Cl. 5.4.6 - Coefficient of thermal expansion = 12 × 10^{-6} per °C.

From BS 5400 Pt4 Table 3 : E_{c} = 31 kN/mm^{2} for f_{cu} = 40N/mm^{2}

Hence restrained temperature stresses per °C = 31 × 10^{3} × 12 × 10^{-6} = 0.372 N/mm^{2}

__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) __Positive temperature difference__

Force F to restrain temperature strain :

0.372 × 10^{3} × [ 150 × ( 3.0 + 5.05 ) + (195 × 1.5) + (195 × 1.05)] × 10^{-3} = 634.2 kN

Taking moments about centroid of section to determine required moment M to restrain
curvature due to temperature strain :

0.372 × 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} = 111.5kNm

b) __Reverse temperature difference__

Force F to restrain temperature strain :

- 0.372 × 10^{3} × [ 130 × ( 1.8 + 2.5 + 1.5 + 1.9 ) + 163 × ( 0.9 + 0.75 )] × 10^{-3} = - 472.4kN

Taking moments about centroid of section to determine required moment M to restrain
curvature due to temperature strain :

- 0.372 × 10^{3} × [130 × ( 1.8 × 260 + 2.5 × 282 - 1.5 × 260 - 1.9 × 282 ) + 163 × ( 0.9 × 141 - 0.75 × 141 )] × 10^{-6} = -13.34kNm

Note: Sign convention is compressive stresses are positive.

Dead + Superimposed Dead Loading (per metre width of deck)

SLS = Serviceability Limit State

ULS = Ultimate Limit State

Design SLS moment = ∑(γ_{fL} × M) = [(1.0 × 16.3)+(1.2 × 2.4)] × 12^{2} / 8 = **345 kNm**

Design ULS moment = γ_{f3} × ∑(γ_{fL} × M) = 1.1 × [{(1.15 × 16.3)+(1.75 × 2.4)} × 12^{2} / 8] = **454 kNm**

Live Loading (per metre width of deck)

Nominal HA mid span moment = 17.5 × 12.0^{2} / 8 + 33.0 × 12.0 / 4 = **414kNm**

The maximum moment for the HB vehicle occurs at point X in the diagram below with the vehicle positioned as shown.

(CG = position of the centre of gravity of the three 75kN wheel loads)

Nominal HB moment at X = 99.4 × 5.3 - 75 × 1.8 = **392kNm**

Combination 1 Loading

Design HA SLS moment = γ_{fL} × M = 1.2 × 414 = 497 kNm

Design HB SLS moment = γ_{fL} × M = 1.1 × 392 = 431 kNm < 497 kNm ∴ HA critical

Total Design SLS Moment (Dead + Live) = 345 + 497 = **842 kNm**

Design HA ULS moment = γ_{f3} × γ_{fL} × M = 1.1 × 1.5 × 414 = 683 kNm

Design HB ULS moment = γ_{f3} × γ_{fL} × M = 1.1 × 1.3 × 392 = 561 kNm < 683 kNm

∴ HA loading critical

Total Design ULS Moment (Dead + Live) = 454 + 683 = **1137 kNm**

Combination 3 Loading

Design HA SLS moment = γ_{fL} × M = 1.0 × 414 = 414 kNm

Design HB SLS moment = γ_{fL} × M = 1.0 × 392 = 392 kNm <414 kNm ∴ HA loading critical

Design SLS Moment (Dead + Live) = 345 + 414 = **759 kNm**

Design HA ULS moment = γ_{f3} × γ_{fL} × M = 1.1 × 1.25 × 414 = 569 kNm

Design HB ULS moment = γ_{f3} × γ_{fL} × M = 1.1 × 1.1 × 392 = 474 kNm < 569 kNm

∴ HA loading critical

Design ULS Moment (Dead + Live) = 454 + 569 = **1023 kNm**

Ultimate Capacity of Deck Slab

Ultimate Design Moment = **1137 kNm**

BS 5400 Pt 4 cl. 5.1.2.1

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

cl. 5.4.2

Use clause 5.3.2 for the resistance moments in slabs.

cl. 5.3.2.3

**Try 32mm dia. reinforcement at 125mm centres:**

Nominal cover to reinforcement in deck soffit = 60mm

d = 650 - 60 -32/2 = 574

A_{s} = Π16^{2} × 1000 / 125 = 6434mm^{2}/m

f_{y} = 500N/mm^{2}

f_{cu} = 40N/mm^{2}

z = [1 - ({1.1f_{y}A_{s}}/{f_{cu}bd})]d

z = [1 - ({1.1 × 500 × 6434}/{40 × 1000 × 574})]d = 0.85d < 0.95 d

∴ z = 0.85 × 574 = **488mm**

M_{uSteel} = 0.87f_{y}A_{s}z = 0.87 × 500 × 6434 × 488 × 10^{-6} = 1366 kNm/m

M_{uConcrete} = 0.15f_{cu}bd^{2} = 0.15 × 40 × 1000 × 574^{2} × 10^{-6} = 1977 kNm/m > 1366

∴ **M _{u} = 1366 kNm/m** > 1137kNm/m ∴ OK.

Check Serviceability Limit State

**Combination 1** SLS Design Moment = **842 kNm** (345DL + 497LL)

Determine depth '**X**' to neutral axis of cracked section:

cl. 4.3.2.1 Table 3

Youngs Modulus for concrete for short term loading = E_{c} = 31 kN/mm^{2}

cl. 4.3.2.2

Youngs Modulus for steel reinforcement = E_{s} = 200 kN/mm^{2}

**Case 1**) When the bridge has just opened (when only a small amount of creep has occurred):

Modular Ratio = E_{s} / E_{c} = 200 / 31 = 6.45

Taking first moments of area about the neutral axis:

1000 × **X**^{2} / 2 = 6.45 × 6434 × (574 - **X**)

500**X**^{2} + 41510**X** - 23.83×10^{6} = 0

**X** = 177 mm

Second Moment of Area of cracked section:

I_{xx} = 1000×177^{3} / 3 + 6.45×6434×(574-177)^{2} = 8.39×10^{9} mm^{4}

Cl 4.1.1.3

Max compressive stress in concrete = 842×10^{6} × 177 / 8.39×10^{9} = 17.8 N/mm^{2}

Table 2

Allowable compressive stress = 0.5f_{cu} = 20 N/mm^{2} > 17.8 ∴ OK

**Case 2**) When creep and shrinkage in the bridge are substantially complete:

cl. 4.3.2.1(b)

Youngs Modulus for concrete for long term loading = E_{c}/2 = 15.5 kN/mm^{2}

Hence Modified E_{c} for (345DL + 497LL) = (345 × 15.5 + 497 × 31) / 842 = 24.65 kN/mm^{2}

cl. 4.3.2.2

Youngs Modulus for steel reinforcement = E_{s} = 200 kN/mm^{2}

Modular Ratio = E_{s} / E_{c} = 200 / 24.65 = 8.1

Taking first moments of area about the neutral axis:

1000 × **X**^{2} / 2 = 8.1 × 6434 × (574 - **X**)

500**X**^{2} + 52115**X** - 30×10^{6} = 0

**X** = 198 mm

Second Moment of Area of cracked section:

I_{xx} = 1000×198^{3} / 3 + 8.1×6434×(574-198)^{2} = 9.96×10^{9} mm^{4}

Cl 4.1.1.3

Max compressive stress in concrete = 842×10^{6} × 198 / 9.96×10^{9} = 16.7 N/mm^{2}

Table 2

Allowable compressive stress = 0.5f_{cu} = 20 N/mm^{2} > 16.7 ∴ OK

Tensile stress in reinforcement = 842×10^{6} × (574 - 198) × 8.1 / 9.96×10^{9} = 257.5 N/mm^{2}

Table 2

Allowable tensile stress = 0.75f_{y} = 375 N/mm^{2} > 257.5 ∴ OK

Crack Control:

Strain in reinforcement = ∈_{1} = 257.5 / 200000 = 0.00129

Table 13

Notional surface for crack calculation = 35mm cover to reinforcement

5.8.8.2

a_{cr} = √[(125/2)^{2} + (16+35)^{2}] - 16 = 65

Allow for stiffening effect of concrete:

eqn 25

∈_{m} = ∈_{1} - [{3.8b_{t}h(a'-d_{c})} / {∈_{s}A_{s}(h-d_{c})}] × [(1-M_{q}/M_{g})×10^{-9}]

∈_{m} = ∈_{1} - [{3.8×1000×650×(625 - 198)} / {0.00129×6434×(650 - 198)}] × [(1 - 497/345)×10^{-9}]

∈_{m} = ∈_{1} - [-0.00012] but not greater than ∈_{1} Hence no stiffening effect

∈_{m} = ∈_{1} = 0.00129 × (625 - 198) / (574 - 198) = 0.00146

Design crack width = 3 × 65 × 0.00146 / [1 + 2 × (65 - 35) / (650 - 198)] = 0.25 mm

eqn 24

Design crack width = 3a_{cr}∈_{m} / [1+2(a_{cr}-c_{nom})/(h-d_{c})]

Table 1

Maximum allowable crack width = 0.25 mm ∴ OK

**Combination 3** SLS Design Moment = **759 kNm** (345DL + 414LL)

Determine depth '**X**' to neutral axis of cracked section:

cl. 4.3.2.1 Table 3

Youngs Modulus for concrete for short term loading = E_{c} = 31 kN/mm^{2}

cl. 4.3.2.1(b)

Youngs Modulus for concrete for long term loading = E_{c}/2 = 15.5 kN/mm^{2}

Hence Modified E_{c} for (345DL + 414LL) = (345 × 15.5 + 414 × 31) / 759 = 23.95 kN/mm^{2}

cl. 4.3.2.2

Youngs Modulus for steel reinforcement = E_{s} = 200 kN/mm^{2}

Modular Ratio = E_{s} / E_{c} = 200 / 23.95 = 8.35

Taking first moments of area about the neutral axis:

1000 × **X**^{2} / 2 = 8.35 × 6434 × (574 - **X**)

500**X**^{2} + 53724**X** - 30.84×10^{6}

**X** = 200 mm

Second Moment of Area of cracked section:

I_{xx} = 1000×200^{3} / 3 + 8.35×6434×(574-200)^{2} = 10.18×10^{9} mm^{4}>

Cl 4.1.1.3

Max compressive bending stress in concrete = 759×10^{6} × 200 / 10.18×10^{9} = 14.9 N/mm^{2}

Max compressive stress due to positive temperature difference = γ_{fL} × 2.31 = 0.8 × 2.31 = 1.8 N/mm^{2}

Total compressive stress in concrete = 14.9 + 1.8 = 16.7 N/mm^{2}

Table 2

Allowable compressive stress = 0.5f_{cu} = 20 N/mm^{2} > 16.7 ∴ OK

Tensile stress in reinforcement = 759×10^{6} × (574 - 200) × 8.35 / 10.18×10^{9} = 232.8 N/mm^{2}

Tensile stress due to reverse temperature difference =

γ_{fL} × 8.35×[{(1.43+0.06)×(130-60-16)/130}-0.06] = 0.8 × 4.7 = 3.8 N/mm^{2}

Total tensile stress in reinforcement = 232.8 + 3.8 = 237 N/mm^{2}

Table 2

Allowable tensile stress = 0.75f_{y} = 375 N/mm^{2} > 237 ∴ OK

**Hence B32 bars at 125 centres are adequate for the mid span.**

Shear Design

Shear is designed for ultimate limit state.

cl. 5.4.4

V = shear force due to ultimate loads.

Maximum Dead Load V = γ_{f3} × 12 × (1.15 × 16.3 + 1.75 × 2.4) / 2

Maximum Dead Load V = 1.1 × 137.7 kN = 151 kN

Determine shear effects at distance d away from the support. **Try 32mm dia. bars at 125 c/c : d = 574mm**

Maximum HA V = γ_{f3} × 1.5 × (12 × 17.5 / 2 + 33 × 11.426 / 12 - 17.5 × 0.574)

Maximum HA V = 1.1 × 190 kN = 209 kN

Maximum HB V = γ_{f3} × 1.3 × 75 × (11.426 + 9.626 + 3.626 + 1.826) / 12

Maximum HB V = 1.1 × 215 kN = 237 kN > 209 ∴ HB loading critical

Maximum V = 151 + 237 = 388 kN

Shear stress = V / bd = 388×10^{3} / (1000 × 574) = 0.68 N/mm^{2}

Design for no shear reinforcement condition then ξ_{s}v_{c} > 0.68 N/mm^{2}

Table 9

ξ_{s} = (500/d)^{1/4} = (500/574)^{1/4} = 0.97

Table 8

v_{c} = 0.27/γ_{m}(100A_{s}/b_{w}d)^{1/3}(f_{cu})^{1/3}

v_{c} = (0.27 / 1.25) × [100 × 6434 / (1000 × 574)]^{1/3} × (40)^{1/3} = 0.77 N/mm^{2}

ξ_{s}v√ = 0.97 × 0.77 = 0.75 N/mm^{2} > 0.68 ∴ OK

cl 5.3.3.1

Check that the maximum allowable shear stress is not exceeded:

Maximum allowable shear stress = 0.75√f_{cu} or 4.75 N/mm^{2}

0.75√f_{cu} = 0.75√40 = 4.74 > 0.68 ∴ OK

**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. 5.8.4.1

Minimum area of reinforcement = 0.15% of b_{a}d = 0.15 × 1000 × 574 / 100 = 861 mm^{2}/m ∴ use B12 bars at 125 centres (A_{s} = 905 mm^{2}/m) for distribution reinforcement.

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