7. Design Example - Reinforced Concrete Deck Slab

1. Problem Definition

Live LoadHL-93

Deck Concretef’c = 4 ksiwc = 150 pcf

Mild Steel (Non-Prestressed)fy = 60 ksiEs = 29,000 ksi

DimensionsThickness = 8.0 in.(LRFD 9.7.11 & 13.7.3.1.2)

Cover = 2.0 in. (Top)(LRFD 5.12.3) = 1.0 in. (Bottom)

Future Wearing Surface Allowance: FWS = 30 psf

7-1

Figure 1: Cross section of bridge showing girder layout, barriers, and 8 in. thick deck.

The LRFD Specifications contain provisions for designing the deck of a multi-girder bridge by using either refined or approximate methods. Refined methods consist of modeling the deck and girders using finite element techniques. General guidance is provided by the specifications to assist in formulating a suitable model. Most of the detail, however, is left up to the engineer. Due to the complexities involved in creating and validating the results of a finite element model, in practice, it is more likely that one of the approximate methods will be used. In this case, two procedures are given in the specifications: the empirical method and the strip method. Both methods will be demonstrated in this design example.

Design of the overhang is discussed briefly, but calculations are not presented.

2. The Empirical Method

No analysis is performed with the empirical method. If the criteria for using the empirical method are met, then a prescribed amount of reinforcement is provided in each direction of the top and bottom mats of steel (4 layers total), and the design is complete for the interior portion of the deck. However, the overhang must still be designed using other means.

LRFD 9.7.2.4 lists the conditions that must be satisfied before using the empirical method. Those applicable to the structure in this example are:

Diaphragms are used throughout the cross-section at lines of support

Supporting elements are concrete and/or steel

Deck is fully cast-in-place and water cured

Deck is of uniform depth

Ratio of effective length to design depth does not exceed 18.0 and is not less than 6.0

Core depth of the slab is not less than 4.0 in.

Effective length does not exceed 13.5 ft.

Slab thickness is not less than 7.0 in.

Length of overhang is at least 5.0 times the slab depth, or 3.0 times the slab depth and a structurally continuous barrier is made composite with the slab

The specified concrete strength is not less than 4.0 ksi

The deck is composite with the supporting elements

Assuming all of the above criteria are met, the following amount of reinforcement must be provided:

Bottom Layer, each way: 0.27 in.2 / ft.Use #5 bars @ 13.5 in. spacing - As prov'd = 0.276 in. 2 / ft.

Top Layer, each way: 0.18 in.2 / ft.Use #4 bars @ 13 in. spacing - As prov'd = 0.185 in. 2 / ft.)

Each mat of steel must be placed as close to the outside surface as permitted by cover and spacing. Spacing of reinforcement must not exceed 18 in. The outermost layer of reinforcement must be placed in the direction of the effective length. Reinforcing steel must at least Grade 60.

If skew exceeds 25, the specified quantity of reinforcement must be doubled in the end zones of the deck (LRFD 9.7.2.5). The end zone is taken as the longitudinal distance equal to the effective length of the slab.

3. The Strip Method

The strip method consists of approximating the behavior of a deck using transverse strips. A unit width strip of the deck running of from out to out of the bridge cross section is modeled as a continuous beam with the girders acting as rigid, knife-edge supports. Different strip widths are used for the design of the reinforcement, depending on which portion of the deck is being designed. For a cast-in-place concrete deck on longitudinal girders, three zones are considered: overhang, positive moment, and negative moment.

The equivalent strip method of deck design is based on the following: A transverse strip of the deck is assumed to support the truck axle loads. The strip is assumed to be supported on rigid supports at the center of the girders. The width

of the strip for different load effects is determined using the equations in LRFD 4.6.2.1. Dead load moments are calculated on a per foot width of deck. For live load effects, the truck axle loads are moved laterally to produce the moment

envelopes (refer to LRFD 3.6.1.3). One or more axles may be placed side by side on the deck (representing axles from trucks in different traffic lanes) and move them transversely across the deck to maximize the moments (LRFD 4.6.2.1.6). Multiple presence factors and the dynamic load allowance are included. The total moment is divided by the strip distribution width to determine the live load per unit width.

Alternatively, the live load moment per unit width of the deck to be determined using LRFD Table A4.1-1 (method used in this design example).

The loads transmitted to the bridge deck during vehicular collision with the railing system are determined.

Designs factored moments are then determined using the appropriate load factors for different limit states.

The reinforcement is designed to resist the applied loads using conventional principles of reinforced concrete design.

Generally, the LRFD Specifications require that four limit states be investigated in the design of a structure or structural component: service, fatigue and fracture, strength, and extreme event. With regard to the design of a reinforced concrete deck, the following limit states are pertinent:

Service Limit States (LRFD 9.5.2)

Check control of cracking by distribution of reinforcement.

Fatigue and Fracture Limit State (LRFD 9.5.3)

Fatigue need not be investigated for concrete decks in multi-girder applications.

Strength Limit States (LRFD 9.5.4)

Deck may be analyzed as elastic and shall be designed to satisfy provisions of Section 5.

Extreme Event Limit State (LRFD 9.5.5)

Decks shall be designed for force effects transmitted by traffic and combination railings as specified in Section 13.

3.1 Analysis

3.1.1 Dead Load Moments:

Dead loads represent a small fraction of the deck loads. Use of a simplified approach to determine the deck dead load effects is quite appropriate. Traditionally, dead load positive and negative moments in the deck, except for the overhang, for a unit width strip of the deck are calculated using the following approach:

where:

M = dead load positive or negative moment in the deck for a unit width strip (k-ft/ft)w = dead load per unit area of the deck (ksf)l = girder spacing = 9 ft.c = constant, typically taken as 10 or 12. We will use c = 10 for this example.

Self weight of the deck = 8(150)/12 = 100 psf = 0.1 ksf

Unfactored self weight positive or negative moment =

Future wearing surface = 30 psf = 0.03 ksf

Unfactored FWS positive or negative moment =

3.1.1.2 Live Load Moments:

The live load moments (with impact) are determined from the LRFD Table A4-1. The critical section for negative moments may be taken at one third the flange width (14 in., governs), but no more than 15 in. away from the centerline of the girder, as allowed in LRFD 4.6.2.1.6. For 9 ft. span the following live load positive and negative moments (conservatively assuming the negative moment design section at 12 in. from CL of girder) can be obtained from the table:

3.1.2 Factored Design Moments

3.1.2.1 Service Limit State:

Negative Interior Moment: (14 in. from centerline of girder)

Mneg = -(0.81+0.24+3.71) = -4.76 kip-ft. / ft.

Positive Moment:

Mpos = (0.81+0.24+6.29) = 7.34 kip-ft. / ft.

3.1.2.2 Strength Limit State:

Negative Interior Moment: (14 in. from centerline of girder)

Mneg = -(1.25x0.81 + 1.5x0.24 + 1.75x3.71) = -7.87 kip-ft. / ft.

Positive Moment:

Mpos = 1.25x0.81 + 1.5x0.24 + 1.75x6.29 = 12.38 kip-ft. / ft.

3.2 Design

Figure 2: Longitudinal cross section of deck. Transverse mats are the upper-most and lower-most mats.

3.2.1 Negative Moments:

3.2.1.1 Strength Limit State

Mneg = -7.87 kip-ft. / ft.

Try #5 bars at 10 in. spacing

where:

= 0.90 for extreme event limit state - flexure of reinforced concrete members

= depth of compression block

As = (12/10)(0.31 in.2/bar) = 0.372 in.2

b = 12 in.

d = 8 – 2.5 - 0.625/2 = 5.19 in.

Provided moment capacity: Mn = 8.29 kip-ft. / ft. > Mneg = 7.87 kip-ft. / ft. O.K.

3.2.1.2 Service Limit State

3.2.1.2.1 Calculate Maximum Spacing of Tension Reinforcement

The provisions of AASHTO LRFD Article 5.7.3.4 regarding the maximum spacing of tension reinforcement apply if the tension in the cross-section exceeds 80 percent of the modulus of rupture, f r, at Service I Limit Sate.

(LRFD 5.4.2.6)

0.8fr = 0.8(0.48) = 0.38 ksi

Section modulus for the gross cross-section,

The negative moment at Service Limit State = -4.76 kip-ft./ft.

Tensile stress at the bottom fiber is given by,

> 0.38 ksi

Therefore, the maximum spacing, s, of the tension reinforcement should satisfy the following:

(LRFD Eq. 5.7.3.4-1)

where:

, for Class 2 exposure condition (increased concern of appearance and/or corrosion)dc = concrete cover measured from extreme tension fiber to center of extreme flexural reinforcement =

2.5” (clear cover) + 0.625 (diameter of No. 5 bar)/2 = 2.81 in.

fs = tensile stress in steel reinforcement (ksi)

h = overall thickness (in) = 8 in.

Calculate stress in reinforcement:

The stress in the steel reinforcing bars due to the fatigue load is calculated using cracked section analysis based on elastic theory. The elastic theory assumes:

1. Plane sections before bending remain plane after bending.

2. Tensile strength of concrete is ignored.

3. Stress-strain relationships for concrete and steel are linear elastic.

Figure 3 depicts the slab cross-section and the resulting stresses, strains, and forces acting on the section using the elastic theory.

The stress in the reinforcement is calculated as follows:

where:

M = -4.76 kip-ft./ft.

As = No. 5 at 10” o.c. = 0.31/10*12 = 0.372 in.2/ ft.

ds = 8 – 2.5 – 0.625/2 = 5.19 in.

n = modular ratio = Es / Ec

n = 29,000 / 3,830 = 7.57

But n shall be rounded to the nearest integer and shall be greater than 6.

Therefore, n = 8 (LRFD 5.7.1)

M

b

d s

c

s

fc

fs

NeutralAxis

kds

13kds

jd = (1 - )dk3s s

Elevation Section Strain Stress Resultant Forces

C

T

Figure 3: Reinforced concrete rectangular beam section at service load

j = 1 - 0.265/3 = 0.912

Ms = 4.76 kip-ft./ft.

Therefore,

Provided No. 5 at 10” is not adequate for crack control. N.G.Reduce the spacing of No. 5 bars to 7” o.c.The revised maximum spacing of the reinforcement is 7.2 in. o.c. > provided O.K.

Therefore, for negative interior moments: Use #5 bars @ 7 IN spacing (As prov'd = 0.53 in. 2 / ft.)

3.2.2 Positive Moments:

3.2.2.1 Strength Limit State

M+ = 12.38 kip-ft. / ft.

Try #5 bars at 8 in. spacing

As = (12/8)(0.31 in.2/bar) = 0.465 in.2

b = 12 in.

a = = 0.684 in.

d = 8 – 1 – 0.625/2 = 6.69 in.

Mn = = 13.3 kip-ft. / ft.

Provided moment capacity: Mn = 13.3 kip-ft. / ft. > Mu = 12.38 kip-ft. / ft. O.K.

3.2.2.2 Check crack control requirements: (LRFD 5.7.3.4)

3.2.2.2.1 Calculate Maximum Spacing of Tension Reinforcement

The provisions of AASHTO LRFD Article 5.7.3.4 regarding the maximum spacing of tension reinforcement apply if the tension in the cross-section exceeds 80 percent of the modulus of rupture, f r, at Service I Limit Sate.

(LRFD 5.4.2.6)

0.8fr = 0.8(0.48) = 0.38 ksi

Section modulus for the gross cross-section,

The positive moment at Service Limit State = 7.34 kip-ft./ft.

Tensile stress at the bottom fiber is given by,

> 0.38 ksi

Therefore, the maximum spacing, s, of the tension reinforcement should satisfy the following:

(LRFD Eq. 5.7.3.4-1)

where:

, for Class 2 exposure condition (increased concern of appearance and/or corrosion)dc = concrete cover measured from extreme tension fiber to center of extreme flexural reinforcement =

1.0” (clear cover) + 0.625 (diameter of No. 5 bar)/2 = 1.313 in.

fs = tensile stress in steel reinforcement (ksi) = 31 ksi (see section 3.2.1.2.1 for the calculation procedure)

h = overall thickness (in) = 8 in.

Therefore,

< 8 in. provided. O.K.

Therefore, for positive moments: Use #5 bars @ 8 in. spacing (As prov'd = 0.465 in. 2 / ft.)

3.2.2.3 Distribution Reinforcement

Reinforcement must be placed in the secondary direction in the bottom of the slab as a percentage of the primary reinforcement for positive moment. For primary reinforcement that is perpendicular to traffic, the required percentage is:

(LRFD 9.7.3.2)

where S = distance between flange tips, plus flange overhang, taken as distance from the extreme flange tip to the face of the web, disregarding any fillets. (LRFD 9.7.2.3)

= 108 - 6 = 102 in. = 8.5 ft.

Thus, the required percentage is:

Therefore, for distribution reinforcement, use 67% of positive moment steel.

As = 0.67(0.47 in.2 / ft.) = 0.31 in.2 / ft. - Use #5 @ 12 in. (As prov'd = 0.310 in. 2 / ft.)

3.2.2.4 Shrinkage and Temperature Reinforcement:

Provide minimum reinforcement for top layer of longitudinal reinforcement:

As 0.11 Ag / fy (LRFD 5.10.8.2-1)

As 0.11 [8.0(12.0)] / 60 = 0.176 in.2 / ft. - Use #4 @ 13.5 in. (As prov'd = 0.178 in. 2 / ft.)

3.2.3 Summary and Comparison of Reinforcement Requirements:

Total reinforcement per square foot of deck, as computed above:

Empirical method 2[0.276 + 0.185] = 0.922 in.2 / ft. (- 38%)

Traditional method 0.53 + 0.465 + 0.310 + 0.178 = 1.483 in.2 / ft. (+ 61%)

4. Overhang Design

Bridge barriers are designed for resisting accidental impact of a standard Test Vehicle. LRFD Article 13.7.2 specifies various levels of Test Vehicles for different applications. For the majority of applications on high speed highways, the Test Vehicle, TL-4 is specified.

Barriers are designed by the yield line theory which assumes development of nominal moment strength of the barrier wall in the transverse and/or longitudinal direction. Article A13.3.1 provides equations for calculating the critical barrier wall length over which the yield line mechanism occurs, L c and the nominal barrier resistance to transverse loads, Rw, for two cases: (a) impact within a barrier wall segment and (b) impact at end of barrier wall segment or at a joint.

For sustaining the assumed yield line mechanism and containing the damage due to accidental vehicle impact in the barrier, the deck overhang must be designed to resist the loads caused by the Test Vehicle impact (Rw ≥ Ft ). LRFD Table A13.2.1-1 specifies the barrier design forces due to Test Vehicle, TL-4:

Design Forces and DesignationsFt Transverse Force 54 KIPFL Longitudinal Force 18 KIPFv Vertical Force Down 18 KIPLt and LL 3.5 FTLv 18 FTHe min (Height of impact above deck) 32 INH Minimum Height of Barrier 32 IN

Figure 4: Design forces for bridge barrier

Figure 5: Case 1 – Yield line mechanism for impact within barrier wall segment

Figure 6: Case 2 – Yield line mechanism for impact at end of barrier wall segment or at a joint

4.1 Design Cases:

Bridge deck overhangs are designed for the effects of three separate load cases. (LRFD A13.4.1)

Design Case 1: Dead loads and the transverse and longitudinal forces specified in LRFD Article A13.2. Load factor, load and resistance factors = 1.0. – extreme event limit state.

Design Case 2: Dead loads and the vertical forces specified in LRFD Article A13.2 Load factor, load and resistance factors = 1.0. – extreme event limit state. Typically does not govern for concrete barriers.

Design Case 3: Strength I Limit State: 1.25DC + 1.5 DW + 1.75 (LL+IM).

For Design Case 3, the live load consists of a design truck axel with the wheel closest to the barrier located at a distance 1.0 ft. from the face of the barrier. The multiple presence factor, m = 1.2, of LRFD Article 3.6.1.1.2 applies.

For deck overhangs not exceeding 6.0 ft., LRFD Article 3.6.1.3.4 allows the use of a uniform live load. Fro structurally continuous barriers, a uniform live load of 1.0 KLF

intensity is applied at a distance of 1.0 FT. from the face of the barrier. For structurally discontinuous barriers, the design truck axle load should be used.

For decks spanning transversely across multiple girders, the effective deck width is taken as specified in LRFD Article 4.6.2.1.3 and the deck is analyzed as a continuous beam supported by the girders to calculate the design moments along the entire width of the bridge. This design case typically only governs for widely spaced girders, approximately 12 FT. or more, which allow the use of wider overhangs.

This example will focus on design case 1.

The flexural resistance of the deck, Ms, acting concurrently with a tensile force T = Ft should exceed the factored moments caused by dead loads and a concentrated moment equal to the flexural resistance, Mc,

of the parapet at its base applied at the edge of the deck. This is to allow the assumed yield line failure mechanism to develop in the barrier.(LRFD A13.4.2)

The deck design moment, Ms, is distributed over a distance Lc whereas the axial tension, T, is distributed over a distance Lc+2H and Lc+H for the yield line cases 1 and 2 respectively.

4.2 Strength of Barrier:

The standard IDOT (Illinois Department of Transportation) “F-Shape” concrete barrier is used for this example. Figures 5.1 and 5.2 depict the barrier geometry and reinforcement.

LRFD A13.3.1 provides equations for calculating the nominal resistance of a barrier wall, Rw. For brevity, detailed calculations for Rw which involve calculation of the nominal flexural resistance of the barrier wall about the horizontal and the vertical axes, Mc and Mw, are omitted. The Rw for the barrier is equal to 61.6 kip (governed by yield line case 2) which is greater than Ft = 54 kip. OK

4.3 Flexural Design of Deck Overhang:

At the inside face of the barrier:

MDC = (8/12)*(0.150)*(1.5)2 / 2 = 0.06 kip-ft. / ft.Mbarrier = (0.450)*(1.5)2 / 2 = 0.34 kip-ft. / ft.Mc = 13.9 kip-ft. / ft.

The design forces for the deck (at the inside face of barrier) are:

M = MDC + Mbarrier + Mc = 0.06 + 0.34 + 13.9 = 14.30 kip-ft. / ft.P = T1 (yield line case 1) = 6.94 kip / ft., assumed to act at the centroid on the deckCalculate the reinforcement needed at the top of the deck in the transverse direction (empirical design):

Figure 6: Longitudinal and Transverse Flexural Resistance of Barrier

Figure 5.1: Barrier geometry Figure 5.2: Barrier reinforcement

8”

The nominal flexural resistance, Mn, of a section subject to moment, M, and axial tension, P, is given by:

Where,

C = Total compressive forcea = depth of the Whitney stress blockT = Tensile force in the flexural tension reinforcement = C + P

The reinforcement provided in the top of the slab is No. 4 bars spaced 13.0 in. o.c. (Empirical Method),

As = 0.185 in.2 / ft.

T = 0.185x60 = 11.1 kip / ft.

C = 11.1 - 6.94 = 4.16 kip / ft.

a = 4.16/(0.85x12x4) = 0.10 in.

c = a/0.85 = 0.10/0.85 = 0.12 in.

de = 8 – 2.5 – 0.5/2 = 5.25 in.

Mn = 2.53 < M = 14.30 kip-ft. / ft. NG

Provide additional No. 7 at 13 in. o.c. (alternating with No. 4 at 13 o.c.) in the overhang portion of the deck.

As = (0.20+0.60)/13*(12) = 0.74 in.2 / ft.

T = 0.74x60 = 44.4 kip / ft.

C = 44.4 - 6.94 = 37.46 kip / ft.

a = 37.46/(0.85x12x4) = 0.92 in.

c = a/0.85 = 0.92/0.85 = 1.08 in.

de = 8 – 2.5 – 0.875/2 = 5.06 in.

Mn = 15.0 > M = 14.30 kip-ft. / ft. OK

Check the maximum reinforcement requirement of LRFD 5.7.3.3.1.

c/de = 1.08/5.06 = 0.21 < 0.42 OK

One option to reduce the amount of steel would be to thicken the deck in the overhang region. Several additional sections should then be checked to determine where the additional steel in the overhang zone can be stopped.

At the inside face of the barrier, the impact forces are distributed over a distance Lc for the transverse bending moment and Lc+2H or Lc+H for axial force. This effective width increases at sections away from the impact location. It may be reasonable to assume a dispersion of 30 to 45 degrees to calculate the increased effective width of the deck.