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Bridges & Structures

LRFD Steel Girder SuperStructure Design Example

Concrete Deck Design Example Design Step 2

Table of Contents

Design Step 2.1 - Obtain Design Criteria
Design Step 2.2 - Determine Minimum Slab Thickness
Design Step 2.3 - Determine Minimum Overhang Thickness
Design Step 2.4 - Select Slab and Overhang Thickness
Design Step 2.5 - Compute Dead Load Effects
Design Step 2.6 - Compute Live Load Effects
Design Step 2.7 - Compute Factored Positive and Negative Design Moments
Design Step 2.8 - Design for Positive Flexure in Deck
Design Step 2.9 - Check for Positive Flexure Cracking under Service Limit State
Design Step 2.10 - Design for Negative Flexure in Deck
Design Step 2.11 - Check for Negative Flexure Cracking under Service Limit State
Design Step 2.12 - Design for Flexure in Deck Overhang
Design Step 2.13 - Check for Cracking in Overhang under Service Limit State
Design Step 2.14 - Compute Overhang Cut-off Length Requirement
Design Step 2.15 - Compute Overhang Development Length
Design Step 2.16 - Design Bottom Longitudinal Distribution Reinforcement
Design Step 2.17 - Design Top Longitudinal Distribution Reinforcement
Design Step 2.18 - Design Longitudinal Reinforcement over Piers
Design Step 2.19 - Draw Schematic of Final Concrete Deck Design

Design Step 2.1 - Obtain Design Criteria

The first design step for a concrete bridge deck is to choose the correct design criteria. The following concrete deck design criteria are obtained from the typical superstructure cross section shown in Figure 2-1 and from the referenced articles and tables in the AASHTO LRFD Bridge Design Specifications (through 2002 interims).

Refer to Design Step 1 for introductory information about this design example. Additional information is presented about the design assumptions, methodology, and criteria for the entire bridge, including the concrete deck.

The next step is to decide which deck design method will be used. In this example, the equivalent strip method will be used. For the equivalent strip method analysis, the girders act as supports, and the deck acts as a simple or continuous beam spanning from support to support. The empirical method could be used for the positive and negative moment interior regions since the cross section meets all the requirements given in S9.7.2.4. However, the empirical method could not be used to design the overhang as stated in S9.7.2.2.

S4.6.2

MathCad tip logo Overhang Width
The overhang width is generally determined such that the moments and shears in the exterior girder are similar to those in the interior girder. In addition, the overhang is set such that the positive and negative moments in the deck slab are balanced. A common rule of thumb is to make the overhang approximately 0.35 to 0.5 times the girder spacing.

Superstructure cross section consisting of a concrete deck with Jersey barriers on 5 steel girders which are spaced at 9 foot 9 inches each for a total of 39 foot 0 inches and with a 3 foot 11 and one quarter inch deck overhang on each side for a total deck width of 46 foot 10 and one half inches. The barriers are 3 foot 6 inches high and 1 foot 5 and one quarter inches wide. There are 2 lanes at 12 foot 0 inches each and 2 shoulders at 10 foot zero inches each.

Figure 2-1 Superstructure Cross Section

The following units are defined for use in this design example:

Formula: K = 1000lb   Formula: kcf = numerator (K) divided by denominator ( feet cubed )   Formula: ksi = numerator (K) divided by denominator ( inches squared )

Deck properties:

Girder spacing: Formula: S = 9 point 75 feet  
Number of girders: Formula: N = 5  
Deck top cover: Formula: Cover subscript t = 2 point 5 inches STable 5.12.3-1
Deck bottom cover: Formula: Cover subscript b = 1 point 0 inches STable 5.12.3-1
Concrete density: Formula: W subscript c = 0 point 150kcf STable 3.5.1-1
Concrete 28-day compressive strength:   Formula: f prime subscript c = 4 point 0ksi S5.4.2.1
Reinforcement strength: Formula: f subscript y = 60ksi S5.4.3 & S6.10.3.7
Future wearing surface: Formula: W subscript fws = 0 point 140kcf STable 3.5.1-1

Parapet properties:

Weight per foot: Formula: W subscript par = 0 point 53 Kips per foot  
Width at base: Formula: w subscript base = 1 point 4375 feet  
Moment capacity at base*: Formula: M subscript co = 28 point 21 Kips foot per foot  
Parapet height: Formula: H subscript par = 3 point 5 feet  
Critical length of yield line failure pattern*:   Formula: L subscript c = 12 point 84 feet (calculated in Design Step 2.12)   SA13.3.1
Total transverse resistance of the parapet*: Formula: R subscript w = 117 point 40K (calculated in Design Step 2.12)   SA13.3.1

* Based on parapet properties not included in this design example. See Publication Number FHWA HI-95-017, Load and Resistance Factor Design for Highway Bridges, Participant Notebook, Volume II (Version 3.01), for the method used to compute the parapet properties.

Deck top cover - The concrete top cover is set at 2.5 inches since the bridge deck may be exposed to deicing salts and/or tire stud or chain wear. This includes the 1/2 inch integral wearing surface that is required.

STable 5.12.3-1

Deck bottom cover - The concrete bottom cover is set at 1.0 inch since the bridge deck will use reinforcement that is smaller than a #11 bar.

STable 5.12.3-1

Concrete 28-day compressive strength - The compressive strength for decks shall not be less than 4.0 KSI. Also, type "AE" concrete should be specified when the deck will be exposed to deicing salts or the freeze-thaw cycle. "AE" concrete has a compressive strength of 4.0 KSI.

S5.4.2.1

STable C5.4.2.1-1

Future wearing surface density - The future wearing surface density is 0.140 KCF. A 2.5 inch thickness will be assumed.

STable 3.5.1-1

Design Step 2.2 - Determine Minimum Slab Thickness

The concrete deck depth cannot be less than 7.0 inches, excluding any provision for grinding, grooving, and sacrificial surface.

S9.7.1.1

Design Step 2.3 - Determine Minimum Overhang Thickness

For concrete deck overhangs supporting concrete parapets or barriers, the minimum deck overhang thickness is:

S13.7.3.1.2

Formula: t subscript o = 8 point 0 inches

Design Step 2.4 - Select Slab and Overhang Thickness

Once the minimum slab and overhang thicknesses are computed, they can be increased as needed based on client standards and design computations. The following slab and overhang thicknesses will be assumed for this design example:

Formula: t subscript s = 8 point 5 inches and Formula: t subscript o = 9 point 0 inches

Design Step 2.5 - Compute Dead Load Effects

The next step is to compute the dead load moments. The dead load moments for the deck slab, parapets, and future wearing surface are tabulated in Table 2-1. The tabulated moments are presented for tenth points for Bays 1 through 4 for a 1-foot strip. The tenth points are based on the equivalent span and not the center-to-center beam spacing.

STable 3.5.1-1

After the dead load moments are computed for the slab, parapets, and future wearing surface, the correct load factors must be identified. The load factors for dead loads are:

STable 3.4.1-2

For slab and parapet:

Maximum Formula: gamma subscript pDCmax = 1 point 25

Minimum Formula: gamma subscript pDCmin = 0 point 90

For future wearing surface:

Maximum Formula: gamma subscript pDWmax = 1 point 50

Minimum Formula: gamma subscript pDWmin = 0 point 65

DISTANCE SLAB DEAD LOAD PARAPET DEAD LOAD FWS DEAD LOAD
BAY 1 BAY 2 BAY 3 BAY 4 BAY 1 BAY 2 BAY 3 BAY 4 BAY 1 BAY 2 BAY 3 BAY 4
1.0 -0.71 -0.72 -0.71 -0.74 0.43 -0.23 0.47 -1.66 -0.24 -0.18 -0.24 -0.06
0.9 -0.30 -0.31 -0.30 -0.33 0.22 -0.16 0.40 -1.45 -0.11 -0.07 -0.12 0.04
0.8 0.01 0.01 0.02 -0.01 0.02 -0.09 0.33 -1.24 0.00 0.01 -0.02 0.11
0.7 0.24 0.24 0.24 0.22 -0.19 -0.02 0.26 -1.03 0.08 0.07 0.05 0.15
0.6 0.37 0.38 0.38 0.36 -0.40 0.05 0.19 -0.82 0.14 0.10 0.09 0.17
0.5 0.41 0.42 0.42 0.41 -0.61 0.12 0.12 -0.61 0.17 0.11 0.11 0.17
0.4 0.36 0.38 0.38 0.37 -0.82 0.19 0.05 -0.40 0.17 0.09 0.10 0.14
0.3 0.22 0.24 0.24 0.24 -1.03 0.26 -0.02 -0.19 0.15 0.05 0.07 0.08
0.2 -0.01 0.02 0.01 0.01 -1.24 0.33 -0.09 0.02 0.11 -0.02 0.01 0.00
0.1 -0.33 -0.30 -0.31 -0.30 -1.45 0.40 -0.16 0.22 0.04 -0.12 -0.07 -0.11
0.0 -0.74 -0.71 -0.72 -0.71 -1.66 0.47 -0.23 0.43 -0.06 -0.24 -0.18 -0.24

Table 2-1 Unfactored Dead Load Moments (K-FT/FT)

Design Step 2.6 - Compute Live Load Effects

Before the live load effects can be computed, the following basic parameters must be defined:

The minimum distance from the center of design vehicle wheel to the inside face of parapet = 1 foot

S3.6.1.3.1

The minimum distance between the wheels of two adjacent design vehicles = 4 feet

S3.6.1.3.1

Dynamic load allowance, IM Formula: IM = 0 point 33

STable 3.6.2.1-1

Load factor for live load - Strength I Formula: gamma subscript LL = 1 point 75

STable 3.4.1-1

Multiple presence factor, m:

With one lane loaded, m = 1.20
With two lanes loaded, m = 1.00
With three lanes loaded, m = 0.85

STable 3.6.1.1.2-1

Fatigue does not need to be investigated for concrete deck design.

S9.5.3 & S5.5.3.1

Resistance factors for flexure:

Strength limit state Formula: phi subscript str = 0 point 90

S5.5.4.2

Service limit state Formula: phi subscript serv = 1 point 00

S1.3.2.1

Extreme event limit state Formula: phi subscript ext = 1 point 00

S1.3.2.1

Based on the above information and based on S4.6.2.1, the live load effects for one and two trucks are tabulated in Table 2-2. The live load effects are given for tenth points for Bays 1 through 4. Multiple presence factors are included, but dynamic load allowance is excluded.

DISTANCE SINGLE TRUCK(MULTIPLE PRESENCE FACTOR OF 1.20 INCLUDED) TWO TRUCKS(MULTIPLE PRESENCE FACTOR OF 1.00 INCLUDED)
MAX. MOMENT MIN. MOMENT MAX. MOMENT MIN. MOMENT
BAY 1 BAY 2 BAY 3 BAY 4 BAY 1 BAY 2 BAY 3 BAY 4 BAY 1 BAY 2 BAY 3 BAY 4 BAY 1 BAY 2 BAY 3 BAY 4
1.0 4.55 5.82 4.07 3.62 -28.51 -27.12 -28.37 -0.44 2.28 -2.87 2.28 2.66 -29.39 -27.94 -28.83 -0.27
0.9 5.68 6.01 7.84 17.03 -15.11 -15.84 -19.10 -2.08 2.06 7.19 8.04 12.49 -18.32 -17.37 -14.44 -1.26
0.8 18.10 18.35 20.68 30.43 -13.48 -13.63 -16.39 -3.70 4.50 15.14 16.98 22.32 -8.18 -8.01 -2.10 -2.25
0.7 27.13 27.20 25.39 36.64 -11.84 -11.41 -13.68 -5.33 14.00 20.48 19.15 26.46 -7.19 -5.74 -1.44 -3.24
0.6 32.48 28.00 29.28 36.62 -10.22 -9.20 -10.97 -6.96 21.21 19.58 21.30 25.84 -6.20 -4.62 -2.68 -4.22
0.5 31.20 28.26 28.14 31.10 -8.59 -8.57 -8.27 -8.59 21.01 21.72 21.64 20.93 -5.21 -3.51 -3.92 -5.21
0.4 36.76 29.09 28.14 26.14 -6.96 -11.38 -9.20 -10.22 25.93 21.19 19.41 17.96 -4.23 -2.40 -5.17 -6.20
0.3 36.44 25.56 27.37 16.19 -5.33 -14.18 -11.42 -11.84 26.35 19.26 16.71 10.48 -3.24 -1.29 -6.41 -7.19
0.2 30.53 20.88 18.22 11.66 -3.70 -17.00 -13.63 -13.48 22.36 17.10 7.73 4.59 -2.71 -8.53 -8.02 -8.18
0.1 22.94 7.73 6.22 5.56 -14.45 -19.81 -15.85 -15.11 17.44 7.98 7.32 2.30 -12.09 -16.92 -17.38 -18.33
0.0 5.62 4.07 6.04 4.55 -25.75 -28.38 -27.13 -28.51 4.36 2.04 -2.92 2.55 -21.47 -29.36 -27.92 -29.40

Table 2-2 Unfactored Live Load Moments (Excluding Dynamic Load Allowance) (K-FT)

Design Step 2.7 - Compute Factored Positive and Negative Design Moments

For this example, the design moments will be computed two different ways. For Method A, the live load portion of the factored design moments will be computed based on the values presented in Table 2-2. Table 2-2 represents a continuous beam analysis of the example deck using a finite element analysis program. For Method B, the live load portion of the factored design moments will be computed using STable A4.1-1. In STable A4.1-1, moments per unit width include dynamic load allowance and multiple presence factors. The values are tabulated using the equivalent strip method for various bridge cross sections. The values in STable A4.1-1 may be slightly higher than the values from a deck analysis based on the actual number of beams and the actual overhang length. The maximum live load moment is obtained from the table based on the girder spacing. For girder spacings between the values listed in the table, interpolation can be used to get the moment.

STable A4.1-1

Based on Design Step 1, the load modifier eta (η) is 1.0 and will not be shown throughout the design example. Refer to Design Step 1 for a discussion of eta.

S1.3.2.1

Factored Positive Design Moment Using Table 2-2 - Method A

Factored positive live load moment:

The positive, negative, and overhang moment equivalent strip equations are presented in Figure 2-2 below.

Figure showing equivalent strip equations for various parts of the deck consisting of 5 steel girders concrete deck and Jersey barriers. Overhang moment between barrier and fascia girder equals 45 point 0 plus 10 point 0X. Positive moment one half way between girders equals 26 point 0 plus 6 point 6S. Negative moment at center line of girders equals 48 point 0 plus 3 point 0S.

STable 4.6.2.1.3-1

Figure 2-2 Equivalent Strip Equations for Various Parts of the Deck

The width of the equivalent strip for positive moment is:

STable 4.6.2.1.3-1

Formula: w subscript posstripa = 26 point 0 + 6 point 6S. Equation not used

For Formula: S = 9 point 75feet

Formula: w subscript posstripa = 90 point 35 inches or Formula: w subscript posstripa = 7 point 53 feet

Based on Table 2-2, the maximum unfactored positive live load moment is 36.76 K-ft, located at 0.4S in Bay 1 for a single truck. The maximum factored positive live load moment is:

Formula: Mu subscript posliveA = gamma subscript LL times ( 1 + IM) times numerator (36 point 76K feet ) divided by denominator (w subscript posstripa)

Formula: Mu subscript posliveA = 11 point 36 Kips foot per foot

Factored positive dead load moment:

Based on Table 2-1, the maximum unfactored slab, parapet, and future wearing surface positive dead load moment occurs in Bay 2 at a distance of 0.4S. The maximum factored positive dead load moment is as follows:

Formula: Mu subscript posdead = gamma subscript pDCmax times ( 0 point 38 times Kips foot per foot ) + gamma subscript pDCmax times ( 0 point 19 times Kips foot per foot ) + gamma subscript pDWmax times ( 0 point 09 times Kips foot per foot )

Formula: Mu subscript posdead = 0 point 85 Kips foot per foot

The total factored positive design moment for Method A is:

Formula: Mu subscript postotalA = Mu subscript posliveA + Mu subscript posdead

Formula: Mu subscript postotalA = 12 point 21 Kips foot per foot

It should be noted that the total maximum factored positive moment is comprised of the maximum factored positive live load moment in Bay 1 at 0.4S and the maximum factored positive dead load moment in Bay 2 at 0.4S. Summing the factored moments in different bays gives a conservative result. The exact way to compute the maximum total factored design moment is by summing the dead and live load moments at each tenth point per bay. However, the method presented here is a simpler and slightly conservative method of finding the maximum total factored moment.

Factored Positive Design Moment Using STable A4.1-1 - Method B

Factored positive live load moment:

For a girder spacing of 9'-9", the maximum unfactored positive live load moment is 6.74 K-ft/ft.

STable A4.1-1

This moment is on a per foot basis and includes dynamic load allowance. The maximum factored positive live load moment is:

Formula: Mu subscript posliveB = gamma subscript LL times 6 point 74 Kips foot per foot

Formula: Mu subscript posliveB = 11 point 80 Kips foot per foot

Factored positive dead load moment:

The factored positive dead load moment for Method B is the same as that for Method A:

Formula: Mu subscript posdead = 0 point 85 Kips foot per foot

The total factored positive design moment for Method B is:

Formula: Mu subscript postotalB = Mu subscript posliveB + Mu subscript posdead

Formula: Mu subscript postotalB = 12 point 64 Kips foot per foot

Comparing Methods A and B, the difference between the total factored design moment for the two methods is:

Formula: numerator (Mu subscript postotalB minus Mu subscript postotalA) divided by denominator (Mu subscript postotalB) = 3 point 4 %

MathCad tip logo Method A or Method B
It can be seen that the tabulated values based on STable A4.1-1 (Method B) are slightly greater than the computed live load values using a finite element analysis program (Method A). For real world deck design, Method B would be preferred over Method A due to the amount of time that would be saved by not having to develop a finite element model. Since the time was spent to develop the finite element model for this deck design, the Method A values will be used.

Factored Negative Design Moment Using Table 2-2 - Method A

Factored negative live load moment:

The deck design section for a steel beam for negative moments and shear forces is taken as one-quarter of the top flange width from the centerline of the web.

S4.6.2.1.6

Figure 2-3 Location of Design Section: Girder cross section. Width of top flange designated b sub f and design section located one forth b sub f from centerline of web.

S4.6.2.1.6

Figure 2-3 Location of Design Section

Assume

Formula: b subscript f = 1 point 0 feet

Formula: numerator (1) divided by denominator (4) b subscript f = 0 point 25 feet

The width of the equivalent strip for negative moment is:

STable 4.6.2.1.3-1

Formula: w subscript negstripa = 48 point 0 + 3 point 0S. Equation not used

Formula: w subscript negstripa = 77 point 25 inches or Formula: w subscript negstripa = 6 point 44 feet

Based on Table 2-2, the maximum unfactored negative live load moment is -29.40 K-ft, located at 0.0S in Bay 4 for two trucks. The maximum factored negative live load moment is:

Formula: Mu subscript negliveA = gamma subscript LL times ( 1 + IM) times numerator ( minus 29 point 40K feet ) divided by denominator (w subscript negstripa)

Formula: Mu subscript negliveA = minus 10 point 63 Kips foot per foot

Factored negative dead load moment:

From Table 2-1, the maximum unfactored negative dead load moment occurs in Bay 4 at a distance of 1.0S. The maximum factored negative dead load moment is as follows:

Formula: Mu subscript negdead = gamma subscript pDCmax times ( minus 0 point 74 times Kips foot per foot ) + gamma subscript pDCmax times ( minus 1 point 66 times Kips foot per foot ) + gamma subscript pDWmax times ( minus 0 point 06 times Kips foot per foot )

Formula: Mu subscript negdead = minus 3 point 09 Kips foot per foot

The total factored negative design moment for Method A is:

Formula: Mu subscript negtotalA = Mu subscript negliveA + Mu subscript negdead

Formula: Mu subscript negtotalA = minus 13 point 72 Kips foot per foot

Factored Negative Design Moment Using STable A4.1-1 - Method B

Factored negative live load moment:

For a girder spacing of 9'-9" and a 3" distance from the centerline of girder to the design section, the maximum unfactored negative live load moment is 6.65 K-ft/ft.

STable A4.1-1

If the distance from the centerline of the girder to the design section does not match one of the distances given in the table, the design moment can be obtained by interpolation. As stated earlier, these moments are on a per foot basis and include dynamic load allowance.

The maximum factored negative live load moment is:

Formula: Mu subscript negliveB = gamma subscript LL times minus 6 point 65 Kips foot per foot

Formula: Mu subscript negliveB = minus 11 point 64 Kips foot per foot

Factored negative dead load moment:

The factored negative dead load moment for Method B is the same as that for Method A:

Formula: Mu subscript negdead = minus 3 point 09 Kips foot per foot

The total factored negative design moment for Method B is:

Formula: Mu subscript negtotalB = Mu subscript negliveB + Mu subscript negdead

Formula: Mu subscript negtotalB = minus 14 point 73 Kips foot per foot

Comparing Methods A and B, the difference between the total factored design moment for the two methods is:

Formula: numerator (Mu subscript negtotalB minus Mu subscript negtotalA) divided by denominator (Mu subscript negtotalB) = 6 point 8 %

MathCad tip logo Method A or Method B

It can be seen that the tabulated values based on STable A4.1-1 (Method B) are slightly greater than the computed live load values using a finite element analysis program (Method A). For real world deck design, Method B would be preferred over Method A due to the amount of time that would be saved by not having to develop a finite element model. Since the time was spent to develop the finite element model for this deck design, the Method A values will be used.

Design Step 2.8 - Design for Positive Flexure in Deck

The first step in designing the positive flexure steel is to assume a bar size. From this bar size, the required area of steel (As) can be calculated. Once the required area of steel is known, the required bar spacing can be calculated.

Figure 2-4 Reinforcing Steel for Positive Flexure in Deck: This is a deck overhang with parapet cross section showing transverse and longitudinal bars and designating bottom transverse bars as reinforcing steel for positive flexure in deck.

Figure 2-4 Reinforcing Steel for Positive Flexure in Deck

Assume #5 bars:

Formula: bar_diam = 0 point 625 inches

Formula: bar_area = 0 point 31 inches squared

Effective depth, de = total slab thickness - bottom cover - 1/2 bar diameter - top integral wearing surface

Formula: d subscript e = t subscript s minus Cover subscript b minus numerator (bar_diam) divided by denominator (2) minus 0 point 5 inches

Formula: d subscript e = 6 point 69 inches

Solve for the required amount of reinforcing steel, as follows:

Formula: phi subscript f = 0 point 90

S5.5.4.2.1

Formula: b = 12 inches

Formula: Rn = numerator (Mu subscript postotalA times 12 inches ) divided by denominator (( phi subscript f times b times d subscript e squared )) Formula: Rn = 0 point 30 Kips per square inch

Formula: rho = 0 point 85 ( numerator (f prime subscript c) divided by denominator (f subscript y) ) times left bracket 1 point 0 minus square root of (1 point 0 minus numerator (( 2 times Rn)) divided by denominator (( 0 point 85 times f prime subscript c ))) right bracket

Formula: rho = 0 point 00530

Note: The above two equations are derived formulas that can be found in most reinforced concrete textbooks.

Formula: A subscript s = rho times numerator (b) divided by denominator ( feet ) times d subscript e Formula: A subscript s = 0 point 43 square inches per foot

Required bar spacing = Formula: numerator (bar_area) divided by denominator (A subscript s) = 8 point 7 inches

Use #5 bars @ Formula: bar_space = 8 point 0 inches

Once the bar size and spacing are known, the maximum reinforcement limit must be checked.

S5.7.3.3.1

Formula: T = bar_area times f subscript y Formula: T = 18 point 60 K

Formula: a = numerator (T) divided by denominator (0 point 85 times f prime subscript c times bar_space) Formula: a = 0 point 68 inches

Formula: beta subscript 1 = 0 point 85

S5.7.2.2

Formula: c = numerator (a) divided by denominator ( beta subscript 1) Formula: c = 0 point 80 inches

S5.7.2.2

Formula: numerator (c) divided by denominator (d subscript e) = 0 point 12 where Formula: numerator (c) divided by denominator (d subscript e) less than or equal to 0 point 42

S5.7.3.3.1

Formula: 0 point 12 less than or equal to 0 point 42 OK

Design Step 2.9 - Check for Positive Flexure Cracking under Service Limit State

The control of cracking by distribution of reinforcement must be checked.

S5.7.3.4

For members in severe exposure conditions: Formula: Z = 130 numerator (K) divided by denominator ( inches )

Thickness of clear cover used to compute dc should not be greater than 2 inches:

Formula: d subscript c = 1 inches + numerator (bar_diam) divided by denominator (2)

Formula: d subscript c = 1 point 31 inches

Concrete area with centroid the same as transverse bar and bounded by the cross section and line parallel to neutral axis:

Formula: A subscript c = 2 times ( d subscript c ) times bar_space

Formula: A subscript c = 21 point 00 inches squared

The equation that gives the allowable reinforcement service load stress for crack control is:

Formula: f subscript sa = numerator (Z) divided by denominator (( d subscript c times A subscript c ) superscript numerator (1) divided by denominator (3)) where Formula: f subscript sa less than or equal to 0 point 6 times f subscript y

Formula: f subscript sa = 43 point 04 ksi Formula: 0 point 6f subscript y = 36 point 00 ksi

Use Formula: f subscript sa = 36 point 00ksi

Figure 2-5 Bottom Transverse Reinforcement. 8 and 1/2 slab cross section. Four Number 5 bars spaced at 8 inches center to center and located 1 and 15/16 inches from centerline of bars to bottom face of slab. Reinforcement is Number 5 bars, diameter equals zero point 625 inch, cross-sectional area equals zero point 31.

Figure 2-5 Bottom Transverse Reinforcement

Formula: E subscript s = 29000ksi

S5.4.3.2

Formula: E subscript c = 3640ksi

S5.4.2.4

Formula: n = numerator (E subscript s) divided by denominator (E subscript c) Formula: n = 7 point 97

Use Formula: n = 8

Service positive live load moment:

Based on Table 2-2, the maximum unfactored positive live load moment is 36.76 K-ft, located at 0.4S in Bay 1 for a single truck. The maximum service positive live load moment is computed as follows:

Formula: gamma subscript LL = 1 point 0

Formula: Mu subscript posliveA = gamma subscript LL times ( 1 + IM) times numerator (36 point 76K feet ) divided by denominator (w subscript posstripa)

Formula: Mu subscript posliveA = 6 point 49 Kips foot per foot

Service positive dead load moment:

From Table 2-1, the maximum unfactored slab, parapet, and future wearing surface positive dead load moment occurs in Bay 2 at a distance of 0.4S. The maximum service positive dead load moment is computed as follows:

Formula: gamma subscript pDCserv = 1 point 0

STable 3.4.1-1

Formula: gamma subscript pDWserv = 1 point 0

STable 3.4.1-1

Formula: Mu subscript posdead = gamma subscript pDCserv times ( 0 point 38 times Kips foot per foot ) + gamma subscript pDCserv times ( 0 point 19 times Kips foot per foot ) + gamma subscript pDWserv times ( 0 point 09 times Kips foot per foot )

Formula: Mu subscript posdead = 0 point 66 Kips foot per foot

The total service positive design moment is:

Formula: Mu subscript postotalA = Mu subscript posliveA + Mu subscript posdead

Formula: Mu subscript postotalA = 7 point 15 Kips foot per foot

To solve for the actual stress in the reinforcement, the transformed moment of inertia and the distance from the neutral axis to the centroid of the reinforcement must be computed:

Formula: d subscript e = 6 point 69 inches Formula: A subscript s = 0 point 465 square inches per foot Formula: n = 8

Formula: rho = numerator (A subscript s) divided by denominator ( numerator (b) divided by denominator ( feet ) times d subscript e) Formula: rho = 0 point 00579

Formula: k = square root of (( rho times n ) squared + ( 2 times rho times n )) minus rho times n

Formula: k = 0 point 262

Formula: k times d subscript e = 1 point 75 inches

Figure 2-6 Crack control for Positive Reinforcement under Live Load. Slab cross section with one Number 5 bar at 8 point zero inch spacing. Bar is located one point three one inches from the bottom of the slab. The slab is eight and 1/2 inches thick including a zero point five zero inch wearing surface at the top. The neutral axis of the section is located four point nine four inches above the centerline of the bar.

Figure 2-6 Crack Control for Positive Reinforcement under Live Loads

Once kde is known, the transformed moment of inertia can be computed:

Formula: d subscript e = 6 point 69 inches

Formula: A subscript s = 0 point 465 square inches per foot

Formula: I subscript t = numerator (1) divided by denominator (3) times ( 12 inches per foot ) times ( k times d subscript e ) cubed + n times A subscript s times ( d subscript e minus k times d subscript e ) squared

Formula: I subscript t = 112 point 22 numerator ( inches superscript 4) divided by denominator ( feet )

Now, the actual stress in the reinforcement can be computed:

Formula: Mu subscript postotalA = 7 point 15 K times numerator ( feet ) divided by denominator ( feet ) Formula: y = d subscript e minus k times d subscript e Formula: y = 4 point 94 inches

Formula: f subscript s = numerator (n times ( Mu subscript postotalA times 12 inches per foot times y )) divided by denominator (I subscript t)

Formula: f subscript s = 30 point 23 ksi Formula: f subscript sa greater than f subscript s OK

Design Step 2.10 - Design for Negative Flexure in Deck

The negative flexure reinforcing steel design is similar to the positive flexure reinforcing steel design.

S4.6.2.1

Figure 2-7 Reinforcing Steel for Negative Flexure in Deck: This is a deck overhang with parapet cross section showing transverse and longitudinal bars and designating top transverse bars in first bay as reinforcing steel for negative flexure in deck.

Figure 2-7 Reinforcing Steel for Negative Flexure in Deck

Assume #5 bars:

Formula: bar_diam = 0 point 625 inches

Formula: bar_area = 0 point 31 inches squared

Effective depth, de = total slab thickness - top cover - 1/2 bar diameter

Formula: d subscript e = t subscript s minus Cover subscript t minus numerator (bar_diam) divided by denominator (2) Formula: d subscript e = 5 point 69 inches

Solve for the required amount of reinforcing steel, as follows:

Formula: phi subscript f = 0 point 90

S5.5.4.2.1

Formula: b = 12 inches

Formula: Rn = numerator ( minus Mu subscript negtotalA times 12 inches ) divided by denominator (( phi subscript f times b times d subscript e squared )) Formula: Rn = 0 point 47 Kips per square inch

Formula: rho = 0 point 85 ( numerator (f prime subscript c) divided by denominator (f subscript y) ) times left bracket 1 point 0 minus square root of (1 point 0 minus numerator (( 2 times Rn)) divided by denominator (( 0 point 85 times f prime subscript c ))) right bracket

Formula: rho = 0 point 00849

Formula: A subscript s = rho times numerator (b) divided by denominator ( feet ) times d subscript e Formula: A subscript s = 0 point 58 square inches per foot

Required bar spacing = Formula: numerator (bar_area) divided by denominator (A subscript s) = 6 point 4 inches

Use #5 bars @ Formula: bar_space = 6 point 0 inches

Once the bar size and spacing are known, the maximum reinforcement limit must be checked.

S5.7.3.3.1

Formula: T = bar_area times f subscript y Formula: T = 18 point 60 K

Formula: a = numerator (T) divided by denominator (0 point 85 times f prime subscript c times bar_space) Formula: a = 0 point 91 inches

Formula: beta subscript 1 = 0 point 85

S5.7.2.2

Formula: c = numerator (a) divided by denominator ( beta subscript 1) Formula: c = 1 point 07 inches

S5.7.2.2

Formula: numerator (c) divided by denominator (d subscript e) = 0 point 19 where Formula: numerator (c) divided by denominator (d subscript e) less than or equal to 0 point 42

S5.7.3.3.1

Formula: 0 point 19 less than or equal to 0 point 42 OK

Design Step 2.11 - Check for Negative Flexure Cracking under Service Limit State

Similar to the positive flexure reinforcement, the control of cracking by distribution of reinforcement must be checked.

S5.7.3.4

Formula: Z = 130 numerator (K) divided by denominator ( inches )

Note: clear cover is greater than 2.0 inches; therefore, use clear cover equals 2.0 inches.

S5.7.3.4

Formula: d subscript c = 2 inches + numerator (bar_diam) divided by denominator (2) Formula: d subscript c = 2 point 31 inches

Formula: A subscript c = 2 times ( d subscript c ) times bar_space Formula: A subscript c = 27 point 75 inches squared

Formula: f subscript sa = numerator (Z) divided by denominator (( d subscript c times A subscript c ) superscript numerator (1) divided by denominator (3)) where Formula: f subscript sa less than or equal to 0 point 6 times f subscript y

Formula: f subscript sa = 32 point 47 ksi Formula: 0 point 6f subscript y = 36 point 00 ksi

Use Formula: f subscript sa = 32 point 47ksi

Service negative live load moment:

From Table 2-2, the maximum unfactored negative live load moment is -29.40 K-ft, located at 0.0S in Bay 4 for two trucks. The maximum service negative live load moment is:

Formula: gamma subscript LL = 1 point 0

STable 3.4.1-1

Formula: Mu subscript negliveA = gamma subscript LL times ( 1 + IM) times numerator ( minus 29 point 40K feet ) divided by denominator (w subscript negstripa)

Formula: Mu subscript negliveA = minus 6 point 07 Kips foot per foot

Service negative dead load moment:

From Table 2-1, the maximum unfactored negative dead load moment occurs in Bay 4 at a distance of 1.0S. The maximum service negative dead load moment is computed as follows:

Formula: gamma subscript pDCservice = 1 point 0 Formula: gamma subscript pDWservice = 1 point 0

STable 3.4.1-1

Formula: Mu subscript negdead = gamma subscript pDCservice times ( minus 0 point 74 times Kips foot per foot minus 1 point 66 Kips foot per foot ) + gamma subscript pDWservice times ( minus 0 point 06 times Kips foot per foot )

Formula: Mu subscript negdead = minus 2 point 46 Kips foot per foot

The total service negative design moment is:

Formula: Mu subscript negtotalA = Mu subscript negliveA + Mu subscript negdead

Formula: Mu subscript negtotalA = minus 8 point 53 Kips foot per foot

Formula: d subscript e = 5 point 69 inches Formula: A subscript s = 0 point 62 square inches per foot Formula: n = 8

Formula: rho = numerator (A subscript s) divided by denominator ( numerator (b) divided by denominator ( feet ) times d subscript e) Formula: rho = 0 point 00908

Formula: k = square root of (( rho times n ) squared + ( 2 times rho times n )) minus rho times n

Formula: k = 0 point 315

Formula: k times d subscript e = 1 point 79 inches

Figure 2-8 Crack Control for Negative Reinforcement under Live Loads. Eight point five zero inch slab cross section with one Number 5 bar at six point zero spacing. Neutral axis of slab section is located one point seven nine inches above bottom of slab Centerline of bar is located two point eight one inches from top of slab.

Figure 2-8 Crack Control for Negative Reinforcement under Live Loads

Once kde is known, the transformed moment of inertia can be computed:

Formula: d subscript e = 5 point 69 inches

Formula: A subscript s = 0 point 62 square inches per foot

Formula: I subscript t = numerator (1) divided by denominator (3) times ( 12 inches per foot ) times ( k times d subscript e ) cubed + n times A subscript s times ( d subscript e minus k times d subscript e ) squared

Formula: I subscript t = 98 point 38 numerator ( inches superscript 4) divided by denominator ( feet )

Now, the actual stress in the reinforcement can be computed:

Formula: Mu subscript negtotalA = minus 8 point 53 K times numerator ( feet ) divided by denominator ( feet ) Formula: y = d subscript e minus k times d subscript e Formula: y = 3 point 90 inches

Formula: f subscript s = numerator (n times ( minus Mu subscript negtotalA times 12 inches per foot times y )) divided by denominator (I subscript t)

Formula: f subscript s = 32 point 44 ksi Formula: f subscript sa greater than f subscript s OK

Design Step 2.12 - Design for Flexure in Deck Overhang

Bridge deck overhangs must be designed to satisfy three different design cases. In the first design case, the overhang must be designed for horizontal (transverse and longitudinal) vehicular collision forces. For the second design case, the overhang must be designed to resist the vertical collision force. Finally, for the third design case, the overhang must be designed for dead and live loads. For Design Cases 1 and 2, the design forces are for the extreme event limit state. For Design Case 3, the design forces are for the strength limit state. Also, the deck overhang region must be designed to have a resistance larger than the actual resistance of the concrete parapet.

SA13.4.1

CA13.3.1

Figure 2-9 Deck Overhang Dimensions and Live Loading: This is a deck overhang with parapet cross section. Overhang deck thickness is 9 inches and bay one deck thickness is 8 and one half inches. Outside face of parapet is 3 feet 11 and one forth inches from centerline of girder. Parapet is i foot 5 and one forth inches thick. Wheel load is located 1 foot from inside edge of parapet and one foot 6 inches form centerline of girder. Parapet center of gravity, C period G period, is 6 point 16 inches from outside face of parapet and edge of deck. Overhang design section is 3 inches from centerline of girder on overhang side and Bay one design section is 3 inches on Bay one side.

Figure 2-9 Deck Overhang Dimensions and Live Loading

Figure 2-10 Reinforcing Steel for Flexure in Deck Overhang: This is a deck overhang with parapet cross section showing transverse and longitudinal bars and designating top transverse bars in overhang and over girder as reinforcing steel for flexure in deck overhang.

Figure 2-10 Reinforcing Steel for Flexure in Deck Overhang

Design Case 1 - Design Overhang for Horizontal Vehicular Collision Force

SA13.4.1

The horizontal vehicular collision force must be checked at the inside face of the parapet, at the design section in the overhang, and at the design section in the first bay.

Case 1A - Check at Inside Face of Parapet

The overhang must be designed for the vehicular collision plus dead load moment acting concurrently with the axial tension force from vehicular collision.

For the extreme event limit state:

Formula: phi subscript ext = 1 point 0

S1.3.2.1

Formula: gamma subscript pDC = 1 point 25

STable 3.4.1-2

Formula: M subscript co = 28 point 21 K times numerator ( feet ) divided by denominator ( feet ) (see parapet properties)

Formula: M subscript DCdeck = gamma subscript pDC times numerator (left bracket ( numerator (9 inches ) divided by denominator (12 inches per foot) ) times ( 0 point 150kcf) times ( 1 point 4375 feet ) squared right bracket) divided by denominator (2)

Formula: M subscript DCdeck = 0 point 15 Kips foot per foot

Formula: M subscript DCpar = gamma subscript pDC times W subscript par times ( 1 point 4375 feet minus numerator (6 point 16 inches ) divided by denominator (12 inches per foot) )

Formula: M subscript DCpar = 0 point 61 Kips foot per foot

Formula: Mu subscript total = M subscript co + M subscript DCdeck + M subscript DCpar

Formula: Mu subscript total = 28 point 97 Kips foot per foot

The axial tensile force is:

SA13.4.2

Formula: T = numerator (R subscript w) divided by denominator (L subscript c + 2H subscript par)

Before the axial tensile force can be calculated, the terms Lc and Rw need to be defined.

Lc is the critical wall length over which the yield line mechanism occurs:

SA13.3.1

Formula: L subscript c = numerator (L subscript t) divided by denominator (2) + square root of (( numerator (L subscript t) divided by denominator (2) ) squared + numerator (8 times H times ( M subscript b + M subscript w times H )) divided by denominator (M subscript c)) . Equation not used

Since the parapet is not designed in this design example, the variables involved in this calculation are given below:

Formula: L subscript t = 4feet longitudinal length of distribution of impact force Ft

SATable 13.2-1

Formula: M subscript b = 0 K feet * additional flexural resistance of beam in addition to Mw, if any, at top of wall *

Formula: M subscript c = 16 point 00 Kips foot per foot flexural resistance of the wall about an axis parallel to the longitudinal axis of the bridge

Formula: M subscript w = 18 point 52 K feet * flexural resistance of the wall about its vertical axis

Formula: H = 3 point 50feet height of parapet

* Based on parapet properties not included in this design example. See Publication Number FHWA HI-95-017, Load and Resistance Factor Design for Highway Bridges, Participant Notebook, Volume II (Version 3.01), for the method used to compute the parapet properties.

Lc is then:

Formula: L subscript c = numerator (L subscript t) divided by denominator (2) + square root of (( numerator (L subscript t) divided by denominator (2) ) squared + numerator (8 times H times ( M subscript b + M subscript w times H )) divided by denominator (M subscript c))

Formula: L subscript c = 12 point 84feet

Rw is the total transverse resistance of the railing and is calculated using the following equation for impacts within a wall segment:

SA13.3.1

Formula: R subscript w = ( numerator (2) divided by denominator (2 times L subscript c minus L subscript t) ) times ( 8 times M subscript b + 8M subscript w times H + numerator (M subscript c times L subscript c squared ) divided by denominator (H) )

Formula: R subscript w = 117 point 36K

use Formula: R subscript w = 117 point 40K

Now, the axial tensile force is:

SA13.4.2

Formula: T = 5 point 92 Kips per foot

The overhang slab thickness is: Formula: t subscript o = 9 point 0 inches

For #5 bars: Formula: bar_diam = 0 point 625 inches

Formula: d subscript e = t subscript o minus Cover subscript t minus numerator (bar_diam) divided by denominator (2) Formula: d subscript e = 6 point 19 inches

The required area of reinforcing steel is computed as follows:

Formula: b = 12 inches

Formula: Rn = numerator (Mu subscript total times 12 inches ) divided by denominator (( phi subscript ext times b times d subscript e squared )) Formula: Rn = 0 point 76 Kips per square inch

Formula: rho = 0 point 85 ( numerator (f prime subscript c) divided by denominator (f subscript y) ) times left bracket 1 point 0 minus square root of (1 point 0 minus numerator (( 2 times Rn)) divided by denominator (( 0 point 85 times f prime subscript c ))) right bracket

Formula: rho = 0 point 0145

Formula: A subscript s = rho times numerator (b) divided by denominator ( feet ) times d subscript e Formula: A subscript s = 1 point 07 square inches per foot

Use Formula: A subscript s = 1 point 24 square inches per foot (2 - #5 bars bundled at 6.0 in)

Once the required area of steel is known, the depth of the compression block must be checked:

Formula: T subscript a = A subscript s times f subscript y Formula: T subscript a = 74 point 40 Kips per foot

Formula: C = T subscript a minus T Formula: C = 68 point 48 Kips per foot Use Formula: C = 68 point 48K

Formula: a = numerator (C) divided by denominator (0 point 85 times f prime subscript c times b)

Formula: a = 1 point 68 inches

Formula: M subscript n = T subscript a times ( d subscript e minus numerator (a) divided by denominator (2) ) minus T times ( numerator (d subscript e) divided by denominator (2) minus numerator (a) divided by denominator (2) ) Formula: M subscript n = 32 point 05 Kips foot per foot

Formula: M subscript r = phi subscript ext times M subscript n Formula: M subscript r = 32 point 05 Kips foot per foot

Formula: M subscript r greater than or equal to Mu subscript total OK

Formula: c = numerator (a) divided by denominator ( beta subscript 1) Formula: c = 1 point 97 inches

S5.7.2.2

Formula: numerator (c) divided by denominator (d subscript e) = 0 point 32 where Formula: numerator (c) divided by denominator (d subscript e) less than or equal to 0 point 42

S5.7.3.3.1

Formula: 0 point 32 less than or equal to 0 point 42 OK

Case 1B - Check at Design Section in Overhang

The collision forces are distributed over a distance Lc for moment and Lc + 2H for axial force. When the design section is moved to 1/4bf away from the girder centerline in the overhang, the distribution length will increase. This example assumes a distribution length increase based on a 30 degree angle from the face of the parapet.

Figure 2-11 Assumed Distribution of Collision Moment Load in the Overhang: This is a plan and elevation view of a concrete New Jersey shaped barrier on a concrete deck supported by steel beams with a 3 foot 11 and 1 fourth inch overhang measured from centerline of fascia beam. The Distribution length L sub c equals 12 point 84 feet plus an additional 2 point 6 feet for 2H.

Figure 2-11 Assumed Distribution of Collision Moment Load in the Overhang

For the extreme event limit state:

Formula: phi subscript ext = 1 point 0

S1.3.2.1

Formula: gamma subscript pDC = 1 point 25

STable 3.4.1-2

Formula: gamma subscript pDW = 1 point 50

STable 3.4.1-2

Formula: L subscript c = 12 point 84 feet (see parapet properties)

Formula: M subscript co = 28 point 21 Kips foot per foot (see parapet properties)

Formula: M subscript cB = numerator (M subscript co times L subscript c) divided by denominator (L subscript c + 2 times 1 point 30 feet ) Formula: M subscript cB = 23 point 46 Kips foot per foot

Factored dead load moment:

Formula: M subscript DCdeck = gamma subscript pDC times numerator (left bracket ( numerator (9 point 0 inches ) divided by denominator (12 inches per foot) ) times ( W subscript c ) times ( 3 point 6875 feet ) squared right bracket) divided by denominator (2)

Formula: M subscript DCdeck = 0 point 96 Kips foot per foot

Formula: M subscript DCpar = gamma subscript pDC times W subscript par times ( 3 point 6875 feet minus numerator (6 point 16 inches ) divided by denominator (12 inches per foot) )

Formula: M subscript DCpar = 2 point 10 Kips foot per foot

Formula: M subscript DWfws = gamma subscript pDW times numerator (left bracket ( numerator (2 point 5 inches ) divided by denominator (12 inches per foot) ) times ( W subscript fws ) times ( 3 point 6875 feet minus 1 point 4375 feet ) squared right bracket) divided by denominator (2)

Formula: M subscript DWfws = 0 point 11 Kips foot per foot

Formula: Mu subscript total = M subscript cB + M subscript DCdeck + M subscript DCpar + M subscript DWfws

Formula: Mu subscript total = 26 point 63 K times numerator ( feet ) divided by denominator ( feet )

The axial tensile force is:

SA13.4.2

Formula: T = numerator (R subscript w) divided by denominator (L subscript c + 2H subscript par + 2 times ( 1 point 30 feet ))

Formula: T = 5 point 23 Kips per foot

The overhang slab thickness is: Formula: t subscript o = 9 point 0 inches

For #5 bars: Formula: bar_diam = 0 point 625 inches

Formula: d subscript e = t subscript o minus Cover subscript t minus numerator (bar_diam) divided by denominator (2) Formula: d subscript e = 6 point 19 inches

The required area of reinforcing steel is computed as follows:

Formula: b = 12 inches

Formula: Rn = numerator (Mu subscript total times 12 inches ) divided by denominator (( phi subscript ext times b times d subscript e squared )) Formula: Rn = 0 point 70 Kips per square inch

Formula: rho = 0 point 85 ( numerator (f prime subscript c) divided by denominator (f subscript y) ) times left bracket 1 point 0 minus square root of (1 point 0 minus numerator (( 2 times Rn)) divided by denominator (( 0 point 85 times f prime subscript c ))) right bracket

Formula: rho = 0 point 0131

Formula: A subscript s = rho times numerator (b) divided by denominator ( feet ) times d subscript e Formula: A subscript s = 0 point 97 square inches per foot

The above required reinforcing steel is less than the reinforcing steel required for Case 1A.

Case 1C - Check at Design Section in First Span

The total collision moment can be treated as shown in Figure 2-12. The moment ratio, M2/M1, can be calculated for the design strip. One way to approximate this moment is to set it equal to the ratio of the moments produced by the parapet self-weight at the 0.0S points of the first and second bay. The collision moment per unit width can then be determined by using the increased distribution length based on the 30 degree angle distribution (see Figure 2-11). The dead load moments at this section can be obtained directly from Table 2-1.

Figure 2-12 Assumed Distribution of the Collision Moment Across the Width of the Deck. Moment distribution diagram showing constant negative Moment M1 extending from inside edge of New Jersey Barrier to facia beam then diagram moment decreases uniformly to a maximum positive Moment M2 at the first interior support.

Figure 2-12 Assumed Distribution of the Collision Moment Across the Width of the Deck

Collision moment at exterior girder:

Formula: M subscript co = minus 28 point 21 Kips foot per foot Formula: M subscript 1 = M subscript co

Parapet self-weight moment at Girder 1 (0.0S in Bay 1):

Formula: Par subscript 1 = minus 1 point 66 Kips foot per foot

Parapet self-weight moment at Girder 2 (0.0S in Bay 2):

Formula: Par subscript 2 = 0 point 47 Kips foot per foot

Collision moment at 1/4bf in Bay 1:

Formula: M subscript 2 = M subscript 1 times ( numerator (Par subscript 2) divided by denominator (Par subscript 1) ) Formula: M subscript 2 = 7 point 99 Kips foot per foot

By interpolation for a design section at 1/4bf in Bay 1, the total collision moment is:

Formula: M subscript cM2M1 = M subscript co + 0 point 25 feet times numerator (( minus M subscript co + M subscript 2 )) divided by denominator (9 point 75 feet )

Formula: M subscript cM2M1 = minus 27 point 28 Kips foot per foot

As in Case 1B, the 30 degree angle distribution will be used:

Formula: phi subscript ext = 1 point 0

S1.3.2.1

Formula: gamma subscript pDC = 1 point 25

STable 3.4.1-2

Formula: gamma subscript pDW = 1 point 50

STable 3.4.1-2

Formula: M subscript cM2M1 = minus 27 point 28 Kips foot per foot

Formula: M subscript cC = numerator (M subscript cM2M1 times L subscript c) divided by denominator (L subscript c + 2 times ( 1 point 59 feet )) Formula: M subscript cC = minus 21 point 87 Kips foot per foot

Factored dead load moment (from Table 2-1):

Formula: M subscript DCdeck = gamma subscript pDC times ( minus 0 point 74 Kips foot per foot )

Formula: M subscript DCdeck = minus 0 point 93 K times numerator ( feet ) divided by denominator ( feet )

Formula: M subscript DCpar = gamma subscript pDC times ( minus 1 point 66 Kips foot per foot )

Formula: M subscript DCpar = minus 2 point 08 Kips foot per foot

Formula: M subscript DWfws = gamma subscript pDW times ( minus 0 point 06 Kips foot per foot )

Formula: M subscript DWfws = minus 0 point 09 Kips foot per foot

Formula: Mu subscript total = M subscript cC + M subscript DCdeck + M subscript DCpar + M subscript DWfws

Formula: Mu subscript total = minus 24 point 96 Kips foot per foot

The axial tensile force is:

SA13.4.2

Formula: T = numerator (R subscript w) divided by denominator (L subscript c + 2H subscript par + 2 times ( 1 point 59 feet ))

Formula: T = 5 point 10 Kips per foot

Use a slab thickness equal to: Formula: t subscript s = 8 point 50 inches

For #5 bars: Formula: bar_diam = 0 point 625 inches

Formula: d subscript e = t subscript s minus Cover subscript t minus numerator (bar_diam) divided by denominator (2) Formula: d subscript e = 5 point 69 inches

The required area of reinforcing steel is computed as follows:

Formula: b = 12 inches

Formula: Rn = numerator ( minus Mu subscript total times 12 inches ) divided by denominator (( phi subscript ext times b times d subscript e squared )) Formula: Rn = 0 point 77 Kips per square inch

Formula: rho = 0 point 85 ( numerator (f prime subscript c) divided by denominator (f subscript y) ) times left bracket 1 point 0 minus square root of (1 point 0 minus numerator (( 2 times Rn)) divided by denominator (( 0 point 85 times f prime subscript c ))) right bracket

Formula: rho = 0 point 0148

Formula: A subscript s = rho times numerator (b) divided by denominator ( feet ) times d subscript e Formula: A subscript s = 1 point 01 square inches per foot

The above required reinforcing steel is less than the reinforcing steel required for Case 1A.

Design Case 2 - Design Overhang for Vertical Collision Force

SA13.4.1

For concrete parapets, the case of vertical collision force never controls. Therefore, this procedure does not need to be considered in this design example.

Design Case 3 - Design Overhang for Dead Load and Live Load

SA13.4.1

Case 3A - Check at Design Section in Overhang

The resistance factor for the strength limit state for flexure and tension in concrete is:

S5.5.4.2.1

Formula: phi subscript str = 0 point 90

The equivalent strip for live load on an overhang is:

STable 4.6.2.1.3-1

Formula: w subscript overstrip = 45 point 0 + 10 point 0 times X. Equation not used

For Formula: X = 1 point 25feet

Formula: w subscript overstrip = 45 point 0 + 10 point 0X

Formula: w subscript overstrip = 57 point 50inches or Formula: w subscript overstrip = 4 point 79 feet

Use a multiple presence factor of 1.20 for one lane loaded.

STable 3.6.1.1.2-1

Use a dynamic load allowance of 0.33.

STable 3.6.2.1-1

Design factored overhang moment:

Formula: gamma subscript LL = 1 point 75

STable 3.4.1-1

Formula: gamma subscript pDC = 1 point 25

STable 3.4.1-2

Formula: gamma subscript pDW = 1 point 50

STable 3.4.1-2

Formula: M subscript DCdeck = gamma subscript pDC times numerator (left bracket ( numerator (9 point 0 inches ) divided by denominator (12 inches per foot) ) times ( W subscript c ) times ( 3 point 6875 feet ) squared right bracket) divided by denominator (2)

Formula: M subscript DCdeck = 0 point 96 Kips foot per foot

Formula: M subscript DCpar = gamma subscript pDC times W subscript par times ( 3 point 6875 feet minus numerator (6 point 16 inches ) divided by denominator (12 inches per foot) )

Formula: M subscript DCpar = 2 point 10 Kips foot per foot

Formula: M subscript DWfws = gamma subscript pDW times W subscript fws times numerator (( numerator (2 point 5 inches ) divided by denominator (12 inches per foot) ) times ( 3 point 6875 feet minus 1 point 4375 feet ) squared ) divided by denominator (2)

Formula: M subscript DWfws = 0 point 11 Kips foot per foot

Formula: M subscript LL = gamma subscript LL times ( 1 + IM) times ( 1 point 20) times ( numerator (16K) divided by denominator (w subscript overstrip) ) times 1 point 25 feet

Formula: M subscript LL = 11 point 66 Kips foot per foot

Formula: Mu subscript total = M subscript DCdeck + M subscript DCpar + M subscript DWfws + M subscript LL

Formula: Mu subscript total = 14 point 83 Kips foot per foot

Calculate the required area of steel:

For #5 bars: Formula: bar_diam = 0 point 625 inches

Formula: d subscript e = t subscript o minus Cover subscript t minus numerator (bar_diam) divided by denominator (2)

Formula: d subscript e = 6 point 19 inches

Formula: b = 12 inches

Formula: Rn = numerator (Mu subscript total times 12 inches ) divided by denominator (( phi subscript str times b times d subscript e squared )) Formula: Rn = 0 point 43 Kips per square inch

Formula: rho = 0 point 85 ( numerator (f prime subscript c) divided by denominator (f subscript y) ) times left bracket 1 point 0 minus square root of (1 point 0 minus numerator (( 2 times Rn)) divided by denominator (( 0 point 85 times f prime subscript c ))) right bracket

Formula: rho = 0 point 00770

Formula: A subscript s = rho times numerator (b) divided by denominator ( feet ) times d subscript e Formula: A subscript s = 0 point 57 square inches per foot

The above required reinforcing steel is less than the reinforcing steel required for Cases 1A, 1B, and 1C.

Case 3B - Check at Design Section in First Span

Use a slab thickness equal to: Formula: t subscript s = 8 point 50 inches

The dead and live load moments are taken from Tables 2-1 and 2-2. The maximum negative live load moment occurs in Bay 4. Since the negative live load moment is produced by a load on the overhang, compute the equivalent strip based on a moment arm to the centerline of girder.

Design factored moment:

Formula: gamma subscript LL = 1 point 75

STable 3.4.1-1

Formula: gamma subscript pDC = 1 point 25

STable 3.4.1-2

Formula: gamma subscript pDW = 1 point 50

STable 3.4.1-2

Formula: w subscript overstrip = 45 point 0 + 10 point 0 times X. Equation not used

For Formula: X = 1 point 50feet

Formula: w subscript overstrip = 45 point 0 + 10 point 0X

Formula: w subscript overstrip = 60 point 00inches or Formula: w subscript overstrip = 5 point 00 feet

Formula: M subscript DCdeck = gamma subscript pDC times ( minus 0 point 74 Kips foot per foot )

Formula: M subscript DCdeck = minus 0 point 93 Kips foot per foot

Formula: M subscript DCpar = gamma subscript pDC times ( minus 1 point 66 Kips foot per foot )

Formula: M subscript DCpar = minus 2 point 08 Kips foot per foot

Formula: M subscript DWfws = gamma subscript pDW times ( minus 0 point 06 Kips foot per foot )

Formula: M subscript DWfws = minus 0 point 09 Kips foot per foot

Formula: M subscript LL = gamma subscript LL times ( 1 + IM) times numerator (( minus 29 point 40K feet )) divided by denominator (w subscript overstrip)

Formula: M subscript LL = minus 13 point 69 Kips foot per foot

Formula: Mu subscript total = M subscript DCdeck + M subscript DCpar + M subscript DWfws + M subscript LL

Formula: Mu subscript total = minus 16 point 78 Kips foot per foot

Calculate the required area of steel:

For #5 bars: Formula: bar_diam = 0 point 625 inches

Formula: d subscript e = t subscript s minus Cover subscript t minus numerator (bar_diam) divided by denominator (2)

Formula: d subscript e = 5 point 69 inches

Formula: b = 12 inches

Formula: Rn = numerator ( minus Mu subscript total times 12 inches ) divided by denominator (( phi subscript str times b times d subscript e squared )) Formula: Rn = 0 point 58 Kips per square inch

Formula: rho = 0 point 85 ( numerator (f prime subscript c) divided by denominator (f subscript y) ) times left bracket 1 point 0 minus square root of (1 point 0 minus numerator (( 2 times Rn)) divided by denominator (( 0 point 85 times f prime subscript c ))) right bracket

Formula: rho = 0 point 0106

Formula: A subscript s = rho times numerator (b) divided by denominator ( feet ) times d subscript e Formula: A subscript s = 0 point 72 square inches per foot

The above required reinforcing steel is less than the reinforcing steel required for Cases 1A, 1B, and 1C.

The required area of reinforcing steel in the overhang is the largest of that required for Cases 1A, 1B, 1C, 3A, and 3B.

Case 1A controls with: Formula: A subscript s = 1 point 24 square inches per foot

The negative flexure reinforcement provided from the design in Steps 2.10 and 2.11 is:

#5 bars at 6.0 inches: Formula: bar_diam = 0 point 625 inches

Formula: bar_area = 0 point 31 inches squared

Formula: A subscript sneg = numerator (bar_area) divided by denominator ( feet ) times ( numerator (12 inches ) divided by denominator (6 inches ) )

Formula: A subscript sneg = 0 point 62 square inches per foot

Formula: 0 point 62 square inches per foot less than 1 point 24 square inches per foot

Since the area of reinforcing steel required in the overhang is greater than the area of reinforcing steel required in the negative moment regions, reinforcement must be added in the overhang area to satisfy the design requirements.

Bundle one #5 bar to each negative flexure reinforcing bar in the overhang area.

The new area of reinforcing steel is now:

Formula: A subscript s = 2 times ( 0 point 31 times square inches per foot ) times ( numerator (12 inches ) divided by denominator (6 inches ) )

Formula: A subscript s = 1 point 24 square inches per foot

Once the required area of reinforcing steel is known, the depth of the compression block must be checked. The ratio of c/de is more critical at the minimum deck thickness, so c/de will be checked in Bay 1 where the deck thickness is 8.5 inches.

Formula: d subscript emin = t subscript s minus Cover subscript t minus numerator (bar_diam) divided by denominator (2)

Formula: d subscript emin = 5 point 69 inches

Formula: T = A subscript s times f subscript y Formula: T = 74 point 40 Kips per foot Use Formula: T = 74 point 40K

Formula: a = numerator (T) divided by denominator (0 point 85 times f prime subscript c times b)

Formula: a = 1 point 82 inches

Formula: c = numerator (a) divided by denominator ( beta subscript 1) Formula: c = 2 point 15 inches

S5.7.2.2

Formula: numerator (c) divided by denominator (d subscript emin) = 0 point 38 where Formula: numerator (c) divided by denominator (d subscript e) less than or equal to 0 point 42

S5.7.3.3.1

Formula: 0 point 38 less than or equal to 0 point 42 OK

Design Step 2.13 - Check for Cracking in Overhang under Service Limit State

Cracking in the overhang must be checked for the controlling service load (similar to Design Steps 2.9 and 2.11). In most deck overhang design cases, cracking does not control. Therefore, the computations for the cracking check are not shown in this deck overhang design example.

Design Step 2.14 - Compute Overhang Cut-off Length Requirement

The next step is to compute the cut-off location of the additional #5 bars in the first bay. This is done by determining the location where both the dead and live load moments, as well as the dead and collision load moments, are less than or equal to the resistance provided by #5 bars at 6 inch spacing (negative flexure steel design reinforcement).

Compute the nominal negative moment resistance based on #5 bars at 6 inch spacing:

Formula: bar_diam = 0 point 625 inches

Formula: bar_area = 0 point 31 inches squared

Formula: A subscript s = numerator (bar_area) divided by denominator ( feet ) times ( numerator (12 inches ) divided by denominator (6 inches ) )

Formula: A subscript s = 0 point 62 square inches per foot

Formula: d subscript e = t subscript s minus Cover subscript t minus numerator (bar_diam) divided by denominator (2)

Formula: d subscript e = 5 point 69 inches

Formula: T = A subscript s times f subscript y Formula: T = 37 point 20 Kips per foot Use Formula: T = 37 point 20K

Formula: a = numerator (T) divided by denominator (0 point 85 times f prime subscript c times b) Formula: a = 0 point 91 inches

Formula: M subscript n = A subscript s times f subscript y times ( d subscript e minus numerator (a) divided by denominator (2) )

Formula: M subscript n = 16 point 22 Kips foot per foot

Compute the nominal flexural resistance for negative flexure, as follows:

Formula: M subscript r = phi subscript f times M subscript n

Formula: M subscript r = 14 point 60 Kips foot per foot

Based on the nominal flexural resistance and on interpolation of the factored design moments, the theoretical cut-off point for the additional #5 bar is 3.75 feet from the centerline of the fascia girder.

The additional cut-off length (or the distance the reinforcement must extend beyond the theoretical cut-off point) is the maximum of:

S5.11.1.2

The effective depth of the member: Formula: d subscript e = 5 point 69 inches

15 times the nominal bar diameter: Formula: 15 times 0 point 625 inches = 9 point 38 inches

1/20 of the clear span: Formula: numerator (1) divided by denominator (20) times ( 9 point 75 feet times 12 inches per foot ) = 5 point 85 inches

Use Formula: cut_off = 9 point 5 inches

The total required length past the centerline of the fascia girder into the first bay is:

Formula: cut_off subscript total = 3 point 75 feet times 12 inches per foot + cut_off

Formula: cut_off subscript total = 54 point 50 inches

Design Step 2.15 - Compute Overhang Development Length

Formula: d subscript b = 0 point 625inches

Formula: A subscript b = 0 point 31inches squared

Formula: f prime subscript c = 4 point 0ksi

Formula: f subscript y = 60ksi

The basic development length is the larger of the following:

S5.11.2.1.1

Formula: numerator (1 point 25 times A subscript b times f subscript y) divided by denominator ( square root of (f prime subscript c) ) = 11 point 63inches or Formula: 0 point 4 times d subscript b times f subscript y = 15 point 00inches or 12 inches

Use Formula: I subscript d = 15 point 00 inches

The following modification factors must be applied:

S5.11.2

Epoxy coated bars: 1 point 2

S5.11.2.1.2

Bundled bars: 1 point 2

S5.11.2.3

Spacing > 6 inches with more than 3 inches of clear cover in direction of spacing:

0 point 8

S5.11.2.1.3

Formula: I subscript d = 15 point 00 inches times ( 1 point 2) times ( 1 point 2) times ( 0 point 8)

Formula: I subscript d = 17 point 28 inches Use Formula: I subscript d = 18 point 00 inches

The required length past the centerline of the fascia girder is:

Formula: 3 point 0 inches + I subscript d = 21 point 00 inches

Formula: 21 point 00 inches less than 54 point 50 inches provided

Figure 2-13 Length of Overhang Negative Moment Reinforcement: This is a deck overhang with parapet cross section. Number 5 at 6 inch parenthesis bundled bars parenthesis extend from facia to 54 and one half inches into bay one. Bay one design section located 3 inches from girder centerline into bay one then 18 inch development length equals 21 inches.

Figure 2-13 Length of Overhang Negative Moment Reinforcement

Design Step 2.16 - Design Bottom Longitudinal Distribution Reinforcement

The bottom longitudinal distribution reinforcement is calculated based on whether the primary reinforcement is parallel or perpendicular to traffic.

S9.7.3.2

Figure 2-14 Bottom Longitudinal Distribution Reinforcement: This is a deck overhang with parapet cross section showing transverse and longitudinal bars and designating bottom longitudinal bars as bottom longitudinal distribution reinforcement.

Figure 2-14 Bottom Longitudinal Distribution Reinforcement

For this design example, the primary reinforcement is perpendicular to traffic.

Formula: S subscript e = 9 point 25feet

Formula: A subscript sbotpercent = numerator (220) divided by denominator ( square root of (S subscript e) ) where Formula: A subscript sbotlong less than or equal to 67 %

Formula: A subscript sbotpercent = 72 point 3%

Use Formula: A subscript sbotpercent = 67 %

For this design example, #5 bars at 8 inches were used to resist the primary positive moment.

Formula: bar_diam = 0 point 625 inches

Formula: bar_area = 0 point 31 inches squared

Formula: A subscript s_ft = bar_area times ( numerator (12 inches ) divided by denominator (8 inches ) )

Formula: A subscript s_ft = 0 point 465 square inches per foot

Formula: A subscript sbotlong = A subscript sbotpercent times A subscript s_ft

Formula: A subscript sbotlong = 0 point 31 square inches per foot

Calculate the required spacing using #5 bars:

Formula: spacing = numerator (bar_area) divided by denominator (A subscript sbotlong)

Formula: spacing = 1 point 00 feet or Formula: spacing = 11 point 94 inches

Use Formula: spacing = 10 inches

Use #5 bars at 10 inch spacing for the bottom longitudinal reinforcement.

Design Step 2.17 - Design Top Longitudinal Distribution Reinforcement

Figure 2-15 Top Longitudinal Distribution Reinforcement: This is a deck overhang with parapet cross section showing transverse and longitudinal bars and designating top longitudinal bars as top longitudinal distribution reinforcement

Figure 2-15 Top Longitudinal Distribution Reinforcement

The top longitudinal temperature and shrinkage reinforcement must satisfy:

S5.10.8.2

Formula: A subscript s greater than or equal to 0 point 11 numerator (A subscript g) divided by denominator (f subscript y) . Equation not used

Formula: A subscript g = 8 point 5 inches times ( 12 point 0 inches per foot ) Formula: A subscript g = 102 point 00 square inches per foot

Formula: 0 point 11 numerator (A subscript g) divided by denominator (f subscript y) = 0 point 19 square inches per foot

When using the above equation, the calculated area of reinforcing steel must be equally distributed on both concrete faces. In addition, the maximum spacing of the temperature and shrinkage reinforcement must be the smaller of 3.0 times the deck thickness or 18.0 inches.

The amount of steel required for the top longitudinal reinforcement is:

Formula: A subscript sreq = numerator (0 point 19 times square inches per foot) divided by denominator (2) Formula: A subscript sreq = 0 point 10 square inches per foot

Check #4 bars at 10 inch spacing:

Formula: A subscript sact = 0 point 20 times square inches per foot times ( numerator (12 inches ) divided by denominator (10 inches ) ) Formula: A subscript sact = 0 point 24 square inches per foot

Formula: 0 point 24 square inches per foot greater than 0 point 10 square inches per foot OK

Use #4 bars at 10 inch spacing for the top longitudinal temperature and shrinkage reinforcement.

Design Step 2.18 - Design Longitudinal Reinforcement over Piers

If the superstructure is comprised of simple span precast girders made continuous for live load, the top longitudinal reinforcement should be designed according to S5.14.1.2.7. For continuous steel girder superstructures, design the top longitudinal reinforcement according to S6.10.3.7. For this design example, continuous steel girders are used.

Figure 2-16 Longitudinal Reinforcement:over Piers: This is a deck overhang with parapet cross section showing transverse and longitudinal bars and designating top and bottom longitudinal bars as longitudinal reinforcement over piers.

Figure 2-16 Longitudinal Reinforcement over Piers

The total longitudinal reinforcement should not be less than 1 percent of the total slab cross-sectional area. These bars must have a specified minimum yield strength of at least 60 ksi. Also, the bar size cannot be larger than a #6 bar.

S6.10.3.7

Deck cross section:

Formula: A subscript deck = numerator (8 point 5 inches times 12 inches ) divided by denominator ( feet )

Formula: A subscript deck = 102 point 00 square inches per foot

Formula: A subscript s_1_percent = 0 point 01 times A subscript deck

Formula: A subscript s_1_percent = 1 point 02 square inches per foot

Two-thirds of the required longitudinal reinforcement should be placed uniformly in the top layer of the deck, and the remaining portion should be placed uniformly in the bottom layer. For both rows, the spacing should not exceed 6 inches.

S6.10.3.7

Formula: ( numerator (2) divided by denominator (3) ) times A subscript s_1_percent = 0 point 68 square inches per foot Formula: ( numerator (1) divided by denominator (3) ) times A subscript s_1_percent = 0 point 34 square inches per foot

Use #5 bars at 5 inch spacing in the top layer.

Formula: A subscript sprovided = 0 point 31 square inches per foot times ( numerator (12 inches ) divided by denominator (5 inches ) )

Formula: A subscript sprovided = 0 point 74 square inches per foot > 0 point 68 square inches per foot OK

Use #5 bars at 5 inch spacing in the bottom layer to satisfy the maximum spacing requirement of 6 inches.

Formula: A subscript sprovided = 0 point 31 square inches per foot times ( numerator (12 inches ) divided by denominator (5 inches ) )

Formula: A subscript sprovided = 0 point 74 square inches per foot > 0 point 34 square inches per foot OK

Design Step 2.19 - Draw Schematic of Final Concrete Deck Design

Figure 2-17 Superstructure Positive Moment Deck Reinforcement: This is a deck overhang with parapet cross section showing and designating transverse and longitudinal bars. Overhang deck thicknes is 9 inches. Interior bay deck thickness is 8 and one half inches.Top transverse overhang bars are Number 5 at 6 inches parenthesis bundled bars parenthesis with Number 5 at 6 inch top transverse negative flexure bars. Clearance for all transverse bars is 2 and on half inches. Bottom transverse bars are number 5 at 8 inches. Clearance for bottom transverse bars is one inch in interior bays. Top longitudinsal bars are Number 4 at 10 inches. Bottom longitudinal bars are Number 5 at 10inches. Number 5 at 6 inch top overhang transverse bars extend 54 and one half inches into bay one.

Figure 2-17 Superstructure Positive Moment Deck Reinforcement

Figure 2-118 Superstructure Negative Moment Deck Reinforcement: This is a deck overhang with parapet cross section showing and designating transverse and longitudinal bars. Overhang deck thicknes is 9 inches. Interior bay deck thickness is 8 and one half inches. Top transverse overhang bars are Number 5 at 6 inches parenthesis bundled bars parenthesis with Number 5 at 6 inch top transverse negative flexure bars. Clearance for all transverse bars is 2 and one half inches. Bottom transverse bars are number 5 at 8 inches. Clearance for bottom transverse bars is one inch in interior bays. Top longitudinsal bars are Number 5 at 5 inches. Bottom longitudinal bars are Number 5 at 5 inches. Number 5 at 6 inch top overhang transverse bars extend 54 and one half inches into bay one.

Figure 2-18 Superstructure Negative Moment Deck Reinforcement

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Updated: 07/18/2013
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