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Bridges & Structures

LRFD Steel Girder SuperStructure Design Example

Abutment and Wingwall Design Example Design Step 7

Table of Contents

Design Step 7.1 - Obtain Design Criteria
Design Step 7.2 - Select Optimum Abutment Type
Design Step 7.3 - Select Preliminary Abutment Dimensions
Design Step 7.4 - Compute Dead Load Effects
Design Step 7.5 - Compute Live Load Effects
Design Step 7.6 - Compute Other Load Effects
Design Step 7.7 - Analyze and Combine Force Effects
Design Step 7.8 - Check Stability and Safety Requirements
Design Step 7.9 - Design Abutment Backwall
Design Step 7.10 - Design Abutment Stem
Design Step 7.11 - Design Abutment Footing
Design Step 7.12 - Draw Schematic of Final Abutment Design
Design Step 7.2 - Select Optimum Wingwall Type
Design Step 7.3 - Select Preliminary Wingwall Dimensions
Design Step 7.4 - Compute Dead Load Effects
Design Step 7.5 - Compute Live Load Effects
Design Step 7.6 - Compute Other Load Effects
Design Step 7.7 - Analyze and Combine Force Effects
Design Step 7.9 - Design Wingwall Stem
Design Step 7.12 - Draw Schematic of Final Wingwall Design

Design Step 7.1 - Obtain Design Criteria

This abutment and wingwall design example is based on AASHTO LRFD Bridge Design Specifications (through 2002 interims). The design methods presented throughout the example are meant to be the most widely used in general bridge engineering practice. The example covers the abutment backwall, stem, and footing design, using pile loads from Design Step P, Pile Foundation Design Example. The wingwall design focuses on the wingwall stem only. All applicable loads that apply to the abutment and wingwall are either taken from design software or calculated herein.

The wingwall design utilizes the same flowchart as the abutment. Design Step 7.1 is shared by both the abutment and wingwall. After Design Step 7.1, Design Steps 7.2 through 7.12 are for the abutment. For the wingwall, any Design Steps from 7.2 through 7.12 that apply to the wingwall follow at the end of the abutment design steps. For example, there are two Design Steps 7.2 - one for the abutment and one for the wingwall (after Design Step 7.12 of the abutment).

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 abutments and wingwalls.

In order to begin the design, the abutment and wingwall properties as well as information about the superstructure that the abutment supports is required.

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

          Formula: K = 1000lb       Formula: kcf = Kips per cubic foot       Formula: ksi = Kips per square inch       Formula: ksf = Kips per square foot       Formula: kIf = Kips per foot

It should be noted that the superstructure loads and plate girder dimensions used in this design step are based on the first trial of the girder design.

Material properties:
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 & SC5.4.2.1 & STable C5.4.2.1-1
Reinforcement strength: Formula: f subscript y = 60ksi S5.4.3

Reinforcing steel cover requirements:

Backwall back cover: Formula: Cover subscript b = 2 point 5 inches STable 5.12.3-1
Stem back cover: Formula: Cover subscript s = 2 point 5 inches STable 5.12.3-1
Footing top cover: Formula: Cover subscript feet = 2 point 0 inches STable 5.12.3-1
Footing bottom cover: Formula: Cover subscript fb = 3 point 0 inches STable 5.12.3-1

Backwall back cover - Assuming that the backwall will be subject to deicing salts, the cover is set at 2.5 inches.

STable 5.12.3-1

Stem cover - The stem cover is set at 2.5 inches. This will allow the vertical flexure reinforcement in the stem to be lapped with the vertical back face reinforcement in the backwall. Also, it is assumed that the stem may be exposed to deicing salts due to the abutment having an expansion joint.

STable 5.12.3-1

Footing top cover - The footing top cover is set at 2.0 inches.

STable 5.12.3-1

Footing bottom cover - Since the footing bottom is cast directly against the earth, the footing bottom cover is set at 3.0 inches.

STable 5.12.3-1

Relevant superstructure data:

Girder spacing:  Formula: S = 9 point 75 feet
Number of girders: Formula: N = 5
Span length: Formula: L subscript span = 120 feet
Parapet height: Formula: H subscript par = 3 point 5 feet
Parapet weight (each): Formula: W subscript par = 0 point 53 Kips per foot
Out-to-out deck width: Formula: W subscript deck = 46 point 875 feet

Superstructure data - The above superstructure information was taken from Design Steps 1 and 2.

Abutment and wingwall height S2.3.3.2

Abutment stem height: Formula: h subscript stem = 22 feet
Wingwall stem design height: Formula: h subscript wwstem = 20 point 75 feet       use height at 3/4 point

Abutment and wingwall length S11.6.1.4

Abutment length: Formula: L subscript abut = 46 point 875 feet
Wingwall length: Formula: L subscript wing = 20 point 5 feet

Design Step 7.2 - Select Optimum Abutment Type

Selecting the optimal abutment type depends on the site conditions, cost considerations, superstructure geometry, and aesthetics. The most common abutment types include cantilever, gravity, counterfort, mechanically-stabilized earth, stub, semi-stub or shelf, open or spill through, and integral or semi-integral. For this design example, a full-depth reinforced concrete cantilever abutment was chosen because it is the most economical for the site conditions. For a concrete cantilever abutment, the overturning forces are balanced by the vertical earth load on the abutment heel. Concrete cantilever abutments are the typical abutment type used for most bridge designs and is considered optimal for this abutment design example.

S11.2

Elevation view of a reinforced concrete cantilever abutment. The toe is shown to the right and the heel is shown to the left.

Figure 7-1 Full-Depth Reinforced Concrete Cantilever Abutment

Design Step 7.3 - Select Preliminary Abutment Dimensions

Since AASHTO does not have standards for the abutment backwall, stem, or footing maximum or minimum dimensions, the designer should base the preliminary abutment dimensions on state specific standards, previous designs, and past experience. The abutment stem, however, must be wide enough to allow for the minimum displacement requirements. The minimum support length is calculated in Design Step 7.6.

S4.7.4.4

The following figure shows the preliminary dimensions for this abutment design example.

The figure shows a cross section of the entire abutment giving preliminary dimensions. The abutment toe is to the right and the heel is to the left. The top width not including the paving notch is 1 foot 3 inches. The paving notch is 9 inches wide and 1 foot 4 inches deep. The depth to the corbel along the front face of backwall is 2 feet 0 inches. The corbel depth is 4 inches deep by 4 inches wide. The entire backwall depth is 7 feet and 0 inches. The beam seat is sloped 5 percent toward the front of the abutment. The abutment stem width is 3 feet 6 inches. The stem height is 15 feet 0 inches. The entire footing width is 10 feet 3 inches. The heel width from the back edge of footing to the back of stem is 4 feet 0 inches. The toe width from the front edge of footing to the front face of stem is 2 feet 9 inches. The footing thickness is 2 feet 6 inches.

Figure 7-2 Preliminary Abutment Dimensions

For sealed expansion joints, slope the top surface of the abutment (excluding bearing seats) a minimum of 5% towards the edge.

S2.5.2.1.2

Design Step 7.4 - Compute Dead Load Effects

Once the preliminary abutment dimensions are selected, the corresponding dead loads can be computed. Along with the abutment dead loads, the superstructure dead loads must be computed. The superstructure dead loads acting on the abutment will be given based on the superstructure output from the software used to design the superstructure. The dead loads are calculated on a per foot basis. Also, the dead loads are calculated assuming the beam seat is level.

S3.5.1

The superstructure dead load reactions per bearing were obtained from trial one of the steel grider design and are as follows.

Fascia girder:

                              Formula: R subscript DCfascia = 69 point 25K       Formula: R subscript Dwfascia = 11 point 24K

Interior girder:

                                Formula: R subscript DCinterior = 73 point 51K       Formula: R subscript DWinterior = 11 point 24K

As previously stated, the superstructure dead load reactions must be converted into a load applied to a one-foot strip of abutment. This is accomplished by adding the two fascia girder dead load reactions with the three interior girder dead load reactions and then dividing by the abutment length.

                        Formula: R subscript DCtot = numerator (( 2 times R subscript DCfascia ) + ( 3 times R subscript DCinterior )) divided by denominator (L subscript abut)

                        Formula: R subscript DCtot = 7 point 66 Kips per foot

                          Formula: R subscript DWtot = numerator (( 2 times R subscript Dwfascia ) + ( 3R subscript DWinterior )) divided by denominator (L subscript abut)

                          Formula: R subscript DWtot = 1 point 20 Kips per foot

Backwall dead load:

                      Formula: DL subscript bw = left bracket ( 1 point 25 feet times 1 point 33 feet ) + ( 2 point 0 feet times 0 point 67 feet ) + left bracket ( numerator ( 2 feet + 1 point 67 feet) divided by denominator ( 2 )) times 0 point 33 feet right bracket + ( 4 point 67 feet times 1 point 67 feet ) right bracket times W subscript c

                      Formula: DL subscript bw = 1 point 71 Kips per foot

Stem dead load:

                          Formula: DL subscript stem = 15 feet times 3 point 5 feet times W subscript c

                          Formula: DL subscript stem = 7 point 88 Kips per foot

Footing dead load:

                      Formula: DL subscript ftg = 10 point 25 feet times 2 point 5 feet times W subscript c

                      Formula: DL subscript ftg = 3 point 84 Kips per foot

Earth dead load:

                Formula: gamma subscript s = 0 point 120kcf       use average of loose and compacted gravel

STable 3.5.1-1

                          Formula: DL subscript earth = 22 feet times 4 feet times gamma subscript s

                          Formula: DL subscript earth = 10 point 56 Kips per foot

Design Step 7.5 - Compute Live Load Effects

The live load effects were also obtained from trial one of the girder design. The reactions for one girder are given as unfactored, without impact, and without distribution factors. The given reactions are converted into one loaded lane and then converted into a per foot load.

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

STable 3.6.2.1-1

Multiple presence factor, m (1 lane)       Formula: m subscript 1 = 1 point 20

STable 3.6.1.1.2-1

Multiple presence factor, m (2 lanes)       Formula: m subscript 2 = 1 point 00

STable 3.6.1.1.2-1

Multiple presence factor, m (3 lanes)       Formula: m subscript 3 = 0 point 85

STable 3.6.1.1.2-1

For this design example, the backwall live load is computed by placing three design truck axles along the abutment and calculating the load on a per foot basis including impact and the multiple presence factor. This load is applied to the entire length of abutment backwall and is assumed to act at the front top corner (bridge side) of the backwall. This load is not applied, however, when designing the abutment stem or footing.

                        Formula: R subscript LLbw = numerator (left bracket 6 times ( 16 K) times ( 1 + IM) + 3 times ( 0 point 64 times kIf) times ( 2 point 0 feet ) right bracket) divided by denominator (L subscript abut)

                        Formula: R subscript LLbw = 2 point 81 Kips per foot

The following loads are obtained from girder design software output for one lane loaded and they are applied at the beam seat or top of abutment stem for the stem design.

                            Formula: V subscript vehmax = 64 point 90K       Based on first trial of girder design

                            Formula: V subscript lanemax = 33 point 25K       Based on first trial of girder design

                          Formula: V subscript vehmin = minus 7 point 28K       Based on first trial of girder design

                            Formula: V subscript lanemin = minus 5 point 15K       Based on first trial of girder design

The controlling maximum and minimum live loads are for three lanes loaded. The loads are multiplied by dynamic load allowance and the multiple presence factor.

Maximum unfactored live load used for abutment stem design:

                        Formula: r subscript LLmax = V subscript vehmax times ( 1 + IM) + V subscript lanemax

                        Formula: r subscript LLmax = 119 point 57 K       for one lane

                          Formula: R subscript LLmax = numerator (3 times m subscript 3 times r subscript LLmax) divided by denominator (L subscript abut)

                          Formula: R subscript LLmax = 6 point 50 Kips per foot

Minimum unfactored live load representing uplift used for abutment stem design:

                        Formula: r subscript LLmin = V subscript vehmin times ( 1 + IM) + V subscript lanemin

                        Formula: r subscript LLmin = minus 14 point 83 K       for one lane

                          Formula: R subscript LLmin = numerator (3 times m subscript 3 times r subscript LLmin) divided by denominator (L subscript abut)

                        Formula: R subscript LLmin = minus 0 point 81 Kips per foot

The following loads are applied at the beam seat or top of abutment stem for the footing design. The loads do not include dynamic load allowance, but do include the multiple presence factor.

S3.6.2.1

Maximum unfactored live load used for abutment footing design:

                          Formula: r subscript LLmax1 = V subscript vehmax + V subscript lanemax

                          Formula: r subscript LLmax1 = 98 point 15 K       for one lane loaded

                            Formula: R subscript LLmax1 = numerator (3 times m subscript 3 times r subscript LLmax1) divided by denominator (L subscript abut)

                            Formula: R subscript LLmax1 = 5 point 34 Kips per foot

Minimum unfactored vehicle load used for abutment footing design:

                          Formula: r subscript LLmin1 = V subscript vehmin + V subscript lanemin

                          Formula: r subscript LLmin1 = minus 12 point 43 K       for one lane loaded

                          Formula: R subscript LLmin1 = numerator (3 times m subscript 3 times r subscript LLmin1) divided by denominator (L subscript abut)

                          Formula: R subscript LLmin1 = minus 0 point 68 Kips per foot

Design Step 7.6 - Compute Other Load Effects

Other load effects that need to be computed include braking force, wind loads, earthquake loads, earth pressure, live load surcharge, and temperature loads.

Braking Force

Since the abutment has expansion bearings, the braking force does not apply at the abutment. The entire braking force is resisted by the fixed bearings located at the pier. Braking force calculations are provided in Design Step 8.

Wind Load on Superstructure

S3.8.1.2

When calculating the superstructure wind load, the total depth from the top of the barrier to the bottom of the girder is required. Included in this depth is any haunch and/or depth due to the bridge deck cross slope. Once the total depth is known, the wind area can be calculated and the wind pressure can be applied.

The total depth is:

                    Formula: h subscript par = 42 inches

                      Formula: t subscript deck = 9 inches       overhang deck thickness

                        Formula: t subscript csIope = 0 inches       assume no cross slope for design

                      Formula: t subscript topflg = 0 inches       top flange embedded in haunch; therefore, ignore top flange thickness

                      Formula: d subscript web = 66 inches       based on first trial of girder design

                      Formula: t subscript botflg = 2 point 25 inches       use maximum bottom flange thickness, based on first trial of girder design

                          Formula: t subscript haunch = 3 point 5 inches

                    Formula: D subscript tot = numerator (h subscript par + t subscript deck + t subscript csIope + t subscript topflg + d subscript web + t subscript botflg + t subscript haunch) divided by denominator (12 inches per foot)

                    Formula: D subscript tot = 10 point 23 feet

The wind load on the abutment from the superstructure will be from one-half of one span length or:

                      Formula: L subscript wind = 60 feet

The wind area is:

                          Formula: A subscript wsuper = D subscript tot times L subscript wind

                          Formula: A subscript wsuper = 613 point 75 feet squared

Since the abutment is less than 30 feet in height, the design wind velocity, VDZ, does not have to be adjusted and is equal to the base wind velocity.

S3.8.1.1

                  Formula: V subscript B = 100       mph

                    Formula: V subscript DZ = V subscript B

From this, the design wind pressure is equal to the base wind pressure:

S3.8.1.2.1

                  Formula: P subscript D = P subscript B times ( numerator (V subscript DZ) divided by denominator (V subscript B) ) squared . Equation not used       or       Formula: P subscript D = P subscript B times ( numerator (100mph) divided by denominator (100mph) ) squared . Equation not used

                  Formula: P subscript D = P subscript B. Equation not used

Also, the total wind loading on girders must be greater than or equal to 0.30 klf:

S3.8.1.2.1

                            Formula: Wind subscript total = 0 point 050ksf times D subscript tot

                            Formula: Wind subscript total = 0 point 51 Kips per foot       , which is greater than 0.30 klf

The wind load from the superstructure acting on the abutment depends on the attack angle of the wind. Two wind load calculations are provided for two different wind attack angles. All wind loads are tabulated in Table 7-1 for the various attack angles. The attack angle is measured from a line perpendicular to the girder longitudinal axis. The wind pressure can be applied to either superstructure face. The base wind pressures for the superstructure for various attack angles are given in STable 3.8.1.2.2-1. Since the abutment has expansion bearings, the longitudinal component of the wind load on superstructure will not be resisted by the abutment and is not required to be calculated. The fixed pier will resist the longitudinal wind component.

S3.8.1.2.2

Plan and elevation view showing how the superstructure wind load is applied to the abutment. The elevation view shows the superstructure depth as 10 point 23 feet. The plan view shows skew angle between the centerline of bearings and the centerline of girders at 90 degrees. The superstructure wind angle or wind attack angle is shown as measured from a line perpendicular from the girders.

Figure 7-3 Application of Superstructure Wind Load on Abutment

For a wind attack angle of 0 degrees, the superstructure wind loads acting on the abutment are:

STable 3.8.1.2.2-1

                                      Formula: WS subscript supertrans0 = A subscript wsuper times 0 point 050ksf

                                      Formula: WS subscript supertrans0 = 30 point 69 K

                                      Formula: WS subscript superlong0 = A subscript wsuper times 0 point 000ksf

                                    Formula: WS subscript superlong0 = 0 point 00 K       not applicable due to expansion bearings at abutment

For a wind attack angle of 60 degrees, the superstructure wind loads acting on the abutment are:

STable 3.8.1.2.2-1

                                        Formula: WS subscript supertrans60 = A subscript wsuper times 0 point 017ksf

                                        Formula: WS subscript supertrans60 = 10 point 43 K

                                      Formula: WS subscript superlong60 = A subscript wsuper times 0 point 019ksf

                                      Formula: WS subscript superlong60 = 11 point 66 K       not applicable due to expansion bearings at abutment

  Abutment Design Wind Loads from Superstructure
Wind Attack Angle Bridge Transverse Axis Bridge * Longitudinal Axis
Degrees Kips Kips
0 30.69 0.00
15 27.01 3.68
30 25.16 7.37
45 20.25 9.82
60 10.43 11.66
* Provided but not applicable due to expansion bearings at abutment.

Table 7-1 Abutment Design Wind Loads from Superstructure for Various Wind Attack Angles

Wind Load on Abutment (Substructure)

S3.8.1.2.3

The wind loads acting on the exposed portion of the abutment front and end elevations are calculated from a base wind pressure of 0.040 ksf. These loads act simultaneously with the superstructure wind loads.

Since all wind loads acting on the abutment front face decrease the maximum longitudinal moment, all abutment front face wind loads will be conservatively ignored.

The abutment exposed end elevation wind area is:

                              Formula: A subscript wsubend = ( 3 point 5 feet ) times ( 22 feet )

                              Formula: A subscript wsubend = 77 point 00 feet squared

Two wind load calculations for the abutment end elevation are shown below for a wind attack angle of zero and sixty degrees. All other wind attack angles do not control and are not shown.

For a wind attack angle of 0 degrees, the wind loads acting on the abutment end elevation are:

                                          Formula: WS subscript subtransend0 = A subscript wsubend times ( 0 point 040 times ksf times cos( 0 times deg))

                                        Formula: WS subscript subtransend0 = 3 point 08 K

                                        Formula: WS subscript sublongend0 = A subscript wsubend times ( 0 point 040ksf times sin( 0deg))

                                        Formula: WS subscript sublongend0 = 0 point 00 K

For a wind attack angle of 60 degrees, the wind loads acting on the abutment end elevation are:

                                          Formula: WS subscript subtransend60 = A subscript wsubend times ( 0 point 040 times ksf times cos( 60 times deg))

                                          Formula: WS subscript subtransend60 = 1 point 54 K

                                          Formula: WS subscript sublongend60 = A subscript wsubend times ( 0 point 040ksf times sin( 60deg))

                                          Formula: WS subscript sublongend60 = 2 point 67 K

Wind Load on Vehicles

S3.8.1.3

The wind load applied to vehicles is given as 0.10 klf acting normal to and 6.0 feet above the roadway. For wind loads that are not normal to the roadway, the Specifications give a table of wind components on live load. For normal and skewed wind pressures on vehicles, the wind load is calculated by multiplying the wind component by the length of structure over which it acts. An example calculation is provided and Table 7-2 shows all the vehicle wind loads for the various wind attack angles. As with the superstructure wind load, the longitudinal wind load on vehicles is not resisted by the abutment due to expansion bearings. The calculation for longitudinal vehicular wind loads is not required but is provided in this design example.

For a wind attack angle of 0 degrees, the vehicular wind loads are:

                      Formula: L subscript wind = 60 feet

                            Formula: WL subscript trans0 = L subscript wind times ( 0 point 1 times kIf)

STable 3.8.1.3-1

                            Formula: WL subscript trans0 = 6 point 00 K

                            Formula: WL subscript long0 = L subscript wind times ( 0 point 000kIf)

STable 3.8.1.3-1

                            Formula: WL subscript long0 = 0 point 00 K       not applicable due to expansion bearings at abutment

  Design Vehicular Wind Loads
Wind Attack Angle Bridge Transverse Axis Bridge * Longitudinal Axis
Degrees Kips Kips
0 6.00 0.00
15 5.28 0.72
30 4.92 1.44
45 3.96 1.92
60 2.04 2.28
* Provided but not applicable due to expansion bearings at abutment.

Table 7-2 Design Vehicular Wind Loads for Various Wind Attack Angles Vertical Wind Load

S3.8.2

The vertical wind load is calculated by multiplying a 0.020 ksf vertical wind pressure by the out-to-out bridge deck width. It is applied to the windward quarter-point of the deck only for limit states that do not include wind on live load. Also, the wind attack angle must be zero degrees for the vertical wind load to apply.

                      Formula: W subscript vert = 0 point 020ksf times W subscript deck       Formula: W subscript vert = 0 point 94 Kips per foot       acts vertically upward

Earthquake Load

S3.10

This design example assumes that the structure is located in Seismic Zone I with an acceleration coefficient of 0.02 and a Soil Type I. For Seismic Zone I, no seismic analysis is required except designing for the minimum connection force between the superstructure and substructure and the minimum bridge seat requirements.

S4.7.4.1

S3.10.9

S4.7.4.4

The horizontal connection force in the restrained direction is 0.1 times the vertical reaction due to the tributary permanent load and the tributary live loads assumed to exist during an earthquake. In addition, since all abutment bearings are restrained in the transverse direction, the tributary permanent load can be taken as the reaction at the bearing. Also, γEQ is assumed to be zero. Therefore, no tributary live loads will be considered. This transverse load is calculate and used to design the bearing anchor bolts and is mentioned here for reference only. Refer to Design Step 6 for bearing and anchor bolt design and the calculation of the horizontal connection force.

S3.10.9.2

SC3.10.9.2

S3.4.1

From S4.7.4.3, for Seismic Zone I, no seismic analysis is required. Therefore, the minimum displacement requirement must be obtained from a percentage of the empirical seat width. The percentage of the minimum support length, N, is based on Seismic Zone I, an acceleration coefficient of 0.02, and Soil Type I. From the above information, 50 percent or greater of the minimum support length is required.

S4.7.4.4

STable 4.7.4.4-1

Minimum support length required:

                Formula: N = ( 8 + 0 point 02L + 0 point 08H) times ( 1 + 0 point 000125S squared ) . Equation not used

S4.7.4.4

              Formula: L = 240feet       Formula: H = 22feet       Formula: S = 0       deg

                Formula: N = ( 8 + 0 point 02L + 0 point 08H) times ( 1 + 0 point 000125S squared )

                Formula: N = 14 point 56inches       Use       Formula: N = 15inches

Since the selected preliminary abutment dimensions in Design Step 7.3 leave 18 inches as a support length, this design example will use 100 percent of the minimum support length.

STable 4.7.4.4-1

Elevation view of abutment beam seat showing the minimum support length required. This is the distance from the end of beam to the front face of abutment. The "N" value is given as 15 inches. The distance from the end of beam to the front face of backwall is 3 inches.

Figure 7-4 Minimum Support Length Required

Earth Loads

S3.11

The earth loads that need to be investigated for this design example include loads due to basic lateral earth pressure, loads due to uniform surcharge, and live load surcharge loads.

S3.11.5

S3.11.6

The water table is considered to be below the bottom of footing for this design example. Therefore, the effect of hydrostatic water pressure does not need to be added to the earth pressure. Hydrostatic water pressure should be avoided if possible in all abutment and retaining wall design cases through the design of an appropriate drainage system. Some ways that can reduce or eliminate hydrostatic water pressure include the use of pipe drains, gravel drains, perforated drains, geosynthetic drains, or backfilling with crushed rock. It should be noted that the use of weep holes, or drains at the wall face, do not assure fully drained conditions.

S3.11.3

S11.6.6

Loads due to basic lateral earth pressure:

S3.11.5

To obtain the lateral loads due to basic earth pressure, the earth pressure (p) must first be calculated from the following equation.

S3.11.5.1

              Formula: p = k subscript a times gamma subscript s times z. Equation not used

Bottom of backwall lateral earth load:

                Formula: k subscript a = 0 point 3       obtained from geotechnical information

                Formula: gamma subscript s = 0 point 120kcf       use average of loose and compacted gravel

STable 3.5.1-1

              Formula: z = 7 feet       backwall height

              Formula: p = k subscript a times gamma subscript s times z

              Formula: p = 0 point 25 ksf

Elevation view of abutment showing the backwall design earth pressure. The backwall depth from top of backwall to bottom of backwall is 7 feet 0 inches. The lateral earth pressure on the backwall, p, is maximum at the base of the backwall and zero at the top of backwall. The horizontal load due to p, R sub EHbw, acts at 2 feet 4 inches up from the base of the backwall acting in the direction towards the toe.

Figure 7-5 Backwall Design Earth Pressure

Once the lateral earth pressure is calculated, the lateral load due to the earth pressure can be calculated. This load acts at a distance of H/3 from the bottom of the section being investigated.

S3.11.5.1

SC3.11.5.1

                      Formula: h subscript bkwII = 7 feet

                        Formula: R subscript EHbw = ( numerator (1) divided by denominator (2) ) times p times h subscript bkwII

                        Formula: R subscript EHbw = 0 point 88 Kips per foot

Bottom of abutment stem lateral earth load:

                Formula: k subscript a = 0 point 3       obtained from geotechnical information

                Formula: gamma subscript s = 0 point 120kcf       use average of loose and compacted gravel

STable 3.5.1-1

              Formula: z = 22 feet       height used for maximum moment at bottom of abutment stem

              Formula: p = k subscript a times gamma subscript s times z

              Formula: p = 0 point 79 ksf

Elevation view of abutment showing the abutment design earth pressure from the top of backwall to bottom of stem. The backwall depth from top of backwall to bottom of backwall is 7 feet 0 inches. The stem depth from top of stem to top of footing is 15 feet 0 inches. The lateral earth pressure on the stem, p, is maximum at the base of the stem and zero at the top of backwall. The horizontal load, R subscript EHstem due to p acts at 7 feet 4 inches up from the base of the stem acting in the direction towards the toe.

Figure 7-6 Abutment Stem Design Earth Pressure

Once the lateral earth pressure is calculated, the lateral load due to the earth pressure can be calculated. This load acts at a distance of H/3 from the bottom of the section being investigated.

S3.11.5.1

SC3.11.5.1

                            Formula: R subscript EHstem = ( numerator (1) divided by denominator (2) ) times p times h subscript stem

                            Formula: R subscript EHstem = 8 point 71 Kips per foot

Bottom of footing lateral earth load:

                Formula: k subscript a = 0 point 3       obtained from geotechnical information

                Formula: gamma subscript s = 0 point 120kcf       use average of loose and compacted gravel

STable 3.5.1-1

              Formula: z = 24 point 5 feet       height from top of backwall to bottom of footing

              Formula: p = k subscript a times gamma subscript s times z

              Formula: p = 0 point 88 ksf

Elevation view of abutment showing the design earth pressure from the top of backwall to the bottom of footing. The backwall depth from top of backwall to bottom of backwall is 7 feet 0 inches. The stem depth from top of stem to top of footing is 15 feet 0 inches. The footing thickness is 2 feet 6 inches. The lateral earth pressure on the stem, p, is maximum at the bottom of the footing and zero at the top of backwall. The horizontal load, R subscript EHftg due to p acts at 8 feet 2 inches up from the bottom of the footing acting in the direction towards the toe.

Figure 7-7 Bottom of Footing Design Earth Load

Once the lateral earth pressure is calculated, the lateral load due to the earth pressure can be calculated. This load acts at a distance of H/3 from the bottom of the section being investigated.

S3.11.5.1

SC3.11.5.1

                  Formula: t subscript ftg = 2 point 5 feet

                        Formula: R subscript EHftg = ( numerator (1) divided by denominator (2) ) times p times ( h subscript stem + t subscript ftg )

                        Formula: R subscript EHftg = 10 point 80 Kips per foot

Loads due to uniform surcharge:

S3.11.6.1

Since an approach slab and roadway will cover the abutment backfill material, no uniform surcharge load will be applied.

Loads due to live load surcharge:

S3.11.6.4

Loads due to live load surcharge must be applied when a vehicular live load acts on the backfill surface behind the back face within one-half the wall height. The horizontal pressure increase due to live load surcharge is estimated based on the following equation:

                  Formula: Delta subscript p = k times gamma subscript s times h subscript eq. Equation not used

Bottom of backwall live load surcharge load:

              Formula: k = k subscript a

                Formula: gamma subscript s = 0 point 120kcf       use average of loose and compacted gravel

STable 3.5.1-1

                  Formula: h subscript eq = 3 point 6 feet       equivalent height of soil for vehicular loading based on 7ft backwall height (interpolate between 4 and 3 in the Table)

STable 3.11.6.4-1

                  Formula: Delta subscript p = k times gamma subscript s times h subscript eq

                  Formula: Delta subscript p = 0 point 130 ksf

The lateral load due to the live load surcharge is:

                        Formula: R subscript LSbw = Delta subscript p times h subscript bkwII

                        Formula: R subscript LSbw = 0 point 91 Kips per foot

Bottom of abutment stem live load surcharge load:

              Formula: k = k subscript a

                Formula: gamma subscript s = 0 point 120kcf       use average of loose and compacted gravel

STable 3.5.1-1

                  Formula: h subscript eq = 2 feet       equivalent height of soil for vehicular loading based on stem height

STable 3.11.6.4-1

                  Formula: Delta subscript p = k times gamma subscript s times h subscript eq

                  Formula: Delta subscript p = 0 point 072 ksf

The lateral load due to the live load surcharge is:

                            Formula: R subscript LSstem = Delta subscript p times h subscript stem

                          Formula: R subscript LSstem = 1 point 58 Kips per foot

Bottom of footing live load surcharge load:

              Formula: k = k subscript a

                Formula: gamma subscript s = 0 point 120kcf       use average of loose and compacted gravel

STable 3.5.1-1

                  Formula: h subscript eq = 2 feet       equivalent height of soil for vehicular loading

STable 3.11.6.4-1

                  Formula: Delta subscript p = k times gamma subscript s times h subscript eq

                  Formula: Delta subscript p = 0 point 072 ksf

The lateral load due to the live load surcharge is:

                        Formula: R subscript LSftg = Delta subscript p times ( h subscript stem + t subscript ftg )

                        Formula: R subscript LSftg = 1 point 76 Kips per foot

Since one edge of the approach slab will be supported by the abutment, a reduction of live load surcharge could be taken into account. For this design example, a surcharge reduction is not accounted for.

S3.11.6.5

Loads due to temperature:

S3.12

For this abutment design example, two horizontal temperature loads need to be calculated: load due to temperature rise and load due to temperature fall. To calculate these loads, the steel girder setting temperature is required. Also, the temperature range, as well as the thermal coefficient of expansion for steel, is needed. The expansion or contraction can then be calculated. Using the expansion or contraction, the thermal loads can be calculated based on the neoprene bearing properties.

S3.12.2.2

STable 3.12.2.1-1

S6.4.1

S14.6.3.1

              Formula: epsilon = 6 point 5 times 10 superscript - 6       (in/in/oF)

S6.4.1

                  Formula: t subscript set = 68       oF       assumed steel girder setting temperature

For this design example, assume a moderate climate. The temperature range is then 0 oF to 120 oF

STable 3.12.2.1-1

Expansion calculation:

                      Formula: Delta subscript exp = epsilon times Delta subscript t times ( L subscript span times 12 inches per foot ). Equation not used

                      Formula: Delta t subscript rise = 120 minus t subscript set

                      Formula: Delta t subscript rise = 52       oF

                      Formula: Delta subscript exp = epsilon times Delta t subscript rise times ( L subscript span times 12 inches per foot )

                    Formula: Delta subscript exp = 0 point 487 inches

Contraction calculation:

                        Formula: Delta subscript contr = epsilon times Delta subscript t times ( L subscript span times 12 inches per foot ). Equation not used

                    Formula: Delta t subscript fall = t subscript set minus 0

                    Formula: Delta t subscript fall = 68       oF

                        Formula: Delta subscript contr = epsilon times Delta t subscript fall times ( L subscript span times 12 inches per foot )

                        Formula: Delta subscript contr = 0 point 636 inches

Once the expansion and contraction is known, the loads due to temperature can be calculated based on the following equation:

                  Formula: H subscript u = G times A times numerator ( Delta ) divided by denominator (h subscript rt) . Equation not used

S14.6.3.1

Before the loads due to temperature rise and fall can be calculated, the neoprene bearing properties are needed (see Design Step 6). If the bearing pad design is not complete at the time the temperature loads are being calculated, the temperature loads can be estimated by assuming bearing pad properties that are larger than expected from the bearing pad design. The bearing pad properties for this design example are:

                Formula: G = 0 point 095ksi       shear modulus

STable 14.7.5.2-1

                Formula: A = 14 inches times 15 inches       area of the bearing pad in plan view

                Formula: A = 210 point 00 inches squared

                  Formula: h subscript rt = 3 point 5 inches       elastomer thickness (not including steel reinforcement)

Load due to temperature rise:

                        Formula: H subscript urise = G times A times numerator ( Delta subscript exp) divided by denominator (h subscript rt)

                      Formula: H subscript urise = 2 point 77 K       per bearing

Now, multiply Hurise by five bearings and divide by the abutment length to get the total load due to temperature rise:

                            Formula: H subscript urisetot = numerator (H subscript urise times 5) divided by denominator (L subscript abut)

                          Formula: H subscript urisetot = 0 point 30 Kips per foot

Load due to temperature fall:

                      Formula: H subscript ufall = G times A times numerator ( Delta subscript contr) divided by denominator (h subscript rt)

                      Formula: H subscript ufall = 3 point 63 K

Now, multiply Hufall by five bearings and divide by the abutment length to get the total load due to temperature fall:

                          Formula: H subscript ufalltot = numerator (H subscript ufall times 5) divided by denominator (L subscript abut)

                          Formula: H subscript ufalltot = 0 point 39 Kips per foot

Design Step 7.7 - Analyze and Combine Force Effects

There are three critical locations where the force effects need to be combined and analyzed for an abutment design. They are the base or bottom of the backwall, the bottom of stem or top of footing, and the bottom of footing. For the backwall and stem design, transverse horizontal loads do not need to be considered due to the high moment of inertia about that axis, but at the bottom of footing, the transverse horizontal loads will need to be considered for the footing and pile design, although they are still minimal.

Bottom of Abutment Backwall

In order to analyze and combine the force effects, the abutment backwall dimensions, the appropriate loads, and the application location of the loads are needed. The small moment that is created by the top of the backwall corbel concrete will be neglected in this design example.

Elevation view of abutment showing the backwall design earth pressure. The backwall depth from top of backwall to bottom of backwall is 7 feet 0 inches. The lateral earth pressure on the backwall, p, is maximum at the base of the backwall and zero at the top of backwall. The horizontal load due to p, R sub EHbw, acts at 2 feet 4 inches up from the base of the backwall acting in the direction towards the toe.

Figure 7-8 Abutment Backwall Dimensions and Loading

The following limit states will be investigated for the backwall analysis. The load factor for future wearing surface is given, but the load due to future wearing surface on the abutment backwall will be ignored since its effects are negligible. Also, limit states that are not shown either do not control or are not applicable. In addition, Strength III and Strength V limit states are included but generally will not control for an abutment with expansion bearings. Strength III or Strength V may control for abutments supporting fixed bearings.

  Load Factors
  Strength I Strength III Strength V Service I
Loads γmax γmin γmax γmin γmax γmin γmax γmin
DC 1.25 0.90 1.25 0.90 1.25 0.90 1.00 1.00
DW 1.50 0.65 1.50 0.65 1.50 0.65 1.00 1.00
LL 1.75 1.75 --- --- 1.35 1.35 1.00 1.00
EH 1.50 0.90 1.50 0.90 1.50 0.90 1.00 1.00
LS 1.75 1.75 --- --- 1.35 1.35 1.00 1.00

STable 3.4.1-1

STable 3.4.1-2

Table 7-3 Applicable Abutment Backwall Limit States with the Corresponding Load Factors

The loads that are required from Design Steps 7.4, 7.5, and 7.6 include:

                      Formula: DL subscript bw = 1 point 71 Kips per foot       Formula: R subscript EHbw = 0 point 88 Kips per foot

                        Formula: R subscript LLbw = 2 point 81 Kips per foot       Formula: R subscript LSbw = 0 point 91 Kips per foot

Abutment backwall Strength I force effects:

The following load factors will be used to calculate the force effects for Strength I. Note that eta (η), the product of ductility, redundancy, and operational importance factors, is not shown. Eta is discussed in detail in Design Step 1. For all portions of this design example, eta is taken as 1.0, and will not be shown.

                    Formula: gamma subscript DC = 1 point 25

STable 3.4.1-2

                  Formula: gamma subscript LL = 1 point 75

STable 3.4.1-1

                    Formula: gamma subscript EH = 1 point 50

STable 3.4.1-2

                  Formula: gamma subscript LS = 1 point 75

STable 3.4.1-1

The factored vertical force at the base of the backwall is:

                          Formula: F subscript vbwstrI = gamma subscript DC times DL subscript bw + gamma subscript LL times R subscript LLbw

                          Formula: F subscript vbwstrI = 7 point 05 Kips per foot

The factored longitudinal shear force at the base of the backwall is:

                          Formula: V subscript ubwstrI = ( gamma subscript EH times R subscript EHbw ) + ( gamma subscript LS times R subscript LSbw )

                          Formula: V subscript ubwstrI = 2 point 91 Kips per foot

The factored moment at the base of the backwall is:

                          Formula: M subscript ubwstrI = ( gamma subscript LL times R subscript LLbw times 1 point 17 feet ) + ( gamma subscript EH times R subscript EHbw times 2 point 33 feet ) + ( gamma subscript LS times R subscript LSbw times 3 point 50 feet )

                          Formula: M subscript ubwstrI = 14 point 38 Kips foot per foot

Abutment backwall Strength III force effects:

The following load factors will be used to calculate the force effects for Strength III:

                    Formula: gamma subscript DC = 1 point 25

STable 3.4.1-2

                  Formula: gamma subscript LL = 0 point 00

STable 3.4.1-1

                    Formula: gamma subscript EH = 1 point 50

STable 3.4.1-2

                  Formula: gamma subscript LS = 0 point 00

STable 3.4.1-1

The factored vertical force at the base of the backwall is:

                            Formula: F subscript vbwstrIII = gamma subscript DC times DL subscript bw + gamma subscript LL times R subscript LLbw

                            Formula: F subscript vbwstrIII = 2 point 14 Kips per foot

The factored longitudinal shear force at the base of the backwall is:

                            Formula: V subscript ubwstrIII = ( gamma subscript EH times R subscript EHbw ) + ( gamma subscript LS times R subscript LSbw )

                            Formula: V subscript ubwstrIII = 1 point 32 Kips per foot

The factored moment at the base of the backwall is:

                            Formula: M subscript ubwstrIII = ( gamma subscript LL times R subscript LLbw times 1 point 17 feet ) + ( gamma subscript EH times R subscript EHbw times 2 point 33 feet ) + ( gamma subscript LS times R subscript LSbw times 3 point 50 feet )

                            Formula: M subscript ubwstrIII = 3 point 08 Kips foot per foot

Abutment backwall Strength V force effects:

The following load factors will be used to calculate the force effects for Strength V:

                    Formula: gamma subscript DC = 1 point 25

STable 3.4.1-2

                  Formula: gamma subscript LL = 1 point 35

STable 3.4.1-1

                    Formula: gamma subscript EH = 1 point 50

STable 3.4.1-2

                  Formula: gamma subscript LS = 1 point 35

STable 3.4.1-1

The factored vertical force at the base of the backwall is:

                            Formula: F subscript vbwstrV = gamma subscript DC times DL subscript bw + gamma subscript LL times R subscript LLbw

                            Formula: F subscript vbwstrV = 5 point 93 Kips per foot

The factored longitudinal shear force at the base of the backwall is:

                            Formula: V subscript ubwstrV = ( gamma subscript EH times R subscript EHbw ) + ( gamma subscript LS times R subscript LSbw )

                            Formula: V subscript ubwstrV = 2 point 55 Kips per foot

The factored moment at the base of the backwall is:

                            Formula: M subscript ubwstrV = ( gamma subscript LL times R subscript LLbw times 1 point 17 feet ) + ( gamma subscript EH times R subscript EHbw times 2 point 33 feet ) + ( gamma subscript LS times R subscript LSbw times 3 point 50 feet )

                            Formula: M subscript ubwstrV = 11 point 80 Kips foot per foot

Abutment backwall Service I force effects:

The following load factors will be used to calculate the force effects for Service I:

                    Formula: gamma subscript DC = 1 point 0

STable 3.4.1-2

                  Formula: gamma subscript LL = 1 point 0

STable 3.4.1-1

                    Formula: gamma subscript EH = 1 point 0

STable 3.4.1-2

                  Formula: gamma subscript LS = 1 point 0

STable 3.4.1-1

The factored vertical force at the base of the backwall is:

                            Formula: F subscript vbwservI = gamma subscript DC times DL subscript bw + gamma subscript LL times R subscript LLbw

                            Formula: F subscript vbwservI = 4 point 52 Kips per foot

The factored longitudinal shear force at the base of the backwall is:

                              Formula: V subscript ubwservI = ( gamma subscript EH times R subscript EHbw ) + ( gamma subscript LS times R subscript LSbw )

                              Formula: V subscript ubwservI = 1 point 79 Kips per foot

The factored moment at the base of the backwall is:

                              Formula: M subscript ubwservI = ( gamma subscript LL times R subscript LLbw times 1 point 17 feet ) + ( gamma subscript EH times R subscript EHbw times 2 point 33 feet ) + ( gamma subscript LS times R subscript LSbw times 3 point 50 feet )

                              Formula: M subscript ubwservI = 8 point 51 Kips foot per foot

The maximum factored backwall vertical force, shear force, and moment for the strength limit state are:

                            Formula: F subscript vbwmax = max ( F subscript vbwstrI , F subscript vbwstrIII , F subscript vbwstrV )

                            Formula: F subscript vbwmax = 7 point 05 Kips per foot

                            Formula: V subscript ubwmax = max ( V subscript ubwstrI , V subscript ubwstrIII , V subscript ubwstrV )

                            Formula: V subscript ubwmax = 2 point 91 Kips per foot

                            Formula: M subscript ubwmax = max ( M subscript ubwstrI , M subscript ubwstrIII , M subscript ubwstrV )

                            Formula: M subscript ubwmax = 14 point 38 Kips foot per foot

Bottom of Abutment Stem

The combination of force effects for the bottom of abutment stem are similar to the backwall with the addition of the superstructure dead and live loads.

Elevation view of abutment section 1 foot thick showing the loads that are applied when forces are desired at the bottom of stem. The backwall dead load, DL subscript bw, acts vertically downward at 11 and three eights inches from the back face of the backwall. The stem dead load, DL subscript stem, acts vertically downward at the center of the 3 feet 6 inch stem thickness. The superstructure dead load, R subscript DCtot, acts vertically downward at the beam seat and 7 and one half inches in from the front face of abutment. The superstructure wearing surface load, R subscript DWtot, acts at the same location as R subscript DCtot. Other vertical loads that act at the beam seat and 7 and one half inches in from the front face of abutment include: the maximum and minimum live load, R subscript LLmax and R subscript LLmin. The thermal load, H subscript ufalltot acts horizontally at the beam seat along the centerline of bearings, which is 7 feet down from the top of backwall and 15 feet 0 inches up from the bottom of stem. The horizontal earth load, R subscript EHstem, acts 7 feet 4 inches up from the bottom of the stem toward the toe. The horizontal live load surcharge, R subscript LSstem, acts at the midpoint of the stem plus backwall depth toward the toe.

Figure 7-9 Abutment Stem Dimensions and Loading

The force effects for the stem will be combined for the same limit states as the backwall. The loads and load factors are also similar to the backwall with the addition of wind on structure, wind on live load, and thermal effects. As with the backwall, the extreme event limit states will not be investigated.

  Load Factors
  Strength I Strength III Strength V Service I
Loads γmax γmin γmax γmin γmax γmin γmax γmin
DC 1.25 0.90 1.25 0.90 1.25 0.90 1.00 1.00
DW 1.50 0.65 1.50 0.65 1.50 0.65 1.00 1.00
LL 1.75 1.75 --- --- 1.35 1.35 1.00 1.00
EH 1.50 0.90 1.50 0.90 1.50 0.90 1.00 1.00
LS 1.75 1.75 --- --- 1.35 1.35 1.00 1.00
WS --- --- 1.40 1.40 0.40 0.40 0.30 0.30
WL --- --- --- --- 1.00 1.00 1.00 1.00
TU 0.50 0.50 0.50 0.50 0.50 0.50 1.00 1.00

STable 3.4.1-1

STable 3.4.1-2

Table 7-4 Applicable Abutment Stem Limit States with the Corresponding Load Factors

The loads that are required from Design Steps 7.4, 7.5 and 7.6 include:

                      Formula: DL subscript bw = 1 point 71 Kips per foot       Formula: R subscript LLmax = 6 point 50 Kips per foot

                          Formula: DL subscript stem = 7 point 88 Kips per foot       Formula: R subscript EHstem = 8 point 71 Kips per foot

                        Formula: R subscript DCtot = 7 point 66 Kips per foot       Formula: R subscript LSstem = 1 point 58 Kips per foot

                          Formula: R subscript DWtot = 1 point 20 Kips per foot       Formula: H subscript ufalltot = 0 point 39 Kips per foot

Abutment stem Strength I force effects:

The following load factors will be used to calculate the controlling force effects for Strength I:

                    Formula: gamma subscript DC = 1 point 25

STable 3.4.1-2

                    Formula: gamma subscript DW = 1 point 50

STable 3.4.1-2

                  Formula: gamma subscript LL = 1 point 75

STable 3.4.1-1

                    Formula: gamma subscript EH = 1 point 50

STable 3.4.1-2

                  Formula: gamma subscript LS = 1 point 75

STable 3.4.1-1

                    Formula: gamma subscript TU = 0 point 50       use contraction temperature force

STable 3.4.1-1

The factored vertical force at the base of the abutment stem is:

                              Formula: F subscript vstemstrI = ( gamma subscript DC times DL subscript bw ) + ( gamma subscript DC times DL subscript stem ) + ( gamma subscript DC times R subscript DCtot ) + ( gamma subscript DW times R subscript DWtot ) + ( gamma subscript LL times R subscript LLmax )

                              Formula: F subscript vstemstrI = 34 point 74 Kips per foot

The factored longitudinal shear force at the base of the stem is:

                              Formula: V subscript ustemstrI = ( gamma subscript EH times R subscript EHstem ) + ( gamma subscript LS times R subscript LSstem ) + ( gamma subscript TU times H subscript ufalltot )

                              Formula: V subscript ustemstrI = 16 point 03 Kips per foot

The factored moment about the bridge transverse axis at the base of the abutment stem is:

                              Formula: M subscript ustemstrI = ( gamma subscript DC times DL subscript bw times 0 point 80 feet ) + ( gamma subscript DC times R subscript DCtot times 1 point 13 feet ) + ( gamma subscript DW times R subscript DWtot times 1 point 13 feet ) + ( gamma subscript LL times R subscript LLmax times 1 point 13 feet ) + ( gamma subscript EH times R subscript EHstem times 7 point 33 feet ) + ( gamma subscript LS times R subscript LSstem times 11 point 00 feet ) + ( gamma subscript TU times H subscript ufalltot times 15 feet )

                              Formula: M subscript ustemstrI = 156 point 61 Kips foot per foot

Abutment stem Strength III force effects:

The following load factors will be used to calculate the force effects for Strength III:

                    Formula: gamma subscript DC = 1 point 25

STable 3.4.1-2

                    Formula: gamma subscript DW = 1 point 50

STable 3.4.1-2

                    Formula: gamma subscript EH = 1 point 50

STable 3.4.1-2

                    Formula: gamma subscript WS = 1 point 40       all longitudinal wind loads ignored

STable 3.4.1-1

                    Formula: gamma subscript TU = 0 point 50       use contraction temperature force

STable 3.4.1-1

The factored vertical force at the base of the abutment stem is:

                              Formula: F subscript vstemstrIII = ( gamma subscript DC times DL subscript bw ) + ( gamma subscript DC times DL subscript stem ) + ( gamma subscript DC times R subscript DCtot ) + ( gamma subscript DW times R subscript DWtot )

                              Formula: F subscript vstemstrIII = 23 point 36 Kips per foot

The factored longitudinal shear force at the base of the stem is:

                                Formula: V subscript ustemstrIII = ( gamma subscript EH times R subscript EHstem ) + ( gamma subscript TU times H subscript ufalltot )

                              Formula: V subscript ustemstrIII = 13 point 26 Kips per foot

The factored moment about the bridge transverse axis at the base of the abutment stem is:

                                Formula: M subscript ustemstrIII = ( gamma subscript DC times DL subscript bw times 0 point 80 feet ) + ( gamma subscript DC times R subscript DCtot times 1 point 13 feet ) + ( gamma subscript DW times R subscript DWtot times 1 point 13 feet ) + ( gamma subscript EH times R subscript EHstem times 7 point 33 feet ) + ( gamma subscript TU times H subscript ufalltot times 15 feet )

                                Formula: M subscript ustemstrIII = 113 point 25 Kips foot per foot

Abutment stem Strength V force effects:

The following load factors will be used to calculate the force effects for Strength V:

                    Formula: gamma subscript DC = 1 point 25

STable 3.4.1-2

                    Formula: gamma subscript DW = 1 point 50

STable 3.4.1-2

                  Formula: gamma subscript LL = 1 point 35

STable 3.4.1-1

                    Formula: gamma subscript EH = 1 point 50

STable 3.4.1-2

                  Formula: gamma subscript LS = 1 point 35

STable 3.4.1-1

                    Formula: gamma subscript WS = 0 point 40       all longitudinal wind loads ignored

STable 3.4.1-1

                    Formula: gamma subscript WL = 1 point 00       only applicable for wind angle of 0 degrees

STable 3.4.1-1

                    Formula: gamma subscript TU = 0 point 50       use contraction temperature force

STable 3.4.1-1

The factored vertical force at the base of the abutment stem is:

                              Formula: F subscript vstemstrV = ( gamma subscript DC times DL subscript bw ) + ( gamma subscript DC times DL subscript stem ) + ( gamma subscript DC times R subscript DCtot ) + ( gamma subscript DW times R subscript DWtot ) + ( gamma subscript LL times R subscript LLmax )

                              Formula: F subscript vstemstrV = 32 point 14 Kips per foot

The factored longitudinal shear force at the base of the stem is:

                                Formula: V subscript ustemstrV = ( gamma subscript EH times R subscript EHstem ) + ( gamma subscript LS times R subscript LSstem ) + ( gamma subscript TU times H subscript ufalltot )

                              Formula: V subscript ustemstrV = 15 point 40 Kips per foot

The factored moment about the bridge transverse axis at the base of the abutment stem is:

                                Formula: M subscript ustemstrV = ( gamma subscript DC times DL subscript bw times 0 point 80 feet ) + ( gamma subscript DC times R subscript DCtot times 1 point 13 feet ) + ( gamma subscript DW times R subscript DWtot times 1 point 13 feet ) + ( gamma subscript LL times R subscript LLmax times 1 point 13 feet ) + ( gamma subscript EH times R subscript EHstem times 7 point 33 feet ) + ( gamma subscript LS times R subscript LSstem times 11 point 00 feet ) + ( gamma subscript TU times H subscript ufalltot times 15 feet )

                                Formula: M subscript ustemstrV = 146 point 70 Kips foot per foot

Abutment stem Service I force effects:

The following load factors will be used to calculate the force effects for Service I:

                    Formula: gamma subscript DC = 1 point 00

STable 3.4.1-2

                    Formula: gamma subscript DW = 1 point 00

STable 3.4.1-2

                  Formula: gamma subscript LL = 1 point 00

STable 3.4.1-1

                    Formula: gamma subscript EH = 1 point 00

STable 3.4.1-2

                  Formula: gamma subscript LS = 1 point 00

STable 3.4.1-1

                    Formula: gamma subscript WS = 0 point 30       use for wind on stem end face for controlling wind at 60 degrees

STable 3.4.1-1

                    Formula: gamma subscript WL = 1 point 00       only applicable for wind angle of 0 degrees

STable 3.4.1-1

                    Formula: gamma subscript TU = 1 point 00       use contraction temperature force

STable 3.4.1-1

The factored vertical force at the base of the abutment stem is:

                                Formula: F subscript vstemservI = ( gamma subscript DC times DL subscript bw ) + ( gamma subscript DC times DL subscript stem ) + ( gamma subscript DC times R subscript DCtot ) + ( gamma subscript DW times R subscript DWtot ) + ( gamma subscript LL times R subscript LLmax )

                                Formula: F subscript vstemservI = 24 point 95 Kips per foot

The factored longitudinal shear force at the base of the stem is:

                                Formula: V subscript ustemservI = ( gamma subscript EH times R subscript EHstem ) + ( gamma subscript LS times R subscript LSstem ) + ( gamma subscript TU times H subscript ufalltot )

                                Formula: V subscript ustemservI = 10 point 68 Kips per foot

The factored moment about the bridge transverse axis at the base of the abutment stem is:

                                  Formula: M subscript ustemservI = ( gamma subscript DC times DL subscript bw times 0 point 80 feet ) + ( gamma subscript DC times R subscript DCtot times 1 point 13 feet ) + ( gamma subscript DW times R subscript DWtot times 1 point 13 feet ) + ( gamma subscript LL times R subscript LLmax times 1 point 13 feet ) + ( gamma subscript EH times R subscript EHstem times 7 point 33 feet ) + ( gamma subscript LS times R subscript LSstem times 11 point 00 feet ) + ( gamma subscript TU times H subscript ufalltot times 15 feet )

                                Formula: M subscript ustemservI = 105 point 82 Kips foot per foot

The maximum factored abutment stem vertical force, shear force, and moment for the strength limit state are:

                              Formula: F subscript vstemmax = max ( F subscript vstemstrI , F subscript vstemstrIII , F subscript vstemstrV )

                              Formula: F subscript vstemmax = 34 point 74 Kips per foot

                                Formula: V subscript ustemmax = max ( V subscript ustemstrI , V subscript ustemstrIII , V subscript ustemstrV )

                              Formula: V subscript ustemmax = 16 point 03 Kips per foot

                                Formula: M subscript ustemmax = max ( M subscript ustemstrI , M subscript ustemstrIII , M subscript ustemstrV )

                                Formula: M subscript ustemmax = 156 point 61 Kips foot per foot

Bottom of Abutment Footing

The combination of force effects for the bottom of abutment footing are similar to the backwall and stem with the addition of the earth load on the abutment heel. Also, dynamic load allowance must be removed from the live load portion of the force effects for foundation components that are completely below the ground level.

S3.6.2.1

Elevation view of abutment section 1 foot thick showing the loads that are applied when forces are desired at the bottom of footing. The backwall dead load, DL subscript bw, acts vertically downward at 11 and three eights inches from the back face of the backwall. The stem dead load, DL subscript stem, acts vertically downward at the center of the 3 feet 6 inch stem thickness. The vertical earth dead load, DL subscript earth, acts at the center of footing heel. The superstructure dead load, R subscript DCtot, acts vertically downward at the beam seat and 7 and one half inches in from the front face of abutment. The superstructure wearing surface load, R subscript DWtot, acts at the same location as R subscript DCtot. Other vertical loads that act at the beam seat and 7 and one half inches in from the front face of abutment include: the maximum and minimum live load, R subscript LLmax and R subscript LLmin. The thermal load, H subscript ufalltot acts horizontally at the beam seat along the centerline of bearings, which is 7 feet down from the top of backwall and 15 feet 0 inches plus 2 feet 6 inches up from the bottom of footing. The horizontal earth load, R subscript EHftg, acts 8 feet 2 inches up from the bottom of the footing toward the toe. The horizontal live load surcharge, R subscript LSftg, acts at the midpoint of the footing plus stem plus backwall depth toward the toe. The entire footing width is 10 feet 3 inches. The footing toe width is 2 feet 9 inches.

Figure 7-10 Abutment Footing Dimensions and Loading

The force effects for the bottom of footing will be combined for the same limit states as the backwall and stem. The loads and load factors are also similar with the addition of vertical earth load.

  Load Factors
  Strength I Strength III Strength V Service I
Loads γmax γmin γmax γmin γmax γmin γmax γmin
DC 1.25 0.90 1.25 0.90 1.25 0.90 1.00 1.00
DW 1.50 0.65 1.50 0.65 1.50 0.65 1.00 1.00
LL 1.75 1.75 --- --- 1.35 1.35 1.00 1.00
EH 1.50 0.90 1.50 0.90 1.50 0.90 1.00 1.00
EV 1.35 1.00 1.35 1.00 1.35 1.00 1.00 1.00
LS 1.75 1.75 --- --- 1.35 1.35 1.00 1.00
WS --- --- 1.40 1.40 0.40 0.40 0.30 0.30
WL --- --- --- --- 1.00 1.00 1.00 1.00
TU 0.50 0.50 0.50 0.50 0.50 0.50 1.00 1.00

STable 3.4.1-1

STable 3.4.1-2

Table 7-5 Applicable Abutment Footing Limit States with the Corresponding Load Factors

The loads that are required from Design Steps 7.4, 7.5, and 7.6 include:

                      Formula: DL subscript bw = 1 point 71 Kips per foot       Formula: R subscript EHftg = 10 point 80 Kips per foot

                          Formula: DL subscript stem = 7 point 88 Kips per foot       Formula: R subscript LSftg = 1 point 76 Kips per foot

                      Formula: DL subscript ftg = 3 point 84 Kips per foot       Formula: H subscript ufalltot = 0 point 39 Kips per foot

                          Formula: DL subscript earth = 10 point 56 Kips per foot       Formula: WS subscript supertrans0 = 30 point 69 K

                        Formula: R subscript DCtot = 7 point 66 Kips per foot       Formula: WS subscript subtransend0 = 3 point 08 K

                          Formula: R subscript DWtot = 1 point 20 Kips per foot       Formula: WL subscript trans0 = 6 point 00 K

                          Formula: R subscript LLmin1 = minus 0 point 68 Kips per foot       Formula: WS subscript subtransend60 = 1 point 54 K

                            Formula: R subscript LLmax1 = 5 point 34 Kips per foot       Formula: WS subscript sublongend60 = 2 point 67 K

Abutment bottom of footing Strength I force effects using the maximum load factors:

The following load factors will be used to calculate the controlling force effects for Strength I:

                    Formula: gamma subscript DC = 1 point 25

STable 3.4.1-2

                    Formula: gamma subscript DW = 1 point 50

STable 3.4.1-2

                  Formula: gamma subscript LL = 1 point 75

STable 3.4.1-1

                    Formula: gamma subscript EH = 1 point 50

STable 3.4.1-2

                    Formula: gamma subscript EV = 1 point 35       use maximum value to maximize the pile loads

STable 3.4.1-2

                  Formula: gamma subscript LS = 1 point 75

STable 3.4.1-1

                    Formula: gamma subscript TU = 0 point 50       use contraction temperature force

STable 3.4.1-1

The factored vertical force at the bottom of footing is:

                          Formula: F subscript vftgstrI = ( gamma subscript DC times DL subscript bw ) + ( gamma subscript DC times DL subscript stem ) + ( gamma subscript DC times DL subscript ftg ) + ( gamma subscript EV times DL subscript earth ) + ( gamma subscript DC times R subscript DCtot ) + ( gamma subscript DW times R subscript DWtot ) + ( gamma subscript LL times R subscript LLmax1 )

                          Formula: F subscript vftgstrI = 51 point 76 Kips per foot

The factored longitudinal horizontal force at the bottom of footing is:

                            Formula: F subscript IonftgstrI = left bracket ( gamma subscript EH times R subscript EHftg ) + ( gamma subscript LS times R subscript LSftg ) + ( gamma subscript TU times H subscript ufalltot ) right bracket

                            Formula: F subscript lonftgstrI = 19 point 49 Kips per foot

The factored transverse horizontal force at the bottom of footing is:

                            Formula: F subscript traftgstrI = 0 Kips per foot       The load factors for the loads that produce transverse horizontal forces are zero for Strength I.

The factored moment about the bridge transverse axis at the bottom of footing is:

M subscript IonftgstrI = left bracket gamma subscript DC times ( DL subscript bw ) times ( minus 0 point 177 feet ) right bracket + left bracket gamma subscript DC times (DL subscript stem) times 0 point 625 feet right bracket + left bracket gamma subscript EV times DL subscript earth times ( minus 3 point 125 feet right bracket + (gamma subscript DC times R subscript DCtot times 1 point 75 feet) + ( gamma subscript DW times R subscript DWtot times 1 point 75 feet ) + ( gamma subscript LL times R subscript LLmax1 times 1 point 75 feet ) + ( gamma subscript EH times R subscript EHftg times 8 point 17 feet ) + ( gamma subscript LS times R subscript LSftg times 12 point 25 feet ) + ( gamma subscript TU times H subscript ufalltot times 17 point 5 feet )

                              Formula: M subscript lonftgstrI = 171 point 09 Kips foot per foot

The factored moment about the bridge longitudinal axis at the bottom of footing is:

                              Formula: M subscript traftgstrI = 0 Kips foot per foot       The load factors for the loads that produce transverse horizontal forces are zero for Strength I.

Abutment bottom of footing Strength I force effects using the minimum load factors:

The following load factors will be used to calculate the controlling force effects for Strength I:

                    Formula: gamma subscript DC = 0 point 90

STable 3.4.1-2

                    Formula: gamma subscript DW = 0 point 65

STable 3.4.1-2

                  Formula: gamma subscript LL = 1 point 75

STable 3.4.1-1

                    Formula: gamma subscript EH = 0 point 90

STable 3.4.1-2

                    Formula: gamma subscript EV = 1 point 00       use minimum value to minimize the pile loads

STable 3.4.1-2

                  Formula: gamma subscript LS = 1 point 75

STable 3.4.1-1

                    Formula: gamma subscript TU = 0 point 50       use contraction temperature force

STable 3.4.1-1

The factored vertical force at the bottom of footing is:

                                Formula: F subscript vftgstrImin = ( gamma subscript DC times DL subscript bw ) + ( gamma subscript DC times DL subscript stem ) + ( gamma subscript DC times DL subscript ftg ) + ( gamma subscript EV times DL subscript earth ) + ( gamma subscript DC times R subscript DCtot ) + ( gamma subscript DW times R subscript DWtot ) + ( gamma subscript LL times R subscript LLmin1 )

                              Formula: F subscript vftgstrImin = 29 point 14 Kips per foot

The factored longitudinal horizontal force at the bottom of footing is:

                                  Formula: F subscript lonftgstrImin = ( gamma subscript EH times R subscript EHftg ) + ( gamma subscript LS times R subscript LSftg ) + ( gamma subscript TU times H subscript ufalltot )

                                  Formula: F subscript lonftgstrImin = 13 point 00 Kips per foot

The factored transverse horizontal force at the bottom of footing is:

                                  Formula: F subscript traftgstrImin = 0 Kips per foot       The load factors for the loads that produce transverse horizontal forces are zero for Strength I.

The factored moment about the bridge transverse axis at the bottom of footing is:

M subscript lonftgstrImin = left bracket gamma subscript DC times ( DL subscript bw ) times ( minus 0 point 177 feet ) right bracket + left bracket gamma subscript DC times ( DL subscript stem ) times ( 0 point 625 feet ) right bracket + left bracket gamma subscript EV times DL subscript earth times ( minus 3 point 125 feet ) right bracket + ( gamma subscript DC times R subscript DCtot times 1 subscript 75 feet ) + ( gamma subscript DW times R subscript DWtot times 1 point 75 feet ) + ( gamma subscript LL times R subscript LLmin1 times 1 point 75 feet ) + ( gamma subsrcipt EH times R subscript EHftg times 8 point 17 feet ) + ( gamma subscript LS times R subscript LSftg times 12 point 25 feet ) + ( gamma subscript TU times H subscript ufalltot times 17 point 5 feet )

                                  Formula: M subscript lonftgstrImin = 103 point 16 Kips foot per foot

The factored moment about the bridge longitudinal axis at the bottom of footing is:

                                  Formula: M subscript traftgstrImin = 0 Kips per foot       The load factors for the loads that produce transverse horizontal forces are zero for Strength I.

Abutment bottom of footing Strength III force effects:

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Load Combinations

There are numerous load factor combinations for each limit state as can be seen from STables 3.4.1-1 and 3.4.1-2. It is possible to check one limit state, such as Strength I, over and over again using many different load factor combinations to obtain the controlling factored effects. The engineer should use engineering judgement when selecting the most appropriate load factor for each individual load within a limit state.

For the Strength III force effects below, the horizontal earth load is factored by the maximum load factor while the vertical earth load is factored by the minimum factor to maximize the overturning moment.

S3.4.1

The following load factors will be used to calculate the force effects for Strength III:

                    Formula: gamma subscript DC = 1 point 25

STable 3.4.1-2

                    Formula: gamma subscript DW = 1 point 50

STable 3.4.1-2

                    Formula: gamma subscript EH = 1 point 50

STable 3.4.1-2

                    Formula: gamma subscript EV = 1 point 00       use minimum value to maximize the longitudinal moment

STable 3.4.1-2

                    Formula: gamma subscript WS = 1 point 40       use a wind angle of 0 degrees

STable 3.4.1-1

                    Formula: gamma subscript TU = 0 point 50       use contraction temperature force

STable 3.4.1-1

Vertical wind load will be ignored since the moment of inertia about the abutment longitudinal axis is so large.

The factored vertical force at the bottom of footing is:

                            Formula: F subscript vftgstrIII = ( gamma subscript DC times DL subscript bw ) + ( gamma subscript DC times DL subscript stem ) + ( gamma subscript DC times DL subscript ftg ) + ( gamma subscript EV times DL subscript earth ) + ( gamma subscript DC times R subscript DCtot ) + ( gamma subscript DW times R subscript DWtot )

                          Formula: F subscript vftgstrIII = 38 point 72 Kips per foot

The factored longitudinal horizontal force at the bottom of footing is:

                              Formula: F subscript lonftgstrIII = ( gamma subscript EH times R subscript EHftg ) + ( gamma subscript TU times H subscript ufalltot )

                              Formula: F subscript lonftgstrIII = 16 point 40 Kips per foot

The factored transverse horizontal force at the bottom of footing is:

                              Formula: F subscript traftgstrIII = gamma subscript WS times ( numerator (WS subscript supertrans0) divided by denominator (L subscript abut) + numerator (WS subscript subtransend0) divided by denominator (L subscript abut) )

                              Formula: F subscript traftgstrIII = 1 point 01 Kips per foot

The factored moment about the bridge transverse axis at the bottom of footing is:

M subscript lonftgstrIII = left bracket gamma subscript DC times ( DL subscript bw ) times ( minus 0 point 177 feet ) right bracket + left bracket gamma subscript DC times (DL subscript stem ) times 0 point 625 feet right bracket + left bracket gamma subscript EV times DL subscript earth times ( minus 3 point 125 feet ) right bracket + ( gamma subscript DC times R subscript DCtot times 1 point 75 feet) + ( gamma subscript DW times R subscript DWtot times 1 point 75 feet) + ( gamma subscript WS ( numerator( WS subscript sublongend0) divided by denonmiator (L subscript abut) ) times 10 feet right bracket + gamma subscript EH times R subscript EHftg times 8 point 17 feet )+ ( gamma subscript TU times H subscript ufalltot times 17 point 5 feet )

                            Formula: M subscript lonftgstrIII = 128 point 47 Kips foot per foot

The factored moment about the bridge longitudinal axis at the bottom of footing is:

                            Formula: M subscript traftgstrIII = gamma subscript WS times ( numerator (WS subscript supertrans0) divided by denominator (L subscript abut) times 17 point 5 feet ) + gamma subscript WS times ( numerator (WS subscript subtransend0) divided by denominator (L subscript abut) times 10 feet )

                            Formula: M subscript traftgstrIII = 16 point 96 Kips foot per foot

Abutment bottom of footing Strength V force effects:

The following load factors will be used to calculate the force effects for Strength V:

                    Formula: gamma subscript DC = 1 point 25

STable 3.4.1-2

                    Formula: gamma subscript DW = 1 point 50

STable 3.4.1-2

                  Formula: gamma subscript LL = 1 point 35

STable 3.4.1-1

                    Formula: gamma subscript EH = 1 point 50

STable 3.4.1-2

                    Formula: gamma subscript EV = 1 point 00       use minimum value to maximize the longitudinal moment

STable 3.4.1-2

                  Formula: gamma subscript LS = 1 point 35

STable 3.4.1-1

                    Formula: gamma subscript WS = 0 point 40       use a wind angle of 0 degrees

STable 3.4.1-1

                    Formula: gamma subscript WL = 1 point 00

STable 3.4.1-1

                    Formula: gamma subscript TU = 0 point 50       use contraction temperature force

STable 3.4.1-1

The factored vertical force at the bottom of footing is:

                            Formula: F subscript vftgstrV = ( gamma subscript DC times DL subscript bw ) + ( gamma subscript DC times DL subscript stem ) + ( gamma subscript DC times DL subscript ftg ) + ( gamma subscript EV times DL subscript earth ) + ( gamma subscript DC times R subscript DCtot ) + ( gamma subscript DW times R subscript DWtot ) + ( gamma subscript LL times R subscript LLmax1 )

                          Formula: F subscript vftgstrV = 45 point 93 Kips per foot

The factored longitudinal horizontal force at the bottom of footing is:

                              Formula: F subscript lonftgstrV = ( gamma subscript EH times R subscript EHftg ) + ( gamma subscript LS times R subscript LSftg ) + ( gamma subscript WS times numerator (WS subscript sublongend0) divided by denominator (L subscript abut) ) + ( gamma subscript TU times H subscript ufalltot )

                              Formula: F subscript lonftgstrV = 18 point 78 Kips per foot

The factored transverse shear force at the bottom of footing is:

                                Formula: F subscript traftgstrV = gamma subscript WS times ( numerator (WS subscript supertrans0) divided by denominator (L subscript abut) + numerator (WS subscript subtransend0) divided by denominator (L subscript abut) ) + gamma subscript WL times ( numerator (WL subscript trans0) divided by denominator (L subscript abut) )

                                Formula: F subscript traftgstrV = 0 point 42 Kips per foot

The factored moment about the bridge transverse axis at the bottom of footing is:

M subscript IonftgstrV = left bracket gamma subscript DC times ( DL subscript bw ) times ( minus 0 point 177 feet ) right bracket + left bracket gamma subscript DC ( DL subscript stem ) times 0 point 625 feet right bracket + left bracket gamma subscript EV times DL subscript earth times ( minus 3 point 125 feet ) right bracket + (gamma subscript DC times R subscript DCtot times 1 point 75 feet ) + ( gamma subscript DW times R subscript DWtot times 1 point 75 feet ) + ( gamma subscript LL times R subscript LLmax1 times 1 point 75 feet ) + ( gamma subscript WS times numerator (WS subscript subIongend0) divided by denominator (L subscript abut) times 10 feet ) + ( gamma subscript EH times R subscript EHftg times 8 point 17 feet ) + ( gamma subscript LS times R subscript LSftg times 12 point 25 feet ) + ( gamma subscript TU times H subscript ufaIItot times 17 point 5 feet )

                              Formula: M subscript lonftgstrV = 170 point 26 Kips foot per foot

The factored moment about the bridge longitudinal axis at the bottom of footing is:

                              Formula: M subscript traftgstrV = gamma subscript WS times ( numerator (WS subscript supertrans0) divided by denominator (L subscript abut) times 17 point 5 feet ) + gamma subscript WS times ( numerator (WS subscript subtransend0) divided by denominator (L subscript abut) times 10 feet ) + gamma subscript WL times ( numerator (WL subscript trans0) divided by denominator (L subscript abut) times 30 point 5 feet )

                              Formula: M subscript traftgstrV = 8 point 75 Kips foot per foot

Abutment bottom of footing Service I force effects for wind at 0 degrees and maximum live load:

The following load factors will be used to calculate the force effects for Service I:

                    Formula: gamma subscript DC = 1 point 00

STable 3.4.1-2

                    Formula: gamma subscript DW = 1 point 00

STable 3.4.1-2

                  Formula: gamma subscript LL = 1 point 00

STable 3.4.1-1

                    Formula: gamma subscript EH = 1 point 00

STable 3.4.1-2

                    Formula: gamma subscript EV = 1 point 00

STable 3.4.1-2

                  Formula: gamma subscript LS = 1 point 00

STable 3.4.1-1

                    Formula: gamma subscript WS = 0 point 30       use wind at 0 degrees

STable 3.4.1-1

                    Formula: gamma subscript WL = 1 point 00

STable 3.4.1-1

                    Formula: gamma subscript TU = 1 point 00       use contraction temperature force

STable 3.4.1-1

The factored vertical force at the bottom of footing is:

                            Formula: F subscript vftgservI = ( gamma subscript DC times DL subscript bw ) + ( gamma subscript DC times DL subscript stem ) + ( gamma subscript DC times DL subscript ftg ) + ( gamma subscript EV times DL subscript earth ) + ( gamma subscript DC times R subscript DCtot ) + ( gamma subscript DW times R subscript DWtot ) + ( gamma subscript LL times R subscript LLmax1 )

                            Formula: F subscript vftgservI = 38 point 19 Kips per foot

The factored longitudinal shear force at the bottom of footing is:

                                Formula: F subscript lonftgservI = ( gamma subscript EH times R subscript EHftg ) + ( gamma subscript LS times R subscript LSftg ) + ( gamma subscript WS times numerator (WS subscript sublongend0) divided by denominator (L subscript abut) ) + ( gamma subscript TU times H subscript ufalltot )

                                Formula: F subscript lonftgservI = 12 point 96 Kips per foot

The factored transverse shear force at the bottom of footing is:

                              Formula: F subscript traftgservI = gamma subscript WS times ( numerator (WS subscript supertrans0) divided by denominator (L subscript abut) ) + gamma subscript WS times ( numerator (WS subscript subtransend0) divided by denominator (L subscript abut) ) + gamma subscript WL times ( numerator (WL subscript trans0) divided by denominator (L subscript abut) )

                              Formula: F subscript traftgservI = 0 point 34 Kips per foot

The factored moment about the bridge transverse axis at the bottom of footing is:

M subscript IonftgservI = left bracket gamma subscript DC times ( DL subscript bw ) times ( minus 0 point 177 feet ) right bracket + left bracket gamma subscript DC times ( DL subscript stem ) times 0 point 625 feet right bracket + left bracket gamma subscript EV times DL subscript earth ( minus 3 point 125 feet ) right bracket + (gamma subscript DC times R subscript DCtot times 1 point 75 feet ) + ( gamma subscript DW times R subscript DWtot times 1 point 75 feet ) + ( gamma subscript LL times R subscript LLmax1 times 1 point 75 feet ) + ( gamma subscript WS times numerator (WS subscript subIongend0) divided by denominator (L subscript abut) times 10 feet ) + ( gamma subscript EH times R subscript EHftg times 8 point 17 feet ) + ( gamma subscript LS times R subscript LSftg times 12 point 25 feet ) + ( gamma subscript TU times H subscript ufaIItot times 17 point 5 feet )

                                Formula: M subscript lonftgservI = 113 point 12 Kips foot per foot

The factored moment about the bridge longitudinal axis at the bottom of footing is:

                                Formula: M subscript traftgservI = gamma subscript WS times ( numerator (WS subscript supertrans0) divided by denominator (L subscript abut) times 17 point 5 feet ) + gamma subscript WS times ( numerator (WS subscript subtransend0) divided by denominator (L subscript abut) times 10 feet ) + gamma subscript WL times ( numerator (WL subscript trans0) divided by denominator (L subscript abut) times 30 point 5 feet )

                                Formula: M subscript traftgservI = 7 point 54 Kips foot per foot

Abutment bottom of footing Service I force effects for wind at 60 degrees and minimum live load:

The following load factors will be used to calculate the force effects for Service I:

                    Formula: gamma subscript DC = 1 point 00

STable 3.4.1-2

                    Formula: gamma subscript DW = 1 point 00

STable 3.4.1-2

                  Formula: gamma subscript LL = 1 point 00

STable 3.4.1-1

                    Formula: gamma subscript EH = 1 point 00

STable 3.4.1-2

                    Formula: gamma subscript EV = 1 point 00

STable 3.4.1-2

                  Formula: gamma subscript LS = 1 point 00

STable 3.4.1-1

                    Formula: gamma subscript WS = 0 point 30       use for wind on stem end face for wind at 60 degrees

STable 3.4.1-1

                    Formula: gamma subscript WL = 1 point 00       only applicable for wind angle of 0 degrees

STable 3.4.1-1

                    Formula: gamma subscript TU = 1 point 00       use contraction temperature force

STable 3.4.1-1

The factored vertical force at the bottom of footing is:

                                  Formula: F subscript vftgservImin = ( gamma subscript DC times DL subscript bw ) + ( gamma subscript DC times DL subscript stem ) + ( gamma subscript DC times DL subscript ftg ) + ( gamma subscript EV times DL subscript earth ) + ( gamma subscript DC times R subscript DCtot ) + ( gamma subscript DW times R subscript DWtot ) + ( gamma subscript LL times R subscript LLmin1 )

                                  Formula: F subscript vftgservImin = 32 point 17 Kips per foot

The factored longitudinal shear force at the bottom of footing is:

                                      Formula: F subscript lonftgservImin = ( gamma subscript EH times R subscript EHftg ) + ( gamma subscript LS times R subscript LSftg ) + ( gamma subscript WS times numerator (WS subscript sublongend60) divided by denominator (L subscript abut) ) + ( gamma subscript TU times H subscript ufalltot )

                                    Formula: F subscript lonftgservImin = 12 point 97 Kips per foot

The factored transverse shear force at the bottom of footing is:

                                    Formula: F subscript traftgservImin = gamma subscript WS times ( numerator ( WS subscript supertrans60 ) divided by denominator (L subscript abut) + numerator ( WS subscript subtransend60 ) divided by denominator ( L subscript abut) )subscript abut | d60 | )

                                    Formula: F subscript traftgservImin = 0 point 08 Kips per foot

The factored moment about the bridge transverse axis at the bottom of footing is:

M subscript IonftgservImin = left bracket gamma subscript DC times ( DL subscript bw ) times ( minus 0 point 177 feet ) + left bracket gamma subscript DC times ( DL subscript stem ) times 0 point 625 feet right bracket + left bracket gamma subscript EV times DL subscript earth times ( minus 3 point 125 feet ) right bracket + ( gamma subscript DC times R subscript DCtot times 1 point 75 feet ) + ( gamma subscript DW times R subscript DWtot times 1 point 75 feet ) + ( gamma subscript LL times R subscript LLmax1 times 1 point 75 feet ) + ( gamma subscript WS times numerator (WS subscript subIongend60) divided by denominator (L subscript abut) times 10 feet ) + ( gamma subscript EH times R subscript EHftg times 8 point 17 feet ) + ( gamma subscript LS times R subscript LSftg times 12 point 25 feet ) + ( gamma subscript TU times H subscript ufaIItot times 17 point 5 feet )

                                      Formula: M subscript lonftgservImin = 113 point 29 Kips foot per foot

The factored moment about the bridge longitudinal axis at the bottom of footing is:

                                      Formula: M subscript traftgservImin = gamma subscript WS times ( numerator (WS subscript supertrans60) divided by denominator (L subscript abut) times 17 point 5 feet ) + gamma subscript WS times ( numerator (WS subscript sublongend60) divided by denominator (L subscript abut) times 10 feet )

                                      Formula: M subscript traftgservImin = 1 point 34 Kips foot per foot

The following table summarizes the combined forces at the bottom of footing that were calculated above. The forces were calculated at the center of the bottom of footing. The values shown in the table were multiplied by the abutment length to obtain the total effect. These forces are required for the geotechnical engineer to design the pile foundation. It should be noted that Design Step P was based on preliminary pile foundation design forces. In an actual design, the geotechnical engineer would need to revisit the pile foundation design calculations and update the results based on the final design bottom of footing forces given below.

Limit State Vertical Force
(K)
Long. Moment
(K-ft)
Trans. Moment
(K-ft)
Lateral Load
(Long. Direction)
(K)
Lateral Load
(Trans. Direction)
(K)
Strength I Max/Final 2426 8020 0 913 0
Strength I Min/Final 1366 4836 0 610 0
Strength III Max/Final 1815 6022 795 769 47
Service I Max/Final 1790 5302 353 607 16
Service I Min/Final 1508 5310 63 608 4

Table 7-6 Pile Foundation Design Forces

Design Step 7.8 - Check Stability and Safety Requirements

For abutment footings supported by piles, the stability and safety requirements deal with the amount of settlement that will occur to the substructure. For this design example, 1.5 inches of horizontal movement is acceptable and 0.5 inches of vertical settlement is acceptable. Design Step P verifies that less than the allowable horizontal and vertical displacements will take place using the pile size and layout described in Design Step P.

S10.7.2.2 & C11.5.2

Design Step 7.9 - Design Abutment Backwall

It is recommended that Pier Design Step 8.8 is reviewed prior to beginning the abutment design. Design Step 8.8 reviews the design philosophy used to design the structural components of the pier and is applicable for the abutment as well.

Design for flexure:

Assume #5 bars:

                              Formula: bar_diam = 0 point 625 inches

                            Formula: bar_area = 0 point 31 inches squared

First, the minimum reinforcement requirements will be calculated. The tensile reinforcement provided must be enough to develop a factored flexural resistance at least equal to the lesser of 1.2 times the cracking strength or 1.33 times the factored moment from the applicable strength load combinations.

S5.7.3.3.2

The cracking strength is calculated by:

SEquation

5.7.3.6.2-2

                  Formula: M subscript cr = numerator (f subscript r times I subscript g) divided by denominator (y subscript t) . Equation not used

Cross section of abutment backwall showing the cracking moment dimensions. The front face of the backwall is at the top of the cross section and the rear face is at the bottom of the cross section. The cross section is 1 foot and 0 inches wide by 1 foot and 8 inches in length. The centroidal axis is located 10 inches from the front face. The centroid of the reinforcing steel is 2 and thirteen-sixteenths inches from the rear face of the backwall. The reinforcing steel consists of number 5 bars at 8 point 0 inches spacing.

Figure 7-11 Abutment Backwall Cracking Moment Dimensions

              Formula: f subscript r = 0 point 24 times square root of (f prime subscript c)

S5.4.2.6

              Formula: f subscript r = 0 point 48ksi

                Formula: I subscript g = numerator (1) divided by denominator (12) ( 12 inches ) ( 20 inches ches ) cubed

                Formula: I subscript g = 8000 inches superscript 4

                Formula: y subscript t = 10 inches

                  Formula: M subscript cr = numerator ( numerator (f subscript r times I subscript g) divided by denominator (y subscript t)) divided by denominator ( feet )

                  Formula: M subscript cr = 32 point 00 Kips foot per foot

                          Formula: 1 point 2 times M subscript cr = 38 point 40 Kips foot per foot

1.33 times the factored controlling backwall moment is:

                            Formula: M subscript ubwmax = 14 point 38 Kips foot per foot

                                      Formula: 1 point 33 times M subscript ubwmax = 19 point 13 Kips foot per foot

Since 1.33 times the controlling factored backwall moment controls the minimum reinforcement requirements, use:

                            Formula: M subscript ubwdes = 1 point 33 times M subscript ubwmax

                            Formula: M subscript ubwdes = 19 point 13 Kips foot per foot

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

                  Formula: t subscript bw = 20 inches       backwall thickness

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

                Formula: d subscript e = 17 point 19 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 (M subscript ubwdes times 12 inches ) divided by denominator (( phi subscript f times b times d subscript e squared ))       Formula: Rn = 0 point 072 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 00121

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 25 square inches per foot

                                                                    Formula: numerator (bar_area) divided by denominator (A subscript s) = 14 point 9 inches

Required bar spacing =

Use #5 bars @       Formula: bar_space = 9 point 0 inches

                  Formula: A subscript s = bar_area times ( numerator (12 inches ) divided by denominator (bar_space) )       Formula: A subscript s = 0 point 41 inches squared       per foot

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

S5.7.3.3.1

              Formula: T = A subscript s times f subscript y       Formula: T = 24 point 80 K

              Formula: a = numerator (T) divided by denominator (0 point 85 times f prime subscript c times b)       Formula: a = 0 point 61 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 72 inches

S5.7.2.2

                  Formula: numerator (c) divided by denominator (d subscript e) = 0 point 04       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 04 less than or equal to 0 point 42       OK

The backwall flexure reinforcement bar spacing was set at 9.0 inches so that it could lap with the flexure reinforcement in the stem. Originally, the backwall bars were set at 12.0 inches. After completing the stem design, the backwall design was updated to match the stem flexure reinforcement bar spacing.

Check crack control:

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

S5.7.3.4

Since this design example assumes that the backwall will be exposed to deicing salts, use:       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 = 2 point 5 inches + numerator (bar_diam) divided by denominator (2)

                                                                                              Formula: d subscript c = 2 point 81 inches

use       Formula: d subscript c = 2 point 0 inches + numerator (bar_diam) divided by denominator (2)

                                                                                              Formula: d subscript c = 2 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 = 41 point 63 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 = 28 point 37 ksi       Formula: 0 point 6f subscript y = 36 point 00 ksi

Use       Formula: f subscript sa = 28 point 37ksi

                  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 backwall total load moment:

                              Formula: M subscript ubwservI = 8 point 51 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 = 17 point 19 inches       Formula: A subscript s = 0 point 413 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 002

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

              Formula: k = 0 point 164

                    Formula: k times d subscript e = 2 point 81 inches

Cross section of abutment backwall showing the dimensions for the crack control check. The front face of backwall is at the top and the rear face is at the bottom. The cross section width is 12 point 0 inches. The backwall length is 20 point 0 inches. The distance from the front face of backwall to the neutral axis is 2 point 81 inches. The distance from the rear face of backwall to the centroid of the vertical backwall reinforcement is 2 point 81 inches. The distance from the neutral axis to the centroid of vertical reinforcement is 14 point 37 inches.

Figure 7-12 Abutment Backwall Crack Control Check

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

                Formula: d subscript e = 17 point 19 inches

                  Formula: A subscript s = 0 point 413 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 = 771 point 73 numerator ( inches superscript 4) divided by denominator ( feet )

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

              Formula: y = d subscript e minus k times d subscript e       Formula: y = 14 point 37 inches

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

                Formula: f subscript s = 15 point 22 ksi       Formula: f subscript sa greater than f subscript s       OK

Design for shear:

The factored longitudinal shear force at the base of the backwall is:

                            Formula: V subscript ubwmax = 2 point 91 Kips per foot

The nominal shear resistance is the lesser of:

S5.8.3.3

                    Formula: V subscript n1 = V subscript c + V subscript s. Equation not used

or

                    Formula: V subscript n2 = 0 point 25 times f prime subscript c times b subscript v times d subscript v. Equation not used

where:

                  Formula: V subscript c = 0 point 0316 times beta times square root of (f prime subscript c) times b subscript v times d subscript v. Equation not used

and

                  Formula: V subscript s = numerator (A subscript v times f subscript y times d subscript v times ( cot theta + cot alpha ) times sin alpha ) divided by denominator (s) . Equation not used       neglect for this abutment design

Before the nominal shear resistance can be calculated, all the variables used in the above equations need to be defined.

                Formula: beta = 2 point 0

S5.8.3.4.1

                Formula: b subscript v = 12inches

                Formula: d subscript v = max ( d subscript e minus numerator (a) divided by denominator (2) , 0 point 9 times d subscript e , 0 point 72 times h ). Equation not used

S5.8.2.9

where:

              Formula: h = 20inches

                Formula: d subscript v = 16 point 80inches

Now, Vn1 and Vn2 can be calculated:

For       Formula: f prime subscript c = 4 point 0ksi

                    Formula: V subscript n1 = 0 point 0316 times beta times square root of (f prime subscript c) times b subscript v times d subscript v

                    Formula: V subscript n1 = 25 point 48       Kips per foot

                    Formula: V subscript n2 = 0 point 25 times f prime subscript c times b subscript v times d subscript v

                    Formula: V subscript n2 = 201 point 60       Kips per foot

Use:       Formula: V subscript n = 25 point 48 Kips per foot

The factored shear resistance is then:

                  Formula: phi subscript v = 0 point 90

S5.5.4.2.1

                Formula: V subscript r = phi subscript v times V subscript n

                Formula: V subscript r = 22 point 93 Kips per foot

                Formula: V subscript r greater than V subscript ubwmax       OK

Shrinkage and temperature reinforcement:

S5.10.8

For members less than 48.0 inches thick, the area of reinforcement in each direction shall not be spaced greater than 12.0 inches and satisfy the lesser of:

S5.10.8.2

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

                      or

                    Formula: Sigma A subscript b = 0 point 0015A subscript g. Equation not used

                  Formula: A subscript g = ( 20 inches ) times ( 12 inches per foot )       Formula: A subscript g = 240 point 0       square inches per foot

                Formula: f subscript y = 60ksi

                            Formula: 0 point 11 times numerator (A subscript g) divided by denominator (f subscript y) = 0 point 44       square inches per foot

                      or

                              Formula: 0 point 0015A subscript g = 0 point 36       square inches per foot

As must be greater than or equal to 0.36in2/ft

The above steel must be distributed equally on both faces of the backwall.

Try 1 horizontal # 4 bar for each face of the backwall at 12.0 inch spacing:

                              Formula: bar_diam = 0 point 500 inches

                            Formula: bar_area = 0 point 20 inches squared

                  Formula: A subscript s = 2 times numerator (bar_area) divided by denominator ( feet )       Formula: A subscript s = 0 point 40 square inches per foot

                          Formula: 0 point 40 square inches per foot greater than or equal to 0 point 36 square inches per foot       OK

Based on the backwall design, #5 bars at 9.0 inch spacing will be used for the back face flexure reinforcement. The same bar size and spacing will be used for the front face vertical reinforcement. The horizontal temperature and shrinkage reinforcement will consist of #4 bars at 12.0 inch spacing for the front and back faces.

Design Step 7.10 - Design Abutment Stem

Design for flexure:

Assume #9 bars:

                              Formula: bar_diam = 1 point 128 inches

                            Formula: bar_area = 1 point 00 inches squared

                Formula: f subscript y = 60ksi

As with the backwall, the minimum reinforcement requirements will be calculated for the stem. The tensile reinforcement provided must be enough to develop a factored flexural resistance at least equal to the lesser of 1.2 times the cracking strength or 1.33 times the factored moment from the applicable strength load combinations.

S5.7.3.3.2

The cracking strength is calculated by:

SEquation

5.7.3.6.2-2

                  Formula: M subscript cr = numerator (f subscript r times I subscript g) divided by denominator (y subscript t) . Equation not used

Cross section of abutment stem showing the cracking moment dimensions. The front face of the stem is at the top of the cross section and the rear face is at the bottom of the cross section. The cross section is 1 foot and 0 inches wide by 3 feet 6 inches in length. The centroidal axis is located 1 foot 9 inches from the front face. The centroid of the vertical reinforcing steel is 3 and one sixteenth inches from the rear face of the stem. The reinforcing steel consists of number 9 bars at 9 point 0 inches spacing.

Figure 7-13 Abutment stem Cracking Moment Dimensions

              Formula: f subscript r = 0 point 24 times square root of (f prime subscript c)

S5.4.2.6

              Formula: f subscript r = 0 point 48ksi

                Formula: I subscript g = numerator (1) divided by denominator (12) ( 12 inches ) ( 42 inches ches ) cubed

                Formula: I subscript g = 74088 inches superscript 4

                Formula: y subscript t = 21 inches

                  Formula: M subscript cr = numerator ( numerator (f subscript r times I subscript g) divided by denominator (y subscript t)) divided by denominator ( feet )

                  Formula: M subscript cr = 141 point 12 Kips foot per foot

                          Formula: 1 point 2 times M subscript cr = 169 point 34 Kips foot per foot

1.33 times the factored controlling stem moment is:

                                Formula: M subscript ustemmax = 156 point 61 Kips foot per foot

                                        Formula: 1 point 33 times M subscript ustemmax = 208 point 29 Kips foot per foot

1.2 times the cracking moment controls the minimum reinforcement requirements. 1.2 times the cracking moment is also greater than the controlling applied factored moment, therefore, use 1.2 times the cracking moment for design.

                                Formula: M subscript ustemdes = 1 point 2 times M subscript cr

                              Formula: M subscript ustemdes = 169 point 34 Kips foot per foot

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

thickness of stem:       Formula: t subscript stem = 42 inches

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

                Formula: d subscript e = 38 point 94 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: f prime subscript c = 4 point 0ksi

                  Formula: Rn = numerator (M subscript ustemdes times 12 inches ) divided by denominator (( phi subscript f times b times d subscript e squared ))       Formula: Rn = 0 point 124 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 00211

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 98 square inches per foot

Required bar spacing = Formula: numerator (bar_area) divided by denominator (A subscript s) = 12 point 2 inches

Use #9 bars @       Formula: bar_space = 9 point 0 inches

                  Formula: A subscript s = bar_area times ( numerator (12 inches ) divided by denominator (bar_space) )       Formula: A subscript s = 1 point 33 inches squared       per foot

Now, the maximum reinforcement limit must be checked. This check could be skipped since the calculated factored design moment is less than 1.2 times the cracking moment.

S5.7.3.3.1

              Formula: T = A subscript s times f subscript y       Formula: T = 80 point 00 K

              Formula: a = numerator (T) divided by denominator (0 point 85 times f prime subscript c times b)       Formula: a = 1 point 96 inches

                  Formula: beta subscript 1 = 0 point 85

S5.7.2.2

              Formula: c = numerator (a) divided by denominator ( beta subscript 1)       Formula: c = 2 point 31 inches

S5.7.2.2

                  Formula: numerator (c) divided by denominator (d subscript e) = 0 point 06       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 06 less than or equal to 0 point 42       OK

Check crack control:

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

S5.7.3.4

Since this design example assumes that the abutment stem will be exposed to deicing salts, use:       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 = 2 point 5 inches + numerator (bar_diam) divided by denominator (2)

                                                                                              Formula: d subscript c = 3 point 06 inches

use       Formula: d subscript c = 2 point 0 inches + numerator (bar_diam) divided by denominator (2)

                                                                                              Formula: d subscript c = 2 point 56 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 = 46 point 15 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 = 26 point 48 ksi       Formula: 0 point 6f subscript y = 36 point 00 ksi

Use       Formula: f subscript sa = 26 point 48ksi

                  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

Stem factored service moment:

                                Formula: M subscript ustemservI = 105 point 82 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 = 38 point 94 inches       Formula: A subscript s = 1 point 33 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 00285

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

              Formula: k = 0 point 192

                    Formula: k times d subscript e = 7 point 47 inches

Cross section of abutment stem showing the dimensions for the crack control check. The front face of abutment stem is at the top and the rear face is at the bottom. The cross section width is 12 inches. The abutment stem length is 42 point 0 inches. The distance from the front face of stem to the neutral axis is 7 point 47 inches. The distance from the rear face of stem to the centroid of the vertical stem reinforcement is 3 point 06 inches. The distance from the neutral axis to the centroid of vertical reinforcement is 31 point 47 inches. The vertical reinforcement consists of number 9 bars at 9 point 0 inches spacing.

Figure 7-14 Abutment Stem Crack Control Check

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

                Formula: d subscript e = 38 point 94 inches

                  Formula: A subscript s = 1 point 330 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 = 12202 point 09 numerator ( inches superscript 4) divided by denominator ( feet )

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

              Formula: y = d subscript e minus k times d subscript e       Formula: y = 31 point 47 inches

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

                Formula: f subscript s = 26 point 20 ksi       Formula: f subscript sa greater than f subscript s       OK

Design for shear:

The factored longitudinal shear force at the base of the stem is:

                              Formula: V subscript ustemmax = 16 point 03 Kips per foot

The nominal shear resistance is the lesser of:

S5.8.3.3

                    Formula: V subscript n1 = V subscript c + V subscript s. Equation not used

or

                    Formula: V subscript n2 = 0 point 25 times f prime subscript c times b subscript v times d subscript v. Equation not used

where:

                  Formula: V subscript c = 0 point 0316 times beta times square root of (f prime subscript c) times b subscript v times d subscript v. Equation not used

and

                  Formula: V subscript s = numerator (A subscript v times f subscript y times d subscript v times ( cot theta + cot alpha ) times sin alpha ) divided by denominator (s) . Equation not used       neglect for this abutment design

Before the nominal shear resistance can be calculated, all the variables used in the above equations need to be defined.

                Formula: beta = 2 point 0

S5.8.3.4.1

                Formula: b subscript v = 12inches

                Formula: d subscript v = max ( d subscript e minus numerator (a) divided by denominator (2) , 0 point 9 times d subscript e , 0 point 72 times h ). Equation not used

S5.8.2.9

where:

              Formula: h = 42inches

                Formula: d subscript v = 38 point 2inches

Now, Vn1 and Vn2 can be calculated:

For       Formula: f prime subscript c = 4 point 0ksi

                    Formula: V subscript n1 = 0 point 0316 times beta times square root of (f prime subscript c) times b subscript v times d subscript v

                    Formula: V subscript n1 = 57 point 94       Kips per foot

                    Formula: V subscript n2 = 0 point 25 times f prime subscript c times b subscript v times d subscript v

                    Formula: V subscript n2 = 458 point 40       Kips per foot

use       Formula: V subscript n = 57 point 94 Kips per foot

The factored shear resistance is then:

                  Formula: phi subscript v = 0 point 90

S5.5.4.2.1

                Formula: V subscript r = phi subscript v times V subscript n

                Formula: V subscript r = 52 point 15 Kips per foot

                Formula: V subscript r greater than V subscript ustemmax       OK

Shrinkage and temperature reinforcement:

S5.10.8

For members less than 48.0 inches thick, the area of reinforcement in each direction shall not be spaced greater than 12.0 inches and satisfy the lesser of:

S5.10.8.2

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

                      or

                    Formula: Sigma A subscript b = 0 point 0015A subscript g. Equation not used

                  Formula: A subscript g = ( 42 inches ) times ( 12 inches per foot )       Formula: A subscript g = 504 point 0       square inches per foot

                Formula: f subscript y = 60ksi

                            Formula: 0 point 11 times numerator (A subscript g) divided by denominator (f subscript y) = 0 point 92       square inches per foot

                      or

                              Formula: 0 point 0015A subscript g = 0 point 76       square inches per foot

As must be greater than or equal to 0.76in2/ft

The above steel must be distributed equally on both faces of the stem.

Try 1 horizontal # 5 bar for each face of the stem at 9.0 inch spacing:

                              Formula: bar_diam = 0 point 625 inches

                            Formula: bar_area = 0 point 31 inches squared

                  Formula: A subscript s = 2 times numerator (bar_area times ( numerator (12 inches ) divided by denominator (9 inches ) )) divided by denominator ( feet )       Formula: A subscript s = 0 point 83 square inches per foot

                          Formula: 0 point 83 square inches per foot greater than or equal to 0 point 76 square inches per foot       OK

Based on the abutment stem design, #9 bars at 9.0 inch spacing will be used for the back face flexure reinforcement. The same bar size and spacing will be used for the front face vertical reinforcement to reduce design steps. The horizontal temperature and shrinkage reinforcement will consist of #5 bars at 9.0 inch spacing for the front and back faces.

Design Step 7.11 - Design Abutment Footing

The abutment footing is designed for flexure in the heel and toe, one-way and two-way shear action, and the control of cracking by the distribution of reinforcement. For footings supported by pile foundations, the footing and pile foundation designs are interdependent and should be designed concurrently to be more efficient. Refer to Design Step P for the pile foundation design.

S5.13.3

S5.7.3.4

The following figures show the assumed footing dimensions and pile locations within the footing.

Plan view of abutment footing showing footing dimensions and pile layout. The footing and pile group longitudinal axis and the footing length is parallel to the girders or stations ahead direction. The footing and pile group transverse axis and the footing width is perpendicular to the girders. The footing length is 10 feet 3 inches. The footing width is 46 feet 10 and one half inches. There are two rows of piles. The centroidal axis of the back row is 1 foot 3 inches from the heel back edge. The centroidal axis of the front row is 1 foot 3 inches measured along the footing bottom. The distance between centroidal axes of the front and back row of piles is 7 feet 9 inches measured along the bottom of the footing. The distance from the left or right edge of footing to the centroidal axis of the nearest pile in the front and back row is 1 foot 5 and one quarter inches. The pile spacing within each row is 7 feet 4 inches. There are 7 piles per row. The strong axis of the piles are oriented parallel with the longitudinal direction of the footing. The piles are numbered from back row to ahead row from the right side of footing to left. The back rightmost pile is number 1. The front rightmost pile is number 2. The back second from the rightmost pile is number three. The front second from rightmost pile is pile number 4. The pile numbering sequence continues for the entire pile pattern until pile 14, which is the front leftmost pile. The distance from the back of the footing to the transverse axis of the pile group is 5 feet 1 and one half inches. The distance from the back of the footing to the abutment stem centerline transverse axis is 5 feet 9 inches. The distance from the right edge of footing to the longitudinal pile and footing axis is 23 feet 5 and one quarter inches. The back row of piles are vertical and the front row of piles are battered toward to toe.

Figure 7-15 Abutment Footing and Pile Foundation Plan View

Elevation view of abutment footing and pile foundation layout. The abutment footing heel is to the left and the toe is to the right. The abutment footing heel width is 4 feet 0 inches from the left face of the heel to the back face of the abutment stem. The abutment stem width is 3 feet 6 inches wide. The abutment footing toe width is 2 feet 9 inches measured from the front face of stem to the front face of the toe. The footing thickness is 2 feet 6 inches. The entire footing width is 10 feet 3 inches. There are two rows of piles with the front row battered at a 3 vertical to 1 horizontal batter. The centroidal axis of the back row is 1 foot 3 inches from the heel edge. The distance from the front edge of footing to the centroidal axis of the front row of piles is 1 foot 3 inches measured along the bottom of footing.

Figure 7-16 Abutment Footing and Pile Foundation Elevation View

Design for flexure:

The flexure reinforcement must be designed at two critical sections for abutment footings. The two sections include the back and front face of the stem. The moments at the abutment faces are calculated from the pile reactions.

S5.13.3.4

For the abutment front face, the following moment arm will be used:

Elevation view of abutment footing showing the abutment toe critical flexure section location. The footing toe is to the right. The toe critical flexure section is located along the front face of abutment stem. The distance from the abutment front face to the centroidal axis of the front row of piles measured along the bottom of footing is 1 foot 6 inches. The front row of piles is battered towards the front of toe.

Figure 7-17 Abutment Toe Critical Flexure Section

The controlling moment on the critical section occurs when the pile loads on the front row of piles are maximized. From Tables P-17 to P-20, the front row pile loads are maximized for Strength I using the maximum load factors at the final construction condition and are summarized below.

                  Formula: P subscript 2 = 315 point 3K       Formula: P subscript 10 = 335 point 4K

                  Formula: P subscript 4 = 330 point 5K       Formula: P subscript 12 = 330 point 7K

                  Formula: P subscript 6 = 336 point 1K       Formula: P subscript 14 = 315 point 6K

                  Formula: P subscript 8 = 339 point 9K

Since the above pile loads are already factored, no load factors need to be applied and the total factored moment is as follows:

                      Formula: M subscript utoe = 1 point 5 feet times ( P subscript 2 + P subscript 4 + P subscript 6 + P subscript 8 + P subscript 10 + P subscript 12 + P subscript 14 )

                      Formula: M subscript utoe = 3455 point 25 K feet

The moment on a per foot basis is then:

                          Formula: M subscript utoeft = numerator (M subscript utoe) divided by denominator (L subscript abut)

                        Formula: M subscript utoeft = 73 point 71 K times numerator ( feet ) divided by denominator ( feet )

Once the maximum moment at the critical section is known, the same procedure that was used for the backwall and stem to calculate the flexure reinforcement must be followed. The footing toe flexure reinforcement is located longitudinally in the bottom of the footing since the bottom of footing is in tension at the critical toe section. These bars will extend from the back of the heel to the front of the toe taking into account the clear cover:

Assume #8 bars:

                              Formula: bar_diam = 1 point 000 inches

                            Formula: bar_area = 0 point 79 inches squared

                Formula: f subscript y = 60ksi

The footing toe critical section minimum tensile reinforcement requirements will be calculated. The tensile reinforcement provided must be enough to develop a factored flexural resistance at least equal to the lesser of 1.2 times the cracking strength or 1.33 times the factored moment from the applicable strength load combinations.

S5.7.3.3.2

The cracking strength is calculated by:

SEquation

5.7.3.6.2-2

                  Formula: M subscript cr = numerator (f subscript r times I subscript g) divided by denominator (y subscript t) . Equation not used

Cross section of abutment footing toe showing the cracking moment dimensions. The top of the abutment footing is at the top of the cross section and the bottom of footing is at the bottom of the cross section. The cross section is 1 foot and 0 inches wide by 2 feet 6 inches in depth. The centroidal axis is located 1 foot 3 inches from the top of footing. The centroid of the flexure reinforcing steel is 3 and one half inches from the bottom of the footing. The flexure reinforcing steel consists of number 8 bars at 12 point 0 inches spacing.

Figure 7-18 Abutment Footing Toe Cracking Moment Dimensions

              Formula: f subscript r = 0 point 24 times square root of (f prime subscript c)

S5.4.2.6

              Formula: f subscript r = 0 point 48ksi

                Formula: I subscript g = numerator (1) divided by denominator (12) ( 12 inches ) ( 30 inches ches ) cubed

                Formula: I subscript g = 27000 inches superscript 4

                Formula: y subscript t = 15 inches

                  Formula: M subscript cr = numerator ( numerator (f subscript r times I subscript g) divided by denominator (y subscript t)) divided by denominator ( feet )

                  Formula: M subscript cr = 72 point 00 Kips foot per foot

                          Formula: 1 point 2 times M subscript cr = 86 point 40 Kips foot per foot

1.33 times the factored controlling stem moment is:

                        Formula: M subscript utoeft = 73 point 71 Kips foot per foot

                                  Formula: 1 point 33 times M subscript utoeft = 98 point 04 Kips foot per foot

1.2 times the cracking moment controls the minimum reinforcement requirements. 1.2 times the cracking moment is also greater than the factored footing toe moment. Therefore, use 1.2 times the cracking moment to design the toe flexure reinforcement.

                              Formula: M subscript ufttoedes = 1 point 2 times M subscript cr

                              Formula: M subscript ufttoedes = 86 point 40 Kips foot per foot

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

thickness of footing:       Formula: t subscript ftg = 30 inches

                          Formula: Cover subscript fb = 3 point 00 inches

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

                Formula: d subscript e = 26 point 50 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: f prime subscript c = 4 point 0ksi

                  Formula: Rn = numerator (M subscript ufttoedes times 12 inches ) divided by denominator (( phi subscript f times b times d subscript e squared ))       Formula: Rn = 0 point 137 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 00233

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 74 square inches per foot

Required bar spacing = Formula: numerator (bar_area) divided by denominator (A subscript s) = 12 point 8 inches

Use #8 bars @       Formula: bar_space = 12 point 0 inches

                  Formula: A subscript s = bar_area times ( numerator (12 inches ) divided by denominator (bar_space) )       Formula: A subscript s = 0 point 79 inches squared       per foot

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

S5.7.3.3.1

              Formula: T = A subscript s times f subscript y       Formula: T = 47 point 40 K

              Formula: a = numerator (T) divided by denominator (0 point 85 times f prime subscript c times b)       Formula: a = 1 point 16 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 37 inches

S5.7.2.2

                  Formula: numerator (c) divided by denominator (d subscript e) = 0 point 05       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 05 less than or equal to 0 point 42       OK

Check crack control:

The control of cracking by distribution of reinforcement must be checked for the abutment toe.

S5.7.3.4

Since the footing is buried, moderate exposure will be assumed, use:       Formula: Z = 170 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 = 3 point 0 inches + numerator (bar_diam) divided by denominator (2)

                                                                                              Formula: d subscript c = 3 point 50 inches

use       Formula: d subscript c = 2 point 0 inches + numerator (bar_diam) divided by denominator (2)

                                                                                              Formula: d subscript c = 2 point 50 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 = 60 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 = 32 point 00 ksi       Formula: 0 point 6f subscript y = 36 point 00 ksi

Use       Formula: f subscript sa = 32 point 00ksi

                  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

The pile loads used to compute the controlling footing toe moment for the Service I limit state are again taken from Design Step P, Tables P-17 through P-20.

                  Formula: P subscript 2 = 221 point 5K       Formula: P subscript 10 = 234 point 9K

                  Formula: P subscript 4 = 232 point 2K       Formula: P subscript 12 = 230 point 8K

                  Formula: P subscript 6 = 235 point 9K       Formula: P subscript 14 = 219 point 3K

                  Formula: P subscript 8 = 238 point 6K

The footing toe service moment is then calculated by:

                              Formula: M subscript utoeservI = 1 point 5 feet times ( P subscript 2 + P subscript 4 + P subscript 6 + P subscript 8 + P subscript 10 + P subscript 12 + P subscript 14 )

                              Formula: M subscript utoeservI = 2419 point 80 K feet

The moment on a per foot basis is then:

                                Formula: M subscript utoeftservI = numerator (M subscript utoeservI) divided by denominator (L subscript abut)

                                Formula: M subscript utoeftservI = 51 point 62 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 = 26 point 50 inches       Formula: A subscript s = times superscript 79 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 00248

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

              Formula: k = 0 point 180

                    Formula: k times d subscript e = 4 point 78 inches

Cross section of abutment footing toe showing the dimensions for the crack control check. The top of abutment footing toe is at the top and the bottom of footing toe is at the bottom. The cross section width is 12 inches. The abutment footing toe depth is 30 point 0 inches. The distance from the top of toe to the neutral axis is 4 point 78 inches. The distance from the bottom of footing toe to the centroid of the toe flexure reinforcement is 3 point 50 inches. The distance from the neutral axis to the centroid of flexure reinforcement is 21 point 72 inches. The flexure reinforcement consists of number 8 bars at 12 point 0 inches spacing.

Figure 7-19 Abutment Footing Toe Crack Control Check

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

                Formula: d subscript e = 26 point 50 inches

                  Formula: A subscript s = 0 point 790 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 = 3418 point 37 numerator ( inches superscript 4) divided by denominator ( feet )

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

              Formula: y = d subscript e minus k times d subscript e       Formula: y = 21 point 72 inches

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

                Formula: f subscript s = 31 point 48 ksi       Formula: f subscript sa greater than f subscript s       OK

For the abutment back face flexure design, the following moment arm will be used:

Elevation view of abutment footing showing the abutment heel critical flexure section. The critical flexure section for the heel is located at the back face of the stem. The heel is to the left and the toe is to the right. The distance from the abutment heel left face to the back face of the abutment stem is 4 feet 0 inches. The distance from the heel face to the centroidal axis of the back row of piles is 1 foot 3 inches. The distance from the back row of piles to the critical section for flexure is 2 feet 9 inches. The footing thickness is 2 feet 6 inches. The entire footing width is 10 feet 3 inches.

Figure 7-20 Abutment Heel Critical Flexure Section

The controlling moment on the critical section occurs when the pile loads on the back row of piles are minimized. From Tables P-17 to P-20, the back row pile loads are minimized for Strength I using the minimum load factors at the final construction condition and are summarized below. Piles in tension are shown as having negative pile loads.

                  Formula: P subscript 1 = minus 15 point 3K       Formula: P subscript 9 = minus 14 point 5K

                  Formula: P subscript 3 = minus 14 point 8K       Formula: P subscript 11 = minus 14 point 8K

                  Formula: P subscript 5 = minus 14 point 5K       Formula: P subscript 13 = minus 15 point 3K

                  Formula: P subscript 7 = minus 14 point 4K

Since the above pile loads are already factored, no load factors need to be applied and the total factored moment is as follows:

                        Formula: M subscript uheeI = 2 point 75 feet times ( P subscript 1 + P subscript 3 + P subscript 5 + P subscript 7 + P subscript 9 + P subscript 11 + P subscript 13 )

                        Formula: M subscript uheeI = minus 284 point 90 K feet

The moment on a per foot basis is then:

                          Formula: M subscript uheeIft = numerator (M subscript uheeI) divided by denominator (L subscript abut)

                          Formula: M subscript uheeIft = minus 6 point 08 K times numerator ( feet ) divided by denominator ( feet )

Once the moment at the critical section is known, the same procedure that was used for the toe must be followed. The flexure reinforcement for the footing heel is placed longitudinally along the top of the footing since the top of the footing heel is in tension at the critical heel section. The bars will extend from the back of the heel to the front of the toe taking into account the concrete cover.

Assume #5 bars:

                              Formula: bar_diam = 0 point 625 inches

                            Formula: bar_area = 0 point 31 inches squared

                Formula: f subscript y = 60ksi

The footing heel critical section minimum tensile reinforcement requirements will be calculated. The tensile reinforcement provided must be enough to develop a factored flexural resistance at least equal to the lesser of 1.2 times the cracking strength or 1.33 times the factored moment from the applicable strength load combinations.

S5.7.3.3.2

The cracking strength is calculated by:

SEquation 5.7.3.6.2-2

                  Formula: M subscript cr = numerator (f subscript r times I subscript g) divided by denominator (y subscript t) . Equation not used

Cross section of abutment footing heel showing the cracking moment dimensions. The top of the footing heel is at the top of the cross section and the bottom of the footing heel is at the bottom of the cross section. The cross section is 1 foot and 0 inches wide by 2 feet 6 inches in depth. The centroidal axis is located 1 foot 3 inches from the bottom of the footing. The centroid of the flexure reinforcing steel is 2 and five sixteenths inches from the top of the footing heel. The reinforcing steel consists of number 5 bars at 12 point 0 inches spacing.

Figure 7-21 Abutment Footing Heel Cracking Moment Dimensions

              Formula: f subscript r = 0 point 24 times square root of (f prime subscript c)

S5.4.2.6

              Formula: f subscript r = 0 point 48ksi

                Formula: I subscript g = numerator (1) divided by denominator (12) ( 12 inches ) ( 30 inches ches ) cubed

                Formula: I subscript g = 27000 inches superscript 4

                Formula: y subscript t = 15 inches

                  Formula: M subscript cr = numerator ( numerator (- f subscript r times I subscript g) divided by denominator (y subscript t)) divided by denominator ( feet )

                  Formula: M subscript cr = minus 72 point 00 Kips foot per foot

                          Formula: 1 point 2 times M subscript cr = minus 86 point 40 Kips foot per foot

1.33 times the factored controlling heel moment is:

                                    Formula: 1 point 33 times M subscript uheeIft = minus 8 point 08 Kips foot per foot

1.33 times the factored controlling heel moment controls the minimum reinforcement requirements. Use 1.33 times the factored controlling heel moment to design the heel flexure reinforcement.

                                Formula: M subscript uftheeldes = 1 point 33 times ( minus M subscript uheeIft )

                                Formula: M subscript uftheeldes = 8 point 08 Kips foot per foot

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

thickness of footing:       Formula: t subscript ftg = 30 inches

                          Formula: Cover subscript feet = 2 point 00 inches

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

                Formula: d subscript e = 27 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: f prime subscript c = 4 point 0ksi

                  Formula: Rn = numerator (M subscript uftheeldes times 12 inches ) divided by denominator (( phi subscript f times b times d subscript e squared ))       Formula: Rn = 0 point 012 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 00020

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 06 square inches per foot

Required bar spacing = Formula: numerator (bar_area) divided by denominator (A subscript s) = 57 point 2 inches

Use #5 bars @       Formula: bar_space = 12 point 0 inches

                  Formula: A subscript s = bar_area times ( numerator (12 inches ) divided by denominator (bar_space) )       Formula: A subscript s = 0 point 31 inches squared       per foot

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

S5.7.3.3.1

              Formula: T = A subscript s 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 b)       Formula: a = 0 point 46 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 54 inches

S5.7.2.2

                  Formula: numerator (c) divided by denominator (d subscript e) = 0 point 02       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 05 less than or equal to 0 point 42       OK

The crack control check for the footing heel critical section will not be carried out. The calculations are similar to that of the abutment backwall, stem, and footing toe.

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Preliminary Design

A quick way to come up with a design section that will probably work for all design checks is to just check the crack control requirements for LRFD. It has been the designer's experience that in many footing designs, the crack control requirements control the footing design. The above is true for LRFD because LFD allows a certain percentage of overstress for the service cases due to the low probability that the loads combined for each service case will actually occur simultaneously.

Shrinkage and temperature reinforcement:

S5.10.8

For members less than 48.0 inches thick, the area of reinforcement in each direction shall not be spaced greater than 12.0 inches and satisfy the lesser of:

S5.10.8.2

                  Formula: A subscript s greater than or equal to 0 point 11 numerator (A subscript g) divided by denominator (fy) . Equation not used       or       Formula: Sigma A subscript b = 0 point 0015A subscript g. Equation not used

                  Formula: A subscript g = ( 30 inches ) times ( 12 inches per foot )       Formula: A subscript g = 360 point 00       square inches per foot

                Formula: f subscript y = 60ksi

                            Formula: 0 point 11 times numerator (A subscript g) divided by denominator (f subscript y) = 0 point 66       square inches per foot

                      or

                              Formula: 0 point 0015A subscript g = 0 point 54       square inches per foot

The total combined amount of reinforcing steel on the top and bottom transverse faces must be greater than or equal to 0.54 in2/ft.

For one face only:

                      Formula: A subscript stop = numerator (0 point 54 square inches per foot) divided by denominator (2)       Formula: A subscript stop = 0 point 27 square inches per foot

Try 1 # 5 bar at 12.0 inch spacing for one face:

                              Formula: bar_diam = 0 point 625 inches

                            Formula: bar_area = 0 point 31 inches squared

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

                          Formula: 0 point 31 square inches per foot greater than or equal to 0 point 27 square inches per foot       OK

Based on the abutment footing flexure design, #8 bars at 12.0 inch spacing are required for the bottom longitudinal flexure reinforcement. #5 bars at 12.0 inch spacing are required for the top longitudinal flexure reinforcement. In the footing transverse direction, the shrinkage and temperature reinforcement calculations require #5 bars at 12.0 inch spacing for the top and bottom mats.

Design for shear:

S5.13.3.6

Shear design in abutment footings consists of having adequate resistance against one-way action and two-way action. For both one-way and two-way actions, the design shear is taken at a critical section. For abutments, one-way action is checked in the toe and heel. The factored shear force at the critical section is computed by cutting the footing at the critical section and summing the pile loads or portions of pile loads that are outside the critical section. Two-way action in abutment footings supported by piles is generally checked taking a critical perimeter around individual piles or around a group of piles when the critical perimeter of individual piles overlap.

For one way action in the abutment footing toe, the critical section is taken as the larger of:

S5.13.3.6.1 & S5.8.3.2

                      0 point 5 times d subscript v times cot theta       or       d subscript v

              Formula: theta = 45deg

The term dv is calculated the same as it is for the backwall and stem:

                Formula: d subscript v = max ( d subscript e minus numerator (a) divided by denominator (2) , 0 point 9 times d subscript e , 0 point 72 times h ). Equation not used

S5.8.2.9

where:

                  Formula: d subscript e = 26 point 50inches       taken from footing toe strength flexure design

              Formula: a = 1 point 16inches       taken from footing toe strength flexure design

              Formula: h = 30inches

                Formula: d subscript v = 25 point 92

Now the critical section can be calculated:

                                    Formula: 0 point 5 times d subscript v times cot ( theta ) = 12 point 96inches       or       Formula: d subscript v = 25 point 92inches

use       Formula: d subscript v = 25 point 92inches

Elevation view showing the abutment footing toe one way action critical section. The toe is to the right. The distance from the front face of abutment stem to the face of the abutment toe is 2 feet 9 inches. The distance from the front face of toe to the centroidal axis of the front row of piles is 1 foot 3 inches measured along the bottom of footing. The front row of piles is battered. The distance from the front face of abutment stem to the abutment toe critical one way shear section is 25 point 92 inches.

Figure 7-22 Abutment Toe One-way Action Critical Section

Since the front row of piles are all inside the critical section, the factored shear outside the critical section is zero and does not have to be checked. However, the manner in which the design shear force would be calculated if the front row of piles were outside the critical section is shown below. Note that this check is not required and does not apply since the front row of piles are all inside the critical section.

The pile loads used to compute the controlling footing toe shear force are for the Strength I limit state using the maximum load factors at the final construction stage. They are taken from Design Step P, Tables P-17 through P-20 and are as follows:

                  Formula: P subscript 2 = 314 point 5K       Formula: P subscript 10 = 335 point 3K

                  Formula: P subscript 4 = 330 point 3K       Formula: P subscript 12 = 330 point 5K

                  Formula: P subscript 6 = 336 point 0K       Formula: P subscript 14 = 314 point 8K

                  Formula: P subscript 8 = 339 point 9K

The factored one-way shear force at the abutment footing toe critical section on a per foot basis is then:

                          Formula: V subscript uftgtoe = numerator (( P subscript 2 + P subscript 4 + P subscript 6 + P subscript 8 + P subscript 10 + P subscript 12 + P subscript 14 )) divided by denominator (L subscript abut)

                          Formula: V subscript uftgtoe = 49 point 09 Kips per foot

The nominal shear resistance is the lesser of:

S5.8.3.3

                    Formula: V subscript n1 = V subscript c + V subscript s. Equation not used

or

                    Formula: V subscript n2 = 0 point 25 times f prime subscript c times b subscript v times d subscript v. Equation not used

where:

                  Formula: V subscript c = 0 point 0316 times beta times square root of (f prime subscript c) times b subscript v times d subscript v. Equation not used

and

                  Formula: V subscript s = numerator (A subscript v times f subscript y times d subscript v times ( cot theta + cot alpha ) times sin alpha ) divided by denominator (s) . Equation not used       neglect for this abutment design

Before the nominal shear resistance can be calculated, all the variables used in the above equations need to be defined.

                Formula: beta = 2 point 0

S5.8.3.4.1

                Formula: b subscript v = 12inches

                Formula: d subscript v = 25 point 92inches

S5.8.2.9

Now, Vn1 and Vn2 can be calculated:

For       Formula: f prime subscript c = 4 point 0ksi

                    Formula: V subscript n1 = 0 point 0316 times beta times square root of (f prime subscript c) times b subscript v times d subscript v

                    Formula: V subscript n1 = 39 point 32       Kips per foot

                    Formula: V subscript n2 = 0 point 25 times f prime subscript c times b subscript v times d subscript v

                    Formula: V subscript n2 = 311 point 04       Kips per foot

Use:       Formula: V subscript n = 39 point 32 Kips per foot

The factored shear resistance is then:

                  Formula: phi subscript v = 0 point 90

S5.5.4.2.1

                Formula: V subscript r = phi subscript v times V subscript n

                Formula: V subscript r = 35 point 39 Kips per foot

                Formula: V subscript r less than V subscript uftgtoe       N.G.

If the front row of piles were outside the critical section, the one-way shear for the abutment footing toe would fail. The footing depth would have to be increased or the piles would have to be redesigned to reduce the shear force outside the critical section. Again, the above design shear force and resistance are just shown to illustrate the toe one-way shear check if the pile loads were outside the critical section.

For one way action in the abutment footing heel, the critical section is taken at the abutment face for heels that are in tension on the top face of the heel. For heels that are in compression on the top face, the critical section is calculated according to S5.8.2.9. The maximum factored abutment footing heel shear occurs when the heel is in tension on the top face. Therefore, the critical section is taken at the stem back face.

S5.13.3.6.1 & C5.13.3.6.1

The term dv is calculated the same as it is for the abutment toe:

                Formula: d subscript v = max ( d subscript e minus numerator (a) divided by denominator (2) , 0 point 9 times d subscript e , 0 point 72 times h ). Equation not used

S5.8.2.9

where:

                  Formula: d subscript e = 26 point 50inches       use the same effective depth as the toe - conservative

              Formula: a = 1 point 16inches       use the same stress block depth as the toe - conservative

              Formula: h = 30inches

                Formula: d subscript v = 25 point 92inches

Elevation view showing the abutment footing heel one way action critical section. The abutment heel is to the left. The distance from the left edge of the abutment heel to the back face of the abutment stem is 4 feet 0 inches. The distance from the left edge of abuemnt heel to the centroidal axis of the back row of piles is 1 foot 3 inches. The distance from the left edge of heel to the abutment heel critical shear section is 4 feet 0 inches. The critical section is taken at the back face of abutment stem.

Figure 7-23 Abutment Heel One-way Action Critical Section

Since the back row of piles are all outside the critical section, the factored shear is computed by summing all the back row pile loads.

The pile loads used to compute the controlling footing heel shear force are for the Strength I limit state using the minimum load factors at the final construction stage. They are taken from Design Step P, Tables P-17 through P-20 and are as follows:

                  Formula: P subscript 1 = minus 15 point 3K       Formula: P subscript 9 = minus 14 point 5K

                  Formula: P subscript 3 = minus 14 point 8K       Formula: P subscript 11 = minus 14 point 8K

                  Formula: P subscript 5 = minus 14 point 5K       Formula: P subscript 13 = minus 15 point 3K

                  Formula: P subscript 7 = minus 14 point 4K

The factored one-way shear force at the abutment footing heel critical section on a per foot basis is then:

                            Formula: V subscript uftgheeI = numerator (( P subscript 1 + P subscript 3 + P subscript 5 + P subscript 7 + P subscript 9 + P subscript 11 + P subscript 13 )) divided by denominator (L subscript abut)

                            Formula: V subscript uftgheeI = minus 2 point 21 Kips per foot

The nominal shear resistance is the lesser of:

S5.8.3.3

                    Formula: V subscript n1 = V subscript c + V subscript s. Equation not used

or

                    Formula: V subscript n2 = 0 point 25 times f prime subscript c times b subscript v times d subscript v. Equation not used

where:

                  Formula: V subscript c = 0 point 0316 times beta times square root of (f prime subscript c) times b subscript v times d subscript v. Equation not used

and

                  Formula: V subscript s = numerator (A subscript v times f subscript y times d subscript v times ( cot theta + cot alpha ) times sin alpha ) divided by denominator (s) . Equation not used       neglect for this abutment design

Before the nominal shear resistance can be calculated, all the variables used in the above equations need to be defined.

                Formula: beta = 2 point 0

S5.8.3.4.1

                Formula: b subscript v = 12inches

                Formula: d subscript v = 25 point 92inches

S5.8.2.9

Now, Vn1 and Vn2 can be calculated:

For       Formula: f prime subscript c = 4 point 0ksi

                    Formula: V subscript n1 = 0 point 0316 times beta times square root of (f prime subscript c) times b subscript v times d subscript v

                    Formula: V subscript n1 = 39 point 32       Kips per foot

                    Formula: V subscript n2 = 0 point 25 times f prime subscript c times b subscript v times d subscript v

                    Formula: V subscript n2 = 311 point 04       Kips per foot

Use:       Formula: V subscript n = 39 point 32 Kips per foot

The factored shear resistance is then:

                  Formula: phi subscript v = 0 point 90

S5.5.4.2.1

                Formula: V subscript r = phi subscript v times V subscript n

                Formula: V subscript r = 35 point 39 Kips per foot

                Formula: V subscript r greater than V subscript uftgheeI       OK

For two-way action, the pile critical perimeter, bo, is located a minimum of 0.5dv from the perimeter of the pile. If portions of the critical perimeter are located off the footing, that portion of the critical perimeter is limited by the footing edge.

Two-way action should be checked for the maximum loaded pile, or pile # 8 (see Design Step P - Tables P-17 through P-20). The effective shear depth, dv, is the same as that used for the one-way shear check for the footing toe.

S5.13.3.6.1

                            Formula: V subscript utwoway = P subscript 8

                            Formula: V subscript utwoway = 339 point 90 K

                Formula: d subscript v = 25 point 92inches

                        Formula: 0 point 5 times d subscript v = 12 point 96inches

Plan view of abutment footing showing the pile two way action critical perimeter for pile number 8 which is in the front row of piles. The pile size is an HP 12 by 53. The footing heel is to the left and the footing toe is to the right. The critical perimeter is 37 point 92 inches along the front face of footing toe and 33 point 86 inches starting at the footing toe back measured towards the footing heel. The clear distance from the edge of the pile to the critical perimeter is 12 point 96 inches on all sides except the side of the pile that is adjacent to the footing toe front face. The lines representing the stem back face, backwall back face, backwall front face, and stem front face are also shown. The stem front face is just inside the critical perimeter.

Figure 7-24 Pile Two-way Action Critical Perimeter

In the above figure, it can be seen that the critical perimeter is approximately at the face of the stem. In fact, the critical perimeter overlaps the front face of the stem by approximately 0.07 inches. Since the overlap is minimal, ignore the overlap and assume the critical perimeter and the front face of the stem are aligned at the same plane.

Two-way action or punching shear resistance for sections without transverse reinforcement can then be calculated as follows:

S5.13.3.6.3

                  Formula: V subscript n = ( 0 point 063 + numerator (0 point 126) divided by denominator ( beta subscript c) ) times square root of (f prime subscript c) times b subscript o times d subscript v less than or equal to 0 point 126 times square root of (f prime subscript c) times b subscript o times d subscript v . Equation not used

                  Formula: beta subscript c = numerator (37 point 92 inches ) divided by denominator (33 point 86 inches )       ratio of long to short side of critical perimeter

                  Formula: beta subscript c = 1 point 12

                  Formula: b subscript o = 2 times ( 33 point 86 + 37 point 92)inches

                Formula: b subscript o = 143 point 56inches

                                                              Formula: ( 0 point 063 + numerator (0 point 126) divided by denominator ( beta subscript c) ) times square root of (f prime subscript c) times b subscript o times d subscript v = 1306 point 17K

                                          Formula: 0 point 126 times square root of (f prime subscript c) times b subscript o times d subscript v = 937 point 71K

use       Formula: V subscript n = 937 point 71K

The factored punching shear resistance is then:

                  Formula: phi subscript v = 0 point 90

S5.5.4.2.1

                Formula: V subscript r = phi subscript v times V subscript n

                Formula: V subscript r = 843 point 94 K

                Formula: V subscript r greater than V subscript utwoway       OK

Design Step 7.12 - Draw Schematic of Final Abutment Design

Elevation view of final abutment design. The backwall top width is 1 foot 3 inches. The paving notch width is 9 inches and the paving notch depth is 1 foot 4 inches. The depth along the front of backwall down to the corbel is 2 feet 0 inches. The total backwall height is 7 feet 0 inches. The stem height from the bottom of backwall to top of footing is 15 feet 0 inches. The stem thickness is 3 feet 6 inches. The footing width is 10 feet 3 inches. The footing thickness is 2 feet 6 inches. The distance from the footing toe to the front face of stem is 2 feet 9 inches. The back row of piles is 1 foot 3 inches from the heel face. The front row of piles is 1 foot 3 inches from the toe face measured along the bottom of footing. The front row of piles is battered 3 vertical to 1 horizontal. Each row of piles is embedded into the footing 1 foot 0 inches. The backwall back and front face vertical reinforcement bars are number 5 bars at 9 point 0 inches spacing. The horizontal or temperature and shrinkage reinforcement bars in the backwall consist of number 4 bars at 12 point 0 inches vertical spacing for the back and front faces of the backwall. The stem back and front face vertical reinforcement bars consist of number 9 bars at 9 point 0 inch spacing. The horizontal or temperature and shrinkage reinforcement in the back and front faces of the stem consist of number 5 bars at 9 point 0 inches vertical spacing. The footing longitudinal reinforcement on the top of the footing or heel reinforcement consists of number 5 bars at 12 point 0 inches spacing. The footing longitudinal reinforcement on the bottom of the footing or toe reinforcement consists of number 8 bars at 12 point 0 inches spacing. The top and bottom of footing transverse reinforcement bars consist of number 5 bars at 12 point 0 inches spacing. The stem back face vertical bars protruding from the footing into the stem have a splice length of 8 feet 0 inches. The backfill material is to be acceptable material.

Figure 7-25 Final Abutment Design

Design Step 7.2 - Select Optimum Wingwall Type

Selecting the most optimal wingwall type depends on the site conditions, cost considerations, and aesthetics. Wingwalls can be integral or independent. Wingwall classifications include most of the abutment types listed in the abutment section. For this design example, a reinforced concrete cantilever wingwall was chosen. The wingwall is skewed at a 45 degree angle from the front face of the abutment stem.

S11.2

Elevation view of reinforced concrete cantilever wingwall. The wingwall heel is to the left and the toe is to the right.

Figure 7- 26 Reinforced Concrete Cantilever Wingwall

Design Step 7.3 - Select Preliminary Wingwall Dimensions

The designer should base the preliminary wingwall dimensions on state specific standards, previous designs, and past experience.

The following figure shows the preliminary dimensions for the wingwall.

Elevation view of preliminary wingwall dimensions looking at the front face of the wingwall. Adjacent to the abutment, the wingwall height from the top of wingwall to top of footing is 22 feet 0 inches. The wingwall stem height at the free end of the wingwall is 17 feet 0 inches. The wingwall length is 20 feet 6 inches. The top width of the wingwall stem is is 1 foot 6 inches. The top of the wingwall stem adjacent to the abutment is flat for a 6 inch length of wingwall and then the top of the wingwall slopes downward at a rate of 4 horizontal to 1 vertical to the free end. The back face of the wingwall stem is battered at a rate of 1 horizontal to 12 vertical.

Figure 7-27 Preliminary Wingwall Dimensions

Design Step 7.4 - Compute Dead Load Effects

Once the preliminary wingwall dimensions are selected, the corresponding dead loads can be computed. The dead loads are calculated on a per foot basis. For sloped wingwalls, the design section is generally taken at a distance of one-third down from the high end of the wingwall.

S3.5.1

Design section stem height:

Distance from start of slope at high end of stem:

                          Formula: numerator (( 20 point 5 feet )) divided by denominator (3) = 6 point 83 feet

Amount wingwall stem drops per foot:

                                  Formula: numerator (( 22 feet minus 17 feet )) divided by denominator (20 feet ) = 0 point 25 numerator ( feet ) divided by denominator ( feet )

The wingwall design height is then:

                      Formula: H subscript wing = 22 feet minus left bracket ( 6 point 83 feet minus 0 point 5 feet ) times 0 point 25 numerator ( feet ) divided by denominator ( feet ) right bracket

                      Formula: H subscript wing = 20 point 42 feet

Use       Formula: H subscript wing = 20 point 50 feet

Elevation view of wingwall design section. The design section is taken at 6 feet 10 inches from the wingwall end adjacent to the abutment. The wingwall top width is 1 foot 6 inches. The wingwall stem design height is 20 feet 6 inches. The wingwall back face batter is 1 horizontal to 12 vertical. The wingwall stem width at the base is 3 feet 2 and one half inches.

Figure 7-28 Wingwall Design Section

Wingwall stem:

                              Formula: DL subscript wwstem = left bracket ( numerator (1 point 5 feet + 3 point 21 feet ) divided by denominator (2) ) times 20 point 50 feet right bracket times W subscript c

                              Formula: DL subscript wwstem = 7 point 24 Kips per foot

Design Step 7.5 - Compute Live Load Effects

Since the wingwall does not support a parapet, the only live load effects are from live load surcharge. The effects from live load surcharge are computed in Design Step 7.6.

Design Step 7.6 - Compute Other Load Effects

Other load effects that need to be computed include: wind loads, earthquake loads, earth pressure, live load surcharge, and temperature loads.

Wind Load on Wingwall

S3.8.1.2.3

The wind loads acting on the exposed portion of the wingwall front and end elevations are calculated from a base wind pressure of 0.040 KSF. In the wingwall final state, the wind loads acting on the wingwall will only decrease the overturning moment and will be ignored for this design example. For the wingwall temporary state, the wind loads acting on the wingwall should be investigated. Also, any wind loads that produce a transverse shear or moment in the wingwall footing are ignored. The reason for this is due to the fact that the majority of force effects required to produce a transverse shear or moment will also reduce the maximum overturning moment.

Earthquake Load

S3.10

This design example assumes that the structure is located in seismic zone I with an acceleration coefficient of 0.02. For seismic zone I, no seismic analysis is required.

Earth Loads

S3.11

The earth loads that need to be investigated for this design example include: loads due to basic lateral earth pressure, loads due to uniform surcharge, and live load surcharge loads.

S3.11.5

S3.11.6

Loads due to basic lateral earth pressure:

S3.11.5

To obtain the lateral loads due to basic earth pressure, the earth pressure (p) must first be calculated from the following equation.

S3.11.5.1

              Formula: p = k subscript a times gamma subscript s times z. Equation not used

Bottom of wingwall stem lateral earth load:

                Formula: k subscript a = 0 point 3       obtained from geotechnical information

                Formula: gamma subscript s = 0 point 120kcf       use average of loose and compacted gravel

STable 3.5.1-1

              Formula: z = 20 point 5 feet + 2 point 0 feet       Depth below the surface of the earth

              Formula: p = k subscript a times gamma subscript s times z

              Formula: p = 0 point 81 ksf

Elevation view of wingwall showing the design earth pressure from the top of wingwall to the bottom of wingwall stem. The wingwall depth from top of wingwall to bottom of wingwall is 20 feet 6 inches. The lateral earth pressure on the stem, p, is maximum at the bottom of the stem and zero at the top of grade. The distance from the top of grade to the bottom of wingwall stem is 22 feet 6 inches. The horizontal load, R subscript EHstem due to p acts at 7 feet 6 inches up from the bottom of the wingwall stem acting in the direction towards the toe. The backfill material is sloped at an angle of 18 point 5 degrees from the horizontal up and away from the back face of the wingwall.

Figure 7-29 Wingwall Stem Design Earth Pressure

Once the lateral earth pressure is calculated, the lateral load due to the earth pressure can be calculated. This load acts at a distance of H/3 from the bottom of the section being investigated. For cases where the ground line is sloped, H is taken as the height from the top of earth to the bottom of the section being investigated.

S3.11.5.1

SC3.11.5.1

                          Formula: h subscript wwstem = 20 point 5 feet + 2 point 0 feet

                            Formula: R subscript EHstem = ( numerator (1) divided by denominator (2) ) times p times h subscript wwstem

                            Formula: R subscript EHstem = 9 point 11 Kips per foot

Since the ground line is sloped, REHstem, must be broken down into horizontal and vertical components as follows:

                                  Formula: R subscript EHstemhoriz = R subscript EHstem times cos( 18 point 5deg)

                                  Formula: R subscript EHstemhoriz = 8 point 64 Kips per foot

                                  Formula: R subscript EHstemvert = R subscript EHstem times sin( 18 point 5deg)

                                  Formula: R subscript EHstemvert = 2 point 89 Kips per foot

Loads due to uniform surcharge:

S3.11.6.1

Since an approach slab and roadway will cover the abutment backfill material, no uniform surcharge load will be applied.

Loads due to live load surcharge:

S3.11.6.4

Loads due to live load surcharge must be applied when a vehicular live load acts on the backfill surface behind the backface within one-half the wall height. Since the distance from the wingwall back face to the edge of traffic is greater than one foot, the equivalent height of fill is constant. The horizontal pressure increase due to live load surcharge is estimated based on the following equation:

                  Formula: Delta subscript p = k times gamma subscript s times h subscript eq. Equation not used

Bottom of wingwall stem live load surcharge load:

              Formula: k = k subscript a

                Formula: gamma subscript s = 0 point 120kcf       use average of loose and compacted gravel

STable 3.5.1-1

                  Formula: h subscript eq = 2 point 0 feet       equivalent height of soil for vehicular loading

STable 3.11.6.4-1

                  Formula: Delta subscript p = k times gamma subscript s times h subscript eq

                  Formula: Delta subscript p = 0 point 072 ksf

The lateral load due to the live load surcharge is:

                            Formula: R subscript LSstem = Delta subscript p times h subscript wwstem

                          Formula: R subscript LSstem = 1 point 62 Kips per foot

Since the ground line is sloped, RLSstem, must be broken down into horizontal and vertical components as follows:

                                  Formula: R subscript LSstemhoriz = R subscript LSstem times cos( 18 point 5deg)

                                  Formula: R subscript LSstemhoriz = 1 point 54 Kips per foot

                                  Formula: R subscript LSstemvert = R subscript LSstem times sin( 18 point 5deg)

                                  Formula: R subscript LSstemvert = 0 point 51 Kips per foot

Loads due to temperature:

S3.12

Temperature loads are not applicable for the wingwall design.

Design Step 7.7 - Analyze and Combine Force Effects

There are two critical locations where the force effects need to be combined and analyzed for design. They include: the bottom of stem or top of footing and the bottom of footing. For the stem design, transverse horizontal loads do not need be considered due to the high moment of inertia about that axis, but at the bottom of footing, the transverse horizontal loads will need to be considered for the footing and pile design. Note that the footing design calculations for wingwalls are similar to abutments. Therefore, the wingwall footing design calculations will not be shown.

Bottom of Wingwall Stem

The combination of force effects for the bottom of the wingwall stem includes:

Elevation view of wingwall section 1 foot thick showing the loads that are applied when forces are desired at the bottom of wingwall stem. The wingwall dead load, DL subscript wwstem, acts vertically downward at the midpoint of the stem base. The stem base is 3 feet 2 and one half inches wide. The horizontal component of the earth load, R subscript EHstemhorizontal, acts 7 feet 6 inches up from the bottom of the stem toward the toe. The vertical component of the earth load, R subscript EHstemvert, acts vertically downward at the back face of the heel. The horizontal component of the live load surcharge, R subscript LSstemhorizontal, acts at the midpoint of the depth between the finshed grade and the bottom of stem toward the toe. The vertical component of the live load surcharge, R subscript LSstemvert, acts vertically downward at the back face of the heel. The angle between the finished grade and the horizontal is 18 point 5 degrees.

Figure 7-30 Wingwall Stem Dimensions and Loading

The force effects for the wingwall stem will be combined for the following limit states.

  Load Factors
  Strength I Strength III Strength V Service I
Loads γmax γmin γmax γmin γmax γmin γmax γmin
DC 1.25 0.90 1.25 0.90 1.25 0.90 1.00 1.00
DW 1.50 0.65 1.50 0.65 1.50 0.65 1.00 1.00
LL 1.75 1.75 --- --- 1.35 1.35 1.00 1.00
EH 1.50 0.90 1.50 0.90 1.50 0.90 1.00 1.00
LS 1.75 1.75 --- --- 1.35 1.35 1.00 1.00

STable 3.4.1-1

STable 3.4.1-2

Table 7-7 Applicable Wingwall Stem Limit States with the Corresponding Load Factors

The loads that are required to combine force effects at the base of the wingwall stem include:

                              Formula: DL subscript wwstem = 7 point 24 Kips per foot

                                  Formula: R subscript EHstemhoriz = 8 point 64 Kips per foot

                                  Formula: R subscript LSstemhoriz = 1 point 54 Kips per foot

Wingwall stem Strength I force effects:

The following load factors will be used to calculate the controlling force effects for Strength I:

                    Formula: gamma subscript DC = 1 point 25

STable 3.4.1-2

                    Formula: gamma subscript EH = 1 point 50

STable 3.4.1-2

                  Formula: gamma subscript LS = 1 point 75

STable 3.4.1-1

The factored vertical force at the base of the wingwall stem is:

                            Formula: F subscript vstmstrI = gamma subscript DC times DL subscript wwstem

                          Formula: F subscript vstmstrI = 9 point 05 Kips per foot

The factored longitudinal shear force at the base of the wingwall stem is:

                            Formula: V subscript ustmstrI = ( gamma subscript EH times R subscript EHstemhoriz ) + ( gamma subscript LS times R subscript LSstemhoriz )

                            Formula: V subscript ustmstrI = 15 point 65 Kips per foot

The factored moment about the bridge transverse axis at the base of the wingwall stem is:

                            Formula: M subscript ustmstrI = ( gamma subscript EH times R subscript EHstemhoriz times 7 point 5 feet ) + ( gamma subscript LS times R subscript LSstemhoriz times 11 point 25 feet )

                            Formula: M subscript ustmstrI = 127 point 46 Kips foot per foot

Wingwall stem Strength III force effects:

The following load factors will be used to calculate the force effects for Strength III:

                    Formula: gamma subscript DC = 1 point 25

STable 3.4.1-2

                    Formula: gamma subscript EH = 1 point 50

STable 3.4.1-2

The factored vertical force at the base of the wingwall stem is:

                            Formula: F subscript vstmstrIII = gamma subscript DC times DL subscript wwstem

                            Formula: F subscript vstmstrIII = 9 point 05 Kips per foot

The factored longitudinal shear force at the base of the abutment stem is:

                              Formula: V subscript ustmstrIII = gamma subscript EH times R subscript EHstemhoriz

                              Formula: V subscript ustmstrIII = 12 point 96 Kips per foot

The factored longitudinal moment at the base of the wingwall stem is:

                              Formula: M subscript ustmstrIII = gamma subscript EH times R subscript EHstemhoriz times 7 point 5 feet

                              Formula: M subscript ustmstrIII = 97 point 22 Kips foot per foot

Wingwall stem Strength V force effects:

The following load factors will be used to calculate the force effects for Strength V:

                    Formula: gamma subscript DC = 1 point 25

STable 3.4.1-2

                    Formula: gamma subscript EH = 1 point 50

STable 3.4.1-2

                  Formula: gamma subscript LS = 1 point 35

STable 3.4.1-1

The factored vertical force at the base of the wingwall stem is:

                            Formula: F subscript vstmstrV = gamma subscript DC times DL subscript wwstem

                            Formula: F subscript vstmstrV = 9 point 05 Kips per foot

The factored longitudinal shear force at the base of the wingwall stem is:

                              Formula: V subscript ustmstrV = ( gamma subscript EH times R subscript EHstemhoriz ) + ( gamma subscript LS times R subscript LSstemhoriz )

                              Formula: V subscript ustmstrV = 15 point 04 Kips per foot

The factored longitudinal moment at the base of the wingwall stem is:

                              Formula: M subscript ustmstrV = ( gamma subscript EH times R subscript EHstemhoriz times 7 point 5 feet ) + ( gamma subscript LS times R subscript LSstemhoriz times 11 point 25 feet )

                              Formula: M subscript ustmstrV = 120 point 55 Kips foot per foot

Wingwall stem Service I force effects:

The following load factors will be used to calculate the force effects for Service I:

                    Formula: gamma subscript DC = 1 point 00

STable 3.4.1-2

                    Formula: gamma subscript EH = 1 point 00

STable 3.4.1-2

                  Formula: gamma subscript LS = 1 point 00

STable 3.4.1-1

The factored vertical force at the base of the wingwall stem is:

                              Formula: F subscript vstmservI = gamma subscript DC times DL subscript wwstem

                              Formula: F subscript vstmservI = 7 point 24 Kips per foot

The factored longitudinal shear force at the base of the wingwall stem is:

                              Formula: V subscript ustmservI = ( gamma subscript EH times R subscript EHstemhoriz ) + ( gamma subscript LS times R subscript LSstemhoriz )

                              Formula: V subscript ustmservI = 10 point 18 Kips per foot

The factored longitudinal moment at the base of the wingwall stem is:

                              Formula: M subscript ustmservI = ( gamma subscript EH times R subscript EHstemhoriz times 7 point 5 feet ) + ( gamma subscript LS times R subscript LSstemhoriz times 11 point 25 feet )

                              Formula: M subscript ustmservI = 82 point 10 Kips foot per foot

The maximum factored wingwall stem vertical force, shear force, and moment for the strength limit state are:

                                      Formula: F subscript vertstemmax = max ( F subscript vstmstrI , F subscript vstmstrIII , F subscript vstmstrV )

                                    Formula: F subscript vertstemmax = 9 point 05 Kips per foot

                                      Formula: V subscript uwwstemmax = max ( V subscript ustmstrI , V subscript ustmstrIII , V subscript ustmstrV )

                                      Formula: V subscript uwwstemmax = 15 point 65 Kips per foot

                                      Formula: M subscript uwwstemmax = max ( M subscript ustmstrI , M subscript ustmstrIII , M subscript ustmstrV )

                                      Formula: M subscript uwwstemmax = 127 point 46 Kips foot per foot

Design Step 7.9 - Design Wingwall Stem

Design for flexure:

Assume #9 bars:

                              Formula: bar_diam = 1 point 128 inches

                            Formula: bar_area = 1 point 00 inches squared

First, the minimum reinforcement requirements will be calculated. The tensile reinforcement provided must be enough to develop a factored flexural resistance at least equal to the lesser of 1.2 times the cracking strength or 1.33 times the factored moment from the applicable strength load combinations.

S5.7.3.3.2

The cracking strength is calculated by:

SEquation

5.7.3.6.2-2

                  Formula: M subscript cr = numerator (f subscript r times I subscript g) divided by denominator (y subscript t) . Equation not used

Cross section of wingwall stem showing the cracking moment dimensions. The front face of the wingwall stem is at the top of the cross section and the rear face is at the bottom of the cross section. The cross section is 1 foot and 0 inches wide by 3 feet 2 and one half inches in length. The centroidal axis is located 1 foot 7 and one quarter inches from the front face. The centroid of the vertical reinforcing steel is 3 and one sixteenth inches from the rear face of the wingwall stem. The reinforcing steel consists of number 9 bars at 9 point 0 inches spacing.

Figure 7-31 Wingwall Cracking Moment Dimensions

              Formula: f subscript r = 0 point 24 times square root of (f prime subscript c)

S5.4.2.6

              Formula: f subscript r = 0 point 48ksi

                Formula: I subscript g = numerator (1) divided by denominator (12) ( 12 inches ) ( 38 point 5 inches ches ) cubed       Formula: I subscript g = 57067 inches superscript 4

                Formula: y subscript t = 19 point 25 inches

                  Formula: M subscript cr = numerator ( numerator (f subscript r times I subscript g) divided by denominator (y subscript t)) divided by denominator ( feet )       Formula: M subscript cr = 118 point 58 Kips foot per foot

                          Formula: 1 point 2 times M subscript cr = 142 point 30 Kips foot per foot

1.33 times the factored controlling backwall moment is:

                                    Formula: M subscript uwwstemmax = 127 point 46 Kips foot per foot

                                              Formula: 1 point 33 times M subscript uwwstemmax = 169 point 53 Kips foot per foot

1.2 times the cracking moment controls the minimum reinforcement requirements. 1.2 times the cracking moment is also greater than the factored wingwall stem moment. Therefore, use 1.2 times the cracking moment to design the wingwall stem flexure reinforcement.

                                  Formula: M subscript uwwstemdes = 1 point 2 times M subscript cr

                                  Formula: M subscript uwwstemdes = 142 point 30 Kips foot per foot

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

                  Formula: t subscript bw = 38 point 5 inches       wingwall thickness at base

                        Formula: Cover subscript s = 2 point 50 inches

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

                Formula: d subscript e = 35 point 44 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: f prime subscript c = 4 point 0ksi

                Formula: f subscript y = 60ksi

                  Formula: Rn = numerator (M subscript uwwstemdes times 12 inches ) divided by denominator (( phi subscript f times b times d subscript e squared ))       Formula: Rn = 0 point 13 ksi

                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 00214

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 91 square inches per foot

Required bar spacing = Formula: numerator (bar_area) divided by denominator (A subscript s) = 13 point 2 inches

Use #9 bars @       Formula: bar_space = 9 point 0 inches       to match the abutment stem vertical bar spacing

                  Formula: A subscript s = bar_area times ( numerator (12 inches ) divided by denominator (bar_space) )       Formula: A subscript s = 1 point 33 inches squared       per foot

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

S5.7.3.3.1

              Formula: T = A subscript s times f subscript y       Formula: T = 80 point 00 K

              Formula: a = numerator (T) divided by denominator (0 point 85 times f prime subscript c times b)       Formula: a = 1 point 96 inches

                  Formula: beta subscript 1 = 0 point 85

S5.7.2.2

              Formula: c = numerator (a) divided by denominator ( beta subscript 1)       Formula: c = 2 point 31 inches

S5.7.2.2

                  Formula: numerator (c) divided by denominator (d subscript e) = 0 point 07       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 07 less than or equal to 0 point 42       OK

Check crack control:

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

S5.7.3.4

Since this design example assumes that the wingwall will be exposed to deicing salts, use:       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 = 2 point 5 inches + numerator (bar_diam) divided by denominator (2)

                                                                                              Formula: d subscript c = 3 point 06 inches

use       Formula: d subscript c = 2 point 0 inches + numerator (bar_diam) divided by denominator (2)

                                                                                              Formula: d subscript c = 2 point 56 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 = 46 point 15 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 = 26 point 48 ksi       Formula: 0 point 6f subscript y = 36 point 00 ksi

Use       Formula: f subscript sa = 26 point 48ksi

                  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 backwall total load moment:

                              Formula: M subscript ustmservI = 82 point 10 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 = 35 point 44 inches       Formula: A subscript s = numerator (bar_area) divided by denominator ( feet ) times ( numerator (12 inches ) divided by denominator (bar_space) )       Formula: A subscript s = 1 point 33 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 00314

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

              Formula: k = 0 point 200

                    Formula: k times d subscript e = 7 point 10 inches

Cross section of wingwall stem at the base showing the dimensions for the crack control check. The front face of wingwall stem is at the top and the rear face is at the bottom. The cross section width is 12 inches. The wingwall stem length is 38 point 50 inches. The distance from the front face of stem to the neutral axis is 7 point 10 inches. The distance from the rear face of stem to the centroid of the vertical stem reinforcement is 3 point 06 inches. The distance from the neutral axis to the centroid of vertical reinforcement is 28 point 34 inches. The vertical reinforcement consists of number 9 bars at 9 point 0 inches spacing.

Figure 7-32 Wingwall Crack Control Check

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

                Formula: d subscript e = 35 point 44 inches

                  Formula: A subscript s = 1 point 33 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 = 9996 point 22 numerator ( inches superscript 4) divided by denominator ( feet )

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

              Formula: y = d subscript e minus k times d subscript e       Formula: y = 28 point 34 inches

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

                Formula: f subscript s = 22 point 34 ksi       Formula: f subscript sa greater than f subscript s       OK

Design for shear:

The factored longitudinal shear force at the base of the wingwall is:

                                    Formula: V subscript uwwstemdes = V subscript uwwstemmax

                                  Formula: V subscript uwwstemdes = 15 point 65 Kips per foot

The nominal shear resistance is the lesser of:

S5.8.3.3

                    Formula: V subscript n1 = V subscript c + V subscript s. Equation not used

or

                    Formula: V subscript n2 = 0 point 25 times f prime subscript c times b subscript v times d subscript v. Equation not used

where:

                  Formula: V subscript c = 0 point 0316 times beta times square root of (f prime subscript c) times b subscript v times d subscript v. Equation not used

and

                  Formula: V subscript s = numerator (A subscript v times f subscript y times d subscript v times ( cot theta + cot alpha ) times sin alpha ) divided by denominator (s) . Equation not used       neglect for this wingwall design

Before the nominal shear resistance can be calculated, all the variables used in the above equations need to be defined.

                Formula: beta = 2 point 0

S5.8.3.4.1

                Formula: b subscript v = 12inches

                Formula: d subscript v = max ( d subscript e minus numerator (a) divided by denominator (2) , 0 point 9 times d subscript e , 0 point 72 times h ). Equation not used

S5.8.2.9

where:

                Formula: d subscript e = 35 point 44 inches

              Formula: a = 1 point 96 inches

              Formula: h = 38 point 50inches

Therefore:

                Formula: d subscript v = 34 point 46inches

Now, Vn1 and Vn2 can be calculated:

For       Formula: f prime subscript c = 4 point 0ksi

                    Formula: V subscript n1 = 0 point 0316 times beta times square root of (f prime subscript c) times b subscript v times d subscript v

                    Formula: V subscript n1 = 52 point 27       Kips per foot

                    Formula: V subscript n2 = 0 point 25 times f prime subscript c times b subscript v times d subscript v

                    Formula: V subscript n2 = 413 point 52       Kips per foot

Use:       Formula: V subscript n = 52 point 27 Kips per foot

The factored shear resistance is then:

                  Formula: phi subscript v = 0 point 90

S5.5.4.2.1

                Formula: V subscript r = phi subscript v times V subscript n

                Formula: V subscript r = 47 point 04 Kips per foot

                Formula: V subscript r greater than V subscript uwwstemdes       OK

Shrinkage and temperature reinforcement:

S5.10.8

For members less than 48.0 inches thick, the area of reinforcement in each direction shall not be spaced greater than 12.0 inches and satisfy the lesser of:

S5.10.8.2

                  Formula: A subscript s greater than or equal to 0 point 11 numerator (A subscript g) divided by denominator (fy) . Equation not used       or       Formula: Sigma A subscript b = 0 point 0015A subscript g. Equation not used

                  Formula: A subscript g = ( 38 point 5 inches ) times ( 12 inches per foot )       Formula: A subscript g = 462 point 0       square inches per foot

                Formula: f subscript y = 60ksi

                            Formula: 0 point 11 times numerator (A subscript g) divided by denominator (f subscript y) = 0 point 85       square inches per foot

                      or

                              Formula: 0 point 0015A subscript g = 0 point 69       square inches per foot

As must be greater than or equal to 0.69in2/ft

The above steel must be distributed equally on both faces of the wingwall.

Try 1 horizontal # 5 bar for each face of the wingwall at 9.0 inch spacing:

                              Formula: bar_diam = 0 point 625 inches

                            Formula: bar_area = 0 point 31 inches squared

                  Formula: A subscript s = 2 times numerator (bar_area) divided by denominator ( feet ) times ( numerator (12 inches ) divided by denominator (9 inches ) )

                  Formula: A subscript s = 0 point 83 square inches per foot

                          Formula: 0 point 83 square inches per foot greater than or equal to 0 point 69 square inches per foot       OK

Based on the wingwall design, #9 bars at 9.0 inch spacing will be used for the back face flexure reinforcement. Use # 5 bars at 9.0 inch spacing for the front face vertical reinforcement. The horizontal temperature and shrinkage reinforcement will consist of #5 bars at 9.0 inch spacing for the front and back faces.

Design Step 7.12 - Draw Schematic of Final Wingwall Design

Elevation view of final wingwall design. The wingwall stem top width is 1 foot 6 inches. The wingwall stem design height is 20 feet 6 inches. The back face batter is 1 horizontal to 12 vertical. The design stem width at the base of stem is 3 feet 2 and one half inches. The reinforcement cover on the front and back faces of the stem is 2 and one half inches. The stem back face vertical reinforcement bars consist of number 9 bars at 9 point 0 inches spacing. The back face vertical reinforcing bars protruding from the footing into the stem have a lap length of 8 feet 0 inches. The stem front face vertical reinforcement bars consist of number 5 bars at 9 point 0 inches spacing. The horizontal reinforcement bars in the wingwall stem or temperature and shrinkage reinforcement consist of number 5 bars at 9 point 0 inch vertical spacing. The backfill material should be acceptable material.

Figure 7-33 Final Wingwall Design

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Updated: 07/18/2013
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