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Bridges & Structures

LRFD Steel Girder SuperStructure Design Example

Pier Design Example Design Step 8

Table of Contents

Design Step 8.1 - Obtain Design Criteria
Design Step 8.2 - Select Optimum Pier Type
Design Step 8.3 - Select Preliminary Pier Dimensions
Design Step 8.4 - Compute Dead Load Effects
Design Step 8.5 - Compute Live Load Effects
Design Step 8.6 - Compute Other Load Effects
Design Step 8.7 - Analyze and Combine Force Effects
Design Step 8.8 - Design Pier Cap
Design Step 8.9 - Design Pier Column
Design Step 8.10 - Design Pier Piles
Design Step 8.11 - Design Pier Footing
Design Step 8.12 - Final Pier Schematic

Design Step 8.1 - Obtain Design Criteria

This pier 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 first design step is to identify the appropriate design criteria. This includes, but is not limited to, defining material properties, identifying relevant superstructure information, determining the required pier height, and determining the bottom of footing elevation.

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 pier.

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

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 CTable5.4.2.1-1
Reinforcement strength: Formula: f subscript y = 60 point 0ksi S5.4.3

Concrete 28-day compressive strength - For all components of this pier design example, 4.0 ksi is used for the 28-day compressive strength. However, per the Specifications, 2.4 ksi could be used for the pier footing.

C5.4.2.1

Reinforcing steel cover requirements (assume non-epoxy rebars):

Pier cap: Formula: Cover subscript cp = 2 point 5 inches STable 5.12.3-1
Pier column: Formula: Cover subscript co = 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

Pier cap and column cover - Since no joint exists in the deck at the pier, a 2-inch cover could be used with the assumption that the pier is not subject to deicing salts. However, it is assumed here that the pier can be subjected to a deicing salt spray from nearby vehicles. Therefore, the cover is set at 2.5 inches.

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
 Deck overhang: Formula: DOH = 3 point 9375 feet
 Span length:  Formula: L subscript span = 120 point 0 feet
Parapet height:  Formula: H subscript par = 3 point 5 feet
Deck overhang thickness:  Formula: t subscript o = 9 point 0 inches
Haunch thickness: Formula: H subscript hnch = 3 point 5 inches       (includes top flange)
Web depth: Formula: D subscript o = 66 point 0 inches       (based on 1st trial section)
Bot. flange thickness: Formula: t subscript bf = 2 point 25 inches       (maximum thickness)
Bearing height: Formula: H subscript brng = 5 point 0 inches
Superstructure Depth:  Formula: H subscript super = H subscript par + ( numerator (t subscript o + H subscript hnch + D subscript o + t subscript bf) divided by denominator (12 inches per foot) )

 Formula: H subscript super = 10 point 23 feet

                                                                 

Superstructure data - The above superstructure data is important because it sets the width of the pier cap and defines the depth and length of the superstructure needed for computation of wind loads.

S3.8

Pier height - Guidance on determining the appropriate pier height can be found in the AASHTO publication A Policy on Geometric Design of Highways and Streets. It will be assumed here that adequate vertical clearance is provided given a ground line that is two feet above the top of the footing and the pier dimensions given in Design Step 8.3.

S2.3.3.2

Bottom of Footing Elevation - The bottom of footing elevation may depend on the potential for scour (not applicable in this example) and/or the geotechnical properties of the soil and/or rock. However, as a minimum, it should be at or below the frost depth for a given geographic region. In this example, it is assumed that the two feet of soil above the footing plus the footing thickness provides sufficient depth below the ground line for frost protection of the structure.

S10.6.1.2

Design Step 8.2 - Select Optimum Pier Type

Selecting the most optimal pier type depends on site conditions, cost considerations, superstructure geometry, and aesthetics. The most common pier types are single column (i.e., "hammerhead"), solid wall type, and bent type (multi-column or pile bent). For this design example, a single column (hammerhead) pier was chosen. A typical hammerhead pier is shown in Figure 8-1.

S11.2

This figure shows a front elevation view of a typical hammerhead pier.

Figure 8-1 Typical Hammerhead Pier

Design Step 8.3 - Select Preliminary Pier Dimensions

Since the Specifications do not have standards regarding maximum or minimum dimensions for a pier cap, column, or footing, the designer should base the preliminary pier dimensions on state specific standards, previous designs, and past experience. The pier cap, however, must be wide enough to accommodate the bearing.

Figures 8-2 and 8-3 show the preliminary dimensions selected for this pier design example.

This figure shows a front elevation view of the hammerhead pier used in this design example. The overall width of the pier cap is dimensioned as 46 feet 6 inches. The pier cap overhang is dimensioned as 15 feet 6 inches. The depth of the end of the pier cap is dimensioned as 5 feet and the depth of the tapered portion of the pier cap overhang is 6 feet. The height of the pier column is dimensioned as 15 feet, and the width of the pier column is dimensioned as 15 feet 6 inches. The footing thickness is dimensioned as 3 feet 6 inches, and the footing width is dimensioned as 23 feet.

Figure 8-2 Preliminary Pier Dimensions - Front Elevation

This figure shows a side, or end, elevation view of the pier used in this design example. The pier cap thickness is dimensioned as 5 feet. The depth of the end of the pier cap is dimensioned as 5 feet. The depth of the tapered portion of the pier cap overhang is dimensioned as 6 feet. The thickness of the pier column is dimensioned as 4 feet 6 inches. The width of the footing is dimensioned as 12 feet.

Figure 8-3 Preliminary Pier Dimensions - End Elevation

Design Step 8.4 - Compute Dead Load Effects

Once the preliminary pier dimensions are selected, the corresponding dead loads can be computed. The pier dead loads must then be combined with the superstructure dead loads. The superstructure dead loads shown below are obtained from the superstructure analysis/design software. Based on the properties defined in Design Step 3 (Steel Girder Design), any number of commercially available software programs can be used to obtain these loads. For this design example, the AASHTO Opis software was used, and the values shown below correspond to the first design iteration.

S3.5.1

Exterior girder dead load reactions (DC and DW):

                      Formula: R subscript DCE = 253 point 70K       Formula: R subscript DWE = 39 point 20K

Interior girder dead load reactions (DC and DW):

                      Formula: R subscript DCI = 269 point 10K       Formula: R subscript DWI = 39 point 20K

Pier cap dead load:

Overhang:       Formula: DL subscript ovrhg = ( 5 feet times 5 feet times 15 point 5 feet ) times W subscript c + numerator (1) divided by denominator (2) times ( 6 feet times 5 feet times 15 point 5 feet ) times W subscript c

                                                    Formula: DL subscript ovrhg = 93 point 00 K

Interior:       Formula: DL subscript int = ( 11 feet times 5 feet times 15 point 5 feet ) times W subscript c

                                                Formula: DL subscript int = 127 point 88 K

Total:       Formula: DL subscript cap = 2 times DL subscript ovrhg + DL subscript in t

                                                  Formula: DL subscript cap = 313 point 88 K

Pier column dead load:

                                                Formula: DL subscript col = ( 15 point 5 feet times 4 point 5 feet times 15 feet ) times W subscript c

                                                Formula: DL subscript col = 156 point 94 K

Pier footing dead load:

                                                Formula: DL subscript ftg = ( 3 point 5 feet times 23 feet times 12 feet )W subscript c

                                                Formula: DL subscript ftg = 144 point 90 K

In addition to the above dead loads, the weight of the soil on top of the footing must be computed. The two-foot height of soil above the footing was previously defined. Assuming a unit weight of soil at 0.120 kcf :

STable 3.5.1-1

                                                Formula: EV subscript ftg = 0 point 120kcf times ( 2 feet ) times ( 23 feet times 12 feet minus 15 point 5 feet times 4 point 5 feet )

                                                Formula: EV subscript ftg = 49 point 50 K

Design Step 8.5 - Compute Live Load Effects

For the pier in this design example, the maximum live load effects in the pier cap, column and footing are based on either one, two or three lanes loaded (whichever results in the worst force effect). Figure 8-4 illustrates the lane positions when three lanes are loaded.

The positioning shown in Figure 8-4 is arrived at by first determining the number of design lanes, which is the integer part of the ratio of the clear roadway width divided by 12 feet per lane. Then the lane loading, which occupies ten feet of the lane, and the HL-93 truck loading, which has a six-foot wheel spacing and a two-foot clearance to the edge of the lane, are positioned within each lane to maximize the force effects in each of the respective pier components.

S3.6.1.1.1

S3.6.1.2.1

S3.6.1.2.4

S3.6.1.3.1

This figure shows a front elevation view of the pier with a cross section of the superstructure and the positioning of the live load in the three design lanes. The curb to curb distance is dimensioned as 44 feet. Each of the three lanes are dimensioned as 12 feet. Each lane has a 10 foot uniformly distributed transverse load along with two concentrated loads 6 feet apart. Each concentrated load is labeled P. The lanes are labeled A B and C from left to right. The three lanes are moved as far to the right as possible, with the right edge of Lane C at the right curb line. Within each lane, the uniformly distributed load is moved to the right edge of the lane with one of the concentrated loads two feet from the right edge and the other concentrated load 6 feet from the first. A 2 foot dimension is given in each lane that shows both the distance from the right edge of each lane to a concentrated load and the distance from the left end of the uniformly distributed load to the left edge of the lane. The five steel girders are shown and are numbered 1 thru 5 from left to right. The center to center spacing of the girders is dimensioned as 9 feet 9 inches. A 2 foot 6 inch dimension is given for the distance from the centerline of Beam 5 to the right most curb. A 2 foot dimension is shown for the distance from the centerline of Beam 4 to the face of the pier column. The centerline of the pier is indicated and it aligns with the centerline of Beam 3.

Figure 8-4 Pier Live Loading

The unfactored girder reactions for lane load and truck load are obtained from the superstructure analysis/design software. These reactions do not include dynamic load allowance and are given on a per lane basis (i.e., distribution factor = 1.0). Also, the reactions do not include the ten percent reduction permitted by the Specifications for interior pier reactions that result from longitudinally loading the superstructure with a truck pair in conjunction with lane loading. The value of these reactions from the first design iteration are as follows:

S3.6.1.3.1

                        Formula: R subscript truck = 124 point 50K

                      Formula: R subscript lane = 97 point 40K

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

STable 3.6.2.1-1

The values of the unfactored concentrated loads which represent the girder truck load reaction per wheel line in Figure 8-4 are:

                        Formula: P subscript wheeI = numerator (R subscript truck) divided by denominator (2) times ( 1 + IM) times ( 0 point 90)

                        Formula: P subscript wheeI = 74 point 51 K

The value of the unfactored uniformly distributed load which represents the girder lane load reaction in Figure 8-4 is computed next. This load is transversely distributed over ten feet and is not subject to dynamic load allowance.

S3.6.2.1

                        Formula: W subscript lane = numerator (R subscript lane) divided by denominator (10 feet ) times ( 0 point 90)       Formula: W subscript lane = 8 point 77 Kips per foot

The next step is to compute the reactions due to the above loads at each of the five bearing locations. This is generally carried out by assuming the deck is pinned (i.e., discontinuous) at the interior girder locations but continuous over the exterior girders. Solving for the reactions is then elementary. The computations for the reactions with only Lane C loaded are illustrated below as an example. The subscripts indicate the bearing location and the lane loaded to obtain the respective reaction:

                      Formula: R subscript 5_c = numerator (P subscript wheeI times ( 4 point 25 feet + 10 point 25 feet ) + W subscript lane times 10 feet times 7 point 25 feet ) divided by denominator (9 point 75 feet )

                      Formula: R subscript 5_c = 176 point 00 K

                      Formula: R subscript 4_c = P subscript wheeI times 2 + W subscript lane times 10 feet minus R subscript 5_c

                      Formula: R subscript 4_c = 60 point 69 K

The reactions at bearings 1, 2 and 3 with only Lane C loaded are zero. Calculations similar to those above yield the following live load reactions with the remaining lanes loaded (for simplicity, it is assumed that Lane B's loading is resisted entirely, and equally, by bearings 3 and 4):

                      Formula: R subscript 5_a = 0 point 0K       Formula: R subscript 5_b = 0 point 0K

                      Formula: R subscript 4_a = 0 point 0K       Formula: R subscript 4_b = 118 point 36K

                      Formula: R subscript 3_a = 70 point 96K       Formula: R subscript 3_b = 118 point 36K       Formula: R subscript 3_c = 0 point 0K

                      Formula: R subscript 2_a = 161 point 59K       Formula: R subscript 2_b = 0 point 0K       Formula: R subscript 2_c = 0 point 0K

                      Formula: R subscript 1_a = 4 point 19K       Formula: R subscript 1_b = 0 point 0K       Formula: R subscript 1_c = 0 point 0K

Design Step 8.6 - Compute Other Load Effects

Other load effects that will be considered for this pier design include braking force, wind loads, temperature loads, and earthquake loads.

Braking Force

S3.6.4

Since expansion bearings exist at the abutments, the entire longitudinal braking force is resisted by the pier.

The braking force per lane is the greater of:

25 percent of the axle weights of the design truck or tandem

5 percent of the axle weights of the design truck plus lane load

5 percent of the axle weights of the design tandem plus lane load

The total braking force is computed based on the number of design lanes in the same direction. It is assumed in this example that this bridge is likely to become one-directional in the future. Therefore, any and all design lanes may be used to compute the governing braking force. Also, braking forces are not increased for dynamic load allowance. The calculation of the braking force for a single traffic lane follows:

S3.6.1.1.1

S3.6.2.1

25 percent of the design truck:

                          Formula: BRK subscript trk = 0 point 25 times ( 32K + 32K + 8K)

                        Formula: BRK subscript trk = 18 point 00 K

25 percent of the design tandem:

                          Formula: BRK subscript tan = 0 point 25 times ( 25K + 25K)

                          Formula: BRK subscript tan = 12 point 50 K

5 percent of the axle weights of the design truck plus lane load:

                                Formula: BRK subscript trk_Ian = 0 point 05 times left bracket ( 32K + 32K + 8K) + ( 0 point 64 Kips per foot times 240 feet )right bracket

                                Formula: BRK subscript trk_Ian = 11 point 28 K

5 percent of the axle weights of the design tandem plus lane load:

                                Formula: BRK subscript tan_Ian = 0 point 05 times left bracket ( 25K + 25K) + ( 0 point 64 Kips per foot times 240 feet )right bracket

                                Formula: BRK subscript tan_Ian = 10 point 18 K

Use       Formula: BRK = max ( BRK subscript trk , BRK subscript tan , BRK subscript trk_Ian , BRK subscript tan_Ian )

                                Formula: BRK = 18 point 00 K

The Specifications state that the braking force is applied at a distance of six feet above the roadway surface. However, since the bearings are assumed incapable of transmitting longitudinal moment, the braking force will be applied at the bearing elevation (i.e., five inches above the top of the pier cap). This force may be applied in either horizontal direction (back or ahead station) to cause the maximum force effects. Additionally, the total braking force is typically assumed equally distributed among the bearings:

S3.6.4

                          Formula: BRK subscript brg = numerator (BRK) divided by denominator (5)

                          Formula: BRK subscript brg = 3 point 60 K

Wind Load from Superstructure

S3.8.1.2

Prior to calculating the wind load on the superstructure, the structure must be checked for aeroelastic instability. If the span length to width or depth ratio is greater than 30, the structure is considered wind-sensitive and design wind loads should be based on wind tunnel studies.

S3.8.3

                      Formula: L subscript span = 120 feet

                      Formula: Width = 47 feet

                        Formula: Depth = H subscript super minus H subscript par

                      Formula: Depth = 6 point 73 feet

                        Formula: numerator (L subscript span) divided by denominator (Width) = 2 point 55       OK       Formula: numerator (L subscript span) divided by denominator (Depth) = 17 point 83       OK

Since the span length to width and depth ratios are both less than 30, the structure does not need to be investigated for aeroelastic instability.

To compute the wind load on the superstructure, the area of the superstructure exposed to the wind must be defined. For this example, the exposed area is the total superstructure depth multiplied by length tributary to the pier. Due to expansion bearings at the abutment, the transverse length tributary to the pier is not the same as the longitudinal length.

S3.8.1.1

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

The total depth was previously computed in Section 8.1 and is as follows:

                        Formula: H subscript super = 10 point 23 feet

For this two-span bridge example, the tributary length for wind load on the pier in the transverse direction is one-half the total length of the bridge:

                        Formula: L subscript windT = numerator (240) divided by denominator (2) feet       Formula: L subscript windT = 120 feet

In the longitudinal direction, the tributary length is the entire bridge length due to the expansion bearings at the abutments:

                        Formula: L subscript windL = 240 feet

The transverse wind area is:

                              Formula: A subscript wsuperT = H subscript super times L subscript windT

                            Formula: A subscript wsuperT = 1228 feet squared

The longitudinal wind area is:

                            Formula: A subscript wsuperL = H subscript super times L subscript windL

                            Formula: A subscript wsuperL = 2455 feet squared

Since the superstructure is approximately 30 feet above low ground level, the design wind velocity, VB, does not have to be adjusted. Therefore:

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

or

                    Formula: P subscript D = P subscript B times ( numerator (100mph) divided by denominator (100mph) ) squared

                    Formula: P subscript D = P subscript B

Also, the minimum transverse normal 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 H subscript super

                            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 pier depends on the angle of wind direction, or attack angle of the wind. The attack angle is taken as measured from a line perpendicular to the girder longitudinal axis (see Figure 8-5). The base wind pressures for the superstructure for various attack angles are given in STable 3.8.1.2.2-1.

S3.8.1.2.2

This figure shows a plan view and, directly underneath it, a side elevation view of the pier supporting the superstructure. Cut lines are used to indicate that only a portion of the entire superstructure length is being shown. The plan view does not show the bridge deck, so the bridge girders can be seen and are shown as solid lines. The pier cap is visible as well. The wind attack angle is shown in the plan view as the angle between a line drawn perpendicular to the fascia girder and an inclined arrow that has its head at the base of the line. The arrow represents the applied wind force. The side elevation view dimensions the previously defined superstructure depth for wind loads of 10.23 feet.

Figure 8-5 Application of Wind Load

Two wind load calculations are illustrated below for two different wind attack angles. The wind loads for all Specifications required attack angles are tabulated in Table 8-1.

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

                                  Formula: WS subscript suptrns0 = A subscript wsuperT times 0 point 050ksf

STable 3.8.1.2.2-1

                                Formula: WS subscript suptrns0 = 61 point 38 K

                                Formula: WS subscript supIng0 = A subscript wsuperL times 0 point 00ksf

                                Formula: WS subscript supIng0 = 0 point 00 K

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

                                  Formula: WS subscript suptrns60 = A subscript wsuperT times 0 point 017ksf

STable 3.8.1.2.2-1

                                  Formula: WS subscript suptrns60 = 20 point 87 K

                                  Formula: WS subscript supIng60 = A subscript wsuperL times 0 point 019ksf

                                  Formula: WS subscript supIng60 = 46 point 65 K

Pier Design Wind Loads from Superstructure
Wind Attack Angle
Degrees
Bridge Transverse Axis
KIPS
Bridge Longitudinal Axis
Kips
0
61.38
0.00
15
54.01
14.73
30
50.33
29.46
45
40.51
39.28
60
20.87
46.65

Table 8-1 Pier Design Wind Loads from Superstructure for Various Wind Attack Angles

The total longitudinal wind load shown above for a given attack angle is assumed to be divided equally among the bearings. In addition, the load at each bearing is assumed to be applied at the top of the bearing (i.e., five inches above the pier cap). These assumptions are consistent with those used in determining the bearing forces due to the longitudinal braking force.

The transverse wind loads shown in Table 8-1 for a given attack angle are also assumed to be equally divided among the bearings and applied at the top of each bearing. However, as shown in Figure 8-6, the transverse load also applies a moment to the pier cap. This moment, which acts about the centerline of the pier cap, induces vertical loads at the bearings as illustrated in Figure 8-6. The computations for these vertical forces with an attack angle of zero are presented below.

This figure shows a cross section of the superstructure at the pier. This includes the bridge deck, five beams and the pier cap. The out to out deck width of 46 feet 10 and a half inches is shown along with the 9 foot 9 inch center to center beam spacing. A uniformly distributed transverse load (shown with arrows) is applied over the depth of the superstructure at the right hand side of the figure. This force acts from right to left. The forces at the bottoms of the beams that are required to resist the applied transverse wind load are shown and are also represented as arrows. These include horizontal and vertical forces at the bottom of each beam. The horizontal forces act from left to right and are present at each beam. The vertical forces are present at the two outer beams on the left and right of the centerline. The middle beam does not have a vertical load. The vertical arrows on the two left most beams are shown in the upward sense, and the arrows on the two right most beams are shown in the downward sense. The arrows on the outer most beams are shown larger than the arrows on the next innermost beams.

Figure 8-6 Transverse Wind Load Reactions at Pier Bearings from Wind on Superstructure

                              Formula: M subscript trns0 = WS subscript suptrns0 times ( numerator (H subscript super) divided by denominator (2) )

                              Formula: M subscript trns0 = 313 point 91 K feet

                              Formula: I subscript girders = 2 times ( 19 point 5 feet ) squared + 2 times ( 9 point 75 feet ) squared

                              Formula: I subscript girders = 950 point 63 feet squared

                                          Formula: RWS1_5 subscript trns0 = numerator (M subscript trns0 times 19 point 5 feet ) divided by denominator (I subscript girders)

                                          Formula: RWS1_5 subscript trns0 = 6 point 44 K

The reactions at bearings 1 and 5 are equal but opposite in direction. Similarly for bearings 2 and 4:

                                        Formula: RWS2_4 subscript trns0 = numerator (M subscript trns0 times 9 point 75 feet ) divided by denominator (I subscript girders)

                                        Formula: RWS2_4 subscript trns0 = 3 point 22 K

Finally, by inspection:

                                    Formula: RWS3 subscript trns0 = 0 point 0K

The vertical reactions at the bearings due to transverse wind on the superstructure at attack angles other than zero are computed as above using the appropriate transverse load from Table 8-1. Alternatively, the reactions for other attack angles can be obtained simply by multiplying the reactions obtained above by the ratio of the transverse load at the angle of interest to the transverse load at an attack angle of zero (i.e., 61.38K).

Vertical Wind Load

S3.8.2

The vertical (upward) wind load is calculated by multiplying a 0.020 ksf vertical wind pressure by the out-to-out bridge deck width. It is applied at 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.

From previous definitions:

                      Formula: Width = 47 point 00 feet

                        Formula: L subscript windT = 120 point 00 feet

The total vertical wind load is then:

                          Formula: WS subscript vert = times superscript 02ksf times ( Width) times ( L subscript windT )

                          Formula: WS subscript vert = 112 point 80 K

This load causes a moment about the pier centerline. The value of this moment is:

                                  Formula: M subscript wind_vert = WS subscript vert times numerator (Width) divided by denominator (4)       Formula: M subscript wind_vert = 1325 K feet

The reactions at the bearings are computed as follows:

                                Formula: RWS subscript vert1 = numerator ( minus WS subscript vert) divided by denominator (5) + numerator (M subscript wind_vert times 19 point 5 feet ) divided by denominator (I subscript girders)

                                Formula: RWS subscript vert2 = numerator ( minus WS subscript vert) divided by denominator (5) + numerator (M subscript wind_vert times 9 point 75 feet ) divided by denominator (I subscript girders)

                                Formula: RWS subscript vert3 = numerator ( minus WS subscript vert) divided by denominator (5)

                                Formula: RWS subscript vert4 = numerator ( minus WS subscript vert) divided by denominator (5) minus numerator (M subscript wind_vert times 9 point 75 feet ) divided by denominator (I subscript girders)

                                Formula: RWS subscript vert5 = numerator ( minus WS subscript vert) divided by denominator (5) minus numerator (M subscript wind_vert times 19 point 5 feet ) divided by denominator (I subscript girders)

The above computations lead to the following values:

                                  Formula: RWS subscript vert1 = 4 point 63 K

                                  Formula: RWS subscript vert2 = minus 8 point 97 K       (vertically upward)

                                  Formula: RWS subscript vert3 = minus 22 point 56 K       (vertically upward)

                                  Formula: RWS subscript vert4 = minus 36 point 15 K       (vertically upward)

                                  Formula: RWS subscript vert5 = minus 49 point 75 K       (vertically upward)

Wind Load on Vehicles

S3.8.1.3

The representation of wind pressure acting on vehicular traffic is given by the Specifications as a uniformly distributed load. Based on the skew angle, this load can act transversely, or both transversely and longitudinally. Furthermore, this load is to be applied at a distance of six feet above the roadway surface. The magnitude of this load with a wind attack angle of zero is 0.10 klf. For wind attack angles other than zero, STable 3.8.1.3-1 gives values for the longitudinal and transverse components. For the transverse and longitudinal loadings, the total force in each respective direction is calculated by multiplying the appropriate component by the length of structure tributary to the pier. Similar to the superstructure wind loading, the longitudinal length tributary to the pier differs from the transverse length.

                        Formula: L subscript windT = 120 point 00 feet       Formula: L subscript windL = 240 point 00 feet

An example calculation is illustrated below using a wind attack angle of 30 degrees:

                                  Formula: WL subscript trans30 = L subscript windT times ( 0 point 082 times kIf)

STable 3.8.1.3-1

                                  Formula: WL subscript trans30 = 9 point 84 K

                                  Formula: WL subscript long30 = L subscript windL times ( 0 point 024kIf)

STable 3.8.1.3-1

                                  Formula: WL subscript long30 = 5 point 76 K

Table 8-2 contains the total transverse and longitudinal loads due to wind load on vehicular traffic at each Specifications required attack angle.

Design Vehicular Wind Loads
Wind Attack Angle
Degrees
Bridge Transverse Axis
Kips
Bridge Longitudinal Axis
Kips
0
12.00
0.00
15
10.56
2.88
30
9.84
5.76
45
7.92
7.68
60
4.08
9.12

Table 8-2 Design Vehicular Wind Loads for Various Wind Attack Angles

The vehicular live loads shown in Table 8-2 are applied to the bearings in the same manner as the wind load from the superstructure. That is, the total transverse and longitudinal load is equally distributed to each bearing and applied at the the top of the bearing (five inches above the top of the pier cap). In addition, the transverse load acting six feet above the roadway applies a moment to the pier cap. This moment induces vertical reactions at the bearings. The values of these vertical reactions for a zero degree attack angle are given below. The computations for these reactions are not shown but are carried out as shown in the subsection "Wind Load from Superstructure." The only difference is that the moment arm used for calculating the moment is equal to (Hsuper - Hpar + 6.0 feet).

                                        Formula: RWL1_5 subscript trns0 = 3 point 13K

                                        Formula: RWL2_4 subscript trns0 = 1 point 57K

                                    Formula: RWL3 subscript trns0 = 0 point 0K

Wind Load on Substructure

S3.8.1.2.3

The Specifications state that the wind loads acting directly on substructure units shall be calculated from a base wind pressure of 0.040 ksf. It is interpreted herein that this pressure should be applied to the projected area of the pier that is normal to the wind direction. This is illustrated in Figure 8-7. The resulting force is then the product of 0.040 ksf and the projected area. For nonzero wind attack angles, this force is resolved into components applied to the front and end elevations of the pier, respectively. These loads act simultaneously with the superstructure wind loads.

This figure shows two pier caps side by side in plan view, oriented with the long dimension of the pier caps in a vertical plane. Shown on the pier cap to the left are vertical and horizontal lines that intersect at the middle of the pier cap. Also shown on the pier cap to the left is an arrow that represents the wind load. This arrow is at a counterclockwise acute angle to the vertical line and the head of the arrow is at the intersection of the horizontal and vertical lines. Shown on the pier cap to the right are two lines that represent the projected face of the pier that is perpendicular to the wind load. These are shown in the figure as follows: A line is drawn at a counterclockwise acute angle from the upper left corner of the pier cap to an intersection with a line that is drawn from the lower left corner of the pier cap. These lines are at 90 degrees to each other. The line that is drawn from the lower left corner of the pier cap and intersects the line drawn from the upper left corner is the first portion of the projected face. The remaining portion of the projected face is defined by drawing a line from the upper right corner of the pier cap to the line drawn from the upper left corner of the pier cap. The line drawn from the upper right corner of the pier cap is perpendicular to the line drawn from the upper left corner of the pier cap. This is the remaining portion of the projected face.

Figure 8-7 Projected Area for Wind Pressure on Pier

What follows is an example of the calculation of the wind loads acting directly on the pier for a wind attack angle of 30 degrees. For simplicity, the tapers of the pier cap overhangs will be considered solid (this is conservative and helpful for wind angles other than zero degrees). The column height exposed to wind is the distance from the ground line (which is two feet above the footing) to the bottom of the pier cap.

Component areas of the pier cap:

                                          Formula: A subscript cap1 = ( 11 point 0 feet ) times ( 5 point 0 feet )       Formula: A subscript cap1 = 55 point 00 feet squared

                                          Formula: A subscript cap2 = ( 11 point 0 feet ) times ( 46 point 5 feet )       Formula: A subscript cap2 = 511 point 50 feet squared

Projected area of pier cap:

                                          Formula: AP subscript cap = A subscript cap1 times cos( 30 times deg) + A subscript cap2 times sin( 30 times deg)

                                          Formula: AP subscript cap = 303 point 38 feet squared

Component areas of the pier column:

                                        Formula: A subscript col1 = ( 15 feet minus 2 feet ) times ( 4 point 5 feet )       Formula: A subscript col1 = 58 point 50 feet squared

                                        Formula: A subscript col2 = ( 15 feet minus 2 feet ) times ( 15 point 5 feet )       Formula: A subscript col2 = 201 point 50 feet squared

Projected area of pier column:

                                        Formula: AP subscript col = A subscript col1 times cos( 30 times deg) + A subscript col2 times sin( 30 times deg)

                                        Formula: AP subscript col = 151 point 41 feet squared

The total wind force is then:

                                              Formula: WS subscript sub30 = 0 point 040ksf times ( AP subscript cap + AP subscript col )

                                              Formula: WS subscript sub30 = 18 point 19 K

The transverse and longitudinal force components are:

                                                  Formula: WS subscript sub30T = WS subscript sub30 times cos( 30 times deg)

                                                  Formula: WS subscript sub30L = WS subscript sub30 times sin( 30 times deg)

                                                  Formula: WS subscript sub30T = 15 point 75 K

                                                Formula: WS subscript sub30L = 9 point 10 K

The point of application of these loads will be the centroid of the loaded area of each face, respectively. This point will be approximated here as 17 feet above the top of the footing for both the transverse and longitudinal directions.

The wind attack angles for the pier must match the wind attack angles used for the superstructure. Table 8-3 shows the pier wind loads for the various attack angles.

Wind Loads Applied Directly to Pier
Wind Attack Angle
Degrees
APcap
ft2
APcol
ft2
Total Wind Load
Kips
Trans. Force
Kips
Long. Force
Kips
0
55.00
58.50
4.54
4.54
0.00
15
185.51
108.66
11.77
11.37
3.05
30
303.38
151.41
18.19
15.75
9.10
45
400.58
183.85
23.38
16.53
16.53
60
470.47
203.75
26.97
13.49
23.36

Table 8-3 Design Wind Loads Applied Directly to Pier for Various Wind Attack Angles Earthquake Load

S3.10

It is assumed in this design example that the structure is located in Seismic Zone I with an acceleration coefficient of 0.02. For Seismic Zone I, a seismic analysis is not required. However, the Specifications require a minimum design force for the check of the superstructure to substructure connection. Also, at locations of expansion bearings, a minimum bridge seat must be provided.

S4.7.4.1

S3.10.9

S4.7.4.4

Since the bearings at the pier are fixed both longitudinally and transversely, minimum bridge seat requirements for seismic loads are not applicable. Also, since the bearing design is carried out in Design Step 6, the calculations for the check of the connection will not be shown here. Therefore, the earthquake provisions as identified in the above paragraph will have no impact on the overall pier design and will not be discussed further.

Temperature Loading (Superimposed Deformations)

S3.12

In general, uniform thermal expansion and contraction of the superstructure can impose longitudinal forces on the substructure units. These forces can arise from restraint of free movement at the bearings. Additionally, the physical locations and number of substructure units can cause or influence these forces.

S3.12.2

STable 3.12.2.1-1

In this particular structure, with a single pier centered between two abutments that have identical bearing types, theoretically no force will develop at the pier from thermal movement of the superstructure. However, seldom are ideal conditions achieved in a physical structure. Therefore, it is considered good practice to include an approximate thermal loading even when theory indicates the absence of any such force.

For the purpose of this design example, a total force of 20 kips will be assumed. This force acts in the longitudinal direction of the bridge (either back or ahead station) and is equally divided among the bearings. Also, the forces at each bearing from this load will be applied at the top of the bearing (i.e., five inches above the pier cap).

                                  Formula: TU subscript 1 = 4 point 0K

                                  Formula: TU subscript 2 = 4 point 0K

                                  Formula: TU subscript 3 = 4 point 0K

                                  Formula: TU subscript 4 = 4 point 0K

                                  Formula: TU subscript 5 = 4 point 0K

Design Step 8.7 - Analyze and Combine Force Effects

The first step within this design step will be to summarize the loads acting on the pier at the bearing locations. This is done in Tables 8-4 through 8-15 shown below. Tables 8-4 through 8-8 summarize the vertical loads, Tables 8-9 through 8-12 summarize the horizontal longitudinal loads, and Tables 8-13 through 8-15 summarize the horizontal transverse loads. These loads along with the pier self-weight loads, which are shown after the tables, need to be factored and combined to obtain total design forces to be resisted in the pier cap, column and footing.

It will be noted here that loads applied due to braking and temperature can act either ahead or back station. Also, wind loads can act on either side of the structure and with positive or negative skew angles. This must be kept in mind when considering the signs of the forces in the tables below. The tables assume a particular direction for illustration only.

Superstructure Dead Load
Wearing Surface Dead Load
Bearing
Variable Name
Reaction (Kips)
Variable Name
Reaction (Kips)
1
RDCE
253.70
RDWE
39.20
2
RDCI
269.10
RDWI
39.20
3
RDCI
269.10
RDWI
39.20
4
RDCI
269.10
RDWI
39.20
5
RDCE
253.70
RDWE
39.20

Table 8-4 Unfactored Vertical Bearing Reactions from Superstructure Dead Load

Vehicular Live Load **
Lane A
Lane B
Lane C
Bearing
Variable Name
Reaction (Kips)
Variable Name
Reaction (Kips)
Variable Name
Reaction (Kips)
1
R1_a
4.19
R1_b
0.00
R1_c
0.00
2
R2_a
161.59
R2_b
0.00
R2_c
0.00
3
R3_a
70.96
R3_b
118.36
R3_c
0.00
4
R4_a
0.00
R4_b
118.36
R4_c
60.69
5
R5_a
0.00
R5_b
0.00
R5_c
176.00

**Note: Live load reactions include impact on truck loading.

Table 8-5 Unfactored Vertical Bearing Reactions from Live Load

Reactions from Transverse Wind Load on Superstructure (kips)
Wind Attack Angle (degrees)
Bearing
0
15
30
45
60
1
6.44
5.67
5.28
4.25
2.19
2
3.22
2.83
2.64
2.12
1.09
3
0.00
0.00
0.00
0.00
0.00
4
-3.22
-2.83
-2.64
-2.12
-1.09
5
-6.44
-5.67
-5.28
-4.25
-2.19

Table 8-6 Unfactored Vertical Bearing Reactions from Wind on Superstructure

Reactions from Transverse Wind Load on Vehicular Live Load (kips)
Wind Attack Angle (degrees)
Bearing
0
15
30
45
60
1
3.13
2.76
2.57
2.07
1.07
2
1.57
1.38
1.28
1.03
0.53
3
0.00
0.00
0.00
0.00
0.00
4
-1.57
-1.38
-1.28
-1.03
-0.53
5
-3.13
-2.76
-2.57
-2.07
-1.07

Table 8-7 Unfactored Vertical Bearing Reactions from Wind on Live Load

Vertical Wind Load on Superstructure
Bearing
Variable Name
Reaction (Kips)
1
RWSvert1
4.63
2
RWSvert2
-8.97
3
RWSvert3
-22.56
4
RWSvert4
-36.15
5
RWSvert5
-49.75

Table 8-8 Unfactored Vertical Bearing Reactions from Vertical Wind on Superstructure

Braking Load **
Temperature Loading
Bearing
Variable Name
Reaction (Kips)
Variable Name
Reaction (Kips)
1
BRKbrg
3.60
TU1
4.00
2
BRKbrg
3.60
TU2
4.00
3
BRKbrg
3.60
TU3
4.00
4
BRKbrg
3.60
TU4
4.00
5
BRKbrg
3.60
TU5
4.00

**Note: Values shown are for a single lane loaded

Table 8-9 Unfactored Horizontal Longitudinal Bearing Reactions from Braking and Temperature

Longitudinal Wind Loads from Superstructure (kips)
Wind Attack Angle (degrees)
Bearing
0
15
30
45
60
1
0.00
2.95
5.89
7.86
9.33
2
0.00
2.95
5.89
7.86
9.33
3
0.00
2.95
5.89
7.86
9.33
4
0.00
2.95
5.89
7.86
9.33
5
0.00
2.95
5.89
7.86
9.33
Total =
0.00
14.73
29.46
39.28
46.65

Table 8-10 Unfactored Horizontal Longitudinal Bearing Reactions from Wind on Superstructure

Longitudinal Wind Loads from Vehicular Live Load (kips)
Wind Attack Angle (degrees)
Bearing
0
15
30
45
60
1
0.00
0.58
1.15
1.54
1.82
2
0.00
0.58
1.15
1.54
1.82
3
0.00
0.58
1.15
1.54
1.82
4
0.00
0.58
1.15
1.54
1.82
5
0.00
0.58
1.15
1.54
1.82
Total =
0.00
2.88
5.76
7.68
9.12

Table 8-11 Unfactored Horizontal Longitudinal Bearing Reactions from Wind on Live Load

Longitudinal Substructure Wind Loads Applied Directly to Pier (kips)
Wind Attack Angle (degrees)
0
15
30
45
60
0.00
3.05
9.10
16.53
23.36

Table 8-12 Unfactored Horizontal Longitudinal Loads from Wind Directly on Pier

Transverse Wind Loads from Superstructure
Wind Attack Angle
Bearing
0
15
30
45
60
1
12.28
10.80
10.07
8.10
4.17
2
12.28
10.80
10.07
8.10
4.17
3
12.28
10.80
10.07
8.10
4.17
4
12.28
10.80
10.07
8.10
4.17
5
12.28
10.80
10.07
8.10
4.17
Total =
61.38
54.01
50.33
40.51
20.87

Table 8-13 Unfactored Horizontal Transverse Bearing Reactions from Wind on Superstructure

Transverse Wind Loads from Vehicular Live Load (kips)
Wind Attack Angle (degrees)
Bearing
0
15
30
45
60
1
2.40
2.11
1.97
1.58
0.82
2
2.40
2.11
1.97
1.58
0.82
3
2.40
2.11
1.97
1.58
0.82
4
2.40
2.11
1.97
1.58
0.82
5
2.40
2.11
1.97
1.58
0.82
Total =
12.00
10.56
9.84
7.92
4.08

Table 8-14 Unfactored Horizontal Transverse Bearing Reactions from Wind on Live Load

Transverse Substructure Wind Loads Applied Directly to Pier (kips)
Wind Attack Angle (degrees)
0
15
30
45
60
4.54
11.37
15.75
16.53
13.49

Table 8-15 Unfactored Horizontal Transverse Loads from Wind Directly on Pier

In addition to all the loads tabulated above, the pier self-weight must be considered when determining the final design forces. Additionally for the footing and pile designs, the weight of the earth on top of the footing must be considered. These loads were previously calculated and are shown below:

                          Formula: DL subscript cap = 313 point 88 K       Formula: DL subscript ftg = 144 point 90 K

                        Formula: DL subscript col = 156 point 94 K       Formula: EV subscript ftg = 49 point 50 K

In the AASHTO LRFD design philosophy, the applied loads are factored by statistically calibrated load factors. In addition to these factors, one must be aware of two additional sets of factors which may further modify the applied loads.

STable 3.4.1-1

STable 3.4.1-2

The first set of additional factors applies to all force effects and are represented by the Greek letter η (eta) in the Specifications. These factors are related to the ductility, redundancy, and operational importance of the structure. A single, combined eta is required for every structure. These factors and their application are discussed in detail in Design Step 1.1. In this design example, all eta factors are taken equal to one.

S1.3.2.1

The other set of factors mentioned in the first paragraph above applies only to the live load force effects and are dependent upon the number of loaded lanes. These factors are termed multiple presence factors by the Specifications. These factors for this bridge are shown as follows:

STable 3.6.1.1.2-1

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

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

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

Table 8-16 contains the applicable limit states and corresponding load factors that will be used for this pier design. Limit states not shown either do not control the design or are not applicable. The load factors shown in Table 8-16 are the standard load factors assigned by the Specifications and are exclusive of multiple presence and eta factors.

It is important to note here that the maximum load factors shown in Table 8-16 for uniform temperature loading (TU) apply only for deformations, and the minimum load factors apply for all other effects. Since the force effects from the uniform temperature loading are considered in this pier design, the minimum load factors will be used.

S3.4.1

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
BR
1.75
1.75
---
---
1.35
1.35
1.00
1.00
TU
1.20
0.50
1.20
0.50
1.20
0.50
1.20
1.00
WS
---
---
1.40
1.40
0.40
0.40
0.30
0.30
WL
---
---
---
---
1.00
1.00
1.00
1.00
EV
1.35
1.00
1.35
1.00
1.35
1.00
1.00
1.00

Table 8-16 Load Factors and Applicable Pier Limit States

STable 3.4.1-1

STable 3.4.1-2

The loads discussed and tabulated previously can now be factored by the appropriate load factors and combined to determine the governing limit states in the pier cap, column, footing and piles. For this design example, the governing limit states for the pier components were determined from a commercially available pier design computer program. Design calculations will be carried out for the governing limit states only.

Pier Cap Force Effects

The controlling limit states for the design of the pier cap are Strength I (for moment, shear and torsion) and Service I ( for crack control). The critical design location is where the cap meets the column, or 15.5 feet from the end of the cap. This is the location of maximum moment, shear, and torsion. The reactions at the two outermost bearings (numbered 4 and 5 in Figure 8-4), along with the self-weight of the cap overhang, cause the force effects at the critical section. In the following calculations, note that the number of lanes loaded to achieve the maximum moment is different than that used to obtain the maximum shear and torsion.

For Strength I, the factored vertical and horizontal forces at the bearings and corresponding force effects at the critical section are shown below. Also shown are the moment arms to the critical section.

Flexure from vertical loads (reference Tables 8-4 and 8-5):

                                            Formula: FV4 subscript cap_fIexstr1 = 1 point 25 times R subscript DCI + 1 point 50 times R subscript DWI + 1 point 75 times R subscript 4_c times m subscript 1

                                          Formula: FV4 subscript cap_fIexstr1 = 522 point 62 K

                                  Formula: ArmV4 subscript cap = 2 point 0 feet       (see Figure 8-4)

                                            Formula: FV5 subscript cap_fIexstr1 = 1 point 25 times R subscript DCE + 1 point 50 times R subscript DWE + 1 point 75 times R subscript 5_c times m subscript 1

                                          Formula: FV5 subscript cap_fIexstr1 = 745 point 52 K

                                  Formula: ArmV5 subscript cap = 11 point 75 feet

                                    Formula: Mu subscript cap_str1 = FV4 subscript cap_fIexstr1 times ArmV4 subscript cap + FV5 subscript cap_fIexstr1 times ArmV5 subscript cap + 1 point 25 times DL subscript ovrhg times ( numerator (15 point 5) divided by denominator (2) feet )

                                    Formula: Mu subscript cap_str1 = 10706 feet K

Shear from vertical loads (reference Tables 8-4 and 8-5):

                                              Formula: FV4 subscript cap_shearstr1 = 1 point 25 times R subscript DCI + 1 point 50 times R subscript DWI + 1 point 75 times ( R subscript 4_c + R subscript 4_b ) times m subscript 2

                                              Formula: FV4 subscript cap_shearstr1 = 708 point 51 K

                                              Formula: FV5 subscript cap_shearstr1 = 1 point 25 times R subscript DCE + 1 point 50 times R subscript DWE + 1 point 75 times ( R subscript 5_c + R subscript 5_b ) times m subscript 2

                                              Formula: FV5 subscript cap_shearstr1 = 683 point 92 K

                                    Formula: Vu subscript cap_str1 = FV4 subscript cap_shearstr1 + FV5 subscript cap_shearstr1 + 1 point 25 times DL subscript ovrhg

                                  Formula: Vu subscript cap_str1 = 1509 K

Torsion from horizontal loads (reference Table 8-9):

                                          Formula: FH4 subscript cap_torstr1 = 2 times ( 1 point 75 times BRK subscript brg times m subscript 2 ) + 0 point 50 times TU subscript 4

                                          Formula: FH4 subscript cap_torstr1 = 14 point 60 K

                                          Formula: FH5 subscript cap_torstr1 = 2 times ( 1 point 75 times BRK subscript brg times m subscript 2 ) + 0 point 50 times TU subscript 5

                                          Formula: FH5 subscript cap_torstr1 = 14 point 60 K

                                Formula: ArmH subscript cap = numerator (11) divided by denominator (2) feet + numerator (H subscript brng) divided by denominator (12 inches per foot)

                                Formula: ArmH subscript cap = 5 point 92 feet

                                  Formula: Tu subscript cap_str1 = ( FH4 subscript cap_torstr1 + FH5 subscript cap_torstr1 ) times ArmH subscript cap

                                  Formula: Tu subscript cap_str1 = 172 point 77 feet K

The applied torsion would be larger than the value just calculated if the vertical loads at the bearings are not coincident with the centerline of the pier cap. Some state agencies mandate a minimum eccentricity to account for this possibility. However, AASHTO does not. Therefore, no eccentricity of vertical loads is considered in this design example.

For Service I, the factored vertical forces at the bearings and corresponding force effects at the critical section are shown next. First, variables for transverse wind load on the structure and on the live load with an attack angle of zero degrees will be defined. Force effects from vertical wind load on the structure are not applicable since the Service I limit state includes wind on live load.

S3.8.2

                                        Formula: RWS5 subscript trans0 = 6 point 44K       Formula: RWL5 subscript trans0 = 3 point 13K

                                        Formula: RWS4 subscript trans0 = 3 point 22K       Formula: RWL4 subscript trans0 = 1 point 57K

Flexure from vertical loads (reference Tables 8-4 and 8-5):

                                            Formula: FV4 subscript cap_fIexser1 = 1 point 00 times R subscript DCI + 1 point 00 times R subscript DWI + 1 point 00 times R subscript 4_c times m subscript 1 + 0 point 30 times RWS4 subscript trans0 + 1 point 00 times RWL4 subscript trans0

                                            Formula: FV4 subscript cap_fIexser1 = 383 point 66 K

                              Formula: Arm4V = 2 point 0 feet       (see Figure 8-4)

                                            Formula: FV5 subscript cap_fIexser1 = 1 point 00 times R subscript DCE + 1 point 00 times R subscript DWE + 1 point 00 times R subscript 5_c times m subscript 1 + 0 point 30 times RWS5 subscript trans0 + 1 point 00 times RWL5 subscript trans0

                                            Formula: FV5 subscript cap_fIexser1 = 509 point 16 K

                              Formula: Arm5V = 11 point 75 feet

                                    Formula: Mu subscript cap_ser1 = FV4 subscript cap_fIexser1 times Arm4V + FV5 subscript cap_fIexser1 times Arm5V + 1 point 00 times DL subscript ovrhg times ( numerator (15 point 5) divided by denominator (2) feet )

                                    Formula: Mu subscript cap_ser1 = 7471 feet K

Pier Column Force Effects

The controlling limit states for the design of the pier column are Strength I (for biaxial bending with axial load), Strength III (for transverse shear) and Strength V (for longitudinal shear). The critical design location is where the column meets the footing, or at the column base. The governing force effects for Strength I are achieved by excluding the future wearing surface, applying minimum load factors on the structure dead load, and loading only Lane B and Lane C with live load. Transverse and longitudinal shears are maximized with wind attack angles of zero and 60 degrees, respectively.

For Strength I, the factored vertical forces and corresponding moments at the critical section are shown below.

Axial force (reference Tables 8-4 and 8-5):

                                        Formula: Ax subscript DL_super = 0 point 90 times ( 2 times R subscript DCE + 3 times R subscript DCI )

                                        Formula: Ax subscript DL_super = 1183 point 23 K

                                      Formula: Ax subscript DL_sub = 0 point 90 times ( DL subscript cap + DL subscript col )

                                      Formula: Ax subscript DL_sub = 423 point 73 K

                              Formula: Ax subscript LL = 1 point 75 ( R subscript 3_b + R subscript 4_b + R subscript 4_c + R subscript 5_c ) times m subscript 2

                              Formula: Ax subscript LL = 828 point 46 K

                              Formula: Ax subscript col = Ax subscript DL_super + Ax subscript DL_sub + Ax subscript LL

                              Formula: Ax subscript col = 2435 K

Transverse moment (reference Table 8-5):

                                        Formula: ArmV4 subscript col = 9 point 75 feet       Formula: ArmV5 subscript col = 19 point 5 feet

                                  Formula: Mut subscript col = 1 point 75 ( R subscript 4_b + R subscript 4_c ) times m subscript 2 times ArmV4 subscript col + 1 point 75 times ( R subscript 5_c ) times m subscript 2 times ArmV5 subscript col

                                  Formula: Mut subscript col = 9061 feet K

Longitudinal moment (reference Table 8-9):

                                              Formula: ArmH subscript col_sup = 15 feet + 11 feet + numerator (H subscript brng) divided by denominator (12 inches per foot)

                                            Formula: ArmH subscript col_sup = 26 point 42 feet

                                  Mul subscript col = 5 times ( 1 point 75 times BRK subscript brg times ArmH subscript col_sup ) times 2 times m subscript 2 + 0 point 50 ( TU subscript 1 + TU subscript 2 + TU subscript 3 + TU subscript 4 + TU subscript 5 ) times ArmH subscript col_su

                                  Formula: MuI subscript col = 1928 feet K

For Strength III, the factored transverse shear in the column is:

                                          Formula: WS subscript suptrns0 = 61 point 38 K       Formula: WS subscript sub0T = 4 point 54K

                                  Formula: Vut subscript col = 1 point 40 ( WS subscript suptrns0 + WS subscript sub0T )

                                Formula: Vut subscript col = 92 point 28 K

For Strength V, the factored longitudinal shear in the column is (reference Table 8-9):

                                            Formula: WS subscript supIng60 = 46 point 65 K       Formula: WS subscript sub60L = 23 point 36K

                                        Formula: WL subscript long60 = 9 point 12K

                                Vul subscript col = 0 point 40 ( WS subscript suplng60 + WS subscript sub60L ) + 1 point 00 times WL subscript long60 + 0 point 50 times ( TU subscript 1 + TU subscript 2 + TU subscript 3 + TU subscript 4 + TU subscript 5 ) + 1 point 35 times (5 times BRK subscript brg ) times 3 times m subscript 3

                                Formula: VuI subscript col = 109 point 09 K

Pier Pile Force Effects

The foundation system for the pier is a reinforced concrete footing on steel H-piles. The force effects in the piles cannot be determined without a pile layout. The pile layout depends upon the pile capacity and affects the footing design. The pile layout used for this pier foundation is shown in Design Step 8.10 (Figure 8-11).

Based on the pile layout shown in Figure 8-11, the controlling limit states for the pile design are Strength I (for maximum pile load), Strength III (for minimum pile load), and Strength V (for maximum horizontal loading of the pile group).

The force effects in the piles for the above-mentioned limit states are not given. The reason for this is discussed in Design Step 8.10.

Pier Footing Force Effects

The controlling limit states for the design of the pier footing are Strength I (for flexure, punching shear at the column, and punching shear at the maximum loaded pile), Strength IV (for one-way shear), and Service I ( for crack control). There is not a single critical design location in the footing where all of the force effects just mentioned are checked. Rather, the force effects act at different locations in the footing and must be checked at their respective locations. For example, the punching shear checks are carried out using critical perimeters around the column and maximum loaded pile, while the flexure and one-way shear checks are carried out on a vertical face of the footing either parallel or perpendicular to the bridge longitudinal axis.

The Strength I limit state controls for the punching shear check at the column. The factored axial load and corresponding factored biaxial moments at the base of the column are obtained in a manner similar to that for the Strength I force effects in the pier column. However, in this case the future wearing surface is now included, maximum factors are applied to all the dead load components, and all three lanes are loaded with live load. This results in the following bottom of column forces:

                                  Formula: Ax subscript col_punch = 3583K

                                    Formula: Mut subscript col_punch = 5287 feet K

                                  Formula: MuI subscript col_punch = 2756 feet K

Factored force effects for the remaining limit states discussed above are not shown. The reason for this is discussed in Design Step 8.11.

Design Step 8.8 - Design Pier Cap

Prior to carrying out the actual design of the pier cap, a brief discussion is in order regarding the design philosophy that will be used for the design of the structural components of this pier.

When a structural member meets the definition of a deep component, the Specifications recommends, although does not mandate, that a strut-and-tie model be used to determine force effects and required reinforcing. Specifications Commentary C5.6.3.1 indicates that a strut-and-tie model properly accounts for nonlinear strain distribution, nonuniform shear distribution, and the mechanical interaction of Vu, Tu and Mu. Use of strut-and-tie models for the design of reinforced concrete members is new to the LRFD Specification.

S5.2

S5.6.3.1

Traditionally, piers have been designed using conventional methods of strength of materials regardless of member dimensions. In this approach, it is assumed that longitudinal strains vary linearly over the depth of the member and the shear distribution remains uniform. Furthermore, separate designs are carried out for Vu and Mu at different locations along the member.

C5.6.3.1

For the purpose of this design example, all structural components, regardless of dimensions, will be designed in accordance with the conventional strength of materials assumptions described above. This approach is currently standard engineering practice.

The design of the pier cap will now proceed.

As stated in Design Step 8.7, the critical section in the pier cap is where the cap meets the column, or 15.5' from the end of the cap. The governing force effects and their corresponding limit states were determined to be:

Strength I

                                    Formula: Mu subscript cap_str1 = 10706 feet K

                                  Formula: Vu subscript cap_str1 = 1509 K

                                  Formula: Tu subscript cap_str1 = 172 point 77 feet K

Service I

                                    Formula: Mu subscript cap_ser1 = 7471 feet K

A preliminary estimate of the required section size and reinforcement is shown in Figure 8-8.

This figure shows a fully reinforced cross section of the pier cap at the face of the pier column. The pier cap dimensions are shown as 5 feet wide by 11 feet deep. The effective depth of the section is dimensioned as 125.97 inches from the bottom of the section to the centroid of the flexural tension steel. Inner and outer hoops of number 5 reinforcing bars at a 9 inch spacing is shown. This is the transverse shear reinforcement. All longitudinal bars are shown inside the transverse shear reinforcing. A two and one half inch clear cover is shown from any face of the pier cap to the outer hoop. The longitudinal steel shown is two rows of 10 number 11 bars at the top of the section with a 3 inch clear space between the two rows. The remaining longitudinal bars are shown as number 8 bars at an eight inch center to center spacing along both side faces of the section and number 8 bars at a 12 inch center to center spacing along the bottom face of the section.

Figure 8-8 Preliminary Pier Cap Design

Design for Flexure (Strength I)

Assume #11 bars:

                                    Formula: bar_diam11 = 1 point 41 inches

                                    Formula: bar_area11 = 1 point 56 inches squared

                  Formula: f subscript y = 60ksi

The minimum reinforcement requirements will be calculated for the cap. 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 as follows:

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

S5.4.2.6

              Formula: f subscript r = 0 point 48ksi

                Formula: I subscript g = numerator (1) divided by denominator (12) ( 60 inches ) ( 132 inches ches ) cubed

                Formula: I subscript g = 11499840 inches superscript 4

                Formula: y subscript t = 66 inches

                  Formula: M subscript cr = numerator (f subscript r times I subscript g) divided by denominator (y subscript t) times numerator (1) divided by denominator (12 inches per foot)

                  Formula: M subscript cr = 6970 feet K

                          Formula: 1 point 2 times M subscript cr = 8364 feet K

By inspection, the applied moment from the Strength I limit state exceeds 120 percent of the cracking moment. Therefore, providing steel sufficient to resist the applied moment automatically satisfies the minimum reinforcement check.

The effective depth (de) of the section shown in Figure 8-8 is computed as follows:

                          Formula: Cover subscript cp = 2 point 50 inches

                  Formula: d subscript e = 132 inches minus ( Cover subscript cp + point 625 inches + 1 point 41 inches + numerator (3) divided by denominator (2) inches )

                Formula: d subscript e = 125 point 97 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 = 60 inches

                Formula: f subscript c = 4 point 0ksi

                                Formula: Mu subscript cap_str1 = 10706 feet K

                  Formula: Rn = numerator (Mu subscript cap_str1 times 12 inches per foot) divided by denominator (( phi subscript f times b times d subscript e squared ))       Formula: Rn = 0 point 15 ksi

                Formula: rho = 0 point 85 ( numerator (f 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 subscript c ))) right bracket

              Formula: rho = 0 point 00256

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

                  Formula: A subscript s = rho times b times d subscript e       Formula: A subscript s = 19 point 32 inches squared

The area of steel provided is:

                          Formula: A subscript s_cap = 20 times ( bar_area11)

                        Formula: A subscript s_cap = 31 point 20 inches squared

                        Formula: A subscript s_cap greater than or equal to A subscript s       OK

The reinforcement area provided must now be checked to ensure that the section is not overreinforced:

S5.7.3.3.1

              Formula: T = A subscript s_cap times f subscript y       Formula: T = 1872 point 00 K

              Formula: a = numerator (T) divided by denominator (0 point 85 times f subscript c times b)       Formula: a = 9 point 18 inches

                  Formula: beta subscript 1 = 0 point 85

S5.7.2.2

              Formula: c = numerator (a) divided by denominator ( beta subscript 1)       Formula: c = 10 point 80 inches

S5.7.2.2

                  Formula: numerator (c) divided by denominator (d subscript e) = 0 point 09       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 09 less than or equal to 0 point 42       OK

Design for Flexure (Service I)

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

S5.7.3.4

Since this design example assumes that the pier cap will be exposed to deicing salts, use:

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

The distance from the extreme tension fiber to the center of the closest bar, using a maximum cover dimension of 2 inches, is:

                Formula: d subscript c = 2 point 0 inches + 0 point 625 inches + numerator (bar_diam11) divided by denominator (2)

                Formula: d subscript c = 3 point 33 inches

The area of concrete having the same centroid as the principal tensile reinforcement and bounded by the surfaces of the cross-section and a straight line parallel to the neutral axis, divided by the number of bars, is:

                  Formula: A subscript c = numerator (2 times ( d subscript c + numerator (bar_diam11) divided by denominator (2) + numerator (3 inches ) divided by denominator (2) ) times b) divided by denominator (20)

                  Formula: A subscript c = 33 point 21 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

S5.7.3.4

                  Formula: f subscript sa = 27 point 08 ksi       Formula: 0 point 6f subscript y = 36 point 00 ksi

Use       Formula: f subscript sa = 27 point 08ksi

                  Formula: E subscript s = 29000ksi

S5.4.3.2

                  Formula: E subscript c = 3640ksi

SEquation C5.4.2.4-1

              Formula: n = numerator (E subscript s) divided by denominator (E subscript c)       Formula: n = 7 point 97       Use       Formula: n = 8

The factored service moment in the cap is:

                                Formula: Mu subscript cap_ser1 = 7471 feet K

To solve for the actual stress in the reinforcement, the distance from the neutral axis to the centroid of the reinforcement (see Figure 8-9) and the transformed moment of inertia must be computed:

              Formula: n = 8

                Formula: d subscript e = 125 point 97 inches       Formula: A subscript s_cap = 31 point 20 inches squared

                Formula: rho = numerator (A subscript s_cap) divided by denominator (b times d subscript e)       Formula: rho = 0 point 00413

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

              Formula: k = 0 point 226

                    Formula: k times d subscript e = 28 point 48 inches

This figure is a cross section of the pier cap at the face of the pier column. Dimensions of the pier cap are shown and dimensioned as 60 inches wide and 132 inches deep. A bold horizontal line that represents the centroidal axis of the cracked section is shown and is dimensioned as being 28.48 inches above the bottom of the pier cap. The distance from the centroidal axis to the center of the flexural tension steel is shown and dimensioned as 97.49 inches.

Figure 8-9 Pier Cap Under Service Loads

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

                Formula: d subscript e = 125 point 97 inches

                        Formula: A subscript s_cap = 31 point 20 inches squared

              Formula: I subscript t = numerator (1) divided by denominator (3) times ( 60 inches ) times ( k times d subscript e ) cubed + n times A subscript s_cap times ( d subscript e minus k times d subscript e ) squared

              Formula: I subscript t = 2834038 inches superscript 4

Now, the actual stress in the reinforcement is computed:

                                Formula: Mu subscript cap_ser1 = 7471 feet K

              Formula: y = d subscript e minus k times d subscript e       Formula: y = 97 point 49 inches

                Formula: f subscript s = numerator (( Mu subscript cap_ser1 times 12 inches per foot times y )) divided by denominator (I subscript t) times n

                Formula: f subscript s = 24 point 67 ksi       Formula: f subscript sa = 27 point 08 ksi

                  Formula: f subscript sa greater than f subscript s       OK

Design for Flexure (Skin Reinforcement)

S5.7.3.4

In addition to the above check for crack control, additional longitudinal steel must be provided along the side faces of concrete members deeper than three feet. This additional steel is referred to in the Specifications as longitudinal skin reinforcement. This is also a crack control check. However, this check is carried out using the effective depth (de) and the required longitudinal tension steel in place of specific applied factored loads.

Figure 8-8 shows longitudinal skin reinforcement (#8 bars spaced at 8" on center) over the entire depth of the pier cap at the critical section. The Specifications require this steel only over a distance de/2 from the nearest flexural tension reinforcement. However, the reinforcing bar arrangement shown in Figure 8-8 is considered good engineering practice. This includes the placement of reinforcing steel along the bottom face of the pier cap as well, which some state agencies mandate.

The calculations shown below are for the critical section in the pier cap. The skin reinforcement necessary at this section is adequate for the entire pier cap.

                Formula: d subscript e = 125 point 97 inches       Formula: A subscript s_cap = 31 point 20 inches squared

                              Formula: bar_area8 = 0 point 79 inches squared

                  Formula: A subscript sk greater than or equal to 0 point 012 times ( d subscript e minus 30 )       and       Formula: A subscript sk less than or equal to numerator (A subscript s) divided by denominator (4)

SEquation 5.7.3.4-4

                  Formula: A subscript sk = 0 point 012 times ( 125 point 97 minus 30) square inches per foot

                  Formula: A subscript sk = 1 point 15 square inches per foot       (each side face)

                                  Formula: ( numerator (31 point 2) divided by denominator (4) ) times square inches per foot = 7 point 80 square inches per foot

                  Formula: A subscript sk less than or equal to 7 point 8 square inches per foot       OK

Spacing of the skin reinforcement:

                        Formula: S subscript Ask = min ( numerator (d subscript e) divided by denominator (6) , 12 inches ches )

                      Formula: S subscript Ask = 12 point 00 inches

Verify that #8 bars at 8" on center is adequate:

                                                  Formula: bar_area8 times ( numerator (12) divided by denominator (8) ) times numerator (1) divided by denominator ( feet ) = 1 point 18 square inches per foot

                              Formula: 1 point 18 square inches per foot greater than or equal to A subscript sk       OK

Design for Shear and Torsion (Strength I)

S5.8

The shear and torsion force effects were computed previously and are:

                              Formula: Vu subscript cap_str1 = 1509 K

                              Formula: Tu subscript cap_str1 = 172 point 77 feet K

The presence of torsion affects the total required amount of both longitudinal and transverse reinforcing steel. However, if the applied torsion is less than one-quarter of the factored torsional cracking moment, then the Specifications allow the applied torsion to be ignored. This computation is shown as follows:

S5.8.2.1

                Formula: phi subscript t = 0 point 90

S5.5.4.2.1

                    Formula: A subscript cp = ( 60 inches ) times ( 132 inches )

                  Formula: A subscript cp = 7920 inches squared

                  Formula: P subscript c = 2 times ( 60 inches + 132 inches )

                  Formula: P subscript c = 384 point 00 inches       Formula: square root of (4) ( 1 ksi) = 2 point 00 ksi

                  Formula: T subscript cr = times superscript 125( 2ksi) times numerator (( A subscript cp ) squared ) divided by denominator (P subscript c) times ( numerator (1) divided by denominator (12 inches per foot) )

                  Formula: T subscript cr = 3403 feet K

                                Formula: 0 point 25 times phi subscript t times T subscript cr = 765 point 70 feet K

                              Formula: Tu subscript cap_str1 less than 0 point 25 times phi subscript t times T subscript cr

Based on the above check, torsion will be neglected and will not be discussed further. The shear check of the critical cap section will now proceed.

The nominal shear resistance of the critical section is a combination of the nominal resistance of the concrete and the nominal resistance of the steel. This value is then compared to a computed upper-bound value and the lesser of the two controls. These calculations are illustrated below:

S5.8.3.3

                Formula: b subscript v = 60 inches       Formula: h = 132 inches

                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 )

S5.8.2.9

                Formula: d subscript v = 121 point 38 inches

                Formula: beta = 2 point 0       Formula: theta = 45deg

S5.8.3.4.1

The nominal concrete shear strength is:

                              Formula: square root of (4) ( 1 ksi) = 2 point 00 ksi

                  Formula: V subscript c = 0 point 0316 times beta times ( 2ksi) times b subscript v times d subscript v

S5.8.3.3

                  Formula: V subscript c = 920 point 52 K

Note that unless one-half of the product of Vc and the phi-factor for shear is greater than Vu, then transverse reinforcement must be provided. Therefore, when Vc is less than Vu, as in this case, transverse reinforcement is automatically required.

S5.8.2.4

The nominal steel shear strength is (using vertical stirrups, theta equal to 45 degrees):

                  Formula: A subscript v = 1 point 24 inches squared       (4 legs of #5 bars)

              Formula: s = 9 inches

S5.8.3.3

                  Formula: V subscript s = numerator (A subscript v times f subscript y times d subscript v) divided by denominator (s)

                  Formula: V subscript s = 1003 K

The nominal shear strength of the critical section is the lesser of the following two values:

                    Formula: V subscript n1 = V subscript c + V subscript s       Formula: V subscript n1 = 1924 K       (controls)

S5.8.3.3

                    Formula: V subscript n2 = 0 point 25 times f subscript c times b subscript v times d subscript v       Formula: V subscript n2 = 7283 K

Define Vn as follows:

                  Formula: V subscript n = V subscript n1       Formula: V subscript n = 1924 K

The factored shear resistance is:

                  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 = 1732 K

                Formula: V subscript r greater than Vu subscript cap_str1   OK

The shear check is not complete until the provided transverse steel is compared to the Specifications requirements regarding minimum quantity and maximum spacing.

Minimum quantity required:

S5.8.2.5

                                Formula: square root of (4) ( 1 ksi) = 2 point 00 ksi

                          Formula: A subscript v_min = 0 point 0316 times ( 2 point 0ksi) times numerator (b subscript v times s) divided by denominator (f subscript y)

                          Formula: A subscript v_min = 0 point 57 inches squared

                  Formula: A subscript v greater than A subscript v_min       OK

Maximum spacing allowed:

S5.8.2.7

                                  Formula: Vu subscript cap_str1 = 1509 K

                              Formula: v subscript u_stress = numerator (Vu subscript cap_str1) divided by denominator (( phi subscript v ) times ( b subscript v ) times ( d subscript v ))

S5.8.2.9

                              Formula: v subscript u_stress = 0 point 23 ksi

                              Formula: 0 point 125 times f prime subscript c = 0 point 50 ksi

                              Formula: v subscript u_stress less than 0 point 50ksi

                            Formula: s minus stress = 0 point 8 times ( d subscript v )

                            Formula: s minus stress = 97 point 10 inches

                      Formula: s subscript max = min ( s minus stress , 24 inches ches )

                      Formula: s subscript max = 24 point 00 inches

                Formula: s less than or equal to s subscript max       OK

Design Step 8.9 - Design Pier Column

As stated in Design Step 8.7, the critical section in the pier column is where the column meets the footing, or at the column base. The governing force effects and their corresponding limit states were determined to be:

Strength I

                                      Formula: Ax subscript col = 2435 K

                                        Formula: Mut subscript col = 9061 feet K

                                      Formula: MuI subscript col = 1928 feet K

Strength III

                                      Formula: Vut subscript col = 92 point 28 K

Strength V

                                      Formula: VuI subscript col = 109 point 09 K

A preliminary estimate of the required section size and reinforcement is shown in Figure 8-10.

This figure shows a fully reinforced cross section of the pier column. The column is dimensioned as 15 feet 6 inches wide by 4 feet 6 inches thick. Number 10 reinforcing bars are shown around the column perimeter enclosed within a number 4 hoop that is called out to be spaced at 12 inches center to center. A two and one half inch clear cover is shown from any face of the pier column to the hoop.

Figure 8-10 Preliminary Pier Column Design

Design for Axial Load and Biaxial Bending (Strength I):

S5.7.4

The preliminary column reinforcing is show in Figure 8-10 and corresponds to #10 bars equally spaced around the column perimeter. The Specifications prescribes limits (both maximum and minimum) on the amount of reinforcing steel in a column. These checks are performed on the preliminary column as follows:

S5.7.4.2

                              Formula: Num_bars = 76       Formula: bar_area10 = 1 point 27 inches squared

                        Formula: A subscript s_col = ( Num_bars) times ( bar_area10)

                        Formula: A subscript s_col = 96 point 52 inches squared

                        Formula: A subscript g_col = ( 4 point 5 feet ) times ( 15 point 5 feet ) times numerator (144 inches squared ) divided by denominator (1 feet squared )

                        Formula: A subscript g_col = 10044 inches squared

                        Formula: numerator (A subscript s_col) divided by denominator (A subscript g_col) = 0 point 0096             Formula: 0 point 0096 less than or equal to 0 point 08       OK

                              Formula: numerator (A subscript s_col times f subscript y) divided by denominator (A subscript g_col times f subscript c) = 0 point 144       Formula: 0 point 144 greater than or equal to 0 point 135       OK

The column slenderness ratio (Klu/r) about each axis of the column is computed below in order to assess slenderness effects. Note that the Specifications only permit the following approximate evaluation of slenderness effects when the slenderness ratio is below 100.

S5.7.4.3

S5.7.4.1

For this pier, the unbraced lengths (lux,luy) used in computing the slenderness ratio about each axis is the full pier height. This is the height from the top of the footing to the top of the pier cap (26 feet). The effective length factors, Kx and Ky, are both taken equal to 2.1. This assumes that the superstructure has no effect on restraining the pier from buckling. In essence, the pier is considered a free-standing cantilever.

CTable4.6.2.5-1

For simplicity in the calculations that follow, let lu=lux=luy and Kcol=Kx=Ky. This is conservative for the transverse direction for this structure, and the designer may select a lower value. The radius of gyration (r) about each axis can then be computed as follows:

                  Formula: I subscript xx = numerator (1) divided by denominator (12) times ( 186 inches ) times ( 54 inches ches ) cubed       Formula: I subscript xx = 2440692 inches superscript 4

                  Formula: I subscript yy = numerator (1) divided by denominator (12) times ( 54 inches ) times ( 186 inches ches ) cubed       Formula: I subscript yy = 28956852 inches superscript 4

                  Formula: r subscript xx = square root of ( numerator (I subscript xx) divided by denominator (A subscript g_col))

                  Formula: r subscript xx = 15 point 59 inches

                    Formula: r subscript yy = square root of ( numerator (I subscript yy) divided by denominator (A subscript g_col))

                    Formula: r subscript yy = 53 point 69 inches

The slenderness ratio for each axis now follows:

                      Formula: K subscript col = 2 point 1       Formula: I subscript u = 312 inches

                            Formula: numerator (( K subscript col times I subscript u )) divided by denominator (r subscript xx) = 42 point 03       Formula: 42 point 03 less than 100       OK

                            Formula: numerator (( K subscript col times I subscript u )) divided by denominator (r subscript yy) = 12 point 20       Formula: 12 point 20 less than 100       OK

The Specifications permits slenderness effects to be ignored when the slenderness ratio is less than 22 for members not braced against sidesway. It is assumed in this example that the pier is not braced against sidesway in either its longitudinal or transverse directions. Therefore, slenderness will be considered for the pier longitudinal direction only (i.e., about the "X-X" axis).

S5.7.4.3

In computing the amplification factor that is applied to the longitudinal moment, which is the end result of the slenderness effect, the column stiffness (EI) about the "X-X" axis must be defined. In doing so, the ratio of the maximum factored moment due to permanent load to the maximum factored moment due to total load must be identified (βd).

S4.5.3.2.2b

S5.7.4.3

From Design Step 8.7, it can be seen that the only force effects contributing to the longitudinal moment are the live load braking force and the temperature force. Neither of these are permanent or long-term loads. Therefore, βd is taken equal to zero for this design.

The column stiffness is taken as the greater of the following two calculations:

                  Formula: E subscript c = 3640 ksi       Formula: I subscript xx = 2440692 inches superscript 4

                  Formula: E subscript s = 29000 ksi       Formula: I subscript s = 44997 inches superscript 4

                  Formula: EI subscript 1 = numerator (( E subscript c times I subscript xx )) divided by denominator (5) + E subscript s times I subscript s

                  Formula: EI subscript 1 = 3 point 08 times 10 superscript 9 K inches squared

                  Formula: EI subscript 2 = numerator (E subscript c times I subscript xx) divided by denominator (2 point 5)

                  Formula: EI subscript 2 = 3 point 55 times 10 superscript 9 K inches squared       (controls)

The final parameter necessary for the calculation of the amplification factor is the phi-factor for compression. This value is defined as follows:

S5.5.4.2.1

                      Formula: phi subscript axial = 0 point 75

It is worth noting at this point that when axial load is present in addition to flexure, the Specifications permit the value of phi to be increased linearly to the value for flexure (0.90) as the factored axial load decreases from ten percent of the gross concrete strength to zero. However, certain equations in the Specification still require the use of the phi factor for axial compression (0.75) even when the increase just described is permitted. Therefore, for the sake of clarity in this example, if phi may be increased it will be labeled separately from Φaxial identified above.

S5.5.4.2.1

                      Formula: Ax subscript col = 2435 K

                                                Formula: ( 0 point 10) times ( f prime subscript c ) times ( A subscript g_col ) = 4018 K

Since the factored axial load in the column is less than ten percent of the gross concrete strength, the phi-factor will be modified and separately labeled as follows:

                              Formula: phi subscript Low_axial = 0 point 90 minus 0 point 15 times left bracket numerator (Ax subscript col) divided by denominator (( ( 0 point 10)) times ( f subscript c ) times ( A subscript g_col )) right bracket

                              Formula: phi subscript Low_axial = 0 point 81

The longitudinal moment magnification factor will now be calculated as follows:

S4.5.3.2.2b

                  Formula: P subscript e = numerator (pi 2 times ( EI subscript 2 )) divided by denominator (( K subscript col times I subscript u ) squared )       Formula: P subscript e = 81701 K

                  Formula: delta subscript s = numerator (1) divided by denominator (1 minus ( numerator (Ax subscript col) divided by denominator ( phi subscript axial times P subscript e) ))       Formula: delta subscript s = 1 point 04

The final design forces at the base of the column for the Strength I limit state will be redefined as follows:

                          Formula: P subscript u_col = Ax subscript col       Formula: P subscript u_col = 2435 K

                      Formula: M subscript ux = MuI subscript col times delta subscript s       Formula: M subscript ux = 2008 feet K

                      Formula: M subscript uy = Mut subscript col       Formula: M subscript uy = 9061 feet K

The assessment of the resistance of a compression member with biaxial flexure for strength limit states is dependent upon the magnitude of the factored axial load. This value determines which of two equations provided by the Specification are used.

S5.7.4.5

If the factored axial load is less than ten percent of the gross concrete strength multiplied by the phi-factor for compression members (Φaxial), then the Specifications require that a linear interaction equation for only the moments is satisfied (SEquation 5.7.4.5-3). Otherwise, an axial load resistance (Prxy) is computed based on the reciprocal load method (SEquation 5.7.4.5-1). In this method, axial resistances of the column are computed (using ΦLow_axial if applicable) with each moment acting separately (i.e., Prx with Mux, Pry with Muy). These are used along with the theoretical maximum possible axial resistance (Po multiplied by Φaxial) to obtain the factored axial resistance of the biaxially loaded column.

Regardless of which of the two equations mentioned in the above paragraph controls, commercially available software is generally used to obtain the moment and axial load resistances.

For this pier design, the procedure as discussed above is carried out as follows:

                                                              Formula: ( 0 point 10) times ( phi subscript axial ) times ( f prime subscript c ) times ( A subscript g_col ) = 3013 K

                        Formula: P subscript u_col less than 3013K

Therefore, SEquation 5.7.4.5-3 will be used.

                    Formula: M subscript ux = 2008 feet K       Formula: M subscript uy = 9061 feet K

                  Formula: M subscript rx = 10440 feet K       Formula: M subscript ry = 36113 feet K

                                Formula: numerator (M subscript ux) divided by denominator (M subscript rx) + numerator (M subscript uy) divided by denominator (M subscript ry) = 0 point 44       Formula: 0 point 44 less than or equal to 1 point 0       OK

The factored flexural resistances shown above, Mrx and Mry, were obtained by the use of commercial software. These values are the flexural capacities about each respective axis assuming that no axial load is present. Consistent with this, the phi-factor for flexure (0.90) was used in obtaining the factored resistance from the factored nominal strength.

Although the column has a fairly large excess flexural capacity, a more optimal design will not be pursued per the discussion following the column shear check.

Design for Shear (Strength III and Strength V)

S5.8

The maximum factored transverse and longitudinal shear forces were derived in Design Step 8.7 and are as follows:

                      Formula: Vut subscript col = 92 point 28 K       (Strength III)

                      Formula: VuI subscript col = 109 point 09 K       (Strength V)

These maximum shear forces do not act concurrently. Although a factored longitudinal shear force is present in Strength III and a factored transverse shear force is present in Strength V, they both are small relative to their concurrent factored shear. Therefore, separate shear designs can be carried out for the longitudinal and transverse directions using only the maximum shear force in that direction.

For the pier column of this example, the maximum factored shear in either direction is less than one-half of the factored resistance of the concrete. Therefore, shear reinforcement is not required. This is demonstrated for the transverse direction as follows:

S5.8.2.4

                Formula: b subscript v = 54 point inches       Formula: h = 186 inches

S5.8.3.3

                Formula: d subscript v = ( 0 point 72) times ( h)

S5.8.2.9

                Formula: d subscript v = 133 point 92 inches

The above calculation for dv is simple to use for columns and generally results in a conservative estimate of the shear capacity.

                Formula: beta = 2 point 0       Formula: theta = 45deg

S5.8.3.4.1

The nominal concrete shear strength is:

                              Formula: square root of (4) ( 1 ksi) = 2 point 00 ksi

                  Formula: V subscript c = 0 point 0316 times beta times ( 2ksi) times b subscript v times d subscript v

S5.8.3.3

                  Formula: V subscript c = 914 point 08 K

The nominal shear strength of the column is the lesser of the following two values:

                    Formula: V subscript n1 = V subscript c       Formula: V subscript n1 = 914 point 08 K       (controls)

S5.8.3.3

                    Formula: V subscript n2 = 0 point 25 times f subscript c times b subscript v times d subscript v       Formula: V subscript n2 = 7232 K

Define Vn as follows:

                  Formula: V subscript n = 914 point 08K

The factored shear resistance is:

                  Formula: phi subscript v = 0 point 90

S5.5.4.2.1

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

S5.8.2.1

                Formula: V subscript r = 822 point 67 K

                  Formula: numerator (V subscript r) divided by denominator (2) = 411 point 34 K

                  Formula: numerator (V subscript r) divided by denominator (2) greater than Vut subscript col       OK

It has just been demonstrated that transverse steel is not required to resist the applied factored shear forces. However, transverse confinement steel in the form of hoops, ties or spirals is required for compression members. In general, the transverse steel requirements for shear and confinement must both be satisfied per the Specifications.

S5.7.4.6

S5.10.6

It is worth noting that although the preceding design checks for shear and flexure show the column to be overdesigned, a more optimal column size will not be pursued. The reason for this is twofold: First, in this design example, the requirements of the pier cap dictate the column dimensions (a reduction in the column width will increase the moment in the pier cap, while good engineering practice generally prescribes a column thickness 6 to 12 inches less than that of the pier cap). Secondly, a short, squat column such as the column in this design example generally has a relatively large excess capacity even when only minimally reinforced.

Transfer of Force at Base of Column

S5.13.3.8

The provisions for the transfer of forces and moments from the column to the footing are new to the AASHTO LRFD Specifications. Although similar provisions have existed in the ACI Building Code for some time, these provisions are absent from the AASHTO Standard Specifications. In general, standard engineering practice for bridge piers automatically satisfies most, if not all, of these requirements.

In this design example, and consistent with standard engineering practice, all steel reinforcing bars in the column extend into, and are developed, in the footing (see Figure 8-13). This automatically satisfies the following requirements for reinforcement across the interface of the column and footing: A minimum reinforcement area of 0.5 percent of the gross area of the supported member, a minimum of four bars, and any tensile force must be resisted by the reinforcement. Additionally, with all of the column reinforcement extended into the footing, along with the fact that the column and footing have the same compressive strength, a bearing check at the base of the column and the top of the footing is not applicable.

In addition to the above, the Specifications requires that the transfer of lateral forces from the pier to the footing be in accordance with the shear-transfer provisions of S5.8.4. With the standard detailing practices for bridge piers previously mentioned (i.e., all column reinforcement extended and developed in the footing), along with identical design compressive strengths for the column and footing, this requirement is generally satisfied. However, for the sake of completeness, this check will be carried out as follows:

                  Formula: A subscript cv = A subscript g_col       Formula: A subscript cv = 10044 inches squared

S5.8.4.1

                  Formula: A subscript vf = A subscript s_col       Formula: A subscript vf = 96 point 52 inches squared

                  Formula: c subscript cv = 0 point 100ksi       Formula: lamda = 1 point 00

S5.8.4.2

                Formula: mu = 1 point 0 times lamda       Formula: mu = 1 point 00

                Formula: f subscript y = 60 ksi       Formula: f prime subscript c = 4 point 0 ksi

                Formula: phi subscript v = 0 point 90

S5.5.4.2.1

The nominal shear-friction capacity is the smallest of the following three equations (conservatively ignore permanent axial compression):

S5.8.4.1

                      Formula: V subscript nsf1 = c subscript cv times A subscript cv + mu times A subscript vf times f subscript y       Formula: V subscript nsf1 = 6796 K

                      Formula: V subscript nsf2 = 0 point 2 times f prime subscript c times A subscript cv       Formula: V subscript nsf2 = 8035 K

                      Formula: V subscript nsf3 = 0 point 8 times A subscript cv times ( 1 ksi)       Formula: V subscript nsf3 = 8035 K

Define the nominal shear-friction capacity as follows:

                      Formula: V subscript nsf = V subscript nsf1       Formula: V subscript nsf = 6796 K

The maximum applied shear was previously identified from the Strength V limit state:

                        Formula: VuI subscript col = 109 point 09 K

It then follows:

                                Formula: phi subscript v times ( V subscript nsf ) = 6116 K

                                Formula: phi subscript v times ( V subscript nsf ) greater than or equal to VuI subscript col       OK

As can be seen, a large excess capacity exists for this check. This is partially due to the fact that the column itself is overdesigned in general (this was discussed previously). However, the horizontal forces generally encountered with common bridges are typically small relative to the shear-friction capacity of the column (assuming all reinforcing bars are extended into the footing). In addition, the presence of a shear-key, along with the permanent axial compression from the bridge dead load, further increase the shear-friction capacity at the column/footing interface beyond that shown above. This may account for the absence of this check in both the Standard Specifications and in standard practice.

MathCad tip logo

Transfer of Force at Column Base

For common bridges with standard detailing of bridge piers and the same design compressive strength of the column and the footing, S5.13.3.8 can be considered satisfied.

Design Step 8.10 - Design Pier Piles

The foundation system for the pier is a reinforced concrete footing on steel H-piles. The force effects in the piles cannot be determined without a pile layout. The pile layout depends upon the pile capacity and affects the footing design. The pile layout used for this pier foundation is shown in Figure 8-11.

S10.7

Based on the given pile layout, the controlling limit states for the pile design were given in Design Step 8.7. However, pile loads were not provided. The reason for this is that the pile design will not be performed in this design step. The abutment foundation system, discussed in Design Step 7, is identical to that of the pier, and the pile design procedure is carried out in its entirety there. Although individual pile loads may vary between the abutment and the pier, the design procedure is similar. The pile layout shown in Figure 8-11 is used only to demonstrate the aspects of the footing design that are unique to the pier. This is discussed in the next design step.

The figure shows a plan view of the pier footing, pile layout, and pier shaft. The footing width is 23 feet. The footing length is 12 feet. The X X axis is through the footing centroid and parallel with the footing width face. The Y Y axis is through the footing centroid and parallel with the footing length face. The pile pattern is numbered from 1 to 20. Number one is ahead station leftmost pile. The back station rightmost pile is numbered as 20. The pile numbering scheme goes from left to right and top to bottom. There are 4 rows of 5 piles per row at equal spacing in both directions. The spacing along the X X axis is 5 feet. The spacing along the Y Y axis is 3 feet. The distance from the left and right edges of footing to the centerline of the nearest pile is 1 foot 6 inches. The distance from the back and ahead edges of the footing to the centerline of the nearest pile is 1 foot 6 inches. The piles are oriented so that the strong axis of the pile is parallel with the X X axis. In other words, the pile flanges are parallel with the X X axis and the pile web is parallel with the Y Y axis. The distance from the ahead station face of the pier shaft to the ahead station edge of the footing is 3 foot 9 inches. This distance is the same for the back station direction. The distance from the right edge of pier shaft to the right edge of footing is 3 feet 9 inches. This is the same for the left side.

Figure 8-11 Pier Pile Layout

Design Step 8.11 - Design Pier Footing

In Design Step 8.7, the governing limit states were identified for the design of the pier footing. However, the factored force effects were only given for the Strength I check of punching shear at the column. The reason for this is that most of the design checks for the pier footing are performed similarly to those of the abutment footing in Design Step 7. Therefore, only the aspects of the footing design that are unique to the pier footing will be discussed in this design step. This includes the punching (or two-way) shear check at the column and a brief discussion regarding estimating the applied factored shear and moment per foot width of the footing when adjacent pile loads differ.

The factored force effects from Design Step 8.7 for the punching shear check at the column are:

                                      Formula: Ax subscript col_punch = 3583K

                                        Formula: Mut subscript col_punch = 5287 feet K

                                      Formula: MuI subscript col_punch = 2756 feet K

It should be noted that in Design Step 8.5, the live load reactions at the bearings include dynamic load allowance on the truck loads. These live load force effects are part of the factored axial load and transverse moment shown above. However, the Specifications do not require dynamic load allowance for foundation components that are entirely below ground level. Therefore, the resulting pile loads will be somewhat larger (by about four percent) than necessary for the following design check. For the sake of clarity and simplicity in Design Step 8.5, a separate set of live load reactions with dynamic load allowance excluded was not provided.

S3.6.2.1

The longitudinal moment given above must be magnified to account for slenderness of the column (see Design Step 8.9). The computed magnification factor and final factored forces are:

                                  Formula: delta subscript s_punch = 1 point 06

                                    Formula: P subscript u_punch = Ax subscript col_punch       Formula: P subscript u_punch = 3583 K

                                      Formula: M subscript ux_punch = ( MuI subscript col_punch ) times ( delta subscript s_punch )

                                      Formula: M subscript ux_punch = 2921 feet K

                                      Formula: M subscript uy_punch = Mut subscript col_punch       Formula: M subscript uy_punch = 5287 feet K

With the applied factored loads determined, the next step in the column punching shear check is to define the critical perimeter, bo. The Specifications require that this perimeter be minimized, but need not be closer than dv/2 to the perimeter of the concentrated load area. In this case, the concentrated load area is the area of the column on the footing as seen in plan.

S5.13.3.6.1

The effective shear depth, dv, must be defined in order to determine bo and the punching (or two-way) shear resistance. Actually, an average effective shear depth should be used since the two-way shear area includes both the "X-X" and "Y-Y" sides of the footing. In other words, dex is not equal to dey, therefore dvx will not be equal to dvy. This is illustrated as follows assuming a 3'-6" footing with #9 reinforcing bars at 6" on center in both directions in the bottom of the footing:

S5.13.3.6.3

                              Formula: bar_area9 = 1 point 00 inches squared       Formula: bar_diam9 = 1 point 128 inches

                  Formula: b subscript ftg = 12 inches       Formula: h subscript ftg = 42 inches

                        Formula: A subscript s_ftg = 2 times ( bar_area9)

                      Formula: A subscript s_ftg = 2 point 00 inches squared       (per foot width)

Effective depth for each axis:

                            Formula: Cover subscript ftg = 3 inches

                  Formula: d subscript ey = 42 inches minus Cover subscript ftg minus numerator (bar_diam9) divided by denominator (2)

                  Formula: d subscript ey = 38 point 44 inches

                  Formula: d subscript ex = 42 inches minus Cover subscript ftg minus bar_diam9 minus numerator (bar_diam9) divided by denominator (2)

                  Formula: d subscript ex = 37 point 31 inches

Effective shear depth for each axis:

                  Formula: T subscript ftg = A subscript s_ftg times f subscript y

                  Formula: T subscript ftg = 120 point 00 K

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

                  Formula: a subscript ftg = 2 point 94 inches

                  Formula: d subscript vx = max ( d subscript ex minus numerator (a subscript ftg) divided by denominator (2) , 0 point 9 times d subscript ex , 0 point 72 times h subscript ftg )

S5.8.2.9

                  Formula: d subscript vx = 35 point 84 inches

                  Formula: d subscript vy = max ( d subscript ey minus numerator (a subscript ftg) divided by denominator (2) , 0 point 9 times d subscript ey , 0 point 72 times h subscript ftg )

S5.8.2.9

                  Formula: d subscript vy = 36 point 97 inches

Average effective shear depth:

                        Formula: d subscript v_avg = numerator (( d subscript vx + d subscript vy )) divided by denominator (2)

                        Formula: d subscript v_avg = 36 point 40 inches

With the average effective shear depth determined, the critical perimeter can be calculated as follows:

                    Formula: b subscript col = 186 inches       Formula: t subscript col = 54 inches

                  Formula: b subscript 0 = 2 left bracket b subscript col + 2 times ( numerator (d subscript v_avg) divided by denominator (2) ) right bracket + 2 times left bracket t subscript col + 2 times ( numerator (d subscript v_avg) divided by denominator (2) ) right bracket

                Formula: b subscript 0 = 625 point 61 inches

The factored shear resistance to punching shear is the smaller of the following two computed values:

S5.13.3.6.3

                  Formula: beta subscript c = numerator (b subscript col) divided by denominator (t subscript col)       Formula: beta subscript c = 3 point 44

                              Formula: square root of (4) ( 1 ksi) = 2 point 00 ksi

                                Formula: V subscript n_punch1 = ( 0 point 063 + numerator (0 point 126) divided by denominator ( beta subscript c) ) times 2ksi times ( b subscript 0 ) times ( d subscript v_avg )

                              Formula: V subscript n_punch1 = 4535 K

                                Formula: V subscript n_punch2 = 0 point 126 times ( 2ksi) times ( b subscript 0 ) times ( d subscript v_avg )

                              Formula: V subscript n_punch2 = 5739 K

Define Vn_punch as follows:

                              Formula: V subscript n_punch = V subscript n_punch1

                Formula: phi subscript v = 0 point 90

                            Formula: V subscript r_punch = phi subscript v times ( V subscript n_punch )

                            Formula: V subscript r_punch = 4082 K

With the factored shear resistance determined, the applied factored punching shear load will be computed. This value is obtained by summing the loads in the piles that are outside of the critical perimeter. As can be seen in Figure 8-12, this includes Piles 1 through 5, 6, 10,11, 15, and 16 through 20. These piles are entirely outside of the critical perimeter. If part of a pile is inside the critical perimeter, then only the portion of the pile load outside the critical perimeter is used for the punching shear check.

S5.13.3.6.1

This figure shows a plan view of the pier footing. A horizontal line labeled X X intersects a vertical line labeled Y Y at the centroid of the footing. The cross sections of all twenty (four rows of five) of the steel H piles are shown and are numbered consecutively from left to right with pile number 1 in the upper left and pile number 20 in the lower right. The outline of the pier column perimeter is shown with a fine line and it can be seen that piles 7 thru 9 and 12 thru 14 are within the pier column perimeter. A distance dimensioned as d subscript v divided by 2 is shown and this is the distance from all sides of the pier column perimeter to a dashed bold line that is called out to be the critical perimeter for column punching shear. The figure shows that piles 1 thru 5, 6,10,11,15, and 16 thru 20 are outside the critical perimeter. Also shown is the positive sense of the footing moments about each axis. A positive moment about the horizontal axis (X X) would put compression in piles 1 thru 10 and tension in piles 11 thru 20. A positive moment about the vertical axis (Y Y) would put piles 1,2,6,7,11,12, and 16 and 17 in compression and piles 4,5,9,10,14,15, and 19 and 20 in tension.

Figure 8-12 Critical Perimeter for Column Punching Shear

The following properties of the pile group are needed to determine the pile loads (reference Figures 8-11 and 8-12):

                              Formula: n subscript piles = 20

                            Formula: I subscript p_xx = 5 times left bracket ( 1 point 5 feet ) squared + ( 4 point 5 feet ) squared right bracket times 2

                            Formula: I subscript p_xx = 225 feet squared

                              Formula: I subscript p_yy = 4 times left bracket ( 5 feet ) squared + ( 10 feet ) squared right bracket times 2

                            Formula: I subscript p_yy = 1000 feet squared

The following illustrates the pile load in Pile 1:

                            Formula: P subscript 1 = numerator (P subscript u_punch) divided by denominator (n subscript piles) + numerator (M subscript ux_punch times ( 4 point 5 feet )) divided by denominator (I subscript p_xx) + numerator (M subscript uy_punch times ( 10 feet )) divided by denominator (I subscript p_yy)

                            Formula: P subscript 1 = 290 point 45 K

Similar calculations for the other piles outside of the critical perimeter yield the following:

                            Formula: P subscript 2 = 263 point 47K       Formula: P subscript 3 = 237 point 03K

                            Formula: P subscript 4 = 210 point 6K       Formula: P subscript 5 = 184 point 16K

                            Formula: P subscript 6 = 251 point 31K       Formula: P subscript 10 = 145 point 57K

                              Formula: P subscript 11 = 212 point 73K       Formula: P subscript 15 = 106 point 99K

                              Formula: P subscript 16 = 174 point 14K       Formula: P subscript 17 = 147 point 71K

                              Formula: P subscript 18 = 121 point 27K       Formula: P subscript 19 = 94 point 84K

                              Formula: P subscript 20 = 68 point 40K

The total applied factored shear used for the punching shear check is:

                                        Formula: V subscript u_punch = P subscript 1 + P subscript 2 + P subscript 3 + P subscript 4 + P subscript 5 + P subscript 6 + P subscript 10 + P subscript 11 + P subscript 15 + P subscript 16 + P subscript 17 + P subscript 18 + P subscript 19 + P subscript 20

                                      Formula: V subscript u_punch = 2509 K

                                      Formula: V subscript u_punch less than or equal to V subscript r_punch       OK

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Alternate Punching Shear Load Calculation

An alternate method for carrying out the column punching shear check is to simply use the applied factored axial load to obtain equal pile loads in all of the piles. This is only valid for the case where the piles outside of the critical perimeter are symmetric about both axes. The applied factored shear on the critical section is obtained as above (i.e., the sum of the piles located outside of the critical perimeter). This approach yields the same value for Vu_punch as wasderived above. This is illustrated as follows:

                                              Formula: V subscript u_punch_aIt = ( numerator (P subscript u_punch) divided by denominator (n subscript piles) ) times 14

                                              Formula: V subscript u_punch_aIt = 2508 K

                                              V subscript u_punch_aIt = V subscript u_punch

It has just been shown that the factored axial load alone is sufficient for the punching shear check at the column. However, consideration of the factored axial load along with the corresponding applied factored moments is necessary for other footing design checks such as punching shear at the maximum loaded pile, one-way shear, and flexure. This applies to the abutment footing in Design Step 7 as well. However, what is unique to the pier footing is that significant moments act about both axes. What follows is a demonstration, using the pile forces previously computed, of an estimation of the applied factored load on a per-foot basis acting on each footing face. The following estimations are based on the outer row of piles in each direction, respectively. Once these estimates are obtained, the appropriate footing design checks are the same as those for the abutment footing.

                                    Formula: L subscript ftg_xx = 23 feet       Formula: L subscript ftg_yy = 12 feet

Estimation of applied factored load per foot in the "X" direction:

                                              Formula: R subscript estimate_xx = numerator (2 times ( P subscript 1 + P subscript 2 ) + P subscript 3) divided by denominator (L subscript ftg_xx)

                                              Formula: R subscript estimate_xx = 58 point 47 Kips per foot

Estimation of applied factored load per foot in the "Y" direction:

                                              Formula: R subscript estimate_yy = numerator (2 times ( P subscript 1 + P subscript 6 )) divided by denominator (L subscript ftg_yy)

                                              Formula: R subscript estimate_yy = 90 point 29 Kips per foot

Design Step 8.12 - Final Pier Schematic

Figure 8-13 shows the final pier dimensions along with the required reinforcement in the pier cap and column.

This figure shows a front elevation view of the pier with final pier dimensions and reinforcement called out. The final pier dimensions are the same as the preliminary pier dimensions described in Figure 8 dash 2 and will not be described again here. The final reinforcing in the pier cap is called out to be number 5 hoops at 9 inch horizontal spacing, which are represented as vertical lines in the pier cap, 2 rows of 10 number 11 bars, which are shown as two dark horizontal lines at the top of the pier cap, number 8 bars at 8 inch vertical spacing over the depth of the pier cap, which are shown as light horizontal lines, and number 8 bars that conform to the shape of the bottom of the pier cap. These bars are called out to be at a 12 inch spacing along the bottom face of the pier cap but only a single bar is shown due to the orientation of the figure. The final reinforcing in the pier column is called out to be 76 number 10 bars equally spaced around the column perimeter. This is shown as two vertical lines adjacent to each end of the pier column separated with a horizontal line with arrows on each end. These bars are shown extending up into the pier cap and down into the pier footing. These bars are shown with hooks at the ends where they are extended into the footing. The other reinforcing in the pier column is called out to be number 4 hoops at 12 inch vertical spacing, which is represented as horizontal lines in the pier column. The vertical range of these hoops is shown as being from the middle of the footing to just inside the pier cap. A top and bottom mat of reinforcement in each direction is shown adjacent to the top and bottom faces of the pier footing but no reinforcement sizes are called out. This is due to the fact that the design of the pier footing was not performed in this design step. Similarly with the piles, they are shown in the figure but their size is not identified.

Figure 8-13 Final Pier Design

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Updated: 07/18/2013
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