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Bridges & Structures

LRFD Steel Girder SuperStructure Design Example

Pile Foundation Design Example Design Step P

Table of Contents

Design Step P.1 - Define Subsurface Conditions and Any Geometric Constraints
Design Step P.2 - Determine Applicable Loads and Load  Combinations
Design Step P.3 - Factor Loads for Each Combination
Design Step P.4 - Verify Need for a Pile Foundation
Design Step P.5 - Select Suitable Pile Type and Size
Design Step P.6 - Determine Nominal Axial Structural Resistance for Selected Pile Type / Size
Design Step P.7 - Determine Nominal Axial Geotechnical esistance for Selected Pile Type / Size
Design Step P.8 - Determine Factored Axial Structural Resistance for Single Pile
Design Step P.9 - Determine Factored Axial Geotechnical Resistance for Single Pile
Design Step P.10 - Check Drivability of Pile
Design Step P.11 - Do Preliminary Pile Layout Based on Factored Loads and Overturning Moments
Design Step P.12 - Evaluate Pile Head Fixity
Design Step P.13 - Perform Pile Soil Interaction Analysis
Design Step P.14 - Check Geotechnical Axial Capacity
Design Step P.15 - Check Structural Axial Capacity in lower portion of pile)
Design Step P.16 - Check Structural Axial Capacity in Combined Bending and Axial Load (upper portion of pile)
Design Step P.17 - Check Structural Shear Capacity
Design Step P.18 - Check Maximum Horizontal and Vertical Deflection of Pile Group at Beam Seats Using Service Load Case
Design Step P.19 - Additional Miscellaneous Design Issues
References

Design Step P.1 - Define Subsurface Conditions and Any Geometric Constraints

This task involves determining the location and extent of soil and rock materials beneath the proposed abutment and determining engineering design properties for each of those materials. It also includes identification of any specific subsurface conditions that may impact the performance of the structure. The design of the foundation system needs to address any identified issues.

A subsurface investigation was conducted at the site. Two test borings were drilled at each substructure unit. Soils were sampled at 3 foot intervals using a split spoon sampler in accordance with ASTM D-1586. Rock was continuously sampled with an N series core barrel in accordance with ASTM D-2113.

For Abutment 1, one boring was drilled at each side of the abutment. These borings are illustrated graphically in Section A1 below.

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 Pile Foundation Design.

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

                      Formula: PCF = pounds per cubic foot       Formula: ksi = numerator (1000 times lb) divided by denominator ( inches squared )       Formula: TSF = tons per square foot       Formula: kiIo = 1000

                    Formula: KSF = numerator (1000 times lb) divided by denominator ( feet squared )       Formula: K = 1000 times lb       Formula: psi = pounds per square inch       Formula: PSF = pounds per square foot

This figure shows that there are two types of soil. The first layer is a loose silty sand (sm) and the second layer is a hard grey sandstone. The bottom of footing elevation is 101 point 00.

Figure P-1 Section A1 - Subsurface Conditions at Abutment 1

Evaluation of Section A1 indicates that subsurface conditions are relatively uniform beneath the proposed abutment consisting of essentially 2 materials.

Loose silty sand was encountered in the top 35 feet of each boring. This material is non-plastic and contains about 15% fine material. Below a depth of about 5' the soil is saturated.

Rock was encountered at about elevation 70 in both borings. The rock consists of a hard gray sandstone. Fractures are tight with no infilling and occur at a spacing of 1-3'; primarily along bedding planes which are horizontal. Slight weathering was observed in the upper 1' foot of the rock but the remainder of the rock is unweathered.

Special Geotechnical Considerations:

The loose fine sandy soils could be subject to liquefaction under seismic loading. Liquefaction is a function of the anticipated maximum earthquake magnitude and the soil properties. If liquefaction is a problem, the soils can not be relied upon to provide lateral support to deep foundation systems. For this example it is assumed that the potential for liquefaction has been evaluated and has been found to be negligible. (Note: Seed and Idriss (NCEER-97-0022) provides more up to date material for evaluation of liquefaction)

C10.5.4, SAppendix A10

The weight of the approach embankment will cause compression of the loose soil horizon. The granular material should compress essentially elastically with little or no long term consolidation. However, since the full height abutment will likely be placed prior to completion of the approach embankment in the vicinity of the abutment, soil compression beneath the abutment must be accounted for in foundation design. For shallow foundations, this compression will result in settlement and rotation of the footing. For deep foundations this compression could result in negative skin friction (downdrag) loads on the foundation elements; particularly in the back row of piles.

S10.7.1.4, C10.7.1.4

Development of Parameters for Design:

Layer 1 - Soil

Depth:

Assuming a bottom of footing elevation of 101 FT and a top of rock elevation of 70 FT as described above:

                                  Formula: 101 feet minus 70 feet = 31 feet

Unit Weight ( Υ ):

Consider relevant published data when selecting design parameters. For unit weights of in-situ soil materials, a good reference is NAVFAC DM7.1-22. Based on this reference, general and local experience, and the above description of the soil horizon as loose silty sand, the unit weights were selected as follows:

C10.4.1

Dry unit weight:       Formula: Y subscript dry = 90 PCF

Wet unit weight:       Formula: Y subscript wet = 110 PCF

Unit weight of water:       Formula: Y subscript water = 62 point 4 PCF

Effective unit weight:       Formula: Y subscript eff = Y subscript wet minus Y subscript water

                                                          Formula: Y subscript eff = 47 point 6 PCF

Angle of internal friction ( φ ):

The angle of internal friction can be estimated based on correlation to Standard Penetration Test (SPT) N values. The raw SPT N-values determined in the test borings must be corrected for overburden pressure as follows:

Formula: N subscript corr = ( 0 point 77 times log ( numerator (20) divided by denominator ( sigma prime subscript nu) )) times N

SEquation 10.7.2.3.3-4

where:

Corrected SPT blow count (Blows/FT)

Note: The formula above is generally considered valid for values of σ' > 0.25 TSF (Bowles 1977):       N subscript corr

SPT blow count (Blows/FT):       N

Vertical effective stress at bottom of sample (TSF):       sigma prime subscript nu Formula: sigma prime subscript nu = numerator ( Sigma ( h subscript i times Y subscript effi )) divided by denominator (2000)

where:

Thickness of soil layer i above point being considered (FT):       h subscript i

Effective unit weight of soil layer i (PCF):       Y subscript effi

Number of soil layer under consideration:       i

This formula is implemented for each of the borings below. Wet unit weight is used for the soil above the water table and effective unit weight is used for the soil below the water table.

Depth to Top of Sample (FT)
Depth to Bottom of Sample (FT)
Ueff i (PCF)
sn' (TSF)
N Blows/Ft (BPF)
Ncorr Blows/Ft (BPF)
Boring A1-1
0
1.5
110
0.0825
5
9
3
4.5
110
0.2475
5
7
6
7.5
47.6
0.3189
4
6
9
10.5
47.6
0.3903
3
4
12
13.5
47.6
0.4617
5
6
15
16.5
47.6
0.5331
6
7
18
19.5
47.6
0.6045
3
4
21
22.5
47.6
0.6759
3
3
24
25.5
47.6
0.7473
6
7
27
28.5
47.6
0.8187
9
10
30
31.5
47.6
0.8901
12
12
33
34.5
47.6
0.9615
14
14
Boring A1-2
0
1.5
110
0.0825
2
4
3
4.5
110
0.2475
3
4
6
7.5
47.6
0.3189
5
7
9
10.5
47.6
0.3903
6
8
12
13.5
47.6
0.4617
8
10
15
16.5
47.6
0.5331
4
5
18
19.5
47.6
0.6045
6
7
21
22.5
47.6
0.6759
9
10
24
25.5
47.6
0.7473
10
11
27
28.5
47.6
0.8187
10
11
30
31.5
47.6
0.8901
11
11
33
34.5
47.6
0.9615
13
13

Table P-1 Calculation of Corrected SPT Blow Count

Find average values for zone between bottom of footing and top of rock. This means ignoring the first two values of each boring.

                Formula: N = 7 point 35            BPF

                Formula: N subscript corr = 8 point 3       BPF

The correlation published in FHWA-HI-96-033 Page 4-17 (after Bowles, 1977) is used to determine the angle of internal friction. This correlation is reproduced below.

Description
Very Loose
Loose
Medium
Dense
Very Dense
Ncorr =
0-4
4-10
10-30
30-50
>50
jf =
25-30o
27-32o
30-35o
35-40o
38-43o
a =
0.5
0.5
0.25
0.15
0
b =
27.5
27.5
30
33
40.5

Table P-2 Correlation

This correlation can be expressed numerically as: Formula: phi prime subscript f = a times N subscript corr + b

where:

a and b are as listed in Table P-2.

              Formula: a = 0 point 5

                      Formula: N subscript corr = 8 point 3

              Formula: b = 27 point 5

Thus

                Formula: phi prime subscript f = a times N subscript corr + b

                Formula: phi prime subscript f = 31 point 65       o       say       Formula: phi prime subscript f = 31       o

Modulus of elasticity (E):

Estimating E0 from description

STable 10.6.2.2.3b-1

Loose Fine Sand E0=       80 - 120 TSF

Estimating E0 from Ncorr

Note, in Table 10.6.2.2.3b-1 N1 is equivalent to Ncorr

Clean fine to medium sands and slightly silty sands

                  Formula: E subscript 0 = 7 times N subscript 1

STable 10.6.2.2.3b-1

                  Formula: E subscript 0 = 7 times N subscript corr

                  Formula: E subscript 0 = 58 point 1       TSF

Based on above, use:

                  Formula: E subscript 0 = 60 times TSF

                  Formula: E subscript 0 = 0 point 833 ksi

Poisons Ratio ( ν ):

Estimating ν from description

STable 10.6.2.2.3b-1

Loose Fine Sand:       Formula: nu = 0 point 25

Shear Modulus (G):

From Elastic Theory:

                  Formula: G subscript 0 = numerator (E subscript 0) divided by denominator (2 times ( 1 + nu ))

                  Formula: G subscript 0 = 24TSF

                  Formula: G subscript 0 = 0 point 33 ksi

Coefficient of variation of subgrade reaction (k):

As per FHWA-HI-96-033, Table 9-13:

This is used for lateral analysis of deep foundation elements

Submerged Loose Sand

              Formula: k = 5430 times numerator (kiIo times neWton) divided by denominator (m cubed )       Formula: k = 20 times psi

Layer 2 - Rock:

Depth:

Rock is encountered at elevation 70 and extends a minimum of 25 FT beyond this point.

Unit Weight ( Υ ):

Determined from unconfined compression tests on samples of intact rock core as listed below:

Boring No.
Depth (FT)
U (PCF)
A1-1
72.5
152
A1-1
75.1
154
A1-2
71.9
145
A1-2
76.3
153
P1-1
81.2
161
P1-2
71.8
142
A2-1
76.3
145
A2-2
73.7
151
Average U
150.375

Table P-3 Unit Weight

                      Formula: Y subscript ave = 150 point 375 PCF

Unconfined Compressive Strength (q):

Determined from unconfined compression tests on samples of intact rock core as listed below:

Boring No.
Depth (FT)
qu (PSI)
A1-1
72.5
12930
A1-1
75.1
10450
A1-2
71.9
6450
A1-2
76.3
12980
P1-1
81.2
14060
P1-2
71.8
6700
A2-1
76.3
13420
A2-2
73.7
14890
Average qu
11485

Table P-4 Unconfined Compressive Strength

                        Formula: q subscript uave = 11485 times psi

Modulus of elasticity (E):

STable 10.6.2.2.3d-2

This is to be used for prediction of deep foundation response

For sandstone, Average:       Formula: E subscript 0 = 153000 times TSF

                                                                  Formula: E subscript 0 = 2125 ksi

Poisons Ratio ( ν ):

STable 10.6.2.2.3d-1

This is to be used for prediction of pile tip response

For sandstone, Average:       Formula: nu subscript ave = 0 point 2

Shear Modulus (G):

From elastic theory

                  Formula: G subscript 0 = numerator (E subscript 0) divided by denominator (2 times ( 1 + nu subscript ave ))

                  Formula: G subscript 0 = 63750TSF

                  Formula: G subscript 0 = 885 point 417 ksi

Rock Mass Quality:

Rock mass quality is used to correct the intact rock strength and intact modulus values for the effects of existing discontinuities in the rock mass. This is done through empirical correlations using parameters determined during core drilling.

Data from the test borings is summarized below:

Depth (FT)
Run Length (FT)
Recovery (%)
RQD (%)
Boring A1-1
35
5
100
80
40
5
96
94
45
5
100
96
50
5
98
92
55
5
98
90
Boring A1-2
35
5
98
90
40
5
100
80
45
5
100
96
50
5
96
90
55
5
98
96
Averages
98.4
90.4

Table P-5 Rock Mass Quality

Design Step P.2 - Determine Applicable Loads and Load Combinations

Loads and load combinations are determined elsewhere in the design process. The critical load cases for evaluation of foundation design are summarized below:

  1. The load combination that produces the maximum vertical load on the foundation system. This will typically be a Strength I and a Service I load case with the maximum load factors applied.

  2. The load combination that produces the maximum overturning on the foundation which will tend to lift a spread footing off the bearing stratum or place deep foundation elements in tension.

  3. The load combination that produces the maximum lateral load. If several combinations produce the same horizontal load, select the one with the minimum vertical load as this will be critical for evaluation of spread footing sliding or response of battered deep foundations. In some cases, particularly deep foundations employing all vertical elements, the highest lateral load and associated highest vertical load should also be evaluated as this case may produce higher foundation element stress and deflections due to combined axial load and bending in the foundation elements.

Design Step P.3 - Factor Loads for Each Combination

It is extremely important to understand where the loads are being applied with respect to foundation design. In this case the loads were developed based on an assumed 10' 3" wide by 46' 10 1/2" long footing that is offset behind the bearings a distance of 1' 9". The loads are provided at the horizontal centroid of the assumed footing and at the bottom of that footing. A diagram showing the location and direction of the applied loads is provided below.

The figure shows a footing with a length of 46 feet 10 and one half inches and a footing width of 10 feet 3 inches. The distance from centerline of bearings to the centerline of footing is 1 foot 9 inches. The distance from the centerline of bearing to the edge of footing is 3 feet 4 and one half inches. There are 3 loads and 2 moments applied to the footing. There is a vertical load, a longitudinal load, a transverse load, a moment in the transverse axis and a moment in the longitudinal axis.

Figure P-2 Application of Loads

LIMIT STATE AXIAL FORCE Pvert (K) LONG MOMENT Mlong (K-FT) TRANS MOMENT Mtrans (K-FT) LATERAL LOAD (IN LONG. DIR.) Plong (K) LATERAL LOAD (IN TRANS. DIR.) Ptrans (K)
Maximum Vertical Load STR-I MAX/FIN 2253 7693 0 855 0
SER-I MAX/FIN 1791 4774 162 571 10
Maximum Overturning STR-I MIN/FIN 1860 7291 0 855 0
SER-I MIN/FIN 1791 4709 162 568 10
Maximum Lateral Load STR-III MAX/FIN 1815 6374 508 787 37
SER-I MAX/FIN 1791 4774 162 571 10

Table P-6 Summary of Factored Loads

It should be noted that the calculations performed in Design Step P are based on preliminary pile foundation design forces. In an actual design, the geotechnical engineer would need to revisit the pile foundation design calculations and update the results based on the final design bottom of booting forces given at the end of Design Step 7.7.

Design Step P.4 - Verify Need for a Pile Foundation

Evaluate a spread footing design:

Check vertical capacity:

Presumptive Bearing Capacity for loose sand with silt (SM)

Presumptive bearing capacity       Formula: SM = 1 point 5 times TSF

STable 10.6.2.3.1-1

                                                                        Formula: SM = 3 KSF

Presumptive bearing capacity is a service limit state, thus compare against maximum service load.

S10.5.2

From Design Step P.3, the Maximum service load is Formula: P subscript vert = 1791 K

The Required area:       Formula: A = 597 feet squared

The length of the footing is controlled by the length of the abutment step required to support the steel beams and the approach roadway. This is determined from previous geometry calculations.

Maximum possible length of footing       Formula: L = 46 point 875 feet

Preliminary minimum required width       Formula: B subscript min = 12 point 736 feet

Excessive loss of contact:

This is a strength limit state thus use strength loads for the case of maximum overturning which is STR I Min.

S10.5.3

Determine the maximum eccentricity eB in the direction parallel to the width of the footing (B)

                 Formula: e subscript B = numerator (M subscript long) divided by denominator (P subscript vert)

From the loads obtained in Design Step P.3,       Formula: M subscript long = 7291 K feet

                                                                                                      Formula: P subscript vert = 1860 K

                  Formula: e subscript B = numerator (M subscript long) divided by denominator (P subscript vert)

                  Formula: e subscript B = 3 point 92 feet

To prevent excessive loss of contact eB must be less than B/4.

S10.6.3.1.5

Width of the footing:       Formula: B subscript i = 10 point 25 feet

                  Formula: numerator (B subscript i) divided by denominator (4) = 2 point 563 feet

In order to resolve the bearing pressure and eccentricity issue, the footing will have to be widened and the centroid shifted toward the toe. This can be accomplished by adding width to the toe of the footing. Note that the issue could also be resolved by adding width to the heel of the footing, which would increase the weight of soil that resists overturning. This would require recalculation of the loads and was not pursued here.

In order to satisfy bearing pressure and eccentricity concerns, the footing width is increased incrementally until the following two criteria are met:

                  Formula: e subscript B less than numerator (B subscript i) divided by denominator (4)       Based on Strength Loads

            B prime       >       Formula: B subscript min = 12 point 736 feet       Based on Service Loads

Where B' is the effective footing width under eccentric load

                Formula: B prime = B subscript i minus 2 times e subscript B

SEquation 10.6.3.1.5-1

For the Strength Load case:

Footing width B (FT)
Distance from heel to Centroid of footing (FT)
Distance from heel to centroid of load (FT)
eB (FT)
B/4 (FT)
10.25
5.13
9.05
3.92
2.56
11.00
5.50
9.05
3.55
2.75
12.00
6.00
9.05
3.05
3.00
13.00
6.50
9.05
2.55
3.25
14.00
7.00
9.05
2.05
3.50
15.00
7.50
9.05
1.55
3.75
16.00
8.00
9.05
1.05
4.00
17.00
8.50
9.05
0.55
4.25

Table P-7 Excessive Loss of Contact - Strength

For the Strength Load Case, the condition was satisfed first when the width of the footing B = 13.00 FT

For the Service Load Case

                  Formula: e subscript B = numerator (M subscript long) divided by denominator (P subscript vert)

From the loads obtained from Design Step P.3,       Formula: M subscript long = 4774 K feet

                                                                                                        Formula: P subscript vert = 1791 K

                  Formula: e subscript B = numerator (M subscript long) divided by denominator (P subscript vert)

                  Formula: e subscript B = 2 point 67 feet

Footing width B (FT)
Distance from heel to Centroid of footing (FT)
Distance from heel to centroid of load (FT)
eB (FT)
B' (FT)
10.25
5.13
7.80
2.67
4.91
11.00
5.50
7.80
2.30
6.41
12.00
6.00
7.80
1.80
8.41
13.00
6.50
7.80
1.30
10.41
14.00
7.00
7.80
0.80
12.41
15.00
7.50
7.80
0.30
14.41
16.00
8.00
7.80
-0.21
16.41

Table P-8 Presumptive Bearing Pressure - Service

For the Service Load Case, the condition was satisfed first when the width of the footing B = 15.00 FT

The first width to satisfy both conditions is 15.00 FT. Which would require the toe of the footing to be extended:

                  Formula: Delta B = 15 feet minus B subscript i

                  Formula: Delta B = 4 point 75 feet

This increase may not be possible because it may interfere with roadway drainage, roadside utilities, or the shoulder pavement structure. However, assume this is not the case and investigate potential settlement of such a footing.

Settlement is a service limit state check.

For the granular subsoils, settlement should be esentially elastic thus Settlement (S0) is computed from:

                Formula: S subscript 0 = numerator (q subscript 0 times ( 1 minus nu 2 ) times A superscript 0 point 5) divided by denominator (E subscript s times beta subscript z)

SEquation 10.6.2.2.3b-1

Assume the footing is fully loaded, thus q0 is the presumptive bearing capacity and effective loaded area is as calculated above

Average bearing pressure on loaded area:       Formula: q subscript 0 = SM

                                                                                                  Formula: q subscript 0 = 1 point 5TSF

Effective are of footing: Formula: A = L prime times B prime

Length of footing

                                Formula: L prime = L       Formula: L prime = 46 point 875 feet

Width of the footing

                                Formula: B prime = B subscript min       Formula: B prime = 12 point 736 feet

Therfore, the Effective Area is

                                Formula: A = L prime times B prime       Formula: A = 597 feet squared

Modulus of elasticity of soil, from Design Step P.1:       Formula: E subscript s = 60 times TSF

Poisson's ratio of soil, from Design Step P.1:       Formula: nu = 0 point 25

Shape factor for rigid footing:       beta subscript z       at       Formula: numerator (L prime) divided by denominator (B prime) = 3 point 681

From Table 10.6.2.2.3b-2 for rigid footing:

L'/B' bz
3 1.15
5 1.24

STable 10.6.2.2.3b-2

Table P-9 Rigid Footing

By interpolation, at       Formula: numerator (L prime) divided by denominator (B prime) = 3 point 681       Formula: beta subscript z = 1 point 18

                  Formula: S subscript 0 = numerator (q subscript 0 times ( 1 minus nu 2 ) times A superscript 0 point 5) divided by denominator (E subscript s times beta subscript z)

                  Formula: S subscript 0 = 0 point 49 feet       Formula: S subscript 0 = 5 point 8 inches

Note: This computation assumes an infinite depth of the compressible layer. Other computation methods that allow for the rigid base (NAVFAC DM-7.1-211) indicate the difference between assuming an infinite compressible layer and a rigid base at a depth equal to 3 times the footing width (H/B = 3) below the footing can be estimated by computing the ratio between appropriate influence factors (I) as follows:

As per NAVFAC DM7.1-212, and DM7.1-213:

I for rigid circular area over infinite halfspace:       Formula: I subscript inf = 0 point 79

I for rigid circular area over stiff base at H/B of 3:       Formula: I subscript sb = 0 point 64

The influence value determined above is for a Poisson's ratio of 0.33. A Poisson's ration of 0.25 is used for the soil. This difference is small for the purposes of estimating elastic settlement.

Ratio of I values:

                  Formula: numerator (I subscript sb) divided by denominator (I subscript inf) = 0 point 810127

Since I is directly proportional to settlement, this ratio can be multiplied by S0 to arrive at a more realistic prediction of settlement of this footing.

                  Formula: S prime subscript 0 = S subscript 0 times numerator (I subscript sb) divided by denominator (I subscript inf)

                  Formula: S prime subscript 0 = 4 point 718 inches

This settlement will occur as load is applied to the footing and may involve some rotation of the footing due to eccentricities of the applied load. Since most of the loads will be applied after construction of the abutment (backfill, superstructure, deck) this will result in unacceptable displacement.

The structural engineer has determined that the structure can accommodate up to 1.5" of horizontal displacement and up to 0.5" vertical displacement. Given the magnitude of the predicted displacements, it is unlikely this requirement can be met. Thus, a deep foundation system or some form of ground improvement is required.

Note that the above calculation did not account for the weight of the approach embankment fill and the effect that this will have on the elastic settlement. Consideration of this would increase the settlement making the decision to abandon a spread footing foundation even more decisive.

Design Step P.5 - Select Suitable Pile Type and Size

It will be assumed that for the purposes of this example, ground improvement methods such as vibro-flotation, vibro replacement, dynamic deep compaction, and others have been ruled out as impractical or too costly. It is further assumed that drilled shaft foundations have been shown to be more costly than driven pile foundations under the existing subsurface conditions (granular, water bearing strata). Thus a driven pile foundation will be designed.

Of the available driven pile types, a steel H-pile end bearing on rock is selected for this application for the following reasons.

  1. It is a low displacement pile which will minimize friction in the overlying soils.

  2. It can be driven to high capacities on and into the top weathered portion of the rock.

  3. It is relatively stiff in bending thus lateral deflections will be less than for comparably sized concrete or timber piles.

  4. Soils have not been shown to be corrosive thus steel loss is not an issue.

To determine the optimum pile size for this application, consideration is given to the following:

1) Pile diameter:

H-Piles range in size from 8 to 14 inch width. Since pile spacing is controlled by the greater of 30 inches or 2.5 times the pile diameter (D); pile sizes 12 inches and under will result in the same minimum spacing of 30 inches. Thus for preliminary analysis assume a 12 inch H-Pile.

2) Absolute Minimum Spacing:

Per referenced article, spacing is to be no less than:       2 point 5 times D

S10.7.1.5

Where the pile diameter:       Formula: D = 12 inches

                              Formula: 2 point 5 times D = 30 inches

3) Minimum pile spacing to reduce group effects:

As per FHWA-HI-96-033, Section 9.8.1.1:

Axial group effects for end bearing piles on hard rock are likely to be negligible thus axial group capacity is not a consideration. However, note that the FHWA driven pile manual recommends a minimum c-c spacing of 3D or 1 meter in granular soils to optimize group capacity and minimize installation problems. The designer's experience has shown 3D to be a more practical limit that will help avoid problems during construction.

Lateral group effects are controlled by pile spacing in the direction of loading and perpendicular to the direction of loading.

From Reese and Wang, 1991, Figure 5.3 (personal communication):

For spacing perpendicular to the direction of loading 3D results in no significant group impacts.

As per FHWA-HI-96-033, Section 9.8.4 & NACVFAC DM7.2-241:

For spacing in the direction of loading, various model studies indicate that group efficiency is very low at 3D spacing, moderate at about 5D spacing and near 100% for spacings over about 8D. Thus it is desirable to maintain at least 5D spacing in the direction of the load and preferable to maintain 8D spacing.

Maximum pile spacing

Spacing the piles more than 10 feet c-c results in higher bending moments in the pile cap between each pile and negative bending moments over the top of each pile that may result in additional steel reinforcing or thicker pile caps. Thus it is desirable to keep the pile spacing less than 10 feet c-c.

4) Edge clearance

Referenced section indicates minimum cover:       Formula: cover subscript min = 9 inches

S10.7.1.5

Thus for a 12 inch pile, minimum distance from edge of footing to center of pile:

                                  Formula: dist subscript min = cover subscript min + numerator (D) divided by denominator (2)

                                  Formula: dist subscript min = 1 point 25 feet

5) Maximum pile cap dimensions

The length of the pile cap in the direction perpendicular to the centerline (L) is limited to the width of the abutment. Thus:

From Design Step P.4:

                              Formula: L subscript max = L

                              Formula: L subscript max = 46 point 875 feet

The width of the pile cap in the direction parallel to the centerline of the bridge (B) can generally be made wider as required. Initial loadings were developed assuming a width of 10.25 FT thus use this dimension as a starting point.

                        Formula: B = 10 point 25 feet

Determine the maximum and minimum number of piles that can be placed beneath the cap (See sketch below for definition of variables)

This figure shows a footing with a length of L and a width of B. The pile spacing in the B direction is defined as SB and the pile spacing in the L direction is defined as SL. The number of pile spaces is the B direction is defined as NB and number of pile spaces is the L direction is defined as NL.

Figure P-3 Plan View of Pile Cap

In B direction:

        S subscript B       is defined as: Width of the pile cap - 2 times the edge distance

                  Formula: S subscript B = B minus 2 times dist subscript min

                  Formula: S subscript B = 7 point 75 feet

Max number of spaces at 5D spacing (NB)

                  Formula: N subscript B less than numerator (S subscript B) divided by denominator (5 times D)

                    Formula: numerator (S subscript B) divided by denominator (5 times D) = 1 point 55

                  Formula: N subscript B less than 1 point 55

Minimum number of spaces at 10' each (NB)

                  Formula: N subscript B greater than numerator (S subscript B) divided by denominator (10 feet )

                      Formula: numerator (S subscript B) divided by denominator (10 feet ) = 0 point 775

                  Formula: N subscript B greater than 0 point 775

Since the number of spaces has to be an integer

                  Formula: N subscript B = 1

Which results in two rows of piles in the B direction.

In L direction:

        S subscript L       is defined as: Width of the pile cap - 2 times the edge distance

                  Formula: S subscript L = L minus 2 times dist subscript min

                  Formula: S subscript L = 44 point 375 feet

Max number of spaces at 3D spacing (NL)

                  Formula: N subscript L less than numerator (S subscript L) divided by denominator (3 times D)

                    Formula: numerator (S subscript L) divided by denominator (3 times D) = 14 point 792

                  Formula: N subscript L less than 14 point 792

Minimum number of spaces at 10' each (NL)

                  Formula: N subscript L greater than numerator (S subscript L) divided by denominator (10 feet )

                      Formula: numerator (S subscript L) divided by denominator (10 feet ) = 4 point 438

                  Formula: N subscript L greater than 4 point 438

Since the number of spaces has to be an integer

                  Formula: N subscript L = 5       to 14

Which results in 6 to 15 rows of piles in the L direction.

Determine maximum axial load acting on piles

Using factored loads and diagram below, determine reactions on the front and back pile rows:

This figure shows a section view of the pile cap. There is a vertical load, P vertical, and a Moment, M longitudinal, applied to the pile cap. The positive X direction goes from left to right and the positive Z direction goes from top to bottom. There are two reactions R back and R front. The spacing in the B direction is SB which is equal to 7 point 75 feet.

Figure P-4 Section View of Pile Cap

Summing the forces in the z-direction and the moments about point B:

                    Formula: Sigma F subscript z = 0 Formula: Sigma F subscript z = P subscript vert minus R subscript BACK minus R subscript FRONT

                      Formula: Sigma M subscript B = 0 Formula: Sigma M subscript B = minus M subscript long minus numerator (S subscript B) divided by denominator (2) times P subscript vert + S subscript B times R subscript FRONT

For STR I max, from Table P.6:

                        Formula: P subscript vert subscript 1 = 2253 K       Formula: M subscript long subscript 1 = 7963 K feet

                              Formula: R subscript FRONT subscript 1 = 2119 K       Formula: R subscript BACK subscript 1 = 134 K

For STR I min, from Table P.6:

                        Formula: P subscript vert subscript 2 = 1860 K       Formula: M subscript long subscript 2 = 7291 K feet

                              Formula: R subscript FRONT subscript 2 = 1871 K       Formula: R subscript BACK subscript 2 = minus 11 K

Max axial load on front row of piles:

                            Formula: R subscript FRONT = max ( R subscript FRONT subscript 1 , R subscript FRONT subscript 2 )

                            Formula: R subscript FRONT = 2119 K

Since the front row can have 6 - 15 piles,

Max anticipated factored pile load can range between:

                              Formula: R subscript FRONT6 = numerator (R subscript FRONT) divided by denominator (6)       Formula: R subscript FRONT15 = numerator (R subscript FRONT) divided by denominator (15)

and

                              Formula: R subscript FRONT6 = 353 point 167 K       Formula: R subscript FRONT15 = 141 point 267 K

Assuming the following:

Axial pile resistance is controlled by structural resistance

SEquation 6.9.2.1-1 and SEquation 6.9.4.1-1

Structural resistance Formula: P subscript r = phi subscript c times F subscript y times A subscript s

NOTE: λ in equation 6.9.4.1-1 is assumed to be zero (because unbraced length is zero) resulting in the simplified equation shown above.

                  Formula: phi subscript c = 0 point 6

S6.5.4.2

                  Formula: F subscript y = 36 ksi

NOTE: Grade 36 steel is assumed at this stage even though most H-pile sections are available in higher grades at little or no cost differential. The need for using a higher strength steel will be investigated in future design steps

Compute required pile area to resist the anticipated maximum factored pile load. The required steel area can range between:

                                      Formula: numerator (( numerator (R subscript FRONT6) divided by denominator ( phi subscript c) )) divided by denominator (F subscript y) = 16 point 35 inches squared       and       Formula: numerator (( numerator (R subscript FRONT15) divided by denominator ( phi subscript c) )) divided by denominator (F subscript y) = 6 point 54 inches squared

For preliminary layout and design, select: HP 12x53

Properties of HP 12x53:

                  Formula: A subscript s = 15 point 5 inches squared

              Formula: d = 11 point 78 inches

                Formula: b subscript f = 12 point 045 inches

              Formula: t subscript f = 0 point 435 inches

                Formula: t subscript w = 0 point 435 inches

                  Formula: I subscript xx = 393 inches superscript 4

                  Formula: I subscript yy = 127 inches superscript 4

                Formula: Z subscript x = 74 inches cubed

Note: Plastic section modulus is used to evaluate nominal moment capacity

                Formula: Z subscript y = 32 point 2 inches cubed

                  Formula: E subscript s = 29000 ksi

Design Step P.6 - Determine Nominal Axial Structural Resistance for Selected Pile Type / Size

Ultimate axial compressive resistance is determined in accordance with either equation 6.9.4.1-1 or 6.9.4.1-2. The selection of equation is based on the computation of l in equation 6.9.4.1-3 which accounts for buckling of unbraced sections. Since the pile will be fully embedded in soil, the unbraced length is zero and therefore l is zero. Based on this this, use equation 6.9.4.1-1 to calculate the nominal compressive resistance.

S6.9.4.1

Formula: P subscript n = 0 point 66 superscript lamda times F subscript y times A subscript s

SEquation

6.9.4.1-1

where:

                Formula: F subscript y = 36 ksi

                  Formula: A subscript s = 15 point 5 inches squared

                Formula: lamda = 0

Therefore:

                  Formula: P subscript n = 0 point 66 superscript lamda times F subscript y times A subscript s

                  Formula: P subscript n = 558 K

Design Step P.7 - Determine Nominal Axial Geotechnical Resistance for Selected Pile Type / Size

Geotechnical axial resistance for a pile end bearing on rock is determined by the CGS method outlined in 10.7.3.5

Nominal unit bearing resistance of pile point, qp Formula: q subscript p = 3 times q subscript u K subscript sp times d

SEquation 10.7.3.5-1

for which: Formula: K subscript sp = numerator (3 + numerator (s subscript d) divided by denominator (D)) divided by denominator (10 times ( 1 + 300 numerator (t subscript d) divided by denominator (s subscript d) ) superscript 0 point 5)

SEquation 10.7.3.5-2

Formula: d = 1 + 0 point 4 times numerator (H subscript s) divided by denominator (D subscript s)

                                                              Formula: d less than 3 point 4

where:

Average compressive strength of rock core:

From Design Step P.1:       Formula: q subscript u = q subscript uave

                                                                                        Formula: q subscript u = 11485 psi

Spacing of discontinuities:

Based on high observed RQD in Design Step P.1 and description of rock:       Formula: s subscript d = 1 feet

Width of discontinuities:

Joints are tight as per discussion in Design Step P.1:       Formula: t subscript d = 0 feet

Pile width:

HP 12x53 used:       Formula: D = 1 feet

Depth of embedment of pile socketed into rock:

Pile is end bearing on rock:       Formula: H subscript s = 0 feet

Diameter of socket:

Assumed but does not matter since Hs = 0:       Formula: D subscript s = 1 feet

so:

                    Formula: K subscript sp = numerator (3 + numerator (s subscript d) divided by denominator (D)) divided by denominator (10 times ( 1 + 300 numerator (t subscript d) divided by denominator (s subscript d) ) superscript 0 point 5)

                    Formula: K subscript sp = 0 point 4

and:

              Formula: d = 1 + 0 point 4 times numerator (H subscript s) divided by denominator (D subscript s)

              Formula: d = 1

Thus:

                  Formula: q subscript p = 3 times q subscript u K subscript sp times d

                  Formula: q subscript p = 1985 KSF

Nominal geotechnical resistance (Qp): Formula: Q subscript p = q subscript p times A subscript p

SEquation 10.7.3.2-3

where:

Nominal unit bearing resistance as defined above:       Formula: q subscript p = 1985 KSF

Area of the pile tip:

Area determined assuming a plug develops between flanges of the H-Pile. This will be the case if the pile is driven into the upper weathered portion of the rock.       Formula: A subscript p = 1 feet squared

Therefore:

                  Formula: Q subscript p = q subscript p times A subscript p

                  Formula: Q subscript p = 1985 K

Design Step P.8 - Determine Factored Axial Structural Resistance for Single Pile

Factored Structural Resistance (Pr):

                Formula: P subscript r = phi subscript c times P subscript n

SEquation 6.9.2.1

where:

Resistance factor for H-pile in compression, no damage anticipated:       Formula: phi subscript c = 0 point 6

S6.5.4.2

Nominal resistance as computed in Design Step P.6:       Formula: P subscript n = 558 K

Therefore:

                Formula: P subscript r = 334 point 8 K

Design Step P.9 - Determine Factored Axial Geotechnical Resistance for Single Pile

Factored Geotechnical Resistance (QR): Formula: Q subscript R = phi subscript qp times Q subscript p

SEquation 10.7.3.2-2

Note: remainder of equation not included since piles are point bearing and skin friction is zero.

where:

Resistance factor, end bearing on rock (CGS method): Formula: phi subscript qp = 0 point 5 times lamda subscript v

STable 10.5.5-2

Factor to account for method controlling pile installation:

For this porject, stress wave measurements will be specified on 2% of the piles (a minimum of one per substructure unit) and the capacity will be verified by CAPWAP analysis. Thus:       Formula: lamda subscript v = 1 point 0

STable 10.5.5-2

and therefore:

                    Formula: phi subscript qp = 0 point 5 times lamda subscript v       Formula: phi subscript qp = 0 point 5

Nominal resistance as computed in Design Step P.7:       Formula: Q subscript p = 1985 K

Therefore:

                  Formula: Q subscript r = phi subscript qp times Q subscript p

                Formula: Q subscript r = 992 K

Note: This is greater than the structural capacity, thus structural capacity controls.

Design Step P.10 - Check Drivability of Pile

Pile drivability is checked using the computer program WEAP. The analysis proceeds by selecting a suitable sized hammer. Determining the maximum pile stress and driving resistance (BPF) at several levels of ultimate capacity and plotting a bearing graph relating these variables. The bearing graph is then entered at the driving resistance to be specified for the job (in this case absolute refusal of 20 BPI or 240 BPF will be used) and the ultimate capacity and driving stress correlating to that driving resistance is read.

If the ultimate capacity is not sufficient, a bigger hammer is specified and the analysis is repeated.

If the driving stress exceeds the permitted driving stress for the pile, a smaller hammer is specified and the analysis is repeated.

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Drivability of Piles

If a suitable hammer can not be found that allows driving the piile to the required ultimate capacity without exceeding the permissible driving stress, modification to the recommended pile type are necessary. These may include:

  • Specifying a heavier pile section
  • Specifying a higher yield stress for the pile steel
  • Reducing the factored resistance of the pile

Develop input parameters for WEAP

Driving lengths of piles

The finished pile will likely be 32-33 feet long which includes a 1 foot projection into the pile cap and up to 1' of penetration of the pile tip into the weathered rock. Therefore assume that 35' long piles will be ordered to allow for some variation in subsurface conditions and minimize pile wasted during cut off.

Distribution and magnitude of side friction

This pile will be primarily end bearing but some skin friction in the overlying sand will develop during driving. This skin friction can be quickly computed using the FHWA computer program DRIVEN 1.0. The soil profile determined in Step P.1 is input and an HP12x53 pile selected. The pile top is set at 4 foot depth to account for that portion of soil that will be excavated for pile cap construction. No driving strength loss is assumed since the H-Pile is a low displacement pile and excess pore pressure should dissipate rapidly in the loose sand. Summary output from the program is provided below.

The figure shows the output from Driven 1 point 0.

Figure P-5 DRIVEN 1.0 Output

This graph shows the capacity with units of kips along the horizontal axis and depth with units of feet along the vertical axis. The graph shows the skin friction capacity, the end bearing capacity and the total capacity.

Figure P-6 Bearing Capacity

From this analysis, the side friction during driving will vary in a triangular distribution, and will be about:

                  Formula: Q subscript s = 50 K

The distribution will start 4 feet below the top of the pile which is:

                      Formula: numerator (4 feet ) divided by denominator (35 feet ) = 11 %       below the top of the pile.

The desired factored resistance was determined in Design Step P.8 and is controlled by structural resistance of the pile. This value is:

                Formula: P subscript r = 334 point 8 K

The ultimate resistance that must be achieved during wave equation analysis will be this value divided by the appropriate resistance factor for wave equation analysis + the estimated side friction.

NOTE: Side friction is added here because downdrag is expected to reduce or reverse the skin friction in the final condition. Therefore, sufficient point capacity must be developed during driving to adequately resist all applied loads plus the downdrag.

                Formula: phi = 0 point 65 times lamda subscript v

STable 10.5.5-2

From Design Step P.9:

                  Formula: lamda subscript v = 1

Thus:

                  Formula: phi = 0 point 65

and

                    Formula: Q subscript P = numerator (P subscript r) divided by denominator ( phi )

                    Formula: Q subscript P = 515 K

At this Ultimate point resistance the percent side friction is:

                            Formula: numerator (Q subscript s) divided by denominator (Q subscript s + Q subscript P) = 9 %

and the resistance required by wave equation analysis is:

                      Formula: Q subscript req = Q subscript s + Q subscript P

                      Formula: Q subscript req = 565 K

Soil parameters (use Case damping factors):

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Damping Factors

Case damping factors are used here because of experience with similar jobs. In general, Smith damping factors are preferred. In this case, the Smith damping factors would likely give very similar results to what is computed using the selected Case damping factors.

The parameters for loose sand and hard sandstone were estimated based on local experience with similar soils.

Loose Sand

Skin Damping:       Formula: S subscript D = 0 point 2       DIM

Skin Quake:       Formula: S subscript Q = 0 point 1 inches

Toe Damping:       Formula: T subscript D = 0 point 15       DIM

Toe Quake:       Formula: T subscript Q = 0 point 1 inches

Use skin damping and skin quake for pile shaft.

Hard Sandstone

Skin Damping:       Formula: S subscript D = 0 point 05       DIM

Skin Quake:       Formula: S subscript Q = 0 point 1 inches

Toe Damping:       Formula: T subscript D = 0 point 05       DIM

Toe Quake:       Formula: T subscript Q = 0 point 05 inches

Use toe damping and toe quake for pile toe.

Hammer Selection:

As a rule of thumb, start out with a rated energy of 2000 ft-lbs times the steel area of the pile.

Area:       Formula: A subscript s = 15 point 5inches squared       from Design Step P.5

Rated Energy:       Formula: E subscript r = ( 2000 feet times lb) times A subscript s

                                              Formula: E subscript r = 31000 feet times lb

Select open ended diesel common to area

DELMAG 12-32 (ID=37) rated at:       31 point 33 feet K

Helmet weight:       2 point 15 kip

Hammer Cushion Properties:

Area:       283 point 5 inches squared

Elastic Modulus:       280 KSI

Thickness:       2 inches

COR:       0 point 8

Hammer Efficiency:       72 %

Permissible Driving Stress:

Driving Stress,       Formula: S subscript d less than 0 point 9 times phi times F subscript y

S10.7.1.16

Note that the equation above was modified to yield stress rather than load.

where:

Resistance factor for driving:       Formula: phi = 1 point 0

S6.5.4

Steel yield stress, from Design Step P.5:       Formula: F subscript y = 36 ksi

                        Formula: 0 point 9 times phi times F subscript y = 32 point 4 ksi

              Formula: S subscript d less than 32 point 4 ksi

Summary of Wave Equations Analysis:

This figure shows the wave equation analysis.

Figure P-7 Wave Equation Analysis

at refusal the pile has an ultimate capacity of       Formula: Q subscript ult = 660 K

at refusal the driving stress in the pile is       Formula: S subscript d_act = 42 ksi

Check:

The ultimate capacity exceeds that required

                        Formula: Q subscript ult greater than Q subscript req

                        Formula: Q subscript ult = 660 K       >       Formula: Q subscript req = 565 K       OK

The permissible driving stress exceeds the actual value

                      Formula: S subscript D greater than S subscript d_act

                      Formula: S subscript d = 32 point 4 ksi       >       Formula: S subscript d_act = 42 ksi

This condition is not satisfied - no good.

Try reducing hammer energy

DELMAG D 12 (ID=3) rated at       23 point 59 feet kip

Hammer Cushion Properties same as before

Summary of Wave Equations Analysis:

This figure shows the wave equation analysis.

Figure P-8 Wave Equation Analysis

at refusal the pile has an ultimate capacity of       Formula: Q subscript ult = 520 K

at refusal the driving stress in the pile is       Formula: S subscript d_act = 34 ksi

Check:

The ultimate capacity exceeds that required

                        Formula: Q subscript ult greater than Q subscript req

                        Formula: Q subscript ult = 520 K       >       Formula: Q subscript req = 565 K

This condition is not satisfied - no good

The permissible driving stress exceeds the actual value

                      Formula: S subscript D greater than S subscript d_act

                      Formula: S subscript d = 32 point 4 ksi       >       Formula: S subscript d_act = 34 ksi

This condition is not satisfied - no good.

A decision must be made at this point:

Is pile drivable to minimum of Ultimate Geotechnical Axial Resistance or Ultimate Structural Resistance without pile damage?

Based on above analysis, no hammer can possibly drive this pile to the required capacity without exceeding the permissible driving stress.

There are 2 approaches to resolving this problem

1)       Reduce the factored resistance of the pile to a value that can be achieved without over stressing the pile.

Based on the above bearing graph and allowing for some tolerance in the driving stress (requiring the contractor to select a driving system that produces exactly 32.4 KSI in the pile is unreasonable) a reasonable driven capacity is estimated. Using a minimum driving stress of 29 KSI (0.8 Fy) the penetration resistance is about 100 BPF and the ultimate capacity would be:

                                Formula: Q subscript ult = 420 K

This value includes skin friction during driving which was set in the program to be 9% of the ultimate resistance. Therefore, point resistance at this driving stress would be:

                              Formula: Q subscript p = 91 % times Q subscript ult       Formula: Q subscript p = 382 point 2 K

and:

                            Formula: phi = 0 point 65

                              Formula: Q subscript R = phi times Q subscript p       Formula: Q subscript R = 248 point 43 K

2)       Increase the yield strength of the pile without increasing the previously computed factored resistance

Using grade 50 steel

Driving Stress:       Formula: S subscript d less than 0 point 9 times phi times F subscript y

S10.7.1.16

(Equation modified to yield stress instead of load)

where:

Resistance factor for driving:       Formula: phi = 1

S6.5.4

Steel yield stress:       Formula: F subscript y = 50 ksi

                                        Formula: 0 point 9 times phi times F subscript y = 45 ksi

                        S subscript d       <       45 ksi

Since option 2 involves little or no additional cost and option 1 will result in significant increase in cost due to required additional piles, select option 2

In this case The Delmag 12-32 produced acceptable driving results.

It can be seen from the results of the wave equation analysis that the driving stress times the pile area is about equal to the mobilized pile capacity. Thus, if the factored structural resistance determined in step P.8 is used as the final design pile resistance, then the ultimate required dynamic capacity determined above is valid and the driving stress associated with this capacity can be estimated by:

Driving Stress       Formula: S subscript d = numerator (Q subscript ult) divided by denominator (A subscript s)

where:

Ultimate required capacity as previously determined by wave equation analysis:       Formula: Q subscript ult = 565 K

Pile area, from Design Step 9.5:       Formula: A subscript s = 15 point 5 inches squared

Driving Stress       Formula: S subscript d = numerator (Q subscript ult) divided by denominator (A subscript s)

                                                                  Formula: S subscript d = 36 point 5 ksi

Thus, so long as the contractor selects a hammer that will produce a driving stress between about 37 and 45 KSI at refusal, an acceptable driven capacity should be achieved during construction.

Using a minimum driving stress of       Formula: S subscript d_min = 37 ksi

                            Formula: Q subscript ult = S subscript d_min times A subscript s

                            Formula: Q subscript ult = 573 point 5 K

                              Formula: Q subscript p = Q subscript ult minus Q subscript s

                                  Formula: Q subscript s = 50 K       As defined previously

                              Formula: Q subscript p = 523 point 5 K

Again, side friction is subtracted from the ultimate capacity since it will be present during driving but will not be present in the final condition. Resistance is based on the point resistance achieved during driving the pile to refusal.

and the minimum driven resistance is

                              Formula: Q subscript R = phi times Q subscript p

                                Formula: phi = 0 point 65

                                  Formula: Q subscript p = 523 point 5 K

                              Formula: Q subscript R = phi times Q subscript p

                              Formula: Q subscript R = 340 point 275 K

Recompute structural resistance based on higher yield steel, as in Design Step P.6

                              Formula: P subscript n = 0 point 66 superscript lamda times F subscript y times A subscript s

SEquation 6.9.4.1-1

where

Nominal compressive resistance:       P subscript n

                            Formula: F subscript y = 50 ksi

                              Formula: A subscript s = 15 point 5 inches squared

                          Formula: lamda = 0

                              Formula: P subscript n = 775 K

The factored axial structural resistance, as in Design Step P.8 is:

                            Formula: P subscript r = phi subscript c times P subscript n

SEquation 6.9.2.1-1

                              Formula: phi subscript c = 0 point 6

                            Formula: P subscript r = 465 K

Driven capacity controls

Thus final axial resistance of driven pile:

                                Formula: Q = Q subscript R

                                Formula: Q = 340 K

Design Step P.11 - Do Preliminary Pile Layout Based on Factored Loads and Overturning Moments

The purpose of this step is to produce a suitable pile layout beneath the pile cap that results in predicted factored axial loads in any of the piles that are less than the final factored resistance for the selected piles. A brief evaluation of lateral resistance is also included but lateral resistance is more fully investigated in step P.13

The minimum number of piles to support the maximum factored vertical load is:

                Formula: N = numerator (P subscript vert) divided by denominator (Q subscript R)

where:

The maximum factored vertical load on the abutment, from Design Step P.3, Load Case STR I max:       Formula: P subscript vert = 2253 K

The final controlling factored resistance for the selected pile type, from Design Step P.10:       Formula: Q subscript R = 340 K

                Formula: P subscript f = Q subscript R

                Formula: N = numerator (P subscript vert) divided by denominator (P subscript f)

                Formula: N = 6 point 6       Piles

Additional piles will be required to resist the over turning moment.

From Design Step P.5, the maximum load that needed to be supported by each row of piles was calculated.

                            Formula: R subscript FRONT = 2119 K       Formula: R subscript BACK = 134 K

The required number of piles in the front row is determined as above.

                            Formula: N subscript FRONT = numerator (R subscript FRONT) divided by denominator (P subscript f)

                            Formula: N subscript FRONT = 6 point 2       Piles

Additional load in the corner pile will come from the lateral moment but this is small, so start with 7 piles in the front row.

                            Formula: N subscript FRONT = 7       Piles

This results in a pile spacing of:

c-c spacing of piles:       Formula: s = numerator (S subscript L) divided by denominator (N subscript FRONT)

where:

The length of footing available for piles, from Design Step P.5:       Formula: S subscript L = 44 point 375 feet

c-c spacing of piles:       Formula: s = numerator (S subscript L) divided by denominator (N subscript FRONT minus 1)

                                                          Formula: s = 7 point 396 feet

Set c-c spacing of piles = 7' 4"

This is approaching the maximum pile spacing identified in Step 5 thus set the back row of piles to the same spacing. This will result in the back row of piles being under utilized for axial loads. However, the additional piles are expected to be necessary to help handle lateral loads and to resist downdrag loads that will be applied to the back row only. Further, a load case in which the longitudinal loads such as temperature and braking loads are reversed will increase the loads on the back row.

Thus, the final preliminary layout is diagramed below

This figure shows a pile footing with a length of 46 feet and a width of 10 feet 3 inches. The pile spacing in the width direction is 7 feet 9 inches and the pile spacing in the length direction is 7 feet 4 inches. The number of pile spaces is the width direction is one and number of pile spaces is the length direction is 6. Therefore, the total pile spacing in the width direction is 7 feet 9 inches and the total pile spacing in the length direction is 44 feet 90 inches. The edge distance in the length direction is 1 foot 5 and one quarter inches and the edge distance in the width direction is 1 foot 3 inches.

Figure P-9 Plan View of Pile Cap

The spreadsheet below is used to calculate individual pile loads using the following formula: Formula: P = numerator (F prime subscript z) divided by denominator (N) + M prime subscript x times numerator (x prime) divided by denominator (I subscript yy) + M prime subscript y times numerator (y prime) divided by denominator (I subscript xx)

where:

Vertical load and moments applied at the centroid of the pile group:       F prime subscript z , M prime subscript x , M prime subscript y

Distance from centroid of pile group to pile in the x and y directions:       x prime , superscript y prime

Moment of inertia of the pile group about the y and x axis respectively:       I subscript yy , I subscript xx

Calculation of Individual Pile Loads on an Eccentrically Loaded Footing:

Input Applied Loads:

At x = 0, y = 0

                  Formula: F subscript z = minus 2253 K

                  Formula: M subscript x = 0 K feet

                  Formula: M subscript y = 7693 K feet

The coordinate system for the following calculations is provided in Figure P.10: The coordinate system in shown in the figure and the positive x direction is show going from left to right. The positive z direction is shown going from bottom to top and the positive y direction is shown going into the page.

Figure P-10 Coordinate System

Table P-10 is used to calculate the vertical load and moments, and the moment of inertia of the pile group.

Input Pile Location
Calculated Values
Pile Number
x
y
x'
y'
x' 2
y' 2
Pile load
1
-3.875
-22
-3.875
-22
15.01563
484
-19.1221
2
3.875
-22
3.875
-22
15.01563
484
-302.735
3
-3.875
-14.6667
-3.875
-14.6667
15.01563
215.111
-19.1221
4
3.875
-14.6667
3.875
-14.6667
15.01563
215.111
-302.735
5
-3.875
-7.33333
-3.875
-7.33333
15.01563
53.7778
-19.1221
6
3.875
-7.33333
3.875
-7.33333
15.01563
53.7778
-302.735
7
-3.875
0
-3.875
0
15.01563
0
-19.1221
8
3.875
0
3.875
0
15.01563
0
-302.735
9
-3.875
7.33333
-3.875
7.333333
15.01563
53.7778
-19.1221
10
3.875
7.33333
3.875
7.333333
15.01563
53.7778
-302.735
11
-3.875
14.6667
-3.875
14.66667
15.01563
215.111
-19.1221
12
3.875
14.6667
3.875
14.66667
15.01563
215.111
-302.735
13
-3.875
22
-3.875
22
15.01563
484
-19.1221
14
3.875
22
3.875
22
15.01563
484
-302.735
Table P-10 Pile Calculations

Sum of the distances in the x direction is zero.

Sum of the distances in the y direction is zero.

Centroids:

                Formula: y subscript c = 0 inches

                Formula: x subscript c = 0 inches

Moment of Inertia about the y axis:       Formula: I subscript yy = 210 point 2188 inches squared

Moment of Inertia about the x axis:       Formula: I subscript xx = 3011 point 556 inches squared

Resolved loads at Centroid:

                  Formula: F prime subscript z = F subscript z

                      Formula: F prime subscript z = minus 2253 K

                  Formula: M prime subscript x = minus F prime subscript z times y subscript c + M subscript x

                      Formula: M prime subscript x = 0 K feet

                  Formula: M prime subscript y = minus F prime subscript z times x subscript c + M subscript y

                      Formula: M prime subscript y = 7693 K feet

Summary of individual pile loads for all load cases:

This table was generated by inserting each load case in the spreadsheet above and recording the resulting pile loads for that load combination.

Load Case
STR-I MAX/FIN
SER-I MAX/FIN
STR-I MIN/FIN
SER-I MIN/FIN
STR-III MAX/FIN
SER-I MAX/FIN
Fz =
-2253
-1791
-1860
-1791
-1815
-1791
Mx =
0
162
0
162
508
162
My =
7693
4774
7291
4709
6374
4774
Pile No.
1
-19.1
-41.1
1.5
-42.3
-15.9
-41.1
2
-302.7
-217.1
-267.3
-215.9
-250.8
-217.1
3
-19.1
-40.7
1.5
-41.9
-14.6
-40.7
4
-302.7
-216.7
-267.3
-215.5
-249.6
-216.7
5
-19.1
-40.3
1.5
-41.5
-13.4
-40.3
6
-302.7
-216.3
-267.3
-215.1
-248.4
-216.3
7
-19.1
-39.9
1.5
-41.1
-12.1
-39.9
8
-302.7
-215.9
-267.3
-214.7
-247.1
-215.9
9
-19.1
-39.5
1.5
-40.7
-10.9
-39.5
10
-302.7
-215.5
-267.3
-214.3
-245.9
-215.5
11
-19.1
-39.1
1.5
-40.3
-9.7
-39.1
12
-302.7
-215.1
-267.3
-213.9
-244.7
-215.1
13
-19.1
-38.7
1.5
-39.9
-8.4
-38.7
14
-302.7
-214.7
-267.3
-213.5
-243.4
-214.7
Maximum
-302.7
-217.1
-267.3
-215.9
-250.8
-217.1
Minimum
-19.1
-38.7
1.5
-39.9
-8.4
-38.7

Table P-11 Individual Loads for All Load Cases

Pile loads range between -302.7 K in compression and 1.5 K in tension for all load cases.

The maximum compressive load is reasonably close to the factored resistance for the selected pile and the tension load is minimized thus this is a reasonable layout with respect to axial load.

Evaluate lateral loads:

If all piles are vertical they can all be assumed to take an equal portion of the applied horizontal load since group effects have been minimized by keeping the pile spacing large enough.

The controlling criterion with respect to horizontal loads on vertical piles is usually deflection which is a service load case. Looking at the maximum horizontal loads in section P.3, it can be seen that the transverse loads are relatively small and can be ignored for the purposes of this step. The maximum longitudinal service load is:

                      Formula: P subscript long = 571 K

Number of piles:       Formula: N subscript pile = 14

Thus, load per pile:       Formula: P = numerator (P subscript long) divided by denominator (N subscript pile)

                                                  Formula: P = 40 point 8 K

MathCad tip logo

Lateral Capacity

The design chart used below to estimate the lateral capacity of steel H-Piles is one of many methods available to the designer. Brohms method can be used to estimate ultimate capacity (strength limit state) and various published elastic solutions may be used to estimate deflection (service limit state). Pressumptive allowable lateral capacities based on the designer's experience (service limit state) may be used or a preliminary P-y analysis using COM624 may be performed at this point to assist in initial pile group layout

Based on the design chart below, the maximum service load per pile for an assumed 1.5" deflection (38mm) is: Formula: 92KN = 20 point 6K

From PennDOT DM4 Appendix F-20:

This figure shows the force in units of kilo Newton's along the vertical axis and deflection in units of millimeters along the horizontal axis.

Figure P-11 Maximum Service Load Per Pile

Notes on chart:

Solid lines represent load vs deflection for full depth loose saturated sand

I values are moment of inertia for pile about axis perpendicular to applied load (shown in mm4 x 108)

For HP 12 x 53

                          Formula: I subscript xx = 393 inches superscript 4

                        Formula: I subscript xx = 1 point 636 times 10 superscript 8 mm superscript 4

Load in KN is applied at ground surface and pile head is assumed to be 50% fixed

Thus, there probably will not be sufficient lateral load capacity with 14 vertical piles. To resolve this, it will be necessary to add more piles or batter some of the piles. Since at least twice as many piles would be required to handle the anticipated horizontal loads, battering the piles makes more sense.

Investigate battering front row of piles at 1:3 (back row of piles not battered due to lack of vertical load and potential for downdrag)

Total vertical load on front row for each of the load cases is computed by summing the individual pile loads computed above.

From Design Step P.3:

Load Case STR-I MAX/FIN SER-I MAX/FIN STR-I MIN/FIN SER-I MIN/FIN STR-III MAX/FIN SER-I MAX/FIN
Total vertical load on front row of piles (kips) 2119.1 1511.5 1870.8 1503.1 1730.0 1511.5
Batter =0.333333333            
Available resisting force due to horizontal component of axial pile load = Batter x vertical load on front row (kips) 706.4 503.8 623.6 501.0 576.7 503.8
Plong =(kips) 855.0 571.0 855.0 568.0 787.0 571.0
Remaining force to be handled by bending of pile = Plong - available horizontal force (kips) 148.6 67.2 231.4 67.0 210.3 67.2
Force per pile (kips) 10.6 4.8 16.5 4.8 15.0 4.8

Table P-12 Vertical Load on Front Row of Piles for Each Load Case         

The remaining force per pile to be handled in bending is in the reasonable range thus this may be a workable configuration but it must be confirmed by interaction analysis. Thus proceed to next step with a 14 pile group with the front row battered at 3V:1H.

Design Step P.12 - Evaluate Pile Head Fixity

The performance of the pile group and the resulting pile stresses are greatly influenced by the degree to which piles are fixed against rotation at the pile head. This fixity is provided by the pile cap and is a function of the embedment of the pile into the cap, the geometry of the pile group, the stiffness of the pile cap, and the deflection. Each of these is evaluated below.

S10.7.3.8

Embedment

Research has shown that a pile needs to be embedded 2-3 times its diameter into the pile cap in order to develop full fixity. These piles will be embedded the minimum of 1 foot since the thickness of the pile cap is expected to be only 2.5 feet. Embedding the piles 2 feet into a 2.5 thick cap places the tops of the piles near the top layer of reinforcing and increases the probability of the pile punching through the top of the cap under load. Thus full pile head fixity will likely not develop regardless of other factors.

S10.7.1.5

Group geometry

In the transverse direction, there will be 7 rows of piles that when deflected force the pile cap to remain level. This condition will result in full fixity of the pile head pending evaluation of other factors. In the longitudinal direction there will be only 2 rows of piles which should be sufficient to enforce fixity pending evaluation of other factors. However, if the front row of piles is battered and the back row of piles is left vertical, the pile cap will tend to rotate backwards as it deflects. This could conceivably result in a moment applied to the pile heads greater than that required to fix the head (i.e. greater than 100% fixity) This backwards rotation of the pile cap is accounted for in the group analysis so it does not need to be considered here.

Pile cap stiffness

Flexing of the pile cap due to applied loads and moments tends to reduce the fixity at the head of the pile. In this case the pile cap is expected to be relatively thin so this effect becomes important. The stiffness of the pile cap is accounted for in the group interaction analysis so this does not effect the evaluation of fixity.

Deflection

The fixity of a pile is reduced at large deflections due to cracking of the concrete at the bottom of the pile cap. For the vertical pile group deflections are expected to be large but for the battered group deflections are likely to be small.

Conclusion

Since the group analysis will account for the group geometry and the stiffness of the pile cap, the remaining factors of embedment and deflection need to be accounted for. Both of these indicate that pile head fixity is likely to be somewhere between 25 and 75% with the higher values for the battered group. To be conservative, the group will be analyzed with 0 and 100% fixity to determine the critical conditions for pile stress (usually 100% fixity) and deflection (0 % fixity)

Design Step P.13 - Perform Pile Soil Interaction Analysis

Group interaction analysis will be performed using the computer program FB-Pier developed by FHWA, FloridaDOT and University of Florida. This program is available from the Bridge Software Institute associated with the University of Florida. Version 3 of the program is used in this example.

S10.7.3.11

In order to properly use the program, a few additional soil and pile cap properties need to be established. These are:

1)       The location and thickness of the abutment stem. This controls the relative stiffness of the pile cap.

2)       The location and distribution of applied loads.

3)       The axial response of the soil and rock (T-z and Q-z)

4)       The lateral response of soil (P-y)

5)       The torsional response of the soil and rock (T- q)

6)       Other miscellaneous considerations

Each is evaluated below:

Location and thickness of stem.

Previous analysis has developed a preliminary stem thickness of 3.5 feet located 2.75' from the toe of the footing. The stem is 15' tall thus the footing will be thickened to 15' in this zone as shown on the sketch below:

This figure shows a pile footing with a length of 46 feet and a width of 10 feet 3 inches. The pile spacing in the width direction is 7 feet 9 inches and the pile spacing in the length direction is 7 feet 4 inches. The number of pile spaces is the width direction is one and number of pile spaces is the length direction is 6. Therefore, the total pile spacing in the width direction is 7 feet 9 inches and the total pile spacing in the length direction is 44 feet 90 inches. The edge distance in the length direction is 1 foot 5 and one quarter inches and the edge distance in the width direction is 1 foot 3 inches. The width of the footing zone thickened to account for the stem is 3 point 5 feet in the width direction. The distance from the edge of footing to the footing zone thickened to account for the stem is 2 point 75 feet in the width direction.

Figure P-12 Location and Thickness of Stem

Location of applied loads

The loads as supplied so far were resolved to a point at the center of the footing and the bottom of the pile cap. The loads actually consist of numerous loads due to earth pressure, superstructure, self weight etc. that are distributed over the proposed structure. To simplify the analysis, only the pile cap will be modeled in FB-Pier. The supplied loads will be divided by 3 and applied to the pile cap at 3 locations along the length of the stem at the centerline of the pile group. Since the cap will be modeled as a membrane element at an elevation that corresponds to the base of the pile cap and the loads were supplied at the base of the pile cap, no additional changes to the supplied loads and moments are required. The assumed locations of the applied loads are shown above.

The magnitude of loads and moments are computed from those provided in section P.3 as shown below. The terminology and sign convention has been converted to that used in FB-Pier. The coordinate system used is a right handed system as shown in the sketch above with Z pointing down.

Note the loads at each point provided below are in Kip-FT units

LIMIT STATE
FB-Pier Load Case
Fz (K)
My (K-FT)
Mx (K-FT)
Fx (K)
Fy (K)
STR-I MAX/FIN
1
751.0
-2564.3
0.0
285.0
0.0
SER-I MAX/FIN
2
597.0
-1591.3
54.0
190.3
3.3
STR-I MIN/FIN
3
620.0
-2430.3
0.0
285.0
0.0
SER-I MIN/FIN
4
597.0
-1569.7
54.0
189.3
3.3
STR-III MAX/FIN
5
605.0
-2124.7
169.3
262.3
12.3
SER-I MAX/FIN
6
597.0
-1591.3
54.0
190.3
3.3

Table P-13 Loads for Each Limit State

The axial response of the soil and rock (T-z and Q-z)

Since the piles will be point bearing, friction response of the soil will be small compared to the point resistance and can be ignored. However, for cases that develop tension in the piles, frictional response of the soil will be the only thing that resists that tension. Therefore, two cases will need to be run, one with the frictional response set to zero by specifying a custom T-z curve and the second with the friction response set to the default for a driven pile in granular material.

Point response of the pile bearing on rock (Q-z) will be a function of the elastic properties of the rock and will be input as a custom Q-z curve as defined below.

This figure shows the Q-z curve. The Q is shown along the vertical axis and z is shown along the horizontal axis. Point 1 is shown at the origin and point 2 is at Q max and Z at Q max.

Figure P-13 Q-z Curve

From Design Step P.7:

                      Formula: Q subscript max = 1985 K

z @ Qmax is estimated using the methods for a drilled shaft socketed in rock.

z @ Qmax: Formula: rho subscript base = numerator ( Sigma P subscript i times I subscript p) divided by denominator (D subscript s times E subscript r)

CEquation 10.8.3.5-2

where:

Load at top of socket:       Formula: Sigma P subscript i = 992 times ton

Since Formula: numerator (H subscript s) divided by denominator (D subscript s) = 0

CFigure 10.8.3.5-1

Influence coefficient (DIM):       Formula: I subscript p = 1 point 1       DIM

Diameter of socket, for HP12 pile:       Formula: D subscript s = 1 feet

Modulus of elasticity of rock mass (TSF):       E subscript r

where:

Modulus modification ratio based on RQD, from Design Step P.1:       Formula: RQD = 90 point 4 %

                                                                                                Formula: K subscript e = 0 point 74

CFigure 10.8.3.5-3

Modulus of elasticity of intact rock, from Design Step P.1:       Formula: E subscript i = 153000 times TSF

Thus,

                Formula: E subscript r = K subscript e times E subscript i

CEquation 10.8.3.5-3

                Formula: E subscript r = 113220TSF

and:

                        Formula: rho subscript base = numerator ( Sigma P subscript i times I subscript p) divided by denominator (D subscript s times E subscript r)

                        Formula: rho subscript base = 0 point 009638 feet

                        Formula: rho subscript base = 0 point 11565 inches

Thus Q-z curve is defined by the following points

(refer to the sketch above for location of the points)

Point Q (kips) z (IN)
1
0
0.00
2
1985
0.12
3
1985
2.00

Table P-14 Q-z Curve Points

The lateral response of soil and rock (P-y)

For Soil, use built in P-y curve for sand (Reese) with

                Formula: phi prime subscript f = 31       o

                      Formula: Y subscript wet = 110 PCF

              Formula: k = 20 psi

Assume pile will drive into top weathered portion of rock estimated to be 1' thick.

The embedment of the pile into the rock will provide some amount of lateral restraint at the pile tip. The response of the rock will be relatively stiff compared to the soil. To simulate this response, use the built in P-y curve for a stiff clay above the water table since the shape of this curve is closest to actual rock response. Input parameters for this curve are estimated below:

Shear strength

Average qu, Design Step P.1:       Formula: q subscript uave = 11485 psi

                                                                      Formula: q subscript uave = 1653840 PSF

Arbitrarily reduce to 10% of this value to account for weathering

10% of Average qu:       Formula: 10 % times q subscript u = 165384 PSF

Shear strength, 1/2 qu:       Formula: numerator (1) divided by denominator (2) times ( 10 % q subscript u ) = 82692 PSF

Say shear strength:       Formula: q subscript u = 80000 times PSF

Unit weight

Average Υ, Design Step P.1:       Formula: Y subscript ave = 150 PCF

Strain at 50% ultimate shear strength (ε50)

                    Formula: epsilon subscript 50 = 0 point 002

This is based on experience with similar rocks or it can be determined from the results of the unconfined tests if stress and strain data was recorded during the test.

The torsional response of the soil and rock (T- q)

From Design Step P.1:

                Formula: phi prime subscript f = 31       o

                      Formula: Y subscript wet = 110 PCF

                  Formula: G subscript o = 0 point 33 ksi

From Design Step P.10:

                      Formula: T subscript max = 417 times PSF

Note: Τmax calculated as the total skin friction calculated by DRIVEN analysis divided by surface area of pile embedded in soil during that analysis. This represents an average value along the length of the pile and is not truly representative of the torsional response of the pile. However, a more sophisticated analysis is not warranted since torsional response of the piles will be minimal in a multi pile group that is not subject to significant eccentric horizontal loading.

Miscellaneous other considerations

Modulus of elasticity of concrete in pile cap

Assume pile cap is constructed of concrete with       Formula: f prime subscript c = 3000       psi

Then, modulus of elasticity of concrete

                  Formula: E subscript c = 57000 times f prime subscript c superscript 0 point 5

                  Formula: E subscript c = 3122019       psi

                  Formula: E subscript c = 3122 point 019 ksi

Poisson's ratio for concrete

Assume:

                  Formula: nu subscript c = 0 point 2

Pile lengths

Since top of rock is level and front row of piles is battered, front row of piles will be slightly longer than back row so set up front row as a second pile set.

Back row of piles:       Formula: L subscript back = 32 feet

Batter       Formula: Btr = 0 point 3333       (3V:1H)

Front row of piles:       Formula: L subscript front = 33 point 73 feet

Group Interaction

c-c spacing in direction of load:       Formula: s subscript Ioad = 7 point 75 times D

c-c spacing in direction perpendicular to load:       Formula: s subscript perp_Ioad = 7 point 33 times D

The C-C spacing in direction of load is almost 8D and since it gets larger with depth due to the batter on the front row, there should be no horizontal group effects.

The C-C spacing in both directions is greater than 3D thus there should be no horizontal or vertical group effects.

Therefore set all group interaction factors to 1.0

Deflection measurement location

See previous design sections for geometry of abutment

The critical point for evaluation of deflections is at the bearing locations which are 17.5 feet above the bottom of the pile cap as modeled. To account for pile cap rotations in the computation of displacement, add a 17.5' tall column to the center of the footing. This is a stick only with nominal properties and sees no load due to the way the problem is modeled.

Results of Analyses

Four runs were made with different combinations of pile head fixity and considering frictional resistance from the soil. These are expected to bracket the extremes of behavior of the pile group. The results of the four runs are summarized in the table below.

The results in Table P-15 are summarized from the FB-Pier Output files

Run # Units 1 2 3 4
Pile head condition   Fixed Pinned Fixed Pinned
Soil Friction   No No Yes Yes
Strength Limit State  
Maximum Axial load Kip 340 332 340 332
Pile number and LC   Pile 8 LC1 Pile 8 LC1 Pile 8 LC1 Pile 8 LC1
   
Maximum Tension Kip 0.06 1.45 15.3 2.25
Pile number and LC   Pile 7 LC3 Pile 7 LC3 Pile 1 LC3 Pile 13 LC3
           
Max combined load  
Axial kip 288 289 336 290
M2 kip-ft 0 0 0 0
M3 kip-ft 107 100 26 97
Pile number and LC   Pile 8 LC3 Pile 8 LC3 Pile 6 LC1 Pile 8 LC3
Depth FT 8 8 0 8
   
Max V2 Kips 18.1 18.2 15.9 18.1
Pile number and LC   Pile 7 LC3 Pile 7 LC3 Pile 7 LC3 Pile 7 LC3
   
Max V3 Kips 3.4 3 3.3 3
Pile number and LC   Pile 2 LC5 Pile 13 LC5 Pile 2 LC5 Pile 13 LC5
Service Limit State
Max X Displacement IN 0.481 0.489 0.46 0.474
Max Vertical Displacement IN 0.133 0.122 0.123 0.108
Load Case   LC6 LC6 LC6 LC6
Max Y displacement IN 0.02 0.053 0.02 0.053
Load Case   LC6 LC6 LC6 LC6

Table P-15 Results

View of model

This figure shows that the piles are embedded in loose sand and are taken down to rock. The beam seat elevation is the location where the displacement measurement is taken.

Figure P-14 Model

Design Step P.14 - Check Geotechnical Axial Capacity

From the FB-Pier analyses, and Design Step P.13:

Max factored axial pile load:       Formula: P subscript max_a = 340 K

Max factored tension pile load:       Formula: P subscript max_t = 15 point 3 K

These occurred when the pile was assumed to be fully fixed in the pile cap and when soil friction was considered

The maximum factored geotechnical axial resistance, from Design Step P.10 is:

                  Formula: Q subscript r = 340 K       (Controlled by drivability considerations)

The ultimate geotechnical tension resistance can be taken as the reverse of what was computed in step P.10 using driven

S10.7.3.7.2

                  Formula: Q subscript s = 50 point 39 K

Factored resistance: Formula: Q subscript R = phi subscript u times Q subscript s

where:

The ultimate shaft resistance in compression:       Q subscript s

Resistance factor for tension loading:       Formula: phi subscript u = 0 point 4

STable 10.5.5-2

The computer program Driven employs the Nordlund method to compute shaft friction. No resistance factor is provided for the Nordlund method applied to granular soils but the method is similar to the b method and has similar reliability.

thus:

                  Formula: Q subscript R = phi subscript u times Q subscript s

                  Formula: Q subscript R = 20 K

The geotechnical resistance in compression and tension exceeds the maximum factored compressive and tensile pile loads. Thus geotechnical resistance is adequate.

S10.5.3

Design Step P.15 - Check Structural Axial Capacity (in lower portion of pile)

From the FB-Pier analyses, and Design Step P.13:

Max factored axial pile load:       Formula: P subscript max_a = 340 K

Max factored tension pile load:       Formula: P subscript max_t = 15 point 3 K

These occurred when the pile was assumed to be fully fixed in the pile cap and when soil friction was considered

The maximum factored structural axial resistance in the lower portion of the pile, from Design Step P.10 is:

                Formula: P subscript r = 465 K

This is also applicable to tension.

The factored structural resistance far exceeds the maximum factored loads. Thus, the piles are adequately sized to transmit axial loads.

Design Step P.16 - Check Structural Axial Capacity in Combined Bending and Axial Load (upper portion of pile)

The equation to use to evaluate combined axial load and bending is determined by the ratio:

            numerator (P subscript u) divided by denominator (P subscript r)

S6.9.2.2

where:

Axial compressive load:       P subscript u

Factored compressive resistance:       Formula: P subscript r = phi subscript c times P subscript n

where:

From Design Step P.10:       Formula: P subscript n = 775 K

For combined axial and bending (undamaged section of pile):       Formula: phi subscript c = 0 point 7

S6.5.4.2

so:

                Formula: P subscript r = phi subscript c times P subscript n

                Formula: P subscript r = 542 point 5 K

From Design Step P.13, maximum combined loadings range from:

                          Formula: P subscript u_min = 288 K       to       Formula: P subscript u_max = 336 K

so:

                          Formula: numerator (P subscript u_min) divided by denominator (P subscript r) = 0 point 531       to       Formula: numerator (P subscript u_max) divided by denominator (P subscript r) = 0 point 619

Since these are both greater than 0.2

The combined loading must satisfy:

                                                    Formula: numerator (P subscript u) divided by denominator (P subscript r) + numerator (8) divided by denominator (9) times ( numerator (M subscript ux) divided by denominator (M subscript rx) + numerator (M subscript uy) divided by denominator (M subscript ry) ) less than or equal to 1 point 0

SEquation 6.9.2.2-2

where:

P subscript u , P subscript r       are as defined above

Factored flexural moment about the x axis, from Design Step P.13:       M subscript ux

Factored flexural moment about the y axis, from Design Step P.13:       M subscript uy

Factored flexural resistance about the x axis:       M subscript rx

Factored flexural resistance about the y axis:       M subscript ry

Flexural resistance is: Formula: M subscript r = phi subscript f times M subscript n

S6.10.4-1

where:

The resistance factor for combined bending and axial load in piles:       Formula: phi subscript f = 1 point 0

S6.5.4.2

Mn is computed in accordance with the provisions of Section 6.12.

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Deep Foundations Surrounded by Soil

In most cases where deep foundations are completely surrounded by soil, lateral support from even the weakest soil is sufficient such that the unbraced length can be considered zero. When the unbraced length is zero, the buckling considerations of section 6.10.4 generally result in no reduction of the ultimate bending stress and Mn=the plastic moment or Mn=fy*Z where Z is the plastic section modulus. Note that the plastic section modulus is used in LRFD design, not the elastic section modulus. The evaluation of buckling criteria on the following pages is presented for completeness.

For bending about the x axis, the provisions of Section 6.10.4 apply as follows:

S6.12.2.2.1

Criteria from Section 6.10.4.1.2:

                        Formula: numerator (2 times D subscript cp) divided by denominator (t subscript w) less than or equal to 3 point 76 times ( numerator (E) divided by denominator (F subscript yc) ) superscript 0 point 5

SEquation 6.10.4.1.2-1

for HP 12 x 53 Grade 50 piles: Formula: D subscript cp = numerator (d minus 2t subscript f) divided by denominator (2)

                                                                        Formula: D subscript cp = 5 point 455 inches

From Design Step P.5:       Formula: t subscript w = 0 point 435 inches

Modulus of Elasticity:       Formula: E = 29000 ksi

As in Design Step P.10:       Formula: F subscript yc = 50 ksi

Check:

                            Formula: numerator (2 times D subscript cp) divided by denominator (t subscript w) = 25 point 08       Formula: 3 point 76 times ( numerator (E) divided by denominator (F subscript yc) ) superscript 0 point 5 = 90 point 553

Therefore:

                            Formula: numerator (2 times D subscript cp) divided by denominator (t subscript w) less than or equal to 3 point 76 times ( numerator (E) divided by denominator (F subscript yc) ) superscript 0 point 5       is satisfied.

Criteria from Section 6.10.4.1.3:

                    Formula: numerator (b subscript f) divided by denominator (2 times t subscript f) less than or equal to 0 point 382 times ( numerator (E) divided by denominator (F subscript yc) ) superscript 0 point 5

SEquation 6.10.4.1.3-1

where:

                    Formula: b subscript f = 12 point 045 inches

                  Formula: t subscript f = 0 point 435 inches

Check:

                        Formula: numerator (b subscript f) divided by denominator (2 times t subscript f) = 13 point 845       Formula: 0 point 382 times ( numerator (E) divided by denominator (F subscript yc) ) superscript 0 point 5 = 9 point 2

Therefore:

                        Formula: numerator (b subscript f) divided by denominator (2 times t subscript f) less than or equal to 0 point 382 times ( numerator (E) divided by denominator (F subscript yc) ) superscript 0 point 5       NOT SATISFIED

Proceed with criteria of Section 6.10.4.1.4:

Criteria from Section 6.10.4.1.4:

                        Formula: numerator (b subscript f) divided by denominator (2 times t subscript f) less than or equal to 12

Check:

                        Formula: numerator (b subscript f) divided by denominator (2 times t subscript f) = 13 point 845       >       12       Condition is NOT SATISFIED

However, this criteria is intended for welded sections to prevent distortion of the flange during welding. Since this is a rolled section, this practical limit does not apply.

S6.10.4.1.4

Therefore proceed to bracing requirements of Section 6.10.4.1.9:

Criteria from Section 6.10.4.1.9:

The pile is laterally braced along its entire length by the adjacent soil thus the unbraced length (Lb) is zero and this condition is always satisfied.

S6.10.4.1.9

Proceed to noncompact section flange flexural resistance of Section 6.10.4.2.4

For compression flange: Formula: F subscript n = R subscript b times R subscript h times F subscript cr

SEquation 6.10.4.2.4a-2

where:

                          Formula: F subscript cr = numerator (1 point 904 times E) divided by denominator (( numerator (b subscript f) divided by denominator (2 times t subscript f) ) squared times ( numerator (2 times D subscript cp) divided by denominator (t subscript w) ) superscript 0 point 5)

                          Formula: F subscript cr = 57 point 52 ksi

but:

                F subscript cr       cannot exceed       Formula: F subscript yc = 50 ksi

so:

                      Formula: F subscript cr = 50 ksi

and:

Hybrid factor as specified in Section 6.10.4.3.1 for a homogeneous section:       Formula: R subscript h = 1 point 0

S6.10.4.3.1a

Load shedding factor specified in Section 6.10.4.3.2:       R subscript b

Check:

                            Formula: numerator (2 times D subscript cp) divided by denominator (t subscript w) less than or equal to lamda subscript b times ( numerator (E) divided by denominator (f subscript c) ) superscript 0 point 5

SEquation 6.10.4.3.2a-1

where:

Since       Formula: D subscript cp less than or equal to numerator (d) divided by denominator (2)

                                      Formula: lamda subscript b = 5 point 76

Compressive stress in the flange due to factored loads. Since this condition will be critical when fc is the largest, assume fc = the maximum possible stress which is the yield stress of the steel.

                        Formula: f subscript c = F subscript yc       Formula: f subscript c = 50 ksi

Check:

                              Formula: numerator (2D subscript cp) divided by denominator (t subscript w) = 25 point 08

                                        Formula: lamda subscript b times ( numerator (E) divided by denominator (f subscript c) ) superscript 0 point 5 = 138 point 719

Therefore:

                                Formula: numerator (2 times D subscript cp) divided by denominator (t subscript w) less than or equal to lamda subscript b times ( numerator (E) divided by denominator (f subscript c) ) superscript 0 point 5       is satisfied

thus:

                          Formula: R subscript b = 1 point 0

so:

                          Formula: F subscript n = R subscript b times R subscript h times F subscript cr

                          Formula: F subscript n = 50 ksi

For tension flange: Formula: F subscript n = R subscript b times R subscript h times F subscript yt

SEquation 6.10.4.2b-1

where:

                      Formula: R subscript h = 1

                      Formula: R subscript b = 1       for tension flange

S6.10.4.3.2b

                      Formula: F subscript yt = 50 ksi

Therefore:

                      Formula: F subscript n = R subscript b times R subscript h times F subscript yt

                      Formula: F subscript n = 50 ksi

Since the nominal plastic stress in all components of the pile is equal to the yield stress, The nominal moment capacity may be computed as the plastic moment. Formula: M subscript n = M subscript p

                      Formula: M subscript p = F subscript y times Z subscript x

where:

The plastic section modulus about the x axis, from Design Step P.5:       Formula: Z subscript x = 74 inches cubed

so:

                        Formula: M subscript nx = M subscript p

                      Formula: M subscript p = F subscript y times Z subscript x       Formula: M subscript p = 3700 K inches

                        Formula: M subscript nx = 3700 K inches

                        Formula: M subscript nx = 308 point 3 K feet

For bending about the y axis, provisions of Section 6.12.2.2.1 apply.

                        Formula: M subscript ny = M subscript p

SEquation 6.12.2.2.1-1

                      Formula: M subscript p = F subscript y times Z subscript y

                    Formula: Z subscript y = 32 point 2 inches cubed

From Design Step P.5.

                        Formula: M subscript ny = F subscript y times Z subscript y

                        Formula: M subscript ny = 1610 K inches

                        Formula: M subscript ny = 134 point 2 K feet

Check using alternate method from Section C6.12.2.2.1 Formula: M subscript ny = 1 point 5 times F subscript y times S subscript y

CEquation 6.12.2.2.1

where:

The elastic section modulus about the y axis:       Formula: S subscript y = 21 point 1 inches cubed

                        Formula: M subscript ny = 1 point 5 times F subscript y times S subscript y

                        Formula: M subscript ny = 1582 point 5 K inches       close to that computed above

Use       Formula: M subscript ny = F subscript y times Z subscript y       Formula: M subscript ny = 1610 K inches

The factored moment resistances are now determined as: Formula: M subscript r = phi subscript f times M subscript n

SEquation 6.10.4-1

so:

                          Formula: M subscript rx = phi subscript f times M subscript nx

                          Formula: M subscript rx = 308 point 3 K feet

and:

                          Formula: M subscript ry = phi subscript f times M subscript ny

                          Formula: M subscript ry = 134 point 2 K feet

From the maximum combined loads from Design Step P.13:

The interaction equation is now applied to the maximum combined loading conditions determined in the 4 FB-Pier analyses as follows

                                                                Formula: numerator (P subscript u) divided by denominator (542 point 5) + numerator (8) divided by denominator (9) times ( numerator (M subscript ux) divided by denominator (308 point 3) + numerator (M subscript uy) divided by denominator (134 point 2) ) less than or equal to 1 point 0

SEquation 6.9.2.2-2

FB-Pier Run # Pu (kips) Mux (kip-ft) Muy (kip-ft) Results of interaction equation
1
288
107
0
0.84
2
289
100
0
0.82
3
336
26
0
0.69
4
290
97
0
0.81

Table P-16 Results of Interaction Equation

All conditions satisfy the interaction equation thus piles are acceptable under combined loading.

Design Step P.17 - Check Structural Shear Capacity

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Pile Capacity

The capacity of the pile section to resist the maximum applied shear force is usually not critical for steel pile sections placed in groups such that high overturning moments are not required to be resisted by the pile. However, in foundation systems consisting of concrete foundation elements arranged as a single element or a single row of elements supporting a tall laterally loaded pier or supporting a column subject to a large eccentric vertical load, this can become the controlling criteria. It is checked here for completeness.

The nominal shear capacity of the pile section is computed as for an unstiffened web of a steel beam. Formula: V subscript n = C times V subscript p

SEquation 6.10.7.2-1

where: Formula: V subscript p = 0 point 58 times F subscript yw times D times subscript Z subscript y

SEquation 6.10.7.2-2

                        Formula: F subscript yw = 50 ksi

From Design Step P.5

                    Formula: D = 11 point 78 inches

                    Formula: t subscript w = 0 point 435 inches

so:

                      Formula: V subscript p = 0 point 58 times F subscript yw times D times subscript Z subscript y

                      Formula: V subscript p = 148 point 6 K

and

C is determined based on criteria in Section 6.10.7.3.3a with k = 5

compute:       Formula: numerator (D) divided by denominator (t subscript w) = 27 point 08

compute:       1 point 1 times ( numerator (E k) divided by denominator (F subscript yw) ) superscript 0 point 5

SEquation 6.10.7.3.3a-5

                                                                      Formula: 1 point 1 times ( numerator (29000 times 5) divided by denominator (50) ) superscript 0 point 5 = 59 point 237

Check:       Formula: 27 point 08 less than 59 point 237

thus:       Formula: C = 1 point 0

so:

                      Formula: V subscript n = C times V subscript p

                      Formula: V subscript n = 148 point 6 K

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

Resistance factor for shear:       Formula: phi subscript v = 1 point 0

S6.5.4.2

so:

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

                    Formula: V subscript r = 148 point 6 K

From Design Step P.13, the maximum factored shear in any pile in the FB-Pier analysis was 18.2 K.

Thus, piles are acceptable for shear.

Design Step P.18 - Check Maximum Horizontal and Vertical Deflection of Pile Group at Beam Seats Using Service Load Case

Displacements were determined in the interaction analysis with FB-Pier

It can be seen from the results that the horizontal displacements at the beam seat elevation are slightly higher for the cases of pinned head piles. This is expected and the difference is usually much greater. In this case, the battered piles in the front row resist the majority of the lateral load so pile head fixity is not critical to performance of the foundation system.

From Design Step P.13:

The maximum horizontal deflection observed is       Formula: Delta subscript h = 0 point 489 inches

The maximum vertical deflection observed is       Formula: Delta subscript v = 0 point 133 inches

The structural engineer has determined allowable deflections as

The maximum horizontal deflection allowed is       Formula: Delta subscript h_aII = 1 point 5 inches

S10.7.2.2

The maximum vertical deflection allowed is       Formula: Delta subscript v_aII = 0 point 5 inches

S10.7.2.3.1

Thus deflections are within tolerances and Service limit states are satisfied.

S10.7.2.4 and S10.7.2.3.1

Design Step P.19 - Additional Miscellaneous Design Issues

Downdrag

S10.7.1.4

As indicated in step P.1 elastic settlement of the loose sand will occur after construction of the pile foundation and abutment as the backfill behind the abutment is placed and the approach embankment is constructed.

Compute Settlement for consideration of Downdrag

Figure P-15 shows the location and dimensions of rectangles used to simulate approach embankment loading. The 150' length was arbitrarily selected as representative of the length beyond which additional influence from the approach embankment at the abutment location is not significant. The final approach embankment geometry relative to existing grade may decrease or increase this value. However, use of 150' is considered a reasonable upper bound.

This figure shows the plan view of the approach embankment. Area R3 and R4 are closest to the abutment. Area R1 is in the upper left portion of the plan, Area R2 is in the lower left, Area R3 is in the upper right and Area R4 is in the lower right. Area R1 and R2 have a length of 150 feet and a width of 30 feet. Area R3 and R4 have a length of 9 feet and a width of 30 feet.

Figure P-15 Plan View of Approach Embankment

Compute settlement at back edge of pile cap (Point A)

Depth of layer =       31 feet

                Formula: nu = 0 point 25

                  Formula: E subscript s = 60 times TSF

h of fill =       21 feet

Υ of fill =       130 pcf

                  Formula: q subscript 0 = 1 point 365 times TSF

At point A include influence from R1 and R2

                Formula: B = 30 feet Formula: numerator (L) divided by denominator (B) = 5 Formula: numerator (H) divided by denominator (B) = 1 point 0333333

Note: Influence factors from NAVFAC DM7 are used here because they allow proper consideration of a layer of finite thickness underlain by a rigid base. The influence values in AASHTO assume an infinite elastic halfspace. Also note that the influence values in NAVFAC are for use with a different form of the elastic settlement equation than the one contained in AASHTO. The influence values published in NAVFAC must be used with the settlement equation in NAVFAC as presented below.

From NAVFAC DM7.1-213:

              Formula: I = 0 point 16       for ν = 0.33

NAVFAC DM7.1-211:

                          Formula: S subscript 0_R1 = numerator (q subscript 0 times ( 1 minus nu 2 ) B times I) divided by denominator (E subscript s)

                        Formula: S subscript 0_R1 = 0 point 102375 feet

For two rectangles:

                              Formula: S subscript 0_R1R2 = 2 times S subscript 0_R1

                              Formula: S subscript 0_R1R2 = 2 point 457 inches

Compute settlement at front row of piles (Point B)

To simulate this case; the corner of R1 and R2 are shifted forward to be coincident with point B, and the settlement due to the approach fill weight will be equal to that computed for Point A. However, the weight of the approach embankment above the heel of the footing will be supported by the pile foundation and will not contribute to elastic settlement. Thus the settlement at point B can be computed by subtracting the influence of rectangles R3 and R4 from the settlement computed for rectangles R1 and R2 alone.

Contribution of R3 and R4 only

                Formula: B = 9 feet Formula: numerator (L) divided by denominator (B) = 3 point 333333 Formula: numerator (H) divided by denominator (B) = 3 point 444444

From NAVFAC DM7.1-213

              Formula: I = 0 point 45       for ν = 0.33

                          Formula: S subscript 0_R3 = numerator (q subscript 0 times ( 1 minus nu 2 ) B times I) divided by denominator (E subscript s)

                              Formula: S subscript 0_R3R4 = 2 times S subscript 0_R3

                              Formula: S subscript 0_R3R4 = 0 point 172758 feet       (for two rectangles) Formula: R1 + R2 minus ( R3 + R4) = S subscript 0_R1R2 minus S subscript 0_R3R4

                                                    Formula: S subscript 0_R1R2 minus S subscript 0_R3R4 = 0 point 031992 feet

                                                    Formula: S subscript 0_R1R2 minus S subscript 0_R3R4 = 0 point 383906 inches

This is not sufficient settlement to mobilize downdrag on the front row of piles as per FHWA HI-96-033, Section 9.9.1

Sufficient settlement to mobilize downdrag forces is expected at the back row of piles but not at the front row of piles. This is because the loading producing the settlement is transmitted to the soil starting at the back edge of the footing. Evaluation of downdrag loads is required for the back row of piles but not the front row. Since the back row of piles is lightly loaded and vertical, they can probably handle the downdrag load without any special details. To verify this, the following conservative approach is used.

The maximum possible downdrag force per pile is equal to the ultimate tension capacity computed in step P.14. This conservatively assumes that downdrag is mobilized along the entire length of the pile and is not reduced by the live load portion of the axial load.

S10.7.1.4

and

C10.7.1.4

                  Formula: Q subscript s = 50 point 39 K

Since downdrag is a load, it is factored in accordance with Section 3.4.1-2.

STable 3.4.1-2

                    Formula: phi subscript dd = 1 point 8       (maximum)

Maximum factored drag load per pile

                        Formula: Q subscript dd = phi subscript dd times Q subscript s

                        Formula: Q subscript dd = 90 point 7 K

From FB-Pier analysis, the maximum factored Axial load on back row of piles is 23.85 K.

Note: higher loads were observed for service load cases.

If the factored downdrag is added to the maximum observed factored pile load on the back row, the total factored load is:

                        114 point 4K

This is well below the factored resistance computed in Design Step P.10

                    Formula: Q = 340 K

Thus downdrag loads can be safely supported by the back row of piles as designed.

Battered Piles

S10.7.1.6

This bridge is not in seismic zones 3 or 4 thus battered piles are OK

No downdrag is expected at the front row of piles thus batter of front row is OK

Protection Against Deterioration

S10.7.1.8

Design Step P.1 determined soils and ground water were non corrosive thus no special protection scheme or sacrificial steel is required.

Uplift and Pile to Pile Cap Connection

S10.7.1.9

The FB-Pier analysis showed some of the piles in the back row to be in tension under some of the strength limit states. None of the service limit states showed piles in tension.

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Pile Capacity

To adequately transfer the tension load from the pile to the pile cap, a special connection detail involving reinforcing passing through a hole in the pile web or shear studs would be normally required.

However, The cases run in FB-Pier that used no skin friction effectively simulate the case of a pile pulling out of the bottom of the footing under tension load. Review of these runs indicate that the pile could be pulled completely out of the bottom of the footing thus design of a tension connection should be included in the design of the pile cap.

From Design Step P.13, the maximum factored tension force is

                  15 point 3K

The pile cap connection should be designed to resist this force.

Evaluation of the Pile Group Design

Does Pile Foundation Meet all Applicable Criteria?

Design Steps P.14 through P.19 indicate that all the applicable criteria are met

Is Pile System Optimized?

Determine if the pile system could be improved to reduce cost

Maximum factored axial load is:

                                Formula: numerator (P subscript max_a) divided by denominator (Q subscript r) = 100 %       of resistance

Maximum factored combined load is:

From Table P.16, the maximum results of the interaction equation yields:       84% of resistance

Some of the front row are not fully loaded due to flexing of the relatively thin pile cap but the front row can be considered optimized.

The back row of piles is severely under utilized for the loads investigated.

However, load cases in which the longitudinal forces are reversed will result in higher loads on the back row of piles. These loads will not exceed the loads on the front row since some longitudinal loads can not be reversed (earth pressure). Still, it may be possible to eliminate every other pile in the back row and still meet all criteria.

A brief evaluation of this possibility using FB-Pier indicates that removing 3 piles from the back row could cause the combined bending and axial stress in the front row of piles to exceed that allowed by the interaction equation. This is because elimination of the piles in the back row causes more of the horizontal loads to be absorbed by the front piles which produces higher bending moments in these piles.

Based on the above, the design is optimized to the greatest extent practical

Summary of Final Design Recommendations

Final Pile Cap Layout

All Piles are HP 12 x 53 Grade 50

All dimensions shown at bottom of pile cap

This figure shows a pile footing with a length of 46 feet and a width of 10 feet 3 inches. The pile spacing in the width direction is 7 feet 9 inches and the pile spacing in the length direction is 7 feet 4 inches. The number of pile spaces is the width direction is one and number of pile spaces is the length direction is 6. Therefore, the total pile spacing in the width direction is 7 feet 9 inches and the total pile spacing in the length direction is 44 feet 90 inches. The edge distance in the length direction is 1 foot 5 and one quarter inches and the edge distance in the width direction is 1 foot 3 inches. The width of the footing zone thickened to account for the stem is 3 point 5 feet in the width direction. The distance from the edge of footing to the footing zone thickened to account for the stem is 2 point 75 feet in the width direction. The front row of piles are battered with a 1 horizontal to 3 vertical slope.

Figure P-16 Final Pile Cap Layout

Design considerations for design of pile cap

Piles to be embedded 1' into pile cap

Piles to have bar through web or shear stud to transfer 15 Kip tension load to cap

For structural design of the cap, the factored axial load per pile is summarized in tables below.

From FB-Pier File FHWA_bat_fix_noskin.out

CASE: Fixed Pile Heads No Skin Friction

FB-Pier Load Case
LC1
LC2
LC3
LC4
LC5
LC6
Limit State
STR-I MAX/FIN
SER-I MAX/FIN
STR-I MIN/FIN
SER-I MIN/FIN
STR-III MAX/FIN
SER-I MAX/FIN
Pile Number
1
-6.9
-35.0
0.1
-36.1
0.0
-35.0
2
-315.3
-221.5
-273.3
-220.3
-275.2
-221.5
3
-12.7
-40.1
0.1
-41.3
0.0
-40.1
4
-330.5
-232.2
-287.1
-230.9
-284.7
-232.2
5
-15.4
-41.9
0.1
-43.1
0.0
-41.9
6
-336.1
-235.9
-292.4
-234.7
-286.3
-235.9
7
-17.0
-42.8
0.1
-44.0
0.0
-42.8
8
-339.9
-238.6
-292.4
-237.3
-286.2
-238.6
9
-15.3
-40.7
0.1
-41.9
0.0
-40.7
10
-335.4
-234.9
-288.4
-233.7
-279.3
-234.9
11
-12.6
-37.8
0.1
-38.9
0.0
-37.8
12
-330.7
-230.8
-287.8
-229.5
-272.6
-230.8
13
-6.7
-31.3
0.1
-32.4
0.0
-31.3
14
-315.6
-219.3
-274.1
-218.1
-256.4
-219.3

Table P-17 Factored Axial Load per Pile Fixed Pile Heads - No Skin Friction          

From FB-Pier File FHWA_bat_pin_noskin.out

CASE: Pinned Pile Heads No Skin Friction

FB-Pier Load Case
LC1
LC2
LC3
LC4
LC5
LC6
Limit State
STR-I MAX/FIN
SER-I MAX/FIN
STR-I MIN/FIN
SER-I MIN/FIN
STR-III MAX/FIN
SER-I MAX/FIN
Pile Number
1
-13.7
-36.6
1.4
-37.8
-14.6
-36.6
2
-308.4
-219.2
-274.5
-218.0
-254.3
-219.2
3
-19.5
-41.7
1.4
-42.9
-16.8
-41.7
4
-323.8
-230.3
-288.4
-229.0
-267.1
-230.3
5
-22.2
-43.4
1.4
-44.6
-16.7
-43.4
6
-329.5
-234.3
-293.7
-233.1
-271.7
-234.3
7
-23.8
-44.2
1.4
-45.5
-15.8
-44.2
8
-332.2
-237.0
-294.0
-235.7
-272.1
-237.0
9
-22.1
-42.1
1.4
-43.3
-12.4
-42.1
10
-327.8
-233.6
-290.0
-232.4
-268.2
-233.6
11
-19.3
-39.0
1.4
-40.2
-7.9
-39.0
12
-324.3
-230.0
-289.0
-228.7
-266.7
-230.0
13
-13.3
-32.4
1.4
-33.6
-0.8
-32.4
14
-309.1
-218.8
-275.3
-217.5
-253.8
-218.8

Table P-18 Factored Axial Load per Pile Pinned Pile Heads - No Skin Friction          

From FB-Pier File FHWA_bat_fix_skin.out

CASE: Fixed Pile Heads Skin Friction

FB-Pier Load Case
LC1
LC2
LC3
LC4
LC5
LC6
Limit State
STR-I MAX/FIN
SER-I MAX/FIN
STR-I MIN/FIN
SER-I MIN/FIN
STR-III MAX/FIN
SER-I MAX/FIN
Pile Number
1
-5.6
-34.0
15.3
-35.1
-1.6
-34.0
2
-314.5
-220.6
-287.0
-219.4
-271.7
-220.6
3
-11.8
-39.3
14.8
-40.5
-1.4
-39.3
4
-330.3
-231.8
-301.2
-230.5
-282.8
-231.8
5
-14.6
-41.2
14.5
-42.4
0.2
-41.2
6
-336.0
-235.7
-306.4
-234.4
-285.7
-235.7
7
-16.2
-42.2
14.4
-43.4
0.8
-42.2
8
-339.9
-238.4
-309.3
-237.1
-286.7
-238.4
9
-14.5
-40.0
14.5
-41.2
1.8
-40.0
10
-335.3
-234.6
-305.1
-233.4
-280.7
-234.6
11
-11.6
-37.0
14.8
-38.2
2.9
-37.0
12
-330.5
-230.5
-301.4
-229.2
-274.9
-230.5
13
-5.5
-30.3
15.3
-31.4
4.4
-30.3
14
-314.8
-218.5
-287.3
-217.3
-259.3
-218.5

Table P-19 Factored Axial Load per Pile Fixed Pile Heads - Skin Friction          

From FB-Pier File FHWA_bat_fix_skin.out

CASE: Pinned Pile Heads Skin Friction

FB-Pier Load Case
LC1
LC2
LC3
LC4
LC5
LC6
Limit State
STR-I MAX/FIN
SER-I MAX/FIN
STR-I MIN/FIN
SER-I MIN/FIN
STR-III MAX/FIN
SER-I MAX/FIN
Pile Number
1
-13.3
-36.2
2.1
-37.3
-14.3
-36.2
2
-307.5
-218.0
-274.6
-216.8
-254.4
-218.0
3
-19.5
-41.6
1.4
-42.7
-16.7
-41.6
4
-323.5
-229.6
-289.2
-228.3
-267.7
-229.6
5
-22.3
-43.3
1.1
-44.5
-16.7
-43.3
6
-329.4
-233.8
-294.7
-232.5
-272.6
-233.8
7
-23.9
-44.2
0.9
-45.5
-15.9
-44.2
8
-332.0
-236.4
-295.0
-235.1
-273.0
-236.4
9
-22.1
-42.0
1.1
-43.2
-12.5
-42.0
10
-327.4
-232.9
-290.9
-231.6
-268.9
-232.9
11
-19.2
-38.8
1.5
-40.0
-7.9
-38.8
12
-324.0
-229.3
-289.9
-228.0
-267.4
-229.3
13
-13.0
-32.0
2.2
-33.2
-0.7
-32.0
14
-308.2
-217.6
-275.6
-216.3
-253.9
-217.6

Table P-20 Factored Axial Load per Pile Pined Pile Heads - Skin Friction          

Absolute maximum from above:       15 point 319

Absolute minimum from above:       minus 339 point 9

FB-Pier may be used to print out all stresses in each element of the pile cap as a check on manual methods if desired.

Notes to be placed on Final Drawing

Maximum Factored Axial Pile Load = 340K

Required Factored Axial Resistance = 340K

Piles to be driven to absolute refusal defined as a penetration resistance of 20 Blows Per Inch (BPI) using a hammer and driving system components that produces a driving stress between 37 and 45 KSI at refusal. Driving stress to be estimated using wave equation analysis of the selected hammer.

Verify capacity and driving system performance by performing stress wave measurements on a minimum of 2 piles in each substructure. One test shall be on a vertical pile and the other shall be on a battered pile.

Perform a CAPWAP analysis of each dynamically tested pile. The CAPWAP analysis shall confirm the following:

Driving stress is in the range specified above.

The ultimate pile point capacity (after subtracting modeled skin friction) is greater than:

                  Formula: Q subscript p = 523 K

This is based on a resistance factor (Φ) of 0.65 for piles tested dynamically.

References:

FHWA HI-96-033 Design and Construction of Driven Pile Foundations, Hannigan, P.J., Gobel, G.G, Thedean, G., Likins, G.E., and Rausche, F. for FHWA, December 1996, Volume 1 and 2

NAVFAC DM7 Design Manual 7; Volume 1 - Soil Mechanics; Volume 2 - Foundations and Earth Structures, Department of the Navy, Naval Facilities Engineering Command, May 1982.

PADOT DM4 Design Manual Part 4, Pennsylvania Department of Transportation Publication 15M, April 2000

Reese and Wang (1991) Unpublished paper presenting group efficiencies of pile groups subject to horizontal loads in diferent directions and at different spacings.

NCEER-97-0022 Proceedings of the NCEER Workshop on Evaluation of Liquefaction Resistance of Soils, Edited by T.L. Youd, I.M. Idriss. Summary Report, 1997. MCEER Publication NCEER-97-0022

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Updated: 07/18/2013
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