bridge lrfd theory deep foundations

7/29/2019 Bridge Lrfd Theory Deep Foundations

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LRFD Theory forGeotechnical Design

Topic 3 Part A

Deep Foundations

Session 3

LRFD for Highway Bridge Substructures

and Earth Retaining Structures

Course No. 130082A


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A. State the performance limits thatshould be evaluated when designinga deep foundation

B. Be able to select a deep foundationtype

C. Be able to select the appropriateresistance factor for eachperformance limit evaluated

Learning Outcomes


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Deep Foundation

Performance Limits

A


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Start F.1 F.2 F.3 F.4 F.5

F.6 F.7 F.8 F.9 F.10

D.1

F.11

D.2

F.12

F.13

F.14

F.15

F.16

F.17

F.18D.3D.4End

Detailed Flow ChartRM page 3.3.6

1. Decide deep foundation type2. Select resistance factor

3. Compute resistances4. Layout foundation group and

analyze at the strength limit state5. Check the service limit state

Deep Foundation DesignProcess


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Structuralresistance

Axial geotechnicalresistance

Driven resistance

Structuralresistance

Axial geotechnicalresistance

Driven Piles Drilled Shafts

Strength Limit State Checks


6/172

Structural Axial Failure


7/172

Structural Flexure Failure


8/172

Structural Shear Failure


9/172

Axial Geotechnical Resistance


10/172

Pile damage

Driven Resistance


11/172

Driven Performance Limit


12/172

Driven Performance Limit


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Global StabilityVerticalDisplacement

Horizontal

Displacement


Global StabilityVerticalDisplacement

Horizontal

Displacement

Service Limit State Checks


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Global Stability


15/172

x

z

Displacement


16/172

Same

Determining Resistance

Determining Deflection

Different

Comparison of load and resistance

Specific separation of resistance anddeflection

LRFD Differences from ASD


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Deep foundation type

selectionMethod of support

Bearing material depthLoad type, direction and magnitude

Constructability

Cost

B


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Deep Foundation Material

Driven X X X X X X X

Drilled or Bored -- X -- X X -- X

Jacked / Special X -- -- X X -- X

Pres

tresse

dC

oncre

te

Pos

t-tension

Cocnre

te

Pre-cast

Concre

te

Cast-in-p

lace

Concre

te

Stee

l

Woo

d

Spec

ialty

/Compos

ites

Drilled Shafts

Driven Piles

Deep Foundation Types


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End Bearing Side Friction Combined

Method of Support


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Driven LowDisplacement Piles


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Driven HighDisplacement Piles


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Drilled Shafts


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Depth to Bearing/ Scour


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Permanent/ Transient/ Cyclic

Horizontal or Vertical

Load Type and Direction


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Wood is better for transientresistance than permanent

Steel pile better cyclic resistance

High horizontal loads better resistedby stiffer piles or shafts

Load Type and Direction


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Deep foundationtype

Typical range ofnominal (ultimate)resistance (kips)

Typicallength (feet)

Timber pile 75 200 20 40

Concrete pile 200 2,000 20 150

Steel H-pile 200 1,000 20 160

Pipe pile 175 2,500 20 100

Drilled shaft 750 10,000 20 160

Load Magnitude


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Obstructions/ Rock

Use low displacement

steel piles-or-Drilled shafts

Constructability


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Equipment access

Low headroom requires pile splicingEquipment size a function of pile/shaft size


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Wrap Up

1. Decide deep foundation type

2. Select resistance factor3. Compute resistances

4. Layout foundation group and analyzeat the strength limit state

5. Check the service limit state


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Selection of

Resistance factorsStrength limit state

Structural Resistance

Geotechnical Resistance

Driven Resistance (piles only)

Service limit state

Resistance factor = 1.0(except global stability)

C


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Axial compression

Combined axial and flexure

Shear

LRFD

Specifications

Concrete Section 5

Steel Section 6

Wood Section 8

Methods for determiningstructural resistance


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Concrete (5.5.4.2.1)Axial Comp. = 0.75Flexure = 0.9

Shear = 0.9

Steel (6.5.4.2)Axial = 0.5-0.6Combined

Axial= 0.7-0.8Flexure = 1.0Shear = 1.0

Timber (8.5.2.2)Compression = 0.9Tension = 0.8Flexure = 0.85

Shear = 0.75

LRFD

Specifications

Structural resistance factors


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Field methods

Static load test

Dynamic load test (PDA) Driving Formulae

Static analysis methods

Determining GeotechnicalResistance of Piles


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Determining GeotechnicalResistance of Piles


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Load

Settlemen

t

Pile top settlement

Static Load Test


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Dynamic Load Test (PDA)


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Driving Formulas



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Geotechnical ResistanceFactors for Piles

Method Site Variability

Static LoadTest

Low 0.8 0.9

Medium 0.7 0.9

High 0.55 0.8

AASHTO Table 10.5.5.2.2-2

Site Variability Defined in NCHRP

Report 507 Range of Values of Resistance

Factors Depends on Number ofStatic Load Tests



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Method Dynamic Test w/Signal Matching (e.g.,PDA + CAPWAP)

0.65

AASHTO Table 10.5.5.2.2-1 & 3

Test 1% to 50% of ProductionPiles, Depending on SiteVariability and Number of Piles

Driven Site Variability Defined in NCHRP

Report 507



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Method

Wave Equation only 0.4

FHWA-Modified Gates 0.4

ENR 0.1


Geotechnical Safety


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Geotechnical SafetyFactors for Piles

Basis for Design and Typeof Construction Control

Increasing Design/Construction Control

Subsurface Exploration

Static Calculation

Dynamic Formula

Wave Equation

CAPWAPAnalysis

Static Load Test

Factor of Safety (FS) 3.50

X

X

X

X

X

X

X

X

1.90

X

X

X

X

2.00

X

X

X

X

2.25

X

X

X

2.75


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Computation of StaticGeotechnical Resistance

AASHTO10.7.3.7.5-2

RP

RS

RR= Rn

Rn = qpRp + qsRs

RP = AP qP

RS = AS qs


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Static Analysis Methods

a method

b methodl method

Nordlund -Thurman

method

SPT-method

CPT-method

a method

b methodSide friction inRock

Tip Resistance in

Rock



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Pile Group ResistanceStatic Geotechnical Resistance

Take lesser of



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Geotechnical ResistanceFactors Pile Static

Analysis MethodsMethod Comp Ten

- Method 0.35 0.25

- Method 0.25 0.20

- Method 0.40 0.30

Nordlund-Thurman 0.45 0.35

SPT 0.30 0.25

CPT 0.50 0.40

Group 0.60 0.50


Driven Pile Time


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Driven Pile TimeDependant Effects

Setup Relaxation

RP

RS

RP

RS

RP

RS

RP

RS


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Side Resistance

Tip Resistance

Total Resistance

A

B

CD

RP

RS

RR= Rn = qpRp + qsRs

Displacement

Resistan

ce

Drilled Shaft Resistance


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For cohesivesoils useequivalent pier

approach

For cohesionlesssoils, use groupefficiency factorapproach

Drilled Shaft Group Resistance

Rn group = x Rn single

where: = 0. 65 at c-c spacingof 2.5 diameters = 1.0 at c-c spacing of

6 diameters

G t h i l R i t F t


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Geotechnical Resistance FactorsDrilled Shafts

Method Comp Ten

Shafts in Clay

- Method (side) 0.45 0.35

Total stress (tip) 0.40 --Shafts in Sand

- Method (side) 0.55 0.45

ONeill & Reese (tip) 0.50 --Group (sand or clay) 0.55 0.45


Geotechnical Resistance Facto s


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Geotechnical Resistance FactorsDrilled Shafts

Method Comp Ten

Shafts in Interm. Geomatls (IGMs)

ONeill & Reese (side) 0.60

ONeill & Reese (tip) 0.55 --Shafts in Rock

Side (H&K, O&R) 0.55 0.40

Side (C&K) 0.50 0.40Tip (CGS, PMT, O&R) 0.50 --

Load Test (all matls)


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AxialGeotechnicalResistance of aDrilled Shaft in

Clay

ReferenceManual3.3.7.5

Example 9

50

2.5

Stiff Clay

Su = 1500 psf

E = 200 ksf = 125 pcf

e50 = 0.007

Drilled Shaft

fc = 4 ksi

Ec = 3600 ksi


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Determine Unit

Side Resistance

qs = Su

To find a, checkSu/pa = 1.5 / 2.12Su/pa = 0.7 < 1.5

So

= 0.55

qs = 0.55 x 1500 psf

qs = 0.825 ksf

50

2.5

Stiff Clay

Su = 1500 psf

E = 200 ksf = 125 pcf

e50 = 0.007

Drilled Shaft

fc = 4 ksi

Ec = 3600 ksi


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Determine ExclusionZones

Per AASHTO 10.8.3.5.1bTop 5' non contributingBottom 1 diameter (2.5')

non contributing

Ls= 50 5- 2.5 = 42.5

2.5

50

2.5

5

42.5


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As = D LsAs = (2.5)(42.5)

As = 334 ft2

Rs = qs AsRs = (0.825 ksf)(334 ft

2)

Rs = 275 kips

2.5

50

Rs=275kips


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Point Resistance

qp = Nc Su

Nc = 6(1 + 0.2 (Z/D)) < 9Nc = 6(1 + 0.2 (50/2.5))Nc = 30not less than 9 thus

Nc = 9

qp = 9 (1.5 ksf)q

p

= 13.5 ksf 2.5

50

Rs=275kips


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Point Resistance

Rp = qp Ap

Ap = D2/4

Ap = (2.5)2/4

Ap = 4.9 ft2

Rp = 13.5 ksf (4.9 ft2)

Rp = 66 kips

Rs=275kips

Rp = 66 kips


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Combining Side andPoint Resistance

RR= qs Rs + qp Rp

qs = 0.45

qps = 0.4

RR= 0.45 (275) + 0.4 (66)RR= 150 kips

Rs

=275kips

Rp = 66 kips


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Combining Side andPoint Resistance

1.0 2.0

1.0

Dzt/ D (%)

Rsd

/Rs

0 5.0 10.0Dz

t/ D (%)

1.0

Rpd

/Rp

Dzt/ D (%)

00

0


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Check Relative Stiffness

SR

= (Z/D) (Esoil

/Eshaft

) < 0.01

SR= (50/2.5) (1.39 ksi

l/3600 ksi)

SR = 0.008 < 0.01

If

Shaft can be considered rigid

Shaft can be considered rigid


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0

50100

150

200

250

300

350

0 0.5 1 1.5 2

Displacement (in)

Develo

ped

Resistance

(kips)

Developed Side Resistance Developed Base Resistance

Developed Nominal Resistance

Rs = 256 kips

RP

= 38 kips

0.3


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RR= qs Rs + qp RpRR= 0.45 (256) + 0.4 (38)

RR= 131 kips


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Driven Resistance

Drivehead

Ground

Surface

Ram

Cushion

Soft Layer

Dense

Layer

(a) (b) (c)Permanent Set

(d)

Pile

elastic

elastic

vo

c

c

cc

CompressiveForce Pulse

(Incident)

Compressive

Force Pulse

(Attenuated)

Compressive

Force Pulse

Tensile or

Compressive

Force Pulse

(Reflected)

Comp Str

k i

Tens Str

k i


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Wave

EquationResults

ksi

30

20

10

Ult Cap

200

400

600

800

kips

0 160 320 480 Blows/ft

4.0

8.0

12.0

16.0

ft

Stroke

ksi


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Concrete piles, = 1.00

AASHTOArticle 5.5.4.2.1

Steel piles, = 1.00AASHTOArticle 6.5.4.2

Timber piles, = 1.15

AASHTOArticle 8.5.2.2

Driven Resistance Factors


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Participant Workbook

Page 3.3A.22


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67/172

Method Qn

(kips)

Qr(kips)

# ofPiles

a-method 0.4 550 220 17

PDA on 5% 0.65 550

Gates Formula 0.4 550

Structural Resistance 0.6 775 8465

17220

11358


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Comparison to ASDService Load = 2794 kips

Method FSQn

(kips)

Qr(kips)

# ofPiles

a-method 3.5 550 15718

(17)

PDA on 5% 2.25 550 24412

(11)

Gates Formula 3.5 550 157 18(17)

Structural Resistance3

(0.33fy)775 256

11(8)


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Wrap Up

1. Decide deep foundation type

2. Select resistance factor

3. Compute resistances

4. Layout foundation group and analyze atthe strength limit state



70/172


Page 3.3A.25


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1. Geotechnical resistance

2. Structural resistance3. Driven resistance

1. Horizontal deflection2. Vertical deflection (settlement)

3. Global stability

Exercise 1: List the three strength limitstate checks for driven piles

Exercise 2: List the three service limit

state checks for drilled shafts


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Exercise 3: Match thedeep foundation type to

the condition.

Condition1) Deep granular material

2) Loose random filloverlying rock3) Large horizontal loads

TypeA)Steel H-Pile

B) Closed endpipe

C) Large diameterdrilled shaft

A

B

C


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74/172

A. State the performance limits thatshould be evaluated when designinga deep foundation

B. Be able to select a deep foundationtype

C. Be able to select the appropriate

resistance factor for eachperformance limit evaluated

Learning Outcomes

Session 3


75/172

LRFD Theory for

Geotechnical Design

Topic 3 Part B

Deep Foundations

Session 3



Course No. 130082A


76/172

D. Apply the rigid cap method toevaluate the strength limit state

checks

Learning Outcome


77/172


78/172

X

Y

Z

Centroid ofPile group

Rigid Cap Model


79/172

X

Y

Z

Mx

MyFz

-xi yi

Pi

n

1i

2

i

iy

n

1i

2

i

ixzi

x

xM

y

yM

n

FP

Distribution of Axial Loads

Distribution of


80/172

X

Y

Z

FxHi

Distribution ofHorizontal Loads

i l


81/172

Ht

Qt

Mty

P

y

y

PropertiesA, E, I

Horizontal Response

C d l


82/172

P-y Curve development

Typical requiredsoil parameters

Su

f

k

50

k coefficient of variation of subgrade reaction

50 - strain at 50% of ultimate strength

l f i l l


83/172

10.1 k

1740 k8000 in-k

Depth,

ft

50

40

30

20

10

-0.2 0 0.2 0.4 0.6 0.8 0 20 40 60 80 -60 -40 -20 0 20

Deflection,

in.Moment,

in. -kx102Shear,

k

0.84 8640

65.5

P-y Results for Single Element


84/172

Pil H d Fi it


85/172

xx

Service Limit State

Moment Moment

Strength Limit State

Pile Head Fixity

G Eff t


86/172

Fx

H1H2

Group Effects

P I t ti Eff t


87/172

P

yPm * P

P

P-y Interaction Effects

O t t f lti l l d


88/172

Output for multiple loads

AppliedHorizontal

Load

ResultingDeflection

MaximumMoment

3.00E+01ad


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-2.00E+03

-1.50E+03

-1.00E+03

-5.00E+02

0.00E+000 0.5 1

0.00E+00

1.00E+01

2.00E+01

0 0.5 1HorizontalLoa

(kip

s)

Deflection (in)

Maxim

um

Moment(

in-kips)

C t P M d li


90/172

Computer P-y Modeling

Horizontal Loads,


91/172

FxH1H2

x x

M1M2

Horizontal Loads,Pile Moment

Wh W A G i


92/172


3. Compute resistances4. Layout foundation group and analyze at

the strength limit stateCompute load effects in piles using rigid

cap methodCompare load effects to factoredresistances for piles


Where We Are Going

Guided Walk Through


93/172


Page 3.3B.7


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12-0

4-6

5-0

5-0

6-0

15-0

3-6

46-6

15-6

23-0

15-6


95/172

18 60 1860 60 60

18

18

36

36

36

HP 12x53 Centroid


96/172

Fz

Fy

Fx

-My

Mx

Applied LoadsStrength V load case

Fx

= 38.4 kipsFy = 109.1 kipsFz = 3594.0 kips

Mx = 3196.5 k-ftMy = -8331.9 k-ft


97/172

n

1i

2

i

iy

n

1i

2

i

ixzi

x

xM

y

yM

n

FP

Example calculation, pile 9:

Fz = 3594.0 kips Mx = 3196.5 k-ftn = 20 piles y

i= 18 in (1.5 ft)

xi2 = 1000 ft2 My = -8331.9 k-ft

yi2 = 225 ft2 xi = 60 in (5 ft)

P9

= 243 kips


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160 k

Y

X

202 k 244 k 285 k 327 k

118 k 159 k 201 k 243 k 284 k

75 k 117 k 158 k 200 k 242 k

32 k 74 k 116 k 157 k 199 k


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y

y assumed to be 0.15

Fy


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-200

-400

-600

108

6

4

2

00.1 0.2

Lo

ad(kips)

Max.

Momen

t(k-in)

Deflection (in)

0.1

5

in

7.2 kips

5.9 kips4.5 kips

-340 k-in-390 k-in-450 k-in


101/172

Row Pm Hy Mmax1 0.35 4.5 kips -340 k-in

2 0.35 4.5 kips -340 k-in

3 0.5 5.9 kips -390 k-in

4 0.7 7.2 kips -450 k-in

Sum of Hy forces times piles per column =(22.1 kips/column) (5 columns) = 110.5 kips

110.5 kips close to 109.1 kips


102/172

x

x assumed to be 0.05

Fx


103/172

-33

-66

-100

2.0

1.5

1.0

0.5

00.025 0.075

Lo

ad(kips)

Max.

Momen

t(k-in)

Deflection (in)

0.0

5

in

2.2 kips2.0 kips1.8 kips

-75 k-in-80 k-in-90 k-in

2.5


104/172

Column Pm Hx Mmax

1 0.7 1.8 kips -75 k-in

2 0.7 1.8 kips -75 k-in

3 0.7 1.8 kips -75 k-in

4 0.85 2.0 kips -80 k-in

5 1.0 2.2 kips -90 k-in

Sum of Hx forces times piles per row =(9.6 kips/row) (4 rows) = 38.4 kips

38.4 kips = 38.4 kips


105/172

100

200

300

Dep

th(in)

Shear (kips)-2 0 2 4 6 8


106/172

For load case Strength V:

Max. shear = 7.2 kips in y-direction(Piles 1, 2, 3, 4, 5 at the top of pile)

Max. axial load (Pile 5) = 326 kipsMin. axial load (Pile 16) = 32 kips

Maximum combined loading (Pile 5)Axial load = 326 kipsMoment (x-direction) = -37.5 kip-ft

Moment (y-direction) = -7.5 kip-ft

(no uplift)

Where We Are Going


107/172


3. Compute resistances4. Layout foundation group and analyze at

the strength limit stateCompute load effects in piles using

rigid cap methodCompare load effects to factoredresistances for piles


Where We Are Going


108/172

Loose

Silty sand

Hard Clay

Driven

HP 12 x 534

35

>100

f= 31o

sat = 110 pcf

Su = 8000 psfsat = 125 pcf

OCR = 10


109/172


Axial compression

As = 15.5 in2

(after corrosion loss)Fy = 50 ksil = 0 in

Pn

AASHTOArticles 6.9.4.1-1, 10.7.3.12.1

Pn = 0.66lFyAs = 0.66

0(50)(15.5)

Pn = 775 kips


110/172

Structural ResistanceFlexure Resistance

zx = 74 in

3

zy = 32.2 in3

Fy = 50 ksi

Mny

Mnx = (50 ksi)(74 in3) =Mny = (50 ksi)(32.2 in

3) =3700 k-in1610 k-in

y

xMnx


111/172


Shear ResistanceD = 11.78 intw = 0.435 in

Fy = 50 ksiC = 1.0

Vny

AASHTO Articles 6.10.7.2-1,6.10.7.2-2,6.10.7.3.3a

Vp = (0.58)(50 ksi)(11.78 in)(0.435 in)

VpC = 149(1.0) = 149 kips

y

x


112/172

Axial Compression= 0.6 for Pr

Combined Compression and Flexure= 0.7 for Pr, 1.0 for Mr

Shear= 1.0 for Vr


113/172

Geotechnical Resistance Axial compression

Use the beta method fro axial resistance in sand

and clay.

qs = b'v and qp = Nt'v

ForSand


114/172

Sand

0.28

ForSand


115/172

Sand

28

ForClay


116/172

Clay

1.5


117/172

Tip resistance in clay

qp = 9Su

Depth

(ft)

Average

'v (ksf)

Cum.side

friction

(kips)

Qp

=q

pA

p

(kips)

TotalResistance

(kips)0 0 0 0 0

5 0.12 0.67 6.6 7.3

00 500 1000 1500 2000

Resistance (kips)AxialGeotechnical


118/172

0

20

40

60

80

100

120

140

0 500 1000 1500 2000

Depth(

ft)

Side Friction

Point Resistance

Total Resistance

Resistance

vs. Depth


119/172


120/172

Steps to perform drivability analysis: Estimate total soil resistance and

distribution

Select hammer

Model driving system and soil resistance

Run wave equation analysis

00 500 1000 1500 2000

Resistance (kips)

Estimate ResistanceDistribution


121/172

0

20

40

60

80

100

120

140

Depth(

ft)

Side Friction

Point Resistance

Total Resistance

Q = dyn Rn

stat = 0.65Rn = 465 kips/0.65

Rn = 715 kips

715 kips

Dest= 70

EB = 10%

20%40%

80%

100%

60%


122/172

Skin quake = 0.1 default per WEAP

manual

Skin damping = 0.2 From WEAP manual

Toe quake = 0.1 1/120 of pile width

Toe damping = 0.15 per FHWA NHI-05-042

page 17-68

Select dynamic properties of soil


123/172


124/172

715 kips

Bigger hammer (Delmag 46-13)


125/172

715 kips

58 ksi


126/172

Evaluate driving stressdr= 0.9 da fy (permissible driving stress)da = 1.0

dr= 0.9 (1.0) 50 ksidr= 45 ksi

45 ksi < 58 ksi (driving stress exceeded)

What is the maximum resistance that canbe developed without exceeding thepermissible driving stress?

45 ksi


127/172

550 kips

45 ksi

17 BPI


128/172

Factored resistance limited by drivingstress (driven resistance)

RR

=dyn

Rn

dyn = 0.65RR= 0.65 (550 kips)RR = 358 kips


129/172

*AASHTO Eqn. 6.9.2.2-2

Axial geotechnical performance ratio =326/465 = 0.7

Axial structural performance ratio =

326/465 = 0.7Combined axial and flexural performanceratio = 0.78*

Driven performance ratio

326 / 358 = 0.91Shear performance ratio =7.2 / 256 = 0.03

00 500 1000 1500 2000

Resistance (kips)EstimateRequired Length


130/172

0

20

40

60

80

100

120

140

Depth

(ft)

Side Friction

Point Resistance

Total Resistance

for Actual

Factored load Q = 326 kips

Q = stat Rnstatstat = 0.25

Rnstat = 326 kips/0.25

Rnstat = 1304 kips

1304 kips

Dest= 91


131/172


132/172


133/172

Fz

Fy

Fx

-My

Mx

Beam seat elevation

Rock

Loose Sand

Applied Loads


134/172

Pile 5

Pile 16

Rigid Cap Results

Shear Results0.0 ft

2.0 ft

4.0 ft

6.0 ft

8.0 ft

10.0 ft

12.0 ft

14.0 ft

16.0 ft

18.0 ft

20.0 ft

22.0 ft

24.0 ft

26.0 ft

28.0 ft

30.0 ft

32.0 ft

-7.

59

kips

1.

87

kips

Shear = 7.2 kipsMoment = - 37.5 k-in

0.0 ft

2.0 ft

4.0 ft

6.0 ft

8.0 ft

10.0 ft

12.0 ft

14.0 ft

16.0 ft

18.0 ft

20.0 ft

22.0 ft

24.0 ft

26.0 ft

28.0 ft

30.0 ft

32.0 ft

-30.

1

kip-ft

15.

3

kip-ft

Moment ResultsAxial Results

0.0 ft

2.0 ft

4.0 ft

6.0 ft

8.0 ft

10.0 ft

12.0 ft

14.0 ft

16.0 ft

18.0 ft

20.0 ft

22.0 ft

24.0 ft

26.0 ft

28.0 ft

30.0 ft

32.0 ft

-324

kips

-18.

2

kips

Max. Axial = 327 kipsMin. Axial = 32 kips


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*AASHTO Eqn 6 9 2 2-2

Axial geotechnical performance ratio =327/465 = 0.7

Axial structural performance ratio =327/465 = 0.7

Combined axial and flexural performanceratio = 0.73*

Driven performance ratio327 / 358 = 0.91

Shear performance ratio =7.59 / 256 = 0.03

(0.7)

(0.78)

(0.91)

(0.03)

(0.7)

00 500 1000 1500 2000

Resistance (kips)Accounting forScour


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0

20

40

60

80

100

120

140

Depth

(ft)

Side Friction

Point Resistance

Total Resistance

Q = 358 kips

Q = stat Rnstatstat = 0.25


Rnstat = 1432 kips

RS scour= 20 kips

1432 kips

Dest= 96

20 kips

Scoured

Accounting for Scour


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Accounting for Scour

Required driven resistance during construction

Q = 358 kips

Q = dyn Rndr RS scourRndr = Q / dyn+ RS scourdyn = 0.65

Rndr= 326 kips/0.65 + 20 kipsRndr= 571 kips

00 500 1000 1500 2000

Resistance (kips)Accounting forDowndrag

Q

358 ki


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0

20

40

60

80

100

120

140

Depth

(ft)

Side Friction

Point Resistance

Total Resistance

Q = 358 kips +

DD DD

DD = 1.8

RS scour= 20 kips

DD = 20 kips

Q = 394 kips


Rnstat = 1576 kips

1576 kips

Dest= 100

20 kips

Settling

Accounting for Downdrag


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ccou g o o d agRequired driven resistance during construction

Q = 358 kips + DD DD

DD = 1.0

Since resistance in downdrag zone determined bysignal matching

Q = 358 kips + 1.0 (20 kips) = 378 kips

Q = dyn Rndr RS downdrag

Rndr = Q / dyn+ RS downdragdyn = 0.65

Rndr= 378 kips/0.65 + 20 kips

Rndr= 602 kips


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00 500 1000 1500 2000

Resistance (kips)Accounting forSet up


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0

20

40

60

80

100

120

140

Depth

(ft)

Side Friction

Point Resistance

Total Resistance

R1dr= 25.6 kips

Rndr= 963 kips

963 kips

Dest= 95

Setup

25.6 kips

End Bearing on Hard Rock


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g

Assume structural resistance is much less thangeotechnical resistance.

Assume potential damage to pile

RR = PnPn = 775 kips

= 0.5 (due to potential for damage)

RR = 0.5 (775 kips) = 388 kips

Estimate length based on depth to rock Control driving to prevent damage


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Participant WorkbookPage 3.3B.29

Given a load case with loadingdirections as depicted in the


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X

Y Z

a. Which pile will havethe highest axial load?

b. Which pile will havethe lowest axial load?c. Which pile will be

subject to the highest

horizontal load?d. Which pile will be

subject to the highestbending moments?

directions as depicted in the

adjacent figure:

Mx

My

Fx

Fz

Fy

1 2

3 4

5D c-c

1

4

2

2

Learning Outcome


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D. Apply the rigid cap method toevaluate the strength limit state

checks

LRFD Th f

Session 3


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LRFD Theory for

Geotechnical DesignTopic 3 Part C

Deep Foundations



Course No. 130082A

Learning Outcome


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E. Be able to perform a rigid capanalysis of a driven pile group at

Service Limit State

Where We Are Going


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3. Compute resistances4. Layout foundation group and

analyze at the strength limit state5. Check the service limit state

Axial Response of a Single Element(Approximate method)


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Qtop

L

ztop

zp

Point bearing onlyztop = zp + Qtop L/ (A E)

Constant side friction onlyztop = zp + Qtop L/ (2 A E)

Linear increasing friction

onlyztop = zp + Qtop L/ (3 A E)

Pile Properties A, E

(Approximate method)

Axial Response of a Group


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Perform Rigid CapAnalysis Driven Pile

E


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Analysis, Driven Pile


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18 60 1860 60 60

18

18

36

36

36

HP 12x53 Centroid


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155/172

128 k

Y

X

159 k 191 k 222 k 254 k

94 k 125 k 157 k 188 k 220 k

60 k 91 k 123 k 154 k 186 k

26 k 57 k 89 k 120 k 152 k


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1 2 3 4

Fy = 86.1 kips / 5 rowsFy = 17.2 kips/row

Assume deflection = 0.11

Pm 0.35 0.35 0.5 0.7

Fy = H1 + H2 + H3 + H4Fy = 3.7+3.7+4.6+5.5Fy = 17.5 kips


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10

8

6

4

2

0 0.1 0.2

Load(kips)

Deflection (in)

0

.11

in

5.5 kips4.6 kips3.7 kips

HP 12x53 in loose sand, fixed x-x axis


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QP

Qtop

L=3

84

in

ztop

zp

Estimate Dzp=0.03 in @ Qp=500 k

2900015.5

3845000.03z top

ztop= 0.46 in= 0.00092(Qtop)

Assume point bearing:

AE

LQ

zz

top

ptop


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Pile head displacements

Dztop, Pile B = 0.00092 (88.8 kips)= 0.082 in.

Dztop, Pile C = 0.00092 (190.6 kips)= 0.175 in.

Dy for both piles is 0.11 in.


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A

B

CD

Initial coordinates, A(72, -333)

Final coordinates, A(72.40, -332.87)

Displacement of A

DyA= 0.40 inDzA= 0.13 in

+y

+z


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0.23 in

0.50 in

0.13 in

FB Pier AnalysisDyA= 0.50 inDzA= 0.13 in

Rigid CapDyA= 0.40 inDzA= 0.13 in

Wrap Up


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3. Compute resistances4. Layout foundation group andanalyze at the strength limit state



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Participant WorkbookPage 3.3C.10


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1 2 3 4

Pm 0.7 0.7 0.7 0.85 1.0

5


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Deflection (in.)

Load(kips)

0.01 0.03 0.05

2.0

1.6

1.2

0.8

0.4

0.0

1.00.850.7

HP 12x53 in loose sand, fixed y-y axis


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Average loads in XZ planePB = (26+60+94+128)/4 = 77 kipsPC = (152+186+220+254)/4 = 203 kips

Horizontal ReactionsDisplacement assumed to be 0.04 inFx = 31.8 kips / 4 rows = 8 kips/row

H1+H2+H3+H4+H5 =1.5+1.5+1.5+1.7+1.8 = 8 kips, OK

Settlement as a Function of Qtop

Dztop = 0 00092Qtop


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Pile Head DisplacementsPile B: Dztop = 0.071 in, Dx = 0.04 inPile C: Dztop = 0.187 in, Dx = 0.04 in

Displaced GeometryzD = 3 (0.129)a = 0.02769o

Final coordinates, A = (138.20, -332.87)

DisplacementDxA= 0.20 in, DzA= 0.13 in


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0.23 in

0.50 in

0.13 in

FB Pier AnalysisDxA= 0.23 inDzA= 0.13 in

Rigid CapDxA= 0.20 inDzA= 0.13 inResults

Learning Outcome


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E. Be able to perform a rigid capanalysis of a driven pile group at

Service Limit State


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bridge lrfd theory deep foundations

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