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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
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Structural Axial Failure
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Structural Flexure Failure
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Structural Shear Failure
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Axial Geotechnical Resistance
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Pile damage
Driven Resistance
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Driven Performance Limit
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Driven Performance Limit
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Global StabilityVerticalDisplacement
Horizontal
Displacement
Driven Piles Drilled Shafts
Global StabilityVerticalDisplacement
Horizontal
Displacement
Service Limit State Checks
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Global Stability
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x
z
Displacement
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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
Geotechnical Resistance
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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
Geotechnical Resistance
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Geotechnical ResistanceFactors for Piles
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
Geotechnical Resistance
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Geotechnical ResistanceFactors for Piles
Method
Wave Equation only 0.4
FHWA-Modified Gates 0.4
ENR 0.1
AASHTO Table 10.5.5.2.2-1
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
Driven Piles Drilled Shafts
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Pile Group ResistanceStatic Geotechnical Resistance
Take lesser of
Geotechnical Resistance
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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
AASHTO Table 10.5.5.2.2-1
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
AASHTO Table 10.5.5.2.3-1
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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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
5. Check the service limit state
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Participant Workbook
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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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
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LRFD Theory for
Geotechnical Design
Topic 3 Part B
Deep Foundations
Session 3
LRFD for Highway Bridge Substructures
and Earth Retaining Structures
Course No. 130082A
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D. Apply the rigid cap method toevaluate the strength limit state
checks
Learning Outcome
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X
Y
Z
Centroid ofPile group
Rigid Cap Model
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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
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X
Y
Z
FxHi
Distribution ofHorizontal Loads
i l
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Ht
Qt
Mty
P
y
y
PropertiesA, E, I
Horizontal Response
C d l
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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
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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
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Pil H d Fi it
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xx
Service Limit State
Moment Moment
Strength Limit State
Pile Head Fixity
G Eff t
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Fx
H1H2
Group Effects
P I t ti Eff t
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P
yPm * P
P
P-y Interaction Effects
O t t f lti l l d
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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
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Computer P-y Modeling
Horizontal Loads,
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FxH1H2
x x
M1M2
Horizontal Loads,Pile Moment
Wh W A G i
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1. Decide deep foundation type2. Select resistance factor
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
5. Check the service limit state
Where We Are Going
Guided Walk Through
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Participant Workbook
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
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18 60 1860 60 60
18
18
36
36
36
HP 12x53 Centroid
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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
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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
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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
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x
x assumed to be 0.05
Fx
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-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
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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
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100
200
300
Dep
th(in)
Shear (kips)-2 0 2 4 6 8
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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
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1. Decide deep foundation type2. Select resistance factor
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
5. Check the service limit state
Where We Are Going
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Loose
Silty sand
Hard Clay
Driven
HP 12 x 534
35
>100
f= 31o
sat = 110 pcf
Su = 8000 psfsat = 125 pcf
OCR = 10
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Structural Resistance
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
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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
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Structural Resistance
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
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Axial Compression= 0.6 for Pr
Combined Compression and Flexure= 0.7 for Pr, 1.0 for Mr
Shear= 1.0 for Vr
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Geotechnical Resistance Axial compression
Use the beta method fro axial resistance in sand
and clay.
qs = b'v and qp = Nt'v
ForSand
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Sand
0.28
ForSand
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Sand
28
ForClay
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Clay
1.5
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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
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0
20
40
60
80
100
120
140
0 500 1000 1500 2000
Depth(
ft)
Side Friction
Point Resistance
Total Resistance
Resistance
vs. Depth
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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
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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%
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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
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715 kips
Bigger hammer (Delmag 46-13)
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715 kips
58 ksi
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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
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550 kips
45 ksi
17 BPI
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Factored resistance limited by drivingstress (driven resistance)
RR
=dyn
Rn
dyn = 0.65RR= 0.65 (550 kips)RR = 358 kips
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*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
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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
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Fz
Fy
Fx
-My
Mx
Beam seat elevation
Rock
Loose Sand
Applied Loads
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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 = 326 kips/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 = 394 kips/0.25
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
LRFD for Highway Bridge Substructures
and Earth Retaining Structures
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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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
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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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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1. Decide deep foundation type2. Select resistance factor
3. Compute resistances4. Layout foundation group andanalyze at the strength limit state
5. Check the service 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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