Session 3
LRFD Theory for
Geotechnical Design
Topic 3 – Part A
Deep Foundations
Course No. 130082A
LRFD for Highway Bridge Substructures
and Earth Retaining Structures
Learning Outcomes
A. State the performance limits that
should be evaluated when designing
a deep foundation
B. Be able to select a deep foundation
type
C. Be able to select the appropriate
resistance factor for each
performance limit evaluated
A
Deep Foundation
Performance Limits
Deep Foundation Design
Process
F.12
F.13
Start
F.1
F.2
F.3
F.4
F.14
F.5
D.1
F.6
F.7
F.8
F.9
D.2
F.10
F.15
F.16
F.11
F.17
End
1.
2.
3.
4.
D.4
D.3
Detailed Flow Chart –
RM page 3.3.6
F.18
Decide deep foundation type
Select resistance factor
Compute resistances
Layout foundation group and
analyze at the strength limit state
5. Check the service limit state
Strength Limit State Checks
Driven Piles
Structural
resistance
Axial geotechnical
resistance
Driven resistance
Drilled Shafts
Structural
resistance
Axial geotechnical
resistance
Structural Axial Failure
Structural Flexure Failure
Structural Shear Failure
Axial Geotechnical Resistance
Driven Resistance
Pile damage
Driven Performance Limit
Driven Performance Limit
Service Limit State Checks
Driven Piles
Global Stability
Vertical
Displacement
Horizontal
Displacement
Drilled Shafts
Global Stability
Vertical
Displacement
Horizontal
Displacement
Global Stability
Displacement
Dx
Dz
LRFD Differences from ASD
Same


Determining Resistance
Determining Deflection
Different


Comparison of load and resistance
Specific separation of resistance and
deflection
B
Deep foundation type
selection
Method of support
Bearing material depth
Load type, direction and magnitude
Constructability
Cost
Deep Foundation Types
Prestressed Concrete
Post-tension Cocnrete
Pre-cast Concrete
Cast-in-place Concrete
Steel
Wood
Specialty / Composites
Deep Foundation Material
Driven
X
X
X
X
X
X
X
Drilled or Bored
--
X
--
X
X
--
X
Jacked / Special
X
--
--
X
X
--
X
Driven Piles
Drilled Shafts
Method of Support
End Bearing
Side Friction
Combined
Driven Low
Displacement Piles
Driven High
Displacement Piles
Drilled Shafts
Depth to Bearing/ Scour
Load Type and Direction
Permanent/ Transient/ Cyclic
Horizontal or Vertical
Load Type and Direction
Wood is better for transient
resistance than permanent
Steel pile better cyclic resistance
High horizontal loads better resisted
by stiffer piles or shafts
Load Magnitude
Deep foundation
type
Typical range of
nominal (ultimate)
resistance (kips)
Typical
length (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
Constructability
Obstructions/ Rock
Use low displacement
steel piles
-orDrilled shafts
Equipment access
Low headroom requires pile splicing
Equipment size a function of pile/shaft size
Wrap Up
1.
2.
3.
4.
Decide deep foundation type
Select resistance factor
Compute resistances
Layout foundation group and analyze
at the strength limit state
5. Check the service limit state
Selection of
Resistance factors
C
Strength limit state
Structural Resistance
 Geotechnical Resistance
 Driven Resistance (piles only)

Service limit state

Resistance factor = 1.0
(except global stability)
Methods for determining
structural resistance
Axial compression
Combined axial and flexure
Shear
Concrete – Section 5
Steel – Section 6
LRFD
Specifications
Wood – Section 8
Structural resistance factors
Concrete (5.5.4.2.1)
Axial Comp. = 0.75
Flexure = 0.9
Shear = 0.9
LRFD
Specifications
Steel (6.5.4.2)
Axial = 0.5-0.6
Combined
Axial= 0.7-0.8
Flexure = 1.0
Shear = 1.0
Timber (8.5.2.2)
Compression = 0.9
Tension = 0.8
Flexure = 0.85
Shear = 0.75
Determining Geotechnical
Resistance of Piles
Field methods
Static load test
 Dynamic load test (PDA)
 Driving Formulae

Static analysis methods
Determining Geotechnical
Resistance of Piles
Static Load Test
Settlement
Load
Pile top settlement
Dynamic Load Test (PDA)
Driving Formulas
Geotechnical Resistance
Factors for Piles
Method
Static Load
Test

Site Variability
Low
Medium
0.8 – 0.9
0.7 – 0.9
High
0.55 – 0.8
 Site Variability Defined in NCHRP
Report 507
 Range of Values of Resistance
Factors Depends on Number of
Static Load Tests
AASHTO Table 10.5.5.2.2-2
Geotechnical Resistance
Factors for Piles
Method
Dynamic Test w/Signal Matching (e.g.,
PDA + CAPWAP)

0.65
 Test 1% to 50% of Production
Piles, Depending on Site
Variability and Number of Piles
Driven
 Site Variability Defined in NCHRP
Report 507
AASHTO Table 10.5.5.2.2-1 & 3
Geotechnical Resistance
Factors for Piles

Method
Wave Equation only
FHWA-Modified Gates
0.4
0.4
ENR
0.1
AASHTO Table 10.5.5.2.2-1
Geotechnical Safety
Factors for Piles
Basis for Design and Type
of Construction Control
Increasing Design/Construction Control
X
X
X
X
X
Static Calculation
X
X
X
X
X
Dynamic Formula
X
X
X
X
X
Subsurface Exploration
Wave Equation
X
CAPWAP Analysis
Static Load Test
Factor of Safety (FS)
3.50
2.75
2.25
X
X
X
2.00
1.90
Computation of Static
Geotechnical Resistance
RR = Rn
Rn = qpRp + qsRs
RP = AP qP
RS
RP
RS = AS qs
AASHTO 10.7.3.7.5-2
Static Analysis Methods
Driven Piles
a method
b method
l method
Nordlund Thurman
method
SPT-method
CPT-method
Drilled Shafts
a method
b method
Side friction in
Rock
Tip Resistance in
Rock
Pile Group Resistance
Static Geotechnical Resistance
Take lesser of
Geotechnical Resistance
Factors Pile Static
Analysis Methods
Comp
Ten
a - Method
0.35
0.25
b - Method
0.25
0.20
l - 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
Method
AASHTO Table 10.5.5.2.2-1
Driven Pile Time
Dependant Effects
Setup
RS
RP
Relaxation
RS
RP
RS
RP
RS
RP
Drilled Shaft Resistance
Total Resistance
Resistance
A
Side Resistance
D
B
C
Tip Resistance
RS
RP
Displacement
RR = Rn = qpRp + qsRs
Drilled Shaft Group Resistance
For cohesive
soils use
equivalent pier
approach
Rn group = h x Rn single
where:
h = 0. 65 at c-c spacing
of
2.5 diameters
h = 1.0 at c-c spacing of
6 diameters
For cohesionless
soils, use group
efficiency factor
approach
Geotechnical Resistance Factors
Drilled Shafts
Comp
Ten
a - Method (side)
0.45
0.35
Total stress (tip)
0.40
--
b - Method (side)
0.55
0.45
O’Neill & Reese (tip)
0.50
--
0.55
0.45
Method
Shafts in Clay
Shafts in Sand
Group (sand or clay)
AASHTO Table 10.5.5.2.3-1
Geotechnical Resistance Factors
Drilled Shafts
Comp
Method
Ten
Shafts in Interm. Geomat’ls (IGMs)
O’Neill & Reese (side)
0.60
O’Neill & Reese (tip)
0.55
--
Side (H&K, O&R)
0.55
0.40
Side (C&K)
0.50
0.40
Tip (CGS, PMT, O&R)
0.50
--
<=0.7
<=0.7
Shafts in Rock
Load Test (all mat’ls)
AASHTO Table 10.5.5.2.3-1
Axial
Geotechnical
Resistance of a
Drilled Shaft in
Clay
Reference
Manual
3.3.7.5
Example 9
Stiff Clay
Su = 1500 psf
E = 200 ksf
 = 125 pcf
e50 = 0.007
Drilled Shaft
f’c = 4 ksi
Ec = 3600 ksi
50’
2.5’
Determine Unit
Side Resistance
qs = a Su
To find a, check
Su/pa = 1.5 / 2.12
Su/pa = 0.7 < 1.5
So
a = 0.55
qs = 0.55 x 1500 psf
qs = 0.825 ksf
Stiff Clay
Su = 1500 psf
E = 200 ksf
 = 125 pcf
e50 = 0.007
Drilled Shaft
f’c = 4 ksi
Ec = 3600 ksi
50’
2.5’
Determine Exclusion
Zones
Per AASHTO 10.8.3.5.1b
Top 5' non contributing
Bottom 1 diameter (2.5')
non contributing
Ls = 50’ – 5’ - 2.5’ = 42.5’
5’
42.5’
50’
2.5’
2.5’
Rs = qs As
Rs = (0.825 ksf)(334 ft2)
Rs = 275 kips
Rs = 275 kips
A s =  D Ls
As =  (2.5’)(42.5’)
As = 334 ft2
50’
2.5’
Point Resistance
Nc = 6(1 + 0.2 (Z/D)) < 9
Nc = 6(1 + 0.2 (50/2.5))
Nc = 30
not less than 9 thus
Nc = 9
qp = 9 (1.5 ksf)
qp = 13.5 ksf
Rs = 275 kips
qp = Nc Su
50’
2.5’
Point Resistance
Ap =  D2/4
Ap =  (2.5’)2/4
Ap = 4.9 ft2
Rp = 13.5 ksf (4.9 ft2)
Rp = 66 kips
Rs = 275 kips
R p = q p Ap
Rp = 66 kips
RR = qs Rs + qp Rp
qs = 0.45
qps = 0.4
Rs = 275 kips
Combining Side and
Point Resistance
RR = 0.45 (275) + 0.4 (66)
RR = 150 kips
Rp = 66 kips
Combining Side and
Point Resistance
1.0
Rsd / Rs
Rpd / Rp
1.0
0
0
1.0
2.0
Dzt / D (%)
0
0
5.0
10.0
Dzt / D (%)
Check Relative Stiffness
If
SR = (Z/D) (Esoil/Eshaft) < 0.01
Shaft can be considered rigid
SR = (50’/2.5’) (1.39 ksil/3600 ksi)
SR = 0.008 < 0.01
Shaft can be considered rigid
Developed Resistance
(kips)
350
Rs = 256 kips
300
250
200
150
RP = 38 kips
100
50
0
0
0.3 0.5
1
1.5
2
Displacement (in)
Developed Side Resistance
Developed Nominal Resistance
Developed Base Resistance
RR = qs Rs + qp Rp
RR = 0.45 (256) + 0.4 (38)
RR = 131 kips
Driven Resistance
vo
Ram
Cushion
elastic
Drivehead
elastic
c
Compressive
Force Pulse
(Incident)
Ground
Surface
Compressive
Force Pulse
(Attenuated)
Pile
Soft Layer
Compressive
Force Pulse
c
Tensile or
Compressive
Force Pulse c
(Reflected)
(a)
Dense
Layer
(b)
(c)
c
Permanent Set
(d)
Comp Str
ksi
Wave
Equation
Results
Tens Str
ksi
30
20
10
Ult Cap
kips
Stroke
800
ft
16.0
600
12.0
400
8.0
200
4.0
0
160
320
480 Blows/ft
Driven Resistance Factors
Concrete piles,  = 1.00

AASHTO Article 5.5.4.2.1
Steel piles,  = 1.00

AASHTO Article 6.5.4.2
Timber piles,  = 1.15

AASHTO Article 8.5.2.2
Participant Workbook
Page 3.3A.22
Fz = 3594.0 kips

Qn
(kips)
a-method
0.4
550
220
17
PDA on 5%
0.65
550
358
11
Gates Formula
0.4
550
220
17
Structural Resistance
0.6
775
465
8
Method
Qr
# of
(kips) Piles
Comparison to ASD
Service Load = 2794 kips
FS
Qn
(kips)
Qr
(kips)
# of
Piles
a-method
3.5
550
157
18
(17)
PDA on 5%
2.25
550
244
12
(11)
Gates Formula
3.5
550
157
18
(17)
3
(0.33fy)
775
256
11
(8)
Method
Structural Resistance
Wrap Up
1.
2.
3.
4.
Decide deep foundation type
Select resistance factor
Compute resistances
Layout foundation group and analyze at
the strength limit state
5. Check the service limit state
Participant Workbook
Page 3.3A.25
Exercise 1: List the three strength limit
state checks for driven piles
1. Geotechnical resistance
2. Structural resistance
3. Driven resistance
Exercise 2: List the three service limit
state checks for drilled shafts
1. Horizontal deflection
2. Vertical deflection (settlement)
3. Global stability
Exercise 3: Match the
deep foundation type to
the condition.
Condition
1) Deep granular material B
2) Loose random fill
A
overlying rock
3) Large horizontal loads C
Type
A)Steel H-Pile
B) Closed end
pipe
C) Large diameter
drilled shaft
Exercise 4: What criteria should be
used to select the geotechnical resistance
factor for a driven pile?
The method used to determine the
ultimate resistance.
Exercise 5: Where would you find the
structural resistance factors for a drilled
shaft?
AASHTO Section 5 – Concrete Design
Learning Outcomes
A. State the performance limits that
should be evaluated when designing
a deep foundation
B. Be able to select a deep foundation
type
C. Be able to select the appropriate
resistance factor for each
performance limit evaluated
Session 3
LRFD Theory for
Geotechnical Design
Topic 3 – Part B
Deep Foundations
Course No. 130082A
LRFD for Highway Bridge Substructures
and Earth Retaining Structures
Learning Outcome
D. Apply the rigid cap method to
evaluate the strength limit state
checks
Where We Are Going …
1.
2.
3.
4.
Decide deep foundation type
Select resistance factor
Compute resistances
Layout foundation group and analyze at
the strength limit state
Compute load effects in piles using rigid
cap method
Compare load effects to factored
resistances for piles
5. Check the service limit state
Rigid Cap Model
Centroid of
Pile group
X
Y
Z
Distribution of Axial Loads
Fz
My
Pi
X
Mx
-xi
Y
yi
Z
My x i
Fz
Mx y i
Pi   n 2  n
2
n  yi
 xi
i1
i1
Distribution of
Horizontal Loads
X
Fx
Hi
Y
Z
Horizontal Response
Qt
Ht
Properties
A, E, I
Mt
y
P
y
y
P-y Curve development
Typical required
soil parameters
Su
f

k
50
k – coefficient of variation of subgrade reaction
50 - strain at 50% of ultimate strength
P-y Results for Single Element
-0.2 0
0.2 0.4 0.6 0.8
0.84
Depth, ft
10.1 k
1740 k
8000 in-k Deflection, Moment,
in.
in. -kx102
0
20 40 60 80 -60 -40 -20
8640
10
20
30
40
50
Shear,
k
65.5
0
20
Reinforced
Concrete Shaft
27
0
45
9
58
2
67
7
11
00
13
00
13
50
13
70
13
80
.4
1.8E+07
1.6E+07
1.4E+07
1.2E+07
1.0E+07
8.0E+06
6.0E+06
4.0E+06
2.0E+06
0.0E+00
79
Stiffness EI (kip-in^2)
Variation of Stiffness (EI)
Moment (in-kip)
Pile Head Fixity
Dx
Dx
Moment
Strength
LimitState
State
Service Limit
Moment
Group Effects
Fx
H2
H1
P-y Interaction Effects
P
P
Pm * P
y
Output for multiple loads
Applied
Horizontal
Load
Resulting
Deflection
Maximum
Moment
Horizontal Load
Maximum
(kips)
Moment (in-kips)
3.00E+01
2.00E+01
1.00E+01
0.00E+00
0.00E+00
-5.00E+02
-1.00E+03
-1.50E+03
-2.00E+03
00
0.5
Deflection (in)
1
Computer P-y Modeling
Horizontal Loads,
Pile Moment
Fx
M2
Dx
Dx
H2
M1
H1
Where We Are Going …
Guided Walk Through…
1.
2.
3.
4.
Decide deep foundation type
Select resistance factor
Compute resistances
Layout foundation group and analyze at
the strength limit state
Compute load effects in piles using rigid
cap method
Compare load effects to factored
resistances for piles
5. Check the service limit state
Participant Workbook
Page 3.3B.7
5’-0”
46’-6”
5’-0”
6’-0”
4’-6”
15’-0”
15’-6”
3’-6”
12’-0”
23’-0”
15’-6”
HP 12x53
Centroid
18”
36”
36”
36”
18”
18”
60”
60”
60”
60”
18”
Applied Loads
Strength V load case
Fz
Fy
-My
Mx
Fx
Fx = 38.4 kips
Fy = 109.1 kips
Fz = 3594.0 kips
Mx = 3196.5 k-ft
My = -8331.9 k-ft
My x i
Fz
Mx y i
Pi   n 2  n
2
n  yi
 xi
i1
i1
Example calculation, pile 9:
Fz = 3594.0 kips
n = 20 piles
 xi2 = 1000 ft2
 yi2 = 225 ft2
Mx = 3196.5 k-ft
yi = 18 in (1.5 ft)
My = -8331.9 k-ft
xi = 60 in (5 ft)
P9 = 243 kips
Y
X
32 k
74 k
116 k
157 k
199 k
75 k
117 k
158 k
200 k
242 k
118 k
159 k
201 k
243 k
284 k
160 k
202 k
244 k
285 k
327 k
Fy
Dy
Dy assumed to be 0.15”
8
6
4
7.2 kips
5.9 kips
4.5 kips
2
Deflection (in)
0
0.1
-200
-400
-600
-340 k-in
-390 k-in
-450 k-in
0.15 in
Max. Moment (k-in)
Load (kips)
10
0.2
Row
1
2
3
4
Pm
0.35
0.35
0.5
0.7
Hy
4.5 kips
4.5 kips
5.9 kips
7.2 kips
Mmax
-340 k-in
-340 k-in
-390 k-in
-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
Dx
Fx
Dx assumed to be 0.05”
2.0
1.5
2.2 kips
2.0 kips
1.8 kips
1.0
0.5
Deflection (in)
0
0.025
-33
-66
-100
-75 k-in
-80 k-in
-90 k-in
0.05 in
Max. Moment (k-in)
Load (kips)
2.5
0.075
Column
1
2
3
4
5
Pm
0.7
0.7
0.7
0.85
1.0
Hx
1.8 kips
1.8 kips
1.8 kips
2.0 kips
2.2 kips
Mmax
-75 k-in
-75 k-in
-75 k-in
-80 k-in
-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
Shear (kips)
-2
Depth (in)
100
200
300
0
2
4
6
8
For load case Strength V:
Max. axial load (Pile 5) = 326 kips
Min. axial load (Pile 16) = 32 kips
(no uplift)
Maximum combined loading (Pile 5)
Axial load = 326 kips
Moment (x-direction) = -37.5 kip-ft
Moment (y-direction) = -7.5 kip-ft
Max. shear = 7.2 kips in y-direction
(Piles 1, 2, 3, 4, 5 at the top of pile)
Where We Are Going …
1.
2.
3.
4.
Decide deep foundation type
Select resistance factor
Compute resistances
Layout foundation group and analyze at
the strength limit state
Compute load effects in piles using
rigid cap method
Compare load effects to factored
resistances for piles
5. Check the service limit state
Driven
HP 12 x 53
4’
f = 31o
sat = 110 pcf
Su = 8000 psf
sat = 125 pcf
OCR = 10
35’
Loose
Silty sand
>100’
Hard Clay
Pn
Structural Resistance –
Axial compression
As = 15.5 in2
(after corrosion loss)
Fy = 50 ksi
l = 0 in
Pn = 0.66lFyAs = 0.660(50)(15.5)
Pn = 775 kips
AASHTO Articles 6.9.4.1-1, 10.7.3.12.1
y
x
Mnx
Structural Resistance –
Flexure Resistance
Mny
zx = 74 in3
zy = 32.2 in3
Fy = 50 ksi
Mnx = (50 ksi)(74 in3) =
Mny = (50 ksi)(32.2 in3) =
3700 k-in
1610 k-in
y
x
Vny
Structural Resistance –
Shear Resistance
D = 11.78 in
tw = 0.435 in
Fy = 50 ksi
C = 1.0
Vp = (0.58)(50 ksi)(11.78 in)(0.435 in)
VpC = 149(1.0) = 149 kips
AASHTO Articles 6.10.7.2-1,6.10.7.2-2,6.10.7.3.3a
Combined Compression and Flexure
 = 0.7 for Pr, 1.0 for Mr
Shear
 = 1.0 for Vr
Axial Compression
 = 0.6 for Pr
Geotechnical Resistance – Axial compression
Use the beta method fro axial resistance in sand
and clay.
qs = b 'v and qp = Nt 'v
For
Sand
0.28
For
Sand
28
For
Clay
1.5
Tip resistance in clay
qp = 9 Su
Depth
(ft)
0
5
Average
'v (ksf)
0
0.12
Cum.
side
friction
(kips)
0
0.67
Qp =
qp Ap
(kips)
0
6.6
Total
Resistance
(kips)
0
7.3
Axial
Geotechnical
Resistance
vs. Depth
0
Resistance (kips)
0
500 1000 1500 2000
20
Depth (ft)
40
60
80
100
Side Friction
120
Point Resistance
Total Resistance 140
Estimate
Required Length
0
Assume
  Q =  Pn
 Pn = 0.6 (775 kips)
 Pn = 465 kips
20
1860 kips
40
Depth (ft)
  Q = stat Rnstat
stat = 0.25
Rnstat = 465 kips/0.25
Rnstat = 1860 kips
Resistance (kips)
0
500 1000 1500 2000
60
80
100
Side Friction
120
Point Resistance
Total Resistance 140
Dest = 108’
Steps to perform drivability analysis:




Estimate total soil resistance and
distribution
Select hammer
Model driving system and soil resistance
Run wave equation analysis
Estimate Resistance
Resistance (kips)
Distribution
0
500 1000 1500 2000
0
  Q = dyn Rn
stat = 0.65
Rn = 465 kips/0.65
Rn = 715 kips
715 kips
20
Depth (ft)
40
20%
40%
60%
80%
100%
Side Friction
Point Resistance
Total Resistance
60
Dest = 70’
80
100
120
140
EB = 10%
Select dynamic properties of soil
Skin quake =
Skin damping =
Toe quake =
Toe damping =
0.1 default per WEAP
manual
0.2 From WEAP manual
0.1 1/120 of pile width
0.15 per FHWA NHI-05-042
page 17-68
Identify pile properties (HP12x53)
As =
Es =
 s=
15.5 in2
300000 ksi
490 pcf
Identify hammer properties
(Delmag 30-23)
Helmet weight =
Cushion Area =
Cushion E =
Cushion Thickness =
2.15
283.5
280
2
kip
in2
ksi
in
715 kips
Bigger hammer (Delmag 46-13)
58 ksi
715 kips
Evaluate driving stress
dr = 0.9 da fy (permissible driving stress)
da = 1.0
dr = 0.9 (1.0) 50 ksi
dr = 45 ksi
45 ksi < 58 ksi (driving stress exceeded)
What is the maximum resistance that can
be developed without exceeding the
permissible driving stress?
45 ksi
550 kips
17 BPI
Factored resistance limited by driving
stress (driven resistance)
RR = dyn Rn
dyn = 0.65
RR= 0.65 (550 kips)
RR = 358 kips
Axial geotechnical performance ratio =
326/465 = 0.7
Axial structural performance ratio =
326/465 = 0.7
Combined axial and flexural performance
ratio = 0.78*
Driven performance ratio
326 / 358 = 0.91
Shear performance ratio =
7.2 / 256 = 0.03
*AASHTO Eqn. 6.9.2.2-2
Estimate
Required Length
for Actual
Factored load
0
20
1304 kips
  Q = 326 kips
40
Depth (ft)
  Q = stat Rnstat
stat = 0.25
Rnstat = 326 kips/0.25
Rnstat = 1304 kips
Resistance (kips)
0
500 1000 1500 2000
60
80
100
Side Friction
120
Point Resistance
Total Resistance 140
Dest = 91’
Wrap Up
1.
2.
3.
4.
Decide deep foundation type
Select resistance factor
Compute resistances
Layout foundation group and analyze at
the strength limit state
Compute load effects in piles using rigid
cap method
Compare load effects to factored
resistances for piles
5. Check the service limit state
Non-linear
Column and
Cap Beam
Flexible
Membrane
Pile Cap
Non-linear Pile
Material
Non-linear Soil
Response
T-z
-
P-y (& P-x)
Q-z
Beam seat elevation
Applied Loads
Loose Sand
Fz
Fy
-My
Mx
Rock
Fx
8.0
ft
8.0
8.0
ft ft
1.87 15.3
kips kip-ft
6.0
ft
6.0
6.0
ft ft
-18.2 kips
4.0
ft
4.0
4.0
ft ft
-324 kips
2.0
ft
2.0
2.0
ft ft
-7.59-30.1
kips kip-ft
0.0
ft
0.0
0.0
ft ft
Axial
Results
Shear
Results
Moment
Results
Pile 5
10.0
ft
10.0
10.0
ft ft
12.0
ft
12.0
12.0
ft ft
14.0
ft
14.0
14.0
ft ft
Pile 16
16.0
ft
16.0
16.0
ft ft
18.0
ft
18.0
18.0
ft ft
20.0
ft
20.0
20.0
ft ft
22.0
ft
22.0
22.0
ft ft
24.0
ft
24.0
24.0
ft ft
26.0
ft
26.0
26.0
ft ft
28.0
ft
28.0
28.0
ft ft
30.0
ft
30.0
30.0
ft ft
32.0
ft
32.0
32.0
ft ft
Rigid Cap Results
Shear = 7.2 kips
Max. Axial
= 327
kips
Moment
= - 37.5
k-in
Min. Axial = 32 kips
Axial geotechnical performance ratio =
(0.7)
327/465 = 0.7
Axial structural performance ratio =
(0.7)
327/465 = 0.7
Combined axial and flexural performance
(0.78)
ratio = 0.73*
Driven performance ratio
327 / 358 = 0.91
(0.91)
Shear performance ratio =
(0.03)
7.59 / 256 = 0.03
*AASHTO Eqn. 6.9.2.2-2
Accounting for
Scour
  Q = stat Rnstat
stat = 0.25
Rnstat = 326 kips/0.25
Rnstat = 1432 kips
RS scour = 20 kips
Depth (ft)
  Q = 358 kips
Resistance (kips)
0
500 1000 1500 2000
0
Scoured
20
20 kips
40
1432 kips
60
80
100
Side Friction
120
Point Resistance
Total Resistance 140
Dest = 96’
Accounting for Scour
Required driven resistance during construction
  Q = 358 kips
  Q = dyn Rndr – RS scour
Rndr =   Q / dyn+ RS scour
dyn = 0.65
Rndr = 326 kips/0.65 + 20 kips
Rndr = 571 kips
Accounting for
Downdrag
DD = 1.8
RS scour = 20 kips
DD = 20 kips
  Q = 394 kips
Rnstat = 394 kips/0.25
Rnstat = 1576 kips
Depth (ft)
  Q = 358 kips +
DD DD
Resistance (kips)
0
500 1000 1500 2000
0
Settling
20
20 kips
40
1576 kips
60
80
100
Side Friction
120
Point Resistance
Total Resistance 140
Dest = 100’
Accounting for Downdrag
Required driven resistance during construction
  Q = 358 kips + DD DD
DD = 1.0
Since resistance in downdrag zone determined by
signal matching
  Q = 358 kips + 1.0 (20 kips) = 378 kips
  Q = dyn Rndr – RS downdrag
Rndr =   Q / dyn+ RS downdrag
dyn = 0.65
Rndr = 378 kips/0.65 + 20 kips
Rndr = 602 kips
Accounting for
Set up
20
40
1432 kips
60
Set up
Depth (ft)
  Q = 358 kips
Rnstat = 358 kips/0.25
Rnstat = 1432 kips
0
Resistance (kips)
0
500 1000 1500 2000
80
100
Side Friction
120
Point Resistance
Total Resistance 140
Dest = 95’
Accounting for Set Up
Required driven resistance during construction
  Q = 358 kips
Rndr =   Q / ( S2) - R1dr S1 / S2 + R1dr
S1 = 1.0 (no strength change expected in layer 1)
S2 = 1.5 (50% strength gain in layer 2)
 = 0.25 (static analysis only)
R1dr = 25.6 kips (resistance in layer 1)
Rndr = 358 kips/(0.25 x1.5) – 25.6 kips (1.0)/1.5 +
25.6 kips
Rndr = 963 kips
Accounting for
Set up
0
Resistance (kips)
0
500 1000 1500 2000
963 kips
20
25.6 kips
40
60
Set up
Depth (ft)
R1dr = 25.6 kips
Rndr = 963 kips
80
100
Side Friction
120
Point Resistance
Total Resistance 140
Dest = 95’
End Bearing on Hard Rock
Assume structural resistance is much less than
geotechnical resistance.
Assume potential damage to pile
RR =  Pn
Pn = 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
Participant Workbook
Page 3.3B.29
Given a load case with loading
directions as depicted in the
Fy
adjacent figure:
My
Fz
a. Which pile will have
the highest axial load? 1 Fx
b. Which pile will have
the lowest axial load? 4 Y
Z
c. Which pile will be
subject to the highest
horizontal load?
2
3
d. Which pile will be
subject to the highest
1
2
bending moments?
X
Mx
4
2
5D c-c
Learning Outcome
D. Apply the rigid cap method to
evaluate the strength limit state
checks
Session 3
LRFD Theory for
Geotechnical Design
Topic 3 – Part C
Deep Foundations
Course No. 130082A
LRFD for Highway Bridge Substructures
and Earth Retaining Structures
Learning Outcome
E. Be able to perform a rigid cap
analysis of a driven pile group at
Service Limit State
Where We Are Going …
1.
2.
3.
4.
Decide deep foundation type
Select resistance factor
Compute resistances
Layout foundation group and
analyze at the strength limit state
5. Check the service limit state
Axial Response of a Single Element
(Approximate method)
Dztop
Point bearing only
Dztop = Dzp + Qtop L/ (A E)
Constant side friction only
L
Qtop
Dztop = Dzp + Qtop L/ (2 A E)
Linear increasing friction
only
Dztop = Dzp + Qtop L/ (3 A E)
Dzp
Pile Properties A, E
Axial Response of a Group
E
Perform Rigid Cap
Analysis, Driven Pile
HP 12x53
Centroid
18”
36”
36”
36”
18”
18”
60”
60”
60”
60”
18”
Applied Loads
Fz
Fy
-My
Mx
Fx
Fx =
Fy =
Fz =
31.8 kips
86.1 kips
2794 kips
Mx = 2547.7 k-ft
My = -6306.9 k-ft
Y
X
26 k
57 k
89 k
120 k
152 k
60 k
91 k
123 k
154 k
186 k
94 k
125 k
157 k
188 k
220 k
128 k
159 k
191 k
222 k
254 k
Mx
Fy
Average load,
PB = 88.8 kips
PB
Fz
PC
Average load,
PC =190.6 kips
Fy = 86.1 kips / 5 rows
Fy = 17.2 kips/row
Assume deflection = 0.11”
1
2
3
4
Pm 0.35 0.35 0.5 0.7
Fy = H1 + H2 + H3 + H4
Fy = 3.7+3.7+4.6+5.5
Fy = 17.5 kips
HP 12x53 in loose sand, fixed x-x axis
8
6
4
5.5 kips
4.6 kips
3.7 kips
2
0
0.1
0.11 in
Load (kips)
10
Deflection (in)
0.2
Qtop
Dztop
Estimate Dzp=0.03 in @ Qp=500 k
L = 384 in
Assume point bearing:
Dzp
QP
Δztop  Δzp 
Δztop
Q topL
AE

500 384 
 0.03 
15.5 29000 
Dztop= 0.46 in
= 0.00092(Qtop)
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.
A
B
D
+y
+z
C
Given coordinates:
A = (72 , -333)
B = (18.11 , Dztop, Pile B)
C = (126.11 , Dztop, Pile C)
D = (72.11 , zD)
1. Find zD by similar triangles
2. Find a of line BC
3. Use trigonometry to find:
DyA, DzA
A
Initial coordinates, A
(72, -333)
B
D
C
Final coordinates, A
(72.40, -332.87)
Displacement of A
DyA = 0.40 in
DzA = 0.13 in
+y
+z
0.50 in
0.13 in
0.23 in
FB Pier Analysis
DyA = 0.50 in
DzA = 0.13 in
Rigid Cap
DyA = 0.40 in
DzA = 0.13 in
Wrap Up
1.
2.
3.
4.
Decide deep foundation type
Select resistance factor
Compute resistances
Layout foundation group and
analyze at the strength limit state
5. Check the service limit state
Participant Workbook
Page 3.3C.10
1
2
3
4
5
Pm 0.7 0.7 0.7 0.85 1.0
HP 12x53 in loose sand, fixed y-y axis
2.0
Load (kips)
1.6
1.2
1.0
0.85
0.7
0.8
0.4
0.0
0.01
0.05
0.03
Deflection (in.)
Average loads in XZ plane
PB = (26+60+94+128)/4 = 77 kips
PC = (152+186+220+254)/4 = 203 kips
Horizontal Reactions
Displacement assumed to be 0.04 in
Fx = 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
Pile Head Displacements
Pile B: Dztop = 0.071 in, Dx = 0.04 in
Pile C: Dztop = 0.187 in, Dx = 0.04 in
Displaced Geometry
zD = 3 (0.129)
a = 0.02769o
Final coordinates, A = (138.20, -332.87)
Displacement
DxA = 0.20 in, DzA = 0.13 in
0.50 in
0.13 in
0.23 in
Results
FB Pier Analysis
DxA = 0.23 in
DzA = 0.13 in
Rigid Cap
DxA = 0.20 in
DzA = 0.13 in
Learning Outcome
E. Be able to perform a rigid cap
analysis of a driven pile group at
Service Limit State