# LRFD Theory - Deep Foundations

Document Sample

```					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
Deep Foundation
A
Performance Limits
Deep Foundation Design
Process
F.12

Detailed Flow Chart –
F.13

F.14
Start   F.1   F.2   F.3   F.4         F.5

D.1          D.2   F.15

RM page 3.3.6
F.6   F.7   F.8   F.9     F.10
F.16
F.11
F.17

End         D.4          D.3   F.18

1. Decide deep foundation type
2. Select resistance factor
3. Compute resistances
4. Layout foundation group and
analyze at the strength limit state
5. Check the service limit state
Strength Limit State Checks

Driven Piles          Drilled Shafts
Structural            Structural
resistance            resistance
Axial geotechnical    Axial geotechnical
resistance            resistance
Driven 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        Drilled Shafts
Global Stability    Global Stability
Vertical            Vertical
Displacement        Displacement
Horizontal          Horizontal
Displacement        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
Deep foundation type
B
selection
Method of support
Bearing material depth
Constructability
Cost
Deep Foundation Types
Deep Foundation Material

Specialty / Composites
Cast-in-place Concrete
Post-tension Cocnrete
Prestressed Concrete

Pre-cast Concrete
Driven Piles

Wood
Steel
Driven              X                     X                          X               X                     X        X      X
Drilled or Bored       --                     X                         --               X                     X        --     X
Jacked / Special        X                   --                          --               X                     X        --     X

Drilled Shafts
Method of Support
End Bearing   Side Friction   Combined
Driven Low
Displacement Piles
Driven High
Displacement Piles
Drilled Shafts
Depth to Bearing/ Scour

Permanent/ Transient/ Cyclic
Horizontal or Vertical

Wood is better for transient
resistance than permanent
Steel pile better cyclic resistance
by stiffer piles or shafts
Typical range of
Deep foundation                           Typical
nominal (ultimate)
type                              length (feet)
resistance (kips)

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
-or-
Drilled shafts
Equipment access

Equipment size a function of pile/shaft size
Wrap Up

1. Decide deep foundation type
2. Select resistance factor
3. Compute resistances
4. Layout foundation group and analyze
at the strength limit state
5. Check the service limit state
Selection of
C
Resistance factors
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)   Steel (6.5.4.2)
Axial Comp. = 0.75     Axial = 0.5-0.6
Flexure = 0.9          Combined
Shear = 0.9               Axial= 0.7-0.8
Flexure = 1.0
Shear = 1.0
Timber (8.5.2.2)
LRFD
Compression = 0.9
Specifications    Tension = 0.8
Flexure = 0.85
Shear = 0.75
Determining Geotechnical
Resistance of Piles

Field methods

 Driving Formulae

Static analysis methods
Determining Geotechnical
Resistance of Piles
Settlement

Pile top settlement
Driving Formulas
Geotechnical Resistance
Factors for Piles
Method      Site Variability       
Low          0.8 – 0.9
Medium         0.7 – 0.9
Test
High          0.55 – 0.8

 Site Variability Defined in NCHRP
Report 507
 Range of Values of Resistance
Factors Depends on Number of
AASHTO Table 10.5.5.2.2-2
Geotechnical Resistance
Factors for Piles
Method                    
Dynamic Test w/Signal Matching (e.g.,
0.65
PDA + CAPWAP)

 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                 0.4
FHWA-Modified Gates                0.4
ENR                                0.1

AASHTO Table 10.5.5.2.2-1
Geotechnical Safety
Factors for Piles
Basis for Design and Type   Increasing Design/Construction Control
of Construction Control
Subsurface Exploration      X       X       X       X       X
Static Calculation         X       X       X       X       X
Dynamic Formula            X
Wave Equation                     X       X       X       X
CAPWAP Analysis                            X               X
Factor of Safety (FS)      3.50    2.75    2.25    2.00   1.90
Computation of Static
Geotechnical Resistance

RR = Rn

Rn = qpRp + qsRs

RP = AP qP
RS
RS = AS qs
RP
AASHTO 10.7.3.7.5-2
Static Analysis Methods
Driven Piles   Drilled Shafts
a method       a method
b method       b method
l method       Side friction in
Nordlund -     Rock
Thurman        Tip Resistance in
method         Rock
SPT-method
CPT-method
Pile Group Resistance
Static Geotechnical Resistance

Take lesser of
Geotechnical Resistance
Factors Pile Static
Analysis Methods
Method         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
AASHTO Table 10.5.5.2.2-1
Driven Pile Time
Dependant Effects
Setup        Relaxation

RS      RS     RS     RS
RP      RP    RP     RP
Drilled Shaft Resistance
Total Resistance

A
Side Resistance    B
Resistance    D
C
Tip Resistance
RS
Displacement
RP
RR = Rn = qpRp + qsRs
Drilled Shaft Group Resistance

For cohesive
soils use
equivalent pier
approach

Rn group =  x Rn single
For cohesionless
where:
soils, use group
 = 0. 65 at c-c spacing
efficiency factor
of   2.5 diameters
approach
 = 1.0 at c-c spacing of
6 diameters
Geotechnical Resistance Factors
Drilled Shafts
Method             Comp   Ten
Shafts in Clay
a - Method (side)          0.45    0.35
Total stress (tip)         0.40     --
Shafts in Sand
b - Method (side)          0.55    0.45
O’Neill & Reese (tip)      0.50     --
Group (sand or clay)         0.55    0.45

AASHTO Table 10.5.5.2.3-1
Geotechnical Resistance Factors
Drilled Shafts
Method              Comp    Ten
Shafts in Interm. Geomat’ls (IGMs)
O’Neill & Reese (side)        0.60
O’Neill & Reese (tip)         0.55      --
Shafts in Rock
Side (H&K, O&R)               0.55    0.40
Side (C&K)                    0.50    0.40
Tip (CGS, PMT, O&R)           0.50      --
Load Test (all mat’ls)         <=0.7    <=0.7

AASHTO Table 10.5.5.2.3-1
Axial
Geotechnical
Resistance of a    Stiff Clay
Drilled Shaft in   Su = 1500 psf
Clay               E = 200 ksf
 = 125 pcf     50’
e50 = 0.007
Reference
Drilled Shaft
Manual             f’c = 4 ksi
3.3.7.5            Ec = 3600 ksi
Example 9                          2.5’
Determine Unit
Side Resistance
Stiff Clay
qs = a Su              Su = 1500 psf
E = 200 ksf
To find a, check        = 125 pcf     50’
Su/pa = 1.5 / 2.12     e50 = 0.007
Su/pa = 0.7 < 1.5
So                     Drilled Shaft
a = 0.55               f’c = 4 ksi
qs = 0.55 x 1500 psf   Ec = 3600 ksi
qs = 0.825 ksf                         2.5’
Determine Exclusion
Zones                           5’

Per AASHTO 10.8.3.5.1b
Top 5' non contributing        42.5’   50’
Bottom 1 diameter (2.5')
non contributing

Ls = 50’ – 5’ - 2.5’ = 42.5’   2.5’
2.5’
A s =  D Ls
As =  (2.5’)(42.5’)
As = 334 ft2

Rs = 275 kips
Rs = qs As                                  50’
Rs = (0.825 ksf)(334 ft2)
Rs = 275 kips

2.5’
Point Resistance

qp = Nc Su

Nc = 6(1 + 0.2 (Z/D)) < 9

Rs = 275 kips
Nc = 6(1 + 0.2 (50/2.5))
50’
Nc = 30
not less than 9 thus
Nc = 9

qp = 9 (1.5 ksf)
qp = 13.5 ksf                               2.5’
Point Resistance

R p = q p Ap

Rs = 275 kips
Ap =  D2/4
Ap =  (2.5’)2/4
Ap = 4.9 ft2

Rp = 13.5 ksf (4.9 ft2)
Rp = 66 kips
Rp = 66 kips
Combining Side and
Point Resistance

RR = qs Rs + qp Rp

Rs = 275 kips
qs = 0.45
qps = 0.4

RR = 0.45 (275) + 0.4 (66)
RR = 150 kips
Rp = 66 kips
Combining Side and
Point Resistance
1.0
1.0

Rpd / Rp
Rsd / Rs

0                                      0
0        1.0    2.0                    0        5.0    10.0
Dzt / D (%)                            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
350
Rs = 256 kips
Developed Resistance

300
250
(kips)

200
150
100              RP = 38 kips
50
0
0   0.3 0.5               1          1.5           2
Displacement (in)
Developed Side Resistance          Developed Base Resistance
Developed Nominal Resistance
RR = qs Rs + qp Rp
RR = 0.45 (256) + 0.4 (38)
RR = 131 kips
Driven Resistance vo
Ram

Cushion
elastic

Compressive
c     Force Pulse
(Incident)

Ground
Surface

Pile                                                  Compressive
Force Pulse
(Attenuated)
Soft Layer                                                         Compressive
c                     Force Pulse
Tensile or
Compressive     c
Force Pulse c
(Reflected)

(a)      Dense      (b)                       (c)
Permanent Set
Layer                                                        (d)
Comp Str                      Tens Str
ksi                            ksi
Wave       30

Equation   20

Results    10

Ult Cap                        Stroke
kips                            ft
800                        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       Qr   # of
Method             
(kips)   (kips) Piles

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
Comparison to ASD
Qn       Qr     # of
Method              FS
(kips)   (kips)   Piles
18
a-method                  3.5       550      157
(17)
12
PDA on 5%                2.25       550      244
(11)
18
Gates Formula             3.5       550      157
(17)
3                          11
Structural Resistance               775      256
(0.33fy)                      (8)
Wrap Up

1. Decide deep foundation type
2. Select resistance factor
3. Compute resistances
4. 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                     Type
1) Deep granular material B   A)Steel H-Pile
2) Loose random fill          B) Closed end
overlying rock            A     pipe
3) Large horizontal loads C   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. Decide deep foundation type
2. Select resistance factor
3. Compute resistances
4. Layout foundation group and analyze at
the strength limit state
Compute load effects in piles using rigid
cap method
resistances for piles
5. Check the service limit state
Rigid Cap Model

Centroid of
Pile group
X

Y

Z
Fz     Pi              My

Mx                                   X

Y            -xi         yi

Z
Fz Mx y i  My x i
Pi   n 2  n
n  yi     xi
2
i1             i1
Distribution of

X
Fx
Hi
Y

Z
Horizontal Response
Qt
Ht     Mt   P
y

Properties                y
A, E, I               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
1740 k
8000 in-k Deflection, Moment,                                  Shear,
in.     in. -kx102                                 k
10.1 k                        -0.2 0   0.2 0.4 0.6 0.8   0   20 40 60 80 -60 -40 -20   0   20

0.84           8640
10
Depth, ft

20                                 65.5

30

40

50
Variation of Stiffness (EI)
1.8E+07
Stiffness EI (kip-in^2)

1.6E+07
1.4E+07
1.2E+07
1.0E+07
8.0E+06
6.0E+06
4.0E+06   Reinforced
2.0E+06   Concrete Shaft
0.0E+00

00

50
00

70

80
.4

0

9

2

7
27

45

58

67
79

13
11

13

13

13
Moment (in-kip)
Dx                 Dx

Moment              Moment

Service Limit State
StrengthLimit State
Group Effects

Fx
H2   H1
P-y Interaction Effects
P             P

Pm * P
y
Applied     Resulting    Maximum
Horizontal   Deflection   Moment
2.00E+01
(kips)

1.00E+01
0.00E+00
0
0        0.5          1
Deflection (in)
Moment (in-kips)

0.00E+00
Maximum

-5.00E+02
-1.00E+03
-1.50E+03
-2.00E+03
Computer P-y Modeling
Pile Moment
Dx    Dx

Fx
H2    H1
M2    M1
Where We Are Going …
Guided Walk Through…

1. Decide deep foundation type
2. Select resistance factor
3. Compute resistances
4. Layout foundation group and analyze at
the strength limit state
Compute load effects in piles using rigid
cap method
resistances for piles
5. Check the service limit state
Participant Workbook
Page 3.3B.7
5’-0”             46’-6”

5’-0”

6’-0”

15’-6”
15’-0”   15’-6”
4’-6”

3’-6”

23’-0”
12’-0”
HP 12x53               Centroid
18”

36”

36”

36”

18”
18”   60”   60”   60”       60”    18”

Fz         Fx = 38.4 kips
Fy              Fy = 109.1 kips
-My   Fz = 3594.0 kips
Mx
Mx = 3196.5 k-ft
My = -8331.9 k-ft
Fx
Fz Mx y i  My x i
Pi   n 2  n
n  yi     xi
2
i1            i1

Example calculation, pile 9:
Fz = 3594.0 kips          Mx = 3196.5 k-ft
n = 20 piles              yi = 18 in (1.5 ft)
 xi2 = 1000 ft2          My = -8331.9 k-ft
 yi2 = 225 ft2           xi = 60 in (5 ft)

P9 = 243 kips
Y
32 k    74 k    116 k   157 k   199 k

75 k    117 k   158 k   200 k   242 k

X   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”
10
8    7.2 kips

6    5.9 kips
4.5 kips
4
2
Deflection (in)
0
Max. Moment (k-in)

0.1                       0.2

0.15 in
-200
-340 k-in
-390 k-in
-400     -450 k-in

-600
Row        Pm         Hy        Mmax
1        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
Dx

Fx

Dx assumed to be 0.05”
2.5    2.2 kips
2.0    2.0 kips

1.8 kips
1.5
1.0
0.5
Deflection (in)
0
Max. Moment (k-in)

0.025                       0.075

0.05 in
-33

-66    -75 k-in
-80 k-in
-90 k-in
-100
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
Shear (kips)
-2   0    2      4      6   8

100
Depth (in)

200

300
Max. axial load (Pile 5) = 326 kips
Min. axial load (Pile 16) = 32 kips   (no uplift)

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. Decide deep foundation type
2. Select resistance factor
3. Compute resistances
4. Layout foundation group and analyze at
the strength limit state
Compute load effects in piles using
rigid cap method
resistances for piles
5. Check the service limit state
Driven
HP 12 x 53
4’
f = 31o                                   Loose
sat = 110 pcf                        Silty sand
35’

Su = 8000 psf                 >100’
sat = 125 pcf
Hard Clay
OCR = 10
Structural Resistance –
Axial compression
Pn
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

Structural Resistance –
Mnx               Flexure Resistance
x
Mny
zx = 74 in3
zy = 32.2 in3
Fy = 50 ksi

Mnx = (50 ksi)(74 in3) =     3700 k-in
Mny = (50 ksi)(32.2 in3) =   1610 k-in
y         Structural Resistance –
Shear Resistance
D = 11.78 in
tw = 0.435 in
x
Fy = 50 ksi
C = 1.0

Vny       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
Cum.
side      Qp =       Total
Depth   Average     friction   qp Ap    Resistance
(ft)   'v (ksf)    (kips)    (kips)     (kips)
0         0           0         0          0
5       0.12        0.67       6.6        7.3
Axial                                 Resistance (kips)
Geotechnical                        0   500 1000 1500 2000
0
Resistance
vs. Depth                      20

40

Depth (ft)
60

80

100
Side Friction    120
Point Resistance
Total Resistance 140
Estimate                                     Resistance (kips)
Required Length                            0   500 1000 1500 2000
0
Assume
  Q =  Pn                          20
 Pn = 0.6 (775 kips)                            1860 kips
 Pn = 465 kips                       40

Depth (ft)
  Q = stat Rnstat                  60
stat = 0.25
Rnstat = 465 kips/0.25                80
Rnstat = 1860 kips
100
Side Friction    120                     Dest = 108’
Point Resistance
Total Resistance 140
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                              715 kips
stat = 0.65                      20
Rn = 465 kips/0.65
Rn = 715 kips                     40

Depth (ft)
20%                   60
40%                             Dest = 70’
60%                   80
80%
100%                  100
Side Friction               120   EB = 10%
Point Resistance
Total Resistance            140
Select dynamic properties of soil
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
Identify pile properties (HP12x53)
As =        15.5 in2
Es =     300000 ksi
 s=         490 pcf
Identify hammer properties
(Delmag 30-23)
Helmet weight =         2.15   kip
Cushion Area =         283.5   in2
Cushion E =              280   ksi
Cushion Thickness =        2   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                                     Resistance (kips)
Required Length                            0   500 1000 1500 2000
0
for Actual
1304 kips
  Q = 326 kips                      40

Depth (ft)
  Q = stat Rnstat                  60
stat = 0.25
Rnstat = 326 kips/0.25                80
Rnstat = 1304 kips
100     Dest = 91’
Side Friction    120
Point Resistance
Total Resistance 140
Wrap Up

1. Decide deep foundation type
2. Select resistance factor
3. Compute resistances
4. Layout foundation group and analyze at
the strength limit state
Compute load effects in piles using rigid
cap method
resistances for piles
5. Check the service limit state
Non-linear
Column and
Cap Beam
Non-linear Soil
Flexible           Response
Membrane              T-z
Pile Cap              -
Non-linear Pile      P-y (& P-x)
Material             Q-z
Beam seat elevation

Fy         -My
Loose Sand
Mx
Rock
Fx
0.0 ft ft
0.0
0.0 ft
2.0 ft ft                                                                   Axial Results
Shear Results
Moment Results
-7.59-30.1 kip-ft
2.0
2.0 ft

1.87 15.3 kip-ft
-18.2 kips
4.0 ft ft
4.0
4.0 ft      -324 kips
kips

kips
6.0 ft ft
6.0
6.0 ft
8.0 ft ft
8.0
8.0 ft                                                                               Pile 5
10.0 ft ft
10.0
10.0 ft
12.0 ft ft
12.0
12.0 ft
14.0 ft ft
14.0
14.0 ft                                                                               Pile 16
16.0 ft ft
16.0
16.0 ft
18.0 ft ft
Rigid Cap Results
18.0
18.0 ft
20.0 ft ft
20.0
20.0 ft
22.0 ft ft
22.0
22.0 ft                                                                    Shear = 7.2 kips
24.0 ft ft
24.0
24.0 ft                                                                   Max. Axial = 327 kips
Moment = - 37.5 k-in
Min. Axial = 32 kips
26.0 ft ft
26.0
26.0 ft
28.0 ft ft
28.0
28.0 ft
30.0 ft ft
30.0
30.0 ft
32.0 ft ft
32.0
32.0 ft
Axial geotechnical performance ratio =
327/465 = 0.7                           (0.7)
Axial structural performance ratio =
327/465 = 0.7                        (0.7)
Combined axial and flexural performance
ratio = 0.73*                         (0.78)
Driven performance ratio
327 / 358 = 0.91                        (0.91)
Shear performance ratio =
7.59 / 256 = 0.03                        (0.03)

*AASHTO Eqn. 6.9.2.2-2
Accounting for                             Resistance (kips)
Scour                                    0   500 1000 1500 2000
0
  Q = 358 kips                            Scoured
20
  Q = stat Rnstat                           20 kips
stat = 0.25                          40
1432 kips

Depth (ft)
Rnstat = 326 kips/0.25
Rnstat = 1432 kips                    60

80
RS scour = 20 kips
100
Dest = 96’
Side Friction    120
Point Resistance
Total Resistance 140
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                             Resistance (kips)
Downdrag                                 0   500 1000 1500 2000
0
  Q = 358 kips +                          Settling
DD DD                         20
20 kips
DD = 1.8                             40
RS scour = 20 kips                            1576 kips

Depth (ft)
DD = 20 kips                          60

  Q = 394 kips                      80
Rnstat = 394 kips/0.25
Rnstat = 1576 kips                    100
Dest = 100’
Side Friction    120
Point Resistance
Total Resistance 140
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                               Resistance (kips)
Set up                                     0   500 1000 1500 2000
0

20
  Q = 358 kips
Rnstat = 358 kips/0.25                40
Rnstat = 1432 kips                               1432 kips

Depth (ft)
60

Set up
80

100
Dest = 95’
Side Friction    120
Point Resistance
Total Resistance 140
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                            Resistance (kips)
Set up                                  0   500 1000 1500 2000
0
963 kips
20
25.6 kips
R1dr = 25.6 kips                40
Rndr = 963 kips

Depth (ft)
60

Set up
80

100
Dest = 95’
Side Friction    120
Point Resistance
Total Resistance 140
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
directions as depicted in the
Fy
Fz
a. Which pile will have                               X
the highest axial load? 1 Fx
Mx
b. Which pile will have
the lowest axial load? 4 Y
Z
c. Which pile will be
subject to the highest
d. Which pile will be               3        4
subject to the highest       1        2
bending moments?        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. Decide deep foundation type
2. Select resistance factor
3. Compute resistances
4. Layout foundation group and
analyze at the strength limit state
5. Check the service limit state
Axial Response of a Single Element
(Approximate method)
Qtop
Dztop
Point bearing only
Dztop = Dzp + Qtop L/ (A E)

Constant side friction only
Dztop = Dzp + Qtop L/ (2 A E)
L

Linear increasing friction
only
Dztop = Dzp + Qtop L/ (3 A E)
Dzp
Pile Properties A, E
Axial Response of a Group
Perform Rigid Cap
E
Analysis, Driven Pile
HP 12x53               Centroid
18”

36”

36”

36”

18”
18”   60”   60”   60”       60”    18”

Fz         Fx =   31.8 kips
Fy =   86.1 kips
Fy        -My   Fz =   2794 kips

Mx        Mx = 2547.7 k-ft
My = -6306.9 k-ft
Fx
Y
26 k    57 k    89 k    120 k   152 k

60 k    91 k    123 k   154 k   186 k

X   94 k    125 k   157 k   188 k   220 k

128 k   159 k   191 k   222 k   254 k
Mx
Fz
Fy

PB = 88.8 kips    PB             PC =190.6 kips
PC
Fy = 86.1 kips / 5 rows
Fy = 17.2 kips/row

Assume deflection = 0.11”

1   2    3   4     Fy = H1 + H2 + H3 + H4
Fy = 3.7+3.7+4.6+5.5
Fy = 17.5 kips
Pm 0.35 0.35 0.5 0.7
HP 12x53 in loose sand, fixed x-x axis

10
8

6       5.5 kips
4.6 kips
4       3.7 kips

0.11 in
2
Deflection (in)
0
0.1                       0.2
Qtop
Dztop
Estimate Dzp=0.03 in @ Qp=500 k
Assume point bearing:
Q topL
L = 384 in
Δz top  Δz p 
AE

Δz top    0.03 
500 384 
15.5 29000 
Dztop= 0.46 in
= 0.00092(Qtop)
Dzp
QP
   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

Given coordinates:
A = (72 , -333)
B         D         B = (18.11 , Dztop, Pile B)
C
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:
+y
DyA, DzA
+z
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   FB Pier Analysis
DyA = 0.50 in
0.13 in   0.23 in
DzA = 0.13 in

Rigid Cap
DyA = 0.40 in
DzA = 0.13 in
Wrap Up

1. Decide deep foundation type
2. Select resistance factor
3. Compute resistances
4. 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

1.6

1.2
1.0
0.85
0.8
0.7
0.4

0.0
0.01     0.03      0.05
Deflection (in.)
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 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   FB Pier Analysis
DxA = 0.23 in
0.13 in    0.23 in
DzA = 0.13 in

Rigid Cap
DxA = 0.20 in
Results        DzA = 0.13 in
Learning Outcome

E. Be able to perform a rigid cap
analysis of a driven pile group at
Service Limit State

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