LRFD Theory - Deep Foundations

Document Sample
LRFD Theory - Deep Foundations Powered By Docstoc
					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
    Load type, direction and magnitude
    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
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
                     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

  Low headroom requires pile splicing
  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
      Static load test
      Dynamic load test (PDA)

      Driving Formulae

    Static analysis methods
Determining Geotechnical
Resistance of Piles
Static Load Test
                        Load
  Settlement




               Pile top settlement
Dynamic Load Test (PDA)
Driving Formulas
Geotechnical Resistance
   Factors for Piles
  Method      Site Variability       
                    Low          0.8 – 0.9
Static Load
                  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
  Static Load Tests
            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
   Static Load Test                                  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
     Drivehead                        elastic
            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
             Service Load = 2794 kips
                                     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
       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     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
Horizontal Loads


                   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)
Pile Head Fixity
    Dx                 Dx




            Moment              Moment




          Service Limit State
         StrengthLimit State
Group Effects



       Fx
                H2   H1
P-y Interaction Effects
      P             P



                    Pm * P
                             y
Output for multiple loads
 Applied     Resulting    Maximum
Horizontal   Deflection   Moment
  Load
                 Horizontal Load   3.00E+01
                                   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
Horizontal Loads,
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
       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”

                           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”
                Applied Loads
                Strength V load case

     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
           Load (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
Load (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
      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. 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
       Compare load effects to factored
        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
Factored load                         20
                                              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
    Compare load effects to factored
     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

Applied Loads               Fz

                      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
Given a load case with loading
directions as depicted in the
adjacent figure:            My
                               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
   horizontal load?        2
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”
                Applied Loads

     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


Average load,                    Average load,
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
Load (kips)




              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
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

                    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
Load (kips)


              1.2
                                              1.0
                                              0.85
              0.8
                                              0.7
              0.4

              0.0
                    0.01     0.03      0.05
                      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   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

				
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
views:30
posted:8/18/2011
language:English
pages:172