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     NPTEL – ADVANCED FOUNDATION ENGINEERING-I

    Module 8

    (Lecture 33)

    PILE FOUNDATIONS

    Topics

    1.1 

    PILE-DRIVING FORMULAS

    1.2 

    NEGATIVE SKIN FRICTION

    Clay Fill over Granular Soil 

    Granular Soil Fill over Clay

    1.3 GROUP PILES

    1.4 

    GROUP EFFICIENCY

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    PILE-DRIVING FORMULAS

    To develop the desired load-carrying capacity, a point bearing pile must penetrate the

    dense soil layer sufficiently or have sufficient contact with a layer of rock. Thisrequirement cannot always be satisfied by driving a pile to a predetermined depth

     because soil profiles vary. For that reason, several equations have been developed to

    calculate the ultimate capacity of a pile during driving. These dynamic equations arewidely used in the field to determine whether the pile as reached satisfactory bearing

    value at the predetermined depth. One of the earliest of these dynamic equations-

    commonly referred to as the  Engineering News Record (ENR) formula-is derived fromthe work-energy theory. That is,

    Energy imparted by the hammer per blow = (pile resistance) (penetration per hammer blow)

    According to the ENR formula, the pile resistance is the ultimate load , expressed as = ℎ+   [8.118]Where

    = ℎ  ℎ  (  ,   . 4  ) ℎ = ℎℎ     ℎ   =     ℎ   =   The pile penetration, S , is usually based on the average value obtained from the last few

    driving blows. In the equations’ original form, the following values of C   were

    recommended.

    For drop hammers: C  = 1 in. (if the units of S  and h are in inches)

    For steam hammers: C  = 0.1 in. (if the units of S  and h are in inches)

    Also, a factor of safety, = 6, was recommended to estimate the allowable pilecapacity. Note that, for single- and double-acting hammers, the term ℎ can be replaced

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     by   (where = ℎ    =    ℎ).Thus

    = +  [8.119]The ENR formula has been revised several times over the years, and other pile-drivingformulas also have been suggested. Some of them are tabulated in table 1.

    The maximum stress developed on a pile during the driving operation can be estimated

    from the pile-driving formulas presented in table 11. To illustrate, we use the modified

    ENR formula:

    = ℎ+ +2+  In this equation, S  equals the average penetration per hammer blow, which can also be

    expressed as

    = 1  [8.120]Where

       ℎ  =   ℎ   ℎ   Table 11 Pile-Driving formulas

     Name Formula

    Modified ENR

    formula = ℎ +   + 2 +  Where

    = ℎ   = 0.1 . ,  ℎ     ℎ   ℎ  = ℎ  ℎ   =    ℎ   ℎ   Typical values for E  

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    Single- and double-acting hammers 0.7-0.85

    Diesel hammers 0.8-0.9

    Drop hammers 0.7-0.9

    Typical values for n 

    Cast iron hammer and concrete pile (without cap) 0.4-0.5

    Wood cushion on steel piles 0.3-0.4

    Wooden pile 0.25-0.3

    Michigan State

    HighwayCommission

    formula (1965)

    = 1.25 +   + 2 +  

    Where

    =  ′  ℎ  ( −. )  = ℎ   = 0.1 . A factor of safety of 6 is recommended.

    Danish formula(Olson and Flaate,

    1967) =

    + 2  

    Where

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    = ℎ   =  ℎ  

    =

      

     

     

     

    ℎ 

     

     

    = ℎ     =   ℎ    

    Pacific Coast

    Uniform Building

    Code formula

    (International

    Conference of

    Building Officials,1982)

    = () + +  +    

    The value of n should be 0.25 for steel piles and 0.1 a for all other

     piles. A factor of safety of 4 is generally recommended.

    Janbu’s formula(Janbu, 1953)

    = ′ 

    Where

    ′ =  1 + 1 + ′  = 0.75 + 0.14  

    ′ =

      

     2

     

    Gates formula

    (Gates, 1957) =  ( − ) If   is in , then S   is  . = 27, = 1,    is in − .

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    If   is in ,  then S   is in mm, = 104.5, = 2.4,    is in − 

    = 0.75 for drop hammer;

    = 0.85 for all other hammer

    Use a factor of safety of 3.

     Navy-McKay

    formula = 1+0.3

     

    Use a factor of safety of 6.

    Thus

    =

      ℎ(1/)+0.1+2

    +  [8.121]

    Different values of  N  may be assumed for a given hammer and pile and  calculated.The driving stress can then be calculated for each value of N  and /  .

      = 100 2 The weight of the pile is

     =

     100 2144

     (80

     )(150

    /

     3) = 8.33

     

    If the weight of the cap is 0.67 , = 8.33 + 0.67 = 9  Again, from for an 11B3 hammer,

    Rated energy= 19.2 − =  = ℎ Weight of ram= 5  Assume that the hammer efficiency is 0.85 and that

    = 0.35. Substituting these values

    in equation (121) yields

    = (0.85)(19.2×2)1+0.1 5+(0.35)2(9)5+9   = 85.371+0.1   Now the following table can be prepared:

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       ()     (2)  /   (/2) 0 0 100 0

    2 142.3 100 1.42

    4 243.9 100 2.44

    6 320.1 100 3.20

    8 379.4 100 3.79

    10 426.9 100 4.27

    12 465.7 100 4.66

    20 569.1 100 5.69

    Both the number of hammer blows per inch and the stress can now be plotted in a graph,

    as shown in figure 8.47. If such a curve is prepared, the number of blows per inch of pile

     penetration corresponding to the allowable pile-driving stress can be easily determined.

    Figure 8.47

    Actual driving stresses in wooden piles are limited to about 0.7  . Similarly, for concreteand steel piles, driving stresses are limited to about 0.6 ′   0.85  , respectively.

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    In most cases, wooden piles are driven with hammer energy of less than 45 − (≈ 60 ∙ ). Driving resistances are limited are limited mostly to 4-5 blows per inch of pile penetration. For concrete and steel piles, the usual  N   values are 6-8 and 12-14,

    respectively.

    Example 11

    A precast concrete pile 12 .× 12 . in cross sections in driven by a hammer. Given:  ℎ  = 30 −   = 0.8 ℎ   = 7.5   ℎ = 80     = 0.4 ℎ     = 550   = 3× 106 /2  Number of blow for last 1 in. of penetration = 8

    Estimate the allowable pile capacity by the

    a.  Modified ENR formula (use = 6)  b.  Danish formula (use

    = 4) 

    c. 

    Gates formula (use = 3) Solution

    Part a

    = (0.8)(30×12 − .)18+0.1

    ×7.5+(0.4)2(12.55)

    7.5+12.55= 607  

    =  = 6076 ≈ 101  Part b

    =   +  2 

     

    Use = 3× 106  /2.

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     2  =  (0.8)(30×12)(80×12)2(12×12)3×106

    1000  /2 = 0.566 .

    =

    (0.8)(30×12)18+0.566

     ≈417

     

    = 4174 ≈ 104  Part c

    =    ( − ) = 2 7 (0.8)(30) [1−18] ≈ 252   = 2523 = 84  NEGATIVE SKIN FRICTION

     Negative skin friction is a downward drag force exerted on the pile by the soilsurrounding it. This action can occur under conditions such as the following:

    1.  If a fill of clay soil is placed over a granular soil layer into which a pile is driven,the fill will gradually consolidate. This consolidation process will exert a

    downward drag force on the pile (figure 8.48a) during the period of

    consolidation.2.  If a fill of granular soil is placed over a layer of soft clay, as shown in figure 8.

    48b, it will induce the process of consolidation in the clay layer and thus exert a

    downward drag on the pile.3.  Lowering of the water table will increase the vertical effective stress on the soil at

    any depth, which will induce consolidation settlement in clay. If a pile is located

    in the clay layer, it will be subjected to a downward drag force.

    In some cases, the downward drag force may be excessive and cause foundation failure.

    This section outlines two tentative methods for the calculation of negative skin friction.

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    Figure 8.48 Negative skin friction

    Clay Fill over Granular Soil (Figure 8.48a)

    Similar to the  method presented in section 12, the negative (downward) skin stress onthe pile is

      = ′ ′    [8.122] Where

    ′=

     ℎ 

     

    =

     = 1

     −  

    ′ =      ℎ  = ′  ′  =   ℎ     =  −      ≈ 0.5− 0.7 Hence the total downward drag force, , on a pile is = ∫ ( ′ ′  )  = ′ ′    2 2 0   [8.123]Where

      = ℎℎ  ℎ   If the fill is above the water table, the effective unit weight, ′  , should be replaced by themoist unit weight.

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    Granular Soil Fill over Clay (figure 8.48b)

    In this case, the evidence indicates that the negative skin stress on the pile may exist from = 0   = 1, which is referred to as the neutral depth (see Vesic, 1977, pp. 25-26, fordiscussion). The neutral depth may be given as (Bowles, 1982):

    1 = (− )1 − 2 + ′  ′  − 2′  ′   [8.124]Where

    ′     ′ =  ℎ  ℎ    ℎ   ,  For end-bearing piles, the neutral depth may be assumed to be located at the pile tip (that

    is, 1 =  −  ).Once the value of 1 is determined, the downward drag force is obtained in the followingmanner. The unit negative skin friction at any depth from = 0   = 1 is  = ′′   [8.125]Where

    ′ =  = 1 −  ′ = ′   + ′ 

    = 0.5

    −0.7

     

    = ∫    = ∫  ′  ′   ′  1010  = ( ′ ′   )1 + 12 ′ ′  2   [8.126] If the soil and the fill are above the water table, the effective unit weights should be

    replaced by moist unit weights. In some cases, the piles can be coated with bitumen in the

    downdrag zone to avoid this problem. Baligh et al. (1978) summarized the results ofseveral field tests that were conducted to evaluate the effectiveness of bitumen coating in

    reducing the negative skin friction. Their results are presented in table 12.

    A limited number of case studies on negative skin friction is available in the literature.

    Bjerrum et al. (1969) reported monitoring of downdrag force on a test pile at Sorenga in

    the harbor of Oslo, Norway (noted as pile G in the original paper). This was alsodiscussed by Wong and The (1995) in terms of the pile being driven to bedrock at 40 m.

    Figure 8.49 a shows the soil profile and the pile. Wong and The (1995) estimated the

    following:

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    Figure 8.49 Negative skin friction on a pile in the harbor of Oslo, Norway [based onBjerrum et al., (1969); and Wong and The (1995)]

    :  ℎ,   = 16 /3   ℎ, ( ) = 18.5 /3 So

    ′  = 18.5− 9.81 = 8.69/3   = 13 :′  ≈ 0.22   ℎ, ′ = 19 − 9.81 = 9.19/3 

    :

    = 40

     

    , = 500 Thus, the maximum downdrag force on the pile can be estimated from equation. (126).

    Since it is a point bearing pile, the magnitude of 1 = 27, so = ()(′ )[  × 2 + (13 − 2) ′  ](1) + 12 ′ (′  )2  

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    Or

    = ( × 0.5)(0.22)[(16×2) + (8.69 × 11)](27) + (27)2(×0.5)(9.19)(0.22)2 = 2348  The measured value of maximum

      was about 2500

      (figure 8. 49b), which is in

    good agreement with the calculated value.

    Table 12 Summary of Case Studies of Bitumen-Coated Piles(After Baligh et al.

    (1978)) 

    Downward drag Test loadings

    Case number 1 2 3 4 5 6 7

    Soil type Fill,sand, and

    clay

    Fill andsilty clay

    Fill andclay

    Sandand

    silty

    clay

    Siltyclay

    Siltyclay

    Sandfill,

    clay,

    and peat

    Pile type

    Pile crosssection (mm)

    Cast-in-

     place

    concrete

    = 530 Steel

     pipe

    = 300 

    Steel

     pipe

    = 500 

    Steel

     pile

    = 760 

    6 RC

     piles

    300× 300 

    6 RC

     piles

    300× 300 

    Precast

    concrete

    380

    × 450 

    Length incontact with

    settling soil

    (m)

    25 26 40 25 7-17 9-16 24

    Installation

    method

    Predriven

    casing

    Enlarged

    tip and

    slurry

    Enlarge

    tip and

    casing

    Driving Driving Driving Driving

    Bitumen

    coating

    Type (pen

    25° )Coating

    thickness

    (mm)

    20/30

    10

    80/100

    1.2

    80/10

    1.2

    60/70

    1.5

    60/70

    1.2

    80-100RC-0

    cutback

    1.2

    43special

    grade

    10

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    Measured

    shaft

    resistance

    Uncoated pile (ton)

    Coated pile(ton)

    Coating

    effectiveness

    (%)

    70-80

    5-7

    92

    120

    10

    92

    300

    15

    95

    180

    3

    98

    31-40

    10-33

    30-80

    31-40

    20-42

    30-80

    160

    Predicted

    downdrag

    Coated pile

    (ton)

    Coating

    Effectiveness(%)

    0.1

    100

    2-11

    91-98

    5

    98

    0-23

    87-100

    Example 12

    Refer to figure 8. 48a;   = 3. The pile is circular in cross section with a diameter of0.5 m. For the fill that is above the water table,   = 17.2 /3    = 36°.Determine the total drag force. Use = 0.7.Solution

    From equation. (123),

    = ′     2 2  

    = (0.5

    )= 1.57 ′ = 1− = 1− 36° = 0.41 

    = (0.7)(36) = 25.2°  = (1.57)(0.41)(17.2)(3)2 25.22 = 23.4 

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    Example 13

    Refer to figure 8. 48b. Here,

      = 2

    ,

     

    = 0.305

    ,

      = 16.5

    /

    3,

    = 34°

    , = 17.2 /3

    ,   = 20. The water table coincides with the topof the clay layer. Determine the downward drag force. Assume = 0.6 .Solution

    The depth of the neutral plane in given in equation (124) as

    1 = − 1 − 2 +   ′  − 2  ′   Note that ′   in equation (124) has been replaced by    because the fill is above the watertable, so

    1 = (20−2)1 (20−2)2 + (16.5)(2)(17.2−9.81) − (2)(16.5)(2)(17.2−9.81) 1 = 242.41 − 8.93; 1 = 11.75  Now, referring to equation (126), we have

    = ( ′   )1 + 12′ ′  2  

    =

     (0.305) = 0.958

     

    ′ = 1− 34° = 0.44 = (0.958)(0.44)(16.5)(2)[(0.6 × 34)](11.75) +

    (11.75)2(0.958)(0.44)(17.2−9.81)[ (0.6×34)]2

    = 60.78 + 79.97 = 140.75  GROUP PILES

    GROUP EFFICIENCY

    In many cases, piles are used in groups, as shown in figure 8.50, to transmit the structural

    load to the soil.  A pile cap is constructed over  group piles. The pile cap can be contact

    with the ground, as in most cases (figure 8.50a), or well above the ground, as in the caseof offshore platforms (figure 8.50b).

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    Figure 8.50 Pile groups

    Determining the load-bearing capacity of group piles is extremely complicated and has

    not yet been fully resolved. When the piles are placed close to each other, a reasonableassumption is that the stresses transmitted by the piles to the soil will overlap (figure 8.50c), reducing the load-bearing capacity of the piles. Ideally, the piles in a group should

     be spaced so that the load-bearing capacity of the group should not be less than the sumof the bearing capacity of the individual piles. In practice, the minimum center-to-center

     pile spacing,

    ,

     2.5

    , and in ordinary situations, is actually about 3

    −3.5

    .

    The efficiency of the load-bearing capacity of a group pile may be defined as

    = ( )   [8.127]Where

    =   

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    () =   −    ℎ    =   −   ℎ  ℎ ℎ   Many structural engineers use a simplified analysis to obtain the group efficiency for

     friction piles,  particularly in sand. This type of analysis can be explained with the aid offigure 8. 50a. Depending on their spacing within the group, the piles may act in one of

    two ways: (1) as a block  with dimensions ×  × , or (2) as individual piles. If the piles act as a block, the frictional capacity is  ≈ (). [Note: = perimeter ofthe cross section of block = 2(1 + 2 − 2) + 4 ,    =  average unit frictionalresistance.] Similarly, for each pile acting individually,  ≈ . (Note: =  perimeter of the cross section of each pile.) Thus

    = ( )  =  [2(1+2−2)+4]12 = 2(1+2−2)+412   [8.128] Hence

    () = 2(1+2−2)+4 12       [8.129]From equation (129), if the center-to-center spacing, ,  s large enough, > 1. In thatcase, the piles will behave as individual piles. Thus, in practice, if < 1,() =     And, if

    ≥1,

    () =    There are several other equations like equation (129) for the group efficiency of friction

     piles. Some of these are given in table 13.

    Feld (1943) suggested a method by which the load capacity of individual piles (friction)

    in a group embedded in sand could be assigned. According to this method, the ultimatecapacity of a pile is reduced by one-sixteenth by each adjacent diagonal or row pile. The

    technique can be explained by referring to figure 8.51, which shows the plan of a group

     pile. For pile type A, there are eight adjacent piles; for pile type B, there are five adjacent

     piles; and for pile type C, there are three adjacent piles. Now the following table can be prepared:

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    Figure 8.51 Feld’s method for estimation of group capacity of friction piles

    Table 13 Equations for Group Efficiency of Friction Piles

    Converse-Labarre equation = 1 − (1 − 1)2 + (2 − 1)19012  ℎ  () = −1(/) 

    Los Angles Group Action

    equation = 1−   12 [1(2 − 1)] + 2(1 − 1)

    + √ 2(1 − 1)(2 − 1)] Seiler and Keeney equation

    (Seiler and Keeney, 1944) = �1 − 111

    7(2 − 1) 1 +

    2

     −2

    1 + 2 − 1+0.3

    1 + 2 ℎ      Pile type No. of

    Piles

     No. of adjacent

     piles/pile

    Reduction

    factor for each pile

    Ultimate

    capacity

    A 1 81

    −8

    16 

    0.5

     

    B 4 51− 5

    16 

    2.75  C 4 3

    1− 316

     3.25  

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     6.5 = () 

    (No. of piles)(

    ) (reduction factor)

    =        Hence

    = ( )  = 6.5 9 = 72% Figure 8.52  shows a comparison of field test results in clay with the theoretical group

    efficiency calculated from the Converse-Labarre equation (table 13). Reported by Brand

    et al. (1972), these tests were conducted in soil for which the details are given in figure 8.

    7 from chapter 3. Other test details includeℎ   = 6    = 150    = 2 × 2    ℎ = 1.5  ℎ   

    Figure 8.52 Variation of group efficiency with / (after Brand et al., 1972)

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    Pile tests were conducted with and without a cap (free-standing group). Note that for/ ≥ 2, the magnitude of  was greater than 1.0. Also for similar values of /  thegroup efficiency was greater with the pile cap than without the cap. Figure 8.53 shows

    the pile group settlement at various stages of the load test.

    Figure 8.53 Variation of group pile settlement at various stages of load (after Brand et al.,

    1972)

    Figure 8.54 Variation of efficiency of pile group in sand (based on Kishida and

    Meyerhof, 1965)

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    Figure 8.55 Behavior of low-set ad high-set pile groups in terms of average skin friction

    (based on Liu et al., 1985)

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    Figure 8. 55 (Continued)

    Figure 8.54 shows the variation of group efficiency ()   3 × 3 group pile in sand(Kishida and Meyerhof, 1965). It can be seen that, for loose and medium sands, themagnitude of group efficiency is larger than one. This is primarily due to the

    densification of sand surrounding the pile.

    Liu et al. (1985) reported the results of field tests on 58 pile groups and 23 single piles

    embedded in granular soil. Test details include

      ℎ, = 8 − 23  , = 125 − 330     =  

     

     

     

     

    ,

    = 2

    −6

     

    Figure 8. 55 shows the behavior of 3 × 3 pile groups with low-set and high-set pile caps

    in terms of average skin friction,  . Figure 8.56  shows the variation of average skinfriction based on the location of a pile in the group.

    Figure 8.56 Average skin friction based on pile location (based on Liu et al., 1985)

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    Based on eh experimental observations of the behavior of group piles in sand to date, the

    following general conclusions may be drawn.

    1.  For driven group piles in  sand with ≥ 3,()  may be taken to be 

    , which includes the frictional and the point bearing capacities of individual

     piles.2.  For bored  group piles in  sand  at conventional spacings ( ≈ 3),() may betaken to be 23  3

    4     (frictional and point bearing capacities of individual

     piles).