Ch3 pilefoundations

JABATAN KEJURUTERAAN INFRASTRUKTUR DAN GEOMATIK

FAKULTI KEJURUTERAAN AWAM DAN ALAM SEKITAR

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3.0 PILE FOUNDATION 3.1 Types of piles and their structural characteristics

1. Steel piles, Figure 3.1 Consist of pipe piles or rolled steel H-section piles The allowable structural capacity of steel piles :

ssall fAQ

Where : As – cross-sectional area of steel fs – allowable stress of steel

Use of additional thickness and epoxy coating are used to avoid corrosion, and typical condition of splicing (sambat) when needed is shown in Figure 3.1.

Figure 3.1 Steel Piles

2. Concrete piles Two categories of concrete piles are (a) precast and (b) cast-in-

situ Precast piles, Figure 3.2:

- prepared with ordinary reinforcement



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- in the shape of square or octagonal

Figure 3.2 Precast piles with ordinary reinforcement

Cast-in-situ or cast-in-place, Figure 3.3 :

- made by driving a steel casing with mandrel into the ground

- upon reaching the desired depth, mandrel is pulled out and the casing remain

- with or without pedestal - uncased piles :

- casing is driven to the desired depth, and filled with fresh concrete later gradually withdrawn - with or without pedestal

- allowable loads : cased pile : ccssall fAfAQ

uncased pile : ccall fAQ

where : As – cross sectional area of steel Ac - cross sectional area of concrete fs – allowable stress of steel fc - allowable stress of concrete

Figure 3.3 Cast in place concrete piles



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3. Timber piles Three classifications are :

o Class A : to carry heavy loads; min butt dia. = 14in

(356mm) o Class B : to carry medium loads; min butt dia. = 12-13in

(305-330mm) o Class C : used as temporary works but permanently for

submerged structure; min butt dia. = 12in (305mm)

Splicing can be done by means of pipe sleeves or metal straps or bolts, Figure 3.4

The allowable load-carrying capacity :

wpall fAQ

Where : Ap – average cross-sectional area of the pile fw – allowable stress for the timber

Figure 3.4 Splicing of timber piles (a) use of pipe sleeves (b) use of

metal straps and bolts 4. Composite piles

Upper and lower portions of composite piles are made of

different material They may in the form of : steel-cast-in-place concrete or

timber-concrete piles



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5. Pile in term of their function support capacity, Figure 3.5:

(a) Bearing pile, (b) friction pile, (c) piles under uplift, (d) piles under lateral loads, (e) batter piles under lateral loads

Figure 3.5

Requirements and conditions for pile foundations, Figure 3.6 :

Figure 3.6 Conditions for use of pile foundations



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- transmit load to the stronger underlying bedrock, 3.6(a) - gradually transmitting the load to the surrounding soil by

means of frictional resistance at the soil-pile interface, 3.6(b) - subjected to horizontal load while supporting the vertical load

transmitted by superstructure, 3.6(c) - built extended into hard stratum under collapsible soil (loess) to

avoid the zone of moisture change that lead to swell and shrink, 3.6(d)

- to resist uplifting forces for basement mats under water table, 3.6(e)

- to resist scouring at the bridge abutments and piers that can lead to possible loss of bearing capacity of soil underneath, 3.6(f)

3.2 Estimating Pile Length, Figure 3.7

Figure 3.7 (a) and (b) Point Bearing Piles; and (c) Friction Piles

Length of pile estimation depending upon the mode of load transfer to the soil ; namely :

o Point Bearing Piles

- the ultimate capacity of the piles depends entirely on the bearing capacity of the hard stratum

- hence the length, L of the pile is fairly well established



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- the ultimate pile load is then; spu QQQ (Figure 3.7a)

where : Qp – load carried at the pile point Qs – load carried by skin friction developed at the side of the pile

- piles can be extended into hard stratum with pu QQ (Figure

3.7b)

o Friction Piles

- if no hard stratum presence, piles are driven through softer soil to specified depths

- resistance to vertical loading, is provided mainly by the skin friction; (in clayey soil is called adhesion)

- the ultimate load is given by : su QQ

o Compaction Piles

- piles are driven in granular soil to achieve proper compaction of

soil close to ground surface - the length depends on :relative density before and after

compaction as well as required depth of compaction



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3.3 Installation of Piles, Figure 3.8

Figure 3.8 Pile driving equipment

Four method used in piles driving are ; drop hammer, single

acting air or steam hammer, double-acting and differential air or steam hammer, and diesel hammer

- drop hammer, Figure 3.8a

o raised by a winch, and allowed to drop at a certain height H

o slow rate of hammer blows

- single acting air or steam hammer, Figure 3.8b o ram is raised by air or steam pressure and then drops by

gravity

- double-acting and differential air or steam hammer, Figure 3.8c



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o ram is raised and pushed downward by air or steam pressure

- diesel hammer, Figure 3.8d

o consist of ram, an anvil block and a fuel-injection system o ram is raised, fuel is injected near the anvil, ram is

released, drops and compresses air-fuel mixture and ignites it

o this causes; pile to be pushed downward and ram raised

Vibratory pile driver, Figure 3.8e; consists of counter-rotating weights that produces centrifugal force that cancel each other but sinusoidal dynamic vertical force produced pushes the pile downward

3.4 Pile Load Transfer Mechanism Frictional resistance, f(z) with depth is given by :

zp

Qf

z

z

Where :

zQ - increase in pile load Δz – increase in depth

P – perimeter of pile

Nature of variation of pile load is as given by Figure 3.9 and Woo and Juang(1970) has obtained actual variation of load transfer by a bored concrete pile in Taiwan as in Figure 3.10



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3.5 Equations for Estimating Pile Capacity Ultimate load-carrying capacity of pile, Qu is :

spu QQQ

Where :

Qp – load-carrying capacity of the pile point Qs – frictional resistance

Point bearing capacity, Qp is :

** ' qcpppp NqcNAqAQ

Where : Ap – cross sectional area of pile tip c – cohesion of the soil supporting the pile tip qp - unit point cohesion

Figure 3.9 Load transfer mechanism for piles

Figure 3.10 Load transfer curves for a concrete bored pile, Woo and

Juang (1975)



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q’ =γ’L – effective vertical stress at the level of the pile tip

L- pile length ** , qc NN - the bearing capacity factors

Frictional resistance, Qs is :

LfpQs

Where : p – perimeter of the pile section ΔL – incremental pile length where, p and f is constant f – unit friction resistance at any depth z There are many other methods for estimating Qp and Qs

3.6 Meyerhof’s Method – Estimation of Qp The value of unit point resistance qp remains constant beyond

the critical embedment ratio, (Lb/D)cr, Figure 3.11

Figure 3.11 Nature of variation of unit point resistance in a

homogeneous sand

Figure 3.12 is the relationship of (Lb/D)cr and Ø(degree) where at Ø = 45°, (Lb/D)cr = 25



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For piles in sand, c=0; but Qp should not exceed Apql,

*' qpppp NqAqAQ and lpqpp qANqAQ *'

The limiting point resistance is :

SI unit : tan50/ *2

ql NmkNq ; or English

tan/

tan1000/

*2

*2

ql

ql

Nftkipq

Nftlbq

Where : Ø – soil friction angle in the bearing stratum

Using SPT method (Meyerhof, 1976):

NDLNmkNq p 400/40/ 2

where N - average SPT number at 10D above and 4D below the pile point.

For piles in clay, with saturated and undrained conditions (Ø=0)

pupucp AcAcNQ 9*

Where : cu – undrained cohesion (undrained shear strength) of the soil below the pile tip

Figure 3.12 Nature of variation of unit point resistance in sand

Figure 3.13 Variation of the

maximum values of *

qN with Ø



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3.7 Vesic’s Method – Estimation of Qp Vesic (1977) proposed value of Qp as :

*'*

NcNAqAQ ocpppp

Where :

'

o - mean normal ground effective stress = '3

21q

Ko

Ko – earth pressure coefficient = 1 – sin Ø ** , NNc - bearing capacity factors (see Table D.6 of Das textbook)

3.8 Janbu’s Method – Estimation of Qp NOT to be covered Janbu (1976) proposed value of Qp as :

** ' qcpp NqcNAQ

Where : **, qc NN - bearing capacity factors, Figure 9.14

Figure 3.14 (a)Meyerhof’s and (b) Janbu’s bearing capacity factors



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3.9 Coyle and Castello’s Method (Estimation of Qp in Sand) NOT TO BE COVERED

Coyle and Castello (1981) proposed value of Qp as :

pqp ANqQ *'

Where : q’ – effective vertical stress at the pile tip *

qN - bearing capacity factor, Figure 3.15



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Figure 3.15 Variation of *

qN with L/D, unit frictional resistance and K

value for piles in sand (Coyle and Castello, 1981)

3.10 Frictional Resistance, Qs in Sand Frictional resistance is, LfpQs

Factors to be kept in mind while estimating unit frictional, f

- the nature of pile installation - unit skin friction increases with depth - at similar depth, bored or jetted piles has a lower unit

skin friction compared to driven piles Approximation of f : (Figure 3.15)

For z = 0 to L’ : tan'

vKf

For z = L’ to L : 'Lzff

Where : K – effective earth coefficient

'

v - effective vertical stress at specified depth

- soil-pile friction angle L’ = 15d

Read text for values of K, fav and Qs between 1976 and 1982

3.11 Frictional Resistance, Qs in Clay Three method of estimating Qs in Clay :

1. Method :

- proposed by Vijayvergia and Focht (1972) - assumption : displacement of soil caused by pile driving results

in a passive lateral pressure at any depth - average unit skin resistance as :



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uvav cf 2

'

Where :

'

v - mean effective vertical stress for entire embedment length,

L

AAA ......321

cu – mean undrained shear strength (Ø=0) - refer to Figure 3.16b

- total frictional resistance is : avs pLfQ

Figure 3.16a Critical embedment ratio and bearing capacity factors

for various soil friction angles, (Meyerhof, 1976).



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Figure 3.16b Variation of with pile embedment length and its

application, (McCleland – 1974).

2. Method :

- unit skin resistance in clayey soil is : ucf

- empirical adhesion factor, Figure 3.17

α



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Figure 3.17 Variation of with undrained cohesion of clay

- total frictional resistance is : LpcLfpQ us

3. Method :

- assumption : excess pore water pressure in normally

consolidated clay for driven pile shall dissipates gradually - thus unit frictional resistance for the pile is :

'

vf

Where : '

v - vertical effective stress = γ’z

RK tan

ØR – drained friction angle of remolded clay K – earth pressure coefficient Where : RK sin1 for normally consolidated clays

OCRK Rsin1 for overly consolidated clays

- total frictional resistance is : LfpQs

3.12 Point Bearing Capacity of Piles Resting on Rock Goodman (1980) has approximate the ultimate unit point

resistance in rock as :

1 Nqq up

Where : 2/45tan2 N

qu – unconfined compression strength of rock - drained angle of friction



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After taking care of scale effect, 5

)(

)(

labu

designu

qq

Table 3.1 is the typical value of qu(lab) for rocks and Table 3.2 the value of angle of friction respectively

Table 3.1 Typical unconfined compressive strength of rocks

Rock type

qu

lb/in2 MN/m2

Sandstone Limestone

Shale Granite

Marble

10,000 – 20,000 15,000 – 30,000

5,000 – 10,000 20,000 – 30,000

8,500 – 10,000

70 – 140 105 – 210

35 – 70 140 – 210

60 – 70

Table 3.2 Typical Values of angle of friction, Ø, of rocks

Rock type Angle of friction, Ø

Sandtone

Limestone Shale

Granite

Marble

27 – 45

30 – 40 10 – 20

40 – 50

25 - 30

Hence, with FS = 3, the allowable point bearing capacity, Qp is :

FS

ANqQ

pdesignu

allp

1)(

)(

Table 3.3 Typical pre-stressed concrete pile in use



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Table 3.4 : Bearing capacity factors for deep foundations, N*c and N*σ, Vesic’s, 1977.



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Table 3.5 Janbu’s bearing capacity factors



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Example 3.1 Given : A square 305 mm x 305 mm concrete pile and 12 m long.

Fully embedded in homogeneus sand layer, γd = 16.8 kN/m3 ,

c=0 and Øavg=35°. The average SPT value near pile tip is 16.

Find : a. Qp using Meyerhof’s, Vesic’s, Janbu’s and SPT method.

b. Qs using LfpQs and )'......(

)'0....(tan

'

'

LLzforff

LzforKf

Lz

v

if K=1.3

and 8.0 .

c. Estimate the load-carrying capacity of pile, Qall if FS=4. d. Qall using Coyle and Costello’s method Solution : a. Meyerhof’s : Because it is a homogeneous soil, Lb=L. For Ø=35°,

(Lb/D)cr =(L/D)cr ≈ 10 (Figure 3-16a). So for this pile, Lb/D = 39.34 >

(Lb/D)cr. Hence, from the same figure 120* qN

kNNqAqAQ qpppp 4.22471206.2010929.0' *

2*2 /25.420135tan12050tan50/ mkNNmkNq ql

*'3.39042010929.0 qplpp NqAkNqAQ

Qp = 390 kN Vesic’s : use 90rrI ; with Ø=35°; 5.79* N so :

kN

NqANAQ popp

9235.796.2013

35sin1210929.0

'3

sin121 **'

Janbu’s : with c=0; use erpolationbyNand q int3.41;..35..;..90' *

kNmkNmNqAQ qpp 5.7733.41/6.2010929.0' 22*

SPT method : NDLNmkNq p 400/40/ 2

kNmqAQ ppp 233934.3916400929.0 2



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Limiting value = kNmNAQ pp 595164000929.0400 2

For design purpose : kNkNkNkN

Qp 5863

3905.773595

b. from sub-topic 3.10 from the note : mmDL 58.4305.01515'

For z = 0 : 0tan;0 '' vv Kf

For z = L’ to L :

2'

23'

/2.53358.0tan94.763.1tan

/94.7658.4/8.16'

mkNKf

mkNmmkNL

v

v

Thus

:

kN

mmmkNmmmkN

LLpfpLff

Q ftzmzz

s

631482149

58.412305.04/2.5358.4305.042

/2.530

''2

22

2058.40

c. thus load carrying capacity of pile, Qu = Qp(avg) + Qs

kNFS

QQkNQandkNQ ult

allsavgp 25.3044

631586;.....631586)(

d. Coyle and Castello’s

3.39305.0

12...;....8.0tan' '*

D

LandpLKANqQQQ vpqspult

For Ø=35° and L/D=39.3; 40* qN K≈1.0

Thus :

kN

mmkN

pLKANqQQQ vpqspult

23181569749

12305.04358.0tan128.160.1

0929.040/6.201

8.0tan'

22

'*

And kNFS

QQ ult

all 6.5794

2318



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Example 3.2

Find :

a. Net point bearing capacity. b. Skin resistance using α, λ and β method if ØR =30°; the

top 10m is normally consolidated clay and the bottom clay layer has OCR=2.

c. Net allowable pile capacity, Qall if FS=4. Solution :

a. Cross section of pile, 222 1295.0406.044

mDAp

kNcNAqAQ ucpppp 55.11610091295.0)2(

*

b. Skin resistance, Qs :

(α method) : LpcLfpQ us

From Figure α vs cu : cu(1)=30kN/m2 α=1.0; cu(2)=100 α=0.5

Thus :

kN

ccLpcLfpQ uuus

2.165820406.01005.010406.0301

20406.010406.0 )2(2)1(1

(λ method) : where )(

' 2 avuvav cf

2)2()1(

)( /7.7630

201001030

30

2010mkN

ccc

uu

avgu

Use the plotted Figure E9.2b, for σ’v vs depth;

2321'

/48.17830

457738.552225mkN

L

AAAv

Given : A driven pile in clay as in Figure E9.2. The pipe pile has outside diameter of 406mm and wall thickness of 6.35mm.



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From Figure λ vs L; λ=0.14 for L=30m; so 2

)(

' /46.467.76248.17814.02 mkNcf avuvav

Hence; kNpLfQ avs 8.177746.4630406.0

(β method) : where ØR =30°; '

vf ; RK tan ; RK sin1

OCRK Rsin1

For z=0-5m :

2'

)()1( /0.132

90030tan30sin1tansin1 mkNf avvRRav

For z=5-10m :

2'

)()2( /9.312

95.1309030tan30sin1tansin1 mkNf avvRRav

For z=10m-30m , OCR=2:

2'

)()3( /43.932

75.32695.130230tan)30sin1(tansin1 mkNOCRf avvRRav

so kNfffpQ avavavs 7.26692043.9359.31513406.02055 )3()2()1(

c. So use α and λ method which produced almost similar results,

kNQs 17182

8.17771.1658

kNFS

QQhencekNQQQ ult

allspult 6.4584

46.1834...;....46.1834171846.116

Example 3.3 Given : An H-pile (size HP 310 x 1.226), length of embedment = 26m, driven through soft clay and rest on sandstone, qu(lab) for sandstone = 76 MN/m2, Ø=28°, FS=5. Find : The allowable point bearing capacity, Qp(all)

Solution : Since 1 Nqq up ; 2/45tan2 N and 5

)(

)(

labu

designu

qq

kN

mmkN

FS

Aq

FS

ANq

FS

AqQ

p

labu

pupp

allp

1825

109.1512

2845tan

5

/1076

12

45tan51

23223

2)(

)(



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EXAMPLE OF FINAL EXAMINATION QUESTION

Q4 The most common function of piles is to transfer a load that cannot be adequately

supported at shallow depths to a depth where adequate support becomes available.

Hence, the piles can also be categorized based on its function/ support capacity.

(a) Briefly describe with relevant sketches the five (5) functions / support

capacity of piles.

(5 marks)

(b) Reinforced concrete piles 18 m long, of square section and width 400 mm

are driven through 8 m of loose fill with unit weight of 13 kN/m3 to

penetrate 10 m into an underlying firm to stiff saturated clay. The

groundwater table is found at a depth of 2 m below ground surface.

(i) Determine the ultimate bearing capacity, Qult, of pile by the given formula,

if the undrained shear strength of the clay increases linearly with depth

from 65 kN/m2 at the top of the clay to 100 kN/m

2 at a depth of 10 m

below the surface of the clay.

Assuming that the unit weight of stiff saturated clay is 17 kN/m3

throughout the layer and the frictional capacity of the loose fill is

negligible.

(10 marks)

(ii) Assuming that it is necessary to provide a number of such piles to carry

the total foundation load, explain the bearing capacity of the pile group is

estimated? Discuss your answer with the help of relevant sketches.

(5 marks)



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ANSWER

Q4 The most common function of piles is to transfer a load that cannot be adequately

supported at shallow depths to a depth where adequate support becomes available.

Hence, the piles can also be categorized based on its function/ support capacity.

(a) Briefly describe with a relevant sketch what are the five (5) function/

support capacity of piles.

(5 marks)

(a) Bearing pile, (b) friction pile, (c) piles under uplift,

(d) piles under lateral loads, (e) batter piles under lateral loads

(b) A reinforced concrete piles 18 m long, of square section and width 400

mm is driven through 8 m of loose fill with unit weight of 13 kN/m3 to

penetrate 10 m into the underlying firm to stiff saturated clay. The

groundwater table is found at a depth of 2 m below ground surface.

(i) Determine the ultimate bearing capacity, Qult of pile by the given

formula, if the undrained shear strength of the clay increases



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linearly with depth from 65 kN/m2 at the top of the clay to 100

kN/m2 at a depth of 10 m below the surface of the clay.

Assuming that the unit weight of firm to stiff saturated clay is 17

kN/m3 throughout the layer and the frictional capacity of the loose

fill is negligible.

Given that:-

qtip = cu Nc (Based on Meyerhof’s equation); uvavgs cf 2')(

(10 marks)

Answer:-

To determine Qp:-

qtip = cu Nc = 100 kN/m2 x 9 = 900 kN/m

2 [1M]

Ap = 0.4 x 0.4 = 0.16 m2 [1M]

Qp = Apqtip = 0.16 x 900 = 144 kN [0.5M]

To determine Qs:-

2/09.8110

2

)10)(04.11714.45(

' mkNv

[1M]

Elevation (m) Effective Vertical Pressure (kN/m2)

0 0

2 26

8 45.14

18 117.04

[1M]

2/5.8210

2

)10)(10065(

mkNcu

[1M]

Based on Figure 1, = 0.185 [1M]

uvavgs cf 2')(

= (0.185)[81.09+2(82.5)]

= 45.53kN/m2 [1M]

As = 4 x 0.4 x 10 = 16 m2 [0.5 M]



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Qs = As. fs = 16 x 45.53 = 728.48 kN [1M]

Qult = Qs + Qp = 728.48 + 144

= 872.48 kN [1M]

(ii) Assuming that it is necessary to provide a number of such piles to

carry the total foundation load, how could the bearing capacity of

the pile group be estimated? Discuss your answer with a relevant

sketch.

(5 marks)

Answer:-

For most practical purposes, the ultimate load of pile group, (QvG)ult, can be estimated

based on the smaller value of the following two values:-

(a) Group Action – block failure (Figure A) of pile group by

breaking into the ground along an imaginary perimeter and bearing

at the base. The ultimate capacity for the group failure can be

estimated from the following relationship:-

(QvG)ult = x n x (Qv)ult

[2M]

(b) Individual Action (Figure B) – if there is no group action

(when the center to center spacing, s, is large enough, >1), in that

case, the piles will behave as individual piles. The total load of the

group can be taken as n times the load of the single pile, in which

(QvG)ult = n x (Qv)ult = (Qv)ult

[2M]



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Figure : (A) Individual action, (B) Group action

[2 x 0.5M = 1M

3.13 Pile Load Test Pile load test arrangement by means of hydraulic jack is shown

in Figure 3.18a Step loads are applied to the pile, so that a small amount of

settlement is allowed to occur Settlement from field test is recorded as in Figure 3.18b Net settlement calculation for any load Q :

- When Q = Q1 : Net settlement, )1()1()1( etnet sss



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- When Q = Q2 : Net settlement, )2()2()2( etnet sss

Where : snet – net settlement se – elastic settlement of the pile itself st – total settlement

The values of Q then plotted against se produces diagram in Figure 3.18c

Figure 3.18 (a) Test arrangement (b) load vs total settlement

(c) load vs net settlement 3.14 Failure criteria of a pile The ultimate failure load for a pile is defined as the load when

the pile plunges or the settlements occur rapidly under sustained load and the amount of settlement exceed the acceptable soil-pile system

Or

Besides it, many engineers define the failure load at the point

of intersection of the initial tangent to the load-settlement curve and the tangent to or the extension of the final portion of the curve.



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Arbitrary settlement limits that the pile is considered to have failed when the pile head has moved 10 percent of the pile end diameter or the gross settlement of 1.5 in. (38 mm) and net settlement of 0.75 in. (19 mm) occurs under two times the design load. (JKR standard)

However, all of these definitions for defining failure are

judgemental. 3.15 Pile Driving Formulas Due to varying soil profiles layers a point bearing pile cannot

always satisfied the capability of penetrating the dense soil to a predetermined depth; therefore several equations have been developed by many to calculate the ultimate capacity of pile during driving.

According to Engineering News Record (ENR), Qu is :

CS

hWQ R

u

Where : WR – weight of the ram h – height of fall of the ram S – penetration of pile per hammer blow (from last few

driving blows) C – a constant (for drop hammers : C = 1 in. ; S and h are in inches) (for steam hammers : C = 0.1 in. ; S and h are in inches) FS = 6 For single and double-acting hammers WRh is replaced by EHE Thus :

CS

EHQ E

u

Example 3.4 A precast concrete pile 12 in. x 12 in. in cross section is driven by a hammer. Given : Maximum rated hammer energy = 30 kip-ft Hammer efficiency = 0.8



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Weight of ram = 7.5 kip Pile length = 80 ft Coefficient of restitution = 0.4 Weight of pile cap = 550 lb Ep = 3 x 106 kip/in2 Number of blows for last 1 in. of penetration = 8 Estimate the allowable pile capacity by the

a. Modified ENR formula (use FS=6) b. Danish formula (use FS = 4) c. Gates formula (use FS = 3)

Solution : a. Weight of pile + cap = kipftlb 55.12550/15080 3

1212

1212

and ftkiphWR 30

kip

inkip

WW

WnW

CS

hEWQ

pR

pRRu 607

55.125.7

55.124.05.7

1.0

12308.02

81

2

kipFS

QQ u

all 1016

607

b.

pp

E

Eu

EA

LEHS

EHQ

2

Use Ep = 3 x 106 lb/in2

And

.566.0

/1000

10312122

128012308.0

22

6in

inkipEA

LEH

pp

E

kipQu 417566.0

12308.0

81

kipQall 1044

417

c. kipSbEHaQ Eu 252log1308.027log

81

kipQall 843

252

3.16 Hiley’s Formula for estimating single RC pile capacity.



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The Hiley’s formula gives the simplest method of calculating the final setting or the ultimate load of a pile while driving depending upon the given parameter. It is usually written as :

PW

PeW

WFS

hBWCs

H

H

L

H

2

2

And qpc CCCC

where : s - Set value /1 blow (mm/blow) C - Temporary compression of pile & soil (mm)

WH - Weight of hammer (kN) h - Drop of hammer (mm) P - Total load (P1 + P2) (kN)

P1 - Weight of pile (kN) P2 - Weight of driving assembly (kN) WL - Pile working load (kN) FS - Factor of safety

e - Coefficient of restitution Cc - Temporary compression coefficient due to pile

head and cap (mm), Table 3.3 Cp, - Temporary compression coefficient due to pile

length (mm), Table 3.3 Cq, - Temporary compression coefficient due to ground

or quake (mm), Table 3.3

Note : (a) This formula was developed by Hiley (1925). The formula

assumes the energy of the falling hammer during pile driving is proportional resisted by the pile. This method is widely considered to be one of the better formulas that intended to be applied to cohesionless, well-drained soils or rock.

(b) Weight of the hammer shall be about 0.5 to 2.0 times of the total pile weight.

(c) The term mass and weight are interchangeably (d) The term Cp and Cq are shown in Figure 3.19 after a pile

set measurement of pile are made.



34

Figure 3.19 : Example graph of pile set

Table 3.6 : Values of Cc, Cp and Cq

Form of compression

Material Easy driving (inch)

Medium driving (inch)

Hard driving (inch)

Very hard driving (inch)

Pile head and cap, Cc

Head of timber pile

0.05 0.10 0.15 0.20

Short dolly in helmet or

driving cap 0.05 0.10 0.15 0.20

3 in/76.2mm packing under

helmet or driving cap

0.07 0.15 0.22 0.30

1 in/25.4mm pad only on

head of reinforced

concrete pile

0.03 0.05 0.07 0.10

Pile length, Cp

Timber pile (E=1,500,000

lb/in2) or (E=10,342,500

kPa)

0.004L 0.008L 0.012L 0.016L

Pre-cast pile (E=2,000,000

lb/in2) or (E=13,790,000

kPa)

0.003L 0.006L 0.009L 0.012

Steel pile for cast in place

(E=30,000,000 lb/in2) or

(E=206,850,000 kPa)

0.003L 0.006L 0.009L 0.012

Quake, Cq Ground surrounding pile and under pile

0.05 0.10 – 0.20

0.15 – 0.25

0.05 – 0.15



35

point

Note : Length, L measure in feet 1 feet = 0.3048 m 1 inch = 25.4 mm

Table 3.7 : Coefficient of restitution, e.

Description Coefficient of restitution, e Piles driven with double acting hammer

- Steel piles without driving cap - Reinforced concrete pile without helmet but with

packing on top of pile - Reinforced concrete piles with short dolly in helmet

and packing - Timber pile

0.5 0.5

0.4

0.4

Piles driven with single acting and drop hammer

- Reinforced concrete piles without helmet but with packing on top of piles

- Steel piles or steel tube of cast in place piles fitted with driving cap and short dolly covered by steel plate

- Reinforced concrete piles with helmet and packing, dolly in good condition

- Timber pile in good condition - Timber pile in poor condition

0.4

0.32

0.25

0.25 0.00

Example 3.5 Using Hiley’s formula calculate the final set of a 200mm X 200mm RC pile. The pile driven with single acting and drop hammer with medium driving. The type of pile is the reinforced concrete pile with helmet and packing, dolly in good condition. Other data and parameters are : Pile working load, = 275 kN Mass of hammer, = 25 kN Factor of safety, FS = 2.0 Pile length, L = 18 m Mass driving assembly, = 2.0 kN Drop of hammer, = 400 mm Hammer efficiency, = 85%



36

Density of concrete, = 24 kN/m3 Solution : Mass of pile, P1 = Concrete density X Area X Length of pile = 24 X (0.2 X 0.2) X 18 = 17.28 kN Total load, P = P1 + P2 = 17.28 + 2.0 = 19.28 kN Value of e = 0.25 (Table 3.7)

mmWFS

hBW

L

H 454.152750.2

4002585.0

592.028.1925

25.028.192522

PW

PeW

H

H

Value of C : Cc = 0.15in X 25.4 = 3.81 mm Cp = 0.006(59ft) = 0.354in X 25.4 = 8.99 mm Cq = 0.10in X 25.4 = 2.54 mm C = Cc + Cp + Cq = 3.81 + 8.99 + 2.54 = 15.34 mm

Using

)(10/8.14

/48.1

592.0454.152

34.15

2

2

SetFinalblowmmSor

blowmms

s

PW

PeW

WFS

hBWCs

H

H

L

H

Example 3.6 Given :



37

A 200mm x 200mm RC square pile. The pile driven with single-acting and drop hammer with hard driving. The type of pile is reinforced concrete pile with helmet and packing, dolly in good condition. Mass of hammer, Wn =25kN Factor of safety, FS =2.0 Pile length, l =24m Mass Driving assembly,P2 =3.0 kN Drop hammer, h =500mm Hammer efficiency, B =85% Set value, S =19mm/10 blow (Figure 3.20)

Figure 3.20

Required : Ultimate load of pile Solution : Mass of pile, P1 = Concrete densityxAreaxlength = 24x(0.2x0.2)x24=23.04kN Total load, P2 = P1 + P2 = 23.04 + 3=26.04kN Value of e = 0.25 (Table 3.7) Cp + Cq = 20mm (Figure 3.20) Cc = 0.22inx25.4=5.59mm Temporary compression, C = 5.59 + 20 = 25.59mm Set value, s = 19mm/10 blow =1.9mm/blow



38

kN

CCCsFS

hBW

qpc

H 52.361

59.252

19.10.2

5002585.0

2

1

5217.004.2625

25.004.262522

PW

PeW

H

H

By using Hiley’s equation :

kNPW

PeW

CCCsFS

hBWW

H

H

cqp

HL 6.1885217.052.361

2

1

2

Therefore, the pile working load must be less than 188.6kN 3.17 Settlement of Piles, Vesics (1969) Settlement of a pile under vertical working load, Qw is :

321 ssss

Where : s – total pile settlement s1 – elastic settlement of pile s2 – settlement caused by the load at the pile tip s3 – settlement caused by the load transmitted along pile shaft

Formulae : - elastic settlement, s1 :

pp

wswp

EA

LQQs

1

Where : Qwp – load carried at the pile point under working condition Qws – load carried by frictional resistance under work load Ap – area of pile cross section L – length of pile Ep – modulus of elasticity of the pile material - nature of unit skin friction (=0.5 or 0.67), Figure 3.21



39

Figure 3.21 Various types of unit friction resistance along pile shaft

- load at pile point, s2 :

wps

s

wpI

E

Dqs 2

2 1

Where : D – width or diameter of pile qwp – point load per unit area = Qwp/Ap Es – modulus of elasticity of soil at or below the pile point μs – Poisson’s ratio of soil Iwp – influence factor = 0.85 Or

p

pwp

Dq

CQs 2

Where : qp – ultimate point resistance of the pile Cp – an empirical coefficient, Table 3.8

Table 3.8 Typical Values of Cp

Soil type Driven Pile Bored Pile

Sand (dense to loose) Clay (stiff to soft) Silt (dense to loose)

0.02-0.04 0.02-0.03 0.03-0.05

0.09-0.18 0.03-0.06 0.09-0.12

- load carried by pile shaft, s3 :

wss

s

ws IE

D

pL

Qs 2

3 1

Where :



40

p – perimeter of the pile L – embedded length of pile

Iws – influence factor = D

L35.02

Or

p

sws

Lq

CQs 3 and Cs (a constant) = pCDL /16.093.0

Cp from Table 3.8 Example 3.7 Given : A pre-stressed concrete pile 21m long, being driven into sand. Working load, Qw = 502 kN. The pile is octagonal in shape with D = 356 mm, see Figure E9.4. Skin resistance, Qs carries 350 kN, and Qp carries the rest. Use Ep = 21 x 106 kN/m2, Es = 25 x 103 kN/m2, μs = 0.35 and ξ = 0.62.

Find : The settlement of the pile. Solution : From Table D3; for D=356mm, Ap=1045cm2, p=1.168mm and Qws=350 kN; so Qwp=502-350=152 kN

Due to material :

mmm

mEA

LQQs

pp

wswp35.300353.0

10211045.0

2135062.0152621

Due to point load :

mmmIE

Dqs wps

s

wp5.150155.085.035.01

1025

356.0

1045.0

1521 2

3

2

2

Due to skin :

With 69.4356.0

2135.0235.02

D

LI ws

p

wp

wpA

Qq



41

And

mmmm

IE

D

pL

Qs wss

s

ws

84.000084.0

69.435.011025

356.0

21168.1

3501 2

3

2

3

Therefore the total settlement is :

mmssss 69.1984.05.1535.3321

3.18 Pullout Resistance of Piles The gross ultimate resistance of a pile subjected to uplifting

force, Figure 3.22 is :

WTT unug

Where : Tug – gross uplift capacity Tun – net uplift capacity W – effective weight of pile

Figure 3.22 Uplift capacity of piles

a. In Clay

Das and Seeley (1982), estimated Tun as :



42

uun cLpT '

Where : L – length of the pile p – perimeter of pile section ’- adhesion coefficient at soil-pile interface cu – undrained cohesion of clay Values of ' :

- for cast-in-situ: (for bore pile)

uc00625.09.0' for cu ≤ 80 kN/m2

4.0' for cu > 80kN/m2

- for pipe piles :

uc0191.0715.0' for cu ≤ 27 kN/m2

2.0' for cu > 27 kN/m2

b. In Sand

Das and Seeley (1975), estimated Tun as :

L

uun dzpfT0

with fu varies by tan'

vuu Kf for (z≤Lcr) such

as in Figure 3.23a Steps in finding Tun in dry soil;

- find relative density and use Fig 3.23c to find Lcr - if L ≤ Lcr then :

tan2

1 2

uun KLpT with values Ku and from Figure

3.23b&c

- if L > Lcr then :



43

crucrucrun LLKLpKLpT tantan2

21 with values Ku and

from Figure 3.23b&c

Where : Ku – uplift coefficient

'

v - effective vertical stress at a depth z

- soil-pile friction

Thus with FS=2 to 3, allowable uplift capacity Tu(all) is :

FS

TT

ug

allu )(

Figure 3.23 (a) Variation of fu (b) Ku (c) Variation of /Ø, (L/D)cr with relative density of sand Dr

Example 3.8



44

Given : A 50 ft long concrete pile embedded in a saturated clay cu=850 lb/ft2. 12 in x 12 in. in cross section. Use FS=4. Find : Allowable pullout capacity, Tun(all) Solution : with cu =850 lb/ft2 ≈ 40.73 kN/m2 645.073.4000625.09.000625.09.0' uc

kipcLpT uun 7.1091000

850645.01450'

And kipFS

T allun 4.274

7.1097.109)(

Example 3.9 Given : A precast concrete pile, with cross section = 350mm x 350mm. Length of pile as 15m. Assume : γsand=15.8 kN/m3,

Øsand=35°, Dr=70%. Find : Pullout capacity if FS=4. Solution : From Figure 3.23; for Ø=35° and Dr=70%

2;.......35351;..1

08.535.05.14;..5.14

u

cr

cr

K

mmLD

L

Hence : for L (15m) > Lcr (5.08m)

kN

LLKLpKLpT crucrucrun

1961

35tan08.515208.58.15435.035tan208.58.15435.0

tantan

2

21

2

21

kNFS

TT

ug

allu 4904

1961)(



45

3.19 Group piles efficiency Converse –Labarre method of estimating pile-group efficiency

developed by Jumikis, 1971 using the following equation :

mn

nmmnEg

90

111

Where : Eg – pile-group efficiency θ – tan-1(d/s), (deg)

n – number of piles in row m – number of rows of piles d – diameter of piles s – spacing of piles, center to center, same unit as pile

diameter.

Example 3.10 Given : A pile group consists of 12 friction piles in cohesive soil, Figure 3.24. Each pies diameter is 300mm and center-to-center spacing is 1 m. By means of a load test, the ultimate load of a single pile was found to be 450 kN. Take SF as 2.0. Required : Design capacity of the pile group, using the Converse-Labarre equation.

4.18

3

1tan 1 ;

710.04390

4133144.181

gE

Allowable bearing capacity of a single pile=450kN/2=225kN Design capacity of the pile group = 0.710(12)(225kN)=1917kN.



46

Figure 3.24

Another method of estimating efficiency of pile group as quoted by Das (2007) as follows :

A pile cap is normally constructed over group piles; either in contact or well above the ground, Figure 3.25 a&b.

In practice, minimum center-to-center pile spacing, d = 2.5D, or 3-3.5D as in ordinary situations; where D - diameter of piles

Thus, the efficiency of a group pile, η is :

u

ug

Q

Q )(

Where : Qg(u) – ultimate load-bearing capacity of the group pile Qu – ultimate load-bearing capacity of each pile without group

effect If group as a block thus :

21

21 422

npn

ddnn



47

For most practical purposes, the ultimate load of pile group, (QvG)ult, can be estimated based on the smaller value of the following two values, Figure 3.27 (a) and (b):-

(a) Group Action – block failure (Figure A) of pile group by

breaking into the ground along an imaginary perimeter and bearing at the base. The ultimate capacity for the group failure can be estimated from the following relationship:-

(QvG)ult = x n x (Qv)ult

(b) Individual Action (Figure B) – if there is no group action (when the center to center spacing, s, is large enough, >1), in

that case, the piles will behave as individual piles. The total load of the group can be taken as n times the load of the single pile, in which

(QvG)ult = n x (Qv)ult = (Qv)ult

Figure 3.25 Pile groups

3.26 timate Capacity of Group Piles in Saturated Clay



48

Figure 3.27 : (A) Individual action, (B) Group action Feld’s Method : in estimating group capacity of friction piles,

Qg(u)

Figure 3.28 Feld’s Method

Table 3.9 Arrangement of Feld’s Method

Pile

type

No. of Piles No. of adjacent

piles

Reduction factor

for each pile

Ultimate capacity

Col.2 x Col.4 A 1 8 1-8/16 # 0.5Qu

B 4 5 1-5/16 2.75Qu

C 4 3 1-3/16 3.25Qu Σ6.5Qu=Qg(u)

Note: 16 # no. of arrow



49

Therefore, efficiency, %729

5.6)(

u

u

u

ug

Q

Q

Q

Q

3.20 Ultimate Capacity of Group Piles in Saturated Clay Figure 3.29 shows a group of pile in saturated clay, steps to

find the ultimate load-bearing capacity Qg(u) are : Find Qu in pile group :

As individual From : spu QQnnQ 21 ; )(9 pupp cAQ and LpcQ us

So : LpccAnnQ upupu )(21 9 (1)

As pile group (dimensions of LgxBgxL): LcBLLcp uggug 2

Point bearing capacity as : *

)(

*

)( cpuggcpuppp NcBLNcAqA

With *

cN from Figure 3.29, thus :

LcBLNcBLQ uggcpuggu 2*

)( (2)

Where :

2211

DdnLg and

2212

DdnBg

The lower value from (1) and (2) is Qg(u)



50

Figure 3.29 Ultimate group piles in clay

Figure 3.30 Variation of *

cN with Lg/Bg and L/Bg



51

Example 3.11 Given : The section of 3 x 4 group pile in a layered saturated clay is shown in Figure 3.31. The piles are square in cross section (350mm x 350mm). The center-to-center spacing d, of the piles is 1220mm. Required : The allowable load-bearing capacity of the pile group. Use FS=4.

Figure 3.31 Group pile in clay soil

If pile act as single pile:

2)2(21)1(1)(21

)(21

9

9

LpcLpccAnn

LpccAnnQ

uupup

upupu

With cu(1)=40 kN/m2;α1=0.86 and cu(2)=70 kN/m2;α2=0.63 thus;

Ap=0.350x0.350=0.093m2, p=4x0.350=1.22m

kN

LpcLpccAnnQ uupupu

04.967702.5388.2096.5843

107022.163.054022.186.070093.0943

9 2)2(21)1(1)(21

If pile as a group :

mD

dnLg 965.3305.022.1142

211

mD

dnBg 745.2305.022.1132

212

70 kN/m2

40 kN/m2 5 m

10 m

1.22 m



52

46.5745.2

15;.....44.1

745.2

965.3

gg

g

B

L

B

L

From Figure 3.29: 6.8* cN

(assuming : that at the end of curve at right hand stays horizontal)

Thus :

kN

LcBLNcBLQ uggcpuggu

1863070020042.136552

1070540745.2965.326.870745.2965.3

2*

)(

Hence, ΣQu=9677 kN,

kNFS

Qall 24194

96779677

3.21 Consolidation Settlement of group pile in clay by mean of 2:1

distribution method.

Figure 3.32

o L=depth of pile embedment

o Qg – total load of superstructure (–) weight of soil

excavated

o Assume load Qg transmitted at depth of 2L/3

from top of pile.

o The load Qg spread out at 2 : 1 horizontal line

from this depth

o Line a-a’ and bb’ are two 2:1 lines

o Stress increased at the middle of each soil layer :

igig

g

izLzB

Qp

o Lg and Bg – the length and width of pile group

o zi – distance from z=0 to the middle of clay layer

o For layer 2 : zi=L1/2; For layer 3 : zi=L1+L2/2

o For layer 4 : zi=L1+L2+L3/2

o Consolidation Settlement, i

i

i

i He

es

)(

)(

1

o Where :

0

0logp

ppCe c

;

Layer 2 : Hi=L1; Layer 3 : Hi=L2; Layer 4 : Hi=L3

o Total consolidation settlement, ig ss



53

Example 3.12 A group of pile in clay is shown in Figure 3.33. Determine the consolidation settlement of the pile groups. All clays are normally consolidated.

Figure 3.33 Pile group in clay soil Solution :

2

)1( /6.515.33.35.32.2

2000mkN

zLzB

Qp

igig

g

2

)2( /52.1493.392.2

2000mkN

zLzB

Qp

igig

g

2

)3( /2.9123.3122.2

2000mkN

zLzB

Qp

igig

g

With

)1(0

)1()1(0

)1(0

1)1(

1 log1 p

pp

e

HCs

c and ;



54

2

)1(0 /8.13481.9185.122.162 mkNp

mmm

p

pp

e

HCs

c4.1621624.0

8.134

6.518.134log

82.01

73.0log

1 )1(0

)1()1(0

)1(0

1)1(

1

2

)2(0 /62.18181.99.18281.918162.162 mkNp

mmm

p

pp

e

HCs

c7.150157.0

62.181

52.1462.181log

7.01

42.0log

1 )2(0

)2()2(0

)2(0

2)2(

2

2

)3(0 /99.20881.919181.99.18262.181 mkNp

mmm

p

pp

e

HCs

c4.50054.0

99.208

2.999.208log

75.01

225.0log

1 )3(0

)3()3(0

)3(0

2)3(

3

Therefore the total settlement : Δsg = 162.4 + 15.7 + 5.4 = 183.5mm

3.22 Elastic settlement of pile group. Vesic (1969) developed the simplest relation of :

Elastic settlement of group pile, sD

Bs

g

eg )(

Bg – width of pile group D – width or diameter of each pile in the group s = s1 + s2 + s3 – total elastic settlement at working load Meyerhof (1976) developed elastic settlement of pile group in

sand and gravel.

Elastic settlement of group pile, .

)(

2

corr

g

egN

IBqins

Where : q=Qg/(LgBg) in ton/ft2 Lg and Bg – length and width of pile group section (ft) Ncor – average of SPT no. at Bg below pile tip (within seat of settlement)

Influence factor, I=1-L/8Bg ≥ 0.5 L – length of pile embedment



55

Example 3.13 (Cumulative)

A reinforced concrete piles 18m long, of square section (diameter) and width 300 mm is driven through 6 m of loose fill with unit weight of 15 kN/m3 to penetrate 12 m into the underlying firm to stiff saturated clay. The groundwater table is found at a depth of 3 m below ground surface.

(i) Determine the ultimate bearing capacity, Qult of pile by the given formula, if the undrained shear strength of the clay increases linearly with depth from 80 kN/m2 at the top of the clay to 120 kN/m2 at a depth of 12 m below the surface of the clay.

Assuming that the unit weight of firm to stiff saturated clay is 18 kN/m3 throughout the layer and the frictional capacity of the loose fill is negligible.

Given that:- qtip = cu Nc (Based on Meyerhof’s equation);

uvavgs cf 2')(

(ii) Evaluate Qa if using total FS=2.5 (iii) Evaluate Qa if using FS = 2 for skin and FS = 3 for tip. 3m 3m 12m



56

To determine Qp:- qtip = cu Nc = 120 kN/m2 x 9 = 1080 kN/m2 Ap = 0.3 x 0.3 = 0.09 m2 Qp = Apqtip = 0.09 x 1080 = 97.2 kN To determine Qs:-

Depth(m) Effective Vertical Pressure (kN/m2)

0 0

3 3x15=45

6 45 + 3(15-9.81) = 60.57

18 60.57 + 12(18-9.81) = 158.85

2/71.109122

)12)(85.15857.60(

' mkNv

2/100122

)12)(12080(

mkNcu

Based on Figure 1, = 0.185 for L=18m

uvavgs cf 2')(

= (0.185)[109.71+2(100)] = 57.3 kN/m2 As = 4 x 0.3 x 12 = 14.4 m2

Qs = As. fs = 14.4 x 57.3 = 825.12 kN Qult = Qs + Qp = 825.12 + 97.2 = 922.32 kN

(ii) Qa = 922.32/2.5 = 368.9kN

(iii) Qa = 825.12/2 + 97.2/3 = 444.96kN



57

3.23 Calculation of single, group pile capacity and settlement from Prakash & Sharma - for sandy soil.

Table used for values of Nq and Ø, Table 3.10.

Table 3.10

Ø 20 25 28 30 32 34 36 38 40 42 45

Nq(driven) 8 12 20 25 35 45 60 80 120 160 230

Nq(drilled) 4 5 8 12 17 22 30 40 60 80 115

Table used for values of Ks for various pile types in sand, Table

3.11 Table 3.11

Pile type Ks

Bored pile Driven H pile Driven displacement pile

0.5 0.5 – 1.0 1.0 – 2.0

For most design purpose δ=2/3Ø (Meyerhof, 1976)

Example 3.14 A closed-ended 12-in (300mm) diameter steel pipe is driven into sand to a 30ft (9m) depth. The water is at ground surface and sand has Ǿ=36° and unit weight (γsat) is 125 lb/ft3 (19.8kN/m3).

Estimate the pipe pile’s allowable load. Solution :

For circular pile :

ftpftft

Ap 14.31,785.04

1 2

2

Nq=60, Table 3.10; Ks=1.0, Table 3.11; 24363

2

3

2

Using the formula of the ultimate capacity :

LpKNAQQQLL

L

vlsqvpfpultv

0

'' tan

Where :



58

BLBBBL satsub

LL

L

vl 2020202/200

'

This is with the assumption of : σ’vl increases with depth up to 20B. Below this depth, σ’vl remains constant. With γsub or γ’ = 125 – 62.5 =62.5 lb/ft3, B=1ft, L=30ft.

Then :

kNkips

lb

BLBBBL subsub

LL

L

vl

25.11125500,12500,12

120301205.621201105.62

2020202/200

'

Thus :

kipsB

LpKNAQQQ

sub

LL

L

vlsqvpfpultv

83.9395.3488.582524tan114.36020785.0

tan0

''

Therefore with FS=3: (Qv)all=(Qv)ult/FS=93.83/3=31kips (137.95kN) Example 3.15 For the pile described in example 3.14, estimate the pile settlement. The pile has ¾ in. wall thickness and is closed at the bottom. Solution : B=12 in. (outside diameter); L=30x12=360 in. (Qv)all=31,000 lb (from Example 3.14)

Area of base 2211312

4in

Pipe inside diameter .5.10)4/32(12 in

Area of steel section 2222 496.26184.01444/5.1012 inft

1. Semiempirical method :



59

From the relation of : (from Example 3.14)

kipsB

LpKNAQQQ

sub

LL

L

vlsqvpfpultv

83.9395.3488.582524tan114.36020785.0

tan0

''

And (Qv)all=(Qv)ult/FS=93.83/3=31kips Assuming allowable loads are actual loads; then

...........)3/95.34....(......4.116.1931

;....6.193/83.58

errorroundoffsomewithkipsorQQ

kipsQQ

allffa

allppa

Due to material :

inEA

LQQS

pp

faspa

s 011.0103496.26

10363.25

1030496.26

36010004.115.06.197

4

6

Vesic (1977) recommends αs = 0.5 for uniform or parabolic

skin friction distribution along pile shaft. Ep = 30x106 psi for steel Ep = 21 x 106 kN/m2 for concrete Due to point :

inBq

QCS

p

pap

p 094.088.5812

1136.1903.0

Cp=0.03 (Table 9.3); qp=Qp/Ap=58.88/113

Cs 054.003.012

36016.093.0.16.093.0 p

fC

B

D

Due to skin :

in

qD

QCS

pf

fas

ps 0033.088.58360

1134.11054.0

Using St=Ss+Sp+Sps=0.011+0.094+0.0033=0.108in(2.7mm)

2. Empirical method :

Using :

)35.3.(134.0

014.012.01030496.26

100036031

100

12

100 6

mmin

EA

LQBS

pp

va

t



60

Example 3.16 Using data of example 3.14, find the allowable bearing capacity based on standard penetration data as given in Figure 3.34.

Solution : (b) Average N value near pile tip, Navg(tip)=(10+12+14)/3=12 (c) Point bearing, Qp

)(/938.0/187530/5.62125 223' tsffttonftlbftftlbv

1 ton = 2000 lb Correction for depth of N values,

02.1938.0/20log77.0 10 NC

Therefore ; 121202.1 NCN N

And tonsBADN pf 1131/785.030124.0/4.0

tonsAN p 7.37785.01244

The lower of these values is Qp=37.7 tons (d) Shaft friction, Qf Average N value along pile shaft, Navg(shaft)= (4+6+6+8+10)/5=6.8 Use σ’v for average depth of L/2=30/2=15ft so σ’v= 0.938/2=0.469tsf

25.1469.0/20log77.0 10 NC Therefore ;

For driven piles :

ppfp ANADBNQ 4/4.0

(Meyerhof,1976)

tsfNfDpfQ sfsf 150/;.. *

(Meyerhof,1976)

Figure 3.34



61

5.88.625.1 NCN N ; )1(17.050/5.850/ tsftsfNf s

So tonsLpfQ sf 1630117.0

(e) Allowable bearing capacity, Qall :

)156..(36...8.359.173/7.53/

7.53167.37

kNkipssaytonsFSQQ

tonsQQQ

ultvallv

fpultv

Pile group sample calculations Settlement of pile group and check on design :

1. Vesic’s Method (1977) : BbSS tG /

2. Meyerhof’s Method (1976) (if SPT N values available) :

N

bpIS

2

where :

5.08

1...........

b

DIand

bb

Qp

fallG



62

Example 3.16 Using data from Example 3.14, calculate the pile group bearing capacity if the piles are placed 4ft center to center and joined at the top by a square pile cap supported by nine piles. Estimate pile group settlement.

Solution : (a) bearing capacity B=1ft; s=4ft; ,9144 ftb ; b=10ft; n=9

kipsQultv 83.93 for a single pile (from empirical method Ex 3.15)

0.3...),...1250(2813

83.939

47.84483.939

FSwithkNkipsQ

kipskipsQnQ

allvG

ultvultvG

(b) settlement B=1ft; ,9144 ftb (square arrangement); n=9 piles;

(Qg)all=281kips; zone of influence, b =9ft below the group base; Navg=(12+14+14)/3≈13; for single pile st=0.134in.(EX.3.14)

1. Vesic’s (1977): inBbSS tG 40.01/9134.0/

2. Meyerhof’s (1976): (N values)

Figure 3.35



63

5.058.0

98

301

81

/74.1/47.399

281 22

b

DI

fttonsftkipsbb

Qp

f

allG

where Df is pile length = 30 ft

So :

in

N

bpIS 93.0

13

958.047.322

47.322

.

)( corr

g

egN

IBqins

.

)(

2

corr

g

egN

IBqins

Example 3.17

Given : A 236-kip(1050kN) of vessel (water tank) is to be supported on a pile foundation in an area where soil investigations indicated soil profile Fig 3.36. Required : Design a pile foundation so that the maximum allowable settlement for the group does not exceed allowable settlement, Sa=0.6in (15mm).

Figure 3.36 Soil profile and soil properties used : N-SPT value;



64

33

3'

/5.625.62125';..../125

;../110;......36;........

ftlbftlb

ftlbsandforstressverticaleffective

sandsand

clayv

Solution : 1. Soil profile as in Figure 3.36 2. Pile dimensions and allowable bearing capacity

- top 4 ft consist of top soil and soft clay – this layer has no contribution to the side frictional resistance.

- Increasing in N values except at 24ft – due to gravel – neglected

- Try 34ft(10.3m) long with 30ft(9.1m) penetration into sand and 12-in(305mm) diameter steel-driven frictional pile

- This pile has 0.75in thickness and is closed at the bottom - Static analysis by utilizing soil strength :

LpKNAQLL

L

vlsqvpultv

0

'' tan and 22785.014/ ftAp

Nq=60 for Ǿ=36° from Table 9.5; perimeter, p=πB=3.14ft Ks=1.0 from Table 9.6; δ=2/3Ǿ=2/3(36°)=24°

Thus :

)5.182()41(1.413

3.123

3.1237.436.79

101690202

/169044024tan114.3

601690785.0

2

2

2

kNorkipssaykipsFS

QQ

kip

ftftlb

ft

ft

lbftQQQ

ultv

allv

fpultv

- Empirical analysis by utilizing standard penetration test (SPT) : Point bearing, Qp: Navg near pile tip = (8+12+14+14)/4=12 σ’v near pile tip = 440+(125-62.5)30=2315lb/ft2=1.15t/ft2 Correction for depth of N values, 0.115.1/20log77.0 10 NC

Therefore ; 12120.1 NCN N



65

And ppfp ANADBNQ 4/4.0

tonsBADNQ pfp 1131/785.030124.0/4.0 > than

tonsAN p 7.37785.01244

Therefore use Qp=38 tons = 76kips Shaft friction, Qf: Navg along shaft = (4+6+6+8+12)/5=7.2 say 7 And tsfNf s 150/

tsftsfNf s 114.050/750/ ; tonpLfQ sf 2.133014.314.0

Therefore :

)3.151..(343

4.102

4.102)2.1338(

kNkipsFS

QQ

kipstonsQQQ

ultv

allv

fpultv

3. Number of piles and their arrangements The number of piles required to support 236kip vessel load :

9.634

236

allv

va

Q

Qn

Try a group of 9 piles (Figure 3.37); Piles at 4ft center-to-center A 10ft x 10ft pile cap is required Assume pile cap = 3ft thick Pile cap width, b = 10ft Outer periphery, ftbb 91101

(see Figure 9.34)

settlementpredictinginusedbewill

kipsQ

kipston

kiptonQ

allf

allp

..........

8.83/22.13

3.253/1

238

Figure 3.37



66

Pile cap weight = (3 x 10 x 10)ft3 x 0.15kip/ft3 = 45 kips Total weight = 236 + 45 = 281 kips Load per pile = 281/9 = 31kips<34kips OK Pile group capa. = 34 x 9 = 306kips>281kips OK 4. Settlement of single pile Semiempirical Method St=Ss+Sp+Sps

Where :

pp

faspa

sEA

LQQS

and

faactualf

paactualp

QkipsQ

QkipsQ

834/318.8

2334/313.25

Ep=30 x 106psi; αs= =0.5

in

EA

LQQS

pp

faspa

s 012.0

10305.10124

1000123085.023

622

; Ap=26.5 in2

p

pap

pqB

QCS

and Cp=0.03; Qpa=23kips; B=12in; Ap=113.09 in2

qp =Qp/Ap=76/113.09=0.672kip/in2; 2209.113124/ inAp

ininkipin

kips

qB

QCS

p

pap

p 086.0/672.012

2303.02

pf

fas

psqD

QCS and Qfa=8kips; Df=30x12in; qp=0.672kips/in2

054.003.012

123016.093.016.093.0

p

f

s CB

DC

in

inkipin

kips

qD

QCS

pf

fas

ps 0018.0/672.01230

8054.02

; Ap=113.09 in2

Therefore : St=Ss+Sp+Sps=0.012in+0.086in+0.0018in=0.0998in Say 0.1in (2.5mm)



67

Empirical Method

)35.3..(134.0014.012.0

/10305.26

/100036031

100

12

100 262

mmin

inlbin

kiplbinkipsin

EA

LQBS

pp

va

t

From the two results consider the larger : settlement for a single pile St=0.134 in. 5. Settlement of pile groups in cohesionless soils With B=1ft; ftb 9 ; n=9 piles; within zone of influence of 9 ft;

Navg=(12+14+14)/3≈13; group load, Qg=281kips; Total settlement of single pile; St=0.134 in;

By Vesic’s : )10(4.0....402.01/9134.0/ mminsayftftinBbSS tG

By Meyerhof’s (SPT) method :

Where :

22 /74.1/47.399

281fttonsftkips

bb

Qp G

58.0

98

3018/1

bDI f

)13(5.013

58.0974.122 mmin

N

IbpSG

The larger is SG=0.5in(13mm) < allowable settlement, Sa=0.6in Therefore OK.. 3.24 Distribution of load in pile groups The load on any particular pile within a group may be computed by using the elastic equation :

22 y

yM

x

xM

n

QQ xy

m

Where : Qm – axial load on any pile m Q – total vertical load acting at the centroid of the pile

group n - number of piles Mx, My - moment with respect to x and y axis respectively



68

x, y - distance from pile to y and x axes respectively Example 3.18 Given : A pile cap consists of 9 pile as in Figure 3.38. A column load of 2250 kN acts vertically on point A. Required : Load on pile 1,6 and 8.

Solution :

22 y

yM

x

xM

n

QQ xy

m

Q=2250kN; n=9 222 616 mmx

222 616 mmy

mkNkNM x .9004.02250

mkNkNM y .5.56225.02250

Load on pile no. 1:

kNm

mmkN

m

mmkNQ 25.306

6

1.900

6

1.5.562

9

2250221

Figure 3.38



69

Load on pile no. 6:

kNm

mkN

m

mmkNQ 75.343

6

0.900

6

1.5.562

9

2250226

Load on pile no. 8:

kNm

mmkN

m

mkNQ 100

6

1.900

6

0.5.562

9

2250226

Figure 3.39



70

Example 3.19 Given : A pile cap with five piles. The pile cap is subjected to a 900 kN vertical load and a moment with respect to the y axis of 190 kN.m, Figure 3.39. Required : Shear and bending moment on section a-a due to the pile reacting under the pile cap. Solution :

Q=1000kN; n=5; mkNM y .190 ; mkNM x .0 ;

222 414 mmx

kN

y

ymkN

m

mmkNQQ 5.247

.0

4

1.190

5

10002242

Shear at a-a : (247.5kN)(2) = 495kN Moment at a-a : (2)(247.5kN)(1m-0.3m) = 173 kN.m (Draw free body diagram of the pile cap and take summation of shear and moment at section a-a)

Example 3.20 Given : A pile group consists of four friction piles in cohesive soil, Figure 3.40. Each pile’s diameter is 300 mm and center-to-center spacing is 0.75m. Required :

(a) Block capacity of the pile group. Use safety factor of 3. (b) Allowable group capacity based on individual pile failure.

Use a factor of safety of 2, along with the Converse-Labarre equation for the pile-group efficiency.

(c) Design capacity of the pile group.



71

Figure 3.40 Solution :

(a) Block capacity: Since c-to-c spacing = 0.75 and < 0.90m; Coyle and Sulaiman, 1970 suggested :

LWNcfLWDQ cg 3.12

D=10.5m W=0.75+0.15+0.15=1.05m L=0.75+0.15+0.15=1.05m f=αc qu=200 kN/m2; c=200/2=100kN/m2; α=0.56 (Figure 3.17)

f=0.56x100=56kN/m2



72

Nc=5.14 (from Table 2.3 for shallow rectangular footing for Ø=0˚- Vesic, 1973)

kN

mkN

LWNcfLWDQ cg

32067.7366.2469

05.105.114.5/1003.15605.105.15.102

3.12

2

Allowable block capacity kN10693

3206

(b) based on individual pile

kNQkNQThus

kNmkNAcNQ

kNmmAfQ

QQQ

allult

tipctip

surfaces

tipsult

3092

618;61864554

644

3.09/100

5549.9565.103.056

2

2*

With : n=2, m=2, θ=tan-1(1/2.5)=21.8˚

758.02290

2122128.211

90

111

mn

nmmnEg

Qall for group (based on individual pile) : kNkNQ allg 937758.04309)(

(c) Design capacity of group is the smaller of two = 937kN

(even using FS=2)



73

3.25 Conventional rigid method Example 3.21 The allowable bearing capacity of vertical pile ( length 12 m and 30 cm in diameter )

against vertical load = 120 kN, against horizontal load = 30 kN dan 65 kN against pull

out load, Figure 3.41.

That pile group will retain vertical load V = 1500 kN, horizontal load H = 300 kN and

momen = 150 kNm at the centroid of the pile group. Design the proper pile lay out to

retain those of external load. For stability control, use this formula (conventional rigid

method):

22

][][

y

yy

x

xx

ne

eVeM

e

eVeM

n

VS

take 1.0 m

Answer

4.2 m

yy

Try this lay out : a b c d

1

ey 4.2 m

xx 2

3

4 a b c d

ex

Figure 3.41

Answer

Number of piles = 1500 / 120 ~ 12 ; 300 / 30 ~ 10 ; 150 / 65 ~ 3

Efficiency take 0.7, so number of pile = 12/0.7 = 16 piles

2 d = 2 x 0.3 = 0.6 m ( minimum length for pile to edge of pile cap )

take 0.6 m

3 d = 3 x 0.3 = 0.9 m ( minimum length for centre to centre of pile )



74

ey1a = ey1b =ey1c =ey1d = ey4a = ey4b = ey4c = ey4d =1.5 m

ey1a2 = 2.25

ey2a = ey2b =ey2c =ey2d =ey3a =ey3b =ey3c =ey3d = 0.5 m

ey2a2 = 0.25

ey2 = 20

Mx only and V positioned at the centroid, formula is simplified to

2

][

x

xn

e

eM

n

VS

Q1a = 1500 / 16 150 x 1.5 / 20 = 93.75 – 11.25 = 82.5 kN < 120 OK

Q1d = 93.75 + 11.25 = 105 kN < 120 OK

Check all the piles !

Check stability to obtain how much the external load imposed to each piles, then each

piles should be compared to allowable bearing capacity.

ex1a = ex2a =ex3a =ex4a = ex1d = ex2d = ex3d = ex4d =1.5 m

ex1a2 = 2.25

ex1b = ex2b =ex3b =ex4b =ex1c =ex2c =ex3c =ex4c = 0.5 m

ex1b2 = 0.25

ex2 = 8x2.25 + 8 x 0.25 = 18 + 2 = 20



75

3.26 DESIGN AND ANALYSIS OF PILE UNDER LATERAL STATIC LOADS (CASE FROM PRAKASH AND SHARMA)

BRINCH HANSEN’S METHOD The ultimate soil reaction at any depth is given by equation (6.3), cqvxxu cKKP

For cohesionless soil, equation becomes: qvxxu KP

Where; vx is the effective vertical overburden pressure at depth x and coefficient qK and Kc is determined from Figure 3.42.

The procedure for calculating ultimate lateral resistance consists of the following steps: 1. Divide the soil profile into a number of layers. 2. Determine σvx and Kq and Kc for each layer and then calculate Pxu for

each layer and plot it with depth. 3. Assume a point of rotation at depth xr below ground and take the moment about the point of application of lateral load Qu (Figure 6.2). 4. If this moment is small or near zero, then xr is the right value. If not, repeat steps (1) through (3) until the moment is near zero. 5. Once xr (the depth of the point of rotation) is known, take moment about the point (center) of rotation and calculate Qu. This method is illustrated in Example 3.22.



76

EXAMPLE 3.22 A 20 - ft (6.0m) long 20 - in. (500mm) - diameter concrete pile is instated into sand that has Ø’ = 30' and γ = 120 lb/ft3 (I920kg/m3). The modulus of elasticity of concrete is 5 x 105 kips/ft2 (24 x 106 kN/m2). The pile is 15 ft (4.5 m) into the ground and 5 ft (1.5 m) above ground. The water table is near ground surface. Calculate the ultimate and the allowable lateral resistance by Brinch Hansen’s method.

Figure 3.42 Coefficients Kq and Kc (Brinch Hansen, 1961)



77

SOLUTION (a) Divide the soil profile in five equal layers, 3 ft long each (Figure 6.8). (b) Determine σvx

σvx = γ’x = (120 – 62.5) x = 0.0575 x kips/ft2 1000

Where x is measured downwards from the ground level. For each of the five soil layers, calculations for σvx and pxu are carried out as

0.6



78

shown in Table 6.1. pxu is plotted with depth in Figure 6.8. The values for pxu at the middle of each layer are shown by a solid dot. (c) Assume the point of rotation at 9.0 ft below ground level and take moment about the point of application of lateral load, Qu. Each layer is 3 ft thick, which

Gives: ∑ M = 0.6 x 3 x 6.5 + 2 x 3 x 9.5 + 3.8 x 3 x 12.5 – 5.9 x 3 x 15.5 - 8 x 3 x 18.5 = 11.7 + 57 + 142.50 - 274.35 - 444 = 211.2 - 71 8.35 = - 507.2 kip-ft/ft width Where : (0.6 - from center point) x (3 – thickness of each layer) x (6.5 – distance from center to Qu) (d) This is not near zero; therefore, carry out a second trial by assuming a point of rotation at 12ft below ground. Then, using the above numbers,



79

∑ M = 11.7 + 57 + 142.50 + 274.35 - 444 = 41.6 kip ft/ft The remainder is now a small number and is closer to zero. Therefore, the point of rotation xr can be taken at 12 ft below ground. (e) Take the moment about the center of rotation to determine Qu: Qu(5 + 12) = 0.6 x 3 x 10.5 + 2 x 3 x 7.5 + 3.8 x 3 x 4.5 + 5.9 x 3 x

1.5 – 8 x 3 x 1.5 = 18.9 + 45 + 51.3 + 26.55-36 = 105.8 Qult = 105.8/17 = 6.2 kips/ft width Qult = 6.2 x B = 6.2 x 1.67 =10.4 kips

(where B = 20 in. = 1.67 ft) Qall = 10.4/2.5 = 4.2 kips using a factor of safety 2.5

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Ch3 pilefoundations