CH3 pile foundations.doc

JABATAN KEJURUTERAAN INFRASTRUKTUR DAN GEOMATIK FAKULTI KEJURUTERAAN AWAM DAN ALAM SEKITAR

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 :

Where : As – cross-sectional area of steelfs – 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 Piles2. Concrete piles

Two categories of concrete piles are (a) precast and (b) cast-in-situ

Precast piles, Figure 3.2:

- prepared with ordinary reinforcement- in the shape of square or octagonal

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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- casing is driven to the desired depth, and filled with fresh concrete later gradually withdrawn- with or without pedestal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 :uncased pile : where : As – cross sectional area of steel

Ac - cross sectional area of concretefs – allowable stress of steelfc - allowable stress of concrete

Figure 3.3 Cast in place concrete piles3. Timber piles

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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 :

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 :

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Figure 3.6 Conditions for use of pile foundations

- 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

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

where :

Qp – load carried at the pile pointQs – load carried by skin friction developed at the side of the pile

- piles can be extended into hard stratum with (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 :

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

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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 3.8ao raised by a winch, and allowed to drop at a certain

height Ho slow rate of hammer blows

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

drops by gravity

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

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

- diesel hammer, Figure 3.8do consist of ram, an anvil block and a fuel-injection

systemo 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 :

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Where :

- increase in pile load Δz – increase in depthP – 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

3.5 Equations for Estimating Pile Capacity

Ultimate load-carrying capacity of pile, Qu is :

Where :

Qp – load-carrying capacity of the pile pointQs – frictional resistance

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Figure 3.9 Load transfer mechanism for piles

Figure 3.10 Load transfer curves for a concrete bored pile, Woo and Juang (1975)


Point bearing capacity, Qp is :

Where :

Ap – cross sectional area of pile tip c – cohesion of the soil supporting the pile tip qp - unit point cohesionq’ =γ’L – effective vertical stress at the level of the

pile tipL- pile length - the bearing capacity factors

Frictional resistance, Qs is :

Where :

p – perimeter of the pile sectionΔL – 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

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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

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

and

The limiting point resistance is :

SI unit : ; or English

Where : Ø – soil friction angle in the bearing stratum

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Using SPT method (Meyerhof, 1976):

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)

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

3.7 Vesic’s Method – Estimation of Qp

Vesic (1977) proposed value of Qp as :

Where :

- mean normal ground effective stress = Ko – earth pressure coefficient = 1 – sin Ø

- 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 :

Where :

- bearing capacity factors, Figure 9.14

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Figure 3.12 Nature of variation of unit point resistance in sand

Figure 3.13 Variation of the maximum values of with Ø


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

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 :

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

- bearing capacity factor, Figure 3.15

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Figure 3.15 Variation of 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, 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’ :For z = L’ to L :

Where :K – effective earth coefficient

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- 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 :

Where :

- mean effective vertical stress for entire embedment length,

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

- total frictional resistance is :

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Figure 3.16a Critical embedment ratio and bearing capacity factors for various soil friction angles, (Meyerhof,

1976).

Figure 3.16b Variation of with pile embedment length and its application, (McCleland – 1974).

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2. Method :

- unit skin resistance in clayey soil is : - empirical adhesion factor, Figure 3.17

Figure 3.17 Variation of with undrained cohesion of clay


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 :

Where :

- vertical effective stress = γ’z

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α


ØR – drained friction angle of remolded clayK – earth pressure coefficient Where : for normally consolidated clays

for overly consolidated clays


3.12 Point Bearing Capacity of Piles Resting on Rock

Goodman (1980) has approximate the ultimate unit point resistance in rock as :

Where :

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

After taking care of scale effect, 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

SandstoneLimestone

ShaleGraniteMarble

10,000 – 20,00015,000 – 30,0005,000 – 10,000

20,000 – 30,0008,500 – 10,000

70 – 140105 – 21035 – 70

140 – 21060 – 70

Table 3.2 Typical Values of angle of friction, Ø, of rocksRock type Angle of friction, ØSandtoneLimestone

ShaleGraniteMarble

27 – 4530 – 4010 – 2040 – 5025 - 30

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

Table 3.3 Typical pre-stressed concrete pile in use

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

Example 3.1

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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 and if K=1.3 and . 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

Qp = 390 kN Vesic’s : use ; with Ø=35°; so :

Janbu’s : with c=0; use

SPT method :

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Limiting value =

For design purpose :

b. from sub-topic 3.10 from the note :

For z = 0 : For z = L’ to L :

Thus :

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

d. Coyle and Castello’s

For Ø=35° and L/D=39.3; K≈1.0Thus :

And

Example 3.2

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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,

b. Skin resistance, Qs :(α method) :

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

(λ method) : where

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

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

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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Hence;

(β method) : where ØR =30°; ; ;

For z=0-5m :

For z=5-10m :

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

so

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

Example 3.3Given : 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 ; and

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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/m2 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/m 3 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);

(10 marks)Answer:-

To determine Qp:-

qtip = cu Nc = 100 kN/m2 x 9 = 900 kN/m2 [1M]

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

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

To determine Qs:-

[1M]

Elevation (m) Effective Vertical Pressure (kN/m2)0 02 268 45.1418 117.04

[1M]

[1M]

Based on Figure 1, l = 0.185 [1M]

= (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 = h 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, h>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 = å (Qv)ult

[2M]

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

[2 x 0.5M = 1M

3.13Pile 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,

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- When Q = Q2 : Net settlement,

Where :

snet – net settlementse – elastic settlement of the pile itselfst – 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

3.14Failure 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.

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

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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.

3.15Pile 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 :

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 :

Example 3.4A precast concrete pile 12 in. x 12 in. in cross section is driven by a hammer. Given :

Maximum rated hammer energy = 30 kip-ftHammer efficiency = 0.8Weight of ram = 7.5 kip

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Pile length = 80 ftCoefficient of restitution = 0.4Weight of pile cap = 550 lbEp = 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 =

and

b.

Use Ep = 3 x 106 lb/in2

And

c.

3.16Hiley’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 :

And

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 restitutionCc - Temporary compression coefficient due to pile

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

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

or quake (mm), Table 3.3Note :

(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.

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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 cap0.05 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

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point

Note :

Length, L measure in feet1 feet = 0.3048 m1 inch = 25.4 mm

Table 3.7 : Coefficient of restitution, e.

Description Coefficient of restitution, ePiles 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

0.50.5

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

0.250.00

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%

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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)

Value of C :Cc = 0.15in X 25.4 = 3.81 mm

Cp = 0.006(59ft) = 0.354in X 25.4 = 8.99 mmCq = 0.10in X 25.4 = 2.54 mm

C = Cc + Cp + Cq = 3.81 + 8.99 + 2.54 = 15.34 mm

Using

Example 3.6

Given :

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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 =25kNFactor of safety, FS =2.0Pile length, l =24mMass Driving assembly,P2=3.0 kN

Drop hammer, h =500mmHammer efficiency, B =85%Set value, S =19mm/10 blow (Figure 3.20)

Figure 3.20Required : 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=26.04kNValue of e = 0.25 (Table 3.7)Cp + Cq = 20mm (Figure 3.20)Cc = 0.22inx25.4=5.59mmTemporary compression, C= 5.59 + 20 = 25.59mmSet value, s = 19mm/10 blow =1.9mm/blow

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By using Hiley’s equation :

Therefore, the pile working load must be less than 188.6kN

3.17Settlement of Piles, Vesics (1969)

Settlement of a pile under vertical working load, Qw is :

Where :s – total pile settlements1 – elastic settlement of piles2 – settlement caused by the load at the pile tips3 – settlement caused by the load transmitted along

pile shaft

Formulae :- elastic settlement, s1 :

Where :Qwp – load carried at the pile point under working

conditionQws – load carried by frictional resistance under work

loadAp – area of pile cross sectionL – length of pileEp – modulus of elasticity of the pile material - nature of unit skin friction (=0.5 or 0.67), Figure 3.21

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Figure 3.21 Various types of unit friction resistance along pile shaft

- load at pile point, s2 :

Where :

D – width or diameter of pileqwp – point load per unit area = Qwp/ApEs – modulus of elasticity of soil at or below the pile

pointμs – Poisson’s ratio of soilIwp – influence factor = 0.85

Or

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

Table 3.8 Typical Values of Cp

Soil type Driven Pile Bored PileSand (dense to loose)Clay (stiff to soft)Silt (dense to loose)

0.02-0.040.02-0.030.03-0.05

0.09-0.180.03-0.060.09-0.12

- load carried by pile shaft, s3 :

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Where :p – perimeter of the pileL – embedded length of pileIws – influence factor = Or

and Cs (a constant) = 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 :

Due to point load :

Due to skin :

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With

And

Therefore the total settlement is :

3.18Pullout Resistance of Piles

The gross ultimate resistance of a pile subjected to uplifting force, Figure 3.22 is :

Where :Tug – gross uplift capacityTun – net uplift capacityW – effective weight of pile

Figure 3.22 Uplift capacity of piles

a. In Clay

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Das and Seeley (1982), estimated Tun as :

Where :

L – length of the pilep – perimeter of pile section’- adhesion coefficient at soil-pile interface

cu – undrained cohesion of clay

Values of :

- for cast-in-situ: (for bore pile) for cu ≤ 80 kN/m2

for cu > 80kN/m2

- for pipe piles : for cu ≤ 27 kN/m2

for cu > 27 kN/m2

b. In Sand

Das and Seeley (1975), estimated Tun as :

with fu varies by 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 :

with values Ku and from Figure 3.23b&c

43


- if L > Lcr then :

with values Ku and from Figure 3.23b&c

Where :

Ku – uplift coefficient - effective vertical stress at a depth z

- soil-pile friction

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

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

And

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%

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

3.19Group piles efficiency

45


Converse –Labarre method of estimating pile-group efficiency developed by Jumikis, 1971 using the following equation :

Where :Eg – pile-group efficiencyθ – tan-1(d/s), (deg)n – number of piles in rowm – number of rows of pilesd – 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.

; Allowable bearing capacity of a single

pile=450kN/2=225kNDesign 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 :

Where :

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

group effect

If group as a block thus :

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 = h x n x (Qv)ult

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

48

Figure 3.25 Pile groups 3.26 ultimate Capacity of Group Piles in Saturated Clay


(when the center to center spacing, s, is large enough, h>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 = å (Qv)ult

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 MethodPile type

No. of Piles No. of adjacent piles

Reduction factor for each pile

Ultimate capacityCol.2 x Col.4

49


A 1 8 1-8/16 # 0.5QuB 4 5 1-5/16 2.75QuC 4 3 1-3/16 3.25Qu

Σ6.5Qu=Qg(u)

Note: 16 # no. of arrow Therefore, efficiency,

3.20ULTIMATE 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 individualFrom : ; and

So : (1)

As pile group (dimensions of LgxBgxL):

Point bearing capacity as :

With from Figure 3.29, thus :

(2)

Where :

and

The lower value from (1) and (2) is Q g(u)

50


Figure 3.29 Ultimate group piles in clay

Figure 3.30 Variation of 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: 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

If pile as a group :

52

70 kN/m2

40 kN/m25 m

10 m

1.22 m


From Figure 3.29: (assuming : that at the end of curve at right hand stays horizontal)Thus :

Hence, ΣQu=9677 kN,

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

Figure 3.32

53

o L=depth of pile embedmento Qg – total load of superstructure (–) weight of soil

excavatedo 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 deptho Line a-a’ and bb’ are two 2:1 lineso Stress increased at the middle of each soil layer :

o Lg and Bg – the length and width of pile groupo zi – distance from z=0 to the middle of clay layero For layer 2 : zi=L1/2; For layer 3 : zi=L1+L2/2o For layer 4 : zi=L1+L2+L3/2

o Consolidation Settlement,

o Where : ;

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

o Total consolidation settlement,


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 :

With and ;

54


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,Bg – width of pile groupD – width or diameter of each pile in the groups = 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,

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 ≥ 0.5L – length of pile embedment

Example 3.13 (Cumulative)

55


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);

(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 03 3x15=456 45 + 3(15-9.81) = 60.5718 60.57 + 12(18-9.81) = 158.85

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

= (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

3.23 Calculation of single, group pile capacity and

settlement from Prakash & Sharma - for sandy soil.

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

Table 3.10Ø 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 3.11Pile type KsBored pileDriven H pileDriven displacement pile

0.50.5 – 1.01.0 – 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 :

Nq=60, Table 3.10; Ks=1.0, Table 3.11; Using the formula of the ultimate capacity :

Where :

58


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 :

Thus :

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 Pipe inside diameterArea of steel section

1. Semiempirical method :

59


From the relation of : (from Example 3.14)

And (Qv)all=(Qv)ult/FS=93.83/3=31kips

Assuming allowable loads are actual loads; then

Due to material :

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 :

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

CsDue to skin :

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

2. Empirical method :

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Using :

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

1 ton = 2000 lbCorrection for depth of N values,

Therefore ; And

The lower of these values is Qp=37.7 tons

(d)Shaft friction, QfAverage N value along pile shaft, Navg(shaft)= (4+6+6+8+10)/5=6.8Use σ’v for average depth of L/2=30/2=15ft so

61

For driven piles :

(Meyerhof,1976)

(Meyerhof,1976)

Figure 3.34


σ’v= 0.938/2=0.469tsf Therefore ;

; So

(e)Allowable bearing capacity, Qall :

Pile group sample calculations

Settlement of pile group and check on design :

1. Vesic’s Method (1977) :

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

where :

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 capacityB=1ft; s=4ft; ; b=10ft; n=9

for a single pile (from empirical method Ex 3.15)

(b) settlementB=1ft; (square arrangement); n=9 piles; (Qg)all=281kips; zone of influence, =9ft below the group base; Navg=(12+14+14)/3≈13; for single pile st=0.134in.(EX.3.14)

1. Vesic’s (1977): 2. Meyerhof’s (1976): (N values)

63

Figure 3.35


where Df is pile length = 30 ft

So :

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


Solution :

1. Soil profile as in Figure 3.362. 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 : and

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 :

- Empirical analysis by utilizing standard penetration test (SPT) :

Point bearing, Qp:Navg near pile tip = (8+12+14+14)/4=12

65


σ’v near pile tip = 440+(125-62.5)30=2315lb/ft2=1.15t/ft2

Correction for depth of N values, Therefore ; And

> than

Therefore use Qp=38 tons = 76kips

Shaft friction, Qf:Navg along shaft = (4+6+6+8+12)/5=7.2 say 7And

; Therefore :

3. Number of piles and their arrangements

The number of piles required to support 236kip vessel load :

Try a group of 9 piles (Figure 3.37); Piles at 4ft center-to-centerA 10ft x 10ft pile cap is requiredAssume pile cap = 3ft thickPile cap width, b = 10ftOuter periphery, (see Figure 9.34)

66

Figure 3.37


Pile cap weight= (3 x 10 x 10)ft3 x 0.15kip/ft3 = 45 kipsTotal weight = 236 + 45 = 281 kipsLoad per pile = 281/9 = 31kips<34kips OKPile group capa. = 34 x 9 = 306kips>281kipsOK

4. Settlement of single pile

Semiempirical MethodSt=Ss+Sp+SpsWhere :

and Ep=30 x 106psi; αs= =0.5

; Ap=26.5 in 2

and Cp=0.03; Qpa=23kips; B=12in; Ap=113.09 in 2

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

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

; Ap=113.09 in 2

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

67


Empirical Method

From the two results consider the larger : settlement for a single pile St=0.134 in.

5. Settlement of pile groups in cohesionless soilsWith B=1ft; ; 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 : By Meyerhof’s (SPT) method : Where :

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 :

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

68


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

Q=2250kN; n=9

Load on pile no. 1:

69

Figure 3.38


Load on pile no. 6:

Load on pile no. 8:

70

Figure 3.39


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; ; ;

Shear at a-a : (247.5kN)(2) = 495kNMoment 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.

71


(c) Design capacity of the pile group.

Figure 3.40Solution :

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

D=10.5mW=0.75+0.15+0.15=1.05mL=0.75+0.15+0.15=1.05mf=αcqu=200 kN/m2; c=200/2=100kN/m2; α=0.56 (Figure 3.17)

72


f=0.56x100=56kN/m2

Nc=5.14 (from Table 2.3 for shallow rectangular footing for

Ø=0˚- Vesic, 1973)

Allowable block capacity

(b) based on individual pile

With :n=2, m=2, θ=tan-1(1/2.5)=21.8˚

Qall for group (based on individual pile) :

(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):

take 1.0 m

Answer 4.2 m yyTry this lay out : a b c d

1 ey 4.2 m xx 2

3 4 a b c d ex

Figure 3.41

AnswerNumber 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 m3 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 mey1a

2 = 2.25

ey2a = ey2b =ey2c =ey2d =ey3a =ey3b =ey3c =ey3d = 0.5 mey2a

2 = 0.25

ey2 = 20

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

Q1a = 1500 / 16 150 x 1.5 / 20 = 93.75 – 11.25 = 82.5 kN < 120 OKQ1d = 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 mex1a

2 = 2.25ex1b = ex2b =ex3b =ex4b =ex1c =ex2c =ex3c =ex4c = 0.5 mex1b

2 = 0.25 ex

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

75


3.26DESIGN 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),

For cohesionless soil, equation becomes:

Where; is the effective vertical overburden pressure at depth x and coefficient 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

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


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.

77

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


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 asshown 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.

78

0.6


(c) Assume the point of rotation at 9.0 ft below ground level and take moment about the point of application of lateral load, Q u. 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 Q u)

(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,

∑ M = 11.7 + 57 + 142.50 + 274.35 - 444 = 41.6 kip ft/ft

79


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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