Principles of Foundation Engineering

Principles of Foundation Engineering

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,..._--- -

J ;-0---I'

\

,

CONVERSION FACTORS FROM ENGLISH TO SI UNITS Len gth:

1 ft 1 ft 1 ft 1 in. 1 in.

1

Area:

Volume:

Force:

in.

1 fl' 1 ft' 1 ft' 1 in:! 1 in' 1 in:!

1 ft" ift' ,1 in] lin'

lIb lIb lib 1 kip 1 U.S. ton lib 1 Ib/n

0.3048 m 30.48 em = 304.8 mm = 0.0254 m = 2.54 cm = 25.4 mm = =

929.03 X Hr'm' 929.03 cm' 929.03 X 10' mm' 6.452 X 10-' m' 6.452 cm' = 645.16 mm'

= = = = =

28.317 X 10-3 m' = 28.317 X 10' cm" = 16.387 X 10-6 m' = 16.387 cm' =

4.448 N = 4.448 X 10-3 kN = 0.4536 kgf = 4.448 kN = 8.896 kN = 0.4536 X 10'\ metric tall = 14.593 N/m =

Stress:

11b/ft' llb/ft' 1 U.S. ton/ft' 1 kip/ft' lIb/in'

Unit weight:

1 1b/ft3 l Ib/in'

Moment:

1 lb-ft l Ib-in.

Energy:

1 ft-Ib

Moment of inertia:

1 in' 1 in' 1 in3 I in3

Section modulus: Hydraulic conductivity:

1 ft/min 1 ft/min 1 ft/min 1 ft/sec 1 fI/sec 1 in/min 1 in./sec 1 in/sec

Coefficient of consolidation:

1 in'/sec 1 in'/sec 1 ft'/sec

47.88 N/m' 0.04788 kN/m' 95.76 kN/m' 47.88 kN/m' = 6.895 kN/m'

= = = =

0.1572 kN/m' 271.43 kN/mJ = 1 .3558 N ' m = 0.11298 N· m =

=

=

= = =

=

1 .3558J

0.4162 X 106 mm' 0.4162 X 10 - 6 m'

0.16387 X 10' mm' 0.16387 X lW'm'

0.3048 m/min 30.48 cm/min = 304.8 mm/min = 0.3048 m/sec = 304.8 mm/sec = 0.0254 m/min = 2.54 cm/sec = 25.4 mm/sec =

=

=

6.452 cm'/sec

= 20.346 X 103 m'/yr = 929.03 cm'/sec

Principles of Foundation Engineering Sixth Edition

Braja

M.

Das

THOMSON

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Principles of Foundation Engineering, Sixth Edition by Braja M. Das

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Geotechnical Properties of Soil

1

Introduction 1 Grain-Size Distribution 2 Size Limits for Soils 5 Weight-Volume Relationships 5 1.5 Relative Density 9 1.6 Atterberg Limits 12 1.7 Soil Classification Systems 13 1.8 Hydraulic Conductivity of Soil 21 1.9 Steady-State Seepage 23 1.10 Effective Stress 25 1.11 Consolidation 27 1.12 Calculation of Primary Consolidation Settlement 32 1.13 Time Rate of Consolidation 33 1.14 Degree of Consolidation Under Ramp Loading 40 1.15 Shear Strength 43 1.16 Unconfined Compression Test 48 1.17 Comments on Friction Angle, '1" 49 1.18 Correlations for Undrained Shear Strength, c" 52 1.19 Sensitivity 53 Problems 54 References 58 1.1 1.2 1.3 1.4

2

Natural Soil Deposits and Subsoil Exploration 2.1

60

Introduction 60

Natural Soil Deposits 60 2.2 Soil Origin 60 vii

viii

Contents

2.3 2.4 2.5 2.6 2.7 2.8 2.9

Residual Soil

61

Gravity Transported Soil 62 Alluvial Deposits 62 Lacustrine Deposits 65 Glacial Deposits 65 Aeolian Soil Deposits 67 Organic Soil 68

Subsurface Exploration

�

68

Purpose of Subsurface Exploration 68 Subsurface Exploration Program 68 Exploratory Borings in the Field 71 Procedures for, Sampling Soil 74 Observation of Water Tables 85 Vane Shear Test 86 Cone Penetration Test 90 Pressuremeter Test (PMT) 97 Dilatometer Test 99 Coring of Rocks 101 Preparation of Boring Logs 105 Geophysical Exploration 105 Subsoil Exploration Report 113 Problems 114 References 119

2.10 2.11 2.12 2.13 2.14 2.15 2.16 2.17 2.18 2.19 2.20 2.21 2.22

Shallow Foundations: Ultimate Bearing Capacity

121

Introduction 121 General Concept 121 Terzaghi's Bearing Capacity T heory 124 Factor of Safety 128 Modification of Bearing Capacity Equations for Water Table 130 The General Bearing Capacity Equation 131 Meyerhofs Bearing Capacity, Shape, Depth, and Inclination Factors 136 Some Comments on Bearing Capacity Factor, Ny, and Shape Factors 138 A Case History for Bearing Capacity Failure 140 Effect of Soil Compressibility 142 Eccentrically Loaded Foundations 146 Ultimate Bearing Capacity under Eccentric Loading-Meyerhofs Theory 148 3.13 Eccentrically Loaded Foundation-Prakash and Saran's Theory 154 3.14 Bearing Capacity of a Continuous Foundation Subjected to Eccentric Inclined Loading 161 Problems 165 References 168

3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.11 3.12

I

Contents

ix

Ultimate Bearing Capacity of Shallow Foundations: Special Cases 1 70 Introduction 170 Foundation Supported by a Soil with a Rigid Base at Shallow Depth Bearing Capacity of Layered Soils: Stronger Soil Underlain by Weaker Soil 177 4.4 Closely Spaced Foundations-Effect on Ultimate Bearing Capacity 185 4.S Bearing Capacity of Foundations on Top of a Slope 188 4.6 Bearing Capacity of Foundations on a Slope 191 4.7 Uplift Capacity of Foundations 193 Problems 199 References 202

4.1 4.2 4.3

Shallow Foundations: Allowable Bearing Capacity and Settlement 203 203

5.1

Introduction

5.2 5.3 5.4 5.5

Stress Due to a Concentrated Load 204 Stress Due to a Circularly Loaded Area 205 Stress below a Rectangular Area 206 Average Vertical Stress Increase Due to a Rectangularly Loaded Area 213 Stress Increase under an Embankment 216

Ve�iical si�e;s increase in a Soil Mass Caused by Foundation Load 5.6

Elastic Settlement 5.7 5.8 5.9 5.10 5.11 5.U 5.13 5.14

204

220

Elastic Settlement Based on the Theory of Elasticity 220 Elastic Settlement of Foundations on Saturated Clay 230 Improved Equation for Elastic Settlement 230 Settlement of Sandy Soil: Use of Strain Influence Factor 236 Range of Material Parameters for Computing Elastic Settlement 240 Settlement of Foundation on Sand Based on Standard Penetration Resistance 241 General Comments on Elastic Settlement Prediction 246 Seismic Bearing Capacity and Settlement in Granular Soil 247

cons�lidation Settlement 5.15 5.16 5.17 5.18

..

252

Primary Consolidation Settlement Relationships 252 Three·Dimensional Effect on Primary Consolidation Settlement Settlement Due to Secondary Consolidation 258 Field Load Test

260

254

170

x

Contents 5.19 Presumptive Bearing Capacity 263 5.20 Tolerable Settlement of Buildings 264 Problems 266 References 269

6

Mat Foundations

272 .

Introduction 272 6.1 Combined Footings 272 6.2 6.3 Common Types of Mat Foundations 275 Bearing Capacity of Mat Foundations 277 6.4 6.5 Differential Settlement of Mats 280 6.6 Field Settlement Observations for Mat Foundations Compensated Foundation 281 6.7 6.S Structural Design of Mat Foundations 285 Problems 304 References 307

7

Lateral Earth Pressure 7.1 7.2

:I !

281

308

Introduction 308 Lateral Earth Pressure at Rest

309

Active Pressure 312 7.3 7.4 7.5 7.6 7.7 7.S 7.9

Rankine Active Earth Pressure 312 A Generalized Case for Rankine Active Pressure 315 Coulomb's Active Earth Pressure 323 Active Earth Pressure for Earthquake Conditions 328 Active Pressure for Wall Rotation about the Top: Braced Cut Active Earth Pressure for Translation of Retaining Wall-Granular Backfill 334 General Comments on Active Earth Pressure 338

Passive Pressure 338 Rankine Passive Earth Pressure 338 Rankine Passive Earth Pressure: Inclined Backfill 344 Coulomb's Passive Earth Pressure 345 Comments on the Failure Surface Assumption for Coulomb's Pressure Calculations 347 Passive Pressure under Earthquake Conditions 348 7.14 Problems 349 References 352

7.10 7.11 7.12 7.13

333

Contents

Retaining Walls 8.1 ""

�--. "'�

Introduction . �.

,�

".�'-'

353 353

Gravity and Cantilever Walls 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9

xi

355

Proportioning Retaining Walls 355 Application of Lateral Earth Pressure Theories to Design Stability of Retaining Walls 358 Check for Overturning 359 Check for Sliding along the Base 361 Check for Bearing Capacity Failure 364 Construction Joints and Drainage from Backfill 374 Some Comments on Design of Retaining Walls 377

356

M;Ch�;'k;Jry-ft�bTiii.;;rR:;'i;;i�i;'iWai[;'· 379 ..

8.10 Soil Reinforcement 379 8.11 Considerations in Soil Reinforcement 380 8.12 General Design Considerations 382 8.13 Retaining Walls with Metallic Strip Reinforcement 383 8.14 Step-by-Step-Design Procedure Using Metallic Strip Reinforcement 8.15 Retaining Walls with Geotextile Reinforcement 395 8.16 Retaining Walls with Geogrid Reinforcement 399 8.17 General Comments 402 Problems 404 References 407

Sheet Pile Walls 9.1 9.2

9.3 9.4 9.5 9.6 9.7

9.8 9.9 9.10 9.11 9.12 9.13 9.14 9.15 9.16

409

Introduction 409 Construction Methods 413 Cantilever Sheet Pile Walls 414 Cantilever Sheet Piling Penetrating Sandy Soils 415 Special Cases for Cantilever Walls Penetrating a Sandy Soil 422 Cantilever Sheet Piling Penetrating Clay 423 Special Cases for Cantilever Walls Penetrating Clay 428 Anchored Sheet-Pile Walls 429 Free Earth Support Method for Penetration of Sandy Soil 430 Design Charts for Free Earth Support Method (Penetration into Sandy Soil) 435 Moment Reduction for Anchored Sheet-Pile Walls 440 Computational Pressure Diagram Method for Penetration into Sandy Soil 443 Field Observations of an Anchored Sheet Pile Wall 447 Free Earth Support Method for Penetration of Clay 448 Anchors 452 Holding Capacity of Anchor Plates in Sand 454

390

xii

Contents 9.17 Holding Capacity of Anchor Plates in Clay 9.18 Ultimate Resistance of Tiebacks 460 Problems 461 References 464

10

11

Braced Cuts

(cp = 0 Condition)

460

466

10.1 Introduction 466 Pressure Envelope for Braced-Cut Design 467 10.2 Pressure Envelope for Cuts in Layered Soil 471 10.3 10.4 Design of Various Components of a Braced Cut 472 10.5 Bottom Heave of a Cut in Clay 482 10.6 Stability of the Bottom of a Cut in Sand 485 10.7 Lateral Yielding of Sheet Piles and Ground Settlement Problems 489 References 490

Pile Foundations

487

491

11.1 11.2 11.3 11.4 11.5 11.6 11.7 11.8 11.9 11.10 11.11 11.12 11.13 11.14 11.15 11.16 11.17 11.18 11.19 11.20 11.21

Introduction 491 Types of Piles and Their Structural Characteristics 493 Estimating Pile Length 502 Installation of Piles 504 Load Transfer Mechanism 508 Equations for Estimating Pile Capacity 509 Meyerhofs Method for Estimating Qp 512 Vesic's Method for Estimating Qp 515 Janbu's Method for Estimating Qp 516 Coyle and Castello's Method for Estimating Qp in Sand 520 Other Correlations for Calculating Qp with SPT and CPT Results Frictional Resistance (Q,) in Sand 524 Frictional (Skin) Resistance in Clay 528 Point-Bearing Capacity of Piles Resting on Rock 531 Pile Load Tests 538 Comparison of Theory with Field Load Test Results 542 Elastic Settlement of Piles 543

Group

Pile;

11.22 11.23 11.24

Group Efficiency 573 Ultimate Capacity of Group Piles in Saturated Clay Elastic Settlement of Group Piles 580

Laterally Loaded Piles 546 Pile-Driving Formulas 562 Pile Capacity For Vibration-Driven Piles Negative Skin Friction 570

521

568

..

573 576

Contents 11.25 Consolidation Settlement of Group Piles 11.26 Piles in Rock 584 Problems 584 References 588

· 12

Drilled-Shaft Foundations

59 1

Introduction 591 Types of Drilled Shafts 592 Construction Procedures 593 Other Design Considerations 598 Load Transfer Mechanism 598 Estimation of Load-Bearing Capacity 599 Drilled Shafts in Granular Soil: Load-Bearing Capacity Drilled Shafts in Clay: Load-Bearing Capacity 613 Settlement of Drilled Shafts at Working Load 620 Lateral Load-Carrying Capacity-Characteristic Load and Moment Method 622 U.11 Drilled Shafts Extending into Rock 631 Problems 635 References 639

U.l U.2 U.3 U.4 U.5 U.6 U.7 U.8 U.9 U.I0

,-

13

Foundations on Difficult Soils 13.1

Introduction

Collapsible Soil 13.2 13.3 13.4 13.5 13.6

13.7 13.8 ' 13.9 13.10 13.11

640

640

640

649

General Nature of Expansive Soils 649 Laboratory Measurement of Swell 650 Classification of Expansive Soil on the Basis of Index Tests Foundation Considerations for Expansive Soils 656 Construction on Expansive Soils' 661

Sanitary Landfills

665

13.U General Nature of Sanitary Landfills 665 13.13 Settlement of Sanitary Landfills 666 Problems 668 References 670

(

/

602

Definition and Types of Collapsible Soil 640 Physical Parameters for Identification 641 Procedure for Calculating Collapse Settlement 645 Foundation Design in Soils Not Susceptible to Wetting 646 Foundation Design in Soils Susceptible to Wetting 648

Expansive Soils

I

581

655

xiii

xiv

Contents

14.;

Soi/lmprovement and Ground Modification 14.1 14.2 14.3 14.4 14.5 14.6 14.7 14.8 14.9 14.10 14.11 14.12 14.13 14.14 14.15 14.16

672

Introduction 672 General Principles of Compaction 673 Correction for Compaction of Soils with Oversized Particles Field Compaction 678 Compaction Control for Clay Hydraulic Barriers 681 Vibroflotation 682 Precompression 688 Sand Drains 696 Prefabricated Vertical Drains 706

676

Lime Stabilization 711 Cement Stabilization 714 F ly-Ash Stabilization 716 Stone Columns 717 Sand Compaction Piles 722 Dynamic Compaction 725 Jet Grouting 727 Problems 729 References 731

Appendix A

735

Answers to Selected Problems Index

740

745

..

Soil mechanics and foundation engineering have rapidly developed during the last fifty years. Intensive research and observation in the field and the laboratory have refined and improved the science of foundation design. Originally published in 1984, this text on the principles of foundation engineering is now in the sixth edi­ tion. The use of this text throughout the world has increased greatly over the years; it also has been translated into several languages. New and improved materials that have been published in varibus geotechnical engineering journals and conference proceedings have been continuously incorporated into each edition of the text. Principles of FULI/.daciall Ellgilleering is intended primarily for undergraduate civil engineering students. The first chapter, on geotechnical properties of soil, re­ views the topics covered in the introductory soil mechanics course, which is a pre­ requisite for the foundation engineering course. The text is comprised of fourteen chapters with examples and problems, one appendix, and an answer section for se­ lected problems. The chapters are mostly devoted to the geotechnical aspects of foundation design. Systeme International (SI) units and English units are used in the text. Because the text introduces civil engineering students to the application of the fundamental concepts of foundation analysis and design, the mathematical deriva­ tions are not always presented . Instead, just the final form of the equation is given. A list of references for further inforI]1ation and study is included at the end of each chapter. Example problems that will help students understand the application of vari­ ous equations and graphs are given in each chapter. A number of practice problems also are given at the end of each chapter. Answers to some of these problems are given at the end of the text. Following is a brief overview of the changes from the fifth edition. •

•

L ,

Seve�al tables were changed to graphical form to make interpolation conve­ nient. Chapter 1 on geotechnical properties of soil has several additional empirical re­ lationships for the compression index. Also included are the time factor and xv

xvi

Preface

•

average degree of consolidation relationships for initial pore water pressure increasi ng and having trapezoidal. sinusoidal. and triangular shapes. . hapte r 2 on natural deposits and subsoil exploratton has an expanded descnp­ lion on natural soil deposits. Also incorporat ed mto thiS chapter are several re­ cently developed correlations between:

-

1t v,

-1

(aj

V(I+ V". + Vs= V lV".+ Ws= W

,-----,

�ht

t'�O

t w,

-1

V,,=wG.,

t

V

i =

1

(n) Unsaturated soil;

Figure 1.3

Vs = I

(c) Saturated soil; Vs=

I

Weight-volume relationships

where Vu = volume of voids V, = volume of soil solids The porosity, 11, is the ratio of the volume of voids to the volume of the soil specimen. or V,. 11:::: (1.5) V where V = total volume of soil

1.4

Weight-Volume Relationships

7

Moreover, V,.

n

=

' v,. = v,. = -::-:-_v."-= V v,+v,. V, v,. -+V, v,

=

e l+e

(1.6)

The degree of saturation, S, is the ratio of the volume of water in the void spaces to the volume of voids, generally expressed as a percentage, or S(%)

=

�

v

Vv

X 100

(1.7)

where v'v = volume of water Note that, for saturated soils, the degree of saturation is 100%. The weight relationships are moisture content, moist unit weight, dry unit weight, and saturated unit weight, often defined as follows: .

MOIsture content = w ( %) = where

Ww X 100 W ,

(1.8)

W, = weight of the soil solids "'" = weight of water ,

. . MOl· st umt welght = 'Y = where

V W

(1.9)

W = total weight of the soil specimen = W, + W".

The weight of air, W.. in the soil mass is assumed to be negligible. Dry unit weight = Yd =

v'

W

(1.10)

When a soil mass is completely saturated (i.e., all the void volume is occupied by water), the moist unit weight of a soil [Eq. (1.9)] becomes equal to the saturated unit weight (y,,, ) . So y = y", if V,. = \1;,.. More useful relations can now be developed by considering a representative soil specimen in which the volume of soil solids is equal to u/lity, as shown in Figure 1.3b. Note that if V, = 1 , then, from Eq. (1.4), Y" = e, and the weight of the soil solids is where G, = specific gravity of soil solids y". = unit weight of water (9.81 kNjm'. or 62.4 Ib/ft')

8

Chapter

1

Geotechnical Properties of Soil

Also, from Eq. (1.8), the weight of water W,.. = wW,. Thus, for the soil specimen un­ der consideration. W.,. = wW, = wG;y.,.. Now, for the general relation for moist unit weight given in Eq. (1.9), W W, + W;, G;yw( 1 + w) = y = -= 1 +e V,+v" V

(1.11)

Similarly, the dry unit weight [Eq. (1.10)] is Yd

G,yw W, W, =-= = -V V,+Vv l+e

(1.12)

From Egs. (1.11) and (1.12), note that

Y I'll = 1 + 'W

(1.13)

If a soil specimen is completely saturated. as shown in Figure 1.3c. then \/� :-::

Also, for this case,

t:

Thus, e = wG,

( for saturated soil only)

(1.14)

The saturated unit weight of soil then becomes G,1., +lIYw' . .... . "", i:.W.. V. = l' +' ' v. , , :y.;,�. ,' .' ., -, +'

(1.15)

Relationships similar to Eqs. (1.11), (1.12), and (1.15) in terms of porosity can also be obtained by considering a representative soil specimen with a unit volume. These relationships are

and

Y = G,y".(l - n) (1 + 10) Yd = ( 1 - n)G,y.,.

y,,, = [(1

- Il

)G,

+ n]y".

(1.16) (1.17) (1.18)

1.5

Table 1.3

Relative Density

9

Specific Gravities of Some Soils G.

Typo of Soil

2.64-2.66 2.67-2.73 2.70-2.9 2.60-2.75 2.65-2.73 1.30-1.9

Quartz sand Silt Clay Chalk Loess Peat

Except for peat and highly organic soils, the general range of the values of spe­ cific gravity of soil solids (G,) found in nature is rather small. Table 1.3 gives some representative values. For practical purposes, a reasonable value can be assumed in lieu of running a test.

Relative Density In granular soils, the degree of compaction in the field can be measured according to the relative density, defined as (1.19) where erna, = void ratio of the soil in the loosest state emin = void ratio in the densest state e = in situ void ratio

The relative density can also be expressed in terms of dry unit weight, or )

D, ( 0/0 =

{

Yd - Yd{m',) 1'd(max) - I'd(min)

-- X 100

}Yd(m,,) Yd

(1.20)

where

= in situ dry unit weight = dry unit weight in the densest state; that is, when the void ratio is em" m ) = dry unit weight in the loosest state; that is, when the void ratio is em" Yd( in Yd

Yd (m",)

The denseness of a granular soil is sometimes related to the soil's relative den­ sity. Table 1.4 gives a general correlation of the denseness and D,. For naturally oc­ curring sands, the magnitudes of em" and em" [Eq. (1.19)] may vary widely. The main reasons for such wide variations are the uniformity coefficient, Cu, and the round­ neSS of the particles.

10

Chapter

1

Geotechnical Properties of Soil

Table 1.4

Denseness of a Granular Soil

Relative density, 0,(%)

Description

0-20 20-40 40-60 60-80 80-100

Very loose Loose Medium Dense Very dense

1.1 A 0.25 ft3 moist soil weighs'3().8 lb. Given G, = 2.7. Determine the a.

Moist unit weight, l'

b. Moisture content, W c.

Dry unit weight, I'd

d. Void ratio, e e.

Porosity, n

f. Degree of saturation, S

Solution

From Eq.(f.?);�·

.

.: "�

Parta,.Moist UnitWeigtit

.'

'.

.

.

Part b: Moisture.Content .

From Eq. (1.8),

Part c: Dry UnitWei, gb . t

From Eq. (1.10),

....

7.5

Part d: Void Ratio From Eq. (1.4),

Relative Density

11

V,

e=V, W, 28.2 = = V; G,,,/w (2.67) (62.4) e=

0.25 - 0.169 0.169 .

=

0.169 ft3

0.479

fl�rte; Porosity From Eq. (1.6),

n From Eq. (1.7),

0.479 e = + = 0.324 e 1 0.479 1+

=

Partf: Degree of Saturation

v" S(%) = v.-x 100 v,. =

V

w

s I

I

=

=

v

= 0.25 - 0.169

v - v,

Woo 111'

=

0.042 0.081

=

0.081 ft'

30.8 - 28.2 = 0'042 ft3 62.4

x

100= 51

•

•

9%

Example 1.2

A soil has a void ratio of 0.72, moisture content = 12%, and G, = 2.72. Deter­

mine the a. Dry unit weight (kN/m3) b. Moist unit weight (kN/m3) Co Weight of water in kN/m' to be added to make the soil saturated

Solution Part a: Dry Unit Weight

From Eq. (1.12),

Yd -

_

3

G,,,/w -_ (2.72 ) 0( 9.81) - 15.S1kNjm _

1 +e

--

1 + .72

12

Chapter 1

Geotechnical Properties of Soil

Part b: Moist Unit Weight

From Eq. (1.11),

'Y

=

G,'Yw(1 + IV) 1 +e

(2.72)(9.81)(1 + 0.12) =

1 + 0.72

=

17.38 kN/m3

Part c: Weight of Water to be Added From Eq. (1.15), 'Y,,,

=

(G, + elY",

Water to be added

1+ e =

'Y,,,

-

'Y

(2.72 + 0.72)(9.81) =

=

1 + 0.72

19.62 - 17.38

=

=

19.62 kN/m',

2.24 kN/m3

•

Example 1.3 The maximum and minimum dry unit weights of a sand are 17.1 kN/m3 and 14.2 kN/m3, respectively. The sand in the field has a relative density of 70% with a moisture content of 8%. Determine the moist unit weight of the sand in the field. Solution

From Eq. (1.20), Dr 0.7

=

'Yd(mioJ ]['Yd(m�)] ['Yd('Ydmax)- -'Yd( min) ----:y:;[ 'Yd - )[ 'Yd ]

'Yd 'Y 'Yd

1.6

=

14.2

17.1

17.1 - 14.2

=

16.11 kN/m3

=

(1 + w )

=

( �)

16.11 1 +

1

= 17.4 kN/m3

•

Atterberg Limits When a clayey soil is mixed with an excessive amount of water. it may flow like a semi­ liqllid. If the soil is gradually dried. it will behave like a plastic. semisolid. or solid mater­ ial, depending on its moisture content. The moisture content, in percent. at which the soil changes from a liquid to a plastic state is defined as the liqllid limit (LL). Similarly. the

1.7 Soil Classification Systems ,

Solid I Semisolid , state state 1----+I ,-

I

,

Volume of the soil-water mixture

tsL

, ,

Plastic state

I

,

'Semiliquid , state ]ncrease of moisture content

I I ....--.. ---.--_ )1

-

,

, , , , , ,

, , ,

hL

tLL

, , '-__----',___--'-_____-"__ -+

Figure 1.4

13

Moisture content

Definition of Atterberg limits

moisture content, in percent, at which the soil changes from a plastic to a semisolid state and from a semisolid to a solid state are defined as the plastic limit (PL) and the shrink­ age limit (SL), respectively. These limits are referred to as Atterberg limits (Figure 1.4): The liquid limit of a soil is determined by Casagrande's liquid device (ASTM Test Designation D-4318) and is defined as the moisture content at which a groove closure of 12.7 mm (1/2 in.) occurs at 25 blows. The plastic limit is defined as the moisture content at which the soil crumbles when rolled into a thread of3.l8 mm (1/8 in.) in diameter (ASTM Test Designa­ tion D-4318). The shrinkage limit is defined as the moisture content at which the soil does not undergo any further change in volume with loss of moisture (ASTM Test Desig­ nation D-427). The difference between the liquid limit and the plastic limit of a soil is defined as the plasticity index (PI), or

•

•

•

PI

1.7

'"

LL

-:-PL

(1.21)

Soil Classification Systems Soil classification systems divide soils into groups and subgroups based on common engineering properties such as the grain-size distribution, liquid limit, and plastic limit. The two major classification systems presently in use are (1) the American Association of State Highway and Transportation Officials (AASHTO) System and (2) the Unified Soil Classification System (also ASTM). The AASHTO system is used mainly for the classification of highway subgrades. It is not used in foundation construction.

1

14

Chapter 1

Geotechnical Properties of Soil

AASHTO System

The AASHTO Soil Classification System was originally proposed by the Highway Research Board's Committee on Classification of Materials for Subgrades and Granular Type Roads (1945). According to the present form of this system, soils can be classified according to eight major groups, A-I through A-S, based on their grain­ size distribution, liquilllimit, and plasticity indices. Soils listed in groups A-I, A-2, and A-3 are coarse-grained materials, and those in groupsA-4,A-5.A-6, and A-7 are fine-grained materials. Peat. muck, and other highly organic soils are classified un­ der A-S. They are identified by visual inspection. The AASHTO classification system (for soils A-I through A-7) is presented in Table 1.5. Note that group A-7 includes two types of soil. For the A-7-5 type, the Table 1.5

AASHTO Soil Classification System Granular materials

General C,. > 3"

Fines classify as ML or MH

OM

Silty gravelJ.l�. h

More than 1 2 % finesc

Fines classify as CL or CH

GC

Less than 5',�;, finesd

Liquid limit less than 50

Sills and Clays

Silts and Clays

blf field sample contained cobbles or boulders, or

SW

Poorly graded sand'

Fines classify as ML or MH

SM

Silty sand�·I1·1

Fines classify as CL or CH

SC

Clayey sandl!·h"

Ino rgan ic

PI > 7 and plots on or above "A" line'

CL

Lean c1C1yk . 1 .1I1

PI < 4 or plots below "A" line'

ML

Siltk . 1 ,m

Organic

Liquid Iimit-oven dried . . ' " " < 0,75 IqUld ! Imlt-not d fie

OL

PI plots on or above "A" line

CH

Fat clayL l . m

PI plots below "A" line

MH

Elastic siltk . l·ru

More than [ 2 % finesd

Sand with Fines

Inorganic

Liquid limit-oven dried . d < 0.75 · 'd I"lIDlt---not d rIe LIqUi

both. add "with cobbles or boulders. or both" to

f

name.

OW-OM well-graded gravel with silt; OW-OC well­ graded gravel with clay; OP-OM poorly graded gravel with silt; OP-GC poorly graded gravel with clay. dSands with 5 to 12% fines require dual symbols:

SW-SM weJl-graded sand with silt; SW-SC wen­ -.j

-

group name.

�Gravels with 5 to 12% fines require dual symbols:

...

Primarily organi..· matter. dark in color, and organic odor

'C,,

graded sand with clay; SP-SM poorly graded sand with silt; SP-SC poorly graded sand with clay.

DfW'DII!

c< �. .. ( I)In

D ,,,)' x

Db!l

if soil contains � 15% sand,add "With sand" to group

S If

Clayey gravelf.� .h

SP

Organic

a Based on the material passing the 75-mm. (3-in) sieve.

ell

C,, � 6 and J .s:; C, .s:; 3e

Liquid limit 50 or more

HighlV organic soils

C,, � 4 and l � C,. � 3"

C,, < 6 and/or l > Cr > 3c

Clean Sand:-

Sands 50% or more of coarse

Fine-grained soils

Clean Gravels Less Ihnn 5"\. finesc

Group nameb

fines classify as CL·ML. use dual symbol GC-GM

or SC·SM. b If fines are

organic, add "with organic fines" to group

name. If soil contains �15% gravel, add "with grave)" to

I

group name. i lf Atterbcrg limits pint in hatched area, soil is a CL-ML, silty clay.

OH PT

Well-grnded smid'

Organic c1ayk,l. nJ.1l Organic silt);· I. III. I'

Organic

:lal·I.�)

Organic silt\;·

I. m.lt

Peat

It If soil contains 1 5 to 29% plus No. 200, add "with sand" or "with gravel," whichever is predominant. I If soil contains ."..30% plus No. 200, predominantly

rolf soil contains ""'30% plus No. 200. predominantly sand. add "sandy" to group name.

gravel, add "gravelly" to group name.

n PI � 4 and plots on or above "A" line.

°PI < 4 or plots below"/l:' linc.

q PI plots below "A" line.

PPI plots on or above "A" line.

-

'"

Group Symhol < 5% lines

� C,' � �C,, C .(' _ -" .l

C,, ;;'= 4 and l � C, ';; J

Gravel % gravel > % sand

'-12% """

C" < 4 and/or I :>

.> 12% line�

.-: S'if. lines

Sand % sand ?:.': % gravel

C,,
J

C" < 6 nnd/or I >

> 1 2% """

C,. > J

�

�C,, ;;"

,-,, "",,


gravel < 15% s..1.rn.I ' % sand � ,;: gra\'d ....... :?; 15% sand


ood � ,,% ,,""

"

< " ", p" " No. 2IMl "t l .5.i9

. '

i

- t· .. . .

.

.

..

:

.

: :

" '

.,

Saturated unit weight 'Ysat

Iz,

1 t t t t Flow of water

=

I-- x ---+/ (aJ

Figure

Saturated unit weight 'Ysat A

=

(b)

1. 12 Calculation of effective stress

Thus. from Eq. (1.34). = cr - u = ( h,y". + hzY,,,) - (h, + h, + h)y". = h ,( y", - y",) - hy", = h ,y' - hy" or 0"

(

cr' = hz y'

- �', y",) = h,(y' - iyw)

(1.36)

Note in Eq. (1.36) that hih, is the hydraulic gradient i. If the hydraulic gradient is very high, so that y' iyw becomes zero, the effective stress will become zero. In other words, there is no contact stress between the 'soil particles, and the soil will break up. This situation is referred to as the quick condition, or failure by heave. So, for heave. -

(1.37) where i" = critical hydraulic gradient. For most sandy soils, i" ranges from 0.9 to 1.1, with an average of about unity. :

;

:.

For the soil profile shown in Figure 1.13, determine the, total vertical stre�:pOte . ' ' water pressure, and effective vertical stress at A, B, and C. . .

Solution

..

.

.

.

The following table can now be prepared.

.

' "

,

'

.

',

1, 1 1

Point

u(kN/m')

(4)(1'.) = (4) ( 14.5) = 58 5 8 + (1' ,)(5) = 58 + (17,2)(5) 60%) (1.61) Figure 1.22 shows a trapezoidal variation of initial excess pore water pressure with two-way drainage. For this case the variation of Tv with U will be the same as shown in Figure 1.21. T,.

(100

Trapezoidal Variation

1.0

1+1.----- Eq (1.60) -----...·� I·-- Eq (1.61) ----+1

Sand

0.8 ...: 9

�

<E u

E ;::: "

Sand

0.6 0.4

o Figure 1.21

IO

20

Average30 degree40of consol50idation.60

U (%)

70

80

90

Plot of time factor against average degree of consolidation (auo = constant)

37

1.13 lime Rate of Consolidation Sand

Figure 1.22 Trapezoidal initial excess pore water-pressure distribution

Sand

z

T Hc = 2H � 1.0

z

Sand

T Hc = 2H �

Sand

(.)

Sand

j.-- .jllo -+l

Figure 1.23 Sinusoidal initial excess pore water-pressure distribution

(b)

l

0.8 0.6

'

.....

For Figure

1.23a

0.4

For Figure 1.23b

2

0.

0 0

10

20

30

40 50 U (%)

60

80

70

90

Figure 1.24 Variation of U with T,­ sinusoidal variation of initial excess pore water-pressure distribution

Sinusoidal Variation This variation is shown in Figures 1.23a and 1.23b. For the initial excess pore water-pressure variation shown in Figure 1.23a,

CJ.u

=

,

CJ.uo sin

Similarly, for the case shown in Figure 1.23b, tJ.u = CJ.uo cos

7TZ

(1 .62)

7TZ

(1 .63)

2H

4H The variations of T" with U for these two cases are shown in Figure 1.24.

38

Chapter 1

Geotechnical Properties of Soil

Sand '

.,

Sand '

, . .. .

... .".;>"'-_ 1�.7, : ; ��Ir;- �'I IRock I+- Juo -+-l '"

1 .0

Sand

Sand

+------1

0.8

;...,-

0.6

0.4

0.2

o ��_,--,_-_.--,_-_,--,_-� 50 60 80 90 40 70 20 30 10 o u (o/c)

"

1.25 Variation of U with T1.-triangular initial excess pore water-pressure distribution Figure

Triangular Variation Figures 1.25 and 1.26 show several types of initial pore water­ pressure variation and the variations of TO' with the average degree of consolidation.

Example 1.7 A laboratory consolidation test on a

lowing results:

normally consolidated clay showed the fol­

Load. Au' (kN/m2)

Void ratio at the end of consolidation. e

140 212

0.92 0.86

The specimen tested was 25.4 mm in thickness and drained on both sides. The time required for the specimen to reach 50% consolidation was 4.5 min.

"

,

,

1. 13 Time Rate of Consolidation Sand . :.

,'"

I--

. : '.

.:1u" -.j . ,"

," " . \ . '. .. . '. ',

Rock

"

"

1.0

39

'"

. .. . : .

' , , ;

'

'

-1------1

0.8

0.6

0.4

,

l

o Figure

with T;,

10

30

20

50

40

60

70

80

90

U {%)

1.26 Triangular initial excess pore water-pressure distribution-variation of U

A similar clay layer in the field 2.8 m thick and drained on both sides, is sub­ 2 . jected. to a simiiar increase in average effective pressure (i.e" Uo. = 140 IcN/m . 2 and Uo + Au' = 212 kN/m ). Determine

a. the expected maximum primary consolidation settlement in the field. b. the length of time required for the total settlement in the fielcl to reach 40 mm. (Assume a uniform initial increase in excess pore water p�essure with depth.)

Solution

Part a

.

For normally consolidated clay [Eq. (1.38)], C, =

el

log

e2

(:D -

=

0.92

log

-

0.86 = 0.333

G!�)

40

Chapter 1

Geotechnical Properties of Soil

From Eq. (1.47), Sc =

212 (0.333) (2.8) C,H, ero + f!,er' = 0.0875 m = 87.5 mm = log log 1 + eo 1 + 0.92 140 Uo

Part b From Eq. (1.56), the average degree of consolidation is

- �(100) - 45.7 Yo u-� S _

c(max)

_

87.5

_

0

The coefficient of consolidation, C," can be calculated from the laboratory test. From Eq. (1.55), Cvt T,, = H'

For 50% consolidation (Figure 1.21), T,. 12.7 mm,so

=

0.197. t = 4.5 min. and H = HJ2 =

H' (0.197) (12.7)' , . = 7.061 mm-/mm C,. = TSO -- = 4.5 t

Again, for field consolidation. U = 45.7%. From Eg. (1 .60)

( ) ( )

' ' T = 'Tf U % = 'Tf 45.7 = 0.164 '4 100 , '4 100

But

or I

=

�,

T H,

o

=

·

1 64

(

2.8 X 1000 2 7.061

)

'

= 45,523 min = 3L6 days

•

Degree of Consolidation Under Ramp Loading The relationships derived for the average degree of consolidation in Section 1.13 as­ sume that the surcharge load per unit area (f!, er) is applied instantly at time I = O. However, in most practical situations. f!,er increases gradually with time to a maxi­ mum value and remains constant thereafter. Figure 1.27 shows f!,er increasing lin­ early with time (t) up to a maximum at time t,. (a condition called ramp loading).

1.14 Degree of Consolidation Under Ramp Loading z

41

Au

.,.., ,-:'."7-:'",-, '7 ;-l!:-: - - - .,.., - -;--,- -;- - - -

" ', Sand Clay

(a) Load per unit area, .do-

t"

Time, t

(b)

Figure 1.27 OneMdimensional consolidation due single ramp loading

tu

For t "" t" the magnitude of ACT remains constant. Olson (1977) considered this phenomenon and presented the average degree of consolidation, U, in the following form: T.

{

2 m= = 1

U = � 1 - - 2: . [l - exp { - M 'Tv)] Tc 7;" m= O M

and for Tv "" T"

}

2 m ='" 1 1 - - 2: -, [exp{M'Tcl - l]exp{ - M'T,) Tc m= O M where m, M, and Tv have the same definiti,on as in Eq. (1.54) and where U

=

T., =

C.t,

H'

(1.64)

(1.65)

(1.66)

Figure 1.28 shows the variation of U with T, for various values of T"., based on the solution given by Eqs. (1.64) and (1.65).

42

Chapter 1

Geotechnical Properties of Soil

0.01

Figure

0. 1 Time

1.28

72 kN/m'

factor. Til

1.0

10

Olson's ramp-loading solution: plot of V vs. T,. (Eqs. 1.64 and 1.65)

- - - - - -,,----

tc =

IS

days

Time, t

.,

FfgUftl 1.29 . Ramp

lo.di!'j�:!:'.?�

1. 15 Shear Strength

Also

43

,

Shear Strength The shear strength of a soil, defined in terms of effective stress, is (1.67) where u' = effective normal stress on plane of shearing

c' ::;; cohesion, or apparent cohesion effective stress angle of friction Equation (1.67) is referred to as the Mohr-Coulomb fai/u,e criterion. The value of c ' for sands and normally consolidated clays is equal to zero. For overcon­ solidated clays,c' > O. For most day-to-day work, the shear strength parameters of a soil (Le., c' and ') are determined by two standard laboratory tests: the direct shear test and the triax­ ,,>' =

ial test. Direct Shear Test

Dry sand can be conveniently tested by direct shear tests. The sand is placed in a shear box that is split into two halves (Figure 1.30a}.First a normal load is applied to the specimen. Then a shear force is applied to the top half of the shear box to cause failure in the sand. The normal and shear stresses at failure are u' =

N

-

A

and R s ::;; -

A

where A = area of the failure plane in soil-that is, the cross-sectional area of the shear box.

44

Chapter 1

Geotechnical Properties of Soif

Shear stress

N

.j.

'. ' . " , ... .. .

.

·.

=

c'

+

' cr

tan

�------------------

::r � : ;'

s

·

"

' .

::

:

':'", . -' '. ,.

.

cpo

s,

--R

sl _ _ _ _ _ _ _

ive normal "'---'-''-'-:---'-,---"7--'"'''''' Effect stress. 0",

CT ' J

u�

O" . �

cr '

(b) (a) Figure 1.30 Direct shear test in sand: (a) schematic diagram of test equipment; (b) plot of test results to obtain the friction angle '. = tan

u'

(1.68)

For sands, the angle of friction usually ranges from 26' to 45', increasing with the relative density of compaction. A general range of the friction angle, 4>', for sands is given in Table 1.8. Triaxial Tests

Triaxial compression tests can be conducted on sands and clays Figure 1.31a shows a schematic diagram of the triaxial test arrangement. Essentially. the test consists of placing a soil specimen confined by a rubber membrane into a lucite chamber and then applying an all-around confining pressure «(J'3) to the specimen by means of the chamber fluid (generally. water or glycerin). An added stress ( f),u) can also be

Porous stone

Lucite chamber

Rubber nembrane

Chamber fluid

Soil specimen

Shear stress

Porous stone

4>'

Base plare Chamber fluid

1 L_-1U3�J._., _---_'-;---__

pore waler pressure device

c

Shematic diagram of triaxial test equipment

0"]

-'-_

uj

Consolidated-drained test

Effective nonnal stress

(b)

(aJ Shear stress

Shear stress

{

Effective stress failure envelope

Tolal stress failure envelope

L__J q, ___J_....J'-____..L_ nonnal

Total

UJ

stress,

1 U3

c

a

LJc----L------L------__-L;o nonnal

uj

u;

Consolidated-undrained [est

u;

Effective

stress,

a'

(c)

Shear stress

T

s=

Total stress failure envelope

(4) = 0)

l L-L-_-'----'--_� ell

Unconsolidated-undrained test

Nonnal stress (lOlal), u

(d) lure

1.31 Triaxial test

45

46

Chapter 1 Geotechnical Properties of Soil

. .....�

.. .

.

: .. . '. .. . .

: ': � .-

. .. .: : " ', .

'

I

:, .

-1.

. .•.

:

. .' " ., , . ; : , -. : ' .- , ,

' '"

> •

"

.

Figure 1.32 Sequence of stress application in triaxial test

applied to the specimen in the axial direction to cause failure (t>u = t>uf at failure). Drainage from the specimen can be allowed or stopped, depending on the condition being tested. For clays, three main types of tests can be conducted with triaxial equipment (see Figure 1.32): 1. Consolidated-drained test (CD test) 2. Consolidated-undrained test (CU test) 3.

Unconsolidated-undrained test (UU test)

Consolidated-Drained Tests:

Step 1. Apply chamber pressure U3' Allow complete drainage, so that the pore water pressure (u = uo) developed is zero. Step 2. Apply a deviator stress t>u slowly. Allow drainage, so that the pore water pressure (u = Ud) developed through the application of t>u is zero. At failure, t>u = t>uf; the total pore water pressure uf = Uo + Ud = O.

So for consolidated-drained tests, at failure,

Major principal effective stress = u3 + t>uf Minor principal effective stress = U3 = uo

=

u, = ui

Changing U3 allows several tests of this type to be conducted on various clay speci­ mens. The shear strength parameters (c' and "" ) can now be determined by plotting Mohr's circle at failure, as shown in Figure 1.31b, and drawing a common tangent to the Mohr's circles. This is the Mohr-Coulomb failure envelope. (Note: For normally consolidated clay, c ' = 0.) At failure, (1.69)

Consolidated-Undrained Tests:

Step 1. Apply chamber pressure U3' Allow complete drainage, so that the pore water pressure (u = uo ) developed is zero.

I

1.15 Shear Strength

47

Step 2. Apply a deviator stress t:.. u. Do not allow drainage, so that the pore water pressure U = Ud * O. At failure, t:..u = t:.. uf; the pore water pres­ sure uf = Uo + Ud = 0 + ud(fl' Hence, at failure,

Major principal total stress = UJ + t:.. uf = u, Minor principal total stress = UJ Major principal effective stress = (''

=

0.68 [0', 70
'. Mathe· matically. the correl3tion can be approximated as (Kulhawy and Mayne. 1990) ch' =

tan-l

where

[

N

6U

12.2 + 2u.3

(�J

]0."

(2.25)

N", = field standard penetration number O'� = effective overburden pressure Pa = atmospheric pressure in the same unit as u� 4>' = soil friction angle

3. Hatanaka and Uchida (1996) provided a simple correlation between 4>' and (NI)", that can be expressed as 4>'

=

V20(N' )60 + 20

(2.26)

The following qualifications should be noted when standard penetration resis­ tance values are used in the preceding correlations to estimate soil parameters: The equations are approximate. Because the soil is not homogeneous, the values of N,o obtained from a given borehole vary widely. 3. In soil deposits that contain large boulders and gravel. standard penetration numbers may be erratic and unreliable. 1.

2.

Although approximate. with correct interpretation the standard penetration test provides a good evaluation of soil properties. The primary sources of error in

� l

2. 13 Procedures for Sampling Soil

83

standard penetration tests are inadequate cleaning of the borehole, careless mea­ surement of the blow count, eccentric hammer strikes on the drill rod, and inade­ quate maintenance of water head in the borehole. The modulus of elasticity of granular soils (E,) is an important parameter in estimating the elastic settlement of foundations. A first order estimation for E, was given by Kulhawy and Mayne (1990) as E, Pa = OIN6f)

{

where

Pa =

01 =

(2.27)

atmospheric pressure (same unit as E,) 5 for sands with fines 10 for clean normally consolidated sand 15 for clean overconsolidated sand

Scraper Bucket

When the soil deposits are sand mixed with pebbles, obtaining samples by split spoon with a spring core catcher may not be possible because the pebbles may pre­ vent the springs from closing. In such cases, a scraper bucket may be used to obtain disturbed representative samples (Figure 2.15a). The scraper bucket has a driving point and can be attached to a drilling rod. The sampler is driven down into the soil and rotated, and the scrapings from the side fall into the bucket. Thin-Waited Tube

Thin-walled tubes are sometimes referred to as Shelby tubes. They are made of seamless steel and are frequently used to obtain undisturbed clayey soils. The most common thin-walled tube samplers have outside diameters of 50.8 mm (2 in.) and 76.2 mm (3 in.). The bottom end of the tube is sharpened. The tubes can be attached

Drill rod

Figure 2. 75

\\\

'- - + I

S

(a)

Sampling devices: (aJ scraper bucket

1\ll

Driving point

c Section at S - S

84

Chapter 2 Natural Soil Deposits and Subsoil Exploration

� i Drill rod (bl

•

Thiin-y,all,ed tube

Drill rod

Water (in)

Vent

Piston (e)

Sample

(d)

Figure 2_ 15 (Continued) (b) thin-walled tube; (e) and (d) piston sampler

2. 14 Observation of Water Tables

85

to drill rods (Figure 2.1Sb). The drill rod with the sampler attached is lowered to the bottom of the borehole. and the sampler is pushed into the soil. The soil sample in­ side the tube is then pulled out. The two ends are sealed, and the sampler is sent to the laboratory for testing. Samples obtained in this manner may be used for consolidation or shear tests. A thin-walled tube with a 50.8-mm (2-in.) outside diameter has an inside diameter of about 47.63 mm (lhn.). The area ratio is AR( % )

= D� D-i Di (100) =

(50.8)2 - (47.63) 2

(47.63) 2

(100)

= 13 75% .

Increasing the diameters of samples increases the cost of obtaining them. Piston Sampler

When undisturbed soil samples are very soft or larger than 76.2 mm (3 in.) in di­ ameter, they tend to fall out of the sampler. Piston samplers are particularly useful under such conditions. There are several types of piston sampler; however, the sampler proposed by Osterberg (1952) is the most useful. (see Figures 2.15c and 2.15d). It consists of a thin-walled tube with a piston. Initially, the piston closes the end of the tube. The sampler is lowered to the bottom of the borehole (Figure 2.15c), and the tube is pushed into the soil hydraulically, past the piston. Then the pres­ sure is released through a hole in the piston rod (Figure 2.15d). To a large extent. the presence of the piston prevents distortion in the sample by not letting the soil squeeze into the sampling tube very fast and by not admitting excess soil. Conse­ quently, samples obtained in this manner are less disturbed than those obtained by Shelby tubes.

2. 14

Observation of Water Tables The presence of a water table near a 'foundation significantly affects the founda­ tion's load-bearing capacity and settlement, among other things. The water level will change seasonally. In many cases, establishing the highest and lowest possible levels of water during the life of a project may become necessary. If water is encountered in a borehole during a field exploration, that fact should be recorded. In soils with high hydraulic conductivity, the level of water in a borehole will stabilize about 24 hours after completion of the boring. The by lowering a chain or tape into depth of the water table can then be recorded . the borehole. In highly impermeable layers, the water level in a borehole may not stabilize for several weeks. In such cases, if accurate water-level measurements are required, a piezometer can be used. A piezometer basically consists of a porous stone or a per­ forated pipe with a plastic standpipe attached to it. Figure 2.16 shows the general placement of a piezometer in a borehole (also see Figure A.l in Appendix A). This procedure will allow periodiC checking until the water level stabilizes.

86

Chapter 2 Natural Sail Deposits and Subsoil Exploration

Piezometer water level

Protective cover

Groundwater level

Standpipe

Bentonite cement grout

Bentonite plug

Filter tip Sand

Figure 2. 16 Casagrande-type piezometer (Courtesy of N. Sivakugan, James Cook University, Australia)

Vane Shear Test The vane shear test (ASTM D-2573) may be used during the drilling operation to de­ termine the in situ undrained shear strength (c.) of clay soils-particularly soft clays. The vane shear apparatus consists of four blades on the end of a rod, as shown in Figure 2.17. The height, H, of the vane is twice the diameter, D. The vane can be either rectangular or tapered (see Figure 2.17). The dimensions of vanes used in the field are given in Table 2.6. The vanes of the apparatus are pushed into the soil at the bottom of a borehole without disturbing the soil appreciably. Torque is applied at the top of the rod to rotate the vanes at a standard rate of 0.1 o/sec. This rotation will induce failure in a soil of cylindrical shape surrounding the vanes. The maximum torque, T, applied to cause failure is measured. Note that T = f(c", H, and D)

(2.28)

.,

2. 15 Vane Shear Test

87

1

•

+

Rectangular vane

Tapered vane

or

where

eu = KT

Figure 2. 17 Geometry of field vane (After ASTM, 2(01)

(2.29)

K = a constant with a magnitude depending on the dimension and shape T is in N· m, c" is in kN/m', and

of the vane

The constant

(2.30)

I

L

88

Chapter 2 Natural Soil Deposits and Subsoil Exploration Table 2.6

ASTM Recommended Dimensions of Field Vanes3

Casing size

Height, H mm {in.)

Diameter. D mm (in.)

38.1 (II) .50.8 (2) 63.5 (2l) 92.1 (31 )

AX BX NX

101.6 mm (4 in.)b

Thickness of blade mm (in.)

1.6 (;\;) 1.6 (;\;) 3.2 (I) 3.2 ell

76.2 (3) 101.6 (4) 127.0 (5) 184.1 (7�)

Diameter of rod mm (in.)

12.7 ." · Gu .,d

. : -.

: �..

.

' '

.. .

,

:.' )...

. ., ': •. .

.

·'

-

Pressure. p •

PI

Guard ..". �,..,. cell

'

(al

.. . .

..'

Figure 2.24

I

1I

I

I

Zone II

I I I

I

�P

I

.iI _ _ �

__

r-�\'I

I

Zone III

I

I

P..

I

I i i1 - - - - - -

I I

Measuring _4-.'-- cell

...( .

Zone I

-----

.

, >.

l-�,.

cell vo u

conduct a

I

Total

!""''---_-+_-+_+_____f-_+ caviIY v"

v" + I'" V" +

I'm

Va

-i-

(b)

V(

2(\/" + \',,)

volume. V

(a) Pressure meter; (b) plot of pressure versus total cavity volume

98

Chapter 2 Natural Soil Deposits and Subsoil Exploration

the cavi totalty.volAfter umetheof thecomplexpanded cavity (V)theis about twiis cdefle theatedvolandumeadvanced of the oriforgi­ nal of the test, probe e tion testingTheat another depth. in theZonegraphical resul t s of the aspressuremeter test are expressed formtheof pressure represents versus vol u me, shown in Figure In the fi g ure, I theThe reloadingstateportion duristateng whi ch thein before soil around borehol e is pushedrepresents back intothethein initial (Le. , the it was dril l i ng). pressure total horizontal stress. Zone II represents aThepseudoel astic zone in whichthethecreep, .cell volsitu u me versus cel l pressure i s practical l y l i near. pressure represents ortheyiellimitd, pressure. The zone marked III is the plastic zone. The pressure represents pressure. pressuremeter soil isThus, determined with the use of the of anmodul theoryTheof expansion infinitelus,yEpthi' cofk cyltheinder. 2.24b.

Po

Pr

PI

(2.56)

where up .1 1 '

= PI =

p., =

t'l

-2

Po l'"

Poisson's ratio (which may be assumed to be The limit pressure is usually obtained by extrapolation and not by direct mea­ surement. Inf-boring order topressuremeters overcome the di(SBPMTs) fficulty ofhave preparing the borehol e to The the proper size, sel al s o been devel o ped. details concerning SBPMTs can be found i n the work of Baguel i n et al . Correlations tests between various soiloparameters andinvestigators. the results obtained from the pressuremeter have been devel ped by various Kul h awy and Mayne proposed that a; where a; preconsolidation pressure. On the basis of the cavity expansion theory, Baguelin et al. proposed that 0.33)

PI

(1978).

(1990) =

=

0.45pI

(2.57)

(1978)

(2.58)

where c" = undrained shear strength of a clay = 1 + In(::'.) Np

2.18 Di/atometer Test

99

Typical values of Np vary between 5 and 12, with an average of about 8.5. Ohya et al. (1982) (see also Kulhawy and Mayne, 1990) correlated Ep with field standard penetration numbers (NIiJ) for sand and clay as follows: Clay: Ep(kN/m2) = 1930N:!o63

(2.59)

Sand: Ep(kN/m2) = 908N&,66

(2.60)

Dilatometer Test The use of the flat-plate dilatometer test (DMT) is relatively recent (Marchetti, 1980; Schmertmann, 1986). The equipment essentially consists of a flat plate measur­ ing 220 mm (length) x 95 mm (width) x 14 mm (thickness) (8.66 in. X 3.74 in. X 0.55 in.). A thin, flat, circular, expandable steel membrane having a diameter of 60 mm (2.36 in.) is located flush at the center on one side of the plate (Figure 2.25a). The dilatometer probe is inserted into the ground with a cone penetrometer testing rig (FIgure 2.25b). Gas and electric lines extend from the surface control box, through the penetrometer rod, and into the blade. At the required depth, high-pressure nitro­ gen gas is used to inflate the membrane. Tho pressure readings are taken: L The pressure A required to "lift off" the membrane. 2. The pressure B at which the membrane expands 1.1 mm (0.4 in.) into the sur­ rounding soil

The A and B readings are corrected as follows (Schmertmann, 1986): Contact stress, Po = 1.05(A + t1A - Zm) - O.05(B - t1B

Expansion stress. PI = B - Zm - t1 B

60

- ..

o J.- 95 mrn -+{ (a)

.

' .

.

. ' '.; .

�::

.

' �.'

-

Zm) (2.61)

(2.62)

.

:

(b)

Figure 2.25 (a) Schematic diagram of a flat-plate dilatometer; (b) dilatometer probe inserted into ground

100

Chapter 2 Natural Soil Deposits and Subsoil Exploration

where

I ... I

seating vacuum pressure required to keep the membrane in contact with its air pressureofrequired inside the membrane to deflect it outward to a center expansion 1.1 mm gauge pressure >leviation from zero when vented to atmospheric pressure The ly conducted at depths 200 to 300 mm apart. The result of a given test itest s usedis normal to determi ne three parameters: . I 'Index, -1. Matena -, 2. Honzontal stress mdex, Ko = 3. Dilatometer modulus, Eo(kNjm') = 34.7 kNjm' - kNjm') where pore water = verticalpressure effective stress gurereported 2.26 shows the results of a dilatometer test conducted in Bangkok soft cl(1980) ay andFiprovi Based on his initial tests, Marchetti ded thebyfolShibuya lowing and correlHanh ations(2001). . IlA = IlB = 2m =

ID =

.

Po

PI Po

Uo

.

Po

Uo

(J"o

( PI

Po

Uo = (J"� in situ

Ko =

( )0-.47 K 1.5

-E.

0.6

OCR = (0.5Ko)Ll6 (for normally consolidated clay)

I

( c�) = (c,: ) (Fa

OC

Ua

NC

(0.5Ko)'-"

(2.63) (2.64) (2.65) (2.66) (2.67)

where Ko coefficient of at-rest earth pressure OCR overconsol i dation ratio = OC overconsolidated soil· NC = normally consolidated soil E, = modulus of elasticity Other relevant correlations using the results of dilatometer tests are as follows: For undrained cohesion in clay (Kamei and Iwasaki, 1995): = =

•

c"

= 0.35 (J"o (0.47Ko) L14

(2.68)

"

2. 19

o

Po. PI

(kN/m2)

300

0.6 0

600 0

KD 3

6

1 01

Coring of Rocks

0

ED (I(N/m') 2,000

4,000 5,000

J

Figure 2.26 A dilatometer test result conducted on soft Bangkok clay (Redrawn from Shibuya and Hanh. 2001) •

For soil friction angle (ML and SP-SM soils) (Ricceri et aI., 2002): KD ' = 31 + 0.236 + 0.066KD ;"

" 2. 19 "

I

=

28 + 14.6 10gKD - 2.1(logKD)'

(2,69a) (2.69b)

For definition of ;1(, see Fig. 1.36. Schmertmann (1986) also provided a correlation between the material index (ID) and the dilatometer modulus (ED) for a determination of the nature of the soil and its unit weight (y). This relationship is shown in Figure 2.27.

Coring of Rocks When a rock layer is encountered during a drilling operation, rock coring may be necessary. To core rocks, a core barrel is attached to a drilling rod. A coring bit is at­ tached to the bottom of the barrel (Fig. 2.28). The cutting elements may be diamond,

102

Chapter 2 Natural Soil Deposits and Subsoil Exploration CJa�..

Silt Silly

Clayey

.

Sandy

-

100 -+---1----+--+-1--++--1--, 50

�

.," �. �.,..... ('.-
', of the sand. Use Eq. (2.47). Refer to Problem 2.11. Using Eq. (2.45 ) , determine the variation of Ihe relative density wilh deplh. In the soil profile shown in Figure n.l3, if the cone penelration resistance (q,) al A (as delermined by an electric friction-cone penelrometer) is lbjin.',estimale s. The undrained cohesion, CII b. The overconsolidation ratio, OCR In a pressuremeter lesl iil a soft saturated clay, the measuring cell volume Vo = 535 em ' . P" = 42 .4 kN/m'. P I = 326.5 kN/m', 110 = 46 em). and VI = 180 em". Assuming Poisson's ratio (JLJ to be 0.5 and using Figure 2.24, calcu­ late the pressuremeter modulus (£,,). A dilatometer test was conducted in a clay deposit. The groundwater lable was located at a depth of 3 m below the surface. At a depth of 8 m below the surface, the contact pressure (1',,) was 280 kNjm' and Ihe expansion slress (1'\) was 350 kNjm'- Determine the following:

Figure P2. 13

L

Point resistance of cone, qc (MN/m2)

90

1 18

Chapter 2 Natural Soil Deposits and Subsoil Exploration H.

Coefficient of at-rest earth pressure, KG

h. Overconsolidation ratio, OCR

2.16

2.17

I

2.18

1

2.19

c. Modulus of elasticity, E, Assume '/2 than to e/>'. If this change is accepted, the values of N" Nq, and Ny for a given soil friction angle will also change from those given in Table 3.1. With a = 45 + e/>'/2, it can be shown that (3.24) and (3.25) Equation (3.25) for N, was originally derived by Prandtl (1921), and Eq. (3.24) for Nq was presented by Reissner (1924). Caquot and Kerisel (1953) and Vesic (1973) gave the relation for Ny as Ny

=

2 (Nq + l ) tan e/>'

(3.26)

3.6 The General Bearing Capacity Equation

133

Table 3.3 Bearing Capacity Factors 'I>' y

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

N.

N,

Ny

4>'

N.

N,

Ny

5.14 5.38 5.63 5.90 6.19 6.49 6.81 7.16 7.53 7.92 8.35 8.80 9.28 9.81 10.37 10.98 11.63 12.34 13.10 13.93 14.83 15.82 16.88 18.05 19.32 20.72

1.00 1.09 1.20 1.31 1.43 1.57 1.72 1.88 2.06 2.25 2.47 2.71 2.97 3.26 3.59 3.94 4.34 4.77 5.26 5.80 6.40 7.07 7.82 8.66 9.60 10.66

0.00 0.07 0.15 0.24 0.34 0.45 0.57 0.71 0.86 1.03 1.22 1.44 1 .69 1 .97 2.29 2.65 3.06 3.53 4.07 4.68 5.39 6.20 7.13 8.20 9.44 10.88

26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43

22.25 23.94 25.80 27.86 30.14 32.67 35.49 38.64 42.16 46.12 50.59 55.63 61.35 67.87 75.31 83.86 93.71 105.11 118.37 133.88 152.10 173.64 199.26 229.93 266.89

11.85 13.20 14.72 16.44 18.40 2Q.63 23.18 26.09 29.44 33.30 37.75 42.92 48.93 55.96 64.20 73.90 85.38 99.02 115.31 134.88 158.51 187.21 222.31 265.51 319.07

12.54 14.47 16.72 19.34 22.40 25.99 30.22 35.19 41.06 48.03 56.31 66.19 78.03 92.25 109.41 130.22 155.55 186.54 224.64 271.76 330.35 403.67 496.01 613.16 762.89

44

45 46 47 48 49 50

Table 3.3 shows the variation of the preceding bearing capacity factors with soil fric­ tion angles. The equations for the shape factors F", Fq" and Fy, were recom­ mended by De Beer (1970) and are

Shape Factors

(3.27) (3.28) and (3.29) where L = length of the foundation ( L > B). The shape factors are empirical relations based on extensive laboratory tests.

I

l

134

Chapter 3

Shallow Foundations: Ultimate Bearing Capacity

(1970) proposed the following equations for the depth (3.30) Fed = 1 + 0.4 ( ;) , Fqd = 1 + 2 tan '(1 sin ')' ; (3.31) � =are1 valid for D,jB 1. For a depth-of-embedment-to­ Equations (3.30) and (3.31) foundation-width ratio greater than unity ( DdB > 1), the equations have to be modified to (3.33) Fc-I = 1 + (OA) tan-' ( ;) F"d = 1 + tan ' (1 - sin')' tan- ' (�) (3.34) and (3.35) F'/l = respectively, The factor tan (Dr!B) is in radians in Eqs, (3.33) and (3.34), Meyerhof (1963) and Hanna and Meyerhof ( 1981) suggested the following inclination factors for use in Eq. (3.23): (3.36) Fci = �Ji = (1 - 900 ) (3.37) Here, f3 = inclination of the load on the foundation with respect to the vertical. Hansen

Depth Factors

factors:

-I

-

0�

"'"

2

1

"

-I

Inclination Factors

tr '

I

::

Example 3.2 A square foundation (B x B) has to be constructed as shown in Figure 3.7. &< sume that 'Y = lb/ft', 'Y,,, = 118 1bjft3, Df and Dl 2 ft. The grOl;s,M: lowable load, Q.u, with FS is lb. The standard penetration resistali�;,' N6fJ values are as follows:

105

= 3 150,000 = 4 ft, Depth (It)

5 10 15 20 25

=

,',

N.. (blowllt)

4 6 6 10 5

�

3.6

135

The General Bearing Capacity Equation

BXB

Figure 3.7 A Square fou.ndation

Determine the size of the footing. Use Eq. (3.23).

SciIUtiOIi

From :E;qs. (2.13) and (2.14), (a) Combining Eqs. (2.26) and (a) yields &'

Thus, Po

=

=

5J0.5 . 0 ) ( [ cr;, 20N�1

P

+ 20

(b)

2000 Ib/ft'. Now the following table can be prepared:

Depth (ft)

N..

5 10 15 20 25

4 6 6 10 5

u ;(lb/ft')

X 105 + 3 (118 - 62.4) = 376.8 376.8 + 5 ( 1 18 - 62.4) = 654.8 654.8 + 5 (118 - 62.4) = 932.8 932.8 + 5 (118 - 62.4) = 1210.8 1210.8 + 5 (118 - 62.4) = 1488.8

2

. (deg) [Eq. (bll

33.6 34.5 33.3 36.0 30.8

Average &' - 33.640 - 34° Next, we have q,1I

- - 150,000 . B' Ib/ft'

_

Q,,, B'

_

From Eq. (3.23) (with c' = 0), we obtain

t. I

(c)

136

Chapter 3 Shallow Foundations: Ultimate Bearing Capacity

For

'

=

34°, from Table 3.3, Nq = 29.44 and Ny = 41.06. Hence,

Fq, = 1 +

B L

tan

Fy, = 1 - 0.4

Fyd = 1

=

1 + tan 34 = 1.67

(Z) � 1 - 0.4

Fqd = 1 + 2 tan

I

'

-- I,

=

0.6

'(1 - sin � ') 2

=

1 + 2 tan 34

(1 - sin 34)2 �

=

1+

�

1 5

and q =

(2) (105) + 2 (118 - 62.4) = 321.2 Ib/ft2

So q,n =

=

(

)

1

[

+

(i) = 0,

F" Fq1 =

1 + 0.2 (B/L) 1

FYI

For 4>' "" 10",

F" Fqs = Fys

1 + 0.2 (B/L) tan' (45 + 4>'/2) 1 + 0.1 (B/L) tan' (45 + 4>'/2) Depth

For 4> = O.

F,d Fqd = Fyd

1 + 0.2 (Df/B) 1

For 4>' "" 10"

F,d Fqd =

1 + 0.2 (Df/B) tan (45 + 4>'/2) 1 + 0.1 (DJiB) tan (45 + 4>'/2)

Fyd

Inclination Equation (3.36) Equation (3.37)

Tabla 3.5 Meyerhof's Bearing Capacity Factor Ny = (N, - 1 ) tan (1.4 4>') ""

0 1 2 3 4 5 6 7 8 9 10 11 12 13

L I ,

IV,

0.00 0.002 0.01 0.02 0.04 0-.07 0.11 0.15 0.21 0.28 0.37 0.47 0.60 0.74

""

14 15 16 17 18 19 20 21 22 23 24 25 26 27

IV,

0.92 1.13 1.38 1.66 2.00 2.40 2.87 3.42 4.07 4.82 5.72 6.77 8.00 9.46

""

28 29 30 31 32 33 34 35 36 37 38 39 40 41

IV,

11.19 13.24 15.67 18.56 22.02 26.17 31.15 37.15 44.43 53.27 64.07 77.33 93.69 113.99

""

42 43

44

45 46 47 48 49 50 51 52 53

IV, 139.32 171.14 211.41 262.74 328.73 414.32 526.44 674.9 1 873.84 1143.93 1516.05 2037.26

1 38

Chapter 3 Shallow Foundations: Ultimate Bearing Capacity

3.8

Some Comments on Bearing Capacity Factor, Ny, and Shape Factors Bearing Capacity Factor, Ny

At the present time, th� bearing capacity factors Nq and N, given by Eqs. (3.24) and (3.25) are well accepted for bearing capacity calculation. However, there are sev­ eral relationships for N, that can be found in the literature. Some of these relation. ships are: Vesic (1973):

N, = 2(Nq + l)tan � '

[Eq. (3.26)]

Meyerhof (1953): N, = (N. - 1)tan(1.4 �')

(see Table 3.5)

Hansen (1970):

Ny = 1.5(Nq - l ) tan �' Michalowski (1997):

A comparison of these Ny values for varying friction angles (�') is shown in Figure 3.8. It can be seen from this figure that Vesic 's and Michalowski's N, values are ap­ proximately the same, and they are the upper limits of all suggested relationships. These values are recommended for use in this text. Shape Factors (Fcv Fop and F,,)

I

The shape facto�s proposed by De Beer (1970) are given in Eqs. (3.27) to (3.29). Similarly, the shape factors suggested by Meyerhof are given in Table 3.4. These are conservative relationships acceptable for design purposes. However, it is interesting to compare the relationships for F." . According to De Beer (1970) [see Eq. (3.29)] F."

=

1 - 0.4

(�)

Similarly, according to Meyerhof (1963) (see Table 3.4) F." = 1 + 0.{

�}

(

an2 45 +

�' )

(for�' > 10')

The two relationships stated above appear to contradict each other. For a given soil friction angle with an increase in B/ L, De Beer's value of F" decreases, whereas the magnitude of F" suggested by Meyerhof increases. More recently, Zhu and

3.8 Some Comments on Bearing Capacity Factor, N" and Shape Factors

139

50

Hansen (1970)

...j ....lJj

_ _

40

'" 9

.:

J!!

?>

Meyerhof (1963) --....--JL 30

'l

.r 20

Michalowski (1997)

'" �

Vesic (1973)

10

o

20

10

40

30

Soil friction ungi\!. 4>' (deg)

Figure 3.8 Variat ion of Ny with soil friction angle, 4>'

Michalowski (2005) evaluated the shape factors based on the elastoplastic model of soil and finite element analysis. They are as follows: F" = 1 + ( 1.8 tan' ' + O.l) Fq,

= 1 + 1 .9 tan''

(�)

O.l

(�y

F" = 1 + (0.6 tan' ' - 0.25)

and F"

=1

+ (1.3 tan' ' - 0.5)

'

(�)

()

(3.38) (3.39) (for ' ." 30°)

L 1.l e-(L(B) (for ' > 30°)

B

(3.40)

(3.41)

Equations (3.38) through (3.41) have been derived based on sound theoretical background and may be used for bearing capacity calculation.

I l.

I

1 40

Chapter 3 Shallow Foundations: Ultimate Bearing Capacity C,

20

o

40

(kN/m') 60

80

100

•

I

I

2

§

;: C

"u

•

•

3 •

•

•

•

•

•

•

•

:1

"

� �; 2

•

•

• •

Figure 3.9 Variation of ell with depth obtained from field vane shear test

A Case History for Bearing Capacity Failure An excellent case of bearing capacity failure of a 6-m (20-ft) diameter concrete silo was provided by Bozozuk (1972). The concrete tower silo was 21 m (70 ft) high and was constructed over soft clay on a ring foundation. Figure 3.9 shows the vari­ ation of the undrained shear strength (c,,) obtained from field vane shear tests at the site. The groundwater table was located at about 0.6 m (2 ft) below the ground surface. On September 30, 1970,just after it was filled to capacity for the first time with corn silage, the concrete tower silo suddenly overturned due to bearing capacity fail­ ure. Figure 3.10 shows the approximate profile of the failure surface in soil. The fail­ ure surface extended to about 7 m (23 ft) below the ground surface. Bozozuk (1972) provided the following average parameters for the soil in the failure zone 'and the foundation: • •

Load per unit area on the foundation when failure occurred = 160 kN/m' Average plasticity index of clay (PI) = 36

3.9 A Case History for Bearing Capacity Failure

141

60"

12

Figure 3.10 Approximate profile of silo failure (Adapted from Bozozuk, 1972) •

•

Average undrained shear strength (c,,) from 0.6 10 7 m depth obtained from field vane shear tests = 27.1 kN/m2 From Figure 3.10, B = 7.2 m and Df = 1.52 m.

We can now calculate the factor of safety against bearing capacity failure. From Eq. (3.23)

qll = c' NcFcsFcdFci + qNcFqsFqdFqi + � yB NyF"IsFYdFyj For '" = 0 condition and vertical loading, c' = c,,, � = 5.14, Nq = 1 , Ny = 0, and F.o; = F,,; = F.,; = O. Also. from Eqs. (3.27), (3.28), (3.30). and (3.31), . 7.2 1 = 1.195 F.o, = 1 + 7.2 5.14 F",

=1

F.od

=

.

( )( )

1 + ( 0.4)

( 1.52 ) = 1.08 7.2

Thus,

q" = (c,,) (5.14) (1.l95)(1.08)(l) + (y)(1.52) Assuming y = 18 kN/m3 q" = 6.63c" + 27.36

I

l .

l

I

(3.42)

142

Chapter 3 Shallow Foundations: Ultimate Bearing Capacity

According to Eqs. (2.34) and (2.35), A

= 1.7 - 0.54 log [PI(%)] For this case, PI = 36 and C,,( VST) = 27.1 kN/m'. So C'«]' }" ' ,

-

'

.'

(3.46)

"

Figure 3.11 shows the variation of Fr, = Fq, [see Eg. (3.46)] with "" and I,. For '" = 0,

144

Chapter 3 Shallow Foundations: Ultimate Bearing Capacity

F" = 0.32 + 0.12

B + 0.60 log I, L

(3.47)

For q,' > 0, 1 - Fqc FfC = FIJC - -:-:---'-:-c Nq tan cp'

(3.48)

Example 3.3 For a shallow founda tion, B = 0.6 m. L = 1.2 m, and Df = 0.6 m. The known soil characteristics are as follows: Soil: q,' = 25 ' c' = 48 kNjm' Y = 1 8 kN/m3 Modulus ofelasticity. E, = 620 kN/m' Poisson's ratio. j.L, = 0.3 Calculate the ultimate bearing capacity.

Solution

From Eq. (3.44),

However,

I .

G = ,

So

2 ( 1 + j.L,)

E,

'

I, =

Now, q' = Thus,

Y

.,--,------:'=-''-- -:---c-::­

2 (1 + j.L,) [c' +

( �) ( Df +

I, =;

E,

= 18 0.6 +

q' tan q,']

�

06

)

= 16.2 kNjm'

620 = 4.29 2 (1 + 0.3 ) [48 + 16.2 tan 25]

3.10

Effect of Soil Compressibility

145

From Eq. (3.45), = Hex{ (3.3 - 0.45 �)cot(45 - �' )]} = Hex{ (3.3 - 0.45 �:�)cot(45 �)]} .= ' = 25°, Nq = 10.66 (see Table 3.3); therefore, = 0.347 - 10.66 0.tan34725 0.216 Now, from Eq. (3.43). + c' From Tabl quentl y. e 3.3, for c/>' 25°. N". 20.72, 10.66, and N, = 10.88. Conse­ 1 (�:)(�) 1 G�:��)(�:�) 1.257 0.6 tan 25 1.233 = 1 BL tan 1.2 - 0.4 L 1 - 0.4·0.1.26 0.8 = 0.4 (IiDr) 1 + 0.4 (0.0.66) = 1.4 = 2 tan - sin. ,(IiDr) 2 tan25{l - Sin25)2 (�:�) = 1.311 1,(0.)

-

1,(a) >

F"

c// +

+

+

25

+

+

o

8m

(

.

25

Fct = F 1

F".

I

,

=

+

Fq,

+

F",

1 +

Fqd

1 +

= 1 +

I

-

F" =

. F" = 1

l

F.,

NcF;.·sF'..·JFcc + qNqF;fJFqdFqr; = = Nq =

-

c/>'

B

=

=

=

=

.p' (1

!'YBN"'IFysF'(dF1C

+

= 1 +

=

.. . . . .

---'''-

qll =

=

_

c/>')"

146

Chapter 3 Shallow Foundations: Ultimate Bearing Capacity

and Thus, q"

3. 1 1

;

(4S) (20,72) (J.257) (L4) ( O.216) + (0.6 x IS)(10.66) (1.233) (1.31l) ;

(0.347) + (t)(lS) (0.6) ( 10.SS) (O.S) ( 1 ) (0.347)

459 kN/m'

,',

_

Eccentrically Loaded Foundations In several instances, as with the base of a retaining wall, foundations are subjected to moments in addition to the vertical load, as shown in Figure 3.12a. In such cases, the distribution of pressure by the foundation on the soil is not uniform. The nominal dis­ tribution of pressure is (3.49) and (3.50)

Q

M . r ·

'

I

-::+I

I

:" ,

',

-" ,

-

' .� . ' . '

.,' . � '"

.. .:. ,

.. . .

I '.

BxL

.

','

I I

I

Q e :

.

'! '

B

. . " .� .

:.

� .: . "

�B'-+:j:::= ' "�', .

: r

For e < 816

L

__ _

-

For e > BI6

--

---

(a) Figure 3.12

(b)

Eccentrically loaded foundalions

I,

3, 1 1 Eccentrically Loaded Foundations

147 '

where Q = total vertical load M = moment on the foundation Figure 3,12b shows a force system equivalent to that shown in Figure 3,12a, The distance e=

M

(3 5 1 )

Q

is the eccentricity, Substituting Eq, (3.51) into Eqs, (3,49) and (3,50) gives (3,52) and ' ,'

q� ,

Q

( 6�.) .

= BL,', l -" . .

(3.53)

B

" > ,

Note that, in these equations, when the eccentricity e becomes B/6, qmi" is zero, For e > B/6, qmi" will be negative, which means that tension will develop. Because soil cannot take any tension. there will then be a separation between the foundation and the soil underlying it. The nature of the pressure distribution on the soil will be as shown in Figure 3.12a, The value of qm"" is then

-

4Q

q = m" 3L(B

(3 54) 2e) The exact distribution of pressure is difficult to estimate. Figure 3,13 shows the nature of failure surface in soil for a surface strip foun­ dation subjected to an eccentric load. The factor of safety for sucli type of loading against bearing capacity failure can be evaluated as FS =

Q"it

(3.55)

Q

where Q"'t = ultimate load-carrying capacity. The following section describes several theories for determining Q"it'

;------------------- ------L

Figure 3.13 Nature of failure surface in soil supporting a strip foun­ dation subjected to eccentric loading (Nole: D,. 0; QuH is ulti­ mate lo ad per unit length of foundation)

=

1 48

Chapter 3 Shallow Foundations: Ultimate Bearing Capacity

3. 12 .

Ultimate Bearing Capacity under Eccentric Loading -Meyerhof's Theory In 1953, Meyerhof proposed a theory that is generally referred to as the effective area method. The following is a step-by-step procedure for determining the ultimate load that the soil can support and the factor of safety against bearing capacity failure: Step 1. Determine the effective dimensions of the foundation (Figure 3.12b): B' = effective width = B - 2e L' = effective length = L

Step 2.

Step 3.

Note that if the eccentricity were in the direction of the length of the foundation, the value of L' would be equal to L - 2e. The value of B' would equal B. The smaller of the two dimensions (i.e., L' and B') is the effective width of the foundation. Use Eq. (3.23) for the ultimate bearing capacity: (3.56) To evaluate F". £',,, and Fy" use Egs. (3.27) through (3.29) with effective length and effeci;ve lVidth dimensions instead of L and B, respectively. To determine £,.,/. £',./. and F..d• use Eqs. (3.30) through (3.35). Do not replace B with B'. The total ultimate load that the foundation can sustain is A' Q"1t = (3.57) ;, (B') (L' ) q

where A' = effective area. Step 4. The factor of safety against bearing capacity failure is Q FS = uit Q

I .•.

Step 5. Check the factor of safety against qm,,' or FS = q:';qm,,' Foundations with Two-Way Eccentricity

Consider a situation in which a foundation is subjected to a vertical ultimate load Q"it and a moment M, as shown in Figures 3.14a and b. For this case, the compo­ nents of the moment M about the x- and y-axes can be determined as M, and M" respectively. (See Figure 3.14.) This condition is eqUivalent to a load Q "1t placed ec­ centrically on the foundation with x = eB and y = eL (Figure 3.14d). Note that M,. ' en = (3.58) Q"1t and (3.59)

3.12

Ultimate Bearing Capacity under Eccentric Loading-Meyerhof's Theory

149

QuI!

�M

(a) ,

. ." . ..

.. . .. . .

f'

I

+Y

T fk� 1 L' - -

i

i

I

I � B --1

+- X

�

f- Quit I

I

(b)

- f- -

M".

M

j Ql,lh -- - , I

-

I

eL

-

(d)

(e)

Figure 3. 14 Analysis of foundation with two-way eccentricity

If Q,,, is needed. it can be obtained from Eg. (3.57): that is. where, from Eg. (3.56), and

A' = effective area '= B'L'

As before, to evaluate F", Fq" and Fy, [Eqs. (3.27) through (3.29)], we use the effective length L' and effective width B' instead of L and B, respectively. To calcu­ late F,d, Fqd, and F,d, we use Egs. (3.30) through (3.35); however, we do not replace B with B'. In determining the effective area A', effective width B' , and effective length L', five possible cases may arise (Highter and Anders, 1985). Case I. eLiL '" i and Figure 3.15, or

eBI

B

'"

i. The effective area for this condition is shown in (3.60)

where

(

BI = B 1.5 -

3e8

B

)

(3.6 1)

1 50

Chapter 3 Shallow Foundations: Ultimate Bearing Capacity

T 1

1-I 8, -\

Effective

L-

Figure 3. 15 Effective area for the case of eJL

�B----I

and eB/B � i

and L, = L

(

1.5

- 3eL L

)

'"

!

(3.62)

The effective length L' is the larger of the two dimensions B, and L,. So the effec­ tive width is

i .

'

;;

B' =

A' L'

(3.63)

Case II. eLiL < 0.5 and 0 < eBIB < �. The effective area for this case, shown in Figure 3.16a, is

(3.64) The magnitudes of L, and L, can be determined from Figure 3.16b. The effective width is

I

-=.::-----

A' L, or L, (whichever is larger)

(3.65)

L' = L, or L, (whichever is larger)

(3.66)

B'

=

-

The effective length is Case III. eLi L < t and 0
' = _/8

JO

30 20 Friction angle. r/>' (deg)

40

N"1!e)

0 0.1

115.80 71.80

0.3 0.4

1 8.50 4.62

0.2

o

40'

41.60

Figure 3.22 Variation of Ny1cl with cb'

Figu�e 3.23 A continuous foundation with load

eccentricity

For t/>' = 35°, from Table 3.3, Nq = 33.3 and Ny = 48.03. Also,

B' = 6

l.

i

-

(2) (0.5)

=

5 it

158

Chapter 3 Shallow Foundations: Ultimate Bearing Capacity

Because Fqs = 1, Fythse=fo1,undation in question is a strip foundation, /L' is zero. Hence, Fqi = FYi = 1 Fqd = 1 2 tan ' (1 -sin",')' = 1 0.255 ( 4) = 1.17 Fy,/ = 1 and = (440)(33.3)(1)(1.17)(1) G)(110)(5)(48.03)(1)(1)(1) ,: 30,351l1?Lft2 Consequently, = (5)(1)(30,351) = 151,7551bjft = = B'

+

D

.

Iif

+

"6

+

q�

Q.1t

( B ' ) ( 1 ) (q;,)

_

75.88 ton/ft

,

,"

•

Example 3.5

Solve Example 3.4 using Eq. (3,73). Sincec' = 0 = [ �'YBNy(,)] �= 0.65 =0083. . ForNy(,)A" 26.35°8. Hence, and ejB = 0.083, Figures 3,21 �d. 3.22. give Nq(,) 4,AA4' = 6[(440)(33.4) ('/,)(110)(6)(26.8)] 141,2401bjft = 70.6Zto�/it·· .' NotMeyere:Thheofsultmetimatheod.loadThiof70. s is due62 ttoonltfhteisconsaboutervat6%ivleowerassumptthanitohnatofobtdeptainhed,using Solution

.

Qui.

B qN,,(,) +

.

=

�

� � ��

'

, ";.;'\';'''" .'.

= 3�..

+

�. ' :: ', '

. ':"S/;'{l;":�!:'

B '

Qui.

, ,

"

:

."

"\:i''- ' .

=

factors ' . �

3. 13

Eccentrically Loaded Foundation-Prakash and Saran's Theory

159

Example 3.6

AAssume squar€ltwo-way foundatieccentri on is shown in Fi g ure wi t h city, and determine the ultimate= load,m and have . 3.24,

"

eL

0.3

,Solution'

We

This

� = �� = 0.1

==

0.15 m.

From Figure 3.i6b, for L1 = (0.85) ( 1.5 ) = 1.275 m

case, andsimilar to that shown in Figure is

esiB =

0.1,

L1

7 := 0.85; and

i = 0.21;

.

i-J,s =

I l.5 m 'I eL - -----,-I I •

L,

3.16a.

= (0.21) (1.5) = 0.315 m

0.5 m

�3

-

.

m

-- l-

,

,

1.5 m

1 !.

es

� = �:� = 0.2

ei.JL = O.2

T

Quit.

Figure 3.24 An eccentrically loaded

foundation

160

Chapter 3 Shallow Foundations: Ultimate Bearing Capacity

(3.64), =! From Eq. (3.66), From Eq. (3.65), From Eq.

A'

( LI

) !(1.275 0.315)(1.5) = 1.193 m2

Lo B =

+

+

1.275 m 1.193 = A' = 1.275 = 0.936 m = 0. L' = L I =

B'

L'

(3.56) with = where = (0.7 )(18) = 12.6 kNJm2 = 22.4. Thus, For = 30°, from Table 3. 3 . = 18. 4 and = 1 (BT:') tan' ' = 1 ( 1:2;5 ) tan 30° = 1.424 ( ') ( 0.1.297536 ) = 0.706. = 1 0.4 B = 1 0.4 (0.7) 1. 135 = 2 tan sind>") 1.5 c'

Note from Eq.

q�JFqsFqdFqi

q;,

q



40

(b)

200

30

I

10 30 20 Soil friction angle. cf:/ (deg)

30 J3 ; 300

" ho

:£ 20

,,�

10

10

0.3

0

0 0

30 20 10 Soil friction angle. c{J' (deg) (e)

40

Figure 3.26 Variation of Hc(.-,) with t/J', e/B. and f3

0

30 10 20 Soil friction angle, cf>' (deg) (d)

40

3.14 Bearing Capacity of a Continuous Foundation Subjected to Eccentric Inclined Loading

1 63

50

80

60

40

(3= 10'

(3 = 0

"i!,i

30

20

20

10

0.3

0

0 0

10

,

10

0

40

30

20

30

40

30

40

Soil friction angle, cP' (deg)

Soil friction angle. cP' (deg)

(a)

(b)

'" ]

30

f3 ::= 20"

20

'' = 34', and c' = O. Use Terzlighi's equation to determine the size of the foundation (B). Assume gen­ eral shear failure. 3.3 Use the general bearing capacity equation [Eq. (3.23)] to solve the following: a. Problem 3.la b. Problem 3.1b c. Problem 3.1c Use bearing capacity, shape, and depth factors given in Section 3.6. 3.4 The applied load on a shallow square foundation makes an angle of IS' with the vertical. Given: B = 5.5 ft, Dr = 4 ft, 'Y = 107 1b/ft3, e/>' = 25', and c' = 350 1b/ft'. Use FS = 4 and dete,mine the gross allowable load. Use Eq. . (3.23). Use the bearing capacity, shape, depth, and inclination factors given in Section 3.6. 3.5 A column foundation (Figure P3.5) is 3 m x 2 m in plan. Given: Dr = 2 m, e/>' = 25', c' = 50 kN/m'. Using Eq. (3.23) and FS = 4, determine the net allowable load [see Eq. (3.15)] the foundation could carry. Use bearing capac­ ity, shape, and depth factors given in �ection 3.6. 3.2

L

166

Chapter 3 Shallow Foundations: Ultimate Bearing Capacity

Figure P3.5

3.6

3.7

For a square foundation that is B X B in plan, Dr = 3 ft; vertical gross allowable load, Q,I\ = 150.000 lb; y = 115 Ib/ft3; cf/ = 40°; c' = 0; and FS = 3. Deter­ mine the size of the foundation. Use Eq. (3.23) and bearing capacity, shape, and depth factors given in Section 3.6. A foundation measuring 8 ft X 8 ft has to be constructed in a granular soil de­ posit. Given: Dr = 5 ft and y = 110 Ib/ft3• Following are the results of a stan­ dard penetration test in that soil. Depth (ft) 5

a.

3.8

10

11 14

I)

l li

20 25

21 24

Use Eq. (2.25) to estimate an average friction angle, 4>', for the soil. Use = 14.7 Ib/in'. b. Using Eq. (3.23). estimate the gross ultimate load the foundation can carry. Use the bearing capacity, shape, and depth factors given in Section 3.6. For the design of a shallow foundation, given the following: Po

Soil:

I

Field standard penetration number, N6fJ

•

Foundation:

4>'

= 20°

c' = 72 kN/m'

Unit weight, y = 17 kN/m3 Modulus of elasticity, E, = 1400 kN/m' Poisson's ratio. 1-', = 0.35

LB == 21 mm

Dr = 1 m 3.9

Calculate the ultimate bearing capacity. Use Eq. (3.43). An eccentrically loaded foundation is shown in Figure P3.9. Use FS of and determine the maximum allowable load that the foundation can carry. Use Meyerhofs effective area method and the bearing capacity, shape, and depth factors given in Section 3.6.

4

Problems

; ' . (Eccentricity ". . , . in one direction :'., ,'.,., only) O.IS m

•

� g:�.

,J

i

.

.

..

. '. ' ' C -

.

:'

\

.

' ' . 'r" '"

Centerline

0

:- CP', = 36"

I .s m � I.S m

,

.

y �' 17 kN/mJ , '

': c' =

QaJl

1 67

.

.. . .

..

Figure P3.9

'r =

1 10 Ib/ftJ

.,:"

.

::

l

e=

0.65 ft

I

6.5 ft �Eccentricity one direction I I only) 'rQ' = 122 Ib/ft' = SOO Ib/ft' if! x8 ft ' cp' = 26" c

'

m

... .

Q,

Centerline

Figure P3. 10

Q" •.

':. .

., . 'r =

.

.

. ..

j+2 ft 1-- 5 ft ---l +i

IOS lb/ftJ , '

.

. .,:

· �� �:.. ..::�·.:.·�.����n���:.r !' are given in Figures 4.3, 4.4, and 4.5, respectively.

172

Chapter 4 Ultimate Bearing Capacity of Shallow Foundations: Special Cases

--r��

L l'""':L, __

",.

c'

Figure 4.2 Failure surface under a rough, continuous foundation with a rigid, rough base located at a shallow depth

�

IO.O{JO -,

5000 2000

HIB =

0.2/; 0.33

1000 5oo 2oo N*,

loo 50 20 10 5

0.7

2

0

10

20 ",. (deg)

30

40

Figure 4.3 Mandel and Salencon's bearing capacity factor N; [Eq. (4.2)J

l

4.2 Foundation Supported by a Soil with a Rigid Base at Shallow Depth

173

10.000 5000

. �

2000 1000 500

/

200 N*q

1.0

100

DIB =

3.0

50 20 10 5

1 .2

20

25

30

35

cb' l deg)

40

45

Figure 4.4 Mandel and Salencon"s bearing capacity factor N; [Eq. (4.2)J

Neglecting the depth factors, the ultimate bearing capacity of rough circular and rectangular foundations on a sand layer (c' = 0) with a rough, rigid base lo­ cated at a shallow depth can be given as (4.3) where F;" F;, = modified shape factors. The shape factors F;;. and F;, are functions of H! B and q,'. On the basis of the work of Meyerhof and Chaplin (1953 J, and simplifying the assumption that, in radial planes. the stresses and shear zones are identical to those in transverse planes. Meyerhof (1974) proposed that

174

Chapter 4 Ultimate Bearing Capacity of Shallow Foundations: Special Cases 10.000 5000

2000 1000

HIB = 0.2

500

D18 = 1.5

200

N*y

1 00 50

20 10 5

2

0.5

Figure 4.5 Mandel and

20

25

30

35

40

45

.p' (deg)

I

Salencon's bearing capacity factor N; [Eq. (4.2)1

(4.4) and

L

F;, = 1

-

m2(�)

(4.5)

where = length of the foundation. The variations of "" and nI, with HIB and .p' are shown in Figure 4.6. For saturated clay (i.e., under the undrained condition. or q, = 0). Eq. (4.2) will simplify to the form (4.6)

•

4.2 Foundation Supported by a Soil with a Rigid Base at Shallow Depth

175

1.0 0.8 0.6 m,

0.4 0.2 0 20

35

30

25

,,/ ideg)

1.0

40

45

0.8 0.6 0.4 0.2

-- - � 0.4

-

------

Q -t-----.---,--,----.---,

20

30

25

35

;

(4.9a)

(,UU)

K. = punching shear coefficient. TIlen. ( 2Df) K, tan 4>, "fI H (4.11) 2c�H q" = q" + B + y I H - 1 + H B The punching shear coefficient. K,. , is a function of q,jql and 4>, , or, specifically.

where

'

-

Note that ql and q2 are the ultimate beafing capacities of a continuous founda­ tion of width under vertical load on the surfaces of homogeneous thick beds of upper and lower soil, or

B

(4.12) and (4.13) where N'(l), Ny(l) = bearing capacity factors for friction angle 4>; (Table 3.3) N,(2) , N y(2) = bearing capacity factors for friction angle 4>, (Table 3.3)

1 80

Chapter 4 Ultimate Bearing Capacity of Shallow Foundations: Special Cases

Observe that, for the top layer to be a stronger soil, q,jql should be less than unity. The variation of K, with q,/q, and ; is shown in Figure 4.8. The variation of c�/c; with q,/ql is shown in Figure 4.9. If the height H is relatively large, then the failure 40

30

� 20

10

�,

30

,



(

=

109.41

(30)(5.14) � Z) 0.081 ql O 5YI B�(1) (0.5)(17.5)(2)(109.41 ) From Figure 4.8, for c'�(Z/0.5YIBN>(I) 0.081 and 40°, the value of K, = 2.5. Equation (4.21) then gives [ 1 + (0.2) (�)}1 Z + ( 1 + �)Y1HZ(1 + 2:t)K, ta� + ylDt q,

=

.

2

=

q. =

=

4c

[1 + (0.2) (0)] (5.14) (30) + ( 1

X =

I

[

1

154.2

+

+

107.4

+ 21

=

=

i

+ 0) (17.5) (1.5) '

(2) ( 1.2) tan 40 + (17.5) ( 1 .2) (2.5) 1.5 2.0

J

i

282.6 kN/m2

184

Chapter 4

Ultimate Bearing Capacity of Shallow Foundations: Special Cases

Again. from Eq.

(4.21),

3.3, for 4>i = 40°. Ny 10904 and Nq = 64.20. (3.28) and (3.29), F.'lll = 1 + ( �}an ',

,

. -'

."

For this problem, Eqs. q"

,,;

( 1 + 0.2 �)5.14CI + ylDf ,I

t.

4.4

Closely Spaced Foundations-Effect on Ultimate Bearing Capacity

We are given the following data: '.

�

B= 1m L · =" 1'.

H =i ni.

5 '"

fJil. '.' .

185

. D} '7 1 m . . ' . · · . Yt... .,;;;. 1·6·..8·'�"T . �./·.'.. ...3' . .

'

;c

. FI.O m Fig\ll'� .4.9 for cJc, = 48/120, = 0.4, �h{Y!ll� ofc;;''l2, "3 = 180 - 24>')

Case II. (Figure 4.lOb) If the center-to-center spacing of the two foundations (x = x, < xtl are such that the Rankine passive zones just overlap. then the magni­

tude of q" will still be given by Eq. (4.29). However, the foundation settlement at ul­ timate load will change (compared to the case of an isolated foundation). Case III. (Figure 4.lOc) This is the case where the center-to-center spacing of the two continuous foundations is x = x, < x,. Note that the triangular wedges in the soil under the foundations make angles of 1 800 - 2' at points dl and d,. The arcs

4.4

Closely Spaced Foundations-Effect on Ultimate Bearing Capacity

1 87

of the logarithmic spirals d Ig, and dI e are tangent to each other at d,. Similarly. the arcs of the logarithmic spirals d,g, and d,e are tangent to each other at d,. For this case, the ultimate bearing capacity of each foundation can be given as (c' = 0)

(4.30) where 1;., 1;, = efficiency ratios. The efficiency ratios are functions of xlB and soil friction angle q,'. The theo­ retical variations of 1;. and 1;, are given in Figure 4.11. 2.0

--

Rough base ---- Along this line two footings act as one

" 1.5

3

2

5

4

xiS

Ca)

3.5

I I

Rough base Along this line two footings act as one

3.0

2

3

xiS

4

(b)

Figure 4. 1 1 Variation of efficiency ratios with xlB and ¢' i

(

5

1 88

Chapter 4 Ultimate Bearing Capacity of Shallow Foundations: Special Cases

�

""

�

.2 "

80 60

"

.§

�� � E

� �

"

.§

or.

Angulur sand cf>' � :

•

Relative density

• •

• •

40 20

•

E

.,

• •

O O,.IB = O .. DrIB 1 ==

0 2

=

3

xlB

4

5

Figure 4.12 Settlement at ultimate load for two closely spaced continuous foundations (Redrawn from Das and LarbiCheriL 1982)

Case IV. (Figure 4.10d): If the spacing of the foundation is further reduced such that x = x, < x). blocking will occur and the pair of foundations will act as a single foundation. The soil between the individual units will form an inverted arch which travels down with the foundation as the load is applied. When the two foundations touch. the zone of arching disappears and the system behaves as a single foundation with a width equal to 2B. The ultimate bearing capacity for this case can be given by Eq. (4.29). with B being replaced hy 2B in the second term. The ultimate bearing capacity of two continuous foundations spaced close to each other may increase since the efficiency ratios are greater than one. However, when the closely spaced foundatIOns are subjected to a similar load per unit area the settlement S, will be larger when compared to that for an isolated foundation. Figure 4.12 shows the laboratory model test results of Das and Larbi-Cherif (1982) as they relate to the settlement at ultimate load for two closely spaced continuous foundations. The settlement decreases with the increase in xlB and becomes practi­ cally constant at xlB ., 4. •

4.5

I

Bearing Capacity of Foundations on Top of a Slope In some instances, shallow foundations need to be constructed on top of a slope. In Figure 4.13, the height of the slope is H, and the slope makes an angle {3 with the horizontal. The edge of the foundation is located at a distance b from the top of the slope. At ultimate load. q" . the failure surface will be as shown in the figure. Meyerhof (1957) developed the following theoretical relation for the ultimate bearing capacity for continuous foundations: (4.31) For purely granular soil. c' = 0, thus, q" = �yBN'YCf

(4.32)

Again, for purely cohesive soil. dJ = 0 (the undrained condition); hence, (4.33) where c = undrained cohesion.

.

I

4.5 Bearing Capacity of Foundations on Top of a Slope

I

' .

: .'

.

.

H

"

ib ; .

'"

"."

: .- .

. .,

189

.

y c'

;' . '-�'"

;;..

Figure 4. 13 Shallow foundation on top of a slope

400 300

Qo

f3

- - ------

200

�

,....

� 100

... ...

...

20° �

--

_

40°

E

.... _

_

...

... ,. ...

... ... -

�

... ...

=

�'''-OO------���'_=�4� �� � �::;;��

__

50 ... ... .......

25

�

===== I =:: :::���� - - - -- - -......... - ...cP'-:-�--40-°

- -- - ----- -

�

-

¢' = 30 � - - -= = = = - - - - - - - - - - - - - - - - ---

40"

10 5 o

2

b B 3

4

5

6

Figure 4. 14 Meyerhofs bearing capacity factor Nyq for granular soil (c' = 0)

The variations of Ny. and N,q defined by Eqs. (4.32) and (4.33) are shown in Figures 4.14 and 4.15, respectively. In using N,. in Eq. (4.33) as given in Figure 4.15, the following points need to be kept in mind: 1. The term

is defined as the stability number.

N, =

yH c

-

2. If B < H, use the curves for N, = O. 3. If B .. H, use the curves for the calculated stability number N,.

I

(4.34)

190

Chapter 4 Ultimate Bearing Capacity of Shallow Foundations: Special Cases

8�

�

Df ---- s =1

---

--�----,

---

--

Dr =0 8

6

2

N.•

=4

o +---�--,,---r---.--� .j o

; forNs = 0; � for /v� > O

Figure 4.15 Meyerhofs bearing capac­ ity factor /V,.q for purely cohesive soil

Example 4.5

Di

and

B

=

1.2 1.2

% �:� =

= 1

= 0.67

4.6 Bearing Capacity of Foundations on a Slope

191

For f3 = 30°, Dr/B = 1 and b/ B = 0.75, Figure 4.15 gives N,q = 6.3. Hence, and

q" = (50) (6.3) = 315 kN/m' 315 _ !b.. - ""4 - 78.8 kNIm' q.1I FS _

_

•

Example 4.6 Figure 4.16 shows a continuous foundation on a slope of a granular soil. Estimate the ultimate bearing capacity. , . .. 6m

'Y

�

.1.

1.5 m

I'

-I'

-I

1 .5 m 1.5 m

16.8 kN/m'

l/J'= 40° c' = 0

Figure 4.16 Foundation on a granular slope Solution

For granular soil ( c ' = 0), from Eq. (4.32), We are given that b/B = 1.5/1.5 = 1, Df/B = 1.5/1.5 = 1, cp' = 40°, and f3 = 30°.. . From Figure 4.14, Nyq = 120. So, q" = l (16.8) (1.5) ( 120) = 1512 kN/m'

4.6

•

Bearing Capacity of Foundations on a Slope A theoretical solution for the ultimate bearing capacity of a shallow foundation lo­ cated on the face of a slope was developed by Meyerhof (1957). Figure 4.17 shows the nature of the plastic zone developed under a rough continuous foundation of width B. In Figure 4.17, abc is an elastic zone. oed is a radial shear zone, and ode is a

192

Chapter4 Ultimate Bearing Capacity of Shallow Foundations: Special Cases

" T' .;

','

" '",

",

"e'

Figure 4.17 Nature of plastic zone under a rough continuous foundation on the face of a slope

8 7

"

6

,

,

D,IB = 0 D/B = I , ,

5

,

,

' ... N . = O , '

, ,

,

, ,

, , ,

,

o 3 2

o

I

20

60 40 f3 (deg)

,

,

,

, , ,

, , , ,

80

Figure 4.18 Variation of tv.." with {J.

(Note: N,

=

yH/c)

mixed shear zone. Based on this solution, the ultimate bearing capacity can be ex­ pressed as q"

=

eN" ., (for purely cohesive soil. that is. '"

and q"

=

=

0)

4 yBN" ., (for granular soil. that is e' = 0)

(4.35) (4.36)

The variations of N", and Nrq., with slope angle {3 are given in Figures 4.18 and 4.19.

4.7 Uplift Capacity of Foundations ,

600 500

'-+

, ,

,,

, , ,

,

400 300

J

200

, , , , ,

,

"

,, [00

, ,

, 4l

=

, ,

45°

' ' ' , 40° " ,, ,

193

,

45° "

50 25

,

,

,,

,

,

"

,

,,

,,

,,

'.

10 5 [ 0 0

4.7

10

20

30 /3 (deg)

40

50 Figure 4.19 Variation of N,q, with /3

Uplift Capacity of Foundations Foundations may be subjected to uplift forces under special circumstances. During the design process for those foundations. it is desirable to provide a sufficient factor of safety against failure by uplift. This section will provide the relationships for the uplift capacity of foundations in granular and cohesive soils. Foundations in Granular Soil (c = 0)

Figure 4.20 shows a shallow continuous foundation that is being subjected to an up­ lift force. At ultimate load, Q", the failure surface in soil will be as shown in the figure. The ultimate load can be expressed in the form of a nondimensional break­ out factor, Fq• Or Fq =

�

,

AyDf

(4.37)

where A = area of the foundation. The breakout factor is a function of the soil friction angle q,' and DtiB. For a given soil friction angle, F, increases with DtiB to a maximum at DtiB = (Dr/B)" and remains constant thereafter. For foundations subjected to uplift, Dr/B :5 ( Dr/B)" is considered a shallow foundation condition. When a foundation has an embedment ratio of Dr/B > (Dr/B),,, it is referred to as a deep foundation.

194

Chapter 4 Ultimate Bearing Capacity of Shallow Foundations: Special Cases

Q"

.

:.

,

'.,

'

I

J I I I I I I I I I I I

(

: : -' -:r'-':"

' "; :.' : .

.. . .

I I I I I I I I I I

(Dr/B)", as determined in Step 2, it is a deep foundation. However, if DriB :5 (Dr/B),,, it is a shallow foundation. Step 5. For DriB > (Dr/B)" F, = F; = 7.56 +

Thus, Q" = A

l.44(n

{[ 7.56 + (�)}" + YDr} 1.44

where A = area of the foundation.

I.2 1.0 0.8 /3' 0.6 0.4 0.2 0 0

0.2

Figure 4.22 Plot of f3' versus a'

I

l .

0.6

0.4 c(

0.8

1.0

(4.47)

198

Chapter 4 Ultimate Bearing Capacity of Shallow Foundations: Special Cases

Step 6. For DJiB :S (D;/B)" Q"

[ + (�)} + }

= A (f3'F,: c" + "'(Dr) = A {f3' 7.56

1.44

,

YDr

(4.48)

The value of f3' can be obtained from the average curve of Figure 4.22. The out­ lined procedure outlined above gives fairly good results for estimating the net ulti­ mate uplift capacity of foundations and agrees reasonably well with the theoretical solution of Merifield et al. (2003).

Example 4.7 Consider a circular foundation in sand. Given for the foundation: diameter, B = 1.5 m and depth of embedment. Df = 1.5 m. Given for the sand: unit weight, Y = 17.4 kN/m3, and friction angle, cp' = 35'. Calculate the ultimate bearing capacity. Solution

Dr/B = 1.5/1.5 = 1 and cp' = 35°. For circular foundation. (DtfB)" it'is a shallow foundation. From Eq. (4.38) �J = I +

{I (i) J (� ) + m

r

=

5. Hence,

K" l an cb'

For cp' = 35°, m = 0.25, and K" = 0.936 (Table 4.2). So

Fq = 1 + 2[1 + (0.25) (1))(1) (0.936 ) ( tan35) = 2.638

So Q" = FqyADr = (2.638)(17.4

I

{( :) J

,'� 40' ' =0

el

H

'

eli?kN/m'

4>, = 0 c, � 30 kN/m'

y,

1m

O.S m

1

�.

�

Figure P4.6

IS kN/m'

4> , ' � 40' ' el = 0

1+1'--- 1 .5 m ---+I-I

.

,

:... :.

'

.

"

::

" '

. ..:. : : .

1', � 16 kN/m" q,2 '= 30" c./ = 0

Figure P4.7

spacing = 6 ft. The soil friction angle, the total load on the elemen­ tal area is (SA)

This elemental load, dp, may be treated as a point load. The increase in vertical stress at point A caused by dP may be evaluated by using Eq. (5.1). Note, however, the need to substitute dP q" dx dy for P and x' + y' for r' in that equation. Thus, =

The stress increase at A caused by dP ::;:::

3q (dx dv)zJ

27T(X·�' + y-., + �

z- ) -''1

5 '

The total stress increase :"cr caused by the entire loaded area at point A may now be obtained by integrating the preceding equation:

2 07

Stress below a Rectangular Area

5.4

'
30

5.0

'\.

O.B

2.0

0.6

1.0

0.4

0.5

� �

0.2

HIB, � 0.2

0 0.01 2 4 6 0.1

10 f3 �

E,

100 Figure 5. 19 Variation of IG

kB,

with f3

1.0 0.95 0.9 .!:-

0.B5

( EO :fi k ) ( �: )

l

O.B 0.75

KF � =

Flexibility factor

0.7 +--r-..-rr-r-'--'''''-,.....,.---.TTT..----rr'''--'-TTn 0.1 1.0 10.0 0.001 2 4 0.01 100

K,

l L

Figure 5.20 Variation of rigidity correction factor h with flexibility factor K,. [Eq. (5.40)J

234

Chapter 5 Shallow Foundations: Allowable Bearing Capacity and Settlement 1 .0

0.95

0.9 .::: 0.85

0.4 0.3 0.2

0.8

0. 1

o

0.75 0. 7 0

5

15

1O

20

D,

n;

Figure 5.21 Variation of embedment correction factor h with Dr/B, [Eq (5.41) ]

Example 5.5 For a shallow foundation supported by a silty clay, as shown in Figtfr� S IS; · Length = L = 10 ft .

"

..

.� . -

.

ot foundation Load per unit area = qo = 5000 Ib/ft' E, = 2.3 X 106 1b/in2 The silty clay soil has the following (>ioperties: H = 12 ft Ms = 0.3 Eo = 1400 Ibjin2 k = 25 lb/in'/ft Estimate the elastic settlement of the foundation. J. tU"l...l.it;;;SS

=

.

Solution

From Eq. (5.36), the equivalent diameter is B' =

�4BL = 7r

,

(4) (5) (10) 7r

=

7.98 ft

L

5.9 Improved Equation for Elastic Settlement

so f3

'w

and

= kB,E.

(5.40);

IF

= '4 + 'Ir

= '4 +

= 7.02 and HIB, = 1.5, the value of 10

4.6 + 10

E•

4.6 + 10

From Eq. (5041),

'

+ ...! k

2

1

2.3 X 106 7 8

(2

1

-

1

3.5 exp(1.22/Ls

-

004)

,

=

E.

q.B,IoIFh ( 1

so, with qo = 5000 Ib/ft', it follows that

i

0)

--''---;--=---;-

-

s

\

.

(�; 1.6) 1 = 0.908 . =1. ( 7 98 3.5 exp( 1.22) (0.3) 0.4] -'s 1.6) =

From Eq. (5.39),

l

0.789

(25)

+

+

-

..

0:69. From I\q"

( ErB )(; )3 = [ ( � ) ][ �.�� J 1400 +

_ S, -

=

1

'Ir

IE

7.02

= 7.9812 = 1.5

H B,

FroD1 I'igu�e 5. 19, for f3

1400 (25)(7.98)

=

235

'

_

/Ls )

2

(5000) (7.98) (0.69) (0.789) (0.908) (1 - 0.3 ) - 0.089 ft - 1.07 D1. (1400) (144) ,

_

_

.

•

236

Chapter 5 Shallow Foundations: Allowable Bearing Capacity and Settlement

5. 10

Settlement of Sandy Soil: Use of Strain Influence Factor The settlement of granular soils can also be evaluated by the use of a semiempirical

strain influence factor proposed by Schmertmann et al. (1978). According to this method, the settlement)s

S,

= C1C,Cq - q) }:; Az ,

"I 0

(5.42)

�

where I, = strain influence factor

C, = a correction factor for the depth of foundation embedment = [q/(q - q)l

J

- 0.5

C, = a correction factor to account for creep in soil = 1 + 0.2 log (time in years/a.! ) q = stress at the level of the foundation q = yDf

The recommended variation of the strain influence factor I, for square (L/ B = 1) or circular foundations and for foundations with LI B ;,: JO is shown in Figure 5.22. The I, diagrams for 1 < L/B. < 10 can be interpolated. The procedure to calculate elastic settlement using Eq. (5.42) is given here (Figure 5.23).

I--- B

J

'1

.J...

" = B/2 ZI

=B

LIB =

1

LIB '" 1 0

;:2

= 48

Figure 5.22 Variation of the strain influence factor. I�

I

5. 10 Settlement of Sandy Soil: Use of Strain Influence Factor

237

E,

"

1.-

_ _ _ _____________

lkpth, : (a)

�

Depth. � (b)

Figure 5.23 Procedure for calculation of S" using the strain influence factor

Step 1. Plot the foundation and the variation of i, with depth to scale (Figure 5.23a).

Step 2. Using the correlation from standard penetration resistance (N6o) or cone penetration resistance (qel, plot the actual variation of E, with &epth (Figure 5.23b). Schmertmann et al. (1978) suggested E, = 3.5q,.

Step 3. Approximate the actual variation of E, into a number of layers of soil having a constant E" such as E,(tl, E'(2 ), . . . . E,(I)." E,(,,) (Figure 5.23b). Step 4. Divide the soil layer from z 0 to Z = Z2 into a number of layers by =

drawing horizontal lines. The number of layers will depend on the break in continuity in the i, and E, diagrams. L

Step 5. Prepare a table (such as Table 5.6) to obtain l-'- az. E, Step 6. Calculate C, and C2• Step 7. Calculate S, from Eq. (5.42).

I

l .

I 238

Chapter 5 Shallow Foundations: Allowable Bearing Capacity and Settlement L

Table 5.6 Calculation of � E, layer No. I

2

,J,z �Z(1) .:.\ ;: w

E,

r

.'.: /z at the

middle of the layer /:( 1 )

E'(I)

I, .l.z E,

1;( I )

--

E.,( l l

I:C�1

£" (2)

- I!

r

;l z J

II

Example 5.6 Figure 5.24 shows a shallow continuous foundation along with the variation o f E, with depth obtained from cone penetration tests (broken line in Figure 5.24b). Given: B = 8 ft, D, = 4 ft, Y = 110 Ib/ft3, and 7j = 25 lb/in'. Estimate the elastic settlement using the strain influence factor method described in Section 5.10:

Solution

Given L/B > 10. Based on this, the I, diagram is plotted in Figure 5.24a: The ap­ proximate variation of E, is shown in Figure 5.24b. The soil below the foundation has been divided into S layers. Now Table 5.7 can be prepared. Since y = 1 1 0 Ib/ft3, q = yDf = (4) ( 1 10) = 440 lb/ft' = 3.06 lb/in'. Given q = 25 lb/in'. Thus, q - q = 25 - 3.06 = 21.94 Jb/in'. Also,

( )

C, = 1 - 0.5 7j

q

_

q

= 1 - 0.5

( ) 3.06 21.94

Assume the time for creep is 10 years. Hence,

C, = 1 + O.2 l0g

G.�)

= 1.4

= 0.93

5. 10 Settlement of Sandy Soil: Use of Strain Influence Factor Table 5.7 Elasijc settlement Calculation

!.aYer Nd;',

,

. •

" Az ,

"" (lit,1

z to the middle ofthe,layer (In.I

, E,

1I"lIn3,·

'; . 7�O ' " 1250

24 72 144 216 312

1 2$9 l

OW

· 2000, . .

11/ at the middle oftbe layer 0.275 0.425 0.417

0.292

0.125

'

o .. . : : ' . . . " .�

.

Layer I

, I I I \

2

, ,

16

I, -, Az E

(1r>'1Jb1

, O:Oi76

0.016 '; 0.632 ,

o:oi� ' 1i , oo!k , "

, 'A

, ."

E,(lb/in' ) 2000

1000

239

\ \ \ I I I I , ,

\ \ \

5

I I I

\ I

, , , ,

32 z (f,)

(a)

(b)

Figure 5.24

Thus, S, = C, C,(/i

-

'" I, q ) "JF:l1z ,

= (0.93) (1.4) (21.94) (0.0886) = 2.53 ;0. I

l

I

•

240

Chapter 5 Shallow Foundations: Allowable Bearing Capacity and Settlement Table 5.8 Elastic Parameters of Various Soils

Modulus of elasticity. Es

Type of soil

Loose sand Medium dense sand Dense sand Sil ty sand Sand and gravel Soft clay Medium clay Stiff clay

5. 1 1

MN/m'

Ib/l"'

10.5-24.0 17.25 -27 .60 34.50-55.20 10.35-17.25 69.00-172.50 4.1-20.7 20.7-41.4 41.4-96.6

1500-3500 2500-4000 5000-8000 1500-2500 10,000-25,000 600-3000 3000-6000 6000-14.000

Poisson's ratio, p. 0.20-0.40 0.25-0.40 0.30-0.45 0.20-0.40 0.15-0.35 0.20-0.50

Range of Material Parameters for Computing Elastic Settlement Relationships for calculating elastic settlement presented in Sections 5.7 through 5.10 involve elastic parameters E, and /-" for soils. Table 5.8 shows some approxi­ mate values of modulus of elasticity and Poisson's ratio for various soils. However, it must be realized that actual estimation of E, is difficult and challenging. The relia­ bility of the elastic settlement calculation primarily uepcIllb on that. As a first approximation, the magnitude of E, in sandy soil can be approxi­ mated according to Eq. (2.27), or E,

- = exNw

{

Po

where Po

(5.43)

= atmospheric pressure = 100 kN/m' ( = 2000 lb/ft' )

ex =

5 for sands with fines 10 for clean normally consolidated sand 15 for clean overconsolidated sand

Schmertmann et al. (1978) further suggested that the following correlations for sand may be used with the strain influence factors described in Section 5.10. E, and

= 2.5q" (for square and circular foundations)

(5.44a)

= 3.5q" (for strip foundations)

(5.44b)

E,

[Note: Any consistent set of units may be used in Eqs. (5.44a) and (5.44b ).] The modulus of elasticity (E,) for clays can, in general. be given as (5.45) where c" = undrained shear strength:

5.12 Settlement of Foundation on Sand Based on Standard Penetration Resistance Table 5.9

241

Range of f3 for Clay [Eq. (5.45)]'

Plasticity index

OCR - 1

OCR _ 2

JJ OCR - 3

OCR - 4

OCR

< 30 30 to 50 > 50

1500-600 600-300 300-150

1380-500 550-270 270-120

1200-580 580-220 220-100

950-380 380-180 180-90

730-300 300-150 150-75

-

5

'Interpolated from Duncan and Buchignani (1976) The parameter {3 is primarily a function of the plasticity index and overconsol­ idation ratio. Table 5.9 provides a general range for {3 based on that proposed by Duncan and Buchignani (1976). In any case, proper judgment should be used in selecting the magnitude of {3. .

·5. 12

Settlement of Foundation on Sand Based on Standard Penetration Resistance Meyerhof's Method Meyerhof (1956) proposed a correlation for the net bearing pressure for foundations with the standard penetration resistance, Nw. The net pressure has been defined as q,,, = q - yD,

where q = stress at the level of the foundation; According to Meyerhofs theory. for 25 mm (1 in.) of estimated maximum set­ tlement. N60 q,,,,(kIp/ft") = 4 (for B •

I

'"

and N60 . q,,,(klp/ft' ) = 6"

B () B-

+1 '

" 4 ft)

(for B > 4 tt)

(5.46a)

(5.46b)

Since the time that Meyerhof proposed his original correlations. researchers have observed that its results are rather conservative. Later, Meyerhof (1965) sug­ gested that the net allowable bearing pressure should be increased by about 50%. Bowles ( 1 977) proposed that the modified form of the bearing equations be ex­ pressed as • 2 _ N,; q,,, (kIp/tt ) F"S, (for B 2.5

and NW q,,,(kip/ft') = 4

( -B B-)

+1 '

:5 4 ft)

F"S, (for B > 4 tt)

(5.47a)

(5.47b)

242

Chapter 5 Shallow Foundations: Allowable Bearing Capacity and Settlement

12

10

-r---,

-f----,.

8

:a

25

"

I

"I"'� �

;::. :.:.., o

�

20

6

15

4

10 2 +---_

5

0 +---,-,-----,--, R o 16 B (fl)

Figure 5.25 Plot of q"JF,,s, against B [Eqs. (5.47a) and (5.47b)]

where

=

Fd = depth factor = 1 + 0.33(Dr/ B) B foundation width, in feet S, = settlement, in inches Figure 5.25 shows a plot of q",JFdS, for various values of B and N60. Thus, ' 2.5q",, (kip/ft ) . (for B '" 4 ft) S,(m.) =

I

and

NoFd

( )

' 4qn,, ( ki P/ft' ) B . (for B > 4 ft) S,( m.) = B+ 1 N60Fd In SI, units, Eqs. (5.47a) and (5.47b) can be written as , q",, ( kN/m ) -

_

and

N60 , q",. (kN/m- ) = 0.08

( )

S, N60 0.05 F" 25

(

) ()

S, B + 0.3 ' F" 25 B

where B is in meters and S, is in mm. Hence, l.25q",,( kN/m' ) , mm) SJ N",Fd

=

(for B

'"

1.22 tn )

(for B > 1.22 m)

(for B '" 1.22 m)

(5.48)

(5.49)

(5.50)

(5.51)

(5.52)

5.12 Settlement of Foundation on Sand Based on Standard Penetration Resistance and

243

)

(

2q,,,(kNjm' ) ' B (for B > 1.22 m) (5.53) Nr,oFd B + 0.3 The Nro referred to in Eqs. (5.46) through (5.53) is the standard penetration resis­ S,(mm)

=

tance between the bottom of the foundation and 2B below the bottom.

Burland and Burbidge's Method Burland and Burbidge (1985) proposed a method of calculating the elastic settle­ ment of sandy soil using the field standard penetration number, Nro. (See Chapter 2.) The method can be summarized as follows: 1_

Variation of Standard Penetration Number with Depth

Obtain the field penetration numbers (Nro) with depth at the location of the foundation. The following adjustments of Nro may be necessary, depending on the field condi tions: For gravel or sandy gravel,

Nro(.) = 1.25 Nro For fine sand or silty sand below the groundwater table and Nro > where

Nro(.) = 15 + O.5(Nro - 15) Nro(.) = adjusted Nro value.

(5.54) 15, (5.55)

2. Determination of Depth of Stress InDuence (z')

In determining the depth of stress int1uence, the following three cases may arise:

- = 1.4(-)".75

Case I. If N", [or Nro('ll is approximately constant with depth, calculate z' from B z' where BR B

. { == 0 3

= reference WIdth

=

(5.56)

BR

BR

I ft ( if B is in ft) " m m) . m ('f I B Is width of the actual foundation

Case II. If Nro [or Nro(.,l is increasing with depth, use Eq. (5.56) to calculate z'.

=

=

Case III. If Nro [or Nro(.)l is decreasing with depth, calculate z ' 2B and z ' distance from the bottom of the foundation to the bottom of the soft soil layer (z"). Use z ' = 2B or z ' z " (whichever is smaller).

=

3. Calculation of Elastic Settlement S,.

The elastic settlement of the foundation, S" can be calculated from ,

(�) J 0.25 (�) (:J,(; l.25

+

(5.57)

244

Chapter 5 Shallow Foundations: Allowable Bearing Capacity and Settlement where

at a, a3 Pa

L

= = = = =

a constant compressibility index correction for the depth of influence atmospheric pressure = 100 kN/m' ( =2000 Ib/ft' ) length of the foundation

For normally consolidated sand,

a, = 0.14

(5.58)

and

a2 =

1.71

[N6() or N6D(a)) 1.4

(5.59)

where N6() or N6()(a) = average value of N6() or N6()(,) in the depth of stress influence. Z" Z" a- = -' 2 -' ", 1 (5.60) .> Z Z

(- )

and

(5.61) in which q,et = net applied stress at the level of the foundation (i.e., the stress at the level of the foundation minus the overburden pressure). For overconsolidated sand with q,,, ", a;, the preconsolidation pressure,

a, = 0.047

(5.62)

and 0.57

(5.63)

For a3, use Eq. (5.57): ' q ::;; qnt!1

(5.64)

For overconsoLidated sand with qnet > u�.

a, = 0.14

(5.65)

For a2, use Eq. (5.63), and for a3, use Eq. (5.60). Finally, use q

' ==

qnel -

O.67u�

(5.66)

A.��� �utid�tiori measuring 1.75 m x 1.75 m isto be constructed over a).:ii�f.�i san4. qiv�D;; l m; Noo is generally increasing with depth; Noo m the aePd,f:;&( d; .· .

5.12 Settlement of Foundation on Sand Based on Standard Penetration Resistan* Sollnion

245

( )0.75

FfOIJl:Eq. (5.56),

B Z' = 1.4 BR Bf(

DepthoiStrtiss influerice,

Fo��.omiallY consolidated sand, a,

2-

a

a3

1.71

= 0.14

(Noo) !.'

=1

[

1.71 O.068 (10) ,.4 =

q ' = qnot = 120 kNlm'

So,

� = (0.14) (0.068) ( 1 ) 0.3

S,

(1.25)

0.25 +

= 0.0118 m = 11.8 mm

Example S.S Solve Example 5.7 using Meyerhof's method. Soiiition From Eq. (5.53),

S, =

2qnot (Noo ) (Fd)

(B B0.3 ) +

'

]

(1.75 1.75 ) ' (1.750.3 ) ' (100 120 ) (1.75 1.75 ) 07

•

246

Chapter 5 Shallow Foundations: Allowable Bearing Capacity and Settlement " Fd

5. 13

S,

= 1 + 0.33(Df/B) = 1 + 0.33(1/1.75) = 1.19 (2) (120) = (10) (1.19)

( 1.751.750.3 ) +

'

= 14.7 mm

•

General Comments on Elastic Settlement Prediction Idealized versions of elastic settlement prediction have been discussed in the pre­ ceding sections of this chapter. Predicting settlements of shallow foundations in sand is prone to uncertainties due to highly erratic density and compressibility vari­ ations and the difficulties associated with sampling and assessing the in situ charac­ teristics of the granular soil deposits (for example, see Briaud and Gibbens, 1994). To illustrate this point. Sivakugan et al. (1998) compiled the settlement records for 79 foundations summarized by Jeyapalan and Boehm (1986) and Papadopoulos (1992) and compared them to those predicted by Schmertmann's strain influence factor method. The comparison between the predicted and observed settlements in shown in Figure 5.26. From this figure. it can be seen that predicted settlements

., :

1 20

�

�'i

l

O (ID O

tOO

� §

0 0

80

0

0

0

C

E

"

! �

.!j

60

"-

40

0;

" "0 "

U '5 �

0

0

20

o fL------.-----,--r ----. 60 o 20 80 40 Observed settlement (mm)

Figure 5.26 Predicted versus observed settlement of a shal­ low foundation or sand. Prediction by strain influence factor method [Adapted from Sivakugan and Johnson

(2004) J

5. 14 Seismic Bearing Capacity and Settlement in Granular Soil

247

Table 5. 10 Staiistical Analysis of Settlement Prediction in Sand-Based on the Study of Sivakugan and Johnson (2004) \.>-

Probability of exceeding 25 mm settlement in the field

Predicted settlement (mm)

Schmertmann et at (1978)

Burland and Burbidge (1985)

1 5 10 15 20 25 30 35 40

0 0 2% 13% 20% 27% 32% 37% 42%

0 3% 15% 25% 34% 42% 49% 55% 61%

Figure 5.27 Failure surface in soil for static bearing capacity analysis. (Note: etA = 45 + '

1.24 1.13 1.00 0.87 0.74 0.61 0047 0.32

=

3D·

' = 35·

1.39 1.26 1.15 1.02 0.92 0.77 0.66 0.55 0.42 0.27

1.57 1.44 1.32 1.18 1.06 0.94 0.84 0.73 0.63 0.50 0.44 0.32

' = 40'

1.75 1 .63 1048 1.35 1 .23 1.10 0.98 0.88 0.79 0.68 0.60 0.50

\: ; ,\,>' A strip foundation

is to be constructed on a sandy soil with B = 4 ft, Df = :i ft, 3, and cP' = 30'. 110 Ib/ft Y= ,

8.

Determine the gross ultimate bearing capacity quE' Assume that ku = 0 and

k. = 0.176.

b. If the design earthquake parameters are V = 1.3 ft/sec and A = 0.32, deter­ mine the seismic settlement of the foundation. Use FS = 3 to obtain the sta­ . tic allowable bearing capacity.

SOlution Part a

FromFigure 5.29, for r/>' = 30', Nq = 16.51 and N

,

.

tan 0 = i

For tan 0 = 0.176, Figure 5.30 gives N Ny

2!!. =

0.4

Thus,

=

kh k" = 0.176

23.76. Also,

_

and

;/q = 0.63

N

NYE = (0.4) (23.76) = 9.5 NqE =

(0.63) (16.51) = 10.4

252

Chapter 5

Shallow Foundations: AJ/owable Bearing Capacity and Settlement

and

quE = qNq£ + hBNYE =

x

(3 110)(10.4) + (�)(110)(4)(9.5) = 55221b/ft2

Part b For the foundation,

3 4 0.75 5.31,5.11,q,' k�30', 0.26 3, q,' DJiB 0.75, 30', (5.69), 0.174Ik�'I-4 ( AVg') Df

-=

B

From Figure for = Also, from Table for From Eq. we have

=

=

FS = and and =

(meters)

V 3

= l . ft = OA m

it follows that SEq =

0.174 (0.32)(9.81) 1°0..32261 -' (0.92) 0.0187 (OA)'

ki,

= the value of � 0.26. the value of tan "A£ =. 0.92�

tan "AE

SEq =

with

-

=

.

m = 0.74 m. .

-

Consolidation Settlement

. '. ::5.;15".

Primary Consolidation Settlement Relationships As mentioned before, consolidation settlement occurs over time in saturated clayey soils subjected to an increased load caused by construction of the foundation. (See Figure On the basis of the one-dimensional consolidation settlement equations given in Chapter 1, we write •

5.32.)

Seep) =

where = vertical strain �e = + eo �e = change of void ratio = f(
(2.5) log( 52.84 + 14.38 ) - 0.0465 m - (0.32) 52.84 1 + 0.8 _

_

= 46.5mm

258

Chapter 5 Shallow Foundations: Allowable Bearing Capacity and S.ttl.ment

Now assuming that the 2:1 method of stress increase (see Figure 5.5) holds good, the area of distribution of stress at the top of the clay layer will have dimensions B'

and

= width = B + Z =

I + (1.5 + 0.5) = 3 m

=

2 + (1.5 + 0.5) = 4 m

L' = width

L+

Z =

The diameter of an equivalent circular area, B,q' can be given as 71"

4

so that B'q =

Also,

�

B2'q

4B'L' 71"

=

= B'L' (4) (3)(4)

='

H, Beq

= 3.91

2.5 = 0.64 3.91

From Figure 5.34, for A = 0.6 and H,IB,q Hence, S,fp) = K"S,fp).o:'.:':�: ·�'.· ���c.�n��iidali�� press�r� = '2� i�ft2

: · f.::, · ;-;' :�" .�·+�.l. :t-}�;::. :\y�, :=s�� :'� .:: .: ;:::.: . y�, .

"

'

"

. '•

.

.

. -

'

'

.

.

.

.

.

eo

'�.,

. ..

Figure P5.6

/-20 f'-1

1 Center line

I V:2H

30 f'

IV:2H Y=

1 I5 Ib/f,' .

. �

.

'

"

'. "

Figure P5.8

Rock

Figure P5.9

5.12 Repeat Problems 5.11 for a foundation of size = 2.1 m x 2.1 m and with qo = 230 kN/m', DJ = 1.5 m, H = 12 m; and soil conditions of /L, = 0.4, E, = 16,000 kN/m', and l' = 18.1 kN/m'. 5.13 Refer to Figure 5.17. A foundation measuring 1.5 m X 3 m is supported by a saturated clay. Given: DJ = 1.2 m, H = 3 m, E, (clay) = 600 kNlm', and qo = 150 kN/m'. Determine the elastic settlement of the foundation.

5.14 For a shallow foundation supported by a silty clay, as shown in Figure 5.18, the

following are given:

268

Chapter 5 Shallow Foundations: Allowable Bearing Capacity and Settlement

Length, L = 8 ft Width, B = 2.5 ft Depth of foundation. Df = 2.5 ft Thickness of foundation. t = 1 ft Load per uuit area, qo =· 3000 lb/ft' , Ef = 2 X 10' lb/jn' The silty clay soil has the following properties: H = 8 ft

fJ..,

= 0.4 Eo = 1250 lb/in'

k = 30 lb/in'/ft

Using Eq. (5.39). estimate the elastic settlement qf the foundation. plan calls for a square foundation measuring 3 m X 3 m, supported by a layer of sand. (See figure 5.18.) Led Df= 1 .5 m, ( = 0.25 m, Eo = 16,000 kN/m2, k = 400 kN/m2/m. fJ., = 0.3, H = 20 m, Ef = 15 X 10' kN/m2, and qo = 150 kN/m'. Calculate the elastic settlement. Use Eq. (5.39). 5.16 SOlve Problems 5.11 with Eq. (5.42). For the correction factor C,., use a time of 5 years for creep and, for the unit weight of soil. use y = 115 lb/ft'. Assum­ ing an I" plot the same as that for a square foundation. 5.17 Solve Problem 5.12 with Eq. (5.42). For the correction factor C,. use a time of 5 years for creep. 5.18 A continuous foundation on a deposit of sand layer is shown in Figure P5.18 along with the variation of the modulus of elasticity of the soil (E ) Assuming y = 115 lb/ft' and C2 = 10 years, calculate the elastic settlement of the foun­ dation using the strain influence factor. 5.15

A

, ,

E,

Sand

=

875 Ib/in'

6 E, = 1 740 Ib/in'

20 - - - - - E, = 1450 Ib/in'

40 - - - - - Depth (ft)

Figure P5. 18

References

269

5.19 Following are the results of standard penetration tests in a granular soil de­

posit.

Depth (It)

W>

5 10 15 20 25

Standard penetration number. Neo 10 12 9 14 16

What will be the net allowable bearing capacity of a foundation planned to be 5 ft X 5 ft? Let Df = 3 ft and the allowable settlement = 1 in., and use the re­ lationships presented in Section 5.12. 5.20 A shallow foundation measuring 1 m x 2 m in plan is to be constructed over a normally consolidated sand layer. Given: Df = 1 m,N60 increases with depth, N60 (in.the depth of stress influence) = 8, and q,,, = IS3 kN/m'. Estimate the elastic settlement using Burland and Burbidge's method. 5.21 Following are the average values of cone penetration resistance in a granular soil deposit: Depth (m) 2 4 6 8

10 15

Cone penetration resistance. qt: (MN/m') 2.1 4.2 5.2 7.3 8.7 14

Assume that y = 16.5 kN/m' and estimate the seismic ultimate bearing ca­ pacity (q"E) for a continuous foundation with B = 1.5 m, Df = 1.0 m, kh = 0.2, and k, = O. Use Eqs. (2.47), (S.67), and (5.68). 5.22 In problem 5.21, if the design earthquake parameters are = 0.35 m/sec and A = 0.3, determine the seismic settlement of the foundation. Assume that FS = 4 for use in obtaining the static allowable bearing capacity. 5.23 Estimate the consolidation settlement of the clay layer shown in Figure PS.6 using the results of Problem 5:6. 5.24 Estimate the consolidation settlement of the clay layer shown in Figure PS.6 using the results of Problem 5.7.

V

References

Ahlvin, R. G. and Ulery. H. H. (1962). Tabulated Vollies ofDetermining the Composite Pattern of Stresses, Strains, and Deflections beneath a Uniform Load on a Homogeneous Half Space. Highway Research Board Bulletin 342, pp. l-13. .

270

Chapter 5 Shallow Foundations: Allowable Bearing Capadty and Settlement American Society forTesting and Materials (2000) .AllIwal Book ofASTM Standards, Vol. 04.08. West Conshohocken. PA. Bjerrum. L. (1963). "Allowable Settlement of Structures:' Proceedings, European Confer­ ence 011 Soil ..\llecflanicJ alld Foundation Engineering. \Viesbaden. Germany, Vol. III. pp. 135-137. Boussinesq. 1. (1883). Application des Potentials a L'Etude de L'Equilibre et dll MOllvement des Solides £lastiques. Gauthier-Villars. Paris. Bowles. 1. E. (1987). "Elastic Foundation Settlement on Sand Deposits." Journal .of Geotechnical Engineering. ASCE. Vol. 113. No. 8. pp. 846-860. Bowles. 1. E. ( 1 977). Foul/dation Analysis and'"Design. 2d ed.. McGraw-HilI. New York. Briaud. 1. L.. and Gibbens. R. M. (1994). "Predicted and Measured Behavior of Five Spread Footings on Sand," Geolechnical Special Publications No. 41, American Society of Civil Engineers. Burland. 1. B., and B urbidge. M. C. (1985). "Settlement of Foundations on Sand and Gravel." Proceedings. [nstitllle of Ci"il Engineers, Part I. Vol. 7. pp. 1325-1381. Christian. 1. T.. and Carrier. W. D. (19781. "Janbu. Bjerrum. and Kjaernsli's Chart Reinter­ preted:' Canadian Geotechnical Journal, Vol. 15. pp. 124-128. Das. B. (1997). Advanced Soil ,)

Plan

•

t

•

,- - - ,I , I

'. _ - '

• •

_OM, I '

•

.. .... "': _ _ :_ -! _ -: _ .!- _ -+ _ J '

•

,

__•

,

'_ O M '

'_ O M '

,

' '_ O M '

�---:r---;r---: •

r--�T---n---T

I 11 I J! I �------'. -:---"---1 �------1 ,I ]l II I II II '- _ _ _1 1... _ _ _1 1.. _ _ _1

'_OM'

, I ,_ _ _ ,

(c) Section

•

I II 11 jl I II I � _ _ _ � _ _ _I � _ _ _l

(d)

, ,_ _ _ ,,

•

'_OM'

r---)r---lf

-

,

• • • .... . . . • • • • ,. . . ., ,. . . ., ,. . . .,

_ _

r--t--f---i

+

,

_OM'

' ,

J

•

Plan

•

Plan

(e)

Figure 6.4 Common types of mat foundation

load·bearing capacities. but that will have to support high column or wall loads. Un­ der some conditions, spread footings would have to cover more than half the build­ ing area, and mat foundations might be more economical. Several types of mat foundations are used currently. Some of the common ones are shown schematically in Figure 6.4 and include the following: 1. Flat plate (Figure 6.4a). The mat is of uniform thickness. 2. Flat plate thickened under columns (Figure 6.4b). 3. Beams and slab (Figure 6.4c). The beams run both ways, and the columns are lo­

cated at the intersection of the beams.

4. Flat plates with pedestals (Figure 6.4d). 5. Slab with basement walls as a part of the mat (Figure 6.4e). The walls act as stiff­

eners for the mat.

Mats may be supported by piles, which help reduce the settlement of a struc­ ture built over highly compressible soil. Where the water table is high, mats are of­ ten placed over piles to control buoyancy. Figure 6.5 shows the difference between the depth Dr and the width B of isolated foundations and mat foundations. .

.,

6.4 Bearing Capacity of Mat Foundations

I I

277

I

H Drf 'B

vI I I·

I

B

Figure 6.5 Comparison of isolated foundation and mat foundation (8 = width, Dr = depth)

,I

Bearing Capacity of Mat Foundations

gross ultimate bearing capacity

The of a mat foundation can be determined by the same equation used for shallow foundations (see Section 3.6), or [Eq. (3.23)]

B

(Chapter 3 gives the proper values of the bearing capacity factors. as well as the shape depth. and load inclination factors.) The term in Eq. (3.23) is the smallest di­ mension of the mat. The of a mat foundation is

net "liinu"e capacity

[Eq. (3.14)j

allowable

A suitable factor of safety should be used to calculate the net bear­ ing capacity. For mats on clay, the factor of safety should not be less than 3 under dead load or maximum live load. However. under the most extreme conditions, the factor of safety should be at least 1.75 to 2. For mats constructed over sand, a factor of safety of 3 should normally be used. Under most working conditions, the factor of safety against bearing capacity failure of mats on sand is very large. For saturated clays with ", = 0 and a vertical loading condition,Eq. (3.23) gives (6.8)

where = undrained cohesion. = 5.14, N., = I , and From Eqs. (3.27) and (3.30). for '" = 0,

c"

(Note: N,

F =1 "

and

+

B (NN,q) L

=1

+

,( B ) (_I_) L

5.14

=1

+

Ny

= 0.)

0.195B L

278

Chapter 6 Mat Foundations

(6.8) yields qu = 5.14cu( 1 + 0.195B)( O.4 DBf) + q (6.9)

Substitution of the preceding shape and depth factors into Eq. 1+

L

Hence, the net ultimate bearing capacity is

q.,,(u) q" - q = 5.14c". (1 :-.0.195B)( + O.4BDr) . . 3. the net allowable soil bearing capacity becomes qUI'OI) = 1.713c" ( 1 + 0.195B) ( 1 + OABDr ) q'OI(,n) = "-"FS =

L

(6.10)

1

For FS =

(6.11)

-L --

The net allowable bearing capacity for mats constructed over granular soil de­ posits can be adequately determined from the standard penetration resistance num. bers. From Eq. for shallow foundations.

(5.51). q",,(kN/m") = ,

IV",

(l.OK

(B + 0.3)' (5,) R

where

F" 25

[Eq.

(5.51)]

IVB60 = width standard penetration resistance (m) F" 1 + 0.3 3( Dr/B) ';; 1 .33. = settlement, (mm) =

=

5,

When the width B is large, the preceding equation can be approximated as

q.,,(kN/m') ;O� F"G;) IV ( r) (mm) ] = 60[ 1 + 0.33 D ][ S, 0.08 B 25 .;; 16.63IV60[ ,(�m) ] In English units, Eq. (6.12) may be expressed as q.,,(au) (kip/ft') = O.25IV,{1 + O.33(�)Jrs,(in.)] .;; 0.33IV60[S,(in.)] =

(6.12)

5

(6.13)

6.4

1

Figure 6.6

1

Bearing Capacity of Mat Foundations

279

Unit weight = -y

Q

Defmition of net pressure on soil caused by a mat foundation

Generally, shallow foundations are designed for a maximum settlement of 25 mm (1 in.) and a differential settlement of about 19 mm (0.75 in.). However, the width of the raft foundations are larger than those of the isolated spread footings. As shown in Table 5.3, the depth of significant stress increase in the soil below a foundation depends on the width of the foundation. Hence, for a raft foundation, the depth of the zone of influence is likely to be much larger than that of a spread footing. Thus, the loose soil pockets under a raft may be more evenly dis­ tributed, resulting in a smaller differential settlement. Accordingly, the customary as­ sumption is that, for a maximum raft settlement of 50 mm (2 in.), the differential settlement would be 1 9 mm (0.75 in.). Using this logic and conservatively assuming that Fd = 1. we can respectively approximate Eqs. (6.12) and (6.13) as (6.14) and (6.15) The net allowable pressure applied on a foundation (see Figure 6.6) may be expressed as q=

Q D A-y r

(6.16)

where Q = dead weight of the structure and the live load A = area of the raft

In all cases, q should be less than or equal to allowable q,,,.

Example 6.1 Deteimine the net ultimate bearing capacity of a mat foundation measuring 45 ft X 30 ft on a saturated clay with Cu = 1950 Ib/ft', '" = 0, and Dr = 6.5 ft.

280

Chapter 6 Mat Foundations Solution

(6.10), qnot(u) = 5.14Cu[ (0.1�5B ) ][ 1 + 0.4�J 0.19! X 3b )J [1 ( 0.4:06.5 ) J = (5. 1 4) (1950{ 1 + ( 5

.r

From Eq.

1 +

+

= 12,307 Ib/ft2

•

Example 6.2 What will be the net allowable bearing capacity of a mat foundation with dimeitsions m x m constructed over a sand deposit? Here,DJ = m, the allowable set­ tlement is mm, and the average penetration number N60 =

of15 2510 210. From Eq. (6.12), qnott,lI) :�[ 1 0.33(;) J(:;) .. 16.63N60(:; ) Solution

=

or

0.33 2 J(25 ) 1 + qnot(aU) - JQ.. 0.08[ 10 25 _

6.5

+

x

-

_

133.25 kN/m2

Differential Settlement of Mats

1988,

336

In the American Concrete Institute Committee suggested a method for calculating the differential settlement of mat foundations. According to this method, the rigidity factor K, is calculated as

(6.17) where = modulus of elasticity of the material used in the structure = modulus of elasticity of the soil = width of foundation = moment of inertia of the structure per unit length at right angles to B The term can be expressed as

E'E,. lbB

E'lb

. E'Ib = E'(IF + 2,lb;+ 2,��)

(6.18)

, 6. 7

Compensated Foundation

281

where E'Ib = flexural rigidity of the superstructure and foundation per unit length at right angles to B lE' I b = flexural rigidity of the framed members at right angles to B l(E'ah'/12) = flexural rigidity of the shear walls a = shear wall thickness h = shear wall height E'IF = flexibility of the foundation

Based on the value of K" the ratio (8) of the differential settlement to the total set­ tlement can be estimated in the following manner: 1. If K, > 0.5, it can be treated as a rigid mat, and 8 = O. 2. If K, = 0.5, then 8 = 0.1. 3. If K, = 0, then /) = 0.35 for square mats (B/L = 1) and 8 = 0.5 for long foundations (B/ L = 0).

6.6

Field Settlement Observations for Mat Foundations Several field settlement observations for mat foundations are currently available in the literature. In this section, we compare the observed settlements for some mat foundations constructed over granular soil deposits with those obtained from Eqs. (6.12) and (6.13). Meyerhof (\965) compiled the observed maximum settlements for mat foun­ dations constructed on sand and gravel, as listed in Table 6.1. In Eq. (6.12), if the depth factor, \ + 0.33 (Df/B), is assumed to be approximately unity, then S,(mm )

=

2qn"(,II) 6Q N

(6.19)

From the values of qn,, (,II) and N60 given in Columns 6 and 5, respectively, of Table 6.1, the magnitUdes of S, were calculated and are given in Column 8. Column 9 of Table 6.1 gives the ratios of calculated to measured values of S,. These ratios vary from about 0.79 to 3.39. Thus, calculating the net allowable bearing capacity with the use of Eq. (6.12) or (6.13) will yield safe and conservative values. 6. 7

Compensated Foundation Figure 6.6 and Eq. (6.16) indicate that the net pressure increase in the soil under a mat foundation can be reduced by increasing the depth Df of the mat. This approach is generally referred to as the compensated foundation design and is extremely use­ ful when structures are to be built on very soft clays. In this design, a deeper base­ ment is made below the higher portion of the superstructure, so that the net pressure increase in soil at any depth is relatively uniform. (See Figure 6.7.) From Eq. (6.16) and Figure 6.6, the net average applied pressure on soil is Q yDf q=A-

N

�

Table 6. 1

Settlement of Mat Foundations on Sand and Gravel (Based on Meyerhof, 1965)

Case No.

Structure

Reference

B m (ft)

Average No.

kN/m' (kip/It')

Observed maximum settlement, S. mm (in.)

(1)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

(9)

1

T. Edison Sao Paulo, Brazil

Rios and Silva (1948)

229.8 (4.8)

15.24 (0.6)

30.64 (1.21 )

2.01

Banco do Brazil Sao Paulo, Brazil

Rios and Silva (1948); Vargas (1961)

22.86 (75)

15

2

18.29 (60)

239.4 (5.0)

27.94

26.6 (l.05)

0.95

3

Iparanga Sao Paulo, Brazil

Vargas (1948)

9.14 (30)

304.4 (6.4)

35.56 ( 1.4)

67.64 (2.66)

l.9

4

c.B.I., Esplanda Sao Paulo, Brazil

Vargas (1961)

14.63 (48)

22

383.0 (8.0)

27.94 (1.1)

34.82 (1.37)

1 .25 .

5

Riscala Sao Paulo, Brazil

Vargas (1948)

3.96 (13)

20

229.8 (4.8)

12.7 (0.5)

22.98 (0.9)

1.81

6

Thyssen DUsseldorf, Germany

Schultze (1962)

22.55 (74)

239.4 (5)

24.13 (0.95)

19.15 (0.75)

0.79

7

Ministry DUsseldorf, Germany

Schultze (1962)

15.85 (52)

20

220.2 (4.6)

20.32 (0.8)

22.02 (0.87)

l.08

Chimney Cologne, Germany

Schultze (1962)

20.42 (67)

10

172.4 (3.6)

10.16 (0.4)

34.48 (l.36)

3.39

8

..

I� 1.....----t----...... ---j 1 I •

283

Figure 6.7 Compensated foun­ dation

For no increase in the net pressure on soil below a mat foundation, q should be zero. Thus, Q Dr = (6.20) A-y

­

This relation for Dr is usually referred to as the depth of a fully compensated

foundation.

The. factor of safety against bearing capacity failure for partially compensated foundations (Le., Dr < Q/ Al') may be given as (6.21 ) where q,,,v, cf>(4)VJ;bod (0.85)(4)(Y3000) (96 + 2d)d Vu

shear strength of concrete 11;, = factored shear strength

=

=

- -.- - - - - - - - - - - - - - - - - - - - - -

I 1

I , , , , I I

'

:

!!..

2

24 in.

24 + d (in.)

1 2 in.-+''''I'�- 24 in.---->I "

f (in.)

- - - -- - - - - - - - - - - - - - - - - - --

�1.---- 36 +

I

�

� ,I

-

Figure 6. 14 Critical perimeter column

300

Chapter 6 Mat Foundations

f

l

d 24ti i 24 + d (in.) : d 2 in.dhl ·d- i 4 d 1-I-

E 1 6 ft -----.J

A

C;

I I I I

•

1500 kN

400kN

F

-I I'

0.25 m

306

500 kN

H 8m

E

>I-

1

T

O.25 m

7m

1200kN

8 m

4.25 m

_ _ _ _ _ _ _

1500 kN

Figure P6.B

450kN

' 1500 kN

4.25 m

'

C

I I I I I I I I I I

1500 kN

x

y • I I 'B 1 500kN

450kN

1

I I L I I I I I I I I I I I I I I I I I I

t

7m ________

I I

-

.----

+ .1"

1200 kN

7m 350kN

J 8m 0.25 m

1

0.25 m

..-

x'

Figure H.B

References

307

1 1

6.11 From the plate load test (plate dimensions ft x ft) in the field, the coeffi­ cient of subgrade reaction of a sandy soil is determined to be 55 Ib/in3• What will be the value of the coefficient of subgrade reaction on the same soil for a foundation with dimensions of 25 ft x 25 ft? 6.U Refer to Problem If the full-sized foundation had dimensions of 70 ft x 30 ft, what will be the value of the coefficient of subgrade reaction? 6.13 The subgrade reaction of a sandy soil obtained from the plate load test (plate dimensions m x 0.7 m) is kN/m3• What will be the value of k on the same soil for a foundation measuring 5 m X 3.5 m?

6.11.

1

18

American Concrete Institute (1995). ACI Standard Building Code Requirements for Rein­ forced Concrete, ACI 318-95, Farmington Hills, MI. American Concrete Institute Committee 336 (1988). "Suggested Design Procedures for Com­ bined Footings and Mats," Journal of the American Concrete Institute, Vol. 63, No. 10, pp. l041-1077. Hetenyi, M. (1946). Beams ofElastic Foundations, University of Michigan Press, Ann Arbor, MI. Meyerhof, G. G. (1965). "Shallow Foundations," Joumal of the Soil Mechanics and Founda­ tions Division, American Society of Civil Engineers,Vol. 91, No. SM2, pp. 21-31. Rios. L., and Silva, F. P. (1948). "Foundations in Downtown Sao Paulo (Brazil)," Proceedings, Second International Conference on Soil Mechanics and Foundation Engineering, Rot­ terdam, Vol. 4, p. 69. Schultze, E. (1962). " Probleme bei der Auswertung von Setzungsmessungen:" Proceedings, Baugrundtagung, Essen. Germany, p. 343. Terzaghi, K. (1955). "Evaluation of the Coefficient of Subgrade Reactions:' GeotecJmique. I n ­ stitute of Engineers, London, Vol. 5. No. 4, pp. 197-226. Vargas, M. (1948). "Building Settlement Observations in Sao Paulo." Proceedings Second In­ lernational Conference on Soil Mechanics and Foundation Engineering, Rotterdam. Vol. 4, p. 13. Vargas, M. (1961). "Foundations of Tall Buildings on Sand in Sao Paulo (Brazil)," Proceed­ ings, Fifth International Conference on Soil Mechanics and Foundation Engineering.

Paris, Vol. 1, p. 841. Vesic,A. S. (1961). "Bending of Beams Resting on Isotropic Solid," JOllrnal afthe Engineering Mechanics Division, American Society of Civil Engineers. Vol. 87. No. EM2, pp. 35-53.

7. 1

Introduction Vertical or near-vertical slopes of soil are supported by retaining walls, cantilever sheet-pile walls, sheet-pile bulkheads, braced cuts, and other, similar structures. The proper design of those structures requires an estimation of lateral earth pressure, which is a function of several factors, such as (a) the type and amount of wall move­ ment, (b) the shear strength parameters of the soil. (c) the unit weight Qf the soil, and (d) the drainage conditions in the backfill. Figure 7.1 shows a retaining wall of height H. For similar types of backfill. a.

The wall may be restrained from moving (Figure 7.1a). The lateral earth pres­ sure on the wall at any depth is called the at-rest earth pressure. b. The wall may tilt away from the soil that is retained (Figure 7.1b). With sufficient wall tilt, a triangular soil wedge behind the wall will fail. The lateral pressure for this condition is referred to as active earth pressure. Co The wall may be pushed into the soil that is retained (Figure 7.1c). With suffi­ cient wall movement, a soil wedge will fail. The lateral pressure for this condi­ tion is referred to as passive earth pressure.

ui, Height = H

(a)

(al-reSI)

+ dH -I I+-

I rr-----..,-­ I \ \ \ \ \ \ \ \ \

Height = H

Soil failure wedge

Height = H

Ibl

Figure 7. 1 Nature of lateral earth pressure on a retaining wall

308

uir

- dH -I I+-

:

I

1 I I I I I I I I

(passIve)

Soil failure wedge

(e)

7.2 Lateral Earth Pressure at Rest

I I I I I

---,-...:- - - -- - -

( �)p

=

_

ui, (pas.�j\'e)

( �)

for loo,e 1 sand to 0.05 : for soft clay : 0.0 I

AHH

309

=

a

I

for loo,e sand to 0.04 for soft clay

0.001

l ( Aff)

I I

i ( �)p

+-_---''-- _-'---__L-_-'__''-" _

+

AHH

Figure 7.2 Nature of variation of lateral earth pressure at a certain depth

Figure 7.2 shows the nature of variation of the lateral pressure, IT;" at a certain depth of the wall with the magnitude of wall movement. In the sections that follow, we will discuss various relationships to determine the at-rest, active, and passive pressures on a retaining wall. It is assumed that the reader has studied lateral earth pressure in the past, so this chapter will serve as a review.

7.2

Lateral Earth Pressure at Rest Consider a vertical wall of height H, as shown in Figure 7.3, retaining a soil having a unit weight of y. A uniformly distributed load, q/unit area, is also applied at the ground surface. The shear strength of the soil is s

where

=

' c

+ (I' tan cf/

c' = cohesion



7.3

Rankine Active Earth Pressure

-I

313

2c'{if,;

p.

H

Eq. (7. 12)

1+1'-- CT�K. ---+l'1 -I -2c'{if,; I­ (c) Figure 7.5

(Continued)

Mohrcont-Coulinuesombto ifnacriluereaseenvel, the ocorperedefispondinednbyg Mohr' thteanequatcf/sciriclone eventually wil just touch the Thitheshoricirclzonte, marked ihnenthequal e figure,s a;represent sttoheasfathiluere condition in the soil mass;The a l s t r es s t . ref e rred anes) iinntFihegsoiurel 7.mass5a. wil then make angles of (45 with the horiEquatz(ontfaiiloaulnre, as(1pl.shown s the opripe:ncipal stresses for a Mohr's circle that touches the Mohr-Coulomb69)fairelluareteenvel UI U3 tan2(45 �' ) tan(45 �' ) ForMajothrepriMohr'ncipals cistrrcless,e UIin Figure 7.5b. andMinor principal stress, U3 Thus, =. tan-(45 ') tan(45 ' ) i = 30°,

Forthe top soil layer, so tan.-( ) tan' Similarly, for the botto!l' soil layer, and tan-'( ) Becauseatofic pressure the presencehaveoftothbee watcalceulrattaebldeseparat , the effeelyct.ive lateral pressure and tAthe hydrost At At thm,is depth. for the top soil layer.kN/m' kN/m' Similarly, for the bottom soil layer, kN/m' Atz 6m, = + kN/m' ' + kN/m i s zero f r o m t o m. At 6 The hydrostatic pressure Figure The force perkN/m2unit .leThengthpressure distribution diagram is plotted in Area + Area Area + Area + kN/m + Thecan dibestadetnceermiof tnhede libyne oftakiactngionthofe moment the resulstaabout nt fromthethebotbottotmomofoftthhee walwalll (point in Figure or (�) ( + �) + (%) + G) + + + 44.15 cf>

45 - 2i =

K,,(I ) =

(45 - 15)

=

cf>, = 36°.

3"1

36 45 - 2 = 0.26

Ka(,) =

Z = 0, 0"0 = 0, u:1 = 0 Z =3 iTo = yz ( 16) (3) = 48 =

a;, = K""1iTO = (* ) (48) = 16

iT;, = K"miTo = (0.26) (48) = 12.48

=

iTo

(y"" - y",) (3 ) = ( 1 6 ) (3 ) + ( I Y - Y.81 ) (3)

(y)(3)

27.57

= 48

=

75.57

(7; = Ko(2) iTO = (0.26) (75.57) = 19.65

3y",

=

3(9.81) = 29.43 7.9b.

2

1

Po =

(E)

LI

=

!(3) (16)

=

24 + 37.44

0

+

Z =0

3

+

Z =3

Z=

m;u

=

4

(3) (12.48)

+

10.76

�(3) (19.65 - 12.48) + t(3)(29.43 ) 44.15 = 116.35

7.9a).

(37.44)

(24) 3

z=

=

(10.76)

+ (44.15)

1 16.35

96

56 .16

10.76

1 16.35

= 1.779 m

•

I

7.4 A Generalized Case for Rankine Active Pressure

321

Example 7.2 \ Y

A

6-m-high retaining wall is to support a soil with unit weight y = 17.4 kN/m3, soil friction angle q,' = 26°, and cohesion c' = 14.36 kN/m2 Determine the Rankine active force per unit length of the wall both before and after the tensile ctack oc­ curs, and determine the line of action of the resultant in both cases. Solution

For q,' = 26°,

( �' )

K, = tan' 45 -

= tan2(45

YK. = 0.625

-

13) = 0.39

lT� = yHKa - 2c'YK. From Figure 7.5c, at z = O,O'� = - 2e'YK. = - 2(14.36) (0.625) = - 17.95 kNjm2 and at z = 6 m. O'� = (17.4) (6) (0.39) - 2 (14.36) (0.625) = 40.72 - 1 7.95 = 22.77 kN/m'

,

Active Force before the Tensile Crack Appeared: Eq.

(7. 1 0)

I ' P" = zyH-K" - 2c ,;, !(6) (40.72) - (6) (17.95) = 122.16 - 107.7 = 14.46 kN/m

'H V K,

The line of action of the resultant can be determined by taking tl,e moment of the area of the pressure diagrams about the bottom of the wall, or �

Thus,

P;z

= (122.16) (�) - ( 107.7) (�)

-_

z-

244.32 - 323.1 14.46

--

_

5.45 m

Active Force after the Tensile Crack Appeared: Eq.

Z,

=

(7.9)

2(14.36) 2e' = yYK. ( 17.4) (0.625)

2.64 m

Using Eq. (7.11) gives .�

Po = ! (H - z. l ( yH K, - 2e'YK. ) = 4 (6 - 2.64) (22.77) = 38.25 kNjm

322

Chapter 7 Lateral Earth Pressure

Figure 7.5c indicates that the force P, = 38.25 kN/m is the area of the hatched triangle. Hence, the line of action of the resultant will be located at a height z = (H - zJ/3 above the bottom of the wall, or -

Z =

6 - 2.64 = I.II m 3

Example 7.3 Refer to the retaining wall in Figure Wall: Backfill:

7.7.

-

The backfill is granular soil. Given:

H = 10 ft e

=

a =

+10° 15°

"-

,,

Solution

From Eq. (7.25) = 2!�H'K a I From Table 7.4, for a = 0°,,13 = 90°, "" = 30°, and 8' = �",. =20°, Ko = 0.297. Hence, pa

Po = ! (16.5) (4.6)'(O.297) = 51.85 kN/m

•

7.5

\

Coulombls Active Earth Pressure

327

Table 7.5 Values of K, [from Eq. (7.26)J for D' = 4//2

'" (deg)

.p' (deg)

90

85

80

0

28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 28

0.3264 0.3137 0.3014 0.2896 0.2782 0.2671 0.2564 0.2461 0.2362 0.2265 0.2172 0.2081 0.1994 0.1909 0.1828 03477 0.3337 0.3202 0.3072 0.2946

0.3629 0.3502 0.3379 0.3260 0.3145 0.3033 0.2925 0.2820 0.2718 0.2620 0.2524 0.2431 0.2341 0.2253 0.2168 0.3879 0.3737 0.3601 0.3470 0.3342 0.3:> 19 0.3101 0.2986 0.2874 0.2767 0.2662 0.2561 0.2463 0.2368 0.2276 0.4187 0.4026 0.3872 0.3723 0.3580 0.3442 0.3309 0.3181 0.3058 0.2938 0.2823 0.2712 0.2604 0.2500 0.2400 0.4594

5

10

15

O.2R25

0.2709 0.2596 0.2488 0.2383 0.2282 0.2185 0.2090 0.1999 0.1911 0.3743 0.3584 0.3432 0.3286 0.3145 0.3011 0.2881 0.2757 0.2637 0.2522 0.2412 0.2305 0.2202 0.2103 0.2007 004095

{J (deg)

75

70

0.4034 0.3907 0.3784 0.3665 0.3549 0.3436 0.3327 0.3221 0.3118 0.3017 0.2920 0.2825 0.2732 0.2642 0.2554 0.4327 0.4185 0.4048 0.3915 0.3787

0.4490 0.4363 0.4241 0.4121 004005 0.3892 0.3782 0.3675 0.3571 0.3469 0.3370 0.3273 0.3179 0.3087 0.2997 0.4837 0.4694 0.4556 0.4422 0.4292

0.3541 0.3424 0.3310 0.3199 0.3092 0.2988 0.2887 0.2788 0.2693 0.4688 0.4525 0.4368 0.4217 0.4071 0.3930 0.3793 0.3662 0.3534 0.3411 0.3292 0.3176 0.3064 0.2956 0.2850 0.5159

0.4043 0.3924 0.3808 0.3695 0.3585 0.3478 0.3374 0.3273 0.3174 0.5261 0.5096 0.4936 0.4782 0.4633 0.4489 0.4350 0.4215 0.4084 0.3957 0.3833 0.3714 0.3597 0.3484 0.3375 0.5812

0.5011 0.4886 0.4764 004645 0.4529 0.4415 004305 004197 0.4092 0.3990 0.3890 0.3792 0.3696 0.3602 0.3511 05425 0.5282 0.5144 0.5009 0.4878 0.4750 0.4626 0.4505 0.4387 0.4272 0.4160 0.4050 0.3944 0.3840 0.3738 0.5928 0.5761 0.5599 0.5442 0.5290 0.5143 0.5000 0.4862 0.4727 0.4597 0.4470 0.4346 0.4226 0.4109 0.3995 0.6579

0.3662

0..J166

65

0.5616 0.5492 0.5371 0.5253 0.5137 0.5025 0.4915 0.4807 0.4702 0.4599 0.4498 0.4400 0.4304 0.4209 0.4177 0.6115 0.5972 0.5833 0.5698 0.5566 05437 0.5312 0.5190 0.5070 0.4954 0.4840 0.4729 0.4620 0.4514 0.4410 0.6719 0.6549 0.6385 0.6225 0.6071 0.5920 0.5775 0.5633 0.5495 0.5361 0.5230 0.5103 0.4979 0.4858 0.4740 0.7498 (continued)

328

Chapter 7 Lateral Earth Pressure Table 7.5 (Continued)

fl (dog)

.j�lJ

'" (dog)

' (dog)

20

29 30 31 32 33 34 35 36 37 38 39 40 41 42 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42

•

-

90

85

80

75

70

65

0.3908 0.3730 0.3560 0.3398 0.3244 0.3097 0.2956 0.2821 0.2692 0.2569 0.2450 0.2336 0.2227 0.2122 0.4614 0.4374 0.4150 0.3941 0.3744 (1..:i559 0.3384 0.3218 0.3061 0.2911 0.2769 0.2633 0.2504 0.2381 0.2263

0.4402 0.4220 0.4046 0.3880 0.3721 0.3568 0.3422 0.3282 0.3147 0.3017 0.2893 0.2773 0.2657 0.2546 0.5188 0.4940 0.4708 0.4491 0.4286 O..J093 0.3910 0.3736 0.3571 0.3413 0.3263 0.3120 0.2982 0.2851 0.2725

0.4964 0.4777 0.4598 0.4427 0.4262 0.4105 0.3953 0.3807 0.3667 0.3531 0.3401 0.3275 0.3153 0.3035 0.5844 0.5586 0.5345 0.5 1 1 9 0.4906 0.-1704 0.4513 0.4331 0.4157 0.3991 0.3833 0.3681 0.3535 0.3395 0.3261

0.5611 0.5419 0.5235 0.5059 0.4889 0.4726 0.4569 0.4417 0.4271 0.4130 0.3993 0.3861 0.3733 0.3609 0.6608 0.6339 0.6087 0.5851 0.5628

0.6373 0.6175 0.5985 0.5803 0.5627 0.5458 0.5295 0.5138 0.4985 0.4838 0.4695 0.4557 0.4423 0.4293 0.7514 0.7232 0.6968 0.6720 0.6486 0.6264 0.6052 0.5851 0.5658 0.5474 0.5297 0.5127 0.4963 0.4805 0.4653

0.7284 0.7080 0.6884 0.6695 0.6513 0.6338 0.6168 0.6004 0.5846 0.5692 0.5543 0.5399 0.5258 0.5122 0.8613 0.8313 0.8034 0.7772 0.7524 0.7289 0.7066 0.6853 0.6649 0.6453 0.6266 0.6085 0.5912 0.5744 0.5582

054 j 7

0.5216 0.5025 0.4842 0.4668 0.4500 0.4340 0.4185 0.4037 0.3894

I

Active Earth Pressure for Earthquake Conditions Coulomb's active earth pressure theory (see Section 7.5) can be extended to take into account the forces caused by an earthquake. Figure 7.11 shows a condition of active pressure with a granular backfill (c' = 0). Note that the forces acting on the soil failure wedge in Figure 7.11 are essentially the same as those shown in Figure 7.10a with the addition of k"W and kvW in the horizontal and vertical direction respectively; k" and k, may be defined as

klz k,.

= =

horizontal earthquake acceleration component acceleration due to gravity, g

(7.27)

vertical earthquake acceleration component acceleration due to gravity, g

(7.28)

�

i I

7.6 Active Earth Pressure for Earthquake Conditions

329

\.

0 ' I 1

'kvW

Granular backfill

y c' = 0 cp'

,

W

H

Figure 7, 1 1 Derivation of Eq. (7.29)

As in Section 7.5, the relation for the active force per unit length of the wall (F.,) can be determined as (7.29) where K.. = active earth pressure coefficient sin' (cf>' + 13 - 8') . = --���--���� ����� sin (cf/ + 8') sin (cf/ - IJ' + 1 ' ; 0 (13 f3 2 8) cos 8' sin sin . sin (13 - 8' .;. 1J' ) sin ( O! + f3) _

_

«)J'

[

8 ' = tan- 1

[

k"

( 1 - k,, ) . Note that for no earthquake condition k" = 0, k,. 0, and =

(7.30)

]

(7.31) 8' = 0

Hence K" = Ka [as given by Eq. (7.26)]. Some values of K., for f3 = 900 and k, = 0 are given in Table 7.6. Equation (7.29) is usually referred to as the MOllollobe-Okabe solution. Unlike the case shown in Figure 7.lOa, the resultant earth pressure in this situation, as

Table 7.6 Values of K", [Eq. (7.30)] for f3 = 90' and k,. = 0 4>'(deg) kh

0.1 0.2 0.3 0.4 0.5 0.1 0.2 0.3 0.4 0.5 0.1 0.2 0.3 004 0.5 0.1 0.2 0.3 0.4 0.5 0.1 0.2 0.3 0.4 0.5 0.1 0.2 0.3 0.4 0.5 0.1 0.2 0.3 0.4 0.5 0.1 0.2 0.3 0.4 0.5 0.1 0.2 0.3 0.4 0.5 330

28

30

35

40

45

0.427 0.508 0.611 0.753 1.005 0.457 0.554 0.690 0.942

0.397 0.473 0.569 0.697 0.890 0.423 0.514 0.635 0.825

10

0.497 0.623 0.856

0.457 0.570 0.748

0.328 0.396 0.478 0.581 0.716 0.347 0.424 0.522 0.653 0.855 0.371 0.461 0.585 0.780

4>'(2

0

ci>, '2

5

0.396 0.485 0.604 0.778 1 .115

0.368 0.452 0.563 0.718 0.972

0.306 0.380 0.474 0.599 0.774

0.268 0.382 0.400 0.488 0.596 0.282 0.349 0.431 0.535 0.673 0.299 0.375 0.472 0.604 0.809 0.253 0.319 0.402 0.508 0.648

0.537 0.699 1.025

0.497 0.640 0.881

4>/2

10

0.472 0.616 0.908

0.433 0.562 0.780

00412 0.526 0.690 0.962 0.352 0.454 0.602 0.857

let>

0.342 0.438 0.568 0.752 0.285 0.371 0.487 0.656 0.944

0.217 0.270 0.334 0.409 0.500 0.227 0.285 0.356 0.442 0.551 0.238 0.303 0.383 0.486 0.624 0.207 0.267 0.340 0.433 0.522 0.218 0.283 0.367 0.475 0.620 0.230 0.303 0.400 0.531 0.722

0

�

0.327 0.418 0.541 0.722 1.034

0.271 0.350 0.455 0.600 0.812

0.224 0.294 0.386 0.509 0.679

10

0.472 0.625 0.942

0.434 0.570 0.807

0.354 0.463 0.624 0.909

0.290 0.381 0.509 0.699 1.037

0.237 0.317 0.423 0.573 0.800

Il'(deg)

a(deg)

0

0

0

5

0

0.4"8

0.396

n.3�6

0.268

•

,

7.6 Active Earth Pressure for Earthquake Conditions

331

Rgure 7.12 Determining the line of action of Pal!

calculated by Eq. does not act at a distance of HI3 from the bottom of the wall. The following procedure may be used to obtain the location of the resultant force P",.:

(7.29)

Step 1. Calculate P", by using Eq. Step 2. Calculate p" by using Eq. (7.25) Step 3. Calculate

(7.29)

(7.32)

Step 4. Assume that Po acts at a distance of H/3 from the bottom of the wall (Figure Step 5. Assume that /J;p., acts at a distance of O.6H from the bottom of the wall (Figure Step 6. Calculate the location of the resultant as

7.12) 7.12)

(O.6H) (/J;P.,) z =

+ (�)

cP.)

�lImpJe 7.5

. �efer to Figure 7.13. For k, = 0 and kh = O.3, determine:

'iL ? , , . b;'The location of the resultant, Z, from the bottom of the wali

(7.33)

332

Chapter 7 Lateral Earth Pressure

cP' = Y = a' =

350

105 Ib/f,' 1 7.5°

1 0 ft

Figure 7. 13

Solution

Part a

From Eq. (6.29).

Here, 'Y = 105 Ib/ft', H =10 ft, and k, = O. As 8' = '/2, we can use Table 7.6 to determine K... For kh = 0.3, K.. = 0.474, so p..

= !(105) (10)2(1 - 0) (0.474) = 2488.S lb/ft

Part b

From Eq. (6.25), p.tI

= !.vH2K Q 21

From Eq. (7.26) with 8' = 17S,{3 = 900, and a = O°, K, = 0.246 (Table 7.5f , P, = W05) (10)2(0.246) = 1292 1b/ft AP.. = p.. - P, = 2488.5 - 1292 = 1 196.5 lb/ft From Eq. (7.33), _

z =

=

(0.6H) (AP",) + (H/3 ) (P,, ) . [(0.6) ( 10)](1 196.5) + ( 10/3) (1292) = 4.62 ft 2488.5

•

I I I I I I I I I

7. 7

Active Pressure for Wall Rotation about the Top: Braced Cut

333

Strm

I I I I I I I I I I

I I I I I I

I

(aJ

A

I I I I

(b)

Figure 7. 14 Nature of yielding of walls: (a) retaining wall; (b) braced cut

.' 7;7

Active Pressure for Wall Rotation about the Top: Braced Cut �L:l'lions. \'..... ha\ L "cell that a retaining wall rotates ahout its bot­ tom. (See Figure 7.14a.) With sufficient yielding of the walL the lateral earth pres­ sure is approximately equal to that obtaineLi by Rankine's theory or Coulomb's theory. In contrast to retaining walls. braced cuts show a different type of wall yield­ ing. (See Figure 7.14b.) In this case. deformation of the wall gradually increases with the depth of excavation. The variation of the amount of deformation depends on several factors. such as the type of soil. the depth of excavation. and the workman­ ship involved. However. with very little wall yielding at the top of the cut. the lateral earth pressure will be close to the at-rest pressure. At the bottom of the wall. with a much larger degree of yielding. the lateral earth pressure will be substantially lower than the Rankine active earth pressure. As a result. the distribution of lateral earth pressure will vary substantially in comparison to the linear distribution assumed in the case of retaining walls. The total lateral force per unit length of the wall, P,,, imposed on a wall may be evaluated theoretically by using Terzaghi's (1943) general wedge theory, (See Figure 7.15.) The failure surface is assumed to be the arc of a logarithmic spiral, de­ fined as I n the pr' (7.51) cos a- Vcos' a cos' cf>' As in the case of the active force, the resultant force, P", is inclined at an angle a with the horizontal and intersects the wall at a distance H/3 from the bottom of the wall.The values of Kp (the passive earth pressure coefficient) for various values of a and cf>' are given in Table 7.9. Kp = cos a--�-:r=;====:;==: -

-

Tabla 7,9 Passive Earth Pressure Coefficient, Kp [from Eq. 7.51)J

.p' (degl-> .!a (deg)

0 5 10

15

20 25

28

30

32

34

36

38

2.770 2.715 2.551 2.284 1 918 1.434

3.000 2.943 2.775 2.502

3 255

3.537 3 476

3.852 3.788 3.598

4.204 4.136

.

2.132 1.664

.

3.196 3 022 2.740 2.362 1.894 .

.

3.295 3.003

2.612 2.135

3.293 2.886 2.394

3.937

3.615 3.189

2.676

40

4.599 4.527 4.316 3.977

3.526 2.987

.:

7. 12 Coulomb's Passive Earth Pressure

345

Table 7. 10 Values of K�

,..

' (degl 15

20

25

30

a

(deg)

0.025

0 5 10 15 0 5 10 15 0 5 10 15 0 5 10 15

1 .764 1.716 1.564 1.251 2.111 2.067 1.932 1.696 2.542 2.499 2.368 2.147 3.087 3.042 2.907 2.684

tfJyz

0.050

0.100

0.500

1.829 1.783 1.641 1.370 2.182 2.140 2.010 1 .786 2.621 2.578 2.450 2.236 3.173 3.129 2.996 2.777

1.959 1.917 1.788 1.561 2.325 2.285 2.162 1.956 2.778 2.737 2.614 2.409 3.346 3.303 3.174 2.961

3.002 2.971 2.880 2.732 3.468 3.435 3.339 3.183 4.034 3.999 3.895 3.726 4.732 4.674 4.579 4.394

'

c' q, Soil -

If the backfill of the frictionless vertical retaining wall is a c' - soil (see Figure 7.8), then (Mazindrani and Ganjali. 1997) u; = yzKp = yzK� cos a

I

where

K'p -

1 cos' 4>'

,,�, 0 + ,

+

(;;)=.' ,;' .'

(7.52)

)

The variation of K� with 4>',a, and c'fyz is given in Table 7.10 (Mazindrani and Ganjali, 1997).

Coulomb's Passive Earth Pressure Coulomb (1776) also presented an analysis for determining the passive earth pres­ sure (Le., when the wall moves into the soil mass) for walls possessing friction (5' = angle of wall friction) and retaining a granular backfill material similar to that discussed in Section 7.5.

346

Chapter 7 Lateral Earth Pressure Passive force

� + 8' 1 c'

'

=0

R

w 9, + '

(a)

(b)

Figure 7.24 Coulomb's passive pressure

. To understand the determination of Coulomb's passive force, P", consider the wall shown in Figure 7.24a.As in the case of active pressure, Coulomb assumed that the potential failure surface in soil is a plane. For a trial failure wedge of soil, such as ABC., the forces per unit length of the wall acting on the wedge are 1. The weight of the wedge, W 2. The resultant, R, of the normal and shear forces On the plane BCI, and 3. The passive force, P"

Figure 7.24 shows the force triangle at equilibrium for the trial wedge ABCI• From this force triangle, the value of Pp can be determined,because the direction of all three forces and the magnitude of one force are known. Similar force triangles for several trial wedges, such as ABC., ABC2, ABC" . , can be constructed, and the corresponding values of p" can be determined. The top part of Figure 7.24 shows the nature of variation of the p" values for different wedges.The minimum value ofp" in this diagram is Coulombs passive force, mathe­ matically expressed as .

.

(7.54)

7. 13

Comments on the Failure Surface Assumption for Coulomb's Pressure Calculations Table 7. 1 1 Values of Kp [from Eq. (7.55)J for /3

�.

t/>' (deg)

0

5

15 20 25 30 35 40

1.698 2.040 2.464 3.000 3.690 4.600

1.900 2.313 2.830 3.506 4.390 5.590

8'

=

90' and Q

=

347

0'

(deg)

10

15

20

2.130 2.636 3.286 4.143 5.310 6.946

2.405 3.030 3.855 4.977 6.854 8.870

2.735 3.525 4.597 6.105 8.324 11.772

where

;!:;� �;�11��b'���V� P�����,j1fm:l!

•

.••.�1�;{';�y,$�ff�n.�.�;;6:)El:.�. .

:

" ' ', "" ,

,

. . , . ."

· · ���;�;�;�;2�'�t��J.J· .

(7.55)

The values of the passive pressure coefficient, Kp, for various values of , c,

I L

_____

I i

I

363

Use of a dead man anchor

-- - Base slab increase

I I

I

Figure 8.8 Alternatives for increasing the factor of safety with respect to sliding

Combining Eqs. (8.7), (8.8), and (8.9) yields FS (,,1IJm!!1 -

A minimum factor of safety of

=

1.5

(2 V) tan/}' + Be; + E;, p.a cos a:

(8. 1 0 )

against sliding is generally required. In many cases, the passive force Pp is ignored in calculating the factor of safety with respect to sliding. In general, we can write /}' = k} cP, and c; k,c, . In most cases, k} and k, are in the range from t to �. Thus. (}'. V) tan (k} B/6, the design should be reproportioned and calculations redone. The relationships pertaining to the ultimate bearing capacity of a shallow foundation were discussed in Chapter 3. Recall that [Eq. (3.56)]. (8.1 9) where q = "I,D

B' = B - 2e F,d = 1 + 0.4

D

B'

Fqd = 1 + 2 tan ¢,( 1 - sin¢ll '

.

...

F,d = 1 FCI = F1/1 =

( £)' 1

_

900

�

366

Chapter 8 Retaining Walls

F'Ii .pc =

( _,(Po-.p,), ) 1 -

4>20

cosa IV

tan

Note that the shape faCtors Fe" F." . and Fy given in Chapter 3 are all equal to unity, because they can be treated as a continuous foundation. For this reason, the shape factors are not shown in Eq. (8. 1 9). Once the ultimate bearing capacity of the soil has been calculated by using Eq. (8.19). the factor of safety against bearing capacity failure can be determined:

,

FS (h1 '11+-

Figure 8. 10

-

6 [,

=

2.75 ft

"Y2 = 107 Ib/f,]

---I

q,;

180 c; = 900 Ib/f,2 =

Calcula'ion of stability of a retaining wall

Factor of Safety against Overturning

The following table can now be prepared for determination of the resisting moment. Weight Section (kipjftl

. 1

2 3

4

(1.5) (18) (0.15)

=

�(1.0) (18)(O.15)

4.05

=

1.35

Moment about C Ikip-ftjftl

4 + �(1) = 4.�7

6.3

Mo =

+

�

=

H' (8.06) (21.-3-81) 58.6 203.58.617

Ph)" =

FS(ovort","ing) =

IMR = Mo

23.29

5.75

(12.5) (2.75) (0.15) = 5.15� �.25 8 19 06 C \ · )(� ) (0.117) = 13.01 =4 + .2.5 12.5 p. = 1.42 l:V = 24.986

(Note." Yro.cr,,, = 150 1b/ft]) 'The overturning moment, Mo, is

So

Moment arm from C (ftl

=

9.5

32.23

123.6

17.75 :!;203.17 = l:M.

. /

kip ft

= 3.47 > 2-0K

368

Chapter 8 Retaining Walls Factor of Safety against Sliding

From Eq. (8.11),

.

FS(sliding) =

(�V) tan (k,C/>2) + Bk,C2 + Pp P" COsa

Let k, = k, = ¥3 and Pp = O. So

[� ] G )

(�V)tan c/>; + B C; = FS(sliding) ----"£n D=---= -- -----'---'­ COSa

�

[� ]

(24.986) tan (18) + (12.5) (0.9) = ------��-=-----�--8.06 = 1,59 > 1.5-0K Factor of Safety against Bearing Capacity Failure

Combining Eqs. (8.13). (8.14). and (8.15) yields e =

B 2

�MR - �Mo = 6.25 �V

_

= 0.464 ft

�V(


i = 18°,Nc = 13.1, Nq = 5.26, and Ny = 4.07,

q = "'I2D = (4) (0.107) = 0.428 kip/ft'

B' = B '- 2e = 12.5 - (2) (0.464) = 11.572 ft P'd = 1 + 0.4

(�)

( �) sinc/>2) 2 ( D )

= 1 + 0.4

Fqd = 1 + 2 tan ci>,( 1

-

11. 72 B'

= 1.138

= 1 + (0.31)

( 4 ) 11.572

,':'

= 1.1P7.

8.7

1/1

_ -

tan

_ l(PaCOScr. ) ( IV

So, Fd = Fq, = Fl'. =

Hence,

(

1

-

_

-

8.06 24.986

( l - --W)

)

369

- 17.88 . _

,

17.88 2 , = 0.642

.!t cf>2

) ( 2

=

q" = (0.9 ) (13.1 ) ( 1.138) (0.642) +

Check for Bearing Capacity Failure

+

1

_

17'88 , 18 ,

)2

= .0

1'.�."

(0.428 )(5.26) ( 1.107) (0.642) '

!(0.107) (11.572) (4.07) (1) (0)

:'

-

'

= 10.21 kip/ft>

q" FS (be'ria,,,p,,''Y) = = qtoe

10.21 = 4.18 > 3-0K 2 .44

, '. , ' ,;

Example 8.2 Consider the retaining wall given in Example 8.1. Calculate the factor of safety with respect to bearing capacity using Eq. (8.21). Solution

From Eq. (8.21) qu = C ' Nc(ti)Fcd + qNq(ei)Fqtl + �Y2BN"f(ei)

From ,Example 8.1, IjI = 17.88'. Note the change of notation. In Figures 3.26 through 3.28, (3 = 1/1. Also, e/B = 0.464/12.5 = 0.037. By interpollition; ior "', = 18', e/B = 0.037 and IjI = 17.88'. N,«,) = 7.5 Nq(,,) = 4 N,(,,). = 0

So,

q" = (0.9)(7.5) (1.138) + (0:428)(4) (1.107) = 9.58 kip/ft> FS(b"riag"p,d'Yi =

9.58 = 3.93 > 3-0 K. 2.44

•

370

Chapter 8 Retaining Walls

Example 8.3

concrete gravity retaining wall is shown in Figure 8.11. Determine: The factor factor ofof safety safety against against sloverturning The iding The pressure on the soil at the toe and heel Unit weight of concrete Use Rankine's earth pressure theory. 150Ib/ft'.) H' 15 + 2.5 17.5 ft Ka tan2 (45 �; ) tan2 (45 3�) = � �(H')'K" t(121)(17.5)'G) 61761b/ft 6.176 kip/ft Since" 0 6.176 kip/ft Ph

A

a. b. c.

I

(Note:

Solution

=

=

=

P" =

_

=

=

=

=

-

=

= P" =

1',.

, 1

�

0

Not to scale

I I I I

15 ft

® I:

I I I I I I

1

2.5 ft

cP;

-

20'. y, =

121 10/ft'. c; = 1000 Ib/ft'

Figure B. 1 1

=

'Y,

=

8.7

, �.

Check far Bearing Capacity Failure

371

Part a: Factor of Safety against Overturning The following table can now be prepared to obtain IMR• At..

(from Weight figure 8'; 1) (klp/ftl

Moment abOUt C

(klpCft/ft)

Moment arm from C (It)

I

�

1.605 1 .25 + (0.8) 1.783 1.25 + 0.8 + 0.75 = 2.8 9.45 , 5.25 1.25 + 0.8 + 1.5 + -- 5.3 3Jc,31>.: 3 .3 1 5.15 (10.3) (2.5) (y,) = 3.863 ., 19.89' I 2 2(5.25)(15) (0.121) = 4.764 1.25 + 0.8 + 1.5 + 3(5.25) 7.05 �3.59 , (1.5)(15)(0.121) 2.723 1.25 + 0.8 + 1.5 + 5.25 + 0.75 9.55 26.0. 21.531 121.84 ,o M.

1 2 3

2(0.8)(15) (y,) 0.9 (1.5) (15)(y,) = 3.375 I 2(5.25)(15)(yJ = 5.906

4 5 6

=

�

=

•

=

=

=

=

=

The 'overturning moment

Mo =

�' p. = C�·5 ) '. (See Table 3.3.) In Eg. (8.44a), Li is the effective length; that is, Li = L, - 2e

(8.44b)

where e = eccentricity given by L, MR - Mo e=-lV 2

in which lV = WI + W, . . . + qa' . From Eg. 8.24, the vertical stress at Z =

(8.44c)

H

is (8.45)

So the factor of safety against bearing capacity failure is •

(bearing capacity) = -,uo

FS

. . q,h

(H)

(8.46)

Generally, minimum values of FS(o\'crturning) = 3, FS (!>liding) = 3. and FS(t�earingcllpaciIY failure) = 3 to 5 are recommended.

"

8. 14 Step·by·Step·Design Procedure Using Metallic Strip Reinforcement

393

Example 8.6 A 30-ft-high retaining wall with galvanized steel-strip reinforcement in a granular backfill has to be constructed. Referring to Figure 8.23, given: Granular backJill: Foundation soil:

i

1' 1

i 1'2

ci

= =

36° 105 lb/ft3

=

28° 110 lb/ft3 1000 lblft'

= =

Galvanized steel reinforcement:

SHSy

Width of strip, W = 3 in. = 2 ft center-to-center = 3 ft center-to-center fy = 35,000 Ib/in2 � = 20°

=3

Required FS(B) = 3 Required FS(p)

Check for the external and internal stability. Assume the corrosion rate of the gal­ vanized steel to be 0.001 in./year and the life span of the structure to be 50 years. Solution

Internal Stability Check

lie thickness: Maximum tie force. Tmax = (T�lmax) 5v5H

so

( - �i)SySH

Tmox = 1't H tan 2 45

From Eq. (8.39), for tie break,

or t=

[

(

(105) (30) tan' 45 -

C )

)

]

36 (2) (3) (3) 2

3 ft (35,000 x 144Ib/ft') 2

= 0.0117 ft = 0.14 in.

394

Chapter 8 Retaining Walls

50 yr, th n

tIhfethacte ratuale ofthicorckness.rosionofis the tieisn.wiJyrl beand the life span of thie structure is e . a Refer to Eq. woulFord bethenough is case, iT� and iTo so tan( ) tan� Now i and the fol owing table can be prepared. (Note: 0.001

I,

I

+ (0.001) (50) = 0.19 n.

= 0.14

So tie thickness of 0.2 in. •

(8.38).

TIe length:

(H - z)

L=

45

i

+2

� = 20'.)

+

= "'/JzKa = YIZ, Y K..:: SCc:PJ..c:..c:.. IZ_ _ F--, aSVSH --,_..:.:. 2WYIZ

FSCP) = 3, H = 30 ft, W = 3 n.,

lie length L (ft) [Eq. (8.38)]

z(ft)

5 10 15 20 25

38.45 35.89 33.34 30.79

28.25 25.7

30

SoExteusrenala Stability Refer to Figure For this case, using Eq. tie length of L

=

40 ft.

Check

Check for overturning:

8.26.

(8.42)

W,Xl

WI

=

y, HL = ( 105) (30) (40)

=

126,000

Xl = 20 Pa =

ft i iT� H

o

dz =

lb

trlKaH2 = (�) ( 105) (0.26) (30)2

30 z' = - = 10 ft 3 FSco"" urni,g) = Check for sliding:

=

12,2&S lb/ft ,

.

,

(126,000) (20) = 20.5 > 3-O K (12,285) ( 10)

From Eq. tan(ki )

FSC'tiding) =

..

(8.43)

W,

Pa

126,000 =

tan [(t) 12.285

3-OK

8. 15

Retaining Walls with Geotextile Reinforcement

395

30 ft f-- W, --I

y,

�

L � 40 ft

1 1 0 Ib/ft'

28' c; = IOQQ Iblft' "" �

'.

. . .e

.:: ' . ·

':::' " ,-'" . :

L. Mr Mo . =- -

2.

liV·

, ,: , ',

" '< " " " ', " ,-�=- ---lr--Geotextile

....L-'-r..,-C>--"---

-

_ _

s,. -1__ Geotexlile

__

In situ soil

_ _

1'2;¢2;C2

Figure B.27

Retaining wall with geotextile reinforcement

Internal Stability

Step 1.

Determine the active pressure distribution on the wall from the formula whereRankine active pressure coefficient tan'( - ;/2)-.j

",\., !.,

. I

..... ,

•

r

.:'_,.:,. �

_.;',

�_•

• '

(b) Extensible reinforcement

Location of potential failure surface (After Transportation Research Board,

I(n8.1probl 8.6, use 23.58 kN/m' (150 Ib/ft'). Also, in Eq. 1),Foruseemsthe8.cant1 thirough 2/3 and in Figure PR1, let the following data beWalgilvdien:mensiolens:ver retH ain18infgt,xwal, l shown 18 in.,x, 30 4 ft,x, 6 ft.xs 2.75 4f t . a 10° Soil properties: 117Ib/ft"i 34°, 110lb/ft" 18°,c', SOOlb/ft'

S.l

k, = k, =

Y,o",," =

Pp = O.

=

D=

y, =

= =

=

=

y, =

in" X3 =

=

=

2 =

=

ft.

405

Problems

1'1 C; = 0

4>;

I-- x, -+ -1'+1.1--x,

X4

1',

4>i c',

8.2

Calcultayt.e the factor of safety with respect to overturning, sliding, and bearing capaci Repeat Probl e m 8. 1 wi t h t h e f o l o wi n g: Wall dimensions: H2.75 ft22, Dft, xI4 ft1, 2a in.5°,x, 27 in.,x, 4.5 ft,x, 8 ft,x, Soil properties: YI1000l11blf0t'lblft', 36°, 120 lb/ft', 15°, Repeat Probl e m 8. 1 wi t h t h e f o l o wi n g: Wall dimensions: x,H 0.6.85 m,D 1.0.53m,m,ax, 0°0.6 m, x, 0.8 m, x. 2 m, Soil properties: Y30kN/m' I 18.08 kN/m" 36°, 19.65 kN/m', , 15°,c, iin.diFing,guregivenP8.t4h.eCalfolcoulwiatnegthdate faa:ctor of safety ty rettaoinovert ing walurnil insgshown wiAWaltgravihlrespect and sl dimensions: 0.H75 m,6 m,xx. I 0.80.m,6 m,x, 2mm,x, 2 m,x. 0.5 m,x, D 1. 5 Soil properties: 40kN/m' YI 16.5 kN/m" 32°, 18 kN/m', 22°, Use the Rankine active earth pressure in your calculation. =

=

8.3

= =

=

8.4

Figure PB, 1

=

=

=

ro,

=

=

=

i =

=

i =

=

=

i

=

,

=

=

=

=

y, =

c, =

=

=

y, =

=

=

=

y, =

Xl =

=

=

=

=

, =

=

=

c, =

406

Chapter 8 Retaining Walls

H

y,

cj = 0

,�,';,

"",�,��: , :, .,' .

>, ';

�.��\):�:;,:�:f,�:: ," ." ,�>! . .

121.64 ici-tym� ' 0'7 = 4c + (yL, + 'Y'�) ;;: 24a.3!ikN;m� . ' . .···,�;·�,,:i:�,,· S ' �ep f. '. ';' : ri �ur�d�tX����'�1�� C�a�'���'Q diaWn' �s� :f)�'}i' StepB. . R�l� J,5D�"'etical "'1.5(2.13) .�. 3�!I.l Step 6.

,

'>

0'6 = 'Ie - (yL,

�:; �� � ...

•

,

+

y'i,)

.

Again, fr(jDl. Eq.

,- .>'

;"

,

�

"

,

���lJ

(9.46),

So

41)' 52.2(0.41 + 1.78) - 127.64(0, 2 = = 114.32 - 10.73 103.59 tN-m/m Themininl11mrequired section modulus (assllDling t!tafO',u = 172;5 MNJm�)�,.,> " '·· · i,;,; 10359 k#tm,lm" 1J-' x iij-;,� �otth,!� f,;�l1":'�R · ' " ;" C.:�5' ,. f! ' 172.5 x !Ql·kN/m2 . ' Mmax =

,

0

"

. ..•

. .

. .

.

.

.

.

.

.

, .•

.0 '

.

., 0

0.

�:.�

.

�.:

Special Cases for Cantilever Walls Penetrating Clay

9.5,

As in Section relationships for special cases for cantilever walls penetrating clay may also be derived. Referring to Figure we can write

9.12,

0', = yLKa 0'6

4c yL 0'7 = 4c + yL D

Al

and

L ,

_

=

-

1 - 2:1 - 2:I LU2' L'Ka

D(4c

- yL) - �yL2 Ka

4c

(9.47) (9.48) (9.49) (9.50) ...•

I

(9.51) I

I .

t

9.8

I

'0

Anchored Sheet-Pile

Walls

429

Sand y '

E.

P, , '"

".

L,

Clay 'Ysat

L,

c

D

' = 0

1

1

Figure S.12

clay

"7

Sheet-pile wall penetrating

The theoretical depth of penetration, D, can be calculated [in a manner similar to the calculation of Eq. (9.44») as (9.52) where ZI = 3' The magnitude of the maximum moment in the wall is _

L

(9.53)

(9.54)

P where z' = I

-

0'6

hL'Ka

= 4

C

-

YL

'

(9.55)

Anchored Sheet-Pile WaJls When the height of the backfill material behind a cantilever sheet-pile wall exceeds about m (=20 tying the wall near the top to anchor plates, anchor walls, or an­ chor piles becomes more economical. This type of construction is referred to as an­ chored sheet-pile wall or an anchored bulkhead. Anchors minimize the depth of penetration required by the sheet piles and also reduce the cross-sectional area and weight of the sheet piles needed for construction. However, the tie rods and anchors must be carefully designed.

6

ft),

430

Chapter 9

Sheet Pile Walls

Anchor tie rod ' . Water _ _ 1: _ _ _ _ _ ' ,

table

_ _ _ _ _ _

.!. _

_ _ _ _ _ _ _ _

I I I I I I I I

e=-I",in'lec--t,-I

Dredg

,

I

1 D

1

, , , , , ,

Sheet pile simply supported

---------------------

(a)

Anchor tie rod Water table

_ _

:!

_ _ _ _

) j

_ _ _ _ _ _

.1

_ _ _ _ _ _ __ _

I I

Detlection ,

Dredge line

D

1

I I I , \

.... \ \ \

Point of inflection

Sheet pile fixed at lower end (b)

Nature of variation of deflection and moment for anchored sheet piles: (a) free earth supprt method; (b) fixed earth support method

Figure 9.13

The two basic methods of designing anchored sheet-pile walls are (a) the free earth support method and (b) the fIXed earth support method. Figure 9.13 shows the

assumed nature of deflection of the sheet piles for the two methods. The free earth support method involves a minimum penetration depth. Below the dredge line, no pivot point exists for the static system. The nature of the variation of the bending moment with depth for both methods is also shown in Figure 9.13. Note that Dfn:e earlh < Dfixed earth

9.9

Free Earth Support Method for Penetration of Sandy Soil Figure 9.14 shows an anchor sheet-pile wall with a granular soil backfill; the wall has been driven into a granular soil. The lie rod connecting the sheet pile and the anchor is located at a depth I, below the top of the sheet-pile wall.

Free Earth Support Method for Penetration of Sandy Soil

9.9

T

(

I-

L,

Water

___ !f_��� _ _ _ _

431

A 0'

Anchor ti'e rod . I, .--\-...;.= .: ===-f--'-+....;.� F

�i. _

Water

C_ _ _t�b!e_ _�_ _ _

l�

i_ _

Sand

y, '

z

"--�--r--- P

Sand

Dredge line

Sand

D

Figure 9.14 Anchored sheet�pile wall penetrating sand

The diagram of the net pressure distribution above the dredge line is similar to that shown in Figure 9.7. At depth z = L" ") = yL,Ko, and at z = L" "2 = (yL, + y'L,) Ko' Below the dredge line, the net pre.ssure will be zero at z = L, + L, L,. The relation for L, is given by Eq. (9.6), or

L, +

+

At Z =

....

L,

L3

L, + + +

L., the net pressure is given by "8 = y' (Kp - K,J L,

(9.56)

Note that the slope of the line DEF is 1 vertical to y' (Kp - K,) horizontal. For equilibrium of the sheet pile, � horizontal forces = 0, and � moment about 0' = O. (Note: Point 0' is located at the level of the tie rod.) Summing the forces in the horizontal direction (per unit length of the wall) gives Area of the pressure diagram ACDE - area of EBF - F = 0 where F = tension in the tie rod unit length of the wall, or.

/

p

- 1"'L. - F = 0

432

Chapter 9

Sheet Pile Walls

or (9.57) where P = area of the pressure diagram ACDE. Now, taking the moment about point 0' gives -P[(LI + L , + L,) - (z + II)l + H'Y'( Kp - K,)lL�(l, + L , + L3 + �L4) = 0

or

Equation (9.58) may be solved by trial and error to determine the theoretical depth. L4:

The theoretical depth is increased by about 30-40% for actual construction, or Dac,ua' =

1.3 to 1.4Dlheorelicai

(9.59)

The step-by-step procedure in Section 9.4 indicated that a factor of safety can be applied to Kp at the beginning [i.e., Kp(d,,;gol = Kp/FSl. If that is done, there is no need to increase the theoretical depth by 30 to 40%. This approach is often more conservative. The maximum theoretical moment to which the sheet pile will be subjected occurs at a depth between z = L I and z = LI + L2 • The depth z for zero shear and hence maximum moment may be evaluated from (9.60) Once the value of z is determined, the magnitude of the maximum moment is easily obtained. Sometimes, the dredge line slopes at an angle (3 with respect to the horizontal, as shown in Figure 9.15a. In that case, the passive pressure coefficient will not be equal to tan'(45 + "" /2). The variations of Kp (Coulomb's passive earth-pressure analysis for a wall friction angle of zero) with {3 for "" = 30° and 35° are shown in Figure 9.15b. With these values of Kp' the procedure described in this section may be used to determine the depth of penetration, D.

9.9

Free Earth Support Method for Penetration of Sandy Soif

433

4

3

_ _ _ _ _ _ y._ _ _ _ _ _

_ _ y. _ _ _

'

� ��

�•�.;: f.�,:".l,

�

' ffir ,�!��:�f�� �� ==,:�Yl�:� ;�·.;. g�'��":$'·& ��:,;,l,;�'.:�·\,¥:� "'..: , , . ", . . ',J.: pe(e�� theoretical �d ai:tU� #p�'''- = 0). For the definition of y, y' , LI, and L" see Figure 9.28. 2. The nondimensional wall height is (9.77)

.!...

3. The flexibility number is p [see Eq. (9.65) or Eq. (9.66)] 4. Md = design moment Mmox = maximum theoretical moment

450

Chapter 9

Sheet Pile Walls 1.0

Log p � -3 . 1

0.8

Md Mm�x

_======�---"':

a � 0. 8 0 .7-

0.6

0.6--

0.4 1.0

Log p� -2.6

0.8

M, Mma'l.

06 .

� 0.8 --=::::::::====::;;-;;:0.70.6--

0.4 1 .0

Log p � -2.0

0.8

Md M�,

0.6

0.4

0.70.6 -0

0.5

1.0

Stability number. Sn

1.5

1.75

Figure 9.29 Plot of MJ/Mmu against stability number for sheet-pile wall penetrating clay (After Rowe, 1957)

The procedure for moment reduction, using Figure 9.29, is as follows: Step 1. Step 2. Step 3. Step 4. Step 5.

Obtain H' = L] + L, + D","m' Determine a = (L] + L,) IH'. Determine Sn [from Eq. (9.76)]. For the magnitudes of a and S" obtained in Steps 2 and 3, determine MdlMmox for various values of log p from Figure 9.29, and plot MdlMm" against log p. Follow Steps 1 through 9 as outlined for the case of moment reduction of sheet-pile walls penetrating granular soil. (See Section 9.11.)

9. 14

-

.

Free Earth Support Method for Penetration of Clay

451

�xalfiple 9.7 '

t/l == the a

35°, and c = 41 kN/m2. . r.. , " 2QkN/m3; ' ;: ' .., pefemlirle the theoretical depth of embedmeni of the sheet-pile wall. tl'l �@l'e 9.28,·let Ll

.,- ,

:ji; ·.Otcutilie

. , . ,.

"

���:2:\ .'
' = 30°, and y = 110 Ib/ft3 • The anchor plates are made of con­

crete and have a thickness of 3 in. Using Ovesen and Stromann 's method, cal­ culate the ultimate holding capacity of each anchor. Take y","",,, = 150 Ib/ft'. 9.17 A single anchor slab is shown in Figure P9.17. Here, H = O.� m, h = 0.3 m, y = 17 kN/m3, and 4>' = 32°. Calculate the ultimate holding capacity of the anchor slab if the width B is (a) 0.3 m, (b) 0.6 m, and (c) = 0.9 m. (Note: center-to-center spacing, S ' = "' . ) Use the empirical correlation given in Section 9.15 [Eq. (9.82»)'

References

Casagrande, L. (1973). "Comments on Convenlional Design of Retaining Structures," Journal ofche Soil Mechanics and Foundations Division, ASCE, Vol. 99, No. SM2. pp. 181-198. Das. B. M . . and Seeley. O. R. (1975). "Load-Displacement Relationships for Vertical Anchor Plates," Jou.rnal of the Geotechnical Engineering Division. American Society of Civil Engineers. VoL 101. No. on. pp. 71 1-715.

References

465

Ghaly, A. M. (1997). "Load-Displacement Prediction for Horizontally Loaded Vertical Plates." Journal of Geotechnical and Geoenvironmental Engineering, ASCE, Vol. 123, No. 1. pp. 74-76. Hagerty, D. 1., and Nofal, M. M. (1992). "Design Aids: Anchored Bulkheads in Sand," Cana­ dian Geotechnical Journal, Vol. 29, No. 5, pp. 789-795. Littlejohn. G. S. (1970). "Soil Anchors," Proceedings, Conference on Ground Engineering, In­ stitute of Civil Engineers, London, pp. 33-44. Mackenzie, T. R. (1955). Strength. ofDeadman Anchors in Clay, M.S. Thesis, Princeton University. Princeton, N. 1. Nataraj,M.S., and Hoadley,P.G. (1984). "Design of Anchored Bulkheads in Sand," Journal of Geotechnical Engineering, American Society of Civil Engineers, Vol. 110, No. GT4, pp. 505-515. Ovesen, N. K., and Stromann, H. (1972). "Design Methods for Vertical Anchor Slabs in Sand," Proceedings, Specialty Conference on Performance of Earth and Earth-Supported Struc­ tures. American Society of Civil Engineers, Vol. 2.1, pp. 1481-1500. Rowe, P. W. (1952). "Anchored Sheet Pile Walls," Proceedings, Institute ofCivil Engineers, Vol. 1 ,

Part 1 , pp. 27-70. Rowe, P. W. (1957). "Sheet Pile Walls in Clay," Proceedings, Institute ofCivil Engineers, Vol. 7, pp.654-692. Schroeder, W. L.. and Roumillac, P. (1983). "Anchored Bulkheads with Sloping Dredge Lines," Journal of Geotechnical Engineering. American Society of Civil Engineers, Vol. 109. No. 6, pp. 845-851. Tschebotarioff, G. P. (1973). Foundations, Retaining and Earth Structures, 2nd ed., McGraw­ Hill, New York. Tsinker, G. P. (1983). "Anchored Street Pile Bulkheads: Design Practice," Journal of Geotech­ nical Engineering. American Society of Civil Engineers, Vol. 109. No. GT8. pp. 1021-1038.

10

Brac�d Ciits

10. 1

Introduction Sometimes construction work requires ground excavations with vertical or near­ vertical faces-for example, basements of buildings in developed areas or under­ ground transportation facilities at shallow depths below the ground surface (a cut-and-cover type of construction). The vertical faces of the cuts need to be pro­ tected by temporary bracing systems to avoid failure that may be accompanied by considerable settlement or by bearing capacity failure of nearby foundations. Figure 1 0.1 shows two types of braced cut commonly used in construction work. One type uses the soldier beam (Figure 10.la), which is driven into the ground before excavation and is a vertical steel or timber beam. Laggings, which are hori­ zontal timber planks, are placed between soldier beams as the excavation proceeds. When the excavation reaches the desired depth, wales and struts (horizontal steel beams) are installed. The struts are compression members. Figure 10.1b shows an­ other type of braced excavation. In this case, interlocking sheet piles are driven into the soil before excavation. Wales and struts are inserted immediately after excava­ tion reaches the appropriate depth. Figure 10.2 shows the braced-cut construction used for the Chicago subway in 1940. Timber lagging, timber struts, and steel wales were used. Figure 10.3 shows a braced cut made during the construction of the Washington, DC, metro in 1974. In this cut, timber lagging, steel H-soldier piles, steel wales, and pipe struts were used. To design braced excavations (i.e., to select wales, struts, sheet piles, and sol­ dier beams), an engineer must estimate the lateral earth pressure to which the braced cuts will be subjected. The theoretical aspects of the lateral earth pressure on a braced cut were discussed in Section 7.7. The total active force per unit length of the wall (Pa) was calculated using the general wedge theory. However, that analysis does not provide the relationships required for estimating the variation of lateral pressure with depth, which is a function of several factors, such as the type of soil, the experience of the construction crew, the type of construction equipment used, and so forth. For that reason, empirical pressure envelopes developed from field ob­ servations are used for the design of braced cuts. This procedure is discussed in the next section.

466

i

I I

10.2 Pressure Envelope for Braced·Cut Design •

Wale

467

Strut

�'-'c-,---- Soldier --""";>1 ' . beam

Wedge Plan

Elevation

Ca) Wale

l� - - ""' - - -

t,4

Strut

Strut

I

------,

YA 1+---- Sheet ----+II pi le

Elevation

Figure 10. 1

Cb)

Plan

Types of braced cut: (a) use of soldier beams; (b) use of sheet piles

Pressure Envelope for Braced-Cut Design As mentioned in Section 10.1, the lateral earth pressure in a braced cut is dependent on the type of soil, construction method, and type of equipment used. The lateral earth pressure changes from place to place. Each strut shOUld also be designed for

468

Chapter 10 Braced Cuts

Figure 10.2

B. Peck)

Braced cut in Chicago subway construction, January 1940 (Courtesy of Ralph

to which-piressure t may bediasubjgramsectetd.hatTherefore, thpese braced cuthe pressure s should tbehedesimaxignedmumusilonadg apparent are envel o of al l t measured-pressure strut lodiadsagramin tathe fisecteld.ioFingfrom ure strutshowsloads.thIne mettdihaisgrams hfiodgure,fordetlobtetermiPIai'nnP"inedgPtfrom h3,eP4,apparent strut loads.The apparent horizon­ tal pressure can then be calculatbeed tashe measured 0"1 (dPl I � , ) ( d;P, , d3) 0"3 (dP,3 d, ) a

10.4

• • •

=

U2 = =

a�

=

( S)

+

(s)

+ 2'

(s)

2' + 2'

(S)

P,

(�' i') +

...

10.2

Pressure Envelope for Braced-Cut Design

469

Figure 10.3 Braced cut in the construction of Washington, DC. metro. May 1974 (Courtesy of Ralph B. Peck)

�I

T P,

P,

1, d,

P,

f

d, ...

P,

J f

JI

f-

d,l2

+-----

+

d1/2 dl2

-} - - - - - ---d,t2

+

di2

-}-----

d,t2

�?t��

CT,

--- --

'---

- - .-----' -

- --

'--

Figure 10.4 Procedure for calculating apparent-pressure diagram from mea­ sured strut loads

470

Chapter 10 Braced Cuts

where 0" 1 ; 0"2, O":h 0"4 = apparent pressures s = center-to-center spacing of the struts

Using the procedure just described for strut loads observed from the Berlin subway cut, Munich subway cut, and New York subway cut, Peck (1969) provided the envelope of apparent-lateral-pressure diagrams for design of cuts in sand. This envelope is illustrated in Figure 10.5, in which 0".

= 0.65yHK.

(10.1)

where y = unit weight H = height of the cut K. = Rankine active pressure coefficient = tan' (45 - '/2) ' = effective friction angle of sand Cuts in Clay

In a similar manner,Peck ( 1 969) also provided the envelopes of apparent-lateral­ pressure diagrams for cuts in soft ro medium rlay and in su/f clay. The pressure envelope for soft to medium clay is shown in Figure 10.6 and is applicable to the condition yH >4 c

where c = undrained cohesion = 0). The pressure, 0"•• is the larger of

(

I

0".

[ (�)]

=.YH l -

and 0". = 0.3yH

.

(10.2)

where y = unit weight of clay. The pressure envelope for cuts in stiff clay is shown in Figure 10.7, in which 0".

= 0.2yH to O.4yH

(with an average of O.3yH)

(10.3)

is applicable to the condition yH/c .. 4. When using the pressure envelopes just described. keep the following points in mind:

Pressure Envelope for Cuts in Layered Soil

10.3

T '-----

L

0.25 H

H

�

T c:--...

0.25 H

t +

-.....

a.

0.5 H a.

0.75 H

Figure 10.5 Peck's (1%9) apparent-pressure envelope for cuts in sand

1

0.25 H

'-......

471

......

au

1 �./

./

Figure 10.7 Peck's (1969) apparent-pressure envelope for cuts in stiff clay

Peck's (1969) apparent-pressure envelope for cuts in soft to medium clay Figure 10.6

1. They apply to excavations having depths greater than about 6 m ( = 20 ft). 2. They are based on the assumption that the water table is below the bottom of the c u t. 3. Sand is assumed to be drained with zero pore water pressure. 4. Clay is assumed to be undrained and pore water pressure is not considered.

10.3

Pressure Envelope for Cuts in Layered Soil Sometimes, layers of both sand and clay are encountered when a braced cut is being constructed. In this case, Peck (1943) proposed that an equivalent value of cohesion (c/> = 0) should be determined according to the formula (see Figure 1O.8a). o

Cay

-

_

�[ KOH2 tan "" 2H 'Ys

s

s

'lis

+

(Hi '- H )n,qllJ s

(lOA)

where H = total height of the cut y, = unit weight of sand H, = height of the sand layer K, = a lateral earth pressure coefficient for the sand layer ( = 1 ) c/>; = effective angle of friction of sand q" = unconfined compression strength of clay ,, ' = a coefficient of progressive failure (ranging from 0.5 to 1.0; average value 0.75)

472

Chapter

10

Braced Cuts

H

H

• •

•

(b) .

Layered soils in braced cuts

Figure 10.B

The average unit weight of the layers may be expressed as Yo =

1

H

[y,H, + (H - Hshcl

(10.5)

where Yc = saturated unit weight of clay layer. Once the average values of cohesion and unit weight are determined, the pres­ sure envelopes in clay can be used to design the cuts. Similarly, when several clay layers are encountered in the cut (Figure 10.8b), the average undrained cohesion becomes (10.6)

where CI, C"

, Cn

= undrained cohesion in layers 1, 2, . . . , n

HI' H" . . . , Hn = thickness of layers 1, 2. . . . , n •

•

•

The average unit weight is now

(10.7)

10.4

Design of Various Components of a Braced Cut Struts

In construction work, struts should have a minimum vertical spacing of about 2.75 m (9 ft) or more. Struts are horizontal columns subject to bending. The load-carrying capacity of columns depends on their slenderness ratio, which can be reduced by

10.4

Design of Various Components of a Braced Cut

473

providing vertical and horizontal supports at intermediate points. For wide cuts, splicing the struts may be necessary. For braced cuts in clayey soils, the depth of the first strut below the ground surface should be less than the depth of tensile crack, z" From Eq. (7.8), u� = yzKa - 2C/� where Ka = coefficient of Rankine active pressure. For determining the depth of tensile crack, u'Q = 0 = yzCKQ - 2c" V1( Ka

or

2c' z = -, �y

With 4> = 0, Ka = tan'(45 - 4>/2} = 1, so 2c z =y ,

A simplified conservative procedure may be used to determine the strut loads. Although this procedure will vary, depending on the engineers involved in the project, the following is a step-by-step outline of the general methodology (see Figure 10.9): Slep 1.

Slep 2. Slep 3.

Draw the pressure envelope for the braced cut. (See Figures 10.5, 10.6, and 10.7.) Also. show the proposed strut levels. Figure 10.9a shows a pressure envelope for a sandy soil; however, it could also be for a clay. The strut levels are marked A, B, C, and D. The sheet piles (or soldier beams) are assumed to be hinged at the strut levels, ex­ cept for the top and bottom ones. In Figure 10.9a, the hinges are at the level of struts B and C. (Many designers also assume the sheet piles or soldier beams to be hinged at all strut levels except for the top.) Determine the reactions for the two simple cantilever beams (top and bottom) and all the simple beams between. In Figure 10.9b, these reactions are A, BI> B2, CI> C" and D. The strut loads in the figure may be calculated via the formulas PA = (A ) (s) PB = (B, Pc = (C,

,
(b)

Figure 10.9 Determination of strut loads: (a) section and plan of the cut; (b) method for determining strut loads

where

PA, PH, Pc, PD = loads to be taken by the individual struts at lev­ els A, B, C, and D, respectively A, B" Bz, C" Cz, D = reactions calculated in Step 2 (note the unit: forcejunit length of the braced cut) s = horizontal spacing of the struts (see plan in Fig­ ure 10.9a)

Step 4. Knowing the strut loads at each level and the intermediate bracing conditions allows selection of the proper sections from the steel construction manual.

.,

10.4 Design of Various Components of a Braced Cut

475

Sheet Piles

The following steps are involved in designing the sheet piles: Step 1; For each of the sections shown in Figure 10.9b, determine the maxi­ Step 2. Step 3.

mum bending moment. Determine the maximum value of the maximum bending moments ( Mmox) obtained in Step 1. Note that the unit of this moment will be, for example, kN • m/m (lb-ft/ft) length of the wall. Obtain the required section modulus of the sheet piles, namely, s

Step 4.

=

Mmax UaU

(10.9)

where IT,ll = allowable flexural stress of the sheet pile material. Choose a sheet pile having a section modulus greater than or equal to the required section modulus from a table such as Table 9.1.

Wales

Wales may be treated as continuous horizontal members if they are spliced prop­ erly. Conservatively, they may also be treated as though they are pinned at the struts. For the section shown in Figure 10.9a, the maximum moments for the wales (assum­ ing that they are pinned at the struts) are, (A) (s') At level A, Mm" = ":""":' 8-'-':' =

At level B,

lv/max

At level C,

Mmax =

( BI

+

B,)s'

(e l

+

C,)s'

8

8

and At level D, Mmax =

( D) (s' ) 8

where A, BI, Bz, e" e" and D are the reactions under the struts per unit length of the wall (see Step 2 of strut design). Now determine the section modulu� of the wales: s

= Mmrrx Uall

The wales are sometimes fastened to the sheet piles at points that satisfy the lateral support requirements.

476

Chapter

10

Braced Cuts

Example 10.1 The cross section of a tong braced cut is shown in Figure 1O.lOa. a. Draw the earth-pressure envelope. b. Determine the strut loads at levels A, B, and C. c.

Determine the section modulus of the sheet pile section required.

d. Determine a design section modulus for the wales at level B.

(Note: The struts are placed at 3 m, center to center, in the plan.) Use CT,,, = 170 x 103 kN/m2

Solution Part a

We are given that 'Y = 18 kN/m', c 'YH

�=

=

35 kN/m', and H = 7 m. So,

(18) (7) = 3.6 < 4 35

Thus, the pressure envelope will be like the one in Figure 10.7. The envelope is plotted in Figure 1O.lOa with maximum pressure intensity, CT., equal to 0.3'YH = 0.3(18) (7) = 37.8 kN/m2• Part b

To caiclliate the strut loads, examine Figure 10. lOb. Taking the moment about B10 we have I ¥B, = O, and A (2.5 ) :or

G) C·;5 ) +

Also, I verti�al forces =

A

O. Thus,

= 54.02 kN/m

W.75) (37.8) + (37.8) (1.75)

or 33.08 + 66.15 So, Due to symmetry,

-

A = B,

B, = 45.2 kN/m B, = 452 kN/m

and

=

C = 54.02 kN/m

A + B,

=0·

10.4 Design of Various Components of a Braced Cut

1�'----- 6 m --------�-1

i+- 1 .75 m

, , ,

A

1.75 m ---+l

i+- 1.75 m

I

1.75 m ---.j

- -

37.8 kN/m'

37.8 kN/m'

, 2 1 .6 ,

j.-- I m -I'

_I'

2.5 m ----+I

B,

B,

-IC

I+--- 2.5 m

I m ->l

(b) Determination of reaction 43.23 kN

43.23 kN

x � Ll96 m �

I-

-+

__

-I

�

__ __ __

BI B , r'

__ __

__ __

-7L-

+-��

__ __

10.8 kN

45.2 kN

45.2 kN

(e) Shear force diagram Rs.ure 10, 10 Analysis of a braced cut

477

478

Chapter 10 Braced Cuts

Hence, the strut loads at the levels indicated by the subscripts are P" =

54.02

X

horizontal spacing, s =

X

(45.2 + 45.2)3 271.2 and Pc = 54.02 X 3 = 162.06 kN At the left side of Figure 10. lOb, for the maximum moment, the shear force should be zero. The nature of the variation of the shear force is shown in Figure 1O.1Oc. = (B1

PH

+ B,)3

54.02 3 = 162.06 kN

=

=

kN

Part c

The location of point E can be given as

45.2 = 1 .196 m 37.8 37.8 37.8 1 Magnitude of moment at A = ;- ( 1 ) ( -- 1 ) ( - ) 1 .75 3 = 3.6 kN-m/meter of wall x

Also,

=

reaction at B I

=

1

X

_

and

(45.2 1.196) - (37.8 1.196)e·�6) = 54.06 27.03 27.03 kN-mJmeter of wall Because the loading on the left and right sections of Figure 1O.10b are same, the magnitudes of the moments at F and C (see Figure 1O.1Oc) will be respectively. Hence, the maximum moment is same as those at and 27.the0The 3 kN-mJmeter of wall. section modulus of the sheet piles is thus 27.03 kN-m 15.9 10-sm3/m of . Mm" S = 170 x 10' kNJm' Magnitude of moment at E =

X

X

-

=

the

E

=

A,

0'",

=

x

the waD

Part d

The reaction at level B has been calculated in Part b. Hence,

(45.2 + 45.2)3' = 101.7 kN-m 8 8 101.7 1 101.7 Section modulus. S = -;;:;;;;- = ( 70 1000)

Mmax =

and

(BI + B,)s'

=

= 0.598 X

X

10-3 m'

•

10.4 Design of Various Components of a Braced Cut

( i-.

479

Example 1 0.2 Refer to the braced cut shown in Figure 10.11, for which 'Y = 112 Ibjft3,.p' = 32°, lind c' = O. The struts are located 12 ft on center in the plan. Draw the earth­ pressure envelope and determine the strut loads at levels A, B, and C. Solution

thi$ case, the earth-pressure envelope shown in figure 10.5 is applicable, Hence,

For

( �')

Ko = tan2 45 From Equation

(10.1)

-

( 3D

= tan2 45

-

= 0.307

f7o "" 0.65 'YHKo = (0.65) (112) (27) (0.307) = 603.44 lbjft'

Figure 10.12ashows the pressure envelope. Refer to figure 10.12b and calculateBJ:

�MB, = 0 A

=

(603.44) (15) 9

(�)

. = 7543 Ib(ft

B, = (603.44) ( 15) - 7543 = 1 508.6 Ib./ft

=

1509 lb/ft

Now, refer to Figure 1O.12e and calculate B,:

�MB, =

0

kl'--- 16 fl ---+1'1 T

6 fl

A

.,.'

9 fl B 9 fl

Sand y c=O B/Vz B" = VzB'

Bjerrum and Eide (1956) compiled a number of case records for the bottom heave of cuts in clay. Chang (2000) used those records to calculate FS by means of Eq. (10.14): his findings are summarized in Table 10. 1. It can be seen from this table that the actual field observations agree well with the calculated factors of safety.

Table 10.1

Calculated Factors of Safety for Selected Case Records Compiled by Bjerrum and Eide (1956) and Calculated by Chang (2000)

Site

Pumping station, Fomebu, Oslo Storehouse, Drammen Sewerage tank, Drammen Excavation, Grey Wedels Plass, Oslo Pumping station, Jembanetorget, Oslo Storehouse, Freia, Oslo Subway, Chicago

(m)

81L

(m)

HI8

(kN/m')

'Y

(kN/m')

c

(kN/m'l

q

FS [Eq. (10.14)]

5.0

1.0

3.0

0.6

17.5

7.5

0

1.05

Total failure

4.8

0

2.4

0.5

19.0

12

15

1.05

Total failure

5.5

0.69

3.5

0.64

18.0

10

10

0.92

Total failure

5.8 0.72

4.5

0.78

18.0

14

10

1.07

Total failure

8.5

0.70

6.3

0.74

19.0

22

0

1.26

Partial failure

5.0 16

0 0

5.0 11.3

1.00 0.70

19.0 19.0

16 35

0 0

1.10 1.00

Partial failure Near failure

8

H

Type of failure

484

Chapter 10 Braced Cuts

c

H

c-

Ip

= 0

b

a

".

a'

�

Figure 10.15 Force on the buried length of sheet pile

Equation (10.14) is recommended for use in this test. In most cases, a factor of safety of about 1.5 is recommended. In homogeneous clay. if FS becomes less than 1.5, the sheet pile is driven deeper. (See Figure 10.15.) Usually, the depth d is kept less than or equal to B/2, in which caSe the force P per unit length of the buried sheet pile (aa' and bb') may be expressed as (U. S. Department of the Navy. 197 1 ) E

= 0.7("1HI}

and

--, l.4cH.- '!reB)

for 4 :> .0.47B .

(10.15)

(10.16)

10.4

Refer to the braced cut described in Example 10.1 (see F'i21Jlre'10; that the ]tmgth of the cut is 25 m .and it hard stratum is located below the bottom of the cut. Calculate the factor of safety against Use Eq. (10.14) Solution

From Eq. (10.14),

FS

=

(

)

0.2B" cH 5.14c 1 + ---y- + If" yH + q

, .,

'"

h ,' "

70.6 Stability of the Bottom of a Cut in Sand

i ;-

'Yith T 7, 4 m,

B

485

6

y'z = ,Vi

Stability of the Bottom of a Cut in Sand The bottom of a cut in sand is generally stable. When the water table is encountered, the bottom of the cut is stable as long as the water level inside the excavation is higher than the groundwater level. In case dewatering is needed (see Figure 10.16), the factor of safety against piping should be checked. [Piping is another term for failure by heave, as defined in Section 1 .10: see Eq. (1.37).] Piping may occur when a high hydraulic gradient is created by water flowing into the excavation. To check the

_ ; ....

"

." ,'.

\ \ ,

' ..i" Flow of ..... _ � water

Impervious layer

Figure 10.16

sand

Stability of the bottom of a cut in

486

Chapter 10 Braced Cuts

, , , , , I I I I I I

. I

, , , , ,, ' , \ \ \

\

�-:{�:��_�1:f�,�j��;5������!� ���,��::.�Jf);��;���}.f�;��3�� ��::j�t�Xt:�·?/i.�;(�� J',

Imper\'ious layer

Figure 10. 17 Determining the factor of safety against piping by drawing a flow net

factor of safety. draw now nets and determine the maximum exit gradient lim",,,;"] that will Occur at points A and H. Figure 10.17 shows such a !low net, for which the maximum exit gradient is h lmax(exit)

.

=

Nd --;;-

h

= Nda

(10.17)

where a = length of the flow element at A (or B) Nd = number of drops (Note: in Figure 10.17. Nd = 8; see also Section 1.9)

I

The factor of safety against piping may be expressed as i FS = . tt

�(exit)

(10.18)

where itt = critical hydraulic gradient. The relationship for itt was given in Chapter 1 as

G

1 e + 1 -

i = -'-cr

The magnitude of itt varies between 0.9 and 1.1 in most soils, with an average of about 1 . A factor of safety of about 1.5 is desirable.

10.7

On

i

;

inal round surface

Lateral Yielding of Sheet Piles and Ground Settlement x'

I--

"

487

I

Deflecled shape of sheet pile

z ++-+z'

Figure 70. 78

10. 7

I

H

Lateral yielding of sheet pile and ground settlement

Lateral Yielding of Sheet Piles and Ground Settlement In braced cuts, some lateral movement of sheet pile walls may be expected. (See Figure 10.18.) The amount of lateral yield (8,,) depends on several factors, the most im­ portant of which is the elapsed time between excavation and the placement of wales and struts. As discussed before. in several instances the sheet piles (or the soldier, piles as the case may be) are driven to a certain depth below the bottom of the excavation.The reason is to reduce the lateral yielding of the walls during the last stages of excava­ tion. Lateral yielding of the walls will cause the ground surface surrounding the cut to settle. The degree of lateral yielding, however, depends mostly on the type of soil below the bottom of the cut.If clay below the cut extends to a great depth and yHIe is less than about 6, extension of the sheet piles or soldier piles below the bottom of the cut will help considerably in reducing the lateral yield of the walls. However. under similar circumstances, if yHlc is about 8, the extension of sheet piles into the clay below the cut does not help greatly. In such circumstances, we may expect a great degree of wall yielding that could result in the total collapse of the bracing systems. If a hard layer of soil lies below a clay layer at the bottom of the cut. the piles should be embedded in the stiffer layer. This action will greatly re­ duce lateral yield. The lateral yielding of walls will generally induce ground settlement. 8v• around a braced cut. Such settlement is generally referred to as ground loss. On the basis of several field observations. Peck (1969) provided curves for predicting ground settlement in various types of soil. (See Figure 10.19.) The magnitude of ground loss varies extensively; however, the figure may be used as a general guide.

488

Chapter

10

Braced Cuts 3 A - Sand and soft clay and average workmanship

B - Very soft to soft clay. Limited in depth tJ.elow base of excavation

2

C - Very soft to soft clay_ Great depth below excavation

8,.

H (%)

C B A o �-----.----�--� 3 2 4 Distance from the hraced \vnll H

Figure 10.19 Variation of ground settlement with distance (After Peck. 1969)

Moormann (2004) recently analyzed about 153 case histories dealing mainly with the excavation in soft clay (that is, undrained shear strength, c ".. 75 kN/mZ). Following is a summary of his analysis relating to 8Y(moxl' x', 8H(m"l' and z' (see Figure 10.18). •

Maximum Vertical Movement [8 Y(moxll 8Y(m,,)/H = 0.1 to 10.1 % with an average of 1.07% (soft clay) 8 Y(m"l/H = 0 to 0.9% with an average of 0.18% (stiff clay) 8Y(m,,)/H = 0 to 2.43% with an average of 0.33% (non-cohesive soils)

•

•

•

Location of 8V(moxl' that is x ' (Figure 10.18) For 70% of all case histories considered, x' ".. 0.5H. However, in soft clays, x' may be as much as 2H. Maximum Horizontal Deflection of Sheet Piles, 8H(m"l

For 40% of excavation in soft clay, 0.5% ".. 8H(m"JiH ".. 1 %. The average value of 8H(m""JiH is about 0.87%. In stiff clays, the average value of 8H(moxJi H is about 0.25%. In non-cohesive soils, 8H(m"l/H is about 0.27% of the average. Location of 8H(mox ,' that is z' (Figure 10.18) For deep excavation of soft and stiff cohesive soils, z'/ H is about 0.5 to 1.0.

•

Problems

489

10.1 Refer to the braced cut shown in Figure P10.l. Given: y = 17 kN/m'. q,' = 35°, and c ' = O. The struts are located at 3 m center-to-center in the 10.2

10.3 10.4 10.5

plan. Draw the earth-pressure envelope and determine the strut loads at lev­ els A, B, and C. For the braced cut described in Problem 10.1, determine the following: a. The sheet-pile section modulus b. The section modulus of the wales at level B Assume that j Figure 11.3

•

•

Precast piles with ordinary reinforcement

Advan tages: a. Can be subjected to hard driving b. Corrosion resistant Co Can be easily combined with a concrete superstructure Disadvantages: a. Difficult to achieve proper cutoff b. Difficult to transport

Precast piles can also be prestressed by the use of high-strength steel pre­ stressing cables. The ultimate strength of these cables is about 1800 MNjm2 (=260 ksi). During casting of the piles, the cables are pretensioned to about 900 to 1300 MNjm'( =130 to 190 ksi ) . and concrete is poured around them. After curing, the cables are cut. producing a compressive force on the pile section. Table 11.3 gives additional information about prestressed eoncrete piles with square and octagonal cross sections. Some general facts about precast prestressed piles are as follows: • • •

Usual length: 10 m to 45 m (30 ft to 150 tt) Maximum length: 60 m (200 ft) Maximum load: 7500 kN to 8500 kN (1700 kip to 1900 kip)

The advantages and disadvantages are the same as those of precast piles. Cast-ill-situ, or cast-ill-place, piles are built by making a hole in the ground and then filling it with concrete. Various types of cast-in-place concrete piles are currently used in construction, and most of them haye been patented by their manufacturers. These piles may be divided into two broad categories: (a) cased and (b) uncased. Both types may have a pedestal at the bottom. Cased piles are made by driving a steel casing into the ground with the help of a mandrel placed inside the casing. When the pile reaches the proper depth the mandrel is withdrawn and the casing is filled with concrete. Figures l1.4a. l1Ab. l 1.4c, and ll.4d show some examples of cased piles without a pedestal. Figure 11.4e

Chapter 1 1 Pile Foundations

498

Typical Prestressed Concrete Pile in Use (SI Units)

Table 1 1.3a

Design be_U

pile shape-

D (mml

Area of cross section (em'l

Perimeter (mml

Number of strands 11.1·mm 12.7·mm ·diameter diameter

Minimum effective prestress force (kNI

cal!adtv (kNI

Strength of co_ (MN/m'J

Section modulus (m' x 1 0�1

S

254

645

1016

4

4

312

2.737

0

254

536

838

4

4

258

1.186

34.5

41.4

462

555 962

556

778

S

305

929

1219

5

6

449

4.719

801

0

305

768

1016

4

5

369

3.097

662

795

S

3:;6

1265

1422

6

8

610

7.489

1091

1310

0

356

1045

1168

5

7

503

4.916

901

1082

S

406

1652

1626

8

11

796

1 1 .192

1425

1710

0

406

1368

1346

7

9

658

7.341

1180

1416

S

457

2090

1829

10

13

1010

15.928

1803

2163

0

457

1729

1524

8

11

836

10.455

1491

1790

S

508

2581

2032

12

16

1245

2 1 .844

2226

2672

0

508

2136

1677

10

14

1032

14.355

1842

2239

S

559

3123

2235

15

20

1508

29.087

2694

3232

0

559

2587

1854

12

16

1250

19.107

2231

2678

S

610

3658

2438

18

23

1793

37.756

3155

3786

0

610

3078

2032

15

19

1486

34.794

2655

3186

;'S =

T 1 D

square

section; 0

.---.---

=

octagonal �t!ction

I I I

Prestressed strand

I

\\lire spiral

� I

I I

---_._---

Wire spi ral Prestressed strand

,,- .- ....

..

.. \

-lI

.' .......... .fL. .... ;'

T 1 D

shows a cased pile with a pedestal. The pedestal is an expanded concrete bulb that is formed by dropping a hammer on fresh concrete. Some general facts about cased cast-in-place piles are as follows: • • • • •

Usual length: 5 m to 15 m (15 ft to 50 ft) Maximum length; 30 m to 40 m (100 ft to 130 ft) Usual load: 200 kN to 500 kN (45 kip to 115 kip) Approximate maximum load: 800 kN (180 kip) Advantages: B. Relatively cheap b. Allow for inspection before pouring concrete c. Easy to extend

Types of Piles and Their Structural Characteristics

11.2

499

Table 1 1.3b Typical Prestressed Concrete Pile in Use (English Units) I

r-

Pile shape-

D

(in.)

10 10 12 12 14 14 16 16 18 18 20 20 22 22 24 24

S 0

S

0

S

0

S

0

S

0

S

0

S

0

S

0

Area of cross section (in') loo

Perimeter (in.)

40 33 48 40 56 46 64 53 72 60 80

83 144 119 196 162 256 212 324 268 4oo 331 484 401 576 477

a s = square section; 0

Number of strands I-in h-In diameter diameter

4 4 5 4 6 5

4 4 6 5

8

7 11 9 13 11 16 14 20 16 23 19

8

7 10 8 12 10 15 12 18 15

66

88 73 96

80

70 58 101 83 137 113 179 148 227 188 280 234 339 281 403 334

Section modulus Un')

167 109 288 189 457 300 683

Design bearing c.�.city (ki�) Strength of Concrete

5000 pal

448

972 638 1333 876 1775 1166 2304 2123

125 104 180 149 245 203 320 265 405 336 500 414 605 502 710 596

6000 pal

175 125 216 178 295 243 385 318 486 402 600

503 727 602 851 716

= octagonal section

Monotube or Union Metal Pile

Western Cased Pile

Thin. fluted, tapered steel casing driven without mandrel

Thin metal casing

Raymond Step-Taper Pile Corrugated thin cylindrical casing Maximum usual length: 30 m ( 1 00 ft)

(a)

Seamless Pile or Annco Pile

Frankl Cased Pedestal Pile

Thin metal casing

Straight steel pile casing

(c) .

"

Western UncaSed Pile wllhout Pedestal

Frankl Uncased Pedestal Pile Maximum usual length: 30 m-40 m ( 1 00 ft-130 ft)

Maximum usual length: 15 m-20 m (50 ft-65 ft)

Maximum usual length: 30 m-40 m ( 1 00 ft-130 ft)

(e)

Maximum usual length: 30 m-40 m (100 ft-130ft)

Maximum usual length: 40 m ( 1 30 ft)

(b)

Maximum usual length: 30 m-40 m (100 ft-130 ft)

(d)

Minimum effective prestreu force (kip)

(I)

Figure 11.4 Cast-in-place concrete piles

(g)

500

Chapter 1 1 Pile Foundations

Disadvantages: Difficult to splice after concreting b. Thin casings may be damaged during driving • Allowable load: (11.2) Q,II = A,I, + A,/, where A, = area of cross section of steel A , = area of cross section of concrete I, = allowable stress of steel I, = allowable stress of concrete Figures 11.4f and l1.4g are two types of uncased pile, one with a pedestal and other without.The uncased piles are made by first driving the casing to the desired the depth and then filling it with fresh concrete. The casing is then gradually withdrawn. Following are some general facts about uncased cast-in-place concrete piles: • Usual length: 5 m to 15 m (15 ft to 50 ft) • Maximum length: 30 m to 40 m (100 ft to 130 ft) • Usual load: 300 kN to 500 kN (67 kip to 115 kip) • Approximate maximum load: 700 kN (160 kip) • Advantages: 3. Initially economical b. Can be finished at any elevation • Disadvantages: 8. Voids may be created if concrete is placed rapidly b. Difficult to splice after concreting c. In soft soils, the sides of the hole may cave in, squeezing the concrete • Allowable load: (11.3) Q,II = A '/e where Ae = area of cross section of concrete Ie = allowable stress of concrete •

3.

Timber Piles

Timber piles are tree trunks that have had their branches and bark carefully trimmed off. The maximum length of most timber piles is 10 to 20 m (30 to 65 ft). To qualify for use as a pile, the timber should be straight, sound, and without any de­ fects. The American Society of Civil Engineers' Manual 01 Practice, No. 17 (1959), divided timber piles into three classes: 1. Class A piles carry heavy loads. The minimum diameter of the butt should be 356 mm (14 in.). 2. Class B piles are used to carry medium loads. The minimum butt diameter should be 305 to 330 mm (12 to 13 in.). 3. Class C piles are used in temporary construction ·work. They can be used per­ manently for structures when the entire pile is below the water table. The mini­ mum butt diameter should be 305 mm (12 in.).

In any case, a pile tip should not have a diameter less than 150 mm (6 in.).

1 1.2 Types of Piles and Their Structural Characteristics

501

Timber piles cannot withstand hard driving stress; therefore, the pile capacity is generally limited. Steel shoes may be used to avoid damage at the pile tip (bottom). The tops of timber piles may also be damaged during the driving operation. The crushing of the wooden fibers caused by the impact of the hammer is referred to as brooming. To avoid damage to the top of the pile, a metal band or a cap may be used. Splicing of timber piles should be avoided, particularly when they are ex­ pected to carry a tensile load or a lateral load. However, if splicing is necessary, it can be done by using pipe sleeves (see Figure 11.5a) or metal straps and bolts (see Figure 11.5b). The length of the sleeve should be at least five times the diameter of the pile. The butting ends should be cut square so that full contact can be main­ tained. The spliced portions should be carefully trimmed so that they fit tightly to the inside of the pipe sleeve. In the case of metal straps and bolts, the butting ends should also be cut square. The sides of the spliced portion should be trimmed plane for putting the straps on. Timber piles can stay undamaged indefinitely if they are surrounded by satu­ rated soiL However, in a marine environment, timber piles are subject to attack by various organisms and can be damaged extensively in a few months. When located above the water table, the piles are subject to attack by insects. The life of the piles may be increased by treating them with preservatives such as creosote. The allowable load-carrying capacity of wooden piles is (11.4) where = average area of cross s�ction of the pile f... = allowatile stress on the timber

Ap

-

I !i IT -

' I I ,l l \

= -

�

( II

Metal sleeve E�ds cut square

-l��WI

(al

II

,t· ! "

�

'=- Ends cut square

I I !. I I 1 /, 1 1 1 \I It (bl

Metal strap

--

Metal strap

Splicing of timber piles: (a) use of pipe sleeves: (b) use of metal straps and bolts Figure 11.5

502

Chapter 1 1 Pile Foundations

The following allowable stresses are for pressure-treated round timber piles made from Pacific Coast Douglas fir and Southern pine used in hydraulic structures (ASCE, 1993): Pacific Coast Douglas Fir • • • •

Compression parallel to grain: 6.04 MN/m' (875 Ib/in.') Bending: 1 l.7 MN/m' (1700 Ib/in.') Horizontal shear: 0.66 MN/m2 (95 Ib/in.') Compression perpendicular to grain: l.31 MN/m' (190 lb/in.')

Southern Pine

Compression parallel to grain: 5.7 MN/m' (825 Ib/in 2) Bending: 1 l.4 MN/m' (1650 Ib/in 2) Horizontal shear: 0.62 MN/m' (90 Ib/in 2 ) Compression perpendicular to grain: l.41 MN/m' (205 Ib/in.'l The usual length of wooden piles is 5 m to 1 5 m (15 ft to 50 ft). The maximum length is about 30 m to 40 m (100 ft to 130 ft). The usual load carried by wooden piles is 300 kN to 500 kN (67 kip to 115 kip). •

•

•

•

Composite Piles

The upper and lower portions of composiTe piles are made of different materials. For example, composite piles may be malk o[ steel anu concrete or timber and concrete. Steel-and-concrete piles consist of a lower portion of steel and an upper portion of cast-in-place concrete. This type of pile is used when the length of the pile required for adequate bearing exceeds the capacity of simple cast-in-place concrete piles. Timber-and-concrete piles usually consi�t of a lower portion of timber pile below the permanent water table and an upper portion of concrete. In any case, forming proper joints between two dissimilar materials is difficult. and for that reason, com­ posite piles are not widely used.

11.3

Estimating Pile Length Selecting the type of pile to be used and estimating its necessary length are fairly difficult tasks that require good judgment. In addition to being broken down into the classification given in Section 11.2, piles can be divided into three major cate­ gories, depending on their lengths and the mechanisms of load transfer to the soil: (a) point bearing piles, (b) friction piles, and (c) compaction piles. Point Bearing Piles

If soil-boring records establish the presence of bedrock or rocklike material at a sile within a reasonable depth, piles can be extended to the rock surface. (See Figure 11.6a.) In this case, the ultimate capacity of the piles depends entirely on the load-bearing ca­ pacity of the underlying material; thus, the piles are called point bearing piles. In most of these cases, the necessary length of the pile can be fairly well established.

II .

1 1.3 Estimating Pile Length Q"

Q"

Q"

�

�

.! Weak soil

L

q

503

i

Q,

Weak soil

:

i

Q,

Weak soil

' Strong

" Lb . soil " 1. ' layer Qp Q,, � Qp

Lb = depth of penetration into bearing stratum (b)

Figure 11.6

(a) and (b) Point bearing piles; (c) friction piles

If, instead of bedrock, a fairly compact and hard stratum of soil is encountered at a reasonable depth, piles can be extended a few meters into the hard stratum. (See Figure 1 1 .6b.) Piles with pedestals can be constructed on the bed of the hard stratum. and the ultimate pile load may be expressed as Q" = Qp Q, (11.5) where

+

Qp = load carried at the pile point Q, = load carried by skin friction developed at the side of the pile (caused by shear. ing resistance between the soil and the pile) If Q, is very small. (11.6) In this case, the required pile length may be estimated accurately if proper subsoil exploration records are available. Friction Piles

When no layer of rock or rocklike material is present at a reasonable depth at a site, point bearing piles become very long and uneconomical. In this type of subsoil. piles are driven through the softer material to Specified depths. (See Figure 11.6c.) The ul­ timate load of the piles may be expressed by Eq. (11.5). However. if the value of Qp is relatively small. then Q" = Q, (11.7) These piles are called friction piles. because most of their resistance is derived from skin friction. However. the term friction pile. although used often in the

504

Chapter 1 1 Pile Foundations

literature, is a misnomer: In clayey soils, the resistance to applied load is also caused by adhesion. The lengths of friction piles depend on the shear strength of the soil, the ap­ plied load, and the pile size. To determine the necessary lengths of these piles, an en­ gineer needs a good understanding of soil-pile interaction, good judgment, and experience. Theoretical procedures for calculating the load-bearing capacity of piles are presented later in the chapter.

, I

I

Compaction Piles

Under certain circumstances. piles are driven in granular soils to achieve proper compaction of soil close to the ground surface. These piles are called compaction piles. The lengths of compaction piles depend on factors such as (a) the relative density of the soil before compaction, (b) the desired relative density of the soil after compaction, and (c) the required depth of compaction. These piles are gen­ erally short: however, some field tests are necessary to determine a reasonable length.

1 1.4

Installation of Piles Most piles are driven into the ground by means of hammers or \'ibratory drivers. In special circumstances. piles can also be inserted by jetting or partial al/gering. The types of hammer used for pile driving include (a) the drop hammer, (b) the single­ acting air or steam hammer, (c) the double-acting and differential air or steam ham­ mer, and (d) the diesel hammer. In the driving operation, a cap is attached to the top of the pile. A cushion may be used between the pile and the cap. The cushion has the effect of reducing the impact force and spreading it over a longer time; however, the use of the cushion is optional. A hammer cushion is placed on the pile cap. The ham­ mer drops on the cushion. Figure 1 1.7 illustrates various hammers. A drop hammer (see Figure 11.7a) is raised by a winch and allowed to drop from a certain height H. It is the oldest type of hammer used for pile driving. The main disadvantage of the drop hammer is its slow rate of blows. The principle of the single-acting air or steam hammer is shown in Figure 1 1 . 7b. The striking part, or ram, is raised by air or steam pressure and then drops by gravity. Figure 11.7c shows the operation of the double-acting and differ­ ential air or steam hammer. Air or steam is used both to raise the ram and to push it downward, thereby increasing the impact velocity of the ram. The diesel hammer (see Figure 11.7d) consists essentially of a ram, an anvil block, and a fuel-injection system. First the ram is raised and fuel is injected near the anvil. Then the ram is re­ leased. When the ram drops, it compresses the air-fuel mixture, which ignites. This action, in effect, pushes the pile downward and raises the ram. Diesel hammers work well under hard driving conditions. In soft soils, the downward movement of the pile is rather large, and the upward movement of the ram is small. This differ­ ential may not be sufficient to ignite the air-fuel system, so the ram may have to be lifted manually.

I

,

I .

1 1.4 Installation of Piles

505

+- Exhaust

t-�,�. -+-- Intake

t

Ram

Hammer cushion

Ram

Hammer cushion

Pile cap Pile cushion

Pile cap Pile cushion

Pile

Pile

(aJ Figure 1 1.7

hammer.

(bJ

Pile-driving equipment: (aJ drop hammer; (b) single-acting air or steam

The principles of operation of a vibratory pile driver are shown in Figure 11.7e. This driver consists essentially of two counterrotating weights. The horizontal com­ ponents of the centrifugal force generated as a result of rotating masses cancel each other. As a result, a sinusoidal dynamic vertical force is produced on the pile and helps drive the pile downward. Figure 11.7f is a photograph of a vibratory pile driver. Figure 1 1 .8 shows a pile­ driving operation in the field. Jetting is a technique that is sometimes used in pile driving when the pile needs to penetrate a thin layer of hard soil (such as sand and gravel) overlying a layer of softer soil. In this technique, water is discharged at the pile point by means of a pipe 50 to 75 mm (2 to 3 in.) in diameter to wash and loosen the sand and gravel. Piles driven at an angle to the vertical, typically 14 to 20', are referred to as batter piles. Batter piles are used in group piles when higher lateral load-bearing ca­ pacity is required. Piles also may be advanced by partial augering, with power augers (see Chapter 2) used to predrill holes part of the way. The piles can then be inserted into the holes and driven to the desired depth. Piles may be divided into two categories based on the nature of their place­ ment: displacement piles and nondisplacement piles. Driven piles are displace­ ment piles, because they move some soil laterally; hence, there is a tendency for

506

Chapter 1 1 Pile Foundations

t+--

+--

Exhaust and Intake Cylinder Exhaust and Intake Ram Ram Anvil

Hammer cushion

Pile cushion

Pile cap Pile cushion

Hammer cushion

Pile

Pile

Pile cap

Id)

(e)

Static weight

1+--- Pile

(e)

(f)

Figure 11.7 (continued) Pile·driving equipment: (c) double-acting and differential air or

steam hammer; (d) diesel hammer; (e) vibratory pile driver; (f) photograph of a vibratory pile driver (Courtesy of Michael W. O'Neill, University of Houston)

l

1 1.4 Installation of Piles

507

A pile-driving operation in the field (Courtesy of E. C. Shin, University of Incheon. Korea) Flgu,e 11.B

densification of soil surrounding them. Concrete piles and closed-ended pipe piles are high-displacement piles. However, steel H-piles displace less soil later­ ally during driving, so they are low-displacement piles. In contrast, bored piles are nondisplacement piles because their placement causes very little change in the state of stress in the soil.

508

Chapter 1 1 Pile Foundations

11.5

Load Transfer Mechanism The load transfer mechanism from a pile to the soil is complicated. To understand it, consider a pile of length as shown in Figure l1.9a. The load on the pile is gradually increased from zero to Q(, = 0) at the ground surface. Part of this load will be resisted by the side friction developed along the shaft, QI' and part by the soil below the tip of

L,

�I·----- Q" ------�·I

r------r---r--� Q ..

I.

.

L-,,.--J - - - - - - - - - - - - - - - - -

Q,

t

1+ Q,-+j • -- Q, --+I I+--- Qp ---+1 ·1....

(a)

(b)

Unit frictional resistance

z = O ..-----�

Zone II

Pile tip Zone I

, = L '--./

( c) Figure 11.9

Qp (d)

Load Iransfer mechanism for piles

(e)

...-

�one

1 1. 6

Equations for Estimating Pile Capacity

509

the pile, Q,. Now, how are QI and Q, related to the total load? If measurements are made to obtain the load carried by the pile shaft, Q(,), at any depth z, the nature of the variation found will be like that shown in curve 1 of Figure 11.9b. The frictional resistance per unit area at any depth z may be determined as t. Q(,) f(,) = (11.8) (p) (t.z)

where p = perimeter of the cross section of the pile. Figure 11.9c shows the varia­ tion of f(,) with depth. If the load Q at the ground surface is gradually increased, maximum frictional resistance along the pile shaft will be fully mobilized when the relative displacement between the soil and the pile is about 5 to 10 mm (0.2 to 0.3 in.), irrespective of the pile size and length L. However, the maximum point resistance Q, = Qp will not be mobilized until the tip of the pile has moved about 10 to 25% of the pile width (or di­ ameter). (The lower limit applies to driven piles and the upper limit to bored piles). At ultimate load (Figure 11.9d and curve 2 in Figure 11.9b), Q(,-O) = Qu' Thus, QI

=

Q,

and Q, = Qp The preceding explanation indicates that Q, (or the unit skin friction,.t along the pile shaft) is developed at a much smaller pile displacement compared with the point resistance, Qp. At ultimate load, the failure surface in the soil at the pile tip (a bearing capacity failure caused by Qp) is like that shown in Figure 11.ge. Note that pile foundations are deep foundations and that the soil fails mostly in a punching mode, as illustrated pre­ viously in Figures 3.1c and 3.3. That is, a triangular zone, I, is developed at the pile tip, which is pushed downward without producing any other visible slip surface. In dense sands and stiff clayey soils, a radial shear zone, II, may partially develop. Hence. the load displacement curves of piles will resemble those shown in Figure 3.1c.

Equations for Estimating Pile Capacity The ultimate load-carrying capacity Qu of a pile is given by the equation where

Qu = Qp + Q,

(11.9)

Qp = load-carrying capacity of the pile point Q, = frictional resistance (skin friction) derived from the soil-pile interface (see Figure 11.10) Numerous published studies cover the determination of the values of Qp and Q,. Excellent reviews of many of these investigations have been provided by Vesic (1977), Meyerhof (1976), and Coyle and Castello (1981). These studies afford an insight into the problem of determining the ultimate pile capacity.

510

Chapter

11

Pile Foundations Q"

....--Steel ,--- Soil plug

i

I-- D ->I (b) Open-Ended Pipe Pile Section

Tl 1

••t--- Steel , ,

, . _ Soil plug ,

d,

Qp L = length of embedment L" = length of embedment in bearing stratum

I'

( 11.22)

N� , N� = bearing capacity factors

Note that Eq. (11.20) is a modification of Eq. (11.13) with N* = --..:'...-u (1 + 2K.)

( I U3)

Note also that N� in Eq. (11.20) may be expressed as According to Vesic's theory.

N! = (N; - I )cot cf>'

(11.24)

N� = f(1,,)

(11.25)

where I" = reduced rigidity index for the soil. However, I, where I, = rigidity index =

E, = /L, = G, = A. =

2(1

+

q

E, + , tan cf>' )

/L, ) ( C'

G, C, + q, tan cf> ,

(11.26)

(11.27)

modulus of elasticity of soil Poisson's ratio of soil shear modulus of soil average volumatic strain in the plastic zone below the pile point

When the volume does not change (e.g., for dense sand or saturated clay), A. = 0, sO

(11.28)

516

Chapter 1 7 Pile Foundations

Table 11.5 gives the values of N� and N� for various values of the soil friction angle ¢' and Iw For ¢ = 0 (an undrained condition), 7T

4 3

N*c = -(In Irr + 1 ) + - + 1

(11.29)

2

The values of I, can be estimated from laboratory consolidation and triaxial tests corresponding to the proper stress levels. However, for preliminary use, the fol­ lowing values are recommended: Type of soil

Sand Silts and clays (drained condition) Clays (undrained condition)

/,

70-150 50-100 100-200

On the basis of cone penetration tests in the field, Baldi et a!. (1981) gave the following correlations for I,: I ,. and

=

I

,

300 F,( % )

=

(for mechanical cone penetration)

(11.30a)

. · cone penetratlOn (for electnc )

(11.30b)

170 F,( % )

For the definition of F" see Eq. (2.41).

Janbu's Method for Estimating Qp lanbu (1976) proposed calculating Qp as follows: (11.31) Note that Eq. (11.31) has the same form as Eq. (11.13). The bearing capacity factors Nt and NS are calculated by assuming a failure surface in soil at the pile tip similar to that shown in Figure 11.14. The bearing capacity relationships are then N� = (tan ¢' + VI + tan' ¢')' ( e'·· ..n .' )

(11.32a)

(the angle rt' is defined in the figure) and W;

=

(N; t

-

.,

1 ) cot ¢'

from Eq. (11.32a)

i

. _,

(11.32b)

•

'-

Table 1 1.5 Bearing Capacity Factors N: and N� Based on the Theory of Expansion of Cavities

.p'

10

0

6.97 1.00 7.34 1.13 7.72 1.27 8.12 1.43 8.54 1 .60 8.99 1.79 9.45 1.99 9.94 2.22 10.45 2.47 10.99 2.74 . 11.55 3.04 12.14 3.36 12.76 3.71 13.41 4.09 14.08 4.51 14.79 4.96 15.53 5.45

2 3 4 5 6 7 8 9 10 II

12 13 14 15 16 en

-.I �

20

7.90 1.00 8.37 1.15 8.87 1.31 9.40 1.49 9.96 1 .70 10.56 1.92 11.19 2.18 11.85 2.46 12.55 2.76 13.29 3.11 14.08 3.48 14.90 3.90 15.77 4.35 16.69 4.85 17.65 5.40 18.66 6.00 19.73 6.66

40

8.82 1.00 9.42 1.16 j().06 1.35 10.74 1.56 1 1.47 1.80 12.25 2.07 13.08 2.37 13.96 2.71 14.90 3.09 15.91 3.52 16.97 3.99 18.10 4.52 19.30 5.10 20.57 5.75 21.92 6.47 23.35 7.26 24.86 8.13

/"

60

9.36 1.00 10.04 1.18 10.77 1.38 1 1.55 1 .61 12.40 1.87 13.30 2.16 14.26 2.50 15.30 2.88 16.41 3.31 17.59 3.79 18.86 4.32 20.20 4.93 21.64 5.60 23.17 6.35 24.80 7.18 26.53 8.11 28.37 9.14

80

100

200

300

�.75 1.00 I tJ.49 1.18 1 1 .28 1.39 12.14 1 .64

10.04 1.00 10.83 1.19 11.69 1.41 12.61 1.66 13.61 1.95 14.69 2.28 15.85 2.67 17.10 3.10 18.45 3.59 19.90 4.15 21.46 4.78 23.13 5.50 24.92 6.30 26.84 7.20 28.89 8.20 31.08 9.33 33.43 10.58

10.97 1.00 11.92 1.21 12.96 1.45 14.10 1.74 15.34 2.07 16.69 2.46 18.17 2.91 1 9.77 3.43 21.51 4.02 23.39 4.70 25.43 5.48 27.64 6.37 30.03 7.38 32.60 8.53 35.38 9.82 38.37 11.28 41.58 12.92

11.51 1.00 12.57 1.22 13.73 1.48 15.00 1 .79 16.40 2.15 17.94 2.57 19.62 3.06 12.46 3.63 23.46 4.30 25.64 5.06 28.02 5.94 30.61 6.95 33.41 8.10 36.46 9.42 39.75 10.91 43.32 12.61 47.17 14.53

I .l07

1 .91 1 4.07 2.23 15.14 2.59 16.30 3.00 17.54 3.46 1 8.87 3.99 21l.29 4.58 21.81 5.24

23.44

5.98 25.18 6.81 27., =Ko = 1 - sin ,/>' to 1.4Ko = 1.4(1 - sin cj>') = Ko = 1 - sin cj>' to 1.8Ko = 1.8 (1 - sin cj>')

The values of 8' from various investigations appear to be in the range from 0.5' to 0.8c/>'. Coyle and Castello (1981), in conjunction with the material presented in Sec­ tion 11.10, proposed that Q, = f"pL

=

(KiT� tan 8')pL

(11.40)

where •

... .

iT� = average effective overburden pressure 8' = soil-pile friction angle = 0.8' The lateral earth pressure coefficient K, which was determined from field observa­ tions, is shown in Figure 11.19. Thus, if that figure is used, Q, = KiT�tan(0.8')pL

(11.41)

526

Chapter 1 1 Pile Foundations Earth pressure coettlclenl. K

5 1 .0 2 O +-�--�-L-L-L���rr�r-TL��

0. 15 0.2

5 10 '5 1 5 .S e

25 30

35 36 �----����--

Figure 1 1. 19 Variation of K with LjD (Redrawn after Coyle and Castello, 1981)

Correlation with Standard Penetration Test Results

Meyerhof (1976) indicated that the average unit frictional resistance, /"., for high­ displacement driven piles may be obtained from average corrected standard pene­ tration resistance values as (11.42) where (N60) = average value of standard penetration resistance Po = atmospheric pressure (=100 kN/m2 or 2000 Ib/ft2 )

For low-displacement driven piles i" = O.OlPo (N60)

(11.43)

Q., = pL/".

(11.44)

Thus,

Correlation with Cone Penetration Test Results

In Section 1 1.11, the Dutch method for calculating pile tip capacity Qp using cone penetration test results was described. In conjunction with using that method, Not­ tingham and Schmertmann (1975) and Schmertmann (1978) provided correlations

1 1 . 12

Frictional Resistance (a,) in Sand

527

3.0

r '--- ..

Schmertmann (1978); Nottingham and Schmertmann ( 1 975) Steel \ pile "

2.0 -.

, , , , \ '\

'.

Timber pile

\,

..... . " '... , Concrete pile, .... ... .

1 .0

, ... ...

.,;=:.:-.... .

'-::.::.7' -- ---------- - - - - - - - - ­ -._.- . _ . _

.

_._. _ . _._.

O +------.--r---�--_. 20 40 JO 30 o zlD

Figure 11.20

etrometer

Variation of a' with embedment ratio for pile in sand: electric cone pen­ Schmertmann ( 1978): Nottingham and SchmeI1mann ( (975)

2.0 1.5 Steel �1:1

Timber

1.0 0.5 0 +------.--,---.--. 40 20 30 10 o

z/D

Figure 11.21

penetrometer

Variation of a' with embedment ratio for piles in sand: mechanical cone

for estimating Q, using the frictional resistance (f,) obtained during cone penetra­ tion tests. According to this method (11.45) The variations of a ' with z/D for electric cone and mechanical cone penetrometers are shown in Figures 11.20 and 11.21, respectively. We have (11.46) Q, = 'i,p( llL)f = 'i,p(llL)a 'f,

528

Chapter 1 1 Pile Foundations

1 1. 13

Frictional (Skin) Resistance in Clay Estimating the frictional (or skin) resistance of piles in clay is almost as difficult a task as estimating that in sand (see Section 11.12), due to the presence of several variables that cannot easily be qUflntified. Several methods for obtaining the unit frictional re­ sistance of piles are described in the literature. We examine some of them next. A Method

This method, proposed by Vijayvergiya and Focht (1972), is based on the assump­ tion that the displacement of soil caused by pile driving results in a passive lateral pressure at any depth and that the average unit skin resistance is fa. = A(uk +-, �

where

.

'

:22u)'

(11.47)

o:� = mean effective vertical stress for the entire embedment length c" = mean undrained shear strength ('" = 0)

The value of A changes with the depth of penetration of the pile. (See Table 11.7.) Thus, the total frictional resistance may be calculated as Q, = pLJ" Care should be taken in obtaining the values of o:� and c" in layered soil. Figure 11.22 helps explain the reason. Figure 11.22a shows a pile penetrating three layers of clay. According to Figure 11.22b, the mean value of Cu is (c,, ( I )L I + c,,(2)L2 + . . . )/L. Table 11.7

Variation of A with pile embedment length, L

Embedment Length. L (m)

0 5 10 15 20 25 30 35 40 50 60 70 80 90

A

0.5 0.336 0.245 0.200 0.173 0.150 0.136 0.132 0.127 0.1 18 0.113 0.110 0.110 0.110

1 1. 13 Frictional (Skin) Resistance in Clay

I

.

'-

Vertical

Undrained cohesion, ell

.

= A]

1.,

L

1-

!

L,

(a)

_L

Figure 11.22

stress. u�

= Az

Cu(3)

L,

-

529

___

= AJ

Depth (c)

Depth (b)

Application of A method in layered soil

Similarly, Figure 1 1.22c shows the plot of the variation of effective stress with depth. The mean effective stress is u'

_

o

where A ,. A,. A3• a

•

.

•

=

A _ ",,_+ -'CA.:c2,+--,A - = 3 --,+"'

-

L

(11.48)

= areas of the vertical effective stress diagrams.

Method

According to the Cl method, the unit skin resistance in clayey soils can be repre­ sented by the equation (11.49) where a empirical adhesion factor. The approximate variation of the value of a is shown in Figure 11.23, where O'� is the vertical effective stress. 1'hjs variation of a with cu!O'� was obtained by Randolph and Murphy (1985). With it, we have =

(11.50)

fJ Method When piles are driven into saturated clays, the pore water pressure in the soil around the piles increases. The excess pore water pressure in normally consolidated clays may be four to six times CU' However, within a month or so, this pressure gradually dissipates.

530

Chapter 1 1 Pile Foundations

1.2 1 .0 0.8 t!

Randolph and Murphy (1985)

0.6 0.4 0.2 o

0.5

1.0

1.5

2.0 2.5

3.0 3.5 4.0

Figure 1 1.23 Variation of a with cju�

Hence, the unit frictional resistance for the pile can be determined on the basis of the effective stress parameters of the clay in a remolded state (c' '" 0 ) . Thus. at any depth.

(11.51) where a�

=

vertical effective stress

f3 = K tan Q>/, Q>R

K

=

=

drained friction angle of remolded clay earth pressure coefficient

(11.52)

Conservatively, the magnitude of K is the earth pressure coefficient at rest, or K = 1 - sin Q>R (for normally consolidated clays) (1153) , and K = (1 - sin Q>R) YOCR (for overconsolidated clays) (11.54) where OCR = overconsolidation ratio. Combining Eqs. (11.51), (11.52), (11.53), and (11.54), for normally consoli­ dated clays yields I

=

(1

-

sin Q>j,) tan Q>Ra�

(11.55)

and for overconsolidated clays,

(11.56) I = (1 - sin Q>R)tan Q>/,YOCR a� With the value ofI determined, the total frictional resistance may be evaluated as Q,

=

lip ilL

j

,

.

1 1. 14

Point Bearing Capacity of Piles Resting on Rock

531

1.5

�

Nottingham and Schmertmann (1975); Schmertmann (1978) 1 .25

l>--

1.0

-�

, , , , ,

0.75

, ,

,

,

0.5

,

'

........

....

... , ... ....

'-­

0.25

-- -

- --

-- -

- ... -

0 0

�

Concrete and � t im �be �r � pi les �

�______ _

0.5

1.5

;�cl�ii�"' ''' ''' ·

2.0

Figure 11.24 Variation of a' with Mp, for piles in clay (p, = atmosphic pressure =100 kN/m' or 2000 Ib/ft')

Correlation with Cone Penetration Test Results

Nottingham and Schmertmann (1975) and Schmertmann (1978) found the correla­ tion for unit skin friction in clay (with = 0) to be I = a'l, (11.57) The variation of a' with the frictional resistance I, is shown in Figure 11.24. Thus, Q, = IIp( 6. L) = Ia'l,p( 6.L ) (11.58)

Point Bearing Capacity of Piles Resting on Rock Sometimes piles are driven to an underlying layer of rock. In such cases, the engi­ neer must evaluate the bearing capacity of the rock. The ultimate unit point resis­ tance in rock (Goodman, 1980) is approximately (11.59) where N� = tan2 ( 45 + '/2) qu = unconfined compression strength of rock ,-)-,

-

p�is 5Q ft, (L) IOIlg and 16 in. x 16 in. cross section. = 110 Ib/ft3 and ,p' = 30°. Calculate ' point load, Q;. hy using ,,

A

B.

"--

.

:

"

,'

method (Section 1 1.7) method (Section 11.8). Use = = 50 ' Co Janbu's method (Section 11.9). Use 'I) = 90°

b.

thelil

t!i@i\

Point Bearing Capacity of Piles Resting on Rock

1 1. 14

533

Solution Part a

From Eq. (11.15), Qp = Apq'N: = ApyLN.,*

=

For '!>' =

300, N.,* 55 (Figure 11.12), so 16 x 16 1101000 x 50 . . klp/ft (55) - 537.8 kip Qp 16 x 12 ft Again, from Eq. (11.17) qp = (0.5pa N;, tan q,' ) Ap = (0.4) (2000) (55) (tan 30) � : = 56,452 1b = 56.45 kip

(

_

-

2

)(

2

)

.

_

G ��)

Part b From Eqs. (11.20), (11.21), and (11.22), with c' =

0, 1 + 2(I - sin q,') q '" Qp = Apu'oNo = Ap 'N 3 For q,' = 30° and /" = 50. the value of N:, is about 36 (Table 11.5), so

[

•

Qp = Part c

( 12

)[

1 6 X I6 , · ft x

12

I

+ 2( I - Sin 3U ) 3

From Eq. (11.31) with c' = O. For q,' =

300 and Qp =

1)'

](

]

)

· 1l0 X 50 · /ft' ( 36) = 234 7 k'Ip kIp

1000

•

Qp = Apq'N:;

= 900, the value of N;

C� : �� )( 11�0�050 ft'

=

18.4 (Table 11.6)

kiP/ft'

) (18.4)

= 179.9 kip

•

Example 1 1 .2

For the pile described in Example 1 1.1 a.

Determine the frictional resistance. Q,. Use Eqs. (11.14), (11.38), and (11.39). Given: K = 1.3 and Ii' = O.Sq,'. b. Using the results of Example 1 1.1 and Part a of this problem, estimate tIie al­ lowable load-carrying capacity of the pile. Given: FS = 4.

534

Chapter 1 1 Pile Foundations Solution Part a

From Eg. (11.37), L

•

=

15D = 15

( )

16 ft = 20 ft 12

From Eg. (11.38), at Z = 0, ' = 30° and LjD = 37.5, N; Thu� Q"

=

(

1 10 x 50 1000

+ (0.25)

(

G�)

=

:

.'

'

25 (Figure 11 .15) and K = 0.25 (Figure li,j9),.: .

) ( )

k'tpjft2 (25)

110 x 50 1000 x 2

2) ( )

1 6 x 16 ft 12 x 12

tan(O.B X 30)

= 244.4 + B 1.62 = 326.02 kip Q,1l =

:

.

4 x 16 (50) 12

Q" 326.02 = -- = 81.5 kIp FS 4 •

.

..

Example 1 1 .4 A driven-pipe pile in clay is shown in Figure 1 1.25a. The pipe has an outside di­ ameter of 406 mm and a wall thickness of 6.35 mm. a . Calculate the net point bearing capacity. Use Eq. (11.19). b. Calculate the skin resistance (1) by using Eqs. (11.49) and (1 1.50) ( a method). (2) by using Eq. (11.47) ( A method). and (3) by using Eq. (11.51) (/3 method). For all clay layers. c/>k = 30°. The top 10 m of clay is normally consolidated. The bottom clay layer has an OCR of 2. c. Estimate the net allowable pile capacity. Use FS = 4.

Solution

The area of cross section of the pile. including the soil inside the pile. is Ap =

:D' = �( �.406)' = 0.1295 m'

Part a: Calculation of Net Point Bearing Capacity

From Eq. (11.19).

Qp = Apqp = ApN �c"(2) = ( 0.1 295)(9)(100) = 116.55 kN

536

Chapter 1 1 Pi/e Foundations

U;(kN/m2) Saturated clay

2 CUI J ' = 30 kN/m : .l = 1 8 kN/m y

j 326.75

30 m

(b)

(a)

Estimation of the load bearing capacity of a driven-pipe pile

Figure 1 1.25

Part b: Calculation of Skin Resistance

(1) Using Eqs. (1 1.49) and ( 1 1 .50), we have Q, = 'i.ac"pIiL The variation of vertical effective stress with depth is shown in Figure fl.�b. " . : -�: Now the following table can be prepared: '

Depth (m)

Average depth (m)

0-5

2.5

5-10

7.5

10-30

20

AveragD vertical effective stress. u� (kN/mZ)

0 + 90 = 45 2 90 + 130.95 = 1 10.5 2 130.95 + 326.75 = 228.85 2

rr. c.

c. (kN/mZ)

(kN/m')

30

0.67

30

0.27

.. ().?

100

0.44

0.725

,

{Figute , 1�' ,

a'

. OJ; :

Thus, = [(0.6) (30) (5) + (0.9) (30) (5)

+ (0.725 ) (100) (20)](1T X 0.406) = 2136 kN

,'_.

1 1. 14

c.([) ( 10 ) + c"(2)(20) 30

Point Bearing Capacity of Piles Resting on Rock

=

(30) (10) + (100) (20) 30

=

537

76 .

TQ obtain the average value of iT�, the diagram for vertical effe'Cti1�e tion with depth is plotted in Figure 1 l.25b. From Eq. (11.48), u�

=

A} + A2 + A3 L

=

225 + 552.38 + 4577 . =1 30

From Table 1 1 .7, the magnitude of ,\ is 0.136. So fav

=

0.136[178.48 + (2) (76.7)] = 45.14 lCN/m?!'"

aenee,

(3) The top

:z:

Q, = pLfav = 71" (0.406) (30) (45.14)

=

1727 � . j;1�!�)}:

layer of clay (10 m) is normally consolidated, om Eq. (11.55), we have

= 0-5 iii, fr

f.v(}) = (1 - sin cPR) tan cPR iT� =

(1 - sin 300)(tan 300)

Similarly, for z = 5-10 m. .

f"(2) = (1 - sm 3OO) (tan 30°)

For z = 10-30 m from Eq. (11.56),

(

(

0

� 90 ) = 13.0

90 + 130.95 2

)

=

I.M'.;....

31.9 kNjm2

f.v = (1 - sin cP,,)tan cPRVOCR iT�

For OCR = 2, .

(

'- 130.95 + 326.75 2

f.v(3) = (1 - s m 300) (tan 300 ) v 2

So, Q,

= =

•

)

. . .. . ' . = 93.43 kNjD.1� ·

p[f.v( I) (5) + f,,(,, ( 5) + f.v(3) (20 )] (71" ) (0.406)[(13) (5) + (31 :9) (5) + (93.43) (20)]

,�

=

Part c: Calculation of Net Ultimate Capacity, Qu

We have

Q,(.vomgo) =

2136 + 1727 + 2670 3

=

2178 kN

2670 kN

' . :'

538

Chapter 1 1 Pile Foundations

Thus, Q. = Qp + Q, = 1 16.55 + 2178 = 2294.55

and

Q.

QaU =

FS

=

2294.55 4

kN

= 574 kN .

Example 1 1 .5

.

',' : ,,

- "'

-: _-_: "> ':,'--::-:' ·:+'�-· s

Acollcrete pile 305 mm x 305 mm in cross section is driyeitto a dep�({f� !D belo�the ground surface in a saturated clay sOil,A s.Jl!p!l1iifY of �¢ v�ij�.9·9f , , as follilwS: ' frictional resistance Ie obtained from a cone penetratiiiiitestis ,,

.

.

.•

"

Depth (m)

Friction resistance, f" (kg/o...')

0-6 6-12 12-20

0.35 0.56 0.72

"

-


5, so the pile is a long one. Because Eq. (11.75) takes the form T=

E�p =

_

x,(z) = A x

and it follows that

At Z = 0, x, = 8 mm Qg =

=

QgT3

M. ",,; g;' :. :

' ..

I Ep p

0.008 m and Ax = 2.435 (see Table 11.13), so

(0.008) (207 X 106) ( 123 (2.435 ) ( 1.163)

X

10-6)

= 53.59 kN

1

,

1 1 . 18

561

Laterally Loaded Piles

This magnitude of Qg is based on the limiting displacement conditiQri Qilly. However, the magnitude of Qg based on the moment capacity of the pile " alsO needs to be determined. For Mg = 0, Eq. (11.77) becomes '. '

M,(z) = AmQgT

,

According to Table 11.13, the maximum value of Am at any 4eplll;is,().1ti.:: " ;. ; " The maximum allowable moment that the pile can carry is M,,-. � �

�

, ",

"' ,

::'

" , . ,' . '

JIJ:1� .'

•. . .. . ..

;;

,'

LetFy = 248,000 kN/m2 . From Table l1.1a,lp = 123 X 10-6m4lUiadl i;i,'��i�1; ': :;:c;r;l!;�'r;:�,:��{ so 6 Ip 123 X 10= 968.5 X 10-6m3 = 0. 54 .

(i) ( � )

Now, Qg =

M,(mul (968.5 X 10-6) (248.000) , " = 268.2.kN = AmT (0.772) (1.16)

Because Qg = 268.2 kN > 53.59 kN, the deflection criteria applf H�ri�'" •

Q, = 53.59 kN.

Example 1 1 .8

Solve Example 11.7 by Broms's method. Assume that the pile is flexible and is free headed. Let the yield stress of the pile material, Fy = 248 MN/m2 ; t)te unit weight of soil, l' = 18 kN/m3; and the soil friction angle cP' = 35'. Solution

We check for bending failure. From Eq. (11.90), M, = SF"

From Table 11.1 a,

Ip d, 2

S=-=

Also,

,. M. =

[

]

123 X 10-6 0.254 2

123 X 1O-6 (248 _ 0.2)4 2

x

103) = 240.2 kN-m

562

Chapter 1 1

Pile Foundations

and

From Figure l1.34a, for My/D4yKp = 868.8, the magnitude of Qu,.,!KpD3y (for a . free-headed pile with ejD = 0) is about 140, so

(

Qu(,) = 140KpD3y = 140 tan2 45 +

�}0.254)\18) = 152.4

3

kN

Next, we check for pile head deflection. From Eq. (11.91), 7J

so

- 4 - \J,I nh Eplp

12,000 - 0.86 m - 1 (207 X 106) (123 X 10-6) _

7JL = (0.86) (25) = 21.5

From Figure l1.35a, for TJL = 21.5, ejL = 0 (free-headed pile): thus, .

x

;,L

2 5 o (EpI )3i (nh ) /5

and

.

=

0.15

(by interpolation)

Xo(E/p)3/5(nh )2/5 Q = . g ' . . O.15L .. (0.008)[(207 X 106) (123 X W- 6)] 3/5 (12,0Q0) 2/5 = = 40.2 kN (0.15)(25)

.

Hence,Q, = 4O.2kN « 1S2.4 kN).

1 1. 19

'

.i

"// ·

··

Pile-Driving Formulas To develop the desired load-carrying capacity, a point bearing pile must penetrate the dense soil layer sufficiently or have sufficient contact with a layer of rock. This require­ ment cannot always be satisfied by driving a pile to a predetermined depth, because soil profiles vary. For that reason, several equations have been developed to calculate the ultimate capacity of a pile during driving.These dynamic equations are'widely used

1 1. 19 Pile-Driving Formulas

563

in the field to determine whether a pile has reached a satisfactory bearing value at the predetermined depth. One of the earliest such equations-commonly referred to as the Engineering News (EN) Record formula-is derived from the work---energy theory. That is, Energy imparted by the hammer per blow = (pile resistance)(penetration per hammer blow) According to the EN formula, the pile resistance is the ultimate load Q", expressed as (11.107) where WR h S C

= = = =

weight of the ram height of fall of the ram penetration of pile per hammer blow a constant

The pile penetration, S, is usually based on the average value obtained from the last few driving blows. In the equation's original form, the following values of C were recommended: For drop hammers. C

=

For steam hammers, C

=

{25.4

mm if S and h are in mm 1 in. if S and h are in inches

{

2.54 mm if S and h are in mm

0.1 in. if S and h are in inches

Also, a factor of safety FS = 6 was recommended for estimating the allowable pile capacity. Note that, for single- and double-acting hammers, the term WRh can be re­ placed by EHE, where E is the efficiency of the hammer and HE is the rated energy of the hammer. Thus, (11.108) The EN formula has been revised several times over the years, and other pile-driving formulas also have been suggested. Some of them are tabulated in Table 11.15.

564

Chapter 7 7 Pile Foundations Table 11. 15 Pile-Driving Formulas

Formula

Name

Modified EN formula

2 EWR" WR + n Wp

Q = ,, S + C

where

WR + Wp E = efficiency of hammer

C = 2.54 mm if the units of S and h are

in mm C = 0.1 in. if the units of S and h are in in. Wp

=

weight of the pile

n = coefficient of restitution between

the ram and the pile cap

Michigan State Highway Commission formula (1965)

Typical values for E Single- and double-acting hammers Diesel hammers Drop hammers

0.7-0.85 0.8-0.9 0.7-0.9

Typical values for n Cast-iron hammer and concrete piles (without cap) Wood cushion on steel piles 'Wooden piles

0.4-0.5 0.3-0.4 0.25-0.3

I .-) ' - EH � F

WR +

Q = " S+C

where

n:!lt"

WR + Wp

(!

HE = manufacturer's maximum rated

hammer energy (Ib-in.)

E = efficiency of hammer

C = 0.1 in. A factor of safety of 6 is recommended.

Danish formula (Olson and Flaate, 1967)

Q = "

--)-;==== EHE

EHEL -2A,Ep

S +

where

Pacific Coast Uniform Building Code formula (International Conference of Building Officials, 1982)

E = efficiency of hammer HE = rated hammer energy Ep = modulud of elasticity of the pile material L = length of the pile Ap = cross-sectional area of the pile

(EHE)

(

WR + tiWp

--

WR + Wp

S+

)

Q"L

AEp

The value of n should be 0.25 for steel piles and 0.1 for all other piles. A factor of safety of 4 is gener­ ally recommended.

1 1. 19 Pile-Driving Formulas Table 11.15

565

(Continued)

Name

Formula

Janbu's formula (Janbu, 1953)

Q" =

EHE

K'S "

( )

K;, = Cd 1 +

where

1+

Cd = 0.75 + 0.14 A

'

=

(

EHEL

A"EpS'

)

�J

(:;')

Gates's formula (Gates, 1957)

Q" = aVEHE(b log S) If Q" is in kips, then S is in in.,a = 27, b = 1, and HE is in kip-ft. If Q" is in kN, then S is in mm, a = 104.5, b = 2.4, and HE is in kN-m. E = 0.75 for drop hammer; E = 0.85 for all other hammers Use a factor of safety of 3.

Navy-McKay formula

Q"

=

(

£HE

S 1 + 0.3

-

:') W

Use a factor of safety of 6.

TI,e maximum stress developed on a pile during the driving operation can be esti­ mated from the pile-driving formulas presented in Table 11.15. To illustrate, we use the modified EN formula: EWRh WR + n'Wp WR + Wp

Q,, = S + C

In this equation, S is the average penetration per hammer blow, which can also be expressed as (11.109)

where S is in inches N = number of hammer blows per inch f penetration Thus, (11.110)

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

566

Chapter 1 1

Pile Foundations

This procedure can be demonstrated with a set of numerical values. Suppose that a prestressed concrete pile ft in length has to be driven by a hammer. The pile sides measure in. From Table for this pile.

10

8011.3b.

Ap =

100 ( 144 ) (80 )(150 8.33 0.67 8.33 0.67 9 19.5 2 0.85 0.35. (11. 1 10) ) 2 12) 0 . 3 5)2(9 ]rS ( 1 - 85.37 . - [(0.85)(19. 1 1 in'

The weight of the pile i5

A pLYe =

If the weight of the cap is

I OD in'

•

ft

Ib;W) =

kip, then

=

+

Wp =

For the hammer. let

kip

kip

kip-ft = HIi = WRh

Rated energy =

Weight of ram = kip

and that II =

Assume that the hammer efficiency is ues into Eq. yields Q"

_

x

IV +

Substituting these val­

+

-

U. I

5

.J. 9

-

N

Now the following table can be prepared: N

0 2 4 6

8

10 12 20

Qu (kip!

Ap (in'!

q.u/Ap (kip/in'!

0 142.3 243.9 320.1 379.4 426.9 465.7 569.1

100 100 100 100 100 100 100 100

0 1.42 2.44 3.20 3.79 4.27 4.66 5.69

-

kIp

0.1

Both the number of hammer blows per inch and the stress can be plotted in a graph, as shown in Figure If such a curve is prepared, the number of blows per inch of pile penetration corresponding to the allowable pile-driving stress can easily be determined. Actual driving stresses in wooden piles are limited to about Similarly, for concrete and steel piles, driving stresses are limited to about and respectively. In most cases, wooden piles are driven with a hammer energy of less than kN-m kip-ft). Driving resistances are limited mostly to to blows per inch of pile penetration. For concrete and steel piles. the usual values of N are to 8 and to respectively.

11.38.

60 (=45 12 14.

0.0.6/;7/.,. 0.85/,., 4 5 6

1 1. 19 Pile-Driving Formulas

567

6 I

L

5 4

C

N

'a.

;g 3 �

!'; cl

2

O +----r-----,--� o 4 20 12 8 16

Number of blows lin. (N)

Figure 1 1.38

blows/in.

Plot of stress versus

EXample 1 1 .9 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 Weight of ram = 7.5 kip Pile length = 80 ft Coefficient of restitution 0.4 Weight of pile cap = 550 Ib Ep = 3 x 11? kip/in' Number of blows for last 1 in. of penetration = 8 Estimate the allowable pile capacity by the

=

a.

.

.

== = 6)

M()dified EN formula (use FS b.Danish formula (use FS 4) Co Gates formula (use FS 3) Solution Part a

•

EWRh WR + n2Wp = Q,, S + C WR + Wp Weight of pile + cap = G� x �� x 80) (150 Ib/tt3) + 550 = 12,550 Ib 12.55 kip =

568

Chapter 1 1 Pile Foundations

Given: WRh = 30 kip-ft. (0.8)(30 x l2 kip-in.) 7.5 + (0.4)'(12.55) . x Q" = = 607 kip !8 + 0.1 7.5 + 12.55 607 Q . Q," = FS" = Ii = 101 kip Part b

Use Ep = 3 x 106 Ib/in'. (0.8)(30 x 12)(80 x 12) 3 106 kip/in' 2 ( 12 x 12)

Q" = ,

Q,II

=

I�OO

) - 0.566 ' _

m.

(0.8)(30 x 12) . = 417 kip I 8 + 0.566 417

4 = 104 kip

Part c

Qu = aYEHE(b log S) 22 = 84 kip Q.II =

�

-

=

27Y(0.8)(30)[1 - log (1) 1 = 252 kip .

•

Pile Capacity For Vibration-Driven Piles The principles of vibratory pile drivers (Figure 1 1.7e) were discussed briefly in Sec­ tion 1 1.4. As mentioned there, the driver essentially consists of two counterrotating weights. The amplitude of the centrifugal driving force generated by a vibratory hammer can be given as (11.111) F, = mew' where m = total eccentric rotating mass e = distance between the center of each rotating mass and the center of rotation w '= operating circular frequency

7 7.20 Pile Capacity For Vibration-Driven Piles

( L

569

Vibratory hammers typically include an isolated, bias weight that can range from 4 to 40 kN. The bias weight is isolated from oscillation by springs, so it acts as a net downward load helping the driving efficiency by increasing the penetration rate of the pile. The use of vibratory pile drivers began in the early 1930s. Installing piles with vibratory drivers produces less noise and damage to the pile, compared with impact driving. However, because of a limited understanding of the relationships between the load, the rate of penetration, and the bearing capacity of piles, this method has not gained popularity in the United States. Vibratory pile drivers are patented. Some examples are the Bodine Resonant Driver (BRD), the Vibro Driver of the McKiernan-Terry Corporation, and theVibro Driver of the L. B. Foster Company. Davisson (1970) provided a relationship for es­ timating the ultimate pile capacity in granular soil: In SI units,

Q,, ( kN) = (vp m/s)

+

(SLm/cycle) (fHz)

(11.112)

In English units, 550(Hp) + 22,OOO (vp ft/s) Qu (lb) = (vpft/s) + (SL ft/cycle) (fHz)

(11.113)

where Hp = horsepower delivered to the pile vp = final rate of pile penetration SL = loss factor f = frequency, in Hz The loss factor SL for various types of granular soils is as follows (Bowles, 1996): Closed-End Pipe Piles • • •

Loose sand: 0.244 x 10-3 m/cycle (0.0008 ft/cycle) Medium dense sand: 0.762 X 10-3 m/cycle (0.0025 ft/cycle) Dense sand: 2.438 X 10-3 m/cycle (0.008 it/cycle)

H-Piles

Loose sand: -0.213 x 10-3 m/cycle( -0.0007 it/cycle) lO-j m/cycle (0.0025 ft/cycle) Dense sand: 2.134 x 10-3 m/cycle (0',007 it/cycle) In 2000,Feng and Deschamps provided the following relationship for the ulti­ mate capacity of vibrodriven piles in granular soil: •

•

•

. Medium dense sand: 0.762 x

(11.114)

•

570

Chapter 1 1

Pile Foundations

Here, Fe = centrifugal force WB = bias weight vp = final rate of pile penetration c = speed of light [�.8 x 1010 mlmin (5.91 x 1010 ft/min) ]

OCR

=

overconsolidation ratio

L£ = embedded length of pile L = pile length

Example 1 1 . 1 0

Consider a 20-m-Iong steel pile driven by a Bodine Resonant Driver (Section � 310 x 125) in a medium dense sand. If Hp = 350 horsepower, vp = 0.0016 mis, and f = 115 Hz, calculate the ultimate pile capacity, Q". Solution

From

Eq.

(ll.112), Q" =

0.746Hp + 98vp vp + Sd

For an HP pile in medium dense sand, SL = 0.762 X 10-3 m/cycle. So (0.746) (350) + (98) (0.0016 ) kN = Q. 0.0016 + (0.762 x 10-3) ( 115) = 2928

.·.'.1....1;21 ".,'

Negative Skin Friction Negative skin friction is a downward drag force exerted on a pile by the soil sur­ rounding it. Such a force can exist under the following conditions, among others: If a fill of clay soil is placed over a granular soil layer into which a pile is dri­ ven, the fill will gradually consolidate. The consolidation process will exert a downward drag force on the pile (see Figure 11.39a) during the period of consolidation. 2. If a fill of granular soil is placed over a layer of soft clay, as shown in Figure l1.39b, it will induce the process of consolidation in the clay layer and thus exert a down­ ward drag on the pile. 3. Lowering of the water table will increase the vertical effective stress on the soil at any depth, which will induce consolidation settlement in clay. If a pile is lo­ cated in the clay layer, it will be subjected to a downward drag force.

I.

In some cases, the downward drag force may be excessive and cause founda­ tion failure. This section outlines two tentative methods for the calculation of nega­ tive skin friction.

I, .

1 1.21 Negative Skin Friction

T

571

T

i

i

. . . Clay H fill

Sand H ' ml

Sand

(a)

Figure 1 1.39

(b)

Negative skin friction

Clay Fill over Granular Soil (Figure 1 1.39a}

Similar to the f3 method presented in Section the negative (downward) skin stress on the pile is fn = K'U� tan 0' where K' = earth pressure coefficient = Ko = 1 - sin cf/ u� = vertical effective stress at any depth z: = y�z = effective unit weight of fill 8' = soil-pile friction angle = O. - . cf/ Hence, the total downward drag force on a pile is

1l.l3,

(11.115)

YJ

5 07

Qn =

fHI pK'Yf t n 8')z pK'yJH} tan 8' o

" a

(

dz =

--' --' -

2

(11.116)

height of the fill. If the fill is above the water table, the effective unit HYJ,r =should be replaced by the moist unit weight.

where weight,

Granular Soil Fill over Clay (Figure 1 1.39b}

In this case, the evidence indicates that the negative skin stress on the pile may exist from z = 0 to z = L" which is referred to as the (See Vesic, pp. The neutral depth may be given as (Bowles,

25-26.)

"

where and y' respectively.

YJ

neutral1982)depth. 2yjH[ (L - HI) [L - Hr YfHr " + , L, - L 2 Y J Y,

=

_

1

_

1977.

(11.117)

effective unit weights of the fill and the underlying clay layer.

572

Chapter 1 1 Pile Foundations

For end-bearing piles, the neutral depth may be assumed to be located at the pile tip (i.e., = - HI)' Once the value of is determined, the downward drag force is obtained in the following manner: The unit negative skin friction at any depth from z = 0 to Z = is tan 0' (11.118) fll =

L, L

L,

L,

K'a�

where

K' = Ko = 1 -' sin ,/>' (}'� = y'rH, + y z 8' = 0.5-0.7' Qn = fPfn dz fpK'(Y(H, + y'z) tan ?' dz LTPK'y' tan 8' = (pK'y'rH, tan 8')L1 + 2 =

(11.119)

If the soil and the fill are above the water table. the effective unit weights should be replaced by moist unit weights. In some cases, the piles can be coated with bitumen in the downdrag zone to avoid this problem. A limited number of case studies of negative skin friction is a\'ailable in the lit­ erature. Bjerrum et a!. (1969) reported monitoring the downdrag force on a test pile at Sorenga in the harbor of Oslo, Norway (noted as pile G in the original paper). The study of Bjerrum et a!. (1969) was also discussed by Wong andTeh (1995) in terms of the pile being driven to bedrock at 40 m. Figure 11.40a shows the soil profile and the pile. Wong and Teh estimated the following quantities: • Fill: Moist unit weight, = 16 kN/m' Saturated unit weight, y"'(f) = 18.5 kN/m' So

y,

y( = 18.5 - 9.81 = 8.69 kN/m3

and

K' L

H,

=

13 m

tan 8' = 0.22 Saturated effective unit weight,y' = 19 - 9.81 = 9.19 kNjm3 • Pile: = 40 m Diameter, D = 500 m Thus, the maximum down drag force on the pile can be estimated from Eq. (11 .119). Since in this case the pile is a point bearing pile. the magnitude of = 27 m, and •

L,

Clay:

L;py' (K' tan 8') . x 2 + (13 - 2)Yij(Lrl + Qn = (p) (K: tan 8')lYr 2

1 1.22 Group Efficiency ( ,

2m.

. .. �

.

/

-

-------

"It =

Axial force in pj1e (kL'l)

16 kN/mJ

Fill

-

---------

Groundwater Fill table 1m 'YSllI {f) = 18.5 kN/m3

1

2000 3000 o 1000 0 +------'---'---' -

10

r-f--:

•

L

�

\

-

-

'\

I � ,-

(aj

- ,

,

..

40

L

� "'

'" \ \ \ \ \ \ \ \ \ \ \ ,

30

Clay

,

, \ •

•

40 m

Pile D = 500 mm

573

(b)

Figure 1 1.40 Negative skin friction on a pile in the harbor of Oslo, Norway

[Based on Bjerrum et al. (1969) and Wong and Teh (1995)J

or

0.5)(0.22)[(16 2) (8.69 11)](27) (27)'( 0.52)(9.19)(0.22) 2348 2500 11.40b).

Q" = (1T X

=

X

+

X

+

1T X

kN

The measured value of the maximum Q" was about is in good agreement with the calculated value.

kN (Figure

which

Group Piles 1 1.22

Group Efficiency In most cases, piles are used in groups, as shown in Figure to transmit the structural load to the soil. A pile cap is constructed over group piles. The cap can be in contact with the ground, as in most cases (see Figure a), or well above the ground, as in the case of offshore platforms (see Figure Determining the load-bearing capacity of group piles is extremely complicated and has not yet been fully resolved. When the piles are placed close to each other, a

11.41, 11.11.41b)41 .

574

Chapter 1 1

Pile Foundations

Section

Pile

cap

�

L

1

Phm

I+.

I

:. I

B,

1

d ---+- d 4 L"

---_

.. _ (a)

(c)

.

- - - - -

.

: ;. .

-

.

- - - - -

L

,T I

:td

�I

�: ld I

-

- - - -

(bl Number of piles in group =

(Note: Lx'? B".J L� = (1/1 1 Jd -. 2( DI2 ) B" = (lJ2 I)d + 2(DI'!.)

1

II J X 1I�

-

';"'*

Figure 1 1.41

Group piles

reasonable assumption is that the stresses transmitted by the piles to the soil will overlap (see Figure l1.41c), reducing the load-bearing capacity of the piles. Ideally, the piles in a group should be spaced so that the load-bearing capacity of the group is not less than the sum of the bearing capacity of the individual piles. In practice, the minimum center­ to-center pile spacing, d, is 2.5D and, in ordinary situations, is actually about 3 to 3.SD.

1 1.22 Group Efficiency

575

The efficiency of the load-bearing capacity of a group pile may be defined as (11.120)

where 1/ = group efficiency Qg(U) = ultimate load-bearing capacity of the group pile Qu = ultimate load-bearing capacity of each pile without the group effect

Many structural engineers use a simplified analysis to obtain the group effi­ ciency forfriction piles, particularly in sand. This type of analysis can be explained with the aid of Figure 11.41a. Depending on their spacing within the group, the piles may act in one of two ways: (1) as a block, with dimensions L g x Bg x L, or (2) as individ­ ual piles. If the piles act as a block, the frictional capacity is ["p,L = Qg(u)' [Note: Pc = perimeter of the cross section of block = 2(n, + n, - 2 ) d + 4D, and [" = average unit frictional resistance.] Similarly; for each pile acting individually, Qu = pL[". (Note: p = perimeter of the cross section of each pile.) Thus, Qg(u) [,,[2 ( n, + n2 - 2)d + 4D]L 1/ = -- = n,n2pL[" � Qu 2(n, + nz - 2)d + 4D =

Hence, _

Qg(u) -

[

J

2(11, + n, - 2)d + 4D � Qu . pll,Il,

(11.121)

(11.122)

From Eq. (11.122), if the center-to-center spacing d is large enough, 1/ > 1. In that case, the piles will behave as individual piles. Thus, in practice, if 1/ < 1, then and if 1/

.,

1, then

Qg(u) = 1/� Qu Qg(u) = � Qu

There are several other equations like Eq. (11.122) for calculating the group efficiency of friction piles. Some of these are given in Table 1 1.16. Figure 11.42 shows the variation of the group efficiency 1/ for a 3 x 3 group pile in sand (Kishida and Meyerhof, 1965). lt can be seen that, for loose and medium sands, the magnitude of the group efficiency can be larger than unity. This is due pri­ marily to the densification of sand surrounding the pile.

576

Chapter 1 1 Pile Foundations Table 1 1. 16

Equations for Group Efficiency of Friction Piles Equation

Name

Converse-Labarre equation

1/ =

Los Angeles Group Action equation

1

1/ =

rrdll\112

11,(11,

+

Seiler-Keeney equation (Seiler and Keeney, 1944)

_D __ [11, (11,

-

{I

-

-

1)

1 ) + V'2(II,

-

1 ) (11,

-

1) ]

[7()1� IJ:: : ::: = �]}

-

+

where d is in ft

;11,

° 11,

3

dJ' = 10°

>.

". , "

�-$���� f

I

Y I

�

� v

c. �

40'

� l

4·

4'r1: �--$---$1-4+

I

-

1" I

!

-

I

II

J I

45'

0 0

2

4

6

d

D

1 1�23

8

Figure 1 1.42 Variation of efficiency of pile groups in sand (Based on Kishida and Meyerhof, 1965)

Ultimate Capacity of Group Piles in Saturated Clay Figure 1 1 .43 shows a group pile in saturated clay. Using the figure, one can estimate the ultimate load-bearing capacity of group piles in the following manner: Step 1. Determine }; Q" = 1l,I1,(Qp + Q,). From Eq. (1 1 .19), Q"

where c"''''

=

=

A"[9c,,(,,,J

undrained cohesion of the clay at the pile tip.

1 1.23 Ultimate Capacity of Group Piles in Saturated Clay

577

(

L

T 1 BB

I

O

--O� -

Lg Bg Cu /pI N;

"¢ --O-- D I

.

-�.-cr.�.�.�' . b o -9 0 .d I

__

_-

I-

L.R = 20° for all clays, which are normally consolidated. 11.11 A steel pile (H-section; HP14 X 102; see Table l1.1b) is driven into a layer of sandstone. The length of the pile is 62 ft. Fol1owing are the properties of the sandstone: unconfined compression strength = qU(i,b) = 11,400 Ib/in.' and an­ gle of friction = 36°. Using a factor of safety of 3, estimate the al10wable point load that can be carried by the pile. Use [quid,,;,") = q"ll,b/5]. 11.12 A concrete pile is 50 ft long and has a cross section o( 16 in. X 16 in. The pile is embedded in a sand having y = 117 Ib/ftl and = 3r. The allowable working load is 180 kip. If 110 kip are contributed by the frictional resistance

t/>'

586

Chapter 1 1

Pile Foundations

•

piand[Elq.e. 70Gi(11.vkien:6p3)areJ. from3 t1h0e6Ipoib/innt.',load, det5 ermi10'nleb/tihne.',elastic 0.set38,tleandment of0.th57e Sol0.305vemProbl0.em30511.m,12alwilotwabl h theefoworl owikinngg:lloeadngth of338pilkN,e cont12 m,ripibutleiocrossn ofsferctictioionnal resista0.nce3, andto wo�ki0.ng6l[oEadq. (11.24063)kN,]. 21 10' kN/m2, 30,000 kN/m2, Aemhedded 30-m longin concret e pile ti.sl305f mm9200305kN/m'mm, tihnecross sectationground and islefuvell y, a sand deposi moment splacement ofapitelethheade allowabl12 mm;e lateral 21load,10' kN/m' ; andgroundO. ltehvele al.loUse21.wabl0t00heedielkNjm' . cal c ul at t h e stic sol'sumettionhmetod. Assume hod. that the pile is flexible and Solfreeveheaded. ProblemLet11.th14e bysoiaBrom 3 l uni t wei g ht , Y 16 kN/m : t21heMNjm soil fri'c.tion angle, q,A' ste30°el H-: andpileth(esectyieioldnstHP13 ress of th1e00)piliesmatdriveenriabyl, a hammer. The2 kipmaxi. andmtumhe energy i s 40 ki p f t . t h e wei g ht of t h e ram i s 1 ratlengtedhhammer rest i t u t i o n 0. 3 5, of t h e pi l e i s 90 f t . Al s o. we have coef f i c i e nt of ofatbleotwshe weifor gthhte oflastthienchpile cappenetr2.at4iokinp. hammerand efficie30ncy 1060.8Ib5./inumber n . ' . Est i m piSolle,capaci ty using Eq. (ll. ltDS)he .modi TakefiFSed 6. formula. (See Tahle 11.15). Use c SolFiFSgureve Probl11.39aemshows 11.16 ausipinlge.tLethe Danis20h fom,rmul(api(lSeeediTablameteer)11.15).450Usemm,FS ne thtehetowattal edownward fom,rce0.Yonl5ilq,t;h'iIe'17.pil5e.kN/m' Assume. andthatq,tth'l e fil25'.is loDetcateermid above r table anddrthaatg Redo 9 assumi19.8ngkN/ml that t.hIfe twathe eotrhterablquant e coinitciiedsesremaiwithntthheetsame. op of twhathe filwoulProbl anddtehbematt11.he1downward t. lThe. water tO.a121Ibl5q,efbicoiljf' tn'­, Refq,;, e, r to28°,Figure 11.1239b.andLet drag(pil60eforcedifta,metYon',Ietr)he105pi1l8eIb?jfinAssume cionde'thsewipithlet.hAssume e top of the cl0.a5yq,l;a"yer." Determine the total downward drag force E" =

11.13

X

M"

X

E, =

=

X

11.14

M"

�

=

Ep =

=

=

�

=

=

X

=

E, =

X

11" =

Mg =

Q"

=

=

11.16

X

Ep =

=

Fy ipi", =

11.15

=

Fy =

X

=

=

of

1 1. 1 7

11.18 11.19

= 10.

Prnhlcm 1 1 . 1 (, using = 4.

=

EN

L=

4

X

E" = =

D

=

=

=

8' =

11.20

11.21

Y,,""II) =

Hf =

=

ft,

8'

.- - - - - .- - - - -e: , I , I

•

I I I I

•

=

I I I I

e: I

----_._---I- d --->I

Figure 11.22

L = D

8' = Y,,'{d,y) =

=

=

= 3. Hf =

Problems

c"

c"

.

c"

=

550 Ib/ft'. y = 1 15 1b/ft'

=

Clay 875 Iblft'. y = 120 lblft'

587

Clay

Clay = 1200 Iblft'. l' = 124 1blft'

Figure Pl l.26

11.22 The plan of a group pile (friction pile) in sand is shown in Figure Pl1.22. The

piles are circular in cross section and have an outside diameter of 460 mm. The center-to-center spacings of the piles (d) are 920 mm. Determine the effi­ ciency of pile group by a. using Eq. ( 1 1.121 ) b. using the Los Angles Group Action equation (Table 1 1.16). 11.23 Refer to Problem 11.22. If the center-to-center pile spacings are increased to 1200 mm, what will be the group efficiencies? (Solve parts a and b.) 11.24 The plan of a group pile is shown in Figure Pll.22. Assume that the piles are embedded in a saturated homogeneous clay having a c" = 95.8 kN/m2 Given: diameter of piles (D) = 406 mm, center-to-center spacing of piles = 700 mm. and length of piles = 18.5 m. Find the allowable load-carrying capacity of the pile group. Use FS = 3. (Note: 1',,, = 18 kN/m3 and the groundwater table is located at a depth of 23 m below the ground surface.) 11.25 Redo Problem 11.24 with the following: center-to-center spacing of piles = 30 in., length of piles = 45 ft. D = 12 in. c" = 860 1b/ft', 1',,, = 122.4 Ib/ft" and FS = 3. (Note: The ground water table coincides with the ground surface.) 11.26 The section of a 3 x 4 group pile in a layered saturated clay is shown in Figure Pll.26. The piles are square in croS& section (14 in. X 14 in.). The center-to­ center spacing (d) of the piles is 40 in. Determine the allowable load-bearing capacity of the pile group. Use FS = 4. 11.27 Figure PIl.27 shows a group pile in clay. Determine the consolidation settle­ ment of the group. Use the 2:1 method of estimate the average effective stress in the clay layers. .

588

Chapter 1 1

Pile Foundations 1335 kN

Groundwater table

y=

Sand ' '

1 5.72 kN/m'

Sand '

l'�at = 18.55 kN/m3

Figure P11.27

American Petroleum Institute (API). Recommended Practice of Planning, Designing, and Construction of Fixed Offshore Platforms, Rerort No. API-RF-2A, Dallas, 115 pp. American Society of Civil Engineers (1959). "Timber Piles and Construction Timbers," Man­ lIal of Practice, No. 17. American Society of Civil Engineers. New York. American Society of Civil Engineers (1993). Design of Pile FOllndations ( Technical Engineer­ ing and Design Guides as Adapted from rhe us. Army Corps of Engineers, No. 1). Americ� Society of Civil Engineers. New York. Bjerrum. L.. Johannessen. I. 1.. and Eide. O. (1969). "Reduction of Skin Friction on Steel Piles to Rock,'" Proceedings, Seventh International Conference on Soil Mechanics and Foun· dation Engineering. Mexico City. VoL 2. pp. 27-34. /'

-

References

589

Bowles, J. E. (1982). Foundation Analysis and Design, McGraw-Hill. New York. Bowles,J. E. (1996). Foundation Analysis and Design, McGraw-Hill, New York. Briaud,J. L., Moore. B. H and Mitchell, G. B. (1989). "Analysis of Pile Load Test at Lock and Dam 26." Proceedings, Foundation Engineering: Current Principles and Practices, American Society of Civil Engineers, Vol. 2. pp. 925-942. Briaud, J. L .. and Tucker. L. M. (1984). "Coefficient of Variation of In-Situ Tests in Sand:' Proceedings, Symposium on Probabilistic Characterization of Soil Properties, Atlanta. pp. 119-139. Briaud, J. L.,Tucker, L., Lytton. R. L., and Coyle. H. M. (1985). Behavior ofPiles and Pile Groups, Report No. FHWA/RD-83/038, Federal Highway Administration, Washington, DC. Broms, B. B. (1965). "Design of Laterally Loaded Piles," Journal of the Soil Mechanics and Foundations Division, American Society of Civil Engineers, Vol. 91, No. SM3, pp. 79-99. Bustamante, M., and GianeseHi, L. (1982). "Pile Bearing Capacity Prediction by Means of Static Penetrometer CPT," Second European Symposium on Penetration Testing, VoL 2, pp. 493-500. Balkema, Amsterdam. Coyle, H. M., and Castello, R. R. (1981). "New Design Correlations for Piles in Sand," Journal of the Geotechnical Engineering Division, American Society of Civil Engineers, Vol. 107, No. GTI, pp. 965-986. Davisson, M. T. (1970). "BRD Vibratory Driving Formula," Foundation Facts, Vol. VI, No. 1 , pp. 9-11. Davisson, M. T., and Gill, H. L. (1963). "Laterally Loaded Piles in a Layered Soil System," Journal ofthe Soil Mechanics and Foundations Division, American Society of Civil En­ gineers, Vol. 89. No. SM3, pp. 63-94. DeRuiter, 1.. and Beringen. F. L. (1979). "Pile Foundations for Large North Sea Structures." .V/ariue Gt'otel'i1lJ% gy. Vo1. 3. No. 3 . pp. 267- 314. Feng. Z.. and Deschamps. R.J. (2000). "A Study of the Factors InfluenCing the Penetration and Capacity of Vibratory Driven Piles." Soils and Foundations. Vol. 40. No. 3. pp. 43-54. Gates. M. (1957). " Empirical Formula for Predicting Pile Bearing Capacity." Civil Engineer­ ing, American Society of Civil Engineers. Vol. 27. No.3. pp. 65-66. Goodman. R. E. (1980). lmroduction to Rock Mechanics, Wiley, New York. Guang-Yu. Z. (1988). "Wave Equation Applications for Piles in Soft Ground," Proceedings, Third internatiolJal Conference on the Application ofStress-Wave Theory to Piles (B. H. Fellenius. ed.). Ottawa, Ontario, Canada. pp. 831-836. International Conference of Building Officials (1982). "Uniform Building Code," Whittier, CA. Janbu, N. (1953). An Energy Analysis ofPile Driving with the Use 0/ Dimensionless Parame­ lers, Norwegian Geotechnical Institute, Oslo, Publication No. 3. Janbu. N. (1976). "Static Bearing Capacity of Friction Piles." Proceedings, Sixth European COllference all Soil Mechanics and Foundation Engineering, Vol. 1.2, pp. 479-482. Kishida. H., and Meyerhof, G. G. (1965). "Bearing Capacity of Pile Groups under Eccentric Loads in Sand:' Proceedings. Sixth jmernational Conference on Soil Mechanics and Foundation Engineering, Montreal. Vol. 2, pp. 270-274. Matlock. H .. and Reese. L. C. (1960). "Generalized Solution for Laterally Loaded Piles:' lour­ nal of the Soil Mechanics and FOlmdatio.ns Division, American Society of Civil Engi­ neers, Vol. 86. No. SM5, Part I. pp. 63-91. McClelland. B. (1974). "Design of Deep Penetration Piles for Ocean Structures," Journal of the Geotechnical Engineering DiJ,,'ision, American Society of Civil Engineers. Vol. 100. No. GTI. pp. 709-747. Meyerhof, G. G. (1976). "Bearing Capacity and Settlement of Pile Foundations." loumal of the Geotechnical Engineering Division. American Society of Civil Engineers. Vol. 102. No. GT3, pp. 197-228. .•

590

Chapter 1 1 Pife Foundations

MeyerhoL G. G. (1995). "Behavior of Pile Foundations under Special Loading Conditions: 1994 R. M. Hardy Keynote Address," Canadian Geotechnical Journal, Vol. 32, No. 2. pp. 204-222. Michigan State Highway Commission (1965). A Pe/formance llll'esligalioll of Pile Driving Hammers and Piles, Lansing. MI. 338 pp. Nottingham. L. C, and Schmertmann, 1 H. (1975). An lnresligarion of Pile Capacily Design Procedures, Resean;h Report No. D629. Department of Civil Engineering, University of Florida, Gainesville. FL. Olson. R. E .. and Flaate, K. S. (1967). "Pile Driving Formulas for Friction Piles in Sand." Jour­ nal of the Soil Mechanics and Foundmiolls Division, American Society of Civil Engi­ neers. Vol. 93. No. SM6. pp. 279-296. Randolph, M. E. and Murphy. B. S. (1 �85). "Shaft Capacity of Driven Piles in Clay." Proceed­ ings, Offshore Technology Conference, Vo!' l. Houston. pp. 371-378. Schmertmann, 1. H. (1978). Guidelines for Cone Penetration Test: Pelformance and Design. Report FHWA-TS-78-209. Federal Highway Administration, Washington. DC Seiler, l E, and Keeney, W. D. (1944). "The Efficiency of Piles in Groups." Wood Preserving News, Vol. 22. No. 1 1 (November). Skov, R.. and Denver, H. (1988). "Time Dependence of Bearing Capacity of Piles." Proceed­ ings. Third Incernational Conference on Application of Stress ltl-'ave Theory to Piles,

I

Ottawa. Canada, pp. 879-889. Svinkin. M. (1996). Discussion on "Setup and Relaxation in Glacial Sand:' lournal of Geo­ lechnical Engineering. ASCE. Vo! ' 22. pp. 3 19-321 . Vesic. A. S. ( 1961). "Bending of Beams Resting on Isotropic Elastic Solids." IOllrnal ofIhe En­ gineering j\1echanics Division, American Sodety of Civil Engineers. Vol. 87. No. EM2, pp. 35-53. Vesic. A. S. (1970). " Tests on Instrumental Pilcs-Ogcechct! Ri\"er Si t !.!:' jOllmal of the Soil Mechanics and Foundations Division, American Society of Civil Engineers, Vol. 96. No. SM2. pp. 561-584. Vesic,A. S. (1977). Design of Pile Foundations, National Cooperative Highway Research Pro­ gram Synthesis o f Practice No. 42, Transportation Research Board, Washington, DC. Vijayvergiya. V. N., and Focht, 1 A., Jr. (1972). A New Way 10 Predict Capacily of Piles in Clay, Offshore Technology Conference Paper 1718. Fourth Offshore Technology Confer­ ence. Houston. Wong, K. S., and Teh, C I. (1995). "Negative Skin Friction on Piles in Layered Soil Deposit," Journal of Geotechnical and Geoenvironmental Engineering. American Society of Civil Engineers, Vol. 121, No. 6. pp. 457-465. Woodward, R. 1, Gardner, W. S., and Greer, D. M. (1972). Drilled Pier Foundalions, McGraw­ Hill, New York.

I

Introduction The terms caisson, pier, drilled shaft, and drilled pier are often used interchangeably in foundation engineering; all refer to a cast-in-place pile generally having a diameter of about 750 mm (=2.5 ft) or more, with or without steel reinforcement and with or with­ out an enlarged bottom. Sometimes the diameter can be as small as 305 mm (= 1 ft). To avoid confusion, we use the term drilled shaft for a hole drilled or exca­ vated to the bottom of a structure's foundation and then filled with concrete. De­ pending on the soil conditions, casings may be used to prevent the soil around the hole from caving in during construction. The diameter of the shaft is usually large enough for a person to enter for inspection. The use of drilled-shaft foundations has several advantages: 1. A single drilled shaft may be used instead of a group of piles and the pile cap. 2. Constructing drilled shafts in deposits of dense sand and gravel is easier than driving piles. 3. Drilled shafts may be constructed before grading operations are completed. 4. When piles are driven by a hammer. the ground vibration may cause damage to nearby structures. The use of drilled shafts avoids this problem. S. Piles driven into clay soils may produce ground heaving and cause previously driven piles to move laterally. This does not occur during the construction of drilled shafts. 6. There is no hammer noise during the construction of drilled shafts; there is dur­ ing pile driving. 7. Because the base of a drilled shaft can be enlarged, it provides great resistance to the uplifting load. S. The surface over which the base of the drilled shaft is constructed can be visu­ ally inspected. 9. The construction of drilled shafts generally utilizes mobile equipment, which, under proper soil conditions. may prove to be more economical than methods of constructing pile foundations. 10. Drilled shafts have high resistance to lateral loads. There are also a couple of drawbacks to the use of drilled-shaft construction. For one thing, the concreting operation may be delayed by bad weather and always needs 591

592

Chapter 12 Drilled-Shaft Foundations

close supervision. For another, as in the case of braced cuts. deep excavations for drilled shafts may induce substantial ground loss and damage to nearby structures.

12.2

Types of Drilled Shafts Drilled shafts are clasSified according to the ways in which they are designed to transfer the structural load to the substratum. Figure 12.1a shows a drilled straight shaft. It extends through the upper layer(s) of poor soil, and its tip rests oil. a strong load-bearing soil layer or rock. The shaft can be cased with steel shell or pipe when required (as it is with cased, cast-in-place concrete piles; see Figure 1 1.4). For such shafts, the resistance to the applied load may develop from end bearing and also from side friction at the shaft perimeter and soil interface. A belled shaft (see Figures 12.1b and c) consists of a straight shaft with a bell at the bottom, which rests on good bearing soil. The bell can be constructed in the shape of a dome (see Figure 12.1b), or it can be angled (see Figure 12.1c). For angled bells, the underreaming tools that are commercially available can make 30 to 45° angles with the vertical. For the majority of drilled shafts constructed in the United States. the entire load-carrying capacity is assigned to the end bearing only. However, under certain circumstances, the end-bearing capacity and the side friction are taken into account. In Europe, both the side frictional resistance and the end-bearing capacity are always taken into account. Straight shafts can also be extended into an underlying rock layer. (See fig­ ure 12.1d.) In the calculation of the load-bearing capacity of such shafts. the end bearing and the shear stress developed along the shaft perimeter and rock interface can be taken into account. .

. , .. .

,•

'"

.. ' ,

I

\

,

' .•: ! ;

,

..

,

.

' ':'' : : :, � " : ':

. .. :,. ' . .. . � .. �

'

(

.

..

J ' ( I

.

",

:'.

�

•

. ' � ..

.

'

,

: ',, " ; ',"

,

( .

(. . .. .

�

..

f ., , , ,, '.

r

.

:

::.

·• .

,': , ' . ,

:

::

'

.: :

'

•

' ,

�

'':."

' . - . . .

• ·

·

.

•

,

, .

:' .:

'

'

.. ..

.

45° or "

·

30·

·

.

Good bearing soil (b)

•

.

·

.

'

'•

,

.

.

•.. •

•

.

, . , , •

.

:

•

" ,,'

.

,

. '4 '"

�.

"

..

"

"

,',

�

�

• ' "

... �.

I

.

': : �.: :

. "

(

r

'

;

.

,, ,

.. , ', , ' ' , I " : ,' "

.

,

,

'

"

,

·

·

:

.

. . ... "

.

(al

•..

..

·

' ; :',

t ",

. ;', � .;

II.

. : . .

,

·

,'

:

�

' ... �

. -

-

: ' ,

' ' '''

..

• "

: :,

..

'

.�

1 ..

;,

..

. . , ',

• A

•

�.. .

,

'

'.

; � .

' Soft ., , . .. . '1 . .. SOI

"

. "

" �" � . ,

�

(

• J

p

:

'

. �

' ", , .'" .

�.�

. • ,'

,

I

. ,: , ". .. : :':' ' .. ) . . :': .

'4

"

.,

.

:-

.

'

. , ' .

. ,

,

Good bearing soil (el

Figure 12. 1 Types of drilled shaft: (a) straight shaft: (b) and (e) belled shaft: (d) straight shaft socketed into rock

12.3 Construction Procedures

593

Construction Procedures The most common construction procedure used in the United States involves rotary drilling. There are three major types of construction methods: the dry method, the casing method, and the wet method. Dry Method of Construction

This method is employed in soils and rocks that are above the water table and that will not cave in when the hole is drilled to its full depth. The sequence of construc­ tion, shown in Figure 12.2, is as follows: Step 1. The excavation is completed (and belled if desired), using proper drilling Step 2. Step 3. Step 4.

tools, and the spoils from the hole are deposited nearby. (See Figure 12.2a.) Concrete is then poured into the cylindrical hole. (See Figure 12.2b.) If desired, a rebar cage is placed in the upper portion of the shaft. (See Figure 12.2c.) Concreting is then completed, and the drilled shaft will be as shown in Figure 12.2d.

Casing Method of Construction

This method is used in soils Or rocks in which caving Or excessive deformation is likely to occur when the borehole is excavated. The sequence of construction is shown in Figure 12.3 and may be explained as follows: Step 1. Step 2.

Step 3. Step 4. Step 5. Step 6. Step 7. Step 8.

The excavation procedure is initiated as in the case of the dry method of construction. (See Figure 12.3a.) When the caving soil is encountered, bentonite slurry is introduced into the borehole. (See Figure 12.3b.) Drilling is continued until the ex­ cavation goes past the caving soil and a layer of impermeable soil or rock is encountered. A casing is then introduced into the hole. (See Figure 12.3c.) The slurry is bailed out of the casing with a submersible pump. (See Figure 12.3d.) A smaller drill that can pass through the casing is introduced into the hole, and excavation continues. (See Figure 12.3e.) If needed, the base of the excavated hole can then be enlarged, using an underreamer: (See Figure 12.3f.) If reinforcing steel is needed, the rebar cage needs to extend the full length of the excavation. Concrete is then poured into the excavation and the casing is gradually pulled out. (See Figure 12.3g.) Figure 12.3h shows the comI'leted drilled shaft.

Wet Method of Construction

-

This method is sometimes referred to as the slurry displacement method. Slurry is used to keep the borehole open during the entire depth of excavation. (See Figure 12.4.) Following are the steps involved in the wet method of construction:

594

Chapter 12 Drilled-Shaft Foundations

Surface casing. if required

::::: :: Competent, :::::: : : : noncaving : = = : = : soil

(a)

Competent. : = = : = = noncaving ::::: : : : soil

(e)

chute

:: :: : Competent.

noncaving : = ",=,=:: = = = = = = = = = = soil (b)

Competent, noncaving soil

(d)

12.2 Dry method of construction: (a) initiating drilling; (b) starting concrete pour; (c) placing rebar cage; (d) completed shaft (After O'Neill and Reese, 1999)

Figure

12.3 Construction Procedures

��

= = = = = --

====== - �::==

: Cohesive soil ::

: Cohesive soil

: = : Cohesive soil = = = = == = = = = = = = = :

----

------

=

-

, Caving soil

= = = = == = = = == = = = : (c)

.,..'

(b)

: Cohesive soil

Caving soil

: ::: Cohesive soil

;;;'I)rillin� slurry

: = : Cohesive soil = = = = = = = = = == = = : :

(a)

::: Cohesive soil

595

: :: : Cohesive soil := (d)

Figure 12.3 Casing method of construction: (a) initiating drilling; (b) drilling with slurry; (c) introducing casing; (d) casing is sealed and slurry is being removed from interior o f cas­ ing; (e) drilling below casing: (f) underreaming; (g) removing casing; (h) completed shaft (After O'Neill and Reese. 1999)

596

Chapter 12 Drilled-Shaft Foundations

: Competent soil

Competent soil -

Caving soil ,

Caving soil

:: :: : Competent soil

:: Competent soil

=======

- - - - - - - - - - - -' is in degrees.The variation of w with L/Db is given in Figure 12.7. The frictional resistance at ultimate load, Q" developed in a drilled shaft may be calculated from the relation given in Eq. (12.15), in which p = shaft perimeter = 7rD, (12.22) f = unit frictional (or skin) resistance = K lT� tan ii' where K = earth pressure coefficient Ko = 1 - sin 4>' lT� = effective vertical stress at any depth z Thus, =

(12.23)

The value of lT� will increase to a depth of about lSD, and will remain constant thereafter, as shown in Figure 11.18. For cast-in-pile concrete and good construction techniques, a rough interface develops and, hence, ii'/4>' may be taken to be one. With poor slurry construction, ii'/4>' 0.7 to 0.8. =

604

Chapter 12 Drilled·Shaft Foundations

i=

I

5�fl noncontributing zone (cohesive soil only)

t D,

L J;P !lL,

i=i

t

--- t =N t )-----\',

Noncontributing zones: Length = D., (cohesive soil only)

u

i

No side-load transfer permiued on perimeter of bell

Figure 12.8

Development of Eq. ( 1 2.25)

An appropriate factor of safety should be applied to the ultimate load to ob­ tain the net allowable load, or Q,II(net)

=

Qp(not) + Q,

FS

(12.24)

Load·Bearing Capacity Based on Settlement

On the basis of a database of 41 loading tests, Reese and O'Neill (1989) proposed a method for calculating the load-bearing capacity of drilled shafts that is based on settlement. The method is applicable to the following ranges: 1. Shaft diameter: D, = 0.52 to 1.2 m (1.7 to 3.93 it) 2.

Bell depth: L = 4.7 to 30.5 m (15.4 to 100 ft)

3. Field standard penetration resistance: N(iJ = 5 to 60 4. Concrete slump = 100 to 225 mm (4 to 9 in.)

Reese and O'Neill's procedure (see Figure 12.8) gives Qu(ne') =

N

z,fiptl Li + qpAP

i=1

(12.25)

12.7 Drilled Shafts in Granular Soil: Load-Bearing Capacity

605

where (

f, = ultimate unit shearing resistance in layer i p = perimeter of the shaft = 1TD, qp = unit point resistance Ap = area of the base = (1T/4)D� Following are the relationships for determining Qu(nol) from Eq. (12.25) for sandy soils. We have (12.26) where U�'i = vertical effective stress at the middle of layer i f3 = 1.5 - 0.135z?·5 (0.25 "" f3 "" 1.2) Zi = depth to the middle of layer i (ft) The point bearing capacity is

(12.27)

.. ·· · (tr�;iii)· 'i� : ' 90�ii!{f2 ··� ;i6{' �'·;� _.\ .:;:..•..•..•�....l�, : .. .�;..·.·>!..··..., · 'fp . . Pt f .. ·."", .',. "i.� �.> · ll' . t ." .;. .-6 ;' '>:�

•.

;'

.·r'::.

:

�

. . ::.•...;• .• . . : .•...:

.

.

(12.28) ,

where N"" = mean uncorrected standard penetration number within a distance of 2Db below the base of the drilled shaft. If Db is equal to or greater than 50 in., exces­ sive settlement may occur. In that case. Cfp may be replaced by (for Db ;;. 50 in. )

(12.29)

In SI units, Eqs. (12.26) through (12.29) will be of the form ' fi

=

f3U�'i ""

f3 = 1.5 - 0.244z? 5

192 kN/m'

(0.25 "" f3 "" 1.2)

(12.30) (12.31)

(where Zi is in m) and

Cfp

(kN/m' )

=

57.5Nr,o "" 4310 kNjm'

(forDb < 1.27 m)

(12.32)

(12.33)

.....

Based on the desired level of settlement, Figures 12.9 and 12.1 0 may now be used to calculate the allowable load. Q,n(n D" then a* = 0 for 1 diameter above the top of the bell and for the peripheral area of the bell itself) ott = 0.55 elsewhere.

(12.46)

where CUb = average undrained cohesion within 2Db below the base. If Db is large, excessive settlement will occur at the ultimate load per unit area, qp' as given by Eq. (12.46). Thus, for D., > 1.91 m (75 in.), qp may be replaced by

(12.47) where F, �

2.5

in which t/J, = 0.0071 + O.0021 and t/J, =

,,; 1

( 1 2 .48)

(�J ,,; 0.Dl5

(12.49)

t/J, Db (in.) +

t/J,

0.45( CUb)0,5 (0.5 .. t/J, "; 1.5) t

(12.50)

kip/ft' In SI units, Eqs. (12.46), (12.48), (12.49), and (12.50) can be expressed as

(12.51) (12.52) t/J,

=

2.78 X lO-4 + 8.26

x

1O-5

( �J ,,; 5.9

x

10-'

(12.53)

616

Chapter 12 Drilled-Shaft Foundations 1.2

1.0 --i
'-' "" "-

� <E � k

�

"0 �

!f"

� C

g

"0 � 0

:.;', " ,: ' :. ",, ": " .:.

2.5 It

. Drilled Shafts in Clay: Load-Bearing Capacity .

: " .

.. " "

c"

� 800 Ib/lt'

Clay It

J+--' .'

.' . �. ' . : , . , ' . ,' " .'. . ;.. .' " : :

.. , cr�; ·. .

20 It

Figure 12. 19 ,

:

4 It

c"

� 1200 Ibllt'

-I

A drilled shaft in layered clay

Solution Part a .

From Eq. (12.45),

Fro�Figure 12.19,

ALl = 12 - 5 = 7 ft AL, = (20 - 12) - D, = (20 - 12) - 2.5 = 5.5 ft . cull) ",; 800 lb/ft' Cu(2) = 12.o0 1b/ft2

I:!(!I)C�, !,!iPA Li = ka1cu(i)Pt:.. L;

= (0.55) (800) ( 'IT x 2.5) (7) + (0.55) (1200) ('IT X 2.5) (5.5) = 52,700 lb = 52.7 kip

619

620

Chapter 12 Drilled-Shaft Foundations

Again, from Eq_ qp

=

(12.46),

6CUb( 1 + 0.2 �J = (6) (3000{ 1 + 0.2eo: 5 ) J = 40�500 Ib/ft'

= 40.5 kip/ft'

•

A check reveals that

= 9Cub = (9) (3000) = 27,000 Ib/ft' So we use qp = 27 kip/ft': qp

qpAp

=

qp

Hence. Q"

=

27 kip/ft' < 40.5 kipift'

(:D�) = (27{ (:)(4)'J

= latc,,{,)ptJ.Li +

q" A p

=

339.3 kip

= 52.7 + 339.3 = 392 kip

Part b

We have

1 ) - 0.0 I 67 - 1.67

Allowable settlement 0.5 = ? ( S) ( 2 D ,

12.16

___

_

_

Yo

0

The trend line shown in Figure indicates that. for a normalized settlement of 1.67%, the nonnalized side load is about 0.89. Thus, the side load is (0.89) (lfipllLi) = (0.89) (52.7) = 46.9 kip Again, Allowable settlement

Db

=

(

05 = 0.0104 = 1.04% (4) 12)

The trend line shown in Figure 12.17 indicates that, for a normalized settl�ment . of 1.04%, the nonnalized end bearing is about 0.57, so Base load = (0.57) ( qpAp) = (0.57) (339.3) = 193.4 kip Thus, the total load is • Q = 46.9 + 193.4 = 240.3 kip

12.9

Settlement of Drilled Shafts at Working Load The settlement of drilled shafts at working load is calculated in a manner similar to that outlined in Section 1 1.17. In many cases, the load carried by shaft resistance is

small compared with the load carried at the base. In such cases. the contribution of SJ may be ignored. (Note that in Egs. and the term D should be re­ placed by Db for drilled shafts.)

(11.64) (11.65)

I

I

621

72.9 Settlement of Drilled Shafts at Working Load

Example 12.6

Qw

For the drilled shaft of Example 12.4, estimate the elastic settlement at working loads (Le., = 1005 kN). Use Eqs. (11.63), (11.65), and (11.66). Take t; T' 0.65; Ep = 21 X 106 kN/m2, E, = 14,000 kN/m2, /L, = 0.3, and = 250 kN. ", Splution

Qwp

From Eq. (11.63),

.,'

.

Now

,

so S,(I) =

Q",

= 1005 - 250 = 755 kN

[250 + (0.65) (755))( 11)

From Eq. (11.65),

("47T

)

= 0.00022 m = 0.22 mm

X 1.52 (21 X 106)

s,(2 ) =

QD\l;pCp

-

bqp

From Table 11.12, for stiff clay, Cp = 0.04; also, qp = c"lblN! =

( 105) (9) = 945 kN/m2

Hence,

(250) (0.04)

= S,(2) = (1.5 ) (945) = 0.0071 m 7.1 mm Again, from Eqs. (11.66) and (11.67), S,(3)

where

(L = 2 + 0.35 'I) 15;

( Qw.)(DE,,) (1 pL

-

2 /L,)Iw•

IPs-1.5 = 2.95· 1 .5 ) 1 - [ 7551.5) (11) ] ( 14,00 ( - 0.32 ) (2.95 ) - 0.0042 m - 4.2 mm 0

I", = 2 + 0.35

-

=

S,(3) - ( 7T x

L

11

_

_

.

"

622

Chapter 12 Drilled-Shaft Foundations

The total settlement is S, = S,( I)

" 12. 1()

+

S,(2) +

s'(3) = 0.22 + 7.1 + 4.2

=

11.52 mm

•

Lateral Load-Carrying Capacity- Characteristic Load and Moment Method Several methods for analyzing the lateral load-carrying capacity of piles. as well as the load-carrying capacity of drilled sh�fts, were presented in Section 11.18; there­ fore, they will not be repeated here. In 1994, D uncan et al. developed a characteristic load method for estimating the lateral load capacity for drilled shafts that is fairly simple to use. We describe this method next. According to the characteristic load method, the characteristic load Qe and moment Me form the basis for the dimensionless relationship that can be given by the following correlations: Characteristic Load

Q,

=

,

7.34D; (E/?/)

( )OJ,8

Characteristic Moment

Me = 3.86D;

(EpR/)

c

"

EpR/

( � )0.46 c

Ep /

(for clay)

(12.55)

(for sand)

(12.56)

(for clay)

(12.57)

(for sand)

(12.58)

In these equations, D, = diameter of drilled shafts Ep = modulus of elasticity of drilled shafts RJ = ratio of moment of inertia of drilled shaft section to moment of inertia of a solid section (Note: RJ = 1 for uncracked shaft without central void) = y' effective unit weight of sand ' = effective soil friction angle (degrees) Kp = Rankine passive pressure coefficient = tan2(45 + '/2) Deflection Due to Load Q. Applied at the Ground Line

Figures 12.20 and 12.21 give the plot of QgIQ, versus xol D, for drilled shafts in sand and clay due to the load Qg applied at the ground surface. Note that x" is the ground \

l .

12. 10 Lateral Load-Carrying Capacity 0.050

0.050

0.045

0.045

623

(

Q, Q, - Free head

0.030

0.Dl5

/ / / I I

, , ,

, ,

,

,

,

,

,

-

,

,

o

Figure 12.20

-

,

,

,

-

.,.' D,

"

x"

0.005 0.010 0.020

"

"

"

Q

Q,

'

,-

,

-,

,/

0.030

/ M , ,­

Me

Q

0.0 1 5

Q, . d - Fixe

M,

0.0 1 3 3 0.0197 0.0289

0.0024 0.0048 0.0074

-, - Free Q: Q, 0.0065 0.009 1 0.0135

0.05

Plot of -s and

, ,,,,

,,

, ,-

/ .,.

0.10

M,

0.15

Xl) versus Ii in clay M, , Mg

line deflection. If the magnitudes of Qg and Q, are known, the ratio Q,IQ, can be cal­ culated. The figure can then be used to estimate the corresponding value of xolD, and, hence,xoo Deflection Due to Moment Applied at the Ground Line

Figures 12.20 and 12.21 give the variation plot of MgIM, with xjD, for drilled shafts in sand and clay due to an applied moment Mg at the ground line. Again, Xo is the ground line deflection. If the magnitudes of Mg, M" and D, are known, the value of Xo can be calculated with the use of the figure.

624

Chapter 12 Drilled-Shaft Foundations 0.015

0.015

0.010

0.0 1 0

Q, Q� - Free head

oilai' Q, ,.

0.005

-..:. Q .. - Free

.t"

D,

0.005 0.010 0.020

0.05

u

Figure 12.21

0.00 1 3 0.002 1 0.0033

Q

Mg

Qc

Mc

Plot of -g and

Qg ,

Q - Fixed 0.0028 0.0049 0.0079

0. 1 0

0.005

M, M,. 0.0009 0.001 9 0.0032

0. 15

versus - In sand Ds Xu .

Deflection Due to Load Applied Above the Ground Line

When a load Q is applied above the ground line, it induces both a load Qg = Q and a moment Mg = Qe at the ground line, as shown in Figure 12.22a. A superposition solution can now be used to obtain the ground line deflection. The step-by-step pro­ cedure is as follows (refer to Figure 12.22b): Step 1. Calculate Qg and Mg. Step 2. Calculate the deflection XoO that would be caused by the load Qg acting

alone.

Step 3. Calculate the deflection XoM that would be caused by the moment act­ Step 4. Step 5. Step 6. Step 7.

ing alone. Determine the value of a load QgM that would cause the same deflec­ tion as the moment (i.e., XoM). Determine the value of a moment MgQ that would cause the same de­ flection as the load (i.e., x Q). Calculate (Qg + QgM )/Q,. and determine xoOM/D,. Calculate (Mg + MgQ )/Me and determine XnMQ/D,. o

12. 10 Lateral Load-Carrying Capacity

625

/

(a)

Step 6 - - - - - -- - - -

Step 4

+- - - - - - -

Q, Q.-

lstep4 :I ; ',, ' ,I ,'" I I

'T'

Step 2

, t'� I

- - +- - -

i

' ... ... ... ...

,

...

...... ' Step 7 ......... fs�e-p;-----�-r I

l

I

't T I

t

:eoQM

0:

-o(- - - - - - -o(- - - - -

,

S�ep!.. _ ___ _

Steip 5 I "oI\-_-+-_ I �_ .. --_l.-.---.c--t I I I I I I I ' " I Step 2 ' ,Y y Y t I I I I "

/

,, '

r ,,' ,I Step 3 I, ,." _t , ,

"

I

t

----�----�--....

M, Me ,

:

I

...... - - - - - - -

Mg

+ MgQ Me

:8

'

M8Q M

_ C

/

XoQ

0;

XoM

0;

X"

D,

D,

XoMQ

(b)

Figure 12.22

Slep 8.

Superposition of deflection due to load and moment

Calculate the combined defl�ction: xo(oombi",d) =

O.5(XoQM + XoMQ )

(12.59)

Maximum Moment in Drilled Shaft Due to Ground Line Load Only

Figure 12.23 shows the plot of Qg/Q, with MmaJM, for fixed- and free-headed drilled shafts due only to the application of a ground line load Q., For fixed-headed

626

Chapter 12 Drilled-Shaft Foundations 0.045

0.020

Free 0.0 1 5

Fixed 0.030

Q., Q,

Q, 0.0 1 0 Q,

Clay)

(Sand) 0.015 ---

Clay

0.005

- - - - - - Sand

O ��--�---'----r---�-+ O o

Figure 12.23

0.005

" . .

variatIon 0

f

0.0 1 5

0.010

Q , ' Mm" - With � Q, ,

shafts, the maximum moment in the shaft, Mm", occurs at the ground line. For this condition, if Q" M" and Qg are known, the magnitude of Mm" can be easily calcu­ lated. Maximum Moment Due to Load and Moment at Ground Line

If a load Qg and a moment Mg are applied at the ground line, the maximum moment in the drilled shaft can be determined in the following manner: Using the procedure described before, calculate xo(,omb;n
From Figure 12.20, for .toQID,

=

Qg QgM + Q, Q,

=

�

0.0 25

=

0.0025

0.0025, the value of MgQIM, =

=

0.0013, so

0.004 + 0.002 = 0.006

From Figure 12.20, for {Qg + QgM)IQ, = 0.006, the value of xoQMID, Hence, XoQM

Thus, we have Mg Me

+

=

(0.0046) (1)

�,

MQ

=

=

0.0046 m

0.000675 + 0.0013

=

=

=

0.0046.

4.6 mm

0.00198

From Figure 12.20, for (Mg + MgQ)1Me = 0.00198, the value OfXoMQID, = 0.0041. Hence, XoMQ

= (0.0041) (1 )

=

0.0041 m = 4.1 mm

630

Chapter 12 Drilled-Shaft Foundations

Consequently. Part b

X,, « omoin,d) =

0.5( x"Q.11 + X".110 ) = (O.5} (4.6 + 4.1 )

=

4.35 mm

From Eq. (12.60).

so 0.00435 m =

(2.43) ( 150) . ( 1.62) (200) , T-' + T (22 X 106) (0.049) (22 x 100) (0.049) _

or 0.00435 m = 338 X 10-0 T' + 300.6 X 10-6 T'

and it follows that T

=

2.05 m

From Eq. (12.61). M,

=

A.., Q, T + B.., M.e = A..,( 150 ) (2.05) + B..,(200)

=

307.5A.., + 200 Bm

Now the fOllowing table can be prepared: Am (Figure 12.24)

z

T

o

o

Bm

(Figure 12.24)

1.0 0.98 0.95 0.9 0.845 0.8 0.73

0.36 0.52 0.63 0.75 0.765 0.75

0.4 0.6 0.8 1.0 1.J 1.25

So the maximum moment is 399.4 kN-m

Hence,

Part c

z

=

=

400 kN-m and occurs at ziT = 1.

( I ) ( T) = ( 1 ) (2.05 m) = 2.05 m

The maximum tensile stress is (400) U tensile =

200 306.7 349.9 373.7 399.6 395.2 376.6

( &) .

0.049

=

4081.6 kN/m'

12. 1 1

Part

Drilled Shafts Extending into Rock

631

d

We have Etfi, Cu

=

(22 x 106) (1)

100

=

2.2 x lOS

By interpolation, for (EtfiI)/c" = 2.2 X lOS, the value of (LID,)..,;. - 8.5: SO L = (8.5)(1)

=;

8.S m

•

Drilled Shafts Extending into Rock In Section 12.1, we noted that drilled shafts can be extended into rock. In the cur­ rent section, we describe the principles of analysis of the load-bearing capacity of such drilled shafts, based on the procedure developed by Reese and O'Neill (1988, 1989). Figure 12.26 shows a drilled shaft whose depth of embedment in rock is equal to L. In the design process to be recommended, it is assumed that there is ei­ ther side resistance between the shaft and the rock or point resistance at the bottom, but not both. Following is a step-by-step procedure for estimating the ultimate

bearing capacity:

ql' ,,;

unit pOint bearing

Figure 12.26 Drilled shaft socketed into rock

632

Chapter 12 Drilled-Shaft Foundations

Step 1. Calculate the ultimate unit side resistance as f (lb/in' )

=

2.5q?,5 '" 0.15q"

(12.62)

where q" = unconfined compression strength of a rock core of NW size or larger, or of the drilled shaft concrete, whichever is smaller (in lb/in'). In SI units, Eq (12.62) can be expressed as f(kN/m' )

=

6.564q?,.5 (kN/m' ) '" 0.15q" (kN/m' )

(12.63)

Step 2. Calculate the ultimate capacity based on side resistance only, or Q" = 1TD,Lf

(12.64)

Step 3. Calculate the settlement s, of the shaft at the top of the rock socket, or

(12.65) where

s,("

=

sdb)

=

elastic compression of the drilled shaft within the socket, assum­ ing no side resistance settlement of the base

However.

s, ) (.
136.9 I b/in'

x

12) ( 136.9)]

Step 3. From Eqs. (12.65), (12.66), and (12.67).

l�OO

= 2787 kip

" Jr s = -- + ---'-"-'-­ e

QL AcE.

Q DsE mass

From Eq. (12.69), for RDO = 80% E mass =

E core

(0.0266)(80) - 1.66 = 0.468

E""", = 0.468E _ = (0.468) (0.36 x 106) = 0.168 x 10" Ib/in2

15 ft

1

Drilled shaft Figure

12.27 Drilled shaft extending into rock

I

i

Problems

635

12.1 A drilled shaft is shown in Figure P12.l. Determine the net allowable point

bearing capacity. Given

Db = 6 ft 'Y, = 100 lb/ft' D, = 3.5 ft 'Y, = 112 lb/ft' q,' = 38° L, = 18 ft L, = 10 ft c" = 720 lb/ft' Factor of safety = 3 Use Eq. (12.20). 12.2 Redo Problem 12.1, this time using Eq. (12.16). Let E, = 400po ' 12.3 For the drilled shaft described in Problem 12.1, what skin resistance would de­ velop in the top 18 ft, which are in clay? Use Eqs. (12.42) and (12.44). 12.4 Redo Problem 12.1 with the following: 'Y, = 17.8 kN/m' Db = 1.75 m D, = 1 m 'Y, = 18.2 kN/m 3 q,' = 32° L, = 4 m c" = 32 kN/m' L, = 2.5 m Factor of safety = 4 12.5 Redo Problem 12.4 using Eq. (12.16). Let E, = 600po '

636

Chapter 12 Drilled-Shaft Foundations

Silty clay

Sand y,

4>' c' = 0

L,

1

Figure P12. 1

Clay

Clay Figure P12.7

U.6 For the drilled shaft described in Problem 12.4, what skin friction would de­ velop in the top 4 m? a. Use Eqs. (12.42) and (12.44). b. Use Eq. (12.45). U.7 Figure P12.7 shows a drilled shaft without a bell. Assume the following values: L, = 20 ft L, = 15 ft D, = 5 ft

C.,( !)

= 1000 Ib/ft2

C.,(,) =

1800 lb/ft'

Determine: a. The net ultimate point bearing capacity [use Eqs. (12.39) and (12.40)] b. The ultimate skin friction [use Eqs. (12.42) and (12.44)] c. The working load Qw (factor of safety = 4)

!, -

Problems

637

(

Medium sand

Average standard penetration number (N60) with 2Db below the drilled shaf, = 23

Figure P12.9

U.S Repeat Problem 12.7 with the following data:

L, = 25 ft L, = 10 ft D, = 3.5 ft

Cut')

=

Cut') =

1200 Ibjft' 2000 Ibjft'

Use Eqs. (12.45) and ( 12.46).

U.9 A drilled shaft in a medium sand is shown in Figure P12.9. Using the method

proposed by Reese and O'Neill, determine the following: a. The net allowable point resistance for a base movement of 25 mm b. The shaft frictional resistance for a base movement of 25 mm c. The total load that can be carried by the drilled shaft for a total base move­ ment of 25 mm Assume the following values: L = 14 m L, = 12.5 m D, = 1 m Db = 2 m

l' =

cf>'

D,

= =

19 kNjm1 36° 65% (medium sand)

= 3 ft. Db = 4.5 ft. l' = 1 1 8 Ibjft3, and 8. SL < 10, and PS > 1.5 (critical)

Oakshanamanthy See Figure 13.13b and Raman (1973) Raman (1967)

PI > 32 and SI > 40 (very high) 23 ,,; PI ,,; 32 and 30 ,,; SI ,,; 40 (high) 12 ,,; PI ,,; 23 and IS "; SI ,,; 30 (medium) PI < 12 and SI < 15 (low)

Sowers and Sowers SL < 10 and PI > 30 (high) 10 " SL " 1 2 and 15 ,,; PI " 30 ( moderate ) (1970) SL > 12 and PI < 15 (low) Van Der Merwe

Snethen (1984)

Chen (1988) McKeen (1992)

Weston (1980)

Based on plasticity chart Based on PI and SI

Little swell will occur when Wo results in LI of 0.25 Based on PI, percentage of clay 130 (very high) and 91 ,,; EI ,,; 130 (high) 51 ,,; EI ,,; 90 (medium) and 21 ,,; EI " 50 (low) o ,,; EI ,,; 20 (very low)

Based on oedometer test on com· pacted specimen with degree of saturation close to 50% and surcharge of 6.9 kPa

LL > 60, PI > 35, T,,, > 4, and SP > 1.5 (high) 30 ,,; LL ,,; 60,25 ,,; PI ,,; 35, 1.5 ,,; T," ,,; 4, and 0.5 ,,; SP ,,; 1.5 (medium) LL < 30. PI < 25, T,,, < 1.5, and SP < 0.5 (low) '" 35 (very high) and 20 ,,; PI ,,; 55 (high) 10 ,,; PI ,,; 35 (medium) and PI ,,; 15 (low)

PS is representative for field condition and can be used without ornat! but accuracy will be reduced

PI

Based on PI

Figure 13.13d

Based on measurements of soil water content, suction. and change in volume on drying

Vijayvergiya and log SP = (1/12) (0.44 LL - w, + 5.5) Ghazzaly (1973) Nayak and Christensen (1974)

Based on LS, SL, and PS Remolded sample (Pa(m,,) and w"",) soaked under 6.9 kPa surcharge

See Figure 13.13c

(1964)

Uniform Building Code, 1968

Based on oedometer test using compacted specimen, percent­ age of clay 50.8

50.8

Remarks

.

50.8 to 101.6

Building damping movement

Joints:

>101.6

Building independent of movement

Foundation drilled shaft:

Footings should be small and deep. consistent with the soilbearing capacity. Mats should resist bending. Slabs should be designed to resist bending and should be independent of grade beams. Walls on a mat should be as flexible as the mat. There should be no vertical rigid connections. Brickwork should be strengthened with tie bars or bands.

Contacts between structural units should be avoided. or flexible. waterproof material may he inserted in the joints. Walls: Walls or rectangular building Flexible units should heave as a unit. Unit construction Steel frame Foundations: Cellular foundations allow slight Three point soil expansion to reduce swelling Cellular pressure. Adjustable jacks can be lacks inconvenient to owners. Three­ point loading allows motion without duress. Clear Flexible

SmaUest-diameter and widely spaced shafts compatible with the load should be placed. Straight shaft BeU bottom Clearance should be aUowed under grade beams. Suspended floor: Floor should be suspended on grade beams 305 to 460 mm above the soil.

, After Gromko. 1974

!

13. 1 1

Construction on Expansive Soils

. .

663

Figure 13. 18 Waffle slab Dead load, D

T

Active .

zone, Z

1

Drilled shafts with bells lu)

Figure 13. 19 (a) Construction of drilled shafts with bells and grade beam; (b) definition of parameters in Eq. (13.14)

In most cases, the value of cJ>�, varies between 10 and 20°. An average value of the zero horizontal swell pressure must be determined in the laboratory. In the absence of laboratory results, crT tan cJ>�, may be considered equal to the undrained shear strength of clay, C," in the active zone. The belled portion of the drilled shaft will act as an anchor to resist the uplift­ ing force. Ignoring the weight of the drilled shaft, we have Qnot

=

U

-

D

(13.15)

where Qn"

D

=

=

Now,

net uplift load dead load cuN,

Qnel = FS

where Cu =

N, =

(1T)(Db - D, ) '4

2

2

undrained cohesion of the clay in which the bell is located bearing capacity factor

(13.16)

664

Chapter 13 Foundations on Difficult Soils

FS

= factor of safety Db = diameter of the bell of the drilled shaft

Combining Eqs. (13.15) and (13.16) gives

(13.17)

Conservatively, from Table 3.3 and Eq. (3.27), N, = N""';p)F"

=

( �:�)

NC("dP, 1

+

(

= 5.14 1 +

A drilled-shaft design is examined in Example 13.3.

�)

5. 4

=

6.14

Example 1 3.3 Figure 13.20 shows a drilled shaft with a bell. The depth of the active zone is 5 m . The zero swell pressure of the swelling clay (O' T) i s 450 kN/m2• For the drilled shaft, the dead load (D) is 600 kN and the live load is 300 kN.Assume ;,. = 12°. a. Determine the diameter of the bell, Db ' b Check the bearing capacity of the drilled shaft assuming zero uplift force. ..

Solution

Part a: Determining the Bell Diameter, Db

The uplift force, Eq. (13.14), is U

=

1TD,ZO'r tan ;,.

Dead load + live load = 900 kN

1

2m

I 2, = 8665 leN/m

.

(US')

Net bearing capacity of the soil under the bell = qu(n.') 7' C�

{= i7@��/m'

= (450)(6:14

Hence, the factor of safety against bearing capacity failure is FS

=

2763 866.5

=

3.19 > 3-0�

, .

.

•

Sanitary Landfills

. ..13. 12 ,

General Nature of Sanitary Landfills Sanitary landfills provide a way to dispose of refuse on land without endangering public health. Sanitary landfills are used in almost all countries, to varying degrees of success. The refuse disposed of in sanitary landfills may contain organic, wood,

666

Chapter 13 Foundations on Difficult Soils

Soil cover Excavation for soil

Figure 13.21 Schematic diagram of a sanitary landfill in progress

Original ground surface

paper, and fibrous wastes, or demolition wastes such as bricks and stones. The refuse is dumped and compacted at frequent intervals and is then covered with a layer of soil, as shown in Figure 13.21. In the compacted state, the average unit weight of the refuse may vary between 5 and 10 kN/mJ(32 to 64 Ib/ft3). A typical city in the United States. with a population of 1 million. generates about 3.8 x 106 m3 (=135 x 106ft3 ) of compacted landfill material per year. As property values continue to increase in densely populated areas, con­ structing structures over sanitary landfills becomes more and more tempting. In some instances. a visual site inspection may not be enough to detect an old sanitary landfill. However. construction of foundations over sanitary landfills is generally problematic because of poisonous gases (e.g., methane), excessive settlement, and low inherent bearing capacity.

Settlement of Sanitary Landfills Sanitary landfills undergo large continuous settlements over a long time. Yen and Scanlon (1975) documented the settlement of several landfill sites in California. After completion of the landfill, the settlement rate (Figure 13.22) may be ex­ pressed as (13.18) where m =

HI

=

settlement rate maximum height of the sanitary landfill

On the basis of several field observations, Yen and Scanlon determined the follow­ ing empirical correlations for the settlement rate: m =

0.0268 - 0.0116 log II 0.038 - 0.0155 log I I

m =

0.0433 - 0.0183 log II

m =

(for fill heights ranging from 12 to 24 m ) (13.19) ( for fill heights ranging from 24 to 30 m) (13.20) (13.21) (for fill heights larger than 30 m)

l

13. 13 Settlement of Sanitary Landfills

667

Height of sanitary landfill

aHf

� , ,

i+-+-- t l�

'--L.-'-

_

__

at

, ,

--''--_�_... + Ie Ie: = til + Ie

t = It

_ _ _

Time. I

Figure 13.22 Settlement of sanitary landfills

In these equations, m

is in m/mo.

I, is the median fill age, in manths The median fill age may be defined from Figure 13.22 as

Ie I, = I - 2

(13.22)

where

I = time from the beginning of the landfill Ie = time for completion of the landfill Equations (13.19), (13.20), and (13.21) are based on field data from landfills for which Ie varied from 70 to 82 months. To get an idea of the approximate length of time required for a sanitary landfill to undergo complete settlement, consider Eq. (13.19). For a fill 12 m high and for I, = 72 months, m

= 0.0268 - 0.0116 log I,

so log I, =

0.0268 - m 0.0116

If m = 0 (zero settlement rate), log I, = 2.31, or I, = 200 months. Thus, settlement will continue for I, - IJ2 = 200 - 36 = . 164 months (=14 years) after completion of the fill-a fairly long time. This calculation emphasizes the need to pay close at­ tention to the settlement of foundations constructed on sanitary landfills. A comparison of Eqs. (13.19) through (13.21) for rates of settlement shows that the value of m increases with the height of the fill. However, for fill heights greater than about 30 m, the rate of settlement should not be much different from that obtained from Eq. (13.21). The reason is that the decomposition of organic

668

Chapter 13 Foundations on Difficult Soils

matter close to the surface is mainly the result of an anaerobic environment. For deeper fills, the decomposition' is slower. Hence, for fill heights greater than about 30 m, the rate of settlement does not exceed that for fills that are about 30 m in height. In English units, Eqs. (13.19)-(13.21) can be expressed in the following form: (for fill heights 40 to 80 ft) m = 0.088 - 0.038 log I, (13.23)

0.125 - 0.051 log I, (13.24) (for fill heights 80 to 100 ft) (for fill heights greater than 100 ft) (13.25) m = 0.142 - 0.06 log I, Sowers (1973) also proposed a formula for calculating the settlement of a san­ m =

itary landfill, namely,

,

, , . ' aHt 4Hj =

1+e

--

(t")' t' ,

'

(13.26)

log -

where

Hf

e =

=

a =

1 " , 1' , =

AHf

=

height of the fill void ratio a coefficient for settlement times (see Figure 13.22) settlement between times I' and I"

The coefficients a fall between a =

0.0ge (for conditions favorable to decomposition)

(13.27)

and

0.03e (for conditions unfavorable to decomposition) (13.28) Equation (13.26) is similar to the equation for secondary consolidation settlement [Equation (5.78)]. a =

13.1 For leossial soil, given G, 2.74. Plot a graph of "Yd (kN/m3) versus liquid limit to identify the zone in which the soil is likely to collapse on saturation. If a soil has a liquid limit of 27, G, = 2.74, and "Yd 14.5 kN/m3, is collapse likely to occur? 13.2 A collapsible soil layer in the field has a thickness of 3 m. The average effective overburden pressure on the soil layer is 62 kN/m2.An undisturbed specimen of this soil was subjected to a double oedometer test. The preconsolidation pres­ sure of the specimen as determined from the soaked specimen was 84 kN/m'. Is the soil in the field normally consolidated or preconsolidated? =

=

Problems

669

13.3 An expansive soil has an active-zone thickness of 10 ft. The natural moisture

content of the soil is 20% and its liquid limit is 50. Calculate the free surface swell of the expansive soil upon saturation. 13.4 An expansive soil profile has an active-zone thickness of 12 ft. A shallow fourrdation is to be constructed at a depth of 4 ft below the ground surface. Based on the swelling pressure test, the following are given. Depth from ground surface (It)

4 6 8 10 12

Swell under overburden and estimated foundation surcharge pressure. SwU} (%)

4.75 2.75 1.5 0.6 0.0

Estimate the total possible swell under the foundation. 13.5 Refer to Problem 13.4. If the allowable total swell is 1 in., what would be the necessary undercut? 13.6 Repeat Problem 13.4 with the following: active zone thickness = 6 m; depth of shallow foundation = 1.5 m. Depth from ground surface (m)

1.5 2� 3� 4.0 5.0 6.0

Swell under overburden and estimated foundation surcharge pressure, Sw(ll (%)

5.5 �I 1.5 0.75 0.4 0.0

13.7 Refer to Problem 13.6. If the allowable total swell is 30 mm, what would be

the necessary undercut?

13.8 Refer to Figure 13.19b. For the drilled shaft with bell, given:

Thickness of active zone, Z = 9 m Dead load = 1500 kN Live load = 300 kN Diameter of the shaft, D, = 1 m Zero swell pressure for the clay in the active zone = 600 kN(m' Average angle of plinth-soil friction, ",�, = 20° Average undrained cohesion of the 'clay around the bell = 150 kN(m' Determine the diameter of the bell, Db' A factor of safety aD against uplift is required with the assumption that dead load plus live load is equal to zero. 13.9 Refer to Problem 13.8. If an additional requirement is that the factor of safety against uplift is at least 4 with the dead load on (live load = 0), what should be the diameter of the bell?

670

Chapter 13 Foundations on Difficult Soils

References Abduljauwad. S. N.. and AI-Sulaimani. G. J. (1993). "Determination of Swell Potential of Al­ Qatif Clay," Geotechnical Testing Journal, American Society for Testing and Materi­ als,VoI. 16, No. 4, pp. 469-484. Altmeyer, W. T. (1955). "Discussion of Engineering Properties of Expansive Clays,"Journal of the Soil Mechanics 'tmd Foundations Division, American Society of Civil Engineers, Vol. 81, No. SM2, pp. 17-19. Benites. L.A. (1968). "Geotechnical Properties of the Soils Affected by Piping near the Ben­ son Area, Cochise County, Arizona," M. S. Thesis, University of Arizona, Theson. Chen, F. H. (1988). Foundations on Expansive Soils, Elsevier, Amsterdam. Clemence, S. P., and FinbaIT,A. 0. (1981). "Design Considerations for Collapsible Soils," Journal ofrhe Geotechnical Engineering Division, American Society of Civil Engineers, Vol. 107. No. GT3, pp.305-317. Clevenger, W. (1958). "Experience with Loess as Foundation Material," Transactions, American Society of Civil Engineers, Vol. 123, pp. 151-170. Dakshanamanthy. V" and Raman, V. (J 973). "A Simple Method of Identifying an Expansive Soil." Soils and Foundations. Vol. 13. No. 1 , pp. 97-104. Denisov. N. Y. (1951). The Engineering Properties of Loess and Loess Loams, Gosstroiiz­ dat. Moscow. Fcda. J. (1964). "Colloidal Activity. Shrinking and Swelling of Some Clays," Proceedings, Soil A1echanics Seminar. Loda. IUinois. pp. 531-546. Gihbs. H. J. (1961). Properties Which Dil'ide Loose alld Dellse Uncel11el1ted Soils, Earth Labo­ ratory Report EM-658,Bureau of Reclamation, U.S. Department of the Interior, Wash­ ington. DC. Gromko. G. J. (1974). "Review of Expansive Soils." Journal of the Geotechllical Engineering Division, American Society of Civil Engineers, Vol. 100, No. GT6, pp. 667-687. Handy, R. L. (1973). "Collapsible Loess in Iowa," Proceedings, Soil Science Society ofAmer­ ica, Vol. 37, pp. 281-284. Holtz, W. G. (1959). "Expansive Clays-Properties and Problems," Journal of the Colorado School ofMines, Vol. 54, No.4, pp. 89-125. Holtz, W. G., and Hilf, J. W. (1961). "Settlement of Soil Foundations Due to Saturation," Pro­ ceedings, Fifth International Conference on Soil Mechanics and Foundation Engineer­ ing, Paris, Vol. !, 1961, pp. 673-679.

I

Houston, W. N., and Houston, S. L. (1989). "State-of-the-Practice Mitigation Measures for Collapsible Soil Sites," Proceedings, Foundation Engineering: Current Principles and Practices, American Society of Civil Engineers, Vol. 1 , pp. 161-175. Jennings, J. E., and Knight, K. (1975). "A Guide to Construction on or with Materials Exhibit­ ing Additional Settlements Due to 'Collapse' of Grain Structure," Proceedings, Sixth Regional Conference for Africa on Soil Mechanics and Foundation Engineering, Johan­ nesburg, pp. 99-105. Lutenegger, A. J. (1986). "Dynamic Compaction in Friable Loess," Journal of Geotechnical Engineering, American Society of Civil Engineers, Vol. 1 12, No. GT6, pp. 663-667. Lutenegger, A. 1. and Saber, R. T. (1988). "Determination of Collapse Potential of Soils," Geotechnical Testing Journal, American Society for Testing and Materials. Vol. 1 1 . No.3, pp. 173-178. McKeen, R. G. (1992). "A Model for Predicting Expansive Soil Behavior." Proceedings, Sev­ el1th International Conference on Expansive Soils, Dallas, Vol. I . pp. 1-6. Nayak. N. V., and Christensen, R. W. (1974). "Swell Characteristics of Compacted Expansive Soils," Clay and Clay Minerals, Vol. 19. pp. 251-261.

I

t.

l

References

671

O'Neill, M. W., and Poormoayed, N. (1980). "Methodology for Foundations On Expansive Clays," Journal of the Geotechnical Engineering Division, American Society of Civil Engineers, Vol. 106, No. GTI2, pp. 1345-1367. Peck, R. R, Hanson, W. E., and Thornburn,T. R (1974). Foundation Engineering, W!ley, New York. Priklonskii, V. A. (1952). Gruntovedenic, Vtoraya Chast', Gosgeolzdat, Moscow. Raman, V. (1967). "Identification of Expansive Soils from the Plasticity Index and the Shrink­ age Index Data," The Indian Engineer, Vol. 11, No. 1, pp. 17-22. Seed, H. R, Woodward, R. l, Jr., and Lundgren, R. (1962). "Prediction of Swelling Potential for Compacted Clay�" Journal ofthe Soil Mechanics and Foundations Division, Amer­ ican Society of Civil Engineers, Vol. 88, No. SM3, pp. 53-87. Semkin, V. v., Ermoshin, V. M., and Okishev, N. D. (1986). "Chemical Stabilization of Loess Soils in Uzbekistan," Soil Mechanics and Foundation Engineering (trans. from Russ­ ian), Vol. 23, No. 5, pp. 196-199. Snethen, D. R. (1984). "Evaluation of Expedient Methods for Identification and Classifica­ tion of Potentially Expansive Soils," Proceedings, Fifth International Conference on Ex­ pansive Soils, Adelaide, pp. 22-26. Snethen, D. R., Johnson, L. D., and Patrick, D. M. (1977). An Evaluation ofExpedient Method­ ology for Identification of Potentially Expansive Soils, Report No. FHWA-RD-77-94, U.S. Army Engineers Waterways Experiment Station, Vicksburg, MS. Sowers, G. F. (1973). "Settlement of Waste Disposal Fills," Proceedings, Eighth International Conference on Soil Mechanics and Foundation Engineering, Moscow, pp. 207-210. Sowers, G. B., and Sowers, G. F. (1970). Introductory Soil Mechanics and Foundations, 3d ed. Macmillan, New York. Sridharan, A. (2005). "On Swelling Behaviour of Clays," Proceedings, International Confer­ ence on Problematic Soils, North Cypru� Vol. 2, pp. 499-516. Sridharan,A.. Rao.A. S.. and Sivapullaiah, P. V. (1986). "Swelling Pressure of Clays." Geotechni­ cal Testing Journal. American Society for Testing and Materials, Vol. 9. No. 1. pp. 2-1--33. Uniform Building Code (1968). UBC Standard No. 29-2. Van Der Merwe, D. H. (1964), "The Prediction of Heave from the Plasticity Index and Per­ centage Clay Fraction of Soils." Civil Engineer in South Africa, Vol. 6, No. 6, pp. 103-106. Vijayvergiya. V. N.. and Ghazzaly. O. l. (1973). "Prediction of Swelling Potential of Natural Clays;' Proceedings. Third International Research and Engineering Conference on Ex­ pansive Clays. pp. 227-234. Weston, D. l (1980). "Expansive Roadbed Treatment for Southern Africa," Proceedings, Fourth International COllference on Expansive Soils, Vol. 1, pp. 339-360. Yen, R C, and Scanlon, R (1975). "Sanitary Landfill Settlement Rates," Journal of the Geot­ echnical Engineering Division. American Society of Civil Engineers, Vol. 101, No. GT5, pp. 475-487.

14

Soil Improvement and Ground Modification

14. 1

Introduction The soil at a construction site may not always be totally suitable for supporting structures such as buildings. bridges. highways. and dams. For example. in granular soil deposits. the in sitlt soil may be very loose and indicate a large elastic settlement. In such a case, the soil needs to be densified to increase its unit weight and thus its shear strength. Sometimes the top layers of soil are undesirable and must be removed and re­ placed with better soil on \\'hich the structural foundation can he huilt. 1l1e soil used as fill should be well compacted to sustain the desired structural load. Compacted fills may also be required in low-lying areas to raise the ground elevation for con­ struction of the foundation. Soft saturated clay layers are often encountered at shallow depths below foun­ dations. Depending on the structural load and the depth of the layers, unusually large consolidation settlement may occur. Special soil-improvement techniques are required to minimize settlement. In Chapter 13, we mentioned that the properties of expansive soils could be al­ tered substantially by adding stabilizing agents such as lime. Improving in situ soils by using additives is usually referred to as stabilization. Various teChniques are used to 1. Reduce the settlement of structures 2. Improve the shear strength of soil and thus increase the bearing capacity of shallow foundations 3. Increase the factor of safety against possible slope failure of embankments and earth dams 4. Reduce the shrinkage and swelling of soils

This chapter discusses some of the general principles of soil improvement, such as compaction, vibroflotation, precompression, sand drains, wick drains. stabilization by admixtures,jet grouting, and deep mixing, as well as the use of stone columns and sand compaction piles in weak clay to construct foundations.

672

14.2

General Principles of Compaction

673

General Principles of Compaction If a small amount of water is added to a soil that is then compacted, the soil will have a certain unit weight. If the moisture content of the same soil is gradually increased and the energy of compaction is the same, the dry unit weight of the soil will gradually increase. The reason is that water acts as a lubricant between the soil particles, and under compaction it helps rearrange the solid particles into a denser state. The increase in dry unit weight with increase of moisture content for a soil will reach a limiting value beyond which the further addition of water to the soil will result in a reduction in dry unit weight. The moisture content at which the maximum dry unit weight is obtained is referred to as the optimum moisture content.

The standard laboratory tests used to evaluate maximum dry unit weights and optimum moisture contents for various soils are The Standard Proctor test (ASTM designation D-698) The Modified Proctor test (ASTM designation D-1557)

•

•

The soil is compacted in a mold in several layers by a hamm,?r. The moisture content of the soil, W, is changed, and the dry unit weight, 1'd, of compaction for each test is deter­ mined. The maximum dry unit weight of compaction and the corresponding optimum moisture content are determined by plotting a graph of 1'd against w (%). The standard specifications for the two types of Proctor test are given in Tables 14.1 and 14.2. Table 14. 1

Specifications for Standard Proctor Test (Based on ASTM Designation 698)

Item

Method A

Method B

Method C

Diameter of mold Volume of mold Weight of hammer Height of hammer drop Number of hammer blows per layer of soil Number of layers of compaction Energy of compaction

101.6 mm (4 in.) 944 cm'('\; ft') 2.5 kg (5.5 lb) 304.8 mm (12 in.)

101.6 mm (4 in.) 944 cm'('\; ftl) 2.5 kg (5.5 lb) 304.8 mm (12 in.)

152.4 mm (6 in.) 2124 cm'(0.075 fI') 2.5 kg (5.5 lb) 304.8 mm (12 in.)

25

25

56

3

3

3

Soil to be used

600 kN' m/m' 600 kN· m/m' 600 kN ' m/m' (12,400 ft' lb/ft') ( 12,400 ft 'lb/ft') (12,400 ft · lb/ft') Portion passing �-in. Portion passing �-in. Portion passing No. 4 (19.0-mm) sieve. (9.5-mm) sieve. (4.57-mm) sieve. May be used if more May be used if soil May be used if 20% or than 20% by weight of retaint:!d on No. 4 sieve less by weight of material is more than 20% and material is retained on is retained on No. 4 sieve. 20% or less by weight 9.5 mm a in.) sieve is retained on 9.5-rnm and less than 30% by a-in.) sieve. weight is retained on 19.oo-mm ()-in.) sieve.

674

Chapter 14 Soil Improvement and Ground Modification Table 14.2

Specifications for Modified Proctor Test (Based on ASTM Designation 1557)

Item

Method A

Method B

Method C

Diameter of mold Volume of mold Weight of hammer Height of hammer drop Number of hammer blows per layer of soil Number of layers of compaction Energy of compaction Soil to be used

101.6 mm (4 in.) 944 em' (to ft') 4.51 kg (10 lb) 457.2 mm (18 in.)

101.6 mm (4 in.) 944 em' (:I;; ft') 4.54 kg (10 lb) 457.2 mm (18 in.)

152.4 mm (6 in.) 2124 em' (0.075 ft') 4.54 kg (10 lb) 457.2 mm (18 in.)

25

25

56

5

5

5

2700 kN'm/m' 2700 kN' m/m' 2700 kN· mlm' (56.000 ft 'lb/ft') (56.000 ft' lb/ft') (56,000 ft · lb/ft') Portion passing Portion passing 19.0-mm Portion passing 9.5-mm (i-in.) sieve. (j ·in.) sieve. No.4 (4.57-mm) May be used if more sieve. May be used May be used if soil if 20% or less by retained on No. 4 than 20% by weight sieve is m ore than weight of material of material is retained is retained on 20% and 20% or on 9.5-mm (i-in.) No.4 sieve. sieve and less than less by weight is re� tained on 9.5-mm 30% by weight is (� -in.) sieve. retained on 19-mm d-in.)sieve.

Figure 14.1 shows a plot of 'Yd against w (%) for a clayey silt obtained from stan­ dard and modified Proctor tests (method A). The following conclusions may be drawn: 1. The maximum dry unit weight and the optimum moisture content depend on

the degree of compaction. 2. The higher the energy of compaction, the higher is the maximum dry unit weight. 3. The higher the energy of compaction, the lower is the optimum moisture content. 4. No portion of the compaction curve can lie to the right of the zero-air-void line. The zero-air-void dry unit weight, 'Y"", at a given moisture content is the theo­ retical maximum value of 'Yd, which means that all the void spaces of the com­ pacted soil are filled with water, or 'Y'"

=

'Yw -:-'--"'1 - + w G,

(14.1)

where 'Y... =

unit weight of water G, = specific gravity of the soil solids w = moisture content of the soil

S. The maximum dry unit weight of compaction and the corresponding optimum moisture content will vary from soil to soil.

I

L

14.2 General Principles of Compaction

675

24

..

22

�

Zero-air-void curve

20

(G, = 2.7)

3

� 18 .i .. ·u � 16 '2

� 0 =

14 •

12

Standard Proctor test

... Modified Proctor test Figure 14. 1

10 0

5

10

15

20

25

Moisture content, W (%)

Standard and modified Proctor compaction curves for a clayey silt (method A)

Using the results of laboratory compaction (1'" versus 10). specifications may be written for the compaction of a given soil in the field. In most cases, the contrac· tor is required to achieve a relative compaction of 90% or more on the basis of a specific laboratory test (either the standard or the modified Proctor compaction test). The relative compaction is defined as

RC '"

1'd(",ld) 'Yd(max)

(14.2)

Chapter 1 introduced the concept of relative density (for the compaction of granular soils), defined as

where Yd = 1'd(m,,) =

Yd( min) =

dry unit weight of compaction in the field maximum dry unit weight of compaction as determined in the laboratory minimum dry unit weight of compaction as determined in the laboratory

676

Chapter 14 Soil Improvement and Ground Modification

For granular soils in the field, the degree of compaction obtained is often measured in terms of relative density. Comparing the expressions for relative density and rela­ tive compaction reveals that

RC =

A 1 - Dr ( 1

- A)

(14.3)

Yd(mill)

where .A = --. Yt/{max)

Lee and Singh (1971) reviewed 47 different soils and. on the basis of their re­ view, presented the following correlation:

Dr ( % )

=

(RC �---: --,:--'80)

0.2

(14.4)

Gurtug and Sridharan (2004) analyzed a number of laboratory-compaction test results on fine-grained (cohesive) soils. Based on this study. the following correla­ tions were developed:

=

11""", "

Yd(me>J

: 1 .�5 '· O.J8 Iog ( CT ) ; ( PL )

[0.5

+ 0.173 10g(CE )] Yd(PL)

Yd(m"J = 22.26 - 0.28 1.

0

o •

o 1 0-8

•

l::..

A

•

Soil A, Standard Proctor Soil A, Modified Proctor Soil B, Standard Proctor Soil B, Modified Proctor Soil C, Standard Proctor Soil C. Modified Proctor

•

10-9 +----,---,----.,.---'1 40 70 90 80 60 50 1 00

Degree of saluration ('K)

Figure 14. 1 1 Effect of degree of saturation on hydraulic conductivity of Wisconsin A. B. and C soils (After Othman and Luettich. 1 994)

of the grain-size distribution of in siCl/ soil marked Zone 1 in Figure 14.15 is most suit­ able for compaction by vibrotlotation. Soils that contain excessive amounts of fine sand and silt-size particles are difficult [0 compact; for such soils. considerable effort is needed to reach the proper relative density of compaction. Zone 2 in Figure 14.15 is the approximate lower limit of grain-size distribution for compaction by vibrotlota­ tion. Soil deposits whose grain-size distribution falls into Zone 3 contain appreciable amounts of gravel. For these soils, the rate of probe penetration may be rather slow, so compaction by vibrotlotation might prove to be uneconomical in the long run. The grain-size distribution of the backfill material is one of the factors that control the rate of densification. Brown (1977) defined a quantity called suitability number for rating a backfill material. The suitability number is given by the formula (14.11) where D50, D,o, and DiO are the diameters (in mm) through which 50%,20%,and 10%, respectively, of the material is passing. The smaller the value of SN, the more desirable is the backfill material. Following is a backfill rating system proposed by Brown (1977): Range of SN

0-10 10-20 20-30 30-50 '>50

Rating as backfill

Excellent Good Fair Poor Unsuitable

14.6 Vibroflotation

687

Power supply

Follow-up

4Vibrating unit

Cylinder of compacted material. added from the suface to compensate for the loss of volume caused by the increase in density of the compacted soil B

Cylinder of compacted material, produced by a single vibroflot compaction

Figure 14. 12 Vibroflotation unit ( After Brown, 1977)

An excellent case study that evaluated the benefits of vibroflotation was pre­ sented by Basore and Boitano (1969). Densification of granular subsoil was neces­ sary for the construction of a three-story office building at the Treasure Island Naval Station in San Francisco, California. The top 9 m (=30 ft) of soil at the site was loose to medium-dense sand fill that had to be compacted. Figure 14.16a shows the nature of the layout of the vibroflotation points. Sixteen compaction points were arranged in groups of four, with 1.22 m (4 ft), 1.52 m (5 ft), 1.83 m (6 ttl, and 2.44 m (8 tt) spacing.

688

Chapter 14 Soil Improvement and Ground Modification

,1

Slep I Figure 14. 13

Slep 2

Slep 3

SIep 4

Compaction by the vibrofio,"Iion process (After Brown. 1977)

Prior to compaction. standard penetration tests were conducted at the centers of groups of three compaction points. After the completion of compaction by vi­ broflotation. the variation of the standard penetration resiswnce with depth was determined at the same points. Figure 14.16b shows the variation of standard penetration resistance, N60, with depth before and after compaction for vibroflotation point spacings S' = 1.22 m (4 ft) and 2.44 m (8 ft). From this figure, the following general conclusions can be drawn: o

o o

For any given S', the magnitude of N60 after compaction decreases with an in­ crease in depth. An increase in N60 indicates an increase in the relative density of sand. The degree of compaction decreases with the increase in S'. At S' = 1.22 m (4 ft), the degree of compaction at any depth is the largest. However.at S' = 2.44 m (8 fi), the vibroflotation had practically no effect in compacting soil.

During the past 30 to 35 years. the vibroflotation technique has been used suc­ cessfully on large projects to compact granular subsoils. thereby controlling struc­ tural settlement. 14. 7

Precompression When highly compressible. normally consolidated clayey soil layers lie at a limited depth and large consolidation settlements are expected as the result of the construc­ tion of large buildings, highway embankments. or earth dams. precompression of soil may be used to minimize postconstruction settlement. The principles of precompres­ sion are best explained by reference to Figure 14.17. Here, the proposed structural

689

14.7 Precompression Table 14.4

Types of Vibrating Units' 100-HP electric and hydraulic motors

30-HP electric motors

(a) Vibrating tip Length Diameter Weight Maximum movement when free Centrifugal force

2.1 m (7 ft) 406 mm (16 in.) 18 kN (4000 lb) 12.5 mm (0.49 in.) 160 kN (18 ton)

1.86 m (6.11 ft) 381 mm (15 in.) 1 8 kN (4000 lb) 7.6 mm (0.3 in.) 90 kN (10 ton)

(b) Eccentric Weight Offset Length Speed

1.16 kN (260 Ib) 38 mm (1.5 in.) 610 mm (24 in.) 1800 rpm

0.76 kN (170 Ib) 32 mm (1.25 in.) 387 mm (15.3 in.) 1800 rpm

(c) Pump Operating flow rate Pressure

0-1.6 m'/min (0-400 gal/min) 690-1035 kN/m' (1oo-150 Ib/in')

0-6 m'/min (0-150 gal/min) 690-1035 kN/m' (loo-150 Ib/in')

(d) Lower follow-up pipe and extensions Diameter Weight

305 mm (12 in.) 3.65 kN/m (250 Ib/ft)

305 mm (12 in.) 3.65 kN/m (250 Ib/ft)

' After Brown (1977)

load per unit area is Ll.O"(p), and the . thickness of the clay layer undergoing consolidation is H,. The maximum primary consolidation settlement caused by the structural load is then

(14.12)

690

Chapter 14 Soi/ /mprovement and Ground Modification Probe

Zone of influence for each probe Figure 14. 14

Nature of probe spacing for vibroflotation

100

80

�

� c il·1 --... .,.... ;...-" , ,," )0

Clay

'l i....-_ �... II...

, '1 I . • L J , L .J , " ,r , 1 L L , I .'

I .1

-,\n--\. ¥ �

.

Smeared :.-�one Sand -drain

..." "

"

....-'1 ..

'-I

Figure 14.22 Schematic diagram of a sand drain

Surcharge per unit area

Surcharge per unit area

Au,pJ + liu(,)

'------ Time (a)

�--�-----+ TIme " (h)

Figure 14.23 Nature of application of surcharge

(see Figure 14.23a), or (b) the surcharge is applied in the form of a ramp load (see Fig­ ure 14.23b). When the entire surcharge is applied instantaneously (Barron, 1948), (14.19) where (14.20) in which

de 'e n==-

s=

2rw r, -

'w

'w

(14 .21) (14.22)

14.8 Sand Drains

699

and kh = hydraulic conductivity of clay in the horizontal direction in the unsmeared zone k, = horizontal hydraulic conductivity in the smeared zone T, = nondimensional time factor for radial drainage only =

C",

= coefficient of consolidation for radial drainage

�

(14.23)

Cv ,

d,

(14.24)

For a no-smear case, r, = rw and kh = k" so S = 1 and Eq. (14.20) becomes

(14.25) Table 14.5 gives the values of U, for various values of T, and n If the surcharge is applied in the form of a ramp and there is no smear, then (Olson, 1977)

1

U,

T, - [I - exp ( - AT,)] A ----- ---'-'= T"

(14.26)

and (14.27) where T" = and

�

Cv '

d,

(see Figure 14.23b for the definition of t,) 2 A=­ m

(14.28)

(14.29)

Average Degree of Consolidation Due to Vertical Drainage Only

Using Figure 14.23a, for instantaneous application of a surcharge, we may obtain the average degree of consolidation due to vertical drainage only from Eqs. (1.60) and (1.61). We have

7r[Uv(%)]' (torU,

Tv = 4"

100

= O to 60%)

[Eq. (1.60)]

\

,

I

700

Chapter 14 Soil Improvement and Ground Modification Table 14.5

Variation of U, for Various Values of T, and n, No-Smear Case [Eqs. (I4.19) and (14.25)]

Degree of consolidation U, (%)

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44

5

0 0.0012 0.0024 0.0036 0.0048 0.0060 0.0072 0.0085 0.0098 0.0110 0.0123 9·0136 0.0150 0.0163 0.0177 0.0190 0.0204 0.0218

0.0232

0.0247 0.0261 0.0276 0.0291 0.0306 0.0321 0.0337 0.0353 0.0368 0.0385 0.0401 0.0418 0.0434 0.0452 0.0469 0.0486 0.0504 0.0522 0.0541 0.0560 0.0579 0.0598 0.0618 0.0638 0.0658 0.0679

lime factor Tr for value of n (= ,./rw) 10

15

0 0.0020 0.0040 0.0060 0.0081 0.0101 0.0122 0.0143 0.0165 0.0186 0.D208 0.0230 0.0252 0.0275 0.0298 0.0321 0.0344 0.0368 0.0392 0.0416 0.0440 0.0465 0.0490 0.0516 0.0541 0.0568 0.0594 0.0621 0.0648 0.0676 0.0704 0.0732 0.0761 0.0790 0.0820 0.0850 0.0881 0.0912 0.0943 0.0975 0.1008 0.1041 0.1075 0.1109 0.1 144

0 0.0025 0.0050 0.0075 0.0101 0.0126 0.0153 0.0179 0.0206 0.0232 0.0260 0.0287 0.0315 0.0343 0.0372 0.0401 0.0430 0.0459 0.04R9 0.0519 0.0550 0.0581 0.0612 0.0644 0.0676 0.0709 0.0742 0.0776 0.0810 0.0844 0.0879 0.0914 0.0950 0.0987 0.1024 0.1062 0.1 100 0.1139 0.1 1 78 0.1218 0.1259 0.1300 0.1342 0.1385 0.1429

a

20

0.0028 0.0057 0.0086 0.0115 0.0145 0.0174 0.D205 0.0235 0.0266 0.0297 0.0328 0.0360 0.0392 0.0425 0.0458 0.0491 0.0525 0.0559

0.0594 0.0629 0.0664 0.0700 0.0736 0.0773 0.0811 0.0848 0.0887 0.0926 0.0965 0.1005 0.1045 0.1087 0.1128 0.1171 0.1214 0.1257 0.1302 0.1347 0.1393 0.1439 0.1487 0.1535 0.1584 0.1634

25

0 0.0031 0.0063 0.0094 0.0126 0.0159 0.0191 0.0225 0.0258 0.0292 0.0326 0.0360 0.0395 0.0431 0.0467 0.0503 0.0539 0.0576 0.0614 0.0652 0.0690 0.0729 0.0769 0.0808 0.0849 0.0890 0.0931 0.0973 0.1016 0.1059 0.1103 0.1148 0.1193 0.1239 0.1285 0.1332 0.1380 0.1429 0.1479 0.1529 0.1580 0.1632 0.1685 0.1739 0.1793 ( Continued)

I

I

l ..

t

l ,

14.8 Sand Drains Table 14.5 -,

, ...-

( Continued)

Degre8 of consolidation U, (%)

/'

701

5

Time factor T, for value of n (= '.trw) 10

15

20

25 0.1849

45

0.0700

0.1180

0.1473

0.1684

46

0.0721

0.1216

0.1518

0.1736

0.1906

47

0.0743

0.1253

0.1564

0.1789

0.1964

48

0.0766

0.1290

0.1611

0.1842

0.2023

49

0.0788

0.1329

0.1659

0.1897

0.2083

50

0.0811

0.1368

0.1708

0.1953

0.2144

51

0.0835

0.1407

0.1758

0.2020

0.2206

52

0.0859

0.1448

0.1809

0.2068

0.2270

53

0.0884

0.1490

0.1860

0.2127

0.2335

54

0.0909

0.1532

0.1913

0.2188

0.2402

55

0.0935

0.1575

0.1968

0.2250

0.2470

56

0.0961

0.1620

0.2023

0.2313

0.2539

57

0.0988

0.1665

0.2080

0.2378

0.2610 0.2683

58

0.1016

0.1712

0.2138

0.2444

59

0.1044

0.1759

0.2197

0.2512

0.2758

60

0.1073

0.1808

0.2258

0.2582

0.2834

61

0.1102

0.1858

0.2320

0.2653

0.2912

62

0.1133

0.1909

0.2384

0.2726

0.2993

63

0.1164

0.1962

0.2450

0.2801

0.3075

64

0.1196

0.2016

0.2517

0.2878

0.3160

65

0. 1 229

0.2071

0.2587

0.2958

0.3247

66

U. 1263

U.2128

0.2658

U.3039

0.3337

67

0.1298

0.2187

0.2732

0.3124

0.3429

68

0.1334

0.2248

0.2808

0.3210

0.3524

69

0.1371

0.2311

0.2886

0.3300

0.3623

70

0.1409

0.2375

0.2967

0.3392

0.3724

71

0.1449

0.2442

0.3050

0.3488

0.3829

72

0.1490

0.2512

0.3134

0.3586

0. 3937

73

0.1533

0.2583

0.3226

0.3689

0.4050

74

0.1577

0.2658

0.3319

0.3795

0.4167

75

0.1623

0.2735

0.3416

0.3906

0.4288

76

0.1671

0.2816

0.3517

0.4021

0.4414

77

0.1720

0.2900

0.3621

0.4141

0.4546

78

0.1773

0.2988

0.3731

0.4266

0.4683

79

0.1827

0.3079

0.3846

0.4397

0.4827

80

0.1884

0.3175

0.3966

0.4534

0.4978

81

0.1944

0.3277

0.4090

0.4679

0.5137

82

0.2007

0.3383

0.4225

0.4831

0.5304

83

0.2074

0.3496

0.4366

0.4992

0.5481

84

0.2146

0.3616

0.4516

0.5163

0.5668

85

0.2221

0.3743

0.4675

0.5345

0.5868

86

0.2302

0.3879

0.4845

0.5539

0.6081

87

0.2388

0.4025

0.5027

0.5748

0.6311

88

0.2482

0.4183

0.5225

0.5974

0.6558

89

0.2584

0.4355

0.5439

0.6219

0.6827

90

0.2696

0.4543

0.5674

0.6487

0.7 122

( COnlinued)

702

Chapter 14 Soil Improvement and Ground Modification Table 14.5 (Continued)

Time factor T, for value of n (= '.trw)

Degree of consolidation U, (%)

5

10

15

20

25

91 92 93 94 95 96 97 98 99

0.2819 0.2957 0.3113 0.3293 0.3507 0.3768 0.4105 0.4580 0.5391

0.4751 0.4983 0.5247 0.5551 0.5910 0.6351 0.6918 0.7718 0.9086

0.5933 0.6224 0.6553 0.6932 0.7382 0.7932 0.8640 0.9640 1.1347

0.6784 0.7116 0.7492 0.7927 0.8440 0.9069 0.9879 1.1022 1.2974

0.7448 0.7812 0.8225 0.8702 0.9266 0.9956 1.0846 1.2100 1.4244

and T,

=

1 .781 - 0.933 log ( 100 - U,.(%»

(for U, > 60% ) [Eq. (1.61)]

where U,. = average degree of consolidation due to vertical drainage only, and

= H'

C,.l2

T,. C,. =

[Eq. (1.55)]

where coefficient of consolidation for vertical drainage. For the case of ramp loading, as shown in Figure 14.23b, the variation of U,. with T, can be expressed as (Olson, 1977): (14.30)

U, where

1r

2 (2m' + 1 ) m' = 0, 1, 2, . . . M

=

=

1-

1 ' , k T M, [exp(-M TJ - l ]exp(-M T ) 2

'

,

(14.31)

(14.32) where H = length of maximum vertical drainage path. Figure 14.24 shows the vari­ ation of Uv( % ) with T; and T,. Aboshi and Monden (1963) provided details on the field performance of 2700 sand drains used to construct the Toya Quay Wall on reclaimed land in Japan. Re­ sults of this study were summarized by Johnson (1970b) and this analysis is recom­ mended for study.

1

l

14.8 Sand Drains 0

•

10

�

C 0 .'"

;:;

�

§ U

� 0

20 30 40

50 60

�"" 70 Q 80 go 100 0.01

0.1

Time Factor, Tv

10

1.0

Figure 14.24 Variation of U, with T, and 7; [Eqs. (14.30) and (14.31)]

14.1, with the ad.;lition of sot�.e :��.;I.;.gr�ijl1li£�! " Example ,m, d� = 3 m, Cv = C"'t). and (S�e. Fiigu·1·e 14.23a.) Also assume LUauu,"

,.

n

=

d,

.

2rw = 2

3:

X

0.1

.

=

15

703

704

Chapter 74 Soil Improvement and Ground Modification

Again. From Table 14.5 for n U,.,

=

T, =

C"t,

d; =

15 and T,

=

(0.36) (9) ' = 0.36 (3)

0.36, the value of U, is about 77%, Hence,

= 1 - ( 1 - U, ) (1 - U,) = 0.924 = 92.4%

Now, from Figure 14.18, for ila�!a:, ilai/ila� = 0.12. Hence, ilain

=

=

=

1 - ( 1 - 0.67 ) { 1 - 0.77)

0.548 and U" .,

(115) (0.12)

=

=

92.4%, the value of

13.8 kN/m'

_

Example 14.3 Suppose that, for the sand drain project of Figure 14.21. the clay is normally con­ solidated. We are given the following data: Clay: H,. = 15 ft (two-way drainage) C, = 0.31 eo = 1.1 Effective overburden pressure at the middle of the clay layer =

1000 1b/ft' C, 0.115 ft'/day Sand drain: rw = 0.3 ft d, = 6 ft =

C,

=

C'"

A surcharge is applied as shown in Figure 14.25. Assume this to be a no-smear case. Calculate the degree of consolidation 30 days after the surcharge is first applied. Also, determine the consolidation settlement at that time due to the surcharg� Solution

From Eq. (14.32), (0.115 tt'/day) (60 )

and T"

= , =

C'" H

e�y ( )

=

0.123

(0.115) (30) ' = 0.061 15 2

l t

14.8 Sand Drains

705

Surcharge (lb/ft')

2000 = 6.UCPJ +

:lulf)

- - - - - - - - - - - - -r-----

"'----- --'----+ Time .,,60 - days = tc

Figure 14.25 drain project

Ramp load for a sand

Using Figure 14.24 for T, = 0.123 and Tv = 0.061, we have Uv '" 9%. For the sand

drall!,

n

=

d, 6 ,,, = (2) (0.3) 2-

C"t, d,

T" = -,-. and

C"t, d,

Again, from Eq. (14.26),

10

(0.115) (60) ( 6)'

=

(0.115) (30) (6) '-

=

=

, = T, = -

=

0.192

0.096

1 T, - )1 - exp( - AT,) ]

U, = ---''''------­ T" Also, for the no-smear case, m =

3n' -, 1 n' -2-- ln (n ) 4n n -1

and

A so

U,

=

=

2

=

3(10)' - 1 102 In (10) 4 (10) 2 1.,- - 1 n'

2 1.:;78

- = -- = m

=

1.267

1 0.096 - --[1 - exp ( - 1.267 x 0.096) ] 1.267 0.192

=

0.03 = 3%

1.578

706

Chapter 14 Soil Improvement and Ground Modification

From Eq. (14.18), V'J

= 1 - (1 - V, ) ( l - V,,)

=

1 - (1 - 0.03) (1 - 0.09) = 0.117 = 11.7%

The total primary settlement is thus

[

-.: CeHe log U� + Lluip), + AUf S,(p) 1 + eo 0"0 -

_

and the settlement after 30 days is Se(p,DvJ

14.9

(

(0.31) ( 15) 1000 + 2000 log 1000 1 + 1.1 =

(1.056) (0.117) (12)

)

=

]

_ -

1.056 it

1.48 in.

.

'

Prefabricated Vertical Drains Prefabricated vertical drains (PVDs). also referred to as wick or strip drains, were originally developed as a substitute for the commonly used sand drain, With the ad­ vent of materials science. these drains began to be manufactured from synthetic polymers such as polypropylene and high-density polyethylene, PVDs are normally manufactured with a corrugated or channeled synthetic core enclosed by a geotex­ tile filter, as shown schematically in Figure 14.26. Installation rates reported in the literature are on the order of 0.1 to 0.3 mis, eXcluding equipment mobilization and setup time. PVDs have been used extensively in the past for expedient consolidation of low-permeability soils under surface surcharge. The main advantage of PVDs over sand drains is that they do not require drilling; thus, installation is much faster. Figures 14.27a and b are photographs of the installation of PVDs in the field.

l core

Geo�extiJe _ fabric

--: i·.::

(a) Stone columns in a triangular pattern: (b) stress concentration

due to change in stiffness

When a uniform stress by means of a fill operation is applied to an area with stone c'olumns to induce consolidation. a stress concentration occurs due to the change in the stiffness between the stone columns and the surrounding soil. (See Figure 14.J5b.) The stress concentration factor is defined as n

CT,

'

,

= (J:.

.

where IT; = effective stress in the stone column IT; = effective stress in the subgrade soil The relationships for IT; and IT; are

[

:

(14.42)

-)a-J

IT; = IT- -]--(-'"-"-_l + and

=

/L,IT'

( 1 4.43)

(1 4.44)

where

' a =

J.LS! Me

average effective vertical stress

= stress concentration coefficients

The improvement in the soil owing to the stone columns may be expressed as 5d:)

-' = f.l l.."

S,

(14.45)

720

Chapter 14 Soil Improvement and Ground Modification

where 5,«)

=

5,. =

settlement of the treated soil total settlement of the untreated soil

Load-Bearing Capacity of Stone Columns

If a foundation is constructed over a stone column as shown in Figure 14.36, failure will occur by bulging of the column at ultimate load. The bulging will occur within a length of 2.5D to 3D, measured from the top of the stone column, where D is the diameter of the column. Hughes et aL ('1975) provided an approximate relationship for the ultimate bearing capacity of stone columns, which can be given as (14.46)

T

, , , ,

I I I I

Clay c"

, ,

,

\

L

\

I Figure

14.36

I

1-

l:h:C! ring. capacity t)r ;.;tone colulllll

,

14, 1 3

'

C.

c.r; (b'

......:.

.:.

:.�

Srone Columns

721

lli�:!r-Timt:

14.12

731

(Jays)

Figure P. 14. 12

For a sand drain pro,iect (Figure 14.2 1 ) . the following are given: Clay: Normally consolidated H, = 5.5 m (one-way drainage) C, = 0.3 eo

=

0.76

Cv = 0.015 m2jday Effective overburden pressure at the middle of clav. layer = 80 kNjm' Sand drain: r .,. = 0.07 m r w = rf d, = 2.5 C,. = C"

m

A surcharge is applied as shown in Figure PI4.12. Calculate' the degree of consolida­ tion and the consolidation settlement 50 days after the beginning of the surcharge ap­ plication.

Abashi, H., lchimoto. E .. and Harada, K. (1979). "The Compozer-a Method to Improve Characteristics of Soft Clay by Inclusion of Large Diameter Sand Column," Proceed­ ings, International Conference on Soil Reinforcement, Reinforced Earth and Other Tech­ niques, Vol. I, Paris, pp. 21 1-216.

Abashi, H., and Monden, H. (1963). "Determination of the Horizontal Coefficient of Consol­ idation of an Alluvial Clay," Proceedings, Fourth Australia-New Zealand Conference on Soil Mechanics and Foundation Engineering, pp. 159-164. American Society for Testing and Materials (1997). Annllal Book 0/ Standards, Vol. 04.08, West Conshohocken, PA. Bachus, R. c., and Barksdale, R. D. (1989). "Design Methodology for Foundations on Stone Columns," Proceedings, Foundation Engineering: C!lrrem Principles and Practices American Society of Civil Engineers, Vol. 1, pp. 244-257. Barron, R. A. (1948). "Consolidation of Fine-Grained Soils by Drain Wells," Transactions. American Society of Civil Engineers, Vol. 113, pp. 718-754. Basore, C. E., and Boitano, 1. D. (1969). " Sand Densification by Piles and Vibroflotation," Journal of the Soil Nlechanics and FOIlndations Dit'isivn, American Society of Civil En­ gineers, Vol. 95, No. SM6, pp. 1303-1323.

732

Chapter

14 SoU Improvement and Ground Modification Brown, R. E. (1977). "Vibroflotation Compaction of Cohesionless Soils," IOll/'llll1 of the Geo· technical Engineering Division, American Society of Civil Engineers, Vol. 103. No, GT12, pp. 1 437-145 1. Burke, G. K (2004). "Jet Grouting Systems: Advantages and Disadvantages." Proceedings, GeoSupport 2004: Drilled Shafts, Micr"pilillg. Deep Mixing, Remedial Methods, and Special Foundation Systems, American Society of Civil Engineers, pp. 875-886. Christoulas. S.. Bouckovalas, G., and Giannaros, C. (2000). "An Experimental Study on Model Stone Columns," Soils and Foundations, Vol. 40, No. 6, pp. 11-22. D'Appolonia, D. J.. Whitman, R. V., and D'Appolonia, E. (1969). "Sand Compaction with Vibratory Rollers:' Journal oftlte Soil J.\1echanics and Foulldations Didsioll, American Society of Civil Engineers, Vol. 95, No. SMI, pp. 263-284. Gurtug, Y" and Sridharan, A. (2004). "Compaction Behaviour and Prediction of Its Charac­ teristics of Fine Grained Soils with Particular Reference to Compaction Energy." Soils and Foundations, Vol. 44, No. 5, pp. 27-36. Hughes, J. M. 0., and Withers, N. J. (1974). "Reinforcing of Soft Cohesive Soil with Stone Columns," Ground Engineering, Vol. 7, pp. 42-49. Hughes, J. M. 0" Withers, N, J., and Greenwood. D. A. (1975). "A Field Trial of Reinforcing Ef­ fects of Stone Columns in Soil," Geotechniqlle, Vol. 25, No. 1 . pp. 31-34. Ichimoto,A. (1981). "Construction and Design of Sand Compaction Piles," Soil Improvement, General Civil Engineering Laboratory (in Japanese), Vol. 5, pp. 37-45. Johnson, S. 1. (1970a). "Precompression for Improving Foundation Soils." 10llmal of the Soil Mechanics and Fo undations Division, American Society of Civil Engineers. Vol. 96, No. SMI. pp. 1 14-144. Johnson, S. 1. (1970b). "Foundation Precompression with Vertical Sand Drains:' ivumal of the Soil ft.,fechallics and Foundations Divisioll, American Society of Civil Engineers: Vol. 96, No. S1'-n, pp. 145-175. Lee, K. L.. and Singh. A. (1971 ) . .. Relative Density and Relative Compaction:' Journal oftlie Soil Mechanics and Foundations Dil'l:'ion, American Society of Ci\'il Engineers, Vol. 97, No. SM7, pp, 1049-1052. Leonards, G. A.. Cutter, W. A .. and Holtz. R. D. ( 1 980), "Dynamic Compaction of Granu­ lar Soils," Journal of Geoleclm ical Engineering Division, ASCE. Vol. 96. No. GTl . pp. 73-110. Mattes, N. S., and Poulos, H. G. (1969). "Settlement of Single Compressible Pile," lournal of the Soil Mechanics and Foulldations Division, ASCE, Vol. 95, No. SM1, pp. 189-208, Mitchell, J. K. (1970). "In-Place Treatment of Foundation Soils." lournal of tire Soil Mechall­ ics and Foundations Division, American Society of Civil Engineers. Vol. 96, No. SM1, pp.73-110. Mitchell, J. K, and Freitag, D. R, (1959). "A Review and Evaluation of Soil-Cement Pave­ ments," Journal of the Soil Mechanics and FOllndations Din:'iioll, American Society of Civil Engineers, Vol. 85, 1'0. SM6, pp. 49-73. Mitchell, J. K, and Huber. T. R. ( 1985). "Performance of a Stone Column Foundation." lour­ nal of Geolrchnical Engilleering. American Society of Civil Engineers. Vol. i l l. No. GT2. pp. 205-223. �:l urayama, 5. (1962). "An Analysis of Vibro�('ompozer Method on Cohesive Soils;' Con� SlfflCliol1 ill A1echallization (in Japanese). No. 150. pp. lO-15, Ohta . S., and Shibazaki. M. (1982), " A Uniqu� Underpinning of Snil Specification Utilizing Super-High Pressure Liquid Jet:' Proceedill.t..:,\. COlIli:r!!/II 'c on C)'rolllilig in Geotechnical J"�ngineering. New Orleans. Louisiana, ()!qm. R. E. (1977). "Consolidation unde r Time�D(' l'clldt:l't { (I,;. �'llg:' JOllf'lW! of Georer:lmi­ col Eng!'"l.'ering Dh'isioll. ASCE. Vol. 102. 0:0, CiT1 . rT. .:;� '\:1.

I

l

[

I

References

733

( >tllman. i\:l. A . . and Lut.:: ltich. S. !\'L ( YlJ..l) . .. ( \lmpdl'! il)!) ( 'outrol Cri{�ria for Clay Hydrauhc Barriers:' 7iwISPOI'Wfi()JI Re.w:arch Rn:(I! 'd. ' .:" 1 _ln2. National RL'search Counci l, \Va;-:;hinglon.

PUrim, C. 1.

DC. pp. 28-35.

and Rodriguez. J. A.

[cdliliCiil Journal. Vol.

I

( 992 J • •• Dl'>;i,r :l t,'1 2. NO. :1 . Pl'. 7()6-�O=:.

:)\'!l,!illk: COlHpadiou:' Calladiall

ShihuVil. S .. and Hanh. L. '[ (200 1 ). " Esl il11iJting Undrain�d Shear

S tre ngt h

Ueu­

of Soft Clay

Grount..l IJllProv�d by PreioHding with PVD--Cast:; I-iistl'ry in Bangkok," Soils atld Foundations. Vol. 4 1 . No.4.

\Vt:bh . .1. P. . and Burke. G. K . (

pro 95-1 0 1 . 19'1 l ). " Jet Grouting-U�co.; for Soil lmprovt:'IHt:nt,"

PmceedillgJ.

(i('ofcc/lllica/ Ellgil1l'!!ring C/ I/lgrcss, American Society of Civil Enginec:rs, Vol.

334-345.

\Velsh, 1. P.. Ruhright, R. M . . a n d ('oom ba. D. B.

(1986). "Jet

1.

pp.

Grouting for support of Struc­

tures," presented at the Spring Convention of the Am�ric,.ln Society of Civil Engineers,

Seanle. Wash ington .

Yeung. A. T. ( 1 997). " Design CUI'VI:!S for Prdabricated Vertical Drai ns:' '/ourn(/I oI Ceo(ec/llli­

cal lind Geoenvirollmenw/ Engineering, Vol. 123, No. 8, pp. 755-759.

l

4ppendix A

Field Instruments TIJis appendix provides brief descriptions of some common instruments used ill the field in geotechnical engineering projects: {h� piezometer. earth pressure cell, load cell, cone pressuremeler. dilatometer. and inclinometer.

Piezometer

The piezometer measures water kvels and por� water pre�surc. The sinlplest piezome­ ter is the Casagrande piezometer (shown schematically in Figure 2. (6) which is in­ stalled i n a borehole of 75 to 150 11101 di ame te r. Figure A.I shows a photOgraph of the

Figure A. 1 Components of Casagrandt:-lypc piezometer (Courtesy of N. Sivakugan.James Cook Univer­ sity, Australia) 735

736

Appendix A

components of the Ca'''gnlnde piezometer. 1be device wnsists of a plastic riser pipe joined to a filter tip that is placed in sand. A bentonite seal is placed abO\'e the sand to isolate the pore water pressure at the filter tip. TIle annular space between the riser pipe and the borehole is backfilled with a bentonite-cement grout to prevent vertical migration of water. Although they are simple. reliable. and inexpensive, the standpipe piezometers are slow to reflect the change in pore water pressures and require some­ one at the site to monitor their operation. TIle other types are pneumatic piezometers, vibrating wire piezometers, and hydraulic piezometers. Piezometers are used to moni­ tor pore water pressures to determine safe rates of fill or excavation: to monitor pore pressure dissipation in ground improvement techniques such as PVDs. sand drains, or dynamic compaction: to monitor pore pressures to check the performance of em­ bankments, landfills. and tailings dams: and to monitor water drawdown during pump­ ing tests. Its nonmetallic construction makes the piezometer suitable in any adverse environment for long-term operations in most soils. Earth Pressure Cell

An earth pressure cell (Figure A.2) measures the total earth pressure, which includes the effective stress and the pore water pressure. It is made by welding two circular stainless-steel diaphragm plates around their periphery. TIle space between the plates is filled with a de-aired fluid. The cell is installed with its sensitive surface in direct contact with the soil. Any change in earth pressure is transmitted to the lluid inside the cell. which is measured by a pressure transducer that is similar to a diaphragm­ type piezometer. Earth pressure cells can be used to measure stresses within em­ bankments and suhgrades. foundation bearing pressures, and contact pressures on retaining walls. tunnel linings. railroad bases. piers, and bridge abutments.

I

Figure A.2 Earth Fn::ssun: L'nh crsity. Austr
749

S-WJ\'t�. 1 1 1

smear zone, 697, 69K

Q

Resistivity. 1 12-113 Retaining wall:

Jnde,Y

Sliding. retaining wall, 36 l-363

Soil classification systems. 13-18

Soil compressibility [actor, hcaring capacity. J 42-J 44 Spaci ng. boring. 7 1 Specific gravity, 9

Spl i t-spoon sampler, 75-86 Spring. con: catcher. 77

I ,f

! I I

'

'

750

Index

Stability check, retaining wall: bearing capacity. 364-366 overturning, 359-361 sliding, 361-363 Stability number. 189 Stabilization: cement, 714-716 fly ash, 716 hme, 716-717 pozzolanic reaction, 7 11, 713-714 Standard penetration number: correlation, consistency 'of clay. 78-79 correlation, friction angle. 82 correlation, overconsolidation ratio, 79 correlation, relative density, 8 1 correlation, sand, 80 Static penetration test. 90 Stone column: allowable bearing capacity, 720-721 equivalent tri;,mgular pattern, 718 genera!, 717 stress concentration factor, 7 1 7. 719 Strain influence factoe 236 Stress: circular load, 205-206 concentrated lOad, 204 embankment load . 2 1 6-2 1 8 n.:cumgular load. 206-21� Structural dt:!sign. mat: approximate flexible method. 289-295 conventional rigid method. 286-289 Subgrade reaction coefficient, 291. 292-293 Suiwbilitv number. vibroflotation.686 Swell pressure test. 651-653 Swell lest, unrestrained. 651 Swelling index, 31 T Tensile crack. 314 Terminal moraine, 65 Thermal bonding. geotcxtile, 381 Tie failure. retaining wall. 388-389 Tie force. retaining wall. 388 Time factor. 34 lime:: rall: of cOi15olitlation. 33-3R Tolerable s!!ttlement. shallow fuundatioll. 241-244

Trapezoidal footing. �73-275 Triaxial test: consolidated drained,46 consolidated undrained, 46-47 unconsolidated undrained. 47 U Ultimate bearing capacity, Terzaghi, 126-127 Unconfined compression strength, 48 Unconfined compression test, 48-49 Undrained cohesion. 47 Unified classification system, 15-20 Uniformity coefficient, 3 Unit weight: dry, 7 , 8 moist, 7, 8 saturated. 7.8 Unrestrained swell test, 651 Uplift capacity. shallow foundation. 1 9.'-198 V Vane shear test. 86-90 Velocity. P-wave, 106 Vertical stress. average. 213-215 Vibratory roller. 679 Vibroflotation: backfill suitability

number, 686 construction method. 382. 383 effective range. backfill, 688-686 vibratory unit. 682 Virgin.compression curve. 30 Void ratio.5 Volum!!, coefficient of compressibility. 34 W Wame slab, 661, 663 Wash boring. 73 Water table, effect on bearing capacity, 130-131 \V