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PROBLEMS IN SOIL MECHANICS AI{D FOT]NDATION ENGINEERING A'M'LE'(ndia); [ForB.E.(Civil);M'E'(Civil); Examinationsl U.P.S.C'andotherC-ompetitive

DEBASHIS MOITRA

Departmentof civil Engineerilq BengalEngineeringCollege DeemedUniversity Howrah'

\|\\ ,j r u" \ {

i i ,

DHANpA_rLtcATto hls(p)' LrD.84! F_uB_ FIRSTFLOOR,.6ZI+ IT,INONAS HOUSd DARYAGA+TJ, NEWDELHI-J1OOO2 PHONES:3274073

Note: This book or part thereof may not be reproduced in any form or translated without the written permission of the Author and the Publisher.

OTHERUSEFT'LBOOKS 1. AdvanceTheoryofStuctures

N.C. Sinha

2. Concrete Testing Manual

MI. Gambhir

3. Fundamentalsof Limit Analysis of Structures

V.K. Manicka Selvam

4. Modern method of Structural Analvsis

V.K. Manicka Selvam

5. Multistorey Building & Yield Line Analvsis of Slabs

V.K. Manicka Selvam

6. Energy Methods in Structural Mechanics

V.K. Maniclca Selvam

7. Analysis of Skucture in Earth Quake Region

V.K. Manbka Selvam

8. Dock and Harbour Engineering

S.P. Bindra

9. Foundation Design Manual

N.V.Nayak

Preface This book is primarily intended for the undergraduatestudentsof Civil Engineering. However, it will be helpful also to the diploma-level students, A.M.I.E. students,and, in some cases,even to the post-graduatestudentsof Soil Mechanics and Foundation Engineering.

FirstEdition1993 Reprint :

1998

(

A thorough understandingof the basic principles of a subject like Soil Mechanics calls for lhe solution of a large number of numerical problems. In the presentbook a briefinfoduction to the contentsofeach chapterhas been given, which is followed by a number of worked-out examples and quite a few practice problems. For a better understandingof the topics and students are required to solve all the problems by themselves.Effort hasbeenmade to explain the basic principles underlying the solution of the problems so tlat the students may develop the habit of having a logical insight into the numerical problems while solving them. Commentsand 5rrggestionsregardingthe book, from the studentsaswell as the teachers,will be highly appreciated.

Price:Rs.60.00 Calcutta, 9, March 1993

Ptfulishedby Prittted at

Ish Kapur for Dhanpat Rai Publications (p) Ltd. : A.P. Of1.sc.t. Navecn Shahdara.Delhi- | t(X)32.

DEBASHISMOITRA

1 WEIGHT.VOLUME REI.ATIONSHIPS

CONTENTS Clwpter

Page

"{

Weight-VolumeRelationships

,/,

Index Propertiesand Soil Classification

24

,/. I

Capillarity and Permeability .

49

'g..r' lz,

Seepageand Flow-nets

1

.

81

StessDistribution

to7

Consolidation

133

Compaction

165

Nr

Shcar Strength

181

'9J/

Earth Pressure

2r3

10.

Stability of Slopes

?54

L1.

Bearing Capacity

?33

12.

Deep Foundations

310

/J

,€.

o

Matter may exist in naturein threedifferent states,viz., 1.1 Introduction: solid, liquid and gaseous.A soil massin its naturalstatemay consistof all ' three phases.The basic ingredient is the solid grains which form the soil skeleton,while the intermittent void spacesare filled up by either air, or water, or both. Thus, a soil massin its natural statemay be considereda three-phase system. 1.2 Soil Mass as a Three-phase System : In a soil mass in its natural state,tle three phases,viz., solid, liquid andgas,are completely intermingled with one another. However, if one can determine the individual volumes of solid grains, liquid (i.e., water) and gas (i.e., air) presentin a certain volume

: -----Water- ----:

Fis.1.1

I

ofa soil, the entire soil mass can be represelted by a schematicdiagram, as shown in Fig. 1.1, where the volume of each constituent part is shown as a fraction of the total volume. The cross-scctional area of the soil mass fo taken to be unity, so ttat, the volume of each constituent part is numerically equal to ib beight shown in the diagram. Again, the mass of each part may be obtained by multiplying its volume by the corresponding density.

t

Thenotations used inthe diagram are defined below:

t

V = total volume of the soil mass

I I

I

'. \

Problems in SoilMechonics and Fonndation Engineering

2

particlegin the soil % = volume of solid in the soii = voids of volume V, V- = vslspe of water presentin the voids V, = volurne of air presentin the voids

s RelationshiP Weight-Vofume i.e.,

v t=ixrWva

...(1.4)

dry soils) to 1007o(for fully The value of s may vary from oVo (for saturatedsoils). "--is defined as the ratio of the ("tSp"t it'rc gravity of sotids(G".or G) : It to the mass of an equal volume of mass of a given lrotume of solicl grains water, measuredat the sametemperarure'

17 = total mass of the soil !7" = rnassofthe solid Particles W- = mass of water presentin the voids' The massof air presentin the voids is negligible' Vu=V"+Vn Thus,

G =Mny : -

1.c.,

The fundamental physical properties which 1.3 Basic Defrnitions : below : govern the engineeringperformanceofa soil are defined

grains M" = massof anyvolurneVofsolid M. = massof water of volume V' then in the C'G'S' systen If this volume V is arbitrarily taken as unity' of solid grains (y') and dersity the to Lqu't M" and M. become **..i".iry density of water (1.) respectively' Thus'

(i)Voidratio(e):Thevoidratioofasoilisdefinedastheratioofvolurne of voids to the volume of solids'

massolunitvolunggllglids- Ts O massof unitvolumeof water Y-

and,

V =V r+V,

0r,

V=Vr+Vo+Vn

vu

i.e.,

"=v,

where,

...(1.1)

Thevoidratioisadimensionlessparameter,thenumericalvalueofwhich with increasing degree of compactnessof the soil' decreases -aefineAas the ratio of the volume of voids to the 1i4 f-rsity (n): ttis as a percentage' total volume of the soil mass.It is generallyexpressed i.e.,

fu= + x rooe,o

...(r.2)

1' However' as lhe The void ratio of a soil may be greateror less than a soi| mass,its porosity volume ofvoids is alwayslessthalrthe totalvolume of is always lessthan 100%. is defined as (ili) Water content(w) : The water content of a soil mass expressedas always is tne ratlo of the rnassof *.i"t to the massof solids' It a percentage.

i.e., ,/

w... *=frxlooVo

"'(1'3)

,/ (s) : The degreeof saturation of a soil mass is 4i{ O"gr"" of saturation of voids. It is always defin-eias tf,e ratio of volume'of water ro tbe volume expressedas a Perc€ntage'

...(1.s)

T"= G'Y'

or'

as the ratio of the mass of (vi) Mass spectftcgravity (G,,) : It is defined volume of water' measuredat a siven volume of soil to theLiti'of tn equal the sametemPerafure. i.e.,

I ;

M

Y

M*

\n

...(1.6)

where " --'(vit\ Y= unitweightof thesoilmass' of,thetotal Butka"nrityl, unit weight(v): It is ogrineo15n;-ratio KN/m ' gm/ccor t^n- or ,o.,, of u soil to its totalr olume.Its unit is

w

l.e.r,

\=T

...(1.7)

as the massof soil solids per (viii) Unit weiglt of solids(Yr):It is defined unit volume of solids. 1.e.,

w, Y " =%

...(1.8)

a soil mass is defiried as the (ix) Dry density (17) : The dry density of volume of ttre soil mass massof soil solids per unit of the total

Problems in SoilMechanics and Foundation Engineering

4

ws \d=V

i.e.,

...(1.e)

The difference between 1" and y7 should be clearly understood.The dry density of a fufly or partly saturatedsoil is nothing but its bulk density in the dry state.The dry density ofa soil dependson its degreeofcompactness, and hence, on its.void ratio. But $e gnit weight of solids depends only on the properties of iie minerals presentin it and is independentof the statein which the soil exists. (x) Saturated unit'weight (y.",) : When a soil mass is fully saturated,its bulk density is tenrred as the saturatedunit weight of the soil. (xi) Submergeddensity (y.u6): The submergeddensity of a soil massis clefinecl as the subnerged weight of the soil per unit of its total volume. 1.4 Functional Relationships : In order to assessthe engineering performanceandbehaviourofa soil, itis requiredto evaluatethefundamental properties enumeratedin fut' 1.3.While some of theseproperties (e'g', w, G, y etc.) can be easily determinedfrom laboratorytests,someothers(e'g', q s, y" etc.) cannot be evaluated directly. However, all of these properties are interdependent.Hence, if mathematical relationships between two or mor€ such properties can be developedthen the direct determination of a few of them will lead to the indirect detenninationof the others.Thus, the functional relationships have an important role to play in Soil Mechanics. The most important relationshipsare establishedbelow :

vu

"= v" = Vv + V " , o r , V "- V

But,

vu

vr/v

"' e = v - v"= (v:W .'.e=

=

v,/v

considerAlternative prool: The samerelationshipsmay alsobe deduced (b)' and (a) 1'2 in Fig' shown ing the schematicdiagrarnof a soil massas

(1+e)

Fig.1.2

vv We know that,

-Vr.

n

[ . =

+l

V r ,= e . V r . Let us considera soil masshavingtrnitvolume of solids'

= T ? ; t'"J e =

u

e

" = i = 1 . " n =

Again,

= +, or vu n'v

= Cqnsideringa soil masshavir:ga totalvolume V l, - n' V . .= l ' n = n , o r , % = V - V , = |

v,

...e=Vs

V,

vr/v, vu o r ' ng v J v " = Wm=

ys

Now,

...(1.10)

n = T

(b)

(o)

L - n

Again, by definition,

...(1.11) L + e

V

v . =r - "

r-i

. " n =

. ' .% = l , o t , V r = € ' I = € ' .'. Totalvolumeofthesoil, V = V, r V" = | 1 s

(i) Relation betweene and n : By definitnn,

5

Weight -Volume Relat ionslriPs

-,5

n l - n

newion betweene, G, w ands :

With referenceto Fig 1.1'

, = w% ' t " Vn'\n

u

*

*

Problems in Soil Meclnnics and Foundation Engineering

G=!,

Y" = G'Y.

Of,

ln

vn

Vn'ln

Ws

By definition,

tu

,9e G

Wn+W, Vr+V,

Vn.yn + V".Gyn _

_ t v

Vr+V"

Vu+V"

(1.13) = From eqn. weBet,Tsar

Iw

(1.13)we get,fd = From.eqn.

_ (s.e + G)/e ., = G r s€ .., ttt' tw (l+e)/e

V = -

W

v

Wn+W"

v

Wn+W" ...(1.13)

t | 9

(iv) Expressionfor y.", :

Olt

Y

w"

wnr w' l

or,

(V, + GV')/V, - -F;T-q4 tn

or, (G+e)/e

tw

r

- . r t

(l+e\/e

G+e tw

l+e

w"

l a = 1 |i , o r , V = j ,ld

...(ii)

From (i) and (ii) we gel,

For a safurat'edsoil, V,n= l/, Vy.y- r V"'G^ln V, + G.V" = --------------;=-'Y. Tsar - -------i7,rlV, Vu+1,

l+G.(I/el

...(i)

Y = - -

Again,

W n + W " vn"(n + %'Y" = W= Bydefinition,ysar i Vr+V, ffi

t+l/e

'

We know that,

G+se

4 . r ,

#.r*

= -GTn l;"

l+e

v' = - . v l + e

-

.u

u = fi|

(vi) Retation between y and y4 :

t\r

l+l/e

+#

Foradrysoil,s=0

tw

Dividing fle numeratoranddenominator by V, ,we get, VJV, + G.V"/V, s + G/e 1+V"/V,

...(1.ls)

Eqns.(1.14) and (1.15) may also be derivedfrom,eqn.(1.13) as follows : Forasaturatedsoils=1.

Vn + G.V" E

Iw

G^tn

Vn.\n + %.y" Vr+V"

v = - = - = -

ot,

G/e | + l/e

\d = TTe

0f,

The bulk density ofa three-phasesoil systernis given by,

'

Vr,+V"

(V, + V")/V, 'w

(iii) Relatian between y, G, s and e :

=

V".G\n

%'y" Vr+V,

V

s.e = w.G

W V

...(1.14)

G.Vs,/Vv

V"/V"

'

G + e l-Jl'Yw

(v) Expression for y1 :

vJV,

s G/e

.f

Ysar=

of'

= = vr.Gr" y" .G (vr/v,) ' c

= VJV" c =

Weight -Vo htmeRelationshtps

tw

-

Y= Yd =

Wn+W,

(.

=w'

I

a Wr\

W " - ' t , = l t * W . " /l ' r o - ( l + w ) . y a \ v

T;-;

...(1.16)

(vii) Relation between y*5 and y* : A soil is said to be submergedwhen it lies below the ground water table. Such a soil is firlly saturated.Now, accordingto Archimcdes' principlc, when

8

Problems in Soil Meclnnics and Foundation Engineering

an object is submergedin a liquid, it undergoesan apparentreductionin mass, the amount of such reduction being equal to the rnassof the liquid displaced by the object. Consider a soil mass, having a volume V and mass I,Iz,which is fully submerged in water. Volume of water displacedby the soil

From theconsiderationof degreeof saturation,a soil sample

(i) Completely dry (s = 0)

-t

(ii) firlly saturated(s = 1) Unless otherwisementionedin the problem, a soil sampleshould always be taken to be partially saturated.

= V(Y."r - Y-) The apparentdensity or submergeddensity of the soil is given by,

V(Y."r - Y,r) W' Ysub=V = V

Methpd 12'Given ' lT,w, C I

==+Required ' t : [Ta, ' .s,A;l'

l

As e and z are mutually dependent on each other, effectively three unknown parametershave to be determinedfrom the given data. Select the appropriate equationswhich may servethis purpose. The value of y7 can be determinedfrom :

...(r.r7)

Y .,. ' d - l + w Here,

Two differentmetlods :ru

9

(iii) partially saturated(0 < s < 1)

Apparentmassofthesoil, W' = W - V -,{n = V.ysat - V.,{n

Ysub=Ysat-Yw

Solution: may be :

= V

Mass of displacedwater = V . \n

or,

Weight-Volume Relationships

""t"Toyedto solvethe numerical

problems in this chapter. They are : Method I : Solution using mathematical relationships : This process is somewhat mechanical, one has to mernorise all the equations deduced in fut. 1.4 and should select the appropriate equation/s while solving a given problem. However, in most of the casesthis method can yield the desiredresult fairly quickly. Method II : Solutionfrom first principles : In this method the solution is obtained using only the basic definitions with referenceto a three-phasediagram of the soil massunder consideration. This method always allows the student to have an insight into the problem. However, in some casesthe solution becomesa little complicated and more time-consuming than method I. After going tlrough lhe worked out examples, quite a few of which r'llustratethe use of both of tlese methods,one should be able to realise as to which method of solution suits better to a particular type of problem. It may be pointed out that, the methods may also be used in conjunction with one another. Problem 1.1. A soil sample has a unit weight of 1.9 gm/cc and a water content of l2%.If the specific gravity of solids be 2.65, determinethe dry density, degree ofsaturation, void ratio and porosity ofthe soil.

y = unitweightof thesoil = 1.9gm/cc lr = water content = l2%o = 0.t2

\d = T#n=

r'6e6gm/cc

In order to solve for the other two unknowns,viz., s and e; two equations are required. Evidently, the following equationswill serve the purpose : vrG = s€t or re = (0.12)(2.65) = 0.318 Again,

or, or,

G+se l + e

v' = _ . l n

r.n= f41l@)tr.ol l * e l+e=

| . ) ' 1.56,or,e=0.56

The expressionof y7 may also be used.

'{a= of'

G'tn

y-l s,

=(f?P, 1.6s6

OT,

1.696+1.696e=2.65

or,

"=ffi=o'56

...(t

Problems in SoilMechanics and Foundation Engineermg

10

-. = 9 ! 1 9 = 0 . 5 6 8= 5 6 . 8 v o 0.56

From (i),

e " = Ti; Answer.

0.56

=

, . ;s.

= n0.36= 36vo

Dry density = 1'696 gm/cc' void ratio = 0'56 Degree of saturation = 56'87o,Porosity = 36Vo

wn w=-w

Now,

s

wn ; l

Void ratio,

"=2=ffi=os6

Porosity,

=36vo " =+ =ffi x roovo

= rrr, r'n

volume of 300 Problem It2-'F'nundisturbed specimenof soil has a its weight hours' 24 for 105'C at oven in drying After +66got' "" tJ*.igh. reducedto-+sog*.oeterrninethevoidratio,porosity,degreeofsaturation and water conteut. Assume G = 2'70' Solution:

Wn = o.lZgm

0r,

= l'I2gm Totalmassofthesample, W = Wo + W4 rYr " =

Volume of solids,

v' t ' =

Volume of water,

W"

T wn

Weight ofwaterevaporated,

1

Vo=V-(%+V)=0.092cc = .'. Volume of voids, V, = Vo t Vn = 0.12+ 0'fp.2 0'2l2cc vn o.r2 =

.'. Volume of air,

Degree of saturation,

t=

fr

=

ffix

'1"'n't' t I cir"n,fr wg5 cf+ Required ' "

fuid, weight of the dry sample,

= o'12 - 0.12cc

v' =Yl! t- l'rz . 9 = 0.589cc

Total volurne of soil,

Methodl:

After drying itt oven,thewater presentln m€' soti"ffitatts becomescomPletelYdrY. W = 498 gln Now, weight of the moist sample,

I -W' - 0" .' -3"7 7* c r ,, = = c\"= (2f5)(l)

l*

,, =+ =#F = r'6nsm'/cc

Dry densitY,

Method II: Letusconsidera'specimenofthegivensoilinwhichthemassofsolid 1'3' grains = 1 gm. The tnree-phasediagrari of the soil is shown in Fig'

11

Weight-Vol umeReIat ionshiPs

r o o % 5o 6 . 6 7 o

and the soil

Wa = 456 gn' W-='W -Wa= 498 -456 = 42gm'

= 0'0921= 9'21% Watercontent,w - Y 456 wd +G'r'u \d=T;e Dry densitY, \d =

But

wa -456 = r.szpm/cc v 300 G'tn = L5z l + e

\h (0.092cc1 Vw (0'12ccl V ( 0 . 5E 9 c c )

1.s2(r + e) = (2.7)(r) +1.52=2.7 L.S?z e -- 0.78 Void ratio= 0.78

or, or'

:-------Woter-- : ----: - -_:_-_-_-_-_-_-_----_-

of,

VJ (1'12gml

Again,porosity

Vs

(0.377rc)

, -

e TT;

=

From eqn. (1.12), t+G= s€t

F i g .1 . 3 .

0.78 = 0.438= 43'8vo ,ft or,

, = I9

,

LZ

Problems in Soil Mechanics and Foundation Engineering

Weight-VolumeRelationships

Ort

(o.oe2r\ (2.7\ "=ff=0.319=3l.9flo

Problem !J. A saturatedsoil sample,weighing 178 gm, has a volume of 96 cc. If the specific gravity of soil solids be 2.67, determinethe void ratio, water content and unit weight of the soil.

A

Method II : With referenceto the three-phasediagrarn shown in Fig' 1.4,

V-=--:==42cc

Volume of solids,

v' s - w " - w ' Gln

AS6'

= 168'8e cc

'

0f'

V'=V-V'

Ot,

Vu = 3N - 168.89= 131.11cc " =

vu 131.11 = o'78 = 16s€, ,"

vu ,=T=

s= w=

Vn

fi

Again,

131.11= O . 4 3 7 = 4 3 | 7 V o 3Of

=

Wo W=

42

trfu

2'67-+ | x e\(1'o) = 1'954 r + e ) 1.854+L,854e=2.67+e 0.85k = 0.816 e = 0.955

0r,

V=300cc

Total volume,

tu, = l]|.u

But,

Y"

,?in

Required :

y,",={ =y9 =1.854gm/cr v 9 6

yw

=

of'

Given,W, VEe+

Unit weight of the soil,

w...

Volume of water,

Volume of voids,

Solution:

wn=498-456=42gm

Weight of water,

13

= o32= 327o

42 =9'2lc/o ^5t=0'0921

(0L0255) - 0.358= 35,.8vo * =-X -

..u t Prcblery{. A ftrlly saturatedsoil samplehas d volume of 28 cc. The sample was drled in oven and tle weight of the dry soil pat was found to be 48.86 gm. Determine the void ratio, moisture content, saturateddensity and dry densityof the soil mass.Given G =2,68.

. Solution: Given' F % e;l=+

A schematicrepresentationof the given soil is shown in Fig. 1,5. Here, total volume

V=?3cc

Volumeofdrysoil, % = { 1 3 1 . 1c1 )c

Required :

T#"c=18.23cn

Assuming that there was no changein void ratio during ovcn-drying, volumeofwaterevaporated,Vn= V - % = QA - L8.23)cc 7,'9.77cc

w (4 9 8 g m l Sotid

Void ratio,

v,

Ws{ 4 5 6 9 m )

= Fig.r.4

Vn

- v " v , l r = - = -

o11

ffi

= o'536

l'.'v"=vnT

t4

Problems in Soil Mechnnics and Foundation Engineering Weightofwater,

Wn = V*'\*

= (9.77)(1.0,

Wt )Cs =1.7889n/cr 'td=i=,"5

= 9.77 grn Moisture eontent,

g'77

wn

|v=fi=ffi=0.2=20%b

i rral weightof thesoil, W = Wn + 17" = (9.77 + 48.86)gm = 58.63grn density, t*, = Saturated {

=

#

Drydensity, ro =Y= #

= 2.09 gm/cE

= r.745sm/cc

V v= 9 ' 7 7 c c

h(a=9.779m W= 5 S ' 6 3 9 m Ws=4E.869m

Vs=18'23cc

FiB.1.5 . Problem l.rf,. An undisturbed sample of saturatedclay has a volume of 16.5 cc and weighs 35.1 gm. On oven-drying,the weight of the sample reduces to 29.5 gm. Determine the void ratio, moisture content, dry density and the specificgravity of solids. Solution :

Method I:

Given : Vn we+ Weight of thesaturated sample, Weightof thedry sample,

tF,*l,d, c-l Required

W = 35.1gm Wa = 29.5gm

.'. Weightofwalcrevaporated, Wn - W - Wa = (35.1- 29.5\gm

tu, = l]].t* = q, t" , 2.127 i

But.

2.127+2.127e=G+e

or, or,

G=l.l27e+2.127 G'tn I a=

tu,={-i#

-2.ryism/*

...(i)

V r

Again,

l + e

G , 1 1.788 = - ----:r + e = G 1.788e+ 1.788

or,

...(ii)

FromQ) and(ii) we get, e + 2.127 1.788e+ 1.788= L.127 O.66te= 0.339 or, e = 0.51 or, From(i)we get,G = (1.27)(0.51)+ 2.127= 2-r Now, l+G = s€

= r8.evo - = E = gr'}lu -*0.18e

oI'

A thlee-phasediagrarn of the given soil is shown in Fig' 1'6' Here, wet weight of the sample, W = 35-t gm

Method II :

Dry weight of the samPle,

Wd = 29.5 gm

Weightofwater,

Wn = W - Wa = (35"1 - 29'5)gm = 5.6gm

Volume of water

Vn = V, = 5.6 cc

V = 16.5cc V, = V - Vu = (16.5- 5.6)cc = 10'9cc Volumeofsolids, Total volurne

Void ratio,

" = ? =# = o s l ,

Moisture content,

* = V = # = o . r 8 e- r 8 . e %

= 5.6grn

Nou,,

15

Weiglx -Volume Relatians hiPs

Problems in Soil Mechanics and Fonndation Engineering

r7

Weig ht -Vo lame Relat ionship s

was 0.54,dc&rminethc moisturccontent,dry density,bulk density,degree of saturrtionrnd specificgrrvityof solids.

Vy15 =. 5 c c

sotriior: GiveE ,W@+ V * 1&5cc W - 3629m

Totalvolume Totalweigh!

Wa - 3%gm

Dry wcfhl

t -V -#

Bulk density, Fig. 1.6

Ws 29.5 \a = V = ,rj = l.19gm/cc

Dry density,

Required ,F yr,r, ", c I

Dry density,

wd

lo-i

- Le6gm/cn

- 326 - I . 7 6 g m / c c 1g5

Weightof watercvaporated,Wn = W - W"

y" =

Unitweightof solids,

f

=ffi

= Z.7ogm,/v

='# = 2.70 of solids, c = Specificgravity * / Problen i!/ m. initial void ratio of an inorganicclay is foundto be 0.65,while the specificgravity of solidsis 2.68.Determinethe dry density andsaturateddensityof thesoil.AIsodetermineitsbulk densityandmoisture content,if thesoil is 5A%saturated. + Required: Solution: Given' |TZJ Saturated densityof thesoil,

lu, = f]f.U

to'=9o- = ff#i -

ffiff(l)

(t) = 2'o,gm/cc

= L62sm/cc

Whenthesoil is 50%saturated, its bulk density G + se 2.68 + (0.5)(0.65)

Y - ffi'Y-

Moisturecontent, w Now,

0f'

or,

Moisture contenl at SOVosaturation, (0.5) (0.65) .te

yd =

*

e

-(o'll-)!?'7r) -0.55 =55c,o 0.54

,/ Probleqgf. A sample of silty clay has a void ratio of 0.8. The soil is allowed to absorbwater and its saturateddensity was found to be 1.92gmlcc. Determine the water content of the saturated sample. Method I:

It is assumed &at the void ratio of the soil absorption of water. The saturateddensity is given by, Ysat-

, w'e;W=0.12=127o

/ Problem \1 The volume and weight of a partially saturatedclay After drying in an ovenat 105'C sampleare 185cc and362gmrespectively. for 24iho!rs,its weightreducedto 326gm.If thenaturalvcid ratioof thesoil

36 =llVo 326=0.11

I+G - S€,

"-I9

Solution :

= 1'82gm/cc

- f wn r=

G'=l=, 1 . 7 6- , 1 + 0.54 G * (1.76)(1.54)= 2.71

Again,

= ('iltH]]) Dry density,

= (362 - 326)gm - 36gm

9*trl

1+u.o

or,

did not change due to

G + e [J'Yr

_ r.v2 G-(1.92r(1.8)-0.8 - 2.656

I

Problems in Soil Meclnnics and Foundation Engineering

Weight -Vo lume Relat ionships

ttG = s€, we, g€t,

Now, using the relation

se

| + w = 1q, 2.Ew

(1) (0.8)

w=A=ffi=0-30

4.32w=l+w

ort

Required water contenl = 30Vo Fig. 1.7 shows the three-phasediagram of the given soil. Let the weight rf solids be unity. kt lr be the moisture contellt of the saturatedsoil. Method II :

Now, ru =

W #,

ot, Wn = w'W" = w'l

= w gm

or,

Note : Try' to solvq / the problem assumingthe volume of solids to be unity. Problern L/. The bulk density and dry density of a partially saturated soil are 1.9{gm/cc and 1.80gm/cc respectively.The specific gravity of solids is 2.68. Determine the void ratio, moisture coirtent and degreeof saturation of the soil. Solution:

Volnrne 0f waler, Vw = wcc Now, void ratio

w=0.30=3OVo

e = 0.8

*Y = o.t s

v

We have,

t d -

Here,

ya = 1.80gm/cc, y = 1.95gm/cr

l + w

105

0(,

1.80 = ;L

%-*=#=fr=r.x,..

l + w

| + w = 1.95/1.80= L.0833 w = 0 . 0 8 3 3= 8 . 3 3 9 / o

Or'

Total volume of the soil,

of'

V=Vs+Vn

=1.?5w+w=2.?5wcc

Again, we have,

yd=

G^t* I + e

=q?9 1.80 r+e=ffi=r.cl

of,

2'25wcc

e = 0.49

Ort

vtG =se

Now,

1 2 5 v er r

O f ,

J = -

(2.68) r,,C = f(0.0833) f

/

Fig.1.7 Totalweightofthesoil,= WW n + W d = . But,

(1 + w)gm

-0.456=45.69o

Problem l$.The density of a partially saturatedsoil was found to be 1.88 gm/cc. If t[e moisture @ntent and void ratio of the soil be 24.8Voand 0.76 respectively, determine the specific gravity of solids, and the degrec of saturation.

W l + w ysar=f = LZS* yot = 1.92 gm/cc

"

Solution: We have rnd,

G+se

T= 1*.:'Y,"

...(i)

ttfr = Se

...(ii)

Weight-Volune Reht ionships Problemsin SoilMeclunics ard Founfution Engineering

397.58gmof drysoilis obtained from

Substitutingfor se in eqn(i), we get G+t*G - -laz-'ln

Y

Volume ofmoistsoil tobeused = 247.'ll cc. ,l = yd(1 + w) Now, bulk density

G(l + w)

Y ' 1-97-'rn

0r'

= (1.605)(1+ 0.105)= I.773gm/w Totalweightofmoistsoilrequired= y x V = (1.773)(247.71) gm = 439.19gm

1.88={fffirtl ( 1 . S 8 )( 1 . 7 6 ) .t -_W - 2 . 6 5 - , ;

ol'

##*

= 247.71 cc of moist soil

(ii)

/

A given soil masshasa moishrre contcnt of 10.SVoand Problcn 1.{( a void ratio of 0.67. Thc specific gravity of soil solids is 2.68. It is required to conslruct three cylindrical test specimens of diameter 3.75 cm and height 7.5 cm from this soil mass.Each specimenshould have a moisrure content of l57o and a dry dcnsity of 1.6 gm/cc. Determine :

Weight of water presentin this soil

= (439.19- 397.58)sn = 41.6tgm - 59.64 gm Weight of water finally required .'. Weight of water to be added

= 19.03gm Volume of water to be added

(i) the quantity of the given soil to be uscd for this purpose (ii) quantity of water to be mixed with iL Solution : (i) Volume of each specimen - olh

= (59.64 - a1.61)gm = 18.03 cc

Ans : 439.19gm of given soil is to be taken and 18.03 cc of water is to be added to it.

=_f.#:rf (7.s)cc Total volume of three specimens,V - (3) (82.83) = 248.49 cc

Wa = V x ld

Weightofdry soilrequired,

[

= (248.4e)(1.6) = 397.588ln Moisturecontentof finishedspecimens, w a lSVo But,

w 6r=], wd

or, Wn-w

,Wd

Weight of water in the specimens,W. = (0.15) (397.58) - 59'64 8m Now, dry density of the given soil mass, Grn

ta = 1fi

=

(2.68) (1)

ffi#

= I1'605 sft/cn

i.e., 1.605 grn of dry soil is obtained from 1 cc of moist soil

"-*"]

E)(ERCISEI f A soil sample has a porosity of.35Vo.Thesoil is 7SVosafiiratedand J.l. the specific gravity of solids is 2.68. Determine its void ratio, dry density, bulk dercity and moisture content. [Ans : e = 0.54,ld - L.74gm/cc,l = 2.0 gm,/cc,w -'l57ol 1.2. The mass specific gravity of a soil is 1.95, while the specific gravity of soil solids is 2.7. If the moisture content of the soil be 22To, determine the following : (i) Void ratio (ii) porosity {iii) degreeof saturarion(iv) dry density (v) saturateddensity. , [Aor : (i) 0.69 (ii) 4leb (iii) f]6% (iv) r.597 gmlcc (v) 2.00 gm/w I The saturatedand dry densitiesof a soil are 1.93 gm/cc and 1.47 Vl. gm/cc respectively. Determine the porosity and the specific gravity of the solidSris. [Ans : n = 45.9Vo,G=z^721 l\9, A partially saturatedsoil samplehas a natural moisture content of l7%band a bulk density of 2.05 gro/cn.If the specific gravity of soil solids be 2.66, detennine the void ratio, degreeofsaturationand dry density ofthe soil. What will be the bulk densiw of the soil if it is : (i) Fully saturated

)

,t

A

Problemsin SoilMechanicsand FoundationEngineering

Weight -Volume Rela t ion slips

(ii) 6O%saturated? [Ans : Part | 1s = O.52,s = 8'77o,\ a = 1.75 gm/@Part2 : (i) 2.09 gmlcc / /

23

1.12. In problem 1.11,what will be thewater contentand bulk density of the soil if, without undergoingany change in the void ratio, the soil becornes: (i) Fully saturated

(ii) 1'9s gm/cc l

l"/. An undisturbedsoil samplehas a volume of 50 cc and weighs 96'5 gm. On oven-drying, the weight reduces to 83.2 gm' Determine the water content, void ratio and degreeof saturationof the soil. Given, G = 2.65' =72%7 [Ans:w =l6Vo'e =O'59,s I Lfr. The bulk density and dry density of a soil are 1.95 gm/cc and 1.58 gtn/&'..spectively. Assuming G" = 2'68, determine the porosity, water content and degreeof saturation of the soil. =89.2o/ol [Ans: n =4l7o,w =23Vo,s 1.7. A cylindrical sampleof saturatedclay,7.6 cm high and 3'8 cm in diameter,weighs 149.6gm. The samplewas dried in an oven at 105"C for 24 hours, and its weight reduced by 16.9 gm. Determine the dry delsity, void ratio, moisture content and specific gravity of solids. = = = [Ans : 1a = 1.54 gml cc, e 0.74, w 12.7Vo,G 7'68]

(ii)807o saturated [Ans : (i) 2270;2.04gm/cc,(ii) 17.7Vo,L97gnlccl 1.13. A 4 m high embankrnent, with a top width of 5 m and side slopes of 1 : 1, has to be constructedby compactingsoil froln a nearbybqrrow pit. The unit weight and naturalmoisturecontentof the soil are 1.8 tlmr ancl8%, respectively.Detenninethevolume of earthto be excavatedfrorn the borrow pit and the quantity of water to be added to it tbr every krn of finished embankment, if the required dry density and moisture content of the etnbarrkrnent soil be 1.82grn/cc and l87a respeclively. Given, G = 2.j0. [Ans : Vol. of excuvation= 39304m3 ; Vol. of water = 6552 m3]

1.8. Thc moisture contelt a-ndbulk density of a partially saturatedsilt ' respectively. The sample was kept in an sample werc l87o and 19.6 ttft oven at 105' C for 15 minutes, resulting in a partial evaporatiou of the pore water. The bulk density of the sample reducedto 18.3 kN/m'. Assuming the void ratio to rernain unchanged, determine the final water content of the sample. what would have been its bulk density if the sample was kept in the oven for 24hours ? [Ans : 107o,16.6 kN/m3] 1.9. An embankment was constructedwith a clayey soil at a moisture content of 127o.Just after construction, the degree of saturation of the soil was found tobe 55To,The soil absorbedwater during the monsoon and its degreeof saturationincreasedto9O7a Determine the water content of the soil at this stage. What will be the degree of saturation if the moisture content reducestoSVo mthe dry season? Given, G =2.68. lAns:19.67o,27'9%ol 1.10. The natural moisture content of a soil massis 117o,while its void ratio is 0.63. Assuming thc void ratio to remain unchanged, determine the quantity ofwater to be addedto 1 m' of this soil in order to double its moisture ContenL Given, specificgravity of solids =2.72. [Ans : 183.3 kg]

I

1.11. The in-situ density of a soil mass is to be determined by the cote-cutter method. The height and diameter of the core are 13 cm and 10 cm respectively. The core, wien full of soil, weighs 3155 gm, while the self-weight of the empty core is 150 gm. The natural moisture content and the specific gravity of solids are IZlp and 2.66 respectively. Detennine the bulk density, dry density and void ratio ofthe soil. = [Ans : y= 1.87 gmlcc,ya = 1.67gm/cc, e 0.591

I, ,4t)

25

Index Properties and Soil Classificatian Wr = empty weight of PYcnometer. Wz = weight of pycnometerand dry soil' % = weight of pycnometer,soil and water' I4/c= weight of pycnometer filled with water. Now, weight of soil solids = Wz -Wt

2

and, weight of an equal volume of water = (Wa - W) - (Ws * Wz)

INDEX PROPERTIES,ANDSOIL CI.ASSIFICATION

G =

Wc-Wt-W3+W2

wz-wr

...(2.r)

This is determinedin the laboratory by the 2.3 Particle Size Distribution: rnechanicalanalysis,which consistsot

Various physical and engineeringpropertieswitb the 2.1 Introduction: help of which a soil can be properly identified and classifiedare called the index properties.Such propertiescan be broadly divided into the following two categories:

(a) Dry mechanical analysis or sieve analysis: In this method the sample is sieved through a set of sievesof gradually diminishing opening sizes. The percent finer correspondingto each sieve size is determined and thc resulls are plotted on a semilog graph paper to obtain the particle size distribution curye. However, tlis method is applicable only to lhe coarser fractionsofsoils and not to the silt and clay frictions as sieveshaving open sizesless than 0.075 mm are practicallyimpossibleto manufacture. (b) Wet mechanical analysisor lrydrometer analysis:- The percentage of tiner tiactions (i.e.,silt and clay) in a soil canbe analysedindirectly using a hydrometer.The rnethod is basedon Stokes' law which statesthat the terminal velocity of a falling spherein a liquid is given by

(a) Soi/ grain properties: These are the properties pertaining to individual solidgrainsandremainunaffectedbythe stateinwhich a particular soil exists in nature. The most important soil grain properties are the specific gravity and the particle size distribution. (b) SoiI aggregate properti€s: These properties control the behaviour of the soil in actual field. The most important aggragateproperties are: (i) for cohesionlesssoils: the relative density (ii) for cohesivesoils: the consistency,which dependson the moisfure content and which can be measured by either tie Atterberg limits or tht: unconfined compressivestrength.

, = t"irut' ,t

The specificgravity of a soil can be detcrtninedby 2.2 Specific Gravity: a pycnom€ter(i.e., a specificgravity bottle of 500 ml capacity).Fig. 2.1 givcs a schematic representationof the process.Irt,

...(2.2)

where, y" andy- arethe unit weightsof the sphereandtheliquid respectively D = diameterof the sphere p = absoluleviscosity of the liquid nl,l

Fig. 2.2 shows the sketch of a hydrorneter. After irnrnersing the hydrorneterin the rneasuringcylinder containingthe soil-watersuspension; I

readingsaretakenat ;, 1, 2, 4, 8, 15,30, 60, 120,and 1440minutes.Lrt 11 a bc thereadingofhydrometerat time r. The particlesizeand thecorresponding value of percentfiner are obtainedfrom the foilowing equations:

WT ( EmptY Bot)

Wt YIZ W3 ( B o t . * S o i l + W q i e r ) lBot.*DrySoit) { Bot + Woter)

D =\@.

...(2.3)

liig ) |

(A

26

Problems in Soil Meclnnics and Foundation Engineering

and, where,

/v=

Y

s

V (r1 + C^ - rn) x IA}a/o W"'y-

*-'' D = particlesize in mm

...(2.4)

analysis, then the percent finer, N , of the particle size D rrun, with respect to the total quantity of sarnple,is given by'

.|y'' = N "

= unit weiglrt of soil solids = G" . y_

Is V tu,

27

Index Properties and Soil Classification

w.

-T

= unit weight of distilled water at the room temperature

995

t = time interval in sec r1 = reading of hydrometer in suspensionat time t

1000

Z, = distancefrom the surfaceofsuspensionto the centreofgravity of hydrometer bulb at time /, which can be determined from :

I

I

10 0 5

p = viscosity of water al room temperaturein gm-sec,/cm2

ya\ r(, Z,=Hr+;lh-;l * ^ / where, V1 = volume or lyoroor.t.'),n ."

...(2.7)

w

L

W . LL. e v e l Immersion I n i i i a tW . L l

...(2.s)

A = areaof cross-sectionof measuringcylinder in cm2 Hr = distancebetweenthe surfaceofsuspensionand the neck of

t

j l+l lz r t l l, -L

'i V h /2A -T

bulb, in cm

I

lr = length of the bulb in crn

I

The distance fl rnay be rneasuredby a scale. However, a better proposition is to determine.F/1 from the following e.quation:

Hr= where,

(ra+I)-11 r4

x L

...(2.6)

r,t = differencebetweenthe maximum and minimum calibration markson lhe stemof hydrometer L

= lengtb of calibration( - length of stem)

In eqn.(2.4), f{ = percent finer. V = Volume of suspensionin'cc I7, = weight of dry soil takenin gm r- = readingof hydrometerin distilledwaler at roorn temperature Cm = Ireniscus correction If t{2,be the weight of dry soil passing through the 75 p sieve during sieve analysis,which is subsequentlyused for bydrometeranalysis,and if I{2,be the total weight of sample taken for combined dry and wet mechanical

Fig.2.2

Fig. 2.3 showstypical particle size 2.3.1 Particte sizeDistribution curv& distribution curyesfor varioustypesof soils.CurvesA, B and C represeuta uniform soil, a well gradedsoil and a gap gradedsoil respectively' With referenceto the particlesize distributioncurve of a given soil, the following two factorsare helpful tbr defining tbe gradatiottof the soil: (i) Uniformity Co-efficient: = .^u D

g Dto

...(2.8)

(ii) Co-efficientof Curvature: (Dro)2

"=Dto"Doo

...(2.e)

Problemsin Soil Mechnni.csand FoundatianEngineering

28

29

IndexPropertiesand SoilClassificatian 100 90

A

Yd = in-situ dry densitYof the soil.

80

t

70 I

On thebasisof thc relative density,coarse-grrinedsoils areclassifiedasloose, medium or denseas follows:

I

60

I E

50

1!'

aul

(s)

30 20

(r

z u-

J 0

r0

F8,2.3 where, Dfi, Dpand D6grepresenttheparticlesizesin mm,corresponding to l0%o,307o and 6O7ofrnet respectively' When

Cu 15, the soil is uniform Cu = 5 to 15, the soil is medium graded.

is mcdium

t . n o 1' , thesoilis dense. If the water content of a thick soil-water mixture 2.5. Aficrbcrg Limits: is gradually reduced,the mixture passesfrom a liquid stateto a plastic state, then to a semi-solid state and finally to a solid state. The water contents corresponding to the transition from onestate to another are called Attefterg limits or consistency limits. These limits are determined by arbitrary but sbndardised tests. In order to classify fine-grained soils on the basis of their consistency limits, the following indices are used: ...(2.12) Io = w1 - wo (D PlasticitYIndex,

Cu > 15, the soil is well graded. Again, for a well gradedsoil, the value of C" should lie between I' and

tn" soitis loose f ,

f . n p . J, o" *il

s

t0

o.oot0.002o.oo5o.0l 0.02 0.0s0'l 02 05 o'El P A R T I I LS E I Z E( m m ) - - *

If OsRes

d\ptiditY

Index,

,

R., = where,

o "t*

-

€max - €min

= natural void ratio in the field.

The relative density of a soil may also be determined from: Ya - Yddn Ydmax

where,

u

Ydmax

- Ydmin

Ydmax = maximum dry density of the soil Ydmin = minimum dry density of the soil

wl-wp

Wl-Wn

|9l-Wn

Ip

wI-wp

...(2.13) ...(2.14)

w1 t wO and ltz stand for the liquid limit, plastic limit end the na0ral water content of the soil.

(iv) Flow Index (I): It is defincd as the slopeof the w vs. loglg JVcurve obtained from the liquid limit test. wl -r=w7= ...(215) i.e., 'II, - , lqls N2/N1

€min = void ratio at the denseststate

n^ o =

whete,

...(2.10)

emax = void ratio of the soil in its loosest slate e

(iiD Consistency Index, I"

e

wr-wP

Ip

'

3. It is a measureof the degree of compactnessof a 2.4. Relative l)ensity: cobesionlesssoil in the state in which it exists in the field. It is defined as,

wn-wP

t i = T

wbere,

..(2.rr)

N1 and N2 are the number of blows corresponding to the water contents w1 and ul.

(v) Toughnessindex, ,r -

...(2.16)

?

(vi)ActivityNumber,, ffi Soils can be classified accordingto various indices, as follows:

...(2.17)

30

Problems in Soil Mechanics and Foundatian Engineering

a)

Clqssification according to tle plasticity index:

Plasticin Index

Degreeof Plasticity

Typeof Soil

0

Nou - plastic

Sand

17

Highly plastic

Clay

(b) Classilication according to tlrc liquidity index: A soil for which the liquidit i-solid or solid state. The soil is very stiff if { = 0 (i.e., w, = wp) and very soft if .I1= I (i.e. wn = w) Soils having I1> | arein the liquid state.For most soils, bowever,I lies between 0 and 1. Accordingly, the soils are classified as follows: I1

Consistency

0.0 - 0.25

stiff

0.25- 0.50

Medium to soft

0.50- 0.75

Soft

0.75- 1.00

Very soft

(") Clottrft"ofion orrordiog ,, The activity nurnber of a soil representsthe tendency of a soil to swell or shrink due to absorption or evaporationof water. The classification is as follows:

Index Properties and Soil Classification

In order to detennine the shrinkagelirnit, a sampleof soil having a high rnoisture content is filled up in a mould of known volume. The mould containingthe sampleis then kept in the oven at 105'C for 24 hours.After taking it out from the oven, the weight of the dry soil pat is taken and its volume is rneasuredby the mercurydisplacementmethod. Fig.2.a@)an 7.25

Active

2.5.1 Determimtion of Shrinktge Limit: The shrinkage limit of a soil is defined as the water content below which a reduction in the water content does not result in a decreasein the total volume of the soil. This is the minimum water content at which a soil can still be saturated.

c) Drystste

Fig.2.4

!. -u s = -

Activity Number

31

wn Wd

At the initial stage, weight of water = Wo -Wa, Weight of water evaporatedupto shrinkagelimit = (Vg - V)yn

W*=(Wo-Wi-(Vo-Viy* - Wi - (Vo- V) t* *^" _(Wo w d Method II: Let

WhenG is lotown: % = volume of solids

...(2.18)

Problcms k Soil Mechanics and Foundation Engineering

V,-+

hdex Properties ond Soil Classification A

wd c4*

-

w n - ( v a - % ) r - (n' - b*)'"

But,

-

-

Va'ln l?s

ws-

G wd/G

.\"

- VTa ' l n

l

w

1

1

silt

so

50

v1'

"/" OF S I L T

Particle sbe (mm)

/

< 0.002

Fig.2.s

0.002ro 0.075

the soil is then detenninedaccordingto the narneof the segrnentin which the inleisectionpoint lies.

Sand: (i) Fine sand 0.075 ta 0.425 (ii) Mediumsand 0.425 to 2.0 (iii) Coanesand 2.0 tCI 4.75 Gravel

€ o e

...(2.20)

2.5. Cbsslficetlot Bercd on Prrticlc Sizc : Soilsrrc classifiedas clay, silt sandend gnvcl on thc brsis of tteir particlc sizes.IS:1498 - 1970 recommendstbc following clessification:

Clay

o^^*

...(z.re)

G

t-e

Soil Type

o o

10

40

o
\

(2.69)(t nt v, , 4 e o= t + 0 . 2 f r = z l 4 7 g m / c n

2.86

I z Z e r oa i r v o i d s

.--q-:
250L block Hence the designedg4onpof piles is safe from the considerationof failure. ,/ e'rcn footing founded at a depth of 1'5 m below G'L' Pr.oblen n.d a dense in a 19.5 thick stratum of normally consolidated clay underlain by diaand m L2 piles oflength 16 sand layer, is to be supportedby a'groupof carried be to load gross The fonnation. rneter 400 mm arrangid in a squari The piles are by the pile group (including the self-weight of pile cap) is 350 t' level. The ground rhe at is located rable *uter ;;"""d at r.2 m "/". tn. are: soil foundation propertiesofthe w =32Vo, G=2.67, L.L= 4tVo consolidation settlementof the pile group' probable the Estimate Solution:WithrefererrcetoFig.L2.4,theloadfromthepi|egroupis point, assumedto be transmitted to the foundation soil at the lower one-third )"t" tZ = 8 m below the pile cap and 8+ 1'5 = 9'5 m below i.e., at a depth of = G.L. The tirickness of the clay layer undergoing consolidation settlement and m m, 3 3 thickness of sub-layers three 10 m. Let us divide this zone into 4 nr resPectivelY. The settlementof eachsub-layermay be obtainedfrom:

p, = H' f; Now. we have,

'tor,oo 0 +

w G = se, or' e =

eo= v c -

Ao

wG s

and,

- 1'00 = 0'90 t/m3 Ysub= 1'90

= (4 + 3 tan30")2= 32.86rt 350 ^ o = for = 3 L s' -6 10.65t/mz

(300)(0.27e) ,-- e.e-tlq4l = 14'32cn' r'",= f1 frffi'' losto tr of thesecondsub-layer: Settlement os = (0.90)(1.5 + 8.0 + 3'0 + 3'O/2) - I7'6t/m'

,,

A2 = (4 + 2 x 4.5 x tan30")2= 84'57ri

oo=*=#h=4.r4t/m2 (300)(0.27e) ,^- 12'61]!-! = 5'57cm losto P",= fr'dffi' tr Settlementof the third sub-laYer:

lo = ffi

(II - 10) = 0.009(41 - t0) = 0'279 0.0@09

?#r*=Ti..,'s1

= (B + FI, tan30';2

A3 = (4 + 2 x 8 x tan30')2 = 175'?3m2

ryP=08s4

Ysar=

Assurningtheloadtobedispersedalorrgstraight|inesinclinedtothe horizontal at 60", the area over which tf e grossload is distributed at the rniddle of thc first layer, At = Q + 2H/2'tan30') (B + 2H/2'tan30')

o6 = (0.90) (1.5 + 8.0 + 6.0 + 4'0/2\ = 15'75t/r]

o6

AgaiIr,

33t

Pile Foundations

(1)= l'eotzm3

Settlenrentof the fint sub-laYer: oo = initial overburdenpressureat the middle of the layer = \' z = (0.90)(1.5 + 8.0 + 3.02) -- g'st/^z Ditnetrsionsof the block of piles, L = B = 3 s + d = 3 ( 1 . 2 \+ 0 . 4 = 4 m

= r.997t/mz

15.75+ L99J - 3.12cm (400)(0.279)' roglo 'pca 1SJS 1 + 0.g54 = + Totalsetflement,Pc P., * Pc, Pr. = 14.32+ 5.5'7+ 3.12 = 23 cm. EXERCISE12 of an RCCpile drivcn l2.l. Determincthesafeloadcarryingcapac:ity by a drophammerweighing3 t andhavinga freefall of 1.5rn, if theaverage [Ans'20'3t] p.n.nrtion for thelastfiveblowsbe 12mm'

332

Problems in Soil Mechnttics and Fottndation Engineering

of 10 rn 12.2. An RCC pile having a diarneterof 400 rnrn and a length free tall of a height with kN, 30 weighi*g natruner is bei.g driven with i Orop recorded been has blows few last the for penetration of 1.2 rn. The average co-efficient or as 9 rnrn. If the ettlciency of the hammer be 7O% aud the using lnoditied restitution 0.50, detenninettre safe load the pile can carry a factor of = Assume Hiley's fornula. Given,unit weight of RCC 24 kN/m'' 200 kN] Ans' I safetyof 3.0. into a 12.3. A22 m lorrg pile having a diameterof 500 mm is driven 5.6 of strength compressive deep straturn of sofl clay having ai unconfined to a respect with pile the of capacity t/#. Detennine the staticload bearing tl 40 tAns' facror of safety of 2.5.

=16. friction angle = = 10, Nv = 4' [Ans. 279 kN] Nq = ?.6, q Nc 20', for an RCC pile of 500 mm 12.8. Deterrninethe ultimate load capacityof sub-soil conditions are The colutnu' a of diarneter supporting the tboting skt:tchcdin Fig. 12.5. Given' = 0'9 adlrcsiontactor tbr soft clay = 0'7 silt claYeY and that t'or for 0 = 30' is 9'5' The water table is factorNu capacity bcaring Vt'sit"s neglected' [Ans' 232 t] lrx nlcrl rtt il gr(raldcpth.Skin friction iir sandmay be

Soft CtoY

l2.4.Aconcretepileof30cndiameterisernbedd-edinastratunrofsoft clay straturnis clay lraving 1 = 1.7 t/rn3, Qu= 4'2 t/mz'Thethickness of lhe g m and the pild penetratesthrough a distance of 1.2 m into the underlying = 36"' Detennine the sat'e straturnof de;rsesand,havilrg Y = 1'85 t'lm3 and Q of 3' safety of load carrying capacityof the pile with a lactor capacity faclor = bearing Vesic's = 36', Given, O O.gOQ and for Q [Ans.32.3 t] Nq=23,c[=1,K"=1. is driven 12.5. A stnoothsteelpile of 8 m length and 400 rnm diameter properties: into a cohesionlesssoil masshaving the following = 30' Y"ar= 1.8t,zrn3' Q = 0'60 Qand Vesic's The water table is locatedat the ground level' If 6 the safecapacityof = determine 9'5' be 30" bearingcapacity faciorNn for 0 = 0'7. Ks Given, [Ans' 12'1 t] of 2.5. rhe pil! with a iactor of sifety at a 12.6. A 12 m long pile having a diameterof 300 mm is cast-ih-situ site where the sub-soilconsistsof the tbllowing strata: = 10kN'/m2 StratumI: thiclness =5 m, Y' = 10kN,/m3, 0=30" c = = kN'/m2 Stratun II: thiclness= 16 m, Y'= 9 kN'/m3, 0 0', c 60 Detenninethesafeloadonthepilewithafactorofsafetyof2.0.Assume ieasonablevalues for all other data' of 500 mm is 12.7. A 16 m long bored concretepile having a diameter properties following the having silt ernbeddedin a saturatedstratum of sandy 'Yru,= 19'5kN'/m3' c = rlkN'/m2'

0 = 2o'

with a factor of Detennilte the safe load canying capacity of the pile safety of 3.0. Given, = O'75 adhesionfactor = 0'85 pressure ofearth co-efficient

JJJ

PiIe Foundations

10m

Sitt Ctoyey 1 y = 1 . Et /5n F ) , c= 6 l l m 2

I

J

2.0m

T

I

i I I

Ssnd (t=1.75t/m3,@=30o) Fig.12.5

6 rows wi-th a 12.g. A pile group consists of 42 piles anan$ed in pile is 22 rn long Each centre-to_centiespacing of 1,5 rn in each direction. using: pile the of and 500 mm in diameter.Find oul tbe group capacity (i) Convene-Labane formula (ii) tns Angeles formula' Given, load bearing capacity of each pile = 78 t' q [Ans. (i) 2142 t (ii) 2624 12.10. A pile group consistingof 25 piles anangedin a sqlare fonnation are L5 m and is to support a iaft iooting. The length and diameterof eachpile soil is 300 mm respectively,wiile their spacingis 85ocmc/c Thg-foynfation Determine y 1'85 VT'' a normally consotiAatedclay having c = 5 t/mt and = F" = 3'g' the safe load bearing ""p""ity of thi pile group' Take cr 0'85 and [Ans' 527 t] placed 12.11. A multistoried building is to be supportedby a raft footing piles 96 of consists raft on a pile foundation. The pile group supporting the water c/c'The m of 2'0 of 26'm length and 400 mm diameter,with a spacing table is located near tle ground surfaceand the propertiesof the foundation soil are as follows:

334

Problems in Soil Meclwnics snd Foundation Engineering Y " . r = 2 . 0 t / r n 3 c, = 3 . 6 1 / m 2 , O = 0 ' .

The adhesionfactor may be taken as 0.95. Determinethe capacityof the pile group with a factor of safetyof 3.0. 12.12. Designa pile groupto supporra raft footing of 8 m x 12 m size and carrying a gross load of 760 t. The self weight of the pile cap rnay be assumedas 20o/oof tlre gross load on footing. The subsoil consists of a homogeneouslayer of soft clay, extendingto a great depth and having the following properties: y' = 0.85 t/nr3, qu = 5.7 t/m2 Design the pile group with a factor of safety of 3 againstshearfailure. Given. a = 0.85. 12.13. It is required to drive a group of piles in order to support a raft footinqof 10 m x 10 m plan area,and subject to a gross pressureintensity of 15 Vm". The subsoil consists of a 12 m deep layer of soft clay (y = 1.8 t/rn3 , qu = 4.5 Vm2) which is underlainby a densesand layer (y = 2 tlnr3 , 0 = 35'). The raft is founded at 1.5 m below G.L. In order to utilize the bearing resistanceofthe sand layer, each pile should penetratethrough it at least 4 D. The adhesionfactor for clay = 0.90. Vcsic's bearing capacity factor .lfu for 0 = 35' is 18.7. Design a suitable pile group with a factor of safety of 2.5 againstsbearfailure. Assume that the self weight of pile up = 25Voof pressureintensity on the raft. 12.14. A raft footing is founded at a depth of 3.5 m below G.L. in a ?A rn thick stratum of soft clay having the following properties: y""1= 2.05 t/m3, C, = g.3 The gross load to be carried by the pile group, including the self weight of the pile cap, is 8O0L The group consistsof 81 piles of 400 mm $, arranged in a square formation, and extended to a depth of 12 m below the pile cap. The spacingof the piles is 1.25 m. The water table is located at the ground level. Cornpute the probable consolidation settlement of the pile group.

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