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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'

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,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

<7

Inw plastic

silr

7-17

Medium plastic

Silty clay or
clayey silt

>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<[2.4(c)representthe schematicdiagramsof the initial and
final statesof the samplewhile Fig. 2.4(b)representsthat conespondingto

I

Vo

l_

o) Iniiiot Slste

b} Af S,L

the shrinkage limit. With referenceto thesefigures, the shrinkage limit can
be determinedby the following two methods:
Method I:
Wrcn G is unbwwn :
LetVs andV1be the initial and final volumes of the sample and Wg andW6
be its corresponding weights. By definition, the volume of the soil at
shrinkagelimit is equal to its final volume. I*tWnbe the weight of water at
this stage.The shrinkagelimit is then given by,

Typeof Soil

< 0.75

Inactive

0.75- r.25

Normal

> 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<

lYa

ws

olt

\uu

wd

-

Vd.ln

EO

4.75

ro 80

2.6,1. Tcfrtral Cbssiftution
Systamz Any soil, in its natural state,
consistsofparticlcs ofverious sizes.Onthebasis ofthe percentagesofparticle
sizes, and following ccrtain definite principles, broad classification bf such
mixed soil is possiblc.
Fig. 25 shows thc triangular classification chart of the Mississippi River
Comrnission, USA" It essentiallyconsistsof an equilareraltriangle ABC. The
percentagesof sand,silt and clay (ranging from0%ota L0O7o)ate plotted along
the sides AB, BC and CA respectively. The area of the tiangle is divided into
a number of segments and each segment is given a name. In order to find out
the group to which a given soil belongs, three lines are required to be drawn
from the appropriate points on tbe three sides along the directions shown by
the arrnws. These thrcc lines intcrscct at a single point. The nomenclatureof

This chart is usetul for identifying and classifying
2.?. Plasticity chart:
fine-grained soils. In this chart the ordinateand abscissarepresentthe values
t-rfplasticity index and liquid li[rit respectively.A straightline called A-line,
representedby the equation I p= a'73 (wr- 20), is drawn and the areaunder
the chart is divided into a number of segmen8. ou the chart any fine-grained
soil can be representedby a single point if its consistencylimits are known.
The segrnentin which this point lies determinesthe name of the soil.
Fig.2.6shows a plasticitychart.The meaningof thesymbolsusedinthe

chart are as follows:

M
C

o

L
T
H

Silty soils.
Clayey soils.
Organic soils.
l,ow plasticity
h{edium or intermediateplasticity
High plasticity

Main groups of fine-grained soils are
ML, MI, MH - Silty soils

G


Problems in Soil Mecfuinics and Foundation Engineering

34

Indet Properties and Soil Clossiftcation
I.S, Sieue

50

Diameter
of Grains
(mm)

35

Weight
%
Retained R.etaincd
Gm)

Cumulative To
Retained

Vo
Finer

:;

a.75mT

4.75

9.36

1.87

1.87

98.13

- 4 0

2.40mm

2.&

53.75

10.75

12.62

87.38

1.20mm

t.2n

78.10

t5.62

28.24

71.76

500 p

0.600

83.22

\6.64

44.88

55.r2

425tt

0.425

85.79

t7.16

62.04

37.96

300p

0.300

76.82

15.36

77.4)

22.60

150p

0.150

6't.02

13.40

90.80

9.20

75w

0.075

33.88

6.78

97.58

2.42

o\

OJ
C

- 3 0
>. 29
!:

ii

--+-I

M H /O H

I
ro
7
I

0

30

40

50 54 60

70

Liquid Limlt ("hl *
Fig.2.6
CL, CI, CH
OL, OI, OH

* Clayey soils
- Organ-icsoils.

EXAMPLES
/
The results of a sieve analysis performed on a dry soil
Piroblem*l.
sample weighing 500 gm are given below:

The particle size distribution curve is shown in Fig.2.7.
(ii) The required percentagesobtained from the curve are as follows:
Gravel:

- 1.97o
1.8770

Coarsesand:

98.lVo -927o

Mediurn sand:

92Vo-38/o

= 549'o

Fine sand :

-2.4%o
38%b

= 35.60/o

SiIt:

2.42%;2.4%
Mediumj [oqr

Fi n e
Sond

= 6.lVo

t00
90

to
70

,osf% L
l
e
t

(i) Plot the particlq size distribution curve of the soil'

(ii) Find out the percentageof gravel, coarsesand, medium sand, fine
sand and silt presentin the soil,
(iii) Determine the uniformity co-efticient and the co-efficient of
curvature. Hence comment on the type of soil.

sol
I

u

.E
-

rg_f s
3 o l

r

torl'r* t

Solution: (i) The computationsnecessaryfor plotting the particle size
distribution curve are shown below:

0.010,020.040.060.t0e 04 0.6 |
2
P q r t i r l e S i z e( m m l +
Fig.2.7

lo-{,*r.

s ro-T

t


Problems in SoilMeclutnics and Fortntlation Engineering

36

in a sieveanalysis'
Problem 2,/. 500 grn of dry soil samplewas used
''as
i' the steel
collcctcd
p.ssed firough tie T5 p sieve.and
178.;;;;;;i#i1
rnadeby
was
suspension
*u,-ttktn and a 1 litre
fun, ouTof which 50 grn
cylirrder
measuring
a
it
in
agent-to
^OOiogdistilled water and dispersing
'il" uoiuoit of the hydrometcr was 50 cc' the
navinl a diam,eler-of'6-15-crn'

e'Jll,:Thc
of calibrarion^9L+.*:i"
is.s.t atrdthelength
-steln

fi;d?;iilt,

Index Properties and Soil Classificatian

wc ri 090 a*d l 040 icspectivel y'

rni'iururn a*d rnaximurn rnarksiii-ft s
Ahvdrorueter{estwasthenperforrnedattberoomlenperaturtlof25"Cand
the tbllowirrg readittgswere recorded:
Elcused time (min)
Hydrometer reading

I

2

I

;L

I

4

8

l5

30

As

G =7.67,

A.1 ?-5"f,

Z,

i,i}ti

^!s=2'67gitt/cc''
1o = 0,9971En/cc,

'r l{}-1i i;l:-i-.s'r:r:ict{!2

-

Yr \

t\" o)

=
f,rs'tsl' 2elo6cnz
- so/zs.7o6)

= Ht +
lot.t
Z,=Ht+6'908

or,

...(ii)

Using eqn.(2.6),
H1 _ ( r a + 1 ) - r r ,L

r4

r a = t ' 0 4 0 - 0 ' 9 9 = 0'05,

Here,

L = 9.7 crn,
f 1 + 0 ' 0 5 -r1
H . ' = \l
0.05

(e'7)= 194(1.05- rt)

...(iii)

Agairr, o/ofiner on 50 grn of soil

" =,,k

t

*

r(.

=

Hcre,

or, ly' =

"
x i0'-"'
8.95' --.j
----=
llili- :ir'-/ ctn
p = 8.95 rnillipoises

Here

Z,=Ht+

tB24 1023 1t)201 A l 7 101310i{i ic06 1 C 8 i

o = V t s o o P" { z :

...(i)

Usingeqn.(2.5),

60

containing the
Whrrn the hydrometerwas immersethereading
susptnsiorr,
tilt:
in
prt:serrt
thal
as
age-nt
sanrequarrtityof dispersing
grn/ccand
is$.99'
/l
*'ater
of
ggg.5.
unitweight
lt-?s;C,the
*as found tobe
is
solids 2'67 'The
i1svfs6:osit!is 8.95 millipoises.The specificgravity of soil
meniscus correclion rnay be taken as 0'5'
to each
Find out the diameter of particles settied corresponding
ck

volumetrir':
Negiccl
fir{letv:+ltrcs.
hydroincter reading and the respcctive
expansiondue to temperaturcchange'
The terrrperaturot:orrectioll al-lrjil:r: *isittlrsiriii agent
Solijtion:
neecl-notbe applicd here'
correction
'
t/o
using eqn'
The diameier ald ccnesponding tiner rnay be .
it will
int'olve'd'
are
calculations
as
repetitive
12.:; tnrougn (2.?). Howevei,
forms by sul-)stituting
t" uarrunt"!"ous to reducetheseequationsto simplified
col$tant
rernain
wbich
factors
the values of the
Using eqn. (2.3)

Y"-Y*

i/L
D -- 0'09e1 Y ;

()r,

()r,

ir,lc(rt

2.61
"
2'67 - 0.9971 +f

+ c . - r - )x l o o

x 100
x0'ee71(r1
+ 0'0005- 0'eee5)

N = ? , 1 8 2 ' 8 ( 1 1- 0 . 9 9 9 )

...(iu)

Vofiner on 500 gm of soil takeninitially

N'=N.#=0.3s77N

...(v)

Eqn. (i) through (v) rnay now be usedfor the computations'The results
are tatrulatedbelow.

/ .. ,-2.


Problems in Soil Mech.anics and Fottndation Engineering

38
Time

Hydrometef
reoding

n

60
120
24
480
900
1800
3600

r.024
r.023
r.020
L.Ol7
1.013

1.010
1.006
1.001

p =
t-

L
0.0ss1v
I

0.eee)
(%)

g")

5.O4

0.0625

79.57

?3.6

76.39

27.32

5.820 12.728

0.oM
o.0323

66.84

23:9L

6.402 13,310

0.0233

57.29

20.49

7.\78

14.086

0.1697

44.56

t5.94

7.1ffi

14.668

0.0126

35.01

t2.52

8.536 t5.444

0.0092

22.28

7.97

9.506 16.4L4

0.0067

6.37

2.28

r1.952
5.238 tz.16

Total weigbt of unit volume of suspension
W = W" t W- = 0.0599+ 0.9778= 1.0377grn'
Density of the suspension= 1.0377gm/cc o 1.038gm/cc.

ff='
1y'=

3182.8 0.3s77
x(yrxlV

(mm)

(cm)

(cm)

(sec)

30

Zr=
I{r =
195x H t +
(1.0s6.908
ri

/
Problem 2./. Distilledwaterwasaddedto 60 gm of dry soil to prepare
I suspensionYf t litre. What will be the readingof a hydrometerin the
susperuionat t = 0 sec,if the hydrometercould be immersedat that tirne?
Assume,densityof water= | gmlccandspecificgravity of solids=2.70.
Solution: At t = 0 sec,the solid grainshavenot startedto settle.The
havingconstantdensityat any point
tberefore,is homogeneous,
suspension,
in it.
As G = 2.7o,\" = 2.70Emlcc.

Therefore,readingof the bydrometer= 1038"
Asample of dry soil (G, = 2.68) weighing 125 gm is
Problem W
unitbrrnly disp6rsedin water to tbnn a L litre suspeusionat a temperatureof
28"C.
(!)&etermi[e the unit weigbt of the suspensionirnrnediatelyatler its
prcparation.
(ii)!*cc of the suspeusionwas retnovedfrorn a depthof 20 cm beneath
was allowedto settle for 2.5 min. The dry
tbe t6-psurfaceafterthe suspension
wt:ight of the sample in the suspensiondrawn was found to be 0.398 gm.
Determinea singlepoint on the particlesizedistributioncuryecorresponding
to tbis observatiott. Giveu,at 28"C,viscosityof water= 8'36 millipoisesand
unit weightof water = 0.9963gmkc
Solution:

#=

22'22cc'

(i) Volume of solidsin the suspenrion=

#

= 46.64 cc.

Consideringunit weight of suspensiotl,
Volunreof solids present

= ffi=o.o466cc

Volumc of water presenl

= 1 - 0'0466 = 0'9534 cc

Weight of 0.466 cc of solids = (O.04ffi)(2,68) = 0.1249 gm
Weight of 0.9534cc of water at 28oC= (0.9534)(0.9963)= 0.9499gm'

weight
":T.I;'fiffi
rotar

Total volumesf solidsin the suspension

=

39

Index Properties and Soil Classification

=1.0248
sm.

grn/cc.
Theretbre,unit weight of suspension = 1..0748
(ii)

.'. Volumeof solidsin unit volumeof suspension,

We have, frotn Stokes' law,

" =t'i*J".d

,, = ffi=0.0222cc.
Volume of waterin unit volumeof suspension,
= O'9778cc'
Vn = |' O'O222

ol'

W€ightof solidsin unit volumeof suspension,
W" = (0.O222)(2m) = 0.0599gm.

Let D be the diameter of the particles settled to a depth of 20 cm at r =
2.5 min. with a uniform velocitY v.

Weight of water in unit volume of suspension,
W*= (A.W78)(1) = 0.9778gm'

p =

18p x G
Y"-T.

, = 1=
,r#*

= o.133cmlsec

Problems in SoilMechanics and Foundation Engineering

40

F = 8.36 millipoises =

8.36 x L0'3 =
8,522 x 10-6 gm-sec-,/c-m2
981

=
Ys = 2'68 gfit/c'.c, y,,, 0.9963 gm,/cc

41.

Index Properties and Soil Classificotion

55

\

^ 5 0
o\

(18)(8.522x 10-6) x fi133
2.68 - 0.9963

D =

crn

= 3.48 x 10-3 crn = 0.035 mm
=
Agaiu, at tirne t = 0, weight of solids present in 1 cc of suspension
0.1249 gm.
Weight of solids presentin l0 cc of suspension= l'249 gn'
=
At time t = 2.5 min., weight of solids presentin 10 cc of suspension
0.398 gm

0.398
x 100= 31.86%'
LZ4g
Hence the co-ordinates of the requiredpoint on the particle size

E 4 s \tlU=Ir3"h
o

D = 0.035mm

Water content (7o) |

{

\

L,

I

t

L l O
.J

+
CJ

=

\

35

\
30

lo

7, Irner =

distribution curye are:

I

/'o

zo 25 30
No.of Blows --r'

50

60

Fig'2'8
As the plasticrityindex is greater than17Vo,the soil is higltly plaslic itt
nafure.
indexis lessthan1, the soil is friableat liquid lirnit'
As the toughness
Protrlem 2S--Y^brr^tory testson a soil sarnpleyielded the followittg
results:
Liquid linit

= 547o

Plasticlimit

=25%

Natural moisturecoutent = 29o/t,
o/ofiner than 0.002 rnm = l8o/t,

32'l

(a) Determine the lkluid limit of the soiltsoil be 23To,find out the Plasticity index,
iUj ff 1tr" plastic limit ot the
comment on the nature of the soil.
Hence
flow'index anJ toughnessindex.
Solution:(a)Fromthegivendata,acrrrvebetweenthewatercorrtent

shows
and the number oiblows is plotted on a semi-log graph paper' Fig' 2'8
as
to
25-blows'
corresponding
content
wat€r
The
curve.
this w vs. loglg.lV
437o'
is
sotT
the
obtained from the curve, is 43%o.Hencnthe liquid limit of
-237o =20Va
(b) Plasticity index, Io= w1-wo= 43Vo

Flowindex, ,, = ffi
t,=If,=ft
Toughnessindex,
I

= = 38'687o
= o'sz'

(a) Determine the liquidity index of the soil aild courntenton its
consistency.
(b) Find out the activity nurnberand comrnenton the natureof the soil.

(c) Classify tlre soil with the help of a plasticitychart'
w_ -

Solution:

indcx'11=
(a) Liquidity
-,i

v'P

- ?-5
= o'138
= 29
54 - ,s
andis stiff'
As 0 < I1<0.25, thesoilis in theplasticstate
(b) Activitynumber,A =

% i.00,

^^


Problems in Soil Meclmnics andFoundation Engineering

42

s4-)s
o=--=1.611

of'

.

A s A > |..25, the soil is an activesoil.

Solution: Thc three-phasediagramsof the sample at ils liquid litrlit
and shrinftagelimir are shown in Fig. 2.9(a)and (b) respectivcly.
Let e1and e" be the void ratio of the soil at LL andSI respcctively.tct
the volume of solids be 1 cc.

---Jl e i
l 1 '
( 1 + e 1|)
r f

l
lI

{eL-e5)
-n------nF-

,$,re

:-Woter--::

--- - --I

| .t F=-wJ;-=

r

1

t l

-t_-+_
Ai L.L
(Voidrotlo = el)

AI S.L
( V o r dr o i i o = e s )

ts)

tbl
Fig.2.9

wehave,
Atliquid limit,

w

=

= er cc

.'. Volume of water present = el cc
Weigbtofthiswatcr

= €t x | = et Em

W e i g h t c . fs o l i d s

= V r ' G \ n = 0 ) Q . 6 7 ) = L 6 l gm

el

2.67

+2.67 = 0.6 , or, es= (0.6)(2.67)= L6Az
e" = (0.?5)(2.67)= 0'668
Changein volutne per unit of original volume'
Similarly, at SI,

€t - €, _ 1.602- 0.668
= 0.359
1+1.602
l+et
AY = 0.359V = (0.359)(20)= 7.18cc

Av
v

Hence,final volume at

SL = 20 - 7'18
= 12.82cc

Probfem 2.V'' T\econsistencylimits of a soil sampleare:
../
LL = 52c/o,PL=357o,5L = l7%
to 6. 1 cc
I f a spcimen of this soil shrinks from a volume 10 cc at liquid limit
solids'
of
at plasticlimit, determinethe specificgravity
Solution:I-rrelandesbelhevoidratioconespondingtotheliquid
linrit and plasticlimit.
Let volume of solids be 1 cc.
'
.'. At liquid limit, volutne of water = €l cc
= e/ grn
Weight of water

I

=Vr'G\n=1'G'L=Gce:

Weightofsolids

wn
* =

* = " t

I

w= 527o=A.52

But at liquid limit,

" = +Y,s o r , v r = e . v ,
Vu = et'l

w*
w"

w=ffiVa=0'6.

But,atII,

(c) The plasticity chart is given in Fig.2.6. The point correspondingto
=
wt 540/oand 1, = 29Vois tnarkedin the figure as P. As this point lies irr the
segrnentrnark€d Cl{"the soil belongs to the ClJgroup.
The Atterberglirnits of a given soil are,LL = 60aio,PL
Probfern {/;
= 457oand SI =25a/0.The specificgravity of soil solids is2.67. A sampleof
this soil at liquid limit has a volume of 20 cc. What will be its final volume
if the sarnpleis broughtto its shrinkagelimit?

43

Index Properties and Soil Classification

| = o . s zor,, et

= 0-52G

e s = 0 ' 1 7G

Similarly we obtain,

Now, changein volumeperunitof originalvolunie,
€t - €, -:T;
O.52G- 0-17G = 0'35G
LV

i

But,

=

t *o

=

csrd-

1 . o52G

+-{]dq=o'3e
)