Fundarnentals of
Geotechnical
DIEGo INocENcIo
q,orrEseNn
Civil Engineer
BSCE, LIT
-
Magn a Cum Laude
Sth Place, PICE National Students'Quiz, 1989
Awardee, Most Outstanding Student, 1989
3rd Place, CE BoardNovember
1
989
Review Director & Reviewer in all Subjects
Gillesania Engineering Review Center
ReviewerinMathemat*;;!rffiff;rili;:";::;:;,";ff
;:;
Author of Various Engineering Books
I
E*U"-";
i t"-
I Af"t.lt-
l/. ^
'
*"- .-
-,r4
..,;.1,:,,tl,i1,1itl''ii
i| :;.:,IIii,,I
I_r:!
l1r,
:r.l:, ,:ii:rlr,
:.r:
The cardinal objective of this book is to facilitate preparation
lirr the Civil Engineering Licensure examination Siven- bythe
l'r'crf'essional Reluhtiontommission (FRC)' Since this book
irrcludes comp6fe discussion of the principles in geoiechnical
(,n:{,ineering, ihis may also serve as a guide to the civil engineering
lilldcrgraduates.
The book is divided into 9 chapters. Each chapter presents
tlrc formula s and prtnciples in Geotechnical Engineering, followed
lry illustrative proLlems. Each step in the solution is carefully
,:xplained to ensure that it will be rcadlly understood'
F\ndamentals of
Geotechnic al Engineering
Copyright @ Zoofi
byDiego Inocencio Taparg Gilesania
//l
r@hts re:;crvc,cl. ,Yo p,ttt ct'tltis book nt:tv bc
rcproduced, ,stored in a retrieual system, or
tru1tsferrel, in any lbrm or by any means,
without the prior perntissioi of ihe author.
Most of the materials in this book have been used in my
r.cview classes. The choice of these materials was Suided by their
erl'fcctiveness as tested in my classes.
I wish to thank allmy friends andrelatles who inspired me
in writing my books, especially to my wife Imelda who is very
strpportive to me.
I will appreciate any errors pointed o1)t and will welcome
illly sugSestion for further improvement.
lsnNpTr -8614-33-8
.,
'
'"
'""t
dc,.* iS
n by t I
tc, a
u t I t or.
'': "-
' l
,r,.t "
c-ovct'
,:..:'i.i
f'{,
'
'-'.*
"L't
DIEGO INOCENCIO T. GILLESANI.A
Cebu City, Philippines
i
osqst
\r;. .l
'r. i
l**. *f
rl r''
,
"
-\.._{.:,i:.'S":
Fundamentals of
Geotechnical Engineering
Table of
Contents
my mother lluminada,
my wife Imelda,
and oar Children Kiru Deunice,
Kc,tt Dainicl,
and Karla Denise
To
13 to 54
il
Tabrle
of
Contents
Fundamentals
Geotechnical En gineering
[:unclamentals of
Geotechnical Engineering
Table of
Contents
ilt
lllustrativeProbIems'..'..'.'''......'...'
Illustrative Problems
67 to 84
Illustrative
Problems
."..."''"""""1" """""176 tol90
Chapter 06 - Compressibility of Soil (191 - 2L8)
Setilement from One-Diirensional Prirnary Consolidation " " " " " " " " " " "'
""""
Basic Settlement Formula.'.,.,..'....'...,."'
Normally
of
"
"
""""
"
"
"
Settlement
"
"
"
"
"
"
"
"
"
Consolidation
Primary
Consolidated Fine-Grained Soils
Primary Consolidation Settlement of Overconsolidated """""""""""
Fine-Grained Soils
- Overconsotidation Ratio (OCR)
-
Illustrative Problems
98 to 130
191
191
192
193
...,.::... i;3
...'...'."""""':'
Index.'....
"""""" 193
Skempton........:.'......'....
""""' 194
Rendon-Herreo............'...
""""""""194
Nishida
""194
Index...........
Swell
"""""""""194
Nagarajand Murty
' Settlement from Secondary Consolidation""""""""""""""';"""""""""". 194
-Compression
IV
Table
of
contents
Fundamentals of
Geotechnical Engineering
Table of
Contents
Fundamentals of
Geotechnical Engineering
V
-Calculatio. of Consolidation Settlemen,,
- Time Rate of
Consolidation...................
Illustrative Problems
............. 196
tllustrativeProblem .. ... '.'...
' "': ' "'"""
""""":'zor
''61' rto284
- Bearing Capacity of Soils ('*t ]..l'nlru,
.leA
Definitions ,......'..'..'..."...".'.;:..""'1"""' """"""""r """:""'
Chapter 09
""""""' 4uv
Sr:il"""""
Analys1s"""""':""""';"""":""""""'"':"" """' ?-87
""""""""':"""'""""""""'" 288
Capacity"""
""""""""""""""' 289
Equations""
Various Types of Footing on
Bearing Capacitv
Ultimate Soil Bearing
Terzaghi's Bearing iupu"ity
"""""""':"""
Failure
""""":"""""'',.:...,....;33
Failure'........'.."""""":"'r'r
nol
1""""" a2L
Allowable Bearin! Capacity and Factor of Safety"""
"""""""'291
Gross Allow"bl" B:;ti;; Capacity
"""""'"""':-'."""""""
Net Allowable BearingCapacity""""l':"
iZ1
z2 L
c"Ineri
Shear
Local Shear
llespect to Shear ...,...............
Gross Allowable tsearlng capacity with
292
Effect of Water Table on Se"aring Capacity"""""""""""""'. """""""""""'
Equation)....
".................294
G";"rrt f,"rring Capacity
li.y"*"frzuril;
Hansen's Bearing Capacity
Equation"""
"""""296
UltinrateLoadforShallowFoundationunderEccentricLoad....'.',,'.,.',..'298
penetration rest (sPr) "'""""" """""""'zee
n**l,';;;;;-;r.* ild"'J
Illustrative
Problems
"""""""""""
"""""
307 to 334
'
'opics and
Chapter 10 - Miscellaneous T
Additional Problems (335 - 406)
""""""""""
Piles
' " and n"ep fornautton """""""""""""""":"":':""i""
p,'" i"p".'i,y
i-^iintr"s
outu"""""'
Formula"""'
NavY-McKaY Formula
Eytelwein Formula
"""""""""""'
AASHTO
Engineer News Record
..,..........333
aae
JrJ
"""'
"""""""""""""""""""""""""""" 335
""" 335
(ENR) "':"""
#;;;;N;''i' n"'o'a
Modiried
Danish Fofmula
335
11/
'"1i
vt
I::'|":L
Capacity..,....._-:ThoAl^L^^^r1.,^-,
Theoretical Pile
Fundamentars or
'
Fundamentals of
Geotechnical Engineering
Chapter
"""""""3.38
0l *
Properties
of Soil
Chaptef Ol
Pro erties of Soil
I.I SYMBOLS AND NOTATIONS
e = void ratio
n = porosity
D, = relative density
G = specific gravity of solids (usually in the range 2.67 t
GI = group index
LI = liquidity index
LL
Illustrative
problems.....
362to 406
= liquid limit
MC = moisture content
H = plasticity index
Pl = plastic limit
S = degree of saturation
V = volume of soil mass
Vo = volume of air
Vs = volume of solids
V, = volume of water
W = total weight of soil
W, = weight olsolids
W. = weight of water
Ia,y = dry unit weight
fn = unit weight of soil mass
ys = unit weight of solids
ysat = s&turated unit weight
fw = unit weight of water
I.2
DENSITYAND UNITWEIGHT OF WATER
\
Density of water, p, = 1000 kg/
.
pw = 1
*t
kg/liter_= 1 gram/
Unit weigh t f water, T. = 9.81. kN/ma
cc
0,05)
Chapter
of Soil
2
t.3
0l -
Fundamentals of
Geotechnical Engineering
Properties
of
Chapter
BASTC FORMUiAS
Weight of substance , Ws = ls V" =
G
y* V"
Specific gravity of substance, G,rus = 7"uus/T*
Eq. 1.3
Eq. 1.4
1,4.3 RELATIONSHIP BETWEEN eAND n
e_nand
e
e
1,+
\*n
Eq.1.10
1,4.4 WATER CONTENTOR MOISTURE CONTENT, MC OR W
];1-{
,li+
'l'hc ratio of the weight of water to the weight of the solid particles.
MCar*= wu' xlao%
ws
l-i1
Note:
-
Eq. 1.11
0<MC<m
Phase diagram of soil
I.4.5 DEGREE OF SATURATION,
The following relationships can be made from the phase diagram shown:
Total weight of soil, W = W, + lA1,
Volume of voids, Vo = Vo + l/.
Total volume, V = Vr+ l/,
r.4.t vorD
Eq. 1.9
Note:0<n<1"
PHYSICAL PROPERTIES OF SOIL
Figure 01.1
Properties
of Soil
V
Eq. 1.1
Eq.1.2
0l -
v'
n=
""[:;fl:'"'lli;ru :';-=,:,''
I.4
Furrr.lamentals
Grotechnical Engineering
Eq. 1.5
Eq. 1.6
S
l'he ratio of the volume of water to the volume of voids.
s= v* ,L00
Eq. 1.12
V,,
Eq. 1.7
I)egree of saturation varies from S = 0 for completely dry soil and S = 100% for
ttrtally saturated soil.
RATlo, e
Void ratio is the ratio between the volumes of voids to the volume of solids of
a soil mass. It is usually expressed in percent.
I.4.6 REI.ATIONSHIP BET\YEEN
G, MC, S AND E
GxMC=Sxe
V",
Eq. 1.13
Eq. 1.8
vs
Note:0<e<@
1.4.7 UNIT WEIGHT IOR BULI( UNIT wElGHTl OF SOIL MASS, y-
tnv
t.4.2 POROStfi, n
Porosity is the ratio between the volumes of voids to the total volume of a soil
mass. It is usually expressed in percent.
w
G+Se
f*= -=-f
I+e *
Eq. 1.14
Eq. 1.15
Chapter
0f Soii
4
'
0l * Properties
,-.
1.4.8 DRY UNIT WEIGHT,
Fundamentals of
Geotechnical En gineering
c\GMC ,
tu
=
Eq.1.16
1.+e
Fundamentals of
Geotechnical Engineering
Chapter
For dry soils, S = 0 and MC = 0
Volumeof solid,
G
1,+e
- l+
'lo
Iror saturated soils, S =
L,
Y
= J!-
Weightof water, W*=
nt
.
Eq.1.19
Mrc
Weight of solid,
Eq. 1.25
V
Eq. 1.26
*-Vw
\+e
Eq. 1.27
1
W,= ,LV
I+ e
Weightof soil,W=
y5ay
(1
Eq. 1.28
G,,^l,u
+9p
#1'+ e Vy*
Eq. 1.29
Vu= V*
f,ar=
C+e
Tb
Ot
y' = |sat
-,|ro
,
G-1'
ylOf V = .+V-_
1.+e'*
Eq. 1.30
dfu
Eq.1.2O
1;"tw
1.4"10 SUBMERGED OR BUOYANT UMT WEtG!-tT,
I
Eq. 1.24
L+e
Dry unitweight, yr=
I.4,I
5
Eq.1.1B
-
1.4.9 SATURATED UNIT WEIGHT,
V"=
Volume of water, V*
1+ MC
^f
€V
= 1+
,
e.
Eq. 1.17
w,,=
w,= ^
of Soil
These formulas rnay not be memorized. These can be derived from the
previous formulas.
Volume of voids, V,
'" v
Properties
1.4.12 OTHER FORMULAS
ya
_w, _
-y,i-+;=-Tw
0l -
I.5
To
SPECIFIC GRAVlry OF SOME MINERALS
Mineral
OR 7'
Specific Gravity
Gvosum volcanic ash
2.32
Eq 1.21
Orthoclase
Eq 1.22
Kaolinite
Quartz
2.56
2.61
CRITICAL HYDRAULIC GRADIENT
2.67',
Calcite
2.72
Dolomite
Magnetite
2.87
5.17
Critical hyclraulic gradient is the hydraulic gradierrt that brings a soir
(essentially, Coarse-grained soils) to static liquefaction (quick condition).
I.6
Yb - G-1
.,
lcr- -
Tu
-
1,+e
Eq. 1.23
RELATIVE DENSITY OF GRANULAR SOILS
The relative density, D. expresses the state of compactness
ofa
natural
granular soil.
D,= 'mar '
€mar
orD,=
- €min
,rOO
Eq. 1.31
,
1/Y^in*1 /Ya
1/Y-i. -1/"/.rr^
Eq, 1.32
Chapter
6
of Soil
0l -
Properties
Fundamentals of
Geofechnical Engineering
where;
I'Undamentals of
Ccotechnical Engineering
0l -
Properties
of Soll
1.9 sotL tNDtcEs
e-u, = void ratio of the soil in loosest state.
€s1n = void ratio of the soil ih densest state.
e o void ratio of the soil deposit (in-situ state)
|dmar = dry unit weight in.densest state,
]dmin = dry unit weight in loosest state.
Ta = dry unit weight (in-situ state)
lndex
Definition
Plasticity
PI=LL-PL
Liquidity
MC_ PI,
Shrinkage
I.7
Chapter
DT
SI=PL-SL
DI
Activity of
DESIGNATION OF GRANUISR SOILS
nc-
clay'
- LI
Correlation
Str-ength, conrpressibi I ity,
comoactibilitv. ...
Compressibility and stress rate
Shrinkage potential
Swell potential, and so forth
tvhcrc p = percent of soil finer than 0.002 mm (clay size)
0-15
Very loose
Activitu
15-35
35 -7A
70 -85
85 - 100
Loose
Medium dense
Dense
Very dense
A, < 0.7
0.7<4.,<1,.2
4,"> 7.2
Classification
lnactivc clay
Normal clay
Active clay
I.I O DESCRIPTION OF SOIL EASEP QN LIOUIDIry INDFX
1.8 coNsrsTENcY
The consistency of a cohesive soil is greatly affected by the water content of
the soil. A gradual increase of the water content may transform a dry soil
from solid state to a semisolid state, to a plastic state, and, after further moisture
increase, into a liquid state, The water content at the corresponding junction
points sf these states are known as the shrinkage limit, rhe plastic limit, and, the
liqui d timi t, respectively.
Semisolid state - high strength, brittle
(sudden) fracture is expected
Plastic state - intermediate strength, soil
deforms like a plastic material
LI<U
Consistency is the term used to describe the degree of firmness (e.g., soft,
medium, firm, or hard) of a soil.
0<LL<1
Lrqutd stnte - low strength, soil deforms
like a viscous fluid
LI>1
I.I I
DESCRIPTION OF SOIL BASED ON I'LASTICITY INDEX
Liquid state
Uquid limit, LL
,Plastiq
statq
I
,
Plastic limit, PL
Semisolidrstater
Shrinkage limit, SL
Solid state
Figure 01.2
:
..
PI
0
1(
5-L0
n
-20
20-40
>44
Description
Nonplastic
Slightly plastic
Low plasticity
Medium plasticity
High plasticity
Very hish plasticitv
Chapter
of Soil
I
I.
I2
0l -
Properties
Fundamentals of
Geotechnical Engineering
Fundamentals of
Geotechnical Engineering
Chapter
0l -
Properties
of Soil
9
FALL CONE METHOD TO DETERMINE LIOUID AND PI.ASTIC LIMITS
Fall cone test (cone penetration test) offers more accurate method of
determining both the liquid limit and the plastic limit. In this test, a cone with
apex angle of 30o and total mass of 80 grams'is suspended above, but just in
contact with, the soil sample. The cone is permitted to fall freely for a period
of 5 seconds. The water content .o.r"rpond:ing to a cone penetraiion of 20 mm
s
g
o
defines the liquid limit.
B
Thc Iiquid limit is difficult to achieve in just a single test. In this regard, four
or more tests at different moisturg content is required. The results aie plotted
as water content (ordinate, arithmetic scale) versos penetration (abscissa,
logarithmic scale) and the best-fit straight line (liquid state line; linking the
data points is drarvn (see figure below).' The liquidlimit is read irom thJplot
as the water content on the liquid state line corresponding to a penetration of
20.
Figure 01.4
20 mm.
The plastic limit is found by repeating the test with a cone of similar geometry,
but with a^mass af (tv/12) 240 grams. The liquid state line for this cone will be
below the liquid state line for the SO-gram cone (M1) and parallel to it..
30
Penetration (mm)
lhe plastic limit is given
- Typical
-
40 50
50
60
logarithmic scale
fall cone result
as:
PL=LL- 2^ry9
, lvt)
log"Mt
I.I3
CUP METHOD TO DETERMINE LIOUID.LIMIT
fhe device used in this method consists of a brass cup and a hard
brass cup is dropped onto the base by a cam operated by a crank.
Figure 0'1.3
-
Eq. 1.33
Fall cone apparatus
Figure 01.5
-
Liquid limit device and grooving tool
rubber. The
t
o
:iijl"t
0l -
Properties
Fundamentars or
Geotechnical Fnoineerinn
The soil paste is placed in the cup. a groove is then cut at the center of the soil
pat with the standard grooving tool. By the use of the crank-operated cam, the
cup is lifted and dropped from a height of 10 mrn. The moisture content
required to close a distance of 1,2.7 mm along the bottom of the groov e after 2s
blows is defined as the liquid limit.
since it is difficult to adjust the moisture content to meet the requirbd closure
after.25 blows, at least ihree tests for the same soil are conducted at varying
moisture contents, with the number of blows required to achieve closure
varying between 15 and 35. The results are plotted on a graph paper, with the
moisture content along the vertical axis (algebraic scale) ana tne number of
blows, N, along the
.horizontal axis (logarithmic scale). The graph is
approximated as a straight line (called the
flow curae). The moisture content
corresponding to N = 25 is the liquid limit of the soil. The slope of the flow
line is defined as the flow index and may be written
FIow index,
p1
as:
MCt - MCz
= log(N2
/
N1
Eq. 1.34
)
l-undamentals of
Geotechnical En gineering
I.I4
Chapter
Nls20223036
48 45.5 44.7
MC
43.5
42.3
Properties
of Soil
ll
ONE-POINTMETHOD TO DETERMINE LIQUID LIMIT
'l'lris rnethod may be'used when only one test is run for a
cstablished by the U.S, corps of Engineers
ASTM under designation D-4318
,/
in
soil. This is
1949 and was also adopted by
rr\tanB
Lr = MC"l-1r
Eq.'1.35
l
\2si
where:
N = number of blows in the liquid limit device for a 0.5-in groove closure
MCN = corresponding moisture content
tan p = 0.121 (but note that tan.B is not equal to 0,121 for all soils)
'l'his method yields good results for the number of blows between 20 and 30'
I.I5
SHRINKAGE LIMIT
Soil shrinks as moisture is gradually lost from
where MC1 and MCz are the'moisture contents, in percent, corresponding to
number of blows Nr, and N2, respectively.
0l -
it. With continuing
rrroisture, a stage of equilibrium is reached at which more
loss <lf
loss of
moisture wiil
rcsult in no further volume change. The moisture content, in percent, at which
tlre volume of the soil mass ceases to change,.is defined as tl'e shrinkage liruit'
'l'lre shrinkage
limit is determined as follows. A mass of wet soil, mt, is placed
irr a porcelain dish 44.5 mm in diameter and 12.5 mm high,and then oventlried. The volume of oven-dried soil is determined- by using mercury to
occupy the vacant.spaces caused by shrinkage. The mass of mercury is
rletermined and the volume decrease caused by shrinkage can be calculated
lrom the known density of mercury, The shrinkage limit is calculated from
s
C
a
co
<r=ffi1-ffiz -vl-vznru)
m2
m2
o
=
where:
20
30
40 ,50
Number of blows, N
Figure 01.6
-
Flow curve
r,'rr = rnoSS of wet (saturated) soil
m2= mass of oven-dried soil
Vr = volume of wet soil
Vz = volume of oven-dried, soil
p,, = density of water
Eq. 1.36
Chapter
0l * Properties
t2
of Soil
I.I5.I
SHRINKAGE RATIO
Fundamentals of
Geotechnical Engineering
e.
r&
SPECTFTC GRAVTTY
0l -
Properties
of Soil
t3
Eq. 1.37
VzI
PROBLEM
Eq. 1.38
1SL
sR
OI.I
A s,rmple of saturatecl soil weighs 588 N and has a volume of 0.03 m3.
voids ratio of the soil is 0.75, determine the specific gravity of the solids,
OF SOLTDS
^'l
r-=
If the
5OLUTION
-It*= w,
100
I
-
I.I6
Chapter
ILLUSTRATIVE PROBLEMS
sR= 1 ry'
1.15.2
I undamentals of
Gcotechnical Enginebring
s88
,r=b*
LIOUIDITY INDEX & CONSISTENCY INDEX
y,, = 19,600
Liquidity index (L/) defines the relative consistency of a cohesive soil in the
natural state.
.
Liquidity index, LI
MC_PL
LL_PL
[T,=
G+Se
I f r
Eq. 1.39
-Y,,]
19,600=
where NIC = in situ or natural moisture content. If MC is greater than LL, Ll >
G = 2.75
=
N/m3
!o='=t) r,rro
"r*'
1. + 0.75
7. lf MC<PL,Lt<0.
Consistency index, C, =
##
Eq: 1.40
PROBLEM OI.2
A clay sample has unit weight of 21,.1 kN/m: at moisture content of 9.8%.
When completely saturated with water, its unit weight is 22.58 kN/m3.
l)ctermine the porosity of the soil.
If MC is equal to LL, CI is zero. If lvIC =PI, Cl = 1.
Atterberg's limits are aiso used to asses the potential swell of a given soil.
LL
PI
Potential swell classif icatibil,
<50
<25
Low'
50-60
25-35
Medium
>60
>35
Hieh
I
SOLUTION
-L"l. =
G
+GMC
I+ e
-=.-t*)
zr.r =
G*9(o'oe8)
1r.sr;
1,+e
G=1..959+1,,959e
-
[Y"ut
G+e
= :L+e
22.58 =
T*1
(1..959 +1..959e)+
1+e
e
(e.81)
l4
Chapter
of Soil
0l -
Properties
Fundamentals of
Geotechnical Engineering
2'302 + 2.302e = 7.959 + 2.959e
e
= 0.5221"
=
PROBLEM
Chapter
o522t
0r.5lCE MAY t9991
Solve for
e:
[y. =
C+CMC
---:-y*j
I+
PROBLEM Ot.3 {CE NOVEMBER t9981
17.3 =
A specimen of moist clay has a mass of 183.4 grarns. After oven drying, the
e
mass is reduced to :157.7 grams. What is,the moisture content crf the sample:
lG
SOLUTION
Moisture content,
,r=
Moisturecontent,
Wx
7YMC=
#
Moisture content, MC =
MC=
e
2.65 + 2.65(0.1,8)
1.+e
(e.81)
= 0.7732
S e)
2.6s(0.18) = 5(0.7732)
S = 0.617
100%
Total weight of soil mass, W = 1,83.4grams = W, + W,
Weight of scilid (oven-dried weight), W, = 157.7 grams
Weight of water, W*
W, = 783.4 - 1,57.7 = i5.7 grams
x100%
S
=
6'1.70/o
PROBLEM OI.6
!i,rturated silty clay encountered in a deep excavation is found to have a water
r r rrrtcnt of 28%. Determine unit weight of the clay in kN/m3. Assume G = 2.7.
16.30/o
'OLUTION
PROBLEM OI.4
S = 1 (saturated)
[GMC=Se]
A sample of moist clay is found to have moisture content of 400% and degree
of saturation of 85%, The specific gravity of the solids is2.T6. Determine-the
2.7(0.28) =
e = 0.756
1,(e)
voids ratio of this soil.
fy* =
G+GMC
SOLUTION
. IGMC=Sc]
e ='1,2.99
t5
SOLUTION
1+0,522L
n=0,343=34.3o/o
2'76(4) = 0'85
0l - Propetres
A sample of moist soil has water.content of 18% and moist_ unit weight of 17.3
kN/m3. The specific gravity of solids is 2.65. Compute the degree of
t,rturation of the soil.
e I
lr= "l+e'
n
Itrrrdamentals of
Geotechnical Engineering
,,, =
e
1,+ e
v*1
2!)!p1+
0.756
y* = tg.3't kN/ms
p.at1
l6
Chapter
of Soil
0l -
Properties
Fundamentals of
Geotechnical Engineering
PROBLEM OI.7
A
hand-carved sample of soft saturated clay weighs 350 grams and has a
cc. After oven-drying, it weighs 240 grams. Calculate thel
-
G +GMC
[y,=#y.l
I+
e
a) moisture content in percent.
131.1=
&) specific gravity of solids.
e
-
[Y.at
=
SOLUTION
w* |
wr'
2.67 +2.67(0.1.a)
1,+e
(62.4)
= 0,449
c) porosity in percent.
'
tl
SOLUTION
volurne of 200
following;
IMC=
Chapter0l * Propertrei,
Fundamentals of
Ccotechnical Engineering
G+e
:-I+ e
1*t=
'
Y,rJ
2.67 +0.449
#(62.4)
1+ 0.449
.
pcf
Y*r -- 134,32
MC=W#
PROBLEM OI.9
MC = 0.45833
MC = 45.830/a
A soil sample has a bulk unit weight of 19 .6 kN/ m3 at a water content of 10% .
Assuming G = 2.7; determine the percentage air in the voids (air voids).
Sc]
S = 1 (saturated)
G(0.a5833) = 1 e
G = 2."182 e
[CMC=
G+G MC
[Y,,=#y,,]
1+e
l" t= Wf V = 350/20A
"1,* = "1.75
.
lYoot
gramsf cc
19.6
C+e
= ;.---
l+e
7.7s=
Yu, I
-''---
2.182e + e
-
e
.^
(1)
IG
1.+e
L.75+1,.75e=3.1"82e
e
= 1,,222
.e n=
L+e
=0.55=55%
G=2.182(1..222)=2.6,
)
)
porosity
specific gravity of solids
0t.8 (CE NOVEMBER 1999'
13. Nov 99: A soil sample was compacted. The result of the stanclard proctor
test shows that at 100% compaction, the soil weighs 131.1 pcf with optimum
PROBLEM
moisture content of 14%. What is the saturated unit weight of the soil in pcf?
G = 2.67
SOLUTION
-
2,7 +2.7(0.1"0)
L+e
(e.8i)
= 0.4865
MC*
S el
2.7(0.10) * 5(0.4865)
S = 55.5%
Percentage of air in the voids = 100% - S
Percentage of air in the voids = 100% - 55.5%
Percentage of air in the voids = 44.50/o
PROBLEMOI.IO
'lht' moist weight of 0.2
ft3 of a soil is 23 lb. Tlre moisture content and the
,.l)t't'ific gravity of soil solids are determined in the laboratory to be 11% and
.r 7, respectively.
t8
0l -
Chapter
Properties
Fundamentals of
Geotechnical Engineering
of Soil
Fundamentals of
Geotechnical Engineering
-
a) What is the moist unit weight inlb/ft3?
&) What is the dry unit weight in lb/ ft3?
c) What is the degree of saturation in percent?
soLuTtoN
a) Moist unit weight:
y, = W f V = 23 / 0,2 = 115 lb/ft3
b)
Dry unit weight:
- G +GMC
[y,=#y,]
l+e
,r, = 21 *?.7(0.71)
1'+e
e
G-1.
-:-y*)
L+
Ltu=
e
2'71-1 (9.8u
,^=
'" 1+0.8182'
Y6= 9'23 kN/ms
PROBLEM O I.I2
'l'he mass of a sample of saturated soil is 520 grams. The dry mass, after oven
rlrying is 405 grams. Assuming G = 2.7, calculate the effective unit weight of
the soii mass, in kN/m3?
SOLUTION
IMC=
,
.G
w* |
W,,
Lya=
1.
ya
+ 0.626
W _W,
ws
(62.4)
MC=
= 103.521b/ft3
MC=
S el
IGMC=
2.7(0,11) = 5(0'526)
e
.
405
Se]
whereS= L (saturated)
= 0.7667
G-1
#y.I
L+e
lyr=
PROBLEMOI.II
A specimen of sand has a porosity of 45o/o, and the specific gravity of its solids
is 2.71.. Compute the specific weight of this soil in the submerged state, in
kN/m3.
* -n
[e=
-
21(0.284) = (1)e
s=0.4744=47.440/o
SOLUTION
520
405
MC = 28.4%
c) Degree of saturation
lG
0l .- propettrei q
I
62.+1
= 0.626
=-y*l
L+e
2.7
yd = _
Chapter
2.7 . 1, (9.81)
1+ 0.7667 '
yb = 9.44 kN/m3
'- =
vu
PROBLEM OT.I3
L-n
--l
e=
0.45
1-
0,45
e = 0.8182
-
l.,rboratory tests on a soil sample yielded the following information:
= 2.71,, G,,, = 1-.72, MC = 1,3%. Determine the foilowing properties:
t)
,r) void ratio
l)
r)
degree of saturation
porosity
\
20
Chapter
0l * Properties
Fundamentals of
Geotechnical Engineerin g
of Soil
Geotechnical Engineering
SOLUTION
q)
b)
'
Void ratio:
ly,, = y. Grl
.
[y,,,
=
G+CMC
----
l"l. =
],j
y1:2.7'L(0.13)
c)
lG
MC=
2'67 + 2-'67(0'7583)
1,+e
,r,
S el
=
S
(0.
6065)
s=0.697=69,70/o
= 0.78
Ol.r5 (CE NOVEMBER r9?91
A soil sample was compacted. The result of the standard proctor test shows
llr,rt at L00% compaction, the soil weighs 131.1 pcf with optimum moisture
errntcnt of 1,4%. What is the maximum dry unit weight of the soil (at zero air
vrrirls) in pcf? G = 2.67
PROBLEM
lG MC= s el
s=a.452=45.2a/o
c
[rr
= ---:l*rl
1
0.78
1 _ 0.79
'OLUTION
At zero air voids, Vo= Vr(S = 1)
n=0.438=43.80/o
lc MC=
OI.t4
.G=
Ya= 121,28 Pcf.
PROBLEM
1,,=WfV="t8000/9350
a\
ty,tC = Yt:
'w"
^,n_
1g-15.54
y,l
:-*
L+e
2'67
,,, =
'' L+0.3738'(62.4\
lya
In its natural state, a moist soil has a volume of 9350 cc and weighs 18 kg. The
oven-dried weight of the soil is 15.54 kg. Use G = 2.67.
a) Determine the moisture content in percent.
b) Determine the void ratio in percent.
c) Determine the degree of saturation in percent.
SOLUTION
S el
2.67 (0.14) = 1e
e = 0.3738
,:
PROBLEM
=
2.67 (0'1583)
^. _ ^, ,1 n.
2.71,(0.13) = 5(0,78)
c)
L+e
e*0.6066=60.6601o
ffv*=v,(L.72)
b)
zt
C+CMC
-----:-l* I
1..925
-".'Jio
"-/u'=)r'u'r'
t+c
e
chapter 0l _ Propettrei
Furrdamentals of
O
I. I6
compacted clay weighing t.62 kg weighs 0.88 kg when immersed
(r'uspended) in water. Determine the bulk specific gravity of the clay.
A
SOLUTION
Bulk sp. gr.
15.54
MC=0.1583=15.83%
Bulk sp. gr,
Weightinair
. Weightinair - Weightinwater
1'62
=1.52-4.88
= z.tg
22
0l -
Chapter
of Soil
Properties
Fundamentals of
Geotechnical Engineering
OI.I7
PROBLEM
Fundamentals oF
Geotechnical Engineerin g
Chapter
0l - propertiei
Zg
SOLUTION
A 50 cc of moist clay was obtained by pressing a sharpened hollow cylinder
into the wall of a test pit. The extruded sample had an initial weight of 85
grams. After oven-drying it weighs 60 grams. If G = 2,72, determine the
w'l
IMC=
'
ws'
w
MC=
degree of saturation of the saurple.
^,n
SOLUTION
lvlu
Given: W=85g
-w:
ws
17.5-1,4.2
_
-
1.4.2
MC=0.2324=23.24%
V = 50 cm3; W,= 60 g
t*
trm=Wl
vt
,ttilt =w,I
V
Tn,
y,, = 7.7 g/ cc
.
[y*=
w"
'lMg= w,'
MC =
.
L'l,n
=
e
lG
S = 0.7254
S=
2.72 +2.72(0.4167),.,
7+e
\
-
llt /
2(0.4767) =
= O.*r,
-
S (1,
.2667)
Y"ut
r
A cubic meter of soil in its natural state weighs 17.5 kN. After oven-drying,
the soil weighs 14.2 kN. Assume C = 2.7.
the void ratio of the soil.
the degree of saturation of the soil.
the saturated density of the soil in kN/m3.
)
degree of saturation
G+e
=89.5o/o
Or.t8
a) Calculate
b) Calculate
c) Calculate
72.54o/o
[y*,= ]y*J
1+e
2'7 *!!9
,,0,= _il0.g65
= 1..2667
2.7
PROB|-EM
G+GMC
#t.1
l+e
MC= S el
2.7(o.n)4)= s (0.86s)
1+e
,
= 17.5kN/m3
60
0,41"67
fc MC= S e)
,
1'
*t9.n2a)- 0811
1+e
e = 0.865 = 86.5% ) void ratio
-C+GMC
--'l*l
.,
r./ -_
= 77,5 /
2.7
,
85-60
MC=
-
PROBLEM
O
19.g11
= 18.75 kN/m3
r. t 9 (CE MAY 2OOO|
A soil sample has a moisture content of 30% and degree of saturation of 45%.
I'lrc solids have specific gravity of 2.61.. Determine the dry unit weight of the
soil in kN,/m
SOLUTION
'IGMC=Sel
2.61,(w) = 0.45 e
\
0l -
Chapter
of Soil
24
Properties
Fundamentals
Geotechnical Enqi
e = 1,.74
.C
=
[y,r'y
1+
1
.74
f
(g,8t)
-'
'- '-
= -;-^l
kN/m3
4t
r+e
,,,, =
2i7 *=0:?5
kN/m3
95%:
G+Se
'l^= -=-'lw
with a porosity of 0.387 is 1600 kg/ mr.
void ratio of the soil,
a sand
L+e
0.il(9.8s)
,_ = ?.7.*
,, su
specific gravity of the soil solids.
effective density of the soil, in kg/m3.
1 + 0.85
kN/m3
y,, = 18.60
SOLUTION
Given:
.il
rt)
'
(,
=
= 1,600
n = 0.387
pa
,,
0)
p,t
1-n
LJ
= 7I+C
pw
1
.
c)
Pr,
=
2.67 _
18'8:i
-18'60 x 100%
0I.22
A wet soil sample has a volume of 4.85 x LO-a m3 and weighs 8.5 N. After oven
tlryirrg the weight reduces to 7.5 N. Use G = 2,7, Deterrnine the following:
rr) unit weight,
fr) moisture content,
r
)
degree of saturation.
IOLUTION
G-1
Pr= #p,
L+e
pr
'
.
2.6'1.
=
18.60
Percentage Error = 1,18Y0 more
PROBLEM
9 \-""-/
1,600 =
r 0.63'I rrooot
G=
Percentage Error
kg/m,
r' = 0.631
1o.st;
1 + 0.85
Actual S =
t.ZO
The dry clensity of
a) Calculate the
b) Calculate the
c) Calculate the
,E
Soil zr
C+e
sat
Y,,t = 18.82
O
of
Assumed S = l.:
,
2'61
ya.v = 9,34
PROBLEM
0l - Properties
IOLUTION
y,]
;1
I +c
v.,..
'"') =
Chapter
Fundamentals of
Geotechnical Engineering
1
a)
#(1000)
1, * 0.631 '
[y*
-
fft
8.5
lnt
= 987.12kilm3
4.85 x 10-a
1,,-\7,526 N,/ml = 17.526 kN/mg
PROBLEM OI,ZI
The void ratio of a soil is 0.85. What is the percentage error of the bulk unit
weight if the soil were 95% saturatecl and assumed to be totally saturated?
b)
IMC=
w*
w,.'
MC-
1
8.5
-
7.5
-
." I Smtrt JL Fqqu^r
26
0l -
Chapter
of Soil
Properties
Fundamentals
Geotechnical
0l -
Properties
of Soil
27
0I.24
- A sample of dry sand having a unit weight of I6.50 kN/m3 and a
slrccific gravity of 2.70 is placed in the rain. During the rain the volume of the
n,rrnple remains constant but the degree of saturation increases to 40%.
MC = 0.1333 =13,330/o
c)
Chapter
ILrndamentals of
0cotechnical Engineering
PROBLEM
tiiruation 4
Degree of saturation:
Solving for e:
-
G
if n, = ---
+GMC
-L+e
12.526
_
I
Tu,l
2.7 + 2_7(0.1333)
7+e
,rsD
)r,termine the following:
rr) The voids ratio of the sample in percent.
lr) The unit weight of the sample after being in the rain.
t ) The water content of the sample after being in the rain.
e = 0.7128
SOLUTION
ICMC=Sel
.
PROBLEM
g = (2.2 x 0,1"333)/0.7728
G = 2.70
S=0.505=50.5%
';skN/m3
G
'*':
a) [ya,v = ;:-y*l
'L+
e
21
1.6.5=
0t.23
(CE NOVEMBER 20OOl
1.+
A clay sample has unit weight of 20.06 kN/rns with moisture content of 8.
The saturated unit weight of the sarnple is 21.58 kN/m3. Determine
porosity of the soil.
b)
SOLUTION
G
-
+1,.89e
2.70 + 0.40(0.6053)
1'+e
= 0.45
ln=1.+e'
' I
x 9.81
1+ 0.6053
y,n=17.98 kN/m3
c)
,r.rr',
IGMC=Se]
210 MC = (0. 0)(0.6053)
MC = 0,0897 = 8.97o/o
,
2.2+2.2e=1..89+2.89e
e
=-'l*l
l+e
l1k
#y"J
=
S = 0.40
l
h-, = I+e
1'89+1'89e+c
60,530/o
C+Se
ltr,l
G=L.89
G+e
21.58
=
After being in the rain,
t+e.
c 9(o 082)
"
20.06 =
1o.ar)
I+e
.
= 0.6053
e
lT^=
+GMC
tl/r - -------
e
*9.81'
e
PROBLEM
0I.25
lhe moist unit weights and degrees of satur;ition of
Iollowing table:
0.45
1 + 0,45
rr=0.31 =310/d
v (pcf)
s (%)
705.73.
50
112.67
75
a
soil are given in the
28
Chapter
of Soil
0l -
Properties
Fundamentals
Geotechnical
a) Determine the void ratio of the soil in percent,
&) Determine the specific gravity of the soil solids.
c) Determine the porosity of the soil in percent
IOLUTION
Given W=7346grams
G+Se
[t' = ;;}''
L+e
.
First soil;
.tos.7j
Ws = 1076 grams
V = 792 cm3
'
=
G:
.0.s,
1+e
fu*=w114
Y* = 1346/7g2
1oz.+1
1..694 + 1..694e = G + O,Se
G
* t;694 +,1.194e +
T* = 1.7 grams/cc
Eq. (1)
tu*'=ffl
Second soil:
112,67=Gills,
'Y'*= #*
792
rcz.nt
1.806+1.806e =G+0.75e
G = 1.805 + 1.056e + Eq, (2)
1.694 + 7.194e = 1.806 + 1,056e
e = 0.8'j,2 = 81..2o/o
void ratio
MC=
)
Porosity, n
'
r"orosltv,
'
Porosity,
1346
l
-1076
7076
MC = 0.251.
From Eq. (2)
G = 1.806 + 1.056(0.812) = Z.6O
=1'846grams/cc
w,,,w'
wc=
'wrw,ry+ =
lG=GJ
)
specific gravity of solids
.
lT* =
* '
1.+e
1+e
G=1,.26+1..36e
0-812
1+ 0.812
n=0.448't.=44,81o/o
G+CMC
----:-l+e l*l
, (t)
r.z =
"n_G+C(0.251).-.
r
PROBLEM OI.26
In an experiment of determining the porosity and specific gravity
of solids of a
soil, a soil is dug out from the ground, t"igh"d, ihu.r oiu. dried,
and then
saturated with water. The weight of soil taken from the test hole
is 1346
grams' lts was then oven-dried and weighed 1.026 grams. After
saturating it
t.
Chapter
0l -
Properties
of Soil
29
lvillr water, it weighed 1.462 grams. The volume of the test hole was then
trrt',rsured and found tobe792.cc. Determine the voids ratio of the soil.
SOLUTION
.
Fttnclamentals of
Gcotechnical Engineering
.
[T'at
=
+
(1)
C+e
:-L+e
Yu']
G*t (1)
1+e'
G=L.846+0.846e )(2)
1.846=
[G=G]
1.36 + 1,36e = 1,.846 + 0.846
e=0.946=94.6%
e
30
0l * Properties
Chapter
of Soil
Fundamentals
Geotechnical Engineerin
PROBLEM O1.27 ICE MAYZ00rl
A clay sarnple has the following properties:
PorositY = 0'35 (in situ)
Maximum void ratio = 0.85
Minimum void ratio = 0.42
Specific gravity of solids = 2J2
Moisture content = 62%
Determine the dry unit weight of the soil in its natural state, in pcf,
Chapter
Fundamentals of
Gcotechnical Engineering
0l -
Properties
of Soil
3l
G+CM(
-[r,=fr,J
L+e
1.772
=
2'72 + 2'72(0'43)
L+e
e=1.195
r'r,
i
SOLUTION
tu
-!-1
I - II
=
rr) water content of the soil in percent.
t) void ratio inpercent.
c) effective unit weight of the soil in kN/m3.
(=---0.35
- 0.35
e = 0.538
1
,C
l1/,
I
'
lor) = _
-[+c
Ilrl
-"'
Ya,"
I
SOLUTION
2'72
'*
=
0I.29
A sample of saturated clay was placed in a container and weighed' The
wcight was 6 N. The clay in its container was placed in an oven for 24 hours
rrt 105'C. The weight was reduced to a constant value of 5 N''The weight of
llrc container is 1 N. G = 2.7, Determine the following:
PROBLEM
h2.4\
1 + 0.538 '
y.1,u = LL0.36. pcf
Weight of moist soil (saturated),W = 6 - 1 = 5 N
Weight of dry soil, W' = 5 - 1 = 4 N
a) Water content:
PROBLEM 0r.28 (CE NOVEMBER 2OOr)
r'--wc= ryrl
w"
'o-o
MC= "
A
480 cc soil sample taken from the site weighs 850.5 grams. After o
drying, it weighed 594.4 grams. If the specific gravity of solids is 2.
deterrrrine thc void ratio of the soil.
1y,,,
=
w,
_;_;
V
'V'r f u,
IMC
'
MC =
=0.25
250/o
b) Void ratio;
SOLUTION
,
4
=
lG
MC=
S el
l
2.7(0.25) = (1)e
850,5
e=0.675=67.50/o
480
g/ cc
= 1J72
-
.[vo= #v,]
c-l
t+e
wu'
wr'
c) Effective unit weight:
I
850.5-- 594.4
MC=
.
MC=
0.43
591.4
2.7
-" -1- (9.81)
v"=
' 1+0.675'
Yb= 9.956 kN/m3
32
Chapter
of Soil
0l -
Properties
Fundamentals
Geotechnical E
PROBLEM 0 t.30 (CE NOVEMBER Z00 t
A 480 cc soil sample taken from the site
' weighs 850.5
grams. After ove
drving, it weighed 594.4 grams. If the specific gravity of sorids is 2.7i
,
ll'' =
of Soil
Given: W= 56.7lbs
Ws = 487lbs
V = 0,5 ft3
G = 2.69
a) Density of the in-situ soil
7:
-
tnv
l
850.5
4g0
"{* = 1.772
IMC
=
wu'
lm=
w
56.7
05
Y* = 113.4lbsf1s
glcc
b)Porosity:
l
'w,= W*
W,,
*r*
Moisture content, MC
850.5+594.4
XrtC =
Moisture content,'ws
594.4
MC = 0.a3
Moisture content,
.
G -CMC
__1_;;7,,,1
1Y,,, =
2.72
1t.//z_
1_n _ _
+
2.72(0.43) ,",
,;l_trl
c = 1.195
.
lY=
W
-Wt
*, ='u1;.)"
G+GMC
_T.l
L+e
_____
2.69 +2.69$.1,643)
113.4 =
(62.4)
1-+ e
IGxMC=Sxel
2.72(0.43) = S(1 .1es)
S = 0.979
PROBLEM
Properties
TOLUTtON
W,
,,,
'
0l -
treotechnical Engineering
determine the degree of seituration of the soil.
SOLUTION
Chapter
Frrrrdamentals of
Or.3t
e
= 0.7234
Porositv. n
= L+e
'
0.7234
l'orosrtv,
(CE NOVEMBER 2OO2'
Civen the following characteristics of a soil sample:
Volume = 0.5 cubic ft.
. Mass = 56.7 pound mass
The solids have specific gravity of 2.69. After oven drying, the mass of the soil
was 48.7 pounds.
a) What is the density of the in-situ soil?
b) What is the porosity of the in-situ soil?
c) What is the degree of saturation of the in,situ soil?
Porosity, n =
"l + 0.7234
= 0.4198
41.980/o
c) Degree of saturation
[GMC=Se]
2.69(0.1,643) = S (0.723a)
S = 0.611
S
=
61.10/o
= 0.1643
33
3r&
0l * Properties
Chapter
I.32
The saturatecl unit weight of a soiI is1.9.49 kN/m3, and the specific gravity
PROBLEM
f undamentals of
Gcotechnical Engineering
Fundamentals
Geotechnical Engineeri
of Soil
0l -
Properties
of Soil
35
Unit weight:
a)
O
Chapter
o
b=
the soil solids is 2.7.
a) What is the void ratio clf the soil.
b) What is the dry unit weight of the soil in kN/m3,
c) What is the effective unit weight of the soil in kN/mr.
+I
224
' = 500 - 382
o
[G"'
=
J*
= 1.898 e/cc
]
Pw
c,= i'9i8 =1.8e8
SOLUTION
Ctc
a) [1''.,'=t+e
I",j
fY*=Y*xG*l
7g.4g=''''"g.8r,
l*e
h)
iY,r.u
1
.987 + 1.987c = 2.7 +
e
= 0,7226
=
[y,,
Void ratio
.
G+CMC
[Y,,
=
=
Yu.l
2;6 + 2'6(0'225)
e*0.67g
c)
kN/mr
1'+e
* r.r,
Dry unit weight
.G
[ra =
= y*, - y. ]
vr=19.49-9.87
It=
----L+ e
18.62
ffi^P'8t)
y.,y = 1,5.376
c)
b)
e
G
-:y,,,]
w'=
'
T. = 9.81 x 1.898 = 18.62 kN/ms
y.]
-l+e
,ffi"'u'
Yr= -
9.68 kN/m3
Y7
= 15.2kN/m3
PROBLEM 01.33 (CE MAY 2003'
I'he following data was obtained from laboratory tests for a
iohesive
specinren: moisture content, 7r), was 22.5%; G, = 2.60; and to determine th
approximate unit weight, a sample having a tnass of 224.0 g was placed in a
500 cm3 container with 382 cm3 of water required to fill the container.
a) What is the total unit weight of the soil sample in kN/rn3?
b) What is the void ratio e?
c) What is the dry unit weight of the soil sample in kN/m3?
PROBLEM Ot.34 (CE NOVEMBER 2OO3)
A fully saturated clay sample has a mass of 1526 grams. After oven-drying, its
was reduced to 1,053 grams. The specific gravity soil particles is 2,7.
nrass
rr) Calculate the natural water content of the sample in percent,
ll) Calculate the void ratio in percent.
c) Calculate the porosity in percent.
SOLUTION
SOLUTION
Given:
Civu,n: MC = 22.5".,
S=1
M = 1526 grams
C = 2.60
M. = 1053 grams
Mass,M=22-lgrams
Volun.re, y = 500 - 382 = 118 cc
G=2.7
\
0l -
Chapter
36
of
Properties
fundamentals of
Fundamentals
c) lD,= o e^u"-e
_o
Water content, MC
in
'"u'
IMC=
it
'
I
"
lVt-
Void
ratit-r
Lrrntent of 1,4% and a porosity of 38%. In a laboratory test that.simulates field
r orlditions/ it was found that at its densest state, its void ratio is 85% and at its
Itrosest state its void ratio is 40o/.. Determine the relative of the san<i.
e*1.213=121.30/o
Porosity
t,,= tL
t+e
]
SOLUTION
u
le=
' 1.-n'
1,.273
1
1+1.213
t1 = 0.5481 = 54.81a/o
0.38
1
e
-
0.38
= 0.613
0I,35
A sandy soil has a naturai water content of 275% and bulk unit weight of
kN/m3. The void ratios corresponding to the densest and loosest state are
and 0.87" Assume G = 2.7.
a) Which of the following gives the in situ void ratio of the soil..
I.) Which of the following gives the degree of saturation of the soil
c)
,
" mln
A sample of moist sand taken from the field was found to have a moisture
2,7(0.4492) = (7)e;
PROBLEM
37
PROBLEM 0 r.36 (CE MAY 20001
ICMC=Scl
c)
Properties
of Soil
D,=0.308=30.8%
1,053
-=MC=0.4492*44.920/o
b)
max
0l -
-0.759
^ _ 0.87
"'ag7 -l,s:-
1,526 - 1,053
ivIL =
Chapter
Ccotechnical Engineering
GeotechnicalEng
Soil
19.
"
D,=
in
i
Emax - s
D,=
0.5
max 'mtn
- 0.613
0.85 - 0.40
0.85
-
D, = 0,527
natural state.
Which of thr-, following gives the relative density of the soil.
PROBLEM O 1,37
A test of the density of soil in place was performed by digging a small hole in
tlrc soil, weighing the extracted soil, and measuring the volume of the hole.
'l'he soil (moist) weighed 895 g; the volume of the hole was 426 cm3. After
SOLUTION
r)
17,,,
=
C+C MC
.l-C
7v.z=
t
€
h)
2'7
Y" j
"?](0'275)
1*c
= 0.759
[GMC=Se]
2.7(0,275) = S(0.7se)
5 = 0.978
tlrying, the sample weighed 779 g. O{ the dried soil, 400 g was poured into
1o.st1
a
vcssel in a very loose state. Its volume was subsequently determined to be276
cnr3. That same 400 g was then vibrated and tamped to a volume of 212 cm3.
() = 2.7L. Determine the relative density of the soil.
38
Chaoter
of Soit
0l -
Properties
Fundamentals
Geotechnical Engineeri
l'undamentals.of
Properties
of Soil
39
l,irboratory moisture density test on this soil indicated a maximum dry
SOLUTION
rlr'rrsity of 120 pct at an optimum moisture content of 17%. What is the percent
1/y,t^i,. -1/ya
1lyan;n*1/l'a^*
Ya=W
/
tr = 779
Y,r,,,n*
It,r,.tr
rornpaction of the fill.
SOLUTION
V
/
426 = 1,8286 g,l cc
Percent compaction
= 400 / 21,2
='l 8868 g/cc
Dry density,
]d,in = L,4493 g/
11
i.+,1c')
^
"'-'-r
-t.++vr
cc
,rr=
Dry densiry, u =
I
xc6s
Dry density,
D,=A.8946=89.460/o
I.38
For a sandy soil,
void ratio at D, =
en,.,*
= 0.86,
density of soil
Maximum drydensity
1,20
pcf
+
y,r
A!
478.55
c,,,,,,'= 0,43, and G,
=
2.66
= 1.91 gram/cc
= 119.18 pcf
Percentcompaction =
O
- -Dry
V = 0.0-169 ft3 x (123)(2J43)
V = 478.55 cc
il28r,
I
-
Maximum dry density =
Yrlni.*400/276
I'ROBLEM
0l -
Chapter
0cotechnical Engineering
W
x10O%
Percent compaction = 99,30/o
What is the requi
56%?
O r.4O (CE MAY 2OO'r I
,,oil
A
sample in its natural state has a wet density of 155.1 pcf and a moisture
PnoBLEM
SOLUTION
1o =
-5r-1|
-o "rnln
'ntax
0.56 =
e
0.86
0.86
-
rrrrtt'nt of 36%. After compaction, its maximum dry density is 118.5 pcf.
1
*
I
soii. G = 2.65.
e
TOLUTION
0.43
= A.6'].92
Percent compacrion
PROBLEM 0 r.39 (CE MAY 20001
Fielci clensity test on a conipacter-1
\\'t'ight of rnoist soil
'
)r't,'rrnine the percent compaction of the
fill of sancly clav gives the following results:
om tlrt,holc = I038 grams
Oven-dried weight of tiie soil = 914 grams
Volurne of test hole = 0.0169 ft-r
lr
=
#ffifi|.
G +CMC
tv=
yul
t'
1+ c1.8
l5)'li =
- 2.6!.19QlAL+e
e
= 0.45
62.a1
x 1oo?,
40
0l -
Chapter
of Soil
.G =
ly,,u
;l;
Properties
Fundamentals.
Geotechnical En
ya,u
f a,v
=
1;17,
= 114'04 Pcf
+#
118.5
'OLUTION
Porosity, n=e/(1 +e)
For the required soil:
x10O%
Percent compaction = 96.240/o
n = o.i/ (t + o.z)
n = 0.1.667 ='16.67% (% void)
Percent solid = 1 - n =0.8333 = 83.33%
Volume of solid = 10,000 x 0.8333
Volume of solid = 8,333 m3
I.4I
An embankment for a highway 30 m wide and 1.2 m in compacted thickness
to be constructed from a sandy soil trucked from a borrow pit. The wa
content of the sarrdy soil in the borrow pii is 157o and its void ratio is 0.75.
specification requires the embankment be compacted to a dry unit weight
18.2 kN/m3. Length of embankment is 1.5 km. Assume G = 21. Determ
the volume of borrow material requirecl.
For the borrow soil:
rz
= 0.8/(1+ 0.8)
.n=0.4444=44.44o/o
Percent solid = 100 '44,44 = 55'56%
Volume of solid = Loose volume x %solid
8,333=%oo""X0.5556
SOLUTION
Vroo,"
Volurne of finished embankment = (30X1.2X1500) = 54,000
Dry urrit weight of borrow rnaterial:
= 15,000 m3
m3
Or:
.C=
[1,,
;-y,,]
I +c
)7
-"
v., :
' 1-0.75'(9.8i )
[%oo"u=(1 +e)%aial
Vroo," = (1 + 0.8)(8,333.33)
Vtoo," = 15,000 m3
= 15'135 kN/rn3
Total cost = 15,000 mr x P320/m3
Total cost = P4,800,000.00
Y,i
Volume of borrow rnaterial =
(lf rinishe.ren,bankmer.t
9+a
(t ,t)
ri,
Volume of borrow material =
+}]t
(54,000)
Volume of borrow material = 64,936 m3
L
0l -
Properties
of Soil
4l
A building requires a 10,000-m3 fill at a void ratio of 20%. Material for earth
flll was available from a borrow site at P320 per cubic meter. It was found that
thc average void ratio from the site is 807o, Estimate the total cost of fill.
(b/.4)
Percentcompaction =
PROBLEM O
Chapter
IROBLEM O 1.42
y,l
I --E
2.65
Fundamentals of
Geotechnical Engineering
PROBLEM O
I.43
lrollowing are the results of a shrinkage limit test:
Initial volume of soil in saturated state = 24.6 cc
Final volume of soil in a dry state = 15.9 cc
Initial mass in a saturated state = 44 g
Final mass in a dry state = 30.1 g
n) Determine the dry density of the soil in grm/cc.
b) Determine the void ratio of the soil.
\
42
0l -
Chapter
of Soil
Properties
Fundamentals
Geotechnical Engineer
Fundamentals of
Geotechnical En gineering
c) Determine the shrinkage limit of the soil
d) Determine the shrinkage ratio
e) Determine the specific gravity of the solids
e) tc=
=
pa
b)
p"ut
= M,utf V
p*t
= 44/24.6 = 1.7886 g/ cc
PROBLEM O I.44
'l'he following results were obtained from a
liquid limit test on a clay using the
Casagrande cup device, Use the graph in Figure 01,7. The natural water
content of this clay is 38% and the plastic limit is 21%.
,:-r,1
I+e
[p,r
C "l
= 'l+e
t.2236
Number of blows
a)
G+e
=
;-1+e
1.7886
pru
=
"1.7886(1
0.565(1
h)
cL.
fr
+ *
e)
c)
trl
1.2236(1 + e) +
d)
20
28
32
47.1
42.3
38,6
37.5
52.5
is the liquid limit of this clay?
is the plasticity index of this clay?
is the liquidity index of this clay?
is the flow index?
e
+e)=e
0'565+0.565e=e
tl
e=1.3
il
c) sy=1",-*rl-[', -'r]"I t|l';10-
Note:o., =1srm/
Note:P*=1grm/cc
24'6.-1's'9
- 44:-39'"1
30.1 - 30.1 ft\ = 0.1728 = 77.280/o
Es0
o
Itr
oq5
lii
i
;ir
iir iitilliirili
ii
N
B
:l
I
irl
I
1
d) [sR=-Lmz,
u2
Pu
r 30.1
sR= 1 _
1 15'9
-r sR = 1.893
il
:i,
rii
s
I
SL
What
What
What
What
12
6
Water content (%)
G =7.2236(1+e)
Psat
100
{cc
= 1.2236
.C=
--l1, _17,28
G = 2.813
P,l= 30..1/24.6
oa
100
1..893
M,/V
43
r - srl
c=
a)
of Soil
1
sR
SOLUTION
0l * Properties
Chapter
20
30
ril;
ill
;tr
40 50
Number of blows (logarithmic scale)
Figure 01.7
60
44
Chapter
of Soil
0l -
Properties
Fundamentals
Geotechnical Engineeri
Chaprer
Fundamentals of
Geotechnical Engineerin g
0l
=
propertrei
SOLUTION
E;il.'imEEfrfi;,l'11;1;1ii1i11i;;l
Test Number 9
,
I
Cone oenetration. mm
weioht of wet soil + contarner. o
weiqht ol IJrv :;0rl + uontatner, q
vverqnt o1 uontatner. q
Weiqht of Water, q
Weiqht of Dry Soil, s
Water Content, o/o
illri
ili;i
iirli
ri
l,
c
.9
"v
c
I
E45
E
lli
{l
i
l
lOo/o
:
li
Test Number
;]l
35.62
28.84
30.c1
29.69
12.33
31.42
45.70
33.69
11.74
11,45
1U.52
I
30
40
50
Cone penetration (mm)
water uontent, %
Averaoe. %
Figure 01.8
The liquid limit is read from the graph as the water content on the
liquid state line corresponding to 25 blows.
rl
Thus, LL = 40%
Plasticity index, P/ * LL - PL
Plasticity index, PI = 4A - Z, =
ii
s
iil
Es0
o
c
o
U
,r- MCr-PL
I
ii
.;
b4s
o
DI
il
=
?Q_.)1
U="" 19 "
il
li
190/o
=0.895
iii
;i
Frow indcx,
61
=
J!gr:
20
r49:-
30
PROBLEM
O
I.45
&) Determine the nearest value to the Plastic Limit of the soil.
c) Determine the nearest value to the Liquidity Index of the soil.
t
I
01.8. Plot
40
Cone penetration (mm)
Figure 01.9
Civen the laboratory results of the Atterberg Limits Test in Figure
the water content versus the cone penetration in Figure 01,9.
a) Determine the nearest value to the Liquid Limit of the soil.
I
.ZO
iii iiiilll
50
JC
Natural vvater
Conient
2
Z
31.78
27.18
12.33
30.18
25.ltd
10.52
Weiqht of Container, g
Weioht ol Water. o
Weioht of Drv Soil. q
1
20
)
werqht oI wet sorl + uontatner, q
Weioht of Dry Soil + Container, g
l
Liqrridity index,
2A
Plastic Limit
-"j
Liquidity index,
2
l8
I
B
4
1
16
27.77
25.39
11.t4
30.04
+S
Chapter
of Soil
46
0l -
Properties
Fundamentals of
Geotechnical Engineering
Fundamentals of
Chapter
6eotechnical Engineering
SOLUTION
Liquidity index, LI
I
est Number
)
2
penetralon
werght ol wet soil + container. o
werght of Dry soil + container, o
16
35.62
28.84
10.52
Weight of Contalner, q
Weight of Water, g
Weight of Dry Soil, S
Water Content, %
qI
Ilqt[gl{nitrhnE t$elvieirWilrir qElieh!,,
: i:.t
i:
28
41.26
31.42
29.U9
Oi/t'
6..1/
9.U4r'
1U.6tJ
37%
4U-/o
bUY/o
i,,l,riilm
To
)
b45
o
I
25.76
2 / .1t3
:15.39
2/.23
tu,cz
't2.33
t1 t4
11.45
1.C4
4.60
2.d\
15.t4
29%
14.65
2.38
13.65
31%
1t.40%
li
)./o
17.800/"
17.600/0
rl
ti:
,]
ll
lrl
30
Averaqe,
il
t:
40
50 ..60
80
100
LL is the moisture content corresponding to 20 mm cone penetration.
From the graph shown, LL x 420/o
From the table above, the Plastic Limit is the average of the MC of the
two tests, and is equal to 30%
I
I
I
3r"-
MCn
-
PL
LL_PL
22.47
19.44
12.74
2
3
29
21.29
18.78
ZQ
13.24
21.2t
26.12
18.75
13.20
lt.1a
13.27
Natural water
Content
2
23.20
20.42
12 90
7o
Figure 0,1.10
t):
Cone penetration (mm)
Liquidity index, L/ =
38
Number,
Weiqht of Container. o
vvergnl ot vvater, o
weight of Dry soil, g
Water Content, %
iti
20
.;,,.;
wel 5oI + uonlalner. 0
werght ot Dry soil + conlainer, q
iil
10
timjtii ; "
)
vvergnt oI
liil
iillili
A, [idiiid
Test Number
Number of Blows
vverqnt oT vvet $ott + contatner. o
vvelgnt oI Ury SOtl + Uontatner. q
Weiqht of Container, q
wetght of water, q
Weiqht of Drv Soil. o
Water Content, %
Plastic Limit
:,:lt"),1
tr/,
-30
PROBLEM O1.46 ICE MAY ZOO+I
Given the laboratory results of the Atterberg Limits Test in Figure 01.1.0. Plot
the water content versus the number of blows in Figure 01,LL,
c) Determine the nearest value to the Liquid Limit of the soil.
b) Determine the nearest value to the Plastic Limit of the soiJ.
c) Determine the nearest value to the Liquidity Index of the soil.
Test
:iii
iti
llii;i
*30
B. Flastic Liinit and Natuial Watei.Content
iiii
liil
iiii
,{,
LL d (2o1o
=
2
30.04
A
co
2.0'l
lZ:"24i:
54e/t
/7
I
tss0
o
,t
J1. /U
3O1o
I
33.69
1 1.45
l6
JU.
vvergnr or_ vvater. q
vvergnr or ury soil, q
uonlenI,
Average, %
2
1
4l
33
Naturat water
Content
Plastic Limit
esl Number ,
wergnt ot wet sojl + container, q
vvergnl ot ury sotl + container, g
vverqnt or uontalner. q
I
42
Propertres
15./u
t.t4
12,33
7,02
17:56
17.60
-,
Liquidity index, LI = -L,033
4
IE
36.91
=
0l
2
22.4O
20.1
I
12.95
17.53
14.84
9.bu
16.97
14.36
9.55
Chapter
48
0l * Properties
Fundamentals
Geotechnical Engineeri
of Soil
Chapter
Fundamentals of
Geotechnical Engineering
0l * Properties
of Soil
4g
Flow Curve
3
s
Yso
c
o
o
ts
.9
o-
o
+_
-n-s:35%-;-[i-------
C
B
as
o.-
B
E-
=
50 60 70 80 90
Number of blows
-
100
25 30
zo
logarithmic scale
Number of blows
-
40 50 60 70 80 90
logarithmic scale
Figure 01.1 1
Figure 01.11 (a)
soLUfroN
Test Number )
Number of Blows
Weiqht of Wet Soil + Container, q
Weioht of Drv Soil + Contarner, q
Weioht of Container o
Weiohi of Water. o
\A/6idht
^f
Dn/ Snil
Water Content. %
J6
29
22 47
19 44
12.74
3.03
6.7
21.29
14.74
13.24
2.51
5.54
45.310
45.22o/o
Plastlc Limit
Test Number
)
Wei0ht of Wet Soil + Container. q
Weiqht of Ory Soil + Contarner, q
Weiqht of Container. o
Weioht of Water o
\Aiai^ht
nrv C^il
^f
W6ter Content. 7o
Averaoe.
o/o
3
4
20
21.27
14
26.12
t6.I
)2
13.20
2.52
5.55
13.27
1t)
4.U2
22.80
2.ttl
2.61
I .52
7.24
36.050h
v"
36.970k
36.
Figure 01.10 (a)
20.19
12 95
flow curve in Figure
-
U.6J
45.414/o
Natural Water
Content
17 53
14 A4
o.v,
14.36
9.55
I
2.61
5.34
4.Ui
50.37o/o
54.ZA"h
52.32%
01'.11'
(a),LL = 45.35%
b) From Figure 01.10 (a)
36.97% +36.05%
ut
qo.?o
1
23.20
20.42
12.90
a) From the
c)
,
PL
= 36.510/o
y1
=
^
MCn-PL
- LL- PL
50.37 +54.26
_
^,n,,2
-
IYIV,
MC,= 52.32%
,,_52.32-36.51 _anoa
='1"788
45.35 - 36.51
."#fiii'i:]*rr
.*{iif iS,".1-''' '' - ','"?}nltu
' '':'h
ri+:r'It'
l\
i,
''t
'w
', t
1oo
50
-
!if5i",.or
ProPerties
Fundamentals
Geotechnical
Fundamentals of
Gcotechnical Engineering
PROBLEM 01.47 ICE NOVEMBER t9c)8,MAy 2OOt)
The results of Liquid Limit and Plastic Limit tests are shown in Figure 01.12.
a) Determine the Liquid Limit (LL) of the soil,
b) Determine the Plasticity index
Liquid Limit Test
)A
(s)
a
E
E
24
c
))
o
'F
128.6
o
co
a
c
o
20
t32.6
109.6
c
22
(lJ
o
U
20
?,
1B
/
16
74
o
J
I
I
o
I
oo
t
g
24
b
105.4
116.8
E
E
.o
Soil (q)
l4t.4
I
26
Weight of
Oven Dried
I
I
I
o
I
40 45 50 55 60
65
Moisture Content
1B
U
16
14
r34.6
111.2
136.0
113.4
Figure 01.13
a
40 45 5Q 55
60
Plastic Limit Test
Moisture Content
Moist Soil
Weight of
Oven Dried
(Mc)
r28.6
105.4
22.01o/o
747.4
116.8
2r.06o/o
t32.6
109.6
70.99o/o
134.6
111 2
27.040/o
135.0
113.4
19.93%
Weight of
Figure 01.12
(o)
LL
Soil (s)
SOLUTION
a) Liquid limit:
The liquid limit of the soil is the moisture content corresponding to
mm cone penetration,
From Figure
01,"13,
LL =
520/o
b) Plasticity index:
Plasticity Index, PI = LL * PL
From the plastic limit test, PL is the average of all the tests.
,, = W,,,oi,t -W,lru
ffi
.'
wo,,
p7 = (22.01 + 21".06 + 20.99 + 21.04 + 19.93)
PL = 21,.01.%
PI = 52o/, -
LL=MC=W"
w,
21,.01,o/"
0l -
Properties
of Soil
Liquid Limit Test
Plastic Limit Test
Weight of
Moist Soil
Chapter
= 30.990/o
/
5
5t
5Z
0l -
Chapter
of Soil
Properties
Fundamentals
Geotechnical Engineer
l'undamentals of
Gcotechnical Engineering
PROBLEM 0 t.48 (CE MAY 2000'
PROBLEM Ol.+q
The result of a Starrdard Proctor Test is as follows:
I hc
Water Content
. Weight of moist soil
in
(%)
10
Proctor mold (grams)
1.2
1,,606
14
1,696
1,6
1,757
18
1.,741,
20
1.,651.
1,485
Chapter
0l - Properties
of
E2
Soil rr
$E MAY2003l
following data were obtained from the Atterberg Limits test for
a soil:
Liquid Limit = 41'.0 %
Plastic Limit = 21'.1, %
n) What is the plasticity index of the soil?
a) u tire in situ moisture content of the soil is 30%, what is the liquidity
index of the soil?
c) What wguld
The volume of the mold for this test is L/30 cubic feet
(946,000
be the nature of the soil?
SOLUTTON
a) Plasticity Index:
millimeters).
Pl=LL-PL=41.-211
a) Determine the rnaximum dry unit weight of the soil in grams/cc:
b) Determine the optimum moisture content in percent.
PI =
19.9o/o
b) Liquidity Index:
SOLUTION
tAt-
"'-
lorv
Ll =
W
l+rvIC
w,
V
MC_PL
PI
30 -21,.7
Lt-- 19.9
LI = 0.447
w
(1+ MC)V
V=946cc
c) Nature of soil:
MC(%)
W(grams)
I
'
dry
=w+ MC)V
(-l
10
7,485
1..4271.
"12
r,606
1.5158
14
16
1,,696
1.5726
1,757
1.5993
1t
1.,741
7.5596
20
1.,657
1,.4544
From the table shown, the maximum dry unit weight is 1.5993
grams/cm3 and the optimum moisture content is'I..6a/o
,
Since Ll = 0.447, i.e. 0 < LI < 1, the nature of the soil is PLASTIC
See Section 1.10 in pageT:
54
Chapter
of Soil
0l -
Properties
Fundamentals of
Geotechnical Engi
Ciassification
' of Soil
lrundamentals of
Chaoter 02
(lcotechnical Engineering
55
Chapter Oz
Classification of Soil
,.
I
TEXTURAL CLASSIFTCATION
lrr this classification system, the soils are named after their principal
r()nrponents, such as sandy clay,' silty clay, silty loam, and so on. There are a
rrrrrrrber of ciassification systems developed by different organizations. Shown
I'r'lorv is the one developed by the U.S. Department of Agriculture.(USDA).
llris method is based on the following limits of particle size;
size
size
(ilay size
Sand
: 2.0 to 0.05
Silt
: 0.05
mm in diameter
to 0.002 mm in diameter
: smaller than 0.002 mm in diameter
Percentage of sand'
Figure 02.1
-
USDA Triangular Textural Classification Chart
Chapter 02
56
of Soil
-
Classification
Fundamentals of
Geotechnical Engineering
I
Llndamentals of
(r('otechnical
En
Chapter 02
gineering
- Classification Ea
rt
.€ (,
(./)
E*<--6-SX
d.6.
v
i.
f
Ee
a---
/a\
oo
^rv
s5
l\nx
I \"I\-
=ss
Example: 35% clay, 40% sand, 26% sand (clay loam.)
',',
UNTFTED SOrL CLASSTFTCATTON SYSTEM (USCS'
This system classifies soils into two broad categories:
ttlt
VYYV
a@ao)
=
Coarse-grained soils that are gravelly .ancl sandy in nature with
less than 50% passing through the No. 200 Sieve, the group
symbols start with prefixes of either G or S. G for gravel or
gravelly soil, and S for sand or sandy soil.
50% or more passing through the No. 200
sieve. The groups symbol start with prefixes of M, which stands
for inorganic silt; C for inorganic clay, and O for organic silts and
'
clays. The symbol Pt is used for peat, muck, and other highly
organic soils.
u
lo
W - well graded
\i\/l
P
- poorly graded
- low plasticity (LL < 50)
\il-t
oo
t--Fai=:!-i-
I
ttt
lil
=
<
=
\lt
-
{'l.
@a
rr
rV
,4iAQaE
..e€€$s
P€3EAA
?'iaa<.
o9
;q
a=
s e P-o e Pb
r
IV
/
o
o+<9Y
q.9<i-old
+
F-
A
/
+ ='
;-
I
\il
\/t
\tt
ti
e
=L
-
\/l
=
^
/
\
^
^
V
I
A
Y
=
-
-
<<<<6@60
,A,A,ndd;dd
j=
-d9E
ESsS!:!:
E$59 dd=
s
oooo=
€ E
6 d
=
E
E.==+=
=.
a
q.
=
35==asq
=.E
3c
=.+
kl
6o6@oq==
g6
oil
6^ r
a 9q
5€P
s q 9
:eQ.
oq
qs
L
H
\;a
\i
6e
6<
Other symbols used:
A
IA
A
l{lEE
r?=
2. Fine-grained soil with
/t
\--
s,
aS
1.
""
,t
I
'1 " v I J "J Y.+
qaa
u) !, oc
Qaooooo
() () 7
'0E
n n <
TP€€
!€t
F
a)
o,;664
99init
\/ s O <
//)2 I
o <6X
</1
lt /t
//
t=l ll (+
/l
/ /1 /A
r
t
r'
ll /l
/
v+
/l
tt
e+
t/*
r{
&/{/l/{
r'! r'{ r'* r'v lr', tr', Eo
't{
V
IV^IVA
IVA]V
]VAIV^
Y1Y1
,nooooo
oo60
,," .< ;e ;e:e
+-:SS"<;e.Sd-ES
:s ;s:s
$s$s$$ 3S-:!$$$SS Y1IY1
s;s:ss
,J;dSdd
d
sqsqqq
ddad.aa EqE{Sqs{
qqa
q@ q@ q
Luo6o6o
aaaaaeaa
.rvvv.tv vv.tv.r'1,+., .rv'r+; v'l,vvv,l, .t+v,rvvvv vvvv
a@oOa@
ggHdEEEE 33EF
o
dFBFFFFF EdFF=q €€3e"-eE
._ ;d-:-:
< q-5.5€E
aaaa=-==
e g--*+ a
lr
2.2
A
\z
\
,1
\
ii.
< qQ
H -o
< *-r
rr
oo
Atr'
\ v!/t
\pfr
wn
T-
-
g\
oo
^v
\,
\
l\t\. -"zl
Figure 02.2
;9
ao
l/
qo
-o
*-r
-
-\-\\
//- --l--T--\
v|
s
;s
Percentage of sand
G)
a<
Ag
;s
dd
qo
70605040:
Soil
of
rd
e€
ol
65
5
d o d o@
oo6@Jdild
@ @ c
EQEE66A6
gsqq€=€€
-YE€
ESg8.
H ae q
d
{.=
EE.<.r+==+
Q.q5
=5S3ooo-oi-66E='
,^._-_-a
€<::-e^>
ftr
=a
-q
{ sR? a€:B
ds3
Y d
9.:1
qq
6<
<F
F@
-;c
oS
€
-e
g-a
6
g
- high plasticity (LL > 50)
Figure 02.3
-
de
q
E^qP?
yz9.q
Flow chart for classification of coarse-grained soils
(<50% passing No. 200 sieve) (USCS)
!E
z
I
6
-
Chapter 02
of Soil
58
Fundamentals ofl
Geotechnical Engineering
Classification
chaprer 0z
lundamentals of
(icotechnical Engineering
- Crassifi:l,l}l 59
Table 02.1 - UNIFIED SOIL CLASSIFICATION SYSTEM (USCS)
s
--/
tr:
r
d
az
Eo
,C
3
3.
IV
(Jr
O
?6
(rl
o
qq
-
/\
{
v
?
g
!
;e
€
-\
G6,
o€
*a
EJ
-\
r
/^\
'
,\
/ \
a"g,v8rv
oooooo
IV
//\ \
A
d-
i\
N
v\
;e
+_
I
u ^,U lS""l
|
.no.nGX'Xll
EE;e;e;eS5I
g
5x
B
I
I
L I
i I
6^il
tod
c)o@ormm
r i Q
I6ili;
< g<
< -d-d999
g
<
o
<-a gga - a
oouD===
tro3.4==
5o
=
od
=
oj 5
ao
PE
o
q
nia
6o
;; r'! AI
r'*
*sl
;s ;e ;e
V
:PPT
*-qq
I v * .l v +-I
as==a!
==,
A
se
+i
a-
aQ
a-
oor.:oo-.:oo€:
:-E
frA.
ii+
+N
6 5 aH
,4
,/t ,4
t/! t/{ t/*
v\
tr\
:M
n rv
qS €:e
s
d
l\
tllt
ilil
t r,l
aa@@
ddgq
q q<-<*
::sq
il d<
3a
=3.,
SQ
Figure 02.4
-
9E.E
a.qq
E.=
<
ilE
v
?d.
3;
;)
J\
:!v-
/\
rv
,ao
;e
.<
:rY
:X
rv
:;
,E:
GC
I
o
SP
iX
E
0 ili
6 5
HB
^ool
ddadkl'l
9>o
oo@
== E i;=
I
Poorly graded sands and
gravely sands, little or no
E
Clayey sands, sand-clay
mixiures
ML
lnorganlc srlts, very fine
sands, rock flour, silly or
clayey flne sands
ct
lnoraanrc claYs of low to
med'um plasiicity, gr3ysly
'
clays, sandy ciays, siltv
clays, lean clays
OL
organic silts and organrc
sLlty clays oI low Plastrcrlv
l\,4H
lnolgan c sllls, micaceous or
diatomaceous llne sands or
silts, elastc silts
@
@-- vE
_q
SC
E
=
E
ig
^ ^osl
;addkkl
=
qqggaqq
s o
i.r55
-
6Ee."ooo
Q8rs5-5F
0 0 i i o o o
A5e-':::
,rli
li
,E
a6
o
Atterberg limits plol below "A" line ol
Plasticity index less than 4
A[erbero lim ls Dlol above'A"
PLastrctiindex qreaier tnan 7
Qu
:
(D6olDr0) >
6
in.e ot
CH
9
sE
E9
OH
Flow chart for classification of organic fine-grained soils
(>50% passing No. 200 sieve) (USCS)
llrll
ly Organic
Soi s
Atterbero limits
plotting in hatched
area are borderline
requiring use of
dual symbols
Qc = (0a)'?/(Dro x Dm)
Between
AR
Nol me€tlnq both cfller
s
ni
ajor SW
Pe
<R
6._zo
Allerberg limits plot below "A" line 0r
Plasticiiv index less than 4
AtLerbero llmrts olot above "A" l,n€ 0r
Plasrciii rnder greater than 7
Atterberq limits
plorting ;n halched
area are horderline
classifications
requirinq use of
dual symbols
=
Eso
!)
I
10
PT
lnorganrc clays ol high
olasticilv fal
clat/S
Organrc clays ol mediun to
high plasticily
)eal, milck, and olllor hlgnly
ngan c solls
1& 3
ed
9o
1
:'C
=
sdsaoE
&3
Nol meeting both criteria lor GW
E
Well qraded sands and
sands, little or no
pmvely
Siliy sands, sand"sill
mixluaes
tS
EE:sE{
EE?€tE* EE!ggt* 2
€€qqi
:-:
_:a
A'r3o+;
Clayey gravels, gravel-sand.
clay mixlures
SM
@
,ao ,E*
1
:=
,q
fo
E"9
|
A6e-'=,=
Silty gravels, gravel-sand-silt
mixtures
SW
E3
E,S
ot
fft-a*a*?F' ffa*a***I il +_"+l*3|
3 i B B6laU 3'a $ $u"*l I A a B But*!
r r rda
16; I
I I I i 6.6 r| rI rtj_j.Et
I rI II rddl
I 6.Al
J l.l'l v v,l'1, tr l J v,I"I t'i *t v*I
qqggqaq
GM
,:, I
14A-41 uAAAI
A,4AI
nr
v u
, ^,u ^ool
a6ddkkl
a
J\
!*
sl- ?
Hi
;E
E P tl8
E
Fa
o E
o
XB
E-5
,E*
Poorly graded gravels and
gravel-sand m xtures, litlle or
no nnes
B
';
^
Qs= (Da)'?/(Drox Dm) Between
8g
..
1,
4
E
ES
:X
GP
;
*
B^r+N+N
Qe = (D6s/Dio) >
fines
fnes
u4000@
a
3g
--
and
oravefsand mi{tures, little ol
io
I
S:*S:"F.:oENPO
g3B8H8
/\NNANA\
J\ /\
GW
=
vi
t*
/\^
/\
/ \
o
=
/t\..
*--
6
g
o
,q
Classif ication Criteria
Typical Names
Well.graded qravels
oc
d
,/\
,/\
//\
*
Group
Symbols
M,rlor Divisions
--- =\\ ___\*
\\\-
Visual-lVanual identification, See ASTIVl Designation D24BB
Chapter 02
6A
*
Classification
Fundamentals
Geotechnical Engineeri
of Soil
Fundamentals of
Grotechnical Engineering
Chapter 02
- Classification
6l
'l'lrc following example shows the calculation procedure for sieve andlysis.
LUmUlatlve
Sieve
Diameter
No
(mm)
4
4.76
2.38
x
(J
E
8
,o
10
E
ZU
.b
40
60
d
EO
100
200
mass retained
above each
sleve
(percent finer)
5
45
65
92
5
5U
99.370/.
,3./ t70
85.53%
,2.UU
u.2b
l
115
207
359
474
IdU
u.149
27?
bE6
b.-t
'/49
0.o/4
'1'.)
0.84
0.42
1.52
U,
PAN
IOTAL
Figure 02.5
CL
)
)
)
MH
)
)
OH
CL-ML )
ML
OL
CH
2,3
-
Plasticity chart
Inorganic; LL < 50; PI > 7; Atterberg limits plot on or
above A-line
Inorganic; LL < 50; PI < 4 or Atterberg limits plot below A-line
Organic; (LL - ovendr ied) / (LL * not dried) <'0.75; LL < S0
Inorganic; LL>50; Atterberg limits'plot on or above A-line
Inorganic; Ll > 50; Atterberg Iimits plot below A-line
Organic; (Lt * ovendried)/(LL - not dried) < 0.75; LL > 50
Inorganic, Atterberg limits plot in the hatched zone
PARTTCLE-StZE DtSTRtBUTION CURVE (STEVE ANALYStSI
Sieve analysis consists of shaking the soil sample through a set of sieves that
have progressively smaller openings, These sieves are generally 200 mm in
diamcter.
To conduct a sieve analysis, the soil is first oven-dried and then all lumps;
must be broken into small particles. The soil is then shaken through a stack of
sieves with openings of decreasing size from top to bottom. A pin is placed
below thc stack,
Percent
Mass Retained
(grams)
)
\4
passing
/ 5.Yb
70
54.84%
40.38%
'13./'l"k
781
5.79%
7.76%
795
0.00e/"
795
Sicve #4:
Cumulative mass retained = 5 g
Percent passing = (795 - 5)/795 x 100% = 99.37%
Sicve #8:
Cumulative mass retained = 5 + 45 = 50
Sicve #10:
Percent passing = (795 - 50)/795 x"I0O% = 93.71%
Cumulative mass retained = 50 + 65 = 115
Percent passing = (795 - 115)/795 x 100% = 85.53%
Sieve #20:
Cumulative mass retained = 115 + 92 = 207
Percent passing = (795
-
207)/795 x 100% = 73,96%
A particle-size distribution curve can be used to determine the following four
l),lrameters for a give soil:
2.3.I
EFFECTTVE S|ZE, Dro
- 'Ihis parameter is the diameter in the curve correspondin g to 1.0% finer. The
r'[lcctive size of a granular soil is a good measure to estimate the hydraulic
t'rrrrductivity and drainage through soil.
ln the given example, Dro = 0.17 mm
62
Chapter 02
of Soil
-
Classification
Fundamentals
GeotechnicalE
2.3,2 UNIFORMITY COEFFICIENT,
Grain Size Curve
Dog-
Dro
90
Eq. 2.
80
;{
zo
.c
o
60
o
o
**
0.17
Classifi;,lff 6g
100
where Dro = diameter corresponding to 60% finer.
In the example, Cu =
-
Eeotechnical Engineering
CU
c,1=
chapter 02
Fundamentals of
4
o- ,"
=a
I
odn
o-
v00
0.5
0)
o30
2.3.3 COEFFICIENT OF GRADATION OR COEFFICIENT
OF CURVATURE.
ttt
=0i
c=#*
where D:o = diameter corresponding to 30% finer.
' *?:,
0.sl(0.17)
In the example, C. =
2,3.4 SORTING COEFFICIENT,
Dr =iq
7
?1
10
|1
0
0.001
0.1
0.01
1
Particle Diameter, mm
Figure 02.6
- Particle
distribution curve for the above example
= 0.509
I'hc particle-size distribution curve shows not only the range Of particle sizes
of various-size particles.
rrr,sent in the soil, but also the type of distribution
Poorly graded soil is one where most of the soil grains are the same size.
SO
Sr=
tr;
1/%
where
= diameter corresponding to V5% finer
Dzs = diameter cortesponding to 25% finer
Dzs
20
CC
welt graderl soil is one in which the particle sizes are distributed over a
wide range. A well graded soil has C, greater than about 4 for gravels
and 6 for sands, and C, between i. and 3 for gravels and sands.
Cap graded soil is characterized by two ore more humps in the grading
curve.
l'lre average grain size of the soil is Dso.
Chapter 02
64
of Soil
-
Classification
of
Fundamenta
GeotechnicalE
Grain Size Curve
Chapter 02
-
Classificotion
of Soil
Engineering
l'1p = percentage passing No. 200 sieve
Lt. = liquid limit, pI = plasticity index
90
80
;{
Table 02.2 - Classification of Highway Subgrade Materials for
Granular Materials (MSHTO)
zo
o,
c60
a
o
.
{so
Y
r
ir:i.,.:1:l.f :ii:ii.,::;, ",,, Qrdlularfialefl?ls' :,1,
Grouo classiflcation
40
0)
20
A-1.a
A-1-h
A-3
No.10
No.40
5U
JU max,
50 max
1 mln.
No. 200
15 max.
2b ma)(
10 max.
10
Characteristics of
fraction passing
0
N0.40
Figure 02.7
-
Different types of particle-size distribution curve
A-2-5
35 max.
35 max
40 max,
41 min
A-2.6
A-2-7
.Jh max_
Jit max.
6 max.
NP
Usual types ol
signiflcant constituent
materials
Stone fragments,
gravel, and sand
Fine
sand
According
ln tlir_ systenr, soil is classified into seven major groups: Athrough A-7, soils classified under groups A-1, A-2, and A-3 ire gianula
materials of which 35% ar less of the particres pass through the No. 200 sieve
soils of 'which more than 35% puur throrgh the No.200"sieve are classifiec
under groups A-4, A-5, 4-6, and A-7. These soils are mostly silt and clay_ty
designation, example, A-Z-S(35).
uit", g.orp o. iubgioup
- 35X0.2 + 0.00S(LL -a0)l+
-1SXpr-
10)
Eq.2.4
Silty or clayey gravel and sanQ
Excellent to good
Silt-clay materials
(mors:than 35% of totdl Sample passlng No.:2Q0)
A-6
A-5
A-4
A.I-5.
36 min.
36 min.
36 min
36min
Liouid Iimit
40 max.
41 min
40 max
41min
Plasticity index
10 max
10 max.
Characteristics of fraction oassino No. 40
usual tvoes of sionificant constituent materials
General subqrade ratinq
a
0.0j.(Fz,o
11 rn;p,
N0.10
N0,40
b
(Fzoo
|
A.7-6b
No. 200
parentheru,
min.
Sieve analvsis (oercentaoe oassino)
the test data fit is the correct classification.
soil. This index is written in
11
'
General dassif icatioil
Group classiflcation
To evaluate the quality of a soir as a highway subgrade material, one must also
incorporate a number called the grouplndex with Jhe groups and subgroups of
|
Table 02.3 - Classification of Highway Subgrade Materials for
Silt-Clay Materials (AASHTO)
materials.
To classify the soil using the tables below, one must apply the test data
left to right. By process of elimination, the first grorp iro* the left into which
{ I mln.
40 max
10max. | 10max,
Plasticity index
General subgrade
rating
AASHTO CLASSIFICATION SYSTEM
GI=
A-2.4
max.
Liquid limit
Particle Diameter, mm
the
:'
Sieve analysis
fncrccnlaoe nrssino)
LJU
2,4
i.Ii
(350/" or less of total samole oasslno No. 200)
4"2
c
9i
Genelal :iilirrtl; rf'.i
:;lri;ia(;;rHr(r r',';,.
For A-7-5, Pl s LL - 30
For 4-7-6, Pl > LL - 30
'1
1 min.
Siltv soils
1 1
Clavev soils
Fair to poor
min.
65
Chapter 02
66
-
Classification
Chapter 02
ncntals of
hnical Engineering
Funda
Geotechnical Engi
of Soil
- Classiflcation
-'' ""
;;;,,,
STRATIVE PROBLEMS
/
M OZ.l
/
of the sieve analysis is shown below:
What percentage of the soil is retained in No. 200 sieve?
Wt.rut is the effective grain size of the soil in mm?
€) l)ctermine the uniformity coefficient.
tt,sr,rlt
o
!
s40
'o
E
l)
t)
r-7-6
ts30
E
A-2-6
A-6
A-2-7
A-7-5
\-2-4
\-2-5
A-5
A-4
t0 20 30 40 50 60 70 80 90
-
Diameter
(mm)
Mass Retained
4
4,76
2.38
2.00
0.84
0.42
0.25
0.180
0.1 49
0.074
25
B
10
100
20
40
60
80
Liquid Limit
Figure 02.8
Sieve
No
Range of LL and Pl for soils in groups A-2, A-4,A-5, 4-6, and A-7
100
200
Pan
The first group of F,q.2,4 is the partial group index determined from I
limit. The second term is the partial group index determined from plasti
index.
always zero.
When calculating the GI for soils that belong to groups 4-2-6 and Ause the partial CI {or PI, or
GI = 0.01(Fzoo
BO
110
160
180
220
380
590
110
B5
tolurtoru
. lf Eq. 2.4 yields a negative value for GI, it is taken as 0.
. G/ calculated from F'q.2.4 is rounded-off to the nearest whole number.
r There is no upper limit for GI
r The G/ of soils belonging to groups A-l-a, A-1-b, A-2-4, A-2-5, and A-3
.
/orams)
- 15)(Pf - 10)
Cumulative
Mass Retained
Percent Finer
25
25
98.71Yo
80
105
94.59Yo
BB.92o/o
Sieve
#
Size,
mm
Mass
Retained
4
4.76
8
2.38
10
2
110
215
20
0.84
160
375
80.670/o
40
0.42
180
555
71.39Yo
60
0.25
220
775
60.05%
BO
0.18
380
1155
40.46Yo
100
0.149
590
745
200
0.074
'1'10
855
10.05%
4.3A%
85
940
Pan
Total
1
940
From the table shown:
Percent r€tained in No. 200 sieve = 100% - 4.38%
Percent retained in No. 200 sieve = 95.620/o
a)
\
67
68
Chapter 02
of Soil
-
Classification
F
GeotechnicalEng
b)
Effective grain size, Dro = 0.149 mm
c)
Uniformity coefficient, C, = Doo/Drc
Uniformity coefficient, C, = A.25/0.1,49
ilnentals of
t'chnical Engineering
Chapter 02
l,
* Crassifi:?tiffi
6q
='1.,68
/PROBLEM O2.2
A soil has the following particle-size distribution:
Cravel = 20%
Sand = 10%
Silt = 30%
ClaY
=
4g"1
Classify the soii according to USDA textural classification system.
SOLUTION
90
Modified percentages of sand, silt, and clay:
Modified
%sand* " %sand
I00 -%graael
= 10010
* 20
Modified
%
silt
-%graael
30
100
o/o
clay
'
'4i
-
70 60 s0 40 30 20 10
PercentaEe of sand
0
The lines eorrespond to each percentages interseci on the clay rep;iotr,
thus, the soil is clay.
%silt
=
L0Q
Modified
= 1,2*5%
B0
'l
-
20
1n co/
- Jl J to
o/"cloY
00 -o/ograucl
100
-
t,troBLEM 02.3/
{ l,rssify the following soils by the AASF{TO classification system.
20
Description
Percent finer than No, 10 sieve
Percent finer than No. 40 sieve
Percent finer than No. 200 sieve
Liquid limit
Plasticitv index
Soil A
Soil B
Soil C
B3
100
48
48
28
2A
92
86
20
7A
q
1a
6
Nonplastrc
SOLUTION
To classify a soil according to this table, one must applv the test data
from left to right. By process of elimination, the first group from the
left into which the test data fit is the correct classification
Soil
\
A:
Per"cent passing No. 200 = 20'/,< 35%
TIrc soil is either A-1, A-3, or A-2
7A
Chapter 02
-
Classification
Fundament
of Soil
Geotechnical
Chapter 02
hnical Engineering
E
passing No. 10 = BZ'% > 50, it is not A-1-a
% passing No, 40 = 48 < 50
% passing No. 200 = 20 < 25
Thus, the soil is {,-1.b, with GI = 0
Thus, the soil is A-1-b(0)
-
Classification
of Soil
%
Soil
B:
Percent passing No, 200 =- SOy,, SSy"
The soil is either A-4, A-5, A-6, or A-Z
LL = 70 > 40, it is not A-4
PI = 30 > 10, it is not A-5
is not A-6
tt
'
#:'10:it
PI < LL .30, the soil is A-7-5
Gr = (86 -3s)[0.2 + 0,005(70 _ a0)] + 0.01(86 _ 1sx32 _ 10)
GI
-
l0
t/0
f:,
I;
33.47 use 33
Therefore the soil is ,4-7-5(33)
Soil
C:
0
Percent passing No. 200 = 6%< gS%, it is A-1, A-3, or A_2
% passing No, 200 = 6% < S0
% passing No. 40 = 28% < 30
% passing No, 10 48 < 50
=
Thus, the soil is A-1:a and GI = 0
Thus, the soil is A-1-a(0)
Particle Diameter, mm
Size (mm)
Weight
Retained
0.25
0.1 49
33.1
0.074
45.03
0.052
0.02
0.01
0.004
0.001
Pan
18.96
I
54.51
42,66
1 1,85
4.74
4.74
21.33
Weight
-
Grain size curve
Retained
Accumulated
Weioht
Percent Finer
18.96
18.96
92%
49
33.1 8
52.14
78o/o
o.074
45.03
97 17
59o/o
0.052
54.51
151 .68
36%
0.02
42.66
194.34
18o/o
0.25
The table below shows the laboratory results of the sieve analysis of a sam
Plot the grain size curve of the soil in the attached Figure 02.i0. The soll h
liquid limit of 35'/, and plasticity index of 26%. craisity the soil accordi
a) USCS, b) USDA, and c) AASHTO.
0.1
Figure 02,9
Size (mm)
lPROBLEM O2,4
01
0.1
0.01
11.85
206.1
I
13To
0.004
4.74
210.93
11Yo
0.001
4.74
215.67
9o/o
21.33
237
Pan
Total
237.00
7l
Chapter 02
of Soil
72
-
I
80
70
of Soil
D
O)
60
&
50
o
o
L
40
/
s
o
0)
o-
Classification
Percent sand (2.0 mm to 0.05 mm in diameter) = 100 - 33 =
Percent silt (0.05 mm to 0.002 mm) = 33 - 1'0 - 23%
Percent clay = 1gY
D
90
'6
v)
a
-
6
u)
o
30
0-
20
Deo =
0,08
10
I
:o = 0,02
(,tt
0
0.001
Dn = 0.0022
0,01
c
0)
o
o
I
,
0-
i
L
33%
0.1
Particle Diameter, mm
al
73
b)usDA
100
;s
Chapter 02
Fundamentals of
Geotechnical Engineering
Classification
.10%
05 mm
llt
USCS:
Percent passing No. 200 sieve (0.074 mm) = 59% > 50%
Therefore, the soil is FINE GRAINED.
0.1
0.001
Particle Diameter, mm
0
LL = 35% < 50% (ML, CL, or OL)
From the plasticity chart, with LL * 35o1, and PI = 26%, the soil is CL
r00
90
80
70
60
50
40
30
Percentage of sand
From the chart shown, the soil is sandy loam
I
I
I
I
20
100
677o
74
Chapter 02
-
Classification
Chapter 02
Fundamentals
Geotechnical E
of Soil
-
Classification
of Soil
c) ASSHTO
Percent passing No, 200 sieve (0.074 mm) = 59% > 35%
"Silt-clay materials". Use Table 02,3.
General classif ication
' ',,,,
.' :,:',.
Silt"clayrnaterials
.
(more than35o/tof total sample pasSing No. 200)
A-7
Group classification
A-4
A.5
A-0
A:7"5"
A-7-6b
Sieve analvsrs (oercentaoe oassinol
No
10
No. 40
No. 200
36 min "/
Characlerislrcs of fraction passing No, 40
Liquid limit
PlastioitV index
Usual types 0f sionificant constituent materials
40 max.
10 max
r'
r
r'
36 mln,
41 min, x
40 max.
r
11 min,
36 min.
10 max.
Siitv soils
General suborade ratino
6
b
r'
36 min
r'
41 min.
{
11min.
Clavev soils
Fair to pool
For A-7-5, Pl s LL - 30
For 4-7-6, Pl > 11.30
o."Jv
U)
oc
The soil cannot be A-4 because its P/ = 26% > 1,0%.
The soil cannot be A-5 because its LL = 35% > 41,y".
The soil is 4-6
Solving f or Cl:
G/ = (Izoo * 35X0.2 + 0.005(LL - 40)l + 0,01(Fzoo - 15)(P/ - 10)
Cr= (se - 3s)[0.2 + 0.00s(35 - a0)] + 0.01(5e - 15X26 - 10)
ct=
s70
o 60
,s
a
840
0)
L30
10
0
0.01
0.001
11.24
0.1
Particle Diameter, mm
Thus, the soil is ,4'-6(11)
Figure 02,10
/PRoBLEM Oz.5IcE MAY 2003,
'
size.
&) Determine the nearest value to the coefficient, of uniformity, C,.
c) Classify the soil according to the Unified Classification System, using
Percent Passing,
(or % Finer)
Sieve No
Diameter (mm)
+.76
90
B
2.38
64
10
2.00
58
20
40
0.84
35
0.42
0.25
22
following:
Table 02.1.
Grain size curve
SOLUTION
The table below shows the laboratory results of the sieve analysis of a iimple;
Plot the grain size curve of the soil in the attached Figure 02.1b. Determine the
a) Determine the nearest value to the effective
-
60
100
:t:il,r$ttg
200
4.074
15
ilrr:i
10::
4
%
75
_
76
Chapter 02
of
'rarrrtr'
Soil
a)
-
Classification
damentals
Fundamentals
Geotechnical Engineeri,
of
Chapter 02
I".r."iiritrgineering
Effective Size:
The Effectiuc Size, Dro, is the diameter of the particles of whi
10% of the soil is finer. D16 is an important value in regulatir
flow through soils and can significantly influence the mechani<
behavior of soils,
^-
(Dro
oi
)'
From the graph,
^ :)-:::1 :::
C.=
(0,63)2
Dao
= 0,63 mm
=1.21 (Betweenl&3)
For this problem, Dro = 0.149 mm
Since C, > 6, C, is between 1 & 3, the soil
bI
Hrlr[ 0rvisrons
Coefficienl of rrniformitv:
Group
Symbols
Typical Names
GW
Well-qraded gravels and
gravel-sand mixtur6s, lrttle or
no fines
GP
grave[sand mixtures, little
o
az
I
85 o
o)
c60
,6
Arn
""
I
5S
:n
3
fl
c
40
Gl\,4
E
>4
CC= {Dm)2/(OroxDm) Between 1 & 3
Nol meeting both criteria for GW
or
s
I=
p
&go
Clayey Oravels, gravel-sandclay mixtures
't
2A
Co
uz
I
o
co
Poorly graded sands and
sands, liltle or no
Fffily
,e
E"E
3gi
0
u.ul
u.r \o.r+s \,,
'10 63
'
Particle Diameter, mm
D,ro
Dro
Dn= 0.149
From the grain size curve shown ,
Dsa
= 2.2 mm
=14.8
Pait 3 Classification of soil:
Percent gravel (retained in # 4 sieve) = 700% - 90'/o
Percent gravel (retained in # 4 sieve) * 10%
)zz
==.ig
qo:€
9O 6=
a>E
oo6 i
=
Alterberg limits plot bel0w'A'line 0r
Plaslicily index less than 4
Atlerberg limits plol above "A" line or
Plaslicily index greater than 7
Atterberg iimits
plottinq in hatched
area are borderline
classifications
roquirirg use of
dual symbols
'9.4
e
oi\
10
cq
frE
'6
IH
!
a6
Silly gravels, gravel-sand-sill
mixlures
II
o
0,1,49
"(DodDrc)
:E
Poorly graded gravels and
0
lt
))
C,.= '''
Cu
rE
hp
oo
- '-
Cla6silication Criteria
no fines
bE
s70
9?
SW (WelI graded sand)
(,
,'ll-llll{
o-
is
Dso,
I
ta
soir 77
Dro Doo
0.1.4e(2.2)
Th,e Aaersge Grain Size diameter of the soil is
- ' Classification
-
E
8
SM
Silly sands, sand-sill
mixtures
SC
Clayey sands, sand-clay
mixtures
3
E
.9o
d2
>
1sH
a q9
*&
b-z o
=89
Not meeiing both criteria for SW
Atterberg lrmits plot below "A'line or
Plaslicilyindex less lhan 4
5
=
Alterberg limits plol above "A" line ot
Plaslicily index grealer than 7
Atlerberg limits
plotling in hatched
Br€a ar6 borderline
classiflcalions
requiring lse of
dual symbols
Chapter 02 -.Classification
of Soil
78
Fundamentals
Geotechnical Engineer
Using Figure 02.3
q<
AB
__----l-----
r
Ia:th
3s[3
lr-
lrL
VV'lr
a@o
2t/
;lit,
r'{ vv Y+
rV
^rV A,rt ^
ooooco
aS;s"<EP:e.g
c60@o6
d dd idt d
qqqqqg
<:.Eg gE.E
q +€ € S $
-:i33=
*E=
€€:
;6
a
ddd
=
a
a^\ x
\ /{
\en
A
/l\
/l\
/l\
JJJJ
{I
qcna.aa(,Qoooooo
r€
IIFF
6i3?
=
ii5<
a
JJII
."
ri.
\/{
H
ir E=r
80
\t
l\ii
lF
l,Wn
FiF
< 9 s Po e Po
'r
JIJ
pp
v
ll/l\ \
3
ll
a"r\ x
\ ",
l\=":
Wn \+l pI PJfF lP
e
=
=
rp=H
g8g
isiE
l
lol
l,l
ttt
l,l
I/t
\/l
VJ
o.o
.r€
i *
elfs
€oEi:
6Hs
t
o."
o"
E
=@
ac-
€=
I
I
$+o
/
0)
troo
,,1c.
I
20
I
1
/
10
0
0.001
0.01
0.1
Particle Diameter, mm
EE9.g
EfggE5 $;g;€€€6 EEES
dqe€
EEEEE$$g BE$$ EEEBi- E€EEgE$E qq=
P966
5
i=331ii?
5'F'
ia='+=+== E.s
s+se€e== s
E 5
iidi*'r'gg
<=;-F6:Pddo-e-==;;E
Sqa E4e* '
rqi sI*
EE.g
Ess 8El
;Ee
3s €=
I
Alr
6
o
\{h
!o
q :ll
70
o^^
c0u
I
/ I 9t/
////
/) /t
/,/,"4"4
.4/l/l
el
v*
A/+,4A
rv^,!^v^rv^
"{vv
rv
VA ylyAy1
ylyly1yA
YlYA
€-<9
::€il<
:",<F
What is the classification of soil B?
90
3P
=:p
i. e
io e
= \/
i/ii II
F-- 1F+;
dsddaaaa
)
Grain Size Curve
$s" s$ee$ s-c$$ ss$-*S- $-$-s$$S-E ;S.<Ee;9
e+Ais{
sf,sBsssg EEEts
ssBEssgE
sEBE
.r,rovv+
.rov+
v,l,vr+v+v !P=€
vvvvvv
+vJ+JvJ,r. ;JJ;
@@ao6u)
<=6!
79
100
EE
l\i.
l\- i: \:n
\:s
F
of Soil
a
-s;*ps
EfiEI
PP
<qq
Classification
fr1 What is the classification of soil A?
__,---i----_
'!--\-'"+
at/11/l\\/ x
l\, ,.{
\rn ,
-
llrt,grain-size curves for soils A and B is shown in Figure 02.11. lt is required
tr e lassify the soils according to the Unified Soil Classification System.
rt) Determine the value of the coefficient of uniformity of soil A,
?6
,*
,t\
Chapter 02
PRoBLEM 02.6lCE NOVEMBER 20031
f$..--*-ff--.-$9
*
oo
eppp/l\ps,
tdamentals of
Eeotcchnical Engineering
Figure 02.11
*
Grain size curve
10
Chapter 02
of Soil
80
-
Classification
Fundamentals
Geotechnical Engineeri
chapter 02 -
Fundamentals of
Geotechnical Engineering
Uniformity coefficient, C,=
= S.Z
ffi
Coefficient of curvature, C. =
Coefficient of curvature, C. =
Ctassifi:T,i.J g
l
D.il.'
Dro Doo
0.172
= 0.903
0.10(0.32)
Classification:
Retained on No. 200 sieve = 1.00% (more than 50%)
Coarse-Grained Soil
Passing No. 4 sieve
SOLUTION
-
100% (more than 50%)
Ssnds
Grain Size Curve
6;
C. = 0.903 (not between 1 and 3)
100
C, = 3.2 <
90
Since the soil does not meet both criteria for SW,
80
*
70
oi
'6
60
L
50
o
o
o
o
o
0
Soil B:
Dro = 0.15; Dso = 0.66; Dr,o = 2.1'
Uniformity coefficient, C,,=
?.1.
#
0,15
Uniformity coefficient, C, = 14
40
Coefficient of curvature, r,=
30
2p
m
=
ffi
Coefficient of curvature, C, = 1.38
10
Classification:
Retained on No. 200 sieve = 96% (more than 50%)
0
0.1
Particle Diameter, mm
Coarse-Grained Soil
Passing No. 4 sieve = 90% (more than 50%)
Sands
Soil A:
The soil is either SW, SP, SM, or SC
Prs = 0.10; Dn = A.77
D,,,, = 0.32
Uniformity coefficient, C =
C,,=14>6
C. = 1.38 (between 1 and 3)
:!q
Dro
Therefore, the soil is SW
is SP
-
Chapter 02
82
Fundamentals of
Classification
Geotechn ical Engineering
of Soil
Fundamentals of
Geotechnical Engineering
ut
sA<G
d-
oo
rvl
;e
-----+
-------
o
classifi:?rff g3
Using Figure 02.3 for Soil B;
Using Figure 02.3 for SoilA;
s(,
$I
chapter 0z -
.--tl
+..s
o
il
-----1--'-
/l
V
?
$
;s
aR
E
q6
Ss
o6
,4
oi_
-o
*--E=
Ii
YVv
l
(naa
9t^
a
-0
€€aq€€
+.t ., '1, v
'taa60a@
€q3Y€=
_'_dor.o
'lv
aa
U)
a
t
.a
?
/l
A]V
;s
;e ;e ;e
;e a! aS:s:e
il
ddd dd il il
a66.66 q q :q
., ,
vv+
JJ v,.,v
r!n
j= '1,
g!=E
',3g ,6
r?=
s9:
o9.
il
q
-od
€a
5 {€ <-{ =.
g_o*=
e
agq 9 2_.
ae
F6 f
J9€ 6
x€Pq
a'il-
€q
s
q
6 ?,
q'qqa
€?
?
€q
3
q
\
o
oo
IV
nlv
i,,
oo
IT
-rarI
il
aa-a-a
€.
.9.
<.d=
,f,,,.E
E',4
;
a
t_It
t/
lt
1/l
ltt
n
E€!
AI
ll
ll
9.t
ll
ll
v{
Y1Y1
q3q3
ioi
o
e.
-eissass<*
6@6@6A
ef fE qa
9E
a€ f
q ii43€
e6 !
S
as
a<=
qqgq-qq
E
=
e
F'-
JJJJ
=
(t@a@
PT€€
29,:nt)
r/ s O
<
lv^rv^rv
rv^
oooo
sss.*.e"<saF
oa@@
6il6ilJ6ild.
qq9qq.qqg
vvv+.rv vvv.l,.rvvv
PPPP===€
.e€Eeo€*s
eq€QBS
<<da=
$S= q5
a a=
€€2
;
oQ
=@
F'*
,ll
ll ll /{&4'/v
,4
,:^YAYl
oooqoo
ES;S;<.s
,€5€+++5
=
?ttA
3
o
e9 3 q9
io e Ho
a(n@
o
.r+vrr'r+v vv.r.,
sri.r.
aq. q .9.P
a
==*=i
- -
j,i,J
q,
14
t\
t\ I \n'
I \"
dadOdld.id
EEAQB&*A
A3
-
.PCE
r(
6),
zlA
'/l'l
;€.sle
EEE
il
q
555
v
vvv
!
o!o!F
5Y€
:dQB
o
IA
l\
r9 !=
o
\xl
\/l
\/i
VJ
s
rl
=:
lr ,,J
J v',
oo
TT oT oc
€: ooF
o6)
oc
o< o< ?
1,4
f
I
I
;/t/)
v* v*
v*
t
t{' 9V{y9
t4
,r'
,4
oo
T€
ry
<3.P{€€
qqi.i99
ge€€2P
e€gPqe
ooe-9{
dor
55+
*=
EBab
=a=.
!
i€€==
P
:F
pi8836q
g
d
d
€ d
arggE, -& :B
P
.^
qeq
g
E d
od
il
6
zq
9s €5 .q
-{
.=.{.=.fg
=E.q9q 2.
=.
t9
g
a6
,i
,6:P
qa
9.
F
g
F
ia
I
q
q
,q
:9
_{.
o
a
E
o
AA
,t{'l+
t
E
dPFF
o
dil9e
8BH6
E
(a"rY4
'4
Y1Y1Y1
YlYA
oooooo
e;ss s llesssi{
i.e .s.SESA
ESEEsS
E
SE
aaaaaa a
.t,vvv
vv.LvvV
v ,vv,vv -t,,],., .rvv+
!
ag99t2.e. r !r ,F€
q
L5
\
!
I o-
ln
rI
o
el
82.
(
I
v
3
9Q
o:
E
d<
g
a
!
lrr
vvv
ooo
E5
€ad
=s
9*
i/
@d
d
q
o
5qP Iqd.
;3q eg
e.=
\/
oodd€€€{
aoi.aEd==
d d
= =<
r^
r
lrr/W
PP=S
*ebf
-o)i
\/
*r*rsEEE
ss9
ggq9d.dd.d
r=r5=-=E
<,5
3o
E
-<o a -
\a
\i,
\i
YAY
\"1
t\
\i
IA
PP
<gE
v,,
,1,'t+',+
ooooo
.?
1r
s,
\xl
\r
qq=
q
I
I
'r't
FF=F E
.e€.EE ee ddggEE66 <5q9
J$XX
e.i.i t e€ € €q €€----eQQ€
!
Qq{P
=
2
ETE-g
**EE HH TEBEHHFB PXoo
tr60u
HT&3 :;68 a HEHES€-€,€d oq (-o
r
6@doOe
6
*gs4 cP=. e ilildd;6f;6
EEBE
qq€
dts=
E.rS a 6asaiiaa
wvvv
I
]
I ^o
\P,
rll
aooc)
<PP€=
PEPc)
i
t
L,
l
tQOoo
I
\)
<
?
fi/=/
^
11
ll
a
I
/
aA)'i
y!
y*
vY
v{
PV
Y-V vV
tLt vt
v^rv^lv^rv^
Y.A.YA YAYA
oqoq
d6oo
oooo
sEE*Es
s.es;s
EeEQ ;e:e+-SEsSasX
660@
qEqq
SS
EESqEqEq
ail;il
adao
da
dddad.a6-d
5*;5
',P
,t"l
I
fr
r'\
p
5,
^lv
s5
\,t
\./
{r41, ,/ \
F
1P
iF F
q
q
iro
= Fo q
lr
vv
oo
C
3A
=xoqoo
':.:adE
Il\a\Ll\.n
=
€=
@o
o
o<
/ & /
/l vvil
^ v{
{+
v
v
^rv ^
^
oor6uo9
;sIS.s;s;S;S
:_T
.I
t\o^
\
rl
o<
,- i
I
t\
\
p
fr
\A
l\ss,
\
fr
v'z!!
90
o
p
t,
IV
a
a
EPEE
qq€
E,'
1a
g
o
zU
3
o
84
Chapter 02
or 50il
-
Classification
Fundamentals
Geotechnical Engineeri
Ftrrrclamentals
of
Chapter 03
t,r'otechnical Engineering
-
Flow of Water
85
through Soil
Chapter 03
Flow of Water
throu h Soils
,.I
DARCY'S LAW
J
Section a-a
Outflow
\
Figure 03. 1 - Flow of water through soil
l),rrcy's law governs the flow of water through soils. Darcy (1856) proposed
llr,rt the average flow velocity through soils is proportional to the gradient of
llrt' total head. The velocity of flow is:
\
u=ki
\\
Seepage velocity, a, = uf
Eq.3.1
n
Eq. 3.2
\
tv
l
\
r,'rt'
t=
/,
rr
I
HI
"-
I
=,,hvdraulic
L,/
./
- c6efficient
- porosity
sradient
of permeability or hydraulic conductivity, m/s or m/ day
Chapter 03 - Flow of Water
through soil
86
Fundamentals of'
Geotechnical Engineering
The flow of water is:
Q=kiA
Eq. 3,3
Chapter 03 _
Fundamentals of
Geotechnical Engineering
Fl.*
"r#ltfi gl
3,2.2 FALLING.HEAD TEST
'lhe falling head test is used for fine-grained soils because the flow of water
through these soils is too slow to get reasonable measurement from the
constant-head test
3.2 DETERMINATION OF THE COEFFICIENT OF PERMSABILITY
3,2.
I
CONSTANT.HEAD TEST
The constant-head test is used to determine the coefficient of
coarse-grained soils.
perr":
Porous stone
Soil specimen
Porous stone
q\
\t
V
E
tb€l
H-{=
Figure 03.3
-
Falling-head permeability test
Graduated
flask
.
I
t-_
ft_-_-_.-.
Figure 03,2
- Constant head permeability
VL
'
.
k= ::
tAlt
A(t) - t1)
test
where:
Eq.3.4
d
/rr = head at time
= head at time
where:
/ = duration of water collection
/r- \
lnllr-
I
Eq. 3.5
\hz )
\
= cross-sectional area of the siqndpipe
/r2
V = volume o[ water collected in time f
/l = constant head
A = cross-sr.ctional area of the soil
L : length of soil sample
aL
fr
fz
\
\
I
)
3.2.2 EFFECT OF WATER TEMPERIITURE Ory 6
'I'lre hydraul'ic concluctivitV of solL4s a function
of unrt weight of water, and
tlrrrs, it is affected by r,vater;*<rr1f,erature. The relationship is given by:
\
\
\
Fundamentals
chapter
of
o: -
rowoYltji
Geotechnical Engineering
gq
Ground surface
@
Direction of
seepage
.
Impervious layer
Figure 03.5
*
Flow through permeable layer
With reference to Figure 03.5, the hydraulic gradient is:
Hydraulic gradient, i = sin
ct
Eq. 3.8
3.4 EMPIRICAL RELATIONS FOR HYDRAULIC CONDI.'CTIVITY
3.4.1 HAZEN FORMUIIA (for fairly uniform sand):
k(cm/sec)=c(Dro)2
Eq. 3.9
where c = a constant that varies from 1 to1.5
Dro = effective size,
rhm
....
3.4.2 CASAGRANDE (FOR FINE TO
UM CLEAN SAND}:
k=L.4e2
where
t.
hydraulic conductivity at void
ko.as = k at void ratio of 0,85
-
Eq.3.10
tio
e
Chapter 03
-
Fldw of Water
Fundamentals
through Soil
Geotechnical Engineeri
kT, - vT, lu,T,
kT, |rr, f u,Tz
where:
kT.,,krr= hydraulic conductivities at temperatures Tr andT2, respectively
V"r,,Vr, = viscosity of water at temperatures Tr and T2l respectively
T w'f, rT a,7, - unit weight of water at temperatures Tl and Tz, respectively
3.3 FLOW THROUGH PERMEABLE
LAYERS
I
h
rkl..J..
t,
,l;
H
Permeable layer
F(---
li
L
'
>l
r1
Figure 03.4
-
I
Flow through permeable layer
4
,l
With reference to Figure 03.4, the hydraulic gradient
i.
is:
I
t,l
i
I
Hydraulic gradient, I =
L
cos
!
Eq
0.
3.71
90
Chapter 03
-
Flow of Water
Fundamentals
Geotechnical Engineeri
through Soil
3,4.3 I(OZENY-CARMAN EOUATION
l'undamentals of
Gcotechnical Engineering
Chapter
o, -
t,*",i#lH
H = -z
)_
(kr)rq " k"
.3
Eq. 3.11
Eq. 3.15
HZtzo
where k is the hydraulic conductivity at a void ratio of
e
z-
(k"),q-_T-Tk,t
anci Cr is a constant,
k"z
t.
Azn
3,4.4 SAMARASINHE. HUANG, AND DRNEVICH
3.6 FLOW THROUGH I.AYERS OF AOUIFERS
where C: and
rc
c'
Eq.3.12
1,+ e
are constants to be determined experimentally.
3.5 EOUIVALENT HYDRAULIC CONDUCTIVITY IN STRATIFIED SOIL
Impervious layer
Figure 03.7 * Aquifers in horizontal layers
'
kee (LD
= k., Ht + kz Hz
Flow per unit width, q = k,ri
i= Dt-Dz
L
D. +D"
_z
n= _,
(1)
Figure 03.6
-
2
Equivalent hydraulic conductivity in stratified soil
'The equivalent pcrmeability in the:r-direction is (parallel flow):
(k)*H=Zk,z
(k').q
l
'l'he equivalent permeability in the
z-direction is (normal flow):
"lr=C.'
"'r.+e
i( = Lt-
g
H = k,t 4 * k^zzz* ... *
Eq 3.1 3
k,,,2,,
L \,I
3.14
Figure 03.8
-
Non-
a
Eq.3.16
q2
Chapter 03
-
Flow of Water
Fundamen
through Soil
Geotechnical Engineer[
LL"L"
k"- k"
.
D.,'
Equivalent
-D,'
L
"
q)3
k:
H1+H2
,H _-h,
'kea k'
D. +D. .-.
2
FrowJJ#lf;
i=ht/(fu+Hr1
L
U-
-
hnical Engineering
For the entire system:
Total head = hr
Total length of soil = Hl + H,
k"
Flow per unit width, 4 = k"qia
l-:
chapter 03
Fundamentals of
Totalflow,
I
H.
_
+f-
H"
4k2
kq
q=k,riA
At point A:
-l
3,7 CONTINUITY EOUATION
Head, H =
FOR SIMPLE FLOW PROBLEMS
h2
Length, L = yr
h=hz/yr
Permeability = k1
Flow, qa = kt ia A
Continuity Equation; qt = q
At point
B:
Head, h = ht
Length, L=H1 +y2
is =
h/ (H1 + y2)
Permeability,
\*Yz
krqB =
Flow,
k"q a:
H-t
k.t
.,,
lz
k2
qs =k,, e ia A
Continuity Equation;
qB= qA= q
3.8 HYDRAULIC OF UTELLS
Figure 03.9
Consider the one-dimensional
calculations rnay br.' applied:
flow problem
shown,
The followi
Urrderground water constitutes an i-po)'ispource of water supply, The
$tratum of soil in which this water is present is \own as an aquifer. On the
lrirsis of their hydraulic characteristics, wells are d\ided into two categories;
graoity or water-table utells, and qrtesian or pressu\e wells.If the pressure at
the surface of the surrounding underground water i{ atmospheric, the well is
of the grauity typr; if this pressure is above atmospheric because an impervious
soil stratum overlies the aquifer, the well is artesian.
Assume that the water surrounding a
:;t;.rtic conditions.
The lateral flow
of
/
f
well ha/a horizontal surface under
well requires the
watey.4oward the
94
Chapter 03
-
Flow of Water
Fundamentals
Geotechnical Engi
through Soil
existence of a hydraulic gradient, this gradient being caused by a difference
pressure. To create this difference in pressure, the surface of the surroundi
water assumes the shape of an inverted ,'coneu during pumping of the well,
shown in profile in the figure. This cone is known as the corc of depression, t
cross section of the cone at the water surface is called the circle of irtJluence,
the distance through which the water surface is lowered at the well is
the dra7.L)clo7,Jr. The discharge corresponding to a drawdown of 1 m is cal
specific capacity of the well.
FUndamentals of
Gt:otechnical Engineering
Chapter O:
- FlowoofYlte.f gS
ARTESSIAN UTELL
',8.2
R2
iR,
i
i<--)i
i
i
.;
;&*
1 ..l
3.8.I
GRAVITY WELL
Rz'
R,
Bottom of well
Figure 03.1 1 - Artesian well
Drawdown
Cone of
0epressron
<
1',
.tt
2xkt(h7 -h1)
ln(R2
l' /
u
/
R, )
Eq.3.18
I
Iv
lrr.'rc:
lb, Rr, Rz are in meters
= coefficient of permeability in
Itt.
(
m/hr
Q = discharge in m3/hr.
Figure 03.10
^
-
Gravity well
* luz\
In(R2 / R1 )
I.,
TWO-DIMENSIONAL FLOW OF WATER THROUGH SOILS
f ,q.
r rLow Ners
nk(h"2
Eq 3.1V
-1.----
/
Firt'prage losses through the ground or through ealdh darns and levees and the
rrl,rtod flow pattern and rate of energy loss, or y'l'ssipation of hydrostatic head,
grap-fcal technique known as flow net.
nr(' lrequently estimated by means of a
I'hrw net is a graphical representation of a flow field that satisfies Laplace's
e'lr,rtion and comprises a family of flow lines and equipotential lines.
q6
Chapter 03
-
Flow of Water
Fundamentals
Geotechnical Engi
through Soil
Chapter 03
Fundamentals of
6cotechnical Engineering
-
Flow of
Water oa
soil ' '
through
where:
k = coefficient of permeability
H = head
N/= number of flow channels = Number of flow lines minus one
Na =
Na =
N.
=*
Nd
s*idP
Figure 03.12
-
Flow nets
A flow net must meet the following criteria:
The boundary conditions must be satisfied.
Flow lines must intersect equipotential lines at right angles.
3. 'Ihe area between flow lines and equipotential lines must be curvili
squares. A curvilinear square has the properry that an inscribed
can drawn to touch each side of the square and continuous bisecti
results, in the limit, in a point,
4. The quantity of flow through each flow channel id constant.
5. The head loss between each consecutive equipotential line is constant.
5. A flow line canrrot intersect another flow line.
7. An equipotentiaI line cannot intcrsect another equipotential line.
1,
2.
Flout line is the path followed by a particle
saturated soil mass.
of water as it moves t
Equipotential line is a line connecting points of equal potential energy.
The flow of water through isotropic soil is:
Nr
q=kH
'
,
N.,
number of equipotential (pressure) drops
Number of equipotential lines minus one
is called the shape factor
lf the soil is anisotropic:
N,
r
q='H
,
"lk"k.
V ^
NI
L\d
Eq. 3.20
g8
-
Chapter 03
Flow of Water
Fundamentals of
through Soil
$cotechnical Engineering
ILLUSTRATIVE PROBLEMS
(
- ,,,)'
rr= l,r)
k,
ratio
1.1
0.9
Void
through Soil
1'* e,
*A-
n=5.1
b)
Value of G:
The hydraulic conductivity for normally consolidated clay is given by
following equation:
on
[k =
f3-:-
L+e
]
^
0.302xtOr=Crfr11
1.1s
K
en
= L] _
1+c
is the void ratio, Cr and n are constants to be dctermi
experimentally.
b)
c)
c)
Value of k when
Determine the value of n.
Determine the value of C:.
Estimate the lrydraulic conductivity of the clay at a void ratio of 0.2s.
a)
1.+e
€7"
C,
'1.+e1
G- *_!_
" l+c2
kl
k.L
. cr" -
= 7*rl,o-n
"2
!75=t)=
solve this !
given:'(Refer to Figure Figure 03,13)
Length of specimery L = L0 inches
Diameter of specimen = 2.5 inches
Head difference, lz = 22 inches
Vrrlues are
oll
kt _
*n,
= 0.75
lrrrr,r constant head laboratory permeability test on a fine sand, the following
Value of n:
-
ase
e
k= A.5'14- rO"';#;::
paoBLEM 03.2
SOLUTION
1
Cs=0.39x10-7
k=
a)
99
(1"2222)'= 2.781579
n log (7.2222) = log 2.787579
k (cm/sec)
0302x707
0.12 x'L1r
Flow of Water
0.302x t0-7 _ft.r)r t+o.o
ni;nj- - [o.e] 1.1i1
PROBLEM 03. I
For a normally consolidated clay soil, the following values are given:
where e
*
Chapter 03
1- '''
-.'
/
Water collected in 2 min = 0.044 in3
,/
llrc void ratio of the specimen is 0.34.
/
rr) Determine the conductivity, k, of the soil in in/m/n.
lr) Determine the discharge velocity through the soifl in in/min.
, ) Determine the seepage velocity in in/min.
I
\\
r
oo il:ffil?3,, t'"* orwater
Fundamentals
Geotechnical Enqi
Chapter 03
f undamentals of
Gcotechnical En gineering
Seepage velocity, u, =
-
Flow of Water
through Soil
rot
a
n
n=e/(1+e)
n = 03a/ (, + 0.34)
n = 0.2537
seeoase velocitv.
u.-
o'448x10-z
0.2537
Seepage velocity, a, = 1.765x 10-2 iny'min
PROBLEM 03.3
A soil sample 10 cm in diameter is placed in a tube 1 m long. A constant
supply of water is allowed to flow into one end of the soil at A and the outflow
.rt B is collected by a beaker, as shown in Figure 03,14. The average amount of
water collected is 1 cc for every 1.0 seconds. The tube is inclined as shown,
a) Determine the average velocity of flow through the soil in cm/s,
b) Determine the seepage velocity (velocity through the void spaces) in
Figure 03.13
cmf
s.
c) Determine the coefficient of permeability
of the soil in cm/s.
solve this !
SOLUTION
Discharge, Q = 0.022 in3/min
lQ= kiA)
0.022=r2+rz.st,
10 +'
k = 0.2037 x 10-2 iry'min
Dischargevelocity,
u=ki = I
Disclrarge velocity,,
{).022
= r,,;€
J)
4\1
Discharge velocity, u = 0.448 x
10-2
iry'min
Figure 03.14
t
az
,.,l:ffi13i,;''"*
orwater
Fundamentals
GeotechnicalE
lundamentals of
Chapter 03
Ceotechnical Engineering
-
Flow of Water
through Soil
I03
SOLUTION
a)
w.s. El. 140 m
Average velocity
Q=Volume/time=1/10
Q = 0,1 cc/s
Averagevelocitv,
,=
Average velocity,
,=
Q
A
0'1
-
+(10).
Average velocity, a = A,$LLZZ cm/s
b)
Seepage velocity:
Seepage velocity,
U
u..
n
Figure 03.15
e
1,+e
0.6
n= ___
SOLUTION
+ 4.6
tt = 0.375
"I
Seepage velocity,
Q= kiA
zr..
=
0.375
Seepage velocity, rr. = 0.00339 cm/s
c)
'I
Coefficient of permeability:
IQ=kiA)
,_ h _. (2-0.8)
i*
k = 3.5
0.au27
m/ day
i=h/L=20/110
A = 1.5(350) * 525 m2
Q = 3,5(20/110)(52s)
Q = 334.t m3/day x 1000 Lit/ms
Q=
232
Litlrnin
0.'l = /r (1.2) + (10),
k = 0.001061. = 10.G1x 10a cmls
day/24 hrs x t hr/60 min
solve this !
L1,
1,.2
xl
PROBLEM 03.5
A falling-head permeability test was run on a soil sample 9.6 cm in diameter
cm. The coefficient of
1r'rmeability of the soil was found to be 5 x 10-6 cm/s. The diameter of the
arrtl 10 cm long. The head at the start of the test was 90
el,rrrd pipe was 1 cm.
PROBTEM 03.4 (CE MAY t9c)91
A sand layer having the cross-section area as shown in Figure 03.15 has been
determined to exist for a 350-meter length of the levee. The coefficient of,
permeability of the sand layer is 3.5 m/day. Determine the flow of water into
the ditch in Lit/min?
,r) Deiermine the flow at the start of the test, in cm3/hr.
f,) Determine how much head was lost during the first 30 min
r ) Determine the flow after 30 minutes, in cm3/hr.
t
Chapter 03
a4 :,.}ffi133,;''"*orwater
GeotechnicalE
alEngineering
-
Flow of Water
through Soil
r05
hz= 82.84 cm
Lost of head = 90 - 82.84
Lost of head = 7.76 cm
Using the formula:
k=A(t2o!hf&-'l
- t1)
lnz )
5xL0-6=
f, (e.6)2 (1800
t.'29
h2
99
Figure 03.16
h2
-
=
.90
ln_
+ (1)2 (10)
*
0)
b2
o11zs
r0.0,,,
h2= 82.84 cm
Lostofhead=h-hz
SOLUTION
Part
Lost of head = 7.1.6 cm
a:
h=90cm
Q = kia * S x 10-6(90/
10)
*
(9.6),
Q = 0.003257 cm3/sec
Q = 11.72 cm3/hr
Part
Part
c:
Head,h = 90 -7.16 =
hAl
a = 5 x 10-6(82.84/
lQ =
b:
|
Jtt, a
A -X
A,=
i()2=n/q
Q = k iA = 5 x 1,06(h/1"0) t e.6),
Q= i1o.ooo04608h)
/rr = 90 cm
I:
n, (n
oo82s44=
ln
(h2) =
@
/ 4)(0.00004608h)
4.41686
(9.6),
LEM 03.6
A l,t'rmeable soil is underlain by an impervious layer, as shown in Figure
1 17. For the permeable layer, k = 0.0048 cm/sec. H= 3 m and cr = 5o.
in cubic meter.
/ a)dh
frlr,
l]'=
L
J/tr
*
rr) Calculate the hydraulic gradient?
l,) Calculate for the flow of water per meter width in m3/hr.
r ) Calculate the total amount of water percolated per day per meter width,
?
t=30min=1800s
1800 =
10)
Q = 0.002998 cm3/sec
Q = 10.79 cm3/hr
,
1t', A, dlr
r- -
hz=
82.84
,"
(eo)
-
tn (/r,)
iir
t
06
f^?3lH ?i,;
&i
'''*
of warer
GeotechnicalEn
of
Chapter 03
ical'Engineering
Ground surface
ffi
-
Flow of Water
through Soil
t07
q = 0.000048(0,087 16) (2,99)
q = 0.0000125L m3/ s per meter
q = 0.045 m3/hr per meter
H.
ffi
Hi
EI
E,
F1
c)
h
iii
ii
Figure 03,17
Volume percolated per day:
Volume = 4f
Volume =0.045x24
Volume = 1.08 m3
03.7
tcr flows through the permeable layer as shown in Figure 03.18. Given H =
ft, h = 4.6 ft, L = 120 fl, s. = '1.4o, and, k = 0.0016 ft/,sec. Consider 1 ft width
prpendicular to the figure,
SOLUTION
rr) Calculate the hydraulic gradient in percent?
lr) Calculate for the flow of water per ft width in fl3 /hr.
e) Calculate the total amsunt of water percolated per day per ft width, in
cubic ft.
a)
Hydraulic gradient:
I = Head loss/Length
i=Ltanu/ (L/cosa)
l=sincr
i = sin
b)
5o = 0.087L6
Flow per meter width:
fq =
kiAl
A=3cosg,xL
A=3cos50
A = 2.99 m2
k = 0.0048 cmlsec
k = 0.000048 m/sec
Figure 03.'t8
rt
tos f#:ffi ?3,;''o*orwiter
lii
tr
SOLUTION
m,
&.
a)
fi
chaprer 03 _
of
Fun
Geotechnical
Frowooj*tr:l
nical Engineering
tr) If i = 0.70, determine the total flow 4'in cme/sec.
Hydraulic gradient:
.h
,=
,L
il,
iit
$
.
,=
'
-cos 0
4.6
120
-cos L4o
i = 0,03719
i*
b)
3,7N/o
Flow per unit width
[q
* kiAj
A=Hcoscrxl
A=3.5cos14ox1
A = 3.396 tt2
q = 0.0016(0.0 37'tg) (3.396)
q = 2.02078 x 10+ ft:/sec
q = Q.tuft ftt/hr
c)
PROBLEM O3.8 {CE MAY 2OO3l
Consider the stratified soil deposit shown in Figure 03.19.
where;
1. hydraulic gradient is equal; Lq = ir = iz= ... = in
2. quantity of flow in each layer is added to make the total flow q
Note:.Darcy's equation a = ki and, 4 = aA; kpy ks2, ks,... kar are the
of permeability of the individual layers in the horizontal direction. Consider
unit width.
a) Derive the expression for the equivalent coefficient of permeability in
horizontal direction.
b) Iftherearefour layers,3 mthick each, and km=2x1A3,kuz= L x 106,
- 2 x 10-a, kHe = 1 x 10-3 in cm/sec, determine the equivalent coeffici
of permeability in the horizontal direction.
km
SOLUTION
Pafi
Volume percolated per day:
Volume = 0.72T5(24)
Volume = 17.46its
solve this !
'
Figure 03.19
a:
q = azt = k i a
l unit width perpendicular to the figure:
q= (1xFI)(k,*)i = (1xH)(kut)l+ (1x Hz)(kni)i +(l xHsxkrd,
+ ,,, + (1, H,)(kir.),
Considering
.
k,* =
1
i
(km Hr + kuz Hz + kffi Hs+ ... +.kan H^)
v.
L
lOg
I ro
Chapter 03
-
Flow of Water
Fundamentals
Geotechnical Engineeri
through Soil
Part
I trndamentals of
(icotechr:ical Engineering
Chapter 03
&
Hr= Hz= Hs=
He = 3 m
km=2 x 10-3cm/sec
knz=1x 1O-scm/sec
km= 2 x1,0a cm/sec
kns=7x10:scm/sec
1. .. -.
(kmHr+kuzHz+kuoHs + ... + knn
.
k,* =
t="1-
^req
i
Hn)
+
x
(300)
?7[66[2 x 10-s(300) 1x 10-5(300)+ 2 104
(3oo)l
o.ooo8o2u;'/;10-3
k,q = 8.025 x 104 cm/s
k,uo =
Part III:
i = 0.70
q=k'*iA
Considering 1 cm width:
H=3m x4=12m
Figure 03.20
H = 1200 cm
A=Hx1=(1200X1)
A=L2AA cm2
q = (8.025 x 104 )(0.70)(1,200)
q = 0.674 cm3/s Per cm width
Equivalent
k:
UIJ
= r't
l7l
Krq
@arallel flow)
,K
500_ 159 *
solve this !
PROBLEM 03.9
Figure 03.20 shows layers of soil in a tube that is 100 mm x 100 mm in c
section. Water is supplied to maintain a constant head difference of 400
across the sample. The hydraulic conductivities of the soils in the direction
flow through them are as follows:
Soil
A
B
C
(cm/sec)
L x 10-2
3 x 10'3
4.9 x10a
k
Porosity, tt
k,,
lc"o
L0-2
150 n_:00_
10-3 4.9 xL0'a
3x
= 0.001057 cmf sec
k", = 1.057x10^-3
0.1057 m/secm/s
Flow rate:
lQ= kiAj
25o/.
i = h/r-
g2%
l=400/500=0.80
22%
Q = 0.001052(0.80)(10 x 10)
g = 9.63456 srnt/g
a) Calculate the equivalent k in m/sec.
D) Calculate the rate of water supply in cm3/hr.
c) Calculate the seepage velocity through soil C in m/sec.
e = 904.4 cm3/hr
-
Flow of Water
through Soil
llt
tl2
Chapter 03
-
Flow of Water
through Soil
c)
Funclamen
Geotechnical Enqi
Seepage velocity through soil C;
fa, = o^u.f nf
l)ave = Q/ A
Chapter 03
-
FIow of VTater
through Soil
@
!
(,tt
u,u"=0.08456/(10x10)
uave = 0.0008456 cm/sec
u,u" = 0.08456 m/sec
vs
lrundamentals of
(lcotechnical Engineering
t-
t,avel tL
u"= 0.08456/0.22
u, = 0.384 m/s
PROBLEM 03.
solve this !
IO
Given the stratified soil shown in Figure 03.21. The properties of each soil
as follows:
Coefficient of permeability:
kn
= 6.25 cmf hr, ks = 8.15 crtf hr, ko = 3.60
Thickness:
Figure 03.2'l
cm/hr
":
SOLUTION
H =1..20 m, H: = 0.30 m
FIa = 0.50
m, Hs = 0.40 m
0.31
v
Length:
Lr = 0.8 m, L2= Q.7
Lt = L.5 m, Lo = 0.9 ^
m,
0.s
o
Determine the total flow per meter
b) Determine the equivalent coefficient of permeability
I
-+
o.[
s: i'ii!,,k,.
::' :':-:
+
' o.Bm ' o.7m '
Head,ft=1.8m
a)
t
H"=0.8+0.7+1.5+0.9
Ho=3.9m
-. h.
t,= 7l
L=Lr+Lz+Ls
L=0.8+0.7+1.5+0.9
L=3.9m
.
l=
1.8
-3.9
i = 4.461.5
l.5m
....
v
,....
' o.9m
I
lt3
lt4
Chapter 03
*
Flow of Water
Fundamen
Geoteclrnical Engineer
through Soil
lIo _. H
k^^ k
.H^
b=u
"ee
L@/k)
a)
=
b)
h"ri
a1l
I t5
a2l
.
k'
"z eq
3.9
o.8o o.7o 1.5 0.9
r-25- r75*-eN- to
k2.q= 5.272 cm/hr
k2.r= 0.A5272m/hr
= 0.5(1) = 0.5 m2
42
A canal is cut into a soil with a stratigraphy shou,n in Figure 03.22. Assume
lkrw takes place laterally and vertically through the sides of the canal and
vr'r'tically below the canal. The values of k = k, = k. in each layer are given.
rr) What is the equivalent permeability in the horizontal direction through
the sides of the canal, in cm/day.
/r) What is the equivalent permeability in the vertical directions through the
sides of the canal, in cm/day.
t) Determine the equivalent permeability in the vertical directions below
the bottom of the canal, in cm/day.
1.0 m
1.5 m
qs:
-
3.9
, 0.70 1.5 0.9
^3
6.25i rn r sl5 r57
k3"q =,5.7035 cm/hr
eq
0.80
k3"q = 0.057035
m/hr
ar = 0.4(1) = 0.4 m2
qt = 0.057 035(0.4615X0.4)
ry= 0.010529 m3lhr
/n
J
iiirirlii:;ii'llrrii,!ei
2.0 m
1.2 m
lqt = kt "o i a3l
solve this !
PROBLEM 03. I I
= 4.0527 2(0.4615X0.5)
qz= 0.01217 m3/hr
Solving for
k"r(0.4615)(7.2 x 1)
36
(0.4615X0.3)
= 0.00648 m3/hr
Solving for qz:
fqz= kz"ri
*
0.9
g * 0.04681
t.
through Soil
3.9
L5 ,
o.8o o.7o
(rzs- L75 - 4s -
t
= 4,68:l cm/hr
k1"o = 0.04681" m/hr
a = 0.3 x.1 = 0.3 m2
Qz
Flow of Water
k"r= 0.05269 m/hr
k"r= 5.269 cm/hr
4r:
k1 uq
q1
-
Equivalent coefficient of permeability for all layers:
lq = k"ri al
4.029179
,
k- e9
"
Chapter 03
Total flow, q = qi + q2+ ry
Total flow, q = 0.00648 + 0.01217 + 0.010529
Total flow, q = 0.029179 m3/hr
Total flow:
Solving for
[q
Fundamentals of
Er,otechnical Engineering
3.0 m
ir:
.l.lr;
'i' :,r
,';'
T- :Ii:ifr{li,l:tiit,.rl",,'.il:,ii
a,
/t\
0
t
I
l':i:'
r:,
:'i :,rairi:'
k=0.3x10{cm/s
k=0.8x10'rcm/s
Figure 03.22
t
t
6
f,.}:ifi ?i,; '''*
or water
F
GeotechnicalE
mentals of
echnical Engineering
SOLUTION
k,
Chapter 03
1.0 m
k,@q)
Z1 Zt
i+3+...+r
through Soil
lt7
Z"
kr,,
5./
1.5
2"1,0-6
2.0 m
Flow of Water
Ho
("q)
krt krz
1.5 m
-
1.2
0.3 x
10-a
3
0.8 x 10-3
= 7.18 x 10-6 cm/ s
k,Gqt= A.620352 cm/day
k,@q)
1.2 m
3.0 m
FROBLEM 03. r 2 (NOVEMBER 2OO3)
solve this !
A lt'st is set-up as shown in Figure 03.23. A c1rlindrical motrd 4" in diameter is
frllt'd with silt to height fi = 0.2 ft, whose coefficient of permeability k1 = 3.6 x
a)
Equivalent permeability in horizontal direction though the
of the canal (H, = 3 m)
.7t.
k,(.q)
=
k,r"d =
*11o (rrkr,
1
i
tttZ.a *
+ z2k*2 +
l{}rfrlmin.
"' + znk*r)
10-s) + 1.5(5.2
x L0-6) + 0.5(2 x 10+)l
k,r"ql = 10.6 x 10-6 cm/s
k* ("q) = 0.91584 cm/day
b)
Equivalent permeability in vertical direction though the
the canal (H, = 3 m)
t--Ho: -:-------:------
nz (eg.)
aat
La
L-a..-a_
Ln
kzt krz
,3
k.(uq)
=
krn
---l-=f5
t--l_
2.3 x
10-s
5.2 x
-*--6:5*
10-5
Zx70-6
k.("q) = 5.16 x L0-5 cm/b
k. ("q) = 0.445824 cm/day
c)
Equivalent permeability in vertical direction though the
igure 03.23
of the canal
H"=
1.5 + 1..2+ 3 = 5.7 m
A second coaxial mold is placed on top of, the first mold whose inside
rlr,rrneter is d = L.5" and whose height is Hz = 0.30 ft. Its thickness is negligibte.
llr.'insi\e of this second mold is filted with the same silt, but the annulairing
\
\ \,
I l8
Chaoter 03
rhrouqh
-
Flow of Water
Chapter 03
F
GeotechnicalEn
Soit
outside the small tube and outer tube is filled with sand whose coeffici
permeability is k2 = 2,7 x 10 3 ftlmin .
The test set-up is a permeameter of constant'head. Water is placed in the
and maintained at a level h = 1.25 ft. above the level of the outlet. It ma
considered that,the system consists of a fictitious soil of thickness.H = Hr
and coefficient of permeabitity kf.
hnical Engineering
i=h/L
i=1".25/0.5=2.5
'
A= td2
.1= to,s/12)2
.
A=0.012271t2
Q, = 3.6 x 1 0a (2.5)(0. 01.227)
Q, = L.1045 * 16-s f1slmin
The following general equations may be useful:
kuuq= H / | (ru/ k"i + (Hz/ k,z) + ... + (H,/ k,,)l
kt,
(1 / LI)lknt Hr i knz Hz + ... + kn H,l
Path b:
"q=
Q6=
a) What is the total flow of water in ftr/min?
b) What is the equivalent coefficient of permeability, k1, in ft/min? '
c) What is the total amount of water that percolated after 55 minutes.
k.eiA
t.
n"q -
0'5
-Ti-----12-I-
2.7 x7O-3 3.6 x 10-a
k"o= 7 '5 x 10-a ft/min
.l
SOLUTION
i=h/L
i=
1..25/ 0.5 = 2.5
A= t(D2 _ d2)
.
A=
tU#Y-(f
)?l
A = 0"075 ft2
Qb= (7.s x 10a)(2.5)(0.075)
eu_1.40625 x 10-+ ft3/min
Sand,k=2.7x10'3ft/min
Q= Q,+
Qo
Q = 1.1045 x 10-s + 1.40625 x10'a
Q = 1,.5167 * 16-+ f1lmin
Silt,k=3.6x10''ft/min
Part
b:
e=k!il._
i=h/L
i=1..25/0.5=2.5
A= tD'
A=
t(+)'
A = 0.08727 f0
Pafta:
Q
Path
a:
Qo=kiA
k = 3.6 x 104
r
1,.51.67
x 104
ft3lmin
x L0-a = kf (2.5)(0.08727)
6.9518 x 104 fVmin
1..51.67
ft/min
4=
.
-
Flow of Water
through Soil
llq
t
zo
,t,lffil?3,,
part
t'o*
or water
Chapter 03
Geotechnical
c:
V = Q x t = 1.5167 x 10-4 ft3/min x 55 min
V = 0.00834185 ft3 x (100 cml3.28 ft)r
V = 236.4 cml
PROBLEM
03.I3
Refer to Figure 03.24. Given that Hr = 300 mm, Hz = 500 mm, and h= 600
and ttrat at z = 200 mm, /.t = 500 mm. [t is required to determine h at z -
, The following general equations may be useful:
k,uq= H / | (fu/k,) + (H2/ k,z) + ... + (H,/k*))
kn
= (1' / H)[ kn Ht + knz Hz + ... + knn H, I
"q
a)
b)
c)
What is the ratio k1/ k2?
What is the value of h atz = 600 mm?
If k1 = 5 x 10-6 cm/s, what is the equivalent k of soils 1 and
2?
Part a:
For the entire system of soil:
lq=k,qiAl
rl, H =vh't
Krq
0.9
K
0.3 * !g
kq = kt k2
0.8
_
kq
t_ _
''
0.3k2 +0.5kr
k$z
0.8k1k2
0.3k2 + 0.5k1
,H
Figure 03.24
.
L
0.6
0.8
= 0,75
-
Flow of Water
through soit
l2t
122
Fundamentals of
Chaoter 03 - Flow of Water
through soit
a
'
=
* 0.8k:l2
0.3k1_ + 0.5k1
-
Flow of Water
through Soil
123
0.7143k{H/A.q A = 0.5 kl A
t(
t0.75)A
H = 0.42
o.6k,k"
'
Chapter 03
6cotechnical Engineering
0.6-h*0.42
h=0.18m
h=180mm
0'3kz + 0'5kt
At point A:
Head, H = 0.1
Length of soil, L = 0.2
Part c:
kr=5x10-6cm/s
1rr= la: /1.8
i=a/z
ir = 0.1/0.2
kz= 2.778 x
.
D=E
"el
= 0.5
fi = kt (0.5)A
Qt = 0.5ktA
11
t-
Kcq
But 4r = q (continuity equation)
o.Sk,A
"
o'6k$2
= 0.3k2 + 0.5k1
0.3k2+0.5kr=L.2k2
0'5 kt = g'9 L'
kr = 1.8 kz
kr/ kz = '1"8
B:
Head, H =
0.3k2 + 0.5k,
0.8(5 x 10*6 )Q.778 x
to 6 )
0.3(2178 xt0-5 ) + 0.5(5 x 10-6 )
k", = 3'33 x 10-6 cm/s
A
solve this !
PROBLEM 03.14 IMAY ZOO4I
A confined aquifer underlies an unconfined aquifer as shown in Figure 03.25.
( liven the following: Dt = 59 m, Dz = 41 m, Hr = 45 m, Hz = 33 m, Kr = 35 m/ day,
?
r)
k'"0:
_
0.4667
k2
k"s = 1'2857k2
k'"s = 0'7143kt
qz=k',qiL
q= k"qi L
in
horizontal
Calculate the flow of water from one stream to another per meter width.
llte. kl kz
0.6_ 0.3 +0.3
k'", 1,.8k2 k2
0,6
permeability
l,) Calculate the hydraulic gradient,
q.6=0.3+0.3
ftiei
of
dircction.
L=0.3+0.3
L=0.6
Solve for
cm/s.
l:=27mfday,L=2km.
ri) Calculate the equivalent coefficient
Part b:
At point
-
10-6
0.9k, k,
Figure 03.25
t24
,tl3i,,lil?3,;
Flow of Water
Fundamentals
Geotechnical En gineer:i
Frrrrlamentals of
Chapter 03
(*,otechnical Engineering
SOLUTION
El
38m
w.s.
-,ffk
El 28m
Part
a:
K,, H = ZKlt
K*(33 + 4s) = 35(4s) + 27(33)
K,q = 37.6'1.5 rr{d.ay
Part
lmpervious
&:
- n/l=
Figure 03,26
78/2000 = 0.00e
Part c:
IOLUTION
Q = K*iA
A=ho,ux1=
\tr (1)=56*z
Q = 31.615(0.009X50) = 14.227 m3/day
PROBLEM 03. t s (NoVEMBER 2OOt)
The section of a cofferdam is as shorvn in Figure 03.26. If the coefficient
permeability of the soil is k = 5 x 10-3 m/s, determine the seepage into
ditches per meter length of the cofferdam.
El 0
Impervious
N,
ta=kH
,
N,.
H=38-18
H=20m
I N7= number of flow channels
I
-
Flow of Vfater
throuqh Soil
l2s
126
Chaoter 03
through
-
Flow of Water
Fun
GeotechnicalE
Soit
rrlamentals of
cal Engineering
chapter 03 _
Frow#sYl:; 127
N.r= 3
Na = number of pressure drops
Na=7
'?
q = (5
q
"
10
3)(20):
/
* 0.043 m3/s per rneter
Since there are two identical sides:
4=0.043x2
q = 0.086 m3/s per meter
03.I6
The section of a sheet pile is shown in Figure 03.27. The coefficient
PROBLEM
permeability of the soil is k = 4.2 x 10-3 m/s. Determine the seepage into
downstream side per meter length of the sheet pile.
N.
1q=kH
*
/vd
I
k=4.2x103m/s
Nr=
+
Na=8
H=7 -2=5m
q = 4.2x
103(5)f
4 = 0.0105 m3/s Per meter
4 =fi.S as
PROBLEM 03. I 7
masonry dam shown in Figure 03.28, k = 5 m/ day'
n) Determin" th" ru"puge flow per meter width of dam in liters per minute,.
&) Determine the uplift pressure at A and B in kPa'
cj Determine the upliff force per meter of dam. Assume that the uplift
presspre under the dam varies uniformly'
Irrrr the
Figure 03.27
I
\
\
\\
t
zs :,l3L1!3,; '''* or water
Fundamentals
Geotechnical Engi
Fundamentals of
Geotechnical En gineering
Chapter 03
t.
flow,q=kH:
,Nd
Seepage flow, q=
S1f
S11
9
Seepage flow, q - 40
18m
m3
f
day per meter
=40x$x$x1000
* 27.78 Litlmin
I
1m
{,'
I
.," I
b)
Pressure at A and B:
Pressurehead drop =
A
Pressurehead drop =
f;
Nd
Pressure head drop = 2 m per drop
Pressure head, ht= 18
Pressure head, hs = 18
Figure 03.28
SOLUTION
- 2(1) = 16 m
- 2(8) = 2 m
Pt=l*h=9.8L(1.6)
pa = 156.95kPa
30
ll
m
------ri
PB
pa
c)
1m
= nl, hs = 9.81(2)
= 19.62kPa
Uplift pressure per unit length of dam:
I
t-
t' >z---r'
!
\
'.
)./
'r
u= Pe!!.!_(30)x
u=
a)
Flow per unit width:
Number of pressure drops, Na = 9
Number of flow channels, N7 = 4
1
L56.99:79,62
U = 2,548.7
eg1
kN
Flow of Water
throuqh Soil
*+_'
Seepage
-
t29
130
Chaoter 03
- Flow of Water
throush Soil
Funda
GeotechnicalEn
Fundamentals of
err,otechnical Engineering
Chapter 04
-
Stresses
in
soil l3l
Chapter 04
Stresses in Soil
4.1 INTERGRANULAR
STRESS,
p6 (EFFECTIVE
STRESSI
lll('rgranular or effective stress is the stress resulting from particle-to-particlc
r
r)ntact of soil.
pr.=pr-p",
4.2 PORE WATER
PRESSURE,
p*
Eq.41
{NEUTRAL STRESSI
l\rre water pressure or neutral stress is the stress induced by water-pressLlres,
pru
= lru
hru
Eq.4.2
Nrrte: For soils above water table, p* = 0.
4,3 TOTAL
STRESS, PT
l'he sum of the effective and neutral stresses.
Pr=PttP,u
4.4
STRESS
IN SOIL WITHOUT SEEPAGE
(lonsider the soil layer shown in Figure 04.1.
Eq. 4.3
132
Chaoter 04
in soil
-
Fundamentals of
Geotechnical Engineering
Stresses
4.5
Surcharge, q (kPa)
STRESS
- Soil layers with surcharge
sti;T;fr
I SS
lN SATURATED SOIL wlTH SEEPAGE
4.5.I UPWARD
Figure 04.1
chapter 04 -
SEEPAGE
and without seepage
Total stress, pr i y* hn + q
Neutral stress, p, = 0
Effective stress, p6 =
Total stress, pr =
Effective stress,
O.r
Tsaa
pE =
p"r
- p*
Figure 04.2
h5 + y^ h1 + q
pr -
Hydraulic gradient, i = hlHz
q
At point A:
pr= l*Ht
Pw=f*Ht
i
At point
C:
Total shess, pr = lxtthx * lsatlh2+ y^h1 + Q
Neuhal stress, p, = fwho
At point
otPt=*Tuzhz+^lnhz+\^ht+
4
Pr= Pr'Pu = 0
B:
Pr = lsat z1 * Yp Ht
Effective stress, pE = pr - ?*
'
Soil with upward seepage
ht=ixs1=i(h/Hz)
p,
Pr*Yar.hs+Y*h1 +
-
Pr=Pr-P*=fuzt-\r,ht
P,-f*(4+Hr+hi
At point
C;
Pr = lut Hz + Y*.Ht
Pe=Pr'9,r=yoHz-I*h
P*=T,(Hz+fu+h)
The seepage force per unit volume of soil is:
F
= iY'u
Eq. 4.4
t34
Chapter 04
in Soil
-
Stresses
GeotechnicalEng
4.5,2 DOWN\T/ARD SEEPAGE
ntals of
technical Engineering
Chapter 04
-
Stresses
in
Soir 135
EFFECT OF CAPILLARY RISE TO SOIL STRESS
Eapillary rise in soil is demonstrated on the following figure. A sandy soil is
plrrccd in contact with water. After a certain period, water rises and the
Yariation of the degree of saturation with the height of the soil column caused
by capillary rise is approximately given in the figure.
:wT
1000/o
Degree of
saturation
(o/o)
Variation of degree of
saturation with heiqht
Figure 04.3
Hydraulic gradient, i = h/
hr= i x a= i(h/Hz)
- Soil with downward
Hz i i:
,
seepage
Figure 04.4
-
Capillary rise in soil
'l'ltc degree of saturation is about 100% up to a height ft1. Beyond the height &1,
water can occupy only the smaller voids, hence the degree bf saturation is less
tlran 100%.
At pointA:
Pr=l,Ht
'l'hc approximate height of capillary rise is given by Hazen as:
P*=I*Hr
Pt= Pr-P*=0
At point
.
hr=
,z:
C
,pl
B:
P'r=lxt4+lpHt
Pw=T,(4+Hr-hr)
pE*pr-P**lozr+l,kr
where D16 = effective grain size, e = void ratio, and C = a constant that varies
flom L0 to 50 mm2.
'l'he pore water pressu re, pw, ata point
At point
Eq. 4.5
C:
in the layer of soil fully saturated by
r',rpillary rise is:
pr = lsarH2 + Y* Ht
?.=|,(Hz+H1 -h)
P*=
-Y*h
Eq. a.6
PE=Pr-Pt=YuHz+Y*h
where h is the height of the point under consideration measured from the
glound wa{er table.
I
\
\
136
Chaoter 04
in soit
-
Fundam
Stresses
GeotechnicalE
a partial saturation is caused by capillary action, the pore water
can be approximated as:
If
P* = 'S Yrh
where
S is
the degree of saturation at the point under consideration.
Fundamentals of
Geotechnical Engineering
Chapter 04
-
Stresses
in
Soil 137
ILLUSTRATIVE PROBLEMS
PROBLEM O4.1 ICE NOVEMBER r998)
20. Nov 98: A clay layer 4m thick rests beneath a deposit of submerged sand 8
rn thick. The top of the sand is located 3 m below the surface of a lake. The
rraturated unit weight of sand is 25 kN/cu. m and of clay 20 kN/cu. m.
[)ctermine the total vertical pressure (P) at mid-height of the clay layer:
Consider the soil layer shown in Figure 04.5:
-r
i'
SOLUTION
+
I
h
Bm
Sand, 25 KN/m3
Figure 04.5
lpr = Ly
At pointA:
pr720(2) + 25(8)+ e.81(3)
Total stress, pr = yr hr + yzhz
Pore water stress, pu = -51 Y6 h3
At point
B:
Totalstress,
Pr=yht,+yzh
Pore water stress,
At point
'
r*h + y. h*)
p7 = 269.43kPa
,
pw = 0
C:
Tothl stress, pr = y ht + Y2h + Wh4
PROBLEM O4.2 (CENOVEMBER 2OOOI
A clay layer 25 feet thick is overlain with 50 feet thick of sand (G = 2.71). The
water table is 20 feet below the sand (ground) surface. The saturated unit
wcight of clay is 141 pcf. The sand below the water table has a unit weight of
l2li pcf. The sand above the water table has average moisture content of 20%.
Aftcr drying, the sand was found to have a dry unit weight of 92 pcf.
Dctcrmine the effective stress at the mid-height of the clay layer.
TOLUTION
For the sand above the water table
=
C
--f.
l+c
I
Chapter 04 _ Stresses
, Fundamentals of
, Geotechnical Enqineerinq
ih s.,it l3e
i PROBLEM O4.4 ICE MAY 2OO2)
I rhe soil shown in Figure 04.5 has a void ratio of 0.50 and G = 2.70. ht = 1 .5 m,
[ /rz=3m.
What is the effective unit weight of sand in kN/mr.
I a)
e)
What
is the effective stress at point A in kpa.
I c)
wnat is the critical hydrautic gradient rr:l (for quick condition).
"t
I
.
.
solve this !
t
;Yu*i11'ti,,"',;'Uffi
I
I
I
l'"'r,
I
I
II
I
II
I
I
I
Errec,ive uni, weigh,
:::::,
uni, weish,,
yr)
"'=,!fi'o,
b).,,":::::::]'-^r1.,!,'r?riJ;!,iL,+
Ptor^r
e81(15)
= 77.499 kPa
P, = 9.81(3 + 1.5)
p*=
44.1,45kPa
fp.tt =-p.,t,t.- pu,l
p-* = 77.499
(
,.
-
44.145
P*r=33'354kPa
i
-
Chaoter 04
138
in
Fundamentals
Geotechnical Engineeri
Stresses
Soit
s2=
'
e
-
62.+1
= 0.838
=
lYsana
lL
l+e
G
+GMC
--17-'{*
r+e
-
lsano
"-^-,
I
2.7L + 2.71.(0.2),Ur.n,
1+0.g3g
\--'-l
T*na = 110'4 Pcf
|a h)b"to, *ut", tuo," + (I' Yi' h)above watet table
yr = (141 - 62.4\(12.s) + (128 - 62'4X30) + 110'4(20)
Ye = 5158.5 Psf
Yr =
(I
PROBLEM O4.3 (CE NOVEMBER 2OOll
A !;round profile consists of '2 m of -silty sand underlain by 3 m of -clay:
grJund water table is 3 m below the ground t"t'llu.'-The sqnd has a
i,eight of 1.4 kN/m:. The clay has a unit weight of 16 kN/m3 above the w
tabl; and 20 kN/m? below the water table. Determine the total stress at
bottom of the clay laYer.
solufloN
,i
* |c h, + Y"h,
+M(2)
+
QQ(2) 16(1)
pT = |csat he1
pr=
pr = 84kPa
-
Chaoter 04
l40
in
Fundamentals
Geotechnical Engineeri
Stresses
Soit
c)
Fundamentals of
Geotechnical En gineering
G-1
-.
ll.= - ^l
Stresses
in
soit l4l
Vertical effective pressure at A before lowering of the water table:
a)
-
L+e
21'1.
L, -
-
SOLUTION
Critical hYdraulic gradient:
.
'
Chapter 04
lpo=Zyr,hl
pa = (1.20 - 62.4)(12.s) + (135
pa = 4,350 psf
1+0.5
i, = 1,133
-
After lowering of the water table:
b)
solve this !
PROBLEM 04.5 (CE MAY 2002l,
- 62.4)(50)
Ground surface
]-TI
tr_
zsrt
s0ft
to20%.
a) what is the vertical e{fective pressure at the midheight of the clay
T-
before lowering of the water table, in psf'
I
b) What is the veitical effective pressure_at the midheight of the clay la
after lowering of the water table, in psf'
4 \4r1i;;,i.;r3-"i*r1rr".t**
pr"rs,r." when there is nr: water in the
25ft
layer, in psf.
J
Water table
Ground surface
Solving for y* (20% saturated)
y,,= 116 + 0.2(135 y- = 119.8 pcf
50ft
116)
a = (120 - 62.4) (12.s) + (135
Pa = 5530 Psf
p
c)
I
T_
25ft
Vertical effective pressure at A when there is no water
in the sand layer:
pa = (120
-
Pa = 6,710
]Figqre 04.7
FROBLEM
- 62.4)(25) + 119.8(25)
62-.4)(12.5) +119,8(50)
Psf
0+.6 ICEMAY 2OO3l
Tlrc ground water li:vel in a thick, very fine sand deposit is located 2.0 m
bekrw the ground surface. Above the free ground water line, the sand is
142
Chaoter 04
in soit
-
Stresses
a) What is the total stress in kPa on a horizontal plane A located
b)
c)
mentals of
technical Engineering
F
GeotechnicalE
4.5
below the ground surface?
What is the pore water pressure in kPa'at this plane?
What is the effective vertical stress in kPa in plane A?
SOLUTION
x
d16)
ex
dn
C = 0.20 cm2
dio = 10 Fm
drc=10x10-6m
dn = 10 x
cm
10-a
e=0.4
,
0.20
0.4(10 x 10-4 )
Total stress in A:
pr=20.3(2.5)+20.3(2)
h=500cm
pr=91.35kPa
ft=5m
r
y'
r'
Pore water pressure in A:
p* =
'
9.8L(2.5)
: .
kPa
Ground surface
S=50%;e=0.4
Yr
c)
Effective stress in A:
lpr= pr+ p*)
' pr. = pr -- pu'
\,l
\$1
\i.
,st
pr*91.35-24,525
* 66.825 kPa
pr
solve this !
PROBLEM O4.7 ICE MAY 2O03l
A dense silt layer has the following properties: void ratio = 0.40, el
diameter dto = 1,0 pm, capillary constant C = 0.20 cm2. Free ground water
is 8.0 m below the $round surface,
soil 143
f
|,) Find the vertical effective stress in kpa at 5 m depth. Assume unit weight
of solids = 26.5 kN/m3 and that the soil aboue the capillary action rise
and ground surface is partialty saturated at5Oo/o.
r) Find the vertical effeciive stress at 10 m depth. Assume unit weight of
solids = 26.5 kN/mr and that the soil above the capillary action rise and
ground surface is partially saturated at5A%.
t--
p,, = 24.525
Stresses
in
'OLUTION
a) Capillary rise:
a)
-
Find the height of c4pillary rise the silt. Capillary.rise is give n as h c
=
(e
l
Chapter 04
Aii
*,."
*."f*
diskibution
= 25.1 kN/m:
Y$i *.26.5 kN/m3
144
ChaDter 04
-
Fundamentals
Geotechnical Engineeri
Stresses
in soit
b)
a)
b)
c)
Vertical effective stress at 5 m depth
Solving for yr and y.,,:
First we solve G:
[y,
Fundamentals of
Geotechnical Engineering
Chapter 04
-
Stresses
in
soit 145
Determine the density of the clay in kg/m3.
Determine the total vertical stress at the bottom of the clay layer, in kpa.
Determine the effective vertical stress at the bottom of the clay layer, in
kPa.
- y, G]
26.5=9.81xG
SOLUTION
G = 2.701
''
. = G'Sc
lYr
,*;Y"
Given:
MC=0.57
S=1(saturated)
G = 2.84
I
z'zotir(3?(o'a)
"=
y1= 20.33 kN/m'
[GMC =
.''sr
S e]
2.8aQ.57) = 1(e)
e = 1.61,88
.
[y,,=
C re
lyrut=.r_rV"l
G+Se
#y,,]
I+e
*'(''u'8t)
' = ''to1 + 1.6188 ,r.rr',
y* = 16.7026 kN/m: (saturated
.r,,,
.r.,,
"=
2'Zoi
l9'4 , 9.s1
1+0.4
kN/m3
21'73
=
Yunt
a)
Density, p = y/
b)
Total stress, pr = T,,, h
Total stress, pr = 16.7026(20)
Total stress, pr = 334 kPa
c)
Effective stress = pr - p,,
Effective stress = 334 - 9.81(20)
Effec_tive stress = 137.85 kPa
py= 21.73(2) + 20.33(3)
pr = 104.45 kPa
Pore water stress at A (within capillary rise):
P* = '9'8L(3)
p*= -29.43kPa
l\r=Pr+P*l
tO4.4S=p6+\-29.43)
pr = 133.88 kPa
Vertical effective stress at
p
r = (21.73
pe = 193.48
B:
- 9.81X2) + 21'73(5)
16,702.6 /9,8L
Density, p = 1702.6 kglm3
Vertical effective stress at A:
Total stress at A:
c)
g=
unit weight)
PROBLEM 04.9
t
solve this !
A borehole at a site reveals the soil profile shown in Figure 04.8. Assume G 2.7 for allsoil types.
a) What is the unit weight of the soil in layer 1 in kN/ms?
b) What is the effective stress at a depth of 2 m below the ground surface, in
20'33(3)
kPa
kPa?
c)
in kPa?
PROBLEM O4.8 {CE NOVEMBER 2003)
A 20-rn thick submerged saturated clay layer has Water content of
specific gravity of the iolid particles is 2'84.
What is the effective stress at a depth of 20.6 m below the ground surface,
57%.
146
Chaoter 04
-
Chapter 04
Fundamentals
Geotechnical Engir
Stresses
in soit
icalEngineering
Properties of each layer:
0
*
1
2.0
Layer 2
<*
3.0
Layer 3
Layer (1):
"
Verv.fine wet sand with silt
uy
=
5o1o,g
=
1C'tttC=Sel
4Oo/o
2,7(0.05) = Q.4s
= 0.3375
Fine sand saturated bY
capillary action
Fine sand, w
=
e
.
120lo
G+CMC
lf* = ---:--l*l
I+e
5,4
,*
' =
-
280/o
* 2.7(0.05)
+0.3375
20.7gkN/ms
=
y,n
Soft blue clay, w
2.7
1
)
tg.g'
Part
Layers (2) and (3):
IGMC=Sel
2.7(0.12) = 1(e)
= 0.324
e
.
LT'ut
=
G+e
^l*l
-:-L+e
+032a
2.7
Y-,=frffi(e.81)
Figure 04.8
y"u1= 22.405kN/ms
Layer (4):
SOLUTION
Elevation (m)
0
lG
MC=
- Verv fine wet sand with silt
*_1L7=
3.0
<*
e
Fine sand saturated by
capillary action
Finb sand, w
S el
2.7(0.281 = t1s1
5o/o,S=4Oo/o
2.0
.
[Y'ut
= 0.756
=
= 12%
G+e
L+e
-
,..,
=
t Dc'[
5.4
r,1
2.7
1,
Tsat
b)
Soft blue clay, w =
280/o
-
* 0.7s9
+ o.T56
=19.307
/
/
I
19.g11
kN/rf
Stress at 2 m depth:
Paar = 20.79(2)
Pwtat =.41.58 kPa
p* = -9.87(1)
p* = -9.81"
Stresses
in
Elevation (m)
Layer
-
a
Soil 147
148
Chaoter 04
in soit
-
Fundamentals of
Stresses
Geotechnical Engi
*
51.39 kPa
petr
=
(1g ,gO7
Pefr
+ 22.405(1) + 20.7e(2)
= 238.567 kPa
Pert
PROBLEM 04.IO
*
9.81) (20'6
- 5'4) + (22'405 - 9'81X5'4 -
3)
solve this !
Consider the upward flow of water through a layer of sand in a tank shr
Figure 0a.9. Foi the sand, the following properties are given: e = 0'40' G
a) . Calculate the effective stress at point A
b) Calculate the effective stress at point B
c) Calcutate the upward seepage force per unit volume of soil
2.67 +0.4 .^
#(9.81)
1+ 0.4
y*,=
'
!'*
,,h
lr =
= 21.51
.
kN/m:
--l
172
. 1.5
4
l=
2
i = 0.75
,h'h = ::
l\Ote:r=
a)
-zHz
=Constant
Effective stress at point A:
pr-
(21.51)(1) + 9.81(0.7)
pr = 28.177.kPa
Figure 04.9
-
Stresses
in
SOLUTION
= Ptotat'Pw
P"tt= 41.58 - (-9'81)
Peft
'
Chapter 04
Geotechnical Engineering
p* = 9.81. hA
' h,q,=1+0,7 +h'
soil l49
l50
Chaoter 04
in soit
-
Stresses
ll'
Funda
Geotechnical
tals of
Chapter 04
icalEngineering
_ nn_
h'=0,75(D=4.75
fua=1./ +Q./$
ha = 2'45 m
p. = 9.81.(2.45)
p, = 24.0345kPa
ps=
28.377
-
24.0345
pr = 4.343kPa
b)
Effective stress at point B:
py = 21.51(2) + 9.81(0.7)
pr = 49.887 kPa
pa=9.81,(2+0.7+1.5)
p. = 41..202kPa
[pt=Pr'P,]
pr=
Figure 04.10
49.887
-
41..202
pr = 8.685 kPa
c)
Seepage Force per
unit volume of soil = i 1*
i=h/L=1..s/2=0.7s
Seepage Force per unit volume of soil = (0.75)(9.81)
Seepage Force per unit volume of soil = 7.3575 kN/ma
PROBLEM 04. I I
solve this !
Consider the downward flow of water through a layer of sand in a
shown in Figure 04.10. For the sand, the following properties are give4:
0.48, G = 2.7
a) Determine
b) Determine
c) Determine
the saturated unit weight of sand in kN/m3
the effective stress at point A in kPa
the effective stress at point B in kPa
Stresses
in
-,
-=t=u./5
z
[pt=pr-p.)
-
soit
l5l
.
152
Chaoter 04
in soil
-
Stresses
Geotechnical
h1
h,
1,.2
Soir t 53
L1
2.s
pr = 58.581 -18.639
Pr = 39.942kPa
I = 0.48
LEM O4.t2
Saturated unit weight
-
=
llsat
G+e
:-l+c
r,",=
yr,1
0)
in
[w=pr-p*l
LI
,-
Chapter 04 'Stresses
Engineering
Y,j
ffi{e.ar)
= 21..078 kN/mr
solve this !
soil profile consists of a clay layer underlain by a sand layers as shown
in
lure 38. A tube is inserted into the bottom sand layer ani the water level
rs to 1.2 m above the ground surface.
n) Determine the effective stress at point A
lr) Determine the effective stress at point B.
r') Determine the effective stress at point C.
Effective stress at A:
lhz= i Lzl
hz = 0.48(0,8)
hz = 0.384 m
0.6 + 0.8
lz=
-
Y*t'
18.5 kN/m3
0.384
lz = 1.016 m
Y*t = 19 kN/m3
= 21,078(0.8) + 9.81(0.6)
pr = 22.748kPa
p7
lp, = l* yzl
|ot =
p* = 9.81,(1,.0L6)
p* = 9.967 kPa
lft=
pr - p,1
pe=22.748-9.967
pr.
Figure 04.11
= 12.78kPa
Effective stress at point
B:
g=0.5+2.5-1'.2
lt
=1.9 m
*
21.078(2.5) +' 9.81 (0.6)
pr = 58.581 kPa
p y'
[p* = y* yzl
p* = 9.81'(1.9)
p. = 18.639 kPa
TION
Hydraulic gradient (based at point
U
li = lLt
'
L'
.
i=0.4
2.2
C)
17 kN/m3
154 ffffi"
o4 * stresses
Chapter 04
F
GeotechnicalE
Fn
Gcotechnical
gineering
-
Stresses
in
Soil 155
Effective stress, pE = 33.931 kPa
H
c)
.2.2
?.*
-
fet =
Effective stress, pE = pr - pta
Effective stress, p6 = 116 -75.537
Effective stress, p6 = 40.463 kPa
= 17 kN/m3
PROBLEM 04. I 3
a)
At pointA:
0.4=Ho
'li=Hnl
LA',
2
Ha=0.8m
Total stress, pr = 1.9(2) + 18.5(1)
pr - 56.5 kPa
C:
Pore water stress, pr, = 9.81(5.5 + 2.2)
Pore water stress, p,, = 75.537 kPa
19 kN/m3
1.5 m
|*t
At point
Total stress, p r = 17(3.5) + 19(2)+ 18.5(1)
Total stress, pr ='1"16 kPa
18.5 kN/m3
solve this !
A cut is made in a stifl saturated clay that is underlain by a layer of sand as
rltown in Figure 04.12.
rr) Calculate the total stress at point A as a function of /r.
&) Calculate the neutral stress at point A.
c) Calculate the height of water h, in the cut so that the stability of the
saturated clay is not lost.
To_tal stress,
Pore water stress, pa = 9.81(2 + 0.8)
Pore water stress, p@ = 27.468 kPa
i
Effective stress, p5 = pr - p,u
Effective stress, p6 = 56.5 - 27.468
Effective stress, ps = 29.032kPa
b)
At point
ti=I!-t
' LB
B:
0.4=
Hs
2+7.5
Ha= 1.4m
Total stress, pr = 17(1.s) + 19(2) + 18.5(1)
Total stress, pr = 82kPa
Pbre water stress, p, = 9.81(3.5 + 1.4)
Pore water stress, pe = 48.069 kPa
Effective stress, p6 = pr - p*
Effective stress, ps = 82 - 48.069
Saturated clay
it*t = 19 kN/m3
Figure 04.12
IOLUTION
a)
Total stress at point A:
Pr=)9(7-5)*s.erH
pr /38 + 9.8'th
I
I
I
156
Chaoter 04
-
Chapter 04
Stresses
-
Stresses
in soit
b)
in
Neutral stress at A:
P*=9.81(4.5)
p*= 44.145kPa
c)
For lods of stability, pr
*
0
pr= pr - Pw
0=38+981h-44.1,45
h = 0.626 m
PROBLEM
solve this !
04.I4
For the system shown in Figure 04.13, L = 1.2 m, h = 2'3 m, and y = 1,9'
What is the effective stress at point A where z - 0.7 m. '
_t-
f-
l
llVn
Y*t = 19.5 kN/m3
v
I
Porous stone
I
l-
Energy grad'ient, i =
T
2.3
1,2
'1
h =h,
Lz
hl
0.7
h=1.3417 m
he=0.7+1..9-1.3417
ha = 1..2583 m
=
19.5 kN/m3
Porous stone
Figure 04.13
Total stress at A, pr = -tg.5(0.7) + 9.S1(1.9)
Total stress at A,.pr = 32.289 kPa
p, = f*ht
Pore water stress,
Pore water stress,
pro
Por! water
p. ='l2.3MkPa
stress,
= 9.81(1.2583)
Soil 157
158
'
Chaoter 04
in soir
-
Stresses
Funda
Geotechnical Engi
Effective stress, pr = pr * pu
Effective stress, pr = 32.289 - 12.344
Effective stress, p6 = 19.945 kPa
tals of
Chapter 04
-120
o,
- Stresset
in Soil
nical Engineering
kPa
=
t,o=
40 kPa (-120,40)
| 5g
:.
ov = -300 kPa
ro,=
kPa (-300, -40)
.l
PROBLEM 04. I 5
A soil element shown in Figure 04.14 is subjected to the following
ov = 300 kPa
o, = 120 kPa
r=40kPa
-40
0=20o
s
Sign: Tensile normal stresses are taken as positive. .Shear stresses are
considered positive if they act on opposite faces of the element in such
a way that tend to produce clockwise rotation
!
Calculate the normal and shear stress on plane AB in kPa.
B
lo,
-tt
From the Mohr's circle:
Flgure 04.14
Jl=
SOLUTION
oo
I
300
cr = 40/90
a*23.96
O=180o-40o:c[
tan
*{-
-r
I
40
120
0 = 116.04"
t,
j""*..--'.
= -(?tO - R cos $)
oar = -253.23 kPa
,-,o..4!
l
= -R sin $
tes = -88.49 kPa
iAB
l60
Chaoter 04
in soit
-
Stresses
GeotechnicalEng
mentals of
hnical Engineering
Chapter 04
'
PROBLEM 04. I 6
For the stressed soil element shown in Figure 04,15:
b) Calculate the minor piincipal stress.
c) Calculate the normai and shear stress on the plane AE.
150 psi
6oesi
i
I
'
r
I
so ost
60 psi
B
From the Mohr's circle:
p=
Figure 04.15
.ffi1J6
=62.082
Major principal stress = 120 + R
Major principal stress = 187.08 psi
SOLUTION
Minor principal stress = 120 - R
Minor principal stress = 52.92 psi
tan cr = 60/30
a = 53.4349'
- 90' 26.565'
=
0
0 = 180o
c)
o' = -90 psi
t*, = 60 psi (-90,60)
o, = -L50 psi
ryx*
-60
psi
(-150, -60)
63.4349o
,
Norrnal and shear stress on the plane AE:
Normal stress:
'
oa6 = -(120 - 67.082 cos 26.565')
oAE
= -60 psi
Shearing stress:
rar = -67.082 sin 25.565
t,qs = -30
psi
-
Stresses
in
soit 16l
162
Chaoter 04
in Soil
-
Stresses
Chapter 04
of
F
GeotechnicalE
calEngineering
PROBLEM 04. I 7
For the soil element shown in Figure 04.16, determine the following:
a) Maximum and minimum principal stresses
b) Normal and shear stresses on plane CD
Figure 04.16
a)
SOLUTION
Major and minoi principal stresses:
Minor principal stress = 100 - R
Minor principal stress = 55.279 kPa
Major principal stress = 100 + R
Major principal stress = 144.721kPa
b) Normal and shear stress.on
lan a
a=
o, = -80 kPa
qr*
-40
kPa (-8P,40)
kPa
:"y*=40kPa (-120,40)
oy =
'L20
the plane CD:
= 40f20
63.435
0 = 180' -70 '63.435"
B = 46,565"
Norrnal stress, oco = 100 : R cos
Normal stress,
o66r
=
i0fl-
P
44.72X.
Normal stress, oco = 59.253 kPa
cos 46.565'
- Stresset 163
in Soil
164
Chaoter 04
in soil
-
of
Stresses
Chapter 05
-
Stress
,,r,r,1:,1"J
E ngineering
Geotechnical Engi
l65
Shearing stress, tco = R sin P
Shearing stress, rco = 44'72L sin 46.565"
Slrearing stress, t6p = 32,474kPa
Stress Distribution
in Soil
I
STRESS
CAUSED BYA POINT LOAD
Point Load
a
Ground surface
,z
5,
,
Figure 05.1'- Point load on ground surface
Boussinesq equations for the stresses due to concentrated load are:
.*3Q23-Q*,
'
Eq 5.1
Jr--....:-.:/VR
3Q
lxzz
2rR5
22
t-zvl 1
(2R+z)x2
, ll
" - T;
"''
tot-, LntR.,)-nT(R;}"-^tjl
I
,
_(2R+z)y2 _ ll
!y2, *t-zv[
=
"'-G
lRrl 3 Lrr(R.r-RrC;;f-^tll
q
3Q
R=
,/
Fl7*
Eq 5.2
Eq 5.3
Eq 6.4
166
where
Chapter 05
*
of
Stress Distribution
Chapter 05
-
Stress
Distribution
hnical Engineering
in Soil
l6T
VERTICAL STRESS CAUSED BYA FLEXIBLE STRIP LOAD
(FrNrTE WrDTH AND tNFtNTTE LENGTHI
Q = load
F = Poisson's ratio
Table 05.1 - Values of Boussinesq's Vertical Stress Coefficient, Ne
Ne
tlz
0:47746
0.41032
0.27332
2.50
2.75
3.00
tlz
0.c
0.2
o.75
1.00
1.25
1.50
1.75
2.00
z,t5
0.15645
0.08440
0.04543
't5
4.00
0
4.25
4.50
4.75
1436
00854
0.00528
rlz
Ns
223
5.00
5.25
1151
s.50
0.00014
0.00011
0.00009
0.00105
0:00075
0.00054
0.00040
0.00030
0,00023
0.00018
5.75
6.00
6.25
Ns
0.
6:.75
7.25
I
i
zi
'
r+
bi0-667
:.-
06
005
0,o0004
0.00003
0.00003
0.00002
Figure 05.3
-
lop
-.-,x
6
Flexible strip load
- 22 -(82 / 4)l
m=LI,u.r"[
l- B'{x2
'
' l-,un-'[
' nI lx'B/2)
lx+B/2) lxz + z2 - pz 1 +11+ n2 zz )
5.2 VERTICAL
STRESS
=ouu
CAUSED BYA LINE LOAD
When a line load of infinite length having an intensity of 4 ftN/m) acts
surface of a soil mass, the vertical stress, Ap, inside the soil mass is given
I.+ vTRTICNL
,,4.I
t
2az3
a
filx'
STRESS
CAUSED BYA RECTANGULARLY LOADED AREA
INCREASE IN PRESSURE BELOW THE CORNER
OF RECTANGUI.ARAREA
1.)
+ z-
)-
Tlrc increase in vertical stress below the "corner"
width B and length L is given as:
Lp=q,xI,
of a rectangular area of
Eq 5.7
wltere m = Bf z, n = Lf z, and z = depth of the point
,Ap
'- x
-.
Figure 05.2
-
Thc value of I, can be obtained
I
'---._oY
he
flom Figure
05.4, or from Table 05.2. It can also
computed from the following equation:
Line load on ground surface
Eq.5.8
168
Chaoter 05
in soit
-
Stress Distribution
Fun
Geotechnical
Fundamentals of
chapter 05
6eotechnical Engineering
-
Stress
D,r,r,l;,i.Jl
l69
Table 05.2. Variation.of l. wrth m and n to be used in Eq S.7
1.5
mz
0.25
I,
ll
m
0.1
0,2 .:0;L,I 0,4 - -a0T'
Ub
0.7 :
0.8
0.9
10
q,q047 0.00s2 0.0132 o.oroa - o-0198----d6rr-- o.ozss ---o-ozzo----I h
0.00e2 0.0179 0.049. 0.0328 0.0387 0.0435 O.Oqffi
0.0132 0,0259 0.0374 0.0474 0.0ss9 0n620-- 006s6 o.oz:r---oo%o -l6i d
o.oroe o.o:zg o.oaz+ o.oooe
0.01e8 0.0387 0.0ss9 0.0i11 0 0840 0 0947 o.roil -o.fios -0115[---T
0.0222 0.0435 0.0629 0.0801 0 oetr
0.0242 0,a473 0.0686 0.0873
q!4! 0.0504 0.0731 00931 0.1103 0.1247 O.IOOS. 0.1461 0J537*--Tl1e80.0270 0,0528 0.0766 00911 o.rrseffi
0.0279 0.0s47 0,0794 o.rors - -l2oZ----dJI6d--oi59s . 01684--- 0r?s2
!!?93 0.0573 0.0832 0.1063 0.1263 O.urr 0.ffi
o.olot o.osag o.oaso 0.rogl. o.r3
u.0J06 0.0s99 0.0871 0.1114 0.1324 0.1s03 016s2 afi74 il474 olsss
0 030e 0.0606 0.0a80 0 1126 0 1339 0 1521 o.i67z 0i797----0is9t--0
0.03'11 0.0610 00887 0.1134 0.1350 offi
0.0314 0.0615 0.0895 0.1145 _OI:OOffi
0.10
L--/ *
0,20
n1
l0 to infinlty
0.2a
/<
{,
o' 15
0.30
0.8
040
0.50
0.6
0.80
0.70
0.10
,
0.05
'4.
t
0.80
0.4
4
0.90
1.00
120
0.2
140
U.I
r.60
1.80
0.00
0.1
2.0Q
2.50
1
o.osrs
3.00
m
9.01t6 0.0619 0.0901 0.1153 0.1372 OtSOOffi
400
0.0316 0.0620 0,0901 0,11,54
0.0620 0.0002 orrs+
5.00
Fisure 05.4
-
lnfluence
"1"J ffi:,ftli,l"n [:r[-,.a|
stress increase under
To solve the stress at a point using Eq 5.7, divide the area into rectangles
that the point is at the corners of each rectangle.
qql!
6.00
m
.6
0.20
0t
050
0.60
0.i
0.80
0.90
100
1.24
1.40
160
180
2.00
2
3.00
0.2172 _9,2?6 0.232
4.00
Suppose that
-
lncrease in stress under point A
1.8
2
1
5
5
0.0293 0.0301 0
0.0309 0.0311 00314
0.0316 0 0316 0.0316
0.0573_, 0.0589 0.059S 0.0606 o.ooi ffi
0,0832 0.08s6 0.087i _ ooee-- -ooee7---lod-0.a9a1 0n901 01662
0.1063 0.1094 0.1114
1126 0,1134 0.1145 0.1
1153 0,1154 0.11s4
0.1263 0.
1363 0.1368 0.1372 0.1374 0,1374
0.143i 0.1475 0.1503 o.rSi
0.15/ 0162 0 1652 0.1672 0.1686 0 17a4
4.1717 01718 A17
1714 0.1797 o,terz ous2 01si
4.1847 0.1849 0,185
0.1 836
1874 0.1899 0.1915 0.1937 0 194
0.1954 0.1956 0.1957
0.1851 0.1914 0
1981 0.1999 0.20
4 0.2442 0.2044 0
0
0 1958 0.2A28 0.2073 0 2103 oz
0.2028 0.2102 0.2151 o.Ztat
0,2073 0.2151 0.2202
?237
0.2261
0.2294
02i
4.232 4.2324 0.232s
0.2103 0.2184 0.2237 0.22?4 02299 m
0.2124 - 0.2206 0.2261 O.leg!
151 0.2236 0.2294 0 ]4 0.2361 4.2401 0.242 0.2434 02430 O.zlt1
0.2163 0.225
309 0.235 0.2378 0.242 0
0 2455 0.246
0.10
0.30.
Figure 05.5
o.tzlffi
175
5.00
0.2176
6,00
0.2263
0.2264
A.Xg?
6 0.2395 0.2439 0.2461 0.2479 0.2486
1325 0.2367 4.n97 0.2441 0.2463 A.2482 0.2489
0.2324
0.2489
o
ztc2
it is required to solve the increase in pressure below point
the 4 m by 6 m footing shown in Figure 05.5. The area can be divided into
areas (Ar, Az, Ay and Aa). The values of Ap in the corner of each area is
computed. The increase in pressure is the sum of Lp's of each area.
5.4.2 INCREASE IN PRESSURE BELOW THE CENTER OF
RECTANGUI.AR AREA
'l'lrc increase in vertical stress below the "center" of a rectangular area
of
Il arrd lqngth L is given as:
\
\
\
width
l7O
ChaDter 05
in Soit
-
Stress Distribution
Fundamentals of
Geotechnical Engineering
F
Geotechnical
5.5 VERTICAL
Lp = q,x It
STRESS
chapter 05
-
Stress
o,r",T,1"":l
CAUSEP BYA SOUAREAND CONTINUOUS
FOOTING
L>B
L
-T-I
lt
t)
__1_
rt \
\-
Sqrare
footinE
;
,,
6t2
-:r*
3t2
-i
,,2
-|
Figure 05.6
-
2D
lncrease in stress below the center of a rectangular area
th
The value of /a can be obtained using Eq 5.10 or Table 05'3'
/ll1
ll
tr;fi5;1 ))
where
Eq
mt=L/b, h=zfb,and b*B/2
Table 05.3 - Variation of
la
with filr ahd or to be used in Eq 5.9
- Pressure isobars based on Boussinesq equation for square
and continuous footings used to find the pressures along line 1-1
Figure 05.7
ll
I
172
Chaoter 05
in soil
-
Stress Distribution
Fun
mentals of
Chapter 05
calEngineering
Geotechnical
5.6 APPROXIMATE METHOD FOR RECTANGUI.AR LOADS
or=
-
Stress
ro,r,Tri""il
t
t _i
r\,
ttu*;rlaf
Eq512
Lp=qx15
Table 05.4. Variation of
zlR.
lc
zlR
ls
zlR
0
1.00000
0.99999
0.99994
0.9997S
2.6
2.7
2,8
2.9
5.6
5.7
5.8
0.04599
8.6
0.04445
0.042s9
5.9
3
0.5
0.6
0.91056
0.86381
0.04159
0.04027
0.03900
0.03779
0.03664
0.03553
0.03448
8.7
8.8
8.9
s
0.2
0.3
0.4
0.99S49
0.s9901
0.99246
0.97627
o.94877
8693
0.17537
0.1 6479
0.1 5509
0.14619
0.1 3793
0.1
Dispersion of load for approximate increase in vertical
stress under a rectangle
The approximate increase in vertical stress is:
Lp=
O
(B+z)(L+z)
5.7 VERTICAL STRESS BELO\Y THE CENTER OF A UNIFORMLY L
CIRCULAR AREA
I
!
I
*'lAo
6
Figure 05.9
*
lncrease in stress below the center of a circular area
3.2
3.3
0.13044
3.4
0.11702
0,11104
3.5
3.6
3,7
3.8
o.12347
6
6.1
6.2
6.3
6.4
6.5
0.70063
0.64645
0.59487
4.',|
2
0.54662
4.2
3
0.50203
4.3
,4
0.46118
0.42397
44
o.ov276
7.4
4.5
0.06975
7.5
0.06692
7.6
0.7
0.81141
0.8
0.9
o.75622
1
3.9
q
7
0.39020
0.35964
.8
0.3320"1
4.6
4.7
4.8
I
0.30704
4.9
.6
;
a1
0.1
0550
0.1 0035
0.09556
0.09109
0.08692
0.08303
0.07938
0.07597
.5
Load per unit area = q
with z/R be used in Eq 5.'13
ls
0.04
0.06
0.08
-
Eq 5.13
zlR
0.02
Figure 05.8
ls
2
0.28446
5
2.1
0.26403
5.'1
2.2
2.3
0.24552
0.22873
5.2
2.4
0.21347
2.5
0.1 995S
5.3
5.4
5.5
0.1
6.6
6.7
6.8
6.9
ls
.
0.01994
0.01950
0.0'1906
0.01864
s.2
0.41824
0.01784
o.01746
o2
0.01710
9.4
4.u674
9.1
9.5
0 01639
9.6
9.7
9.8
0.01606
0.01573
0.01542
v.v
0.0151
10
7
0.03347
0.03251
0.03158
0.03070
0.02985
7.1
0.02903
10.1
0.01481
0.01453
7.2
0.02825
10.2
0.01425
7.3
0.02750
0.02678
0 02609
'10.3
0.02542
0.02478
o 02416
10.6
0.01397
0.0 371
0.0 345
0.0 320
0.0 256
10.9
10.4
10.5
1
0.06425
7.7
0 061 74
7.8
0.05937
0.05713
7,9
I
0.02356
0.02299
0.05502
0.05302
8.1
0.02243
11.1
8.2
0.02190
11.2
0.0 205
0.0 184
0.051 12
8.3
0.0 163
8.4
0.02139
0.02089
o.02041
1'1.3
0.04932
0 04761
11 .4
0.01 143
1'1.5
o 01124
8.
ll3
10.7
10.8
11
0.0 272
0,0 249
00 227
174
Chaoter 05
in soil
-
Stress Distribution
Fundamentals
Geotechnica I Engineeri
Frrndamentals of
Chapter 05
-
Stress
,lsrr,?;,1:ll
6r,otechnical Engineering
l7S
'l'hc procedure for obtaining the yalue of N to be used in Eq 5.14:
S.A INTLUTNCE CHART FOR VERTICAL PRESSURE
Newmark (1942) presented an inJluence chart based on Boussinesq's theQ
that can be used to determine the vert'ical liressure at any point below
uniformly loaded flexible area of any shape. This chart is based on Eq 5.
number
The influence value of the chart is 1/N, where N is equal to the
is I =
elements. The figure shows 200 elements; thus, the influence value
0
Eq 5.
Lp = iqN
1.
Determine the depth
z below the
loadecl area where the stress
increase is desired.
2.
Plot the plan of the loaded area with a scale of z equal to tlre unit
length AB of the chart. (i.e. if depth z = 4m, then AB = 4 m)
3.
Place the plan on the influence chart in a manner such that the point
below which the stress is required is located at the center of the chart.
4.
Count the number of elements (M of the chart euclosed by the plan of
the ioaded area. If certain segments are noi fully covered, you can
and N is
vrhdre i is the influence value, q is the pressure on the loaded area,
area.
loaded
of
the
plan
the
by
enclosed
the
chart
nrrmber of elements of
I
element
I element
Figure 05.10
-
Newmark's influence chart for vertiqal pressure based on Boussineq's
estimate what fraction is covered.
l7 6
Chaoter 05
in
-
Stress Distribution
GeotechnicalEn
Soit
Fundamentals of
Geotechnical Engineering
PROBLEM 05, I
1,500
-
Stress
,,rrr,?;ri"J
lll
3(1500) 2.53
2n 2.q55
AP = 53.17 kN
ILLUSTRATIVE PROBLEMS
A concentrated load of
Chapter 05
kN is applied at the ground
surface at po
Part c:
x=1,.5;
whose coordinate is (0, 0, 0).
a) Determine the vertical stress, in kN, at a point 2.5 m directly below A.
b) Determine the vertical stress, in kN, at a point whose coordinate
is (1.5,0,2.s).
c) Determine the vertical stress, in kN, at a point whose coordinate
is (1.5,2,2.5).
y=l;
z=2.5
ii?
n= Jt.i'if
R = 3.5355
.,
3(1s00)
A-n =
-i-_
2n
AP = 20.26
2.s3
3.53S5s
kN
SOLUTION
Point Load
Q
=
PROBLEM 05.2
1500 kN
il
A- line load and a point load acting on the ground surface is shown in Figure
05,11. Determine the increase in vertical stress at point A.
;'*tir
,,rl*q*
r.'f:'-.i
,:,
tP
tii
ia
Part a:
loo
v=0; !=0; z=2.5m
R
= Jo2 +02 +2.52
.R
= 2.5
"r
oo
Figure 0S.i
1
3(1500) 2.53
An = ---I---------/-
21
AP = 11+.e
-*
2.5s
kN
SOLUTION
Using Eq 5.1 for point load and Eq 5.5:
Part
Lp*
b:
x=1.5; y=0; z=2.5
R
= Jt.52 +02 +2.52
R = 2.915 m
(Lp)rin"r;ua
+
(Ap)pointroaa
tr=
*3Q
' n(xz?4"==
+ zz12 2n
I
{
R5
"=
178
-
Chaoter 05
Stress Distribution
in soit
q = 150
Fundamentals
Geotechnical Enoi
Fundarnentals of
Chapter 05
-
Stress
Distribution
Geotechnical Engineering
kN/m
x=3m
z=l-.5m
0=650kN
r---;n= ,lt.zz az *Y5z
R = 2.1656 m
B
2(1so)(1.s)3 3(6so)
r(32 +t.52)z 2n
1.s3
2.1.6565
Lp=2.5465+21..991,
LP = 24'537 kPa
L__
3r, ___,1
PROBLEM 05.3
A reetangular concrete slab, 3 m x 4.5 m shown in Figure 05.12, rests on the
surface o"i a soil mass. The load on the slab is 1620 kN'
a) Determine the soil stress below the slab using Eq 5'7
bi Determine the vertical stress increase at point A using Eq 5'7
.1
iiiii iii'i:ii;ii
1j
J m
influence chart.
h) Determine the vertical stress increase
at point A using
Newma
at point B using
Newma
: r
:,
I .1 . .1 i ;. . .1
j.1r:ri.!.,tlit
1:,.,.
..
.
. i .1 I
;. i
.
1r,1...:..,..r
r'1.,.,i]
. i . . i .1 i . . .,
ijr';:j.,.,.
.1-
i
r.,.11,,::.,,,.,.i,
ji:,;I*Eililiigi .diiiiir#;irir:.:
''.i.,,-i,...1..,.1..,r.i,..r
"-
1!r.rr
r,1.ti.i.J.Jj:'r.J:::
i.r'
::: ::: :::
iiiii:ili:iili:i:.:.:'.;.':.;.;,.:,
::
!Figure 05.12
a)
1.620
Q"=
3(4.s)
= 120 kPa
:ti;ii',,,,.''
.:.:
'..l.r;',r".r.'.:.'.:...
::...::r..::i..::r..'.tr.trl j,tr:
2.1
n
j.
:i:
rii.
1ij....ii1.:tr.
--4
jj,l
I
I
I
Ii
4,5 m
+.s,.
ffiiiiii;rii :ilrii:'
j..f
SOLUTION
,,.'
1..!,...4i,,,1,::riii;riiii6riii:i;:;i:.di
:#i
iiiiriiiiiiiiiii
influence chart.
I
j.,.i,.
.- ....:....:,,.r,1,..
,:lr1lr:irji,jir:'.,. ....:.....r,,:l
Determine the vertical stress increase
.1 i
.; .'+:ir,ri.',,''frtlir,ii;itii:#iiii:#ii;i:i#
'-1,
8)
.1_ ..
.jr'..1.,11:1.jt]1.ji.ji1ir.'r,.,.,;.: :r.,.,:
.ll1:i1ri:ri',ir+ i i
Figure 05.7
r
j..1,j.1,
r
I
I
I
I
I
llq
180
Chaoter 05
-
Stress Distribution
in soit
Fun
Geotechnical
b)
Divide the area into small rectangles such that A is at the co
of each rectangle.
Fundamentals of
Geotechnical Engineering
c)
Chapter 05
-
Stress
Distribution
tgt
q,lt
b=B/2
b=3/2
Lp =
b=L.5
m1
=Lfb
mt = 4.5/1.5
mr=3
nt = z/b
nt = 3/1..5
-nl -a
-/
From Table A5.3,Ir = 0.5254
Lp=120x0.5254
AP = 63.048 kPa
B = 2.25
= 1,5
t
d)
Divide the area into small rectangles such that point B is at the
corner of each rectangle.
z=3m
n=B/z=0.75
n=L/z=0.5
rl
.!
From Table 05.2:
For m =0.7 and n = 0.5,1. = 0.1034
For m = 0.8 and n = 0.5, /, = 0.1 103
N
-'2
0.1034 + 0.1103
| = llilli----Jlji.Jl
=n.1t\'7
B
Lp=q,I,x4
Lp=t20x0.1,07 x4
+3,-;
Ap = 51.36 kPa
Note: Using Eq 5.8 for m = 0.75 and n = 0.5, I, = 0.1071
B=3m
L = 2.5m
z=3m
m=B/z=1,
n-L/z=A.75
182
Chaoter 05
in Soit
-
Stress Distribution
Geotechnical
From Table 05.2:
For m = 1 and n = 0,7, I, =
0.1"491,
For m = 1. and n = 0.8, I, = 0.1598
'2= :::::::::::::
O.1491+ 0.1598
l_
Chapter 0s
Fundamentals of
Geotechnical Engineering
-t-- i----.
= 0.1545
t;
-
ff-
-
stress
D,rrr,?;tJ
-*1
I
,I
q,l"x 2
Lp=120x0.1545x2
Lp =
IJ
I
t
.i
* l3l2*i
ap = 37.08kPa
I
Note: Using Eq 5.8 for m
e)
t:U
'
* L and n = 0'75, L'
o
= '-----i :---=
(B + z)(L+ z)
\- _
1.620
(3+3)(a.s+3)
AP = 36 kPa
At point B:
x=1",5/3=0.58
z=3/3=78
At point
C:
/3 = 0.98
z=4.5/3=1.58
x = 21
From the figure shown below:
At point
B:
LP = o'4q
LP = 0.4(120)
AP = 48 kPa
Atpoint
C:
LP = 0.25q
tp = 0.25(1'20)
AP = 30 kPa
A.1547
cont'nou*
j
'
i-
t'/2
'l
tg3
l
84
ffffi"l.05
8)
-
stress Distribution
Funda
GeotechnicalE
Vertical stress at'A using Newmark's influence chart.
z=3m
From the chart, the number of elements N is about
21..6 x 4 = 86.4
LP = iqv
Ap = 0.005(1.20)(86.4)
Ap = 51.84 kPa (very much close to the answer in part b)
Chapter 05
Fundamentals of
Geotechnical Engineering
h)
-
Stress Distribution
in Soil
Vertical stress at B using Newmark's influence chart.
z=3m
From the chart, the number of elements N is about 3L x 2 = 62
Lp = iqN
Ap = 0.00s(120)(6,2)
Lp =
37.ZkPa
(very much close to the answer in part d)
r85
186
Chaoter 05
in soil
-
Stress Distribution
F
Geotechnical
of
Chapter 05
-
Stress
Engineering
PROBLEM 05.4
A square footing is loaded as shown in Figure 05.13. The center of the
footing is at cooidinate (0, 0, 0). Deterrfline the increabe in vertical stress
point whose coordinate is (3, 0, 4) using pressure isobars based on Boussi
equation for square footings shown in Figure 05.7.
| ',t
I
-1*
I
R
I
-J-
" \-.o,u,u
P=900kN
Figure 05.13
SOLUTION
'q=P/A
q=e00/(2x2)
4 = 225kPa
At point (3,0,4):
x=3m=1.58
z= 4m=28
From the pressure isobars shown below;
Lp = 0.045q
Lp = 0.045(225)
Ap = 10.125 kPa
tooting
l-'o
*i
Distribution t oa
,Ot
inSoit
l88
Chaoter 05
in soil
-
Fundamentals of
Stress Distribution
Geotechnical
PROBLEM 05.5
Determine the increase in vertical pressure at point 2 meters below point
the loaded area shown in Figure 05.L4. The fOundation applies a vertical
of 250 kPa on the soil surfaci.
SOLUTION
N=34.91 x3
N = 104.73
Lp = iqN
Ap = 0.005(25q$04.73)
AP = 130'9L kPa
6cotechnical Engineering
Chapter 05
-
Stress D
'",T'iil
t89
lqO
ChaDter 05
-
Stress Distribution
in soit
F
chapter 06 _
Fundamentals of
Geotechnical
hnical Engineering
compr"':iiy
lg l
Using Eq 5.2 divide the area into three rectangles 2 m x 2 m,
Compressibility
of Soil
z=2m
m=Bfz
m = 2/2=1
u =f lz
ati.-4
--. z/
zrL-
I,
lpaces.
divided into three categories:
1. Immediate settlement - caused by the elastic deformation of d{)'moist,
and saturated soils, without any change in moisttire contenf.
2. Primary consolidation settlement - caused by a volume change in
saturated cohesive soils due to expulsion of water that occupies thJvoid
ril settlement may be
L
From Table 05.2:
.
increase in stress caused by foundation and other loads compresses a soil
laycr. This compression is caused by (t) deformation of soil particles, (2)
relocations of soil particles, and (3) expulsion of water or air from the void
* 0.1752
spaces.
Lp=qxI,x3
Lp a 250(0.1752)(3)
LP = 731.4kPa
3. Secondary consolidation settlement - caused by plastic adjustment of
soil fabrics. It is an additional form of compression thai occurs at
constant effective stress.
i
I
SETTLEMENT FROM ONE DIMENSIONAL PRIMARY CONSOLIDATION
I.I
BASIC SETTLEMENT FORMULA
*"^j-ffi];,ti
I
I
lI
I
€Ht
qJ
.l,
+
I
I
I
Hs
<\
ol
+
I
Original
State
Compressed State
Figure 06.1
(
ffiI1
:E
fl
I
-
Chapter 06
of Soil
tq2
P = lf,(l+
rt -
Fundamen
Geotechnical Enqi
ComPressibilitY
6.I.3
e)
H
compr*:lgX
lg3
PRIMARY CONSOLIDATION SEfiLEMENT OF
OVERCONSOLIDATED FINE.GRAINED SOILS
1+e
H' = H,(1+
TJ
H'= It
chapter o6 _
Fundamentals of
Gcotechnical Engineering
e')
(1
W|tenp1<
p,:
I
+e'\
AH=H C' lorL!u
1.+e
nH=H-H'
1+
H fi+e'\
LH=H-1+e
'
on= 11+e:J'l-+e)
Vlhenpl>
e,,
p
=r#
^H=H#,.sh
- e'\=
1+en
H(e^
^LI=
---:-ll-----1
H-
Eq. 6.3
p,,
*H c' 6nU
\+eo " p,
Eq. 6.4
Le
1+eo
where:
C. = swell index
pc
= pr€consolidation pressure
where:
H = thickness of stratum
= void ratio before the vertical load is applied
e' = void ratio after the vertical load is applied
eo
6,2 OVERCONSOLIDATION RATIO, OCR
ocR=
h
Eq. 6.5
Po
I,2
6.
PRIMARY CONSOLIDATION SETTLEM IENT OF NORMALLY
CONSOLIDATED FINE.GRAINED SOILS
'
N7 =
(_
H:-' log Pt
L+eo -Po
where:
'
H = thickness of stratum
C. = compression index
eo = initial void ratio
p, = initial vertical effective soil stress
py = finalvertical effbctive'soil stress
W= P,+ LP
pc = pr€scohsolidation stress (past maximum vertical effective stress)
p, = overburden effective stress (current vertical effective stress)
lf OCR = 1., the soil normally consolidated soil.
6,3 COMPRESSION INDEX,
CC:
6,3.I SI(EMPTON:
For remolded clav:
C,=0.007(LL-77,)
Eq. 6.6
lg4
:ifj,l"'06
-
comPressibiritY
chapter 06 _
ntals of
hnical Engineering
For undisturbed clay:
l95
-,_co
u"ll-t-tp
Eq.6.13
^Ac
L^=
* _log t2 -Lop,t1
8q,6.14
C. = 0.009(tL ,- 10%)
6,3.2 RENDON-HERREOI
compr*:li[X
c.=0.141 G12(1+e0\238
\G)
= secondary compression index
change in void ratio
h = time for completion of primary settlement
fz = time after completion of primary settlement, where settlement
Co
Ae =
6.3.3 NISHIDA:
All clays
eo
C. = 1.15(e,
-0.27)
is required
= void ratio at the end of primary consolidation
er=eo-Le
H = thickness of clay layer
6.4 SWELL INDEX,
C5:
6.6 CALCULATION OF CONSOLIDATION SETTLEMENT UNDER A
The swell index is smaller in magnitude than the compression index. In
FOUNDATION
caseS,
we have seen that the increase in vertical stress caused by a
limited area decreases by depth. To estimate the onedirnensional settlement of a foundation, we can use the equqtions of this
lt,ction. However, the increase in of str6ss Ap should be the average increase
hr Chapter 05,
load applied over a
C,=+C.tofrC.
ln pressure
6.4.I
NAGARAJ AND MURTY:
below the cent€r of the foundation.
Assuming the pressure increase varies parabolica{ly, the average pressure may
q*6.6463!!t *6
Eq
6.
k'estimated as:
.
rt
I
a-^ * Lp, + 4Lpn,
L9ou, =
+ Lp6
'6
r
$
6.5 SETTLEMENT FROM SECONDARY CONSOLIDATION
Secondary consolidation can be calculated as:
AH, = (lo
"rrr(?)
whereApr = increase in pressure at the top of the layer
Lp^ = increase in pressure at the middle of the layer
Apo = increase in pressure at the bottom of the layer
Eq;6.15
196 :i;51"'06 -
comPressibiritY
Fundamenta
Geotechnical Engi
6,7 TIME RATE OF CONSOLIDATION
Eq.6.17
I
60%,
(Ha,)2
To
= 1..781, - 0.933 Iog (100
-
U%)
Eq.6.18
"Cu
'l'he time factor ?, provides
a useful expression to estimate the settlement in
lhe field from the results of a Iaboratory consolidation.
I'
where
H7, = one-half the thickness of the drainage layer if drainage occurs at
top and bottom of the layer (two-way drainage)
i
/ttot\2
0to60%, Tu='"1 "'"
4\100/
For U=
For lL >
L
lgl
lhe approximate values of time factor T, are:
The time / required to achieve a certain degree of consolidation U is evalua
as a function of the shortest drainage path within the compressible zone
coefficient of consolidation C,,, and the dimensionless time factor 7,.
,- *
l-t-.+
chapter 06 _ compr"r:iiH
Fundamentals of
Geotechnical Engineering
Ha, = thickness of the drainage layer
only (one-utay layer)
..
t
'
t
_
(H
a,
faa)2
1; $r;,,f
if drainage occurs at the top or
I
Tnta
I
' tr
Eq.6,19
2
Also, 'l = -1
t^ tt-2
Eq. 6.20
u1
Table 06.1: Variation of T" with U
I
.:1.l%
I
T"
i,
2
0.00008
3
0.00030
li
lI
4
0.00071
iri
5
0.00 126
6
0.00196
7
0.00283
8
0.00502
I
I
0,00636
i
10
0.00785
11
0,0095
i
1
ti
{
t
UVo
1
ill
:
I
12
0.01 13
IJ
0.0133
14
0.0154
15
0.0177
16
0.020'l
17
0.a227
18
0.0254
i
I
i
t
19
0.0283
20
0.0314
2t
22
0346
o o38o
I
ZJ
0.0415
I
24
0
0452
I
25
0
0491
I
]
0
I
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
UYo.
rTt
0,0531
51
0.0s72
I,
0.204
0.212
0.0615
53
0.221
0,0660
54
0.230
0,0707
0.239
0.0754
s6
0.248',
57
0.257
58
0.267
0,0907
59
0.276
0.0962
60
0,286
0.102
61
0.297
0,107
62
0.307
0.0803
I
0,0855
0.113
0.318
0.119
64
0.329
0j26
65
0.304
0.132
66
0.352
67
0.364
68
0.377
0.138
0.145
.
0.152
40
0.390
0.159
70
0.403
0.417
71
0.417
0.173
72
0,43'1
0.181
73
0.446
0.188
74
0.461
0.'197
l5
0.477
A%,1
76
77
78
79
80
81
82
83
84
8s
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
.,!i
0.493
0.511
0,529
where h = time to reach a consolidati on of Llr%
fz = time to reach a consolidation of Uz%
0.547
0.567
0.588
0.610
'l'lre degree of consolidation at a distance z
at
any
tinle
is:
0.633
0.658
Llr=1'
0.684
0.712
Pwz
n'
ruo
Eq. 6.21
0.742
4.774
0.809
0.848
'l he
average degree of consolidation for the entire depth of layer at any time is:
0,891
0.938
0.993
1.055
1.129
AU
lJ.-
= ?i,
Eq,6.22
^frmal
1.219
1.336
1.500
1.781
infinitv
where p*, = excess pore pressure at time I
p,, o = initial excess pore rtrrater pressure
AHr = settlement of the layer at time /
primary consolidation
.-..;;-..-;. ..
*
Chapter 06
198
-
Fundarnentals
Geotechnical Engineeri
ComPressibilitY
ofsoit
lmmediate settlement
6.8 COEFFICIENT OF CONSOLIDATION
chaptero6-compr*:il$
Fundamentals of
Geotechnical En gineering
lgg
of foundations resting on the ground surface of
an
clastic material of finite thickness is given by:
0.848
t" ar )2
t
"'" -" rH,Root time method, 'Cr"*
'
tgo
Log time method,
consolidation (r,[ curve)
f56 = time for 50% consolidation (1og f curve)
where
6.9
,r=
/e6
=
time tor
90%
COEFFICIENT OF VOLUME COMPRESSIBILIW, M,
"
tlii,=
au
-
iri
(en-e)/
1-+err,
LH;
t^^-"' '?
*#?O
LP
= oB1- P2 L
Eq. 6.28
E)
where:
p = net pressure applied in kPa or psf
B = width or diamgter of foundation in m or feet
U = Poisson's ratio
E = modulus of elasticity of soil in kPa or psf
/J = influence factor (dimensionless)
l-lre influence factor for the corner ofa flexible rectangular footing given as:
1,+eo*
_.1
* Jr * ,a,2
r,=Ll*,r,1'.Wl*,,[,,
,I) |
,,L
r
P+e^
t' ''
o -, = ----!Lau(
2
Eq. 6.2e
,i
rii
fi
Itll
The hydraulic conductivity of the layer for the loading range isr
Table 06.2: lnfluence Factors for Foundations
rlll
tI
k
-
Crmry*
itjhd06.i ""i
r
dfiiArt:l!:
Circle
1
where e, = initial void ratio
' e = final void ratio
'
Lp = rise in Pressure
Rectangle
6.t-0 IMMEDIATE SETTLEMENT
fl
Imrnediate or elastic settlement of foundations occurs directly after applica
of a load, without change in moisture content. This depends on the flexibi
of the foundation and the type of material on which it is resting,
q
1.5
2
3
5
10
1.00
u_64
4.12
1.36
0.56
0.68
0.77
0.89
'1
.53
20
50
1.78
2,10
2.54
2.99
3.57
100
4.01
where rer = length of foundation
/
1.05
1.27
1.49
1.8
2,0
width of foundation.
l'
0.79
0.88
1.07
1.21
1.42
1.70
2.10
2.46
3.00
3.43
2OO
Chapter 06
of Soit
-
Fundamentals of
Geotechnical Engineering
ComPressibility
Table 06.3: Values of Modulus of Elasticity
psi
':rkPa,
I
250 - 500
850 - 2000
1,500 - 4,000
1,725
5,865
10,350
34.500
.
.5.000 *,10,000
I
PROBLEM 06. I
-
13,800
- 27,600
- 69,000
Tvpe of $oit,
'PoisBon's Ratlo
Loose sand
Medium sand
Dense sand
Siltv sand
Soft clav
Medium clav
02-0.4
0.25 - 0.4
'fhe soil profile shown in
Figure 06.2 is to carry a surcharge of 60 kpa applied
nt the ground surface The result of laboratory consolidation test eonducted on
a specimen collected from the middle of the clay layer is also shown.
calculate the settlement in the field caused by primary consolidation due to
surcharge.
60 kN/m2
:.
0.3 * 0.45
0.2 - 0.4
0.15 - 0,25
0.2 - 0.5
Clay
Y*t = 18.2 kN/m3
€o
= 1'1
TOTAL SETTLEMENT OF FOUNDATION
The total settlement of a foundation is the sum of the primary, secondary, and
immediate settlement.
1.12
AHr = AH+ AH,+ AHr
20l
r
- 3,450
Table 06.4: Values of Poisson's Ratio
6.I
compr*:il:X
ILLUSTRATIVE PROBLEMS
fr
Soft clav
Hard clav
Loose sand
Densa sand
Chapter o(: _
I-undamentals of
Geotechnical Engineering
Eq. 6.30
l..t 0
0
d
.t.og
u
1,06
E
o
\
\
1.04
\
,.;I
60
Pressure, p (kN/m?)
Figure 06.2
ao {oo
2OZ
Chapter 06
ofSoit
-
Compressibility
Fundamentals
GeotechnicalE
PROBLEM 06.2 (CEMAY
SOLUTION
1,+e,
,*irtl"g
1.1,
t99ql
If
layer.
Solving for e corresponding to p7
Initial effective stress at midheight of clay
P" = Yb(3)
SOLUTION
p,= (18.2 - 9.81X3)
p, = 25.17 kPa
W= P"+ LP
pf
=25.U + 60
Pr= 85.17 kPa
,1.12
,l.10
I-
,,itt
,iiii
lllli
lll ll
itllrl
iliI
4., l'oB
I
AC
jj
I
p
\
I
r.oo
e
-
I
ii
1.047-
1.44
Unit weight of soil:
F.
ui
@
{.00 ,l
2.4
40
Pressure, p (kNim'z)
Sand above water table:
G +GMC
-[Y' = -_;-
v.=
''
Y,.l
t+e
2.()2+0
..
-:"-'" (9.81)
1+0.5'
y"=77.13 kN/m3
From the graph,
e
= 1.047
_ 1..1-1.047
aH-6--
1+ 1.1
AH = 0.1514 m
AH = 151.4 mm
compr*:il$ ZAI
over the
a 3-m thick land fill (Y = 17.3 kN/m3) is placed
the clay
of
settlement
ground surface, compute the consolidation
(sand) surface.
H= 6m
=
-
with 3 m thick of sand
A 2-m clay layer (e = 0.92, G = 2.72, C, = 1/3) is overlain
m below the ground
1'5
)
is
*ater
table
iuy". 1" ='0.5, G 2.a2.,.t4C = O1 tne
Settlement,6ll=g?P-e
e,
chapter 06
Fundamentals of
Geotechnical Engineerih$
Sand below water table (submerged):
-Lla' = G-1
L+e
=-Trl
2.62 -1.
^,,
tbs=. -.-- - (9.91)
1+0.5'
Yo,
= 10.59
kN/m3
-
204 :lfji:'ot
comPressibiritY
Fundamentals
Geotechnical
chapter o6
tundamentals of
Geotechnical Engineerin g
Clay: (submerged)
.
IIa,
G-7
BH 15
= =-T*.1
L+e
w.=
yu,
= 8.79
(-
kN/m:
L+eo "
H
=2m
eo
= 0.92
Yt
= 17.5 kN/ml
1'
oo
0r
1og2-
aH=H:".
m
4
ffiQ'81)
t1.2 m 0
u
po
10.4 kN/ml
* 8'79(1) + 10.59(1.5) + 17.13(1.5)
50.37 kPa
=
Po
P,
"
pf= P"+ aP
LP =
w = 40o/o
LL = 400/o
y, = 27,3 kN/m3
.g
Yth'
U
8.2 m
Lp = 17.3(3)
LP = 519kPa
o
6
where 1, = unit weight of soil particles
p1= 50.37 + 51.9
pf
-
1.02.27
ty=2 1/3
kPa
1+0.92
Figure 06.3
,,a1'0227
"
50.32
AH = 0.1068 m
AH = 106.7 mm
SOLUTION
Part
a:
Io
=
G-1,
=L+e
n
PROBLEM 06.3 ICE MAy 2OO4, NOVEMBER 2OO3l
Given in Figure 06.3 is the borehole log in a project site. The
building will exert a net stress of 12 Newtons per square centimeter.
a) Determine the buoyant unit weight of the clay
b) Determine the effective vertical stress at the mid-height of the clay layer
c) Determine the average settlernent of the normally consolidaied -cla1
layer. Use compression index C. = 0.009(tt - 10)
u=
-
l*
Ts
l*
27'3
9.81
G = 2.783
G MC=
S e,
whereS = 1
2'783(0'40\ =
e = 1.113
vu
=
rs
2.783,1' (9.81)
1+ 1.113
'
y6= 8.27.8 kN/m3
'
'|
-
compr*:pl[{
ZOS
zo6 firTl",
-
06
compressibirity
Fundamentals
Geotechnical
chapter 06
c:
ap
=
s
compr"':pl$
r
p" = (8
+ (10.4)(7.2) + 17 s(4)
?7.8)14!)
p, = 1g0.42kpa
part
-
hnicil Engineering
+-
-.=r_bsT_
l+eo
" po
C.=0.009(40_10)
I
C' = 0'27
en = 1..1I3
p" = Lg0.42
Pf= Po + LP
p _1?!1:
2
Ap = 720,A00 pa
Ap = 120kpa
i,,
iii
llrll
rrii
,,lir
i,ttll
ill ri
i rrf
ii
i,,"
I
I
LI
pf=L80.42+120
pf= 300.42 kpa
At-1=
g.2 0.27 t^-300.42
r
+i.113'"6l8olt
AH = 0.232 m
AH = 23.2 cm
Part a: (Norrnally consolidated clay)
.
'
i illll
i
x (100 cm/m)2
D'f
' logl--LH=H:"'
L+eo Po
PROBLEM 06.4
A soil profile is shown in Figure
06.4. Auniformly distributed
load, Ap
ground"surra.u
G
=
c.
Xl:::::::::"1T"
*
A,;;
Determine the settrement
of the cray layer
clay is nor.mally.onrollJlr"f,"'
fJ ll"
&) The preconsolidatio
.j rn"
.;;;
'""
ort ;;r'consoridation
C.=0.009(Lt-10)
C.*0.009(50'10)=6.36
-
Compute the initial and final effective stresses at
the midheight of clay:
+ (14)r".u ftsand sat + (|a.y)."na hsand
?o = (Ta)auy h.try
po= (19 - 9.81X4) + (18.5 - 9.81X6) + 16(3)
if:
;;;;;;;iiililXi,#:i:[ il8li:,
Po- 136'9 kPa
P7*Po+ LP
W= \36.9 +
5A
Pt'186.9 kPa
AH=8x 0'36 los186'9
1+0.95 " 136.9
6p1=,0.2m
AHf
/
200 mm
dry
ZOI
208
Chapter 06
-
Compressibility
Fund
Geotechnical
of Soil
Fundamentals of
Geotechnical Engineering
chapter o6 _
compr*:il$
Part b: (Overconsoli dated clay, p. = 210 kPa)
p1* 786.9kPa < p,
(
0r
Iog!
"
1,+eo po
LH = H:"'
C,= 0.36/5
C,= 0.072
0.072loq-1.86.9
AH = tJ-
1+0.95 " 136.9
AH = 0.04 m
AH=40mm
rl
Part
'll
c (Overconsolidated
clay, p,
-
150 kPa)
P7='186'9 kPa> P,
rll
:I
,'l
AH =
H:C,
bgL!bglt * uJ,
\+eo "po
1"+eo "p,
aH =
^ 0.072los-150
E1+0.95 "136.9
LH=
A.01172 + 0.141L
i,t
;l
$
+
E- 0.36Ioe-185.9
1+0.95 " 150
AH = 0.1528 m
AH = 152.8 mm
rfil
ffi
ilr
ild
fi
PROBLEM 06.5
Figure 06.5
It is desired to calculate the consolidation settlement of the 4-m thick clay
shown in Figure 05.5 that will result from the load carried by the
measuring 3 m x 1,5 m in plan. Assume the clay to be normally consoli
and the load of footing results to an increase of pressure of 23.2kPa,12.22
and 6.1 kPa at the top, midheight, and bottom portiory respectively, oI the
layer.
Assume the pressure increase varies parabolically and use Simpson's ru
solve for the average pressure increase.
Lp=
a)
b)
4
Lprop*4Lp^ia+LPaot
Calculate the overburden pressure at the base of the footing, in kPa.
Calculate the initial effective stress at the midheight of the clay layer.
Calculate the consolidation settlement of the clay layer.
'a
a\
'",
\
\'r
.\
SOLUTION
uur,
i,= ro,
q = 14.8(1..5)
4 = 22.2kPa
Part
&:
p, = initial effective stress at midheight of the clay layer
p"= (19 - 9.81X3 + 2) + (18.5 - 9.81X1.5) + 14.8(3 + 1.5)
po= 93.42kPa
ZOq
Chapter 06
2lO
ofsoir
Part
-
Compressibility
F
Geotechnical Engi
ls
of
chapter 06
nical Engineering
rt
-c nrl!_
AH_H"l+eo
" po
C'= 0'36
H=3m
r
fi=
Le =
1'A
po
.
Lpw +4Lp^i6 +Lpuot
6
itt
rI
ap=
tI
23,2+4(12,22)+6,1
Le = 0.251on170
AP = 13'03 kPa
,il
'L25
Ae = 0.0334
py= 93;.42+ 13.03
lil
,
aa
lffi
py* LA6.45kPa
=a
0.36
lds
1+1 "
1"06.45
93.42
AH = 0.0306 m
AH = 30.6 mm
itfll
l|4
Or
C,los'
' "pn/
Po= 125kPa
W= Po+ LP
pt= 125 + 45
p1= 170 kPa
+ a'p
uy=
[i
AH, = 3
'
1+ 0.8
Secondary consolidation settlement
AH, =
'
0'0334
AH, = 0;05564 m
AHp = 55.54 mm
fiil
C,H,"r[9]
\11 ,
i
PROBLEM 06.6
A normally consolidated clay layer, 3 m thick, has the following properties;
'
AHprima.y + AH,".on6,r,
L,Ho=
C.=0.009(50-10)
€o
Total consolidation settlement, AH =
Cc
H Le
' 111,+ eolon\L
" pn = 1,+ eo
C. = 0.009(LL: L0)
illtl
compr*:iig
c:
For normally consolidated clay:
rllil
-
Initial void ratio, e, = 0.8
Compression index, C. = 0.25
Average effective pressure, po = 125 kPa
Expected pressure increase, Ap = 45 kPa
Secondary compression index, C. = 0,02
Time for completion of primary settlement = 1.5 years
What is the total settlement of the clay layer five years after the completion
primary consolidation settlement?
-,-Co
l+ep
eo=
eo
- Le
er=0.8-0.0334
eo
'-
= 0,7666
0.02
"=
T;oiG
C'" = 0.01132
5
AH' = (0.01!32)(3) log j=
Nf,= 0.01776 m
LH,=17.76mm
,
Eq.
6.12:
2tt
2lz :i;ji"'o6 - ComPressibiritY
Fundamen
Geotechnical Engineer
Total consolidation settlement, LH = 55.64 + 17,76
Total consolidation settlement, AH = 73,4 mm
b\
'
/\
l.
'
ilrrl
ii't
lrll
120 kPa
itrll
,
,11,
CompressibilitY
of Soil
2t3
p*o)
p,, = initial exaess pore water pressure
Pw= f*h
P*,= (9'81)(12'232)
P,o = 120 kPa
P*, = 9.81(6)
P*, = 58'86 kPa
u,- = (t_ !!.891 roozo
\ 120)
l)o =
llii
-
u, = [r _!-a.zlrcs"/"
PROBLEM 06.7
A surcharge of 120 kpa is applied on the ground surface on n
the soil pro
shown in Figure 06.6.
a) How high will th.e water rise in the piezometer immediately
after
application of the load?
b) What is the degree of consolidation at point D when h = 6 m?
c) Find /z when the degree of consolid.ation at D is g0%
Chapter 06
Fundamentals of
Gcotechnical Engineering
50'950/o
c) u,=(t-'"
\
ltoo%
P"'" )
lh\
"
8o%=lL-r*'lToo%
I ln,tl
\
iilll,irt
liirll
120i
P*, = 24 kPa
Pw'
11
=
lllflll
lw
,24
litiiiliii
9.87
h = 2.446 m
PROBLEM 06.8
Under a given surcharge, a 5:m thick clay layer has a consolidation settlement
of 305 mm, Assume C, = 0.003 cm2/sec.
a) what is the average degree of consolidation for the clay layer when the
settlement is 75 mm?
b) How long will
SOLUTION
a)
n=M
lut
,n= _120
9.81
h = 1,2.232 m
it
take for
50o/o
consolidation to occur
if the layer is
50%
consolidation to occur
if the layer is
drained at the top only?
c) How long will it-take for
drained on both.ends?
214
* comPressibiritY
o6
:lfj,i"o
Fun
Geotechnical
SOLUTION
a)
Fundamentals of
chapter 06
.
Cu
U=
T, = 0.848 (from Table 06.1)
Ha, = 1/z(3.2) = 1.5 m
Ha' = '160 cm
t = 80 days x24 x3600
t = 6,9L2,000 sec
-'"
30s
U=0.2459=24.5go/o
:
Frorn Eq. 6.16:
,
--
6.912.000 = 0.848
-'"
- *-co
(Hrr)?
'
,il
I
t = 0.197
.i||
il[
b)
From Eq. 6.19:
5uu
!t!- ttob
0.003
| ='1,6,41,6,667 sec
x.ffi
x
$
,1ffi
tha
.
.
',
IU
From Table
06.1,,
f
or
L)
=
S0%,
:
0o
/42
T, = 0.1,92
With single drainage, Ha,*1/z(5) = 2.5 m
With single drainage, Ha, = 250 cm
t = o.1g\
1602
-
= 6,750 sec
fr"r = 1..875 hours
C,
frf
(H a,nt)z
frub
1= 7--\Ha')'
N
(Ha,fa)z
6,912.000
t = 190 days
rll!|
".^?
tbu
cu
C, = 0.00314 crn/sec
From Table 06.1, for lJ = 50o/o, Tu = 0.197
With single drainage, Ha,= 5 m * 500 cm
il
c)
2502
0.003
,,:liyi::"secxffixg
PROBLEM 06. I O
Urrder a given load (surcharge), a 3-m thick clay layer undergoes a 180-mm of
total primary consolidation settlement. If it took 80 days for the first L00-mm
trf settlement
to occur, what is the estimated time for the first 50-mm of
rcttlement?
SOLUTION
L
PROBLEM 06.9
From Eq. 6.20,
thick layer of saturated clay under a surcharge loading*
I907o-?.2-*
primary consolidation
in 80 days with double drainage.
a) Determine the coefficient of consolidation for the preisrre .ange
&) For a 10"cm thick specimen of the said clay, how long wilt it take
undergo 90% consolidation
range?
ZIS
(Har)'
,. -_o
AH,
AH.""
il$
tlrI
Compr.r:il$
IOLUTION
FromEq.6.22:
tt-
lll
-
Geotechnical Engineering
in the laboratory for a similar
rr
2
= u7
'-tz Ur'
Degree oi consolidation @ fr = 80 davs.
,
Degree of consolidation, Ur
=
ll"
= Mt
AH-r"
*180
Degree of consolidation, Ur = 0.5556 = 55.56%
2lb
:ifi,i"'06
* comPressibiritY
F
GeotechnicalE
'50
Degree of consolidation @ tz,
Fundamentals of
Gcotechnical Engineering
l)2= .JJ-
(
180
.
Degree of consolidatio n,
80
l-12
ll
d
ll
il
,ll
'tr|
'tu
Under normal loading condition, a 3.6-m thick clay (normally consolida
has po = 190 kPa and eu ='1.22. A surcharge of 190 kia reduceslts void rati
0.98. The hydraulic conductiviry of the clay for the loading range is 6.1 x
= 0.01034
nf/day )
roo
= (0.286)-13:6f
f6e
= 358.5
'
Coeff. of consolidation
0.01034
days )
Time for 60% consolidation
Fressure (kPa)
.60
a; What is the coefficient of volume compressibility of the clay?
b) What is the coefficient of consolidation of the cliy?
c) How long will it take for this clay layer to reach 60zo consolidation if
Void ratio
Determine the coefficient of volume cornpressibility in m2/kN
Determine the hydraulic concluctivity of the clay for the Ioading range.
ilflt
SOLUTION
SOLUTION
,trd
(Ho')'
,-, .C,
ill
L -
k = Cu muy*
lt)-
C'= 7'(H
-t
rl
Ha, = 3.6 m (single drainage)
,
From Table 06.1-, T, = 0.2{6
C,=ffiofR
*,=
@o_-4/
Lp
L+eaoe
eo
.
To
=
a')z
Gq' 6'27\
0,236 (Table 06.1)
* 0.03/2 = 0.015 m (double drain)
=
Ha,
(Eq.6.zT)
-
f = 3.2 min
(o'9T)1
co= 0.286
@q.6.25)
C, = 0.00002011 mz/min
1..22
r,
\"o_e)/ M
m,,L+ero,
e=0.98
'
*"tu-_ (0.87 -0.72)/(1s0-60)
< , rtaiiiiz
0av" =
,2
eur"
IilD-a
.
= 1.1
'
-1.+7.1
mrlkN )
L+'*tt-
m" = O.OOOI285 m2/l$tr
(1.22-0.98\ /190
m, = 0.0006015
lr = 0.0000201 1 (0.00092S5) (9.81)
Coeff. of compressibility (a)
(%)
Time tor 60% consolidation = 3.2 min
a)
illl
(c)
87
72
150
b)
ffi
(&)
PROBLEM 06. I 2
'l'he following data were obtained from h laboratory consolidation test on a 30
rrrnr thick clay specimen drained on both sides:
m/day.
drained on one side only?
zl7
6.1x 10-s
'' = ------------
C,,
55.562
T
- compr*:il$
0.000601s(9.81)
= 0,2778 = 27.78%
4- r?w
PROBLEM 06. I
chapter 06
1 k=1.83xl0rm/min
\
\
218
:iii,i",.06
-
comPressibiritY
Fundamentals
Geotechnical Enoi
chapter 07
tals of
echnical Engineering
-
shear
t.ilS|fi 219
PROBLEM 06. I 3
A rigid column footing 1.2-m in diameter is constructed on unsaturated
layer. The load on the footing is 170 kN. "Estimate the immediate
Assume the clay has E = 6900 kPa and ir
-
Chapter 07
0.2.
.:
SOLUTION
Shear Strength
of Soil
Using Eq. 6.28:
L,Hi =
1)
pB{t-
7,
E
o=
i
l]lh
lrtlll
ru
i
'
y =
irfli
A
I'hc shear strength of soil may be attributed to three basic components:
1. Frictional resistance to sliding between solid particles
170
t0.2)'
-_------;
P = 150'31 kPa
lrm
B
i',ttH
rl
i ur'u
:,1
'
Q
='i,.2m
From Table 06.2,4= 9.79
iilffi
iilt$fl
I
tll
rl
rlt ruil
urtl
2. Cohesion and adhesion between particles
3. tnterlocking and bridging of solid particles to resist deformation
AHr = 150.31(
1.2\1-0'22 (0'79)
AHr = 0.0198 m
AHi = 19.8 mm
6,s00-
7.I MOHR - COULOMB FAILURE
CRITERIA
A material fails because of a critical combination of normal stress and shearing
rtress, and not from either maximum normal or shear stress alone. This theory
was presented by Mohr. Thus, a failure plane can be expressed as a function
of normal and shearing stress as follows:
lilfin
v= f@)
Eq. 7.1
lior most soil mechanics problem, Coulomb suggested that the shear stress on
the failure pla4e can,be expressed as a linear function of normal stress. This
written as:
re lationship is'known as Mohr-Coulumb failure critetia and can be
rl=c+otan$
where
c=
Eq.7.2
cohesion
$ = angle of internal friction
'l'hese functions are ihown in Figure 7.1. The significance of the failure
cnvelope is as follows, If the normal and shearing stress on a plane in a soil
n,oss ui" suchTthat they plot as point X, shear failure will not occur along that
plane. If it plpts at Y, shear failure will occur along that plane because it plots
t
\
\
22a :lf5l""'
-
shear strensth
Fundamen
GeotechnicalE
Chapter 07
Fundamentals of
* Shear t,t:??fi
Eeotechnical Engineering
221
Z cannot exist because it plots above the
envelope and shear failure would have occurred already.
along that plane. Point
Figure 7.2
-
Applied stresses on soil
(a)
Normal stress
o
Figure 7.3
-
Mohr's circle
(,l
c
6
o
3
tn
Eq. 7.3
.i
Eq. 7.4
I
L
Normal Stress
(b)
Figure 7.1
- Mohr's
failure envelope and Mohr-Coulomb failure criteria
,\.
r_r
ll.Z
Chapter 07
of Soil
-
Shear Strength
ls
Funda
Geotechnical EnEi
Chapter 07
of
hnical Engineering
- sheartrt:??S ZZg
following equations can be delived from Figure 7.5:
Air release
valve
r=otan0=Rsin20
Eq.7.5
R=th(q-<r:) =r,u,
Eq. 7,6
or
SlnQ= #
Water
-ol
€q.7.7
01 +O3
pressure, 03
Platen
6*4go+g/2
flil9
Eq. 7.8
cylinder
Radial grooyes
for drainage
h
#
T
water
Water supDly
I
,2.2 COHESIVE SOIL
To pore
t.
pressure transducer
volume chanqe
device
I
il
Figure 7.4
1,F
- Schematic
of ,a
Plane of'failure
triaxial cell
itil
ill
tr
7.2
il
7.2.1
$
Mohr's Strength
Envelope
TRLN<tAL SHEAR TEST {S|NGLE TEST'
COHESTONLESS SOIL
I
Figure 7.6
-
Single test on cohesive soil
r=c+otan{=Rsin20
.ii
l*',_
t!
I'
o,
.-..>i<*
Figure 7.5
-
Deviator
stress
,i
-1..- "
Single test on cohesionless soil
where:
or = Major principal stress at failure
oz = Minor principal stress at failure
r = Shear stress
c = Cohesion of soil
$ = Anglg of internal friction
e=
an[f{ that the failure
\
\\
Eq. 7.9
plane makes with the major principal plane.
224 :lfflt,.07 - shear strensth
7.3 DRA'NEDANDUr,rffi
F
Geotechnical Engin
tdamentals of
echnical Engineering
Chapter 07
For drained triaxial test o,r and o,3
are taken as the effective principal
For undrained triaxial test, o1 and
o.3 a.re taken as the total principal
str
-
Shear
trr:?g:l
or
225
-oq
--f-
Eq,7,10
o1-o3
-t
-l
- -_ _I__
,.,
Eq.7.l1
- - 01+(Ia
u1_--
Eq l.r2
"'
=
_ o'1*o'3
- -2
sln0=
'
Eq. 7.13
Eq,7.l4
Cz'Ct
ilnfri
c=x'tan0
-
iilut'
.
,
itlrrilil
ililrffil
ffi
x'=x-Cr
Figu rel.7 -Drained and undrained condition
rliffi
c-
where:
{o = drained friction angle
0u = undrained friction angle
7.4
rq. z.tS
pore Water pressure
R1
and-=;il
Eq.7.16
cos d - (C,
- Rr sin $) tan $
Eq.7.l7
c = R2cos S - (C,
- Rz sin {) tan $
Eq.7,18
R1
l
?,5 UNcoNFINED coMPREssIoN TEST (UNIAXIAL,
TRLA)<LAL TEST
{SEi?tEs,
o:=0
03=0
Figure 7.9
-
Unconfined compression test
oi
'=R=+
The following equations can be
derived from Figure 7.g:
qu=2C'
tvlrt,re 4, = unconfined compression strerrgth
\
Fq.7.19
Eq,7,2o
2zo :i;51" 07 - shear strensth
Fundamen
GeotechnicalE
7.6 DIRECT SHEAR TEST
Direct shear test is the,simplest for of shear test. The test equipment cor
of a metal shear box (see Figure 7.10) in which the soil ru*pt"i, placed.
sizes of the sample used are usually 50 mm x 50 mm or tbO mm x 100
across and about 25 mm high. The box is split horizontally into halves.
normal force is applied from the top of the shear box. sheai force is appl
by moving half of the box relative to the other to cause failure in the
Chapter 07
Fundamentals of
6eotechnical Engineering
-
Shear
t.:?3|[ 227
ILLUSTRATIVE PROBLEMS
PROBLEM 07. I
is 50
l)irect shear tests were performed on a dry, sandy soil. The sPecirnen
follows;
rrrm in diameter and 25 mrn in height. Test results were as
sample.
Test No.
Normal Force, N
Shear Force, N
1
243
124
2
268
137
3
352
179
412
210
4
Determine the cohesion and angle of internal friction'
SOLUTION
Cross-sectional area of specimenl
A=
f (oos),
A = 0.001963 m2
Figure 7.10
-
Direct shear test arrangement
Tesl
No.
shear stress,
,
3
Cross - sectional area of specimen
4
= Cross
Resisting shear force
- sectional area of specimen
123.8
136.5
124
63.1
137
179.3
209.8
175
69.8
s1.2
210
106.9
243
268
352
412
2
Normal stress,6 =
Shearing
Stress. kPa
Normal
Stress. kPa
1
Normal force
Force: N
Normal
Force. N
Shear
120
a
100
o
ai
o
Test 4
'":'1.
80
testz
st 1
60
^.."t'
a
0
o
a
40
20
Io
o
so
1oo
150
Normal Stress, kPa
2oo
250
22!B 5ii5f,.o7
* shear strensth
F
GeotechnicalEng
The graph shows a straight line that purrur
cohesion c = 0.
Chapter 07
Fundamentals of
Gcotechnical Engineering
* Shear Strength
of Soil
229
tn*rgt tf,Iffi
MA
Solving for
S using the results of Test
1;
120
tan6= r
flx
o
ui
a
63.1
tan6= _
'
rn'i ii .i,,,p rt line
a,60
Eao
a
20
r'
Aa
+
+
j. . zo kn,
PROBLEM O7.2
3:^f::.1:
as follows:
of four drained direct shear tests on an over consolidated
clay
168
50
100
150
Normal Stress, kPa
itnru
jiffi
{
o
N
80
0)
123.8
,0=27"
iiffit
-/t
100
Test No.
Normal Force, N
Shear Force, N
1
120
114
2
220
179
3
320
239
4
420
286
Size of specimen = 50 mm x 50 mm
Height of specimen = 25 mm
From the graph, we draw the best fit straight line.
Cohesion,c=20kPa
Angle of internal friction:
Using the result of Test 4 and the Y-intercept:
xL4:4':20
tan $ r
Determine the cohesion and angle of internal friction.
SOLUTION
Cross-sectional area'of specimen, A = 0.0S(0.05)
Cross-sectional area of specimen, A = 0.0025 m2
168
O *29,38o
PROBLEM 07.3
A direct shear test is performed on a specimen of dry sand. The shear box is
circular in cross-section with a diameter of 50 mm. The normal force imposed
on the specimen is 250 N. The shears when the shear force is 150 N.
Determine the angle of internal friction of this sand.
SOLUTION
From Eq. 7.2:
q=
c+
c=
o tan $
0' (dry sand is cohesionless)
zilo
:i;51"'07
-
shear strensth
F
Geotechnical Engi
v
-
Shear Strength
of Soil
6eotechnical Engineering
231
07.5 {CE NOVEMBER 19981
in a triaxial test, a specimen of saturated (normally consolidated) clay was
Consolidated under a chamber confining pressure of 80 KiloPascals. The axial
Itrcss on the specimen was then increased through the allowing the drainage
from the ,p".i*".r. The specimen fails when the 1.20 KiloPascals, The pore.
Wirter pressure (tI) at that time was 50 KiloPascals. What is the consolidated
Tf=
,A _
1s0
i1o.os;2
ry= 76,394Pa
=i=#
.
Chapter 07
Fundamentals of
runr.lrained
friction angle (Phi):
o = 127,324pa
'OLUTION
For undrained condition, we use the total principal stresses
76,394 = 0 + (127,224) tan g
tan 6 = 6'6
0 = 30'9e'
r-:1!.---roo
PROBLEM 07.4
i.-
A7-m thick soil has water table
ll" *?l:r
3 m berow the.ground surface. The boil abc
table has degree of saturation of 4s%, void,ratio of the
soil is 0.4 a
the solids have specific gravity o{ 2.20, Tests show'th;ith"
tlu" u.,gru
internal friction of 32o and cohesion of 1-4.6 kpa. what
is the potentiar shr
strength on a horizontar prane at a depth of 2 m below
the groriJr;.ru."z
l.--
l_-
;;i
",
= 80
r-----i40 ------l
._+1--
I
+: = t2o
20
SlnQ=
-
0 = 11'54o
SOLUTION
Potential shear strength, r = c + o tan
0
c = 14,6
6 =32
o = vertical effective stress
"
c
=
lt,h
Y*=
*&Y*
t+e
0.45fi.40\
7'=--ffi(9.81)
2.20 +
Y*= 20'18 kN/ms
o = 20.18(2)
o = 40.36 kpa
Potential shear strength, r = 1.4.6 + 4}.36tan 32o
Potential shear strength, r = 39.g2 kpa
PROBLEM 07.6ICE NOVEMBER
z00tl
Irr a triaxial test, a specimen of saturated (normally consolidated) clay was
consolidated under a chamber confining pressure of 90 KiloPascals. The axial
stress on the specimen was then increased allowing the drainage from the
specimen. The ipecimen fails when the deviator stress is 60 KiloPascals. The
pore water pressure (p*) at that time was 40 KiloPascals. What the
consolidated drained friction angle (phi):
SOLUTION
Total principal stress, o: = 90 kPa
Effective principal stress, o': = c't' Pw
Effective principal stress, o'r = 90 - 40
fffective principal stress, o'3 = 50 kPa
Qeviator
\
\\
stress = 60 kPa
232
-
Chaoter 07
Shear Strength
Fundamen
ofSoit
GeotechnicalE
Fundamentals of
Geotechnical Engineering
chapter 07
-
shear
r.:i?$
1
LJ
r60r
stno=
'80_
30
6 = 22'024'
PROBLEM 07.7 ICE NOVEMBER t9991
A triaxial test on a saturated soil has the following results:
Cell pressure
stress
(kPa)
Deviator
, (kPa)
2oo
400
600
119
143
178
Pore pressure
(kPa)
142.5
275.5
396
a) Determine the drained angle of internal friction of the soil.'
b) Determine the cohesion of the soil in drained condition.
From the figure:
& = 119/2
Rr = 59.5
Rz="143/2
Rz= 77.5
cr=57.5+59.5
Cr ='117
(2= !)\.$ + /t.$
Cz= 195
fr) Drained angle of friction
sin0=(Rr-R)/(Ce-G)
sin$= (71.5-59.5)
SOLUTION
For drained condition:
Effective stress, o'3
Deviator stress
200 =L42.5 = 57.5
119
143
4OO-275.5=t2q5
/ (196-717)
6 = 8,737o
b) Cohesion in drained condition:
b=Ct-a
b = 1.17 - 59.5 sin 8.737"
b = 107.96
c=R1 cos$-d
caRlcos$-btanQ
c = 59.5 cos 8.737o - 107 .96 tan 8:737o
c = 42.22kPa
n,_n,
Zgg
234 :if$it'07
-
shear strensth
Fundamentalsl
Geotechnical Ehgi
PROBLEM 07.8lCE NOVEMBER
t9e9'
1.
Sample 2
Sample
kpa
25 kpa
14
o3
tests. The soil
Plunger stress,
34 kpa
56 kpa
What is the angle of internal friction of the soil in degrees.
ll'$,
01
SOLUTION
SOLUTION
ilni
,.:, ,Rir:)
ilffi
Cz
In triangle
34-14
Rr=
1:
f2p$=
4200
6300
R1=10
0 = 33.69o
Cr*14+10
Cr- 24
o =,lq+,zoo12 + (6,300)2
^Xr=
Ra
a = 7,571,.66
56-2s
In triangle 2:
-T
a=90o-0
cr = 90o -33.69'
= 15.5
Cz=25+15.5
a = 56.31o
Cz= 40,5
R - 7,577.66 tan 33.59'
R = 5,047.76
sin$=
sinQ=
O
R=atan0
Rz
*Rr
Cz
-Cr
15.5
- 10
44.5-24
= 19,47o
t,r:i?S Zgs
A triaxial shear test was performed on a well-drained sand sample. The
rrormaI stress on tlre failure plane and the shear stress on the failure plane, at
f,rilure were determined to b; 6,300 psf and 4,200 pst, respectively.
n) Determine the angle of internal friction of the sand?
&) Determine the angle of the failure plane?
c) Determine the maximum principal stress?
r#
t$
ffi
chapter 07 - shear
PROBLEM 07.9 ICE NOVEMBER 20021
A sample of moist sand,was subjected to a series of triaxial
Cell pressure,
tundamentals of
En gineering
Geotechnical
C=asec0
C = 7 ,571..66 sec 33,69o
C = 9,100
236 [ifj,i"'07
-
shear strensth
F
GeotechnicalE
Angle of failure plane,
Fundamentals of
Geotechnical En gineering
b)
0:
q,+20=180o
56.31o*20=180o
0 = 51.84o
.:.-
s,
lt*
lil
rrRoBLEM 07.10 (cE MAY 2OO4t
The results of, a consolidated-drained triaxial test conducted on
consolidated clay, are as follows:
Cha.mber confiniprg stress = 250 kPa
Deviator stress at fiilu." = 350 kpa
a)
Calculate the angle of friction of the soil sample.
b). Calculate the shear stress on.the failure flane.
c) calculate the effective normal stress on the plane of maximum sheai,
c)
Effective normal stress at the point of maximum shear
o=C
,
o=425kPa
PROBLEM 07. I I
A
consolidated undrained (CU) compression test was conducted
pressure recorded was 54 kPa.
a) Determine the undrained shear strength of the clay
friction angle:
Cell pressure, os = 150 kPa
Principal stress, or = 03 + plunger stress
Principal stress, or = 150 + 160 = 310 kPa
Angle of internal friction
n = 35012
R=175
250 + 175
Q = ,1176
sinQ-175/425
a
of 150 kPa and then incrementally applying loads on the plunger while
kceping the cell pressurd'constant. Failure was observed when the stress
cxerted by the plunger (deviator stress) was 160 kPa and the pore water
a) Undrained
sin
on
saturated clay soil by isotropically consolidating the soil using a cell pressure
SOLUTION
C*
237
24.3'16)
&) Determine the undrained friction angle
the drained friction angle
C=425
Shear Strength
of Soil
c) Determine
a)
-
Shear stress on the failure plane
t=Rsin(90_S)
r ='175 sin (90 t = 159.5 kPa
Maximum principal stress, 01 =C+R
Maximum principal stress, ol * 9,100 + 5,047.76
Maximum principal stiess, o1 = 14,'1.47.75 psf
Chapter 07
.
sin $ =
R/nA
R=230-150
R=80kPa
q = 24.316'
(
159$lO = cen
23g
:i;51"'07
-
shea' strensth
Fundamen
Geotechnical
ls
of
chapter 07
*
Shear
r,':ig$
hnical Engineering
sin $,= 367230
+, ='i0.35'4'
ZEg
UTION
b) Undrained shear strength:
tr=R
\,r
t, = 80 kPa
c) Drained
a
friction angle:
'6
R:=
11S
\a,"
\€-r 20 =. 1360
440,. \ "'
tC
,i-.
o':=150-54
40
-rl
*
-.--.-.--o-Y-
l'
155
o's = 96 kPa
o'r=310-54
-+
oi
270
o'r = 256 kPa
0=68o
1i
In right triangle APC:
0=90o-44o
0
'46o
.R
SlnO=
,AC
-
s6+2s6
*-T= t/o
R =176
sin 46o
256
AC *
-96
R=BO
-
115'
1.59.87
In right triangle AOQ:
sin $,i = 80/176
0A=AC-155
0o = 27.03o
OA = 4.869
.c
tand'oA
c = 4,869
PROBLEM 07.12
In a triaxial test for a soil sample, when the principal stresses are 270 kpa e
40 kPa, the soil fails along a plane making an angle of 6!o with the horizon
What is the cohesion of the soil in kPa?
:
AC
c
tan 45o
= 5.04 kPa
IRoBLEM 07.13'
a triaxial test. The critical state
friction angle of the soil is 28o and the normal effective stress at failure is 200
A cohesionless soil sample is subjected to
kl'a.
n) Determine the critical state shear stress
t
)
Determine the plunger stress
d) Determine thg cell pressure
I
I
I
/I
24O
Chapter 07
of Soit
-
Shear Strength
Funda
Geotechnical
Fundaqentals of
Engineering
c)
SOLUTION
chapter 07
-
shear
t.:ig:? Z4l
Cell pressure, o: = C - R
Cell pressure, oz = 256.54 - 120.44
Cell pressure, o: = L36.1 kPa
PROBLEM 07. I 4
An unconfined compression test was carried out on a saturated clay sample,
Thc maximum load the clay sustained was 127 N and the vertical
displacement is 0.8 mm. The siee of the sample was 38 mm diameter x 76 mm
lqrrlg.
o) Calculate the axial strain of the soil sample
tl) Calculate the major principal stress at failure
c) Calculate the undrained shear strength of the soil sample
*#
i
I
a)
b)
Critical state shear stress, ta = o4 tan Q
Critical state shear stress, ta = 200 tan 28o
Critical state shear stress, ta = 106.34 kPa
!OLUTION
a\
Plunger stress
oA
cosd'oA
oA
_
b)
cos 28o
OA = 226.514
R
xA
tan6=
,OA = da
1,4634
226.51"4
200
R = 120.44 kPa
^oA
cosg
L=
H
-0'8
Axialstrain.e=
-76
Axial strain,
200
R
.AH
s=
Axial strain.
226.514
cos 28o
C = 256.54kPa
-
Plungerstress,or=C+R
Plunger stress, or = 256.54 +'120.44
Plunger stress, or = 375.98 kPa
s = 0,010525 mm/rnm
Sample area (initial), A, =
t
(0.0e4;r
Saniple area (initial), A, = [.QQ]1J{ 6z
Sample area at failure, A
'
=
A
="o
1':€--..-.
o
Sample area at failure,
o = r 9or'r1'*,
L - 0.010526
Sar-nple area at failure,
A = 0,001146
*'
-
^-
0.001146
Major principal stress at faiJure, or = 110,814 Pa
Major principal stress at failure, {or)u = 110.814 kPa
242
Chapter 07
of soil
Note:
-
og
Shear Strength
F
Geotechnical
mentals of
Chapter 08
-
Lateral Earth
Pressure
icalEngineering
243
:0
ax
Lateral Earth
Pressure
h
*
It
From the Mohr's circle shown, tlr = 55.4 kPa
Chapter 04 was focused on the vertical stress caused by the weight of soil and
ttructures above the soil surface. This chapter will focus on the lateral stress
excrted by the soil mass on a structure, such as retaining walls, basement
wrrlls, and bulkheads.
Active. earth pressgre coefficient, Ko - the ratio between the lateral and
vertical principal effective strebses when an earth retaining structure moves
away (by a small amount) from a retained soil.
Passive earth pressure coefficient, Kp - the ratio between the lateral and
vertical principal effective stresses, when an earth retaining structure is
forced against a soil mass.
Figure 08.1
iDS "*j
,
244
8,I
Chaoter 08
pressure
-
Lateral Earth
F
GeotechnicalE
Chapter 08
tals of
* Lateral Earth
.
technical Engineering
Pressure
245
th reference to Figure 08.1:
EARTH PRESSUREATREST
If
pn=
a retaining structure does not move either to the right or to the
initial position, the soil mass will be in a state of elastic equilibrium,
the horizontal strain is zero. The ratio of the horizontal stress to the
stress is called the cofficient of earth pressure at rest, K*
F
KoY
-lzK"y
H
Eq, 8,6
H2
Eq, 8.7
RANKINES THEORY
llr
k=:_A_=1_sin0
where $ is the drained friction angle.
For dense sand backfill:
K"=(1-sinQ).
[#-r)rt
Active
where
= actual compacted dry nit weight of the sand behind the wall
|dmrn = dry nit weight of the sand in the loosest state
yd
case
Passive case
Ftgure 08.2
- Vertical
Coefficient of activepressure:
For fine-grained normally consolidated soils:
K,=
Ko= 4.44 + 0.42(pI%/tA})
face and inclined backfill
,
.o,i-..f,of,lIJo"
Eq. 8.8
.or,*fiJi-*J4
.
Coefficient of passive pressure:
For overconsolidated clays:
&
loverconsolidi ted)
OCR
=
Ke1no.-ullyconsolidated)
m
Preconsolidation pressure
=c
Presenteffectlve overburden pressure
Kr=
.ori*.',f,oJ ,-.*t[.ori - 16Jll.oJ 6
COS
T
Eq. 8.9
246
Chapter 08
-
Lateral Earth
Fundamentals
Pressure
f
Chapter 0B
rrrrdamentals of
-
Laterai Earth
(ir:r:technical Engineering
GeotechnicalEn
4.2.1 RAN|(NE',S THEORY (FOR HORTZONTAL BACKFILLI
Pressure
247
8.3 COULOMB'S THEORY
Normal to
tr
tl{
il
ri
Figure 08.3
- Vertical
face and horizontal backfill
T
,il,:
i,13
- Wall
sloping face (active case)
When i = 0, Eq. 8.8 yields
,&
Kr- 1-sing
n
ft
t
Figure 08.4
Coefficient of active pressure:
1+sin$
r4
.
Coefficient of passive pressure:
When i =
Q,
Eq. 8.9 yields
Kp=
sin$
sin0
-
1+
1.
Eq.8,1
Figure 08.5
- Wall
sloping face (passive case)
llecause of frictional iesistance to sliding at the face of the wall, F, and Fn is
irrclined at an angle of 6 with the normal to the wall, where 6 is the angle of
ruall friction
\
248
Chaoter 08
pressure
8.3.I ACTIVE
-
Lateral Earth
GeotechnicalEng
PRESS URE
COEFFICIENT
cos2
Ko*
0o
(O
-
p)
+ tan,o
=
The effect of wall friction on
&
is small, and is usually neglected.. For 6 =
K,-
"or2
16
cor2
Kr=
,ft
tt,*
chapter oB
- LatTire:il::
- p;
T'hc
14
+ p;
Eq..8.17
inclination of the slip plane to the horizontal is:
tan6,=
'_
_:.6i" d."r s
-tand
Wheni=0and6=0:
llor frictionless wall with vertical back face supporting granulal soil backfill
with horizontal surface (i.e. $ = 0o, i = Qe and p = 0'), Eq. 8.17 yields
Note that this is the same
cor2
16
-
B;
Eq. B.18
cosS./sin($+6)
1+sin0
..
k=_
- 1-sin0
K,=
z+9
3.3.2 PASSIVE PRESSURE COEFF.ICIENT
The inclination of the slip plane to the horizontal is:
tan
Fundamentals of
6eotechnical Engineering
Eq.8.19
with Rankine's value given by Eq. 8.11.
'l'he critical value of 0 is:
.or'ofr.H)'
0=0.r=45r+O/2
Eq. 8.20
For wall with vertical back face supporting granulal soil
horizontal surface (i.e. i = 0o.and 0 = 0"), Eq. 8.14 yields
-.K,=
-sin 0
1+sin0
1
-
Note that this is the same with Rankine's value given by Eq. 8.10.
8,4 RETAINING WALLS
A retaining wall may be defined as a structure whose primary purpose is to
prevent lat"eral morement of earth or some other mateiial. Fo. some special
cases, as in basement walls or bridge abutments, a retaining wall may also
have a function of supporting vertica loads.
:
25O
Chaoter 0B
pressure
-
Lateral Earth
F
GeotechnicalE
ls
of
hnical Engineering
Chapter 08
-
Lateral Earth
Pressure
251
shown in Figure 08'6 (a), is usually built of-plain
conciete. Thislype of wall depends only on its own weight for stability,
and hence, its height is subject to some definite practical limits'
G*"tty t"trining wall,
8.4.I ryPES OF RETAINING WALLS
Semi-gravity wall is in essence a gravity wall that has been given a wider
base"(a toe or heel or both) to increase its stability. some reinforcement
is usually necessary for this type of wall'
T-shaped wall as shown in Figure 08'6 (b) is perhaps the most cgmlol
.ur-,iilu,r", wEll. For this type. of wall, the weight of the earth in the back
of,the stem (the backfiil) contributes to its stability'
(a) Gravity retaining wall
(b) T-shaped retaining wall
L-shaped wall as shown in Figure 08'6 (c) is frequently used^when
property line restrictions forbid the use of a T-shaped wall' On the
oitr". t-ra"a, when it is not feasibie (due to construction limitation) to
excavate for a heel, a rettersed L-shape may serve the need'
Counterfort retaining wall, as shown in Figure 08'6 (d) consists of three
main components: base, stem, and intermittent vertical ribes called
counterforis, which tie the base and the stem together. These ribs,
which acts as tension ties, transform the stem and heel into continuous
and at the
slabs supported on three sides - at two adjacent counterforts
base of t-hl stem.
(d) Counterfort retaining wall
(c) L-shaped retaining wall
BridgeabutmentasshowninFigure0s.6(e)isaretainingwall,generally
shirt and typically accompanied by wing walls'
Deck
(e) Counterfort retaining wall
Figure 08.6
of the
Buttressed wall is constructed by placing the ribs on the front face
compression.
stem where they act in
- Types,of retaining
walls
_tr_r
z>z
Chapter 08
-
Lateral Earth
Chapter 0B
tals of
Pressure
GeotechnicalE
8.5 ACTIVE PRESSURE ON WALL
-
Lateral Earth
Pressure
nical Engineering
2s3
Soil:
pt= KotfiHr
Ft=lzptHt
Surcharge, q (kPa)
P+
= K,zfi Ht
Fn= ptHz
ps
=
Knzf azHz
= 1/z ps Hz
F5
Water:
fu= !''
Hz
Fo=1/zftHz
Total active force, F,
9q
Cohesion
Figure 08.7
-
Surcharge
Total active moment, Mo* Ft Yt +
*, = 1-tll1 t
1+sing
Fz^Az+ Ft
-
Fs
W+
t ct J/,r' t
cz
Fa
yt+
Fs
ls+
!,2
Soil
Commonly assumed active pressure on retaining walls
With reference to Figure
- \ + Fs+ FriFr+Fs+Fo-Fcr-Fcz
8,6 PASSIVE PRESSURE ON WALL
08.7:
(Rankine or Coulomb)
Cohesion:
Pd =
2q lKal
^
Fa=PaxHt
p,z=
,, /1
2c2r$j
Fs2=
p4x
---n"{"-
,,,,/f r, I ll
',Til!{n
P:
H2
I{L]J'
9z
Water
Soil
Pr
Cohesion
Surcharge:
Pr*kaq
F1
pz=
= p1x
Figure 08.8
*
Commonly assumed passive pressure on retaining walls
H1
K,zQ
Ft=DtxHt
With reference to Figure 08'8:
KP
_ 1+sin$
1*cos$
(Rankine or Coulomb)
Fr
lr
254
Chapter 0B
*
Lateral Earth
F
Pressure
Geotechnical
Chapter 0B
Fundamentals of
6eotechnical Engineering
-
Lateral Earth
Pressure
255
Factor of safety against sliding:
Cohesion:
h=2c,[$
Ft=fiH
'
rrs
Resistine forces
Active forces
Eq, 8.21
Soil:
P2= KpYbH
For granular backfill, FS' > 1.5
For cohesive backfill, FS' > 2.0
Ft=1/zatH
Water:
Factor of safety against overturning about the toe:
ft=Y,'H
F, =
1/2p3
H
Total passive resistance, Fp= Fr +
Total passive moment, Mp = Ft
8.7
-
rro
Fz + Fs
!t + Fzlz+ hlt
-
Stabilizine moments
Overturning moments
Eq. 8.22
For granular backfill, FS' > 1.5
For cohesive backfill, FS. > 2.0
FACTORS OF SAFETY
The structural elements of the wall should be
following safety factors are realized:
so
proportioned that
'rhe horizontal components of the lateral forces tends to force the wall to slide
resisting force is povided by, fhe horizontal forces
co*iosed of friction and adhlsion, and by passive resistance of soil in- front
*"ir, The passive resistance is not to be counted on if there is a chance
the life of
"r,f,'a
that the soil in front of the wall may be eroded or excavated during,
the wall.
.r""g iir b*", Tie
The force F at the base of the wall consist of the frictio-n and cohesion.
given by:
F = pN +
"'1.'T
Figure 08.9
-
l
Forces acting on wall
c1B
It is
Eq' B'23
base
where N is the normal reactiohi p is the coefficient' bf friction, cr, is-the
of p
values
assumed
cohesion, and B is the base width of wall. Commonly
and ca are as follows:
(213)tan0
0.5c S cu 30.75c
tanQ>p>
,
Eq'8'24
Eq' 8'?5'
2s6
oC[]?J::"
-
LaterarEarth
Fundamentals of
Geotechnical Engineering
GeotechnicalE
8.8 pREssuRE DrsrRrBUTroN AT snsr
oi wALL
chapter
ot '
*"n!:lJ[
257
Whene> 8/6
The actual bearing pressure on the base of the wall is a cornbination
of no
forces and the effects of moments.
Ro
= LFo
Eq.
RrV=RM_Olvl
e=
B-2 - x
Eq.
Eq.
where:
I
RM = righting or stabilizing moments
OM = overturning moments
Figure 08.11
Note that in computing RM and Rr, the passive resistance is not to
be cou
orr if there,is.a chance that_the soil in front of the wall may be
erode
excavated during the life of the wall.
-
e>B/6
Stress distribution at base of wall when e > 8/6
Considering L m length of wall:
'
Whene38/6
Bl2
8.9 IjTERAL
q*t*
=
2R,,
-;f
PRESSURE ON RETAINING WALLS DUE TO POINT.LOAD
SURCHARGE
Figure 08.10
-
Stress distribution at base of wall when e < 8/6
Considering I,m length of wall:
4^u* = min
R,
(',
*6'\
s l'-
s
Eq. 8.30
)
Flgure 08.12
-
Stress on wall caused by a point load
258 F[l?j::rr
-
LaterarEarth
F
Geotechnical
The lateral stress on the wall induced by a point-load surcharge is given
For m > 0.4
chapter
Fundamentals of
6eotechnical Engineering
o, -
t"i1i,;:lJi:
ZSq
induced by a line-load surcharge is given by:
ffiil
For m> 0.4
Or
4a
- 7.77Q m'n'
H2 (m2 +n21?
*2n
6,' = -J- ---------;---.
nH (m'+n')"
For m < 0.4
Eq.8.33
Fol m 30.4
ox.
=
0.28Q
----;H'
n2
(0.16 +
6r=
n')'
0.2a3q
H
-.--.-;-;.
whgre Q is the point load (kN or lbs), H is the height of wall (m or feet),
where 4 is the line load
a,1rtw
Eq. 8.34
(kN/m,or lbs/ft), H is the height of wall (m or
feet),
is the stress (kPa or psf)
Simpson's one-third rule.
'Ihe force F per unit length of wal1 gau,sef by the strip load can be obtained bY
traPezoidal rule or
approximating the area of the shaded portion using
Simpson's one-third rule.
8.IO LATERAL
8.1
The force F per unit length of wall caused by the point load can be obtained
approximating the area
of the shaded portion using
trapezoidal .rule
PRESSURE ON RETAINING WALLS DUE TO LINE-LOAD
SURCHARGE
Figure 08,13
l
LATERAL PREssUftE
oN RETAINING wALLs DUE To STRIP:LoAD
SURCHARGE
- Stress on wall caused
by a line load
Figure 08.14
- Stress
on'wall caused by a strip load
260
F[:,Pff:08
-
LaterarEarth
Fundamentals
Geotechnical Engineeri
The lateral stress on the wall induced by a strip-road surcharge is given by:
or =
2a
--r
(B -
sin
B
cos 2cr)
Eq. 8.3
Chapter 08
approximating the area
of the shaded portion using
trapezoidal rule
Simpson's one-third rule, or by integration of o, with iimits from 0 to H.
Lateral
Earth
26l
ILLUSTRATIVE PROBLEMS
PROBLEM 08. I
A 5-m tall cantilever
The force F per unit length of wall caused by the strip load can be obtained
-
Ceotechnical Engineering
retaining wall retains soil having the following
properties:
Cohesion, c = 0
Unit weight = 19.8 kN/m3
Angle of internal friction, 0 =
30o
The ground surface behind the walt is inclined at a slope of 3 horizontal to 1
vertical, and the wall has moved sufficiently to develop the active condition,
Use Rankine's Theory and consider 1 m length of wall.
a) Determine the coefficient of active pressure
b) Determine the total active force
c) Determine the overturning mornent on the wall
SOLUTION
a) Coefficient of active pressure
Inclined Backfill (Rankine Theory)
co.2 i -.os2 4
Ko-
cos
I
cosi+J.or2i*.or24
tani = 1/3
i=
1.8,43,5o
.. = cos18.435',(
cos
K,=
A,42
1
Jcos2 18.435"
-.or2
30o
8.435' + Jcos2 1 8.435" - .or2 30o
rnq18.435o
262 F[l!ff:"
-
LaterarEarth
Fundamen
Geotechnical Enoi
b) Active pre$sure
Fo
Fundamentals of
Geotechniqal Eng ineering
Chapter 08
-
Lateral Earth
Pressure
b&c)
= t/2y lQ ll2
F, = lz(19.8) (0.42) (5),
F, = 103.95 kN
'
c)
Overturning moment
OM,= F, cos I (5/3)
OM = 1"03.95 cos 18.435" (5/3)
OM = 164.36 kN-m
PROBLEM 08.2
A
3 m high vertical retaining wall is shown in Figure 08.15
a) What is the effective vertical stress at the base of the wall?
b) What is the total active force acting on the wall?
c) What is the total active moment on the wall?
F*Pr---r i .t
l+-pz --+leP t+l
i
3.-*p:*-.--+i
Soil
1-sin$
,, -- :-----;---:
nal
l+slnQ
,, - 7-sin3Oo
-
lliil36i
^or
K,t*1f3
,, *
1-
sin 26o
1 + sin 26o
Koz= 0,39
pr*
Kar! Hr
n= $/3)(15.\(2)
pt - 1A.467 kPa
Ft
\=lt$l'a6h?)
Fr * 10.467 kN
y+ 1. + 2/3
h*5/3
Figure 08.15
soLUTtoi{
a)
Effective vertical stress at the bottom o.f wall:
pe =(18.9 -9.81X1) +15.7(2)
Pt = 4A.49 kPa
* 1/2fu Ht
pz-
Kaz'T
Ht
:
p, = (0.39)(15.7)(2)
pz-12.26kPa
Fz= pzHz
Fz* 12.26(1)
Fu
= 12.26
kN
az= 0.5
!pr-l
263
264 i[:!j::08
p'z=
Laterar Earth
Fundamentals
GeotechnicalE
Ko2y62H2
(0.39)(18.9 - 9.81)(1)
p'z = 3.545 kPa
Ft = 1/z p'z Hz
width.
on the wall per foot width.
.
!/t = 1/3
Pt
=
SOLUTION
$)
-2c
P+='2Q)J6.u
,: To a:,1{0-.p61r.
'' | :. :t', ,.t1.. :
I lr"ili :r irllir:' r::..1
1tt= -2.498kPa
r
14
rr
-- r12
-- p4
l'l
1
Fs = -2.498(1)
Fa
-
-2.498 kN
= 0.5
y*,a,l2Qpcf
Hz
Yq
Ps=f,Hz
Ps
= 9.81(1)
ps = 9.81,
Fs =
Fs
-
Fs =
Effectlve stress
dlagram
,
kPa
1/z ps
Hz
1/z(9.81)(1)
= Tsat'nlw
ybe 720 - 62.4
ya - 57.6 pcf
^la
4.905 kN
Ys=1/3
1-sinQ
Total active pfessure, F, =Fr + Fz+ Fs+ Fa + Fs
Total active pressure, F, = 10.467 + 12.26 + 1,77 _2.49g +
4.905
Total active pressure, F, = 26.904 kN
Total active moment:
M, = F1 fi + F2lz+
24.552
u _ 7K'=
sin3So
TllliF
+ FaAq +
Fs ys
M,= 10.467(5/3) + 12.26(0.5) + 1.7fi0/q
,:=
&= 1+sin$
Ko= 0.271
fi fi
kN-m
ZOS
b) Determine the location of the resultant pressure from the ground surface
c) Determine the overturning moment caused by the active pressure acting
Fz=rze.Saile)
kN
LateratEarl
a) Calculate the total active pressure acting on the wall in pounds per foot
p'z*
Fs = 1.773
Chapter 0B _
Fundamentals of
Geotechnical Engineering
_2,4s8(O.s) + 4,905(1/g)
Part a:
Considering 1-foot length of wall:
Pressure:
9't=
PROBLEM 08.3 (CE MAY 2OO2)
A vertical retaining warl retains 26-ft deep of soil. The soir has
a dry unit
weight of 110 pcf above water table uni tzo p.r ueto*
*ui".
,rur".
The
ground water table is 10 feet berow the ground surface.
rr-r"
internal
- --: u.rgl"'oi
'
-^'o'friction of the soil is 35o.
T
hHr
p1= 710$.271)(70)
Pt = 298.1' Psf
P2=ftKoHz
P2= 57.6(0.271)(1'6)
Pz= 249.75 psf
Pore
stress
water Resultant
diagram force
-
zbb F[]!f:"
Ps
LaterarEarth
GeotechnicalE
-
Total pressure:
Fr = 1/z pt Ht
!,
= lz(298.1)(1,0)
Fr = 1490.5 lbs
Fz=hHz
Ft=VzpzHz
Ft = 1/z(249.75)(16)
F3 = 1998 lbs
SOLUTION
1-sin$
a) Kr= 7---1+sin$
Fs=1/zptHz
-
Fs
= 7987,21bs
y2(ss8.4)(16)
,, - 1- sin33o
' j_:r-nin rr"
33o
Total force, F= Fr + Fz+ Ft+ Ft
F = 1490.5 + 4769.6 + 1998 + Tg87.z
K,=
y=pg
Location of .F:
y = (2/3)(1.0)
y = 1600(9'81)
Y = 15,696 Nlm3
yt = 20/3 ft,
'
lz='10+16/2
Part
15.696o*/:l
z=26-18.6
Overturning moment = F z = 16,245,3 (7.4)
Overturni ng momen t = 120,2'15,22 ft-lb
yr,:
G
ya7;-T"'
'
l+e
/?) | !76e 6(18) + ress12s.6.r,
+ 7987.2(20.664
z=7.4ft
5.696) (8)'?
Solving for G and
fI= \Y+Fzaz+Fsas+Fatu
1.6,24s.3 y 14e_0:l(29
c
(1
b&c)
Taking moment of forces about the top:
ft
o
F, = /z(0.2948)
F, = 148'1kN
lt = As = 10 + (2/ 3)(1.6)
lt = !+ = 20.667 ft.
Y = 18.6
Y
|
yz= 18 ft.
.
_r
A'2948
Fo- 1/zK,Y H2
F = 1,6,245,?lbs
=
Lateral
Earth
26l
soil is dry.
&) Determine the thrust on the wall in kN per lineal meter if owing to
to a 1evel 3.5 m below the surface.
inadequate drainage, it is waterlogged
"b-ase
of th" wall ,where the thrust acts
.1 n"t".ri i.e ttre nefftrt above fhe
during the waterlogged condition.
F: * 298.1(16)
Fz= 4769.6\bs
F4
-
'
PRoBLEM 08.4 {CE NOVEMBER 2003, MAY 2OA4l
,having a dry density of
soil
cohesionless
a
supports
A retaining wall B m high
The
1600 kg/ri3, angle ,of shearing resistance is 33o and void ratio of 0,68.
wall
Neglect
the:wall.
of
top
the
with
level
ftoriro"iuftnd
soil
surface of the
ir
friction and use Rankine's formula for active pressure of a cohesionless soil'
a) Determine the total earth thrust on the wall in kN per Jinear meter if the
l'u Hz
pt = 62.aQ.6)
Pt = 998,4 psf
.
Chapter 0B
Fundamentals of
Geotecfrnical En gineering
1.5.6e6=
G = 2.688
:'
ff*
tr.rrl
G*7
lo= a-1"'
1,+e'
2.688 - 1 (9.81)
=
Yb
1+ 0.58
y6= 9.857 kN/m3
268 F[:iff:"
-
LaterarEarth
chapter
Fundamentals of
GeotechnicalEng
En
6eotechrrical
ot -
gineering
,r,1i,":lJi: 269
Total thrust, Fr = Ft + F2 + F3 + F4
Total thrust, Fr = 28.34 + 72.88 + 29.421+ 99.33
Total thrust, Fr = 229,97 kN
t
Location:
F7x
! =ZFy
i = 28,34(5.667)
22s.s7
i
+ 7 2.88 (2.25) +
29
.421 (1 .5) + 99. 33 (1 s)
.
=2'251m
PROBLEM 08.5
A frictionless retaining wall is shown in Figure 08.16. Consider 1 rn length of
wall.
a) .Determine the total horizontal passive Pressure on the backfilt at the
bottom of the wall
b) Determine the passive resistance on the backfill
cj Determine thelocation of the resultant passive force from the bottom of
o
Soil
Water
the wall
pr=KryH
p t = 0.2948 (15.696X3.5)
Pt = 16.195 kPa
F r = 1/z(16.1.95X3.5X1)
Fr = 28.34 kN
Q
= 10 kN/m2
lt=4.5+3.5/,3
Fr=
lt
= 5.667 m
1.6.1.95(4.5)(1)
Fz = 72.88 kN
y2 = 1/z(4:5)
yz= 2.25 m
Pz= KnytH
pz
= 0.2948(9.857) (4.5)
Pz= 13,076kPa
F3 = i/z(13.076X4.5X1)
FE
= 29.421 kN
lz = 4.5/3
F+
= Vzy* H2
/:=1:5m
h
Fa
'
= lz(9.81)(4.5)z
= 99.33 kN
la= 4.5/3
!a*1.5m
Figure 08.16
27a Ffffff:"
-
LaterarEarth
F
Geotechnical
Chapter 08
Fundamentals of
Geotechnical Engineering
SOLUTION
Location of
,
-
Lateral Earth
Pressure
F:
F *y=h(2)+Fr(4/3)
512y = 20a.8Q) + 307.2(4 / 3)
y =7.6m
PROBLEM 08.6
a)
Deterr,nine the total active pressure before tensile crack occurs:
&) Determine the total active pressure after tensile crack occurS:
q
r
Pressure diagram
,,
].+sin25
l\-=
'
Kp
-7
-
-sin26o
2.5e
The passive pressure at any point is p KryH +
Ko q + 2c
=
AtH
=
{G
O:
fi=2.56(70)+2(B).m
- 51.2 kPa
Pr
@H=4m
pz= 2.56('LS)(4) + 2.s6(10) + 2@) JTR
pz= 204.8kPa
Pz ^
Figure 08.17
Pt = 153.6 kPa
h = s1.2(4)(1)
Fr = 204.8 kN
= 1/z(L33.6)(4)(1)
Fz = 307.2 kN
Fz
F=F,+p,
F=512kN
soruTloN
,, - 1-sin$
o'-
k*
l;ffi
0.39
p,=KoyHaKo4-2rJ\
15 kN/m2
271
Fundamentals of
Geotechnical Engineering
orl
-^'
Chapter 08
-
Later:al Earth
Pressure
lz(6'64x1'03x1)
r,
='t Qt.gn(q.97\(t)'
F,=76kN
Active pressure after tensile crack occurs:
F, = 1/z(31.97)(4'97X1)
F, = 79'45 kN
PROBLEM 08.7
in Figure 08'18'
Analyze the stability of the frictionless wall shown
Q.
= 20
kPa
Figure 08.18
SOLUTION
,'
.
With reference to Figure 08'19:wall'
Note:9 = distanJe oFitt" fott" from the bottom of
Active Pressure
Pr=QsxKa
,
Kt,
=
1
- sin 25o
;;;29
K'r = 0'40586
273
ztz
F[13i::"
-
LaterarEarth
GeotechnicalEng
At H = 0 (at the ground surface)
po
= 0 + 0.3e(1s)
- Z(10t JdB
p, = -6.64 kN/r11z
AtH = 6:
p, = 0.3e(1 6.5)(6)
P,=
31.97
* o.:s(rs)
kNy'pz
- 2(10)
.fi3e
,
y=6
6.64 6.64 + 3't.97
y=
1..03
m
5-Y=4.97m
Depth of tensile crack = 1.03 m
Active pressure before tensile crack:
Fo
= l/zK^yLp + K,qH
- Zc,tQ
ru
F, = Yz(0.3e)(16.s)(6)r+ 0.3e(1s)(6)
F,=76kN
- 2(10) Jo39
101
-
274 F[:rj::"
LaterarEarth
Fundamentals
Geotechnical Engi
Chapter 08
Fundamentals of
Geotechnical Engineerin g
g = 20(0.40586)
p = 8.L2kPa
*
Lateral Earth
Pressure
Pz= 201
pz= 6.67 kPa
Fr= 6.67(4)(1)
\*p1(2)(1)
rr = 8.12(2X1)
Fr = 16.24 kN
kN
Fz = 26.68
!z=2m
th=4+1=5m
P+
= \sutt
p4 =
Kaz
Ht
1e(+)(2)
pn= 12.67 kPa
Ft= 12.67(4)(1\
Fa
= 50'68 kN
lt=2m
:
= 1tz Koz Ht
ps= (20,9.81X+X4)
Ps
Ps
pore
$'ater ' fdir
stress
stress
Passlve Pressure
Figure.08.19
Ps
=
-
= 13'59 kPa
Fs = %(13.sex4x1)
Fs = 27'18 kN
ys=tm
'Pz
P,
Surcharge
stress
Active Presaurestress
SoiPr
stress
Forces acting on the wall
Fe
pore water
Pt=!,Hz
= 9.81.(4)
po= 39.24kPa
po
* lz(39.2a)$)(1)
Fe=7848kN
Ft
Hl
Y*t1K;nt
yu=tm
pt = 1,9(0.40586)(2)
p3 = 15.42
kN
Total active force:
h=vz(1s.a2)Q)$)
F: = 15.42 kN
h=4+2/3
/e=
pz=
tm
K"=
Active morrient
lvfo=
QsKoz
,,
K,r=
Fo=Ft+Fz+F:+Fa+Fs+Fo
t6,zq1 %.Ae + 15.42+ 50"68 + 27.18 + 78.48
F, = 214.68 kN
i,=
1,
-
1+
t
-
sin 30o
M,
-
+ FzAt + FsW + FtY
16-.24(5) + 2.6.68(2) + 15.42(
Frlr
! l?y'*
-p^uV'
+ 50'68(2)
+ 27.18(\) + 78.a8({ )
sin 30'
M,
= 448.76
kN-m
+)
275
276 F[l?j::'n
-
LaterarEarth
Fundamentals
GeotechnicalEn
..
L+sin30o
1
-
sin 30o
Kpz=3
Pz=(20-9.81X3X4)
pz = 122.28kPa
Fr = 1/z(122.2$$)(1)
Fz = 244.56 kN
az=
*
bj
cj
m
Dete.mine the factor of safety against overturning'
Surcharge
Pe=Po=39,24kPa
Fe
= Fo = 78.48
ye=
t
I---
kN
m
I
Total passive force:
I
I
Fp*F7+fs
Fp
= 323.0+
ktl
I
20ft
Passive momeflts
Mp=Ftyz+Fels
Mp = 430.72 kN-m
Ratio of
-
fo..". -
tP
Fa
l(atro ott,torces =
323.04
214.68
Ratio of forces = 1.5 > 1 (OK)
Ratio of moments =
Ratio of momente
.
YL
SOLUTION
Md
430.72
448.76
Ratio of moments = 0.96
Lateral Earth
Pressure
277
The fill behind the
A solid concrete retaining wall is shown in Figure 08.20.
pressure may be assumed
;rll;;;;it *uight of"110 pcf whoseperactivJsoil
The passive pressure. may b1
;;;;il i"-, n.,ii pressure ft eo ptf foot.
psf per foot' rhq live load
;1il; ;;;;;1";i'to a fluid p'"i"'" ofto300an-additionat,?t
feet of fill.
ffih;;;";"r.,i"a ir,..*rll is equivalent
?
"
wall.
of
length
ft.
1
*.igr.i"r.oi..",u =isp p.f. consider
;;ff;*
'wall in lbs,
oi- Ou,"r-ine ihe total active p."rtui" acting on the
in ft-lb'
toe
the
Determine the overturning moment about
"y62 Kp2 H2
x ^= _
Chapter 08
PROBLEM 08.8ICE MAY 2OO2l
Passive Pressure
pt =
Fundamentals of
Geotechnical Engineering
(
1 (unstable)
7
= 30(2)
Pr = 60 Psf
pz= 30(20)
p2= 500 pst
Active Pressure:
\=
Pr(20)
Fl = (60)(20)
Fr = 1200 lbs
278 Flliff:"
j
LaterarEarth
chapter
mentals of
nical Engineertng
Fundamentals
Geotechnical Enqineeri
ot -
*",,i!::J,,: Zlc)
LVj = Ysoit Vs.il
ttl3=110(20*6*1)
* 13200lbs
W:
x1
=!+1 =5ft
=6ft'
xt=L2-'h(6)=9fr'
RM =Wt xt + Wzxt*
x2=1/2(72)
Wt
RM = s400(5) + 3600(6) + 13200(e)
RM = 167,4,00 ft-lb
Factor of safety against overturning =
Factor of safety against overturning =
Pr
2ft
(20)
1/z
pz
Fz= Vz(600)(20)
Fz =
Fz
W
^^^
ffi
Factor of safety against overturning = 3'22
Oz
Active pressure
diagram,
2O0ll
located 1.2 m
ties for the anchored bulkhead shown in Figure 08.21 are
end of the
The
center'.
on
m
5
iro- an" top of the sheet piling and are spaced'
earth
The,
active
shown.
as
iakecl
;i;; ; ;.;ed to two ancho"red piles
*
meter
per
5
kN./m2
of
pressure
fluid
to
a
fr.rru." may be assumed equivalent m?y
assumed equivalenl to a fluid
.rrrd the maximum passive p."r,,to
.be
the compressive force in the
Caiculate
ineter.
Oi f.Nl*z per
;;;r*.
PROBLEM 08.9 (CE MAY
,l,he
,
= 6000 lbs
Total active pressure, F = h + Fz
Total active pressure, F = 7200lbs
Overturning moment:
y=20/2
Yr = 10
ItM
"i
anchor pile.
ft.
lz * 20/3 ft,
OM = Fr yt + Fzlz
OM = 1200(10) + 6000(2013)
OM = 52,000 ft-lb
Righting moment:
'
,
*yrV1
Wt=150(18x2x1)
W1
Anchor piles
Wr = 5400 tbs
Wz='., Vz
Wz=150(72x2x1)
Wz= 3600lbs
Figure 08.21
Laterar Earth
28o' F[:!ff:o' -
F
Geotechnical
SOLUTION
T.--r
Fundamentals of
Geotechnical Engineering
Chapter 08
*
Lateral Earth
Pressure
281
From the diagram shown:.
0r = tan'' (4/i)
il=
_
1.2m
,to
*T-----T--
tt
.it
rml
75.96o
= tan.l (5/1)
0z= 78.69"
0z
lIF,
=
o1
Fr sin 0r = Fz sin 0z
Fz = 0.989 Fr
lIFs = 0l
Ftcos0r+Fzcos0:=T
Fr cos75.96'+ 0.989 Fr cos 78'69o = 134'3
Fr = 307.6
F, = 112(27)g.a)g)
f, = 364.5 kN
lIMe
=
PROBLEM 08. I O
; r;"";i"*"11
is a shown in
depth d for stability' in
a) Determine the minimum ,"rlrre of embedment
61
FP(3.8)
Fp
kN (comPression)
meters.
tension in the tie rod Per
b) Using the minimum value of d, determine the
= 364.27(2.4)
= 230.2 kN
meter length of Pile.
Check for p,:
vz(p,) (1,.2) (s)
Po
= 230.2
= 76.73kPa
,, Maximum
p,=
[IFa = 0]
T+Fr=Fo
T=364.5-230.2
T = 134.3 kN
65(1..2)
= > 76.72 kpa (OK)
zaz F[:!ff:"
*
LaterarEarth
Fundamen
GeotechnicalEn
SOLU,TION
m
m
T
m
',,
a
Fel
Fp
d
\i
f
2d/3) + 47.O8B(1./3) * 23.544(6 + d)$ + d/21
+,,1452(36 + 12d r d1)(5't 2d/3)
pressures:
j,0fl.64 d2 + 9:68 d3 + 15.696
l
*
23.544(24 + 7d + 0.5d2)
*1.452(130 +24d+ 6Od+8d2+5dt +0.67 dt)
g.n
ez
Note:,The water pressrire may not be included
in the analysis. It ,
cancel out because it appeari at the same
level on Uott, sia", of.
wall.
Active
= 1/z(29.04 d)(d) = 14'52 O'
L4.52 d2 (7 +
.:
,,
pr
283
Fra*Fo'r6-Po2c+Frtf
,
7l
F:
Pressure
a=7 +2d/3
<<;-.
i
Lateral Earth
pt=29.04 dkPa
l:.-\
,,..iP'r
-
Passive pressures:
fi= KpybH3
p = 3(9.68)(d)
,'',L,",
lb
vo
Chapter 08
ndamentals of
6eotechnical Engi6eering
Au
+ 7A.99 d2 - 28678d
- 810.72 *
q
By trial and'error: d = 4.34? m
,j
IIFH = 0l
l=Fot+Fnz+F6-Fp
r * s; .aaa * zl.ia+$
).
h=K'YHt
+ 4.347) + 1.452136 + 12(4.347) + (4 3a7J2)
14.52(4,54712
pt = 0.3(1e.62)(4)
pt = 23.544 kPa
T = 771,77
kN
Fa=lz(23,54a)gx1)
Fn = 4Z.0gg kN
PROBLEM 08. I I
b=4/3_7
b = 1./3
F,z= 23.544(6 + d) x
c=1.*1/z(6+A1
L
the back face of a 6A line load of 75 kN/m is located ata distance of 4 m from
due to the line
wall
the
on
nigr,.*rining wall. Determine the lateral force
^load alone.
c=4+d/2
pz= Koy6H2
SOLUTION
rn- 4/6
pz=0.3(9.68)(6+d)
Pz = 2,9A4$ + d) kPa
Fa3
=
lz[2.!}ag
+
0.4
4o
Stress, o, = *
nH
m = 0.667 >
d)i$ + d) x 1
+ d2)
F,s = 1".452(36 + 12d
f =1+ (2/s)(6+
. f=5+2d/3
d)
Use Eq. 8.33
m2n
1m2
+n212
zs4 F[:ff:" - Lateral Earth
F
GeotechnicalEng
Fundamentals of
Geotechnical Engineering
Chapter 09
*
Bearing CaPacitY
of Soils
285
Chapter Oq
Bearing CapacitY
of Solls
9.I
3.3903
Depth, 2
n=zlH
0
0
0
667
0.3333
0.5000
0.6667
0.8333
1 0000
5.2868
7.6394
7.3339
5.9683
0.1
1
2
e
4
5
6
load
Foundation is that part of a structure which transmits the building
are
site
the
at
directly into the underlying soil. If the soil conditions
',(Ir
sufficientiyStronsanacap"auteofsupportingtherequired.load,tlren
shallow spreud fo6tings or mats can be used to transmit the load'
structure
Footing is a foundation consisting of a small slab for transmitting the
supporting
slabs
individual
be
can
load-to the underlying soil. Footings
or be a long
single column, o, oo*li.,ed to support two or more columns '
it
approaches
i'e''
small'
i$
L
ratio
lengtir
to
B
,t.l"p of concrete slab (width
mat'
a
or
wall,
,".o; trpporting a load bearing
depth to the
shallow foundation is one in which the ratio of the embec'lment
minimum plan dimension, which is usually the width' isD/ts <2'5'
4.5446
3 3S03
F = Area of stress diagram (using trapezoidal rule)
d
F=
i[oo
+2ot+2o2+... +2os+oe],
Embedmentdepth(D/isthedepthbelowthegroundsurfacewherethebase
where d = L m (interval of computed stress)
F
=
1
;
of the foundation rests.
soil can
Ultimate bearing caPacitY (4,) is the maximum pressure that the
t0 + 2(5.2868) + Z(7.6394) + 2(7.333e) + 2(s.9683)
support.
the soil can
Ultimate net bearing capacity (4,n) is the maximum pressure that
support above its current overburden pressure'
+ 2(4.s446) + 3.39031
F = 32.458
kN
is.the working
capacity
Allowable
-(q'\ collapse
";;"r;;;; bearing capacity or,safe bearilgsafety
of the
against
of
inut *i"ti un*ru a margin
a
usually
is
capacity
bearing
Itructtre from shear failure. The all,owable
Or:
F
= !:$8@1r) +
I
F = 32.468
ry-$e1
7.333et5.e683
(f) +
$) + z@a{e
rgle$ero (r) *
kN
DEFINITIONS
(1)
Asefe
(f)
fraction of the uitimate net bearing capacity'
removed to
overburden Pressure 4, is the pressure (effective stress) of the soil
place the footing.
Note: The using d = 0.1. m, F = 33.049 kN
of safety or safety factor (FS) is the ratio of the ultimate lret bearing
maximum
capacity to the allowable bearing capacity or to the aPPlied
Factor
28b :i;ji:,.
oe'
BearinscaPacitY
GeotechnicalEng
ver1r5{ stress. In geotechnical. engineering, rr"t*
and 5 is used to calculate the allowiblu beaiing
capacity.
,
9.2 VARIOUS
orIffiilu""r,
, __J
Isolated footing
.#i:iil Zgl
ez
immediately below and adjacent
failure have been identified:
to the footing. Generally three modes of
General Shear Failure: {continuous failure surface develops between the edge
of the footing and the ground surface, This type of failure is characterized
by heaving at'the ground surface accompanied by tiltirig of the footing, It
occurs in soil of low compressibility such as dense sand or stiff clay.
ffi ffi
L*
Bearins
9.3 BEARTNG CAPACITY ANALYSIS
1..
|M
-
Bearing capacity analysis is the method used to determine the ability of the
soil to-support the required load in a safe manner without gross distortion
resulting from objectionable settlement. The ultimate bearing capacity (4,) is
defined as that pressure causing a shear failure of the supporting soil lying
TYPES OF FOOTING ON SOIL
F:
chaprer 0e
Fundamentals of
Geotechnical Engineering
[*'-J
Wall footing
L-r,J
L-s,J
Figure 09.2
-
General shear failure
Combined footing
of the soil
Tilting
of the
ground
surfate.
at
the
occurs but only slight heave occurs
in
highly
failure
occurs
This
type
of
foundation is not expected.
2.Local Shear Failure: a condition where significant compression
compressible soil and the ultimate bearlng capacity is not well defined.
Flgure 09.1
-
Types of footing
Figure 09.3
- Local
shear failure
288
Chapter 09
ofsoils
-
Bearing Capacity
Fun
GeotechnicalE
3, Punching Shear Failure: a condition that occurs where there is relatively
compression of the soil underlying the footing with neither heaving at
ground surface nor tilting of the foundation. Large settlement is ex
without a clearly defined ultimate bearing capacity. Punching will
low compressible soil if the foundation is located at a considerable
below ground surface.
t
.Depthoffoundationislessthanorequa1toitswidth
. No sliding occurs between foundation and soil (rough Ioundation)
r Soil beneath foundation is homogeneous sqmi-infinite mass
. Mohr- Coulomb model for soil
. General shear failure mode is the governing mode (but not the only
, mode)
. No soil consolidation occurs
r Foundation is very rigid relative to the soil
r Soil above bottom of foundation has no shear stength; is only a
.
surcharge load qgainst the overturning load
Applied load is compressivb and applied vertically to the centroid of the
foundation
No applied moments Present
+ &Y, B N,,
9.5.I
GENERAL SHEAR FAILURE:
s.S.t.l LoNG
FOOTINGS
where:
'
.
'
4, = ultimate bearing capacity
ye = unit weight of soil in kPa or pcf
B = width of footing in meter or feet
c * cohesion of soil in kPa or psf
N, = factor for unit weight of soil
N. = factor of soil cohesion
N, = factor of overburden pressure
q = overburden pressure (effective stress)
K,, K;, & K, = constant
28q
I
In general, the ultimate bearing capacity of soil is given by:
K,cN,+ Krq N,
Bearing CaPacitY
of Soils
Terzaghi's bearing capacity equations are based on the following assumptions:
9.4 ULTIMATE SOIL BEARING CAPACIry
Qu=
-
9.5 TERZAGHI'S BEARING CAPACITY EOUATIONS
r
Figure 09.4 * Punching shear failure
Chapter 09
Fundamentals of
Geotechnical Engineering
Qu=,cNc+4Nq+ lzYrBN,
Eq. 9.2
9.s.t.2 5oUARE FOOTINGS
9.5.
r.3
*
+ qN{ + 0.4 Y. B lV,
Eq. 9.3
q, = 1.3 c N, + qN4 + 0.3 Y, B Ny
Eq, 9.4
4,
1.3 c N,
CTRCUI.AR FOOTINGS
Chapter 09
of Soils
290
-
Bearing Capacity
Geotechnical
where:
y, = unit weight of soil at base of footing in kpa or pcf
B = width of footing in meter or feet
c = cohesibn of soil in kPa or psf
N, = factor for unit weight of soil
N. = factor of soil cohesion
N, = factor of overburden pressure
4 = overbrrrden pressure (effective stress) at base of footing
D;= depth of footing in meter or feet
See
chapter oe
Fundamentals of
Geotechnical Engineering
- Bearins.#i:iil
Zgl
The modified bearing capacity factors N'., N'4, and N'y are calculated using the
same general equations at that
(2/3 tan$) for
for N., Na, and N, but by substituting 6 = tan't
$, and using Table 09.1 or Figure 09.10,
The values of bearing capacity factors for a local shear failure are given
Table 09.2 and Figure 09.11.
in
9.6 ALLOWABLE BEARING CAPACIWAND FACTOR OF SAFETY
Table 09.1 or Figure OS.fO for the values of N,, Ny, and N,
The allowable bearing capacity, 4,, is calculated by dividing the ultimate
bearing capacity, q,,by a Factor of safety, FS. The factor of safety is intended
to compensate for the assumptions made in developing the bearing capacity
equations, soil variability, inaccurate soil data, and uncertainties of loads,
9.5.2 LOCAL SHEAR FAILURE
For local shear failure, it may be assumed that:
9.6.I
r^-z-- 5"
tan
[
Eq.
Qall
= ftanQ
Eq.9:
P 4u
FS
Eq. 9.10
9.6.2 NET ALLO\T/ABIE BEARING CAPACIW:
9,5.2.1 LONG FOOTTNGS (STRtp FOOTTNG)
qu=
GROSSALLOWABLE BEARTNG CAPACITY:
iN',+ qN'q+ lzy,BN',
{all(neu
?
4u nete
4 unet
Eq. 9.11
J3:
Qu
-4
Eq. 9.12
9,5.2,2 SOUARE FOOTTNGS
.
9.6.3 GROSSALLOWABLS BEARTNG CAPACITYWTTH
q, = 1,3aN'. + q N'q + 0.4 y. B N'r
FS
WITH RESPECT
TO SHEAR
The gross allowable bearing capacity using a factor of safety on shear strength
of soll may be computed using the developed cohesion cl and values of N" Na,
and
9,5.2.3 CIRCULAR FOOTINGS
N,,,
derived using.the developed angle of friction
$a.
'f'
q, = 1.3e N', + q N'q + 0.3 y" B N'y
F,q.9.9
Developed cohesion, ca=
c/
F$r
Eq. 9.13
292 :|[jf'
oe
-
BearinscaPacitY
ls
GeotechnicalE
chapter 0e
of
-
Engineering
q
* y(Dl- d*)
+ ya
d*
Bearine.*i:il Zg3
.
Unit weight, yr*'fb
For example, on strip footing,
Qu
=
ca
N, +
qNtt
+
li
y,BN,
7.2 GROUNPWATEN LEVELATTHE EASE OF FOOTING'
Alternatively, if the maximum applied foundation stress, (f).u,,
is
-by
factor of safety can be computed
replacing q,by ffi)^*',
6=_J_u__;
(f,)^u*-q' a < ffi)^*
1
9.7
EFFECT OF UTATER TABLE ON BEARING CAPACITY
The unit weight of soil used in the equations for bearing
capacity are.effec
unit weights. With the rising water iable, the subsoil U".o."r
;ri;;;;
the unit weight.of submerge? soil is greatly reduced. rn"
,"ar.tio,r*"oi
unit weight results in a decrease in the;rtim;te bearing capacity
ortn" ,oir.
Figure 09.6
9.7.I
GROUNDWATER LEVELABOVE BASE OF FOOTING.
-
Groundwater at the base of footing
Overburden Pressure, q *
Unit weight,
y. =
Y
Dl
Yu
OF FOOTING'
9.7.2 GROUNDU/ATER LEVEL BELOWTHE BASE
Oveiburden Pressure, 4 = \ Dr
':l
,
When d, < B
,.
,:,.
y, = yr, (1 +
Figure 09.5
- Groundwater above
base of footing
When d.
)
d,/B) = aPProx.
B
^lral
294 :ifji:'oe
-
Bearins caPacitY
Fundamen
Geotethnical Engineer
chapter oe
Fundamentals of
Geotechnical Engineering
- Bearins.#i:iil Zqs
9.8.3 BEARING CAPACITY FACTORS:
Nq
=
etrtan0 *rr'tz (45o + O/ 2)
Eq.9,18
N.=(Nr-1)cot$
Eq. 9.19
N, = (N, - 1) tan (1.4 $)
Eq. 9.20
9.8.4 MEYERHOF',S FACTORS (shape
s,
depth d, and inclination i factorsf
9.8.+,1 SHAPE AND DEPTH FACTORS
For$=go
.:.,
Figure 09.7
-
,It,l;r.;!.
Groundwater at the base of footing
s,= 1+ g.2L
Eq. 9.21
sr=s'r=1'0
Eq. 9.22
d,=l
+
D,
O.r;
Eq. 9.23
Note: If the problem does not provide or mention about the position of
water table, use q a yDf afid qe = y.
dr= dr=7.A
Eq. 9,24
s.af + ozKp:L
Eq. 9.25
For g > 10"
9.8
MEYERHOF'S EOUATTON (General Bearing Capacity equation,
9.8.I
VERTICAL LOAD:
flu = c Nc s, d, + q
s4
Nr s, dn+ 0.5 y B Nr s, d,
= c Nc s, d, i, + q Nr
1+ OJKef'
;D'
t
d,=l+0.2r|K,
dq= dr= 1 + 0.i
9.8.2 INCLINED LOAD:
Qu
= sr =
su
drh + 05
y B Nr s, d.,r,
_Dr
l-
JI(e
Eq, 9.26
Eq,9,27
Eq, 9,29
296
Chapter 09
of Soils
-
Bearing Capacity
chapter oe _
Fundamentals of
Geotechnical Engineering
9.8.4.2 LOAD INCLINATION FACTORS
9.8.I
SHAPE FACTORS:
,
r ir
Z9l
HANSEN'S FACTORS
9.8.I.I
,,..*,.'d,*
Bearins.*i:l{
sr
..:,.ir.d
s,=O.Zi,eBfL
Eq. 9.33
sr=1+sin$Biqt/L
Eq. 9,3a
= 1 -0.4 (B
rrB)
/ (Lirt)
Eq, 9.35
For failure along base length L:
Eq. 9.29
.
i,=
'
where:
t/
cr
[r_9]
\
s,
12
Eq, 9.30,
6)
Eq. 9.36
- 0.4 ('l- irt) /
Eq. 9.38
(B i.,s)
I
9.8.1.2 BASE AND GROUND INCLINATION FACTORS:
= angle of resultant measured from vertical axis.
HANSEN',S BEARTNG CApACtTy EOUATTON (VES|C'S FACTORS'
2v
=
ii
Bq
= 8r = [1
wnere:
= -c cot 0 + (4 + c cot S) N, s, d, irb, + g.5yg7'tr, s,
v = base (of footing) inclination
B = ground inclination
drrrb, Eq. 9.32
Eq. 9.39
Eq. 9.40
,-Z.7vtan$
2P
6' r+2
In the special case of a horizontal ground surface,
vo
147"
h = .-2vtano
y-
qu
B
Eq.9.37
o'=
9.9
= 0.2 i,e I./
sr=1+sin$Lirt/B
sr = 1
d\
2)
K, = tan2 I 45"++
\,
r/
-
Bo
Eq. 9.41
1.47"
0.5 tan P]s
Eq.9.42
ChaPter 09
ofSoits
zq$
9.8.I.3 DEPTH
.
For
-
Fundamentals
Bearing CaPacitY
Geotechnical Engi
chapter 0e
Fundamentals of
Geotechnical Engineering
- Bearins.17i:;il Zqg
FACTORS:
DytB
d, = 0.4 (Dt/B)
dq
=7
*2
tan $ (1
-
Eq:
Eq. 9.
sin $)2 D7/ B
Center of shaded
area
For D1> B
Eq. 9:
d, = 0,4 arctan (Dt/B)
d.r=
|
+ 2 tan $ (1
- sin
$)2
arctan
(4/B)
-
load (one-way eccentricity)
Footinq under eccentric lo
Footing
Eq. 9
Eq. 9.
dr=1
Figure 09.9
the latger value is
The smaller value of x and y is the effective width (B') and
Using the effective
the effective length (L'), ani the effective area 'is B' x L"
width,
F:q.9.1.6 maY be rewritten as:
Load inclination factors:
Forv=0o
i. = 0.5 -o.s
,q,= [1
JT-n
1
Qu
a
Eq. 9.
Eq. 9.49
- 0.5H / (V + Ac cot $)]5
Ultimate load = qt, (B'
i, = [1 - (0.7 - vo / 450")H
/
(V + Ac cot $)]5
t)
Eq. 9.51
Eq. 9.52
by substituting B', for
The values of the shape and depth factors are computed
B andL'f'br
Forv>0o
= c Nc s, d, + q Nr s, dr-r 0.5 T B' N, q d,
L.
'
Eq. 9.50
g.ItBEARINGCAPACITYFRoMSTANDARDPENETRATIoNTEST|sPTl
where:
Allowable bearing caPacitY:
base
| =* foundalion load normal to the
footing
of
H load parallel to the base
q,= 0.41N.o, p, (kPa)
N.o, =
FOR SHALLOW FOUNDATION UNDER
ECCENTI?|CITYI
(ONE-UUAY
ECCENTRIC LOAD
9.IO UL1IMATE LOAD
Meyerhqf introduced the following procedures
to
calculate the bearing
capacity of footings under eccentric load'
The load is assumed to act at the center of the footing whose effective
dimension is x by y, as shown in Figure 09'9'
Ctrt
cwN
where:
't"N:
'
Eq, 9.53
Eq. 9.54
to drive
standard penetration number (the number of blows required
mm)
the rrrnPl"t an additional300
pressure
cN = correction factor for overburden
urY
= f*l'
^
Io".J
'' , r* 12 (Liao and whitman, 1985)
3oo :iij,i:'oe cN
=
0.77,"t,,
[#)
Bearins caPacitY
]
cN
l-2,6',,)
Fundamentals
Geotechnical
24kPa (Peck, 1e74)
o'ro = effective overburden pressure, kPa
cw = corr€ction factor if the groundwater level is
the base of the footing
Ct"+=
Table 09.1
.
- Bearins.*i:il 301
Shear Failure
Terzaghi's Bearing'Capacity Factors for General
i."SP*+
within a depth B be
1,2 _-2 + 2(D1+B)
chapter 0e
Fundamentals of
Geotechnical Engineering
0
5.70
1.00
t
6.00
1.10
't,,2,
z = depth of the ground water table
3
D7
4
5
= depth of footing *
B = width of footing
cw = 1, it the depth Jf th" grorrldwater level if beyond B from the footing
base.
p, = allowable settlement in mm
1.22
0,01
,
0.04
27.05
14.21
9.84
27
29.24
15.90
11.60
28
31.61
17.81
19.98
13.70
16.18
1.35
0.06
29
6.97
1.49
0.10
30
34.24
37.16
22.46
19.13
7.34
1.64
0.14
31
40.41
25.28
22.65
28.52
26,87
6.62
6
7.73
1.81
0.20
32
44.04
7
8.15
2.04
0.27
33
48.09
32.23
31'.94
8
8.60
2.21
0.35
34
52.64
36.50
38.04
35
41.44
45.41
2.69
0.44
0.56
57.75
36
63.53
47.16
2.98
0.69
t7
70.01
53.80
54.36
65.27
3.29
0.85
38.
77.50
61.55
78.61
9
9.09
10
9.61
:17
l'a2
13
,r
6.30
',26
0.00
10.16
10.76
11.41
2.44
3.63
1.04
39
.
95.03
85.97
70.6tr
81.27
1
140.51
15.31
14
12.11
4.02
1.26
40
95.66
til:S,
12.86
4,45
1.52
41 I |
106.81
93.85
42
119.67
'108.75
171.99
,,16
13.68
4.92
,)t7
14.60
5.45
i','t 8i
6.04
2.18
43
134.58
126.50
211.56
2.59
44,
151.95
172.28
147.74
173.28
261.60
325.34
407.11
16.56
6.70
3.07
45
20
17.69
7.44
3.64
46:
196.22
204.19
21
18.92
8.26
4.31
47
224.55
241.80
512.84
20,27
o 40
5.09
48
258.28
287.85
6.00
49
298.71
344.63
24
21.75
23.36
10.23
650.67
831.99
11.40
7.08
50
347.50
415.14
1072.80
25
25.13
12.72
8.34
tg
.ri
15,12
1.82
2L'
23
302 :l;5f',,'
Bearing Capacity
GeotechnicalEn
looo900
400
iiiii,Nr,ii;i,, .ilri;Niu.lti,:
300
28
17.13
25
iri.4:iii::
6.51
1.30
1.39
1.49
1.59
1.70
0.02
0.04
0.06
0.07
0.10
18.03
18.99
20.03
21.16
22.39
23.72
25.18
26.77
:
:;:lilrLri:
ii
ii8.
.6
,..stl
.,10
5
'
u3
o
.'
9z
(u
':,t,1
':12:,'
,ll,Oll.1
6.74
6.97
7.22
7.47
7.74
8.02
8.32
8.63
8.96
::,.t.1i/"t::ii,,,
9.31
'6{
l,,irlfllli;
o
,l,l,liF-l*i
9.67
10.06
lJ-
o_
oG
rriiiilftiir;l
i,lit;i;dlri:
o.s
r,.|2:l'r''
"""2i''
12.52
c
.C
,,,;1,9,1:l
0)
':;,, 0,'.,'
,
r2$.:,,,,
i,'.,|"4.,'r':,
,,:rl;?$ii;i
Fisure 0e.10
-
rerzashi
,",J::::Ii""].
t",-;",erar
shear Fairure
10.47
10.90
11.36
11.85
12.37
C,)
co
7.47
7.66
1.14
20
z4
2.59
r,:3i'rlll
',:1;il
;z
:N',
6.05
6.54
,27
l
{o g
r,Nlai,
15.53
16.30
0
trii"6-":i:li
;6
,,:l\lli,:;,
0.01
,1 f.,
:l5iil:i
30
26
1.07
oo90
80
70
60
50
ao
';.,:,l,fh9::
1.00
,
:,,,:2";;
7
Bearing CaPacitY
of Soils
5.74
5.90
6.10
6.30
'Q:.',
200
.t
-
Table 09.2 - Terzaghi's Modified Bearing Capacity Factors N'", N'0, and N',
for Local Shear Failure
aoo
700
ooo
5oo
'
Chapter 09
Fundamentals of
Geotechnical Engineering
13.51
14.14
14.80
1.22
1,82
1.94
2.08
2.22
2.38
2.55
2.73
2.92
3.1p
3.36
3.61
3.88
4.17
4.48
4.82
5.20
5.60
0.'13
0.16
0.20
4.24
0.30
30
,
.3i!
32
33
34
35
36
37
38
28.51
8.31
9.03
2.88
3.29
3.76
4.39
4.83
9.82
5.51
'10.69
6.32
7.22
11.67
12.75
8.35
13.97
9.41
15.32
16.85
18.56
20.50
22.70
10.90
12.75
41
30.43
32.53
34.87
37.45
0.67
0.76
0.88
1.03
.42
40.33
25.21
r,4q
1.12
ao
43.54
47.13
51.17
55.73
28.06
31.34
3s.11
39.48
1.35
1.55
1.74
1.97
2,25
47
60.91
44.54
49:30
48
49
66,80
73.55
50.46
50
81.31
65.60
59,25
71.45
85.75
0.35
0.42
0.48
0.57
ag
40
.',,44
45
57.41
14.71
17.22
19.75
22.50
26.25
30.40
36.00
41.70
303
3o4 ,tlijf'oe
-
Bearins CaPacitY
\
Fundamentals of
Geotechnical Engineering
chapter oe _
Fundamentals of
Geotechnical Engineering
Bearins.#i:iil
.too
9(,
ao
70
6(,
Table 09.3
:'g:,,
3(,
2,O
{o
-6
z
a
L
o
o
a
1.00
0.00
5.10
1.00
0:00
.2
5.63
1.20
0.01
5.63
1.20
0.01
4
6.19
1.43
0.04
6.19
1.43
,0.
6.81
1.72
0.11
6.81
1.72
,8
7.53
2.06
o.21
7.53
2.06
0.22
rQr.:
8.34
2.47
0.37
8.34
2.47
0.39
:::ijl':
9.28
2.57
0.60
9.28
2.97
0.63
0.92
10.37
'3"59
0.97
i:il14::,
7
6
5
4
3
tr
'6
2
o
o.
o
o
o)
.E
o
o
co
,t.o
o.8
4.34
1.37
4.34
1.43
'r3,10
5.26
2.00
13.10
5.26
2.08
14.33
6.40
2.37
14.83
6.40
2.95
16.88
7.82
4.07
16.88
7.82
4.13
:,t24';
19.32
9.60
5.72
19.32
9.60
5.75
,26
22.25
11.85
8.00
22.25
11.85
7.94
,iif88,
25.80
14.72
11.19
25.80
14.72
10.94
iiEo
30.14
18.40
15.67
30.14
18.40
15.07
32...
35.49
23.18
22,02
35.49
23.18
20.79
il$i.*:;i:;
42.16
29.44
31.15
42.16
29.44
28,77
50,59
37.75
44.43
50.59
37.75
40.05
61.35
48.S3
64.08
61.35
48.93
56.18
75.32
64.20
93.69
75.32
64.20
79.54
93.71
85.38
139.32
93.71
85.38
ffiffi
li#,fir
li1itiffi
t;:{4:
Soit friction angte, g (deg)
Figure 09.11
-
Terzaghi Bearing capacity factors for Local Shear Failure
0.11
11.63
##i!!*it+
o.6
3.59
10.37
0.05
.
'11.63
:il20-:l!
(0
:,Nn,,,
5.10
CD
Z
lNr;:
iii,Nl;)irir.,
40
-&
Meyerhof & Hansen Bearing-Capacity Coefficients
i.l$iliii,lriili!
5(,
z
.
13.96
165.58
1
118.37
1
15.31
211.41
118.37
115.31
ii46:,r
152.14
158.51
329.74
152.10
158.51
244.65
t4B
199.27
222.31
526.47
199.27
222.31
368.68
,ri50r
266.89
319.07
873.89
266.89
319.07
568.59
3O5
Chapter 09
306
-
Bearing Capacity
Funda
Geotechnical Engineer
of Soils
For intermediate values of $, the value of bearing coefficients may
computed by iinear interpolation.
No=N4r +
lJ," - l'/,,
#(Nor-No,)*
Qz -Qr
For example, to find the value of N, for 4
-
)?.32, using Table 09.1:
For $t = 22:
Nrr = 9.19
For
$2
(0-dr)
*
Bearing CaPacitY
of Soils
307
ILLUSTRATIVE PROBLEMS
PROBLEM 09.1 (CE MAY 20031
A footing 6 m square carries a total load, including its own weight, of 10,000
kN. The Lase of the footing is at a depth of 3 m below the ground surface. The
soil strata at the site consist of a layer of stiff saturated clay 27.5 m thick
overlying dense sand. The average bulk density of the clay isl,920 kg/m3 and
its average shear strength determined from undrained triaxial test is 130
kN/m2 and $ = 0.. Given is Tgrzaghi,s ultimate bearing capacity for square
footings:
q,=
= 23o:
ru* = ro'ze
Chapter 09
Fundamentals of
Geotechnical Engineering
1.3 c N. + y DrNq + 0.40 yB Nr
Determine the grosp foundation pressure in kPa.
b) Determine the net foundation pressure.
q)
Thus, N, =9.19
+ 1t-J'4-2:19 e232-Zz)
22-21
Nq= 9'5228
c)
Calculate the factor of safety of the foundation against complete shear
failure under the undrained condition (both gross and net). side
cohesion on the foundation may be neglected'
SOLUTION
p = 1920 kg/m3
c = 130 kPa
T_
d=0'
Dr=3m
Q,
=
1.3 c N. + 1
Ds No
+ 0.4y B N,
308
Chapter 09
of Soils
Parl
-
Bearing Capacity
Geotechnical En
chaprer 0e _
Fundamentals of
Geotechnical Engineering
Bearins.Hi:i{
3Og
Net:
a'.
Gross foundation pressure,4s,=
Gross foundation pressurE,
4s
Y
10.000
Fsnu,
b(b)
-
4'n"'
4
= 277,8 kP a
n"t
Q,i'nut=
Qun
Qunet*
Part. b:
ff _
-
Net foundation pressure , Qn = tls - y DS
IJnor
y=pg=1920(9.81)
= 18,835.2 N/m3
y = 18.835
kN/ms
Qu-fD1
t=1,019.8 - 18.835(3)
963.295kPa
963.295
221.3
FS.* = 4.35
,
Net foundation pressure , qn = 277 .8 - (18.835X3)
Net fourrdation pressure , q, = 22L.3 kPa
PROBLEM 09.2
A continuous footing is shown in Figure 09.12. Use the Terzaghi's bearing
Part
capacity equation.
c:
q, = 1.3c N. + y D1N, + 0.a0 yB N,
ulVen:
c=130
y=p8
y -= 1e20(9.81)
Y
c*
500 psf
0=25o
= 18,835.2 N/ma
18.835
kN/m3
'
D1=3m
B=6
From Table 09.1, for
Q
= Q'
N. = S.Z
NY=0
q"=1,019.8kPa
Factor of safety:
Cross:
'
'
Fsrror"=
U
qB
FSgror"
Factor of safety = 2
b)
area that the footing can
carry, in psf.
Determine the net allowable bearing capacity with a factor of safety of 2,
in psf,
Deiermine the gross ailowable bearing capacity with a factor of safety of
2 with respect to shear failure, in psf.
,i
{
u
til
4, = 1.3(130)(5.7) + 13.335(3)(x) + 0.40(18.835X6X0)
,
Dt'2ft
B=2.5ft
a) Determine the gross allowable load per unit
c)
Nq = 1'0
'
= 115 Pcf
=
ffi
FSrror, = 3.G7
3I
o :i;if,. oe - Bearins caPacitY
chapter
Fundamentals of
Geotechnical Engineering
(g,)n"t = 1,6,434 Psl
,- \ *-
({all)net
=
Bearins
.*i:iI 3t t
.
(lu)ner
\{all/net
y = 115 pcf
c = 500 psf
0e'
fS
16,434
2
({,rr)n", = 8,217 Psf
0=2s"
Part
c:
Calculate the develoPed cohesion:
c
"FS
-
cd=
cr=
")
ca
Figure 09.12
500
-
= 250 psf
Calculate the developed angle of friction of soil:
.
tanO/=
SOLUTION
Part a:
qu=cN,+4Ns+0.5yBNl
For
O
Nr=8'3
'
q=yDt
q = 1'"15(2)
q 230 psf
-
q, = 500(25.1) + 230(12.7) + 0.5(115X2.5) (8.3)
q, = 16,664 ps(
4..
lu
tlal _
- _jl
t5
-t6,664
Qull
= ---:2
qar= 8,332Psf
Part b:
(q,)^*=q"-q
(q,)n"t=1.6,664-230
FS
-tan 25o
tan$a=:
/.
tan
= 25o
From the diagram shown below:
N- = 25.1
Nq = 127
tand
$a
= 0.23315
From the diagram:
N. = 11.4
N, = 3'7
Nx = 1'05
=c4 N. + { Nq + 0.5 y B Nr
+
+
{"u = (250)(11 .4) 230(3.7) 0,5(11s)(2.5X1.05)
3,852
=
Psf
4,u
qu|t
3t2
:i;si:,,o'
* Bearing Capacity
Fundamen
GeotechnicalEn
chapter oe
Fundamentals of
Geotechnical Engineering
- Bearins.17i:;il 313
PROBLEM 09.3
I
A footing 1. m square carries a total load, including its.own weight, of 59,130
kg. The base of the footing is at a depth of 1 m below the grotnd-surface. The
s&l strata at the site consist of a layer of stiff saturated clay 27.5 m thick
overlying dense sand. The average density of the clay is'l',846k8/ mt' Civen
is Terzaghi's ultimate bearing capacity for square footings: q, = 1'3 c N. + yD1
Ns + 0.40 yB Nr. See figure 09.13.
a) Determine the gross foundation pressure.
b) Determine the overburden pressure.
OOO
900
aoo
?oo
000
500
400
too
200
loo 90
ao
?o
60
30
ao
' c) Determine the ultimate bearing capacity of the soil,
d) As-suming local shear failure, determine the ultimate
the soil.
30
20
Qar
Z
to.
;a
27
6s
z1
= 59,130 kg
6
p = 1846 kg/m3
c = 1605 kg/m2
o3
o
IH2
S=30o
'oa
o
o-
(U
O
o.s
O)
,.trG
P*t
o
*
1965 kg/m3
m
Qu
=
1.3cNc + yD6Nq
+ 0.4
Figure 09.13
SOLUTION
a)
Gross foundation pressure:
'"
0 {3.{
25
59130x 9.81x
'"
4e=
Angle of Friction, g
Load
Area
1(1)
q, = 580 kN/.m2
ffi
18N,},
bearing capacity of
314 :iiiJ:l.
b)
-
oe
Bearins caPacitY
Fundamentals
Geotechnical Engi
OverburdenPressure:
*
Bearins
.#i:il
3l
5
y*
tt uy use Table 09.1 or Figure 09.10 using the modified value
of g as follows:
Or,
4=lDt
y=pg
y = 1.846(9.81)
tan$={tang
y = 181"09.26 N/ms
Y = 18.11 kN/ms
q = 18.1"t(1)
tan $ = $tan30"; 6 *21.05"
i = ra.rri'N7-2
c)
chapter oe
Fundamentals of
Geotechnical Engineering
Ultimate bearing capacity:
4, ='1.3 c N. + yD1 Ne r
c - L,605k9/mz
By interpolation between 0 = 21o and $ = 22" in Table 09'1:
For $ = 21t, N, = L8.92, N, = 8.26, Nr = 4.3t
Ny * 5,09
For $ = 22o,N,=
O.+O
.20.27,Nr=9.19,
y.B Nr.
.
4#
Nq = 8.26
.
W(21.05
N, = 4 31
. 'i\:li'
N. = 18.92
c = 15.745 kNTmz
B=1m
From Table 09.1, 4 = 3g'.
N, = 37 '1-5
Nq = 22'4('
e.L.os -21)
(21 0s
=tl.rrrc
- 21) = 8.gouu
-
21)
= 43!P
N, =19.73
Note: This is case 2, the water table is at the base of footir"rg.
For the third term, we will use ye = yb.
Tt=fsat-^lw
yb= (1965 x 9.81)
|a Yr,
q, =
Qu
d)
N/ms
kN/ma
9810
Cohesion-,c=50kPa
Angle of friction = 10o
Assume local shear failure and use Figure 09.14 to get the bearing capacity
9466.65
= 9,467
1 .3 (1 5 .7 4 5) (37 .1
6) + 8.71 (1) (22.a 6) +
"1
= 1,242 kN/mz
PROBLEM O9.4 ICE MAY 2OO4l
A soil has the following properties:
Unit weight, Y = 19.2 kN/mr
s. 4 1s .
a6n(1) (1s .7 3)
factors.
a) Calculate the net bearing capacity for a strip footing of width 1.25 m dt a
'
Local shear failure:
-1
I
=
depth of 4.5
m.
The Terzighi's ultimate bearing capacity equation is
given by:
LF
J
Qu=cNc+YDrNq+7:18N,
c =t1ts.zts!=10.497
From Table 09.2, for 0 = 30o
N'. = 18.99
shear failure only, calculate- the safe Ue111S pr€ssure on a
footing 6 mlong by 1..25 m wide, using a load factor of 2'5' Given:
b)
' Considering
N'q = 8'3t
N'Y = 4'39
q, = "1.37 N', + q N'q + 0.4 y" B N't
q,, = 13Qs.7 4sX1B. 99) + 18.1 1 (1) (8.3 1) + 0.5(S.SO7X1X4.39)
4, = 555.81 kPa
qu
c)
* c N,lL + 0.3(B / L)l + vD:.,^.
Nq + 1/z v B Ny[1 - 0'2(B/ L)]
' 4'=q'"*fFS+YD1
What is the safe total load of the footing?
3t
b
:iij,i:'oe
-
Bearins caPacitY
Fundamen
Geotechnical
,loo
Chapter 09
Fundamentals of
Geotechnical Engineering
-
Bearing CaPacitY
of Soils
SOLUTION
9(,
Ito
?o
6(,
5l,
4(,
3l,
Part a:
Qu=cN.*yDrNs +lzYBN.t
For local shear
failure:
'
qr= e N',+ yDyN'q + lzYB N't
e=
t,
e= 3(so)=33.33
D1= 4.5 m
Y = 19'2 kN/m3
B=1.25m
,to
z
6
GI
o
5
4
2
6
From the figure, for
I
al
7
O
= 10o
N'.=8
N,r=!.?!
Mr = 0'25
q, = 33.33 (8) + 19.2(4.5)
q,* 437.256kPa
o
o
o
lr-
Part
b:
(1.
94) + r1r119.21(1.2sx0'2s)
o
qu= e N',11+ 0.3(B/L)l +yD1N'r!lhI-.l-ryltl .0.2(B/L)l
q, = 33.33(8X1, + 0.3(1.25/ 6)l + 19:2(4.5Jq:?4)
+ 1/z(19.2) (7.2sx0.2s) [1 + 0.2(1.25 / 6\l
c
4, = 454.05 kPa
'6
o
oo
o)
o
o
o
Qu
net = Qu
'!D7
Qunet= 454.05
- 79.2(4'5)
Q,n*= 367'65kPa
q"=
'"
W
2.5
+ D.z(A.s)
q"=233.46kPa
Part
c
Q= q,x Afts
Q= 233.46(7.25 x 6)
Q = 1750.95 kFll
Soilfriction angte, g (deg)
Figure 09.14
317
3
r
I
:ifs',;,o'
-
Bearing Capacity
Fundamentals
Geotechnical Enoi
PROBLEM 09.5
0e'
Bearing
.#i:il 3tg
SOLUTION
General shear failure for'circular footing;
A circular footing
2.5 m in diameter is shown in Figure 09.15. Assume
shear failure and use a factor of safety of 3. Determine the folrowing:
4u
ft" gross allowable bearing
l)
&) the
' c)
Chapter
Fundamentals of
Geotechnical Engineering
=
!.3c N. + f w, + 0,3 Y. B N,
From Table 09.1, fot 0 = 25o:
capacity
net allowable bearing capacity
the safe load that the fogii.,g ca., car.y
N' = 25'13
Nq=12'72
Nr = 8'ga
c=80kPa
{ = effective vertical stress at base of footing
q = (19.2- 9.81X1.8 - 1.1) + 18.s(1.1)
r
q
'
Ground surface
= 26'923kPa
y, = lb - Ig.2 - 9.81
7,= 9.39 kN/m3
B=2,5nr
y
= 18;5
kN/m3
c=80kPa
q, = 1.3(80X25.x3) + 26.923(12.7 2) + 0'3 (9'39) (2.s)
q, = 3,074.72kPa
0=2s"
I
Dr=1.8m
7*t = 19.2
kN/m3
a)
Gross allowable bearing caPacitY:
o
= {}
f)
(quttor)gror,
3,014.72
({rtto*)g'or"
(gutto*)s.or, = 1,004'9
b)
kPa
Net allowable bearing caPacitY:
Qunet=
Q"-
Q
,"t = 3 ,014 .72 - 26 '923
q, net = 2',987 '8 kPa
4u
({uttow)nrt
=
(iiuuo*)net =
4
uiut
r9
2,987.8
({uuo*)n", = 995.9
Figure 09.'15
kPa
Safe load:
Qartow
= ({ulto*).* x Area of footing
Qalow=995.9xtQ5)'z
Qauow = 4,888.8 kN
(8' 34)
zzo :l;ii:'o'- Bearing Capacity
Fundamentals
Geotechnical
chapter 0e
Fundamentals of
Geotechnical Engineering
- Bearins.Hi:il
32t
PROBLEM 09.6
A sguare footing is shown in Figure 09.16. The footing will
carry a gross load,
of 700 kN, Using a factor of safety of 3, determine the required
value of B.
Assume general shear failure
Q
14u
t 3 = Bzt
984.7 +378.38
1
382
-
700
378.383+984.782=2100
-1.2A7
m
By trial and error, B
=
PROBLEM 09.7
Meyerhof
A rectangular footing 5 ft x 2'5 ft is shown in-Figure 09'17' Use the
load
eccentric
under
equation"for the ulti-mate load for shallow foundation
value of the ultimate load Q, in pounds.
1on"-*ay eccentricity). calculate the
Figure 09.16
SOLUTION
^' -ttu-
4all -
-
3
-
Q
An.
"Ir8
Assuming general shear failure:
q, = 1.3cN. +
.
4&
+
0.4T.BNy
(Eq. 9.3)
From Table 09.1:
N. = 63.53
'
I
,/"
N''= 47'76
N, = 54.36
q = 17.4(1.2)
q = 20.88 kpa
T,=T=17.4kN/ma
q,,
='1.3(0))63-.53) + 2s.s1a7.16) + 0.4(17.4)B(s4.s6)
q',=984.7 +378.38
Figure 09.17
t-r
Location of
load
-,- -.-
x
324 :iijf'oe
* Bearins caPacitY
Fundamentals
GeotechnicalEn
N=18
= 0.62
= 1 (the depth of groundwatrlr table was not specified)
N,,,, = 0.62(1)(18)
N.", = 1L.'16
p, = 2 inches
p., = 50.8 mm
cN
crv
q, = 0.41(L1J.6X50.8)
'
chapter 0e
- Bearins.*i:il 325
PROBLEM 09.9
For this problem:
a,
Fundamentals of
Geotechnical Engineering
= 232.44t pu,
*
11'7Pti
each 72" x 12" and
A rectangular footing is to support two'square columns kips and the other
of'40
load
a
curii*'
spaced 12 feeton centers' One column
carriesaloadofs0tips;Thefootingis2ftthickandits.lengthshouldextend
load' rhe base of
;h";;;i-; of the colu;n carrying the 40-kip gravity.of
;.;'ilil;
concrete
specific
i*i*g it S ft below tn" gto""a su^rface'
.i;t*u
\.7g, respectively. Determine the
and soil above the rr"iiig t"'t. z.+ and
psf.
;&il;i;h;
is 2000
footing tf the lllowabte soil bearing capacity
SOLUTION
101.325 kPa
q, = 33.72 psi
q, = 33.72 psi x 144
q, = 4,855.96 psf
q, = 4,855.96 psf + 2,000
q, = 2.43 tons/sq" ft.
Factor of safety:
Qa
.F.S. =
6n _"lDi
o, = maximum applied foundation stress
5,200
'
100(32)
o, = 1.625 tons/sq. ft.
r.J.
the resultant footing load
For the pressure in the soil to be uniform'
should coinclde with the centroid of the footing'
about the 40-kip load:
Locate the resultant load by taking moment
y Dr= 120(8)
90x=40(0)+s0(12)
YDy-960psf+2000
x = 6'67 feet
y D1= A.49.tons/sq. ft,
2.43
-
1,.625
-_
F.S. = 2.12
-
0.48
Frorn the figure
L/2=2.5
+ 6.67
L = 18.34 feet
To compute the width of footing:
Effective soil Pressure
q, = 2OO0 - (62,4 x 2'$(2) - (62.4x 1.78)(5
.
q, = 1367.254 Pst
Total load = 40 + 50
Total load = 90 kiPs
- 2)
3zo fifi,i:'oe' Bearins capacit,
Fundamentals
GeotechnicalE
Total Ioad = 90,000 lbs
A _ 90'000
nfootins
" 1367.264
Arootins
chapter 0e
Fundamentals of
Geotechnical Engineering
-
Bearins.*i:il 327
SOLUTION
q'= q'-27h
q, = 250 - 23'5(0'72) - 17 'S(1'2 - 0;72)
= 65.82ftz
Qe= 224.536kPa
L xW = 1,8.34xW
18.34xW=65.82
Pr = 560 + 850 = 1410 kN
W = 3,59 feet
Pz=720+ 880 = 1600 kN
P =L410 + 1600 = 3010
PROBLEM 09. IO
A trapezoidal combined_ footing is shown in Figure 09.1g. The base of the
footing is 1.20 m below the ground and the allowable soil bearing pressure at
that point is 250 kPa.
a) Determine the effective allowable bearing pressure.
b) Determine the required area of the footing in square meter,
c) Determine the vaiue of a and b.
kN
Arts = P / q' = 3010 / 224'536
Arte = 13:405 mz
Solving
for
A*e=
s
and b:
a+h
ff(4.S;=13.nou
a'*b=5,958m )Eq.(1)
with the resultant load:
The centroid of the footing must coincide
I
t-4,tm
I
4.5 m
.J-
l_
3010x
Figure 09.18
=
1410(0) + 1600(4.1)
x = 2.-179 m
x5=Q)+ x
xc = 2.379 m
g28 :i;i,i:'oe - Bearins capacitv
Ar$ Xc
=
A1x1
+
!
GeotechnicalEn
1.27,
9) = l/z(a) (4.5) (1.s) +
2b = 9.45 ) Eq. (2)
13. 405 (2.37
a
Fundamentals
% (bX4.
5X3)
From Eq. (1):
b=5.958*a
a+2(5,958-a)=9.45
a=2.45m
tr=J.5m
chapter 0e
Fundarnentals of
Geotechnical Engineering
SOLUTION
Part a: E
'\soilhsoil - Yconc hto"t - Surcharge
zsa
q,=
- 18(0.6) - 24('2) - 4'B
Qe
=
Qa
q, = 245.6
kPa
Part b; See Figure 09'20
Value of
a:
Pr:
Solve for X by taking moment of force about
Pr=760+580
PROBLEM 09. r
I
under service Ioads is uniform.
' a) Determine the effective allowable bearing pressure (allowable bearing
pressure minus the weights of concrete, soil and surcharge).
b) Determine the minimum dimensions of the combined footing.
P2=1,100+890
P: = 1,990 kN
p=p",+p2
r=i":+o it,ggo
P = 3,330
kN
[PX = P1 Y1+ P2x2l
3,330X=1,340(0)+
1,990(5.5)
l
X = 3.317 m
a=X+0.225
a=3.377+0.225
a = 3.542m
load.is,uf fo1m.,.the
Such that the resulting pressure under the service
of the footinS'
resultant service load? must coincide with the centroid
Thus,
a
= L/2
3.542= L/2
L=7.084m
Areaof footing,
At=
Area of footing, O,=
P
n
#
Area of footing. A/ = 13.559 mz
Figure 09.19
-
:
Pr = 1,340 kN
(CE MAY 2OO3l
An exterior column with service dead load = 760 kN and service live load +
580 kN, and an interior column with service dead load = 1,100 kN and service
live load = 890 kN are to be supported on a combined rectangular footing
whose outer end cannot protrude beyond the outer face of the exterior
column, as shown in Figure 09.19. The allowable bearing pressure of the soii is.
290 kPa. The bottom of the footing is 1.80 m below grade and a surcharge of
4.8 kPa is specified on the surface. The footing thickness is 1.20 m. The uni
weight of concrete is 24 kN per cubic meter and the unit weight of soil is 18 kN
per cubic meter. The footing is to be designed such that the resulting pressure
- Bearrns.Ti:|il 3Zq
3Bo :i;if'oe
-
Bearins CaPacitY
Fundamentals of
Geotechnical Engineering
[Ar= L xltrl
Fundamentals of
Geotechnical Engineering
chapter oe _
Bearins.Hi:il
331
PROBLEM OqJZ FE NOVEMBER 20OZl
A line of four piles supporting a pier is shown in Figure 09.21. The vertical
Ioad of 200 kips includes the weight of the rigid pile cap. The piles may be
assumed to be fixed at the depth Hr = 3 ft below the bottom of sea. The height
Hz from the sea bead to the bottom of the pile cap is 17 feet.
a) Determine the axial force on pile D?
b) Determine the shear force on each pile?
13.559 = 7,084 x W
W = 7.97 m
Therefore, the footing dimension is 2m x 7m
c) Detgrmine the maximum moment in pile D, assuming point of
contraflexure at a depth of 10 feet below the pile cap?
u2
*----.._i
0,225 m
Po= 760 kN
Pi=
5BO KN
-l
I
It
Pr
Surcharge = 4.8 kpa
H
lPo
=
1,100 kN
= 890 kN
,1,::
Y
Pt-
1,8 m
t
--]
Figure 09.21
Figure 09.20
332 :i;si:'"oe - Bearins CaPacitY
Fundarnenta,s of
Geotechnical Engineering'
SOLUTION
-
Bearins
[IFrr = 0l
F, = 200 kips
Fn
chapter 0e
Fundamentals of
Geotechnical Engineering
4H=20
H = 5 kips (shear force in each Pile)
= 20 kips
Maximum moment at D:
Mo= H x70
MD = s(10)
Mo = 50 kiP-ft
Point
of
inflection
.
Axial force at D;
20'
n-
'
F
Tr
'
N.- i
* 200 kips
N=4
F
I
l
T = 20(10)
T = 200 kip-ft
r=18ft
I=Zxz
l=(62+182)x2
I = 720 ftz
200 200(18)
-Pn=-+>
720
4
r
Po = 55 kiPs
To solve for the other axial force:
200
4
'L
200(6)
720
Pc = 51.67 kips
^ 200
4
Ps = 48.33
^
trt
-
P,r = 45
Check:
720
kips
200
4
=
200(6)
200(18)
720
kips
'
LFv --.0
Pa+Ps+Pc+Pp-200=0
- 200 = 0 (oK)
45 + 48.33 + 51.67 + 55
.#i:iil
333
334 :iijf,-oe - Bearins caPacitv
Fundamentals of
Geotechnical Engineering
Chapter l0
Fundamentals of
Geotechnical Engineering
Topics &
- Miscellaneous 22E
AdditionalProblems rrr
Chapter l0
I
O.I
I O.
PILES
I.I
AND DEEP FOUNDATION
PILE CAPACTTY FROM DRIVING DATA
(DYNAMIC PILE FORMULASI
,,.,.,,I
AASHTO FORMULA
g,=
Zh(W, ! A,P)
s+0.1
,16,
Eq.
lo.i
Recommended factor of safetY = 6
lO.l.l .Z Navy-McKay Formula
, 'rE, ..1b,
,[r*o.rlLl
w,)
Eq. 10.2
I
Recommended factor of sa fety = 6
10.Lt.3 Eytelwein Formula
^r
Q.=;0ffi7_,ta,
Recommended factor of safetY = 6
Eq. 10.3
n_,
55O
Chapter l0 - Miscellaneous
Topics & Additionalprobtems
Chapter l0
Fundafnentals of
Geotechnical Engineering
F
Geotechnical
- Miscellaneous r'
??7
Topics & Additional
Problems
'
Driven weight > striking weight
Where;
ea = Efficiency of hammer
Recommended values:
Single-acting, €r, = 0.8
Eq. 10.8
Double-acting, el = 0.g5
Drop, en = 0,85
Where:
Q, = ultimate Pile caPacitY, lbs
Wr, = weight of hammer in lbs
h = height of fall, ft
Ar = tdfrt cross-sectiory inz
p = pressure, psi
Er, = rated hammer energy, in.lb
s = average penetration of pile during the last few
blows, inches
W,, = Total weight of pile, pounds
Wr = Weight of ram, pounds
lO.l.l.4 Engineer News
, = urr*irg" penetration during the last five blows
E=energytransferredtothepileoneachstroke'ft-lb
I O. I . 1.5
Recbrd {ENR or Engineering News)
Modified Engineer News Record
Drop hammer
1.25e1,Ey
Qu=
'
Single-acting steam hammer
Driven weight < strit(ing weight
Q,=
s
-r-
Eq. 10.9
0.1
Where:
eh= Efficiency of hammer
Recommended values:
Single-acting, an = 0,8
Double-actih g, en = 0'85
DroP, er, = 0'85
ffi,ro,
n = Coefficient of restitution
Driven weight
>
Recommended values:
striking weight
2wt,
\
Qu=
h
Ibs
WoodPiles, n=0'25
Wood cushion on steel' n = 0'32
Steel-on-steel anvil, n = 0'5
Err = rated hammer energY, in-lb
the last five blows, inches
s = average penetration ii pile during
Wp = Total weight of Pile, Pounds
= Weight of rar4, Pounds
Eq. 10,5
,*o.r(wo,u',,\'
)
lw,,
Double-acting steam hammer
Driyen weight < striking weight
^2E--1
Q, =
;;-
, Ibs
Eq.10.7
lO.l.l.6
Danish Formula
one of the few formulae consiclered to have
a reasonable precision, based on a
be used
of pile load tests, is the Danish formuli, which should
3,
with a factor oi safetY of
,trtirti.ut study
338 fi#:1
F
l%;,ffiT",il?",;;,
.
Vr,
c
""
Fundamentals of
Geotechnical Engineering
GeotechnicatE
s.W"H
= ;---*
S+
,jJ:ifili;,YI?ffHi
Table 10.1
0.5S.
Q, = u.ltimate dy.namic bearing capacity of driven pile
driving,hammer efficiency (normally 1)
l:WH pit"
= weight of hammer
,H = ham,mer drop (note that
Wi H = Hammer energy)
S = Inelastic set of pile, in distance per
hammer blow
S. - Elastic set of pile, in clistance plr
hammer blow
L = Pile length
Table 10.2
Cohesion
1.
6.3
7,8
2
8.5
>3
9
48
96
cl
c
One part is due to fric
called skin friction or shait friction or side
shear ei, and the other is due to
bearing at the base or tip of the pile
e1,.
1o.1.2.2 THE BETA
1.0
B
In B-method, the friction caPacity is estimeited
Table 10.3
Qt, =
o
s
Qf=ac,PL
Eq. 10,13
fi'Ar, = N, (c,,)r At,
Eq.
c,,
s,,
= !J!- os
)')
10.t4
Eq,10.15
a fraction of the averagE
Eq, 10.16
Qt=FP'u PL
Qb
The cr-method determines the adhesion factor,
o,, as the ratio
"" of
fliction factor, f,, to the undrainecl shear strength
" the
(coi.rior,y, ,,.
as
pile).
effective vertical stress (as evaluated halfway dorvn the
Eq. 10.
METHOD
0.56
0.43
METHOD
Q1= skin/shaft friction or side shear (ultimate)
Qr, = end bearing or point resistance (ultimatej
q.
0.83
0.s6 - 0,96
0.34 - 0,83
0.26 - 0.78
Where:
THEALPHA
cr
Average
Range of values
144
IO.I.2 THEORETTCAL PILE CAPACITY
Tl: tlltlTale load capacity e, consists of two parrs.
t0.l.2.l
values of the adhesion factor
24
E = Modulus of elasticity of pile material
Qn
- Typical
(GPa)
A = Pile end area
+
N.
0
Where
Q1
Values of N, for driven piles
LIB.
EffirHL
-- v--7F-
Q,,=
-
sEs
Driven
Drilled
-
=
N,1 (P*t)u
Meyerhof values of
No
Eq. 10.U
Aa
foi driven and drilled piles
20'
250
28
300
320
34',
360
380
40'
42
450
I
t2
20
25
35
45
60
80
120
160
230
30
40
60
80
115
4
5
8
t2
17
22
Table10.4-Typicalvaluesofpfordrivenpilesinsofttomediumclaywithc<96GPa
Pile length, L (m)
?
3
40
Table 10.5
Pile type
-
Timber or
Very soft
concrete
Soft
Cohesion,
c
(GPa)
0
Adhesion,
-12
-23
23 -36
36 :45
45-62
Very soft
0 -12
Soft
12
-24
28-48
48 -96
-12
- 22
22 -34
34 - 34.5
96 - 192
34.5 - 36
Medium stiff
stiff
Verv stiff
r
O.I.A
r?d,l
''
CAPACITY OF PILE GROUP
approximately 4 to 3'5 times the pile
Some piles are installed in groups, spaced
c,q
a concrete
The pilestfunition as a group due to the use- of
The weight of the cap
load-transfer cap encasing all of the. piie heads'
due to the pile qap
,rbnru.r, from ihe grom g"totlp capacity' The capacity
ip'eud footing) is disregarded'
;;;il;"',1*
(GPa)
-12
- 24
24-48
48-96
96 - 192
1.2
Medium stiff
stiff
Verv stiff
Steel
Geotechnical Engi
Recommended values of adhesion & cohesion for piles in clay
Consistency of clay
Chapter l0 - Miscellaneous
Topics&Additional Problems
Fundarnentals of
Geotechnical Engineering
Fundamentals
fi;,?l1 f,;,IXi,T"i,[T,1,i.,
aiu*",", apart.
0
12
d;#.(us
bi3sr+D
0
1.2
w=2s:+D
Burland Formula (for clayf
0 = (1
- sin
g) tan O JdCR
I
Eq, 10.18,
Where:
I
= diameter of width of pile
P = perimeter of cross-scction of pile
L = length of pile, m
= undrained shear strength (cohesion)
(c,,)r = undrained shear strength of soil at base of pile
46 = cross-sectional area of pile base
f, = base resistance
N,. = capacity factor
s11 ot
Qu = unconfined compressive strength
perr = average effective vertical stress at midheight of pile in a layer
N,, = capacity factor
c,,
Figure 10.1 . Pile GrouP
Note: The actual valucs of cr, p, and N,, are uncertain.
pile group is taken as the
For cohesionless (granular) soils, the capacity of a
will be, greater.
sum of the indiviJual capacities, although the actual capacity
In-situ tests should be used to justify any increase
z
42 fi;,?l1 l'.;,Hlij"F,?,iiii,
Fundamentals
Geotechnical
ChaPter l0
Fundamentals of
Geotechnical Engineering
Forcohesivesoils,thegroupcaPacltyi't,["@
the. individual capacitie-s u.,a
1r1'tn" capacity-assuming block action. The blc
action capacity is calcurated assuming *rui *,e piles
iorm a t"rj" pier wl
dimensions are-the group's perimeter.*The broc[ depth,
r, is the"diJtar,ce f,
the surface to the depth of the pile points. fhe widih,
;,'";Ji;;;;;
;,;;
pier are the length and width of the pile gfoup as
measured fronithe outsi
(not centers) of the outermost piles.
Perimeter,
p=2(b+sl1
Area, A, = (b +
depth, the pressure
- Miscellaneous il?4?
r
Topics & Additional
Problems
be negligible' Below the (2/3)L
o.ut,lt a verlicil:horizontql rate of 2:1'
aistrid'tio'' .sPleads.
t'ii i'tilength is disregarded'
The presence of tne rower
See
Figure 10'2'
Eq.10.19
7p1z
Eq. 10,20
The average undrained shear strength, c,,, along
the depth of the piles is used
to calculate the skin friction capacity. The averJge
undrained shear
- '--' strength at
the pile tips, c,,i,, is used to calculate the end-beaiing
capacity.
The g.r1up capacity can be more or less than the sum
capacities pile capacities. The pile group effciency,
nc, is:
.^
IO.I.4
=
of the individual pile
Figure 10.2
Pile Graup
Group capacity, e,*
I individual capacities, e,
Eq. 10.21
IO.Z BRACED
TENSTLE CAPACIry OF PILES
Tension piles are intended to resist
below water level may require.
!3wev.T, tall buildings subjected
pile pull-out. Unlike pires roaded
does not include the tip capacity.
-
upward forces. Basement and buried tanks
tension piles to prevent .O"rti"gto overturning moments arso need to"rffi
resist
in compressioir, the puil-out capacity of piles
The puirout capacity incrudes tn" ir"igi-,i .?
the pile and the shaft resistance'(skin frictlon).
IO.I.5
-
SETTLEMENT OF PILE GROUP
Piles bearing on rock essentiaily'do not settre. piles
in sand experience
minimal settlement. piles in. clay may experience significant
settiing. The
settlement of a pile group can be estimaiecr by ussumi.j
that the support
(used to calculate the group capacity) extends
to a deptir of onry
2/i
block
ofthe pile
CUTS
lines
for( water' sanitarY' and other
Bracing is used when temporary trenches
earth
active
the
in which
are opened in soil. Abriced cut is an excavation
bulkhead' The box'
facing
the
support
to
used
is
pressure from one bulkhead
10'3'
Fisure
in
or suppoit are shown
ffi-il;;;;;;;;;;;;i;ethods
active
th"
points'
|']i:gular
The loacl is transferred to the struts at various
:"
the
as
installed
are
iat*t
develop'- Since struts
pressure distribution
the
to
"ot
due
little
very
deflects
wall
excavation go", do*r,,i;;-"Pp;. t"rt oithe
considerably
is
wall
the
of
part
strut restraint. The p'*"t"*'on the uPper
equations'
;,gh";;-h;;;s predicted by the active earth pressure
Thesoilremovedfromtheexcavationisknownasthespolls'-spoilsshouldbe
cut so that they do not produce a
placed far enough fro*'tf'"tuage of the
surcharge lateral loading'
as the
The bottom of the excavation is referred to
line, and
toe of the excat:ation'
base
of the cut, mudline:, tlredge
s44 fl;[:1
l%;,Hii5"3,i,ii5,i,,
Fundamentals
Geotechnical Enoi
Excavations below the water table should be dewatered prior to cutting.
Fundamentals of
Geotechnical Engineering
,fJ:i$t?;,Y5?ffi#i s+s
IO.2.I BRACED CUTS IN SAND
The analysis of braced cuts is approximate due to the extensive bending of.the
sheeting. For drained. sana, ine pressure distribution is approximately
uniform with depth.
Fma,
= 0.65 KaYH
(b) Tschebotarioff
(a) Pecks pressure diagram
Figure 10.4
*
Pressure diagram for design of bracing system
(a) Box Shoring
IO.Z,Z BRACED CUTS IN STIFF CLAY
For undrained clay, $ = 0', In this case, the lateral pressure- distribution
depends on the a*rug* undrainqd shear strength (cohesion) of .the clay' If
yi/ c <4, the clay is stiff and the pressure distribution is given in Figure L0'5'
Figure '10.5 - Peck's pressure diagram for stiff clay
(b) Close Sheeting
Figure 10.3
-
Shoring of Braced Cuts
P^u,=
0.Z:YH
to 0.4YH
Eq. 10.22
z+
o fl;[l1 f,;,Hl,T"F,[1,1X,i,,
Fundamentals
Geotechnical Engin
Except when the cut is underlain by deep, soft, normaily consolidated clay,
maximum pressure can beapproximated as:
,ff:if;t:;,Y#1ffi:#i t+t
Fundamentals of
Geotechnical Engineering
If 6 s yH / c < 8, the bearing capacity of the soil is probably sufficient to prevent
shearing and upward heive.- Simple braced cuts should not be attempted is
yH/c>8.
prtar =
lr-illrru
( yH,l
'
Eq, 10,2
10.2,4 BRACED CUTS IN MEDIUM CLAY
lf 4 < yH/c < 6, the soft and stiff clay cases should both be evaluated, the case
that resuits in greater pressure shouild be used when designing the bracing.
10.2.3 BRACED CUTS tN SOFT CLAY
lf yHlc ) 6, the clay is soft and the lateral pressure distribution
shown in Figure 10.6.
pmax
=
1O.2.5 ANALYSIS OF STRUT REACTION
Since braced excavations with more than one strut are statically indeterminate,
sirut forces and sheet piling moments may be evaluated by assuming hinged
(r-!t\,,
I
beam action
rH]'
The strut load may be determined by assuming that the vertical members are
hinged at each strut level except the topmost and the bottommost ones.
-!tN
,-'-lb)
;ffi
R:r+R
figure
10.6
-
peck's pressure diagram for sofi clay
For cuis underlain by deep, soft, normally consolidated clays, the maximum
pressu re is:
Pnax=\H-4c
Eq. 10.25
Figure 10,7
-
W
D'etermination of strut loads
348 fiff:1f,;,Hffi"Fffi1fi
GeotechnicalE
IO.3 SLOPE STABILITY
FS,=
The maximum slope for cuts in cohesionless (drained) sand is the angle oj
internal friction or angle of repose, $. In cohesive soils such as clay howevei, the
maximum slope for cuts is more difficult to deteimine.
a-
otan$
+otafi$6
c+
#
c7
progressive.
FS'=
*c'tr
Eq, 10,30
Factor of safety with respect to friction:
tan 6
fa,rJa- ._-
,,
Extremely
_ J slow
___,_ ---_.
movements
_--._.,." in
... soils
Rapid rltvvrrrrgrrlD
movements ut"l
oi';
"":." are called soll creep. r\qyru
intact or nearly intact soil or rock masses are called s/ldes. Rock or soil that'i,
frorn a nearly vertical slope and descends mainly through the air byr/i
9:11.h*:
falling, bouncing, or rolling is called afall.yery soft coheiive soils can fail by,,
tan
Relation of FSr, FS., and
Eq. 10.31
Q7
FS6;
Eq, 10,32
lateral spreading or by mud flows.
The factors to be considered for stability of slope are the cohesion of the soil, c,,
,shear strength, r, soil stratification and its in-place shear strength parameters.
failure
When F. = 1, the slope is in a state of impending
seepage through the slope and the choice of potential slip suiface add up to
the complexity of the problem.
SEEPAGE
10.3,2, STABILTTT OF INFINTE SLOPE WITHOUT
firm soil or rock lies parallel to a
slip surfaces are very long
potential
il;.hd;:rori". material and the when a rock
''
surface is parallel to the
compared to their a"pi-r.,. fhr gcculs
the
the rock' In this.analysis'
slope and there is a tf,in iayer of soil overlying
w"dg"s and the iesisting forces of the downhill
Infinite
IO.3.I
FACTORS OF SAFEry
The primary purpose of analyzing slope stability is to determine the
factor of
safety. In general, factor of safety is the ratio the average shear strength of soii
t,totheaVerageshearstressdevelopeda|ongthepotentialfailuresurface,t7.
T
FS=
Eq. 10.26
1,1
.c=c+otan$
1t1= C,/ + O
tan
O,,t
Eq, 10.29
Factor of safety witt-r resPect to cohesion:
The soil or rock in a slope exist in a state of equilibrium between gravity forces,
tending to move the material down the slope and the internal shearing
resistance of the material. A slope failure occurs when the force tending toi
cause rupture exceeds the resisting force. The overstressing of a slope or'l
reduction in shear strength muy .urie displacements that *uy-b" very slow or,
very rapid and
l0 * Miscellaneous ?dO
J''
Topics&Additional Problems
ChaPter
Fundamentals of
Geotechnical Engineerin g
Fundamentals
l,
slope, analysis is used when a layer of
driving forces of th" ;hiil
wedge is considered'
wedgeg are ignored, ani only the remaining central
Consider 1 m strip
perpendicular to the
paper,
---- --a
Eq. 10.27
Eq, 10.28
Factor of safety with respect to strength:
Figure 10.8 * Analysis of infinite slope
350
ffi::
Fundamentals
Geotechnical Engineering
li;,Hiij,,F,i1,li,i,,
Normal stress;
N,n/
^= Area of base- (t)(tWcosB
cos
o
-
=
yVcos0
L/cosB
/
Fundamentals of
Normal stress:
o = y*tH cos2
$i
Effective stress:
o'
Tw
=
Wsin
base
(1)(L
/
IV sin B _ ytH(1)sin
p
L/cos}
= f'Hcos2
p = (y*t
- 1,)H cos2 p
Eq. 10.40
Tangential stress:
Tangential stress:
"
Eq, 10.39
B
rtH_(1)cosB
-7^*e
o=YHcoszp
Area of
,iJ:iffiX;,ffi??1ilffii ssr
c""i.ir"rri.ritngineering
r=]sarHsinPcosB
B
cos B)
rC =
L/cosp
..
y.u, Fi cos2 B tan
Eq. 10.34
FS,=---+---*t?to
yH cos'B tan
B
r0.3.4
F|NTTE SLOPE
v' tandr
r ---1yru, tan
Eq. 10.42
B
-
----:--
r=yHsinBcos[i
p
Eq. 10.41
WlrH PLANE FAILURE (CULMANN'S METHODI
Eq, !0,35
tan B
Jf FS. = 1, H = critical depth, H.,
i=---l.-,
yH.. cosz p tan B
tanB-tan$
tan
p
'
tan0
tan B
c
rHGZBtr"B
rL -C
1
qg. 10.36
I cos'B(tanB-tang)
Figure 10.9
I0.3.3 STAB.'LIry OF INFINTE SLOPE UO/ITH SEEPAGE
For, soils with seepage and ground water
level coincides
su
rface:
with
the ground
Eq. 10.37
LLI-Lia o IanQd
Eq. 10.38
:
Finite slope with plane failure
Normal stress:
o=
T=c+o'tan$
-
- [sin (B-0)llcos 0
*z,vH i -----:--;L stnp l
Eq. 10.43
Chapter l0 - Miscellaneous
Topics & Additional problems
r2 rE,tz
Fundamentals of
Geotechnical Engineering
Critical angle of slip plane:
I0.3.5
Eq. 10.45 can also be
2
SLOPES WITH WATER IN THE TENSILE CRACI(:
-cos(F-0a))
*Y'Ir
4tLtt
writien
,fJ:if;l?; ffi?[[i:#l rsE
When tensile cracks are developed at the top of the slope and filled with
water, the stability of such slope can be determined in the following manner.
P*Oa
6,=
ca=
t
Fundamentals of
Geotechnical Engineering
r,.B."rOr. ]
N
l\
LF
as
Fl
Z*1w
Cs
4
1
= l.l.l =
yH
-
4
cos (B
sin
B
-
9,1)
Eq. 10.46
cos g7
where m= stability number
Figure 10.10
Based on Eq. 10.46, values of 1/ m for various and
B
10.6.
Qa
are shown in Table
-
Finite slope with water on tensile crack
z. = depth of crack
z, = depth of water in the crack
Table '10.6 - Stability Numbers Based on Culmann,s Analysis (Eq.
i0.a6)
X = length AB = (H - z,) / sin 0
0 = angle of failure plane
tlr= r,rfiight of soil wedge ABCD
10
15
0
45.72
10
Z.btr
5
181.84
'15
56.70
0
30.38
20
123.71
67.89
25
0
q
20
25
30
When
c,1
10
267.93
0
22.69
30
40
50
29.64
80
0
4.77
30
44.00
5
5t9
A
O',r
t0
5.90
476.34
0
5
8.09
15
6.59
10.99
10
9.55
20
7.40
14.16
15
11.42
25
8.37
60
E
40.00
10
18.90
20
13.91
10
88.68
lc
26.51
25
17.36
15
347.27
22.39
18,04
40.06
68.39
JU
0
20
.E
0
5.71
30
0
146.57
5
6.49
15
5.21
8.58
10
7.40
20
5.71
10.42
t5
8.51
25
6.28
12.90
20
9.89
JU
6.93
16.37
25
11.63
21.49
30
13.91
5
27,92
'10
48.86
50'
15
107.48
(
20
417,45
10
4,93
0
1
5
21.27.
20
70
90
30
9.55
0
400
10
4.37
4.77
u _ 4c( sinBcosg
- ."t1P -
)
6, J
lole
= ly*zr2
F2= ty*z,X = t^t,2,(H * z)/sin
F1
Eq. 10.47
e
Components of Wand Fr along AB:
,F=Wsine+F1
cos
Resisting force to F:
R =7X + (Wcos 0
-
Fr sin 0
-
Fz)
tan
Q
Factor of safety with respect to strength:
FS, =
= c and 0,r = 0, then H = I-1,,, from Eq. 10.45:
"" - 7[r
Fr = force due to water in the crack
water,Pressure along AB
F2 = force due to
R
F
cX + (W ios 0.- Fi :in 0 - Fz ) tan 0
Wsin0+F1 cos0
Eq, 10.48
The magnitude of FS, for various trial wedges can be calculated by varying the
value of 0. The minimum value of FS, is the factor of safety of the slope'
zs
4
I0.3.6
Fur'idamentals of
i%;,Hi.,:,","i,1i1;i,,
Geotechnical Engineering
"i;ln:I.
ANALYSIS
OF THE FINITE SLOPES WITH CIRCULAR FAILURE
SURFACES - GENERAL FAILURE SURFACES:
Chapter l0
Fundamentals of
Geotechnicat Engineering
-
Miscellaneous
355
Topics & Additional Problerns
10.3.7 MASS PROCEDURE (HOMOGENEOUS CLAY SOIL WITH $ * 0|
Modes of Failure:
Cenerally, finite slope failure occurs in one of the following diagrams:
1.
Slope Failures
Slope Circle
Firm Base
2.
3.
Shallow Slope Failure
Figure 10.12
Base Failure
LL
iei<---ti
I
i
o.l---1
,/
,1&
-
Stability analysis in homogeneous clay
(Q
= 0)
Wr = (Area of FCDEF) y
Wz= (Area of ABFEA) y
Driving force about O to cause instability:
Mo=WtLr-WzLz
Developerl cohesion along the surface of sliding:
Figure
10.1 1
-
Modes of slope failure
Two major classes of stability analysis procedure:
1.
2,
- the soil that'formed the slope is assumed to be
homogeneous and the mass of soil above the surface of sliding is
Laken as a unit.
Mass Procedure
Method of slices * In this procedure, the nonhomogeneity of the soil and
the pore water pressure can be taken into cons-ideration and the soil
abovc the surface of sliding is divided into a number of vertical
Mn = ca (arc AED)(1) r
Ma=ca120
For equilibrium:
Mn= Mo
cl12Q=WtLr-WzLz
n.=
va
w.L.
'z''z
-w"L"
"1"1
r'0
Factor of safety against sliding:
f$,= t = c,
c4 c4
parallel slices.
Eq. 10.49
For critical circles
cd=yHm arm=
Cd
yH
Eq. 10"50
Chapter l0 - Miscellaneous
Topics & Additional Problems
J56
Fundamentals of
Geotechnical Engineering
For critical height; F5. = 1, H = H,, and
Hr, =
where
ca
=
c,,:
!'-
Eq. 10.51
ym
m = stability number
c, = undrained shear strength
cd
-
Fundamentals of
Geotechnical Engineerin g
,ff:if;L?;H5?lili#l
For saturated clay with 0 = 0o, the Taylot slope stability chart (Figure*L0.13) can.
jafety against slope failure. The Taylor'
be used to determine the factor of
chart makes the following assumptions:
(a) There is no open water outside the slope
l
(b) There is no surcharge or tension cracks
(rj Snuur strength is deiived from cohesion onlv and is constant with depth
(d) Failure takes place as rotation on a circular arc
developed cohesion
FS"=N,
c
Eq. 10,52
y'H
ffi
M'
5h-ear.strength,!
Effective unit weight, 'y'
'/rilil///////tr
I
rl ,l
!t
ti
it ti!i
-JJ
FIRM SASE
I
Factor of safety: fS = N" c/(r' H)
I
SLOPE
CIRCLES
value
of
(
d0
The TaylOr chart shows lhat toe'circle t'ailures Occur i1 slopes steeper than 53o.
For slopes less than 53, slope circle failure, toe circle failure, ot base circle failure
may occur.
I
I
!t
I
O = Orft
I
I
I
For B > 53o, all circles are toe circles. The location of the center of critical circle
can be found using the graph shown in Figure 10'14.
t
ti
Ni
li
r1
I
c
lln
l)
tl
8
0,
\
0.2
t'
.
FIRN1 BASE
.'1
I
m = 3.83
@0=90'
lr
\lot
F:'
4 1.
U,J
,l
0.5
2
x
i
d
ol
b
lt
tt
7
'u.r"l,')
F
o
AqF
,'(
,'2t
ICLES
6oo
q)
'u
UI
E
c(o
-50
.,*'.,rt
{/
50
20
10
SLOPE ANGLE, p,'degrees
Figure 10.13
- Taylor Slope
Stability Chart (6 = g";
ssz
60
to
Fisiure 10.14
-
Location
70
gl[:n"]critical
"r,r"
80
90
center ror p > 53'
3
s8 fi#:1 l%;,Hi,T"F,?,ili,i,,
Chapter
Fundamentals of
Geotechnical Engineering
Consider the following examples on how to use
Example (a): Toe Circle
9=70
d=18/36=0.s
U
(J
CI
(J
From Figure'10.15:
I
E
x' = -0'30
E
lo = 1'6
d
I
o
cu
&
&
zF
E
o.
= 0.3(36)
= 10.8 m to the left of O
Y,= 1'6(36) =57.6m.
E
q
F
z
o
g
Example (b): Base Circle
o
F1
z
F=45o
d=20/20="t
o
z
'
x
From Figure 10.15:
lo = 0'5
Y" = 7'65
6
o0
(J
6
ro
X, = 0.5(20)
Xo= 10 m to the right of O
F
z
)
Y, = 1,65(20) = 33 m
-1
I
Figure 10'15
-
center of criticar circre for homogeneous srope in cohesive soirs,
4 = g.
Example (b): Slope Circle
F=20'
d=6/2A=0.3
From Figure 10.15:
xo= 1'45
.,
lo -11
X, = 1.45(20)
X, = 29 m to the right of O
Y" = 2.2\20) = 44 m
l0-
Miscellaneous
Topics & Additional Problems
2qo
Jil
Figure 10.15.
3
60 ffi fl l%;,Hi,:"i",3[iii,i,,
I0.3.8 METHOD OF
Fundamentals
Geotechnical Engineering
Fundamentals'of
Chapter
SLICES
The method of slices was deveroped in the earry 1920s in sweden and
was
later refined- by Bishop to consider interslice fbrces to some degree.
This
analysis method' can accommodate complex slope.geometries, variable
soil
layering and strengths, variable pore water p."rrirr"
internar
reinforcement, and the infruence of external boundary loads,
"o"Jiiio.,u,
but it is onry
applicable to circular slip surfaces. It accomplishes this by dividing
a ,topl
into a series of vertical slices for analysis, with llmiting uqrihbrir*
clnditions
ttl-r.t =
COS CX," +
$
sin cr,
Eq, 10,54
Note that the IS is present on both sides of Eq. 10.53, Hence, a trial-and-error
soltrtion or a programmable calculator is necessary to find the value of FS.
tan $o = 1sn 6755
,sg,
-o."''
-t)' .'
-ivs
:1.
- ..:..
4.,
,€)'
,e'.
tan
Problems
FS
evaluated for each slice, as shown in Figure L0.16,
iq'
lO- Miscellaneous,r- t
5g I
Topics & Additional
Geotechnical Engineering
'tnlq'
- .:
Qrl
ai G/^
^('
'
411
y',
Layel L
, ,.yr;,
2
Layer
'tz, 62, cz
di,
Force Polygoh
c1'
\
Figure 10.17: Forces acting on the
Layer
3:
y:,
rb:,
c:
Figure 10,16 * Typicar stope stabirily anarysis using the method of
srices
Each slice can have different layering, different strength,
and different pore
water pressure than the.other slices. If the condition of-equilibrium
is satisfied
for each slice, then it is considered that the entire mass is in
inu
force system on a single slice is shown in Figure 10.L7.
"q"iliu.i,r*.
n=p
FS
=
F Gr, +w, tan6)-J/-r'
ryd(n)
n="1.
n=p
f w, ,,n o,
/-r
tt=1
Eq. 10.53
nrh
slice in the Bishop simplified method of slices
a6
z fi;H'l l',;,Hiij"i,?,ili,i,,
Fundamentals of
Geotechnical Engineering
ILLUSTRATIVE PROBLEMS
Chapter l0-Miscellaneous
Topici & Additional Problems
Fundamentals of
Geotechnical En gineering
SOLUTION
PROBLEM
Design load Q, = 360 kN
Ultimate pile load capacity, Q, = Qo x FS
Ultimate pile load capacity, Qu = = 360 x 2 = 720 kN
A
Q,= Qr+
t0.I (MAy ZOO5, NOVEMBER 2OO5l
0.36-m square prestressed concrete pile is to be driven
in a clayey soil as
sfovrin in Figure 10.18. The design capacity of the pile
i, soO r.N,-*ith a factor
of safety of 2.0.
Given:
Qo =
Undrained shear streng th, q,, = 111 kN/mz
nit weight of clayey soil = 1 g.5 kN/ m:
ComputJthe
enct
tearing capacity
Qol
Part (a):
fo Ao= N.
(c,)6 ,46
N'=9
(c,)a' q,/2 =
U
9)
r
-,
505
11'1 /2
= 55.5 kPa
A6 = Q'J$z = 0'1296 m?
pir" in if N, = 9
(b) Compute the skin friction expectea "rto develop along the shaft
of the pile
(c) Compute the length of the pile if frictional constantl 0.76
Qr = 9(55.5)(0.1296)
Q6 = 64.74
kN
=
Part (&):
Qo"',e,
lQt= Q, -
= 360 ktl
Qtl
Qr= 720
-
64,74
Qr = 655'26
{
kN
,i
a = 0.76
c,= q,f2=111/2 = 55.5 kPa
p = perimeter = 0.36 x 4 ='t".44 m
655.26 = 0.76(55,5)(0 .36 x 4)L
L = 10.79 m
I
I
PROBLEM IO.2
A square concrete pile 0,3 m x 0.3 m is required to support a load of 175 kN
with a factor of safety of 3, The soil strafification consists of 5 m of soft gray,
I
J
Figure 10.18
normally consolidated clay (c, = 25 kPa, $ = 260,,]sat = tB kN/m3) underlain by
a deep deposit of overconsolidated clay (c, = 80 kPa, 6 = 24", OCR = 4, T,ut =
18.5 kN/m3, N. = 9). Groundwater level is at 2 m below the ground surface.
Assume the soil above groundwater level is saturated. Assume q. = 1 for soft
clay and 0,5 for stiff clay. See Figure 10.19.
1a; Wfrat is the ultimate load iapacity due to skin friction in soft clay?
(b) What is the valu,e of Lr?
(r) Using the p-method,
what is the value of Q, if Lr =
Assume N, = 9.6' Use p = (1
5
- sin $)(OCR)' tan
$
2."1
m?
z6
4
Fundamentals of
Geotechnical Engineering
fi;[:1 li;,Hii5"ilT,X;i,,
Part (c):
aa=liskNl
Qr= FP*P
Y:iflilffi
: os
t
Soft Clay (OCR = 1, normally consolidated)
B = (1 - sin 26")(1)os tan26'
P = 0'274
Soft Clay
Y*t = 18 kN/3
,*q:if!!?;
'
perr
= average effective vert' stress at midheight of pile
perr
= 18(2) + (18
in soft clay
(Q/)sort cruy
(Q7)sortclav
- 9.81)(0,5)
= 40.095 kPa
= 0'27 4(40.095X1,2X5)
= 65 92 kN
Stlff Clay
I.t
= 18,5
kN/3
Stiff claY (OCR = 4)
F = (1
pert
=
- sin 24")(4)0.5 tan24o = 0.528
dv?td.ge effective vert. stress at midheight of pile
in stiff clay
p"* = 18(2)+ (18 - 9,81)(3) + (18.5 - 9,81X1.05)
p*
Figure 10.19
SOLUTION
Parts (a) and (b):
'
Q,= 175 kN
'Q,=QrxFS
Q,,=175x3=525kN
IQ,,= (Ql)*r,.ror'+ (Qr),tirr.r," + Qr]
.
'
= 69,7 kPa
= 0.528 (69,7) (1'.2) (2'1)
Q).tirr
"r,y
(Q|),tirr.r,v = 92'74 kN
(
Qr= N,t(P*r)a Aa
(p*ia = effective vert. stress at the bottom o{ pile.
(p"rit, =18(2) + (18 - 9.81)(3) + (18.5 - 9,81X2.1)
(Paia = 78'82kPa
Qu= e'6(78.82)(0.3 x 0.3)
Qa = 68'1 kN
(Q/"nr"ruv=QC,,PL
(Qr),nrt.r,y = 1(25X0.3 x 4)(5) = 150 kN
Q,,
(Q/.,irr.r.,y = a c,, P Lt
Q, = 75'5
(Q7)stirrcray
Qr'
= 0'5(80X0.3 x 4)Lr = 48Lr
= N' (c')r'
kN
At'
= 9(80)(0.3 x 0.3) = 64.8 kN
,Qr,
525 = 150 + 48L1+
Lt = 6.46 m
= 65.92 + 92.74+ 6g.1 = 226.76kN
e,= e,/FS=226.26/3
64.8
PROBLEM rO.3 (CE NOVEMBER zOOs)
The pile group shown in Figure 10.20 consists of 12 piles, each 0.4 m in
diameter, arranged in a 3 x 4 matrix. The pile penetrates a soft clay (L1 = 2m,
ca = 20 kPa), a medium dense clay lLz = 6 n, cu2 = 60 kPa), and a stiff clay (L3 =
4micus = 95 kPa). Assume N. =,9 and use s = l for soft and medium dense
clay, o. = 0.5 for stiff clay.
366
Chapter l0
:
Miscellaneous
Topics & Additional Problems
Fundamentals
Geotechnical Enqi
(a)
Determine the capacity of the pile group based on singre pile fail
(b)
Determine the capacity of the pile group based on block failure mode.
Compute the maximum center-to-center. spacing of the piles for 1
mode,
(c)
Chapter t0
Fundamentals of
Geotechnical Engineering
d'
=
\rw
Qr= 741.63
Qr,
= N.
- Misceilaneoui
Topics & Additional
Pt
Wt
is7) (2) + 1 (60) (1.2s7X6) + 0.5 (esx1'2s7) (4)
kN
(c,)1,
Aa = 9(95)(0'1257)
-
107'4735 kN
efficiency.
Q, = Qf +
Q6
= 741.63 +
107 ,4735
= 849'1
kN
Group load capacity = 12 x Qu = X'? x 849'1'
GrouP load caPacitY = 10,189'2 kN
Part Z;Block failure mocle;
3.6+0.4=4m
Perimeter, p=2(4+2.8)
Perimeter, P=13.6tn
C,C C
Area, Au = 4(2.8)
Area, Aa = 11.2 mz
C
^iI
t
occc
Q/= q, cn p Lt *
dt=
+ % cg P
\r_.^_.
1(20x13.6X2) + 1(60)(13.6X6) f 0.5(e5)(13'6X4)
s.z cuzP Lz
Qa = ]Y. (c,fu Au=
Qus= Qf
9(95)(11'2) = 9'576 kN
+ Q6=8',024+9,576
= 17.600kN
Part 3:
,
Figure 10.20
SOLUTION
Part 1: Single pile failure mode:
Perimeter, p = nD = n(0.4) = 1.257 m
A.rea, An=
Q.f
tDr=
= o, Cn p L1 *
t(0.+;z= 0.1.257m2
c\2 c,,2
p Lz + az cn p Lt
Efficiency,
,=
eu
ff:
=
1009u'
e,,q= n e,
n Q., = 10,189'2 kN
Solve for Q,, in terms of sPacing s:
+
Ao = (2s + 0.4)(3s + 0,4)-= 6s2 +,2s 0'76
+
1'6
+
10s
+
(3s
0'a)l
=
p = 2l(2s+ 0.4)
+
v
N
s6B fi;[:1
l'.;,Hi,T"F,1?#,i,,
Fundamentals:
GeotechnicalE
(r) Determine the consolidation settlement
Q,,x= Qt+ Qr,
Qr = &t Cul P Lt + az cuzP Lz + se crt P Lz
Q1= 1(20)(10s + 7.6)(2) + 1(60)(10s + i.5)(6)
+ 0,s(es)(10s + 1.6)(a) .
+ 1.6) = 5900s + 944
590(10s
Qr=
= N. (c,)o Ao = 9(95)(6sz + 2s t 0.16)
=
Qr, 5130s2 + 171.0s + 136.8
=
Q,,s (5900s + 944) + (5130su + 1710s + 136.8)
Q,s = 5130s2 +7610s + 1080.8
,ff:ifilX;,Y]i?,$i:,xi roe
Fundamentals of
Geotechnical Engineering
of the clay layer'
SOLUTION
Qr,
Et. 0
Dry Sand
l
ra
= 17.8
y,
= 10.2
kN/m3
El. 3
[Q""'= n
Q,,1
5130su
+ 7610s + 1,080,8 = 70,789.2
- 9708.4 = 0
5130s2 + 7610s
Et. 5
t-
-7,670 t,,l1Z,OtO12 * 4(5,130)(-9,108.4)
Clay
1.5 m
r,= 9.3 kN/m3
',
e = 0.54,
C. = 0'3
,pJJUL
s=0.783m=783mm
PROBLEM IO.4
The foundation shown in Figure 10.21 is supported by 9 piles. The founda
rests on a sand layer underlain with 3 m thick of normally consolidated clay.
i;;-!
rf
:;:;lillr
-:r-!n:JI
Figure 1O22
Part (a):
Lp - 2400/ (6.5 x 5.5)
oo
oo
oo
Sand
kN/m3
Dry Sand
ya
= 17.8
r
= 10.2
kN/m3
Sand
kN/m3
LP = 67.13kPa
Part (b):
p, = 9.3(L.5). +10.2(2) + 17.8(3)
p,= 87.75kPa
Part (c):
f
Clay
= 9.3 kN/m3
= 054. Cr = 0,3
y,
L+e
Po
pf =po+ Ap = 154.88 kPa
H=3m=3000mm
15a'?9
o'3
aH = 3ooo
, to*
Et. 8
Figure 10.21
(a) Compute the effective siress at the midheight of the clay layer.
(b) Determine the increase in pressure at the midheight of the tlay
Dr
LH = H;"' log!-!-
1+0.54
LH * 744.2mm
layer.
"
87.75
3r
o *,:lt i'.;,Hi,'"i"F,i,i?i,i,,
Fundamentals
Geotechnical Engineeri
Chapter l0
Fundamentals of
Geotechnical En gineering
- Miscellaneous ?a tt
il
Topics & Additional
Problems
PROBLEM IO,5
pile weighin g 1.40 rb/ ft,40 feet long is driven
I rt 1.UseI fta concrete
soil.
factor of safety of 6.
a
Given:
Unconlined compression strength,
N,=9
qu
-
-
LU
on a sand
_
1.25(0.85)(198,000) 2750+(0.5)2(5600)
0,2 +
0.1
2750 + 5500
P, = 348,525 lbs
Po= PufFS
P, = 348,525/6 = 58087.5 lbs
P, = 58.09 kips
2,320 psf
a = 0,76.
(o)
(b)
(c)
Determine the ultimate frictional resistance of
the pile,
Determine the allowable bearing capacity of pile.
Determine the allowable pile capaclty usinj the
Modified Engineering
News Record formura. The average penet"ration
during
blows is 0,2
i.ch.
The hamme.l,
beluec
(CE
MAYZ00sl
the last five
D-12
(;"il;;;;;
weighing 2750lbs and rared energy of 16,500 ft_lb.
Use *n =
0.5.
r0.6
PROBLEM
A braced cut in sand 7 m deep is shown in Figure 10'23' In the plan, the struts
are placed at s = 2.0 m center to center. Use Peck's empirical Pressure
diagram. Determine the load on each strut.
O.SS und
nZ
SAND
'0
SOLIJTION
y
Parts a & b:
Q,= Qr+ er,
Q,,=acuPL+N.(c,)uAr
cu=1/zqu=t/z(2,320)
c, = 1,160 psf
Ql= a c,p
L
Q1= 0.76(1,,L60X1 x 4X40)
Qt= 141,056Ibs = 141.06 kips
,
)
part a
Q, = 141,056 + 9(1,160)(1) = 15"t,4961bs
Q, = Q,,/ FS - 157,496/ 6 = 25,249 lbs
Q, = 25.25 kips) part b
BOTTOM OF CUT
Figure 10.23
SOLUTION
Part
.,
,, _ l-sin$ _ 1-sin30o
c:
_1.25ct,E , Wr*nzWp
s+
0.1
W, +Wo
W, = 2,750|bs
= 40(140) = 5,600 lbs
Et, = 76,500 ft-lb = 198,000 in_lb
Wn
s=0.2in
et=0.85, n=0,5
r,
-'r'
^'- t*rinq TllfrJ6 -"
po
,u)
= 0.65K, y H = 0'65(7 / 3)('tn (7) = 25.78i3 l,Pa
= po s = 25.78 x 2* 51.57 kN/m
= 30"
= 17 kN/ml
i__
te
5
Chapter l0 - Miscellaneous
Topics&Additional Problems
Fundamentals
Geotechnical Engineering.
Fundamentals of
Geotechnical Engineering
Chapter l0
- Miscellaneous 4
2a2
r
Topics & Additional
Problems
With reference to Figure 70.4 (a):
Ps
=
0.65K"YH
I lm
l*
ll
I
H
zm
t+
l'i
lr
=','
-
l
ll-
I
2m
JI
Section
Figure 10.25
SOLUTION
Figure 10.24
With reference to Figure 1,0.4(a), the pressure giug:1* is as shown in
Figure 10.26, The irut loads may be determined by assuming that
In Figure 10.24(a):
llMar = 0l 2Rr = S1.57,rr,r.r,'
lIMa = 0l 2Ril = 51.5713)(0.5);
In Figure 10.2a
Plan
Section
are hinged ai each strut level except the topmost
and bottommost ones,
thI vertical members
Ra = 116.03
Rar = 38.68
kN
kN
(\:
[xMc = 0] ZRaz= 51..57(\Q);
[IMaz = 0] ZRc= 5L.57$)(2);
6
@
Rrz=0
Rc = 206.28
N
kN
Ila = Rsr + Rrz = 38.68+ 0
Rn = 38.68 kN
(b)
PROBLEM IO.7
The elevation and plan of a bracing system for an open cut in sand are shown
in Figure 10,25. Assume /sancr = 110 pcf and $ = 36.. Use peck's empirical
pressure diagram. Calculate the load on each strut,
Figure 10.26
r
Br
4 ffilt
chapter l0 - Misceilaneoui
'
Fundamentals of
Geotechnical Engineerin g
l%;,Hiij",3,Xi,l|,i.,
Topics & Additional
P
p, = 0.65K,yH
K, = (1 - sin 36.)/ (1+ sin 36o) = 0.2596
p, = 4.65 (0,2596X1 1 0X30)
p, = 556.842 psf
'u)=paxs=556.842x10
w = 5568.421b/ft
Figure 70,26(a) [LMnr, =
rrFH
Figure 10.26(b)
Raz
*
rr r{^^=*r?,Zii.frflr^",
or
#:::?iii!:liyl-szz,z1z
3
= Rcz = 5,568.42(3.5) = 19,4g9,47 lbs
Ra=Rar+Rsz
R6 = 37,387.95
Figure 10.26(c)
lbs
sot-uTloN
Rcr = 0
Rc=Rcl*Rce
Part a;
c
l'5 = --'----------------.:
yH cos'
Rc=19,489.47 lbs
B
tan
tan (r
tan p
r
P
+- tan
16
Ro = 5,568.42(.t4)
R
tan 20"
18.639(8) cos2 20o tan 20o
o = 77,957.88Lbs
25o
FS = 1.615
Partb:. H,,=
PROBLEM IO.8
A infinite slope is shown in Figure 70.27. Theshear strength parameters
at the
interface of soil and rock are as follows:
Cohesion,c=l6kN,/m2
(a)
(b)
Angle of shearing resistance, $ = 25o
lt H = I m and F = 20", find the factor of safety against sliding on the
rock surface. Assume no seepage.
If P = 30., find the critical height H. Assume no seepage.
(.) If
there were seepage through the soil, and the
froundwater table
coincide with the ground surfJce, and H = 5 ffr, = 20", *f,ut
B
the factor of safety. Assume p,,r = 1900 kg/mt.
*outJ
Uu
76
1'
1C63g
cos2
:O'itu,',a0" - tan 25')
H", = 10.31 m
Part
c:
y*t = 1900 x 9.81 = 18,639 N/m3
y,ut
= 18'639 kN/m3
- 9'81
Y' = lsat. Y, = 18.639
y'=8,829 kN/m3
C
l-5 = -_-----T--
t
y.rrHcos'PtanB
16
, T'
1+
18.639(5) cos2 20o tan 20o
FS = LJ1,41'
tan$
Yrr, tan B
8.829 tan25o
+-.--
18.639 tan20o
ezs
zt
6 r*lt
'I
Fundamentals of
Geotechnical Engineering
fr;,[i,T"F,[il:,i.,
PROBLEM IO.9
A cut is to be made in a soil that has y 416.5 kNf m3, c
= 15 kN/m2, and q =
26"' The side of the cut srope wiil make an angle of 45. with
the horizontar.
(a) What is the developed angle of friction?
(c)
Fundamentals of
Geotechnical Engineering
Chapter l0
- Miscellaneous a77
Topics&Additional Problems r' '
PROBLEM IO.IO
A cut slope is to be made in soft clay with its sides rising at an angle of
the horizontal as shown in Figure 10.29' Given , c,, = 30'87 tN/l' andy = 17 '14
kN/m3, Stability number, rz1 0,219, Assume the critical circle is a toe circle.
Use Figure 15.1.5 for the values of 0 and a.
(a) betermine the maximum depth up to which the excavation can
What is the critical angle of slip plane?
(e)
SOLUTION
(c)
c = 15 kN/m?
Figure 10.2g
1 - cos(B -0r) )
..i = 1 .,,,(
a, ,,[_51np
L
Figure 10.29
*rO"-.,/
c15
f5=;-;c,t=+=5kN/m2
Ld
tan
FS=
J
Q
tan $,i
$a
c,=
s
)
SOLUTION
tan$a= lan0
FS
= 9.230
c.
30.87
a" Hrr= L = ""'I't ' -'
17.1"4(0.219)
Ym
o-
= 8'22m
I.r/;it";--
I sinBcosg, )
cos (45d
-
9,23")
sin 45o cos 9.23o
H=1.48m
Critical angle of slip plane:
o _ 0* 07
22
Part
Part ,bz From'Figure 10.14, for F = 75o, u = 41.8' and 0 = 53o
*vr[t*cost0-0r.)
+r
= + (16.s)
=
tan26o
45".9.23"
= 27.\'1.5'
be
carried out?
Find the radius, r, of the critical circle.when the factor of safety is equal
to 1..
Find the distance BC.
t = 16.5 kN/m3
d=26'
.
75o to
st
I
*rHt l'r;,Hi,',i"F,Xiii,i,,
GeotechnicalEn
In right triangle AFC:
sin q, = H*l AC
AC = 12.33 m
SOLUTION
[G MC = s
sin(0/2)=CD/r
r = 13.817
Part
c:
BC =
-
ln=
12'3?
/
e
/
(1
+e)l
= 0'4995
tO.tt IcENoVEMBER
o=5x103(0.7273)
p = 0.003636 cm/sec = 0.03636 mm/sec
Part a;
ZOO5I
Part
The apparatus shown.in Figure r0.31 maintains
a constant head of 160 mm.
The soil sample has a hydrauric conductivity
of 5 x 10s cm/se. *rd., moisture
Part
Calculate the time required for the red dye to
pass through tne soii,
assuming that there is-no diffusion, or the ied
dye does not J"utte, a"ii
passes through the soil.
u,=
!
=
T#
= 0'1092 mm/sec
b:
Flowof
content of 18.5%. Use G = 2.7.
(a) Calculate the seepage velocity in.
(&) Calculate the flow o] water.
(c)
(1 +O.4gg5)
i=160/220=0.7273
Seepage velocfty,
PROBLEM
/
n=0.4995
n = 0.333
q. _
H., cot 0
BC = H,^(cot cr - cot B) = g.22(cot 47.g _ cot75")
BC = 6.99 m
Topics&Additional Problems
Iirr;rl!;:ll;,*0,.
li = h/L)
la = kil
ryr
- Miscellaneous r''
?-rc,
is saturated (s = 1)
2
r
H* cot
e]
e
In right triangle ODC:
sin (53'12)
Chapter l0
Fundamentals of
Geotechnical Engineering
Fundamentals
water,
c:
Time, I =
L
as
=
Q= Aa= +(100)'z(0'03636)
0 = 285.57 mm3/sec = 0.28557 cm3/sec
220
0.1092
= 2,01.4.6 sec = 33.6 min
,,
t0.tz .lcE MAY 2OO5l
soils as
A reservoir wlth a 3,400 m2 area is underlain- by layers of stratified
3'2 m'
and
,ho*r, in Figure 10.32. The values af Lt, Lu and Ls are 2 0 m' 1'4 m'
PROBLEM
respectively.
of permeability'
ial What the average vertical coefficient
velociiy of water moving
(actual)
ilnterstitial
the
n"t*r*ine
iai
soil if it has a void ratio of 0'60'
PIug
red
of
through the
(c) Compute the water loss from the reservoir in one year' Assume that
the pbre pressure at the bottom sand layer is zero'
chapter l0
Fundamentals of
Geotechnical En gineering
3Io fi;,!!"; li;,Hi,:,1"F,?,iii,i,,
6.6 -
W
2
zsxlo-?
cm/s
@
,
= 3.2
Kr
= 1,1 x
K, =
@
-> ->
l.!
x
10-7
h = 5 + 3.2 + t.4 + 2 +
= 26.6 m
i=26.6/11.6=2.293
cm/s
Sand Layer
1"5
L=5+3.2+1..4+2=11.6m
cm/s
10'o cm/s
10"7
2.3 x70-'
x (1 m/100 cm) x (3600 s/hr)
i=h/L
K, = 2.4 x 10-6 cm/s
Kv
4 -------------;
Average velocity of flow through soil, u = ki
10-6 cm/s
10-7
3.2
1,.4
3./x1o-7
(k/* = 2.54}gs x 10-7 cm/s
it;i";= s.irs * 1o-6 m/hr
K, = 1.8 x
K, = ).$ x
- Misceilaneoui gg I
Topics & Additional Pr
a = 2,54038 x 10t (2.29Q) = 5.8251 x 10' cm/ s
Interstitial (or seepage) velocity, a,=
-> ->
n= ' = 0'6
L+e \+0.6
ffi
Impermeable Laver
!
=0.375
Interstitial velocity, u, = 5'8251 *19'z / 0'375
Interstitial velocity, c', = 1.55 x L0-6 cm/s
Figure 10.32
soLuTtoN
15m
e
K,
- l.g
x
10'7 cm/s
Kv
= 3.2
,
10'7 cm/s
k = 9.L45 x 10-6 m/hr
i = 2.293
A = 3,400 m2
= g :t 45 x 10"6 (2.293) (3,400)
Q - 0.07A5e2a9
Q = 625 m?/Yeat
tf /ht
=
O.O772g 6249
*'/
h.t
x (24hr / day) x (365 days/year)
@
@
11.6 m
Kv
@
>
@
= 2.3 x
10-7 cm/s
..>.Sand Layer
->. ->
Impermeable Layer
PROBLEM
footings
Two footings rest in a layer of sand 2.7 m thick, The bottoms of the
thick
1.8-m
is
a
layer
sand
the
Beneath
are 0.90 m 6elow the ground surface.
fuf"r.
of
Underneatfi the clay layer is solid rock' Water table is at a depth
"iuy
L,8'm below the ground surface. See Figure L0 33'
(a) ComputJthe stress increase in kPa below footing A (1'5 m-x 1'5 m) at
A
of the clay layer. Assume that the pressure beneath footing
(b)
Y- = L h : H =2+1.4
+3.2= 6.6m
'
(k1')rq "k,,"'
10.l3 {cE MAY 2OO5l
(c)
the center
is spread at an angle of 2 vertical to L horizontal'
De;rmine the.si# of footing B so that the settlement in the clay layer
is the same beneath footings A and B'
Determine the settlement in mm beneath footing A'
382
Chapter l0
-
Miscellaneous
Topics & Additionat problems
Fundamentals
GeotechnicalE
lundamentrllof
Chapter l0
.
- Miscellaneous J('''
2oI
Problems
Topics & Additional
Geotechnical Engineering
Part b:
P,=450kN
In order that the settJement in both footings are equal, the increase in
I
pressure at midheight of clay layer under each footing must be equal.
900 / B'2 = 25.51; B' = 5.94 m
= B + 2.7 = 5.94; B = 3.24m
Size of footing B = 3.24m x 3'24 m
B',
0.9 m
0,9 m
I = 18,5 kN/m3
Part c:
p, = 18.5(1.8) + (20.8 - 9.81X0.9) + (18.8 - 9.81X0.9)
po= 5't.282kPa
0.9 m
1.8 m
y = 18.8 kN/m3
e = 1.03; C. = 0.3
ClaY
Pr=P"+ LP=76'792kPa
'
Cr
Solid Bedrock
Dr
r^-Pf
LH=H -3-logz
1+eo " po
Figure 10.33
aH = 1800
0'3,, to*7-97-7?1.03 " 51,,282
1+
SOLUTION
= 46.65 mm
B' = B +Z(d/z)
B'=B+d
PROBLEM IO.I4
In the profile shown in Figure 10.35, steady vertical seepage is occurring.
Part a;
B'=1.5+2J)
B'= 4,2m
+1.2
Lp = 450/4.22
v
77//m7V7fr77777797777m.
Ap = 25.51kpa
Yt
Soil'A
1 cm/sec
19'64 kN/m3
*
k
=
ililo
ild
ll5
Soil B
k = 0.5 cm/sec
yt = 18.85 kN/ml
Figure 10.34
Soil C
k = 2 cm/sec
rt = 20.42 kN/m3
Figure 10.35
l:
384
(a)
Chapter 10 * Miscellaneous
Topics & Additional Problems
Compute the average vertical coefficient of permeabiflty
and B.
(b) Compute the
(c)
Chapter l0
Fundamentals of
Geotechnical Engineering
Fundamentals of
Geotechnical Engineering
A
Problems
(a) Calculate the equivalent coefficient of permeability in the horizontal
l" fry"., a
direction.
hydraulic gradient for flow through
- Miscellaneous rL'r
2oE
Topics & Additional
(b) Calculate the equivalent coefficient
and. B.
Compute the seepage in soil A per square meter.
direction.
(c) What is the ratio of (ks)"' to
of permeability in the
vertical
(krz)"q?
SOLUTION
SOLUTION
Equivalent K for layers A and B (Normal flow, series):
Ha = 4.2m; Hs = 4.5 m
I H _-h,
' Kut k',
Part a: Parallel flow:
9.7 4.2 4.5
Kr, 1 0.5
[kriruq,
H
=Zklt]
kHGo (12)
kH(eq)
Hydrauliccradient, i= H/L= 11.2- (-3.6))
Hydraulic Gradient, i = 0.SS1T
/
t1.2- (-7.5)l
y1!". p"l square meter of area:
=
Q KiA = 0.006591(0.5517)(1) = 0.O03636 m3ls
Flow^of
Q = 3.635 Qs = Qa =
Qa
:
i
A layered soil is shown in Figure
= 8'025 x 10'4 cm/sec
Partb: Normal flow:
H =rlr
I'kvvq)
k'
1233
kv(rq) 10 -3
'r,
kYt.q)
PROBLEM IO.I5
= (103)(3) + (2 x 10''t)(3) + (10-sX3) + (? * 103)(3)
kri("q) = 0.0008025 cm/sec
Part c: Ratio,
+-+--.iJ
2x1O"a
1o*s
2 x 1o-3
= 3'756 x 10-s cm/sec
/cs1.q1/k v@qs
= 21,37
10.36.
PROBLEM IO.I6
The setup shown in Figure 10.37 is 50 mm wide (perpendicular to the paper).
The flow through the soil is known to be 4'08 liters per minute.
Figure 10.36
386 fi#l1
(a)
Fundamentals of
Geotechnical Engineering
l%;,Hi,'"i"i,?,ii5;*
What is the hydraulic gradient?
(o)
(b) Calculate the equivalent
(c)
Fundamentals of
Geotechnical Engineering
hydraulic conductivity in
Calculate the hydraulic conductivity kz.
(b)
G)
SOLUTION
,iJ:if;:?;,Y:i7liffi1ll
Eez
Determine the.number of pressure dropo'
Determine the seePage loss.,
Determine the total amount of water percolated in one year
SOLUTION
Q = 4.08 Liters/minute = 4,080 cm3/min
i = h/ L = (7,2 - 0.1) / 1.7 = 0.647
A=60x50=3000cm2
[Q=K"riA]
H
Ii:
^14
=zh/k)
4,080 = K"q(0.647)(3000)
K, = 2.1 crn/min
770 40
_=_+_+
2.7 1.8
kz = 4.69
80
k2
---[f---:. \\\\ '.
\\\
50
n
',
..
cm/min
PROBLEM IO.I7
Thc section of a concrete gravity dam with ir.s frow
net diagram is shown in
Figure.10.38; The coefficLnt of pur-"ubility of the
soil
in the
vertical and horizontar directions are 0.25 m/day 1l-,istropig'm1aay,
and o.sia
respcctively. The length of dam perpendicular to
the
prp", i,
FiEure 10.39
f ZO"rn.
Nr
4= ^lK,Kt,,
r,
Number of flow channels, NJ= 4
Number of Pressure droPs, Na = 11
Head,H=10-2=8m
Et2
q
El 0
= 167sp58) (8)(4/:'L) = 1.108 m3/day per meter
m3Ar
Q = qL = 1 108(120) = \32.96 m3/day = 5'54
noprC
In one year: Volum e = 1'32.96 x 365 = r[8,530
m3
PROBLEM IO.I8
IMPERMEABLE LAYER
Figure 10.38
A bulkhead is to be constructed of stet,l-sheet piling and tie rocls as shown tn
Figure 10.40. Consider 1 foot strip only of sheet piling'
BBB fi#:1
(a) Determine the
pile.
Fundamentals of
i%;,Hiiii'F,1ill,in,
Geotechnical Engi
nearest value"to the total active pressure acting on
(b) Determine
(r)
the nearest value to the total passive resistance,
What is the tension in the tie rod?
!1, , o i,3oi
\1;.
p.t= K,4 ct= $/3)(500)= 500/3 Psf
lr
Fr=pr(+0) =6667lbs;
KoYa (10) = 333
F3
Problems
Fq
=
20'
zt = 15'
psf
yz* 33.33'
yz=15'
= (333)(30) = 9,990 lbs;
pt = KoY'(30) = 660 Psf
,.oo
- Miscellaneous rL"
2oo
Topics & Additional
Active Pressure:
pz=
5i,r,i,yd,r;,ibO pcf
Chapter l0
Fundamentals of
Geotechnica I Engineering
!+=1.0'
= t/z(650)(30) = 9,990
'Ot
p,=W(30) = (62.4X30) =L,872.Ps[
zz = 1.67'
zz=20'
tu=25'
F* = 1/z(1,872)(30) = 28,080 lbs
Total active force,
Total active force,
Total active force,
Fo
= Fr +
Fz
t
F3
+
Fa
+
f,
Fo = 6667 + 1,667 + 9,990 + 9,900 + 28,080
1,, = 56,304 lbs
Passive Pressure:
Kp = (1+ sin 30")/(1 - sin 30') = 3
Ps= KpY'(15) = 2,970 Pst
Fs,u, = Vz(?,970)(L5)
Fs,nu, =
SOLUTION
22,2751bs
ls = S',
zs = 30'
Total Passive Resistance = F5 *u" * F,
Total Passive Resistance = 22,275 + 28,080
Total Paspive Resistance = 50,355 lbs
Note: The active anci passive water Pressures for this problem
equal, hence it may be disregarded in the analysis.
Solving for actual passive reaction Fs by taking moments about T:
Fszs= Frh+ Fzzz+ Ft41 FtZs
Fs(30) = 6567 (15) + 1,,6'67 (1'67) + 9,990 (20) + 9,900 (25)
(OK)
Fs = 18,336.3 lbs < F5
*u'
[Ifn=
p*
Figure'10.41
0]
Fs = Fr + Fz + Fe + F+
T = 6567.+ 1,667 + g,gga + 9,900
T = 9,887.7lbs
T+
-
18,336.3
are
3eo fi;,?li
i%;,H;:","il;?H,
Fundamentals
Geotechnical Enoi
PROBLEM IO.I9
After,several years of service, the_retaining
Chapter l0 - Miscellaneous
Topics&Additional Problems
Fundamentals of
Geotechnical EnEineering
SOLUTION
wal shown in Figure ro.+2 waii
t1no1"a
*u tu.,#,,1-d; r"" ;td,
f.:jo':After
}3:"^:::ggjl
:'r i:fll"nl:
T1,' l:"r
:r
level.
perfor.ilg,l
fietd. explorarion, taboratr;y1;;;;;:
conducted on the backfiil soir. The difference in
strength par'ameters were
be negligible in comparison to the originai aJsign-conaitions.
f""9.to
t
moist cohesionless
backfiil
y = 119;4 lblfts
Coy1a1 only 1 ft length of wall perpendicular to the-paper,
(a) Determine the totar active-force on the wilt'before
the waterlog
q,-Jr
condition.
gft
(&) Determine the total active force on the
(c)
wal after the waterrog
condition.
Determine the resurting overturning moment about
point A after the
waterlog condition.
ru
= 107.6 lb/ft3
w=
76.70/o
rrt =
125 57 lb/fF
y'= 63.17 lb/ftr
7ft
Figure 10'43
&
= (1
Part
- sin 31')/(1
+ sin 31") = 0.32
the waterlog condition:
'a: before
K, yHz * 1/z(0,32)(119'4)(15)'? = 4298'4 lbs
Fo =
1/z
Parts & & c:
pr * KuyH * 0.3?(119.4)(8) = 305.664 psf
=t/z{3A5.664)(8)=1222.661bs; yr=9.67.tt
yz= 3'5 ft
F2 = 30b.664(7) = 2139.65lbs;
Fr
p2= Koy'H = 0.32(63.17)(7) = 141'5 Psf
'
Figure 10.42
rrl
495,25lbs;
lzgaL\(7) =
Fa o y*Hz f 2 = 62.4(7)2
/2
= 1528.8
ibs
w=
y+
2'33
ft
= 233 ft
= 1222.66 +'2139.65 + 495.25 + 1528'8 = 5,386'36 Ibs
M, = 1222..66x9.67 + 2139.65 x3,5 + 495.25x2.33 + 7528.8x2.33
M, = 24,028 ft-lb
Fo
r'?o I'
ss
z r*lt l%;,H;T,,;[ii""
Fundamentals of
Geotechnical Engineering,
PROBLEM IO.2O
A 1,8.-m.:{uare footing is shown in Figure 10.44. Use
(o) Determine the overburden
a factor of safety of 3.
pressure at the base of footing.
(b)Determinethegrossal]owablebearingcapacity.
Using the value of 4o obtained above, determine the settlement of the
k)
soil. Assume average standard penetration resistance (spr) of N-varue
of 18 and use a corrJction factor crv = 0.62.
,r =
Nc
18.6 kN/ml
B=1.8m
Situation
part
a:
Part
b:
Figure 10.44
SOLUTION
Overburden pressure, q =
Part
= 1.3 c N, +(i N,7 + 0.4 y,.B N,
0 + 1,8.6 (1,.2) (1,2 .81) + 0.4 (18.6) (1.8) (13. Z)
=
Q,,
q,, = 581 kPa
{,,gross = 581/3 = 193.67 kPa
18.6(1, "Z)
=
22.32
Part
'
c:
=
= 18'05 kN/m3
Totr]_load
R."ft
_ 2.5x10s l--.,
,*" = aoloq ==115.741kpa
Depth for full comPensation:
Total load
4= 6'4 m
kpa
Part
&:
p,
N/mr
yDr=lGit aGL8'0s04 q=115'7qt
Part a:
q,,
c
pn-\Dy=
Qanet
Q, = 1.15 c
N. + yDr Nq
for 0 = 0o' N' = 5'7' N' = 1' N' = 0
2=
= 14.35 kPa
+
q, = 1..15(1.L35X5,4 18'05Dr
c = Suf
28.7
/2
qu=94.06+18.05D7
Settlement from standard penetration test (SpT)
Settlement,
4
S=
0.4"1
N
.inmm
'
Qu net
= 4, - "lD1= 94'06 kPa
q, n"t
=
cor
4 u net
-!!-
- 18.05 Dl=
4 = bearing capacity of soil,
115.74L
Ncor =
D1= 4.68m
in kpa
corr€cted SpT value (N value)
Mor = corr€ction factor x N
193.67
)^ = -:-_-..----
0.ai(0.62 x 18)
= 42.3 mm
Problems
has ariaverage.unconfined compressive strength of 28.7 kPa. Neglect depth
correction factors. Use a shape factor of 1.15. 0 = 0q. The bearing capacity
factors are given in Table 09.1 Page 301.
(a) Whit is the actual foundation pressure?
ia) Wnut should be the raft depth for full compensation?
iri What should be the raft depth for partially compensated foundation
using a factor of safetY of 3?
y = 1840(9.81) = 18050.4
= 31.61; Nq = 17.81
'N' * 13'70
- Miscellaneous il2oa
r
Topics & Additional
PROBLEMIl0.2l
A raft foundation is to be designed for a 36 m x 60 m building with a total
loading of 2.5 x 105 kN, The clay specific weight is-18^49 kg/m3,.and the clay
E[tr
c=0
Chapter l0
Fuhdamentals of
Geotechnical Engineering
94.06
3
-
3l .35 kpa
31.35 kPa
394
Chapter l0 - Miscellaneous
topics f nooirirnul Problems
Fundamentats of
Geotechnical Engineering
PROBLEM I O.Z2
i
SOLUTION
11040
A municipal storage tank is to be supported by a circular raft (mat) foundatiori
placed on the surface of the soil. The diameter of the foundation is 12.2 m:
Lo=-
The maximum load exerted on the soil when the tafrk is full is 11,,040
LP = 94'44kPa
kN. The
following data was taken from a boring log and other soil tests from the
proposed site.
Elev.0.0:
ground surface
well-graded sand and gravel
unit weight: 20,44 kN/m:
allowable bearing pressure: 143 kPa
Elev.1.5:
encountered GWT
soft, brown clay
unit weight: 20.81 kN/m3
compression index: 0,34
unloaded, origipal void ratio: 1.15
coefficient of consoliclatiotr: 0.0093 m2 / day
Elev.4.6:
encountered thick, impervious rock layer
What is the factor of safety in bearing?
(b) If the total clay thickness is 3.1 m, what is the primary settlement?
(.) How long will it take for 80% of the primary settlement to occur? Time
factor of 80% consolidation is 0.567.
(a)
,S:il"ltX; YI?lXffi:#i :os
Fundamentals of
Geotechnical Engineering
'
Qa=
+(1,2.2)'
l43kPa
Et. o
1.5 m
Paft at
FS=
El. 1.5
b
ClaY:
Lp
743
f5=
Sand: y =
3.1 m
r
I
94.44
FS = 1.51
-
I
y =.20"81 kN/m3
Q = 0'34
e=1.15
c, = o,oo93 m'z/day
I
lr,-,uffi
ImPervious rock
Part b:
* g.81)(3'1/2) + 20'44(1.5) = 47 '71' kPa
+
pf = Po Lp = 47 .71 + 94.44 = 142.1"5 kPa
p.=
(20.8L
6P1=
(I)r
51-i!-logz
.
l+e Po
1t1.15 "
47.7L
AH=232.4mm
Part c: Please refer to Page.196:
t-----tt
This is a single-drainage layer. Ha. =-Dl3'1 m
-
qt2
t = 0.567
"''
0.0093
= 586 daYs
PROBLEM IO.Z3
The earth dam shown in Figure 10.45 is to be compacted to a void ratio of 0'78
in place. A borrow pit neaiby contains soil having a void ratio of 1207o, attue"
specific gravity of 2.65, and a moisture content of 15"/''--The total cost of
moving ln" roit from the borrow pit to the dam site is P40.00 Per cu. tn.,,based
on the original volume at the borrow pit.
Chapter l0
3q6
-
Miscellaneous
Fundamentals of
Geotechnical En
Topics & Additional Problems
What is the required loose volume of soil from the borrow pit?
How many tons of water will be transportea wltfr the fill? '
what is the total cost of moving all the required soil to the dam site?
(a)
(lr)
(c)
- Miscellaneous a,,a
il u
Topics & Additional
Problems
PROBLEM IO.Z4
\
A 3,000,000 cubic meter earthen dam is constructed from borrow soil with the
following original properties:
t
'
Lenqth of dam = 400
Chapter l0
Fundamentals of
Geotechnical Engineering
Void ratio
...,...,,.,,.
.....0'70
Degree of saturation ......'...'.....'.....0.40
Solids specific gravity,...'..'...'... " "2'65
220
ft
Figure 1o'45
soLUTroN
_ 20+22o (50) x 4gg = 2,400,000 ft3 (loose volume)
,,
vd.,nr = r
2
Required volume of solids:
V
= V,+ V,,;
V = V,
v
i!vs
=
c; V,,= eV,
* eV'= V'(1 + e)
The borrow soil is compacted to a final specific weight of 17.29 kN/m3 and a
17% moisture content, The trucks used to transport the borrow soil from the
borrow pit to the dam site each have maximum capacities of 10 cubic meters.
(a) What is the weight of solids in 1 m3 of compacted fill?
(b) What is the weight of water in 1 m3 of compacted fill?
(c) Given a fluff factor of 10%, how many trips must the truck make to
complete the'dam?
SOLUTION
For compacted fill:
W*
Part a: Required loose volume from borrow pit:
V = 1,348,314.6 (1 + 1,2) = 2,966,292.1.3 ft3
V = 84,060.6 m3
.
Part
1l:
Volume of water;
MC
'
=
17.29
MC
1.+
ya = 74.78
2,400,000=V,(1+0.78)
V, = 1,348,374.6 lt3
v*
te
"=-----.,!!-=-
*
1+0.17
kN per 1 m3
17.29
- 1'4.78
= 2.51 kN per 1 m3
W, + C V" = (62.4 x 2.65)(1,349 ,314.6)
W., = 222,957,302.256
lbs
= MC x W, = 0.75(222,957,302.256)
W,, = 33,443,595 lbs (2000) = 1.6,222 tons
Wu,
)
Part (b)
*,
V,+ l/.= s1/"+ l/,
v
v"=
' 1+e=3'ooo'ooo
L+0.759
w,
Part (a)
2.65+?.65(0.17.\
G +GMC \,
-L"t_= _T;_,t*t
,rj=jlrr,
lT.Zg _
,r.r,
\ ,
V=
wu'
i
=L705,5'l4mj
For the borrow material:
V = %(1 + e\ = 1,705,51,4(1, + 0.7) = 2,899,375 m1
/
Pa
rt
c:
Cost= 84,060.6xp40
Cost = P3,362,424
Number of trips =
ffid
= 318,931
trips
) Part (c)
B
es
fi;,::1
'
Fundamentals of
Geotechnical Engineering
i%;,Hi,T"E?,i1i,i.,
PROBLEM IO.25
A proctor test is performed on four samples. The mold volume and mass are
r/30 ft\ and 4200 g for each sample. The following data are collected.
Samole
'
mold
soil
Mass of
and
7
2
3
4
pRoBLEM t0.26 ICE NOVEMBER ZOO5'
The fotlowing data were obtained from a triaxial test on
8.2%
6300 g
LA1%
6425 g
1L,7%
6330 g
14.8%
(a) Determine the normal stress at failure plane'
(a) Determine the maximum normal stress (plunger
(a)
ft3
=
Given:
rma,=70kPa;d=28;
c
=30kPa
g44 cc
Mass of
MC (g/
, PY"'.
cc)
1900 q.2% 2.01,27
2100 10.1.% 2,2246
2225 1.1,.7% 2.357
21,30 14.8% 2.256
Pa,y
soil
1.
2
3
4
= pwel (1 + MC)
1.86
2.021
2.1.1
1,965
In right triangle TEC:
cos r[ = 70/ R; R = 70 cos28
R = 79.28 kPa
tan $ = EC/70
EC=70 tan28o =37,22kPa
/
o* =
P,rr
#
= 1.85
=
r.rru B/ cc =
2256
kgtmt
In right triangle OAF:
tan 28o = 30/ OA
QA= 56.42kPa
B/cc= 1850 kgimi
Relative compaction of sample 2:
Relative compaction = p,lz f pa^u
(sample 3)
P,i,n, = 2.71, g/ cc
Relative compaction
-
#
x
100%
In right triangle OTC;
sin 28' = R/OC
OC = 79.28/ sin 28o = L68.87 klra
=
stress) applied on the
soil,
Determine the minimum normal stress (confining pressure)'
SOLUTION
Volume, V = 1./30
a cohesive soil:
Cohesion = 30 kPa
SOLUTION
(a) What is the wet maSs density for sample 4?
(b) What is the dry mass density for sample 1.?
(c) What is the relative compaction of sample 2?
399
Maximum shearing stress at failure plane = 70 kPa
Angle of friction = 28o
Water
content
5100 g
Chapter l0 * Miscellaneous
Topics & Additional Problems
Fundamentals of
Geotechnical Engineering
95.78o/o
AC = OC
- OA = 1.68.87 - 56.42=
112.45 kPa
4a
o ffi lt l'r;,Hi,T"F,ii?i,i,,
Part
a:
.
Fundamentals of
Geotechnical Engineering
Chapter l0
Fundamentals of
Geotechnical Engineering
*
Miscellaneous
Topics & Additional Problems
(c) What is the expected settlement of the clay
Normal stress at failure plane:
o= AE= AC-EC=11.2.45-32,22
layer
overconsolidated with OCR = 2'2?
o'= 75.23 kPa
Part
&r
Maximum normal stress (plung"r
rfr"rr;
or = AC + R = 11.2.45 + ZS.ZS =iSt.zS Uyu
Part
c; Minimum normal stress (confining stress)
' o"3 = AB = AC- R = 112.45 _79.28= 33,t7kpa
SOLUTION
Part a:
/soil Vsoil = 3'200
(18.5)
x
t
(10)'?
D1= 3,20A; Dt* 2.2m
Part b:
PROBLEM IO.Z7
The tank shown in Figure 70.47 has an inside diameter of 10
m and is 6 m
high' The tank is used for storage of oil having specific gravity .r osz. in"
combiped weight of empty tank and the conciete footin! is gzoo
kN. It is
required to excavate an amount of soil such that it will comlensate
to the dead
weight of the tank and concrete
Weight of oil = ]ol Vol = (9.81 x 0'82) '? (i0)'z(6) = 3790'74 kN
Pressureincrease-
w: = 3790.7^
-i{ (10)'
=48'2652kPa
or, Pressure increase =fohoit = (9'81 x 0'82)(6) = 48'2'652kPa
Part c:
= initial effective stress at midheight of clay layer
p" = 77.2(1'4) + 18'5(3'5) = 88'83 kPa
'Lp
= Pressure increase.due to oil = 48'2652kPa
po
"
Note;Theweightofthetankwasnolongerincludedbecause
it was com-pensated by the weight of soil excavated'
OCR
= P'
= 2.2; p, = 195.426 kPa
Po
No
pf
-
te'. pc
= preconsolidation stress
po+ Lp = 88'83 + 48'2652=T37'0952kPa
Since p1< p,:
Figure 10.47
(a) Determine the required depth of footing.
(b) what pressure increase in the soil is eipected when the tank is
filled
with oil?
AH =
g-!' bg?
1,+eo " po
au
o'95
zsoo
1+0.45 "
=
nuYP'=
88'83
18.1e6 mm
PROBLEM I0.28
The section of a retaining wall is shown in Figure 10.48. Consider 1 m length o
wall and use Rankine's active state.
(a) Determine the toial active lateral pressure at the bottom of the wall.
442
(b)
(c)
Chapter l0
-
Miscellaneous
Fundamentals of
Geotechnical Engineering
Topics & Additional probtems
Determine total active force acting
Fundamentals of
Geotechnical Engineer:ing
o;ffi
ps
Determine-the location of the t6tar active force measured
from the
bottom of the wall.
,iJ:if ltX;,ffii?lx?i#l +o s
= 0.271(18 - 9'81)(4.5) = 9.988 kPa
Fs = 149.988)(4.5)(1) = 22'473 kN
/r=1.5m
Cohesion:
P
=2cr,!Q,
pt= 2(5),[02?1 = 5.206 kPa
Fs = 5.206(4.5)(1) = 23,422
'y+
tctrt
= 2'25 m
Water pressure
Y*t':
0,= 351
c
ps= 9.81(4.5) = 44|l'45kPa
18 kN/fni
='s'kP,
Fq=1/z(44.145)(4'5)(1) = 99;33
r:
kN
/s =
1'5.
m
r ,
Active lateral Pressure at the bottom of wall:
p = 24 + 19.512+ 9.988 - 5.206 + 44."145
) Part a
P = 92,44kPa
Total active force, Fo= h + Fz+ Fr - Fq + Fs
) Part b
Total active force, Fo = 240'18 kN
r
SOLUTION
Location from the bottorn:
240,18V = 54(6) + 87.804(2.2s)
4.5 m
+ 99.33(1.5)
t-
n
Lt-l
ft
I
4.5 m
I
Qz
Qq
Soil pressure
For the upper soil layer:
Cohesion
For the lower soil layer:
,. - sin
30o
""'""
K-==l
1 +sin30o r
1.
6,=
1Ji"3Y
l+sm35"
=0.271.
Soil pressure:
P = Kolh
fi
=
(7
h
/ 3)76(4.5) = 24
Part c
Classify the following soils by using the unified soil Classification system'
(a) What is the classification of soil A?
(a) What is the classification of soil B?
(a) What is the classification of soil C?
Soil
Sieve analysis, % finer
Liquid
Plastic
No.4
No.200
limit
limit
A
95
7A
B
15
42
C
60
4
48
32
39
24
16
8
Cu
C,
4.8
3.2
2.9
4.2
yru.
= 1/z(24)(4.s)(1) = 5a Pry
yt=4.5+1.5=6m
= 79.512(4.5X1) = 82.804 kN
SOLUTION
Using Table 2.01 in Page 59:
pz= 0.277(16)(4,5) = 19.512 kpa
Fz
'23.427(2.25)
PROBLEM I0.29
U
p:
)
V =2.713m
I
t_
* 22.473(1.5)
lz *
2,25 m
Soil
A:
%finer than
No'
200
=
7'Oo/o
> 507o (Fine-grained soil)
404
Chapter l0
-
Miscellaneous
Fundamentals of
Geotechnical Engineering
Topics & Additional Problems
* 48 < 50% (Silts and Clays, ML,
PI=LL-PL=48-21=24
LL
CL, or OL)
Fundamentals of
Geotechnical Engineering
Soil
From plasticity chart, since the point plot a.bove A-line,
the soil is CL
,S:iflt:;,Yf?lxl[Ti
+os
C:
% Passing No. 200 = 4 < 50 (Coarse-grained soil)
% Passing No. 4 = 60% (Sands)
Since there is less than 5% fines, the soil is either SW or SP'
Since C, < 6, the soil is SP
PLASTICITY CHART
g
PROBLEM IO.3O
:50
A footing rest in a layer of sand 6 m thick. The bottom of the footing is 1.5 m
below the ground surface. Beneath the sand layer is a 4 m thick clay layer.
undemeath the clay layer is solid rock. water table is at a depth of 4.5 m
E
i*o
u,
ai
ABo
P20
ts
0
<10
d
LrourD LrHrT (LL) (%l
CL
ML
OL
CH
MH
OH
CL *
) Inorganic; LL < 50; PI > 7; Atterberg limits pldt on or above A-line
) lnorganic; LL < 50; PI < 4 or Atterberg limits plot below A-line
) Organic; (LL - ovendried)/(tt - not dried) < 0.75; LL < 50
) Inorganic; LL>Sl;AtterberglimitsplotonoraboveA-line
) Inorganic; LL>-50; Atterberg limits plot below A-line
) Organic; (LL - ovendried)/(Ll - not dried) < 0,75; tL > 50
ML )
below the ground surface. See Figure 10.49. Assume that the Pressure beneath
the footing is spread at an angle of 2 vertical to t horizor\tal. Assume OCR = 1.
(a) what is the vertical effective stress at the midheight of the clay layer?
(a) what is the average pressure increase in the clay layer due to the load
on footing? Use SimPson's rule.
(a) Determin-e the settlement in mm beneath the footing.
Inorganic, Atterberg limits plot in the hatched zone
Soil B:
%finer than No. 200;22% < 50% (Coarse-grained soil)
Dry sand
ta = 14.8 kN/m3
Since there is more than127, fines, the soil is either GM or GC.
-
_________.,g
1.5 m
PI=LL-PL=32-"16=76>7
The point plots above A-line
Therefore the soil tt
Yot
=
Groundwater table
18.5 kN/m3
CLAY:1=19kNim3
e = 1.2; LL = 40olo
[5.,r"rHARr
60
>r
L.
=s0
dr{
fr*o
o
,
Figure'10.49
ALINL,:
/Pt,o.tuw.2o',
330
Mt
CL
ts
uzu
OH
SOLUTION
tr
u
<10
c
"0
.l
Part
t'rr-opr
io
20
30 {0 50 fio 7s 80 g0
LmulD
LrrlI
{LLt {?|l
a: p,- (19 = 9.81)(2)
po
100
= 98.015 kPa
+ (18.5 - 9.81X1'5) + 14'8(4.5)
406
Chapter l0
-
Miscellaneous
Fundamentals of
Topics & Addirional Problems
Geotechnical Engineering
Fundamentals of
Geotechnical Engineering
lndex
APartb: 3p,."= * -4APr,+Ap,
,
At any vertical distance ft
from the base of footing:
T-
. b=3+(7/2h)xl=i+11
d = 7.5 + (1/ 2h) x
)
= r.g +
11
i
\t
f7,'
,Vi
At,=(3+h)(-1.5+h)
Lp=
1,100
245,248
Abtivity of clay, 7
Adhesion, 340
i
(3+h)(1.5+h)
AASHTO classification system, 64
AASHTO Fonnula, 335
Active earth pressure cqefficient, 243,
Allowable bearing caPacitY, 297
gross, 291
net,291
At top of clay, (lt = 4.5):
Lp.. =
1,100
= 24.44kPa
(3+a.5)(1.5+4.5)
1,100
(3+6.5)(1.5+6.5)
='14.47 kPa
At bottom of clay, (ft = 8.5)
Lpu, =
1,100
(3+8.5)(1.5+8.s)
24.44+4(14.47)+9.57
'5 = ^ry:llrv:lyz
Apo,"
Lpou"
= 9.57 kPa
6
= 15.315 kPa
Calcite, 5
Capillary rise, 135
pu
o'27
98.015
nH = qool-I
los
All
= 30.95 mm
4
Burland formula, 340
tH=H C, nrL
1+en "
+7.2
Bearing capacity of soils, 285,287
allowable bearin;; caPacitY, 291
Beta B method, 339
Bishop simplified rnethod, 361
Boussinesq equations, 165
Box shoring, 343
Braced cuts,343
in medium clay,347
in santl,345
in soft ctay, 346
in stiff clay, 345
reaction of,347
Bulk unit weight, 3
Buoyant unit weight,
Part r:
C. = 0.009(LL -10) = 0.27
+ 15.315
98.015
Continuity, 92
Correction factor,299
Coulornb's theory,247
Critical hydraulic graclient,
4
Culmannls method, 351
Cup method,9
Cuts, 343
Base circle, 356
At midheight of clay, (h = 6.5)
4P,,, =
Alpha a method, 338
Angle of internal friction, 219
Consolidation
coefficient ofl, 198
degree of, lg7
time of, 198
Constant-head test, 86
Casagrande, 89
Close sheeting, 343
Coefficient of curvature, 62
Coefficient of gradation, 62
Coefficient of permeabilitY, 85
Cohesion, 219, 340
Compression index, 193
secondary, 195
Compressibiliry coefficient of, 198
Consistency,6
index,
12
Danish formula, 337
Darcy's law, 85
Degree of consolidation, 197
Degree of saturation, 3
Density
water, I
Depth factor, 295,298
Developed angle of friction,292
Dcvelopcd cohesion, 291
Direct shear tesl,226
Dolomite,5
Drained triaxial test, 224
Dredge line,343
Drop hammer, 336
Dry unit weight, 4
Earth pressure At rest.244
Effective size,61
Effective stress, 131
Engineer news record, 33(l
Equipotential line, 95
Eytelwein forrnula, 335
Factors of safety, 254,291,34s
Fall,348
Fall cone method, 8
Falling-head test, 87
Flexible strip load, 167
Flow curve, 10
lndex
Flow intlcx, I0
Fiow line. 9(.>
I'low nrts, 95
Footing, 285
types of, 286
Foundation, 285
Friction skin, 3J8
ncr:rl bealing c.:rp:rcity cquation, 294
Genelal shear failure, 287, ZBg
Ce
Group inclex, 64
Cvp.surn,
5
Hanscn bcaring capaciry equation, 296
I
Fundamentals of
Geotechnical Engineering
Nagaraj, 194
Retaining wall,249
Root time method, 198
Nishida,.194
Normally consoliclated, 192
Kaolinite,
5
Kozc.ny-Carntan equation, 90
Lateral ealth pressr-rre, 241
Line-load, 166
Liquid limir, 6
Liquiditv index, 7, 12
Local shear failurc, 287,290
Log tirne rnerhod, 198
Magnetitc,
5
Mirss proceclurc,354
Meyerhol s cquurion. 29,r
Molu-Coulomb fuilure crireria, 219
Moisture content, 3
Mud flow, 34fl
Nludline, 343
Murty,
194
Submerged unit weight, 4
Surcharge, 252,257
Swell index, 194
Taylor chart, 357
One-point method,
One-way. i96
Saturated unit weight, 4
Samarasinhe, 90
li
Orthoclase, 5
Overburden pressure, 285
Ovcrconsolidated, 193
Oveiconsolidation pressure, 193
Overconsolidation ratio (OCR), 193
Ianz.cn lolmr-ill, 89
Herreo, 194
Hyclraulic concluctivity, 85, 89
on stratifled soil, 90
Hyclrar-rlic graclient, 85, 88
Lrntediate settlement, 198
Inclination factor, 296; 297
I nllrrcncr- cltart, 77 4
Influencc valuc, 174
Intcrgranular stress, 13i
Is()tropic soil. 96
Strut, 344
Remolded clay, 194
Rendon, 194
Nary-McKay fornrr.rla, 335
Neutral stress, 131
Newnrark, 174
lndex iii
Fundamentals of
Geptechnical Engineerin g
Particle size distribr"rtion curve, 60
Passive earth pressure coefficient, 243,
245,249
Peck's pres.sure cliagram, 345, 346
Permeability coefficient of, 85
I,ile, 335
AASHTO formula, 335
Danish formula, 337
Engineer news record, 336
Eytelwein fornrula, 335
Nary-McKay formr,rla, 335
settlement of ,342
tensile capacity of, J4?
tlreorerical crprrcity of, 338
Pile group capacity of ,341
efliciency ,>f , 342
Sieve analysis, 60
Skempton, 193
Slices method of ,354,360
Slide, 348
Slip plane, 352
Slope circle, 354
Slope stabiliry, 348
infinite slope,349
finite slope, 351
circular lailure, 354
Soil creep, 348
Soil indices, 7
Sorting coefficient, 62
Specific gravity
of solids, 12
Plastic lirnit, 6
Plasticity cl.rart, 60
Plasticity inclex, 7
I']oinrload, 165
rario, 166
Pore watcl pressure, 1Jl
Poi.s.son's
of substance,
Porosity, 2
Preconsolidation pressure, 193
Pressure clistribution, 256
5
Rankine's theory,245
Relative density of granqlar soits,
Ram, 337
2
Spoil, 343
Stability number, 352
Standard penetration (SI,T), 299
Steam hammer,336
Stresses in soil
effective, 131
neutra[, 131
Pr"rnching shear faih"rre, 288
QLranz,
Seiondary compression inclex, 195
Scepage velocity, 85
Settlement, 191
immediate, 191,198
overconsolidated, 193
primary consolidation, 191
normally consolidated, 192
secondary consolidation, 191, 194
rota[, 200
Shape factor, 295,297
Shear strength, 219
Shrinkage index, 7
Shrinkage limit, 6, 11
Shrinkage ratio, l2
total,
5
131
with seepage,
133
without seepage,
131
Textural classification, 55
'Ierzaghi's bearing capacity equation,
289
Time factor, 197
Tirre ratb of consolidation, 196
Toe circle, 354
Toral stress, 131
Triaxial test, 222
Tschebotarioff, 345
Two-dimensiona{ flow, 95
Two-way drainage, 196
Ultimate bearing capacity, 2ti8
Unconfined compression test, 225
Undisturbed clay,194
Undrained triaxial test, 224
Uniformity coefficient, 62
Unit weight
buoyant, 4
dry,4
saturated, 4
soil mass, 3
sul>merged, 4
water; 1
Upright, 344
USCS
classificetion, 56
USDA classification, 55
Vesic, 296
Void ratio,
2
Vale,344
\trater content, .9ee moisture content
\X/ells, 93
artesian, 95
gravity, 94
lndex
Fundarnentals of
Geotechnical Engineering