;/
long-term
stresses
fill
N/ N/ um
wa
10 kN
mid-depth
f ol ol lo lo w in in g
c on o n d itit io io n s
I ni ni titi al al ly ly , b ef e f or or e c on o n st st ru ru c i o (ii) m m ed e d ia ia te te l a f e r c on on s u c o n (iii) M a n y e a f e r c on on s u c o n
!I
(J
("IJ~
r'
'c
~0
\1
\I
(1'.._
17
\/
Unit
.f.. E W 1 1 1 1 p / e
~P
Geotechnics
lO j()
(,
Ie'!'
Co
Problem
oedometer
mp
mc
me
saturated
=2.73).
858 4.493 * ea ea c
e ad ad in in g w a
ke
4.108
3.449
2.608
3432
1716 1.676
()
'~~_~L·~~~_~~_.r _w"
0.737
1.48
e r 24hol1l'S
19.8%: (i) (ii) Determine th a n 1 00 00 00- 1155 00 00 kP kP a . (iii) W h
mi o f m , c oe oe f
al
nt
o f c om om p e s b i y ) for h e
me
fo
a t e r n c e m en en t
Solution:
''
n.541
Wf
Vo
6, 6e
e h
n,
,6,H
i.e.
~;,-~>(& ('1'('('55
1.541
6e
3.520
f~fc
6e
\·-(\:5 (Jr"E
19
1.··,;(1
6 , " " 0 .3 .3 5
0.541
g iivv e
0.35
( " " : 1 , \ . ~ F/'"
3432
0.891
by
--
i.e. (j)
ncrearnent
er
til
table).
.~ 'o
ng C a 10
100
Ef ecti ective ve pres pressu sure re
Unit
1)(,
~OQO
10000
id"i id"i:\ :\
GelJlec'/l}1ic's
}I)/
IJ
Co
Problem
oedometer
mp
mc
me
saturated
=2.73).
858 4.493 * ea ea c
e ad ad in in g w a
ke
4.108
3.449
2.608
3432
1716 1.676
()
'~~_~L·~~~_~~_.r _w"
0.737
1.48
e r 24hol1l'S
19.8%: (i) (ii) Determine th a n 1 00 00 00- 1155 00 00 kP kP a . (iii) W h
mi o f m , c oe oe f
al
nt
o f c om om p e s b i y ) for h e
me
fo
a t e r n c e m en en t
Solution:
''
n.541
Wf
Vo
6, 6e
e h
n,
,6,H
i.e.
~;,-~>(& ('1'('('55
1.541
6e
3.520
f~fc
6e
\·-(\:5 (Jr"E
19
1.··,;(1
6 , " " 0 .3 .3 5
0.541
g iivv e
0.35
( " " : 1 , \ . ~ F/'"
3432
0.891
by
--
i.e. (j)
ncrearnent
er
til
table).
.~ 'o
ng C a 10
100
Ef ecti ective ve pres pressu sure re
Unit
1)(,
~OQO
10000
id"i id"i:\ :\
GelJlec'/l}1ic's
}I)/
IJ
111"
:oo--,-~~
l+
0'0
10 kN
1112
111\,
111
N/
cr'l
IkN
0.055
11l"
to\!
500
1.632
an
h er er ef ef oorr e
MN
1112
MN
an
0,577
C.
0.31
log(l S O O / l (;(\s\(;"l
wi
~-,-(!mt
th
fin i /
{Yomj)/
probl
fo
ag
ranges
the Iinear pan
va
curve.
Geotechnics
-]OJ()
clay mp
figure.
Consolidation
following results:
-- ~--~-
,.".'-,.,-., .. ..-.
O'p
.~:.!
6;;(;1
_b
:k
•
_
_
0_
•
-
0.08
0.06
0.66
0.45
101 kPa
51 kP
!-.( .:
IT?,()
fi:~ <.~- C:2~
= - .s:
Unit
-f.
Exampl problems
ruge
C, Of C/
C\
:i(),'
'l
vu
_!!-~ o u ( .-~!~~ -,-e0,0
-1
(J
.u. 1-
I)
Initinl
()."'!'f/~?:l:'l I'
Recompression curve
nitial
O'[Of
/.Ie
(J;,O'
o:.l
D'r
0-,' (log scale
Layer
C 'I II I / .
(m)
Example
roge .5
Geotechnics
]0 ()
(p
2.04 M g
iI
to
Mg/ vertical stress profile 2)
,0,()
fill placement.
1~
(0
to
.0
Ill!,
wh
II
j()olillg plus
we
mo
Solution.~
./.c.
~,r'7 ,U
~,
(illif.f.
mp
.Pa
Jf
')'.21 (.
(/\',
{::i)
;.:/
• ~ , ' .~ " ~_ ~
• ' ~ • , • , .
• ' A ~
~""__A
••
,1
_~
",
I;)
IS
II
;:
70 I)
;J
(...:i·;~··(L,·~
JJ
,;
"-H'
liG
.-'_
72
19 1(-)
f5
10
110 1:32
11
12
13
~'; IH
{',~
fiil
"l ( ., ,, ~
'f
t.:
i ,. ,' !
"j
tr
'i
'\0';
/.r,:c:s
. .rte
,,j
elf' ./
!It
,\c< :.
V'.
(,
.It· ;::h. (,;..5' 0..•
!Ilh
Geotechuics
JOf()
with 1 0 k PL I .
,
,
,
.
'C
C
.
,
tank. 2) Compute the
Solution:
u nd e
. [ /2
'2"'
?/r"
-x:
·3
(-~,?
.)1..:
'10-
Ci
"l
....
t.init
.P
. -.,-
,_
_.
_,._
..
..
"._
Geolec/rl1ic·.i
.. l011J
l~,'l is loaded u n i f o rm l y
\b
by
hl._
tooting :5
SolutIOn:
3,-,
(3
,ojl.t,A
: ,J .
(I'J H,'
\/1':",'(
(r":..c 1(!)J~
",
r'~"
Vi'l
J_t \. .!..
ot th
y"
H,,'
:\
\(' ''(d
~n lv mc mp
r'\\'
o>,""V
«(/~
r·.It.;I·(t.~
!J
IAt'
.,\I('I,y:",
,F
(,
~«(yv)
fY1" t'\
z:
L~
/' ..
0,22
_"
f.'.1
fh
.J
v,::,~'; y.
..{!'r.\,
s:
r=:
'i)
('l2d(H"~vk-,,·k)~1(l.fl1 0:'5
1-,,0
I..../;e
Ikx~
6(
zr
\66
0,25
r''':
0.01
} "I I
.J,
10
Geotechnics
20}O
,.
f ~ -
,"
~_
10111 area
ml
with
kl'
th
de
po
Solution: In
.j-t'·I('
"j ;,.,.
...J
---_)
....
Unit .J
I!
Geotcchnics
]0;0
~.
,.
T"""'"'~''''''~'-''''''!''' J_
.a
15~1¥lllf...
0,26
T · · · · · · r . . . .· : · · - T · · T ~ J T . E l i . .. . .. . .. .. j . .
-+-! ······!-i .;
• • .• •
.................
_ • .• . L
interchangeable
..
0.05 ····'-----~·--·,··i··-·,-·~-j··
1-----~~..
.·
O.OO=~=:::0.01
Area 1_
~lt~em~+~A_C
Z
_ G _ ~ ' " [ = ! ~ j J T ~ + " " _ ' = _ A _ B _ F G _ " " " ~ ~ _ C- _A D _ ' _ !
5
5
5
5
Biz
3
2
LIz
1
1
-20.4
-20.0
0.238
23.8
E.\"({IIIJ)/e
.Pa
17.5
Gcotechnics
Three u n if or m l d is tr ib u te cl lo a cl s of profile shown in h e g u e . U nd i u rb e mp th average 0.50, an me
10
to construction,
we
n so l
on
soil
th
an
bo
the clay layer olllym.1_(JQLthG._g_QnJIQ.Q.L~hqmi5hlJ<:JQ'JdQ~L.'II9'\'ssume clay 0 .8 0 Plan:
.. 10,,
Profile:
Solution:
0.01
Unit ·f
10
Geotechnics
j{ (''c,
l{[)
di
c;
r~,Jclte
J . ( e , ~ ' : ; ,cel'it
fOQclf'cl
lu
C (A l
c"A t I" }
() t! (~ ({ ')
> /1 1 1 (
i(ll,
1)
'f
5",
'/
:,
025
020
"-'i"
ec
If
"j
i"
~ I = l i ~ : ~ ~ ~l j , ]I. J~I ~ ~~3 ; ~J. f :-, i ,fr ,l: ,lJ ;,. ~ ~ ~ ~ ,,_~!~~~!._~~:,~~g~~5'_!~J+ii:,i /;;--(-_': :~:FE
=-?~
t- ne:cD
sr
SD ((/
!l
I-
L"
''''
0-'/'1
Iff. ,G
v'C\
2,;;>
'~
t: '::: J?;;,
r-
.. c""
:z__'~,
,(.-"1
Ix /'
c_,'::,
Uni! -I
f]
Geotechuics
3, ]OiIJ
v,
1 · , 1 1 f,p"
(,
62""
c-
/1
?~, I{
;:,
I:::
,_
I>
,:,
Problem Ch
zz:
Ex
/s
%) f.,.
':0:>
F:)
\'
riA
,u,o\(' z.
.\
·:::..rt:i~{
Ie:·"
_;f...
J)
c~/
'(J:~.
c_,-OJ
01(')
t'f\
ch
//i{
Cr
0-1)
1.5 -2
IcG
00;/ "/ -,
C.V:::
II
II
rn'
.....
;.,~.~~ iL
p,'A-'<
g* b-E_~ 'b
5 G f, ,
· 2 11 .j I(0
0.1
'0.2
0.3
Unit 5. Exampl proble
Page
1.
.E
nsol dation
at o, Uz
Geotechnics
2010
lith
a f e r h e p la ce m en t
t.
ofthe fill.
(~, ~\~:-~ol.::)?(.. o J r ( ': ~{S
fIt
Zll
1-U
.!..L
lY
....'>..,""",.,'".,.":
.·:i~;\' \ i ' : z N ' : ' 9 , ' : : l 'i W ! i ; ;
,'@i/i;/ .,':,t-,:"i
c(t.'2
IS .0)'7.
U""
u,,+u.
(Fig)
(m
; ;: :
1I
:;·;9'.····.·,( i;'i'!'Qij,:
! :; '
0.13
,,10
1.0(
::~1
0'.56:
._......
0.65
10
1.6
0.35
14
\0 0.13 Co
ol
on
.0
o. Uz
Geotechnics
2010
u,
(kPa)
15
u, ""
15
so
in Da
wn
Total 0"
::
(mm) kN/m
0.622 cr'l
1.244
2.468
kN/m2, H o
3.400
3.838
3.970
4.000
81 4.051
1440 4.100
mm
Determine:
(I c,
(ii) (iii)
Unit 5.
pl
th
..Pag
m e plot in log time plot in
/year,
/year,
Geotechnics
..2010
Solution:
Sqrt (time) (min1/2) 10
0.5
::::
Q.)
2.5
Q. Q.
U)
3.5
;.6qu.
-;
4.0 - _
)i:"
lIfO.
0.01
0.0 0.5
1.
Q.
2.0 2.5
Q.)
t/
1..\
.f-Q
-F
4.5
C":·,
-:>
Unit 5.
xa pl proble
Page
Geotechnics
20/0
T im e
wa
ur
S(Y%
154 .5 rnrn, cr'o
kN/m
and
N /m 2
0"
Ho
mr
Determine:
(i)
mis,
Solution:
~.
\_.\.
C,I.,1
i~
.
./
(_
Unit
Geotechnics
2010
P ro b
5. To
settlement
on ol da
.8 :::
0.28 O 'Y O
yO':::
kPa C,
Wh
0.02
me me
As me
mp me
mp
ma
me
ma tp
Solution:
We
e t e me n
ma
o m os::::::
5...!L._ +e
0, = c ~ c S : c I-l I+
me
nd
c on da r
c om p e s o n
me tp
mi /./,
I/i:::
\.
.~
.. -.-~
f -, '
Unit 5.
Geotechnics
2010
Depth
Soil profil :-
(111)
:- :-~__:_- "' . "
li n ed . T h om
Ch nd
C,
clay on ol da me
c,
~"
. 1: 8 . M g / m
clay 10
01'3,
111.
Ho we :-'.",
15 ;t'l
Dense'
.s
20
.,
i·l/. '1
'.
..c: "0';
C on so li da ti o
r au e u .
~ 0.238
om
15
::::>
~s
0.03l
0.477 0.561
1\
::i------------+--"~ 100
Unit 5.
01
04
:~=tJ
Co
OR
Geotechnics
10
1.2
2010
r.
clay
his
fficicnt
cm
(l) m p amount.
ma
me wh
zz:
.----~
O.eM 1~
0.008 0.018
25
o.o.~
10
60
1"-..
80
0.2
0.096 0.)2~ 0160 0.)96
3~
<,
40 45 50
r-.
o.~
0.280
95
1)29
0.342
OA03
OA~rl Oc567
0.~O4
ion
O.S
C.O
60 65 70 75 80
,,'
10
l'
ConsoHdiJlion
Unit 5.
ratfo,
Geotechnics
20/0
of
mp
1.1;
::::
Un
ma
1.52 M g /m :
0.06; 0.858 9.81 m/s".
fill
c.,
"2
""
Mg
Required: (1) II.
me
me
r/
avg ""
Th
mp
50%.
me
Solution:
so
pruli!e
100
50
?O
.....-.,,-' . 4 ~ ~ ~ - - = = = : . . . - - 34
...
mid-depth
--,-----\~-.. ovo... _._~
'.
20
76_5 kPa
Unit 5.
1r;~
~n
174.6kPa
Geotechnics
20
ro
silt SM
TABLE·1 •
(1
•
(2)
.;
(3
s,
Tv • •• •• • , •• • "
' " ~ "" "" " -
"U
__
" _" ~
0,031
.}
0.071
8.26
0.4
0.5 0.6
0.197 0.287
0.86
12.92
0.7
OA03
.20
26.42 37,[7
0.848
1.54
55.59 76.25
0::
1.62 1.7J
0.8 OS
1.00
·10
are into th
av
0,52
0.2
~_ ........, '............ . n '_ ~ _ ~ _
substitu ed lu
•• ~-._ .__,._ --- -----.,., _.- +--.-
,[
.10
(4)
..~
multiplying 1m
( oc )u l
'Xi
. . _ .~ , . . .. . ,. . ~ _. . _ _ . .~ ._ _ , ._ . _ _ _ _ . _. .. , _ ._ . _ . . . _ ._ , ,_ , ._ . _ _ _ . ,. _ .u . .. ._ . ~ _ .~ _
II:
(': ;r,
:i.e'
Unit 5.
Data from TABLE-1
I(
Geotechnics
20
u.,YQ
50(%, • _ ,
(I) Depth
(2)
z/
lJ
(3) U,
• _
L _ _ •
_ _ _ _
c_~·__•
(<1) .6
(k Pa)
i.oo
·-5
98.1
--6.88 -8_75
OX)
-10.63 -[2.5
0_75
0.285
28.0
1.0
0.23
22.6
···-14.38
1.25
0.285
28.0
0.455
41.6
0_70
68.7 98_1
o.s
-16.25 -~[8.[J
--20
U5 2.0
1.00
~\Oll
l(iO
100
prolil{t
'< .. ~;)y!~r
ll..'J%
AI
U"... soc:.t1(on
)1
'if.
!~{)',;;:" (>IH~:~j~~~IlllLl)lJfL
.:'l
uuc w - v '
,11m
how,"; c-milif1m~) ~\lJ 1U b~~ ~i:;_;;;tp[)!nd.
10
.~~
\.\.
\,
/G.b I){~11r)(j~tf:
lever
25
0···
Unit 5.
Geotechnics
2010
Unit 5.
2010
Pr
9. La
ma .5
final
s pe c
87%,
om
de
C,
1,2, n d
oe +5
nt
on ol da
g ra v /yr.
G,
2.65,
CD Me +1.5
l ev e CD
inal seabed level ""
-6
CD
Cc
De
1.0
/y
'" 87% 2.65
Sa
(1) me ye
in
Wh
-4
CD?
No
As me
bvr:)
1)
0:
N/
:2'1
;-
;).1)
.b
y ou r
al ul
on
hi-c< ;¥
l-
") C"
t' " i - - .
c:-'(
,-.J
., _-
Unit 5.
ampl problems
Page
Geotechnics
20JO
\_.?>
~J'-.
~.-~____.-""
tt:s (JellV
eO
(;«('<;.;,
St
'.
.. ..
. . . .. .
(-
.f
2"-:~::"i~=:~'_·'~:'~'~::"_"L.JI 0,7
o ns ol id a i o --:--;
r at io , U z
0.6
.9
.z /-/d
.r
Ck\.Jo\i(.:Vl
&W
\'
Q.
l.'e_
p r E " ~ - u c.
I,SIY'
""f.J([oc
1\
z:
CD
C}
-1
_)
6.712
.'
VY,
L.
'.r.. If
-.
Unit 5. Exam le prob em
..P ge
Geotechnics
..201 (J
1.0
~Ex Problem 1 . S u r h a g e p r o ad in g me
ii.c,
Embankment
7m £
~
Soft clay
6m
:::::
mv:::::
kN/m
y:::
3m /year 0,Q004m /k
permeable Solution:
,\
"L.'l6
~.-.--.~.~-.-~.~
(,
2;,J(c
f.
(;,1
k t M ! : .] . ; ' k ~ ' . - 1 , , : ~ · ~ J
,e
/'/
)'
It :.
Unit
~P
, c- " ~
/'
..l. Geotechuics
f(
Problem
Ver mp mm me
me /year.
Ci
.~_(:U:~j )n /"
uSc: c hc "
~(""efi'·
"'
Solution:
1.12, . / 'It"
Yn t.
J . ) [
O.e-l
( ,) ,
r'"(y!):.
('112.73) 1(.(3/2.2((?) t\ le
I:,trtrl
LJI,}A
/(21?') c/;'/(/
1'''\;, 65
,55 /1':'
~:
\1 11
(:J.x'r:~~>/'-.--""
....t.
{r
::'( :!
';
Unit
mp
.Po
Geotcchnic«
]0 J(
an
cross mp me bl
concrete
permeable
oi
um (longest an
wa
shortest)
exit)
l_
(ii) (iii) (iv)
Ho Determine
th
channels th mb
mp (Nr) a n d e q u ip o te n ti a
th lS
to b e i so tr op ic . ma
two
is denoted
(v)
net, mi drops (Nd) have
by n e D . soil. to
irk fo th
wa
ot
111/s.
+13.5rn
(I)
bO'Jf><\("
dn:t
('o(lSfc,tl/1f!oc/l/n,s
-fld",o
I; sr-dell
II .\te.
h.,··
1-
JleQ.J
1;),$·-15
~~
io$) s:
£t.
dn,\:l
1)1"
15)(
12....
t:v-Jo
{lOl-C41tl4'
LY
;;(1 Iii)
h.~e:s
.:s;j",,·r6+)
+·1.5m tOrn
F'j t)w
Geotechnics
i/11e t1 ¥ v : : . c l , : t ,
clrol....
uplift Gi
me
~=
(Nr)
mb
(i) (ii)
,6
now
10'1
et
(Nd)
adjacent equipotentials
th h) he po
(iv)
wa ze
ur
at h e p o n t
wi
of
clam.
6.
In
10.5 rn
Impervious
mp
ms
.P
Gc'Oiec!lIIlcs
.!() Ii
piling permeability [0-"1 m / w a e r ( re du ce d by mb
coefficient
wa O.Sm.
flow
wn
mb
by
I\j.
Calculate
(ii) h e o ta l
(i)
.J
UJ
[\(1111)/0'
1'oge
On mp
O.50m
~~~t':_Jrn
Geotechnics
Problem me me
12
downstream heave Ca
S)
wn
heave,
given rhat
YSill
fo th
C 1 .1 . ; '
yi ;t_;-
~p_-
/'_0
".
~7
'/:;-6/-
{.Jt;e1t (~5--
Unit
7.
E xa m
j ) o /J /e lJ l
F og e
.j
Geotechnic:
}IJ
Consider
th
mo
if t h BC is
wi th
dashed
twice
75
wi
portion
cross section f ol lo w in g ar (rue:
40111.
now
th
head w.
j""l.
"f.~?
h:.-"J')fi'l" {i:c)(
h~:" ' " 1 5 z:
rl'b(
:»
z:
t5
C;?,
tdei\Jq
VY'1
B<
75'" 7;;"__-:::
T-
78_,
If
.:s.;s
7-j;
-c
66/ 2 S " { ' 7 ;;?~
'''7
r-:"
ru
(,:D,., j,
Un
Geotechnics
tP_B
::01
