Chapter 1 Soil Exploration Symbols for for Soil Soil Exploration C B ĺ STP correction factor for the boreholes diameter. C R ĺ STP correction factor for the rod length. C S ĺ STP correction factor for the sampler type used. cu ĺ Soil’s un-drained cohesion. D f ĺ Depth of the foundation’ foundation’ss invert. E m ĺ
The efficiency of the STP hammer.
N ĺ The “raw value of the STP !as obtained in the field". po ĺ The original vertical stress at a point of interest in the soil mass. S ĺ The number of stories of a building. a building. SPT ĺ ĺ Stands for “Standard #enetration Test. N 60 60 ĺ$orrected STP assuming %&' efficiency in the field. N 70 70 ĺ$orrected STP assuming (&' efficiency in the field. mĺ
$orrection factor for the shear vane test using the clay’s #lasticity )nde* PI .
1
1
*Exploration–01. Find the required number of borings and their depth. !,evised Sept. &"
A four story reinforce reinforced d concrete concrete frame office building building will be built on a site wher wheree the soils are expected to be of average quality and uniformity. The building will have a ! m x "! m footprint and is expected to be supported on spread footing foundations located about 1 m below the ground surfac sur face. e. The sit sitee app appear earss to be in its natural natural con condit dition# ion# wit with h no evidence evidence of pr previ evious ous grading. $edroc% is !&m below the ground surface. 'etermine the required number and depth of the borings. Solution:
/ reinforced concrete building is heavier than a steel framed building of the same si0e. ence2 the design engineer will want soil conditions that are at least average or better. 3rom Table-1 below2 one + boring will be needed for every 4&& m of footprint area. Since the total footprint area is 4& m * 5& m + four 612+&& m 2 use borings.. borings &.(
&.(
Table-+ provides the minimum depth re7uired for the borings2 8 S 9 D 6 8!5" 9 1 6 14 m. :ost design engineers want one boring to go to a slightly greater depth to chec; the ne*t lower stratum’s strength. )n summary2 the e*ploration plan will be 5 borings to a depth of 15 m.
Table-1 Ta ble-1 - Spacing of the exploratory borings borings for for buildings on shallow foundations. shallow foundations.
Subsurface !onditions
Structural footprint Structural footprint Area for Area for Each oring Each oring +
+
!m "
! ft "
#oor 7uality and < or erratic
+ &&
+2&&&
/verage
4& &
42&&&
igh 7uality and uniform
%&&
%2&&&
TableTabl e-" " - #ep epth thss of ex expl plor orat ator oryy bo bori ring ngss for buildings on shallow foundations. shallow foundations. $inimum #epth of orings of orings Subsurface !onditions
!S 6 number of stories and D 6 the anticipated depth of the foundation" !m"
+
!ft"
&.(
9D
+&S
&.(
9D
18S
&.(
9D
1&S
#oor and < or erratic
%S
/verage
8S
igh 7ualit y and uniform
4S
&.(
9D
&.(
9D
&.(
9D
+
*Exploration–02. The sample’s disturbane due to the boring diameter. !,evised Sept. &"
The most common soil and soft roc% sampling tool in the (S is the Standard Split Spoon. Split spoon tubes split longitudinal longitudinally ly into halv halves es and permit ta%ing a soil or soft roc% sample. The tube si)e is designated as an *+. The *+ outside diameter is 'o , -!. mm /0 inches and its inside diameter is 'i , ".2 mm /1&3 inches. This small si)e has the advantage of cheapness# because it is relatively easy to drive into the ground. 4owever# it has the disadvantage of disturbing the natural texture of the soil. 5n soft roc%s# such as young limestone# it will destroy the roc% to such a degree that it may be classified as a 6sand7. A better sampler is the Shelby /or thin&tube sampler. 5t has the same outside diameter of 0 inches /although the trend it to use inches. Compare the degree of sample disturbance of a (S standard split&spoon sampler# versus the two Shelby thin&tube samplers /07 and 7 outside diameters via their area ratio A r /a measure of sample disturbance. Solution:
The area ratio for a +=-standard split-spoon sampler is2 + + + +.& 1.4 D + D o i A !'" + !1&&" 11&' + !1&&" D r 1.4 i The area ratio for a +=-Shelby-tube sampler is2 + + Do i D +.& + !1&&" 14.' !1&&" + Ar !'" Di 1.(8 + 1.(8 +
The area ratio for a 4=-Shelby-tube sampler is2 + + Do i D 4.& + !1&&" .>' !1&&" + Ar !'" Di +.(8 + +.(8 +
$learly2 the 4 ?-D Shelby-tube sampler is the best tool to use.
*Exploration–0!. "orreting the S#T for depth and sampling method. !,evision Sept-&"
A standard penetration test /S8T has been conducted in a loose coarse sand stratum to a depth of 19 ft below the ground surface. The blow counts obtained in the field were as follows: ! ; 9 in , " blows< blow s< 9 &10 in , 9 blows< 10 &1 in , blows. The The tests were conducte conducted d using a (S&style donut hammer in a 9 inch diameter boring with a standard sampler and liner. The effective unit weight of the loose sand stratum is about 2. pcf. 'etermine the corrected S8T if the testing procedure is assumed to only be 9!= efficient . Solution:
The raw S#T value is N 6 % 9 6 15 !that is2 only the last two sets of % penetrations". The @S-style donut hammer efficiency is E m 6 &.582 and the other parameters are obtained from the Tables Ta bles provided on the ne*t page C B 6 1.&82 C S 6 1.&&2 C R 6 &.8. Aith these values2 the S#T corrected to %&' efficiency can use S;empton’s relation2
Em C B CS C R
N N %&
& .58
&.%&
1.&8
1.&&
&.%&
& .8
15
>
Botice that the S#T value is always given as a whole number. That corrected S#T N 60 60 is then corrected for depth. 3or e*ample2 using the Ciao and Ahitman method !1>%"2
N
%&
+2 &&& lb < ft + dept dep t effecti!ee u" effecti! u"it it # ei$ t
N %&
>
+
+2 &&& lb < ft 1& ft >4. pc f
1%
?ther methods for corrections are discussed in *ploration-&5.
S%T &ammer Efficiencies 'adapted from !layton( 1))*+.
Cou"tr%
&'mmer T%pe
Rele'(e )ec'"i(m
&'mmer Efficie"c%
/rgentina
donut
cathead
&.58
Era0il
pin weight
hand dropped
&.(+
$hina
automatic
trip
&.%&
donut
hand dropped
&.88
donut
cat-head
&.8&
$olombia
donut
cat-head
&.8&
Fapan
donut
Tombi trigger
&.( - &.8
donut
cat-head 9 sp. release
&.%8 - &.%(
@G
automatic
trip
&.(4
@S
safety
+-turns on cat-head
&.88 - &.%&
donut
+-turns on cat-head
&.58
donut
cat-head
&.54
Hene0uela He ne0uela
!orrection ,actors for the oring #iameter( Sampling $ethod and oring od ength od ength 'adapted from S/empton( 1)0+.
Correctio" *'ctor
E+uipme"t ,' ,'ri'ble( ri'ble(
,'lue
Eorehole diameter factor C B
%8 I 118 mm !+.8 I 5.8 in"
1.&&
18& mm !% in"
1.&8
+&& mm ! in"
1.18
Standard sampler
1.&&
Sampling method factor C S
Sampler without liner !not recommended" ,od length factor2 C R
8
1.+&
4 I 5 m !1& I 14 ft"
&.(8
5 I % m !14 I +& ft"
&.8
% I 1& !+& I 4& ft"
&.>8
J1& m !J4& ft"
1.&&
*Exploration–0$. Three methods used for S#T depth orretions. !,evision Sept.-&"
A raw value of 2 of 2 , "! was obtained from an S%T an S%T at a depth of 0! feet in a sand stratum that has a unit weight of 1- lb3ft . Correct it only for depth. Solution:
/ny of these three methods will provide acceptable answers. Botice how similar their results are from each other 1. (sing the $a)araa >ethod >ethod /129?: K if p& d 1.8 -ip( < ft '"d 5 N + N corrected 1 + po K
5 N
N corr ected ected
4.+8 po
if p&
t
&.8
1&&& lb < -ip K 5 N corrected terefore N 4.+8 &.8 po
+.(& -ip( < ft
K
corrected
N
1&
N
or
!
1.8 -ip( < ft +
4.+8
&.8!+.(& -ip( < ft "
is in tons < ft
if p
#ere C
NC
+
5!5&"
0. (sing the 8ec% >ethod >ethod /12?": &.(( log +& N
+
4
!+& ft "!148 lb < ft "
but p&
1.8 -ip( < ft
48
+
&
p&
&.(( log 1 >1 is in -N < m >1 8 if p 1& & p&
C N
+
+
4
but
!+& f t "!14 8 lb < ft "
p&
1.48 to"( < ft +.(& -ip( < ft
+
+
+&&& lb < to" &.(( log +&
?
1&
C N
&.>&
1.48 to"( < ft
+
N
?
!5&"!&.>&"
4%
corrected
. The @iao&hitman >ethod /129# as used in Explorationin Exploration-*3 *3## + K in p(f in -N < m or N +2 &&& p(f K 1&& N N with p with p corrected o o po po but
+
p
!1.48 to"( < ft " · o
%
§
¨
>%.1 -N < +m 1 to" < ft
+
1+>.( -N < m
+
¸
%
©
¹ +
?
1&& -N < m 5& + 1+>.( -N < m
N
corrected
(
48
(
*Exploration–0%. S#T orretions under a mat foundation. !,evision Sept.-&"
Correct the S%T values shown below for an energy ratio of 9!= using a high&efficient (S&type donut hammer in a 07&diameter boring. The invert /bottom of the mat foundation is at elevation B-.0 feet.
9+&’
Lround Surface 914.+’ 91&.>’
9 (’
Aater Table
98.+’ invert
91&.&’ Sand Sa nd9 9 gravel
T , .-
N 6+%
95.1’
N 6+8
9&.&’
Soft clay
N 6+5 N 64&
:edium sand -1&.&’
-+&.&’
ard clay
N 641
of
Solution:
S;empton proposed in 1>% the following correction for the sampling methods to the raw SPT value2 assuming that only %&' of the energy of the hammer drives the sampler2
N %& E m C B C S C R N &.%&
where N 60 -value corrected for field procedures assuming %&' efficiency 60 6 S#T N -value E m 6 &.%&
o
C B 6 1.&&
o
C S 6 1.&&
o
efficiency for a high-efficiency @S-style safety hammer borehole diameter correction sampler correction2 6 &.(8 !1&’-14’"
C R 6 &.8 !14’-+&’"o rod length correction2 6 &.>8 !+&’-4&’"2 6 1.& !J4&’" N 6 S%T -value -value recorded in the field by the driller !;nown as the “raw S%T ". ".
The depth correction is2 +
N %& 1
N %&
+2 &&& lb < ft t effecti!e u" it # ei ei $ d e p pt u"it $tt
/t depth of 98.+ feet N %&
!&.%&"!1"!1"!&.(8"!+%" !&.%&"!1"!1"!&.(8"!+%" +& '"d N &.%&
/t 95.1’ N %&
/t9+.&’ N %&
/t -1.&’ N %&
/t -8.&’ /t -1&’ /t -+1’
!&.%&"!1"!1"! .%&"!1"!1"! &.( 8" ! +8" &.%&
! &.% &"!1"!1"!& &"!1"!1"!& .(8" .(8" !+5" &.%&
!&.%&"!1"!1"! .%&"!1"!1"! &. 8" !4&"" !4& &.%&
+&
%&
1> '"d N
%&
1>
1 '"d N
%&
1
%&
+%
%&
N %&
! &.%&"!1"!1"! &.%&"!1"!1"! &.>8"!4&" &.>8"!4&" +> '"d N &.%&
%&
+>
&.%&
48
> ft 1+( %+.5 pcf +
41
11 ft 1+8 %+.5 pcf +
4>
15 ft 1+% %+.5 pcf +
+%
+2 &&& lb < ft 1 ft 1+% %+.5 pcf
45
+
+2 &&& lb < ft +4 ft 1+% %+.5 pc f +
54 '"d N %&
%+.5 pcf
+
+2 &&& lb < ft +% '"d N
+
1+(
+2 &&& lb < ft
!&.%&"!1"!1"! .%& "!1"!1"!&.8"!41" 8"!41" +% '"d N &.%&
!&.%&"!1"!1"!1" !54"
ft +2 && &&& & lb < ft
N %&
N %&
+2 &&& lb < ft
54
+2 &&& lb < ft
44 ft 14& %+.5 pcf
Botice that the depth correction does not affect the deeper layers.
51
45
4>
*Exploration–0&. The Shear 'ane Test determines the in(situ ohesion. !,evision Sept.-&"
A shear vane tester is used to determine an approximate value of the shear strength of clay. The tester has a blade diameter d , .90- inches and a blade height h , ?.0- inches. 5n a field test# the vane required a torque of 1?.! ft-lb to shear the clay sample# which has a plasticity index of "?= / % % 5 6 %. %. 'etermine the un&drained cohesion cu corrected for its plasticity.
"?
T
c u
+
S ª!d <
+"
1(.& ft l b 1(.& ft 4
!d < %"º ¬ ¼
S
« ¬
ª !&.4&+1 ft "
+
+
1% p(f 4
!&.%&5+ ft " !&.4&+1 ft " % ¼
º
»
The plasticity inde* helps correct the raw shear vane test value !EMerrum2 1>(5" through the graph shown above. 3or a plasticity inde* of 5(' read a correction factor P 6 &.&. Therefore2
cu corrected
Pcu
!&.&"!1% p(f " 145
p(f
*Exploration–0). eading a soil boring log. !,evision Sept.-&"
Dead the boring boring log shown below and determine determine## /1 the location location of the phreatic phreatic surface# /0 the depth of the boring and / the number of samples ta%en. Solution:
!1" The phreatic surface !the water table" was "ot encountered in this boring and is noted at the bottom the bottom of the reportN !+" The boring was terminated at +1 feet in depthN and !4" 3ive samples were ta;en. ?nly one sample !O+" was used for laboratory tests !dry density and moisture content". Samples O1 and O4 were complete split-spoon samples. Samples O5 and O8 were incomplete split-spoon samples.
engineering parameters. parameters. *Exploration–0+, -sing a boring log to predit soil engineering !,evision Sept.-&"
(sing the boring log and the S%T the S%T versus versus Soil Soil Engineering %arameters Table Table shown on the next two pages# answer these four questions: /1 Correct the values of the S%T of Sample S&" to a ?!= sampling efficiency with a standard sampling method and a (S&donut hammer at elevation ; 1? feet< /0 Correct the same sample sample S&" for depth assuming the unit weight weight is Ȗ , 109 pcf< / hat are your your estimates for the angle of internal internal friction and unit weight Ȗ /" hat is the elevation /above sea level of the groundwater and the elevation of the bottom of the boring Solution:
!1" The log shows a value of N 6 18 !Sample S-5" at elev elevation ation -1%.8’N -1%.8’N at eleva elevation tion -1(’ it has drop dropped ped a small amount to N 6 15. Botice that the “Cegend portion denotes that the sampler was a + ?.D. split spoon. Therefore2 the sampling correction is2
&.58
EC EC S C , N
1.&
1.&
N (&
&.(&
&.(&
!+" $orrect the same sample S-5 for depth. B B +&& & p ( f +&& & p ( f (& (&
J
&.8
15
| P
1+%
1(
p( f
!4" Ahat are your estimates for the angle of internal friction and unit weight ȖQ The log identifies identifies this level at -1(’ as a “brown “brown and grey fine to medium S/BD. S/BD. @se the Table provided on page +4 to obtain an estimate esti mate of some of the t he engineering parameters for granular soils. ,ead the SPT for medium sandsN sandsN then go to the :edium column column and read the value of “ N 6 to obtain the values 0
6 4+R and Ȗ#et 6 1( -N/m . !5" Ahat are the elevations !above sea level" of the groundwater and of the bottom of the boringQ the boringQ - The boring table.
did not report finding a ground water
- The bottom of the boring was at -4%.8’ from the surface2 or 45(.&’ I 4%.8’ 6 941&.8’.
1+
!orrelation between S%T 7alues and some Engineering %arameters of 8ranular Soils 8ranular Soils #escription
9ery 9e ry loose
oose
$edium
#ense
9ery dense
*.0<
#r
el ati7e density
*
*.1<
*.3<
*.<
S%T
fine
1-"
3-
; - 1<
1 - 3*
'2 ;* ;* +
medium
"-3
4-;
0 - "*
"1 - 4*
coarse
3-
<-)
1* - "<
" - 4<
I
" - "0
"0 - 3*
3* - 34
33 - 30
fine
"; - "0
3* - 3"
3" - 3
3 - 4"
medium
"0 - 3*
3* - 34
33 - 4*
4* - <*
= 4*
> <*
coarse
Jwet
;* - 1*"
0) - 11<
1*0 - 1"0
1*0 -14*
11 - 1
14 - 10
1; - "*
1; - ""
Bote O1 These values are based on tests conducted at depths of about % mN Bote O+ Typical values of relative densities are about &.4 to &.(N values of & or 1.& do not e*ist in natureN Bote O4 The value of the angle of internal friction is based on ? 5 "0@ 1<@#r N 4
!orrelation between S%T 7alues and some Engineering %arameters of !ohesi7e Soils !ohesi7e Soils
14
S%T - 2 ;*
!ompressi7e Strength Bu
#escription
*-"
> "< /%a
,er% (oft 1 (+uee2e( bet#ee" fi"$er( bet#ee" fi"$er( ,er% %ou"$ NC cl'%
3-<
"< - <* /%a
Soft 1 e'(il% deformed b% fi"$er( b% fi"$er( 3ou"$ NC cl'%
-)
<* - 1** /%a
)edium
1* - 1
1** - "** /%a
Stiff 1 &'rd to deform #/fi"$er( Sm'll 4CR 1 '$ed cl'%
1; - 3*
"** - 4** /%a
,er% Stiff 1 ,er% 'rd #/fi"$er( I"cre'(i"$ 4CR 1 older cl'%(
= 3*
= 4** /%a
&'rd 1 Doe( "ot deform #/fi"$er( &i$er 4CR 1 ceme"ted cl'%(
1"0 14; "* "3
**Exploration–0. Find the shear strength of a soil from the "#T *eport. !,evision Sept.-&"
Classify a soil from the data provided by the Cone 8enetration Test /C8T shown below at a depth of 11 m. The clay samples recovered from that depth had Ȗ , 0! %*3m and and % % , p , 0!. Compare your estimate of the shear strength versus the lab test value of --! /%a /%a..
Solution.
,eading the data2 + ( 5&& -P' and + c 11 )P' which results in a f R 4'. 3rom the ne*t chart2 the soil appears to be a (ilt% cl'%.
15
15
/t a depth of 11 m2 the in-situ pressure po for a B$ clay is2 po
J
4
!+& -N < m "!11 m "
++& -P'
2
3rom the N - versus I graph2 for I 6 +& yields an N 1(.8. p p - The un-drained shear strength (u is2
18
18
(u po
+c N -
1%
P ' 112 && &&& & - -P'
++& ++ &
%1% -P' !er(u( l'b
88& -P' !a 1+' error".
1(.8
1%
Chapter 0 8hase Delations of Soil Symbols for for #hase #hase elations of of soils soils eC 8 S C
9oids ratio. Specific gra7ity of the t he solids of a soil.
nC
%orosity.
S C
#egree of saturation.
9C
Total 7olume 'solids water air+.
9 a C
9olume of air.
9 9 C
9olume of 7oids 'water air+.
9 S C
9olume of solids.
9 D C 9olume of water. wC
Dater Da ter content 'also /nown as the moisture content+.
D S C
Deight of solids.
D D C Deight of water. g C C nit weight of the soil. g d C #ry unit weight of the soil. g b C uoyant unit weight of the soil 'same as gF as gF + . g SAT C nit weight weight of a saturated soil. g D C nit weight of water
/asi "onepts /asi "onepts and and Formulas Formulas for for the the #hases of of Soils. Soils. 'A+ 9olumetric elationships: 9olumetric elationships:
1. - 9oi 9oids ratio e
e
, ,
+-1
, S
ranges from & to infinity. Typical Ty pical values of sands are very dense &.5 to very loose 1.& Typical Ty pical values for clays are firm &.4 to very soft 1.8. ". - %o %oro rosi sitty n
"
, , ,
1&&'
+.+
ranges from &' to 1&&'. The porosity provides a measure of the permeability of a soil. The interrelationship of the voids ratio and porosity are given by2 '"d " e " e 1 1 " e 3. - Sa Satu tura rati tion on S S
+-4
S
, 5
1&&'
, ,
+-5 ranges from &' to 1&&'.
'+ Deight elationships: Deight elationships: 4. - Dat Dateer co cont nteent w
5 5
#
5 S
1&&'
+-8
Halues range from &' to over 8&&'N also ;nown as moisture content . <. 6 <. 6 nit nit weight weight of of a a soil Ȗ
5
J
5 S
+-%
5 5 ,
, S , A
, 5
The unit weight may range from being dry to being saturated. bul/ l/ de dens nsit ityy ȡ G to refer to the ratio of mass of the solids and water Some engineers use “bu contained in a unit volume !in )$
J J
+-(
1 #
5 S d
,
The soil is perfectly dry !its moisture is 0ero". ;. - The unit wei ght of water
5 5
J #
Ȗ w #ere J
U $ ! *
m' "
, 5 J
#
%+.5 pcf
1 $ < ml
1 -$ < liter
+-%
>.1 -N < m
4
Bote that the above is for fresh water. Salt water is %5 pcf2 etc. 0. - Saturated unit weight of a soil Ȗ sat sat S
J
5 , , &
SAT
S
A
5
+-
5
). - uoyant uoyant unit unit weight weight of of a a soil Ȗ b J
b
J
K
J
#
J
+->
SAT
1*. - Sp Spec ecif ific ic gra7ity gra7ity of the solids of a soil 8 J S
8 S
+-1&
J #
Typic Typ ical al 9alu 9alues es for for the Specific 8ra7ity of $i nerals in Soils and oc/s and oc/s $ineral
!o m po sitio n
Absolute s peci fic g fic gra7ity 8 s
/nhydrite
$aS?5
+.>&
Earites
EaS?5
5.8&
$alcite2 chal;
$a$?4
+.(1
3eldspar
G/CSi4?
+.%& to +.(&
Lypsum
$aS?5 ++?
+.4&
ematite
3e+?4
8.+&
Gaolinite
/l5Si5?1&!?"
+.%&
:agnetite
3e4?5
8.+&
#b
11.45
Si?+
+.%8
?rganic
1.& or less
S;eletons of plants of plants
+.&&
Cead uart0 !silica" #eat Diatomaceous earth
Hthe Ht herr us usef ul formulas dealing with phase with phase relationships:
Se
#8S J
e
(
1
J dr%
9"it #ei$t rel'tio"(ip(
!1
J
#"8S J
!8S S e"J
#
1
""!1
#"
#
#
#
1 e
e
8S J !1
#"8S J
!1
1 #8S S
S'tur'ted u"it #ei$t( !8S #· § e · § 1 J J ¸ ¸ ¨ ¨ e"J # SAT 1 © ¹ © # ¹ 1 e ª
J "J
8 1
J
SAT
d
J
#
J
K
J
#
"
¬
§
J
"º
¨
¼
( #
©
1 1
#
# #8
·
8J
¸
(
#
( ¹
SAT
Dr% u"it #ei$t( 8 (J 1 J " # J
d
1 #
J "J d
J SAT
#
#
1 e
!1 e" #
§
J
#
SAT
eS J
8S J
¨
© 1
e e
·
¸
e8 ( J
!S
#
#8 ( "
J #
.
+&
+&
*#hases of of soils(01, soils(01, "onert from metri units to S and -S units. !,evision ?ct.-&"
A cohesive soil sample was ta%en from an S%T an S%T and returned to the laboratory in a glass Far. 5t was found to weigh 1"!.- grams. The sample was then placed in a container of 9 , -!! cm and "0 cm of water were added to fill the container. Grom these data# what was the unit weight of the soil in %*3m and pcf Solution.
Botice that the 15&.8 grams is a mass. Therefore2 the ratio of mass to volume is a density ȡ 2 U f
15&.8 $
m , 4
!8&&
1 - $ ¨ 1.+
U $
¨
©
$ f cm
4
5+4" cm
§
J
1.+
· §
$
· § f
f
¸
¸
cm
4
m
¨
©
1 -N
¸ ¨ +¸ 4 sec ¹ © 1& N
¨ >.&%
4 1& $ ¸
¹
· §
©
f ¹
· §
¸ ¹
4
+
1 & cm · ¨
1(.>
1m
¹
§
J
&.++5 lb(
¨ 1(.>
©
4 ¹ ©
· § · §
-N
1&&& N
f
¨ ¸ ¸ ¨
m
1 -N 1 N
¹
· §
©
1m
4
·
¸ ¨ ¸ ¹
© 48.4 ft 4
u" it ( " 115 pc f !9S u"it ¹
-N u" it ( " ! SI u"it m
4
©
*#hases of of soils–02, soils–02, "ompation heed ia the oids ratio. !,evision Sept.- &"
A contractor has compacted the base course for a new road and found that the mean value of the 3 test samples shows w , 1".9=# 8 S , 0.1# and J , 1.0 /2Im . The specifications require that e d
!.!. 4as the contractor complied with the specifications Solution: 8S J
J
1
?
5
#
1
8S J 5 #
1 e
1
J
-N ·
e §
+.1¨ >.1
4
1.+ e
1.(5
1
?
e
&.(5
1
&.15% m
©
1 e
&.
¸
¹
1.(5
-N 4 m
&.(5 3e( 2 te co"tr'ctor '( complied .
*#hases of of soils–0!, soils–0!, 'alue of the moisture 3hen fully saturated. !,evision ?ct.-&"
"J
/1 Show that at saturation the moisture /water content is # .
('t
/0 Show that at saturation the moisture /water content is ##¨ ('t
J
J
5
"J
('t
§
©J
1 d
1
¸
J
S ¹
Solution:
!1" )n a fully saturated soil the relation2 Se 1 or
because S
8S
e
J
J
#
('t
1
ª
¬
re'rr'"$i"$
# ('t 1
" 8S ª
J
8
1
"
ª
"º
" ¬
or
J
1
º
« ¬
terefore
# ('t
"J J
# ('t
"
"
"
»
# ('t !1 ""
S
¼
"
('t
"
"
t ('t ('
J#
#8 (
"
# ('t
but
#8S becomes simply e
"
# ('t
¼
#
"J
('t
#
"
J #
e
!+" /gain2 in a fully saturated soil2 # t ('t ('
? #
, ,
#
J
or
# ('t
J# ¨
© 5 S
¸
©
¹
©J
1 d
1 J
S ¹
, ,
8S ·
·
¸
J
J
#
#
, ,
, S
,S
JS
,S 1 5S §
J
, ,
5 S , S
#¨
©
5 S
¹
§
, S
#¨
J
#
, ,
, S
# ¨
5 S 5 S
§
J
·
, ,
('t
¸
§
J
, ,
, S
·
¸
5 S
¹
5
·
*#hases of of soils–0$, soils–0$, Finding the 3rong data. !,evision ?ct.-&"
A geotechnical laboratory reported these results of five samples ta%en from a single boring. 'etermine which are not correctly reported# if any.
Sample H1: w , !=# Ȗd , 1".2 %*3m # Ȗs , 0? %*3m < clay. Sample H0: w , 0!=# Ȗd , 1
%*3m # Ȗs , 0? %*3m < silt.
Sample H: w , 1!=# Ȗd , 19
%*3m # Ȗs , 09 %*3m < sand.
Sample H": w , 00=# Ȗd , 1?. %*3m # Ȗs , 0 %*3m < silt. Sample H-: w , 00=# Ȗd , 1
%*3m # Ȗs , 0? %*3m < silt.
Solution:
, ,
e
#
#
('t (' t
8 # , S
J
,
S
J
S
§
J
J
,
1 5
S
, ,
, ,
, S
5 S
5 S
¹
©
S
§
J
1
#¨
¸
#¨
5 S
©J
§
J ·
#
S
©
J
#
·
#¨
('t
, , , S
1
, ,
5
J
§
, ,
, S
#¨
¸
S
·
¸
J S ¹
d
The water content is in error if it is greater than the saturated moisture2 that is2 ?
#d
#SAT
1"
#SAT
§ 1 J #¨
1
J ©
©S
1
J
d
>.1 -N < m
·
¹
§
4
·
¸
¨
© 15.>
4&'
1
4&' 844D
#
+(
¹
+"
#
§
>.1 1 -N < m
1.8' ! #
1
4
· ¨
SAT
1
+(
¸
4"
§
4
¨
#SAT
>.1 -N < m 1
1&' 844D
1
©
SAT
5"
+5' ! #
#
-N < m 1 >.1 ·
+&' 5R4N8
4
¨
·
1%
+% §
1(.4
¸
1 +
++.1' ! #
++' 844D
© ¹
8"
#
§
>.1 1 -N < m
4
©
1 ·
1.8'
#
·
¸
5
©
S ¹
, S
¹
++' 5R4N 8 SAT
¨
1
+(
¸
*#hases of of soils–0%, soils–0%, nreasing the saturation of a soil. !,evision Sept.-&"
A soil sample has a unit weight of 1!-.? pcf and a saturation of -!=. hen its saturation is increased to ?-=# its unit weight raises to 110.? pcf. 'etermine the voids ratio e and the specific gravity 8 s of this soil. Solution: J ?
'"d
J
5
8S
Se
1 e 1&8.( pcf
!1"
%+.5!8S &.8 &e " 1 e %+.5 ! 8S &. (8 e " 1 e
11+.( pcf
!+"" !+
Solving e*p e* p licitely for 8 ( in e7uation !1"2 8 (
1&8.( %+.5
1
e &.8&e
,eplace 8 ( in e7uation !+" with the above relation from !1"2 ? ?
11+.( e
&.1 15 5
1
e
'" d
8S
1&8.( +. %(
1
e
%+.5
&.+8e
*#hases of of soils–0&, soils–0&, ,ind Ȗ d d ( ( n( n( S and D w . !,evision Sept.-&" 3
The moist unit weight of a soil is 19.- /2Im . Iiven that the w , 1-= and 8 s , 0.?!# find: /1 'ry unit weight Ȗ d d # /0 The porosity n# / The degree of saturation S saturation S # and 3
/" The mass of water in /g m Im that must be added to reach full saturation.
Solution: J
'"
d
b"
J
6 6
1%. 8 8
!1 9 w"
;B
6 15.4
!1 9 &.18"
m
4
3rom the table of useful relationships2 +.(&
8J 8J
(J # d
e
1
(
1
?
>.1
#
J d
15.4
1.8 e
&.8
?
e "
e
&.8
1 e
1
&.8
Se c " Since S
d "
6
J
sat
#8 (
5%'
1&&'
?
#8 ( e
!L S 9 e" J
&.1 8 1&&
&.8
+.( & 9 &.8 +.(& > . 1
w
6 19e
ass of water
m U
;B
6 1.
19&.8
m
The water to be added can be found from the relation ?
5'
+.( &
J
-m < s ª
« ¸ ¬
$ m < (
1. · - 1%. 8 -N 4< m
¨
>.1
-$ -
¼ ©
m
4
U $
J
12 &&& B
º §
¸
>.1;g
· §
»
+
1 -N
¹
+
6 +2 45&
¨
©
N
¹
- $
m 4
*#hases of of soils–0), soils–0), -se the blo diagram to find the degree of saturation. !,evision Sept.-&"
A soil has an 6in&situ7 /in&place voids ratio eo
1.(2 # N
%&'2 '"d 8S
+.(8 .
hat are the
and S S /*ote: All soils are really 6moist7 except when dry# that is when w , !=. Ȗmoist and
4
7
Set , S 6 1 m !Bote this problem could also be solved by setting , 6 1.& m ".
Solution:
1.(
?
,
The =natural= water content is
#
?
, ,
o
e
1.( ,
,
1
,
S
1.(
+.( m
4
1
, S
5 #
N
&.%& 5
&.%&5
? #
5 (
(
5 ( 8 (
J
, (
?
(
J
,
J
(
J
#
5
4
1 m4 >.1 -N < m
8 (
S
+.(8
+%.> -N
#
#
5 # &.%& 5 ( +%.> 5 5 # ?
J
5 moi(t ,
+.( m
5 # J ?
S , # , ,
§
4
©
18.&
-N m
4
1%.1> · ¨
#
, ,
+%.>
5 (
N 54 . 1( -
&.%&
>¸.1 ¹
1.(
.+'
1%.1>
N 1%..1> - 1%
54.1( -N
*#hases of of soils–0+, soils–0+, Same as #rob(0) but setting the total olume '41 m!. !,evision ?ct.-&"
A soil has an 6in&situ7 /in&place voids ratio eo
1.(2 # N
%&'2 '"d 8S
+.(8 . hat are the
and S S /*ote: All soils are really 6moist7 except when dry# that is when w , !=. Ȗ moist moist and
Solution:
, ,
but , 1 m 4
1.( ?
but e o
Set 9 , 1 m /instead of 9 s , 1 m used in 8hases&!?.
, ,
,
S
,
, S
The =natural= water content is
5 #
# N
5 ( 8 (
, (
J
?
5
,
J
8
(
J
(
S
S
&.%& 5
#
5 ( 4
S
S
&.%& 5
?
&.45 m +.(8 4 -N < m
(
J #
S
&.45 '"d ,
?
+.(, ,
1.(,
(
>.1
>.4> -N
#
#
5 # &.%& >.4> 5 5 # ?
5 ,
1m
5 # J ?
S , # , ,
>.4>
5 (
N 18.& -
J moi(t
5 (
4
§
18.&
-N
m
4
8.%4
#
· ¨
, ,
© >.1
¸
¹
&.%8+
.&'
&.%&
8.%4 - N
8.%4
18.&+ -N
,
&.%8+
*#hases of of soils–0, soils–0, Same as #roblem 5% 3ith a blo diagram. !,evision Sept.-&"
A soil sample sample has a unit weight of 1!-.? pcf and a water content content of -!=. hen its saturation saturation is increased to ?- =# its unit weight raises to 110.? pcf. 'etermine the voids ratio e and the specific gravity Is of the soil. /*$: This is the same problem as 8hase;!9# but solved with a bloc% diagram. Solution:
S e t , J
4
ft t 1 f
1 1 + .( ( .&
+
J 1
1 & 8 .(
o f # ' t e r
+ 1 .&
?
lb ( ' r ( 8 o f # ' t e r lb ( e ' + & . > 1 .> l b
1 1 + .(
?
l b ( ' r + 8 e '
5 S , #
4
5#
+ & . lb
J#
% + . p c f
& .4 4 f ft t 4
5 1
, '
, (
, ,
e
, S ' " d
,
4 1 , !
?
J
8
S
4 & . 1 1 f ft t 1
?
,
#
1 & .5 5 5 & .8 8 %
& . 5 5 5
,
,
!
'
,
J
J
& . & > 1 .> lb + .% 8
& .8 8 % #
S
#
#
& .8 8 %
5S S
& .1 1 1 5 5
! % + .5 "
& .4 4 4
& .5
*#hases of of soils–10, soils–10, /lo diagram for a saturated soil. !,evision Sept.-&"
A saturated soil sample has a unit weight of 100.- pcf and 8 s , 0.?!. Gind J
dr%
# e# n# and w.
Solution:
·
5 ¨
J
#
1
5 S
§
1
, , S , #
#
©
¸
S
¹
5
5 S
+
1 + + .8 lb
5#
$ ! 1 " an d !+ " yi eld s o 1m b in in g e7 u atio n s !1 5# ?
,
+ ( .& lb
5
?
#
?
,
#
> 8 .8 lb
5 S
J#
5S
? S
J
?
5S d r%
,
?
, ,
e
, S
?
" , , ,
+ ( .& lb
8S J #
& .5 4 4
©
1 + + .8 + .( &
5# #
¸
5
·
¹
ft
4
& .8 % ft (
4
!% p cf "
4
& .( % 5
& .5 4 4 & .8 % (
?
#
& .5 4 4
5#
5 S
+( > 8 .8
4&
% + .5 p ¨ cf
% + . p cf 5 > 8 .8 l b + .( & + .5
§
> 8 . p cf 8
> 8 .8 lb 1 ft
1
& .5 4 4
1
& .+ 4
4&
"
5 4 .4 ' #
41
+ .4 '
41
*#hases of of soils–11, soils–11, Find the 3eight of 3ater needed for saturation. !,evision Sept.-&"
'etermine the weight of water /in /2 that must be added to a cubic meter of soil to attain a 2- = 3 degree of saturation# if the dry unit weight is 1?.- /2Im # its moisture is "=# the specific gravity of solids is 0.9- and the soil is entirely made up of a clean quart) sand. Solution:
-N
J
J
1(.8 m
1.+
5
5
?
4
1
1 #
5 5
1.+-N
J
?
d
J
5
S
& . & 5 #5
#
m
!1 . & 5 " 5
S
4
S
S
'"d
1(.8-N 2
5
#
& . ( & - N
S
5
, S
8
J
S
#
J
S
#
4
4
& . & ( m ?
, , '
4
! > . 1 -
#
,
&.&(
e
&.+8(
& . % ( 4
, S
#8
T e e i ( t t ii " $ S
e
A e r e e 7 u i r e e a S
(
5
8 #
S
#5
+ . % 8 S
& . 5 > !&.&5" 1&&
+.%8
& . 5 >
> 8 ' 2 t h e r e e f o o r e e 2
&.>8 &.5>
Se
#
S
& . 1 ( -N "
+ . > - N
! & . 1 ( " !1 ( . 8
' l l rr e ' d % ' ! e 5
& . ( & - N N
#
?
e r
m u ( t ' d d # ' t
+ . + - N
& . + 8 ( m
,
N ,,
4
! > . < m "
+ . % 8 1 - N
N & . ( & - N
#
& . % ( 4 m
N 1 ( . 8 - N
J
5
,
N 1 ( . 8 - N
S
+ 1 . % '
#
4
3
Answer: Add "."0 /2 of water per water per m .
3rong piee piee of data. *#hases of of soils–12, soils–12, dentify the 3rong !,evision Sept.-&"
A proFect engineer receives a laboratory report with tests performed on marine marl calcareous silt. The engineer suspects that one of the measurements is in error. Are the engineers suspicions correct 5f so# which one of these values is wrong# and what should be its correct value u " it # e i$ i$ t o f ( ' m 1 . - N 8 i! e " J p le 5 4 m -N J S +%. u " it # e i$ i$ t o f ( o 4 m 1 li d ( ( # # ' te te r r c o " te " t e ! o id ( ( r r ' ' ti o
5&'
1 .1 +
S d ee $ d $ r r ee e o f ( ' tu r r ' ' ti o "
>8'
Solution:
$hec; the accuracy of 5 out of 8 of the variables using2 1.&% Se #8S ? &.>8 1.1+
Se #8S
&.5 +%.1
# J
1.&%
S
?
Therefore2 these four are correct.
>.1
J #
The only possibly incorrect value is , 4 , , , 1 m , '
but
e
, ,
& ,
?
1.1+
,
S
?
'
4
?5
,
S
&.8+ m
&.&+% m §
J S
+%.1
m S
-N ·
but ,
&.>8,
#
,
©
§
&.5(+
4
¨
#5 S
S
&.8&+ m
4
4
m
4
&.5& ¨
1+.4 -N
¸
-N
¹
5 #
4
+
1.1+, #
,
,
1
, '
&.5(+ m 2
,
. /ssume that 6 1 m4 .
S
#
, S ?
J
· 4
5.> -N ¸
1+.4 ©
m
¹
5
1+.4 -N
5.> -N
1(.+ -N Therefore2 the actual unit weight of the soil is2 -N -N N 1(.+ - z 1.5 1(.+ ?J 4 4 4 5 m m 1m ,
*#hases of of soils–1!, soils–1!, The apparent heapest soil is not6 !,evision Sept.-&" 9
Jou are a 8roFect Engineer on a large earth dam proFect that has a volume of -x1! yd of select fill# compacted such that the final voids ratio in the dam is !.!. Jour boss# the 8roFect >anager delegates to you the important decision of buying the earth fill from one of three suppliers. hich one of the three suppliers is the most economical# and how much will you save
Supplier A
Sells fill at K -.03 yd with e , !.2!
Supplier $
Sells fill at K .213 yd with e , 0.!!
Supplier C
Sells fill at K -.123 yd with e , 1.9!
Solution:
Aithout considering the voids ratio2 it would appear that Supplier E is cheaper than Supplier / by U1.4( 4 per yd . 4
4
Therefore To put 1yd of solids in the dam you would need 1. yd of soil. Therefore 4
4
4
4
3or 1yd of solids from / you would need 1.> yd of fill. 3or 1yd of solids from E you would need 4.& yd of fill. 4
4
3or 1yd of solids from $ you would need +.% yd of fill. The cost of the select fill from each supplier is !rounding off the numbers"
A % 1& B 1& C 1&
1 .>
8
%d
¨
1. 1. 4 .&
%
©
8
%d
¨
%
1. 1.
©
8
%d
8 .+ U · 4
%d §
4
1. 1. + .%
§
4
%d
¨
©
%d
4
¸
¸
|
U 4+2 %&&2 &&&
|
U 4 (2 8&&2 &&&
¹
8 .1 > U · 4
U +(2 >&&2 &&&
¹
4 .> 1 U ·
§
4
¸
|
¹
Ther Th eref efor oree Sup uppl plie ierr A is is th thee ch chea eape pest st by ab abou outt K ". ".? ? >i >ill llio ion n co comp mpar ared ed to Su Supp ppli lier er $.
*#hases of of soils–1$, soils–1$, 7umber of tru loads. !,evision Sept.-&"
$ased on the previous problem /8hases;1# if the fill dumped into the truc% has an e , 1.0# how many truc% loads will you need to fill the dam Assume each truc% carries 1! yd of soil. Solution:
1 e
Set , S
, ,
, ,
, S
1
, ,
4
4
4
+.+ %d of (oil for e'c 1 %d of (olid(. 4 4 of (oil for e'c %d i" ' truc- lo'd 1& %d
?
?
4
%d of (olid( per truc- trip.
5.85 The re7uired volume of solids in the dam is2 %
, (olid(
4
of (oil (oil 8 61& %d of (olid(
4
1 %d of
%
4
+. 1& %d of (olid(
4
1. %d of (oil
Therefore2 !rounding off" %
Number of Tr Trucuctrip(
4
1.+ which means that there there is 1 yd yd of solids per 1.+ yd of voids.
4
+. 61& %d of (olid( 4
of (olid( (olid( < truc- 5.85 %d of
%1%2 &&
trip
*#hases of of soils–1%, soils–1%, 8o3 many tru loads are needed for a pro9et: !,evision Sept.-&"
Jou have been hired as the 8roFect Engineer for a development company in South Glorida to build 91! housing units surrounding four la%es. Since the original ground is low# you will use the limestone excavated from the la%e to fill the land in order to build roads and housing pads. Jour estimated fill requirements are ?!!#!!! m # with a dry density equivalent to a voids ratio e , !."9. The 6in&situ7 limestone extracted from the la%es has an e , !.2# whereas the limestone dumped into the truc%s has an e , !.?1. 4ow many truc%loads will you need# if each truc% carries 1! m Solution:
/ssume , S
1m
4
?
e6
, ,
6
, ,
6
4
,
6 &.5% m in the compacted fill
, , S
1
4
4
4
The re7uired (&&2&&& m of fill have 1.5% m of voids per each 1 m of solids 4
4
Therefore2 the (&&2&&& m of fill have 5(>25&& m of solids 4
4
ach truc; carries 1.(1 m of fill per 1 m solids 4
4
)n order for the truc;s to carry 5(>2&&& m of solids they must carry +&2&&& m of fill 4
Since each truc; carries 1& m of fill2
?
+&2 &&& m The number of truc;-loads 6 4 1& m
4
6 +2&&& truc;-loads. truc;-loads.
*#hases of of soils–1&, soils–1&, "hoose the heapest fill supplier. !,evised Sept.-&"
A large housing development requires the purchase and placement of the fill estimated to be 0!!#!!! cubic yards of lime&roc% compacted at 2-= Standard 8roctor with an H$! of 1!=. Two lime&roc% suppliers offer to fill your order: Company A has a borrow material with an in&situ Ȗ , 11- pcf# w , 0-=# 8 S , 0.?!< Standard 8roctor yields a maximum Ȗ d , 110 pcf< at a cost of K!.0!3yd to excavate# and K !.!3yd to haul. Company $ has a borrow material with an in&situ Ȗ , 10! pcf# w , 0!=# 8 S , 0.?!< Standard 8roctor yields a maximum Ȗ d , 11- pcf< a cost of K!.003yd to excavate# and K !.3yd to haul. /1 hat volume would you need from company A /0 hat volume would you need from company $ / hich would be the cheaper supplier Solution:
!1"
The ;ey idea 1 %d of solids from the borrow pit supplies 1 %d of solids in the fill.
!+"
#it A 5 S 6 >+ lb2 5 5 6 +4 lb
e
, ,
&.585
, S
&.85%
&.4
?
4
4
, 5 6 &.4%> ft 2 , S 6 &.85% ft 2 , ' 6 &.&8 ft
4
4
1.4 %d of (oil co"t'i"( 1.& %d of (olid(. 4
e
4
4
#it B 5 S 6 1&& lb2 5 5 6 +& lb2 , 5 6 &.4+1 ft 2 HS 6 &.8>5 ft 2 , ' 6 &.& ft 4 4 , , &.5&1 1.% %d of (oil co"t'i"( 1.& %d of (olid(. ? &.% , S &.8>5
!4" :aterial needed for fill from company / J &.>8J d 1 &.>8 11+ 1 11( pcf ? # &.1& 5
1&%.5 lb2
5 #
1&.% lb
S
e
?
fill
, ,
&.4(
, S
&.%4
&.8>
?
4
4
1.8> %d of (oil co"t'i"( 1.& %d of (olid( 4
Site A re+uire(
+&&2 &&& %d of
4
1+82 && 1+82 && %d of (olid(
1.8>
J
:aterial needed for fill from company E &.>8J d 1 &.>8 118 1 1+& pcf ?
# e
&.1& , ,
&.48
, S
&.%8
?
&.85 1.&
1&>.1 lb2
5 S 4
4
1.85 %d of (oil co"t'i"( %d of (olid(
4
?
fill
Site B re+uire(
+&&2 &&& %d of 1.85
4
14&2 &&& %d of (olid(
5 #
1&.> lb
4
!5" a" $ost of using $ompany $ompany / Co(t
4
1+82 && 1+82 && · %d U&.8&
¨
A
%d ©
U 11821&&
§
1.4 4
¸
¹
$ost of using $ompany E Co(t B
4
14&2 &&& 14&2 &&& · %d U&.%&
¨
U 14121&&
§
1.% 4
%d % © d
¸
¹
9(i"$ Comp'"% A #ill ('!e 'bout : ;6<000.
*#hases of of soils–1), soils–1), -se a matrix to the find the missing data. !,evision Sept.-&" 3
A contractor obtains prices for "#!!! yd "#!!! yd of compacted 6borrow7 material from three pits: 8it H is K11#!!! cheaper than 8it H0 and K2#!!! cheaper than 8it H1. The fill must be compacted down to a voids ratio of !.?. 8it H1 costs K 9.!!3yd and 8it H costs K -.-!3yd . 8its H0 and H reported their voids ratios as !. and !.2- respectively. respectively. (se a matrix to find# a The missing unit cost ! " for 8it H0< b The missing voids ratio e for 8it H1< c The missing volume of fill 9 required from each pit< and d The amount amount paid by the contractor contractor for each pit. Solution:
/ summary of the data provided is herein shown in matri* form2
The volume of solids , ( contained in the total volume of fill , 6 452&&& %d can be found from2 4 452 &&& +&2 &&& %d 4 of (olid( , &.(, &.( 452 &&& %d ? 1 , , , , , ,
S
S
, 4
/t #it O42
S
S
1 e , 4 ,
S
?
S
1 &.>8
,
4
1.( 4
4
+&2 &&& %d
e
1 4>2 &&& %d of (oil
4
S
The total cost of #it O4 is TC 4
4
4>2 &&& %d
U +152 8&&
U 8.8& < %d
4
, +
/t #it O+
1 e ,
S
?, +
+
,
e
4
1
&.
Eut2 the total cost of #it O+ is TC + &&&
/t #it O1 , 1
Eut2 , 1
4
U %.&& < %d 1 e , 1 e S
1
U 112
4(2 %&& %d of (oil
?
U +152 8&&
TC 4
TC +
U ++82 8&&
, +
4(2 %&& %d
TC +
U ++82 8&&
4
U %.&& < %d
4
+2 &&&
T C+ & &&
TC1
4
1
+
S
The unit cost of #it O+ C +
+&2 &&& %d
4
U %.&& < %d 4 +&2 &&& %d 1
++82 8&&
4
5+2 +8& %d of (oil
+2 4
U %.&& < %d 4 ? 5+2 +8& %d
e 1
1.11
**#hases of of soils–1+, soils–1+, Find the oids ratio of;mu< =a highly organi soil>. !,evision Sept.-&"
Jou have been retained by a local municipality to prepare a study of their 6muc%7 soils. Assume that you %now the dry unit weight of the material /solids Ȗsm and the dry unit weight of the organic solids Ȗso. hat is the unit weight Ȗs of the combined dry organic mineral soil whose organic content is >! /The organic content content is the percentage percentage by weight of the dry organic constituent of the total dry weight of the sample for a given volume. hat is the voids ratio e of this soil if it is %nown that its water content is w and its degree of saturation is S is S Solution:
Set D s 6 1 unit and
Js
6
As 1 6 !Hso 9 Hsm " Hs
!a" /ssume ) o 6 D o for a unit weight of the dry soil Therefore 1 - ) o 6 D m so solids : 6 volume of organic 9 so o
J so
!1 - : o " 6 volume of mineral 9 sm sm solids J sm
The total unit weight is the weight of a unit volume.
Therefore
F
1
J
§
·1
:
S
¨
o
J so
©
6
-:
ª so
«
J
:
-
J
"9
sm
º
»
J
J
9
o
J
¬
¸
o
sm
so
so
¼
sm
¹
§
Hv 6 !b" e 6 Hs
volume of water
§
· ¨
¸
© ¹
weight of water ·
¨
¹ 6
Hs
Hs
6
w
weight of solids ·
J wS
©
¹
Hs §
J #
¨
J wS
©
S
§
¹
¨
·
#
¸S © 5&
#J sm J so Therefore e 6 § ) ¨
1 - )
o
©
J J
·
J (o
o
¸
S #
¬
ª
J
) o
(m
J (o
(o
º
¼
J (m
¹
51
