Shear Strength of Soil

Transcription

11
ShearStrength of Soil
The shear strengthof a soil massis the internal resistanceper unit area that the soil
masscan otTerto resistfailure and sliding along any plane inside it. One must understand the nature of shearingresistancein order to analyzesoil stability problemssuch
as bearing capacity,slope stability,and lateral pressureon earthretainingstructures.
11.1
Mohr -Coulomb Failure Criterion
Mohr ( 1900)presenteda theory for rupture in materials that contended that a material fails becauseof a critical combination of normal stressand shcaringstress,and
not from either maximum normal or shear stressalone. Thus. the functional relationship between normal stressand shear stresson a failure plane can be expressed
in the followins form:
r1: f@)
( 11 .1 )
T h c f a i l u r e e n v e l o p ed e f i n e db y E q . ( 1 1 . 1 ) i s a c u r v e d l i n e . F o r m o s t s o i l m e chanicsproblems, it is sufficientto approximate the shearstresson the failure plane
as a linear function of the normal stress(Coulomb. 1776).This linear function can
be written as
rf=c*crtan$
(11.2)
where c : cohesion
d : angle of internal friction
o : normal stresson the failure plane
r/ : shear strength
The preceding equation is called the Mohr-Coulomb failure criterion.
In saturated soil, the total normal stress at a point is the sum of the effective
stress(n') and pore water pressure(tt), or
o : o ' l t t
311
312
Chapter 11
Table 11. ? Typical Values of Drained Angle
of Friction for Sands and Silts
Soil type
d'(des1
Sand: Rr,tundedgrains
Loose
Medium
Dense
27-30
30-35
35-38
Sand: Angulur grains
Loose
Medium
Dense
30-35
3.5- 4t)
40-45
Gravel with some sand
34,48
Sl/ts
26-35
The effectivestressa' is carried by the soil solids.The Mohr-Coulomb failure criterion, expressedin terms of efTectivestress,will bc clf the form
T | - ,'' * tr' tan <['
(11.3)
where c' : cohesionand @' : friction angle,basedon effeotivestrcss.
T h u s , E , q s .( 1 1 . 2 )a n d ( 1 1 . 3 )a r e e x p r e s s i o n o
s f s h e a rs t r e n g t hb a s e do n t o t a l
stressand effectivestress.The value of c' for sand and inorganic silt is 0. For normally consolidatedclays,c' can be approximated at 0. Overconsolidatedclays have
valuesof c' that are greater than 0. The angle ol friction, ry'',is sometimesrcfcrred to
as the drained angLeo.f.frictbn. Typical values of <f' for some granular soils arc given
i n T a b l e 11 . 1 .
The signilicanccof Eq. ( I 1.3)can be explained by referring to Fig. I L l, which
showsan clemental soil mass.Let the effectivenormal stressand the shear stresson
a
(a)
Figure 11.I Mohr-Coulomb failurecriterion
11'2 rncrination of the prane of Fairurecaused by shear
313
the plane ab be o' and r, respectively.Figure 11.1bshowsthe plot of the failure envelope defined by Eq. (11.3).If the magnitudesof o' and r on planeab are such that
they plot as point,4 in Figure 11.1b,shear failure will not occur along the plane. If
the effective normal stress and the shear stress on plane ab plot as point B (which
falls on the failure envelope), shear failure will occur along that plane. A state of
stresson a plane representedby point C cannot exist,becauseit plots above the failure envelope,and shear failure in a soil would have occurrcd alreadv.
11.2
lnclination of the Plane of Failure caused by shear
As statedby the Mohr-Coulomb failure crittlrion, failure from shearwill occur when
the shear stresson a plane reachesa value given by Eq. (1 1.3).To determine the inclination of the failure plane with the major principal plane, refer to Figure 11.2,
where c\ and oi are, respectively,the major and minor effectiveprincipal stresses.
The failure plane EFmakes an angle d with the major principal plane.To determine
the angle 0 and the relationshipbetweenrrj and ai, refer to Figure 11.3,which is a plot
of the Mohr'.scircle for the state of stressshown in Figure I 1.2 (see chapter 9j. In
Figure 11.3,fghis thc failure envelopedefined by the relationshipr[: c, * o, tan .
$,
T h e r a d i a l l i n e a b d e f i n e st h e m a j o r p r i n c i p a lp l a n e ( C D i n F i g u r e ' 1 1 . 2 )a, n d t h e r a d i a l l i n e a d d e f i n e st h e f a i l u r ep l a n e( E - F i nF i g u r e 1 l . z ) . l t c a n b e s h o w n t h a t L b a d :
2e:90*d'.or
0:45+9
2
( r1 . 4 )
A g a i n . f r o m F i g u r e1 1 . 3 .
ad
(r1.-s)
fa:stna
T u - . f O + O u : c ' c t > r g -' o j t ' i
2
U
Figure | 1.2 Inclination of failure plane
in soil with major principal plane
01,
(11.6a)
a
O'I
Ef1'ectivenormal stress
Figure 71.3 Mohr'scircleand failure envelope
314
Chapter 11 Shear Strength of Soil
Also,
r
aa:
o t - o l
( l 1.6b)
2
Eqs.(11.6a)and(11.6b)into Eq. (11.5),we obtain
Substituting
o\-ot
2
sin@' :
c ' c o t $ '+
o,, _f o,.
tcos6'\
/l+sind'\
/
'i -.'1[r
sind/*"\r
,ino,/
( 11.7)
However,
I + sin@'
1 - sin q5'
^/
d'\
:tan'145*
Z)
and
(b' \
cos d'
/
"
t
*
,
t
t
a
n
\
4
5
+
2)
I
Thus,
. +)
. +) * ,r',^n(ot
o\: o\,un'(+s
(11.8)
An expressionsimilar to Eq. (11.tt)could also be derived using Eq. (11.2)(thal
is, total stressparametersc and @),or
rr)
u t - . t t u n ' ( + . s , r r , , " " ( 0 , - -f )
1t.3
( l l.e)
Laboratory Test for Determination
of Shear Strength Parameters
There are several laboratory methods now available to determine the shear strength
parameters (i.e., q 6, c' , 6') of various soil specimensin the laboratory. They are as
follows:
a.
b.
c.
d.
e.
Direct shear test
Triaxial test
Direct simple shear test
Plane strain triaxial test
Torsional rins shear test
11.4 DirectShear Test
315
The direct shear test and the triaxial test are the two commonly used techniques for
determining the shear strength parameters. These two tests will be described in detail in the sections that follow.
11.4
Direct Shear Test
The direct shear test is the oldest and simplestform of shear test arrangement.A diagram of the direct shear test apparatusis shown in Figure 11.4.The teit equipment
consistsof a metal shearbox in which the soil specimenis placed.The soil rp..i."n,
may be square or circular in plan. The size of the specimensgenerally ur",l is about
5 1 m m x 5 L m m o r 1 0 2 m m x 1 0 2 m m ( 2 i n . x 2 i n . o r 4 i n . x 4 i n . ) a c r o s sa n d a b o u t
2-5mm (1 in.) high. The box is split horizontally into halves.Normal force on the
specimen is applied from the top of the shear box. The normal stresson the specimens can be as great as 10-50kN/m2 (150 lb/in.r). Shear force is applied by moving
one-half of the box relative to the other to causefailure in the soil specimen.
Depending on the equipment, the shear test can be either stresscontrolled or
strain controlled. In stress-controlledtests,the shear force is applied in equal increments until the specimenfails.The failure occursalong the plane of split of the shear
box. After the application of each incremental load, the shear displacementof the
top half of the box is measuredby a horizontal dial gauge.Thc changein the height
of the specimen(and thus the volume changeof the specimen)during the test .un b"
obtained from the readingsof a dial gauge that measuresthe vertical movement of
the upper loading plate.
In strain-controlled tests,a constant rate of shear displacementis applied to
one-half of the box by a motor that acts through gears.The constant rate of shear
displacementis measuredby a horizontal dial gauge.The resistingshearforce of the
soil correspondingto any shear displacementcan be measureclby a horizontal proving ring or load cell. The volume changeof the specimenduring the test is obtained
Normal force
Porousstone
€
Shear
force
Porousstone
Figure 11.4 Diagram of direct shear test arrangement
316
directsheartestequipmcnt(courtesyof Soiltest,Inc.,Lake
Figure 1?.5 Strain-controllcd
Bluff, Illinois)
in a manner similar to that in the stress-controlledtests.Figure 11.5showsa photograph of strain-controlleddirect shear test equipment.
The advantageof the strain-controlledtests is that in the caseof dense sand,
peak shear resistance(that is, at failurc) as well as lessershear resistance(that is, at
a point after failure called ultimate streng,th)can be observed and plotted. In stresscontrolled tests,only the peak shear resistancecan be observed and plotted. Note
that the peak shearresistancein stress-controlledtestscan be only approximatedbecausefailure occurs at a stresslevel somewherebetween the prefailure load increment and the failure load increment.Nevertheless,compared with strain-controlled
tests,stress-controlledtestsprobably model real field situationsbetter.
For a given test, the normal stresscan be calculatedas
o : Normal stress:
Normal force
Cross-sectionalarea of the specimen
(11.10)
The resistins shear stressfor anv shear displacementcan be calculatedas
r : Shearstress:
Resistinsshear force
Cross-sectionalarea of the specimen
(1 1 . 1 1 )
Figure 11.6 shows a typical plot of shear stressand change in the height of the
specimen against shear displacement for dry loose and dense sands.These observations were obtained from a strain-controlled test. The following generalizationscan
11.4 DirectShear Test
317
E
a
Sheardisplacentent
.='c
x
u)T
F
d
F
r )
=
o
a
o
U
Figure 11.6 PloIof shearstressanclchangcin heightof specimen
againstsheardisplacementfor looseanddcnsedry sand(dircctshcartest)
be developedfrom Figure I 1.6regarding the variation of resistingshear stresswith
shear displacement:
1. ln loose sand,the resistingshear stressincreaseswith shear displacementuntil
a failure shear stressof 11is reached.After that, the shear resistanceremains
approximately constant for any further increasein the shear displacement.
2. In densesand, the resistingshear stressincreaseswith shear displacementuntil
it reaches a failure stressof rr. This rlis called the peak shear strength. After
failure stressis attained, the resistingshear stressgradually decreasesas shear
displacementincreasesuntil it finalty reachesa constantvalue called the ultimate .shear.strength.
It is important to note that, in dry sand,
o:(r'
and
c ' :0
Direct shear tests are repeated on similar specimensat various normal stresses.
The normal stressesand the corresponding values of r, obtained from a number of
testsare plotted on a graph from which the shear strength parameters are determined.
318
o'(lb/inz)
40
20
^
150
20
z
/
too
ca
<rl
6' = 42"
0
50
100
150
200
2-50
Efl'ective
norntalstress,
o' (kN/rn:,1
300
Figure 11.7 Determination of shear strcngth parameters for a dry sand using the results of
direct shear tests
Figure 1 1.7 shows such a plot for tests on a dry sand. The equation frlr the average
line obtained from experimental results is
T.f : C' tan S'
(1 1 . 1 2 )
So, the friction angle can be determined as follows:
/ r'\
d' : tan-'[\ -i
I
o /
It is important to note that in slla cemented sandsmay show a c' intercept.
11.5
Drained Direct Shear Test on Saturated Sand and Clay
In the direct shear test arrangement,the shearbox that containsthe soil specimenis
generally kept inside a container that can be filled with water to saturatethe specimen. A drained /estis made on a saturated soil specimen by keeping the rate of loading slow enough so that the excesspore water pressure generated in the soil is completely dissipated by drainage. Pore water from the specimen is drained through two
porous stones.(SeeFigure 11.4.)
Because the hydraulic conductivity of sand is high, the excesspore water pressure generated due to loading (normal and shear) is dissipated quickly. Hence, for
11.5 Drained Direct Shear Test on Saturated Sand and Ctay
319
80
Peak shear strength
;i 400
E
z
P
6
<
o
)oo
a
20
Residualshearstrength
o
0
2
Horizonttl
dclirnnation
(t/')
Figure 11.8 Resultsof zrclraincddirectshcartt:ston an ovcrconsolidated
clay.Nore;Residual shearstrengthin clayis similarto ultinrateshcarstrengthin sand- seeFisure11.6
an ordinary loading rate. esscntiallyfull drainageconditionsexist.The friction angle,
',
f obtaincd from a drained direct shear test of saturatedsand will be the same as
t h a t f o r a s i m i l a rs p c c i m e no f d r y s a n d .
The hydraulic conductivity of clay is very small compared with that of sand.
When a normal load is applicd to a clay soil specimen,a sufficientlength of time must
elapsefor full consolidation- that is, for dissipationof excesspore water pressure.
For this reason,the shearingload must be applied very slowly.The test may last from
two to five days.Figure I I .8 showsthc resultsof a drained dircct sheartest on an overconsolidatedclay.Figure I 1.9showsthe plot of riagainst o' obtained from a number
'e
d
O v e r c o n s o l i d a t ecdl a y
6
c^
-r
I
t"
j
N o r m a l l yc o n s o l i d a t ecdl a y
T J= o ' t a n Q ' ( r ' ' =0 )
I
v
Elfective normal stress.o'
Figure
77,9 Failure envelope for clay obtained from drained direct shear tests
Figure 11.10 lnterfaceof a loundationmaterialand soil
of drained direct shear testson a normally conscllidatedclay and an overconsolidated
clay.Note that the value of c'' : 0 for a normally consolidatedclay
11.6
General Comments on Direct Shear Test
The direct shear test is simple to perform. but it has some inherent shortcomings.
The reliability of the resultsmay be questionedbccausethe soil is not allowed to fail
along the wcakest planc but is forced to fail along the planc of split of the shearbox.
Also, the shear stressdistribution over the shear surfaceof the specimenis not uniform. Despitc theseshortcomings,the direct shear test is the simplestand most economical for a dry or saturatedsandy soil.
In many foundation design problems, one must determine the angle of friction between the soil and the matcrial in which the foundation is constructed(Figure 11.10).The foundatiitn matcrial may be concrete, steel, or wood. The shear
strength along the surfaceo[ contact of the soil and the foundation can be given as
rr':
c ' r+
, ottan6
( 11 . 1 3 )
where c,i,: adhesion
6 - effective angle of friction between the soil and the foundation material
Note that the preceding equation is similar in form to Eq. (11.3).The shear
strength parametersbetween a soil and a foundation material can be conveniently
determined by a direct shear test. This is a great advantageof the direct shear test,
The foundation material can be placed in the bottom part of the direct sheartest box
and then the soil can be placed above it (that is, in the top part of the box), as shown
in Figure I 1.11,and the test can be conductedin the usual manner.
Figure 11.12shows the results of direct shear tests conducted in this manner
with a quartz sand and concretewood, and steel as foundation materials,with o' =
1 0 0 k N / m 2 ( 1 4 . 5l b / i n . 2 ) .
11.6 General Comments on Direct Shear Test
Normal forcc
€
I
Shear
fbrce
Figure 11.11 Direct shear test to determinc interface friction ansle
R e l a t i v ed e n s i t y ,/ ) , .( 7 o )
75
-s0
25
Nornral stress
o ' - 1 0 0k N / n r 2
( |4.5 lb/in:)
I
I
IIe
Figure 11.12 Yariationof tan @'and tan 6 with 1/e.fNote:e: void ratio, o' : 100kN/m2
(14.5lb/in.'z),quartzsand(after Acar, Durgunoglu,and Tumay,1982)]
322
E x a m p l e1 1 . 1
Followingare the resultsof four draineddirect sheartestson an overconsolidated
clay:
Diameter of specimen: 50 mm
Height of specimen: 25 mm
Test
no.
1
2
4
Normal
forca, lV
(Nl
Shear force at
failure, So."p
{N}
150
250
3s0
550
Residualshear
forc6, 5r""16961
(N}'
44.2
56.6
102.9
157.5
199.9
257.6
363.4
r44.5
'SeeFigure11.t3
Determine the relationshipsfor peak shear strength (r) and residual shear
strength(r ).
Solution
/ so \2
: 0.0019634
m2.Now the following
Area of the specimen(A) : (rrl+)l #
)
\ luuu/
tablecanbe prepared:'
Residual
Test
no.
Normal
force, N
(Nl
1
2
J
4
150
250
350
550
Normal
sttesg, (''
(kN/m2)
76.4
127.3
178^3
280.1
Peak shear
force, Souul
'
{Nl
$residuat
tt*:l
, Sreriauat
4
(kN/m'l
(Nl
.. = -sr1*
A
{kN/m'?l
r57.5
199.9
257.6
JOJ.4
80.2
101.8
131.2
185.1
44.2
56.6
102.9
1.44.5
22.5
28.8
52.4
73.6
The variationsof rland r, with c' areplotted in Figure 11.13'From the plots,
we find that
Peak strength:
rdkN/m2) : 40 + o' tan27
Residual strength: r,(kN/m2) : a' tan14.6
(Note: For all overconsolidated clays, the residual
pressedas
t' : o' lan 6"
where dl : effectiveresidual friction angle.)
J
shSrstrength can be ex"
| 1.7 Triaxial Shear Test (Generat)
250
tr )oo
z
.: 150
T/ VS. O'
!a loo
21' = Q'
= 40 kN/nr:
r / .v s . o '
r = 11.6"
50
11.7
t(x)
I 50
200
2-50
Effbctivc
norrralstress,
o' (kN/nt:)
300
350
Figure 11.13
Variationsof z7
andr, with (''
Triaxial Shear Test (General)
T h e t r i a x i a l s h e a rt e s t i s o n e o f t h c m o s t r e l i a b l em c t h o c l sa v a i l a b l ef ' o rd c t e r n . r i r r i n g
shcar strength parameters.It is widcly uscclfor researchanclconvcntional testing.,{
d i a g r a mo f t h e t r i a x i a l t e s t l a y o u t i s s h o w n i n F i g u r e I I . 1 4 .
I n t h i s t e s t ,a s o i l s p e c i m e na b o u t 3 6 m m ( 1 . 4i n .) i n d i a m e t . ear l c l 7 6 n t m ( 3 i n . )
long is gcnerallyused.The specimenis encasecl
by a thin rubber membraneanclplacecl
i n s i d ea p l a s t i cc y l i n d r i c a lc h a m b e rt h a t i s u s u t r l l yf i l l e d w i t h w a t e r o r g l y c c r i n c .T h c
specimenis subjectedto a confiningpressureby compressionof the fluiclin the chamber. (Note: Air is sometimesused as a compressionmecliunt.)To cituscshear failure
in the specimeno
, n e m u s t a p p l y a x i a l s t r e s st h r o u g h a v c r t i c a l l o a d i n gr a m ( s o m e times called deviatorsrress).This strcsscan be appried in one of two ways:
1. Application of dead weights or hydraulic pressurein equal incrementsuntil
the specimenfails. (Axial deformation of the specimenresulting from the load
applied through the ram is measuredby a dial gauge.)
2. Application of axial deformation at a constantrate by meansof a geared or
hydraulic loading press.This is a strain-controlledtest.
The axial load applied by the loading ram correspondingto a given axial cleformation is measuredby a proving ring or load cell attachedto the ram.
connections to measure drainage into or out of the specimen,or to mcasure
pressurein the pore water (as per the test conditions), are also providecl.The following three standard types of triaxial testsare generallyconducted:
1. Consolidated-drainedtest or drained test (CD test)
2. Consolidated-undrainedtest (CU test)
3. Unconsolidated-undrainedtest or undrained test (UU test)
324
Axial load
t
Air relelse
---,^
valve
I
Loading ram
:
: ,,.
Top cap
Porousdisc
S p e c i m e ne n c l o s e d
in a rubbcr
mcmbrane
Pressuregauge
"
*--Flexihle
tuhc
.1lr)..1-..
-'\
-
'...1
'.
Rubhcr'
t
| ilrl
\
''
.,":,,.r,..,.i
i,-- ...' :' :l:,.:.-',.-.-...,'....
l
,
LTo cell pressurcontrol
tr'-',..r:,*l
.il, ...-..-.....,..-'ri,
I
Connectionstirr drainageor I
porc pressuremeasuremenI
Figure 11.14 Diagram of triaxial test equipment (al'terBishop and Bjerrum, 1960)
The general proceduresand implications for each of the tests in saturatedroils are
describedin thc followins sections.
11.8
Con so Ii dated -Dra i n ed Triaxi aI Test
In the CD test, the saturated specimenis first subjectedto an all around confining
pressure,oj, by compressionof the chamber fluid (Figure 11.15a).As confiningpressure is applied, the pore water pressureof the specimenincreasesby u. (if drainage
is prevented). This increasein the pore water pressurecan be expressedas a nondimensional Darameterin the form
u.
B = --:C3
(1r.14)
pore pressureparameter(Skempton,1954).
whereB : Skempton's
For saturatedsoft soils,-Bis approximatelyequalto 1; however,for saturated
stiff soils,the magnitudeof B canbe lessthan 1.Black and Lee (1973)gavethe theo-
11.8 Consolidated-DrainedTriaxial Test
325
lo"'
l"
lo,
I
+
i;' ;':
t
.:
--+,1
oj
:.
i.":.."":
'.1=-9:rt,
I:.:'r:::l':':
i.'r
J
U I
',:, ,i
Luo=Q 1I 1 , ' ' - -
:
U]
,
i
lor
I
t
l"'
1o",,
(a)
(b)
Figure 11.15
Consolidated-drainedtriaxial test:
(a) specimcnundcr chambcr confining
pressure:(b) deviator strcssapplication
Table 11.2 Theoretical Values of B at Comnlete Saturation
Theoretical
value
Type of soil
Normally consolidatedsoft clay
Lightly overconsolidatedsoft clays anclsilts
Overconsolidatcdstil'l'claysand sands
Very densc sandsand very stiff clays at high
confining prcssures
0.9998
0.998u
0.987'7
0.9I30
retical valuesof B for various soils at completc saturation.Thcse valuesare listed in
T a b l e 11 . 2 .
Now, if the connectionto drainageis opened,dissipationof the excesspore water pressure,and thus consolidation,will occur. with time, a, will become equal to
0. In saturatedsoil, the changc in the volume of the specimen(A( ) that takes place
during consolidationcan be obtained from the volume of pore water drained (Figure
1 1.16a).Next, the deviator stress,Ao,1,on the specimenis increasedvery slowly (Figure 11.1-5b).
The drainageconnectionis kept open, and the slow rate of deviator stress
application allows complete dissipationol any pore water pressurethat developedas
a result (Au,1: 0).
A typical plot of the variation of deviator stressagainststrain in loose sancland
n o r m a l l y c o n s o l i d a t e dc l a y i s s h o w n i n F i g u r e 1 1 . 1 6 b F
. i g u r e 1 1 . 1 6 cs h o w sa s i m i l a r
plot for dense sand and overconsolidatedclay. The volume change,A2,7,of specimens that occurs becauseof the application of deviator stressin various soilsis also
s h o w n i n F i g u r e s1 1 . 1 6 da n d 1 1 . 1 6 e .
Becausethe pore water pressuredevelopedduring the test is completely dissipated. we have
total and effective confining stress : ot : o\
and
total and effective axial stressat failure : at t (Aor)r : ot : o\
x l
?l
> < 5
i.t
9 l
o *
O '
Axial strain
Axial strain
c l
o l
a l
a l
x l
;'l
"::
f < 6
Axial strain <
frl
6
al
trI
6 {
U '
59
O '
(d)
(c)
by
caused
triaxialtest:(a) volumechangeof specimen
Figure 11.76'Cbnsolidated-draincd
(b) plot ol deviatorstressagainststrainin the verticaldirection
chamberconfiningpressure:
clay;(c) plot of deviatorstressagainststrainin the
for looscszrndand normallyconsolidated
clay;(d) volumechangein loosesand
verticaldirectionfor densesandandoverconsolidated
(e) volumechangein
clayduringdeviatorstrcssapplication;
and normallyconsolidated
clayduringdeviatorstressapplication
densesandandoverconsolidatcd
In a triaxial test, oi is the major principal effectivestressat failure and o! is the minor principal effcctivestressat failure.
Severaltests on similar specimenscan bc conducted by varying the confining
pressure.With the major and minor principal stressesat failure for each test the
Mohr's circlescan be drawn and the failure envelopescan be obtained.Figure 11.17
shows the type of effective stressfailure envelope obtained for tests on sand and normally consolidated clay. The coordinates of the point of tangency of the failure envelope with a Mohrb circle (that is, point A) give the stresses(normal and shear)on
the failure plane of that test specimen.
Overconsolidation results when a clay is initially consolidatedunder an allaround chamber pressure of o. (: oi.) and is allowed to swell by reducing the chamber pressure to oj (: o1). The failure envelope obtained from drained triaxial tests
of such overconsolidated clay specimens shows two distinct branches (ab and bc in
11.8 Consolidated-DrainedTriaxiat Test
327
Eflective
stressiai lure
envelope
Tr = o'tan 0'
a
o:=o:
ot = o'l
Normal strcss
l.-(Ao,i)/--'-*l
l.+tlo")'--*]
Figure 11.17 Effcclive stressfailure cnvclope I'rom drained tests
on sand and normally
consolidatedclay
Figure ll.1tt). The portion ab has a flatter slopc with a cohesion
intercept, and the
shear strength equation for this branch can be written as
TI : c' * o' tan <f\
(11.1-5)
The portion bc of the failure envelope representsa normallyconsolidated
stageof
s o i l a n d f o l l o w st h c e q u a t i o n r t : ( r , t a n6 , .
A consolidated-drainedtriaxial test on a clayey soil may take several
days to
complete. This amount of time is required becausedeviator stress
must be applied
very slowly to ensure full drainage from the soil specimen.For this
reason,the cD
type of triaxial test is uncommon.
Normally
Overconsolidated consolidated
ala
a
T
c'
+
o3=O3
Figure
11'18
Or=o't
O;.
Normal stress
Effective stress failure envelope for overconsolidated
clav
328
o Undisturbedsoil
. '.-.'.
..:
r Remolded soil
t,
" -5
l0
l5
.10
20
30
P l a s t i c ritnyd c xP. I ( % )
60
t30 l(x)
l-so
Figure 11.19 Yttrialion ol'silr r,6'with plasticity indcx lor a number of soils (after Kcnney,
rese)
Comments on Drained and Residual Friction Angles of Clay
'fhe
drained angle of friction, d;', gencrallydccreascswith the plasticityindex of soil.
This fact is illustrated in Figure I 1. l9 for a number of clays from data reported by
Kenncy (19-59).Although the data are considerablyscattered,the general pattern
seemsto hold. ln Figure I l.tt, the rcsiclualshear strength of clay soil is defined.Also
is shown.
in Example I I .1, thc proccdure to calculateresidual friction angle <75i
Skempton ( 1964)provided the resultsof the vzrriationof the rcsidual angleof
friction, $',, of a number ol'clayey sgils with the clay-sizctraction ('2 pm) present.
The followins tzrbleshowsa summary ol'these rcsults:
Clay-size
fraction
l/.1
Sclset
Wiener Tcgel
Jackfield
Oxford clay
Jari
London clay
Walton'.sWood
Weser-Elbe
Little Belt
Biotite
17.7
22.8
-J).4
41.9
46.5
54.9
67
63.2
7'7.2
100
Residual
frictionangle,
d. (deg1
29.8
25.1
19.I
16.3
Iu.6
16.3
t -1.L
9.3
11.2
7.5
At a very high clay content. @iapproachesthe value of the angle of sliding friction for sheet minerals.For highly plastic sodium montmorillonites, the magnitude
of dl mav be as low as 3 to 4o.
329
Example11.2
For a normallyconsolidatedclay,the resultsof a drainedtriaxial testareasfollows:
Chamberconfiningpressure= 16lb/in.2
Deviator stressat failure : 25lblin.z
a. Find the angleof friction, g'.
b. Determinethe angle0 that the failure planemakeswith the major principal plane.
I
E , f l e c t i v sc t r c s s
l u i l u r ee n v o l o p e
q
o'r= 16lb/inr
A
N o r r n a ls t r e s s
Figure 11.20 Mohr'scircleand failure envelopefor a normallyconsolidatedclay
$olution
For a normallyconsolidated
soil,thefailureenvelope
equationis
rf : o' tan+'
(sincec' = 0)
For thetriaxialtest,theeffective
majorandminorprincipalstresses
at failureare
(rl = (rt = ui * (Aou)r * L6 + 25 = 41lb/in.2
and
o\ : ct:
1 6l b / i n . 2
a. The Mohr's circle and the failure envelopeare showninfuigure11.20,
from which we can write
:
srnp-
AB
oA=
(ry)
(*#)
330
or
sind' :
cri - o'^
01+03
47-16 =
0.438
41+16
Q' * 26o
b.9:45 *
o': 4 5 " + 1?6= 5 8 "
z
a
L
Refer to Example1L.2.
a. Find the normal stresso.' and the shearstressrlon the failure plane.
b. Determinethe effectivenormal stresson the planeof maximumshear
stress.
Solution
a. From Eqs.(9,8)and (9.9),
a' (on the failure plane)
o 1 * o ' 3 . o t - c :
'- cos20
--:-;---i
+ "
and
ot - cr5
r, * J
,rsin20
Subsitutingthe valuesof oi = 41 lb/in'z,a\: I6lblin'2, and 0 : 58"into
the precedingequations,we get
4 l + 1 6 4 I - 1 . 6- , - ,
, j;i
x 58) = 23.61b/in.2
* ?cos(2
o, :
and
::
"t
4I_T6
sin(2x 58) = 11'2lb/in.z
Z:
b. From Eq. (9.9),it can be seenthat the maximumshearstresswill occur
on the planewith 0: 45".From Eq' (9.8).
l
- o\cos2g
* o',
I
o
- , : o'\
) + 'o " )
'
o
,
u
.,o
,.
Substituting0 : 45'into the precedingequation gives
o, :4i9
. rycos
tbtin.z
eo: 28.s
331
E x a m p l e1 1 . 4
The equation of the effective stressfailure envelopefor normally consolidated
clayeysoil is 11: o' tan 25'.A drainedtriaxial test wasconductedwiththe same
soil at a chamberconfiningpressureof 80 kN/m2.Calculatethe deviatorstressat
failure.
Solution
For normally consolidatedclay,c' = 0. Thus,from Eq. (11.8),
.
at: oitun'(+s
+)
6' :25o
oi : 8otun'(+sr
(Loo)r = {r't - o\:
T)
: 197 kN/m2
197* 80 : 117kN/m2
E x a m p l e1 1 . 5
The resultsof two drainedtriaxial testson a saturatedclay are asfollows:
ot : o\:
SPecimen
I:
70 kN/m2
(Ao,r)r : 130kN/m2
ct * 6\ = 160kN/m2
SPecimenII:
(Aoir :223.5kN/m?
Determinethe shearstrengthparametersc' and$' .
Solution
For specimenI, the principalstressesat failure are
c ' t : a z : 7 0k N / m 2
and
o't = or : c3 t (Lrit:
70 + 130= 200kN/d2
Similarly,the principalstresses
at failurefor specimenII are \
n\ = oz:
160kNlm2
and
rr\: ot:
{I: * (Aor), : 160+ 223,5:383.5 kN/m2
givenUy& t 11.8)
Usingtherelationship
, we get
dr'\
o\ * o\tan2( 45 + . l
\
+ zc'tan(4t.
+)
Thus, for specimenI,
:.orun'(+s
- +)
2oo
/
6',\
+2c'tan(45*t)
and for specimen II.
383.5=
+ 2c'tan(4t- y)
.
tootan'z(+s
+)
Solvingthe two precedingequations,we obtain
c' : 20 kN/mz
6' :20o
11.9
Consolidated-UndrainedTriaxial Test
The consolidated-undrainedtest is the most common type of triaxial test. ln this test,
the saturatedsoil specimenis first consolidatedby an all-around chamber fluid presSUre,trj, that results in drainage (Figures 11.21aand 11.21b).After the pore water
pressuregeneratedby the application of confining pressureis dissipated,the deviator stress,Ao,7,on thc specimenis increasedto causcshear failure (Figure fl .2Ic).
During this phaseol the test,the drainageline from thc specimenis kept closed.Becausedrainage is not permitted, the pore water pressure,Au,1,will increase.During
the test, simultaneousmeasurementsof Ao,1and Au,yare made. Thc increasein the
pore water pressure,Atr,1,can be expressedin a nondimensionalform as
A:
Au"
-------:Lo,t
(11.16)
where 7 : Skempton'spore pressureparameter (Skempton, 19-54).
The generalpatterns of variation of Ao,1and Aa,1with axial strain for sandand
clay soils are shown in Figures 1l.2ld through l1.2lg. In loose sand and normally
consolidatedclay, the pore water pressureincreaseswith strain. In dense sand and
overconsolidated clay,the pore water pressure increaseswith strain to a certain limit,
beyond which it decreasesand becomesnegative (with respect to the atmospheric
pressure).This decreaseis becauseof a tendencyof the soil to dilate.
Unlike the consolidated-drained test, the total and effective principal stresses
are not the samein the consolidated-undrainedtest.Becausethe pore water pressure
at failure is measured in this test, the principal stressesmay be analyzed as follows:
11.9 Consolidated-UndrainedTriaxial Test
333
E
A x i a ls t r a i n
( c)
4 - ao
'a
(,c
(b)
lno.,
t "
lo.
t
".:,,'.
-+.:
<l-O1
. ,.t,,.
+l o ,
I
lAo,r
I
(c)
Figure 71.21 Consolidatedundrained test: (a) specirnenunder chamber confining pressure:
(b) volume changein specirnencauscdby conlining pressure;(c) cleviatorstress
application;
(d) deviator stressagainstaxial strain fbr krosesand and normally consolidatedclay; (e)
deviator stressagainstaxial strain for dense s:ind and overconsolidatedclay; (f) variation
of
pore water pressurewith axial strain 1or loosc sand and normally consolidatedclay; (g)
variation of pore wiitcr prcssurr:with axial strain lor clensesand ,,nd'ou"ra,rnroliclated
clay
Major principal stressal f a i l u r e ( t o t a l ) :
o3 -f (L,o,1)1: o1
Major principal stressat failure (effective): o r - ( L , u a ) 1 : o \
Minor principal stressat failure (total):
Aj
Minor principal stressat failure (effective): o 1 - ( L , u ) 1 :
o',
\
Total stress
irilr.a envelope
Tt- 6 tan q
o3
oj
o1
o1
l*l
l.-l
(Lutt)|
(Lu,ll
Nortnal stress
'l't't|ltl
and ef1cctivcstressfailureenvelopcsfor consolidatedundrainedtriaxial
Figure 11.22
that no backpressurcis applied.)
tests.(No1c:'l'hcfigureassumcs
: pore water prcssure at failure. The preceding derivaIn thcse equations, (Arz,1)1
t
h
a
t
tionsshow
art
tr; - tr\ -
tr'.
Testson scvcralsimilar specimenswith varying conllning prcssuresmay be conducted to determinc the shear strength parameters.Figure I1.22 showsthe total and
cffectivestressMohr'.scirclesat failure obtained from consolidated-undrainedtriaxial tcsts in sand and normally consolidtrtcdclay. Note that ,4 and B are two total
stressMohr\ circlesobtaincd from two tests.C and D are the effectivestressMohr's
circlcs correspondingto total stresscirclesA and B, respectively.The diametersof
circlesA and C are the same;similarly,thc diametersof circlesB and D are the same.
In Figure ll.22,the total stressfailure envelopecan be obtained by drawinga
line that touches all the total stressMohr's circles.For sand and normally consolidated clays,this will be approximately a straight line passingthrough the origin and
may be expressedby the equation
ry:otanQ
( 11.17)
where rr : total stress
d : the angle that the total stressfailure envelopemakes with the
normal stressaxis. also known as the consolidated-undrained
angle ttf sht'uring re.sistunce
Equation (11.17)is seldom used for practical considerations.
Again referring to Figure ll.22,we see that the failure envelope that is tangent
to all the effective stress Mohr's circles can be represented by the equation 11 =
o' tan rD',which is the same as that obtained from consolidated-drainedtests (see
F i s u r e1 1 . 1 7 ) .
11.9 Consolidated-UndrainedTriaxial Test
o
1=c+OtanQl
U)
6j
Nurn'ii,,r"r,
Figure 11.23Totalstressfailureenvelopeobtainedfrom consolidatcd-unclrained
testsin
over-consolidatcd
clay
In overconsolidatedclays,the total stressfailure envelopeobtained from cons o l i d a t e d - u n d r a i n ctde s t sw i l l t a k e t h e s h a p cs h o w ni n F i g u r e 11 . 2 3 l.' h e s t r a i g h tl i n e
a ' b ' i s r e p r e s e n t e db y t h e e q u a t i o n
rt:c
*otirnd1
(ll.llt)
a n d t h e s t r a i g h tl i n e b ' c ' f o l l o w s t h e r e l a t i o n s h i pg i v e nb y E q . ( 1 1 . 1 7 ) . ' r h ee f T c c t i v e
stressfailure envclope drawn from the cffectivc stressMohr'.scircleswill be similar
t o t h a t s h o w n i n F i g u r e 11 . 2 3 .
Consolidated-drainedtests on clay soils takc considerabletime. For this rcason, consolidated-undraincdtestscan be conductcd on such soilswith pore pressure
measurementsto obtain the drained shearstrength parameters.Becauseclrainageis
not allowed in these testsduring the appliczrtionof deviator stress,they can be performed quickly.
S k e m p t o n ' .pso r e w a t c r p r e s s u r ep a r a m e t c rZ w a s d e l l n e c li n E q . ( 1 1 . 1 6 ) .A t
f a i l u r e .t h e p a r a m c t e 7
r c a n h c w r i t t c na s
A:
A '' :
(Lur),
---( a o , tf)
(11.1e)
The gcneral range of 7, valuesin most clay soils is as follows:
. Normally consolidatedclays:0.-5to I
. Overconsolidatedclays: -0.5 to 0
T a b l e l l . 3 g i v e st h e v a l u e so f A , f o r s o m e n o r m a l l yc o n s o l i d a l e cd l a y sa s o b t a i n e d
by the Norwegian GeotechnicalInstitute.
Laboratory triaxial testsof Bjerrum and Simons(1960)on oslo cray,weald clay,
and London clay showed that A, becomes approximately zero at an overconsolidation value of about 3 or 4.
336
Table 11.3 Triaxial TestResultsfor SomeNormally ConsolidatedClays
Obtainedby the NorwegianGeotechnicalInstitute*
Seven Sisters,Canada
Sarpborg
Lilla Edet, Sweden
Fredrikstad
Fredrikstad
Lilla Edet, Sweden
Cdta River, Sweden
Gota River, Sweden
Oslo
Trondheim
Drammen
Liquid
limit
Plastic
limit
Liquidity
index
r27
69
68
59
-)/
63
60
60
48
36
35
28
30
22
22
30
27
30
25
20
0.28
0.6ri
t --)z
0.58
0.63
1.58
1.30
1.50
0.87
0.50
1.08
1u
-)-l
Drained
friction angle,
Sensitivity"
@'(degl
19
25.5
26
28.5
27
5
50
5
6
50
12
40
4
2
8
LJ
28.5
24
-{1.)
34
28
Al
0.72
1.03
1.10
0.87
1.00
1.02
1.05
1.05
1.00
0.75
1.18
*After Bjerrumand Simons(1960)
" SeeSection1I . l 5 for the delinitionof sensitivity
E xa mp l e1 1 .6
test on a normally consolidatedclay yielded the folA consolidated-undrained
lowingresults:
ot : 12lblin.2
Deviatorstress,(Ar)t
: 9.1lb/in.2
Porepressur
e. (L,u) 1 = 6.8lb/in.2
friction angleandthe consolidated-drained
Calculatethe consolidated-undrained
frictionangle.
Solution
<Yt: 12lblin'2
or:
c3 + (Ao1)1: 72 + 9'l : 2l'1'lblin'2
From Eq. (11.9),for normally consolidatedclaywith c : 0,
./,d\
o t : u t t a n ' l4 5 + ; l
r/
\
;
i
:
^/
d\
21.1: 12tan'( 45 + a )
z /
\
l-
. / 2 t . 1\ o s - . - l
d : 2ltan-r(.;/
4 s l: 1 6 "
11.10 Unconsolidated-UndrainedTriaxialTest
337
Again,
n5 : oz * (Lu)r = 12 * 6.8 : 5.2lb/in.2
oi : ct - (&uir : 21.I - 6.8 : 14.3lblin.z
From Eq. (11.8),for normally consolidatedclay with c' : 0,
o\:o\t"n,(+s.t)
74.3:s.ztan'z(+s.+)
d'- zltun-'(#)"- +sf: zz.r'
11.10
Unconsol idated- Un dra i ned TriaxiaI Test
I n u n c o n s t t l i d a t c d - u n d r a i n et eds t s d
, r a i n e r g le' r o n rt h e s o i l s p e c i n t e ni s n o t p c r m i t t e d
d u r i n g t h c a p p l i c a t i o no [ c h a n . r b epr r c s s u r eo 1 . T h c t e s t s p e c i m c ni s s h e a r c dt o I ' a i l ure by thc application of deviator strcss,Ao,1.ancl drainage is prevented. Bccause
drainageis ncltallowed at any stage,the tcst can be performed quickly.Becauseof the
applicationof chermberconfiningpressurcrrj. the pore water pressurcin the soil spccimen will incrcaseby 4,..A further increascin the pore water pressurc(Aa,,)will occ u r b e c a u s co l ' t h e d e v i a t o rs t r e s sa p p l i c a t i o n .H e n c c , t h e t o t a l p o r c w a t e r p r e s s u r e
rr in thc spccimenat any stageol cleviatorstressapplication can bc given as
Lt-rt(*
F r o m E q s . ( 1 1 . 1 4 )a n d ( l l . 1 6 ) , u , :
( 11.20)
L,u,1
B o j a n c l A r r , ,: 7 A , r , , ,
s.''
u : Bo3 + AAo,1: Bot + A@, * cr)
(lr.2r)
This test is usually conductedon clay specimensand dependson a very imporl.antstrength concept for cohesivesoils if the soil is fully saturated.The added axial
stressat failure (Arlr)r is practically the same regardlessof the chamber confining
pressure.This propcrty is shown in Figure 1l.24.The failure envelope for the total
stressMohr's circlesbecomesa horizontal line and henceis called a rb : 0 condition.
F r o m E q . ( 1 1 . 9 )w i t h O - 0 , w e g e t
r1:c-c,,
(1I.22)
where c,, is the undrained shear strength and is equal to the radius of the Mohr's
circles.Note that the @ : 0 concept is applicableto only saturatedclaysand silts.
The reason for obtaining the same added axial stress(Loitregardless of the
confining pressurecan be explained as follows. If a clay specimen(no. l) is consoli-
338
Total strcss
M o h r ' sc i r c l e s
at tailure ---
a
FailureenvelopeQ=0
61
o1
63
o1
N,,,.,i,lLl"r,
circlesandfailurcenvekrpe(.D- 0) obtainedfrom unconFigure 11.24TotalstrcssMohr'.s
soil
lriaxiallcstsot"tI'ullvsaturltcclcohcsivc
solidated-undrained
dated at a chamber pressurerr, and then shearcd to failure without drainage,the
total stressconditions at I'erilurccan bc representedby the Mohr's circle P in Figure 11.25.The porc pressuredcvclopcd in the specimenat failure is equal to (Arr,1)r.
Thus, the major and minor principal clTcctivestressesat failure are. respectively,
o\:
:
ltrl + (Arr,i)1j (Ar,,)r (rt
(Ar,,)r
and
0\ -
ro1-
(Att,1)7
Q is the effective stressMohrh circle drerwnwith the preceding principal stresses.
N o t e t h a t t h e d i a m e t e r so f c i r c l e sP a n d Q a r c t h e s a m e .
Totalstrcss
M o h r ' sc i r c l e
at fai lurc
------------1
a
l<-
01
O r O:
O'3
-*-l
(Ao,1)1
F-
--=">l
(Ao,7),
l<__
l<_(LuDl_,_,_"____+l . ^;, = ^;;, ,l
Figure 11.25 The rl : 0 concept
(Ao/,
Normal stress
11.11 Unconfined Compression Test on Saturated Clay
339
Now let us consider another similar clay specimen(no. II) that has been consolidated under a chamber pressure o3 with initial pore pressure equal to zero. If the
chamber pressure is increased by Aoj without drainage, the pore water pressure will
increaseby an amount A4,..For saturatedsoilsunder isotropic stresses,the pore water pressureincreaseis equal to the total stressincrease,so Aa,. : Acr: (B : 1). At
this time, the effective confining pressure is equal to oj * Ao. - LJr,.: o7 * Ao3 Ao-, : o.,. This is the same as the effectiveconfining pressureof specimenno. I before the application of deviator stress.Hence, if specimenno. II is shearedto failure
by increasingthe axial stress,it should fail at the samc deviator stress(Ao,,)rthat was
obtained for specimenno. I. The total stressMohr's circle at failure will be R (see
Figure 11.25).The added pore pressureincreasecausedby the application of (An,1)1
will be (Lu)r.
At failure, the minor principal effectivestressis
[(oj + Arrj)]
[Arr, * (Lu,,)r]: ct
( L u , 1 ) 1: c \
and the major principal effectivestressis
l o j * A a . 1+ ( A o , 7 ) 1 ] [ A r l , + ( A a , 7 ) 1 :] l o . * ( A o , 7 ) 1 ] ( L , u , ) 1
: cr
( L u , 1 ) 1- o \
Thus, the effectivestressMohr's circle will still be Q becausestrength is a function
of efTectivestress.Note that the diamcters of circlesP, Q, and R are all the samc.
Any value of Aoj could have been chosen for testing specimen no. II. In any
case,the deviator stress(Ao,1)/to causefailure would have bcen the same as long as
the soil was fully saturatecland fully undrained during both stagesof the test.
I1.11
Unconfined Compression Test on Saturated Clay
The unconfined compressiontest is a specialtype of unconsolidated-undrainecl
test
that is commonly used for clay spccimens.In this test,the confining pressureoj is 0.
An axial load is rapidly applied to the specimento causefailure. At failure, the total
minor principal stressis zero and the total major principal stressis o, (Figure il.26).
a
_F
I
o.r=0
Total stressMohr's
c i r c l ea t l a i l u r e
6t=4,
Normal stress
Figure 11.26 Unconlinedcompressiontest
Chapter 11
Table 11.4 General Relationship of Consistency and
Unconfined Compression Strength of Clays
ton llt2
Consistency
Very soft
Soft
Medium
stiff
Very stiff
Hard
0-25
25-50
-s0,100
100-200
200-400
>400
0-0.25
0.25-0.-5
0.5-1
1
a
2-4
>4
Becausethe undrained shear strength is independent of the confining pressureas
long as the soil is fully saturatedand fully undrained.we have
cl
4"
': T:,,
(lt.23)
where q,, is the unconJined compression strength. Table 11.4 givcs the approximate
consistenciesof clayson the basisof their unconfinedcompressionstrength.A photograph of unconfined compressiontcst equipment is shown in Figure 11.27.
Theoretically, for similar saturated clay specimens.the unconfined compression testsand the unconsolidated-undrainedtriaxial testsshould yicld the samevalues of ci,.In practice,however,unconfinedcompressiontestson saturatedclaysyield
slightly lower valuesof c,,than those obtained from unconsolidated-undrainedtests.
11.12
Sfress Path
Results of triaxial testscan be representedby diagramscalled stresspaths. A stress
path is a line that connectsa seriesof points, each of which representsa successive
stressstate experiencedby a soil specimenduring the progressof a test. There are
severalways in which a stresspath can be drawn. This section covers one of them.
Lambe (1964) suggesteda type of stress path representation that plots q'
againstp' (wherep' and q' are the coordinatesof the top of the Mohr's circle).Thus,
relationshipsfor p' and q' are as follows:
tn
l
:
.1 *
^f
-
o\ * o\
2
"
oi*o"
___:-*-__r__-:
2
(rr.24)
(i 1.2s)
This type of stress path plot can be explained with the aid of Figure 11.28. Let
us consider a normally consolidated clay specimen subjected to an isotropically
11.12 StressPath
341
Figure 11.27
LJnconfined compression test
equipment (courtesyof Soiltest,
Inc.. Lake Bluff. Illinois)
6:=or
i.+(Ao,1)1-,,-,-----------
l
Figure 77.28 Stresspath - plot of q' against p' for a consolidated-drained triaxial test on a
normally consolidatedclay
consolidated-drainedtriaxial test. At the beginning of the application of deviator
s t r e s so
, \ : ot: 03,So
'n ' :
rr'.' -f o'"'
2
U 1
(rt.26)
U 1
and
o\q
o\
-
: U
(11.27)
For this condition, p' and q' will plot as a point (that is, 1 in Figure 11.28).At some
o t h e r t i m e d u r i n g d e v i a t o r s t r e s s a p p l i c a t i o n , -o i o t * L , o , -1 c j l L , o , 1 ; o \ : o t .
The Mohr's circle marked A in Figure 11.28correspondsto this state of stresson the
soil specimen.The valuesof p' and q'for this stresscondition are
(rri * Aa,) * oi
- t r ' r + Lo , - ( r : +
2
2
c', I o',
P'-
2
Lo ,
2"
(11.28)
and
(o1 *
L,o,1) -
o1
q
Ltt,t
2
(tr.2e)
l f t h e s e v a l u e s opf ' a n d q ' w e r e p l o t t e d i n F i g u r e l l . 2 B , t h e y w o u l d b e r e p r e s e n t e d
by point D' at the top of the Mohr'.scircle. So, if the values of p' and q' at various
stagesof the deviator stressapplication are plotted and these points are joined, a
straight line like 1D will result.The straight line 1D is referred to as the stresspathin
a q'-p' plot fbr a consolidated-drainedtriaxial test. Note that the line ID makes an
angle of 45" with the horizontal. Point D representsthe failure condition of the soil
specimen in the test. Also, we can sec that Mohr'.scirclc B representsthe failure
stresscondition.
For normally consolidated clays, the failure envelope can be given by 11 =
c' t an $' . This is thc line OF in Figure 11.28.(Seealso Figure ll .l'7.) A modified failure envelopecan now be dcfined by line OF'. This modified line is commonly called
the K1 line. The equation of the Kr line can be expressedas
q' - p'tana
where rr : the angle that the modified failure envelope makes with the horizontal.
The relationship between the angles S' and a can be determined by referring
to Figure 17.29,inwhich, for clarity, the Mohr's circle at failure (that is, circle B) and
lines OF and OF' as shown in Figure 11.28have been redrawn. Note that O' is the
center of the Mohr's circle at failure. Now.
DO', :
Iana
gO
1 1 . 1 2 S t r e s sP a t h
343
\
c
a
or=O't
O,o',or/l
l.+(Ao,1),+l
Figure 17.29 Relationshipbetwccn ry''ancla
and thus, wc obtain
ni -qt
,
tana -
2
ct*o't
*--;/.
o't*rt5
o\+o,
(11.31)
Again,
CO'
- stn Q
o(y
0\-:L
sind,:_:__o\-ct
o\
*
tr'1
tr'1 | tr'1
(11.32)
2
C o m p a r i nE
g q s .( 1 1 . 3 1a) n d( 1 1 . 3 2 w
) , e s e et h a r
sin {' : 14no
(11.33)
r ( t a na )
{' - sin
(11.34)
Figure 11.30showsa q'-p' plot for a normalry consolidatedclay specimensubjected to an isotropicallyconsolidated-undrainedtriaxial test.At the beginningof the
s
Total stress
Effective stress
s
6
u
I
a't
o , o ' , o r/ '
o'1
Figure 11.30 Stresspath - plot of q' againstp' for a consolidated-undrainedtriaxial test
on a normally consolidatedclay
0.Thisrelationa p p l i c a t i o n o f d c v i a t o r s t r e s s , o l - t r \- o j . H e n c e , p ' : o r a n d q ' ship is representedby point 1. At some other stageof the deviator stressapplication,
rr\ - tra * Arr,l
Ltr,,
and
o\ :oj-
Lu,1
So
p -
oi*o\
2
: c1 *
Ao.,
1'-
1,u,,
(il.3s)
and
'
,
c\
t
-
c3
2
L.tr,1
2
( 11.36)
The precedingvaluesof p' and q' will plot as point U' in Figure 11.30.Pointssuchas
U" representvaluesof p' and q' as the test progresses.At failure of the soil specimen,
p' - (r1+
(Atr,,),
2-
(Arr,1),
( 11.37)
and
(Lor)r
,
, l 2
( 11.38)
T h e v a l u e so f p ' a n d 4 ' g i v e n b y E q s . ( 1 1 . 3 7 )a n d ( 1 1 . 3 8 )w i l l p l o t a s p o i n t U
Hence, the effective stresspath for a consolidated-undrained test can be given by the
curve IU'U. Note that point U will fall on the modified failure envelope, OF' (see
Figure I1.29), which is inclined at an angle a to the horizontal. Lambe (1964) proposed a technique to evaluate the elastic and consolidation settlements of foundations on clav soils bv using the stresspaths determined in this manner.
11.12 StressPath
For a normally consolidatedclay, the failure envelope is given by the equation
tan $' . The correspondingmodified failure envelope(q,-p, plot) ii given
Tr ,o'
bI gq' t 11.30)asq' : p' tan a. In a simirarmanner,if the f;iru;e enuetopeiJrl :
c' * s' lan$' , the correspondingmodifiedfailure envelopeis a q'-p'ploi that ian
be expressedosqrl : m * p' tan a. Expressa as a function ofdr, and givem asa
function of c' andg' .
o-r o,
z
oj
*r'' cotQ'-l-
A
Norntrl stress
o'r I o'r-------..-----*l
Figure | 1.31 Derivationof a asa functionof d' and n asa function of c, and
$olution
FromFigure11.31,
Sln@ E
AB
AC*
+:z
(ry)
AB
CO+OA
c ' c o t S '*
(*#)
c ,\ c o6s' + z( ' i I o 5/ ) , i4n,
(a)
Q':m+p'tana
(b)
or
ComparingEqs. (a) and (b), we find that
m = c'cos@'
and
tana: sin4'
a = tan-1(sind')
1t.13
VaneShear Test
Fairly reliable results for the undrained shear strength, c,, (S : 0 concept), of very
soft to medium cohesive soils may be obtained directly from vane shear tests. The
shear vane usually consistsof four thin, equal-sizedsteel plates welded to a steel
torque rod (Figure 11.32).First, the vane is pushed into the soil. Then torque is applied at the top of the torque rod to rotate the vane at a uniform speed. A cylinder
of soil of height ft and diameter r/ will resist the torque until the soil fails. The
undrained shear strength of the soil can be calculatedas follows.
If I is the maximum torque applied at the head of the torque rod to causefailure, it should be equal to the sum of the resistingmoment of the shear force along
the side surfaceof the soil cylinder (M.) and the resistingmoment of the shearforce
at each end (M,,) (Figure I I .33):
7':M,+M,,+M,,
'l'wo
cnds
The resistingmoment can be given as
M, - (ndh)t',,
(d12)
\ _ J \ - _ ' Y J
Surlacc
arca
Moment
arm
where d : diameter of the shear vanc
/z : height of thc shear vane
For the calculationof M., investigatorshave assumedseveraltypes of distributicln of shear strength mobilization at thc ends of the soil cylinder:
l. Triangular.Shear strength mobilization is c,,at the periphcry of the soil cylinder and decreaseslincarly to zero at thc center.
2, IJni.form.Shear strength mobilization is constant (that is, c,,)from the periphery to the center of thc soil cylinder.
3. Parabolic.Shear strength mobilization is c,,at the periphery of the soil cylinder
and dccreasesparabolicallyto zero at the center.
These variations in shear strength mobilization are shown in Figure 11.33b.In
general,the torque, I at failure can be expressedas
I a2n r/rI
,:rr',,1 +B;l
"
L
L
cu:
I a2n+ ^a31
lrl
t
.
( 11.41)
a - j
lsTl
L
a
l
(rr.42)
1 1 . 1 3 V a n eS h e a r T e s t
I
I
h
I
k-,1
I
--+l
Figure 11.32
Diagram ol'vane shear
test equlpm(jnt
T r i a n g u l a rn t o b i l i z a t i o n
of shearstrength
I
I
I
1
' , - 4
i
l<-f
U n i f b r mm o b i l i z a t i o n
0f shearstrergth
'i-t
2
Parabolicfbrm of
m o b i l i z a t i o no f
shearstrength
I
I
I
-------------*1---
--,
t
Figure 1 1.33 Derjvation of Eq. (11.42): (a) resisting moment
of shear force; (b) variations
i n s h e a rs t r e n g t hm o b i l i z a t i o n
348
where B : \for triangular mobilization of undrained shear strength
B : I for uniform mobilization of undrained shear strength
B : I for parabolic mobilization of undrained shear strength
Note that Eq. (11.42)is usually referred to as Culding'sequtttion.
Vane shear testscan be conductedin the laboratory and in the field during soil
exploration. The laboratory shear vane has dimensionsof about 1-1mm (j in.) in diameter and 25 mm ( I in.) in height. Figurc I 1.34shows a photograph of laboratory
vane shear test equipment. Figurc I 1.35 shows thc field vanes recommended by
ASTM ( 1994).Table I 1.5givesthe ASTM recommendeddimensionsof field vanes.
A c c o r d i n g t o A S T M ( 1 9 9 4 ) .i f h / d : 2 . t h c n
( 11.43)
and
(11.44)
t
(in )
I n t h e f i e l d .w h c r e c o n s i d c r a b l cv a r i a t i o ni n t h c u n d r a i n e ds h e a rs t r e n g t hc a n
be found with depth, vane sheartcstsare extrcmcly useful.In a short period, one can
establisha reasonablcpattern ol'thc changeof c',,with depth. However,if the claydeposit at a given site is more or lessunif'orm,a few unconsolidated-undrainedtriaxial
Figure 1 1.34 Laboratory vane shear test devicc (courtesy of Soiltest, Inc., Lake Bluff, Illinois)
| 1.13 VaneShear Test
1
,
|
I
I
+s)
t
,
<
I
II
+
:f
ll
.lr"
-+;t
ri++
.ii
+i+*i1.i,..t:i*-
iil
I
Rectangularvane
li
Taperedvane
Figure 11.35 ceometry of field vanes[source: FromAnnual Book o.fASTM standards,
04'08'p. 346.CopyrightO 1994AmericanSocietyfor Testingand Materials.Reprintedwith
p e r m i s s i ol n
Table 11.5 Recommended
Dimensions
of FieldVanes*,,
Thickness
of blade,
mm (in.l
Diameter
of rod
mm (in.)
Diameter,
mm (in.)
Height,
mm (in.)
76.2(3)
101.6(4)
127.0
(s)
16(+)
32(i)
1 a ? / l \
t1.t \t)
NX
3 8 . r( r j )
-50.8
(2)
$5 (2))
1 0 r . 6m m ( 4 i n . ) 1 '
9 2 . 1( 3 ; )
1 8 4 . 1( 7 1 )
3 . 2( * )
1 a 1 t ! \
LL.t \2)
Casingsize
AX
BX
1 6 (*r)
LL.t \1)
( +)
t2.7
*After ASTM. 1994
" Selection of vane size is directly related to the consistencyof the soil being tested; that is,
the softer the soil, the larger the vane diameter should be.
blnside
diameter
tests on undisturbed specimenswill allow a reasonable estimation of soil parameters
for design work. Vane shear tests are also limited by the strength of soils in which
they can be used. The undrained shear strength obtained from a vane shear test also
dependson the rate of application of torque Z
Bjerrum (1974) showed that as the plasticity of soils increases,c,,obtained from
vane shear tests may give results that are unsafe for foundation design. For this reason, he suggestedthe correction
ca(dcsign) :
iCr(uane shear)
(11.4s)
where
i : correction factor : l.l - 0.54 log(P1)
(11.46)
P7 : plasticity index
More recently,Morris and Williams (199a)gave the correlations of ,\ as
i :
i ' l )0 ' 5 7
1 . 1 3 n{ ) { ) n ( +
(for P/ > ,5)
(11.47)
(for LL > 20)
(11.48)
and
1 9.57
n : 7.01e tttts(r'r')
where LL : liquid limit (%).
| 1.14
Other Methods for Determining
Undrained Shear Strength
A modified form of the vane sheartest apparatusis the Torvane(Figure I 1.36),which
is a handheld devicewith a calibratedspring.This instrument can be used for determining c,,for tube specimenscollectedfrom the lield during soil exploration, and it
can be used in the field. The Torvane is pushedinto the soil and then rotated until the
soil fails. The undrained shear strength can be read at the top of the calibrateddial.
Figure 1 1.37shows a pocket penetrometer,which is pushed directly into the soil.
The unconfined compressionstrength (q,,) is measuredby a calibrated spring.This
device can be used both in the laboratorv and in the field.
11.15
Sensitivity and Thixotropy of CIay
For many naturally deposited clay soils, the unconfined compression strength is
greatly reduced when the soils are tested after remolding without any change in
11.15 Sensitivityand Thixotropyof Clay
351
Figure 11.36
Torvane (courtesy of
Soiltcst,Inc., Lake Bluff,
Illinois)
Figure 11.37
Pocket penetrometer
( c o u r t e s yo f S o i l t e s t l. n c . .
Lake Bluff, Illinois)
Mrdiutnquic*
E
l6
"l
..;
;
'5a
Slightly quick
d
q
Very sensitive
Mediurr sensitive
S l i g h t l ys e n s i t i v e
Inscns
itivc
Axlal straln
Figure 11.38 L.lnconfinedcomprcssionstrength
for undisturbcd and rcmoldcd clay
Figure ?1.39 Classificationo[ claysbascdon
scnsitivity
t h e m o i s t u r ec o n t e n t ,a s s h o w n i n F i g u r e I 1 . 3 UT. h i s p r o p c r t y o f c l a y s o i l si s c a l l e d
sensitivity.The degree of sensitivitymay bc defined as thc ratio of the unconfined
comprcssionstrength in an undisturbedstate to thettin a remolded statc, or
4a(undisturhc,J)
^
Jr:-._.-4rr(remolded)
( I 1.4e)
The sensitivityratio of most claysrangesfrom about 1 to 8; however,highly flocculent marine clay depositsmay have sensitivityratios ranging from about 10 to 80.
Some clays turn to viscousfluids upon remolding. These clays are found mostly in
the previously glaciatedareasof North America and Scandinavia.Such claysare referred to as quick clays.Rosenqvist(1953)classifiedclayson the basisof their sensitivity. This general classificationis shown in Figure 1 1.39.
The loss of strength of clay soilsfrom remolding is causedprimarily by the destruction of the clay particle structurethat was developedduring the original process
of sedimentation.
11.15 Sensitivity and Thixotropy of Clay
353
If, however, after remolding, a soil specimenis kept in an undisturbed state (that
is, without any change in the moisture content), it will continue to gain strength with
time. This phenomenon is referred to as thixotropy. Thixotropy is a time-dependent,
reversible processin which materials under constant composition and volume soften
when remolded. This loss of strength is gradually regainedwith time when the materials are allowed to rest. This phenomenon is illustrated in Figure 11.40a.
Most soils,however,are partially thixotropic - that is, part of the strength loss
causedby remolding is never regained with time. The nature of the strength-time
variation for partially thixotropic materials is shown in Figure 11.40b.For soils,the
differencebetween the undisturbedstrength and the strength after thixotropic hardening can be attributed to the destruction of the clay-particlestructure that was developed during the original processof sedimentaticln.
,\.
= 4-a ( u n d i s t u r b e d )
{a(renrolded)
Initial
undi sturbed
strength
{u( undisturbed)
clJ
E
q
€
z
E
6"'&
€
E
E
/---"
$t
€
c
E
d,
./a(re Irolded )
a
6*
,/***
!
c
E
r,b"
€
g
+
E
(b)
Figure
11.40 Behavior of (a) thixotropic
marerial; (b) partially
thixotropic
material
354
Table 11.6 Empirical EquationsRelatcdtt'tc,,ando'6
Remarks
Relationship
Reference
(1957)
Skempton
L rrlVSTl
For normally consolidated
clay
+CO: 0 . 1 1 + 0 . 0 0 3 7 ( P l )
P/ : plasticity index (%)
c,(VSr)- undraincd shear
strength frcnt vane shear tcst
1+:
C h a n d l c r( l 9 t t t t )
(f
',
o.u + o.oo37(p1)
rri : preconsolidationpressurc
Jamiolkowski ct al. ( lt)tt5)
:1 -0.23+0.04
(f
Ladd ct al. (1977\
For lightly overconsolidated
clays
',
;(fo:
Mesri ( 19t39)
Can be used in overconsolidiited soil; accur€rcy+257";
not valid for sensitiveand
fissured clays
o.zz
/.,\
\ - /,,",,,,,,,.,,,,,,
,,,, : ( o c R ) ' i E
/(;,\
\a/,,,,,,,,,",ons.,i(r,,c(r
O C R : o v c r c o n s o l i d a t i o rna t i o
11.16
Empirical Relationships between Undrained Cohesion
(cu)and Effective Overburden Pressure (ob)
Severalempirical relertionshipscan be obscrvcd between c,,and the effcctive overburden pressure (o'a1)in the field. Somc of these relationshipsare summarizedin
Table I 1.6.
r a t i o w a s d c f i n o di n C h a p t e r 1 0 a s
The overconsolidation
OCR
:
o,,
tt6
( 11.s0)
pressure.
whereoj. : preconsolidation
10.8
Its liquid
A soil profile is shownin Figure11.41.The clayis normallyconsolidated.
compression
strength
plastic
limit
Estimate
the
60
and
its
is
25.
unconfined
limit is
from the groundsurface.UseSkemptonb
of the clayat a depthof l0 m measured
from TableI L6 andEqs.(11.45)and(lL46).
relationship
Relationships between Undrained Cohesion (c) and Effective Overburden pressure(crL)
Dry sand
Y = 1 5 . 5k N / m :
Clay
ty - 30Vc
G, = 2.68
Figure 11.41
Solution
For the saturatedclay layer,the void ratio is
: 0.8
e = u)Gs: (2.6BX0.3)
The effectiveunit weight is
(2.68-l)(e.81)
.
/c.-l\
:9.r6kN/mr
Tiroy:\,*"i,,:__
The effectivestressat a depth of 10m from the groundsurfaceis
qb : 37"ond
+ Tylrov: (3X15.5)+ (7X9.16)
= 1 1 0 . 6k2N / m 2
From Table11.6.
ca(vST)
o'p
Ca(VST)
n'Sr=
* 0.11+ 0,0037(P1)
- 25)
0.11+ 0.0037(60
and
ca(vsr): 26.49 kN/m2
From Eqs. (11.45)and (11.46),we ger
C, :
,\cr,y51,
* lL.7 - 0.54log(Pl)1c,,y5.r1
* 2s)J26.49
: 11.7- 0.54lo9(60
: 22.95kN/mz
Sotheunconnned
"T':';;:',;eT;:,
: 45.e
kN/m2
356
11.17
Shear Strength of Unsaturated Cohesive Soils
The equation relating total stress,effectivestress,and pore water pressurefor unsaturatedsoils,can be expressedas
(, - uul
(r':
where o' :
o :
rt. :
u?i,:
X(uu- uu,)
(11.s1)
effective stress
total stress
pore air pressure
pore water pressure
When the expression for o' is substituteclinto the shear strength equation
_
[Eq. ( I 1.3)1,which is bascd on effectivestressparameters,we get
r[ : c' + [rr - u,,* y(u,,- rr,,,)]tan
{'
( r r.s2)
The values of 1 depend primarily on the degree of saturation. with ordinary
triaxial equipment uscd for laboratory testing, it is not possibleto determine accurately the effectivestrcssesin unsaturatedsoil specimens,so the common practice is
to conduct undrained triaxial testson unsaturateclspecimensancl measureonly the
total stress.Figure 11.42showsa total strcssfailure envelopeobtained from a number of undraincd triaxial tests conducted with a given initial clegreeof saturation.
The failure envelopc is gcnerally curved. Higher conlining pr"riur" causeshigher
compressionof the air in void spaces;thus, the sotubility of void air in void water is
increased.For design purposes,the curveclenvclopc is sometimesapproximateil as
a straight f ine, as shown in Figurc 11.42,wtthan equation as follows;
rt:
r'f otanrp
(Note that c and rf in the precedingcquation are empirical constants.)
a
Normal stress(total)
Figure 1 1.42
'fotal
stressfailure envelope for unsaturated cohesivesoils
(11.53)
11.18 Summary and GeneralComments
357
o (lb/in2)
160
1000
320
160
Inorganicclay (CL)
? 8oo
z
Degrceof saturation=
tr 600
Rfl
_
'
; .+00
.<
=
,+0
5, 200
0d
!
500
1000
2(XX)22(X)
Totalnonlalstrcss.
o (kN/rnrt
F i g u r e 1 1 . 4 3Y a r i a t i < r n o
t l f. t c t o t aslt r e s fsa i l u r e e n v e l o p c w i t h c h a n gienoi tfi a d
l egrecol'
s:lturationobtaincdfrttmundrainedtcstsof an inorganicclay(al'te
r Casagranclc
alcl
Hirschfeld,1960)
F i g u r e 1 1 . 4 3s h o w st h e v a r i a t i o n o f t h c t o t a l s t r e s sc n v e l o p e sw i t h c h a n g eo f
t h e i n i t i a l d e g r e eo f s a t u r a t i o no b t a i n c d f r o m u n d r a i n e dt e s t so n a n i n o r g a n i cc l a y .
Note that for these teststhe specimcnswerc prepared with approximately the same
i n i t i a l d r y u n i t w e i g h to f a b o u t 1 6 . 7k N / m r ( 1 0 6I b i m r ) . F o r a g i v c nt o t a l n o r m a l s t r c s s ,
tl.reshear stress needed to cause failure dccreascsas the dcgrce of saturalion inc r e a s e sW
. h e n t h c d e g r e co f s a t u r a t i o nr e a c h e s1 0 0 ' 2 ,t.h e t o t a l s t r e s sl ' a i l u r ce n v e lope becomesa horizontal line thzrtis the samc as with thc r/ : 0 conccpt.
In practical caseswhere a cohesivcsoil dcposit may bccome saturateclbccause
o f r a i n f a l l o r a r i s e i n t h c g r o u n d w a t c rt a b l c , t h e s t r e n g t ho f p a r t i a l l ys a t u r a t c dc l a y
s h o u l d n o t b e u s e d f o r d e s i g nc o n s i d e r a t i o n sI.n s t e a d ,t h e u n s a t u r a t e ds o i l s p e c i mens collectedfrom thc field must bc saturatedin the laboratorv and the undraincd
strength determined.
| 1.18
Summary and General Comments
In this chapter, the shear strengths of grandular and cohesivesoils were examined. Laboratory procedures for determining the shcar strength paramcters were
described.
In textbooks, determination of the shear strength parametersof cohesivesoils
appears to be fairly simple. However, in practice, the proper choice of these parametersfor designand stability checksof various earth, earth-retaining,and earthsupported structures is very difllcult and requires experience and an appropriate
theoretical background in geotechnicalengineering.In this chapter, three types of
strength parameters (consolidated-drained,cctnsolidated-undrainecl,anrJunconsolidated-undrained)were introduced. Their use dependson drainageconditions.
Consolidated-drainedstrength parameterscan be used to determine the longterm stability of structuressuchas earth embankmentsand cut slopes.Consolidatedundrained shearstrengthparameterscan be usedto study stability problems relating
to caseswhere the soil initially is fully consolidatedand then there is rapid loadine.
3s8
Nonnally consolidated
anisotropicclay
. Saturated
ciay , ,r..,
(a)
Figure 11.44 Strengthanisotropyin clay
An excellentexample of this is the stability of slopesof carth dams after rapid drawdown. The unconsolidatcd-undrainedshearstrengthof clayscan be usedto cvaluate
the end-of-constructionstability of saturatedcohesivesoilswith thc assumptionthat
the load causcd by construction has bcen applied rapidly and there has been little
time for drainagcto take place.The bcaring capacityof foundationson soft saturated
claysand the stability of the baseol'cmbankments on soft claysarc cxamplesof this
condition.
The unconsolidated-undrainedshearstrength of some saturatcdclayscan vary
depending on the dircction of load application;this is referred to as anisotrcpywith
respectto strength.Anisotropy is primarily causedby the naturc of the depositionof
the cohesivcsoils,and subsequentconsolidationmakes the clay particlesorient perpendicular to the direction of thc major principal stress.Parallel orientation of the
c l a y p a r t i c l e sc a n c a u s et h e s t r c n g t h o f c l a y t o v a r y w i t h d i r e c t i o n . F i g u r e 1 1 . 4 4 a
showsan elcmcnt of saturatedclay in a deposit with the major principal stressmaking an angle a with respectto the horizontal. For anisotropicclays,the magnitudeof
) {:,,1,,:1r"1,
for overc,,will be a lunction of a. For normally consolidatedclays,c,,1,*-e11,,,
consolidatcd clays,c,,,,,q..)( c,,(,,,,.,.Figure 11.44bshows the directional variation
for c,,1,,;.
The anisotropy with rcspcct to strength for clayscan have an important effect on the load-bearingcapacity of foundations and the stability of earth embankments becausethe direction of thc major principal stressalong the potential failure
surfaceschanges.
The sensitivityof clayswas discusscdin Section 11.15.It is imperativethat sensitive clay depositsare properly identified. For instance,when machine foundations
(which are subjectedto vibratory loading) are constructedover sensitiveclays,the
clay may substantiallylose its load-bearingcapacity.and failure may occur.
Problems
11.1 A direct shear test was conductedon a specimenof dry sand with a normal
s t r e s so f 2 0 0 k N / m 2 . F a i l u r eo c c u r r e cal t a s h e a rs t r e s so f 1 7 5k N / m 2 .T h e s i z e
of the specimentested was 75 mm x 75 mm x 30 mm (height). Determine
the angle of friction, @'.For a normal stressof 1-50kN/m2, what shear force
would be required to cause failure in the specimen?
Problems
359
ll.2
The size of a sand specimen in a direct shear test was ,50mm x 50 mm x 30 mm
(height). It is known that, for the sand, tan g,= 0.65/e(where e : void ratio)
and the specificgravity of solids,G, : 2.68.During the test a normal stress
of 150 kN/m2 was applied. Failure occurred at a shear stressof 110 kN/m2.
What was the massof the sand soecimen?
11.3 Following are the resultsof four drained direct shear testson a normally
consolidatedclay:
Size of specimen : 60 mm x 60 mm
Height of specimen : 30 mm
Test
no.
I
2
3
/1
rt.4
Normal
force
{N}
Shear
force at
failure(N}
200
300
400
500
1.5-5
230
3l0
385
Draw a graph for the shear stressat failure againstthe normal stressand determine the drained angle of friction (r/->')
from the graph.
Following are the resultsof four drained direct shcar testson a normally
consolidatedclay:
S p e c i m e ns i z e : d i a m e t e ro f s p e c i n t e n: 2 i n .
- I in.
hcight of spccir.r.ren
Test
no.
Normal
force
(lb)
I
2
3
60
90
ll0
3'7.5
-5-5
70
lll
u0
A
Shear
force at
failure (tbl
Draw a graph for shear stressat failure againstthe normaI stressand determine the drained angle of friction ({,) from the graph.
11.5 The equation of the effectivestressfailure envelopefor a loose sandysoil was
obtained from a direct shear test as r 1.: c' tan 30o.A drained triaxial test
was conductedwith the samesoil at a chamber confining pressureof 10 lb/in.2
a. Calculate the deviator stressat failure.
b. Estimate the angle that the failure plane makes with the major principal
plane.
c. Determine the normal stressand shear stress(when the specimenfailed)
on a plane that makes an angle of 30'with the rnajor principal plane. Also.
explain why the specimendid not fail along the plane during the test.
11.6 The relationship between the relative density D, and the angle of friction,
@,,
of a sand can be given as Q'" : 28 + 0.18 D, (D,is in %). A drained triaxial
test on the same sand was conducted with a chamber confining pressure of
120 kN/m2. The relative density of compaction was 65%. Calculatethe maior
principal stress at failure.
360
Chapter | 1 Shear Strength of Soil
l1..7 For a normally consolidated clay, the results of a drained triaxial test are as
follows:
Chamber confining pressure: 15 lb/in.2
Deviator stressat failure : 34 Ibiin.2
Determine the soil friction angle,@'.
11.8 For a normally consolidatedclay,it is given that $' :24".In a drained triaxial test, the specimenfailed at a deviator stressof 175 kN/m2. What was
the chamber conlining pressure,ai?
11'9 For a normally consolidatedclay,it is given that $'= 28'' In a drained triaxial
test, the specimenfailed at a deviator stressat 30 lb/in.'What was the chamber confining pressure,oi?
11.10 A consolidated-drainedtriaxial test was conductedon a normally cclnsolidated clay.The resultswere as follows:
( r ' r1 2 5 0 k N i m 2
( A r r , , ) ,: 2 7 5 k N / m r
Determine the following:
a. Angle of friction, r/,'
b. Anglc g that thc failure plane makes with the major principal plane
c. Normal stress,rr', and shear strcss,11,on the failure plane
11.11 The resultsof two drained triaxial tcsts on a saturatedclay are as follows;
c h a m b e rc o n f i n i n gp r c s s u r c: 7 0 k N / m r
S p e c i m e nI :
d e v i a t o rs t r c s sa t f a i l u r e - 2 1 5 k N / m 2
S p e c i m e nI I : c h a m b c rc o n f i n i n gp r e s s u r e: 1 2 0k N / m ' )
deviator stressat failurc : 260 kN/m'
Calculatc the shcar strength parametersof the soil.
11.72 lf a spccimenof clay describedin Problem l 1.11 is tested in a triaxial apparatus with a chamber confining prossureof 20()kN/mr, what will be the major principal stressat failurc'/ Assume full drained condition during the test.
11.13 A sandy soil has a drained angle of friction of 35'. In a drained triaxial test
. hat isthe
o n t h e s a m es o i l . l h ed c v i a t o rs t r c s sa t f a i l u r ei s 2 . 6 9t o n / f t 2 W
chamber confining pressure'?
11.14 A deposit of sand is shown in Figure 11.45.Find the maximum shear resistancein kN/m2 along a horizontal plane located l0 m below the ground
surface.
11.15 A consolidated-undrainedtest on a normally consolidatedclay yielded the
following results:
(rr: l5lb/'n'2
D e v i a t o r s t r e s s(, L o , ) r : 1 1 l b / i n . 2
Pore pressure,(Aa,)1 -- 1.2lblin,1
Calculate the consolidated-undrainedfriction angle and the drained friction
angle.
11.16 Repeat Problem 11.15,using the following values:
o 3 - 1 4 0k N / m 2
( A o , i ) , : 1 2 5k N / m 2
(Lu,,)i: 75kN/m2
Problems
361
Ground water table
Sand
Q'= 42"
e =0.6
Gs =2.67
Figure 11.45
11.17 The shear strength of a normally consolidatedclay can be given by the equation 11- o' tan 31'. A consolidated-undrainedtest was conductedon the
clay.Following are the resultsof the tcst:
C h a m b e rc o n f i n i n gp r e s s u r e: 1 1 2k N / m r
Deviator stressat failure - 100kN/m2
Determine
a. The consolidated-undrainedfriction angle,{
b. The pore water pressuredcveloped in the clay specimenat failure
11.18 For the clay specimendescribedin Problem I L 17,what would have been the
deviator stressat failure if a drained test would have been conductedwith
t h e s a m ec h a m b e rc o n f i n i n gp r e s s u r e( i . e . ,o j : 1 1 2k N / m 2 ) ?
11.19 A silty sand has a consolidated-undrainedfriction angle of 22" and a drained
friction angle of 32" (c' : 0). If a consolidated-undrainedtest on such a soil
is conducted at a chamber confining pressureof 1.2tonlft2, what will be the
major principal stress(total) at failure? Also, calculatethe pore pressure
that will be generatedin the soil specimenat failure.
11.20 Repeat Problem 11.19,using the following values:
4>: t9'
6' - 28'
o.r = f35kN/m2
11.21 The following are the resultsof a consolidated-undrainedtriaxial test in a
clay:
Specimen
no.
a3
(kN/mr)
rr, at failure
(kN/mr)
t92
384
375
636
Draw the total stressMohr's circles and determine the shear strength parameters for consolidatedundrained conditions (i.e.,d and c).
362
11.22 The consolidated-undrained test results of a saturated clay specimen are as
follows:
ut: 9J kNim'
ar at failure : 197kN/m2
What will be the axial stressat failure if a similar specimen is subjected to an
unconfined compressiontest?
11.23 The friction angle,@', of a normally consolidatedclay specimencollected
during field exploration was determined from drained triaxial tests to be 25'.
The unconfined compressionstrength,q,,,of a similar specimenwas found to
be 100kN/m2. Determine the pore water pressureat failure for the unconfined compressiontest.
11.24 Repeat Problem I1.23 using the following values:
$' - 23o
q,, - 120 kN/m'
11.25 The resultsol two consolidated-drainedtriaxial tcsts on a clavevsoil are as
follows:
(r'l(failure)
Test no.
u\ (lb lin.2l
I
2
z7
I L
(lb/in.'zl
4n
Use thc failure envelopeequation givcn in Example I 1.7- that is, 11': v11a
p ' L a na . ( D o n o t p l o t t h c g r a p h . )
a. Findrr anda
b. Find c' and 6' .
11.26 A 1,5-m-thicknormally consolidatedclay laycr is shown in Figure 11.46.The
plasticity index of the clay is 18.Estimate the undrained cohesionas would
bc determined from a vane shear test at a depth of 6 m belclwthe ground
surface.Use Skempton'scquation in Table 11.6.
''""1,3m
*
't'
l,'].
:"
l
.'
' " ,1l .5nl
Clay
Y , a r= 1 9 . 5k N / m r
Figure 11.46
References
363
References
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Shear Strength of Soil

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