# SHEAR STRENGTH OF SOIL

Mechanics

Jul 18, 2012 (6 years and 3 months ago)

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CHAPTER 8
SHEAR STRENGTH OF SOIL
8.1 INTRODUCTION
One of the most important and the most controversial engineering properties of soil is its shear
strengt h or ability to resist sliding along internal surfaces within a mass. The stability of a cut, the
slope of an earth dam, the foundations of structures, the natural slopes of hillsides and other structures
built on soil depend upon the shearing resistance offere d by the soil along the probable surfaces of
slippage. There is hardly a problem in the fiel d of engineering which does not involve the shear
properties of the soil in some manner or the other.
8.2 BASIC CONCEPT OF SHEARING RESISTANCE AND
SHEARING STRENGTH
The basic concept of shearing resistance and shearing strength can be made clear by studying firs t
the basic principles of frictio n between solid bodies. Consider a prismatic block B resting on a
plane surface MN as shown in Fig. 8.1. Block B is subjected to the force Pn which acts at right
angles to the surface MN, and the force Fa that acts tangentiall y to the plane. The normal force Pn
remains constant whereas Fa gradually increases from zero to a value which will produce sliding. If
the tangential force Fa is relatively small, block B wil l remain at rest, and the applied horizontal
force will be balanced by an equal and opposite force Fr on the plane of contact. This resisting force
is developed as a result of roughness characteristics of the bottom of block B and plane surface MN.
The angle 8 formed by the resultant R of the two forces Fr and Pn with the normal to the plane MN
is known as the angle of obliquity.
If the applied horizontal force Fa is gradually increased, the resisting force Fr wil l likewise
increase, always being equal in magnitude and opposite in direction to the applied force. Block B
wil l start sliding along the plane when the force Fa reaches a value which will increase the angle of
obliquit y to a certain maximum value 8 . If block B and plane surface MN are made of the same
253
254
Chapte r 8
M
N
Figur e 8.1 Basi c concep t of shearin g resistanc e and strength.
material, the angl e 8m is equa l to (ft whic h is terme d the angle o f friction, and the valu e tan 0 is
terme d the coefficient of friction. If bloc k B and plane surfac e MN are made of dissimila r materials,
the angl e 8 is terme d the angle of wall friction. The applie d horizonta l force Fa on bloc k B is a
shearin g force and the develope d forc e is frictio n or shearing resistance. The maximu m shearin g
resistanc e whic h the material s are capabl e of developin g is calle d the shearing strength.
If anothe r experimen t is conducte d on the same bloc k wit h a highe r norma l load Pn the
shearin g force Fa wil l correspondingl y be greater. A series of such experiment s woul d show that the
shearin g force Fa is proportiona l to the norma l load Pn, that is
F =P tan
(8.1 )
If A is the overal l contac t area of bloc k B on plan e surfac e M/V, the relationshi p may be
writte n as
F P
shear strength, s = —- = —- tan,
A A
or
s = a tan
(8.2)
8.3 THE COULOM B EQUATIO N
The basi c concep t of frictio n as explaine d in Sect. 8.2 applie s to soil s whic h are purel y granula r in
character. Soil s whic h are not purel y granula r exhibi t an additiona l strengt h whic h is due to the
cohesio n betwee n the particles. It is, therefore, stil l customar y to separat e the shearin g strengt h s of
such soil s int o two components, one due to the cohesio n betwee n the soi l particle s and the other due
to the frictio n betwee n them. The fundamenta l shea r strengt h equatio n propose d by the Frenc h
enginee r Coulom b (1776) is
s = c + (J ta n
(8.3)
Thi s equatio n expresse s the assumptio n that the cohesio n c is independen t of the norma l
pressure cr acting on the plane of failure. At zero norma l pressure, the shear strengt h of the soil is
expresse d as
s = c
(8.4)
Shea r Strengt h of Soi l
255
c
1
Norma l pressure, a
Figur e 8.2 Coulomb's law
Accordin g to Eq. (8.4), the cohesio n of a soi l is define d as the shearin g strengt h at zer o
norma l pressur e on the plan e of rupture.
In Coulomb's equatio n c and 0 ar e empirica l parameters, the value s of whic h for any soi l
depen d upo n severa l factors; the mos t importan t of thes e ar e :
1. Th e pas t histor y of the soil.
2. Th e initia l stat e of the soil, i.e., whethe r it is saturate d or unsaturated.
3. Th e permeabilit y characteristic s of the soil.
4. Th e condition s of drainag e allowe d to tak e plac e durin g the test.
Sinc e c and 0 in Coulomb's Eq. (8.3 ) depen d upo n man y factors, c is terme d as apparent
cohesion and 0 the angl e of shearin g resistance. For cohesionles s soi l c = 0, the n Coulomb's
equatio n become s
s = a tan
(8.5)
The relationshi p betwee n th e variou s parameter s of Coulomb's equatio n i s show n
diagrammaticall y in Fig. 8.2.
8.4 METHODS OF DETERMININ G SHEA R STRENGT H
PARAMETER S
Method s
The shea r strengt h parameter s c and 0 of soil s eithe r in the undisturbe d or remolde d state s may be
determine d by any of the followin g methods:
1. Laboratory methods
(a) Direc t or box shea r tes t
(b) Triaxia l compressio n tes t
2. Field method: Van e shea r tes t or by any othe r indirec t method s
Shear Parameter s of Soil s in-situ
The laborator y or the fiel d metho d tha t has to be chose n in a particula r cas e depend s upo n the typ e
of soi l and the accurac y required. Whereve r the strengt h characteristic s of the soi l in-sit u ar e
required, laborator y test s ma y be use d provide d undisturbe d sample s ca n be extracte d fro m the
256
Chapte r 8
stratum. However, soil s ar e subjec t t o disturbanc e eithe r durin g samplin g or extractio n fro m the
samplin g tube s i n the laborator y eve n thoug h soi l particle s posses s cohesion. I t i s practicall y
impossibl e t o obtai n undisturbe d sample s of cohesionles s soil s and highl y pre-consolidate d cla y
soils. Sof t sensitiv e clay s ar e nearl y alway s remolde d durin g sampling. Laborator y method s may,
therefore, be use d onl y i n suc h cases wher e fairl y good undisturbe d sample s ca n be obtained.
Wher e i t is not possibl e to extrac t undisturbe d sample s fro m the natura l soi l stratum, any one of the
followin g method s ma y hav e to be use d accordin g t o convenienc e and judgmen t :
1. Laborator y test s on remolde d sample s whic h coul d at bes t simulat e fiel d condition s of the
soil.
2. An y suitabl e fiel d test.
The presen t tren d is to rel y mor e on fiel d test s as thes e test s hav e bee n foun d to be mor e
reliabl e tha n eve n the mor e sophisticate d laborator y methods.
Shear Strength Parameter s of Compacted Fills
The strengt h characteristic s of fill s whic h ar e to be constructed, suc h as eart h embankments, ar e
generall y foun d i n a laboratory. Remolde d sample s simulatin g the propose d densit y an d wate r
conten t of the fil l material s ar e mad e i n the laborator y and tested. However, the strengt h
characteristic s of existin g fill s ma y hav e t o be determine d eithe r by laborator y or fiel d method s
keepin g i n vie w the limitation s of eac h method.
8.5 SHEA R TEST APPARATU S
Direc t Shea r Tes t
The origina l for m of apparatu s for the direc t applicatio n of shea r forc e is the shea r box. The box
shea r test, thoug h simpl e in principle, ha s certai n shortcoming s whic h wil l be discusse d late r on.
The apparatu s consist s of a squar e bras s box spli t horizontall y at the leve l of the cente r of the soi l
sample, whic h is hel d betwee n meta l grille s and porou s stone s as show n in Fig. 8.3(a). Vertica l loa d
is applie d to the sampl e as show n in the figur e and is hel d constan t durin g a test. A graduall y
increasin g horizonta l loa d is applie d to the lowe r par t of the box unti l the sampl e fail s in shear. The
shea r loa d at failur e is divide d by the cross-sectiona l are a of the sampl e to giv e the ultimat e shearin g
strength. The vertica l loa d divide d by the are a of the sampl e give s the applie d vertica l stres s <7. Th e
tes t ma y be repeate d wit h a few mor e sample s havin g the same initia l condition s as the firs t sample.
Eac h sampl e i s teste d wit h a differen t vertica l load.
— Norma l loa d
Porou s ston e
Provin g rin g
^^^^^^^^
<x><xxx><xxxp> ^
Shearin g
forc e
Roller s
Figur e 8.3(a ) Constan t rat e of strai n shea r box
Shear Strength of Soil 25 7
Figur e 8.3(b) Strai n controlle d direct shea r apparatu s (Courtesy: Soiltest )
The horizonta l load is applied at a constant rate of strain. The lower half of the box is
mounte d on rollers and is pushed forwar d at a unifor m rate by a motorize d gearing arrangement.
The upper half of the box bears agains t a steel proving ring, the deformatio n of which is shown on
the dial gauge indicatin g the shearing force. To measur e the volume change during consolidatio n
and during the shearing process another dial gauge is mounte d to show the vertical movemen t of
the top platen. The horizonta l displacemen t of the bottom of the box may also be measure d by
anothe r dial gauge which is not shown in the figure. Figur e 8.3(b) shows a photograp h of strai n
controlle d direct shear test apparatus.
Procedur e for Determining Shearing Strength of Soil
In the direct shear test, a sampl e of soil is placed into the shear box. The size of the box normall y
used for clays and sands is 6 x 6 cm and the sampl e is 2 cm thick. A large box of size 30 x 30 cm
wit h sampl e thicknes s of 15 cm is sometime s used for gravell y soils.
The soils used for the test are either undisturbe d samples or remolded. If undisturbed, the
specimen has to be carefull y trimmed and fitte d into the box. If remolded samples are required, the soil
is placed into the box in layers at the required initia l water content and tamped to the required dry density.
Afte r the specimen is placed in the box, and all the other necessar y adjustment s are made, a
known normal load is applied. Then a shearing force is applied. The normal load is held constant
258 Chapte r 8
throughou t the tes t bu t the shearin g forc e i s applie d at a constan t rat e of strai n (whic h wil l be
explaine d late r on). The shearin g displacemen t is recorde d by a dia l gauge.
Dividin g the norma l loa d and the maximu m applie d shearin g forc e by the cross-sectiona l are a of
the specime n at the shea r plan e give s respectivel y the uni t norma l pressur e cran d the shearin g strengt h
s at failur e of the sample. These result s may be plotte d on a shearin g diagra m wher e cri s the absciss a
and s the ordinate. The resul t of a singl e tes t establishe s one poin t on the grap h representin g the
Coulom b formul a for shearin g strength. In orde r to obtai n sufficien t point s to draw the Coulom b graph,
additiona l test s mus t be performe d on othe r specimen s whic h are exac t duplicate s of the first. The
procedur e i n thes e additiona l test s is the same as in the first, excep t tha t a differen t norma l stres s is
applie d eac h time. Normally, the plotte d point s of norma l and shearin g stresse s at failur e of the variou s
specimen s wil l approximat e a straigh t line. Bu t in the case of saturated, highl y cohesiv e cla y soil s in the
undraine d test, the grap h of the relationshi p betwee n the norma l stres s and shearin g strengt h is usuall y
a curve d line, especiall y at low value s of norma l stress. However, it is the usua l practic e to dra w the bes t
straigh t line throug h the tes t point s to establis h the Coulom b Law. The slop e of the line give s the angl e
of shearin g resistanc e and the intercep t on the ordinat e give s the apparen t cohesio n (See. Fig. 8.2).
Triaxial Compression Test
A diagrammati c layou t of a triaxia l tes t apparatu s is show n in Fig. 8.4(a). In the triaxia l compressio n
test, thre e or mor e identica l sample s of soi l ar e subjecte d to uniforml y distribute d flui d pressur e
aroun d the cylindrica l surface. The sampl e is seale d in a watertigh t rubbe r membrane. The n axia l
load is applie d to the soi l sampl e unti l it fails. Althoug h onl y compressiv e loa d is applie d to the soi l
sample, i t fail s by shea r on interna l faces. It is possibl e to determin e the shea r strengt h of the soi l
fro m the applie d load s at failure. Figur e 8.4(b ) give s a photograp h of a triaxia l tes t apparatus.
Direc t shea r test s ar e generall y suitabl e for cohesionles s soil s excep t fin e san d and sil t wherea s the
triaxia l tes t is suitabl e for al l type s of soil s and tests. Undraine d and consolidate d undraine d test s on
cla y sample s can be mad e wit h the box-shea r apparatus. The advantage s of the triaxia l ove r the
direc t shea r tes t are:
1. Th e stres s distributio n acros s the soi l sampl e is mor e unifor m in a triaxia l tes t as compare d
to a direc t shea r test.
2. Th e measuremen t of volum e change s is mor e accurat e in the triaxia l test.
3. Th e complet e stat e of stres s is know n at all stage s durin g the triaxia l test, wherea s onl y the
stresses at failur e ar e know n in the direc t shea r test.
4. I n the cas e of triaxia l shear, the sampl e fail s alon g a plan e on whic h the combinatio n of
norma l stres s and the shea r stres s give s the maximu m angl e of obliquit y of the resultan t
wit h the normal, wherea s in the cas e of direc t shear, the sampl e is sheare d onl y on one
plan e whic h is the horizonta l plan e whic h nee d no t be the plan e of actua l failure.
5. Por e wate r pressure s can be measure d in the cas e of triaxia l shea r test s wherea s it is not
possibl e in direc t shea r tests.
6. Th e triaxia l machin e is mor e adaptable.
1. Th e direc t shea r machin e is simpl e and fas t to operate.
2. A thinne r soi l sampl e is use d in the direc t shea r tes t thu s facilitatin g drainag e of the pore
wate r quickl y fro m a saturate d specimen.
3. Direc t shea r requiremen t is muc h les s expensiv e as compare d to triaxia l equipment.
Shear Strength of Soi l 259
Proving ring
Ram
Cell
Rubbe r membran e
Sample
(a ) Diagrammati c layout
Inlet Outlet
(b ) Multiplex 50- E load
frame triaxia l test
apparatu s (Courtesy:
Soiltest USA)
Figur e 8.4 Triaxia l test
apparatu s
260
Chapte r 8
Origina l sampl e
Failur e wit h
unifor m strain s
(a) Direc t shea r test
Actua l failur e
conditio n
_ Stresse d
zone
Zone wit h
large strain s
(b) Triaxia l shear test
Figur e 8.5 Condition of sampl e during shearin g in direct and triaxia l shear tests
The stres s condition s acros s the soil sampl e in the direc t shear test are very comple x becaus e
of the chang e in the shea r area wit h the increas e in shear displacemen t as the test progresses,
causin g unequa l distributio n of shea r stresse s and norma l stresse s over the potentia l surfac e of
sliding. Fig. 8.5(a) shows the sampl e conditio n befor e and afte r shearin g in a direc t shear box. The
fina l sheare d area A,is less than the origina l area A.
Fig. 8.5(b) shows the stresse d conditio n in a triaxia l specimen. Becaus e of the end restraints, dead
zones (non-stresse d zones ) triangula r in section are forme d at the ends wherea s the stress distributio n
across the sampl e midwa y betwee n the dead zones may be taken as approximatel y uniform.
8.6 STRES S CONDITIO N AT A POINT IN A SOI L MASS
Throug h ever y point in a stresse d body ther e are three plane s at right angles to each other whic h are
uniqu e as compare d to all the other plane s passin g throug h the point, becaus e they are subjecte d onl y to
norma l stresse s wit h no accompanyin g shearin g stresse s acting on the planes. These three plane s are
called principal planes, and the norma l stresse s acting on these planes are principal stresses. Ordinaril y
the three principa l stresse s at a point diffe r in magnitude. They may be designate d as the majo r principa l
stress <TJ, the intermediat e principa l stress o~ 2, an d the mino r principa l stress <Jy Principa l stresse s at a
point in a stresse d body are importan t because, once they are evaluated, the stresse s on any other plane
throug h the point can be determined. Many problems in foundatio n engineerin g can be approximate d by
considerin g onl y two-dimensiona l stress conditions. The influenc e of the intermediat e principa l stress
(J 2 on failur e ma y be considere d as not very significant.
A Two-Dimensiona l Demonstratio n of the Existenc e of Principa l Plane s
Conside r the body (Fig. 8.6(a) ) is subjecte d to a syste m of force s such as Fr F2 F3 and F4 whos e
magnitude s and lines of action are known.
Shear Strength of Soil
261
D
dx
(c)
Figur e 8.6 Stres s at a point in a body in two dimensiona l space
Consider a small prismati c element P. The stresses acting on this element in the direction s
parallel to the arbitraril y chosen axes x and y are shown in Fig. 8.6(b).
Consider a plane AA through the element, making an angle a with the jc-axis. The equilibriu m
condition of the element may be analyze d by considerin g the stresses acting on the faces of the
triangl e ECD (shaded) which is shown to an enlarged scale in Fig. 8.6(c). The normal and shearing
stresses on the faces of the triangl e are also shown.
The uni t stress in compressio n and in shear on the face ED are designated as cran d T respectively.
Expression s for c r and T may be obtained by applying the principle s of statics for the
equilibriu m condition of the body. The sum of all the forces in the jc-directio n is
<J x dx tan a + T dx+ rdx sec a cos a - crdx sec a sin a = 0
The sum of all the forces in the y-directio n is
cr dx + T X dx tan a - T dx sec a sin a - crdx sec a cos a = 0
Solving Eqs. (8.6) and (8.7) for cran d T, we have
(8.6)
(8.7)
a V +G X a -GJ
— H — cos2a + T ™ sm2a
o i •*?
T = —| CT V - crr ) sin2a-irv cos2a
fj \ y • * / -v
(8.8)
(8.9)
By definition, a principa l plane is one on which the shearing stress is equal to zero. Therefore,
whe n i is made equal to zero in Eq. (8.9), the orientatio n of the principa l planes is defined by the
relationshi p
tan2a =
2i,
(8.10)
262
Chapte r 8
Equatio n (8.10) indicate s that ther e are two principa l plane s throug h the poin t P in Fig. 8.6(a)
and that they are at right angle s to each other. By differentiatin g Eq. (8.8) wit h respec t to a, and
equatin g to zero, we have
— = - a.. sin 2a + a r sin 2a + 2t _. cos 2a = 0
da y y
or
tan 2a =
a -GX
(8.11 )
Equatio n (8.11) indicate s the orientatio n of the plane s on whic h the norma l stresse s e r are
maximu m and minimum. Thi s orientatio n coincide s wit h Eq. (8.10). Therefore, it follow s that the
principa l plane s are also plane s on whic h the norma l stresse s are maximu m and minimum.
8.7 STRES S CONDITION S IN SOI L DURIN G TRIAXIA L
COMPRESSIO N TEST
In triaxia l compressio n tes t a cylindrica l specime n is subjecte d to a constan t all-roun d flui d
pressur e whic h is the mino r principa l stres s O"3 sinc e the shea r stress on the surfac e is zero. The two
ends are subjecte d to axia l stres s whic h is the majo r principa l stres s or The stres s conditio n in the
specime n goes on changin g wit h the increas e of the majo r principa l stres s crr It is of interes t to
analyz e the stat e of stres s alon g incline d section s passin g throug h the sampl e at any stres s level (J l
since failur e occur s alon g incline d surfaces.
Conside r the cylindrica l specime n of soi l in Fig. 8.7(a) whic h is subjecte d to principa l
stresse s <7 { an d <7 3 (<7 2 = <T 3 ).
Now CD, a horizonta l plane, is calle d a principa l plane since it is norma l to the principa l stres s
<T J and the shea r stres s is zer o on thi s plane. EF is the othe r principa l plane on whic h the principa l
stres s <7 3 acts. AA is the incline d sectio n on whic h the stat e of stres s is require d to be analyzed.
Conside r as befor e a smal l pris m of soi l shown shade d in Fig. 8.7(a) and the same to an
enlarge d scal e in Fig. 8.7(b). All the stresse s actin g on the pris m are shown. The equilibriu m of the
pris m require s
Horizonta l force s = cr 3 sin a dl - a sin a dl + T cos adl =
(8.12)
A/
- D
E
(a) (b)
Figur e 8.7 Stres s condition in a triaxia l compressio n test specime n
Shea r Strength of Soi l 26 3
£ Vertica l force s = o{ cos a dl - a cos a dl - i sin a dl - 0 (8.13 )
Solvin g Eqs. (8.12 ) and (8.13 ) we have
<7, + <7, <7, — (7-,
cr = — - + — -cos2« (8.14 )
2 2
1
r = -(cr1-<J3)sin2« (8.15 )
Let the resultan t of <ran d Tmak e an angl e 8 wit h the norma l to the incline d plane. One shoul d
remembe r tha t whe n ens less tha n 90°, the shea r stres s Ti s positive, and the angl e S is also positive.
Eqs. (8.14 ) and (8.15 ) may be obtaine d directl y fro m the genera l Eqs. (8.8) and (8.9)
respectivel y by substitutin g the following:
cr =<7.,< T =(T,and T = 0
8.8 RELATIONSHI P BETWEE N THE PRINCIPA L STRESSE S AND
COHESIO N c
If the shearin g resistanc e s of a soi l depend s on bot h frictio n and cohesion, slidin g failur e occur s in
accordanc e wit h the Coulom b Eq. (8.3), that is, whe n
T = s = c+cr t an 0 (8.16 )
Substitutin g for the value s of eran d rfrom Eqs. (8.14 ) and (8.15 ) int o Eqs. (8.16) and solvin g
fo r <7 j w e obtai n
c + <7 3 ta n </>
= <r, + ~ 5 (8.17 )
j -"'v-'^'v-cos^ tftan^
The plan e wit h the leas t resistanc e to shearin g alon g it wil l correspon d to the minimu m valu e
of <7 j whic h can produc e failur e in accordanc e wit h Eq. (8.17). ol wil l be at a minimu m whe n the
denominato r in the secon d membe r of the equatio n is at a maximum, tha t is, whe n
d
— — (si n a cos a - cosz a tan <z> ) = 0
da
Differentiating, and simplifying, we obtai n (writin g a - ac)
«, = 45° + 0/2 (8.18 )
Substitutin g for a in Eq. (8.17 ) and simplifying, we have
CT j = CT 3 tan2 (45° + 0/2) + 2c ta n (45° + 0/2) (8.19 )
or (Tl=v3N0 + 2cN (8.20 )
wher e A^ = tan2 (45° + 0/2) is calle d the flow value.
If the cohesio n c = 0, we have
°i = °IN* (8.21 )
If 0 = 0, we have
< T = < T + 2c (8.22 )
264 Chapte r 8
If the sides of the cylindrical specimen are not acted on by the horizontal pressure <7 3, the load
required to cause failur e is called the unconfme d compressive strength qu. It is obvious that an
unconfme d compression test can be performed onl y on a cohesive soil. According to Eq. (8.20), the
unconfme d compressive strength q is equal to
< T = a — 2r N f 8 71\
ui y « -\] </> (o.Zj)
If 0 = 0, then qu = 2c (8.24a )
or the shear strengt h
s = c = — (8.24b )
Eq. (8.24b) shows one of the simplest ways of determining the shear strengt h of cohesive
soils.
8.9 MOHR CIRCL E OF STRES S
Squaring Eqs. (8.8) and (8.9) and adding, we have
i2 / _^ x2
+ ^ = I " 2 j + *ly (8.25 )
Now, Eq. (8.25) is the equation of a circle whose center has coordinates
and whose radius is — i/(c 7 - cr ) -
2 v v y '
The coordinates of point s on the circle represent the normal and shearing stresses on inclined
planes at a given point. The circle is called the Mohr circle of stress, after Mohr (1 900), who firs t
recognized this usefu l relationship. Mohr's method provides a convenient graphical method for
determinin g
I . The normal and shearing stress on any plane through a point in a stressed body.
2. The orientation of the principal planes if the normal and shear stresses on the surface of the
prismati c element (Fig. 8.6) are known. The relationships are valid regardless of the
mechanical properties of the material s since only the considerations of equilibrium are
involved.
If the surfaces of the element are themselves principal planes, the equation for the Mohr
circle of stress may be writte n as
T + oy -- - = - y -- (8.26)
The center of the circle has coordinates T- 0, and o= (a{ + (T3)/2, and its radius is (<J l - (T3)/2.
Again from Mohr's diagram, the normal and shearing stresses on any plane passing through a point
in a stressed body (Fig. 8.7) may be determined if the principal stresses cr l and (J3 are known. Since
<7 j and O"3 are always known in a cylindrical compression test, Mohr's diagram is a very usefu l tool
to analyze stresses on failur e planes.
Shea r Strength of Soi l
265
8.10 MOHR CIRCL E OF STRES S WHEN A PRISMATIC ELEMENT
IS SUBJECTED TO NORMAL AND SHEA R STRESSE S
Conside r firs t the case of a prismati c elemen t subjecte d to norma l and shea r stresse s as in Fig. 8.8(a).
Sign Convention
1. Compressiv e stresse s are positiv e and tensil e stresse s are negative.
2. Shea r stresse s ar e considere d as positiv e if the y give a clockwis e momen t abou t a poin t
above the stresse d plan e as shown in Fig. 8.8(b), otherwis e negative.
The norma l stresse s are take n as absciss a and the shea r stresse s as ordinates. It is
assume d the norma l stresse s c r , c r an d the shea r stres s r ( T = T ) actin g on the surfac e of
x y xy xy yx
the elemen t ar e known. Two point s Pl and P2 may now be plotte d in Fig. 8.8(b), whos e
coordinate s are
If the point s P} and P2 are joined, the line intersect s the absciss a at poin t C whos e coordinate s
are [(0,+op/2,0].
Minor principal
> ai
plane
(a) A prismatic element subjected to normal and shear stresses
(ax + ay)/2
+ ve
(b) Mohr circle of stres s
Figure 8.8 Mohr stress circle for a general case
266 Chapte r 8
Point O is the origi n of coordinate s for the cente r of the Moh r circl e of stress. Wit h cente r C
a circl e may now be constructe d wit h radiu s
Thi s circl e whic h passe s throug h point s Pl and P2 is calle d the Mohr circle of stress. The
Mohr circl e intersect s the absciss a at two point s E and F . The majo r and mino r principa l stresse s
are ol (= OF) and cr 3 (= OE) respectively.
Determination of Normal and Shear Stresse s on Plane AA [Fig. 8.8(a)]
Point P{ on the circl e of stres s in Fig. 8. 8(b) represent s the stat e of stres s on the vertica l plan e of the
prismati c element; similarl y poin t P2 represent s the stat e of stres s on the horizonta l plan e of the
element. If fro m poin t P{ a line is drawn paralle l to the vertica l plane, it intersect s the circl e at poin t P Q
and if fro m the poin t P2 on the circle, a line is drawn paralle l to the horizonta l plane, thi s line also
intersect s the circl e at poin t P Q . The poin t PQ so obtaine d is calle d the origin o f planes or the pole. If
fro m the pol e PQ a line is drawn paralle l to the plane AA in Fig. 8.8(a) to intersec t the circl e at poin t P3
(Fig. 8.8(b) ) then the coordinate s of the poin t give the norma l stres s cran d the shea r stres s Ton plane
AA as expresse d by equation s 8.8 and 8.9 respectively. Thi s indicate s that a line drawn fro m the pol e P Q
at any angl e a to the cr-axi s intersect s the circl e at coordinate s that represen t the norma l and shea r
stresse s on the plan e incline d at the same angl e to the abscissa.
Major and Mino r Principa l Plane s
The orientation s of the principa l plane s may be obtaine d by joinin g poin t PQ to the point s E and F
in Fi g 8.8(b). PQ F is the directio n of the majo r principa l plan e on whic h the majo r principa l stres s
dj acts; similarl y PQ E is the directio n of the mino r principa l plan e on whic h the mino r principa l
stres s <7 3 acts. It is clea r from the Moh r diagra m that the two plane s PQ E and PQ F intersec t at a righ t
angle, i.e., angl e EP Q F = 90°.
8.1 1 MOHR CIRCL E OF STRES S FOR A CYLINDRICAL SPECIMEN
COMPRESSION TEST
Conside r the case of a cylindrica l specime n of soi l subjecte d to norma l stresses <7 j and <J 3 whic h ar e
the majo r and mino r principa l stresse s respectivel y (Fig. 8.9)
From Eqs. (8.14 ) and (8.15), we may writ e
2 2
Agai n Eq. (8.27 ) is the equatio n of a circl e whos e cente r has coordinate s
<7, + CT, (7, — (J-.
<J = — - - - and T = 0 and whos e radiu s is
/O /-*^T\
(8.27 )
2 2
A circl e wit h radiu s (o{ - cr 3)/2 wit h its cente r C on the absciss a at a distanc e of (al + cr 3)/2
may be constructe d as shown in Fig. 8.9. Thi s is the Moh r circl e of stress. The majo r and mino r
principa l stresses ar e shown in the figur e wherei n cr, = OF and <7 3 = OE.
From Fig. 8.8, we ca n writ e equation s for cf j an d <7 3 and Tma x as follow s
±
Shea r Strengt h of Soi l
267
.A
Figur e 8.9 Mohr stress circl e for a cylindrical specimen
(8.29)
where Tma x is the maximum shear stress equal to the radius of the Mohr circle.
The origin of planes or the pole PQ (Fig. 8.9) may be obtained as before by drawing lines fro m
point s E and F parallel to planes on which the minor and majo r principal stresses act. In this case,
the pole PO lies on the abscissa and coincides with the point E.
The normal stress < J and shear stress T on any arbitrary plane AA making an angle a wit h the
majo r principal plane may be determined as follows.
From the pole P0 draw a line PQ Pl parallel to the plane AA (Fig. 8.9). The coordinates of the
point Pl give the stresses cr and i. From the stress circle we may write
= 2a
cr, + cr,
cr, - cr.
-
(8.30)
Normal stress a
0° 15° 30° 45° 60° 75° 90°
Angle of inclination of plane, a ^
Figure 8.10 Variation of crand r with a
268 Chapte r 8
(j, -cr,
r= 3 sin2# (8.31 )
Equations (8.30) and (8.31) are the same as Eqs. (8.14) and (8.15) respectively.
It is of interest to stud y the variation of the magnitudes of normal and shear stresses with the
inclination of the plane.
Eqs. (8.30) and (8.31 ) are plotted wit h a as the abscissa shown in Fig. 8.10. The following
fact s are clear from these curves:
1. The greatest and least principal stresses are respectivel y the maximum and minimum
normal stresses on any plane through the point in question.
2. The maximum shear stress occurs on planes at 45° to the principal planes.
8.12 MOHR-COULOMB FAILUR E THEOR Y
Various theories relating to the stress condition in engineering materials at the time of failure are
available in the engineering literature. Each of these theories may explain satisfactoril y the
actions of certain kinds of material s at the time they fail, but no one of them is applicabl e to all
materials. The failure of a soil mass is more nearly in accordance with the tenets of the Mohr
theory of failur e than wit h those of any other theory and the interpretation of the triaxial
compression test depends to a large extent on this fact. The Mohr theory is based on the postulat e
that a material will fai l when the shearing stress on the plane along which the failure is presumed
to occur is a unique functio n of the normal stress acting on that plane. The material fail s along the
plane only when the angle between the resultant of the shearing stress and the normal stress is a
maximum, that is, where the combination of normal and shearing stresses produces the
maximu m obliquit y angle 8.
According to Coulomb's Law, the condition of failur e is that the shear stress
T^c + atan^ (8.32 )
In Fig 8.1 l(b) MQN and MQNl are the lines that satisf y Coulomb's condition of failure. If the
stress at a given point withi n a cylindrical specimen under triaxial compression is represented by
Mohr circle 1, it may be noted that every plane through this point has a shearing stress which is
smaller than the shearing strength.
For example, if the plane AA in Fig. 8.1 l(a) is the assumed failur e plane, the normal and shear
stresses on this plane at any intermediat e stage of loading are represented by point b on Mohr circle
1 where the line PQb is parallel to the plane AA. The shearing stress on this plane is ab which is less
than the shearing strength ac at the same normal stress Oa. Under this stress condition there is no
possibilit y of failure. On the other hand it would not be possible to apply the stress condition
represented by Mohr stress circle 2 to this sample because it is not possible for shearing stresses to
be greater than the shearing strength. At the normal stress Of, the shearing stress on plane AA is
shown to be fh which is greater than the shear strength of the material s fg which is not possible.
Mohr circle 3 in the figure is tangent to the shear strength line MQN and MQNj at points e and e{
respectively. On the same plane AA at normal stress Od, the shearing stress de is the same as the
shearing strength de. Failure is therefore imminent on plane AA at the normal stress Od and
shearing stress de. The equation for the shearing stress de is
s = de - de'+ e'e = c + crta n 0 (8.33 )
where 0 is the slope of the line MQN which is the maximum angle of obliquit y on the failure plane.
The value of the obliquit y angle can never exceed <5 m = 0, the angle of shearing resistance, without
the occurrence of failure. The shear strength line MQN which is tangent to Mohr circle 3 is called the
Shear Strength of Soil
269
'i /
Ruptur e
plane Mohr
envelope N
Mohr circle of
ruptur e
(b)
Figur e 8.11 Diagra m presenting Mohr's theory of rupture
Mohr envelope or line of rupture. The Mohr envelope may be assumed as a straight line although it
is curved under certai n conditions. The Mohr circle which is tangentia l to the shear strengt h line is
called the Mohr circle of rupture. Thus the Mohr envelope constitute s a shear diagram and is a
graph of the Coulomb equation for shearing stress. This is called the Mohr-Coulomb Failure
Theory. The principa l objective of a triaxial compressio n test is to establis h the Mohr envelope for
the soil being tested. The cohesion and the angle of shearing resistanc e can be determine d fro m this
envelope. When the cohesion of the soil is zero, that is, when the soil is cohesionless, the Mohr
envelope passes through the origin.
8.13 MOHR DIAGRAM FOR TRIAXIAL COMPRESSIO N TEST AT
FAILUR E
Consider a cylindrica l specime n of soil possessin g both cohesion and frictio n is subjecte d to a
conventiona l triaxial compressio n test. In the conventiona l test the lateral pressur e cr 3 is held
constant and the vertical pressur e <T J is increase d at a constant rate of stress or strai n unti l the
sampl e fails. If cr l is the peak value of the vertical pressur e at which the sampl e fails, the two
principa l stresses that are to be used for plotting the Mohr circle of ruptur e are cr 3 and or In
Fig. 8.12 the values of cr { and <7 3 are plotted on the er-axi s and a circle is drawn wit h (o^ - cr 3) as
diameter. The center of the circl e lies at a distance of (<j { + cr3)/2 fro m the origin. As per Eq. (8.18),
the soil fail s along a plane which makes an angle a, = 45° + 0/2 with the major principa l plane. In
Fig. 8.12 the two lines PQPl and PQP2 (wher e P Q is the origi n of planes) are the conjugat e ruptur e
planes. The two lines MQN and M QN^ drawn tangentia l to the ruptur e circle at point s Pl and P2 are
called Mohr envelopes. If the Mohr envelope can be drawn by some other means, the orientatio n of
the failur e planes may be determined.
The result s of analysi s of triaxial compressio n tests as explaine d in Sect. 8.8 are now
presented in a graphica l form in Fig. 8.12. The various informatio n that can be obtained fro m the
figur e include s
1. The angle of shearing resistanc e 0 = the slope of the Mohr envelope.
270
Chapte r 8
A a
Moh r envelop e
(a, - a3)/2
Figur e 8.1 2 Mohr diagra m for triaxia l test at failur e for c-0 soi l
Ruptur e
plan e
T
c
I
Ruptur e
plan e
0 = 0
0
(a) c = 0
C
(b) 0 = 0
Figur e 8.1 3 Mohr diagra m for soil s with c = 0 and
= 0
2. Th e apparen t cohesio n c = the intercep t of the Moh r envelop e on the T-axis.
3. Th e inclinatio n of the ruptur e plan e = a.
4. Th e angl e betwee n the conjugat e plane s = 2a.
If the soi l is cohesionles s wit h c = 0 the Moh r envelope s pas s throug h the origin, and if the
soi l is purel y cohesiv e wit h 0 = 0 the Moh r envelop e is paralle l to the abscissa. The Moh r envelope s
for thes e two type s of soil s ar e show n in Fig. 8.13.
8.1 4 MOHR DIAGRAM FOR A DIREC T SHEA R TES T AT FAILUR E
In a direc t shea r tes t the sampl e is sheare d alon g a horizonta l plane. Thi s indicate s tha t the failur e
plan e is horizontal. The norma l stres s don thi s plan e is the externa l vertica l loa d divide d by the are a
of the sample. Th e shea r stres s at failur e i s the externa l latera l loa d divide d by the are a of the
sample.
Poin t Pj on the stres s diagra m in Fig. 8.14 represent s the stres s conditio n on the failur e plane.
The coordinate s of the poin t are
norma l stres s = <7, shea r stres s i- s.
Shear Strength of Soi l
271
Minor
Plane of ruptur e
0 t Majo r principa l
plan e
Figur e 8.14 Mohr diagra m for a direct shear test at failur e
If it is assume d that the Mohr envelop e is a straigh t line passin g throug h the origi n
(for cohesionles s soi l or normall y consolidate d clays), it follows tha t the maximu m
obliquit y 8m occur s on the failur e plane and 8m = 0. Therefor e the line OP{ mus t be tangen t
to the Moh r circle, and the circl e may be constructe d as follows:
Draw PjC norma l to OPr Point C which is the intersectio n point of the norma l wit h the
absciss a is the center of the circle. CP { is the radius of the circle. The Mohr circl e may now be
constructe d which gives the major and minor principa l stresse s cr { and <7 3 respectively.
Since the failur e is on the horizonta l plane, the origi n of plane s PQ may be obtaine d by
drawin g a horizonta l line throug h P{ giving PQ. PQF and PQE give the direction s of the major and
minor principa l planes respectively.
Example 8.1
Wha t is the shearin g strengt h of soil along a horizonta l plane at a dept h of 4 m in a deposi t of sand
havin g the followin g properties:
Angl e of interna l friction, 0 = 35°
Dry uni t weight, y d - 17 kN/m3
Specifi c gravity, Gs = 2.7.
Assume the ground water tabl e is at a dept h of 2.5 m fro m the ground surface. Also fin d the
change in shear strengt h when the water tabl e rises to the ground surface.
Solution
The effectiv e vertica l stress at the plane of interes t is
<r'=2.50xy d + l.SO x y b
Given y d = 17 kN/m3 and Gs = 2.7
We haver, = 17-
= — X9.8 1
9A9
or lie = 26.5 - 17 = 9.49 or e = —— = 0.56
Therefore, Yb =
l + e
1 + 0.56
*9.81 = 10.7 kN/m3
272
Chapter 8
Hence c/ = 2.5 x 17 + 1.5 x 10.7 = 58.55 kN/m2
Hence, the shearin g strengt h of the sand is
5 = (/ tan 0 = 58.55 x tan 35° = 41 kN/m2
If the wate r tabl e rises to the groun d surfac e i.e., by a heigh t of 2.5 m, the chang e in the
effectiv e stres s wil l be,
Ao" = y d x 2.5 -Yb* 2.5 = 17 x 2.5 - 10.7 x 2.5 = 15.75 kN/m2 (negative )
Hence the decrease in shear strengt h wil l be,
= Ac/ tan 35° = 15.75 x 0.70 = 11 kN/m2
Exampl e 8.2
Direct shear test s wer e conducte d on a dry sand. The size of the sample s used for the test s was
2 in. x 2 in. x 0.75 in. The test result s obtaine d are given below:
Test No. Norma l load Norma l stres s a Shea r forc e Shea r stres s
(Ib) (Ib/ft 2) a t failur e (Ib) (Ib/ft 2)
1 1 5 54 0 1 2 43 2
2 2 0 72 0 1 8 64 8
3 3 0 108 0 2 3 82 8
4 6 0 216 0 4 7 169 2
5 12 0 432 0 9 3 334 8
Determin e the shea r
4000-
3000-
c/f
C/3
£ 2000 -
C/3
j3
1000-
strengt h parameter s c and 0.
^L
S^ A ^"7 8°
/
/
y
/
/
1000
2000
3000
4000
Normal stress, a Ib/ft 2
Figure Ex. 8.2
5000
Shear Strength of Soil
Solution
273
The failur e shea r stresse s r^ as obtaine d fro m the test s are plotte d agains t the norma l stresse s a, in
Figur e Ex 8.2. The shea r parameter s fro m the grap h are: c = 0, 0 = 37.8°.
Exampl e 8.3
A direc t shea r test, whe n conducte d on a remolde d sampl e of sand, gave the followin g observation s
at the time of failure: Norma l load = 288 N; shea r load = 173 N. The cros s sectiona l area of the
sampl e = 36 cm2.
Determine: (i ) the angl e of interna l friction, (ii) the magnitud e and directio n of the principa l
stresse s in the zone of failure.
Solution
Such problem s can be solve d in two ways, namel y graphicall y and analytically. The analytica l
solutio n has been lef t as an exercis e for the students.
Graphica l Solution
173
(i ) Shea r stres s T = = 4.8 N/cm2 = 48 kN/m 2
36
288
Norma l stres s a = — = 8.0 N / cm2 = 80 kN / m2
36
We kno w one poin t on the Moh r envelope. Plot poin t A (Fig. Ex. 8.3) wit h coordinate s 1-
48 kN/m2, and o= 80 kN/m2. Sinc e cohesio n c = 0 for sand, the Moh r envelop e OM passe s
throug h the origin. Th e slop e of OM give s the angl e of interna l frictio n (j) =31°.
(ii ) I n Fig. Ex. 8.3, draw line AC norma l to the envelop e OM cuttin g the absciss a at poin t C.
Wit h C as center, and AC as radius, draw Moh r circl e Cl whic h cut s the absciss a at point s B
and D, whic h give s
120
80
40
Mohr circle C\
Majo r principal plane
C2
40 F 80 C 120
a, kN/m2
160
200
Figure Ex. 8.3
274 Chapte r 8
majo r principa l stres s = OB = (J l = 163.5 kN/m 2
mino r principa l stres s = OD = <J 3 = 53.5 kN/m 2
Now, ZACB = 2cc = twic e the angl e betwee n the failur e plan e and the majo r principa l
plane. Measuremen t give s
2a= 121° or a- 60.5°
Since in a direc t shea r tes t the failur e plan e is horizontal, the angl e mad e by the majo r
principa l plan e wit h the horizonta l wil l be 60.5°. The mino r principa l plan e shoul d be
drawn at a righ t angl e to the majo r principa l plane.
The direction s of the principa l plane s may also be foun d by locatin g the pol e Po. Po is
obtaine d by drawin g a horizonta l line fro m poin t A whic h is paralle l to the failur e plan e in
the direc t shea r test. Now PE and P(D give the direction s of the majo r and mino r principa l
plane s respectively.
8.15 EFFECTIV E STRESSE S
So far, the discussio n has bee n based on consideratio n of tota l stresses. It is to be note d tha t the
strengt h and deformatio n characteristic s of a soi l can be understoo d bette r by visualizin g it as a
compressibl e skeleto n of soli d particle s enclosin g voids. The void s may completel y be fille d wit h
wate r or partl y wit h wate r and air. Shea r stresses are to be carrie d onl y by the skeleto n of soli d
particles. However, the tota l norma l stresse s on any plan e are, in general, the sum of two
components.
Total norma l stres s = componen t of stres s carrie d by soli d particle s
+ pressur e in the flui d in the voi d space.
Thi s visualizatio n of the distributio n of stresse s betwee n soli d and flui d has two importan t
consequences:
1. Whe n a specime n of soi l is subjecte d to externa l pressure, the volume chang e of the specime n
is not due to the tota l norma l stres s but due to the differenc e betwee n the total norma l stres s
and the pressur e of the flui d in the voi d space. The pressur e in the flui d is the por e pressur e u.
The differenc e whic h is calle d the effectiv e stres s d may now be expresse d as
tf = cr-u (8.34 )
2. The shea r strengt h of soils, as of al l granula r materials, is largel y determine d by the
frictiona l force s arisin g durin g sli p at the contact s betwee n the soi l particles. These ar e
clearl y a functio n of the componen t of norma l stres s carrie d by the soli d skeleto n rathe r
tha n of the tota l norma l stress. For practica l purpose s the shea r strengt h equatio n of
Coulom b is give n by the expressio n
s = c' + (o - U) tan </)' = c' + a' tan </)' (8.35 )
wher e c'= apparen t cohesio n in terms of effectiv e stresse s
0' = angl e of shearin g resistanc e in terms of effectiv e stresse s
< 7 = tota l norma l pressur e to the plan e considere d
u = pore pressure.
The effectiv e stres s parameter s c' and 0' of a give n sampl e of soi l may be determine d
provide d the pore pressur e u develope d durin g the shea r tes t is measured. The por e pressur e u is
develope d whe n the testin g of the soi l is don e unde r undraine d conditions. However, if fre e
Shea r Strength of Soil 27 5
drainag e take s plac e durin g testing, ther e wil l not be any developmen t of por e pressure. In suc h
cases, the tota l stresse s themselve s are effectiv e stresses.
8.16 SHEA R STRENGT H EQUATIO N IN TERM S OF EFFECTIV E
PRINCIPA L STRESSE S
The principa l stresse s may be expresse d eithe r as tota l stresse s or as effectiv e stresse s if the value s
of por e pressur e are known.
If u is the por e pressur e develope d durin g a triaxia l test, we may writ e as befor e
o = o, -u
wher e aj and <5'3 ar e the effectiv e principa l stresses. The equatio n for shea r strengt h in terms
of effectiv e stresse s is
<7,' — <7 o G<— (J-, <J, — (T-.
s = — si n 2a = — si n 2a = —; co s 0 (8.37 )
2 2 2
wher e 2a= 90° + 0'
Coulomb's equatio n in terms of effectiv e stresse s is
s = c''+ (<7-u ) tan 0'
(7, — (J~.
Therefore, — cos<z>' = c' + (er-u) tan0'
Since, cr =
2 2
we hav e — co s (/)' = c' H— l- ta n <f)'
+ — cos(9 0 + 0') tan 0' - u tan 0'
Simplifyin g
<7, - cr, , . . . cr, + or, . , O", -1
. — -.
2 22
1 c' cos\$)' + (<7 3 — «) sin^'
- ws n
or
wher e (ci j - cr 3) indicate s the maximu m deviato r stres s at failure. Eq (8.38 ) may also be
expresse d in a differen t for m as follow s by considerin g effectiv e principa l stresse s
1 , , c' cos^' + a. sin^'
— (<j, - cr, ) , = - - -
2 l 3 f 1-sin
or —
276 Chapte r 8
Simplifying, we hav e
(o[ -o'3)f = (o{ + o'3 ) si n (/)' + 2c' cos 0' (8.39 )
8.1 7 STRESS-CONTROLLE D AND STRAIN-CONTROLLE D TESTS
Direc t shea r test s or triaxia l compressio n test s ma y be carrie d ou t by applyin g stresse s or strain s at
a particularl y know n rate. Whe n the stres s is applie d at a constan t rat e it is calle d a stress-controlled
test an d whe n the strai n i s applie d at a constan t rat e i t i s calle d a strain-controlled test. Th e
differenc e betwee n the two type s of test s ma y be explaine d wit h respec t t o bo x shea r tests.
In th e stress-controlle d tes t [Fig. 8.15(a) ] the latera l loa d Fa whic h induce s shea r i s graduall y
increase d unti l complet e failur e occurs. Thi s ca n be don e by placin g weight s on a hange r or by
fillin g a counterweighte d bucke t of origina l weigh t W at a constan t rate. Th e shearin g
displacement s ar e measure d by mean s of a dia l gaug e G as a functio n of the increasin g loa d F . The
shearin g stres s at an y shearin g displacement, is
wher e A is the cros s sectiona l are a of the sample. A typica l shap e of a stress-strai n curv e of the
stress-controlle d tes t is show n in Fig. 8.15(a).
A typica l arrangemen t of a box-shea r tes t apparatu s for the strain-controlle d tes t is show n in
Fig. 8.15(b). Th e shearin g displacement s ar e induce d an d controlle d i n suc h a manne r tha t the y
occu r at a constan t fixe d rate. Thi s ca n be achieve d by turnin g the whee l eithe r by han d or by mean s
of any electricall y operate d moto r so tha t horizonta l motio n is induce d throug h the wor m gear B.
The dia l gaug e G give s the desire d constan t rat e of displacement. The botto m of box C is mounte d
on frictionles s roller s D. The shearin g resistanc e offere d to thi s displacemen t by the soi l sampl e is
measure d by the provin g rin g E. Th e stress-strai n curve s for thi s typ e of tes t hav e the shap e show n
in Fig. 8.15(b).
Bot h stress-controlle d an d strain-controlle d type s of tes t ar e use d in connectio n wit h al l the
direc t triaxia l an d unconfine d soi l shea r tests. Strain-controlle d test s ar e easie r t o perfor m an d
hav e the advantag e of readil y givin g no t onl y the peak resistanc e as i n Fig. 8.1 5 (b ) bu t also the
ultimat e resistanc e whic h is lowe r tha n the pea k suc h as poin t b in the same figure, wherea s the
stres s controlle d give s onl y the pea k value s bu t no t the smalle r value s afte r the peak i s achieved.
The stress-controlle d tes t i s preferre d onl y in some specia l problem s connecte d wit h research.
8.1 8 TYPE S OF LABORATOR Y TESTS
The laborator y test s on soil s ma y be on
1. Undisturbe d samples, or
2. Remolde d samples.
Further, the test s ma y be conducte d on soil s tha t ar e :
1 . Full y saturated, or
2. Partiall y saturated.
The typ e of tes t to be adopte d depend s upo n how bes t we ca n simulat e the fiel d conditions.
Generall y speaking, the variou s shea r test s for soil s ma y be classifie d as follows:
Shea r Strengt h of Soi l
277
Dial
gauge
Displacement
\_ -
(b) Strai n controlle d
Figur e 8.15 Stres s and strai n controlled box shear tests
1. Unconsolidated-Undraine d Test s (UU)
The sample s are subjecte d to an applied pressur e under condition s in which drainag e is prevented,
and then sheared under condition s of no drainage.
2. Consolidated-Undraine d or Quick Test s (CD)
The sample s are allowe d to consolidat e unde r an applie d pressur e and then sheare d unde r
condition s of no drainage.
3. Consolidated-Draine d or Slow Test s (CD)
The sample s are consolidate d as in the previou s test, but the shearin g is carried out slowl y under
condition s of no excess pressur e in the pore space.
The drainag e conditio n of a sampl e is generall y the decidin g factor in choosin g a particula r
type of test in the laboratory. The purpos e of carryin g out a particula r test is to simulat e fiel d
condition s as far as possible. Becaus e of the high permeabilit y of sand, consolidatio n occur s
relativel y rapidl y and is usuall y complete d during the applicatio n of the load. Tests on sand are
therefor e generally carried out under drained conditions (drained or slow test).
For soils other than sands the choice of test condition s depend s upon the purpos e for which
the shear strengt h is required. The guidin g principl e is that drainag e condition s of the test shoul d
confor m as closely as possibl e to the condition s under which the soils wil l be stressed in the field.
Undraine d or quick test s are generall y used for foundation s on clay soils, since during the
period of constructio n onl y a smal l amoun t of consolidatio n wil l have taken place and consequentl y
the moistur e conten t wil l have undergon e littl e change. For clay slopes or cut s undraine d test s are
used bot h for design and for the investigatio n of failures.
Consolidated-undraine d tests are used wher e change s in moistur e content are expecte d to take
place due to consolidatio n befor e the soil is full y loaded. An importan t exampl e is the conditio n known
as "sudde n drawdown" such as that occur s in an earth dam behind which the water level is lowered at
278 Chapte r 8
a faste r rat e tha n at whic h the materia l of the dam can consolidate. In the consolidated-undraine d test s
used in thi s typ e of problem, the consolidatio n pressure s ar e chose n to represen t the initia l condition s
of the soil, and the shearin g load s correspon d to the stresse s calle d int o pla y by the actio n of sudde n
drawdown.
As alread y stated, draine d test s ar e alway s use d in problem s relatin g to sand y soils. In cla y
soil s draine d test s ar e sometime s use d in investigatin g the stabilit y of an eart h dam, an embankmen t
or a retainin g wal l afte r a considerabl e interva l of time ha s passed.
Ver y fin e sand, silt s and silt y sand s als o hav e poor drainag e qualities. Saturate d soil s of thes e
categorie s ar e likel y to fai l i n the fiel d unde r condition s simila r to thos e unde r whic h consolidate d
quic k test s are made.
Shearing Test Apparatu s for the Variou s Types of Tests
The variou s type s of shea r test s mentione d earlie r ma y be carrie d out eithe r by the box shea r tes t or
the triaxia l compressio n tes t apparatus. Test s tha t ma y be mad e by the two type s of apparatu s are:
Box Shear Test Apparatus
1. Undraine d and consolidated - undraine d test s on cla y sample s only.
2. Draine d or Slo w test s on any soil.
The box shea r tes t apparatu s is not suite d for undraine d or consolidated-undraine d test s on
sample s othe r tha n cla y samples, becaus e the othe r soil s ar e so permeabl e tha t eve n a rapi d increas e
of the stresse s in the sampl e ma y caus e at leas t a noticeabl e chang e of the wate r content.
Triaxial Compression Test Apparatus
Al l type s of test s ca n convenientl y be carrie d ou t in thi s apparatus.
8.1 9 SHEARIN G STRENGT H TESTS ON SAND
Shea r test s on san d ma y be mad e whe n the san d is eithe r in a dr y stat e or in a saturate d state. No tes t
shal l be mad e whe n the soi l is in a mois t stat e as thi s stat e exist s onl y due to apparen t cohesio n
betwee n particle s whic h woul d be destroye d whe n it is saturated. Th e result s of shea r test s on
saturate d sample s ar e almos t identica l wit h thos e on the same san d at equa l relativ e densit y in a dr y
stat e excep t tha t the angl e 0 is likel y to be 1 or 2 degree s smalle r for the saturate d sand.
The usua l typ e of tes t use d for coars e t o mediu m san d is the slo w shea r test. However,
consolidate d undraine d test s ma y be conducte d on fin e sands, sand y silt s etc. whic h do no t allo w
fre e drainag e unde r change d stres s conditions. If the equilibriu m of a larg e bod y of saturate d fin e
san d i n an embankmen t i s disturbe d by rapi d drawdow n of th e surfac e of an adjoinin g bod y of
water, the chang e i n wate r conten t of the fil l lag s behin d the chang e i n stress.
In all the shearing tests on sand, onl y the remolde d samples are use d as it is not practicabl e to
obtai n undisturbe d samples. The soi l sample s ar e to be mad e approximatel y to the same dr y densit y
as it exist s in-sit u and teste d eithe r by direc t shea r or triaxia l compressio n tests.
Test s on soil s ar e generall y carrie d out by the strain-controlle d typ e apparatus. The principa l
advantag e of thi s typ e of tes t on dens e san d is tha t it s peak-shea r resistance, as wel l as the shea r
resistance s smalle r tha n the peak, ca n be observe d and plotted.
Direct Shea r Test
Onl y the draine d or the slo w shea r test s on san d ma y be carrie d out by usin g the box shea r tes t
apparatus. The box is fille d wit h san d to the require d density. The sampl e is sheare d at a constan t
Shea r Strength of Soil
279
vertica l pressur e a. The shear stresses are calculate d at various displacement s of the shear box. The
test is repeated with differen t pressure s <7.
If the sampl e consist s of loose sand, the shearing stress increase s wit h increasin g
displacemen t unti l failur e occurs. If the sand is dense, the shear failur e of the sampl e is preceded by
a decreas e of the shearing stress fro m a peak value to an ultimat e value (also known as residua l
value ) lower than the peak value.
Typica l stress-strai n curves for loose and dense sands are shown in Fig. 8.16(a).
The shear stress of a dense sand increase s fro m 0 to a peak value represente d by point a, and
then graduall y decreases and reaches an ultimat e value represente d by point b. The sampl e of sand
in a dense state is closel y packed and the numbe r of contact point s between the particles are more
tha n in the loose state. The soil grains are in an interlocke d state. As the sampl e is subjecte d to shear
stress, the stress has to overcome the resistanc e offere d by the interlocke d arrangemen t of the
particles. Experimenta l evidenc e indicate s that a significan t percent of the peak strengt h is due to
the interlockin g of the grains. In the process of shearing one grai n tries to slide over the other and
the void rati o of the sampl e which is the lowest at the commencemen t of the test reaches the
maximu m value at point a, in the Fig 8.16(a). The shear stress also reaches the maximu m value at
thi s level. Any furthe r increas e of strai n beyond this point is associated wit h a progressiv e
disintegratio n of the structur e of the sand resultin g in a decreas e in the shear stress. Experienc e
shows that the change in void ratio due to shear depends on both the vertical load and the relative
densit y of the sand. At very low vertical pressure, the void ratio at failur e is larger and at very high
pressur e it is smaller than the initial void ratio, whateve r the relative densit y of the sand may be. At
Peak value
Dense sand
b ultimat e value
Displacemen t
(a) Shear stress vs displacemen t
0
Dense sand
Loose sand
(b) Volume change
Normal stress, a
(c) Shear strengt h vs normal stress
Figur e 8.16 Direct shear test on sand
280 Chapte r 8
Table 8.1 Typical values of 0 and (j) u for granular soils
Types of soi l
Sand: rounded grains
Loose
Mediu m
Dense
Sand: angula r grains
Loose
Mediu m
Dense
Sand y gravel
0 deg
28 to 30
30 to 35
35 to 38
30 to 35
35 to 40
40 to 45
34 to 48
0udeg
26 to 30
30 to 35
33 to 36
intermediate values of pressure, the shearing force causes a decrease in the void ratio of loose sand
and an increase in the void ratio of dense sand. Fig 8.16(b) shows how the volume of dense sand
decreases up to a certain value of horizontal displacement and with furthe r displacement the
volume increases, whereas in the case of loose sand the volume continues to decrease with an
increase in the displacement. In saturated sand a decrease of the void ratio is associated with an
expulsion of pore water, and an increase with an absorption of water. The expansion of a soil due to
shear at a constant value of vertical pressure is called dilatancy. At some intermediate state or
degree of density in the process of shear, the shear displacement does not bring about any change in
volume, that is, density. The density of sand at which no change in volume is brought about upon
the application of shear strains is called the critical density. The porosity and void ratio
corresponding to the critical density are called the critical porosity and the critical void ratio
respectively.
By plotting the shear strengths corresponding to the state of failur e in the differen t shear tests
against the normal pressure a straight line is obtained for loose sand and a slightly curved line for dense
sand [Fig. 8.16(c)]. However, for all practical purposes, the curvature for the dense sand can be
disregarded and an average line may be drawn. The slopes of the lines give the corresponding angles of
frictio n 0 of the sand. The general equation for the lines may be written as
s = <J ta n (f)
For a given sand, the angle 0 increases wit h increasing relative density. For loose sand it is
roughly equal to the angle of repose, defined as the angle between the horizontal and the slope of a
heap produced by pouring clean dry sand from a small height. The angle of frictio n varies with the
shape of the grains. Sand samples containing well graded angular grains give higher values of 0 as
compared to uniformly graded sand wit h rounded grains. The angle of frictio n </> for dense sand at
peak shear stress is higher than that at ultimate shear stress. Table 8.1 gives some typical values of
0 (at peak) and 0 M (at ultimate).
Triaxial Compression Test
Reconstructed sand samples at the required density are used for the tests. The procedure of making
samples should be studied separately (refer to any book on Soil Testing). Tests on sand may be
conducted either in a saturated state or in a dry state. Slow or consolidated undrained tests may be
carried out as required.
Drained or Slow Tests
At least three identical samples having the same initial conditions are to be used. For slow tests
under saturated conditions the drainage valve should always be kept open. Each sample should be
Shear Strength of Soil
281
': v •••:..-y<; A
jv:,:^-V^
-• ' -i .* • ' ' - '' • "'
• v ••*».• '-x ' ' '« '"
• •>" "•.., \."' • ••>'
(a) Dense sand (b) Loose sand
Figure 8.17 Typical shapes of dense and loose sands at failur e
Strai n
(a) Stress-strai n curves for three samples at dense state
Mohr
envelope
(b) Mohr envelope
Figur e 8.18 Mohr envelope for dense sand
282 Chapte r 8
tested under differen t constant all-round pressures for example, 1, 2 and 3 kg/cm2. Each sampl e is
sheared to failur e by increasin g the vertical load at a sufficientl y slow rate to prevent any build up
of excess pore pressures.
At any stage of loading the major principal stress is the all-round pressur e <7 3 plus the
intensit y of deviator stress (o{ - cr3). The actuall y applied stresses are the effective stresses in a slow
test, that is <7 } = a\ and O"3 = <r'3, Dense samples fai l along a clearly defined ruptur e plane whereas
loose sand samples fai l along many planes which resul t in a symmetrica l bulging of the sample. The
compressive strengt h of a sampl e is defined as the differenc e between the major and minor
principal stresses at failur e (GI - <T 3 ),,. Typical shapes of dense and loose sand samples at failure are
shown in Fig. 8.17.
Typical stress-strai n curves for three samples in a dense state and the Mohr circles for these
samples at peak strengt h are shown in Fig. 8.18.
If the experimen t is properl y carried out there will be one common tangent to all these three
circles and this will pass through the origin. This indicates that the Mohr envelope is a straight line
for sand and the sand has no cohesion. The angl e made by the envelope wit h the a-axis is called the
angle of internal friction. The failur e planes for each of these samples are shown in Fig. 8.18(b).
Each of them make an angl e a wit h the horizonta l which is approximatel y equal to
a = 45° + 0/2
From Fig. 8.18(b) an expression for the angl e of internal frictio n may be written as
- (J3 (Tj / <73 - 1
(840 )
{Q-™}
Exampl e 8.4
Determine the magnitud e of the deviator stress if a sample of the same sand with the same void ratio
as given in Ex. 8.3 was tested in a triaxial apparatus with a confinin g pressur e of 60 kN/m2.
Solution
In the case of a triaxial test on an identical sampl e of sand as given in Ex. 8.3, use the same Mohr
envelope OM (Fig. Ex. 8.3). Now the point F on the abscissa gives the confinin g pressur e
<7 3 = 60 kN/m2. A Mohr circle C2 may now be drawn passing through point F and tangentia l to the
Mohr envelope OM. The point E gives the major principa l stress <J } for the triaxial test.
Now cr j = OE = 188 kN/m2, <7 3 = 60 kN/m2
Therefor e al - <7 3 = 188 - 60 = 128 kN/m2 = deviator stress
Exampl e 8.5
A consolidate d drained triaxial test was conducte d on a granular soil. At failur e cr'/o^ = 4.0. The
effectiv e minor principa l stress at failur e was 100 kN/m2. Comput e 0' and the principa l stress
differenc e at failure.
Solution
- j -1 4-1
3 +1 4 + 1
The principal stress differenc e at failur e is
sin<z\$' = —; ~ = ~ = 0.6 or 6' - 37°
cr,7cr 3 + 14 + 1
Shear Strength of Soil 283
,
= <^ —-1 =100(4-l) = 300kN/m 2
^
Exampl e 8.6
A drained triaxial test on sand with cr'3 = 3150 lb/ft2 gave (a\laf^)f = 3.7. Compute (a)
(b) (o- j - 0-3)^, and (c) \$'.
Solution
Therefore, o{ = 3.1 (T f 3 = 3.7 x 3 150 = 1 1,655 lb/ft 2
(b) (<T! - o-3)/ = (0-; - crp/ = 1 1,655 - 3150 = 8505 lb/ft 2
l 3.7-1
OT
Exampl e 8.7
Assume the test specimen in Ex. 8.6 was sheared undrained at the same total cell pressure of
3150 lb/ft 2. The induced excess pore water pressure at failur e u, was equal to 1470 lb/ft2. Compute:
(a) o\f
(b) (cr, - 03)f
(c) 0 in terms of total stress,
(d) the angle of the failure plane a,
Solutio n
(a) and (b): Since the void ratio after consolidation would be the same for this test as for Ex. 8.6,
assume §' is the same.
a'
cr, - a-, ) , = cr( , — L - 1
7 J/ -
As before (
°3/ = ^3/ ~ «/ = 3150 - 1470 = 1680 lb/ft 2
So (ff l - cr3 )f = 1680 (3.7 - 1) = 4536 lb/ft 2
a(f = (ff l -03)f + &'3f = 4536 + 1680 = 6216 lb/ft 2
(c) sin<z> f, , = -1 2. = = 0.59 or 0tn1. = 36.17°
<"tota i 6216 + 1470 total
(d) From Eq. (8.18)
284 Chapte r 8
af = 45° + — = 45° + — = 62.5°
7 2 2
where 0'is taken from Ex. 8.6.
Exampl e 8.8
A saturated specimen of cohesionless sand was tested under drained conditions in a triaxial
compression test apparatus and the sample failed at a deviator stress of 482 kN/m2 and the plane of
failur e made an angle of 60° with the horizontal. Find the magnitudes of the principal stresses.
What would be the magnitudes of the deviator stress and the major principal stress at failur e for
another identical specimen of sand if it is tested under a cell pressure of 200 kN/m2?
Solution
Per Eq. (8.18), the angle of the failur e plane a is expressed as equal to
Since a = 60°, we have 0 = 30°.
From Eq. (8.40), sin ^ = —1
wit h 0 = 30°, and (7, - cr 3 = 482 kN/m2. Substituting we have
o- j - <J 3 482
°"i + ^3 = ~7~ ~ • ono = 964 kN/m2 (a)
1 J sin^z ) si n 30
cr, - cr 3 - 482 kN/m2 (b)
solving (a) and (b) we have
ol = 723 kN/m2, and <J 3 = 241 kN/m2
For the identical sample
0 = 30°, <T 3 = 200 kN/m2
From Eq. (8.40), we have
cr, - 200
Sin30°=^7^
Solving for <T J we have al = 600 kN/m2 and (cr, - cr 3) = 400 kN/m2
8.20 UNCONSOLIDATED-UNDRAINED TEST
Saturate d Cla y
Tests on saturated clay may be carried out either on undisturbed or on remolded soil samples. The
procedure of the test is the same in both cases. A series of samples (at least a minimum of three)
having the same initia l conditions are tested under undrained conditions. With ay the all-round
pressure, acting on a sample under conditions of no drainage, the axial pressure is increased until
failur e occurs at a deviator stress (<7, - (7 3). From the deviator stress, the major principal stress cr, is
determined. If the other samples are tested in the same way but with different values of cr 3, it is
Shear Strength of Soi l
285
foun d that for all types of saturated clay, the deviator stress at failur e (compressiv e strength ) is
entirel y independen t of the magnitud e of cr 3 as shown in Fig. 8.19. The diameter s of all the Mohr
circles are equal and the Mohr envelope is parallel to the cr-axi s indicatin g that the angl e of shearing
resistanc e 0 U = 0. The symbol 0 U represent s the angl e of shearing resistanc e under undraine d
conditions. Thus saturate d clays behave as purel y cohesive material s with the followin g properties:
(8.41)
where cu is the symbol used for cohesion under undraine d conditions. Eq. (8.41) holds true for the
particula r case of an unconfine d compressio n test in which <7 3 = 0. Since this test requires a very
simpl e apparatus, it is ofte n used, especiall y for fiel d work, as a ready means of measurin g the
shearing strengt h of saturate d clay, in this case
q
= —!L, wher e
(8.42)
Effectiv e Stresse s
If during the test, pore-pressure s are measured, the effectiv e principa l stresses may be written as
<j( = CT j - U
(8.43)
where u is the pore water pressur e measure d during the test. The effectiv e deviator stress at failur e
may be written as
Eq. (8.44) shows that the deviator stress is not affecte d by the pore water pressure. As such the
effectiv e stress circle is only shifted fro m the position of the total stress circle as shown in Fig. 8.19.
Partiall y Saturate d Clay
Tests on partiall y saturate d clay may be carried out either on undisturbe d or on remolded soil
samples. All the samples shall have the same initial conditions before the test, i.e., they shoul d possess
the same water content and dry density. The tests are conducte d in the same way as for saturated
samples. Each sampl e is tested under undraine d condition s with differen t all-round pressures o~ 3.
T
Cu
A.
Effectiv e stress circle
Total stress circl e
Figur e 8.19 Mohr circl e for undrained shear test on saturate d clay
286
Chapte r 8
Figur e 8.20 Mohr circl e for undraine d shea r test s on partiall y saturate d clay soil s
Tota l stres s
circl e
Figur e 8.21 Effectiv e stres s circle s for undraine d shea r tests on partiall y saturate d
clay soil s
Mohr circle s for three soi l sample s and the Mohr envelop e are shown in Fig. 8.20. Thoug h all the
sample s had the same initia l conditions, the deviato r stres s increase s wit h the increas e in the all-roun d
pressur e o~ 3 as shown in the figure. Thi s indicate s that the strengt h of the soil increase s wit h increasin g
value s of o~ 3. The degre e of saturatio n also increase s wit h the increas e in o~ 3. The Mohr envelop e whic h
is curve d at lowe r value s of o~ 3 become s almos t paralle l to the o*-axi s as ful l saturatio n is reached. Thus
it is not strictl y possibl e to quot e singl e value s for the parameter s cu and §u for partiall y saturate d
clays, but over any range of norma l pressur e cr; encountere d in a practica l example, the envelop e can
be approximate d by a straigh t line and the approximat e value s of cu and 0H can be used in the analysis.
Effectiv e Stresse s
If the pore pressure s are measure d durin g the test, the effectiv e circle s can be plotte d as shown in
Fig. 8.21 and the parameter s c' and 0' obtained. The envelop e to the Moh r circles, whe n plotte d in
terms of effectiv e stresses, is linear.
Typica l undraine d shear strengt h parameter s for partiall y saturate d compacte d sample s are
shown in Tabl e 8.2.
8.21 UNCONFINE D COMPRESSIO N TEST S
The unconfme d compressio n test is a specia l case of a triaxia l compressio n test in whic h the all -
round pressur e o"3 = 0 (Fig. 8.22). The test s are carrie d out onl y on saturate d sample s whic h can
stand withou t any latera l support. The test, is, therefore, applicabl e to cohesiv e soil s only. The test
Shea r Strength of Soil 28 7
Tabl e 8.2 Probabl e undrained shea r strength parameter s for partially saturate d soil s
Types of soi l
Sand wit h clay binde r
Lean silt y clay
Clay, moderat e plasticit y
Clay, ver y plasti c
cu (tsf )
0.80
0.87
0.93
0.87
4>u
23°
13°

c'(t sf )
0.70
0.45
0.60
0.67
0'
40°
31°
28°
22°
is an undraine d test and is base d on the assumptio n that ther e is no moistur e loss durin g the test. The
unconfme d compressio n test is one of the simples t and quickes t test s used for the determinatio n of
the shea r strengt h of cohesiv e soils. Thes e test s can also be performe d in the fiel d by makin g use of
Figur e 8.22 Unconfine d compressio n test equipmen t (Courtesy: Soiltest )
288 Chapte r 8
Any compressio n testin g apparatu s wit h arrangemen t for strai n contro l ma y be use d for
testin g the sample s . The axia l load u\ ma y be applie d mechanicall y or pneumatically.
Specimen s of heigh t to diamete r rati o of 2 ar e normall y used for the tests. The sampl e fail s
eithe r by shearin g on an incline d plan e (i f the soi l is of brittl e type ) or by bulging. The vertica l stres s
at any stage of loadin g is obtaine d by dividin g the tota l vertica l load by the cross-sectiona l area. The
cross-sectiona l area of the sampl e increase s wit h the increas e in compression. The cross-sectiona l
area A at any stage of loadin g of the sampl e ma y be compute d on the basi c assumptio n tha t the tota l
volum e of the sampl e remain s the same. Tha t is
AO/I Q = A h
wher e A Q, hQ = initia l cross-sectiona l area and heigh t of sampl e respectively.
A,h = cross-sectiona l area and heigh t respectivel y at any stage of loadin g
If Ah is the compressio n of the sample, the strai n is
A/z
£ ~ ~j~~ sinc e A/z = h0- h, we may writ e
AO/Z Q = A(/Z O - A/z )
Therefore, A = -j^- = ^^ = ^ (8.45 )
The averag e vertica l stres s at any stage of loadin g may be writte n as
P P(l-e]
A A ()
(8.46)
wher e P is the vertica l load at the strai n e.
Usin g the relationshi p give n by Eq. (8.46 ) stress-strai n curve s may be plotted. The peak valu e
is take n as the unconfine d compressiv e strengt h qti, tha t is
( f f i ) f = V u (8-47 )
The unconfine d compressio n tes t (UC) is a specia l cas e of the unconsolidated-undraine d
(UU) triaxia l compressio n tes t (TX-AC). The onl y differenc e betwee n the UC test and UU tes t is
that a tota l confinin g pressur e unde r whic h no drainag e was permitte d was applie d in the latte r test.
Becaus e of the absenc e of any confinin g pressur e in the UC test, a prematur e failur e throug h a wea k
zone ma y terminat e an unconfine d compressio n test. For typica l sof t clays, prematur e failur e is not
likel y to decreas e the undraine d shea r strengt h by mor e tha n 5%. Fi g 8.23 shows a compariso n of
undraine d shea r strengt h value s fro m unconfine d compressio n test s an d fro m triaxia l compressio n
test s on soft-Natsushim a cla y fro m Toky o Bay. The propertie s of the soil are:
Natura l moistur e conten t w = 80 to 90%
Liqui d limi t w,= 100 to 110 %
Plasticit y inde x /; = 60%
There is a uniqu e relationshi p betwee n remolde d undraine d shea r strengt h and the liquidit y
index, / , as shown in Fig. 8.24 (afte r Terzagh i et al., 1996). Thi s plot include s sof t clay soi l and sil t
deposit s obtaine d fro m differen t part s of the world.
Shea r Strengt h of Soi l
289
50
40
§30
:?20
10
Natsushima Clay
Ip = 60%
cu = undraine d strengt h
cu(UQ
° = 0.80
0 10 20 30 40 50
cu (TQ, kPa
Figur e 8.23 Relatio n between undraine d shea r strength s from unconfine d
compressio n and triaxial compressio n test s on Natsushima clay (dat a from Hanzawa
and Kishida, 1982)
102
10'
M r v
•0 10°
2
"o
10-
10"
2 3 4
Liquidit y index
Figur e 8.24 Relatio n between undraine d shear strengt h and liquidity index of clays
from around the worl d (afte r Terzagh i et al., 1996)
290 Chapte r 8
Exampl e S.9
Boreholes reveal that a thi n layer of alluvia l sil t exists at a depth of 50 ft below the surface of the
ground. The soil above this level has an average dry uni t weight of 96 lb/ft 3 and an average water
content of 30%. The water table is approximatel y at the surface. Tests on undisturbed samples give
the following data: cu = 1008 lb/ft 2, 0 M = 13°, cd = 861 lb/ft 2, (j) d = 23°. Estimate the shearing
resistance of the sil t on a horizontal plane (a) when the shear stress builds up rapidly, and (b) when
the shear stress builds up very slowly.
Solution
Bul k unit weight yt = yd (1 + w) = 96 x 1.3 = 124.8 lb/ft 3
Submerged uint weight yb = 124.8- 62.4 = 62.4 lb/ft 3
Total normal pressure at 50 ft depth = 50 x 124.8 = 6240 lb/ft 2
Effectiv e pressure at 50 ft depth = 50 x 62.4 = 3120 lb/ft 2
(a) For rapid build-up, use the properties of the undrained state and total pressure.
At a total pressure of 6240 lb/ft 2
shear strength, s = c + crta n </> = 1008 + 6240 tan 13° = 2449 lb/ft 2
(b) For slow build-up, use effective stress properties
At an effectiv e stress of 3120 lb/ft 2,
shear strengt h = 861 + 3120 tan 23° = 2185 lb/ft 2
Exampl e 8.10
When an undrained triaxial compression test was conducted on specimens of clayey silt, the
followin g results were obtained:
Specime n No. 1
cr3(kN/m2) 17 44 56
<T! (kN/m2) 15 7 20 4 22 5
M (kN/m 2) 1 2 2 0 2 2
Determine the values of shear parameter s considering (a) total stresses and (b) effective
stresses.
Solution
(a) Total stresses
For a solution wit h total stresses, draw Mohr circles Cr C2 and C3 for each of the specimens
using the corresponding principal stresses a{ and cr 3.
Draw a Mohr envelope tangent to these circles as shown in Fig. Ex. 8.10. Now from the
figur e
c- 48 kN/m2, 0= 15°
Shear Strength of Soil
291
120
80
40
c = 48 kN/m2
c' = 46kN/m2
40
80 120
a, kN/m2 »
Figure Ex. 8.1 0
200
240
(b) Wit h effectiv e stresse s
The effectiv e principa l stresse s ma y be foun d by subtractin g the pore pressure s u fro m the
tota l principa l stresses as give n below.
Specimen No.
cr'3 = (CT 3 - u) kN/m2
o\ = (CT J - w ) kN/m2
1
5
145
2
24
184
3
34
203
As befor e dra w Moh r circle s C',, C"2 and C"3 for eac h of the specimen s as show n in
Fig. Ex. 8.10. No w fro m the figur e
c' = 46 kN/m 2, \$'= 20°
Exampl e 8.11
A soi l has an unconfine d compressiv e strengt h of 120 kN/m2. In a triaxia l compressio n tes t a
specime n of the same soi l whe n subjecte d to a chambe r pressur e of 40 kN/m 2 faile d at an additiona l
stres s of 160 kN/m2. Determine:
(i) The shea r strengt h parameter s of the soil, (ii ) the angl e mad e by the failur e plan e wit h the
axia l stres s in the triaxia l test.
Solution
Ther e is one unconfine d compressio n tes t resul t and one triaxia l compressio n tes t result. Hence two
Moh r circles, Cp and C2 may be draw n as show n in Fig. Ex. 8.11. For Moh r circl e Cr cr 3 = 0 and
CT j = 120 kN/m 2, and for Moh r circl e C2, O 3 = 40 kN/m 2 and a{ = (40 + 160) = 200 kN/m2. A
commo n tangen t to thes e two circle s is the Moh r envelop e whic h give s
(i) c = 43 kN/m2 and 0 = 19°
(ii ) For the triaxia l tes t specimen, A is the poin t of tangenc y for Moh r circl e C2 and C is the
cente r of circl e C2. The angl e mad e by AC wit h the absciss a is equa l to twic e the angl e betwee n the
failur e plan e and the axi s of the sampl e = 26. From Fig. Ex. 8.11, 26 = 71 ° and 6 = 35.5°. The
angl e mad e by the failur e plan e wit h the e r -axi s is a = 90°-35.5° = 54.5°.
292
Chapter 8
80
120
160 200
o, kN/m2 •
Figure Ex. 8.11
Exampl e 8.12
A cylindrical sample of saturated clay 4 cm in diameter and 8 cm high was tested in an unconfined
compression apparatus. Find the unconfined compression strength, if the specimen failed at an axial
load of 360 N, when the axial deformation was 8 mm. Find the shear strength parameters if the angle
made by the failure plane wit h the horizontal plane was recorded as 50°.
Solution
Per Eq. (8.46), the unconfined compression strength of the soil is given by
where P =
A =
= 12.56 cm2,
= — = 0.1
8
50
100 150
a, kN/m2
0= 10°
200
250
Figure Ex. 8.12
Shear Strength of Soi l 29 3
Therefor e a, = 360(1~°-1) = 25.8 N/ cm2 =258kN/m 2
1 12.56
Now 0 = 2a - 90° (Refe r to Fig. 8.12) wher e a = 50°. Therefor e 0 = 2 x 50 - 90° = 10°.
Draw the Mohr circl e as shown in Fig. Ex. 8.12 (a3 = 0 and o~ j = 258 kN/m2) and fro m the
center C of the circle, draw CA at 2a = 100°. At point A, draw a tangen t to the circle. The tangen t is
the Mohr envelop e whic h gives
c = 106 kN/m2, and 0=10°
Exampl e 8.13
An unconfme d cylindrica l specime n of clay fail s unde r an axial stress of 5040 lb/ft 2. The failur e
plane was incline d at an angl e of 55° to the horizontal. Determin e the shear strengt h parameter s of
the soil.
Solution
From Eq. (8.20),
<rl=(T3N</>+2cjN^, where ^ = tan2 45° +|
since < T = 0, we have
= 2c tan ^45 ° + - , wher e ^ = 5040 lb/ft 2 (a)
From Eq. (8.18), the failur e angl e a is
o
a = 45 + — , sinc e a = 55°, we have
2
From Eq. (a),
c =
2tan45°-4 2tan55°
2
Exampl e 8.14
A cylindrica l sampl e of soil havin g a cohesio n of 80 kN/m2 and an angl e of interna l frictio n of 20°
is subjecte d to a cell pressur e of 100 kN/m2.
Determine: (i) the maximu m deviato r stress ((jj- <7 3 ) at whic h the sampl e wil l fail, and (ii) the
angl e made by the failur e plane wit h the axis of the sample.
Graphical solution
<7 3 = 100 kN/m2, 0 = 20°, and c = 80 kN/m2.
A Mohr circle and the Mohr envelope can be drawn as shown in Fig. Ex. 8.14(a). The circle
cuts the cr-axis at B (= <7 3), and at E (= o^). Now <7 j = 433 kN/m2, and <7 3 = 100 kN/m2.
294
Chapter 8
200
100
100 200 300
a, kN/m 2 » •
(a)
400 450
(b)
Figure Ex. 8.14
(<7, - cr 3) = 433 - 100 = 333 kN/m2.
Analytical solution
Per Eq. (8.20)
or, = a, tan2 45° + — +2ctan 45° + —
1 3 I 2) I 2.
Substitutin g the known values, we have
tan(45° + 0/2) = tan (45° + 10) = tan 55° = 1.428
tan2 (45° + 0/2) = 2.04.
Therefore,
<7, = 100 x 2.04 + 2 x 80 x 1.428 « 433 kN/m2
(CT j - <J 3) = (433 - 100) = 333 kN/m2
If 6 = angle made by failur e planes with the axis of the sample, (Fig. Ex. 8.14(b))
29 = 90 - 0 = 90 - 20 = 70° or 6 = 35°.
Therefore, the angle made by the failure plane with the cr-axi s is
a- 90 -35 = 55°
8.22 CONSOLIDATED-UNDRAINED TEST ON SATURATED CLAY
Normall y Consolidate d Saturate d Cla y
If two clay samples 1 and 2 are.consolidated under ambient pressures of pl and p2 and are then
subjected to undrained triaxial tests without furthe r change in cell pressure, the results may be
expressed by the two Mohr circles C L and C2 respectively as shown in Fig. 8.25(b). The failure
envelope tangential to these circles passes through the origin and its slope is defined by 0 CM, the
angle of shearing resistance in consolidated undrained tests. If the pore pressures are measured the
effectiv e stress Mohr circles C\ and C'2 can also be plotted and the slope of this envelope is 0'cu<
The effectiv e principal stresses are:
Shear Strength of Soil
295
Axial strain Axia l strain
(a) Variation of (a\ - a3) and u wit h axial strain
[- "2 H
(b) Mohr envelop e
Figur e 8.25 Normall y consolidate d clay under undrained triaxial test
p\
P2
P^=Pa
Total stress
circle
Effectiv e
stress circle
(^3)1 (^3) 2 (^3) 3
Figure 8.26 Consolidated-undraine d tests on saturated overconsolidate d clay
296
Chapte r 8
° = ° ~
wher e ul and w 2 are the por e wate r pressure s for the sample s 1 and 2 respectively.
It is an experimenta l fact tha t the envelope s to the tota l and effectiv e stres s circle s are
linear. Fig. 8.25(a ) shows the natur e of the variatio n on the deviato r stres s (<7 j - <7 3 ) and the pore
water pressur e u in the specime n durin g the test wit h the axia l strain. The por e wate r pressur e
build s up durin g shearin g wit h a correspondin g decrease in the volume of the sample.
Overconsolidated Clay
Let a saturate d sampl e 1 be consolidate d unde r an ambien t pressur e pa and then allowe d to swel l
unde r the pressur e pr An undraine d triaxia l test is carrie d out on thi s sampl e unde r the all-roun d
pressur e p\(= <T 31 ). Anothe r sampl e 2 is also consolidate d unde r the same ambien t pressur e pa and
allowe d to swel l unde r the pressur e p2(= <7 32 ). An undraine d triaxia l test is carrie d out on thi s sampl e
unde r the same all-roun d pressur e p2. The two Moh r circle s are plotte d and the Mohr envelop e
tangentia l to the circle s is drawn as shown in Fig. 8.26. The shear strengt h parameter s are cu and 0 CU.
If pore wate r pressur e is measured, effectiv e stres s Moh r circle s may be plotte d as shown in the
figure. The strengt h parameter s for effectiv e stresse s are represente d by c'and §'.
8.23 CONSOLIDATED-DRAINE D SHEAR STRENGTH TEST
In draine d triaxia l test s the soil is firs t consolidate d unde r an ambien t pressur e pa and then subjecte d to
an increasin g deviato r stres s unti l failur e occurs, the rate of strai n being controlle d in such a way that
at no time is ther e any appreciabl e pore-pressur e in the soil. Thus at all times the applie d stresse s are
effective, and whe n the stresse s at failur e ar e plotte d in the usua l manner, the failur e envelop e is
directl y expresse d in terms of effectiv e stresses. For normall y consolidate d clays and for sands the
envelop e is linea r for norma l workin g stresse s and passe s throug h the origi n as shown in Fig. 8.27.
The failur e criterio n for such soil s is therefor e the angl e of shearin g resistanc e in the draine d
conditio n 0d.
The draine d strengt h is
-(o- 1 -ff 3 )/= -
. - sin
(8.48)
Eq. (8.48 ) is obtaine d from Eq. (8.38 )
Figur e 8.27 Draine d tests on normall y consolidate d clay sample s
Shea r Strength of Soi l
297
Peak /O.C. clay
Ultimat e
N.C. clay
Axia l strai n
3
I
D,
X
U
CX
o
u
O.C. clay
Axia l strai n
N.C. clay
(a) Variatio n of (a\ - a3) wit h axia l strai n
O.C. clay
Q
N.C. cla y
Norma l stress, o
(b) Mohr envelop e
Figure 8.28 Drained tests on overconsolidated clays
For overconsolidate d clays, the envelop e intersect s the axi s of zero pressur e at a value cd. The
apparen t cohesio n in the draine d test and the strengt h are give n by the expression.
(8.49)
1-sini
The Moh r envelop e for overconsolidate d clays is not linea r as may be seen in Fig. 8.28(b). An
averag e line is to be drawn withi n the rang e of norma l pressur e cr n. The shear strengt h parameter s cd
and (j) d ar e referre d t o thi s line.
Since the stresse s in a draine d test are effective, it migh t be expecte d that a given (f) d woul d be
equa l to 0' as obtaine d fro m undraine d test s wit h pore-pressur e measurement. In normall y
consolidate d clays and in loose sand s the two angle s of shearin g resistanc e are in fac t closel y equa l
since the rat e of volume chang e in such material s at failur e in the draine d test is approximatel y zer o
and ther e is no volume chang e throughou t an undraine d test on saturate d soils. But in dens e sand s
and heavil y overconsolidated clays ther e is typicall y a considerable rate of positive volume chang e
at failur e in draine d tests, and wor k has to be done not onl y in overcomin g the shearin g resistanc e
of the soils, but also in increasin g the volume of the specime n agains t the ambien t pressure. Yet in
298
Chapte r 8
undraine d test s on the same soils, the volum e chang e is zer o and consequentl y (j) d for dens e sand s
and heavil y overconsolidate d clay s is greate r tha n 0'. Fig. 8.28(a ) shows the natur e of variatio n of
the deviato r stress wit h axia l strain. Durin g the applicatio n of the deviator stress, the volume of the
specime n graduall y reduce s for normall y consolidate d clays. However, overconsolidate d clay s go
throug h some reductio n of volum e initiall y but the n expand.
8.24 POR E PRESSUR E PARAMETER S UNDE R UNDRAINED