11 ShearStrength of Soil
The shearstrength of a soil massis the internal resistanceper unit area that the soil masscan otTerto resistfailure and sliding along any plane insideit. One must under- standthe nature of shearingresistance in order to analyzesoil stabilityproblems such as bearing capacity,slope stability,and lateral pressureon earthretainingstructures.
11.1 Mohr -Coulomb Failure Criterion
Mohr ( 1900)presented a theory for rupture in materialsthat contendedthat a ma- terial fails becauseof a critical combination of normal stressand shcaringstress, and not from either maximum normal or shear stressalone. Thus. the functional rela- tionship between normal stressand shearstress on a failure plane can be expressed in the followins form:
r1: f@) (11. 1) Thc failureenvelope defined by Eq.(11.1) is a curvedline. For most soil me- chanicsproblems, it is sufficientto approximatethe shearstress on 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/ : shearstrength
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'ltt
311 312 Chapter 11 Shear Strength of Soil
Table 11. ? Typical Values of Drained Angle of Friction for Sands and Silts
Soil type d'(des1 Sand: Rr,tundedgrains Loose 27-30 Medium 30-35 Dense 35-38 Sand: Angulur grains Loose 30-35 Medium 3.5- 4t) Dense 40-45 Gravel with some sand 34,48 Sl/ts 26-35
The effectivestress a' is carried by the soil solids.The Mohr-Coulomb failure cri- terion, expressedin terms of efTectivestress, will bc clf the form - T| ,'' * tr' tan <[' (11.3) where c' : cohesionand @' : friction angle,based on effeotivestrcss. Thus, E,qs.(11.2) and (11.3)are expressionsof shearstrength based on total stressand effectivestress. The value of c' for sand and inorganic silt is 0. For nor- mally consolidatedclays, c' can be approximated at 0. Overconsolidatedclays have valuesof c' that are greaterthan 0. The angleol friction, ry'',is sometimesrcfcrred to ' as the drained angLeo.f .frictbn. Typical values of a (a) Figure 11.I Mohr-Coulomb failurecriterion 11'2 rncrinationof the prane of Fairurecaused by shear 313 the plane ab be o' and r, respectively.Figure 11.1bshows the plot of the failure en- velope definedby Eq. (11.3).If the magnitudesof o' and r on planeab are suchthat 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,because it plots abovethe fail- ure envelope,and shearfailure 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 occurwhen the shearstress on a plane reachesa value given by Eq. (1 1.3).To determine the in- clination 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 angled with the major principal plane.To determine the angle0 and the relationshipbetween rrj 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$, . The radialline ab definesthe major principalplane (CD in Figure'11.2),and the ra- dial lineaddefines the failureplane (E-Fin Figure 1l.z).ltcan be shownthat Lbad: 2e:90*d'.or 0:45+9 ( 2 r1.4) Again.from Figure11.3. ad (r1.-s) fa:stna Tu - .fO+ Ou : c' ct>rg'-ojt'i (11.6a) 2 U 01, a O'I Ef1'ectivenormal stress Figure | 1.2 Inclination of failure plane in soil with major principal plane Figure 71.3 Mohr'scircle and failureenvelope 314 Chapter 11 Shear Strength of Soil Also, ot-ol aa:r ( l 1.6b) 2 SubstitutingEqs. (11.6a) and (11.6b) into Eq. (11.5),we obtain o\-ot 2 sin@' : o,, _f o,. c'cot$'+ t- /l+sind'\ / cos6'\ 'i -.'1[r (1 1.7) sind/*"\r ,ino,/ However, I + sin@':tan'145*^/ d'\ 1 - sin q5' Z) and cos d' / (b' \ I "t*,t-tan\45+ 2) Thus, : (11.8) o\ o\,un'(+s. +) * ,r',^n(ot . +) An expressionsimilar to Eq. (11.tt)could alsobe derived usingEq. (11.2)(thal is, total stressparameters c and @),or rr) - . rr,,""(0, -- ( l l.e) ut ttun'(+.s, f) 1t.3 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. Direct shear test b. Triaxial test c. Direct simple shear test d. Plane strain triaxial test e. 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 de- tail in the sections that follow. 11.4 Direct Shear Test The direct sheartest is the oldest and simplestform of sheartest arrangement.A di- agram of the direct sheartest 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 squareor circular in plan. The sizeof the specimensgenerally ur",l is about 51mm x 5Lmm or102mm x 102mm(2in.x2in.or4in. x 4in.) acrossand about 2-5mm (1 in.) high. The box is split horizontally into halves.Normal force on the specimenis applied from the top of the shear box. The normal stresson the speci- mens 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 sheartest can be either stresscontrolled or strain controlled. In stress-controlledtests, the shearforce is applied in equal incre- ments 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 gaugethat measuresthe vertical movement of the upper loading plate. In strain-controlledtests, a constant rate of shear displacementis applied to one-half of the box by a motor that acts through gears.The constantrate of shear displacementis measuredby a horizontal dial gauge.The resistingshear force of the soil correspondingto any sheardisplacement can be measureclby a horizontal prov- ing 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 Chapter 11 Shear Strength of Soil Figure 1?.5 Strain-controllcddirect shear test equipmcnt (courtesy of Soiltest,Inc., Lake Bluff,Illinois) in a manner similar to that in the stress-controlledtests. Figure 11.5 shows a photo- graph of strain-controlleddirect sheartest equipment. The advantageof the strain-controlledtests is that in the caseof densesand, peak shearresistance (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 stress- controlled tests,only the peak shear resistancecan be observedand plotted. Note that the peak shearresistance in stress-controlledtests can be only approximatedbe- causefailure occurs at a stresslevel somewherebetween the prefailure load incre- ment and the failure load increment.Nevertheless, compared with strain-controlled tests,stress-controlled tests probably model real field situationsbetter. For a given test,the normal stresscan be calculatedas Normal force o : Normal stress: (11.10) Cross-sectionalarea of the specimen The resistinsshear stress for anv sheardisplacement can be calculatedas Resistinsshear force r : Shearstress : (11.11) Cross-sectionalarea of the specimen 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 observa- tions were obtained from a strain-controlled test. The following generalizationscan 11.4 DirectShear Test 317 E a Sheardisplacentent .='c x u)T F dF = r) o a o U Figure 11.6 PloIof shearstress ancl changc in heightof specimenagainst shear displace- mentfor looseand dcnse dry sand(dircct shcar test) be developedfrom Figure I 1.6regarding the variation of resistingshear stress with sheardisplacement: 1. ln loose sand,the resistingshear stress increases with sheardisplacement until a failure shearstress of 11is reached.After that, the shearresistance remains approximatelyconstant for any further increasein the sheardisplacement. 2. In densesand, the resistingshear stress increases with sheardisplacement until it reaches a failure stressof rr. This rlis called the peak shear strength. After failure stressis attained,the resistingshear stress gradually decreasesas shear displacementincreases until it finalty reachesa constantvalue called the ulti- mate .shear.strength. It is important to note that, in dry sand, o:(r' and c':0 Direct shear testsare 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 shearstrength parameters are determined. 318 Chapter 17 Shear Strength of Soil o'(lb/inz) 20 40 ^ 150 z 20 / too ca 6' = 42" 0 50 100 150 200 2-50 300 Efl'ectivenorntal stress, o' (kN/rn:,1 Figure 11.7 Determination of shear strcngth parameters for a dry sand using the results of direct sheartests Figure 11.7 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' (11.12) 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 sheartest arrangement,the shearbox that containsthe soil specimenis generallykept inside a container that can be filled with water to saturatethe speci- men. A drained /estis made on a saturated soil specimen by keeping the rate of load- ing slow enough so that the excesspore water pressure generated in the soil is com- pletely dissipated by drainage. Pore water from the specimen is drained through two porous stones.(See Figure 11.4.) Because the hydraulic conductivity of sand is high, the excesspore water pres- sure 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 Residualshear strength 20 o 0 2 Horizonttldclirnnation (t/') Figure 11.8 Resultsof zrclraincd direct shcar tt:st on an ovcrconsolidatedclay. Nore; Resid- ualshear strength in clayis similar to ultinrateshcar strength in sand- seeFisure 1 1.6 an ordinary loading rate.esscntially full drainageconditions exist. The friction angle, ', f obtaincd from a drained direct shear test of saturatedsand will be the same as that for a similarspccimen of dry sand. 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 lastfrom two to five days.Figure I I .8shows thc resultsof a draineddircct sheartest on an over- consolidatedclay. Figure I 1.9shows the plot of riagainst o' obtained from a number 'e d Overconsolidatedclay 6 c^ -r Normallyconsolidated clay TJ= o'tan Q'(r''=0) I t" j vI Elfective normal stress.o' Figure 77,9 Failure envelope for clay obtained from drained direct shear tests Chapter 11 Shear Strength of Soil Figure 11.10 lnterfaceof a loundationmaterial and 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 questionedbccause the soil is not allowedto fail along the wcakestplanc but is forced to fail along the planc of split of the shearbox. Also, the shearstress distribution over the shearsurface of the specimenis not uni- form. Despitc theseshortcomings, the direct sheartest is the simplestand most eco- nomical for a dry or saturatedsandy soil. In many foundation design problems, one must determine the angle of fric- tion between the soil and the matcrial in which the foundation is constructed(Fig- ure 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 givenas rr': c'r,+ ottan6 (11.13) where c,i,: adhesion 6 - effectiveangle 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 sheartest, The foundation material can be placedin the bottom part of the direct sheartest box and then the soil can be placedabove 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 resultsof direct shear testsconducted in this manner with a quartz sand and concretewood, and steel as foundation materials,with o' = 100kN/m2 (14.5lb/in.2). 11.6 General Comments on Direct Shear Test Normal forcc € Shear I fbrce Figure 11.11 Direct sheartest to determinc interfacefriction ansle Relativedensity, /),. (7o) 75 -s0 25 Nornralstress o'- 100kN/nr2 ( | 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), quartz sand (after Acar, Durgunoglu,and Tumay,1982)] 322 Chapter 11 Shear Strength of Soil Example11.1 Followingare the resultsof four draineddirect sheartests on an overconsolidated clay: Diameter of specimen: 50 mm Height of specimen: 25 mm Normal Shear force at Residualshear Test forca, lV failure, So."p forc6, 5r""16961 no. (Nl {N} (N}' 1 150 157.5 44.2 2 250 199.9 56.6 3s0 257.6 102.9 4 550 363.4 r44.5 'SeeFigure 11.t3 Determine the relationshipsfor peak shear strength(r) and residual shear strength(r ). Solution / so \2 Area of the specimen(A) : (rrl+)l : 0.0019634m2. Now the following #luuu ) tablecan be prepared:' \ / Residual $residuat Normal Normal Peak shear .. = -sr1* tt*:l , - Test force, N sttesg, ('' force, Souul' A Sreriauat 4 no. (Nl (kN/m2) {Nl {kN/m'?l (Nl (kN/m'l 1 150 76.4 r57.5 80.2 44.2 22.5 2 250 127.3 199.9 101.8 56.6 28.8 J 350 178^3 257.6 131.2 102.9 52.4 4 550 280.1 JOJ.4 185.1 1.44.5 73.6 The variationsof rland r, with c' areplotted in Figure11.13' From the plots, we find that Peak strength: rdkN/m2) : 40 + o' tan27 Residual strength: r,(kN/m2) : a' tan14.6 J (Note: For all overconsolidated clays, the residual shSrstrength can be ex" pressedas t' : o' lan 6" where dl : effectiveresidual friction angle.) | 1.7 Triaxial Shear Test (Generat) 250 tr )oo z .: 150 T/ VS. O' ! loo a 21' = Q' r/.vs. o' = 40 kN/nr: r = 11.6" Figure 11.13 50 t(x) I 50 200 2-50 300 350 Variationsof z7 Effbctivcnorrral stress, o' (kN/nt:) andr, with ('' 11.7 Triaxial Shear Test (General) The triaxialshear test is one of thc most reliablemcthocls available f'or dctern.rirring shcarstrength parameters. It is widcly uscclfor researchancl convcntional testing. ,{ diagramof the triaxialtest layout is shownin Figure I I .14. In thistest, a soilspecimen about 36 mm ( 1.4in. ) in diamet.eralcl 76 ntm (3 in.) long is gcnerallyused. The specimenis encasecl by a thin rubber membraneancl placecl insidea plasticcylindrical chamber that is usutrllyfilled with wateror glyccrinc.Thc specimenis subjectedto a confiningpressure by compressionof the fluiclin the cham- ber. (Note: Air is sometimesused as a compressionmecliunt.) To cituscshear failure in the specimen,one must apply axialstress through a vcrticalloading ram (some- times called deviatorsrress). This strcsscan be appriedin one of two ways: 1. Application of dead weightsor 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 gearedor hydraulic loading press.This is a strain-controlledtest. The axial load applied by the loading ram correspondingto a given axial cleforma- tion is measuredby a proving ring or load cell attachedto the ram. connections to measuredrainage into or out of the specimen,or to mcasure pressurein the pore water (as per the test conditions), are also providecl.The fol- lowing three standardtypes 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 Chapter 11 ShearStrength of Soil Axial load I Air relelse t ---,^ valve Loading ram : : ,,. Top cap Porousdisc Specimenenclosed in a rubbcr mcmbrane tuhc Pressuregauge " *--Flexihle .1lr)..1-.. '. -'\ Rubhcr' -t | ilrl '...1 \ '' .,":,,.r,..,.i i,-- ...' :' :l:,.:.-',.-.-...,'....- l LTo cell pressurccontrol , .il, ...-..-.....,..-'ri,tr'-',..r:,*l 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 testsin saturatedroils are describedin thc followins sections. 11.8 Con so Ii d ated -D ra i n ed Triaxi aI Test In the CD test, the saturatedspecimen is first subjectedto an all around confining pressure,oj, by compressionof the chamberfluid (Figure 11.15a).As confiningpres- sure is applied, the pore water pressureof the specimenincreases by u. (if drainage is prevented).This increasein the pore water pressurecan be expressedas a non- dimensionalDarameter in the form u. B = --:- (1 r.14) C3 whereB : Skempton'spore pressure parameter (Skempton, 1954). For saturatedsoft soils,-B is approximatelyequal to 1; however,for saturated stiff soils,the magnitudeof B canbe lessthan 1.Black and Lee (1973)gave the theo- 11.8 Consolidated-DrainedTriaxial Test 325 lo"' lo, I + tl" i;' .: ;': I:.:'r:::l':':',:, i.":.." i.'r ,i --+,1 ": '.1=-9:rt, J Luo=Q 1- oj :. U I U] I1,''-- , :i lor t I l"' Figure 11.15 Consolidated-drainedtriaxial test: 1o",, (a) specimcnundcr chambcr confining (a) (b) pressure:(b) deviator strcssapplication Table 11.2 Theoretical Valuesof B at Comnlete Saturation Theoretical Type of soil value Normally consolidatedsoft clay 0.9998 Lightly overconsolidatedsoft claysancl silts 0.998u Overconsolidatcdstil'l' clays and sands 0.987'7 Very denscsands and very stiff claysat high confiningprcssures 0.9I30 retical valuesof B for various soilsat completc saturation.Thcse valuesare listed in Table 11.2. Now, if the connectionto drainageis opened,dissipation of the excesspore wa- ter 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 takesplace during consolidationcan be obtained from the volume of pore water drained (Figure 11.16a). Next, the deviator stress,Ao,1, on the specimenis increasedvery slowly (Fig- ure 11.1-5b). The drainageconnection is kept open,and the slowrate of deviatorstress applicationallows complete dissipation ol any pore water pressurethat developedas a result (Au,1: 0). A typical plot of the variation of deviator stressagainst strain in loosesancl and normally consolidatedclay is shownin Figure 11.16b.Figure 11.16cshows a similar plot for dense sand and overconsolidatedclay. The volume change,A2,7, of speci- mens that occursbecause of the application of deviator stressin various soilsis also shownin Figures11.16d and 11.16e. Becausethe pore water pressuredeveloped during the test is completelydissi- pated. we have total and effective confining stress : ot : o\ and total and effectiveaxial stressat failure : at t (Aor)r : ot : o\ Chapter 11 Shear Strength of Soil xl ?l > - <5 i.t 9l o* O' Axial strain Axial strain ol cl al al ;'lxl f- ":: <6 Axial strain <6 frl al trI 59 6{ O' U' (d) (c) Figure 11.76' Cbnsolidated-draincd triaxial test: (a) volumechange of specimencaused by chamberconfining pressure: (b) plot ol deviatorstress against strain in thevertical direction for looscszrnd and normally consolidated clay; (c) plot of deviatorstress against strain in the verticaldirection for densesand and overconsolidated clay; (d) volumechange in loosesand andnormally consolidated clay during deviator strcss application; (e) volumechange in densesand and overconsolidatcd clay during deviator stress application In a triaxial test, oi is the major principal effectivestress at failure and o! is the mi- nor principal effcctivestress at failure. Severaltests on similar specimenscan bc conductedby 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 nor- mally consolidated clay. The coordinates of the point of tangency of the failure en- velope with a Mohrb circle (that is, point A) give the stresses(normal and shear)on the failure plane of that test specimen. Overconsolidationresults when a clay is initially consolidatedunder an all- around chamber pressure of o. (: oi.) and is allowed to swell by reducing the cham- ber pressure to oj (: o1). The failure envelope obtained from drained triaxial tests of such overconsolidated clay specimensshows 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'romdrained testson sandand 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 Normally Overconsolidated consolidated ala a T +c' o3=O3 Or=o't O;. Normal stress Figure 11'18 Effective stress failure envelope for overconsolidated clav 328 Chapter 11 Shear Strength of Soil o Undisturbedsoil . '.-.'...: r Remolded soil t, " -5 l0 l 5 20 30 .10 60 t30 l(x) l-so Plasticrtyindcx. PI (%) Figure 11.19 Yttrialion ol'silr r,6'with plasticityindcx lor a number of soils(after Kcnney, rese) Comments on Drained and Residual Friction Angles of Clay 'fhe drained angleof friction, d;',gencrally dccreascs with 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 in Example I I .1, thc proccdure to calculateresidual friction angle <75iis shown. Skempton ( 1964)provided the resultsof the vzrriationof the rcsidual angleof friction, $',, of a number ol'clayey sgilswith the clay-sizctraction ('2 pm) present. The followins tzrbleshows a summary ol'thesercsults: Clay-size Residual fraction frictionangle, l/.1 d. (deg1 Sclset 17.7 29.8 Wiener Tcgel 22.8 25.1 Jackfield -J).4 19.I Oxford clay 41.9 16.3 Jari 46.5 Iu.6 London clay 54.9 16.3 Walton'.sWood 67 t -1.L Weser-Elbe 63.2 9.3 Little Belt 7'7.2 11.2 Biotite 100 7.5 At a very high clay content. @iapproaches the value of the angle of sliding fric- tion for sheetminerals. For highly plastic sodium montmorillonites, the magnitude of dl mav be as low as 3 to 4o. 11.8 Consolidated-DrainedTriaxial Test 329 Example11.2 For a normallyconsolidated clay, the resultsof a drainedtriaxial test are as follows: Chamberconfining pressure = 16lb/in.2 Deviator stressat failure : 25lblin.z a. Find the angleof friction, g'. b. Determinethe angle0 that the failure planemakes with the major princi- pal plane. I E,flectivcstrcss luilureenvolope q o'r = 16lb/inr A Norrnalstress Figure 11.20 Mohr'scircle and failureenvelope for a normallyconsolidated clay $olution For a normallyconsolidated soil, the failure envelope equation is rf : o' tan+' (sincec' = 0) For thetriaxial test, the effective major and minor principal stresses at failureare (rl = (rt = ui * (Aou)r * L6 + 25 = 41lb/in.2 and o\ : ct: 16lb/in.2 a. The Mohr's circleand the failure envelopeare showninfuigure 11.20, from which we can write : AB (ry) srnp - oA= (*#) 330 Chapter 11 Shear Strength of Soil or cri - o'^ 47-16 sind' : = 0.438 01+03 41+16 Q' * 26o 16 o':45"+?=58" b.9:45 * a L z Refer to Example1L.2. a. Find the normal stresso.' and the shearstress rlon the failure plane. b. Determinethe effectivenormal stresson the planeof maximumshear stress. Solution a. From Eqs.(9,8) and (9.9), o1 *o'3. ot-c: a' (on the failure plane) --:-;---i + - '- cos20 " and ot - cr5 r, * J ,rsin20 Subsitutingthe valuesof oi = 41 lb/in'z,a\: I6lblin'2, and 0 : 58"into the precedingequations, we get 4l + 16 4I - 1.6 , : j;i -,-, x 58) = 23.61b/in.2 o, * ?cos(2 and 4I_T6 :: sin(2x 58) = 11'2lb/in.z "t Z: b. From Eq. (9.9),it canbe seenthat the maximumshear stress will occur the planewith 0: 45".From Eq' (9.8). on l * o', - o\cos2g'o, o,-')') :o\ +o" I u .,o ,. Substituting0 : 45'into the precedingequation gives :4i9 : o, . rycos eo 28.stbtin.z 71.8 Consolidated-DrainedTriaxial Test 331 Example11.4 The equation of the effective stressfailure envelopefor normally consolidated clayeysoil is 11: o' tan 25'.A drainedtriaxial testwas conductedwith the same soil at a chamberconfining pressure of 80 kN/m2.Calculate the deviatorstress at failure. Solution For normally consolidatedclay, c' = 0. Thus,from Eq. (11.8), at: oitun'(+s. +) 6' :25o oi : 8otun'(+s r : 197 kN/m2 T) (Loo)r = {r't- o\: 197* 80 : 117kN/m2 Example11.5 The resultsof two drainedtriaxial testson a saturatedclay are asfollows: SPecimenI: ot : o\: 70kN/m2 (Ao,r)r: 130kN/m2 SPecimenII: ct * 6\ = 160kN/m2 (Aoir :223.5 kN/m? Determinethe shearstrength parameters c' and$' . Solution For specimenI, the principalstresses at failure are c't: az: 70kN/m2 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 Chapter 11 Shear Strength of Soil Usingthe relationship given Uy& t11.8), we get dr'\ o\ * o\tan2( 45 + + zc'tan(4t . \ .l +) Thus, for specimenI, :.orun'(+s / 6',\ 2oo - +) +2c'tan(45*t) and for specimen II. 383.5= . + 2c'tan(4t - y) tootan'z(+s +) Solvingthe two precedingequations, we obtain 6' :20o c' : 20kN/mz 11.9 Consolidated-UndrainedTriaxial Test The consolidated-undrainedtest is the most common type of triaxial test.ln thistest, the saturatedsoil specimenis first consolidatedby an all-aroundchamber fluid pres- SUre,trj, that resultsin drainage (Figures 11.21aand 11.21b).After the pore water pressuregenerated by the applicationof confiningpressure is dissipated,the devia- tor stress,Ao,7, on thc specimenis increasedto causcshear failure (Figure fl .2Ic). During this phaseol the test,the drainageline from thc specimenis kept closed.Be- causedrainage is not permitted, the pore water pressure,Au,1, will increase.During the test, simultaneousmeasurements of Ao,1and Au,yare made. Thc increasein the pore water pressure,Atr,1, can be expressedin a nondimensionalform as Au" A: ------:- (11.16) Lo,t 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 pressureincreases with strain. In densesand and overconsolidated clay,the pore water pressureincreases with strain to a certain limit, beyond which it decreasesand becomesnegative (with respectto 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. Because the pore water pressure at failure is measured in this test, the principal stressesmay be analyzed as follows: 11.9 Consolidated-UndrainedTriaxial Test 333 E Axialstrain (c) 4a-o 'a (,c (b) lno., t" lo. t ".:,,'. -+.: . ,.t,,. + lo, I lAo,r I (c) Figure 71.21 Consolidatedundrained test:(a) specirnenunder chamber confiningpressure: (b) volume changein specirnencauscd by conlining pressure;(c) cleviatorstress application; (d) deviator stressagainst axial strain fbr krosesand and normally consolidatedclay; (e) de- viator stressagainst axial strain for denses:ind and overconsolidatedclay; (f) variation of pore water pressurewith axial strain 1orloosc sandand normally consolidatedclay; (g) vari- ation of pore wiitcr prcssurr:with axial strain lor clensesand ,,nd'ou"ra,rnroliclated clay Major principal stressal failure(total): o3 -f (L,o,1)1: o1 Major principal stressat failure (effective): or-(L,ua)1:o\ Minor principal stressat failure (total): Aj Minor principal stressat failure (effective): o1 - (L,u)1: o', Chapter 11 Shear Strength of Soil 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 Figure 11.22 and ef1cctivcstress failure envelopcs for consolidatedundrained triaxial tests.(No1c:'l'hc figure assumcs that no backpressurc is applied.) In thcse equations,(Arz,1)1 : pore water prcssureat failure. The preceding deriva- tionsshow that art tr; - tr\ - tr'. Testson scvcralsimilar specimenswith varying conllningprcssures may be con- ducted to determinc the shearstrength parameters. Figure I1.22 showsthe total and cffectivestress Mohr'.s circles at failure obtained from consolidated-undrainedtriax- ial tcsts in sand and normally consolidtrtcdclay. Note that ,4 and B are two total stressMohr\ circlesobtaincd from two tests.C and D are the effectivestress Mohr's circlcs correspondingto total stresscircles A 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 touchesall the total stressMohr's circles.For sand and normally consoli- dated clays,this will be approximatelya straight line passingthrough the origin and may be expressedby the equation ry:otanQ (1 1.17) 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 usedfor practicalconsiderations. Again referring to Figure ll.22,we seethat 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 Fisure11.17). 11.9 Consolidated-UndrainedTriaxial Test o 1=c+OtanQl U) 6j Nurn'ii,,r"r, Figure 11.23Total stress failure envelope obtained from consolidatcd-unclrainedtests in over-consolidatcdclay In overconsolidatedclays, the total stressfailure envelopeobtained from con- solidated-undraincdtests will take the shapcshown in Figure1 1.23. l'he straightline a'b'is representedby the equation rt:c *otirnd1 (ll.llt) and the straightline b'c'follows the relationshipgiven by Eq.(11.17).'rheefTcctive stressfailure envclope drawn from the cffectivc stressMohr'.s circles will be similar to that shownin Figure 11.23. Consolidated-drainedtests on clay soils takc considerabletime. For this rca- son, consolidated-undraincdtests can be conductcdon suchsoils with pore pressure measurementsto obtain the drained shearstrength parameters. Because clrainage is not allowed in these testsduring the appliczrtionof deviator stress,they can be per- formed quickly. Skempton'.spore watcr pressureparametcr Z was dellneclin Eq. (11.16).At failure.the paramcter7 canhc writtcnas (Lur), A: A':' ---- (11.1e) (ao,t)f The gcneral range of 7, valuesin most clay soilsis as follows: . Normally consolidatedclays: 0.-5 to I . Overconsolidatedclays: -0.5 to 0 Table ll.3 givesthe valuesof A, for somenormally consolidaled clays as obtained 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 overconsolida- tion value of about 3 or 4. 336 Chapter 11 Shear Strength of Soil Table 11.3 Triaxial TestResults for SomeNormally ConsolidatedClays Obtainedby the NorwegianGeotechnical Institute* Drained Liquid Plastic Liquidity friction angle, limit limit index Sensitivity" @'(degl Al Seven Sisters,Canada r27 35 0.28 19 0.72 Sarpborg 69 28 0.6ri 5 25.5 1.03 Lilla Edet, Sweden 68 30 t --)z 50 26 1.10 Fredrikstad 59 22 0.58 5 28.5 0.87 Fredrikstad -)/ 22 0.63 6 27 1.00 Lilla Edet, Sweden 63 30 1.58 50 LJ 1.02 Cdta River, Sweden 60 27 1.30 12 28.5 1.05 Gota River, Sweden 60 30 1.50 40 24 1.05 Oslo 48 25 0.87 4 -{1.) 1.00 Trondheim 36 20 0.50 2 34 0.75 Drammen -)-l 1u 1.08 8 28 1.18 *After Bjerrumand Simons (1960) " SeeSection 1 I .l 5 for thedelinition of sensitivity Example11.6 A consolidated-undrainedtest on a normally consolidatedclay yielded the fol- lowingresults: ot : 12lblin.2 Deviatorstress, (Ar)t : 9.1lb/in.2 = Porepressur e. (L,u) 1 6.8lb/in.2 Calculatethe consolidated-undrainedfriction angleand the consolidated-drained frictionangle. Solution 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 In unconsttlidatcd-undrainedtests, drainerge l'ronr the soilspecinten is not pcrmitted during thc applicationo[ chan.rberprcssure o1. Thc testspecimcn is shearcdto I'ail- 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.Because of the applicationof chermberconfining pressurc rrj. the pore water pressurcin the soil spcc- imen will incrcaseby 4,..A further increascin the pore water pressurc(Aa,,) will oc- cur becauscol'the deviatorstress application. Hencc, the total porc water pressure rr in thc spccimenat any stageol cleviatorstress application can bc given as Lt-rt(* L,u,1 ( 11.20) From Eqs. (11.14)and (ll .16),u,: Bo janclArr,, :7A,r,,, s.'' u : Bo3+ AAo,1: Bot + A@, * cr) (lr.2r) This test is usuallyconducted on clay specimensand dependson a very impor- l.antstrength conceptfor 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 envelopefor the total stressMohr's circlesbecomes a horizontal line and henceis called a rb : 0 condition. From Eq. (11.9)with O - 0, we get 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 conceptis applicableto only saturatedclays and silts. The reasonfor obtaining the same added axial stress(Loitregardless of the confining pressurecan be explained as follows. If a clay specimen(no. l) is consoli- 338 Chapter 11 Shear Strength of Soil Total strcss Mohr'scircles a at tailure --- FailureenvelopeQ=0 o1 63 o1 61 N,,,.,i,lLl"r, Figure 11.24Total strcss Mohr'.s circles and failurc envekrpe (.D - 0) obtainedfrom uncon- solidated-undrainedlriaxial lcsts ot"t I'ullv saturltccl cohcsivc soil 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 Fig- ure 11.25. The porc pressuredcvclopcd in the specimenat failure is equal to (Arr,1)r. Thus, the major and minor principal clTcctivestresses at failure are. respectively, : o\: ltrl + (Arr,i)1j (Ar,,)r (rt (Ar,,)r and 0\ - ro1- (Att,1)7 Q is the effectivestress Mohrh circle drerwnwith the precedingprincipal stresses. Note that the diametersof circlesP and Q arc the same. Total strcss Mohr'scircle at fai lurc ------1 a O'3 O r O: 01 Normal stress -*-l --=">l l<- (Ao,1)1 F- (Ao,7), l<__ (Ao/, . ,l l<_(LuDl_,_,_"____+l ^;, = ^;;, Figure 11.25 The rl : 0 concept 11.11 Unconfined Compression Test on Saturated Clay 339 Now let us consideranother similar clay specimen(no. II) that has been con- solidated 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 pressurewill increaseby an amount A4,..For saturatedsoils under isotropic stresses,the pore wa- ter pressureincrease is 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 be- fore the application of deviator stress.Hence, if specimenno. II is shearedto failure by increasingthe axial stress,it should fail at the samcdeviator 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 pressureincrease caused by the applicationof (An,1)1 will be (Lu)r. At failure, the minor principal effectivestress is [(oj + Arrj)] [Arr, * (Lu,,)r]: ct (Lu,1)1: c\ and the major principal effectivestress is - : loj * Aa.1+ (Ao,7)1] [Arl, + (Aa,7)1] lo. * (Ao,7)1] (L,u,)1 : cr (Lu,1)1- o\ Thus, the effectivestress Mohr's circle will still be Q becausestrength is a function of efTectivestress. Note that the diamctersof circlesP, Q, and R are all the samc. Any value of Aoj could have been chosenfor testing specimenno. II. In any case,the deviator stress(Ao,1)/ to causefailure would have bcen the sameas long as the soil was fully saturatecland fully undrainedduring both stagesof the test. I1.11 Unconfined Compression Test on Saturated Clay The unconfinedcompression test is a specialtype of unconsolidated-undrainecltest that is commonly usedfor clay spccimens.In this test,the confiningpressure oj 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 Total stressMohr's I circleat lailure o.r=0 6t=4, Normal stress Figure 11.26 Unconlinedcompression test Chapter 11 Shear Strength of Soil Table 11.4 General Relationship of Consistency and Unconfined Compression Strength of Clays Consistency ton llt2 Very soft 0-25 0-0.25 Soft 25-50 0.25-0.-5 Medium -s0,100 0.5-1 stiff 100-200 1a Very stiff 200-400 2-4 Hard >400 >4 Becausethe undrained shear strength is independentof 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 unconfinedcompression strength. A pho- tograph of unconfinedcompression tcst equipment is shown in Figure 11.27. Theoretically, for similar saturatedclay specimens.the unconfined compres- sion testsand the unconsolidated-undrainedtriaxial testsshould yicld the sameval- uesof ci,.In practice,however, unconfined compression tests on saturatedclays yield slightly lower valuesof c,,than thoseobtained from unconsolidated-undrainedtests. 11.12 Sfress Path Resultsof 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 coversone of them. Lambe (1964) suggesteda type of stresspath 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: o\ * o\ t2nl:" (rr.24) oi*o" .1^f *- ___:-*-__r__-: (i 1.2s) 2 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' againstp' for a consolidated-drained triaxial test on a normally consolidatedclay Chapter 11 Shear Strength of Soil consolidated-drainedtriaxial test. At the beginning of the application of deviator stress,o\ : ot: 03, So -f rr'.' o'"' '2n': U 1 U1 (rt.26) and o\- o\ q- :U (11.27) For this condition,p' and q' will plot as a point (that is, 1 in Figure 11.28).At some othertimeduringdeviatorstressapplication,oi- ot* L,o,1- cjl L,o,1;o\: ot. The Mohr's circle marked A in Figure 11.28corresponds to this stateof stresson the soil specimen.The valuesof p' and q'for this stresscondition are (rri * c', I o', * Aa,) oi -tr'r+ Lo , -(r:+ Lo , P'- (11.28) 2 2 2 2" and (o1 * L,o,1) - o1 Ltt,t q (tr.2e) 2 lf thesevaluesofp'and q'were plottedinFigurell.2B,theywouldberepresented by point D' at the top of the Mohr'.scircle. So, if the valuesof 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 makesan 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 consolidatedclays, the failure envelope can be given by 11= c' t an $' . This is thc line OF in Figure 11 .28. (See also Figure ll .l'7.) A modified fail- ure 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 envelopemakes 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', : gO Iana 11.12 StressPath 343 \ c a or=O't O,o',or/l l.+(Ao,1),+l Figure 17.29 Relationshipbetwccn ry''ancl a and thus, wc obtain ni-qt , 2 o't*rt5 tana - (11.31) ct*o't *--;- o\+o, /. Again, CO' - stn o(y Q 0\-:L sind,: _:__o\-ct (11.32) o\ * tr'1 tr'1 | tr'1 2 ComparingEqs. (11.31) and (11.32), we see thar sin {' : 14no (11.33) r(tan {' - sin a) (11.34) Figure 11.30shows a q'-p' plot for a normalry consolidatedclay specimensub- jected to an isotropicallyconsolidated-undrained triaxial test.At the beginningof the Chapter 11 Shear Strength of Soil s Effective stress Total stress Mohr'scircle Mohr'scircle s 6 u a't I o'1 o, o',or /' Figure 11.30 Stresspath - plot of q' againstp' for a consolidated-undrainedtriaxial test on a normally consolidatedclay applicationofdcviatorstress,ol-tr\-oj.Hence ,p':orandq'- 0.Thisrelation- ship is representedby point 1.At some other stageof the deviator stressapplication, rr\ - tra * Arr,l Ltr,, and o\ :oj- Lu,1 So oi*o\ Ao., p- : c1 * 1,u,, (il.3s) 2 1'- and - , c\ c3 L.tr,1 't22 (11.36) The precedingvalues of p' and q' will plot as point U' in Figure 11.30.Points such as U" representvalues of p' and q' asthe test progresses.At failure of the soil specimen, (Atr,,), p' - (r1+ (Arr,1), (1 1.37) 2- and , (Lor)r ,l- (11.38) 2 The valuesof p'and 4'given by Eqs. (11.37)and (11.38)will plot as point 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) pro- posed a technique to evaluate the elastic and consolidation settlements of founda- tions on clav soilsbv using the stresspaths determined in this manner. 11.12 StressPath For a normally consolidatedclay, the failure envelope is given by the equation Tr ,o' tan $' . The correspondingmodified failure envelope(q,-p, plot) ii given gq' : bI t 11.30)as q' p' tan a. In a simirarmanner, if the f;iru;e enuetopeiJrl : c' * s' lan$' , the correspondingmodified failure envelopeis a q'-p'ploi that ian : be expressedos qrl m * p' tan a. Expressa as a function ofdr, and givem asa function of c' andg' . o-r o, z oj A o'r o'r Norntrl stress *r'' cotQ'-l- I ------..-----*l Figure | 1.31 Derivationof a asa functionof d' and n asa functionof c, and $olution FromFigure 11.31, AB AB (ry) Sln@ E AC* CO+OA c'cotS'* (*#) c,cos + ('i o5),in +: z\z/6' I 4, (a) or Q':m+p'tana (b) ComparingEqs. (a) and (b), we find that m = c'cos@' and tana: sin4' a = tan-1(sind') Chapter 11 Shear Strengthof Soil 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 ap- plied 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 shearstrength of the soil can be calculatedas follows. If I is the maximum torque applied at the head of the torque rod to causefail- ure, 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\-_'YJ Surlacc Moment arca arm where d : diameter of the shearvanc /z : height of thc shearvane For the calculationof M., investigatorshave assumed several types of distribu- ticln of shearstrength mobilization at thc ends of the soil cylinder: l. Triangular.Shear strength mobilization is c,,at the periphcry of the soil cylin- der and decreaseslincarly to zero at thc center. 2, IJni.form.Shear strength mobilization is constant(that is, c,,)from the periph- ery to the center of thc soil cylinder. 3. Parabolic.Shear strength mobilization is c,,at the periphery of the soil cylinder and dccreasesparabolically to zero at the center. Thesevariations in shearstrength 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 (11.41) L "L a-j cu: (rr.42) I a2n ^a31 lrl . + lsTl tLal 11.13 VaneShearTest I h I I I k-,1 --+l Figure 11.32 Diagram ol'vane shear test equlpm(jnt Triangularntobilization of shearstrength Unifbrmmobilization I 0f shearstrergth I '21, -4 'i -t Parabolicfbrm of mobilizationof I shearstrength I I I --, i ------*1--- l<-f t Figure 1 1.33 Derjvation of Eq. (11.42): (a) resisting moment of shear force; (b) variations in shearstrength mobilization 348 Chapter 11 Shear Strength of Soil where B : \for triangular mobilization of undrained shearstrength B : I for uniform mobilization of undrained shearstrength B : I for parabolic mobilization of undrained shearstrength Note that Eq. (11.42)is usuallyreferred to as Culding'sequtttion. Vane sheartests can be conductedin the laboratory and in the field during soil exploration. The laboratory shearvane has dimensionsof about 1-1mm (j in.) in di- ameter 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 1.5gives the ASTM recommendeddimensions of field vanes. Accordingto ASTM (1994).if h/d :2. thcn ( 11.43) and (11.44) t (in ) In the field.whcre considcrablcvariation in thc undrainedshear strength can be found with depth, vaneshear tcsts are extrcmcly useful.In a short period, one can establisha reasonablcpattern ol'thc changeof c',,with depth. However,if the clayde- posit at a given site is more or lessunif'orm, a few unconsolidated-undrainedtriaxial Figure 1 1.34 Laboratory vane sheartest devicc (courtesy of Soiltest,Inc., Lake Bluff, Illinois) | 1.13 VaneShear Test 1,,t< | +s) I I I I +I .lr":f -+;t ri++ ll +i+*i1.i,..t:i*- .ii iil I li Rectangularvane Taperedvane Figure 11.35 ceometry of field vanes[source : FromAnnual Book o.fASTM standards, 04'08'p. 346.Copyright O 1994American Society for Testingand Materials. Reprinted with permissionl Table11.5 RecommendedDimensions of FieldVanes*,, Thickness Diameter Diameter, Height, of blade, of rod Casingsize mm (in.) mm (in.) mm (in.l mm (in.) AX 38.r(r j) 76.2(3) 1 6 (*r) LL.t \1) BX -50.8(2) 101.6(4) 1a? / l\ 16(+) t1.t \t) NX $5 (2)) 127.0(s) 32(i) t2.7(+) 10r.6mm (4in.)1' 92.1(3;) 1a 1 t !\ 184.1(71) 3.2(*) LL.t \2) *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 Chapter 11 Shear Strength of Soil tests on undisturbed specimenswill allow a reasonableestimation 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 rea- son, he suggestedthe correction : ca(dcsign) iCr(uane shear) (11.4s) where i : correction factor : l.l - 0.54log(P1) (11.46) P7 : plasticityindex More recently,Morris and Williams (199a)gave the correlationsof ,\ as i : 1.13n{){)n(i'l) + 0'57 (for P/ > ,5) (11.47) and n : 7.01e tttts(r'r')1 9.57 (for LL > 20) (11.48) where LL : liquid limit (%). | 1.14 Other Methods for Determining Undrained Shear Strength A modified form of the vaneshear test apparatus is the Torvane(Figure I 1.36),which is a handheld devicewith a calibratedspring. This instrument can be usedfor deter- mining c,,for tube specimenscollected from the lield during soil exploration, andit can be usedin the field.The Torvane is pushedinto the soil and then rotated until the soil fails.The undrained shearstrength can be read at the top of the calibrateddial. Figure 11.37 shows a pocket penetrometer,which is pushed directly into the soil. The unconfined compressionstrength (q,,)is measuredby a calibrated spring.This devicecan 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 (courtesyof Soiltest.lnc.. Lake Bluff, Illinois) Chapter 11 Shear Strengthof Soil Mrdiutnquic* E l6 "l ..; Slightly quick ; 'a5d q Very sensitive Mediurr sensitive Slightlysensitive Axlal straln Inscns itivc Figure 11.38 L.lnconfinedcomprcssion stre ngth Figure ?1.39 Classificationo[ claysbascd on for undisturbcd and rcmoldcd clay scnsitivity the moisturecontent, as shownin Figure I 1.3U.This propcrty of clay soilsis called 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:-._.-- (I 1.4e) 4rr(remolded) The sensitivityratio of most claysranges from about 1 to 8; however,highly floc- culent marine clay depositsmay have sensitivityratios ranging from about 10 to 80. Some clays turn to viscousfluids upon remolding. These claysare found mostly in the previouslyglaciated areas of North America and Scandinavia.Such clays are re- ferred to as quick clays.Rosenqvist (1953) classified clays on the basisof their sensi- tivity. This generalclassification is shown in Figure 1 1.39. The lossof strengthof clay soilsfrom remolding is causedprimarily by the de- struction of the clay particle structurethat wasdeveloped during 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 ma- terials are allowed to rest.This phenomenon is illustratedin Figure 11.40a. Most soils,however, are partially thixotropic - that is,part of the strengthloss causedby remolding is never regained with time. The nature of the strength-time variation for partially thixotropic materialsis shown in Figure 11.40b.For soils,the differencebetween the undisturbedstrength and the strengthafter thixotropic hard- ening can be attributed to the destructionof the clay-particlestructure that was de- veloped during the original processof sedimentaticln. \=-,. 4a(undisturbed) {a(renrolded) {u(undisturbed ) Initial und i st urbed strength clJ E q € € € z E /---" c E 6"'& E $t E d, ./a(re Irolded ) g a ! ,/*** c r,b" + 6* E €E (b) Figure 11.40 Behavior of (a) thixotropic marerial; (b) partially thixotropic material 354 Chapter 11 Shear Strength of Soil Table 11.6 Empirical EquationsRelatcd tt't c,, and o'6 Reference Relationship Remarks L rrlVSTl Skempton(1957) +:0.11+0.0037(Pl) For normally consolidated CO clay P/ : plasticityindex (%) c,(VSr)- undraincd shear strengthfrcnt vane sheartcst 1+: be used in overconsoli- Chandlcr(l9tttt) ', o.u+ o.oo37(p1) Can (f diited soil; accur€rcy+257"; rri : preconsolidationpressurc not valid for sensitiveand fissured clays :1 -0.23+0.04 For lightly overconsolidated Jamiolkowski ct al. ( lt)tt5) ', (f clays Mesri ( 19t39) o.zz ;:(fo /.,\ \ - /,,",,,,,,,.,,,,,,,,,, Ladd ct al. (1977\ : (ocR)'iE /(;,\ \a/,,,,,,,,,",ons.,i(r,,c(r OC R : ovcrconsolidationratio 11.16 Empirical Relationships between Undrained Cohesion (cu)and Effective Overburden Pressure (ob) Severalempirical relertionshipscan be obscrvcd between c,,and the effcctiveover- burden pressure(o'a1) in the field. Somc of these relationshipsare summarizedin Table I 1.6. The overconsolidationratio wasdcfinod in Chapter 10as o,, OCR : (11.s0) tt6 whereoj. : preconsolidationpressure. 10.8 A soilprofile is shownin Figure11.41. The clayis normallyconsolidated. Its liquid limit is 60 andits plasticlimit is 25.Estimate the unconfinedcompression strength of the clayat a depthof l0 m measuredfrom the groundsurface. Use Skemptonb relationshipfrom Table I L6 andEqs. (11.45) and (lL46). Relationshipsbetween Undrained Cohesion (c) and Effective Overburden pressure(crL) Dry sand Y= 15.5kN/m: Clay ty - 30Vc G, = 2.68 Figure 11.41 Solution For the saturatedclay layer, the void ratio is e = u)Gs: (2.6BX0.3): 0.8 The effectiveunit weight is . /c.-l\ (2.68-l)(e.81) Tiroy:\,*"i,,:__ :9.r6kN/mr The effectivestress at a depth of 10m from the groundsurface is qb : 37"ond+ Tylrov:(3X15.5) + (7X9.16) = 110.62kN/m2 From Table11.6. ca(vST) * 0.11+ 0,0037(P1) o'p Ca(VST) 0.11+ 0.0037(60- 25) n'Sr= 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 - : 11.7 0.54lo9(60* 2s)J26.49: 22.95 kN/mz Sothe unconnned "T':';;:',;eT;:, : 45.ekN/m2 356 Chapter 11 ShearStrength of Soil 11.17 Shear Strength of Unsaturated Cohesive Soils The equation relating total stress,effective stress, and pore water pressurefor un- saturatedsoils, can be expressedas (r': - (, uul X(uu- uu,) (11.s1) where o' : effectivestress o : total stress rt. : pore air pressure u?i,: pore water pressure When the _ expressionfor o' is substituteclinto the shear strength equation [Eq. ( I 1.3)1, which is bascdon effectivestress parameters, we get : - - r[ c' + [rr u,, * y(u,, rr,,,)]tan{' (r r.s2) The valuesof 1 depend primarily on the degreeof saturation.with ordinary triaxial equipment uscd for laboratory testing,it is not possibleto determine accu- rately the effectivestrcsses in unsaturatedsoil specimens,so the common practiceis to conduct undrained triaxial testson unsaturateclspecimens ancl measure only the total stress.Figure 11.42shows a total strcssfailure envelopeobtained from a num- ber 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 designpurposes, the curveclenvclopc is sometimesapproximateil as a straight fine, as shown in Figurc 11.42,wtthan equation as follows; rt: r'f otanrp (11.53) (Note that c and rf in the precedingcquation are empirical constants.) a Normal stress(total) 'fotal Figure 1 1.42 stressfailure envelope for unsaturated cohesivesoils 11.18 Summary and GeneralComments 357 o (lb/in2) 160 320 1000 160 Inorganicclay (CL) ? 8oo z Degrceof saturation= tr 600 .< Rfl_'! ; .+00 = ,+0 5, 200 0d 500 1000 2(XX)22(X) Totalnonlal strcss. o (kN/rnrt Figure11.43Yariati Figure 11.43shows the variationof thc total stresscnvelopes with changeof the initial degreeof saturationobtaincd from undrainedtests on an inorganicclay. Note that for thesetests the specimcnswerc preparedwith approximatelythe same initialdry unit weightof about 16.7kN/mr ( 106Ibi mr). For a givcntotal normal strcss, tl.reshear stressneeded to causefailure dccreascsas the dcgrce of saturalion in- creases.When thc degrecof saturationreaches 100'2,. the total stressl'ailurc enve- lope becomesa horizontal line thzrtis the samc as with thc r/ : 0 conccpt. In practicalcases where a cohesivcsoil dcposit may bccomesaturatecl bccause of rainfallor a rise in thc groundwatcrtablc, the strengthof partiallysaturatcd clay should not be used for designconsiderations. Instead, the unsaturatedsoil speci- 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 exam- ined. Laboratory proceduresfor determining the shcar strength paramcters were described. In textbooks,determination of the shearstrength parametersof cohesivesoils appears to be fairly simple. However, in practice, the proper choice of these pa- rametersfor designand stability checksof various earth, earth-retaining,and earth- supported structures is very difllcult and requires experienceand an appropriate theoretical background in geotechnicalengineering. In this chapter,three types of strength parameters (consolidated-drained,cctnsolidated-undrainecl, anrJ unconsoli- dated-undrained)were introduced. Their use dependson drainageconditions. Consolidated-drainedstrength parameters can be usedto determine the long- term stability of structuressuch as earth embankmentsand cut slopes.Consolidated- undrainedshear strength parameters can be usedto study stabilityproblems relating to caseswhere the soil initially is fully consolidatedand then there is rapid loadine. 3s8 Chapter 11 Shear Strength of Soil Nonnally consolidated anisotropicclay . Saturated ciay , ,r.., (a) Figure 11.44 Strengthanisotropy in clay An excellentexample of this is the stability of slopesof carth dams after rapid draw- down. The unconsolidatcd-undrainedshear strength of clayscan be usedto cvaluate the end-of-constructionstability of saturatedcohesive soils with 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-undrainedshear strength of somesaturatcd clays can vary dependingon 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 subsequentconsolidation makes the clay particlesorient per- pendicular to the direction of thc major principal stress.Parallel orientation of the clay particlescan causethe strcngthof clay to vary with direction.Figure 11.44a showsan elcmcnt of saturatedclay in a depositwith the major principal stressmak- ing an anglea with respectto the horizontal. For anisotropicclays, the magnitudeof c,,will be a lunction of a. For normally consolidatedclays, c,,1,*-e11,,, ) {:,,1,,:1r"1, for over- consolidatcdclays, c,,,,, q..) ( c,,(,,,,.,. Figure 11.44bshows the directional variation for c,,1,,;.The anisotropy with rcspcct to strength for clayscan have an important ef- fect on the load-bearingcapacity of foundations and the stability of earth embank- ments becausethe direction of thc major principal stressalong the potential failure surfaceschanges. The sensitivityof clayswas discusscdin Section 11.15. It is imperativethat sen- sitiveclay 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 sheartest was conductedon a specimenof dry sandwith a normal stressof 200kN/m2. Failure occurrecl at a shearstress of 175kN/m2. The size 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 shearforce would be required to causefailure 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 150kN/m2 was applied.Failure occurred at a shearstress of 110kN/m2. What was the massof the sand soecimen? 11.3 Following are the resultsof four drained direct sheartests on a normally consolidatedclay: Size of specimen : 60 mm x 60 mm Height of specimen : 30 mm Normal Shear Test force force at no. {N} failure(N} I 200 1.5-5 2 300 230 3 400 3l0 /1 500 385 Draw a graph for the shearstress at failure againstthe normal stressand de- termine the drained angle of friction (r/->')from the graph. rt.4 Following are the resultsof four drained direct shcar testson a normally consolidatedclay: Specimensize: diameterof specinten: 2in. hcight of spccir.r.ren- I in. Normal Shear Test force force at no. (lb) failure (tbl I 60 3'7.5 2 90 -5-5 3 ll0 70 A lll u0 Draw a graph for shearstress at failure againstthe normaI stressand deter- mine the drained angle of friction ({,) from the graph. 11.5 The equation of the effectivestress failure envelopefor a loosesandy soil was obtained from a direct sheartest as r 1.: c' tan 30o.A drained triaxial test wasconducted with the samesoil at a chamberconfining pressure of 10 lb/in.2 a. Calculatethe deviator stressat failure. b. Estimate the angle that the failure plane makes with the major principal plane. c. Determine the normal stressand shearstress (when the specimenfailed) on a plane that makes an angleof 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 relativedensity D, andthe 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 compactionwas 65%. Calculatethe maior principal stressat 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 tri- axial test, the specimenfailed at a deviator stressof 175kN/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 cham- ber confining pressure,oi? 11.10 A consolidated-drainedtriaxial test was conductedon a normally cclnsoli- dated clay.The resultswere as follows: (r'r1 250kNi m2 (Arr,,),: 275kN/mr 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 shearstrcss, 11, on the failure plane 11.11 The resultsof two drained triaxial tcstson a saturatedclay are as follows; SpecimenI: chamberconfining prcssurc : 70 kN/mr deviatorstrcss at failure - 215kN/m2 SpecimenII: chambcrconfining pressure : 120kN/m') deviator stressat failurc : 260 kN/m' Calculatcthe shcar strength parametersof the soil. 11.72 lf a spccimenof clay describedin Problem l 1.11 is testedin a triaxial appa- ratus with a chamber confining prossureof 20()kN/mr, what will be the ma- jor principal stressat failurc'/ Assume full drained condition during the test. 11.13 A sandysoil has a drained angle of friction of 35'. In a drained triaxial test on the samesoil.lhe dcviator strcss at failureis 2.69 ton/ft2. What isthe chamber confining pressure'? 11.14 A deposit of sand is shown in Figure 11.45.Find the maximum shearre- sistancein 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 : Deviator stress,(Lo,) r 11lb/in.2 Pore pressure,(Aa,)1 -- 1.2lblin,1 Calculatethe consolidated-undrainedfriction angle and the drained friction angle. 11.16 Repeat Problem 11.15,using the following values: o3 - 140kN/m2 (Ao,i), : 125kN/m2 (Lu,,)i: 75kN/m2 Problems 361 Ground water table Sand Q'= 42" e =0.6 Gs=2.67 Figure 11.45 11.17 The shearstrength of a normally consolidatedclay can be given by the equa- tion 11- o' tan 31'. A consolidated-undrainedtest was conductedon the clay.Following are the resultsof the tcst: Chamberconfining pressure : 112kN/mr 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 specimendescribed in Problem I L 17,what would have been the deviator stressat failure if a drained test would have been conductedwith the samechamber confining pressure (i.e., oj : 112kN/m2)? 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 conductedat 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 a3 rr, at failure no. (kN/mr) (kN/mr) t92 375 384 636 Draw the total stressMohr's circles and determine the shear strength param- etersfor consolidatedundrained conditions (i.e.,d and c). 362 Chapter 11 Shear Strength of Soil 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 unconfinedcompression strength, q,,, of a similar specimenwas found to be 100kN/m2. Determine the pore water pressureat failure for the uncon- fined compressiontest. 11.24 Repeat Problem I1.23 using the following values: $' - 23o q,, - 120kN/m' 11.25 The resultsol two consolidated-drainedtriaxial tcstson a clavevsoil are as follows: (r'l(failure) Test no. u\ (lb lin.2l (lb/in.'zl I z7 2 IL 4n Use thc failure envelopeequation givcn in Example I 1.7- that is, 11': v11a p' Lana. (Do not plot thc graph.) 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 sheartest at a depth of 6 m belclwthe ground surface.Use Skempton'scquation in Table 11.6. ''""1,- 3m 't'* l,']. :"l .',l.l '"15n Clay Y ,ar= 19.5kN/mr Figure 11.46 References 363 References Acan, Y. B., Drrr