Triaxial Compression

12.8 Consolidated-Drained Triaxial Test 385

Hence,

(12.23)

Once the value of f1 is known, we can obtain c as

(12.24)

A consolidated-drained triaxial test on a clayey soil may take several days to complete.This amount of time is required because deviator stress must be applied very slowly to ensurefull drainage from the soil specimen. For this reason, the CD type of triaxial test is uncommon.

c s1112 s3112 tan2 a45 f12 b

2 tan a45 f12b

f1 2 e tan1 cs1112 s1122s3112 s3122 d 0.5 45 f

Example 12.2

A consolidated-drained triaxial test was conducted on a normally consolidated clay.The results are as follows:

s3 16 lb/in.2 (sd)f 25 lb/in.2

Determine

a. Angle of friction, fb. Angle u that the failure plane makes with the major principal plane

SolutionFor normally consolidated soil, the failure envelope equation is

For the triaxial test, the effective major and minor principal stresses at failure are as follows:

and

Part aThe Mohrs circle and the failure envelope are shown in Figure 12.24. From Eq. (12.19),

or

f 26

sin f s1 s

3

s1 s3

41 1641 16

0.438

s3 s3 16 lb/in.2

s1 s1 s3 1sd 2f 16 25 41 lb/in.2tf s tan f 1because c 0 2

386 Chapter 12: Shear Strength of Soil

Figure 12.24 Mohrs circle and failure envelope for a normally consolidated clay

Part bFrom Eq. (12.4),

u 45 f2

45 262

58

Normal stresss3 16 lb/in2 s1 41 lb/in2

Shea

r stre

ss

O

B

2u

fEffective stress failure envelope

A

s3s3

s1

s1

u

Example 12.3

Refer to Example 12.2.

a. Find the normal stress s and the shear stress tf on the failure plane.b. Determine the effective normal stress on the plane of maximum shear stress.

SolutionPart aFrom Eqs. (10.8) and (10.9),

and

Substituting the values of s1 41 lb/in.2, s3 16 lb/in.2, and u 58 into the precedingequations, we get

s 41 16

2

41 162

cos 12 58 2 23.0 lb/in.2tf

s1 s3

2 sin 2u

s 1on the failure plane 2 s1 s32

s1 s

3

2 cos 2u

12.8 Consolidated-Drained Triaxial Test 387

and

Part bFrom Eq. (10.9), it can be seen that the maximum shear stress will occur on the planewith u 45. From Eq. (10.8),

Substituting u 45 into the preceding equation gives

s 41 16

2

41 162

cos 90 28.5 lb/in.2

s s1 s

3

2s1 s

3

2 cos 2u

tf 41 16

2 sin 12 58 2 11.2 lb/in.2

Example 12.4

The equation of the effective stress failure envelope for normally consolidated clayey soilis tf s tan 30. A drained triaxial test was conducted with the same soil at a chamber-conning pressure of 10 lb/in.2 Calculate the deviator stress at failure.

SolutionFor normally consolidated clay, c 0. Thus, from Eq. (12.8),

So,

1sd 2f s1 s3 30 10 20 lb/in.2s1 10 tan2 a45 302 b 30 lb/in.2f 30

s1 s3 tan

2 a45 f2b

Example 12.5

The results of two drained triaxial tests on a saturated clay follow:

Specimen I:

1sd 2f 130 kN/m2s3 70 kN/m2


Specimen II:

Determine the shear strength parameters.

SolutionRefer to Figure 12.25. For Specimen I, the principal stresses at failure are

s3 70 kN/m2

and

Similarly, the principal stresses at failure for Specimen II are

and

Now, from Eq. (12.23),

Figure 12.25 Effective stress failure envelope and Mohrs circles for Specimens I and II

Normal stress, s (kN/m2)

Shea

r stre

ss (k

N/m2

)

c

383.520016070

f

f1 2 e tan1 cs11I2 s11II2s31I2 s31II2 d 0.5 45 f 2 e tan1 c 200 383.570 160 d 0.5 45 f 20s1 s1 s3 1sd 2f 160 223.5 383.5 kN/m2

s3 s3 160 kN/m2

s1 s1 s3 1sd 2f 70 130 200 kN/m2s3

1sd 2f 223.5 kN/m2s3 160 kN/m2

12.9 Consolidated-Undrained Triaxial Test 389

Again, from Eq. (12.24),

c s11I2 s31I2 tan2 a45 f12 b

2 tan a45 f12b

200 70 tan2 a45 202b

2 tan a45 202b 20 kN/m2

12.9 Consolidated-Undrained Triaxial Test

The consolidated-undrained test is the most common type of triaxial test. In this test, thesaturated soil specimen is rst consolidated by an all-around chamber uid pressure, s3,that results in drainage (Figures 12.26a and 12.26b). After the pore water pressure gener-ated by the application of conning pressure is dissipated, the deviator stress, sd, on thespecimen is increased to cause shear failure (Figure 12.26c). During this phase of the test,the drainage line from the specimen is kept closed. Because drainage is not permitted, thepore water pressure, ud, will increase. During the test, simultaneous measurements ofsd and ud are made. The increase in the pore water pressure, ud, can be expressed ina nondimensional form as

(12.25)

where Skemptons pore pressure parameter (Skempton, 1954).The general patterns of variation of sd and ud with axial strain for sand and clay

soils are shown in Figures 12.26d through 12.26g. In loose sand and normally consolidatedclay, the pore water pressure increases with strain. In dense sand and overconsolidatedclay, the pore water pressure increases with strain to a certain limit, beyond which itdecreases and becomes negative (with respect to the atmospheric pressure). This decreaseis because of a tendency of the soil to dilate.

Unlike the consolidated-drained test, the total and effective principal stresses are notthe same in the consolidated-undrained test. Because the pore water pressure at failure ismeasured in this test, the principal stresses may be analyzed as follows:

In these equations, (ud)f pore water pressure at failure. The preceding derivationsshow that

s1 s3 s1 s

3

Minor principal stress at failure 1effective 2 : s3 1ud 2f s3 Minor principal stress at failure 1total 2 : s3 Major principal stress at failure 1effective 2 : s1 1ud 2f s1 Major principal stress at failure 1total 2 : s3 1sd 2f s1

A

A udsd


Example 12.6

A specimen of saturated sand was consolidated under an all-around pressure of 12 lb/in.2The axial stress was then increased and drainage was prevented. The specimen failedwhen the axial deviator stress reached 9.1 lb/in.2 The pore water pressure at failure was6.8 lb/in.2 Determine

a. Consolidated-undrained angle of shearing resistance, fb. Drained friction angle, f

SolutionPart aFor this case, s3 12 lb/in.2, s1 12 + 9.1 21.1 lb/in.2, and (ud)f 6.8 lb/in.2. Thetotal and effective stress failure envelopes are shown in Figure 12.30. From Eq. (12.27),

Part bFrom Eq. (12.28),

Figure 12.30 Failure envelopes and Mohrs circles for a saturated sand

f

f

BB

A5.2

Shea

r stre

ss (l

b/in2

)

Normal stress (lb/in2)12 14.3 21.1A

Effective stress failure envelope

Total stress failure envelope

f sin1 c s1 s3s1 s3 21ud 2f d sin1 c 21.1 1221.1 12 12 2 16.8 2 d 27.8

f sin1 as1 s3s1 s3

b sin1 a 21.1 1221.1 12

b 16

12.10 Unconsolidated-Undrained Triaxial Test 395

12.10 Unconsolidated-Undrained Triaxial Test

In unconsolidated-undrained tests, drainage from the soil specimen is not permitted during theapplication of chamber pressure s3. The test specimen is sheared to failure by the applicationof deviator stress, sd, and drainage is prevented. Because drainage is not allowed at any stage,the test can be performed quickly. Because of the application of chamber conning pressures3,the pore water pressure in the soil specimen will increase by uc. A further increase in the porewater pressure (ud) will occur because of the deviator stress application. Hence, the total porewater pressure u in the specimen at any stage of deviator stress application can be given as

(12.31)From Eqs. (12.18) and (12.25), uc Bs3 and ud sd, so

(12.32)

This test usually is conducted on clay specimens and depends on a very importantstrength concept for cohesive soils if the soil is fully saturated. The added axial stress at fail-ure (sd)f is practically the same regardless of the chamber conning pressure. This propertyis shown in Figure 12.31. The failure envelope for the total stress Mohrs circles becomes ahorizontal line and hence is called a f 0 condition. From Eq. (12.9) with f 0, we get

(12.33)where cu is the undrained shear strength and is equal to the radius of the Mohrs circles.Note that the f 0 concept is applicable to only saturated clays and silts.

tf c cu

u Bs3 Asd Bs3 A1s1 s3 2A

u uc ud

Example 12.7

Refer to the soil specimen described in Example 12.6. What would be the deviatorstress at failure, (sd)f, if a drained test was conducted with the same chamber all-around pressure (that is, 12 lb/in.2)?

SolutionFrom Eq. (12.8) (with c 0),

12 lb/in.2 and f 27.8 (from Example 12.6). So,

1sd 2f s1 s3 33 12 21 lb/in.2s1 12 tan2 a45 27.82 b 33 lb/in.2

s3

s1 s3 tan

2 a45 f2b

12.18 Stress Path 417

O s, s, or p

Shea

r stre

ss, o

r q

U

UU

ud

F

a

Effective stress Mohrs circle

Total stress Mohrs circle

s1s3

1s3

I

Figure 12.55 Stress pathplot of q against p for a consolidated-undrained triaxial test on a normally consolidated clay

and

(12.63)

The preceding values of p and q will plot as point U in Figure 12.55. Points such as Urepresent values of p and q as the test progresses. At failure of the soil specimen,

(12.64)

and

(12.65)

The values of p and q given by Eqs. (12.64) and (12.65) will plot as point U.Hence, the effective stress path for a consolidated-undrained test can be given by the curveIUU. Note that point U will fall on the modied failure envelope, OF (see Figure 12.54),which is inclined at an angle a to the horizontal. Lambe (1964) proposed a technique toevaluate the elastic and consolidation settlements of foundations on clay soils by using thestress paths determined in this manner.

q 1sd 2f

2

p s3 1sd 2f

2 1ud 2f

q s1 s

3

2sd

2

Example 12.9

For a normally consolidated clay, the failure envelope is given by the equation tf stan f. The corresponding modied failure envelope (q-p plot) is given by Eq. (12.57)as q p tan a. In a similar manner, if the failure envelope is tf c s tan f, thecorresponding modied failure envelope is a q-p plot that can be expressed as q m p tan a. Express a as a function of f, and give m as a function of c and f.


SolutionFrom Figure 12.56,

So,

(a)

or

(b)Comparing Eqs. (a) and (b), we nd that

and

or

a tan11sin F 2tan a sin f

m c cos F

q m p tan a

s1 s3

2 c cos f as1 s3

2b sin f

sin f AB

AC

AB

CO OA

as1 s32

bc cot f as1 s3

2b

Normal stressC

O

Shea

r stre

ss

A

B

ftf c s' tan f

c

s1 s3

2

s1' s3'

2

c cot fs1s3

Figure 12.56 Derivation of a as a function of f and m as a function of c and f

12.19 Summary and General Comments

In this chapter, the shear strengths of granular and cohesive soils were examined. Laboratoryprocedures for determining the shear strength parameters were described.

In textbooks, determination of the shear strength parameters of cohesive soilsappears to be fairly simple. However, in practice, the proper choice of these parameters fordesign and stability checks of various earth, earth-retaining, and earth-supported structures

Problems 419

is very difcult and requires experience and an appropriate theoretical background in geot-echnical engineering. In this chapter, three types of strength parameters (consolidated-drained, consolidated-undrained, and unconsolidated-undrained) were introduced. Theiruse depends on drainage conditions.

Consolidated-drained strength parameters can be used to determine the long-termstability of structures such as earth embankments and cut slopes. Consolidated-undrainedshear strength parameters can be used to study stability problems relating to cases wherethe soil initially is fully consolidated and then there is rapid loading. An excellent exampleof this is the stability of slopes of earth dams after rapid drawdown. The unconsolidated-undrained shear strength of clays can be used to evaluate the end-of-construction stabilityof saturated cohesive soils with the assumption that the load caused by construction hasbeen applied rapidly and there has been little time for drainage to take place. The bearingcapacity of foundations on soft saturated clays and the stability of the base of embank-ments on soft clays are examples of this condition.

The unconsolidated-undrained shear strength of some saturated clays can varydepending on the direction of load application; this is referred to as anisotropy with respectto strength. Anisotropy is caused primarily by the nature of the deposition of the cohesivesoils, and subsequent consolidation makes the clay particles orient perpendicular to thedirection of the major principal stress. Parallel orientation of the clay particles can causethe strength of clay to vary with direction. The anisotropy with respect to strength for clayscan have an important effect on the load-bearing capacity of foundations and the stabilityof earth embankments because the direction of the major principal stress along the poten-tial failure surfaces changes.

The sensitivity of clays was discussed in Section 12.13. It is imperative that sensi-tive clay deposits are properly identied. For instance, when machine foundations (whichare subjected to vibratory loading) are constructed over sensitive clays, the clay may loseits load-bearing capacity substantially, and failure may occur.

Problems

12.1 For a direct shear test on a dry sand, the following are given: Specimen size: 75 mm 75 mm 30 mm (height) Normal stress: 200 kN/m2 Shear stress at failure: 175 kN/m2a. Determine the angle of friction, fb. For a normal stress of 150 kN/m2, what shear force is required to cause failure

in the specimen?12.2 For a dry sand specimen in a direct shear test box, the following are given:

Angle of friction: 38 Size of specimen: 2 in. 2 in. 1.2 in. (height) Normal stress: 20 lb/in.2Determine the shear force required to cause failure.

12.3 The following are the results of four drained, direct shear tests on a normallyconsolidated clay. Given: Size of specimen 60 mm 60 mm Height of specimen 30 mm


Normal Shear Test force force at no. (N) failure (N)

1 200 1552 300 2303 400 3104 500 385

Draw a graph for the shear stress at failure against the normal stress, and determine the drained angle of friction from the graph.

12.4 Repeat Problem 12.3 with the following data. Given specimen size: Diameter 2 in. Height 1 in.

Normal Shear Test force force at no. (lb) failure (lb)

1 60 37.52 90 553 110 704 125 80

12.5 The equation of the effective stress failure envelope for a loose, sandy soil wasobtained from a direct shear test at tf s tan 30. A drained triaxial test wasconducted with the same soil at a chamber conning pressure of 10 lb/in.2.Calculate the deviator stress at failure.

12.6 For the triaxial test described in Problem 12.5:a. Estimate the angle that the failure plane makes with the major principal

plane.b. Determine the normal stress and shear stress (when the specimen failed) on a

plane that makes an angle of 30 with the major principal plane. Also, explainwhy the specimen did not fail along the plane during the test.

12.7 The relationship between the relative density, Dr, and the angle of friction, f, ofa sand can be given as f 25 0.18Dr (Dr is in %). A drained triaxial test onthe same sand was conducted with a chamber-conning pressure of 18 lb/in.2. Therelative density of compaction was 60%. Calculate the major principal stress atfailure.

12.8 For a normally consolidated clay, the results of a drained triaxial test are asfollows. Chamber conning pressure: 15 lb/in.2 Deviator stress at failure: 34 lb/in.2Determine the soil friction angle, f.

12.9 For a normally consolidated clay, f 24. In a drained triaxial test, thespecimen failed at a deviator stress of 175 kN/m2. What was the chamberconning pressure, ?

12.10 For a normally consolidated clay, f 28. In a drained triaxial test, thespecimen failed at a deviator stress of 30 lb/in.2. What was the chamber conningpressure, ?s3

s3

Problems 421

12.11 A consolidated-drained triaxial test was conducted on a normally consolidatedclay. The results were as follows:

Determine:a. Angle of friction, fb. Angle u that the failure plane makes with the major principal planec. Normal stress, s, and shear stress, tf , on the failure plane

12.12 The results of two drained triaxial tests on a saturated clay are given next:Specimen I: Chamber conning pressure 15 lb/in.2

Deviator stress at failure 31.4 lb/in.2Specimen II: Chamber-conning pressure 25 lb/in.2

Deviator stress at failure 47 lb/in.2Calculate the shear strength parameters of the soil.

12.13 If the clay specimen described in Problem 12.12 is tested in a triaxial apparatus with achamber-conning pressure of 25 lb/in.2, what is the major principal stress at failure?

12.14 A sandy soil has a drained angle of friction of 38. In a drained triaxial test on thesame soil, the deviator stress at failure is 175 kN/m2. What is the chamber-conning pressure?

12.15 A consolidated-undrained test on a normally consolidated clay yielded the follow-ing results: s3 15 lb/in.2 Deviator stress: (sd)f 11 lb/in.2

Pore pressure: (ud)f 7.2 lb/in.2Calculate the consolidated-undrained friction angle and the drained friction angle.

12.16 Repeat Problem 12.15 with the following:

12.17 The shear strength of a normally consolidated clay can be given by the equationtf s tan 31. A consolidated-undrained triaxial test was conducted on theclay. Following are the results of the test: Chamber conning pressure 112 kN/m2 Deviator stress at failure 100 kN/m2Determine:a. Consolidated-undrained friction angleb. Pore water pressure developed in the clay specimen at failure

12.18 For the clay specimen described in Problem 12.17, what would have been thedeviator stress at failure if a drained test had been conducted with the samechamber-conning pressure (that is, s3 112 kN/m2)?

12.19 For a normally consolidated clay soil, f 32 and f 22. A consolidated-undrained triaxial test was conducted on this clay soil with a chamber-conning pres-sure of 15 lb/in.2. Determine the deviator stress and the pore water pressure at failure.

12.20 The friction angle, f, of a normally consolidated clay specimen collected duringeld exploration was determined from drained triaxial tests to be 25. Theunconned compression strength, qu, of a similar specimen was found to be 100 kN/m2. Determine the pore water pressure at failure for the unconnedcompression test.

1ud 2f 75 kN/m21sd 2f 125 kN/m2s3 140 kN/m2

1sd 2f 275 kN/m2s3 250 kN/m2


12.21 Repeat Problem 12.20 using the following values.

12.22 The results of two consolidated-drained triaxial tests on a clayey soil are as follows.

Test no. (lb/in.2) (lb/in.2)

1 27 732 12 48

Use the failure envelope equation given in Example 12.9that is, q m ptan a. (Do not plot the graph.)a. Find m and ab. Find c and f

12.23 A 15-m thick normally consolidated clay layer is shown in Figure 12.57. Theplasticity index of the clay is 18. Estimate the undrained cohesion as would bedetermined from a vane shear test at a depth of 8 m below the ground surface. Use Eq. (12.35).

S11failure2S3qu 120 kN/m2f 23

g 16 kN/m3

gsat 18.6 kN/m3

Rock

Groundwater table

15 m

3 m

Dry sand Clay Figure 12.57

ReferencesACAR, Y. B., DURGUNOGLU, H. T., and TUMAY, M. T. (1982). Interface Properties of Sand,

Journal of the Geotechnical Engineering Division, ASCE, Vol. 108, No. GT4, 648654.ARMAN, A., POPLIN, J. K., and AHMAD, N. (1975). Study of Vane Shear, Proceedings,

Conference on In Situ Measurement and Soil Properties, ASCE, Vol. 1, 93120.AMERICAN SOCIETY FOR TESTING AND MATERIALS (2004). Annual Book of ASTM Standards,

Vol. 04.08, Philadelphia, Pa.BISHOP, A. W., and BJERRUM, L. (1960). The Relevance of the Triaxial Test to the Solution of

Stability Problems, Proceedings, Research Conference on Shear Strength of Cohesive Soils,ASCE, 437501.

BJERRUM, L. (1974). Problems of Soil Mechanics and Construction on Soft Clays, NorwegianGeotechnical Institute, Publication No. 110, Oslo.

Triax 1Triax 2Triax 3Triax 4

Triaxial Compression

Documents

Transcript of Triaxial Compression