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12.8 Consolidated-Drained Triaxial Test

385

Hence, f1œ


2 e tan

1

c

œ œ  s1122 s1112 œ s3112



œ s3122

d

0.5

 45° f

(12.23)

Once the value of f1 is known, we can obtain c as sœ1112  sœ3112 tan2 a 45  c¿


2 tan a 45 

fœ1 b 2

fœ1 b 2

(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 ensure full drainage from the soil specimen. For this reason, the CD type of triaxial test is uncommon.

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, f b. Angle u that the failure plane makes with the major principal plane Solution For normally consolidated soil, the failure envelope equation is tf
s¿ tan f¿

1because c¿
0 2

For the triaxial test, the effective major and minor principal stresses at failure are as follows: s1œ
s1
s3  1¢sd 2 f
16  25
41 lb/in.2 and s3œ
s3
16 lb/in.2 Part a The Mohr’s circle and the failure envelope are shown in Figure 12.24. From Eq. (12.19), sin fœ


s1œ  s3œ 41  16
0.438 œ œ
s1  s3 41  16

or f¿
26ⴗ

386 Chapter 12: Shear Strength of Soil s1

Shear stress

u s3

s3

Effective stress failure envelope

f

B s1 2u

O

s3
16 lb/in2

s1
41 lb/in2

A

Normal stress

Figure 12.24 Mohr’s circle and failure envelope for a normally consolidated clay

Part b From Eq. (12.4), u
45 

f¿ 26°
45° 
58ⴗ 2 2

■

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. Solution Part a From Eqs. (10.8) and (10.9), s¿ 1on the failure plane2


s1œ  s3œ s1œ  s3œ  cos 2u 2 2

and tf


sœ1  sœ3 sin 2u 2

Substituting the values of s1
41 lb/in.2, s3
16 lb/in.2, and u
58° into the preceding equations, we get s¿


41  16 41  16  cos 12  58 2
23.0 lb/in.2 2 2

12.8 Consolidated-Drained Triaxial Test

387

and tf


41  16 sin 12  582
11.2 lb/in.2 2

Part b From Eq. (10.9), it can be seen that the maximum shear stress will occur on the plane with u
45°. From Eq. (10.8), s1œ  s3œ s1œ  s3œ  cos 2u 2 2 Substituting u
45° into the preceding equation gives s¿


s¿


41  16 41  16  cos 90
28.5 lb/in.2 2 2

■

Example 12.4 The equation of the effective stress failure envelope for normally consolidated clayey soil is tf
s tan 30. A drained triaxial test was conducted with the same soil at a chamberconfining pressure of 10 lb/in.2 Calculate the deviator stress at failure. Solution For normally consolidated clay, c
0. Thus, from Eq. (12.8), s1œ
s3œ tan2 a 45 

f¿ b 2

f¿
30° s1œ
10 tan2 a 45 

30 b
30 lb/in.2 2

So, 1¢sd 2 f
s1œ  s3œ
30  10
20 lb/in.2

Example 12.5 The results of two drained triaxial tests on a saturated clay follow: Specimen I: s3
70 kN/m2

1¢sd 2 f
130 kN/m2

■

388 Chapter 12: Shear Strength of Soil Specimen II: s3
160 kN/m2

1¢sd 2 f
223.5 kN/m2 Determine the shear strength parameters. Solution Refer to Figure 12.25. For Specimen I, the principal stresses at failure are s3œ
s3
70 kN/m2 and s1œ
s1
s3  1 ¢sd 2 f
70  130
200 kN/m2 Similarly, the principal stresses at failure for Specimen II are s3œ
s3
160 kN/m2 and s1œ
s1
s3  1¢sd 2 f
160  223.5
383.5 kN/m2 Now, from Eq. (12.23),
2 e tan

1

c

œ œ  s11II2 s11I2 œ œ s31I2  s31II2

d

0.5

 45° f
2 e tan1 c

200  383.5 0.5 d  45° f
20° 70  160

Shear stress (kN/m2 )

f1œ

f

c

70

160

200

383.5

Normal stress, s (kN/m2 )

Figure 12.25 Effective stress failure envelope and Mohr’s circles for Specimens I and II

12.9 Consolidated-Undrained Triaxial Test

389

Again, from Eq. (12.24), œ œ  s31I2 tan2 a 45  s11I2

c¿


12.9

f1œ b 2 tan a 45  2

f1œ b 2

200  70 tan2 a 45 


20 b 2

20 2 tan a 45  b 2


20 kN/m2

■

Consolidated-Undrained Triaxial Test The consolidated-undrained test is the most common type of triaxial test. In this test, the saturated soil specimen is first consolidated by an all-around chamber fluid pressure, s3 , that results in drainage (Figures 12.26a and 12.26b). After the pore water pressure generated by the application of confining pressure is dissipated, the deviator stress, sd , on the specimen 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, the pore water pressure, ud , will increase. During the test, simultaneous measurements of sd and ud are made. The increase in the pore water pressure, ud , can be expressed in a nondimensional form as A


¢ud ¢sd

(12.25)

where A
Skempton’s 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 consolidated clay, the pore water pressure increases with strain. In dense sand and overconsolidated clay, the pore water pressure increases with strain to a certain limit, beyond which it decreases and becomes negative (with respect to the atmospheric pressure). This decrease is because of a tendency of the soil to dilate. Unlike the consolidated-drained test, the total and effective principal stresses are not the same in the consolidated-undrained test. Because the pore water pressure at failure is measured in this test, the principal stresses may be analyzed as follows: • • •

Major principal stress at failure 1total2:

s3  1¢sd 2 f
s1

Minor principal stress at failure 1total2:

s3

Major principal stress at failure 1effective2 :

s1  1¢ud 2 f
s1œ

œ • Minor principal stress at failure 1effective2: s3  1¢ud 2 f
s3 In these equations, (ud)f
pore water pressure at failure. The preceding derivations show that

s1  s3
s1œ  s3œ

394 Chapter 12: Shear Strength of Soil

Example 12.6 A specimen of saturated sand was consolidated under an all-around pressure of 12 lb/in.2 The axial stress was then increased and drainage was prevented. The specimen failed when the axial deviator stress reached 9.1 lb/in.2 The pore water pressure at failure was 6.8 lb/in.2 Determine a. Consolidated-undrained angle of shearing resistance, f b. Drained friction angle, f Solution Part a For this case, s3
12 lb/in.2, s1
12 + 9.1
21.1 lb/in.2, and (ud)f
6.8 lb/in.2. The total and effective stress failure envelopes are shown in Figure 12.30. From Eq. (12.27), f
sin 1 a

s1  s3 21.1  12 b
16° b
sin 1 a s1  s3 21.1  12

Part b From Eq. (12.28), f¿
sin 1 c

s1  s3 21.1  12 d
27.8° d
sin 1 c s1  s3  21¢ud 2 f 21.1  12  12 2 16.82

f

Shear stress (lb/in2)

Effective stress failure envelope Total stress failure envelope

f

B B

5.2

A

12

14.3

A

21.1

Normal stress (lb/in2)

Figure 12.30 Failure envelopes and Mohr’s circles for a saturated sand

■

12.10 Unconsolidated-Undrained Triaxial Test

395

Example 12.7 Refer to the soil specimen described in Example 12.6. What would be the deviator stress at failure, (sd )f , if a drained test was conducted with the same chamber allaround pressure (that is, 12 lb/in.2 )? Solution From Eq. (12.8) (with c
0), s1œ
s3œ tan2 a 45 

f¿ b 2

s3œ
12 lb/in.2 and f
27.8° (from Example 12.6). So, s1œ
12 tan2 a 45 

27.8 b
33 lb/in.2 2

1¢sd 2 f
s1œ  s3œ
33  12
21 lb/in.2

12.10

■

Unconsolidated-Undrained Triaxial Test In unconsolidated-undrained tests, drainage from the soil specimen is not permitted during the application of chamber pressure s3 . The test specimen is sheared to failure by the application of 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 confining pressure s3 , the pore water pressure in the soil specimen will increase by uc . A further increase in the pore water pressure (ud) will occur because of the deviator stress application. Hence, the total pore water pressure u in the specimen at any stage of deviator stress application can be given as u
uc  ¢ud

(12.31)

From Eqs. (12.18) and (12.25), uc
Bs3 and ud
Asd , so u
Bs3  A¢sd
Bs3  A1s1  s3 2

(12.32)

This test usually is conducted on clay specimens and depends on a very important strength concept for cohesive soils if the soil is fully saturated. The added axial stress at failure (sd)f is practically the same regardless of the chamber confining pressure. This property is shown in Figure 12.31. The failure envelope for the total stress Mohr’s circles becomes a horizontal line and hence is called a f
0 condition. From Eq. (12.9) with f
0, we get tf
c
cu

(12.33)

where cu is the undrained shear strength and is equal to the radius of the Mohr’s circles. Note that the f
0 concept is applicable to only saturated clays and silts.

12.18 Stress Path

417

Shear stress, or q

F Effective stress Mohr’s circle U

Total stress Mohr’s circle

U U a

O

1

s3 s3

I

s1

s, s, or p

ud

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

and q¿


s1œ  s3œ ¢sd
2 2

(12.63)

The preceding values of p and q will plot as point U in Figure 12.55. Points such as U represent values of p and q as the test progresses. At failure of the soil specimen, p¿
s3 

1¢sd 2 f 2

 1¢ud 2 f

(12.64)

and q¿


1¢sd 2 f 2

(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 curve IU U. Note that point U will fall on the modified failure envelope, OF (see Figure 12.54), which is inclined at an angle a to the horizontal. Lambe (1964) proposed a technique to evaluate the elastic and consolidation settlements of foundations on clay soils by using the stress paths determined in this manner.

Example 12.9 For a normally consolidated clay, the failure envelope is given by the equation tf
s tan f. The corresponding modified 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, the corresponding modified 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.

418 Chapter 12: Shear Strength of Soil Solution From Figure 12.56, s1œ  s3œ b 2 AB AB

sin f¿
s1œ  s3œ AC CO  OA c¿ cot f¿  a b 2 a

So, s1œ  s3œ s1œ  s3œ
c¿ cos f¿  a b sin f¿ 2 2

(a)

q¿
m  p¿ tan a

(b)

or Comparing Eqs. (a) and (b), we find that m
cⴕ cos Fⴕ and tan a
sin f¿ or

a
tan1 1sin Fⴕ2

f Shear stress

tf
c  s' tan f B s1'  s3' 2

c

C O c cot f

s3 s1  s3 2

A

s1

Normal stress

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. Laboratory procedures for determining the shear strength parameters were described. In textbooks, determination of the shear strength parameters of cohesive soils appears to be fairly simple. However, in practice, the proper choice of these parameters for design and stability checks of various earth, earth-retaining, and earth-supported structures

Problems

419

is very difficult and requires experience and an appropriate theoretical background in geotechnical engineering. In this chapter, three types of strength parameters (consolidateddrained, consolidated-undrained, and unconsolidated-undrained) were introduced. Their use depends on drainage conditions. Consolidated-drained strength parameters can be used to determine the long-term stability of structures such as earth embankments and cut slopes. Consolidated-undrained shear strength parameters can be used to study stability problems relating to cases where the soil initially is fully consolidated and then there is rapid loading. An excellent example of this is the stability of slopes of earth dams after rapid drawdown. The unconsolidatedundrained shear strength of clays can be used to evaluate the end-of-construction stability of saturated cohesive soils with the assumption that the load caused by construction has been applied rapidly and there has been little time for drainage to take place. The bearing capacity of foundations on soft saturated clays and the stability of the base of embankments on soft clays are examples of this condition. The unconsolidated-undrained shear strength of some saturated clays can vary depending on the direction of load application; this is referred to as anisotropy with respect to strength. Anisotropy is caused primarily by the nature of the deposition of the cohesive soils, and subsequent consolidation makes the clay particles orient perpendicular to the direction of the major principal stress. Parallel orientation of the clay particles can cause the strength of clay to vary with direction. The anisotropy with respect to strength for clays can have an important effect on the load-bearing capacity of foundations and the stability of earth embankments because the direction of the major principal stress along the potential failure surfaces changes. The sensitivity of clays was discussed in Section 12.13. It is imperative that sensitive clay deposits are properly identified. For instance, when machine foundations (which are subjected to vibratory loading) are constructed over sensitive clays, the clay may lose its 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/m2 a. Determine the angle of friction, f b. 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.2 Determine the shear force required to cause failure. 12.3 The following are the results of four drained, direct shear tests on a normally consolidated clay. Given: • Size of specimen
60 mm  60 mm • Height of specimen
30 mm

420 Chapter 12: Shear Strength of Soil

Test no.

Normal force (N)

Shear force at failure (N)

1 2 3 4

200 300 400 500

155 230 310 385

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. Test no.

Normal force (lb)

Shear force at failure (lb)

1 2 3 4

60 90 110 125

37.5 55 70 80

12.5 The equation of the effective stress failure envelope for a loose, sandy soil was obtained from a direct shear test at tf
s tan 30. A drained triaxial test was conducted with the same soil at a chamber confining 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, explain why 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, of a sand can be given as f
25  0.18Dr (Dr is in %). A drained triaxial test on the same sand was conducted with a chamber-confining pressure of 18 lb/in.2. The relative density of compaction was 60%. Calculate the major principal stress at failure. 12.8 For a normally consolidated clay, the results of a drained triaxial test are as follows. • Chamber confining pressure: 15 lb/in.2 • Deviator stress at failure: 34 lb/in.2 Determine the soil friction angle, f. 12.9 For a normally consolidated clay, f
24. In a drained triaxial test, the specimen failed at a deviator stress of 175 kN/m2. What was the chamber confining pressure, s3œ ? 12.10 For a normally consolidated clay, f
28. In a drained triaxial test, the specimen failed at a deviator stress of 30 lb/in.2. What was the chamber confining pressure, s3œ ?

Problems

421

12.11 A consolidated-drained triaxial test was conducted on a normally consolidated clay. The results were as follows: s3
250 kN/m2 1¢sd 2 f
275 kN/m2 Determine: a. Angle of friction, f b. Angle u that the failure plane makes with the major principal plane c. 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 confining pressure
15 lb/in.2 Deviator stress at failure
31.4 lb/in.2 Specimen II: Chamber-confining pressure
25 lb/in.2 Deviator stress at failure
47 lb/in.2 Calculate 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 a chamber-confining 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 the same soil, the deviator stress at failure is 175 kN/m2. What is the chamberconfining pressure? 12.15 A consolidated-undrained test on a normally consolidated clay yielded the following results: • s3
15 lb/in.2 • Deviator stress: (sd)f
11 lb/in.2 Pore pressure: (ud )f
7.2 lb/in.2 Calculate the consolidated-undrained friction angle and the drained friction angle. 12.16 Repeat Problem 12.15 with the following: s3
140 kN/m2 1¢sd 2 f
125 kN/m2 1¢ud 2 f
75 kN/m2 12.17 The shear strength of a normally consolidated clay can be given by the equation tf
s tan 31. A consolidated-undrained triaxial test was conducted on the clay. Following are the results of the test: • Chamber confining pressure
112 kN/m2 • Deviator stress at failure
100 kN/m2 Determine: a. Consolidated-undrained friction angle b. 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 the deviator stress at failure if a drained test had been conducted with the same chamber-confining pressure (that is, s3
112 kN/m2 )? 12.19 For a normally consolidated clay soil, f
32 and f
22. A consolidatedundrained triaxial test was conducted on this clay soil with a chamber-confining pressure 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 during field exploration was determined from drained triaxial tests to be 25. The unconfined compression strength, qu , of a similar specimen was found to be 100 kN/m2. Determine the pore water pressure at failure for the unconfined compression test.

422 Chapter 12: Shear Strength of Soil 12.21 Repeat Problem 12.20 using the following values. f¿
23° qu
120 kN/m2 12.22 The results of two consolidated-drained triaxial tests on a clayey soil are as follows. Test no.

S3œ (lb/in.2)

œ S11failure2 (lb/in.2)

1 2

27 12

73 48

Use the failure envelope equation given in Example 12.9—that is, q
m  p tan a. (Do not plot the graph.) a. Find m and a b. Find c and f 12.23 A 15-m thick normally consolidated clay layer is shown in Figure 12.57. The plasticity index of the clay is 18. Estimate the undrained cohesion as would be determined from a vane shear test at a depth of 8 m below the ground surface. Use Eq. (12.35). g
16 kN/m3

3m

Groundwater table

15 m gsat
18.6 kN/m3

Dry sand

Clay

Rock

Figure 12.57

References ACAR, 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, 648–654.

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, 93–120.

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, 437–501. BJERRUM, L. (1974). “Problems of Soil Mechanics and Construction on Soft Clays,” Norwegian Geotechnical Institute, Publication No. 110, Oslo.