example 12 - civilittee strength... · 12.11 unconsolidated-undrained triaxial test in...

444 Chapter 12: Shear Strength of Soil

Example 12.2

Following are the results of four drained direct shear tests on an overconsolidated clay:

• Diameter of specimen � 50 mm• Height of specimen � 25 mm

Normal Shear force at Residual shearTest force, N failure, Speak force, Sresidualno. (N) (N) (N)

1 150 157.5 44.22 250 199.9 56.63 350 257.6 102.94 550 363.4 144.5

Determine the relationships for peak shear strength (tf) and residual shear strength (tr).

Solution

Area of the specimen . Now the following

table can be prepared.

Residual shear

Normal Normal Peak shear force, Test force, N stress, S� force, Speak Sresidualno. (N) (kN/m2) (N) (kN/m2) (N) (kN/m2)

1 150 76.4 157.5 80.2 44.2 22.52 250 127.3 199.9 101.8 56.6 28.83 350 178.3 257.6 131.2 102.9 52.44 550 280.1 363.4 185.1 144.5 73.6

The variations of tf and tr with s� are plotted in Figure 12.19. From the plots,we find that

Peak strength: tf (kN/m2) � 40 � S� tan 27Residual strength: tr(kN/m2) � S� tan 14.6

(Note: For all overconsolidated clays, the residual shear strength can be expressed as

where � effective residual friction angle.)frœ

tr � sœ tan frœ

Tr �Sresidual

ATf �

Speak

A

1A2 � 1p/42 a 50

1000b

2

� 0.0019634 m2

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Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.

12.8 Triaxial Shear Test-General 445

Effective normal stress, s� (kN/m2)

Shea

r st

ress

, t (

kN/m

2 )

3500

27� � f�

c� � 40 kN/m2

50 100 150 200 250 300

0

50

100

150

200

250

300

fr� � 14.6�

tr versus s�

tf versus s�

Figure 12.19 Variations of tf and tr with s�

12.8 Triaxial Shear Test-General

The triaxial shear test is one of the most reliable methods available for determining shearstrength parameters. It is used widely for research and conventional testing. A diagram ofthe triaxial test layout is shown in Figure 12.20. Figure 12.21 on page 447 shows a triaxialtest in progress in the laboratory.

In this test, a soil specimen about 36 mm in diameter and 76 mm (3 in.) long gener-ally is used. The specimen is encased by a thin rubber membrane and placed inside a plas-tic cylindrical chamber that usually is filled with water or glycerine. The specimen issubjected to a confining pressure by compression of the fluid in the chamber. (Note: Air issometimes used as a compression medium.) To cause shear failure in the specimen, onemust apply axial stress (sometimes called deviator stress) through a vertical loading ram.This stress can be applied in one of two ways:

1. Application of dead weights or hydraulic pressure in equal increments until thespecimen fails. (Axial deformation of the specimen resulting from the load appliedthrough the ram is measured by a dial gauge.)

2. Application of axial deformation at a constant rate by means of a geared orhydraulic loading press. This is a strain-controlled test.

The axial load applied by the loading ram corresponding to a given axial deformation ismeasured by a proving ring or load cell attached to the ram.

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12.9 Consolidated-Drained Triaxial Test 451

The portion bc of the failure envelope represents a normally consolidated stage of soil andfollows the equation tf � s� tan f�.

If the triaxial test results of two overconsolidated soil specimens are known, themagnitudes of and c� can be determined as follows. From Eq. (12.8), for Specimen 1:

(12.21)

And, for Specimen 2:

(12.22)

or

Hence,

(12.23)

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

(12.24)

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

cœ �

s1112œ � s3112œ tan2a45 �f1

œ

2b

2 tana45 �f1

œ

2b

f1œ

f1œ � 2 e tan�1 cs1112œ � s1122œ

s3112œ � s3122œ d 0.5

� 45° f

s1112œ � s1122œ � 3s3112œ � s3122œ 4 tan2145 � f1œ /22

s1122œ � s3122œ tan2145 � f1œ /22 � 2cœ tan145 � f1

œ /22

s1112œ � s3112œ tan 2145 � f1œ /22 � 2cœ tan145 � f1

œ /22f1

œ

Example 12.3

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

• s3 � 276 kN/m2

• (�sd)f � 276 kN/m2

Determine

a. Angle of friction, f�b. Angle u that the failure plane makes with the major principal plane

SolutionFor normally consolidated soil, the failure envelope equation is

tf � sœ tan fœ 1because cœ � 02



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

and

Part a

The Mohr’s circle and the failure envelope are shown in Figure 12.26. From Eq. (12.19),

or

Part b

From Eq. (12.4),

u � 45 �fœ

2� 45° �

19.45°

2� 54.73°

fœ � 19.45�

sin fœ �s1

œ � s3œ

s1œ � s3

œ �552 � 276

552 � 276� 0.333

s3œ � s3 � 276 kN/m2

s1œ � s1 � s3 � 1¢sd2f � 276 � 276 � 552 kN/m2

Normal stress

s3� � 276 kN/m2 s1� � 552 kN/m2

Shea

r st

ress

O

B

2u

f�Effective stress failure envelope

A

s3�s3�

s1�

s1�

u

Figure 12.26 Mohr’s circle and failure envelope for a normally consolidated clay

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Table 12.3 Triaxial Test Results for Some Normally Consolidated Clays Obtained by the Norwegian Geotechnical Institute*

Drained Liquid Plastic Liquidity friction angle,

Location limit limit index Sensitivitya F� (deg)

Seven Sisters, Canada 127 35 0.28 19 0.72Sarpborg 69 28 0.68 5 25.5 1.03Lilla Edet, Sweden 68 30 1.32 50 26 1.10Fredrikstad 59 22 0.58 5 28.5 0.87Fredrikstad 57 22 0.63 6 27 1.00Lilla Edet, Sweden 63 30 1.58 50 23 1.02Göta River, Sweden 60 27 1.30 12 28.5 1.05Göta River, Sweden 60 30 1.50 40 24 1.05Oslo 48 25 0.87 4 31.5 1.00Trondheim 36 20 0.50 2 34 0.75Drammen 33 18 1.08 8 28 1.18

*After Bjerrum and Simons, 1960. With permission from ASCE.aSee Section 12.14 for the definition of sensitivity.

Af

Example 12.7

A specimen of saturated sand was consolidated under an all-around pressure of 105 kN/m2. The axial stress was then increased and drainage was prevented. Thespecimen failed when the axial deviator stress reached 70 kN/m2. The pore waterpressure at failure was 50 kN/m2. Determine

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

SolutionPart a

For this case, s3 � 105 kN/m2, s1 � 105 � 70 � 175 kN/m2, and (�ud)f � 50 kN/m2.The total and effective stress failure envelopes are shown in Figure 12.32. From Eq. (12.27),

Part b

From Eq. (12.28),

fœ � sin�1 c s1 � s3

s1 � s3 � 21¢ud2f d � sin�1 c 175 � 105

175 � 105 � 1221502 d � 22.9�

f � sin�1as1 � s3

s1 � s3b � sin�1a175 � 105

175 � 105b � 14.5°

12.11 Unconsolidated-Undrained Triaxial Test 461

12.11 Unconsolidated-Undrained Triaxial Test

In unconsolidated-undrained tests, drainage from the soil specimen is not permitted dur-ing the application of chamber pressure s3. The test specimen is sheared to failure by theapplication of deviator stress, �sd, and drainage is prevented. Because drainage is notallowed at any stage, the test can be performed quickly. Because of the application ofchamber confining pressure s3, the pore water pressure in the soil specimen will increaseby uc. A further increase in the pore water pressure (�ud) will occur because of the devia-tor stress application. Hence, the total pore water pressure u in the specimen at any stageof deviator stress application can be given as

(12.31)

From Eqs. (12.18) and (12.25), uc � Bs3 and , 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 atfailure (�sd)f is practically the same regardless of the chamber confining pressure. Thisproperty is shown in Figure 12.33. The failure envelope for the total stress Mohr’s circles

u � Bs3 � A¢sd � Bs3 � A1s1 � s32

¢ud � A¢sd

u � uc � ¢ud

f�

f

B B�

A�55

Shea

r st

ress

(kN

/m2 )

Normal stress (kN/m2)

105 125 175A

Effective stress failure envelope

Total stress failure envelope

Figure 12.32 Failure envelopes and Mohr’s circles for a saturated sand

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Problems 485

Problems12.1 Following data are given for a direct shear test conducted on dry sand:

• Specimen dimensions: 63 mm � 63 mm � 25 mm (height)• Normal stress: 105 kN/m2

• Shear force at failure: 300 Na. Determine the angle of friction, f�b. For a normal stress of 180 kN/m2, what shear force is required to cause failure?

12.2 Consider the specimen in Problem 12.1b.a. What are the principal stresses at failure?b. What is the inclination of the major principal plane with the horizontal?

12.3 For a dry sand specimen in a direct shear test box, the following are given:• Size of specimen: 63.5 mm � 63.5 mm � 31.75 mm (height)• Angle of friction: 33°• Normal stress: 193 kN/m2

Determine the shear force required to cause failure12.4 The following are the results of four drained direct shear tests on undisturbed nor-

mally consolidated clay samples having a diameter of 50 mm. and height of 25 mm.

Normal Shear force atTest no. force (N) failure (N)

1 67 23.32 133 46.63 213 44.64 369 132.3

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

12.5 Repeat Problem 12.4 with the following data. Given: Specimen diameter �50 mm; specimen height � 25 mm.

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

1 250 1392 375 2093 450 2504 540 300

12.6 Consider the clay soil in Problem 12.5. If a drained triaxial test is conducted onthe same soil with a chamber confining pressure of 208 kN/m2, what would be thedeviator stress at failure?

12.7 For the triaxial test on the clay specimen in Problem 12.6,a. What is the inclination of the failure plane with the major principal plane?b. Determine the normal and shear stress on a plane inclined at 30° with the

major principal plane at failure. Also explain why the specimen did not failalong this plane.

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99 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 12 12.1 a. c = 0. From Eq. (12.3): f = tan

22 kN/m 75

)063.0)(1000(300

So, 75 = 105 tan

35.510575tan 1

b. For = 180 kN/m2, f = 180 tan 35.5 = 128.39 kN/m2 Shear force, N 509.52)063.0)(1000)(39.128(S 12.2 The point O (180, 128.4) represents the failure stress conditions on the Mohr-

Coulomb failure envelope. The perpendicular line OC to the failure envelope determines the center, C of the Mohr’s circle. With the center at C, and the radius as OC, the Mohr’s circle is drawn by trial and error such that the circle is tangent to the failure envelope at O. From the graph,


a. 2kN/m 1153 ; 2kN/m 4201 b. The horizontal line OP drawn from O determines the pole P. Therefore, the

orientation or the major principal plane with the horizontal is given by the angle 65 .

12.3 For = 28 lb/in2, f = 28 tan 33 = 18.18 lb/in2 Shear force, lb 113.652)5.2)(18.18(S

12.4 Area of specimen 22 in. 14.3)2(4

A

Test No.

Normal force N (lb)

AN

(lb/in.2)

Shear force S (lb)

AS

f

(lb/in.2)

f1tan

(deg) 1 2 3 4

15 30 48 83

4.77 9.55 15.28 26.43

5.25 10.5 16.8 29.8

1.67 3.34 5.35 9.5

19.29 19.27 19.29 19.77

A graph of f vs. will yield = 19.5º.

12.5 Area of specimen 22 m 00196.0)05.0(4

A

Test No.

Normal force N (N)

AN

(N/m2)

Shear force S (N)

AS

f

(N/m2)

f1tan

(deg) 1 2 3 4

250 375 450 540

79.6 119.4 143.3 171.9

139 209 250 300

44.26 66.56 79.61 95.54

29.07 29.13 29.05 29.06

A graph of f vs. will yield 29º.

12.6 c = 0. From Eq. (12.8): 30 ;2

45tan 231


221 kN/m 600

22945tan208

2kN/m 39220860031(failure)d

12.7 a. From Eq. (12.4): 59.52

29452

45

b. Refer to the figure.

= 196 sin 60º = 169.7 kN/m2 = 404 + r cos 60 = 404 + 196 cos 60 = 502 kN/m2

For failure, f = tan = 502 tan 29 = 278.26 kN/m2. Since the developed shear stress = 169.5 kN/m2 (which is less than 278.26 kN/m2), the specimen did not fail along this plane.

12.8 = 28 + 0.18Dr = 28 + (0.18)(68) = 40.24

2kN/m 697.43224.4045tan150

245tan 22

31

Answer

example 12 - civilittee strength... · 12.11 unconsolidated-undrained triaxial test in...

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