Shear Strength Examples.Pdf

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Example 12.2

Following are the results of four drained direct shear tests on an overconsolidated clay: • Diameter of specimen ϭ 50 mm • Height of specimen ϭ 25 mm

Normal Shear force at Residual shear Test force, N failure, Speak force, Sresidual no. (N) (N) (N) 1 150 157.5 44.2 2 250 199.9 56.6 3 350 257.6 102.9 4 550 363.4 144.5 © Cengage Learning 2014

t t Determine the relationships for peak shear strength ( f) and residual shear strength ( r).

Solution 50 2 Area of the specimen 1A2 ϭ 1p/42 a b ϭ 0.0019634 m2 . Now the following 1000 table can be prepared.

Residual S shear peak S force, T ϭ residual ؍ Normal Normal Peak shearT S؅ f r Test force, N stress, force, Speak A Sresidual A no. (N) (kN/m2) (N) (kN/m2) (N) (kN/m2) 1 150 76.4 157.5 80.2 44.2 22.5 2 250 127.3 199.9 101.8 56.6 28.8 3 350 178.3 257.6 131.2 102.9 52.4 4 550 280.1 363.4 185.1 144.5 73.6 © Cengage Learning 2014

t t sЈ The variations of f and r with are plotted in Figure 12.19. From the plots, we find that t 2 ϭ ؉ S؅ Peak strength: f (kN/m ) 40 tan 27 t 2 ϭ S؅ Residual strength: r(kN/m ) tan 14.6 (Note: For all overconsolidated clays, the residual shear strength can be expressed as

t ϭ sœ fœ r tan r

fœ ϭ where r effective residual friction angle.)

300

250 ) 2 200 (kN/m t , 150 t sЈ ss f versus tre s 100 Shear

Њ ϭ fЈ t sЈ 50 27 r versus cЈ ϭ 40 kN/m2 fЈ ϭ Њ r 14.6 0 0 50 100 150 200 250 300 350

Effective normal stress, sЈ (kN/m2) © Cengage Learning 2014

t t sЈ Figure 12.19 Variations of f and r with

12.8 Triaxial Shear Test-General

The triaxial shear test is one of the most reliable methods available for determining shear strength parameters. It is used widely for research and conventional testing. A diagram of the triaxial test layout is shown in Figure 12.20. Figure 12.21 on page 447 shows a triaxial test 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 plastic cylindrical chamber that usually is filled with water or glycerine. The specimen is subjected to a confining pressure by compression of the fluid in the chamber. (Note: Air is sometimes used as a compression medium.) To cause shear failure in the specimen, one must 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 the specimen fails. (Axial deformation of the specimen resulting from the load applied through the ram is measured by a dial gauge.) 2. Application of axial deformation at a constant rate by means of a geared or hydraulic loading press. This is a strain-controlled test.

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

The portion bc of the failure envelope represents a normally consolidated stage of soil and t ϭ sЈ fЈ follows the equation f tan . If the triaxial test results of two overconsolidated soil specimens are known, the fœ Ј magnitudes of 1 and c can be determined as follows. From Eq. (12.8), for Specimen 1: sœ ϭ sœ 21 ϩ fœ 2 ϩ œ 1 ϩ fœ 2 1112 3112 tan 45 1/2 2c tan 45 1/2 (12.21) And, for Specimen 2: sœ ϭ sœ 21 ϩ fœ 2 ϩ œ 1 ϩ fœ 2 1122 3122 tan 45 1/2 2c tan 45 1/2 (12.22) or sœ Ϫ sœ ϭ 3sœ Ϫ sœ 4 21 ϩ fœ 2 1112 1122 3112 3122 tan 45 1/2 Hence, œ œ s Ϫ s 0.5 1112 1122 fœ ϭ e Ϫ1 c d Ϫ f 1 2 tan sœ Ϫ sœ 45° (12.23) 3112 3122 fœ Ј Once the value of 1 is known, we can obtain c as fœ œ œ 2 1 s 1 2 Ϫ s 1 2 tan a45 ϩ b 1 1 3 1 2 œ ϭ c fœ (12.24) 2 tana45 ϩ 1 b 2 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 slowly to ensure full drainage from the soil specimen. For this reason, the CD type of triaxial test is uncommon.

Example 12.3

A consolidated-drained triaxial test was conducted on a normally consolidated clay. The results are as follows: s ϭ 2 • 3 276 kN/m ⌬s ϭ 2 •( d)f 276 kN/m Determine a. Angle of friction, fЈ b. Angle u that the failure plane makes with the major principal plane Solution For normally consolidated soil, the failure envelope equation is t ϭ sœ fœ 1 œ ϭ 2 f tan because c 0

For the triaxial test, the effective major and minor principal stresses at failure are as follows: sœ ϭ s ϭ s ϩ 1¢s 2 ϭ ϩ ϭ 2 1 1 3 d f 276 276 552 kN/m and sœ ϭ s ϭ 2 3 3 276 kN/m Part a The Mohr’s circle and the failure envelope are shown in Figure 12.26. From Eq. (12.19), sœ Ϫ sœ Ϫ fœ ϭ 1 3 ϭ 552 276 ϭ sin sœ ϩ sœ ϩ 0.333 1 3 552 276 or fœ ϭ 19.45؇ Part b From Eq. (12.4), fœ 19.45° u ϭ 45 ϩ ϭ 45° ϩ ϭ 54.73° 2 2

sЈ 1

sЈ sЈ fЈ ss 3 3 Effective stress failure envelope tre s

Shear B

sЈ 1 2u

O sЈ ϭ 2 sЈ ϭ 2 3 276 kN/m A 1 552 kN/m

Normal stress © Cengage Learning 2014

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

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

Drained Liquid Plastic Liquidity friction angle, a FЈ Location limit limit index Sensitivity (deg) Af Seven Sisters, Canada 127 35 0.28 19 0.72 Sarpborg 69 28 0.68 5 25.5 1.03 Lilla Edet, Sweden 68 30 1.32 50 26 1.10 Fredrikstad 59 22 0.58 5 28.5 0.87 Fredrikstad 57 22 0.63 6 27 1.00 Lilla Edet, Sweden 63 30 1.58 50 23 1.02 Göta River, Sweden 60 27 1.30 12 28.5 1.05 Göta River, Sweden 60 30 1.50 40 24 1.05 Oslo 48 25 0.87 4 31.5 1.00 Trondheim 36 20 0.50 2 34 0.75 Drammen 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.

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. The specimen failed when the axial deviator stress reached 70 kN/m2. The pore water pressure at failure was 50 kN/m2. Determine a. Consolidated-undrained angle of shearing resistance, f b. Drained friction angle, fЈ

Solution Part a s ϭ 2 s ϭ ϩ ϭ 2 ⌬ ϭ 2 For this case, 3 105 kN/m , 1 105 70 175 kN/m , and ( ud)f 50 kN/m . The total and effective stress failure envelopes are shown in Figure 12.32. From Eq. (12.27),

s Ϫ s 175 Ϫ 105 f ϭ Ϫ1 a 1 3 b ϭ Ϫ1 a b Ϸ sin s ϩ s sin ϩ 14.5° 1 3 175 105 Part b From Eq. (12.28),

s Ϫ s 175 Ϫ 105 fœ ϭ Ϫ1 c 1 3 d ϭ Ϫ1 c d ϭ ؇ sin s ϩ s Ϫ 1¢ 2 sin ϩ Ϫ 1 21 2 22.9 1 3 2 ud f 175 105 2 50 12.11 Unconsolidated-Undrained Triaxial Test 461

fЈ

) Effective stress failure envelope 2

Total stress failure envelope f

B BЈ Shear stress (kN/m

55 AЈ 105 125A 175 2

Normal stress (kN/m ) © Cengage Learning 2014

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

12.11 Unconsolidated-Undrained Triaxial Test

In unconsolidated-undrained tests, drainage from the soil specimen is not permitted dur- s ing the application of chamber pressure 3. The test specimen is sheared to failure by the ⌬s application of deviator stress, d, and drainage is prevented. Because drainage is not allowed at any stage, the test can be performed quickly. Because of the application of s chamber confining pressure 3, 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

ϭ ϩ ¢ u uc ud (12.31)

ϭ s ¢ ϭ ¢s From Eqs. (12.18) and (12.25), uc B 3 and ud A d, so

ϭ s ϩ ¢s ϭ s ϩ 1s Ϫ s 2 u B 3 A d B 3 A 1 3 (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 ⌬s failure ( d)f is practically the same regardless of the chamber confining pressure. This property is shown in Figure 12.33. The failure envelope for the total stress Mohr’s circles Problems 485

Problems 12.1 Following data are given for a direct shear test conducted on dry sand: • Specimen dimensions: 63 mm ϫ 63 mm ϫ 25 mm (height) • Normal stress: 105 kN/m2 • Shear force at failure: 300 N a. Determine the angle of friction, fЈ 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 ϫ 63.5 mm ϫ 31.75 mm (height) • Angle of friction: 33° • Normal stress: 193 kN/m2 Determine the shear force required to cause failure 12.4 The following are the results of four drained direct shear tests on undisturbed normally consolidated clay samples having a diameter of 50 mm. and height of 25 mm.

Normal Shear force at Test no. force (N) failure (N) 1 67 23.3 2 133 46.6 3 213 44.6 4 369 132.3 © Cengage Learning 2014

Draw a graph for shear stress at failure against the normal stress and determine the drained angle of friction from the graph. 12.5 Repeat Problem 12.4 with the following data. Given: Specimen diameter ϭ 50 mm; specimen height ϭ 25 mm.

Normal Shear force at Test no. force (N) failure (N) 1 250 139 2 375 209 3 450 250 4 540 300 © Cengage Learning 2014

12.6 Consider the clay soil in Problem 12.5. If a drained triaxial test is conducted on the same soil with a chamber confining pressure of 208 kN/m2, what would be the deviator 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 fail along this plane. Chapter 12

12.1 a. cc = 0. From Eq. (12.3): Wf = Vc tan Ic

300 W 75 kN/m 2 (1000)(0.063) 2

So, 75 = 105 tan Ic

2 2 b. For Vc = 180 kN/m , Wf = 180 tan 35.5q = 128.39 kN/m

Shear force, S (128.39)(1000)(0.063) 2 509.5 N

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,

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 Į | 65q.

2 2 12.3 For Vc = 28 lb/in , Wf = 28 tan 33q = 18.18 lb/in

Shear force, S (18.18)(2.5) 2 113.65 lb

§W · N S 1 f Test Normal V c Shear W Ic tan ¨ ¸ f ¨ V c ¸ No. force N A force S A © ¹ (lb) (lb/in.2) (lb) (lb/in.2) (deg) 1 15 4.77 5.25 1.67 19.29 2 30 9.55 10.5 3.34 19.27 3 48 15.28 16.8 5.35 19.29 4 83 26.43 29.8 9.5 19.77

A graph of Wf vs. Vc will yield Ic = 19.5º.

W N S 1§ f · Test Normal c Shear Ic tan ¨ ¸ V W f ¨ V c ¸ No. force N A force S A © ¹ (N) (N/m2) (N) (N/m2) (deg) 1 250 79.6 139 44.26 29.07 2 375 119.4 209 66.56 29.13 3 450 143.3 250 79.61 29.05 4 540 171.9 300 95.54 29.06

A graph of Wf vs. Vc will yield Ic § 29º.

ǻV V c V c 600 208 392 kN/m2 d (failure) 1 3

Ic 29 12.7 a. From Eq. (12.4): T 45 45 59.5q 2 2

b. Refer to the figure.

2 W = 196 sin 60º = 169.7 kN/m

2 Vc = 404 + r cos 60 = 404 + 196 cos 60 = 502 kN/m

2 For failure, Wf = Vc tan Ic = 502 tan 29 = 278.26 kN/m . 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 Ic = 28 + 0.18Dr = 28 + (0.18)(68) = 40.24q

Answer