444 Chapter : Shear Strength of Soil Example. Following are the results of four drained direct shear tests on an overconsolidated clay: Diameter of specimen 50 mm Height of specimen 5 mm Normal Shear force at Residual shear Test force, N failure, S peak force, S residual no. (N) (N) (N) 50 57.5 44. 50 99.9 56.6 3 350 57.6 0.9 4 550 363.4 44.5 Cengage Learning 04 Determine the relationships for peak shear strength (t f ) and residual shear strength (t r ). Solution Area of the specimen table can be prepared. A p/4 a 50 000 b 0.009634 m. Now the following Residual S shear peak Normal Normal Peak shear force, T Test force, N stress, S force, S peak S r S residual T f A residual A no. (N) (kn/m ) (N) (kn/m ) (N) (kn/m ) 50 76.4 57.5 80. 44..5 50 7.3 99.9 0.8 56.6 8.8 3 350 78.3 57.6 3. 0.9 5.4 4 550 80. 363.4 85. 44.5 73.6 Cengage Learning 04 The variations of t f and t r with s are plotted in Figure.9. From the plots, we find that Peak strength: t f (kn/m ) 40 S tan 7 Residual strength: t r (kn/m ) S tan 4.6 (Note: For all overconsolidated clays, the residual shear strength can be expressed as where f r t r s tan f r effective residual friction angle.) Copyright 0 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
.8 Triaxial Shear Test-General 445 300 50 Shear stress, t (kn/m ) 00 50 00 t f versus s 50 0 0 7 f t r versus s c 40 kn/m f r 4.6 50 00 50 00 50 300 Effective normal stress, s (kn/m ) 350 Cengage Learning 04 Figure.9 Variations of t f and t r with s.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.0. Figure. 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 generally 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:. 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.). 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. Copyright 0 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
.9 Consolidated-Drained Triaxial Test 45 The portion bc of the failure envelope represents a normally consolidated stage of soil and follows the equation t f s tan f. If the triaxial test results of two overconsolidated soil specimens are known, the magnitudes of and c can be determined as follows. From Eq. (.8), for Specimen : And, for Specimen : or Hence, f s s Once the value of f s 3 s 3 s s f e tan c s s 0.5 d 45 f s 3 s 3 is known, we can obtain c as c tan 45 f / c tan45 f / tan 45 f / c tan45 f / s 3s 3 s 3 s 3 4 tan 45 f / tan a45 f tana45 f (.) (.) (.3) (.4) 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. b b Example.3 A consolidated-drained triaxial test was conducted on a normally consolidated clay. The results are as follows: s 3 76 kn/m (s d ) f 76 kn/m 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 t f s tan f because c 0 Copyright 0 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
45 Chapter : Shear Strength of Soil 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.6. From Eq. (.9), or Part b From Eq. (.4), s s s 3 s d f 76 76 55 kn/m sin f s s 3 55 76 s 0.333 s 3 55 76 u 45 f s 3 s 3 76 kn/m f 9.45 45 9.45 54.73 s u Shear stress s 3 s 3 Effective stress failure envelope B f O s u s 3 76 kn/m A s 55 kn/m Normal stress Cengage Learning 04 Figure.6 Mohr s circle and failure envelope for a normally consolidated clay Copyright 0 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
460 Chapter : Shear Strength of Soil Table.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 Sensitivity a F (deg) Seven Sisters, Canada 7 35 0.8 9 0.7 Sarpborg 69 8 0.68 5 5.5.03 Lilla Edet, Sweden 68 30.3 50 6.0 Fredrikstad 59 0.58 5 8.5 0.87 Fredrikstad 57 0.63 6 7.00 Lilla Edet, Sweden 63 30.58 50 3.0 Göta River, Sweden 60 7.30 8.5.05 Göta River, Sweden 60 30.50 40 4.05 Oslo 48 5 0.87 4 3.5.00 Trondheim 36 0 0.50 34 0.75 Drammen 33 8.08 8 8.8 * After Bjerrum and Simons, 960. With permission from ASCE. a See Section.4 for the definition of sensitivity. A f Example.7 A specimen of saturated sand was consolidated under an all-around pressure of 05 kn/m. The axial stress was then increased and drainage was prevented. The specimen failed when the axial deviator stress reached 70 kn/m. The pore water pressure at failure was 50 kn/m. Determine a. Consolidated-undrained angle of shearing resistance, f b. Drained friction angle, f Solution Part a For this case, s 3 05 kn/m, s 05 70 75 kn/m, and (u d ) f 50 kn/m. The total and effective stress failure envelopes are shown in Figure.3. From Eq. (.7), Part b From Eq. (.8), f sin a s s 3 75 05 b sin a b 4.5 s s 3 75 05 s f sin s 3 c d sin 75 05 c d.9 s s 3 u d f 75 05 50
. Unconsolidated-Undrained Triaxial Test 46 f Shear stress (kn/m ) Effective stress failure envelope B Total stress failure envelope B f 55 A 05 5 A 75 Normal stress (kn/m ) Cengage Learning 04 Figure.3 Failure envelopes and Mohr s circles for a saturated sand. Unconsolidated-Undrained Triaxial Test In unconsolidated-undrained tests, drainage from the soil specimen is not permitted during the application of chamber pressure s 3. The test specimen is sheared to failure by the application of deviator stress, s d, 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 s 3, the pore water pressure in the soil specimen will increase by u c. A further increase in the pore water pressure (u d ) 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 u c u d (.3) From Eqs. (.8) and (.5), u c Bs 3 and u d A s d, so u Bs 3 A s d Bs 3 As s 3 (.3) 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 (s d ) f is practically the same regardless of the chamber confining pressure. This property is shown in Figure.33. The failure envelope for the total stress Mohr s circles
Problems 485 Problems Test no. Normal force (N) Shear force at failure (N) 3 4 67 33 3 369 3.3 46.6 44.6 3.3 Cengage Learning 04. Following data are given for a direct shear test conducted on dry sand: Specimen dimensions: 63 mm 63 mm 5 mm (height) Normal stress: 05 kn/m Shear force at failure: 300 N a. Determine the angle of friction, f b. For a normal stress of 80 kn/m, what shear force is required to cause failure?. Consider the specimen in Problem.b. a. What are the principal stresses at failure? b. What is the inclination of the major principal plane with the horizontal?.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 3.75 mm (height) Angle of friction: 33 Normal stress: 93 kn/m Determine the shear force required to cause failure.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 5 mm. Test no. Normal force (N) Shear force at failure (N) 3 4 50 375 450 540 39 09 50 300 Cengage Learning 04 Draw a graph for shear stress at failure against the normal stress and determine the drained angle of friction from the graph..5 Repeat Problem.4 with the following data. Given: Specimen diameter 50 mm; specimen height 5 mm..6 Consider the clay soil in Problem.5. If a drained triaxial test is conducted on the same soil with a chamber confining pressure of 08 kn/m, what would be the deviator stress at failure?.7 For the triaxial test on the clay specimen in Problem.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. a. c = 0. From Eq. (.3): f = tan 300 75 kn/m (000)(0.063) So, 75 = 05 tan tan 75 05 35.5 b. For = 80 kn/m, f = 80 tan 35.5 = 8.39 kn/m Shear force, S (8.39)(000)(0.063) 509.5 N. The point O (80, 8.4) 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, 99 04 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
a. 3 5 kn/m ; 40 kn/m 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..3 For = 8 lb/in, f = 8 tan 33 = 8.8 lb/in Shear force, S (8.8)(.5) 3.65 lb.4 Area of specimen A () 4 3.4 in. Test No. 3 4 Normal force N (lb) 5 30 48 83 N A (lb/in. ) 4.77 9.55 5.8 6.43 Shear force S (lb) 5.5 0.5 6.8 9.8 S f A (lb/in. ).67 3.34 5.35 9.5 tan (deg) 9.9 9.7 9.9 9.77 f A graph of f vs. will yield = 9.5º..5 Area of specimen A (0.05) 4 0.0096 m Test No. 3 4 Normal force N (N) 50 375 450 540 N A (N/m ) 79.6 9.4 43.3 7.9 Shear force S (N) 39 09 50 300 S f A (N/m ) 44.6 66.56 79.6 95.54 tan (deg) 9.07 9.3 9.05 9.06 f A graph of f vs. will yield 9º..6 c = 0. From Eq. (.8): tan 3 45 ; 30 00 04 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
9 08tan 45 600 kn/m d (failure) 3 600 08 39 kn/m 9.7 a. From Eq. (.4): 45 45 59.5 b. Refer to the figure. = 96 sin 60º = 69.7 kn/m = 404 + r cos 60 = 404 + 96 cos 60 = 50 kn/m For failure, f = tan = 50 tan 9 = 78.6 kn/m. Since the developed shear stress = 69.5 kn/m (which is less than 78.6 kn/m ), the specimen did not fail along this plane..8 = 8 + 0.8D r = 8 + (0.8)(68) = 40.4 tan 3 45 50tan 45 40.4 697.43kN/m 0 04 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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