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Shear Strength Examples.Pdf 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 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.) 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 445 300 250 ) 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 plas- tic 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. Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 12.9 Consolidated-Drained Triaxial Test 451 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 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 452 Chapter 12: Shear Strength of Soil 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 u 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 Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. 460 Chapter 12: Shear Strength of Soil 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.
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