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# Craig's Soil Mechanics 7th Edition

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compacted to the in-situ density. Determine the value of the shear strength parameter �0. Normal stress (kN/m2) 50 100 200 300 Shear stress at failure (kN/m2) 36 80 154 235 Would failure occur on a plane within a mass of this sand at a point where the shear stress is 122 kN/m2 and the effective normal stress 246 kN/m2? ESP ESP (a) (b) (c) ESP TSP TSP TSP TSP (us > 0) uf ufuf us 1 2 (σ1 – σ3) 1 2 (σ1 – σ3) 1 2 (σ1 + σ3) 1 2 (σ1 + σ3) 1 2 (σ1 + σ3) 1 2 (σ1 + σ3) 1 2 (σ1 + σ3) 1 2 (σ1 + σ3) 1 2 (σ1 – σ3) ′ ′ ′ ′ ′ ′ Consolidated–undrained test Normally consolidated clay Consolidated–undrained test Overconsolidated clay Drained tests Figure 4.15 Stress paths for triaxial tests. 114 Shear strength Figure 4.16 Hydraulic triaxial apparatus. Figure 4.17 Example 4.1. The values of shear stress at failure are plotted against the corresponding values of normal stress, as shown in Figure 4.17. The failure envelope is the line having the best fit to the plotted points; in this case a straight line through the origin. If the stress scales are the same, the value of �0 can be measured directly and is 38�. The stress state � ¼ 122 kN/m2, �0 ¼ 246 kN/m2 plots below the failure envelope, and therefore would not produce failure. Example 4.2 The results shown in Table 4.2 were obtained at failure in a series of triaxial tests on specimens of a saturated clay initially 38mm in diameter by 76mm long. Determine the values of the shear strength parameters with respect to (a) total stress and (b) effective stress. The principal stress difference at failure in each test is obtained by dividing the axial load by the cross-sectional area of the specimen at failure (Table 4.3). The corrected cross-sectional area is calculated from Equation 4.10. There is, of course, no volume change during an undrained test on a saturated clay. The initial values of length, area and volume for each specimen are: l0 ¼ 76mm; A0 ¼ 1135mm2; V0 ¼ 86� 103 mm3 The Mohr circles at failure and the corresponding failure envelopes for both series of tests are shown in Figure 4.18. In both cases the failure envelope is the line nearest to a common tangent to the Mohr circles. The total stress parameters, representing the undrained strength of the clay, are cu ¼ 85 kN=m2; �u ¼ 0 Table 4.2 Type of test All-round pressure (kN/m2) Axial load (N) Axial deformation (mm) Volume change (ml) (a) Undrained 200 222 9.83 – 400 215 10.06 – 600 226 10.28 – (b) Drained 200 403 10.81 6.6 400 848 12.26 8.2 600 1265 14.17 9.5 Table 4.3 �3 (kN/m 2) �l/l0 �V/V0 Area (mm 2) �1 � �3 (kN/m2) �1 (kN/m2) (a) 200 0.129 – 1304 170 370 400 0.132 – 1309 164 564 600 0.135 – 1312 172 772 (b) 200 0.142 0.077 1222 330 530 400 0.161 0.095 1225 691 1091 600 0.186 0.110 1240 1020 1620 116 Shear strength The effective stress parameters, representing the drained strength of the clay, are c0 ¼ 0; �0 ¼ 27� Example 4.3 The results shown in Table 4.4 were obtained for peak failure in a series of consolidated– undrained triaxial tests, with pore water pressure measurement, on specimens of a saturated clay. Determine the values of the effective stress parameters. Values of effective principal stresses �03 and � 0 1 at failure are calculated by sub- tracting pore water pressure at failure from the total principal stresses as shown in Table 4.5 (all stresses in kN/m2). The Mohr circles in terms of effective stress are drawn in Figure 4.19. In this case the failure envelope is slightly curved and a different value of the secant parameter �0 applies to each circle. For circle (a) the value of �0 is the slope of the line OA, i.e. 35�. For circles (b) and (c) the values are 33� and 31�, respectively. 0 200 400 600 800 1000 1200 1400 1600 200 400 600 σ, σ ′ (kN/m2) τ (kN /m 2 ) Figure 4.18 Example 4.2. Table 4.4 All-round pressure (kN/m2) Principal stress difference (kN/m2) Pore water pressure (kN/m2) 150 192 80 300 341 154 450 504 222 Table 4.5 �3 �1 � 0 3 � 0 1 150 342 70 262 300 641 146 487 450 954 228 732 Shear strength of saturated clays 117 Tangent parameters can be obtained by approximating the curved envelope to a straight line over the stress range relevant to the problem. In Figure 4.19 a linear approximation has been drawn for the range of effective normal stress 200–300 kN/m2, giving parameters c0 ¼ 20 kN/m2 and �0 ¼ 29�. 4.5 THE CRITICAL-STATE CONCEPT The critical-state concept, given by Roscoe et al. [14], represents an idealization of the observed patterns of behaviour of saturated clays in triaxial compression tests. The concept relates the effective stresses and the corresponding specific volume (v ¼ 1þ e) of a clay during shearing under drained or undrained conditions, thus unifying the characteristics of shear strength and deformation. It was demonstrated that a char- acteristic surface exists which limits all possible states of the clay and that all effective stress paths reach or approach a line on that surface which defines the state at which yielding occurs at constant volume under constant effective stress. Stress paths are plotted with respect to principal stress difference (or deviator stress) and average effective principal stress, denoted by q0 and p0, respectively. Thus q0 ¼ ð�01 � �03Þ ð4:14Þ In the triaxial test the intermediate principal stress (�02) is equal to the minor principal stress (�03); therefore, the average principal stress is p0 ¼ 1 3 ð�01 þ 2�03Þ ð4:15Þ By algebraic manipulation it can be shown that ð�01 þ �03Þ ¼ 1 3 ð6p0 þ q0Þ ð4:16Þ 300 200 100 0 100 200 300 400 500 600 700 800 A B C (a) (b) (c) σ′ (kN/m2) τ (k N /m 2 ) Figure 4.19 Example 4.3. 118 Shear strength Effective, stress paths for a consolidated–undrained test and a drained triaxial test (C0A0 and C0B0, respectively) on specimens of a normally consolidated clay are shown in Figure 4.20(a), the coordinate axes being q0 and p0 (Equations 4.14 and 4.15). Each specimen was allowed to consolidate under the same all-round pressure p0c and failure occurs at A0 and B0, respectively, these points lying on or close to a straight line OS0 through the origin, i.e. failure occurs if the stress path reaches this line. If a series of consolidated–undrained tests were carried out on specimens each consoli- dated to a different value of p0c, the stress paths would all have similar shapes to that shown in Figure 4.20(a). The stress paths for a series of drained tests would be straight lines rising from the points representing p0c at a slope of 3 vertical to 1 horizontal (because if there is no change in �03, changes in q 0 and p0 are then in the ratio 3:1). In all these tests the state of stress at failure would lie on or close to the straight line OS0. Figure 4.20 Critical-state concept: normally consolidated clays. The critical-state concept 119 The isotropic consolidation curve (NN) for the normally consolidated clay would have the form shown in Figure 4.20(b), the coordinate axes being v and p0. The volume of the specimen during the application of the principal stress difference in a consolidated–undrained test on a saturated clay remains constant, and therefore the relationship between v and p0 will be represented by a horizontal line starting from the point C on the consolidation curve corresponding to p0c and finishing at the point A 00 representing the value of p0 at failure. During a drained test the volume of the speci- men decreases and the relationship between v and p0 will be represented by a curve CB00. If a series of consolidated–undrained and drained tests were carried out on specimens each consolidated to a different value