1.5 STRESS-PATH METHOD OF SETTLEMENT CALCULATION 1.5 STRESS-PATH METHOD OF SETTLEMENT CALCULATION

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1 Module 6 Lecture 40 Evaluation of Soil Settlement - 6 Topics 1.5 STRESS-PATH METHOD OF SETTLEMENT CALCULATION Definition of Stress Path 1.5. Stress and Strain Path for Consolidated Undrained Undrained Triaxial Tests Calculation of Settlement from Stress Point 1.5 STRESS-PATH METHOD OF SETTLEMENT CALCULATION Lambe (1964) proposed a technique for calculation of settlement in clay which takes into account both the immediate and the primary consolidation settlements. This is called the stress-path method Definition of Stress Path In order to understand what a stress path is, consider a normally consolidated clay specimen subjected to a consolidated drained triaxial test (Figure 6.31a). At any time during the test, the stress condition in the specimen can be represented by a Mohr s circle (Figure 6.31b). Note here that, in a drained test, total stress is equal to effective stress. So, σ 3 = σ 3 (minor principal stress) σ 1 = σ 3 + σ = σ 1 (major principal stress) Dept. of Civil Engg. Indian Institute of Technology, Kanpur 1

2 Figure Definition of stress path At failure, the Mohr s circle will touch a line that is the Mohr-Coulomb failure envelope; this makes an angle with the normal stress axis ( is the soil friction angle). We now consider another concept; without drawing the Mohr s circles, we may represent each one by a point defined by the coordinates p = σ 1+σ 3 (59) And q = σ 1 σ 3 (60) This is shown in Figure 6.31b for the smaller of the Mohr s circles. If the points with p and q coordinates of all the Mohr s circles are joined, this will result in the line AB. This line is called a stress path. The straight line joining the origin and the point B will be defined here as the K f line. The K f line makes an angle α with the normal stress axis. Now, tan α = BC = (σ 1 f σ 3 f )/ OC (σ 1 f +σ 3 f )/ (61) Where σ 1 f and σ 3 f are the effective major and minor principal stresses at failure. Similarly, sin = DC = (σ 1 f σ 3 f )/ OC (σ 1 f +σ 3 f )/ (6) From equations (61 and 6), we obtain tanα = sin (63) Dept. of Civil Engg. Indian Institute of Technology, Kanpur

3 Again let us consider a case where a soil specimen is subjected to an oedometer (one-dimensional consolidation) type of loading (Figure 6.3). For this case, we can write Figure 6.3 Determination of the slope of K o line σ 3 = K o σ 1 (64) Where K o is the at-rest earth pressure coefficient and can be given by the expression (Jaky, 1944) K o = 1 sin (65) For the Mohr s circle shown in Figure 6. 3, the coordinates of point E can be given by q = σ 1 σ 3 p = σ 1+σ 3 = σ 1(1 K o ) = σ 1(1+K o ) 1 q Thus, β = tan p = tan 1 1 K o 1+K o (66) Where, β is the angle that the line OE (K o line) makes with the normal stress axis. For purposes of comparison, the K o line is also shown in Figure 6. 31b. In any particular problem, if a stress path is given in a p vs. q plot, we should be able to determine the values of the major and minor principal stresses for any given point on the stress path. This is demonstrated in Figure 6. 33, in which ABC is an effective stress path. Dept. of Civil Engg. Indian Institute of Technology, Kanpur 3

Figure 6. 33 Determination of major and minor principal stresses for a point on a stress path 1.5.

4 Figure Determination of major and minor principal stresses for a point on a stress path 1.5. Stress and Strain Path for Consolidated Undrained Triaxial Tests Consider a clay specimen consolidated under an isotropic stress σ 3 = σ 3 in a triaxial test. When a deviator stress σ is applied on the specimen and drainage is not permitted there will be an increase in the pore water pressure, u (Figure 6. 34a). Figure Stress path for consolidation undrained triaxial test u = A σ (67) Dept. of Civil Engg. Indian Institute of Technology, Kanpur 4

5 Where A is the pore water pressure parameter (chapter 4). At this time, the effective major and minor principal stresses can be given by: Minor effective principal stress = σ 3 = σ 3 u And Major effective principal stress = σ 1 = σ 1 u = σ 3 + σ u Mohr s circles for the total and effective stress at any time of deviator stress application are shown in Figure 6. 34b. (Mohr s circle no. 1 is for total stress and no. is for effective stress). Point B on the effective stress Mohr s circle has the coordinates p and q. If the deviator stress is increased until failure occurs, the effective-stresses Mohr s circle at failure will be represented by circle No. 3 as shown in Figure 6. 34b, and the effective stress path will be represented by the line ABC The general nature of the effective-stress path will depend on the value of the pore pressure parameter A. this is shown in Figure Calculation of Settlement from Stress Point In the calculation of settlement from stress paths, it is assumed that for normally consolidated clays, the volume change between any two points on a p vs. q plot is independent of the path followed. This is explained in Figure For a soil sample, the volume changes between stress paths AB, GH, CD, and CI, for example, are all the same. However, the axial strains will be different. With this basic assumption, we can now proceed to determine the settlement. Figure Volume change between two points of a p vs. q plot Dept. of Civil Engg. Indian Institute of Technology, Kanpur 5

Consolidated undrained traixial tests on these samples at several confining pressures, σ 3 are conducted, along with a standard one-dimensional consolidated test.

6 Consolidated undrained traixial tests on these samples at several confining pressures, σ 3 are conducted, along with a standard one-dimensional consolidated test. The stress-strain contours are plotted on the basis of the CU triaxial test results. The standard one-dimensional consolidation test results with give us the values of compression index C c. For an example, let Figure represent the stress-strain contours for a given normally consolidated clay sample obtained from an average depth of a clay layer. Also let C c = 0.5 and e o = 0.9. the drained friction angle (determined from CU tests) is From equation (66), Figure β = tan 1 1 K o 1+K o And K o = 1 sin = 1 sin 30 = 0.5. So β = tan = Knowing the value of β we can now plot the K o line in Figure Also note that tan α = sin. since = 30, tan α = 0.5. So α = Let us calculate the settlement in the clay layer for the following conditions (Figure 6. 37): 1. In situ average effective overburden pressure = σ 1 = 75 kn/m.. Total thickness of clay layer = H t = 3 m. Due to the construction of a structure, a increase of the total major and minor principal stresses at an average depth are: Dept. of Civil Engg. Indian Institute of Technology, Kanpur 6

7 σ 1 = 40 kn/m σ 3 = 5 kn/m (assuming that the load is applied instantaneously). The in situ minor principal stress (at-rest pressure) is σ 3 = σ 3 = K o σ 1 = = 37.5 kn/m. So, before loading, p = σ 1+σ 3 q = σ 1 σ 3 = = = 56.5 kn/m = kn/m The stress conditions before loading can now be plotted in Figure from the above values of p and q. This is point A. Since the stress paths are geometrically similar, we can plot BAC, which is the stress path through A. also since the loading is instantaneous (i.e., undrained), the stress conditions in clay, represented by the p vs. q plot immediately after loading, will fall on the stress path BAC. Immediately after loading, σ 1 = = 115 kn/m σ 3 = = 6,5 kn/m So, q = σ 1 σ 3 = σ 1 σ = 6.5 kn/m With this value of q, we locate the point D. at the end of consolidation, σ 1 = σ 1 = 115kN/m σ 3 = σ 3 = 6.5kN/m So, p = σ 1+σ 3 = = kn/m and q = 6.5 kn/m The preceding values of p and q are plotted at point E. FEG is a geometrically similar stress path drawn though E, ADE is the effective stress path that a soil element, at average depth of the clay layer, will follow. AD represents the elastic settlement, and DE represents that consolidation settlement. For elastic settlement (stress path A to D), S e = ε 1 at D ε 1 at A H t = = 0.09m For consolidation settlement (stress path D to E), based on our previous assumption the volumetric strain between D and E is the same as the volumetric strain between A and H is on the K o line. For point A, σ 1 = 75 kn/m ; and for point H, σ 1 = 118 kn/m. So the volumetric strain, ε v, is ε v = e = C c log (118/75) 0.9 log (118/75) = = e o Dept. of Civil Engg. Indian Institute of Technology, Kanpur 7

8 The axial strain ε 1 along a horizontal stress path is about one-third the volumetric strain along the K 0 line, or ε 1 = 1 3 ε v = = So, the consolidation settlement is S c = H t = = m And hence the total settlement is S e + S c = = m Another type of loading condition is also of some interest. Suppose that the stress increase at the average depth of the clay layer was carried out in tow steps: (1) instantaneous load application, resulting in stress increases of σ 1 = 40 kn/m and σ 3 = 5 kn/m (stress path AD), followed by () a gradual load increase, which results in a stress path DI (Figure 6. 37). As before, the undrained shear along stress path AD will produce an axial strain of the volumetric strains for stress paths DI and AH will be the same; so ε v = The axial strain ε 1 for the stress path DI can be given by the relation (based on the theory of elasticity) ε 1 ε v = 1+K o KK o (1 K o ) 1+K (68) Where K = σ 3 /σ 1 for the point I. in this case, σ 3 = 4 kn/m and σ 1 = 13 kn/m. So, K = 4 13 = ε 1 = ε (0.5) = = 1.38 ε v [ ] Or ε 1 = = Hence, the total settlement due to the loading is equal to S = [ ε 1 along AD + (ε 1 along DI)H t = H t = 0.066H t Dept. of Civil Engg. Indian Institute of Technology, Kanpur 8

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

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