Evaluation of Soil Settlement

NPTEL- Advanced Geotechnical Engineering

Module 6 EVALUATION OF SOIL SETTLEMENT (Lectures 35 to 40)

Topics 1.1 INTRODUCTION 1.2 IMMEDIATE SETTLEMENT 1.2.1

Immediate Settlement from Theory of Elasticity 

Settlement due to a concentrated point load at the surface



Settlement at the surface due to a uniformly loaded flexible circular area



Settlement at the surface due to a uniformly loaded flexible rectangular area



1.2.2

Settlement of a flexible load area on an elastic layer of finite thickness

Settlement of rigid footings

1.2.3 Determination of Young’s Modulus 1.2.4 Settlement Prediction in Sand by Empirical Correlation 1.2.5 Calculation of Immediate Settlement in Granular Soil Using Simplified Strain Influence Factor

1.3 PRIMARY CONSOLIDATION SETTLEMENT 1.3.1 One-Dimensional Consolidation Settlement Calculation 

Method A



Method B

1.3.2 Skempton-Bjerrum Modification for Calculation of Consolidation Settlement 1.3.3 Settlement of Overconsolidated Clays 1.3.4 Precompression for Improving Foundation Soils

1.4 SECONDARY CONSOLIDATION SETTLEMENT 1.5 STRESS-PATH METHOD OF SETTLEMENT CALCULATION 1.5.1 Definition of Stress Path 1.5.2

Stress and Strain Path for Consolidated Undrained Triaxial Tests

Dept. of Civil Engg. Indian Institute of Technology, Kanpur

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1.5.3 Calculation of Settlement from Stress Point

PROBLEMS

Dept. of Civil Engg. Indian Institute of Technology, Kanpur

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NPTEL- Advanced Geotechnical Engineering

Module 6 Lecture 35 Evaluation of Soil Settlement -1 Topics 1.1 INTRODUCTION 1.2 IMMEDIATE SETTLEMENT 1.2.1

Immediate Settlement from Theory of Elasticity 

Settlement due to a concentrated point load at the surface



Settlement at the surface due to a uniformly loaded flexible circular area



Settlement at the surface due to a uniformly loaded flexible rectangular area



Summary of elastic settlement at the ground surface (z = 0) due to uniformly distributed vertical loads on flexible areas



Settlement of a flexible load area on an elastic layer of finite thickness

1.1 INTRODUCTION The increase of stress in soil layers due to the load imposed by various structures at the foundation level will always be accompanied by some strain, which will result in the settlement of the structures. The various aspects of settlement calculation are analyzed in this chapter. In general, the total settlement S of a foundation can be given as 𝑆 = 𝑆𝑒 + 𝑆𝑐 + 𝑆𝑠

(1)

Where 𝑆𝑒 = Immediate settlement 𝑆𝑐 = Primary consolidation settlement 𝑆𝑠 = Secondary consolidation settlement The immediate settlement is sometimes referred to as the elastic settlement. In granular soils this is the predominant part of the settlement, whereas in saturated inorganic silts and clays the primary consolidation settlement probably predominates. The secondary consolidation settlement forms the major part of the total settlement in highly organic soils and peats. Dept. of Civil Engg. Indian Institute of Technology, Kanpur

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1.2 IMMEDIATE SETTLEMENT 1.2.1 Immediate Settlement from Theory of Elasticity Settlement due to a concentrated point load at the surface



For elastic settlement due to a concentrated point load (Figure 6. 1), the strain at a depth z can be given in cylindrical coordinates, by

Figure 6.1 Elastic settlement due to a concentrated point load

1

𝑒𝑧 = 𝐸 [𝜎𝑧 − 𝑣 𝜎𝑟 + 𝜎𝜃 ]

(2)

Where E is the Young’s modulus of the soil. The expressions for 𝜎𝑧 , 𝜎𝑟 , and 𝜎𝜃 are given in equations , respectively in earlier modules. Substitution of these in equation (2) and simplification yields 𝑄

𝜖𝑧 = 2𝜋𝐸

3(1+𝑣)𝑟 2 𝑧 (𝑟 2 +𝑧 2 )5/2

−

3+𝑣 1−2𝑣 𝑧

(3)

(𝑟 2 +𝑧 2 )3/2

The settlement at a depth z can be found by integration equation (3): 𝑆𝑒 =

𝑄

𝜖𝑧 𝑑𝑧 = 2𝜋𝐸

(1+𝑣)𝑧 2

2 1−𝑣 2

+ (𝑟 2 +𝑧 2 )1/2 (𝑟 2 +𝑧 2 )3/2

The settlement at the surface can be evaluated by putting z = 0 in the above equation: 𝑄

𝑆𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 = 𝜋𝐸𝑟 (1 − 𝑣 2 )



(4)

Settlement at the surface due to a uniformly loaded flexible circular area

Dept. of Civil Engg. Indian Institute of Technology, Kanpur

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The elastic settlement due to a uniformly loaded circular area (Figure 6.2) can be determined by using the same procedure as discussed for a point load, which involves determination of the strain 𝜖𝑧 from the equation and determination of the settlement by integration with respect to z.

Figure 6.2 Elastic settlement due to a uniformly loaded circular area

1

𝜖𝑧 = 𝐸 [𝜎𝑧 − 𝑣(𝜎𝑟 + 𝜎𝜃 ) Substitution of the relation for 𝜎𝑧 , 𝜎𝑟 , and 𝜎𝜃 in the preceding equation for strain and simplification gives (Ahlvin and Ulery, 1962) where q is the load per unit area. A’ and B’ are nondimensional and are functions of z/b and s/b; their values are given in table 7 and 8 in chapter 3. 𝜖𝑧 = 𝑞

1+𝑣 𝐸

[ 1 − 2𝑣 𝐴′ + 𝐵 ′ ]

(5)

The vertical deflection at a depth z can be obtained by integration of equation 6 as where 𝐼1 = 𝐴′ and b is the radius of the circular loaded area. The numerical values of 𝐼2 (which is a function of z/b and s/b) are given in table 1. 𝑆𝑒 = 𝑞

1+𝑣 𝐸

𝑏

𝑧

𝐼 𝑏 1

+ (1 − 𝑣)𝐼2

(6)

From equation (6) it follows that the settlement at the surface (i. e. , at z = 0) is 𝑆𝑒 = 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑞𝑏

1−𝑣 2 𝐸

𝐼2

(7)

The term 𝐼2 in equation (7) is usually referred to as the influence number. For saturated clays, we may assume 𝑣 = 0.5. so, at the center of the loaded area (i. e. , s/b = 0), 𝐼2 = 2 and 𝑆𝑒 = 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑐𝑒𝑛𝑡𝑒𝑟 =

1.5𝑞𝑏 𝐸

=

0.75𝑞𝐵 𝐸

Dept. of Civil Engg. Indian Institute of Technology, Kanpur

(8)

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Table 1 Values of 𝑰𝟐 (After Ahlvin and Ulery 1962) 𝑠/𝑏 𝑧/𝑏 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.2 1.5 2 2.5 3 4 5 6 7 8 9 10

0 2.0 1.80998 1.63961 1.48806 1.35407 1.23607 1.13238 1.04131 .96125 .89072 .82843 .72410 .60555 .47214 .38518 .32457 .24620 .19805 .16554 .14217 .12448 .11079

0.2 1.97987 1.79018 1.62068 1.470044 1.33802 1.22176 1.11998 1.03037 .95175 .88251 .85005 .71882 ,60233 .47022 ,38403 .32403 .24588 .19785

0.4 1.91751 1.72886 1.56242 1.40979 1.28963 1.17894 1.08350 .99794 .92386 .85856 .80465 .70370 .57246 .44512 .38098 .32184 .24820

0.6 1.80575 1.61961 1.46001 1.32442 1.20822 1.10830 1.02154 .91049 .87928 .82616 .76809 .67937 .57633 .45656 .37608 .31887 .25128

0.8 1.62553 1.44711 1.30614 1.19210 1.09555 1.01312 .94120 .87742 .82136 .77950 .72587 .64814 .55559 .44502 .36940 .31464 .24168

1 1.27319 1.18107 1.09996 1.02740 .96202 .90298 .84917 .80030 .75571 .71495 .67769 .61187 .53138 .43202 .36155 .30969 .23932 .19455 .16326 .14077 .12352 .10989

1.2 .93676 .92670 .90098 .86726 .83042 .79308 .75653 .72143 .68809 .65677 .62701 .57329 .50496 .41702 .35243 .30381 .23668

1.5 .71185 .70888 .70074 .68823 .67238 .65429 .63469 .61442 .59398 .57361 .55364 .51552 .46379 .39242 .33698 .29364 .23164

.09900

2 .51671 .51627 .51382 .50966 .50412 .49728 .48061

.45122 .43013 .39872 .35054 .30913 .27453 .22188 .18450 .15750 .13699 .12112 .10854 .09820

where 𝐵 = 2𝑏 is the diameter of the loaded area. At the edge of the loaded area (𝑖. 𝑒. , 𝑧/𝑏 = 0and s/b = 1), I2 = 1.27 and 𝑆𝑒 = 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑒𝑑𝑔𝑒 = 1.27 0.75

𝑞𝑏 𝐸

= 0.95

𝑞𝑏 𝐸

=

0.475𝑞𝐵

(9)

𝐸

The average surface settlement is 𝑆𝑒 = 𝑠𝑢𝑟𝑓𝑎𝑐𝑒, 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 = 0.85𝑆𝑒 (𝑠𝑢𝑟𝑓𝑎𝑐𝑒, 𝑐𝑒𝑛𝑡𝑒𝑟)

(10)

Settlement at the surface due to a uniformly loaded flexible rectangular area



The elastic deformation in the vertical direction at the corner of a uniformly loaded rectangular area of size 𝐿 × 𝐵 can be obtained by proper integration of the expression for strain. The deformation at a depth z below the corner of the rectangular area can be expressed in the form (Harr, 1966) 𝑠/𝑏 𝑧/𝑏 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.2 1.5 2 2.5 3 4 5 6 7 8 9

3 .33815 .33794 .33726 .33638

4 .25200 .25184 .25162 .25124

5 .20045 .20081 .20072

6 .16626

7 .14315

8 .12576

.16688

.14288

.12512

.33293

.24996

.19982

.16668

.14273

.31877 .31162 .29945 .27740 .25550 .23487 .19908 .17080 .14868 .13097 .11680 .10548

.24386 .24070 .23495 .22418 .21208 .19977 .17640 .15575 .13842 .12404 .11176 .10161

.19673 .19520 ..19053 .18618 .17898 .17154 .15596 .14130 .12792 .11620 .10600 .09702

.16516 .16369 .16199 .15846 .15395 .14919 .13864 .12785 .11778 .10843 .09976 .09234

.14182 .14099 .14058 .13762 .13463 .13119 .12396 .11615 .10836 .10101 .09400 .08784

Dept. of Civil Engg. Indian Institute of Technology, Kanpur

10 .09918

12 .08346

14 .07023

.12493

.09996

.08295

.07123

.12394 .12350 .12281 .12124 .11928 .11694 .11172 .10585 .09990 .09387 .08848 .08298

.09952

.08292

.07104

.09876 .09792 .09700 .09558 .09300 .08915 .08562 .08197 .07800 .07407

.08270 .08196 .08115 .08061 .07864 .07675 .07452 .07210 .06928 .06678

.07064 .07026 .06980 .06897 .06848 .06695 .06522 .06377 .06200 .05976

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NPTEL- Advanced Geotechnical Engineering 10

.09510

.09290

.08980

𝑞𝐵

𝑆𝑒 𝑐𝑜𝑟𝑛𝑒𝑟 = 2𝐸 (1 − 𝑣 2 ) 𝐼3 − 1

Where 𝐼3 = 𝜋 𝐼𝑛

1+𝑚 12 +𝑛 12 +𝑚 1 1+𝑚 12 +𝑛 12 −𝑚 1

.08300

1−2𝑣

.08180

.07710

𝐼4

1−𝑣

+ 𝑚 𝐼𝑛

(11) 1+𝑚 12 +𝑛 12 +1 1+𝑚 12 +𝑛 12 −1

(12) 𝐼4 =

𝑛1 𝜋

𝑡𝑎𝑛−1

𝑚1

(13)

𝑛 1 1+𝑚 12 +𝑛 12

𝐿

𝑚1 = 𝐵

(14)

𝑧

𝑛1 = 𝐵

(15)

Values of 𝐼3 and I4 are given in table 2. For elastic surface settlement at the corner of a rectangular area, substituting 𝑧/𝑏 = 𝑛1 = 0 in equation (11) and make the necessary calculations; thus, 𝑞𝐵

𝑆𝑒 𝑐𝑜𝑟𝑛𝑒𝑟 = 2𝐸 (1 − 𝑣 2 )𝐼3

(16)

The settlement at the surface for the center of a rectangular area (Figure 6.3) can be found by adding the settlement for the corner of four rectangular areas of dimension 𝐿/2 × 𝐵/2. Thus, from equation (11),

Figure 6.3 Determination of settlement at the center of a rectangular area of dimensions 𝐿 × 𝐵

𝑆𝑒 𝑐𝑒𝑛𝑡𝑒𝑟 = 4

𝑞(𝐵/2) 2𝐸

(1 − 𝑣 2 )𝐼3 =

𝑞𝐵 𝐸

(1 − 𝑣 2 )𝐼3

(17)

The average surface settlement can be obtained as Dept. of Civil Engg. Indian Institute of Technology, Kanpur

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𝑆𝑒 𝑎𝑣𝑒𝑟𝑎𝑔𝑒, 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 = 0.848𝑆𝑒 (𝑐𝑒𝑛𝑡𝑒𝑟, 𝑠𝑢𝑟𝑓𝑎𝑐𝑒)

(18)

Summary of elastic settlement at the ground surface (z = 0) due to uniformly



distributed vertical loads on flexible areas For circular areas: 𝑆𝑒 = 𝑞𝐵

1−𝑣 2 2𝐸

𝐼2

Where 𝐵 =Diameter of circular loaded area 𝐼2 = 2 (at center) 𝐼2 = 1.27(at edge) 𝐼2 = 0.85 × 2 = 1.7(average) For rectangular areas, on the basic equations (16) to (18) we can write 𝑆𝑒

𝑞𝐵 𝐸

(1 − 𝑣 2 ) 𝐼5

(19)

Where 𝐼5 = 𝐼3 (at center) 𝐼5 = 12𝐼3 (at edge) 𝐼5 ≈ 0.848𝐼3 (average) Table 3 gives the values of 𝐼5 for various 𝐿/𝐵 ratios.



Settlement of a flexible load area on an elastic layer of finite thickness

For the settlement calculation, it was assumed that the elastic soil layer extends to an infinite depth. However, if the elastic soil layer is underlain by a rigid incompressible base at a depth H (Figure 6.4), the settlement can be approximately calculated as 𝑆𝑒 = 𝑆𝑒(𝑧=0) − 𝑆𝑒(𝑧=𝐻)

(20)

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Figure 6.4 Flexible loaded area over an elastic soil layer of finite thickness

Where 𝑆𝑒(𝑧=0) and 𝑆𝑒(𝑧=𝐻) are the settlements at the surface and at z = H, respectively. Foundations are almost never placed at the ground surface, but at some depth 𝐷𝑓 (Figure 6.5). Hence, a correction needs to be applied to the settlement values calculated on the assumption that the load is applied at the ground surface. Fox (1948) proposed a correction factor for this which is a function of 𝐷𝑓 /𝐵, 𝐿/𝐵 and Poisson’s ratio v. thus,

Figure 6.5 Average immediate settlement for a flexible rectangular loaded area located at a depth 𝐷𝑓 from the ground surface Dept. of Civil Engg. Indian Institute of Technology, Kanpur

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𝑆 ′ 𝑒 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 = 𝐼6 𝑆𝑒 (𝑎𝑣𝑒𝑟𝑎𝑔𝑒)

(21)

Where 𝐼6 = correction factor for foundation depth, 𝐷𝑓 𝑆′𝑒 = corrected elastic settlement of foundation 𝑆𝑒 = elastic settlement of foundation calculated on assumption that load is applied at ground surface By computer programming of the equation proposed by Fox, Bowles (1977) obtained the values of 𝐼6 for various values of 𝐷𝑓 /𝐵 length-to-width ratio of the foundation, and Poisson’s ratio of the soil layer. These values are shown in Figure 6.6. Table 3 Values of 𝑰𝟓 𝐼5

𝐿/𝐵 1 2 3 5 10 20 50 100

Center 1.122 1.532 1.783 2.105 2.544 2.985 3.568 4.010

Corner 0.561 0.766 0.892 1.053 1.272 1.493 1.784 2.005

Average 0.951 1.299 1.512 1.785 2.157 2.531 3.026 3.400

Janbu et al, (1956) proposed a generalized equation for average immediate settlement for uniformly loaded flexible footings in the form 𝑆𝑒 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 = 𝜇1 𝜇0

𝑞𝐵 𝐸

(for𝑣 = 0.5)

(22)

Where 𝜇1 = Correction factor for finite thickness of elastic soil layer, H, as shown in Figure 6.5. 𝜇0 = Correction factor for depth of embankment of footing, 𝐷𝑓 , as shown in Figure 6.5. 𝐵 = Width of rectangular loaded area of diameter of circular loaded area

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Figure 6.6 Correction factor for the depth of embedment of the foundation. (Bowles 1977) Christian and Carrier (1978) made a critical evaluation of equation (22), the details of which will not be presented here. However, they suggested that for 𝑣 = 0.5, equation (22) could be retained for immediate settlement calculations with a modification of the values of 𝜇1 and μ2 . The modified values of 𝜇1 are based on the work of Groud (1972) and those for 𝜇0 are based on the work of Burland (1970). These are shown in Figure 6.7. Christian and Carrier inferred that these values are generally adequate for circular and rectangular footings.

Figure 6.7 Improved chart for use in equation (22). (After Christian and Carrier 1978) Another general method for estimation of immediate settlement is to divide the underlying soil into 𝓃 layers of finite thicknesses (Figure 6.5). It the strain at the middle of each layer can be calculated. The total immediate settlement can be obtained as where ∆𝑧(𝑖) is the thickness of the 𝑖𝑡𝑕 layer and 𝜖𝑧(𝑖) is the vertical strain at the middle of the 𝑖𝑡𝑕 layer. Dept. of Civil Engg. Indian Institute of Technology, Kanpur

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𝑆𝑒 =

𝑖=𝑛 𝑖=1

∆𝑧(𝑖) 𝜖𝑧(𝑖)

(23)

The method of using equation (23) is demonstrated in example 2.

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Module 6 Lecture 36 Evaluation of Soil Settlement -2 Topics 1.2.2 Settlement of rigid footings 1.2.3 Determination of Young’s Modulus 1.2.4 Settlement of rigid footings The immediate surface settlement of a uniformly loaded rigid footing (Figure 6.8) is about 7% less than the average surface settlement of a flexible footing of similar dimensions (Schleicher, (1926). So, based on this simplified conclusion, the following expression can be written as: Equation (24) for circular footing Equation (25) for rectangular footing

Figure 6.8 Immediate settlement of rigid footing

For a uniformly loaded rigid circular footing of radius b (note 𝑏 = 𝐵/2): 𝑆𝑒(𝑧=0) = 𝑞𝑏

1−𝑣 2 𝐸

𝐼7

(24)

Where 𝐼7 = 0.93𝐼2 average settlement = 0.93 1.7 = 1.58. For a uniformly loaded rigid rectangular footing of dimension 𝐿 × 𝐵: 𝑆𝑒(𝑧=0) = 𝑞𝑏

1−𝑣 2 𝐸

𝐼8

Dept. of Civil Engg. Indian Institute of Technology, Kanpur

(25) 1

NPTEL- Advanced Geotechnical Engineering

Where 𝐼8 = 0.93𝐼5 (average settlement). The values of 𝐼8 are given in table 4. Example 1 A square tank is shown in Figure 6.9. Assuming flexible loading conditions, find the average immediate (elastic) settlement of the tank for the following conditions: (a) 𝐷𝑓 = 0, 𝐻 = ∞ (b) 𝐷𝑓 = 1.5, 𝐻 = ∞ (c) 𝐷𝑓 = 1.5, 𝐻 = 10𝑚

Figure 6.9 Solution Part (a): 𝑆𝑒 average = (𝑞𝐵/𝐸)(1 − 𝑣 2 )𝐼5 ; 𝐵 = 3𝑚; 𝐿/𝐵 = 3/3 = 1; and 𝐼5 = 0.951. so, 100 3

𝑆𝑒 average = 21,000 1 − 0. 32 951 = 0.0124𝑚 = 12.4𝑚𝑚 Part (b): From equation (18), 𝑆 ′ 𝑒 average = 𝐼6 𝑆𝑒 average ; 𝐷𝑓 /𝐵 = 1.5/3 = 0.5; and I6 = 0.77 (Figure 6. 6). So, 𝑆′𝑒 = average = 0.77 12.4𝑚𝑚 = 9.55𝑚𝑚 Table 4 Values of 𝑰𝟖

𝐿/𝐵 1 2 3 5 10 20 50 100

𝐼8 0.884 1.208 1.406 1.660 2.006 2.353 2.814 3.162

Part (c): From equation (20), 𝑆𝑒 = 𝑆𝑒(𝑧=0) − 𝑆𝑒(𝑧=𝐻) From equation (11), Dept. of Civil Engg. Indian Institute of Technology, Kanpur

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NPTEL- Advanced Geotechnical Engineering

𝑆𝑒 corner =

𝑞𝐵 2𝐸

(1 − 𝑣 2 ) 𝐼3 −

1−2𝑣 1−𝑣

𝐼4

Determine the values of 𝑆𝑒 below the corner at 𝑧 = 10𝑚 for one loaded area of dimension 1.5 × 1.5𝑚 (Figure 6.10) and then multiply that by 4 to obtain the displacement at the center of the tank at depth 𝑧 = 10𝑚. So for a loaded area of 1.5 × 1.5𝑚.

Figure 6. 10 1.5

𝑚 = 1.5 = 1

10

𝑛 = 1.5 = 6.67

From table 2, 𝐼3 = 0.189 and 𝐼4 = 0.047. so, 100 1.5

𝑆𝑒 = 2

21,000

1 − 0.32

0.189 −

1−0.6 1−0.3

0.047 = 0.00053 𝑚

For the whole loaded area of 3 × 3𝑚, the elastic settlement below the center at a depth of 𝑧 = 10𝑚 is equal to 4 × 0.00053 = 0.00212 𝑚. Thus, 𝑆𝑒 (average) without considering the depth effects is 12.4 − 0.848 0.00212 1000 = 10.6𝑚𝑚. now, 𝑆 ′ 𝑒 average = I6 Se 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 ; 𝑎𝑙𝑠𝑜, I6 = 0.77 from part (b). So, Example 2 For the tank shown in Figure 6. 11, (a) Determine the immediate settlement at the center of the tank by using equation (6). (b) Determine the immediate settlement by using equation (23). Divide the underlying soil into three layers of equal thickness of 3 m.

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Figure 6. 11 Solution Part (a): From equation (6), 𝑆𝑒 =

𝑞(1+𝑣) 𝐸

𝑏

𝑧 𝐼 𝑏 1

+ (1 − 𝑣)𝐼2

From equation (20), 𝑆𝑒 = 𝑆𝑒(𝑧=0) − 𝑆𝑒(𝑧=9 𝑚) For 𝑧/𝑏 = 0 and 𝑠/𝑏 = 0, 𝐼1 = 1, and I2 = 2 (Table 1); so, 𝑆𝑒(𝑧=0) =

100 1+0.3 21,000

1.5

1 − 0.3 2 = 0.013𝑚 = 13𝑚𝑚

For 𝑧/𝑏 = 9/1.5 = 6 and 𝑠/𝑏 = 0, 𝐼1 = 0.01361, and I2 = 0.16554; so 𝑆𝑒(𝑧=9 𝑚) =

100 1+0.3 (1.5) 21,000

6 0.01361 + 1 − 0.3 0.16554 = 0.00183𝑚 = 1.83𝑚𝑚

Hence, 𝑆𝑒 = 13 − 18.3 = 11.17𝑚𝑚 Part (b): From equation (5), 𝜖𝑧 =

𝑞(1+𝑣) 𝐸

[ 1 − 2𝑣 𝐴′ + 𝐵 ′ ]

Layer 1: For, 𝑧/𝑏 = 1.5/1.5 = 1 and 𝑠/𝑏 = 0, 𝐴′ = 0.29289, and B′ = 0.35355: 𝜖𝑧 =

100(1+0.3) 21,000

1 − 0.6 0.29289 + 0.35355 = 0.00291]

Layer 2: For, 𝑧/𝑏 = 4.5/1.5 = 3 and 𝑠/𝑏 = 0, 𝐴′ = 0.05132, and B′ = 0.09487: Dept. of Civil Engg. Indian Institute of Technology, Kanpur

4

NPTEL- Advanced Geotechnical Engineering

𝜖𝑧(2) =

100(1+0.3) 21,000

1 − 0.6 0.05132 + 0.09487 = 0.00071]

Layer 3: For 𝑧/𝑏 = 7.5/1.5 = 5 and 𝑠/𝑏 = 0, 𝐴′ = 0.01942, and B′ = 0.03772: 𝜖𝑧(3) =

100(1+0.3) 21,000

1 − 0.6 0.01942 + 0.03772 = 0.00028]

Similarly, the settlement can be calculated for part c The final stages in the calculation are tabulated below: layer no. i

Layer thickness ∆𝑧(𝑖) 𝑚

Strain at the center of the layer, 𝜖𝑧(𝑖)

𝜖𝑧(𝑖) ∆𝑧(𝑖) , 𝑚

1 2 3

3 3 3

0.00291 0.00071 0.00028

0.00873 0.00213 0.00084 Σ 0.0117 𝑚 = 11.7 𝑚𝑚

1.2.3 Determination of Young’s Modulus The equations derived for calculation of immediate settlement require a value of the Young’s modulus E for the soil layers involved. It is difficult to obtain the correct value of E since it increases with the depth of soil, i.e., the effective overburden pressure. Some approximate recommended values of E and Poissons’s ratio v for granular soils are given in table 5.

Table 5 Recommended values of E and v (Harr. 1966)

Type of soil Sand (coarse) v = 0.15 Sand (medium coarse) 𝑣 = 0.2 Sand (fine grained) 𝑣 = 0.2 Sandy silt 𝑣 = 0.3 𝑡𝑜 0.35

Properties of soil* Φ 𝐸(𝑙𝑏/𝑖𝑛2 ) 𝐸(𝑘𝑁/𝑚2 ) Φ 𝐸(𝑙𝑏/𝑖𝑛2 ) 𝐸(𝑘𝑁/𝑚2 ) Φ 𝐸(𝑙𝑏/𝑖𝑛2 ) 𝐸(𝑘𝑁/𝑚2 ) Φ 𝐸(𝑙𝑏/𝑖𝑛2 ) 𝐸(𝑘𝑁/𝑚2 )

0.41 to 0.5 43 6,550 45,200 40 6,550 45,200 38 5,300 36,600 36 2,000 13,800

Void ratio e 0.51 to 0.6 40 5,700 39.300 38 5,700 39.300 36 4.000 27,600 34 1,700 11,700

0.61 to 0.70 38 4.700 32,400 35 4.700 32,400 32 3,400 23,500 30 1,450 10,000

*Conversion factor: 1 𝑙𝑏/𝑖𝑛2 = 6.9𝑘𝑁/𝑚2 (the value of 𝑘𝑁/𝑚2 have been rounded off). 𝜙 is the drained friction angle. More representative values of E and v can be obtained from triaxial compression tests of undisturbed samples collected from a depth equal to the width of the foundation measured from the bottom of the proposed foundation elevation. However, in cohesionless soils, it is usually the secant modulus from zero up to about half of the maximum deviator stress, 𝑖. 𝑒. , 𝐸 = Δ𝜎/𝑒, as shown in Figure 6. 12. Poissson’s ratio v can be calculated by measuring the axial compressive strain and the lateral strain during the triaxial testing. The deviator stress-strain curve can be approximately represented by a hyperbolic equation (Kondner, 1963): where a and b are constants for a given soil.

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Figure 6.12 Young’s modulus from triaxial test 𝜖

∆𝜎 = 𝑎+𝑏𝜖

(26)

For granular soils. Young’s modulus determined from triaxial test is approximately proportional to 𝜎𝑜𝑛 (Figure 6.12) or where 𝜎𝑜 is the hydrostatic confining pressure. 𝐸 ∝ 𝜎𝑜𝑛

(27)

A reasonable average value of n is about 0.5 (Lambe and Whitman, 1969). However, in practical cases the stresses in soil before loading are not isotropic, as shown in Figure 6.13. So Young’s modulus is approximately proportional to the square root of the mean principal stress (Lambe and Whitman 1969), i.e., where 𝜎𝑣 is the effective overburden pressure before application of the foundation load.

Dept. of Civil Engg. Indian Institute of Technology, Kanpur

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Figure 6.13 Stress conditions in soil before loading 𝐸∝

𝜎𝑣 +𝐾𝑜 𝜎𝑣 +𝐾𝑜 𝜎𝑣 3

=

𝜎𝑣

1+2𝐾𝑜 3

(28)

Due to the difficulty in obtaining undisturbed soil samples in cohesionless soils, a number of investigators have attempted to correlate the equivalent Young’s modulus 𝐸𝑠 with the conventional results obtained during field exploration program for calculation of static compression of sand. These conventional results are standard penetration number N and static dutch cone resistance 𝑞𝑐 . It must be pointed out that 𝐸𝑠 is some equivalent to the constrained modulus (odeometer modulus). Some of these correlations of 𝐸𝑠 with 𝑁 and 𝑞𝑐 are given in table 6 and 7. In-saturated clay soils the undrained Young’s modulus can be given by the relation where 𝛽 varies from about 500 to 1500 (Bjerrum 1972) and 𝑆𝑢 is the undrained cohesion. 𝐸 = 𝛽𝑆𝑢

(29)

Some typical values of 𝛽 determined from large-scale field tests are given in table 8.

Dept. of Civil Engg. Indian Institute of Technology, Kanpur

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Figure 6.14 Relationship between 𝐸/𝑆𝑢 and over consolidation ratio from CU tests on three clays determined from 𝐶𝐾𝑜 𝑈 −type direct shear tests. (After D’Appo;onia, et al. 1971)

Dept. of Civil Engg. Indian Institute of Technology, Kanpur

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Module 6 Lecture 37 Evaluation of Soil Settlement - 3 Topics 1.2.4 Settlement Prediction in Sand by Empirical Correlation 1.2.5 Calculation of Immediate Settlement in Granular Soil Using Simplified Strain Influence Factor 1.2.4 Settlement Prediction in Sand by Empirical Correlation Based on several field load tests, Terzaghi and Peck (1967) suggested that for similar intensities of load q on a footing where 𝑆𝑒 is the settlement of a footing with width B and 𝑠𝑒(1) is the settlement of a smaller footing with width 𝐵1 . The value of 𝐵1 is usually taken as 1 ft. 𝑆𝑒 =

𝑆𝐵

2

𝐵+𝐵1

𝑆𝑒(1)

(30)

Table 6 Young’s modulus for vertical static compression of sand from standard penetration number (After Mitchell and Gardner 1975). Reference Schultze and Meizer (1965)

Relationship* 𝐸𝑠 = 𝑣𝜎𝑜 0.522 𝑘𝑔/𝑐𝑚2 𝑣 = 246.2 log 𝑁 − 26.34 𝜎𝑜 + 375.6 ± 57.6 0 < 𝜎𝑜 < 1.2 𝑘𝑔/𝑐𝑚2 𝜎𝑜 = effective overburden pressure

Soil types Dry sand

Webb (1969)

𝐸𝑠 = 5 𝑁 + 15 ton/𝑓𝑡 2 𝐸𝑠 = 10/3 𝑁 + 5 ton/𝑓𝑡 2

Sand Clayey sand

Farrent (1963)

𝐸𝑠 = 40 + 𝐶 𝑁 − 6 𝑘𝑔/𝑐𝑚2 𝑁 > 15 𝐸𝑠 = 𝐶 𝑁 + 6 𝑘𝑔/𝑐𝑚2 𝑁 > 15

Trofimenkov (1974)

𝐸𝑠 = 350 𝑡𝑜 500 log 𝑁 𝑘𝑔/𝑐𝑚2

Silt with sand to gravel with sand Sand

Basis Penetration tests in field and in test shaft. Compressibility based on 𝑒, 𝑒𝑚𝑎𝑥 , and 𝑒𝑚𝑖𝑛 (Schultze and Moussa. (1961) Screw plate tests

Remarks Correlation coefficient = 0.730 for 77 tests

Below water table

Used in Greece

U.S.S.R. practice

Table 7 Equivalent Young’s modulus for vertical static compression of sand-static cone resistance (After Mitchell and Gardner 1975). Reference Buisman (1940) Trofimenkov

Relationship 𝐸𝑠 = 1.5𝑞𝑐

Soil type Sands

𝐸𝑠 = 2.5𝑞𝑐

Sand

Dept. of Civil Engg. Indian Institute of Technology, Kanpur

Remarks Overpredicts settlements by a factor of about 2 Lower limit 1

NPTEL- Advanced Geotechnical Engineering

(1964) De Beer (1967)

𝐸𝑠 = 100 + 5𝑞𝑐 𝐸𝑠 = 1.5𝑞𝑐

Schultze and Meizer (1965)

𝐸𝑠 =

Bachelier Parez (1965)

𝑣 = 310.1 log 𝑞𝑐 − 382.3𝜎𝑜 ± 60.3 ± 50.3 𝜎𝑜 = effective overburden pressure and 𝐸𝑠 = 𝛼𝑞𝑐 𝛼 = 0.8 𝑡𝑜 0.9 𝛼 = 1.3 𝑡𝑜 1.9 𝛼 = 3.8 𝑡𝑜 5.7 𝛼 = 7.7 𝐸𝑠 = 𝛼𝑞𝑐 𝛼 = 3 𝑡𝑜 12

Thomas (1968)

Webb (1969)

1 𝑣𝜎 0.522 𝑚𝑣 𝑜

1 2 1 𝐸𝑠 = 2

𝐸𝑠 =

𝑞𝑐 + 30 𝑡𝑜𝑛/𝑓𝑡 2 𝑞𝑐 + 15 𝑙𝑏/𝑓𝑡 2

Vesic (1970)

𝐸𝑠 = 2(1 + 𝐷𝑅 3 )𝑞𝑐 𝐷𝑟 = relative density

Schmertmann (1970) Bogdanovic (1973)

𝐸𝑠 = 2𝑞𝑐

Schmertmann (1974)

𝐸𝑠 = 𝛼𝑞𝑐 𝑞𝑐 > 40 𝑘𝑔/𝑐𝑚2 𝛼 = 1.5 20 < 𝑞𝑐 < 40 𝛼 = 1.5 𝑡𝑜 1.8 10 < 𝑞𝑐 < 20 𝛼 = 1.8 𝑡𝑜 2.5 5 < 𝑞𝑐 < 10 𝛼 = 2.5 𝑡𝑜 3.0

𝐸𝑠 = 2. 𝑞𝑐

Dept. of Civil Engg. Indian Institute of Technology, Kanpur

Sand Dry sand

Average Overpredicts settlements by a factor of 2 Based on field and lab penetration tests-compressibility based on 𝑒, 𝑒𝑚𝑎𝑥 and 𝑒𝑚𝑖𝑛 Correlation coefficient = 0.778 for 90 tests valid for σo = 0 𝑡𝑜 0.8 kg/cm2

Pure sand Silty sand Clayey sand Soft clay 3 sands

Based on penetration and compression tests in large chambers. Lower values of 𝛼 at higher values of 𝑞𝑐 : attributed to grain crushing

Sand below water table Clayey sand below water table Sand

Based on screw plate tests: correlated will with settlement of oil tanks

Sand

Based on pile load tests and assumptions concerning state of stress Based on screw plate tests

Sand, sandy gravels Silty saturated sands Clayey silts with silty sand and silty saturated sands with silt NC sands 𝐿/𝐵 = 1 𝑡𝑜 2, axisymmetric NC sands 𝐿/𝐵 ≥ 10, plane strain 2

NPTEL- Advanced Geotechnical Engineering

De Beer (1974)

𝐸𝑠 = 3.5𝑞𝑐 𝐸𝑠 = 1.6𝑞𝑐 − 8 𝐸𝑠 = 1.5𝑞𝑐 , 𝑞𝑐 > 30 𝑘𝑔/𝑐𝑚2 𝐸𝑠 = 3𝑞𝑐 , 𝑞𝑐 < 30 𝑘𝑔/𝑐𝑚2 𝐸𝑠 > 1.5𝑞𝑐 , 𝑜𝑟 𝐸𝑠 = 2𝑞𝑐 𝐸𝑠 = 1.9𝑞𝑐 1 𝐸𝑠 = 2 𝑞𝑐 + 3200 𝑘𝑁/𝑚2 ) 1

𝐸𝑠 = 2 𝑞𝑐 + 1600 𝑘𝑁/𝑚2 ) 𝐸𝑠 = 𝛼𝑞𝑐 , 1.5 < 𝛼 < 2

Trofimenkov (1974)

𝐸𝑠 = 3𝑞𝑐 𝐸𝑠 = 7𝑞𝑐

Sand Sand

Bulgarian practice Greek practice

Sand Sand Fine to medium sand Clayey sands, 𝑃𝐼 < 15% Sand Sands Clays

Italian practice South African practice U. K. practice

U. S. S.R. practice

Table 8 Values of 𝜷 from various case studies of immediate settlement (After Appolonia, H. G. Poulos, and C. C. Ladd 1971). Clay properties Sensitivity Overconsolidation ratio 2 3.5

𝐸𝑓𝑖𝑒𝑙𝑑 , 𝑡𝑜𝑛 /𝑚2

𝛽

Source of 𝑆𝑢 ∗

7,600

1,200

CIU

2.5

990

1,000 1,200

Field vane CIU

100

1.7

880

1,000 1,100

Field vane CIU

14

-

1.5

1,300

1,200 1,700

Field vane Bearing capacity

Portsmouty: Highway embankment

15

10

1.3

3,000

2,000 1,700

Boston: Highway embankment Drammen: Circular load test Kawasaki: Circular load test Venezuela: Oil tanks Maine: Rectangular load test †

24

5

28

10

1.5 1.0 1.4

10,000 13,000 3,200

38

6±3

1.0

2,200

1,600 1,200 1,400 1,100 400

37 33±2

8±2 4

1.0 1.5 to 4.5

5.00 100 to 200

Field vane Bearing capacity Field vane 𝐶𝐾𝑜 𝑈 Field vane 𝐶𝐾𝑜 𝑈 Field vane CIU CIU UU and Bearing capacity

No.

Location of structure

Plasticity index

1

Oslo: Nine-story building Asrum I: Circular load

15 16

100

3

Asrum II: Circular load test

14

4

Mastemyr: Circular load test

5

6

2

7 8 9 10

Dept. of Civil Engg. Indian Institute of Technology, Kanpur

800 80 to 160

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Equation (30) can be rewritten in the form 𝑆𝑒 𝑆𝑒(1)

4

= (1+𝐵

(31)

2 1 /𝐵)

D’Appolonia et al. (1970) compared the above equation with several field experiments conducted by Bjerrum and Eggstad (1963) and Bazaraa (1967). The results of the comparison are shown in Figure 6. 15. It appears that the relationship gives the general trend; however, there appears to be a wide scattering of points.

Figure 6.15 Comparison of field test results with equation (31). (After D. J. D’Appolonia, E. D’Appolonia, and R. F. Brisette, discussion on Settlement of Spread Footings on Sand, J. Soil Mech. Found. Div., ASCE, vol. 96, 1970)

Using the standard penetration resistance obtained from field explorations, Meyerhof (1965) proposed the following relationships for settlement calculations in sand: 𝑆𝑒 =

4𝑞 𝑁

And 𝑆𝑒 =

for 𝐵 ≤ 4 𝑓𝑡 6𝑞

𝐵

𝑁

𝐵+1

2

for 𝐵 > 4 𝑓𝑡

(32a) (32b)

Where 𝑞 = intensity of applied load, kip/ft 2 𝐵 = width of footing, ft 𝑆𝑒 = settlement, in Dept. of Civil Engg. Indian Institute of Technology, Kanpur

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NPTEL- Advanced Geotechnical Engineering

𝑁 = standard penetration number Figure 6.16 shows a comparison of the observed settlements to those obtained through equation (32). It appears that the predicted settlements are rather conservative. Bowles (1977) suggested that for a more reasonable agreement equation (32) can be modified as 𝑆𝑒 =

2.5𝑞

for 𝐵 ≤ 4 𝑓𝑡

𝑁

And 𝑆𝑒 =

4𝑞

𝐵

𝑁

𝐵+1

2

for 𝐵 > 4 𝑓𝑡

(33a) (33b)

In a later work, based on the analysis of the field data of Schultze and Sherif (1973), Meyerhof (1974) gave the following empirical correlations for settlement of shallow foundations:

Figure 6.16 Comparison of observed settlement to that calculated from equation (32). (After Meyerhof 1965) 𝑆𝑒 =

𝑞 𝐵

𝑆𝑒 =

𝑞 𝐵

2𝑁

2𝑁

(for sand and gravel)

(34a)

(for silty sand)

(34b)

Where 𝑆𝑒 = settlement, in Dept. of Civil Engg. Indian Institute of Technology, Kanpur

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NPTEL- Advanced Geotechnical Engineering

𝑞 = intensity of applied load, ton/ft 2 𝐵 = width of footing, in

1.2.5 Calculation of Immediate Settlement in Granular Soil Using Simplified Strain Influence Factor The equation for vertical strain 𝜖𝑧 under the center of a flexible circular load was given in equation (5) as where 𝐼𝑧 is the strain influence factor. 𝜖𝑧 =

𝑞(1+𝑣)

Or 𝐼𝑧 =

𝐸 𝜖𝑧 𝐸 𝑞

[ 1 − 2𝑣 𝐴′ + 𝐵 ′ ] = 1 + 𝑣 [ 1 − 2𝑣 𝐴′ + 𝐵 ′ ]

(35)

Figure 6.17 shows the variation of 𝐼𝑧 with depth based on equation (35) for v equal to 0.4 and 0.5 also. According to this simplified strain-influence factor method, the immediate settlement of a foundation can be calculated as where 𝐶1 is the correction factor for the depth of embedment of foundation, and 𝐶2 is a correction factor to account for the creep n soil. The factors 𝐶1 and 𝐶2 are given by the following equations: 𝑆𝑒 = 𝐶1 𝐶2 𝑞

2𝐵 𝐼𝑧 0 𝐸

𝐶1 = 1 − 0.5

𝑞𝑜

𝑠

∆𝑧

(36) (37)

𝑞

Where 𝑞𝑜 = effective overburden pressure at foundation level 𝑞 = net foundation pressure increase = q1 − q o 𝑞1 = average pressure of foundation against soil 𝐶2 = 1 + 0.2 𝑙𝑜𝑔

𝑡 0.1

(38)

Where t is time, in years. Below is an example for using equation (36) which was given in Schmertmann’s 1970 paper.

Dept. of Civil Engg. Indian Institute of Technology, Kanpur

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Figure 6.17 Theoretical and experimental distribution of vertical strain influence factor below the center of a circular loaded area. (after J. Schmertmann,1970)

Dept. of Civil Engg. Indian Institute of Technology, Kanpur

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Module 6 Lecture 38 Evaluation of Soil Settlement -4 Topics 1.3 PRIMARY CONSOLIDATION SETTLEMENT 1.3.1 One-Dimensional Consolidation Settlement Calculation 

Method A



Method B

1.3.2 Skempton-Bjerrum Modification for Calculation of Consolidation Settlement

1.3

PRIMARY CONSOLIDATION SETTLEMENT

1.3.1 One-Dimensional Consolidation Settlement Calculation the settlement for one-dimensional consolidation can be given by: ∆𝑒

𝑆𝑐 = 1+𝑒 𝐻𝑡 (from chapter 5 equation 76) 𝑜

Where ∆𝑒 = 𝐶𝑐 log ∆𝑒 = 𝐶𝑟 𝑙𝑜𝑔 ∆𝑒 = 𝐶𝑟 𝑙𝑜𝑔

𝜎 ′ 𝑜 +∆𝜎 𝜎′ 𝑜

𝜎′ 𝑜 +∆𝜎 𝜎′ 𝑜

for normally consolidated clays

for overconsolidated clays, σ′ o + ∆σ ≤ σ′ c

𝜎′𝑐 𝜎 ′ 𝑜 + ∆𝜎 + 𝐶𝑐 𝑙𝑜𝑔 for σ′ o < σ′ c < σ′ o + ∆σ 𝜎′𝑜 𝜎′ 𝑐

When a load is applied over a limited area, the increase of pressure due to the applied load will decrease with depth, as shown in Figure 6.18. So, for a more realistic settlement prediction, the following methods may be used.

Dept. of Civil Engg. Indian Institute of Technology, Kanpur

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Figure 6. 18 Calculation of consolidation settlement-method A



Method A 1. Calculate the average effective pressure 𝜎′𝑐 on the clay layer before the application of the load under consideration. 2. Calculate the increase of stress due to the applied load at the top, middle, and the bottom of the clay layer. This can be done by using theories developed in chapter. 3. The average increase of stress in the clay layer can be estimated by Simpson’s rule, ∆𝜎𝑎𝑣 = 16(∆𝜎𝑡 + 4∆𝜎𝑚 + ∆𝜎𝑏 ) (39) Where (∆𝜎𝑡 , ∆𝜎𝑚 , and ∆𝜎𝑏 are stress increases at the top, middle, and bottom of the clay layer, respectively. 3. Using the 𝜎′𝑜 and ∆𝜎𝑎𝑣 calculated above, obtain ∆𝑒 from equations whichever is applicable. 4. Calculate the settlement by using equation.



Method B 1. Better results in settlement calculation may be obtained by dividing a given clay layer into n layers as shown in Figure 6.19. 2. Calculate the effective stress 𝜎′𝑜(𝑖) at the middle of each layer. 3. Calculate the increase of stress at the middle of each layer ∆𝜎𝑖 due to the applied load. 4. Calculate ∆𝑒𝑖 for each layer from equations, whichever is applicable. 5. Total settlement for the entire clay layer can be given by 𝑆𝑐 =

𝑖=𝑛 𝑖=1

∆𝑆𝑐 =

∆𝑒 𝑖 𝑛 𝑖=1 1+𝑒

𝑜

∆𝐻𝑖

Dept. of Civil Engg. Indian Institute of Technology, Kanpur

(40)

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Figure 6.19 Calculation of consolidation settlement-Method B Example 3 A circular foundation 2m in diameter is shown in Figure 6.20. A normally consolidated clay layer 5 m thick is located below the foundation. Determine the consolidation settlement of the clay. Solution Divide the clay layer into five layers each 1 m thick. Calculation of 𝜎′𝑜(𝑖) . The effective stress at the middle of layer 1 is 𝜎′𝑜(1) = 17 1.5 + 19 − 9.8 0.5 + 18.5 − 9.81 0.5 = 34.44 𝑘𝑁/𝑚2 . The effective stress at the middle of the second layer is 𝜎′𝑜(2) = 34.44 + 18.5 − 9.81 1 = 34.44 + 8.69 = 43.13 𝑘𝑁/𝑚2 Similarly 𝜎′𝑜(3) = 43.13 + 8.69 = 51.81𝑘𝑁/𝑚2 𝜎′𝑜(4) = 51.82 + 8.69 = 60.51𝑘𝑁/𝑚2 𝜎′𝑜(5) = 60.51 + 8.69 = 69.2𝑘𝑁/𝑚2 Calculation of ∆𝜎𝑖 . For a circular loaded area, the increase of stress below the center is given by 1

∆𝜎𝑖 = 𝑞 1 − [(𝑏/𝑧)2 +1]3/2 Where b is the radius of the circular foundation, 1 m. hence, 1

∆𝜎1 = 150 1 − [(1/1.5)2 +1]3/2 = 63.59 𝑘𝑁/𝑚2 1

∆𝜎2 = 150 1 − [(1/2.5)2 +1]3/2 = 29.93 𝑘𝑁/𝑚2 Dept. of Civil Engg. Indian Institute of Technology, Kanpur

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∆𝜎3 = 150 1 −

1 [(1/3.5)2 +1]3/2

= 16.66 𝑘𝑁/𝑚2

1

∆𝜎4 = 150 1 − [(1/4.5)2 +1]3/2 = 10.46 𝑘𝑁/𝑚2 1

∆𝜎5 = 150 1 − [(1/5.5)2 +1]3/2 = 7.14 𝑘𝑁/𝑚2

Figure 6.20 Calculation of consolidation settlement 𝑆𝑐 : The steps in the calculation are given in the following table ( Figure 6.21): Layer no.

∆𝐻𝑖 , 𝑚

𝜎′𝑜(𝑖) , 𝑘𝑁/𝑚2

∆𝜎′𝑖 𝑘𝑁/𝑚2

∆𝑒 ∗

1 2 3 4 5

1 1 1 1 1

34.44 43.13 51.82 60.51 69.2

63.59 29.93 16.66 10.46 7.14

0.0727 0.0366 0.0194 0.0111 0.00682

∗ ∆𝑒 = 𝐶𝑐 𝑙𝑜𝑔

𝜎′ 𝑜 (𝑖) +∆𝜎 𝑖 𝜎′ 𝑜 (𝑖)

∆𝑒 ∆𝐻𝑖′ 𝑚 1 + 𝑒𝑜 0.0393 0.0198 0.0105 0.0060 0.0037 Σ = 0.0793

; 𝐶𝑐 = 0.16

So, 𝑆𝑐 = 0.0793 𝑚 = 79.3 𝑚𝑚.

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Figure 6.21

1.3.2 Skempton-Bjerrum Modification for Calculation of Consolidation Settlement In one-dimensional consolidation tests, there is no lateral yield of the soil specimen and the ratio of the minor to minor to major principal effective stresses, 𝐾𝑜 , remains constant. In that case, the increase of pore water pressure due to an increase of vertical stress is equal in magnitude to the latter; or ∆𝑢 = ∆𝜎

(41)

Where ∆𝑢 is the increase of pore water pressure and ∆𝜎 is the increase of vertical stress. However, in reality the final increase of major and minor principal stresses due to a given loading condition at a given point in a clay layer do not maintain a ratio equal to 𝐾𝑜 . The increase of pore water pressure at a point due to a given load is (Figure 6.22).

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Figure 6.22 Development of excess pore water pressure below the center line of a circular loaded area ∆𝑢 = ∆𝜎3 + 𝐴(∆𝜎1 − ∆𝜎3 ) Skempton and Bjerrum (1957) proposed that the vertical compression of a soil element of thickness 𝑑𝑧 due to an increase of pore water pressure ∆𝑢 may be given by 𝑑𝑆𝑐 = 𝑚𝑣 ∆𝑢 𝑑𝑧

(42)

Where 𝑚𝑣 is the coefficient of volume compressibility, or ∆𝜎

𝑑𝑆𝑐 = 𝑚𝑣 ∆𝜎3 + 𝐴 𝜎1 − ∆𝜎3 𝑑𝑧 = 𝑚𝑣 ∆𝜎1 𝐴 + ∆𝜎3 (1 − 𝐴) 𝑑𝑧 1

The preceding equation can be integrated to obtain the total consolidation settlement: 𝑆𝑐 =

𝐻𝑡 0

∆𝜎

𝑚𝑣 ∆𝜎1 𝐴 + ∆𝜎3 (1 − 𝐴) 𝑑𝑧

(43)

1

For conventional one-dimensional consolidation (𝐾𝑜 condition) 𝑆𝑐(𝑜𝑒𝑑 ) =

𝐻𝑡 ∆𝑒 0 1+𝑒𝑜

𝑑𝑧 =

𝐻𝑡 ∆𝑒 1 0 ∆𝜎1 1+𝑒𝑜

∆𝜎1 𝑑𝑧 =

𝐻𝑡 0

𝑚𝑣 ∆𝜎1 𝑑𝑧

(44)

. Thus, Settlement ratio, 𝜌𝑐𝑖𝑟𝑐𝑙𝑒 = 𝑆 =

𝐻𝑡 0

𝑆𝑐 𝑐(𝑜𝑒𝑑 )

𝑚 𝑣 ∆𝜎1 [𝐴+(∆𝜎3 /∆𝜎1 )(1−𝐴)]𝑑𝑧 𝐻𝑡 0

𝑚 𝑣 ∆𝜎1 𝑑𝑧

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= 𝐴 + (1 − 𝐴)

𝐻𝑡 0 𝐻𝑡 0

∆𝜎3 𝑑𝑧 ∆𝜎1 𝑑𝑧

= 𝐴 + (1 − 𝐴)𝑀1 Where 𝑀1 =

𝐻𝑡 0 𝐻𝑡 0

(45)

∆𝜎3 𝑑𝑧

(46)

∆𝜎1 𝑑𝑧

The values of 𝑀1 for the stresses developed below the center of a uniformly loaded circular of diameter B are given in Figure 6. 23. The values of settlement ratio, , 𝜌𝑐𝑖𝑟𝑐𝑙𝑒 , for various values of the pore water pressure parameter A are given in Figure 6. 24.

Figure 6. 23 Variation of 𝑀1 with 𝐻𝑡 /𝐵

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Figure 6. 24 Settlement ratio for circular loading [equation (45)]

For consolidation under the center of a strip load (Scott, 1963), of width B (Figure 6.25).

Figure 6. 25 Excess pore water pressure below the center line of a uniform strip load

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∆𝑢 = ∆𝜎3 + So, 𝑆𝑐 =

𝐻𝑡 0

𝐴−

2

1 3

+

1 2 𝐻𝑡 0

𝑚𝑣 ∆𝑢 𝑑𝑧 =

∆𝜎1 − ∆𝜎3

(for v = 0.5) ∆𝜎

𝑚𝑣 ∆𝜎1 𝑁 + (1 − 𝑁) ∆𝜎3 𝑑𝑧 1

(47) 3

Where 𝑁 =

2

1

1

𝐴−3 +2

Hence, settlement ratio, 𝜌𝑠𝑡𝑟𝑖𝑝 = 𝑆 =

𝐻𝑡 0

𝑆𝑐 𝑐(𝑜𝑒𝑑 )

𝑚 𝑣 ∆𝜎1 [𝑁+(1−𝑁)(∆𝜎3 /∆𝜎1 )]𝑑𝑧 𝐻𝑡 0

𝑚 𝑣 ∆𝜎1 𝑑𝑧

= 𝑁 + (1 − 𝑁)𝑀2 Where 𝑀2 =

𝐻𝑡 0 𝐻𝑡 0

(48)

∆𝜎3 𝑑𝑧

(49)

∆𝜎1 𝑑𝑧

Figure 6. 27 Settlement ratio for strip loading [equation (48)]

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Module 6 Lecture 39 Evaluation of Soil Settlement - 5 Topics 1.3.3 Settlement of Overconsolidated Clays 1.3.4 Precompression for Improving Foundation Soils

1.4 SECONDARY CONSOLIDATION SETTLEMENT

1.3.3 Settlement of Overconsolidated Clays Settlement of structures founded on overconsolidated clay can be calculated by dividing the clay layer into a finite number of layers of smaller thicknesses as outlined in method B. thus, 𝑆𝑐(𝑜𝑒𝑑 ) =

𝐶𝑟 ∆𝐻𝑖 1+𝑒𝑜

𝑙𝑜𝑔

𝜎′ 𝑜 (𝑖) +∆𝜎 𝑖 𝜎′ 𝑜 (𝑖)

(50)

To account for the small departure from one-dimensional consolidation, Leonards (1976) proposed a correction factor, 𝛼: 𝑆𝑐 = 𝛼𝑆𝑐(𝑜𝑒𝑑 )

(51)

The values of the correction factor 𝛼 are given in Figure 6.26b and are a function of the average value of 𝜎′𝑐 /𝜎′𝑜 and 𝐵/𝐻𝑡 (B is the width of the foundation and 𝐻𝑡 is the thickness of the clay layer, as shown in Figure 6.26a). According to Leonards, if 𝐵 > 4𝐻𝑡 , 𝛼 = 1 may be used. Also, if the depth to the top of the clay stratum exceeds twice the width of the loaded area, 𝛼 = 1 should be used in equation (51).

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Figure 6. 26 Settlement ratio on overconsolidated clay. (After Leonards 1976)

1.3.3 Precompression for Improving Foundation Soils when it appears that too much consolidation settlement is likely to occur due to the construction of foundations, it may be desirable to apply some surcharge loading before foundation construction in order to eliminate or reduce the post-construction settlement. Let us consider the case where a given construction will require a permanent uniform loading of intensity 𝜎𝑓 as shown in Figure 6. 27. The total primary consolidation settlement due to loading is estimated to be equal to 𝑆𝑐(𝑓) . To eliminate the expected settlement due to primary consolidation, let apply a total uniform load of intensity𝜎 = 𝜎𝑓 + 𝜎𝑠 . This load will cause a faster rate of settlement of the underlying compressible layer; when a total settlement of 𝑆𝑐(𝑓) has been reached, the surcharge can be removed for actual construction.

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Figure 6.27 Concept of precompression technique

For a quantitative evaluation of the magnitude of 𝜎𝑠 and the time it should be kept on, we need to recognize the nature of the variation of the degree of consolidation at any time after loading for the underlying clay layer, as shown in Figure 6.28. The degree of consolidation 𝑈𝑧 will vary with depth and will be minimum at mid plane, 𝑖. 𝑒. , at 𝑧 = 𝐻. If the average degree of consolidation 𝑈𝑎𝑣 is used as the criterion for surcharge load removal, then after removal of the surcharge the clay close to the mid-plane will continue to settle and the clay close to the pervious layer(s) will tend to swell. This will probably result in a net consolidation settlement.. Using the procedure outlined by Johnson (1970),

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Figure 6.28 Choice of degree of consolidation for calculation of precompression

𝑆𝑐(𝑓) =

𝐻𝑡 1+𝑒𝑜

𝐶𝑐 𝑙𝑜𝑔 𝐻𝑡

And 𝑆𝑐(𝑓+𝑠) =

1+𝑒𝑜

𝜎′ 𝑜 +𝜎 𝑓

(52)

𝜎′ 𝑜

𝐶𝑐 𝑙𝑜𝑔

𝜎′ 𝑜 +𝜎 𝑓 +𝜎𝑠 𝜎′ 𝑠

(53)

Where 𝜎′𝑜 is the initial average in situ effective overburden pressure and 𝑆𝑐(𝑓) and 𝑆𝑐(𝑓+𝑠) are the primary consolidation settlements due to load intensities of 𝜎𝑓 and 𝜎𝑓 + 𝜎𝑠 , respectively. But, 𝑆𝑐(𝑓) = 𝑈(𝑓+𝑠) 𝑆𝑐(𝑓+𝑠)

(54)

Where 𝑈(𝑓+𝑠) is the degree of consolidation due to the loading of 𝜎𝑓 + 𝜎𝑠 . As explained before, this is conservatively taken as the mid-plane (z = H) degree of consolidation. Thus, 𝑆𝑐(𝑓)

𝑈(𝑓+𝑠) = 𝑆

𝑐(𝑓+𝑠)

(55)

Combining equations (52, 53 and 55), 𝑈(𝑓+𝑠) = log

log [1+(𝜎 𝑓 /𝜎′ 𝑜 ) 1+(𝜎 𝑓 /𝜎 ′ 𝑜 )[1+(𝜎𝑠 /𝜎 𝑓 ]

(56)

The values of 𝑈(𝑓+𝑠) for several combination of 𝜎𝑓 /𝜎 𝑜 and 𝜎𝑠 /𝜎𝑓 are given in Figure 6.29. Once 𝑈(𝑓+𝑠) is known, we can evaluate the nondimensional time factor 𝑇𝑣 . (Note that 𝑈(𝑓+𝑠) = 𝑈𝑧 at 𝑧 = 𝐻 of is based on our assumption). For convenience, a plot of 𝑈(𝑓+𝑠) against 𝑇𝑣 is given in Figure 6.30. So the time for surcharge load removal, t, is

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𝑡=

𝑇𝑣 𝐻 2

(57)

𝐶𝑣

Where 𝐶𝑣 is the coefficient of consolidation and H is the length of the maximum drainage path.

Figure 6.29 Variation of 𝑈(𝑓+𝑠) with 𝜎𝑠 /𝜎𝑓 and 𝜎𝑓 /𝜎′0 . (after Johnson 1970)

Figure 6.30 Plot of 𝑈(𝑓+𝑠) against 𝑇𝑣 . (Redrawn after Johnson 1970)

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1.4 SECONDARY CONSOLIDATION SETTLEMENT The coefficient of secondary consolidation 𝐶𝛼 was defined as 𝐶𝛼 =

∆𝐻𝑡 /𝐻𝑡 ∆ log 𝑡

Where t is time and 𝐻𝑡 is the thickness of the clay layer. It has been reasonably established that 𝐶𝛼 decreases with time in a logarithmic manner and is directly proportional to the total thickness of the clay layer at the beginning of secondary consolidation. Thus, secondary consolidation settlement can be given by 𝑡

𝑆𝑠 = 𝐶𝛼 𝐻𝑡𝑠 𝑙𝑜𝑔 𝑡

𝑝

(58)

Where 𝐻𝑡𝑠 = thickness of clay layer at beginning of secondary consolidation = Ht − Sc 𝑡 = time at which secondary compression is required 𝑡𝑝 = time at end of primary consolidation Actual field measurements of secondary settlements are relatively scarce. However, good agreement of measured and estimated settlements have been reported by some observers, e.g., Horn and Lambe (1964), Crawford and Sutherland (1971), and Su and Prysock (1972).

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Module 6 Lecture 40 Evaluation of Soil Settlement - 6 Topics 1.5 STRESS-PATH METHOD OF SETTLEMENT CALCULATION 1.5.1 Definition of Stress Path 1.5.2 Stress and Strain Path for Consolidated Undrained Undrained Triaxial Tests 1.5.3 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.

1.5.1 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, 𝜎3 = 𝜎′3

(minor principal stress)

𝜎1 = 𝜎3 + ∆𝜎 = 𝜎′1

(major principal stress)

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Figure 6. 31 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)

2

And 𝑞′ =

𝜎′ 1 −𝜎′ 3

(60)

2

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 𝐾𝑓 line. The 𝐾𝑓 line makes an angle 𝛼 with the normal stress axis. Now, (𝜎 ′

𝐵𝐶

−𝜎 ′ 3 𝑓 )/2 ′ 1 𝑓 +𝜎 3 𝑓 )/2

tan 𝛼 = 𝑂𝐶 = (𝜎 ′ 1 𝑓 Where 𝜎 ′ 1

𝑓

𝐷𝐶

and 𝜎 ′ 3 (𝜎 ′

𝑓

(61)

are the effective major and minor principal stresses at failure. Similarly,

−𝜎 ′ 3 𝑓 )/2 ′ 1 𝑓 +𝜎 3 𝑓 )/2

sin ∅ = 𝑂𝐶 = (𝜎 ′ 1 𝑓

(62)

From equations (61 and 62), we obtain tanα = sin∅

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(63)

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Again let us consider a case where a soil specimen is subjected to an oedometer (one-dimensional consolidation) type of loading (Figure 6.32). For this case, we can write

Figure 6.32 Determination of the slope of 𝐾𝑜 line

𝜎′3 = 𝐾𝑜 𝜎′1

(64)

Where 𝐾𝑜 is the at-rest earth pressure coefficient and can be given by the expression (Jaky, 1944) 𝐾𝑜 = 1 − sin ∅

(65)

For the Mohr’s circle shown in Figure 6. 32, the coordinates of point E can be given by 𝑞′ =

𝜎′ 1 −𝜎′ 3

𝑝′ =

𝜎′ 1 +𝜎′ 3

2

2

= =

𝜎 ′ 1 (1−𝐾𝑜 ) 2 𝜎 ′ 1 (1+𝐾𝑜 )

Thus, 𝛽 = 𝑡𝑎𝑛−1

2 𝑞′ 𝑝′

= 𝑡𝑎𝑛−1

1−𝐾𝑜 1+𝐾𝑜

(66)

Where , 𝛽 is the angle that the line 𝑂𝐸 (𝐾𝑜 line) makes with the normal stress axis. For purposes of comparison, the 𝐾𝑜 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.

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

1.5.2 Stress and Strain Path for Consolidated Undrained Triaxial Tests Consider a clay specimen consolidated under an isotropic stress 𝜎3 = 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 6. 34 Stress path for consolidation undrained triaxial test

∆𝑢 = 𝐴 ∆𝜎 Dept. of Civil Engg. Indian Institute of Technology, Kanpur

(67) 4

NPTEL- Advanced Geotechnical Engineering

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 = 𝜎3 − ∆𝑢 And Major effective principal stress = 𝜎′1 = 𝜎1 − ∆𝑢 = 𝜎3 + ∆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. 2 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 6. 35.

1.5.3 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 6. 36. 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 6. 36 Volume change between two points of a 𝑝′ 𝑣𝑠. 𝑞′ plot

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Consolidated undrained traixial tests on these samples at several confining pressures, 𝜎3 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 𝐶𝑐 . For an example, let Figure 6. 37 represent the stress-strain contours for a given normally consolidated clay sample obtained from an average depth of a clay layer. Also let 𝐶𝑐 = 0.25 and 𝑒𝑜 = 0.9. the drained friction angle ∅ (determined from CU tests) is 300 . From equation (66),

Figure 6. 37

𝛽 = 𝑡𝑎𝑛−1

1−𝐾𝑜 1+𝐾𝑜

And 𝐾𝑜 = 1 − sin ∅ = 1 − sin 30° = 0.5. So 𝛽 = 𝑡𝑎𝑛−1

1−0.5 1+0.5

= 18.43°

Knowing the value of 𝛽 we can now plot the 𝐾𝑜 line in Figure 6. 37. Also note that tan 𝛼 = 𝑠𝑖𝑛∅. since∅ = 30° , 𝑡𝑎𝑛 𝛼 = 0.5. So 𝛼 = 26.57° . 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/𝑚2 . 2. Total thickness of clay layer = 𝐻𝑡 = 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

NPTEL- Advanced Geotechnical Engineering

∆𝜎1 = 40 𝑘𝑁/𝑚2 ∆𝜎3 = 25 𝑘𝑁/𝑚2 (assuming that the load is applied instantaneously). The in situ minor principal stress (at-rest pressure) is 𝜎3 = 𝜎′3 = 𝐾𝑜 𝜎′1 = 0.5 75 = 37.5 𝑘𝑁/𝑚2 . So, before loading, 𝑝′ =

𝜎′ 1 +𝜎′ 3

𝑞′ =

𝜎′ 1 −𝜎′ 3

2

2

=

75+37.5

=

75−37.5

2

2

= 56.25 𝑘𝑁/𝑚2 = 18.75 𝑘𝑁/𝑚2

The stress conditions before loading can now be plotted in Figure 6. 37 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, 𝜎1 = 75 + 40 = 115 𝑘𝑁/𝑚2 𝜎3 = 37.5 + 25 = 62,5 𝑘𝑁/𝑚2 So, 𝑞 ′ =

𝜎′ 1 −𝜎′ 3 2

=

𝜎1 −𝜎3 115−62.5 2

2

= 26.25 𝑘𝑁/𝑚2

With this value of 𝑞′, we locate the point D. at the end of consolidation, 𝜎1 = 𝜎1 = 115𝑘𝑁/𝑚2 𝜎′3 = 𝜎3 = 62.5𝑘𝑁/𝑚2 So, 𝑝′ =

𝜎′ 1 +𝜎′ 3 2

=

115+62.5 2

= 88.75 𝑘𝑁/𝑚2 and 𝑞 ′ = 26.25 𝑘𝑁/𝑚2

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), 𝑆𝑒 =

𝜖1 at 𝐷 − 𝜖1 at 𝐴 𝐻𝑡 = 0.04 − 0.01 3 = 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 𝐾𝑜 line. For point 𝐴, 𝜎′1 = 75 𝑘𝑁/𝑚2 ; and for point 𝐻, 𝜎′1 = 118 𝑘𝑁/𝑚2 . So the volumetric strain, 𝜖𝑣 , is ∆𝑒

𝜖𝑣 = 1+𝑒 = 𝑜

𝐶𝑐 log ⁡ (118/75) 1+0.9

=

0.9 log ⁡ (118/75) 1.9

= 0.026

Dept. of Civil Engg. Indian Institute of Technology, Kanpur

7

NPTEL- Advanced Geotechnical Engineering

The axial strain 𝜖1 along a horizontal stress path is about one-third the volumetric strain along the 𝐾0 line, or 𝜖1 = 13𝜖𝑣 = 13 0.026 = 0.0087 So, the consolidation settlement is 𝑆𝑐 = 0.0087 𝐻𝑡 = 0.0087 3 = 0.0261 𝑚 And hence the total settlement is 𝑆𝑒 + 𝑆𝑐 = 0.09 + 0.0261 = 0.116 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 ∆𝜎1 = 40 𝑘𝑁/𝑚2 and ∆𝜎3 = 25 𝑘𝑁/𝑚2 (stress path AD), followed by (2) 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 0.03. the volumetric strains for stress paths DI and AH will be the same; so 𝜖𝑣 = 0.026. The axial strain 𝜖1 for the stress path DI can be given by the relation (based on the theory of elasticity) 𝜖1 𝜖𝑣

1+𝐾 −2𝐾𝐾

= (1−𝐾 𝑜) 1+2𝐾𝑜

(68)

𝑜

Where 𝐾 = 𝜎′3 /𝜎′1 for the point I. in this case, 𝜎′3 = 42 𝑘𝑁/𝑚2 and 𝜎′1 = 123 𝑘𝑁/𝑚2 . So, 42 𝐾 = 123 = 0.341

𝜖1 𝜖𝑣

𝜖

1 = 0.026 =

1+0.5−2 0.341 (0.5) 1−0.5 [1+2 0.341 ]

= 1.38

Or 𝜖1 = 0.026 1.38 = 0.036 Hence, the total settlement due to the loading is equal to 𝑆 = [ 𝜖1 𝑎𝑙𝑜𝑛𝑔 𝐴𝐷 + (𝜖1 along 𝐷𝐼)𝐻𝑡 = 0.03 + 0.036 𝐻𝑡 = 0.066𝐻𝑡

Dept. of Civil Engg. Indian Institute of Technology, Kanpur

8