Capillary Rise in Soil

8/20/2019 Capillary Rise in Soil

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NPTEL - ADVANCED FOUNDATION ENGINEERING-1

Module 1

(Lecture 3)

GEOTECHNICAL PROPERTIES OF SOIL AND OF

REINFORCED SOIL

Topics

1.1 CAPILLARY RISE IN SOIL

1.2 CONSOLIDATIONS-GENERAL

1.3 CONSOLIDATION SETTLEMENT CALCULATION

1.4 TIME RATE OF CONSOLIDATION

CAPILLARY RISE IN SOIL

When a capillary tube is placed in water, the water level in the tube rises (figure 1.15a).The rise is caused by the surface tension effect. According to figure 1.15a, the pressure at

any point A in the capillary tube (with respect to the atmospheric pressure) can beexpressed as

= −′ (for ′ = 0 to ℎ) And

= 0 (for z′ ≥ hc)


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Figure 1.15 Capillary rise

In a given soil mass, the interconnected void spaces can behave like a number of capillary

tubes with varying diameters. The surface tension force may cause in the soil to rise

above the water table, as shown in figure 1.15b. The height the capillary rise will dependon the diameter of the capillary tubes. The capillary rise will decrease with the increase

of the tube diameter. Because the capillary tube in soil has variable diameters, the height

of capillary rise will be nonuniformly. The pore water pressure at any point in the zone of

capillary rise in soil cause approximated as

= −′ [1.52]Where

= degree fo saturation of soil [equation (7)] ′ = distance measured above the water table CONSOLIDATION-GENERAL

In the field, when the stress on a saturated clay layer is increased-for exam by the

construction of a foundation-the pore water pressure in the clay increase. Because the

hydraulic conductivity of clays is very small, sometime be required for the excess porewater pressure to dissipate and the stress increase to be transferred to the soil skeleton

gradually. According to figure 1.16 if ∆ a surcharge at the ground surface over a verylarge area, the increase of total structure ∆, at any depth of the clay layer will be equalto ∆, or∆ = ∆


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Figure 1.16 Principles of consolidation

However, at time = 0 (that is, immediately after the stress application), the excess porewater pressure at any depth, ∆, will equal ∆, or∆ = ∆ℎ1 = Δ (at time = 0) Hence the increase of effective stress at time = 0 will beΔ′ = Δ − Δ = 0 Theoretically, at time = ∞, when all the excess pore water pressure in the clay layer hasdissipated as a result of drainage into the sand layers,

Δ = 0 at time = ∞) Then the increase of effective stress in the clay layer is

Δ′ = Δ − Δ = Δ − 0Δ This gradual increase in the effective stress in the claylayer will cause settlement over a

period of time and is referred to as consolidation.

Laboratory tests on undisturbed saturated clay specimens can be conducted (ASTM Test

Designation D-2435) to determine the consolidation settlement caused by variousincremental loadings. The test specimens are usually 2.5 in. (63.5 mm) in diameter and 1

in. (25.4 mm) in height. Specimens are placed inside a ring, with one porous stone at thetop and one at the bottom of the specimen (figure 1.17a). Load on the specimen is then

applied so that the total vertical stress is equal to . Settlement readings for the specimenare taken for 24 hours. After that, the load on the specimen is doubled and settlementreadings are taken. At all times during the test the specimen is kept under water. This

procedure is continued until the desired limit of stress on the clay specimen is reached.


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Figure 1.17 (a) Schematic diagram of consolidation test arrangement; (b) − log curvefor a soft clay from East St. Louis, Illinois

Figure 1.17 continued

Based on the laboratory tests, a graph can be plotted showing the variation of the void

ratio at the end of consolidation against the corresponding vertical stress


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(semilogarithmic graph: on the arithmetic scale and on the log scale). The nature ofvariation of against log for a clay specimen is shown in figure 1.17b. After thedesired consolidation pressure has been reached, the specimen can be gradually unloaded,

which will result in the swelling of the specimen. Figure 1.17b also shows the variation

of the void ratio during the unloading period.

From the − log − curve shown in figure 1.17b, three parameters necessary forcalculating settlement in the field can be determined.

1. The preconsolidation pressure, , is the maximum past effective overburden pressure to which the soil specimen has been subjected. It can be determined byusing a simple graphical procedure as proposed by Casegrande (1936). This

procedure for determining the preconsolidation pressure, with reference to figure

1.17b, involves five steps:

a. Determine the point O on the − log curve that has the sharpest curvature(that is, the smallest radius of curvature).

b.

Draw a horizontal line OA.

c. Draw a line OB that is tangent to the − log curve at O. d. Draw a line OC that bisects the angle AOB,

e. Produce the straight-line portion of the − log curve backward to intersectOC . This is point D. the pressure that corresponds to point is the preconsolidation pressure, .

Natural soil deposits can be normally consolidated or overconsolidated (or

preconsolidated ). If the present effective overburden pressure = is equal to the preconsolidated pressure the soil is normally consolidated. However, if


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= liquid limit = plastic limit Nagaraj and Murthy (1985) provided an empirical relation to calculate , which is asfollows:

kN/m2

↓ log =

1.122−−0.0463 log 0.188

↑ kN/m2

[1.54]

Where

=

void ratio

= effective overburden pressure = void ratio of the soil at liquid limit = (%)100 [1.55]

The U. S. Department of the Navy (1982) also provided generalized relationships

between , and the sensitivity of clayey soils (). This relationship was alsorecommended by Kulhawy and Mayne (1990). The definition of sensitivity is given in

section. Figure 1.18 shows the relationship.

Figure 1.18 Variation of with LI (after U. S. Department of the Navy, 1982)


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2. The compression index, , is the slope of straight-line portion (latter part of theloading curve), or

= 1−2log 2−log 1 = 1−2log 21

[1.56]

where 1and 2 are the void ratios at the end of consolidation under stresses 1and 2,respectively

The compression index, as determined from the laboratory − log curve, will besomewhat different from that encountered in the field. The primary reason is that the soil

remolds to some degree during the field exploration. The nature of variation of the

− log curve in the field for normally consolidated clay is shown in figure 1.19. It isgenerally referred to as the virgin compression curve. The virgin curve approximately

intersects the laboratory curve at a void ratio of 0.42 (Terzaghi and Peck, 1967). Notethat

is the void ratio of the clay in the field. Knowing the values of

and

you can

easily construct the virgin curve and calculate the compression index of the virgin curve by using equation (56).

Figure 1.19 Construction of virgin compression curve for normally consolidated clay


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The value of can vary widely depending on the soil. Skempton (1944) has given amempirical correlation for the compression index in which

= 0.009(

−10) [1.57]

Where

= liquid limit Besides Skempton, other investigators have proposed correlations for the compression

index. Some of these correlations are summarized in table 14.

3. The swelling index, , is the slope of the unloading portion of the − log curve. In figure 1.17b, it can be defined as

= 3−4log 43

[1.58]

In most cases the value of the swelling index () is 14 to 15 of the compression index.Flowing are some representative values of / for natural soil deposits. The swellingindex is also referred to as the recompression index.

Description of soil / Boston Blue clay 0.24-0.33

Chicago clay 0.15-0.3

New Orleans clay 0.15-0.28

St. Lawrence clay 0.05-0.1

Table 14 Correlations for Compression Index

Reference Correlation

Azzouz, Krizek, and Corotis (1976) = 0.01 (Chicago clay) = 0.208 + 0.0083 (Chicago clay) = 0.0115 (organic soils, peat) = 0.0046( − 9) (Brazillian clay)


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Rendon-Herrero (1980) = 0.14112 1 + 2.38

Nagaraj and Murthy (1985) = 0.2343 100

Wroth and Wood (1978) = 0.5 100

Leroueil, Tavenas, and LeBihan (1983)

Note:

= specific gravity of soil solids

= liquid limit = plasticity index = sensitivity = natural moisture content The swelling index determination is important in the estimation of consolidation

settlement of overconsolidated clays. In the field, depending on the pressure increase, an

overconsolidated clay will follow an e-log path , as shown in figure 1.20. Note that point with coordinates of and corresponds to the field condition before any pressure increase. Point corresponds to the preconsolidation pressure () of the clay.Line is approximately parallel to the laboratory unloading cure (Schmertmann,1953). Hence, if you know , , , , and , you can easily construct the fieldconsolidation curve.


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Figure 1.20 Construction of field consolidation curve for over consolidated clay

Nagaraj and Murthy (1985) expressed the swelling index as

= 0.0463 100 [1.59]

It is essential to point out that any of the empirical correlations for and given in thesection are only approximate. It may be valid for a given soil for which the relationshipwas developed but may not hold good for other soils. As an example, figure 1.21 shows

the plots of and with liquid limit for soils from Richmond, Virginia (Martin et al.,1985). For these soils,


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Figure 1.21 Variation of and with liquid limit for soils from Richmond, Virginia(after Martin et al., 1995) = 0.0326( − 43.4) [1.60]

And

= 0.00045( + 11.9) [1.61]The / ratio is about 125; whereas, the typical range is about15 to 110.

CONSOLIDATION SETTLEMENT CALCULATION

The one-dimensional consolidation settlement (caused by an additional load) of a clay

layer (figure 1.22a) having a thickness may be calculated as


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= Δ1+ [1.62]

Figure 1.22 One-dimensional settlement calculation: (b) is for equation (64); (c) is forequations (66 and 68)

Where

= settlement Δ = total change of void ratio caused by the additional load application = the void ratio of the clay before the application of load Note that

Δ1+ = = vertical strain

For normally consolidated clay, the field − log curve will be like the one shown infigure 1.22b. If = initial average effective overburden pressure on the clay layer andΔ = average pressure increase on the clay layer caused by the added load, the change ofvoid ratio caused by the load increase is


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Δ = log +Δ [1.63]

Now, combining equations (62 and 63) yields

=

1+

log+Δ

[1.64]

For overconsolidated clay, the field − log curve will be like the one show figure1.22c. In this case, depending on the value of Δ, two conditions may at. First, if + Δ


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Figure 1.23 (a) Derivation of equation (71); (b) nature of variation of ∆ with time

(∆) =

2(∆) 2 [1.69]

Where

= coefficient of consolidation = =

∆

∆(1+ ) [1.70]

Where

= hydraulic conductivity of the clay ∆ = total change of void ratio caused by a stress increase of ∆p

= average void ratio during consolidation = volume coefficient of compressibility = ∆/[∆(1 + )] Equation (69) can be solved to obtain ∆ as a function of time t with the following boundary conditions:


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1. Because highly permeable sand layers are located at = 0 and = , the excess pore water pressure developed in the clay at those points will be immediatelydissipated. Hence

∆ = 0 at = 0 ∆ = 0 at = = 2 Where

= Length of maximum drainage path (due to two-way drainage condition-thatis, at the top and bottom of the clay)

2. At time = 0,∆ = ∆ = initial excess pore water pressure after the load application With the preceding boundary conditions, equation (69) yields

∆ = ∑ 2(∆) −

2=∞0 [1.71]

Where

= [(2 + 1)]/2 = an integer = 1, 2, … = nondimensional time factor = ()/2 [1.72]

Determining the field value of is difficult. Figure 1.24 provides a first-orderdetermination of using the liquid limit (u. A. Department of the Navy, 1971).The value of ∆ for various depths (that is, = 0 to = 2) at time given time t (thus ) can be calculated from equation (71). The nature of this variation of ∆ is shown in figure 1.23b.


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Figure 1.24 Range of (after U. S. Department of the Navy, 1971)

The average degree of consolidation of the clay layer can be defined as

= [1.73]

Where

= average degree of consolidation = settlement of a clay layer at time after the load application =maximum consolidation settlement that the clay will undergo under given loading

If the initial pore water pressure (∆) distribution is constant with depth asshown in figure 1.25a, the average of consolidation can also be expressed as


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Figure 1.25 Drainage condition for consolidation: (a) two-way drainage; (b) one-way drainage

= = ∫ (∆ )−∫ (∆)20

20

∫ (∆) 20

[1.74]

Or

=

(∆)2−∫ (∆)20(∆)2

= 1

− ∫ (∆)20

2(∆ )

[1.75]

Now, combining equations (71 and 75) we obtain

= = 1 − ∑ 2

2 −2=∞=0

[1.76]

The variation of with can be calculated from equation (76) and is plotted infigure 1.26. Note that equation (76) and thus figure 1.26 are also valid when an

impermeable layer is located at the bottom of the clay layer (figure 1.25b). In that

case, excess pore water pressure dissipation can take place in one direction only.


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Figure 1.26 Plot of time factor against average degree of consolidation (∆ =constant)

The length of the maximum drainage path then is equal to = .The variation of with shown in figure 1.26 can also be approximated by

= 4 %1002 (for = 0 − 60%) [1.77]And

= 1.781 − 0.933 log(100 − %) (for > 60%) [1.78]

Sivaram and Swamee (1977) have also developed an empirical relationship between

and that is valid for U varying from 0 to 100%. It is of the form

= 4

%100

2

�1−%100

5.60.357 [1.79]

In some cases, initial excess pore water pressure may not be constant with depth asshown in figure 1.25. Following are a few cases of those and the solutions for the average

degree of consolidation.


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Trapezoidal Variation Figure 1.27 shows a trapezoidal variation of initial excess pore

water pressure with two-way drainage. For this case the variation of with will be thesame as shown in figure 1.26.

Figure 1.27 Trapezoidal initial excess pore water pressure distribution

Sinusoidal Variation This variation is shown in figures 1.28a and 1.28b. For the initialexcess pore water pressure variation shown in figure 1.28a,

z

Figure 1.28 Sinusoidal initial excess pore water pressure distribution


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∆ = ∆ 2 [1.80]

Similarly, for the case shown in figure 1.28b,

∆ = ∆ 4 [1.81]The variations of with for these two cases are shown in figure 1.29

Figure 1.29 Variation of with − sinusoidal variation of initial excess pore water pressure distribution

Triangular Variation Figures 1.30 and 1.31 show several types of initial pore water

pressure variation and the variation of with the average degree of consolidation.


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z

Figure 1.30 Variation of with − triangular initial excess pore water pressuredistribution

Figure 1.31 triangular initial excess pore water pressure distribution-variation of

with


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Example 9

A laboratory consolidation test on normally consolidated clay showed the following

Load, (kN/m2) Void ratio at the end of consolidation, e 140 0.92

212 0.86

The specimen tested was 25.4 mm in thickness and drained on both sides. The timerequired for the specimen to reach 50% consolidation was 4.5 min.

A similar clay layer in the field, 2.8 m thick and drained on both sides, is subjected to

similar average pressure increase (that is, = 140 kN/m2 and po + ∆p = 212kN/m2).Determine the

a. Expected maximum consolidation settlement in the field b. Length of time required for the total settlement in the field to reach 40 mm

(assume uniform initial excess pore water pressure increase with depth)

Solution

Part a

For normally consolidated clay [equation 56]

= 1−221= 0.92−0.86212

140 = 0.333

From equation (64)

= 1+

+∆ =

(0.333)(2.8)

1+0.92 212

140= 0.0875 m = 87.5 mm

Part b

From equation (73) the average degree of consolidation is

= = 4087.5 (100) = 45.7%

The coefficient of consolidation, , can be calculated from the laboratory test.From equation (72)

= 2


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For 50% consolidation (figure 1.26), = 0.197, = 4.5 min, and = /2 =12.7 mm, so

= 50 2

=(0.197)(12.7)2

4.5= 7.061 mm2/min

Again, for field consolidation, = 45.7%. From equation (77) = 4

%100

2 = 4 45.7

1002 = 0.164

But

= 2 Or

= 2 = 0.1642.8×1000

2 2

7.061= 45,523 min = 31.6 days

Capillary Rise in Soil

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