Soil Mechanics I.handout

Chapter One General Introduction 1

Introduction

S o i l i s t h e ol d e s t an d m os t c om p le x e n g i n e e r i n g m at e r i a l . Our ancestors used soils as a construction material to build burial sites, flood protection and shelters. Western civilization credits the Romans for recognizing the importance of soils in the stability of structures. As early as the 1st century BC Roman engineers paid great attention to the soil types (sand, gravel, etc.) and to the design and construction of solid foundations. However, there were no theoretical bases for design; therefore, experience from trial and error was relied upon.

1.1

Coulomb’s Contribution

C o u lo lo m b ( 1 7 7 3 ) i s c r e d i t e d as as t h e f i r s t p e r s o n t o u s e m e c h a n i c s t o s o lv e s oi l p r ob le m s . He was a member of the French Royal engineers, engineers, who were interested in protecting old fortresses fortresses that fell easily from cannon fire. To protect the fortresses from artillery attack, sloping masses of soil were placed in front of them (Fig. 1.1). The enemy had to tunnel below the soil mass and the fortress to attack. Of course, the enemy then became an easy target.

Figure 1.1: 1.1: Unprotected and protected fortress. The soil mass applies a lateral la teral force to the fortress that could cause it to topple over or slide away from the soil mass. Coulomb Coulomb attempted to t o determine the lateral force so that he could evaluate the stability of the fortress. He postulated that a wedge of soil ABC (Fig. 1.1) would fail along a slip plane BC and this wedge would push the wall out or over topple it as it moves down the slip plane. Movement of the wedge along the slip plane would occur only if the soil resistance along the wedge were overcome. Coulomb assumed that the soil resistance is provided by friction between the soil particles and the 1

Soil Mechanics I: Lecture Notes

problem now becomes one of a wedge sliding on a rough (frictional) plane, which you may have analyzed in your Physics or Mechanics course. Coulomb has tacitly defined a failure criterion for soils. Today, Coulomb's failure criterion and method of analysis still prevail.

1.2

Birth of Soil Mechanics From the early 20 th century, the rapid growth of cities, industry and

commerce required numerous building systems. For example, skyscrapers, large public buildings, dams for electric power generation and reservoirs for water supply and irrigation, tunnels, roads and railroads, port and harbor facilities, bridges, airports and runways, mining activities, hospitals, sanitation systems, drainage systems, towers for communication systems, etc. These building systems require stable and economic foundations and new questions about soils were asked. For example, what is the state of stress in a soil mass, mass, how to design safe and economic foundations, foundations, how much would a building settle and what is the stability of structures structures founded on or within soil ? We continue continue to ask these questions and to try to find answers as new issues has confronted us. Some of these new issues include removing toxic compounds from soil and water, water, designing foundations foundat ions and earth structures to mitigate damage from earthquakes and other natural h azards, and designing systems to protect the environment. To answer these questions, we need ne ed the help of some rational method and, consequently, soil mechanics was born. K a r l T er e r z ag a g h i ( 1 8 8 3 - 1 9 6 3 ) is the undisputed

father

of

soil

mechanics.

The

publication

of

his

book

"Erdbaumechanik" in 1925 laid the foundation for soil mechanics and brought recognition to the importanc i mportance e of soils in engineering activities.

1.3

Soil Mechanics and its application to foundations Soil mechanics also called geotechnique or geotechnics or

geomechanics, is the application of engineering mechanics to the solution of problems dealing with soils as a foundation and a construction material. Engineering mechanics is used to understand and interpret the properties, behavior, and performance of soils. Soil mechanics is a subset of geotechnical engineering, which involves the application of soil mechanics, geology and hydraulics to the analysis and design of geotechnical systems such as dams, embankments, tunnels, canals and waterways, foundations for bridges, roads, buildings, etc. Every application of soil mechanics involves uncertainty 2


problem now becomes one of a wedge sliding on a rough (frictional) plane, which you may have analyzed in your Physics or Mechanics course. Coulomb has tacitly defined a failure criterion for soils. Today, Coulomb's failure criterion and method of analysis still prevail.

1.2

Birth of Soil Mechanics From the early 20 th century, the rapid growth of cities, industry and

commerce required numerous building systems. For example, skyscrapers, large public buildings, dams for electric power generation and reservoirs for water supply and irrigation, tunnels, roads and railroads, port and harbor facilities, bridges, airports and runways, mining activities, hospitals, sanitation systems, drainage systems, towers for communication systems, etc. These building systems require stable and economic foundations and new questions about soils were asked. For example, what is the state of stress in a soil mass, mass, how to design safe and economic foundations, foundations, how much would a building settle and what is the stability of structures structures founded on or within soil ? We continue continue to ask these questions and to try to find answers as new issues has confronted us. Some of these new issues include removing toxic compounds from soil and water, water, designing foundations foundat ions and earth structures to mitigate damage from earthquakes and other natural h azards, and designing systems to protect the environment. To answer these questions, we need ne ed the help of some rational method and, consequently, soil mechanics was born. K a r l T er e r z ag a g h i ( 1 8 8 3 - 1 9 6 3 ) is the undisputed

father

of

soil

mechanics.

The

publication

of

his

book

"Erdbaumechanik" in 1925 laid the foundation for soil mechanics and brought recognition to the importanc i mportance e of soils in engineering activities.

1.3

Soil Mechanics and its application to foundations Soil mechanics also called geotechnique or geotechnics or

geomechanics, is the application of engineering mechanics to the solution of problems dealing with soils as a foundation and a construction material. Engineering mechanics is used to understand and interpret the properties, behavior, and performance of soils. Soil mechanics is a subset of geotechnical engineering, which involves the application of soil mechanics, geology and hydraulics to the analysis and design of geotechnical systems such as dams, embankments, tunnels, canals and waterways, foundations for bridges, roads, buildings, etc. Every application of soil mechanics involves uncertainty 2


because of the variability of soils and their compositions. Thus, engineering mechanics can provide only partial solutions to soil problems. Experience and approximate calculations are essential for the successful application of so il mechanics to practical problems. Many of the calculations that you will learn in this course are approximations. 1.4

Marvels of Civil Engineering: - the hidden truth

Figure 1.2: 1.2: A) Sears tower B) Empire state building C) Taj Mahal D) Hoover dam The work that geotechnical engineers do is often invisible once construction is completed. For example, four structures (Figs. 1.2 A – D) demonstrate marvelous engineering and architectural beauty. However, if the foundations, which are invisible, on which these structures stand were not satisfactorily designed then these structures would not exist. A satisfactory foundation design requires the proper application of soil mechanics principles, principles, accumulated experience and good judgment. The stability and life of any structure, e.g., e.g., buildings, roads, ai rports, dams, natural slopes, power plants, etc., etc., depend on the stability, stability, strength and deformation of soils. If the soil fai ls, structures founded founde d on or within within it will fail or be impaired regardless regard less of how well the structures are designed. de signed. 3


1.5

Geotechnical lessons from past failures All structures that are founded on earth rely on our ability to design safe

and economic foundations. Structural failures do occur due to different reasons. Some failures have been catastrophic and caused severe damage to lives and properties. Failures occur because of inadequate site and soil investigations; unforeseen soil and water conditions; natural hazards; poor engineering

analysis,

design,

construction,

and

quality

control;

post-construction activities; and usage outside the design conditions. When failures are investigated thoroughly, we obtain lessons and information that will guide us to prevent similar types of failures in the future.

4


Chapter Two Physical Characteristics of Soils 2. Introduction The purpose of this chapter is to introduce you to soils. You will learn some basic descriptions of soils and some fundamental physical soil properties that you should retain for future use in this course and in geotechnical engineering practice. Soils, derived from the weathering of rocks, are very complex materials and vary widely. One of the primary tasks of a geotechnical engineer is to collect, classify and investigate the physical properties of soils. In this chapter we will deal with descriptions of soils, tests to determine the physical properties of soils, and soil classification systems. When you complete this chapter you should be able to:



Describe and classify soils.



Determine particle size distribution in a soil mass.



Determine the proportions of the main constituents in a soil.



Determine index properties of soils.

1.6

Definition of Key Terms

S oi ls are materials that are derived from the weathering of rocks. W a te r co n te n t ( w ) is the ratio of the weight of water to the weight of solids. V o i d r a t i o ( e ) is the ratio of the volume of void space to the volume of solids. P o r os i t y ( n ) is the ratio of the volume of void to the total volume of soil. D e g r e e o f s at u r a ti o n ( S ) r is the ratio of volume of water to the volume of void.

B u l k u n i t w e i g h t (  ) is the weight of a soil per unit volume (or weight density).

S at u r ate d u ni t w ei g h t (  sat ) is the weight of a saturated soil per unit volume.

D r y u n i t w e i g h t (  d ) is the weight of a dry soil per unit volume. ' is the weight of soil solids in a submerged soil per E f f e ct i v e u n i t w e i g h t (  )

unit volume.

R e l a t i v e d e n s i t y ( D r ) is an index that quantifies the degree of packing between the loosest and densest state of coarse-grained soils.

E f f e c t i v e p a r t i c le s i z e ( D 1 0 ) is the average particle diameter of the soil a t 10 percentile; that is, 10% of the particles are smaller than this size 5


(diameter).

A v e r ag e par ti cle di am et er ( D 5 0 ) is the average particle diameter of the soil.

L i q u i d li m i t ( w LL ) is the water content at which a soil changes from a plastic state to a solid state.

P l a s t i c l i m i t ( w P L ) is the water content at which a soil changes from a semisolid to a plastic state.

S h r i n k ag e li m i t ( w S L ) is the water content at which the soil changes from a solid to a semisolid state without further change in volume.

1.7

Composition of soils

1.7.1 Soil Formation Soils are formed from the physical and chemical weathering of rocks. Physical weathering involves reduction of size without any change in the original composition of parent rock. The main agents responsible for this process are exfoliation, unloading, erosion, freezing and thawing. Chemical weathering causes both reduction in size and chemical alteration of the original parent rock. The main agents responsible for chemical weathering are hydration,

carbonation

and

oxidation.

Often,

physical

and

chemical

weathering take place in concert. Soils that remain at the site of weathering are called residual soils . These soils retain many of the elements that comprise the parent rock.

A llu v i al s oi ls , also called flu v i al s oi ls , are soils that were transported by rivers and streams. The composition of these soils depends on the environment under which they were transported and is often different from the parent rock. The profile of alluvial soils usually consists of layers of different soils. Much of our construction activities has been and is occurring on alluvial soils. G lacial soi ls are soils that were transported by glaciers (large body of ice moving slowly down a slope). Mar i n e s oi ls are soils deposited in a marine environment. Loess is a wind blown, uniform, fine-grained soil.

6


1.7.2 Soil Types

Figure 2.1: Soil types. Common descriptive terms such as gravels, sands, silts and clays are used to identify specific textures in soils. We will refer to these soil textures as soil types; that is, sand is one soil type and clay is another. T e x t u r e refers to the appearance or feel of a soil. Sands and gravels are grouped together as coarse-grained soils. Clay and silts are fine-grained soils. Coarse-grained soils feel gritty and hard. Fine-grained soils feel smooth. The coarseness of soils is determined from knowing the distribution of particle sizes, which is the primary

means

of

classifying

coarse-grained

soils.

To

characterize

fine-grained soils, we need further information on the types of minerals present and their contents. The response of fine-grained soils to loads, known as the mechanical behavior, depends on the type of predominant minerals present.

1.7.3 Clay Minerals Minerals are crystalline materials and make up the solid constituent of a soil. The mineral particles of fine-grained soils are platy. Minerals are classified according to chemical composition and structure. Most minerals of interest to geotechnical engineering are composed of oxygen and silicon, two of the most abundant elements on earth. Silicates are a group o f minerals with a structural unit called the s i li ca tet r ah edr on . A central silica cation (positively charged ions) is surrounded by four oxygen anion (negatively charged ions) one at each corner of the tetrahedron (Fig. 2.2a). The charge on a single tetrahedron is -4 and to achieve a neutral charge, cations must be added or single tetrahedrons must be linked to each other sharing oxygen ions. Silicate 7


minerals are formed by addition of cations and interaction of tetrahedrons. Silica tetrahedrons combine to form sheets, called s i li ca s he ets , which are thin layers of silica tetrahedrons in which three oxygen ions are shared between adjacent tetrahedrons (Fig. 2.2 b). Silicate sheets may contain other structural units such as alum ina sheets .

Figure 2.2: a) Silica tetrahedron b) Silica sheets c) Aluminum Octahedron d) Alumina sheet Alumina sheets are formed by combination of alumina minerals, which consists of aluminum ion surrounded by six oxygen hydroxyl atoms in an octahedron (Fig. 2.2 c, d). The main groups of crystalline minerals that make up clays are the minerals: kaolinite , illite , and montmorillonite .

8


Figure 2.3: Structure of kaolinite, illite and montmorillonite.

Kaolinite has a structure that consists of one silica sheet and one alumina sheet bonded together into a layer about 0.72 nm thick stacked repeatedly (Fig. 2.3a). The layers are held together by hydrogen bonds. Kaolinite is common in clays in humid tropical regions.

I lli t e consists of repeated layers of one alumina sheet sandwiched by two silicate sheets. The layers, each of thickness 0.96 nm, are held together by potassium ions.

M on tm or i llon i te has similar structure to illite, but the layers are held together by weak van der Waals forces and exchangeable ions. Water can easily enter the bond and separate the layers resulting in swelling. Montmorillonite is often called a swelling or expansive clay.

1.7.4 Surface Forces and Adsorbed Water If we subdivide a body, the ratio of its surface area to its volume increases. For example, a cube of sides 1 cm has a surface area of 6 cm2. If we subdivide this cube into smaller cubes of sides 1 mm, the original volume is unchanged but the surface area increases to 60 cm2. The surface area per unit mass ( s pe ci fi c s u r fac e ) of sands is typically 0.01 m2 per gram, while for clays it is as high as 1000 m2 per gram (montmorillonite). The specific surface of kaolinite ranges from 10 to 20 m2 per gram, while that of illite ranges from 65 to 100 m2 per gram. The surface area of 45 grams of illite is equivalent to the area of a football field. Because of their large surfaces, surface forces significantly influence the behavior of fine-grained soils compared to coarse-grained soils. The s u r face ch ar g es on fine-grained soils are negative (anions). These 9


negative surface charges attract cations and the positively charged side of water molecules from surrounding water. Consequently a thin film or layer of water, called adsorbed water, is bonded to the mineral surface. The thin film or layer of water is known as the diffus e double layer (Fig. 2.4). The largest concentration of cations occurs at the mineral surface and decreases exponentially with distance away from the surface.

Figure 2.4: Diffuse double layer. Drying of most soils (with the exception of gypsum type soils) using an oven for which the standard temperature is 105  50C, can not remove the adsorbed water. The adsorbed water influences the way a soil behaves. For example plasticity (which we will deal with in section 2.5) in soils is attributed to the adsorbed water. Toxic chemicals that seep into the ground contaminate soil and groundwater. The surface chemistry of fine-grained soils is important in understanding the migration, sequestration, re-release, and ultimate removal of toxic compounds from soils. These processes are of importance to geotechnical and geo environmental engineers.

1.7.5 Soil Fabric Soil particles are assumed rigid. During deposition, the mineral particles are arranged into structural frameworks that we call s oi l fab r i c (Fig. 2.5). Each particle is in random contact with several particles. The environment under which deposition occurs influences the structural framework that is formed. In particular the electrochemical environment has the greatest influence on the kind of soil fabric that is formed during deposition. Two common types of soil fabric, flo ccu lat ed and d i s p e r s e d, are formed during soil deposition as schematically shown in Fig. 2.5. A flocculated structure,

10


Figure 2.5: Soil Fabric. formed under a saltwater environment, results when many particles tend to orient parallel to each other. A flocculated structure, formed under a freshwater environment, results when many particles tend to orient perpendicular to each other. A dispersed structure is the result when a majority of the particles orient parallel to each other. Any loading (tectonic or otherwise) during or after deposition pe rmanently alters the soil fabric or structural arrangement in a way that is unique to that particular loading condition. Consequently, the history of loading and changes in the environment is imprinted in the soil fabric. The soil fabric is the brain; it retains the memory of the birth of the soil and subsequen t changes that occur. The spaces between the mineral particles are called voids, which may be filled with liquids (essentially water) and g ases (essentially air). Voids occupy a large proportion of the soil volume. Interconnected voids form the passageway through which water flows in and out of soils. If we change the volume of voids, we will cause the soil to either compress (settle) or expand (dilate). Loads applied by a building, for example, will cause the mineral particles to be forced closer together, reducing the volume of voids and changing the orientation of the structural framework. Consequently, the building settles. The amount of settlement depends on how much we compress the volume of voids. The rate at which the settlement occurs depends on the interconnectivity of the voids. Free water, not the adsorbed water, and/or air trapped in the voids must be forced out for settlement to occur. The decrease in volume, which results in settlement of buildings and other structures, is usually very slow in fine-grained soils and almost ceaseless because of their (fine-grained soils) large surface area compared with coarse-grained soils. The large surface area in fine-grained soils compared to coarse-grained soils provides greater resistance to the flow of 11


water through the voids.

1.7.6 Comparison of Coarse and Fine-Grained Soils for Engineering Use Coarse-grained soils have good load bearing capacities and good drainage qualities, and their strength and volume change characteristics are not significantly affected by change in moisture conditions. They are practically incompressible when dense, but significant volume change can occur when they are loose. Fine-grained soils have poor load bearing capacities compared with coarse-grained soils. Fine grained soils are practically impermeable, and change strength and volume with variations in moisture conditions. The engineering properties of coarse-grained soils are controlled mainly by the grain size of the particles and their structural arrangement. The engineering properties of fine-grained soils are controlled by mineralogical factors rather than grain size. Thin layers of fine-grained soils, even within thick deposits of coarse-grained soils, have been responsible for many geotechnical failures and therefore you need to pay special attention to fine-grained soils.

1.8

Phase Relationships Soil is composed of solids, liquids and gases (Fig. 2.6a). The solid phase

may be mineral, organic matter, or both. We will not deal with the organic matter in this course. The spaces between the solids (soil particles) are called voids and are filled with liquids or gases or both. Water is the predominant liquid and air is the predominant gas. We will use the terms water and air instead of liquids and gases. The soil water is commonly called pore water and plays a very important role in the behavior of soils under load. If all the voids are filled with water, the soil is saturated. Otherwise, the soil is unsaturated. If the voids are filled with air the soil is said to be dry.

12


Figure 2.6: Soil Phases. We can idealize the three phases of soil as shown in Fig. 2.6b. The physical properties of soils are affected by the relative proportions of each of these phases. The total volume of the soil is the sum of the volume of solids (V s), volume of water (V w) , and the volume of air (V a), that is: V  V s  V w  V a  V s  V v

(2.1)

where V v is the volume of voids. The weight of the soil is the sum of the weight of solids (W s), and weight of water (W w) . The weight of air is negligible. Thus, W  W s  W w

(2.2)

The following definitions have been established to describe the proportion of each constituent in soil. Each equation can be represented with different variables. The most popular and convenient ones are given here. You should try to memorize these definitions and equations.

1. W a t e r c on t e n t ( w ) is the ratio, often expressed by percentage, of the weight of water to the weight of solids. w 

W w W s

100%

(2.3)

The water content of a soil is found by weighing a sample of the soil and then placing it in an oven at 110  50C until the weight of the sample remains constant; that is all the absorbed water is driven out. For most soils a constant weight is achieved in about 24 hours. The soil is removed from the oven, cooled, and then weighed to obtain the water content. You will later do 13


an example on how to measure and calculate the water content of a soil. 2. V o i d r a t i o ( e ) is the ratio of the volume of void space to the volume of solids. e 

V v V s

(2.4)

3. S pe ci fi c V olu m e ( V ’) is the volume of soil per unit volume of solids. ' V 

V

1  e

V s

(2.5)

This equation is useful in relating volumes and in the calculation of volumetric strains (chapter 3). 4. P o r o s i t y ( n ) is the ratio of the volume of voids to the total volume. Porosity is usually expressed as a percentage. n

V v V

(2.6)

Porosity and void ratio are related by the expression, n

e

1 e

(2.7)

The proof of Eq. (2.7) is simple and is shown below, n

V v V



V v V v  V s



V v V s V v V s  V s V s



e

1 e

5. S pe ci fi c g r av i ty ( G s ) is the ratio of the weight of the soil solids to the weight of water of e qual volume: Gs 

W s V s  w

(2.8)

where  w = 9.81 kN/m 3 is the unit weight of water. The specific gravity of soils ranges from approximately 2.6 to 2.8. For most problems, Gs can be assumed, with little error, to be equal to 2.7. Two types of containers are used to determine the specific gravity. One is a pycnometer, which is used for coarse-grained soils. The other is a 50 mL density bottle, which is used for fin e-grained soils. The container is weighed and a small quantity of dry soil is placed in it. The mass of the container and 14


the dry soil is determined. De-aired water is added to the soil in the container. The container is then agitated to remove air bubbles. When all air babbles has been removed the container is filled with de-aired water. The mass of container, soil, and de-aired water is determined. The contents of the container are discarded and the container is thoroughly cleaned. De-aired water is added to fill the container and the mass of the container and de-aired water is determined. Let m1 be the mass of the container; m2 be the mass of the container and dry soil; m3 be the mass of the container, soil, and water; and m4 be the mass of the container and water. The mass of dry soil is ms = m2 – m1, the mass of water displaced by the soil particles is m5 = m4 – m3 + ms and Gs = m s/m5 . 6. D e g r e e of s a tu r a t i on ( S ) is the ratio, often expressed as a percentage, of the volume of water to the volume of voids: S 

V w V v



wG s

or

e

Se  wGs

(2.9)

If S = 1 or 100%, the soil is saturated. If S = 0, the soil is dry. It is practicall y impossible to obtain a soil with S = 0. 7. U n i t w e i g h t is the weight of a soil per unit volume. We will use the term

b u lk u n i t w e i g h t ,  , to denote the unit weight of a soil:  

 G  Se   w   s V  1  e 

W

(2.10)

Table 2.1: Typical values of unit weight for soils 3 sat (kN/m )

Soil type Gravel Sand Silt Clay

20 18 18 16

– – – –

22 20 20 22

d

(kN/m3) 15 13 14 14

– – – –

17 16 18 21

S pe ci al C as es (a)

Saturated unit weight (S = 1):

 G s  e   w 1  e  

 sat  

(2.11)

15


(b)

Dry unit weight (S = 0)

 d 

 G     s  w  1 1   V e w  

W s

(c)

(2.12)

Effective or buoyant unit weight is the

weight of a saturated soil, surrounded by water, per unit weight of soil:

 G  1   w  '   sat   w   s  1 e  

(2.13)

8. R e la ti v e D e n s i t y ( D ) r is an index that quantifies the degree of packing between the loosest and densest possible state of coarse-grained soils as determined by experiments: Dr 

emax  e emax  emin

where e max is the maximum void ratio (loosest condition), emin

(2.14)

is the

minimum void ratio (densest condition), and e is the current void ratio. The maximum void ratio is found by dry sand, for example, into a mold of volume (V ) using a funnel. The sand that fills the mold is weighed. If the weight of the sand is W , then by combining Eqs. (2.10) and (2.12) we get e max  Gs  w (V / W )  1 . The minimum void ratio is determined by vibrating the

sand with a weight imposing a vertical stress of 13.8 kPa on top of the sand. Vibration occurs for 8 minutes at a frequency of 3600 Hz and amplitude of 0.064 mm. From the weight of the sand (W 1) and the volume (V 1) occupied by it after vibration, we can calculate the minimum void ratio using emin  G s  w (V 1 / W 1 )  1 . Table 2.2: Description based on relative density

D r (%)

Description

0 – 15 15 – 35 35 – 65 65 – 85 85 – 100

Very loose Loose Medium dense Dense Very dense

The relative density correlates very well with the strength of coarse-grained soils, denser soils being stronger than l ooser soils. A description of sand based 16


on relative density is given in Table 2.2. EXAMPLE 2.1 Prove the following relationships: (a) S 

wG s e

(b)  d 

 1 w

 G  Se  G  (1  w)  w  s w (c)    s 1 e  1  e 

EXAMPLE 2.2 A sample of saturated clay was placed in a container and weighed. The weight was 6 N. The clay in its container was placed for 24 hours at 1050C. The weight reduced to a constant value of 5 N. The weight of the container is 1 N. If Gs = 2.7, determine a) the water content, b) void ratio, c) bulk unit weig ht, d) dry density, and e) effective unit weight.

EXAMPLE 2.3 An embankment for a highway is to be constructed from a soil compacted to a dry unit weight of 18 kN/m 3. The clay has to be trucked to the site from a borrow pit. The bulk unit weight of the soil in the borrow pit is 17 kN/m3 and its natural water content is 5%. Calculate the volume of clay from the borrow pit required for 1 cubic meter of embankment. Assume Gs = 2.7.

S t r at eg y : This problem can be solved in many ways. We will here use two ways. One way is direct; the other a bit longer. In the first way we are going to use the ratio of the dry unit weights of the compacted soil and the borrow pit soil to determine the volume. In the second way, we will use the specific volume. In this case, we need to find the void ratio for the borrow pit clay and the desired void ratio for the embankment. We can then relate the specific volumes of the embankment and the borrow pit clay.

EXAMPLE 2.4 If the borrow soil in Example 2.3 were to be compacted to attain dry unit weight of 18 kN/m3 at a water content of 7%, determine the amount of water 17


required per cubic meter of embankment, assuming no loss of water during transportation.

S t r at eg y : Since water content is related to the weight of solids and not the total weight, we need to use the data given to find the weight of solids.

1.9

Determination of Particle Sizes of Soils

1.9.1 Particle Size of Coarse-Grained Soils The distribution of particle size or average grain diameter of coarse-grained soils – gravels and sands – is obtained by screening a known weight of the soil through a stack of sieves of progressively finer mesh size. A typical stack of sieves is shown in Fig. 2.6. Each sieve is identified by a number that corresponds to the number of square holes per linear inch of mesh. The particle diameter in the screening process, often called sieve analysis, is the maximum particle dimension to pass through the square hole of a particular mesh.

Figure 2.7: Stack of sieves. A known weight of dry soil is placed on the la rgest sieve (the top sieve) and the nest of sieves is then placed on a vibrator, called a sieve shaker, and shaken. The nest of sieves is dismantled, one sieve at a time. T he soil retained on each sieve is weighed and the percentage of soil retained on each sieve is calculated. The results are plotted on a graph of percent of particles finer than a given sieve (not the percent retained) as the ordinate versus the logarithm 18


of the particle sizes as shown in Fig. 2.8. T he resulting plot is called a particle size distribution curve (s) or, simply, the gradation curve (s). Engineers have found it convenient to use a logarithmic scale for particle size because the ratio of particle sizes from the largest to the smallest in a soil can be greater than 104. Let W i be the weight of the soil retained on the i th sieve from the top of the nest of sieves and W be the total soil weight. The percent weight retai ned is:

% Retained on ith seive 

W i W

100

(2.15)

The percent finer is:

% Finer than ith seive  100 

i

 (% Retained on ith seive)

(2.16)

i 1

You can use mass instead of weight. The unit of mass is in grams or kilograms.

Figure 2.8: Typical particle size distribution curves.

1.9.2 Particle Size of Fine-Grained Soils The screening process can not be used for fine-grained soils – silts and clays – because of their extremely small size. The common laboratory method to determine the size distribution of fine-grained soils is the hydrometer test (Fig. 2.9). The hydrometer test involves mixing a small amount of soil into 19


suspension and observing how the suspension settles in time. Larger particles will settle quickly followed by smaller particles. When the hydrometer is lowered into the suspension, it will sink into the suspension until the buoyant force is sufficient to balance the weight of the hydrometer. The calibration of hydrometer is affected by temperature and specific gravity of the suspended solids. You must then apply a correlation factor to your hydrometer reading based on the test temperatures. Typically, a hydrometer test is conducted by taki ng a small quantity of a dry fine-grained soil (approximately 10 grams) and thoroughly mixing it with distilled water to form a paste. The paste is placed in a 1 liter glass cylinder and distilled water is added to bring the level to the 1 liter mark. The glass cylinder is then repeatedly shaken and inverted before being placed in a constant temperature bath. A hydrometer is placed in the glass cylinder and a clock is simultaneously started. At different times, the hydrometer is read. The diameter D of the particle at time t D is calculated from Stokes’s law as: D 

z 18

(G s  1) w t D

(2.17)

Where  is the viscosity of water (10.09 millipoises at 200C), z is the depth,

 w is the unit weight of water, and Gs is the specific gravity. The dashed line in Fig. 2.8 shows a typical particle size distribution for fine-grained soils.

Figure 2.9: Hydrometer in soil water suspension.

1.9.3 Characterization of Soils Based on Particle Size The grading curve is used for textural classification of soils. Various 20


classification systems have evolved over the years to describe soils based on their particle size. The Unified Soil Classification System (USCS) – which we will discuss in detail in section 2.7 – is one of most common methods. The USCS separates soils into two categories. One category is coarse-grained soils that are delineated if more than 50% of the soil is greater than 0.075 mm (No. 200 sieve). The other category is fine-grained soils that are delineated if more than 50% of the soil is finer than 0.075 mm. Coarse-grained soils are subdivided into gravels and sands while fine-grained soils are divided into silts and clays. Each soil type – gravel, sand, silt, and clay – is identified by grain size as shown in table 2.3. The USCS does not differentiate silts from clays. Clays have particle sizes less than 0.002 mm Real soils consist of a mixture of particle sizes. The selection of a soil for a particular use may depend on the assortment of particles it contains. Two coefficients have been defined to provide guidance on distinguishing soils based on the distribution of the particles. One of these is a numerical measure of uniformity, called the u n i f o r m i t y c o e f fi c i e n t , UC , defined as:

UC 

D60 D10

(2.18)

where D60 is the diameter of the soil particles for which 60% of the particles are finer, and D10 is the diameter of the soil particles for which 105 of the particles are finer. Both of these diameters are obtained from the grading curve. The other coefficient is the coefficient of curvature , CC (other terms used are the coefficient of gradation and the coefficient of concavity), defined as:

CC 

( D30 ) 2 D10 D60

(2.19)

where D30 is the diameter of the soil particles for which 30% of the particles are finer. A soil that has a UC < 4 contains particles of uniform sizes (approximately one size). The minimum value of UC is 1 and corresponds to an assemblage of particles of the same size. The gradation curve for a uniform soil is almost vertical (Fig. 2.8). Higher values of UC (> 4) indicate a wider assortment of particle sizes. A soil that has a UC > 4 is described as a well-graded soil and has a flat curve (Fig. 2.8). The CC is between 1 and 3 for 21


well-graded soils. The absence of certain grain sizes, termed gap-graded, is diagnosed by a CC outside of the range 1 to 3 and a sudden change of slope in the particle size distribution cure as shown in Fig. 2.8. Table 2.3: Soil types, descriptions, and average grain sizes according to USCS Soil type

Description

Average grain size

Gravel

Rounded and/or angular bulky hard rock Rounded and/or angular bulky hard rock

Coarse : 75 mm to 19 mm Fine: 19 mm to 4.75 mm Coarse : 4.75 mm to 1.7 mm Medium: 1.7 mm to 0.38 mm Fine: 0.38 mm to 0.075 mm 0.075 mm to 0.002 mm

Sand

Silt Clay

Particles smaller than 0.075 mm, exhibit little or no strength when dried. Particles smaller than 0.002 mm, exhibit significant strength when dried, water reduces strength.

< 0.002 mm

The diameter D10 is called the e f f ec t i v e s i z e of the soil and was established by Allen Hazen (1893) in connection with his work on soil filters. The effective size is particularly important in regulating the flow of water through soils. The higher the D10 value, the coarser the soil and the better the drainage characteristics. The diameter of the finer particle sizes, in particular D15 , has been used to develop criteria for soil filters. Terzaghi and Peck (1948), for example, proposed the following set of criteria for an effective soil filter: D15 ( F ) D85( BS )

 4 ( to prevent filter soil from being washed away)

and D15( F ) D15( BS )

 4 (to ensure a higher rate of flow water)

where F denotes filter and BS is the base soil. The a v e r a g e d i am e t e r of a soil is given as D50. Particle size analysis has many uses in engineering. They are used to select aggregates for concrete, soils for the construction of dams and highways, soils as filters, and material for grouting and chemical injection.

EXAMPLE 2.5 22


A sample of dry coarse-grained material of mass 500 grams was shaken through a nest of sieves and the following results were obtained:

Sieve no.

Opening (mm)

Mass retained (grams)

4 10 20 40 100 200 Pan

4.75 2.00 0.85 0.425 0.15 0.075

0 14.8 98 90.1 181.9 108.8 6.1

(a)

Plot the particle size distribution curve.

(b)

Determine (1) the effective size, (2)

the average particle size, (3) the uniformity coefficient, and (4) the coefficient of curvature. (c)

Determine the textural composition of

the soil (i.e. the amount of gravel, sand, etc.).

S t r at e g y : the best way to solve this type of problem is to make a table to carry out the calculation and then plot a gradation curve. Total mass of dry sample (M) used is 500 grams but on summing the masses of the retained soil in column 2 we obtain 499.7 grams. The reduction in mass is due to losses mainly from a small quantity of soil that gets stuck in the meshes me shes of the sieves. You should use the “after sieving” total mass of 499.7 grams in the calculations.

1.10 Physical States and Index Properties of Fine-grained Soils The physical and mechanical behavior of fine-grained soils is linked to four distinct states – solid, semisolid, plastic, and liquid – in order of increasing water content. content. Let us consider a soil in a liquid state that is al lowed to dry uniformly. If we plot a diagram of volume versus water content as shown in Fig. 2.10, we can locate the original liquid state as point A. as the solid dries its water content reduces and consequently consequently its volume.

23


Figure 2.10: 2.10: Change in soil states as a function of soil volume a nd water content. At point B, the soil becomes so stiff that it can no longer flow as a liquid. liq uid.

limi t ; it is denoted by w The boundary water at point B point B is is called the liqui d limi by w LL LL. As the soil continues contin ues to dry, there there is a range of water content conte nt at which the soil can be molded into any desired de sired shape without rapture. rapture. The soil at this state is said to exhibit plastic behavior – the ability to deform continuously without rapture. But if drying is continued beyond the range of water content for plastic plasti c behavior, behavior, the soil becomes a semisolid. The soil can ca n not be molded now without visible cracks appearing. The water content at which the soil changes from a plastic to a semisolid state is known as the pla p la s t i c li m i t , denoted by w PL PL. The range of the water contents over which the soil d eforms plastically is known as the plasticity index, I P P: I P  w LL  wPL

(2.20)

As the soil continues to dry, dry, it comes to a final state called the solid state. At this state, no further furth er volume change occurs since nearly all the water in the soil has been removed. The water content at which the soil changes from the semisolid to a solid is called the s h r i n k ag e li m i t , denoted by w SL SL. The shrinkage limit is useful for the determination of the swelling and shrinkage capacity of soils. The liquid and plastic limits are called A t t e r b e r g limits (also known as c o n s i s t e n c y limits). The Atterberg limits were originated by a Swedish soil scientist, A. Atterberg (1911). We have changed the states of fine-grained soils by changing the water 24


content. Since engineers are interested in the strength and deformation of materials, we can associate specific strength characteristics to each of the soil states. At one extreme – the solid state – the soil has the la rgest strength and the lowest deformation. A measure of soil strength using the Atterberg limits

liquidi ty index (I is known as the liquidi (I L) and is expressed as: I L 

w  wPL I P

(2.21)

The liquidity index is the ratio of the difference difference in water content between the natural (or in situ) water content of a soil and its plastic limit to its plasticity index. Table 2.4 shows a description of soil strength based on values of I L. Table 2.4: 2.4 : Description of soil strength based on l iquidity index Values of I L

Description of soil strength

IL < 0

Semisolid state – high strength, brittle (sudden) fracture is expected. Plastic state – intermediate strength, soil deformation like a plastic material. Liquid state – low strength, soil deforms like a viscous viscous fluid.

0 < IL < 1 IL > 1

Typical values for the Atterberg limits for soils are shown in Table 2.5. The Atterberg limits depend on the type of predominant mineral in the soil. If montmorillonite is the predominant material, the liquid limit can exceed 100%. Why? Recall that the bond between the layers in montmorillonite is weak and large amounts of water can easily infiltrate the spaces between the layers. In the case of kaolinite, the layers are held relatively tightly and water can not easily infiltrate between the layers in comparison with montmorillonite. Therefore, you can expect the Atterberg limits for kaolinite to be, in general, much lower than either montmorillonite montmorillonite or illite. Table 2.5: 2.5 : Typical Atterberg limits for soils Soil Type

w LL (%)

w P L (%)

I P (%)

Sand Silt Clay

30 – 40 40 – 150

Non plastic 20 – 25 10 – 15 25 – 50 15 – 100

Skempton (1953) showed that for soils with a particular mineralogy, the plasticity index is linearly related to the amount of the clay fraction. He coined a term called activity ( A) A) to describe the importance of the clay 25


fractions on the plasticity index. The equation for A is: A 

I P

Clay fraction (%)

You should recall that the clay fraction is the amount of particles less than 2 µm. Skempton classified clays according to Table 2.6. Table 2.6: Typical Atterberg limits for soils A

Description

< 0.75 0.75 – 1.25 > 1.25

Inactive Normal Active

1.11 Determination of Liquid, Plastic, and Shrinkage Limits 1.11.1

Casagrande Cup Method The liquid limit is determined from an apparatus (Fig. 2.11) that

consists of a semispherical brass cup that is repeatedly dropped onto a hard rubber base from a height of 10 mm by a cam-operated mechanism.

Figure 2.10: Cup apparatus for the determination of liquid limit. The apparatus was discovered by A. Casagrande (1932) and the procedure for the test is called the Casagrande cup method. A dry powder of the soil is mixed with distilled water into a paste and placed in the cup to a thickness of about 12.5 mm. The soil surface is smoothed and a groove is cut into the soil using a standard grooving tool. The crank operating the cam is turned at a rate of 2 revolutions per second and the number of blows required to close the groove over a length of 12.5 mm is counted and recorded. A specimen of soil within the closed portio n is extracted for determination of the water content. The liquid limit is defined as the water 26


content at which the groove cut into the soil will close over a distance of 12.5 mm following 25 blows. This is difficult to achieve in a single test. Four or more tests at different water contents are usually required for terminal (number of blows to close the groove over a distance of 12.5 mm) ranging from 10 to 40. The results are presented in a plot of water content (ordinate, arithmetic scale) versus terminal blows (abscissa, logarithmic scale) as shown in Fig. 2.12.

Best-fit straight line called the liquid state l ine

Figure 2.12: Typical liquid limit results from the Casagrande cup method. The best-fit straight line to the data points, usually called the flow line, is drawn. We will call this line the liquid state line to distinguish it from flow lines used in describing flow of water through soils. The liquid limit is read from the graph as the water content on the liquid state line corresponding to 25 blows. The cup method of determining the liquid limit has many shortcomings. Two of these are: 1. The tendency of soils of low pla sticity to slide and to liquefy with shock in the cup rather than to flow plastically. 2. Sensitivity to operator and to small differences in apparatus.

1.11.2

Plastic Limit Test

The plastic limit is determined by rolling a small clay sample into threads and finding the water content at which threads approximately 3 mm in 27


diameter will just start to crumble. Two or more determinations are made and the average water content is reported as the plastic limit.

1.11.3

Fall Cone Method to Determine Liquid and Plastic

Limits A fall cone test, popular in Europe and Asia, appears to offer a more accurate (less prone to operator’s errors) method of determining both the liquid and plastic limits. In the fall cone test (Fig. 2.13), a cone with apex angle of 300 and total mass of 80 grams is suspended above, but in contact with the soil sample. The cone is permitted to fall freely for a period of 5 seconds. The water content corresponding to a cone penetration of 20 mm defines the liquid limit. The sample preparation is similar to the cup method except that the sample container in the fall cone test has a different shape and size (Fig. 2.13). Four or more tests at different water contents are also required because of the difficulty of achieving the liquid limit from a single test. The results are plotted as water content (ordinate, arithmetic scale) versus penetration (abscissa, logarithmic scale) and the best-fit straight line (liquid state line) linking the data points is drawn (Fig. 2.14). The liquid limit is read from the plot as the water content on the liquid state line corresponding to a penetration of 20 mm. The plastic limit is obtained by repeating the test with a cone of similar geometry, but with a mass of 240 grams. The penetration depth in the soil for the bigger cone mass at given water content will be

Figure 2.13: Fall cone apparatus larger than the smaller cone mass of 80 grams. Thus the liquid state line for 28


the 240 gram cone will be below the liquid state line for the 80 gram cone and parallel to it. The plastic limit is given as: wPL  w LL 

2w log 10 ( M 2 M 1 )

 w LL  4.2w

(2.22)

where w is the separation in terms of water content between the liquid state lines (Fig. 2.14) of the two cones, M 1 is the mass of 80 grams cone, and M 2 is the mass of the 240 gram cone.

80 gram cone Best-fit straight line

△w

240 gram cone

w LL=44%

Figure 2.14: Typical fall cone test results.

1.11.4

Shrinkage Limit The shrinkage limit is determined as follows. A mass of wet soil, m1,

is placed in a porcelain dish 44.5 mm in diameter and 12.5 mm high and then oven-dried. The volume of oven-dried soil is determined by using mercury to occupy the vacant spaces caused by shrinkage. The ma ss of the mercury is determined and the volume decrease caused b y shrinkage can be calculated from the density of mercury. The shrinkage limit is calculated from:

 m1  m2

wSL  



m2



V 1  V 2  w  m2

  100

g 

(2.23)

where m1 is the mass of the wet soil, m2 is the mass of the oven-dried soil, V 1 is the volume of the wet soil, V 2 is the volume of the oven-dried soil, and g is the acceleration due to gravity (9.8 m/s2).

29


EXAMPLE 2.6 A liquid limit test conducted on a soil sample in the cup device gave the following results. Number of blows Water content (%)

10 60.0

19 45.2

23 39.8

27 36.5

40 25.2

Two determinations for the plastic limit gave water contents of 20.3% and 20.8%. Determine (a) the liquid limit and plastic limit, (b) the plasticity index, (c) the liquidity index if the natural water content is 27.4 %, and (d) void ratio at the liquid limit, if Gs = 2.7. If the soil were to be loaded to failure, would you expect a brittle failure?

S t r at eg y : To get the liquid limit, you must make a semi-logarithm plot of water content versus number of blows. Use the data to make your plot, then extract the liquid limit (water content on the liquid state li ne corresponding to 25 blows). Two determinations of the plastic limit were made and the differences in the results are small. So, use the average value of water content as a plastic limit.

EXAMPLE 2.7 The results of a Parameter Penetration (mm) Water content (%)

fall cone test are shown in the table below. 80 gram cone 240 gram cone 5.5 7.8 14. 22 32 8.5 15 21 35 8 39. 44. 52. 60. 67 36. 45. 49. 58. 0 8 5 3 0 1 8 1

Determine (a) the liquid limit, (b) the plastic limit, (c) the plasticity index, and (d) the liquidity index if the natural water content is 36%.

S t r at eg y : Adopt the same strategy as in Example 2.6. Make a semi-logarithm plot of water content versus penetration. Use the data to make your plot, then extract the liquid limit (water content on the liquid state line corresponding to 20 mm). Find the water content difference between the two liquid state lines at any fixed penetration. Use this value to determine the plastic limit.

30


1.12 Soil Classification Schemes A classification scheme provides a method of identifying soils in a particular group that would likely exhibit similar characteristics. Soil classification is used to specify a certain soil type that is best suitable for a given application. There are several classification schemes available. Of the number of classification systems proposed over the past few decades, the Unified Soil Classification System (USCS) and the American Association of State Highway and Transportation Officials (AASHTO) system are the most widely used by current practitioners, particularly in America. Other countries like Japan, Europe and China also posses their own soil classification systems. The AASHTO system classifies soils according to their usefulness in roads and highways, while USCS was originally developed for use in airfield construction but was later modified for general use.

1.12.1

Unified Soil Classification System In this course, we will study the USCS in detail. The USCS uses

symbols for particular size groups. These symbols and their representations are: G – gravel, S – Sand, M – Silt, C – Clay. These are combined with other symbols expressing gradation characteristics – W for well-graded and P for poorly graded – and plasticity characteristics – H for high and L for low, and a symbol O for the presence of Organic material. A typical classification of CL means a clay soil with low plasticity and, while SP means a poorly graded sand. The flowcharts shown in Figs. 2.15 a and b provide systematic means of classifying a soil according to the USCS. Experimental results from soils tested from different parts of the world were plotted on a graph of plasticity index (ordinate) versus liquid limit (abscissa). It was found that clays, silts, and organic soils lie in d istinct regions of the graph. A line de fined by the equation I P  0.73 ( wLL  20) %

(2.24)

called the “A-line,” delineates the boundaries between clays (above the line) and silts and organic soils (below the line) a s shown in Fig. 2.16. A second line, the U-line expressed as I P = 0.9(w LL – 8), defines the upper limit of the correlation between plasticity index and liquid limit. If the results of you soil tests fall above the U-line, you should be suspicious of your results and repeat your tests.

31


Figure 2.15a: Unified soil classification flowchart for coarse-grained soil.

Figure 2.15b: Unified soil classification flowchart for fine-grained soil.

32


CH OH CL

CL-ML

MH

ML & OL

ML

Figure 2.16: Plasticity chart.

1.12.2

Engineering Use Chart You may ask “How do I use a soil classification to select a soil for a

particular type of construction, for example, a dam?” Geotechnical engineers have prepared charts based on experience to assist you in selecting a soil for a particular construction purpose. One that chart is shown in Table 2.7. The numerical values 1 to 9 are ratings with No. 1 the best. The chart should only be used to provide guidance and to a preliminary assessment of the suitability of a soil for a particular use. You should not rely on such descriptions as “excellent” shear strength or “negligible” compressibility to make the final design and construction decisions. You will learn later more reliable method to determine strength and compressibility properties.

EXAMPLE 2.8 Particle size analyses were carried out on two soils – Soil A and Soil B – and the particle size distribution curves are shown in Fig. E2.8 below. The Atterberg limits for the two soils are: Soil A B

w LL

w PL

26 18 Non plastic

33


Figure E2.8: Particle size distribution curves for soil A and soil B. (a) Classify these soils according to the Unified Soil Classification System. (b) Is either of the soils organic? (c) In a preliminary assessment, which of the two soils is a better material for the core of a rolled earth dam?

S t r at eg y : If you examine the flowcharts of Figs. 2.15 a and b, you will notice that you need to identify the various soil types based on texture: for example, the percentage of gravel or sand. Use the particle size distribution curves to extract the different percentages of each soil type and then follow the flowchart. To determine whether your soil is organic or inorganic, plot your Atterberg limits on the plasticity chart and check whether the limits fall within an inorganic or organic soil region.

34


Chapter Three One-Dimensional Flow of Water through Soils 3. Introduction In chapter 2, we have discussed particle sizes and index properties and used these to classify soils. You know that water changes the soil states in fine-grained soils; the greater the water content in the soil the weaker it is (I L increases). Soils are porous materials much like sponges. Water can flow between the i nterconnected voids. Particle sizes and structural arrangement of the particles influence the rate of flow. In table 2.7, you should have noticed that one of the important soil properties is permeability. In this chapter, we will discuss soil permeability by considering one-dimensional flow of water through soils. When you complete this chapter you should be able to:



Determine the rate of flow of water through soils.

3.1

Definitions of key Terms

G r o u n d w a t e r is water under gravity in excess of that required to fill the soil pores.

H e ad ( H ) is the mechanical energy per unit weight. C o e ff i c i e n t o f p e r m e a bi l i t y ( k ) is a proportionality constant to determine the flow velocity of water through soils.

3.2

Groundwater We will be discussing gravitational flow of water under a steady state

condition. You may ask: “What is a steady state condition?” Gravitational flow can only occur if there is a gradient. Flow takes place downhill. The steady state flow occurs if neither the flow nor the pore water pressures change with time. Pore water pressure is the water pressure within the voids. If you dig a hole into a soil mass that has all the voids filled with water (fully saturated), you will observe water in the hole up to a certain level. This water level is called groundwater level or groundwater table and exists under a hydrostatic condition. A hydrostatic condition occurs when the flow is zero. The top of the groundwater level is under atmospheric pressure. We will denote the groundwater table by the symbol ▼.

35


3.3

Head Darcy’s law governs the flow of water through soils. But before we delve

into Darcy’s law, we will discuss an important principle in fluid mechanics – Bernoulli’s principle – which is essential in understanding flow through soils. If you cap one end of a tube, fill the tube with water, and then rest it on your table (Fig. 3.1), the height of water with reference to your table is called the pressure head (hP).

Figure 3.1: Illustration of elevation and p ressure heads. Head refers to the mechanical energy per unit weight. If you raise the tube above the table, the mechanical energy or total head increases. You now have two components of total head – the pressure head (hp) and the elevation head (hz). If water were to flow through the tube with a velocity v , under steady state condition, then we have an additional head due to the velocity given as v

2

2 g . The total head (sometimes called piezometric head), H , according to

Bernoulli’s principle is: H  h z  h p 

v

2

2g

(3.1)

The elevation or potential head is referenced to an arbitrary datum and the total head will change depending on the choice of the datum position. Therefore, it is essential that you identify your datum position in solutions to flow problems. Pressures are defined relative to atmospheric pressure. The velocity of flow through soils is generally small (< 1 cm/s) and we usually neglect the velocity head. The total head in soils is then H  h z  h p  h z 

u

 w

(3.2)

where u =  wh p is the pore water pressure. Consider a cylinder containing a soil mass with water flowing through it 36


at a constant rate as dep icted in Fig. 3.2. If we connect two tubes, A and B, called piezometers, at a distance l apart, the water will rise to different heights in each of the tubes. The height of water in tube B near the exit is lower than A. Why? As the water flows through the soil, energy is dissipated through friction with the soil particles, resulting in a loss of head. The head loss between A and B, assuming decrease in head is positive and our datum is arbitrarily selected at the top of the cylinder, is ΔH = ( h p ) B  (h p ) A .

Figure 3.2: Head loss due to flow of water through soil.

3.4

Darcy’s Law Darcy (1856) proposed that average flow velocity through soils is

proportional to the gradient of the total head. The flow in any direction, j , is v j  k j

dH dx j

(3.3)

where v is the average flow velocity, k is a coefficient of proportionality called the coefficient of permeability or hydraulic conductivity, and dH is the change in total head over a distance dx . The unit of measurement for k is length/time, that is, cm/s. With reference to Fig. 3.2, Darcy’s law becomes v x  k x

 H l

 k x i

(3.4)

where i   H l is the hydraulic gradient. Darcy’s law is valid for all soils if the flow is laminar (Reynolds number < 1). The average velocity, v , calculated from Eq. (3.4) is for the cross-sectional area normal to the direction of flow. Flow through soils, however, happens only through the interconnected voids. The velocity through the void spaces is called the s eepag e v elo ci ty (v s) and is obtained 37


by dividing the average velocity by the porosity of the soil: vs 

k j n

i

(3.5)

The volume rate of flow, qv , or, simply, flow rate is the product of the average velocity and the cross-sectional area: qv  v j A  Ak j i

(3.6)

The unit of measurement for q v is m3 /s or cm3 /s. The conservation of flow (law of continuity) stipulates that the volume rate of inflow (qv )in into a soil element must equal the volume rate of outflow, (qv )out , or, simply, inflow must equal outflow: (qv )in = (qv )out . The coefficient of permeability depends on the soil type, the particle size distribution, the structural arrangement of the grains or void ratio, and the wholeness (homogeneity, layering, fissuring, etc) of the soil mass. Typical value ranges of k z for various soil types are shown in Table 2.7. Table 3.1: Coefficient of permeability for common soil types

k z (cm/s)

Soil type

Clean gravel > 1.0 Clean sands, clean sand and gravel mixtures 1.0 to 10Fine sands, silts, mixtures comprising sands, silts, and clays 10-3 to 10 -7 Homogeneous clays < 10 Homogeneous clays are practically impervious. Two popular uses of “impervious” clays are in dam construction to curtail the flow of water through the dam and as barriers in landfills to prevent migration of effluent to the surrounding area. Clean sands and gravels are pervious and can be used as drainage materials or soil filters.

3.5

Empirical Relationships for k For a homogeneous soil, the coefficient of permeability depends

predominantly on its void ratio. You should recall that the void ratio is dependent on the soil fabric and structural arrangement of the soil grains. A number of empirical relationships have been proposed linking k to void ratio and grain size for coarse-grained soils. Hazen (1930) proposed one of the early relationships as: k z  C  D102 cm/s

(2.31) 38


where C is a constant varying between 0.4 and 1.2 if the unit of measurement of D10 (effective diameter) is mm. Typically, C = 1.0. Other relationships were proposed for coarse and fine-grained soils by Samarasinghe et al. (1982), Kenny et al. (1984), and others. One has to be extremely cautious in using empirical relationships for k because it is very sensitive to changes in void ratio and the wholeness of your soil mass. EXAMPLE 3.1 A soil sample 10 cm in diameter is placed in a tube 1 m long. A constant supply of water is allowed to flow into one end of the soil at A and the outflow at B is collected by a beaker (Fig. E3.1). The average amount of water collected is 1 cm3 for every 10 seconds. The tube is inclined as shown in Fig. 3.1. Determine the (a) hydraulic gradient, (b) flow rate, (c) average velocity, (d) seepage velocity, if e = 0.6, and (e) coefficient of permeability.

Figure E3.1

S t r at eg y : For flow problems, you must define a datum position. So your first task is to define the datum position and then find the difference in total head between A and B. Use the head difference to calculate the hydraulic gradient and use Eqs. (3.4 to 3.6) to solve the problem.

EXAMPLE 3.2 A drainage pipe (Fig. E3.2) became completely blocked during a storm by a plug of sand, 1.5 m long, followed by another plug of a mixture of clays, silts, and sands, 0.5 m long. When the storm was over, the water level above ground was 1 m. The coefficient of permeability of the sand is 2 times that of the mixture of clays, silts, and sands. (a) Plot the variation of pressure, elevation, and total head over the length of the pipe. 39


(b) Calculate the pore water pressure at (1) the center of the sand plug and (2) the center of the mixture of clays, silts, and sands. (c) Find the average hydraulic gradients in the sand and in the mixture of clays, silts and sands.

Figure E3.2a: Illustration of blocked drainage pipe.

S t r at eg y : You need to select a datum. From the information given, you can calculate the total head at A and B. The difference in head is the head loss over both plugs but you do not know how much head is lost in the sand and in the mixture of clays, silts, and sands. The continuity equation provides the key to finding the head loss over each plug.

3.6

Flow Parallel to Soil Layers When the flow is parallel to the soil layer (Fig. 3.2), the hydraulic

gradient is the same at all points. The flow through the soil mass as a whole is equal to the sum of the flow through each of the layers. There is a similarity here with the flow of electricity through resistors in parallel. If we consider a unit width (in the y direction) of flow and use Eq. (3.6), we obtain, qv  Av  (1  H 0 )k x ( eq ) i  (1  z1 )k x1i  (1  z 2 )k x 2 i      (1  z n )k xn i (3.6) where H 0 is the total thickness of the soil mass, k x(eq) is the equivalent permeability in the horizontal ( x ) direction, z 1 to z n are the thickness of the first to the nth layers, and k x1 to k xn are the horizontal permeabilities of first to the nth layer. Solving Eq. (3.6) for k x(eq), we get, 40


k x ( eq ) 

1 H 0

( z1 k x1  z 2 k x 2      z n k xn )

(3.7)

Fig. 3.2: Flow through stratified layers.

3.7

Flow Normal to Soil Layers For flow normal to the soil layers, the head loss in the soil mass is the

sum of the head losses in each layer:

 H  h1  h2      hn

(3.8)

where ΔH is the total head loss, and Δh 1 to Δhn are the head losses in each of the n layers. The velocity in each layer is the same. The analogy to electricity is flow of current through resistors in series. From Darcy’s law, we obtain k z ( eq )

 H H 0

 k z1

h1 z1

 k z 2

h2 z 2

     k zn

hn z n

(3.9)

where k z ) direction and k z(eq) is the equivalent permeability in the vertical ( z1 to th k zn are the vertical permeabilities of the first to the n layer. Solving Eqs. (3.8

and 3.9) leads to k z ( eq ) 

z1 k z1



H 0 z 2

z n

k z 2

k zn

  

(3.10)

Values of k z(eq) are generally less than k x(eq) – sometimes as much as 10 times less.

EXAMPLE 3.3 A canal is cut into a soil with a stratigraphy shown in Fig. E3.3. Assuming flow takes place laterally and vertically through the sides of the canal and vertically below the canal, determine the equivalent permeability in the horizontal and 41


vertical directions. Calculate the ratio o f the equivalent horizontal permeability to the equivalent vertical permeability for flow through the sides of the canal.

Fig. E3.3

S t r at eg y : Use Eqs. (3.7 and 3.10) to find the equivalent horizontal and vertical permeabilities over the depth of the canal (3 m) and then use Eq. (3.10) to find the equivalent vertical permeability below the canal. To make the calculations easier, convert all exponential quantities to a single exponent.

3.8

Determination of the Coefficient of Permeability

3.8.1 Constant-Head Test

Fig. 3.3: A constant-head apparatus. The constant-head test is used to determine the coefficient of permeability of 42


coarse-grained soils. A typical constant-head apparatus is shown in Fig. 3.3. Water is allowed to flow through a cylindrical sample of soil under a constant head (h). The outflow (Q) i s collected in a graduated cylinder at a convenient duration (t ). With reference to Fig. 3.2,

 H  h and i 

 H L



h L

The flow rate through the soil is qv = Q /t , where Q is the total quantity of water collected in the measuring cylinder over time t . From Eq. (3.10), k z 

qv Ai



QL tAh

(3.11)

where k z is the coefficient of permeability in the vertical direction. The viscosity of the fluid, which is a function of temperature, influences the value of k . The experimental value (k T0C) is corrected to a baseline temperature of 200C using k 20 0 C  k T 0C

 T 0C  200 C

 k T C RT 0

(3.12)

where µ is the viscosity of water, T is the temperature in 0C at which the measurement was made, and RT = µT /µ20 is the temperature correction factor that can be calculated from, RT  2.42  0.475 ln(T )

(3.13)

3.8.2 Falling-Head Test The falling-head test is used for fine-grained soils because the flow of water through these soils is too slow to get reasonable measurements from the constant-head test. A compacted soil sample or a sample extracted from the field is placed in a metal or acrylic cylinder (Fig. 3.4).

43


Fig. 3.4: A constant-head apparatus. Porous stones are positioned at the top and bottom faces of the sample to prevent its disintegration and to allow water to percolate through it. Water flows through the sample from a standpipe attached to the top of the cylinder. The head of water (h) changes with time as flow occurs through the soil. At different times, the head of water is recorded. Let dh be the drop in head over a time period dt . The velocity or rate of head l oss in the tube is v

dh dt

and the inflow of water to the soil is

( q v ) in  av   a

dh dt

where a is the cross-sectional area of the tube. We now appeal to Darcy’s law to get the outflow: h ( q v ) out  Aki  AK L

where A is the cross-sectional area, L is the length of the soil sample, and h is the head of water at any time t . The continuity condition requires that (qv )in = (qv )out . Therefore,

a

dh dt

 Ak

h

L

By separating the variables (h and t ) and integrating between that appropriate limits, the last equation becomes, 44


t

h2

Ak 2

dt    aL  t 1

k  k z 

h1

dh h

 h1   A(t 2  t 1 )  h2  aL

ln

(3.14)

EXAMPLE 3.4 A sample of sand, 5 cm in diameter and 15 cm long, was prepared at a porosity of 60% in a constant-head apparatus. The total head was kept constant at 30 cm and the amount of water collected in 5 seconds was 40 cm 3. The test temperature was 200C. Calculate the coefficient of permeability and the seepage velocity.

S t r at eg y : From the data given, you can readily apply Darcy’s law to find k .

EXAMPLE 3.5 The data from a falling-head test on a silty clay are Cross-sectional area of soil = 80 cm Length of soil = 10 cm Initial head = 90 cm Final head = 84 cm

Duration of test = 15 minutes Diameter of tube = 6 mm Temperature = 22 C

Determine k .

S t r at eg y : Since this is a falling-head test, you should use Eq. 3.14. Make sure you are using consistent units.

3.8.3 Pumping Test to Determine the Coefficient of Permeability One common method of determining the coefficient of permeability in the field is by pumping water at a constant flow rate from a well and measuring the decrease in ground water level at observation wells (Fig. 3.5).

45


Fig. 3.5: Layout of a pump test to determine k . The equation, called the simple well formula, is derived using the following assumptions. 1. The pumping well penetrates through the water-bearing stratum and is perforated only at the section that is below the groundwater level. 2. The soil mass is homogeneous, isotropic, and of i nfinite size. 3. Darcy’s law is valid. 4. Flow is radial towards the well. 5. The hydraulic gradient at any point in the water-bearing stratum is constant and is equal to the slope of groundwater surface (Dupuit’s assumptions). Let dz be the drop in total head over a distance dr . Then according to Dupuit’s assumption the hydraulic gradient is i

dz dr

The area of flow at a radial distance r from the center of the pumping well is 46


A  2 r z where z is the thickness of an elemental volume of the pervious soil layer. From Darcy’s law, the flow is: qv  2 r z k

dz dr

We need to rearrange the above equation and integrate it between the limits r 1 and r 2, and h1 and h2: r 2

qv

 r 1

dr r

h2

 2k   zdz h1

Completing the integration leads to: k 

q v ln(r 2 r 1 )

 ( h22  h12 )

(3.15)

With measurements of r 1, r 2, h1, h2, and qv (flow rate of the pump), k can calculated using Eq. (3.15). This test is only practical for coarse-grained soils. Pumping tests lower the groundwater, which then causes stress changes in the soil. Since the groundwater is not lowered uniformly as shown in Fig. 3.5, the stress changes in the soil will not be even. Consequently, pumping tests near existing structures can cause them to settle unevenly. You should consider the possibilities of differential settlement on existing structures when you plan a pumping test. Also, it is sometimes necessary to temporarily lower the groundwater level for construction. The process of lowering the groundwater is called dewatering. EXAMPLE 3.6 A pumping test has been carried out in a soil bed of thickness 15 m and the following measurements were recorded. Rate of pumping was 10.6×10 -3 m3 /s; drawdowns in observation wells located at 15 m and 30 m from the center of the pumping well were 1.6 m and 1.4 m, respectively, from the initial groundwater level. The initial groundwater level was located at 1.9 m below the initial ground level. Determine k .

S t r at eg y : You are given all measurements to directly apply Eq. (3.15) to find k . You should draw a sketch of the pump test to identify the values to be used in Eq. (3.15).

47


CHAPTER FOUR STRESSES, STRAINS, AND ELASTIC DEFORMATIONS OF SOILS 4.0

Introduction An elastic analysis of an isotropic material involves only two constants –

Young’s modulus and Poisson’s ratio – and thus if we assume that soils are isotropic elastic materials then we have a powerful, but simple, analytical tool to predict a soil’s response under loading. We will have to determine only the two elastic constants from our laboratory or field tests. A geotechnical engineer must ensure that a geotechnical structure must not collapse under any anticipated loading condition and that settlement under working load (a fraction of the collapse load) must be within tolerable limits. We would prefer the settlement under working load to be elastic so that no permanent settlement would occur and thus use elastic analysis to calculate the settlement. An important task of a geotechnical engineer is to determine the stresses and strains that are imposed on a soil mass by external loads. It is customary to assume that the strains in the soils are small and this assumption allows us to apply our knowledge of mechanics of elastic bodies to soils. Small strains mean infinitesimal strains. For a realistic description of soils, elastic analysis is not satisfactory. We need soil models that can duplicate the complexity of soil behavior. However, even for complex soil models, an elastic analysis is a first step. In this chapter, we will review some fundamental principles of mechanics and strength of materials and a pply these principles to soils. When you finish this chapter, you should be able to:

 Calculate stresses and strains in soils (assuming elastic behavior) from external loads.

 Calculate elastic settlement. You will use the following principles learned from statics and strength of materials.

 Stresses and strains.  Elasticity – Hooke’s law.

S am pl e P r ac ti cal S i tu at i on Two storage tanks are to be founded on a deep layer of stiff saturated clay. Your client and the mechanical engineer, who is 48


designing the pipe works, need an estimate of the settlement of the tanks when they are completely filled. Because of land restrictions, your client desires that the tanks be as close as possible to each other. If two separate foundations are placed too close to each other, the stresses in the soil induced by each foundation overlap and cause intolerable tilting of the structures and their foundations.

4.1 4.1.1

Stresses and Strains Normal Stresses and Strains The normal stresses (with reference to Fig.4.1) are:  z 

P z xy

,  x 

P x yz

,  y 

Py xz

(4.1)

Figure 4.1: Stresses and displacements due to applied loads. The normal strains (with reference to Fig.4.1) are:  z 

4.2

 z z

 x

,  x 

x

,  y 

 y y

(4.2)

Volumetric Strain

The volumetric strain is:  p   x   y   z

4.2.1

(4.3)

Shear Stresses and Shear Strains

For a shearing force F (Fig. 4.2), the shear stress is:  

F

(4.4)

xy

The simple shear strain, also called engineering shear strain,  zx , is (Fig. 4.2)

 zx  tan 1

 x z

or tan  zx 

 x z

49


Figure 4.2: Shear stresses and shear strains. For small strains, tan  zx   zx and therefore,

 x

 zx 

z

(4.5)

In geotechnical engineering, compressive stresses in soils are assumed to be positive. Soils can not sustain any appreciable tensile stresses and we normally assume that the tensile strength of soils is negligible. Strains can be compressive or tensile.

Idealized Stress-Strain Response and Yielding

4.3 4.3.1

Material Response to Normal Loading and Unloading

The change in vertical stress for a uniaxial loading (Fig.4.3) is:  z 

P

(4.6)

A

Figure 4.3: Forces and displacements on a cylinder. The vertical and radial strains are, respectively,  z 

 z H 0

 r 

r r 0

(4.7) (4.8)

where H 0 is the original length and r 0 is the original radius. The ratio of the 50


radial (or lateral) strain to the vertical strain is called Poisson’s ratio,  , defined as  

  r  z

(4.9)

Typical values of Poisson’s ratio for soils are listed in Table 4.1. Table 4.1: Typical values of Poisson’s ratio for soils. * Description υ Soft 0.35 – 0.40 Medium 0.30 – 0.35 Stiff 0.20 – 0.30 Sand Loose 0.15 – 0.25 Medium 0.25 – 0.30 Dense 0.25 – 0.35 * These values are effective values υ’ (discussion later in this chapter).

Soil Type Clay

Figures 4.4 shows typical responses of linearly and non-linearly elastic materials to normal loading and unloading; and Fig. 4.5 shows a typical response of an elastoplastic material to normal loading and unloading.

Figure 4.4: Linear and non-linear stress-strain curves of an elastic material.

51


Figure 4.5: Idealized stress-strain curves of an elastoplastic material.

4.3.2

Material Response to Shear Forces Shear forces distort materials. A typical response of an elastoplastic

material to simple shear is shown in Fig. 4.6.

Figure 4.6: Shear stress-shear strain response of an e lastoplastic material. 4.3.3

Yield Surface Let us consider a more complex situation than the uniaxial loading of a 52


cylinder (Fig. 4.7a). In this case, we are going to apply increments of vertical and radial stresses. Since we are not applying any shear stresses, the axial stresses and radial stresses are principal stresses.

Figure 4.7: Elastic, yield and elastoplastic stress states.

Let us, for example, set σ 3 to zero and increase σ1. The material will yield at some value of σ1, which we will call (σ1)y and plot it as point A in Fig. 4.7b. If, alternatively, we set σ1=0 and increase σ3, the material will yield at

(σ3)y and is represented by point B in Fig. 4.7b. We can then subject the cylinder to various combinations of σ1 and σ3 and plot the resulting yield points. Linking the yield points results in a curve, AB, which is called the y i eld cu r v e or y i eld s u r fac e . A material subjected to a combination of stresses that lies below the curve will respond elastically (recoverable deformation). If loading is continued beyond the yield stress, the material will respond elastoplastically (permanent deformation occurs). If the material is isotropic, the yield surface will be symmetrical about the σ1, σ3 axes.

4.4 4.4.1

Hooke’s Law Generalizes State of Stress Stress and strains for a linear, isotropic, elastic soil are related through

Hooke’s law. For a generalized state of stress (Fig. 4.8), Hooke’s law is  x   1  y      z  1     xy    0   E  0  yz   0  zx 

  1

   

0 0 0

1 0 0 0

 

0 0 0

2(1   ) 0 0

0 0 0 0 2(1   ) 0

0 0 0 0 0 2(1   )

  x    y    z    xy    yz      zx 

(4.10)

where E is the elastic or Young’s modulus and  is the Poisson’s ratio. Equation (4.10) is called the elastic equations or elastic stress-strain constitutive equations.

53


Figure 4.8: Generalized state of stress. From Eq. (4.10), we have, for example,  zx 

2(1   ) E

 zx 

 zx

(4.11)

G

where, G 

E

(4.12)

2(1   )

is the shear modulus. We will call E , G and  the elastic parameters. Only two of these parameters – either E or G and  – are required to solve problems dealing with isotropic, elastic materials. Typical values of E and G are shown in Table 4.2. Table 4.2: Typical values of E and G. Soil Type Description E (MPa) Clay Soft 1 – 15 Medium 15 – 30 Stiff 30 – 100 Sand Loose 10 – 20 Medium 20 – 40 Dense 40 – 80 4.4.2

G (MPa) 0.5 – 5 5 – 15 15 – 40 5 – 10 10 – 15 15 – 35

Principal Stresses If the stresses applied to a soil are principal stresses, then Hooke’s law

reduces to,  1  1  1    2   E    3  

  1  

    1      2  1    3 

(4.13)

The matrix on the right-hand side of Eq. (4.13) is called the compliance matrix. The inverse of Eq. (4.13) is

54


1      1     

 1  E  2    3  (1   )(1  2 )

    1     2  1      3 

(4.14)

The matrix on the right-hand side is called the s ti ffn es s matrix. 4.4.3

Displacements from Strains and Forces from Stresses The displacements and forces are obtained by integration. For example,

the vertical displacement, z , is



 z   z dz

(4.15)

and the axial force is P z 

  dA z

(4.16)

where dz is the height or thickness of the element and dA is the elemental area.

4.5 4.5.1

Plane Strain and Axisymmetric Conditions Plain Strain Let us consider an element of soil, A, behind a retaining wall (Fig.

4.9). Because the displacement that is likely to occur in the Y direction ( y ) is small compared with the length in this direction, the strain tends to zero; that is,  y 

 y y

0

Figure 4.9: Plane strain condition in a soil element behind a retaining wall. Hooke’s law for a plane strain condition is 55


 1   3 

1   E

1   E

(1   ) 1   3 

(4.17)

(1   ) 3   1 

(4.18)

and  2   ( 1   3 )

(4.19)

In matrix form, Eqs. (4.17) and (4.18) become  1   1   1       1       3 E    1     3 

(4.20)

The inverse of Eq. (4.20) gives E  1   1      2  (1   )(1  2 )  

4.5.2

   1   1      2 

(4.21)

Axisymmetric Condition The other condition that occurs in practical problems is axial

s y m m etr y or the a x i s y m m e t r i c condition where two stresses are equal. Let us consider a water tank or an oil tank founded on a soil mass as illustrated in Fig. 4.10. The radial stress (σr) and the circumferential stress (σθ) on a cylindrical element of soil directly under the center of the tank are equal because of axial symmetry.

Figure 4.10: Axisymmetric condition in a soil element under the center of a tank. Hooke’s law for the axisymmetric condition is:  1   3 

1

 1  2 3 

(4.22)

(1   ) 3  1 

(4.23)

E

1 E

56


or in matrix form,  2   1   1   1  1      ( 1   )  3   2  E 

(4.24)

The inverse of Eq. (4.24) gives E  1   1      3  (1   )(1  2 )  

2   1   1    3 

(4.25)

EXAMPLE 4.1

A rectangular retaining wall moves outward causing a lateral strain of 0.1% and a vertical strain of 0.05% on a soil element located 3 m below ground level. Assuming the soil is a linear, isotropic, elastic material with E = 5000 kPa and  = 0.3, calculate the increase in stresses imposed. If the retaining wall is 6 m

high and the stresses you calculate are the average stresses, determine the lateral force increase per unit length of a wall.

S t r at eg y You will have to make a decision whether to use the plane strain or axisymmetric condition. You are asked to find the increase in stresses, so it is best to write the elastic equations in terms of increment. The retaining wall moves outward, so the late ral strain is tensile (–) while the vertical stress is compressive (+). The increase in lateral force is found by integration of the average lateral stress increase. EXAMPLE 4.2

An oil tank is founded on a layer of medium sand 5 m thick underlain by a deep deposit of dense sand. The geotechnical engineer assumed, based on experience, that the settlement of the tank would occur from settlement in the medium sand. The vertical and lateral stresses at the middle of the medium sand directly under the center of the tank are 50 kPa and 20 kPa, respectively. The values of E and  are 20 MPa and 0.3, respectively. Assuming a linear, isotropic, elastic material behavior, calculate the strains imposed on the medium sand and the vertical settlement.

S t r at eg y You have to decide on the stress conditions on the soil element directly under the center of the tank. Once you make your decision, use the appropriate equations to find the strains and then integrate the vertical strains to calculate the settlement. Draw a diagram illustrating the problem.

4.6

Anisotropic Elastic States Anisotropic materials have different elastic parameters in different 57


directions. Anisotropy in soils results from essentially two causes 1. The manner in which the soil is deposited. This is called structural anisotropy and it is the result of the kind of soil fabric that is formed during deposition. You should recall (Chapter 2) that the soil fabric produced is related to the history of the environment in which the soil is formed. A special type of structural anisotropy occurs when the horizontal plane is a plane of anisotropy. We call this form of structural anisotropy, transverse anisotropy. 2. The difference in stresses in the different directions. This is known as stress induced anisotropy.

T r a n s v e r s e a n i s o t r o p y , also called c r o s s a n i s o t r o py , is the most prevalent type of anisotropy in soils. If we were to load the soil in the vertical direction (Z direction) and repeat the same loading in the horizontal direction, say, the X direction, the soil will respond differently; its stress–strain characteristics and strength would be different in theses directions. However, if we were to load the soil in the Y direction, the soil’s response would be similar to the response obtained in the X direction. The implication is that a soil mass will, in general, respond differently depending on the direction of the load. For transverse anisotropy, the elastic parameters are the same in the lateral directions ( X and Y directions) but are different from the vertical direction. To fully describe anisotropic soil behavior, we need 21 elastic constant s (Love, 1927) but for transverse anisotropy, we need only 5 elastic constants; these are E z, E x,  xx ,  zx , and  zz . The first letter in the double subscripts denotes the direction of loading and the second letter denotes the direction of measurement. For example,  zx means poison’s ratio determined from the ratio of the strain in the lateral direction ( X direction) to the strain in the vertical direction (Z direction) with the load applied in the vertical direction (Z direction). For axisymmetric conditions, the transverse anisotropic, elastic equations are 1  z    E z     zr  r   E z

 2 rz



E r   z   (1   rr )    r  E r 

(4.26)

where the subscript z denotes vertical and r denotes radial. It is well known that  rz  zr  E r E z . 58


EXAMPLE 4.3

Redo Example 4.2 but now the soil under the soil tank is an anisotropic elastic material with E z=20 MPa, E r=25 MPa,  rz =0.15, and  rr =0.3.

S t r at eg y The solution of this problem is a straightforward application of Eq. (4.26)

4.7

Total and Effective Stresses

4.7.1

The Principle of Effective Stress

4.7.2

Effective Stresses Due to Geostatic Stress Field (or Self Weight of Soils)

4.7.3

Effects of Capillarity

4.7.4

Effects of Seepage

4.8

Stresses In Soil From Surface Loads The distribution of surface stresses within a soil is determined by

assuming that the soil is a semi-infinite, homogeneous, linear, isotropic, elastic material. A s em i - i n fi ni te mass is bounded on one side and extends infinitely in all other directions; this is also called an “elastic half space. ” For soils, the horizontal surface is the bounding side. Equations and charts for several types of surface loads based on the above assumptions are presented. 4.8.1

Point Load

Figure 4.11: Point load and vertical load distribution with depth a nd radial 59


distance. Boussinesq (1885) presented a solution for the distribution of stresses for a point load applied on the soil surface. An example of a point load is the vertical load transferred to the soil from an electric power line pole. The increases in stresses on a soil element located at point A (Fig. 4.11a) due to a point load, Q, are 3Q 

    z  2 2 2 z  1  ( r / z )   r     

1

5/ 2

Q 

 3r 2 z 1  2    2 2 2 5 / 2 2 2 2 1 / 2 2  ( r  z ) r  z  z ( r  z ) 

Q

2

  1 z    (r 2  z 2 ) 3 / 2 r 2  z 2  z (r 2  z 2 )1/ 2   

(1  2 )

 rz 

3Q 

 rz 2  2 2 5/2  2  ( r  z ) 

(4.27)

(4.28) (4.29) (4.30)

where  is Poisson’s ratio. Most often, the increase in vertical stress is needed in practice. Equation (4.27) can be written as  z 

Q z

2

(4.31)

I

where I is an influence factor, and 3  

  I  2 2  1  (r / z )  1

5/ 2

(4.32)

The distributions of the increase in vertical stress from Eq. (4.32) reveal that the increase in vertical stress decreases with depth (Fig. 4.11 b) and radial distance (c). 4.8.2

Line Load

60


Figure 4.12: (a) Line load and (b) line load near a retaining wall. With reference to Fig. 4.12a, the increase in stresses due to a line load, Q (force/length), are  z   x   zx 

2Q z 3  ( x 2  z 2 ) 2

(4.33)

2

2Q x z 2

2 2

 ( x  z ) 2Qx z 2

 ( x 2  z 2 ) 2

(4.34) (4.35)

A practical example of line load i s the load from a long brick wall. 4.8.3

Line Load Near a Buried Earth Retaining Structure The increase in lateral stress on a buried earth retaining structure (Fig.

4.12b) due to a line load of intensity Q (force/length) is 2

 x 

4Qa b  H 0 (a 2  b 2 ) 2

(4.36)

The increase in lateral force is P x  4.8.4

2Q  (a 2  1)

(4.37)

Strip Load

61


Figure 4.13: Strip load imposing (a) a uniform stress and (b) a linearly varying stress. (c) Strip load near a retaining wall and (b) lateral force near a retaining wall from a strip load. A strip load is the load transmitted by a structure of finite width and infinite length on a soil surface. Two types of strip loads are common in geotechnical engineering. One is a load that imposes a uniform stress on the soil, for example, the middle section of a long embankment (Fig. 4.13a). The other is a load that induces a triangular stress distribution over an area of width B (Fig. 4.13b). An example of a strip load with a triangular stress distribution is the stress under the side of an embankment. The increases in stresses due to a surface stress qs (force/area) are as follows: (a) Area transmitting a uniform stress (Fig. 4.13a)  z 

qs

 x 

qs





 zx 

  sin  cos(  2 )

(4.38)

  sin  cos(  2 )

(4.39)

qs



sin  sin(  2 )

(4.40)

where qs is the applied surface load. (b) Area transmitting triangular stress (Fig. 4.13b) 62


 z 

 x 

q s  x

    12 sin 2    B 

(4.41)

q s  x

2     z ln R1  12 sin 2   2     B B R2 

 zx 

(4.42)

2  q s  1  cos 2  2     B 

(4.43)

(c) Area transmitting triangular stress (Fig. 4.21c,d)  x 

2q s

(   sin  cos 2 )



(4.44)

The lateral force and its location were derived by Jarquio (1981) and are qs

P x  z 

90

H 0 ( 2   1 )

(4.45)

2

H 0 ( 2   1 )  ( R1  R2 )  57.3 BH 0

(4.46)

2 H 0 ( 2   1 )

where  a  B  1  a  ,  2  tan 1  ,  1  tan    H 0   H 0  R1  ( a  B ) 2 (90   2 ), and R2  a 2 (90   1 ) 4.8.5

Uniformly Loaded Circular Area An example of circular area that transmits stresses to a soil mass in a

circular foundation of an oil or water ta nk. The increase of vertical and radial stresses under a circular area of radius r 0 are 3/ 2     1     q s I c  z  q s 1     1  (r 0 z ) 2    

(4.47)

where 3/2     1     I c  1    1  (r 0 z ) 2    

is an influence factor and  r    

2(1   ) 1 qs  (1  2 )   1 / 2 2 2 2  1  (r 0 z ) 1  (r 0 z ) 







  3/ 2  



(4.48)

The vertical elastic settlement at the surface of due to a circular flexible loaded area is Below center of loaded area :  z 

2

q s D(1   ) E

(4.49)

63


Below edge :  z 

2 q s D (1   2 ) 

(4.50)

E

where D = 2r 0 is the diameter of the loaded area. 4.8.6

Uniformly Loaded Rectangular Area Many

structural

foundations

are

rectangular

or

approximately

rectangular in shape. The increase in stresses below the corner of a rectangular area of width B and length L are  z 

qs 

LB LBz  1 1       tan 1 2 2  zR3 R3  R12 R2 

 x   y 

(4.51)

qs 

LBz  1 LB  2   tan 2  zR3 R1 R3 

(4.52)

qs 

LBz  1 LB  2   tan 2  zR3 R2 R3 

 zx 

q s  B



2  R2



(4.53)

z 2 B  2

(4.54)



R1 R3 

where R1  ( L2  z 2 ) 1 2 , R2  ( B 2  z 2 ) 1 2 , and R3  ( L2  B 2  z 2 ) 1 2 . These equations can be written as  z  q s I z

(4.55)

 x  q s I x

(4.56)

 y  q s I y

(4.57)

 zx  q s I 

(4.58)

where I denotes the influence factor. The influence factor for the vertical stress is 

I z 



2 2   1  2mn m 2  n 2  1  m 2  n 2  2     tan 1  2mn m  n  1   m 2  n 2  m 2 n 2  1  4  m 2  n 2  m 2 n 2  1  m 2  n 2  1 





(4.59)



where m  B z and n  L z . You can program your calculator or use a spreadsheet to find I z. You must be careful in the last term (tan-1) in programming. If m 2  n 2  1  m 2 n 2 , then you have to add  to the quantity in the last term. In general, the vertical stress increase is less than 10% of the surface stress when z > 3B. The vertical elastic settlement at the ground surface under a rectangular surface load is  z 

2

q s B(1   ) E

I s

(4.60) 64


where I s is a settlement influence factor that is a function of the L/B ratio (L is length and B is width). Setting  s  L B , the equations for I s are At center of a rectangle (Giroud, 1968): 2  1  1   s  2 2  I s  ln( s  1   s   s ln     s





At corner of a rectangle (Giroud, 1968): I s 

 1  1   s2  1  ln( s  1   s2   s ln     s 



The above equations can be simplified to the following for  s  1 . At center of a rectangle: I s  0.62 ln( s )  1.12 At corner of a rectangle : I s  0.31ln( s )  0.56 4.8.7

Approximate Method for Rectangular Loads In preliminary analyses of vertical stress increases under the center of

rectangular loads, geotechnical engineers often use an approximate method (sometimes called the 2:1 method). The surface load on an area, B  L , is dispersed at a depth z over an area ( B  z )  ( L  z ) as illustrated in Fig. 4.14. The vertical load increase under the center of the rectangle is  z 

q s BL

( B  z )( L  z )

(4.61)

Figure 4.14: Dispersion of load for approximate increase in vertical stress under a rectangle 4.8.8

Vertical Stress Below Arbitrarily Shaped Area Newmark (1942) developed a chart to determine the increase in

vertical stress due to a uniformly loaded area of any shape. The chart consists of concentric circles divided by radial lines (Fig. 4.15). 65


The area of each segment represents an equal proportion of the applied surface stress at depth z below the surface. If there a re 10 concentric circles (only 9 are shown because the 10 th extends to infinity) and 20 radial lines, the stress on each circle is q s 10 and on each segment is q s (10  20) . The radius to depth ratio of the first (inner) circle is found by setting

 z  0.1q s in Eq. (4.47), that is , 32     1  0.1q s  q s 1   2   1  (r 0 z)  

from which r z  0.27 . For the other circles, substitute the appropriate value for  z ; for example, for the second circle,  z  0.2q s , and find r z . The chart is normalized to the depth; that is, all dimensions are scaled by a factor initially determined for the depth. Every chart should show a scale and an influence factor I N, which for our case is 1 (10  20)  0.005 .

Figure 4.15: Newmark’s chart for increase in vertical stress. The procedure for using Newmark’s chart is as follows 1. Set the scale, shown on the chart, equal to the depth at which the increase in vertical stress is required. We will call this the depth scale. 2. Identify the point on the loaded area below which the stress is required. Let 66


us say this point is point A. 3. Plot the loaded area using the depth scale with point A at the center of the chart. 4. Count the number of segments (N s) covered by the scaled loaded area. If certain segments are not fully covered, you can estimate what fraction is covered. 5. Calculate the increase in vertical stress as  z  q s I N N s . EXAMPLE 4.9

A pole carries a vertical load of 200 kN. Determine the vertical stress increase at a depth 5 m (a) directly below the pole and (b) at a radial distance of 2 m.

S t r at eg y : The first step is to determine the type of surface load. The load carried by the pole can be approximated to a point load. You can then use the equation for the vertical stress increase for a point load. EXAMPLE 4.10

A rectangular concrete slab, 3 m×4.5 m, rests on the surface of a soil mass. The load on the slab is 2025 kN. Determine the vertical stress increase at a depth of 3 m (a) under the center of the slab, point A (Fig. E4.10a), (b) under point B (Fig. E4.10a), and (c) at a distance of 1.5 m from a corner, point C (Fig. E4.10a).

S t r at eg y : The slab is rectangular and the equations for a uniformly loaded rectangular area are for the corner of the area. You should divide the area so that the point of interest is the corner of a rectangle(s). You may have to extend the loaded area if the point of interest is outside it (loaded area). The extension is fictitious so you have to subtract the fictitious increase in stress for the extended area.

67


Figure E4.10a EXAMPLE 4.11

The plan of a foundation of uniform thickness for a building is shown in Fig. 4.11a. Determine the vertical stress increase at a depth of 4 m below the centroid. The foundation applies a vertical stress of 200 kPa on the soil surface.

S t r at eg y You need to locate the centroid of t he foundation, which you can find using the given dimensions. The shape of the foundation does not fit nearly into one of the standard shapes (e.g., rectangles or circles) discussed. The convenient method to use for this (odd) shape foundation is Newmark’s chart.

68


Figu re 4.11a,b

69


CHAPTER FIVE TWO-DIMENSIONAL FLOW OF WATER THROUGH SOILS 5.0

Introduction Many catastrophic failures in geotechnical engineering result from

instability of soil masses due to groundwater flow. Lives are lost, infrastructures are damaged or destroyed, and major economic losses occur. In this chapter, you will study the basic principles of two-dimensional flow of water through soils. The topics covered here will help you to avoid pitfalls in the analyses and design of geotechnical systems where groundwater flow can lead to instability. The emphasis in this chapter is on gaining an understanding of the forces that provoke failures resulting from groundwater flow. You will learn methods to calculate flow, pore water pressure distribution, uplift forces, and seepage stresses. When you complete this chapter, you should be able to:

 Calculate flow under and within earth structures.  Calculate seepage stresses, pore water pressure distribution, uplift forces, hydraulic gradients, and the critical hydraulic gradient.

 Determine the stability of simple geotechnical systems subjected to two dimensional flow of water. You will use the following principles learned from previous chapters and your courses in mechanics.

 Statics.  Hydraulic gradient, flow of water through soils (chapter 3).  Effective stress and seepage (chapter 3).

S am pl e P r ac t i cal S i tu at i on A deep excavation is required for the construction of a building. The soil is silty sand with groundwater level just below ground level. The excavation cannot be made unless the sides are supported. You, a geotechnical engineer, are required to design the retaining structure for the excavation and to recommend a scheme to keep the inside of the excavation dry.

4.9 Definition of Key Terms

Equi potential Line is a line representing constant head. F l ow L i n e is the flow path of a particle of water. F l ow N e t is a graphical representation of a flow field. 70


S eepag e S tr es s is the stress (similar to frictional stress in pipes) imposed on a soil as water flows through it.

S t at i c Li qui fac ti on is the behavior of a soil as a viscous fluid when seepage reduces the effective stress to zero.

4.10 Two-dimensional Flow of Water Through Porous Media The flow of water through soils is de scribed by Laplace’s equation. The popular form of Laplace’s equation for two-dimensional flow of water through soils is  2 H  2 H  k x k z 2  0  x 2  z

(5.1)

where H is the total head and k x and k z are the coefficients of permeability in the X and Z directions. Laplace’s equation expresses the condition that t he changes of hydraulic gradient in one direction are ba lanced by the changes in the other directions. The assumptions in Laplace’s equation are:

 Darcy’s law is valid.  The soil is homogeneous and saturated.  The soil and water are incompressible.  No volume change occurs. If the soil were an isotropic material then k x = k z and Laplace’s equation becomes  2 H  2 H  2 0  x 2  z

(5.2)

In this chapter, we are going to emphasis in an approximate (graphical) solution technique for Laplace’s equation called the flo w net

s k et ch i n g . The flow net sketching technique is simple and flexible and conveys a picture of the flow regime. It is the method of choice among geotechnical engineers. But before we delve into this solution technique, we will establish some key conditions that are needed to understand two-dimensional flow. The solution if Eq. (5.1) depends only on the values of the total head within the flow field in the XZ plane. Let us introduce a velocity potential (  ), which describes the variation of total head in a soil mass as   kH

(5.3)

where k is a generic coefficient of permeability. The velocities of flow in the X and Z directions are 71


v x  k x

 H    x  x

(5.4)

v z  k z

 H    x  z

(5.5)

The inference from Eqs. (5.4) and (5.5) is that the velocity of flow ( v ) is normal to lines of constant total head (also called constant piezometric head or equipotential lines ) as illustrated in Fig. 5.1. The direction of v is in the direction of decreasing total head. The head difference between two equipotential lines is called a potential drop or head loss. If lines are drawn that are tangent to the velocity of flow at every point in the flow field in the XZ plane, we will get a series of lines that are normal to the equipotential lines. These tangential lines are called s tr eam li n es or flo w

lines (Fig. 5.1). A flow line represents the flow path that a particle of water is expected to take in steady state flow.

Figure 5.1: Illustration of flow terms. Since flow lines are normal to equipotential lines, there can be no flow across flow lines. The rate of flow between any two flow lines is constant. The area between two flow lines is called a flow channel (fig. 5.1). Therefore, the rate of flow is constant in a flow channel.

4.11 Flow Net Sketching 5.3.1

Criteria for Sketching Flow Nets A flow net is a graphical representation of a flow field that satisfies

Laplace’s equation and comprises a family of flow lines and equipotential lines. A flow net must meet the following criteria: 1. The boundary conditions must be satisfied. 2. Flow lines must intersect equipotential lines at right angles. 72


3. The area between flow lines and equipotential lines must be curvilinear squares. A curvilinear square has the property that an inscribed circle can be drawn to touch each side of the square and continuous bisection results, in the limit, in a p oint. 4. The quantity of flow through each flow channel is constant. 5. The head loss between each consecutive equipotential line is constant. 6. A flow line cannot intersect another flow line. 7. An equipotential line cannot intersect another equipotential line. An infinite number of flow li nes and equipotential lines can be drawn to satisfy Laplace’s equation. However, only a few are required to obtain an accurate solution. The procedure for constructing a flow net is described next. A few examples of flow nets are shown in Figs. 5.2 to 5.3.

Figure 5.2: Flow net for a sheet pile. 5.3.2

Procedure for Sketching Flow Nets

1. Draw the structure and soil mass to a suitable scale. 2. Identify impermeable and permeable boundaries. The soil-impermeable boundary interfaces are flow lines because water can flow along these interfaces. The soil-permeable boundary interfaces are equipotential lines because the total head is constant along these interfaces. 3. Sketch a series of flow lines (four or five) and then sketch an appropriate number of equipotential lines such that the area between a pair of flow lines and a pa ir of equipotential lines (cell) is approximately a curvilinear 73


square. You would have to adjust the flow lines and equipotential lines to make curvilinear squares. You should check that the average width and the average length of a cell are approximately equal. You should also sketch the entire flow net before making adjustments.

Figure 5.3: Flow net under a dam with a cutoff curtain (sheet pile) on the upstream end.

Figure 5.4: Flow net in the backfill of a retaining wall with a vertical drainage blanket. 5.4 5.4.1

Interpretation of Flow Nets Flow Rate The head loss ( h) between each consecutive pair of equipotential lines

is: h 

 H N d

(5.6)

Where H is the total head loss across the flow domain and N d is the number of equipotential drops (number of equipotential lines minus one). From Darcy’s 74


law, the flow through each flow channel for an i sotropic soil is q  Aki  (b 1) k

h L

 k h

b L

 k

 H b

N d L

(5.7)

where B and L are sides of the curvilinear square. By construction B/L = 1, and therefore the total flow is N f  k hN f q  N f q  k  H N d

(5.8)

where N f is the number of flow channels (number of flow lines minus one). The ratio N /N f d is called the shape factor. Both N f and N d can be fractional. In the case of anisotropic soils (different permeabilities in X and Z directions), the quantity of flow is N f q   H N d

5.4.2

k x k z

(5.9)

Hydraulic Gradient You can find the hydraulic gradient over each square by dividing the

head loss by the length, L, of the cell, that is i

h L

(5.10)

You should notice that L is not constant. Therefore the hydraulic gradient is not constant. The maximum hydraulic gradient occurs where L is a minimum, that is, i max 

h Lmin

(5.11)

Where Lmin is the minimum length of the cells within the flow domain. Usually, Lmin occurs at exit points or around corners. 5.4.3

Static Liquefaction Let us consider an element of soil at a depth z near the downstream

end of a sheet pile wall structure subjected to seepage forces with upward flow. As discussed in chapter 3, the vertical effective stress is: '  i w z  z'   z

(5.12)

If the effective stress becomes zero, the soil loses its strength and behaves like a viscous fluid. The soil state at which the effective stress is zero is called static liquefaction. Various other names such as boiling, quicksand, piping, and heaving are used to describe specific events connected to the static liquefaction state. A structure founded on a soil that statically liquefies will collapse. Liquefaction can also be produced by dynamic events such as 75


earthquake. 5.4.4

Critical Hydraulic Gradient

We can determine the hydraulic gradient that brings a soil mass (essentially, coarse grained soils) to static liquefaction. Solving for i in Eq. (5.12) when

 z'  0 , we get i  icr 

 G  1   w Gs  1  s   1 e  w  1  e   w  '

(5.13)

where i cr is the critical hydraulic gradient, Gs is specific gravity, and e is the void ratio. Since Gs is constant the critical hydraulic gradient is solely a function of the void ratio of the soil. In designing structures that are subjected to steady state seepage, it is absolutely essential to ensure that the critical hydraulic gradient cannot develop. 5.4.5

Pore Water Pressure Distribution The pore water pressure at any point j is calculated as follows:

1. Select a datum (for example, choose the downstream water level as the datum.) 2. Determine the total head at j : H j   H  ( N d ) j h, where ( N d ) j

is the

number of equipotential drops at point j . 3. Subtract the elevation head at point j from the total head H j to get the pressure head. For point j below the datum (recall the datum is assumed to be the downstream water level), let the elevation head hz is – z . The pressure head is then (h p ) j   H  ( N d ) j h  h z

(5.14)

4. The pore water pressure is u j  (h p ) j  w

5.4.6

(5.15)

Uplift Forces Lateral and uplift forces due to groundwater flow can adversely affect

the stability of structures such as dams and weirs. The uplift force per unit length (length is normal to the XZ plane) is found by calculating the pore water pressure at discrete points along the base (in the X direction) and then finding the area under the pore water pressure distribution diagram, that is, n

Pw 

 u x j

j

(5.16)

i 1

where P w is the uplift force per unit length, u j is the average pore water pressure over an interval  x j , and n is the number of intervals. It is 76


convenient to use Simpson’s rule to calculate P w:   n n   x  Pw  u1  u n  2 ui  4 ui  3   i 3 i 4   odd even  





(5.17)

EXAMPLE 5.1

A dam, shown in Fig.E5.1a, retains 10 m of water. A sheet pile wall (cutoff curtain) on the upstream side, which is used to reduce seepage under the dam, penetrates 7 m into a 20.3 m thick silty sand stratum. Below the silty sand is a thick deposit of clay. The average coefficient of permeability of the silty sand is 2.0×10-4 cm/s. Assume that the silty sand is homogeneous and isotropic. (a) Draw the flow net under the dam. (b) Calculate the flow, q. (c) Calculate and draw the pore water pressure distribution at the base of the dam. (d) Determine the uplift force. (e) Determine and draw the pore water pressure distribution on the upstream and downstream faces of the sheet pile wall. (f) Determine the resultant lateral force on the sheet pile wall due to the pore water. (g) Determine the maximum hydraulic gradient. (h) Will piping occur if the void ratio of the silty sand is 0.8. (i) What is the effect of reducing the depth of penetration of the sheet pile wall?

S t r at eg y : Follow the procedures described in section 5.2 to draw the flow net and calculate the required parameters.

FigureE5.1a 77


5.5

Flow Through Earth Dams Flow through earth dams is an important design consideration. We

need to ensure that the pore water pressure at the downstream end of the dam will not lead to t o instability and the exit hydraulic gradient does not lead to piping. The major exercise is to find the top flow line called the phreatic surface (Fig.5.2). The pressure head on the phreatic surface is zero. Casagrande (1937) showed that the phreatic surface can be approximated by a parabola with corrections at the points of entry and exit. The focus of the parabola is at the toe of the dam, point F (Fig.5.2).

Figure 5.5: 5.5: Phreatic surface within an earth dam. The procedure to draw a phreatic surface within an earth dam, with reference to Fig. 5.5, is as a s follows. 1. Draw the structure to scale. 2. Locate a point A point A at at the intersection of a vertical line from the bottom of the upstream face and the water surface, and a point B where the waterline intersects the upstream face. 3. Locate point C , such that BC =0.3 =0.3 AB. AB. 4. Project a vertical line from C to to intersect the base of the dam at a t D. 5. Locate the focus of the basic parabola. The focus is located at the toe of the dam. 6. Calculate the focal distance, f  ( b 2  H 2  b) / 2 , where b is the distance FD and H is is the height of water at the upstream face. 7. Construct Construct the basic parabola from z  2 f ( f  x ) . 8. Sketch in a transition transition section BE . 78


9. Calculate the length of the discharge discharge face, a, using a

b

cos 

 cos  b 2  H 2 cot 2  ;   30 0.

For   30 0 , use Fig.5.7 and (a) measure the distance TF , where T is the intersection intersecti on of the basic parabola with the downstream face; (b) for the known angle  , read the corresponding factor a / L from the chart; chart; and (c) find the the distance a  TF (1  a / L) . 10.

Measure the distance a from the toe of the dam along the

downstream face to point G. 11.

Sketch in a transition section, GK .

12.

Calculate the flow using q  ak sin  tan  , where k is is the coefficient of

permeability permeabi lity.. If the downstream slope has a horizont horizontal al drainage blanket as shown in Fig. 5.3, the flow is calculated using q  2 fk .

Figure 5.6: 5.6: A horizontal drainage blanket at the toe of an earth dam.

a L

 (degrees)

Figure 5.7: 5.7: Correction factor for downstream curve.

79


CHAPTER SIX ONE-DIMENSIONAL CONSOLIDATION SETTLEMENT OF FINE SOILS Definitions of Key Terms

Consolidation is the time-dependent settlement of soils resulting from the expulsion of water from the soil pores.

P r i m a r y C o ns n s o l i d at at i o n is the change in volume of a fine-grained fine-grained soil caused by the expulsion of water from the voids and the transfer of load from the excess pore water pressure to the soil particles.

S e c on d ar y C o n s oli ol i d at i on is the change in volume of a fine-grained soil caused by the adjustment of the soil fabric (internal structure) after primary consolidation has been completed.

D r a i n a g e p a th t h , H dr , is the longest vertical path that a water particle will take dr to reach the drainage surface.

P r e c o n s o l i d a t i o n S t r e s s or past maximum effective stress,  zc' , is the maximum vertical vertical effective stress that a soil was subjected to in the past.

Normally Consolidated Soil is one that has never experienced vertical ' ' effective stresses greater than its current vertical effective stress (  z 0   zc ).

Over consolidated S oil is one that has experienced vertical effective stresses greater than its current vertical effective stress (  z 0   zc ). '

'

Overconsolidation ratio, OCR, is the ratio by which the current vertical '

'

effective stress in the soil was exceeded in the past (OCR =  zc /  z 0 ).

C om o m p r e s s i o n I n d e x , C c c , is the slope of the normal consolidation line on a plot of the logarithm of vertical vertical effective stress versus versus void ratio.

U n l od od ai a i n g / r e l oa oa d i n g i n d e x o r r e c o m p r e s s i o n i n de d e x , C , r r is the average slope of the unloading/reloading curves in a plot of the logarithm of vertical effective stress versus void ratio.

M od u lu s o f v o lu m e c om p r e s s i b i li t y , m v , is the slope of the curve between bet ween two stress points in a plot of vertical effective stress versus vertical strain.

CHAPTER SIX ONE-DIMENSIONAL CONSOLIDATION CONSOLIDATION SETTLEMENT OF FINE SOILS 80


6.0

Introduction In this chapter, we will consider one-dimensional consolidation

settlement of fine-grained soils. We will restrict settlement consolidation to the vertical direction. Under load, all soils will settle, causing settlement of structures founded on or within them. If the settlement is not kept to a tolerable limit, the desired use of the structure may be impaired and the design life of the structure may be reduced. Structures may settle uniformly or nonuniformly. The latter is called differential settlement and is often the crucial design consideration. The total settlement usually consists of three parts – immediate or elastic compression, primary consolidation, and secondary compression. We have considered elastic settlement in chapter 4. Here we will develop the basic concepts of consolidation and show how these concepts can be applied to calculate the consolidation settlement from applied loads. After that, we will formulate the theory of one-dimensional consolidation and use it to predict the time rate of settlement. After you have studied this chapter, you should:

 Have a basic understanding of the consolidation of soils under vertical loads.

 Be able to calculate one-dimensional consolidation settlement and time rate of settlement. You will make use of the following concepts learned from the previous chapters and your courses in mechanics and mathematics.

 Stresses in soils – effective stress and vertical stress increases from surface loads (chapter 4).

 Strains in soils – vertical and volumetric strains (chapter 4).  Elasticity (chapter 4).  Flow of water through soils (chapter 3, Darcy’s law).  Solution of partial differential equations.

S am pl e P r ac ti cal S i t uat i on The foundation for an office building is to be constructed on a sand deposit. Below the sand deposit is a layer of soft clay followed by a thick layer of sand and gravel. The office building is not asymmetrical and the loads are not uniformly distributed. The allowable total settlement is 50 mm and the differential settlement is limited to 6 mm. You are expected to determine the total and differential settlement of the building, the rate of settlement, and the time to achieve 90% consolidation settlement. 81


FigureS1: The leaning tower of Pisa. The leaning tower of Pisa is a classic example of differential settlement (Fig.S1). Construction of the tower started in 1173 and by the end of 1178 when two-thirds of the tower was completed it tilted. Since then the tower has been settling differentially. The foundation of the tower is located about 3 m into a bed of silty sand that is underlain by a 30 m of clay on a deposit of sand. A sand layer approximately 5 m thick intersects the clay. The structure of the tower is intact but its function is impaired by differential settlement.

4.12 Basic Concepts In our development of the various ideas on consolidation settlement, we will assume:

 A homogeneous, saturated soil.  The soil particles and the water to be incompressible.  Vertical flow of water.  The validity of Darcy’s law.  Small strains. We will conduct a simple experiment to establish the basic concepts of the one-dimensional consolidation settlement of fine-grained soils. Let us take a thin, soft, saturated sample of clay and place it between porous stones in a rigid, cylindrical container whose inside wall is frictionless (Fig. 6.1a). The porous stones are used to facilitate drainage of the pore water from the top and bottom faces of the soil. The top half of the soil will drain through the top porous stone and the bottom half will drain through the bottom porous stone. A platen on the top of the top porous stone transmits applied loads to the soil. Expelled water is transported by plastic tubes to a burette. A valve is used to control the flow of the expelled water into the burette. Three pore water transducers are mounted in the side wall of the cylinder to measure the excess 82


pore water pressure near the porous stone at the top (A), at a distance of one-quarter the height (B), and at the midheight of the soil (C). A displacement gauge with its stem on the platen keeps track of the vertical settlement of the soil.

Figure 6.1a: Experimental setup for illustrating basic concepts on consolidation. We will assume that the pore water and the soil particles are incompressible and the initial pore water pressure is zero. The volume of excess pore water that drains from the soil is then a measure of the volume change of the soil resulting from the applied loads. Since the side wall of the container is rigid, no radial displacement can occur. The lateral and the circumferential strains are then equal to zero (  r     0 ), and the volumetric strain (  p   z   r    ) is equal to the vertical strain  z   z H 0 , where  z is the change in height or thickness and H 0 is the initial height or thickness of the soil. 5.3.1

Instantaneous Load Let us now apply a load P to the soil through the load platen and keep

the valve closed. Since no excess pore water can drain from the soil, the change in volume of the soil is zero ( V  0) . The pore water carries the total head. The initial excess pore water pressure in the soil ( u 0 ) is then equal to the change in applied vertical stress,

 z'  P A , where A is the

cross-sectional area of the soil, or more appropriately the change in mean total stress, P  (  z  2 r ) 3 , where  r is the change in radial stress. For our thin soil layer, we will assume that the initial excess pore water pressure will be distributed uniformly with depth so that at every point in the soil layer, the initial excess pore water pressure is equal to the applied stress. For example, if  z  100 kPa , then u 0  100 kPa as shown in Fig. 6.1b. 83


Figure 6.1b: Instantaneous or initial excess pore water pressure when a vertical load is applied. 5.3.2

Consolidation Under a Constant Load: Primary Consolidation

Let us now open the valve and allow the initial excess pore water to drain. The total volume of soil at time t 1 decreases by the amount of excess pore water that drains from it as indicated by the change in volume of water in the burette (Fig. 6.1c). At the top and bottom of the soil sample, the excess pore water pressure is zero because these are drainage boundaries. The decrease of initial excess pore pressure at the middle of the soil (point C) is the slowest because a water particle must travel from the middle of the soil to either the top or bottom boundary to exit the system. You may have noticed that the settlement of the soil ( z ) with time t (Fig. 6.1 c) is not linear. Most of the settlement occurs shortly after the valve was opened. The rate of settlement,  z / t , is also much faster soon after the valve was opened compared with later times. Before the valve was opened, an initial hydraulic head, u 0  w , was created by the applied vertical stress. When the valve was opened, the initial excess pore water was forced out of the soil by this initial hydraulic head. With time, the initial hydraulic decreases and, consequently, smaller amount of excess pore water are forced out. We will call the initial settlement response, soon after the valve was opened, the early time response or primary consolidation. P r i m a r y c o n s o l i d a t i o n is the change in volume of the soil caused by the expulsion of water from the voids and the transfer of load from the excess pore water pressure to the solid particles.

84


Figure 6.1c: Excess pore water pressure distribution and settlement d uring consolidation. 5.3.3

Secondary Compression Primary consolidation ends with u  0 . The later time settlement

response

is

called s e con dar y

com pr e s s i on

or

c r e e p.

Secondary

compression is the change in volume of fine-grained soils caused by the adjustment of the soil fabric (internal structure) after primary consolidation has been completed. The rate of settlement from secondary compression is very slow compared with primary consolidation. We have separated primary consolidation and secondary compression. In reality, the distinction is not clear because secondary compression occurs as part of the primary consolidation phase especially in soft clays. The mechanics of consolidation is still not fully understood and to make estimates of settlement, it is convenient to separate primary consolidation and secondary compression. 5.3.4

Drainage Path The distance of the longest vertical path taken by water to e xit the soil

is called the length of the drainage path. Because we allowed the soil to drain on the top and bottom faces (double drainage) , the length of the drainage path, H dr, is H dr 

H av

2



H 0  H f

4

(6.1)

where H av is the average height and H 0 and H f are the initial and final heights, respectively, under the current loading. If drainage were permitted from only 85


one face of the soil, then H dr  H av . Shorter drainage paths will cause the soil to complete its settlement in a shorter time than longer drainage path. You will see later that, for single drainage, our soil sample will take four times longer to reach a particular settlement than for double drainage. 5.3.5

Rate of Consolidation The rate of consolidation for a homogeneous soil depends on the soil’s

permeability, the thickness, and the length of the drainage path. A soil with a coefficient of permeability lower than that of our current soil will take longer to drain, the initial excess pore water and settlement will proceed at a slower rate. 5.3.6

Effective Stress Changes Since the applied vertical stress (total st ress) remains constant, then

according to the principle of the effective stress (  z'   z  u ), any reduction of initial excess pore water pressure must be balanced by a corresponding increase in vertical effective stress. Increases in vertical stresses lead to soil settlement caused by changes to the soil fabric. As time increases, the initial excess water pressure continues to dissipate and the soil continues to settle (Fig. 6.1c). After some time, usually within 24 hours for many soils tested in laboratory, the initial excess pore water pressure in the middle of the soil reduces to approximately zero and the rate of decrease of the volume of the soil becomes very small. Since the initial excess pore water pressure becomes zero, then from the principle of effective stress all of the applied vertical stress is transferred to the soil; that is, the vertical effective stress is equal to the vertical total stress (  z'   z ). 5.3.7 Void Ratio and Settlement Changes Under a Constant Load The initial volume (specific volume) of a soil is V  1  e 0 (chapter 2), where e0 is the initial void ratio. Any change in volume of the soil ( V ) is equal to the change in void ratio ( e ). We can calculate the volumetric strain from the change in void ratio as

 p 

V V



e 1 e

(6.2)

Since for one-dimensional consolidation,  z   p , we can write a relationship between settlement and the change in the void ratio as 86


 z 

 z H 0



e 1 e

(6.3)

where H0 is the initial height of the soil. We can rewrite Eq. (6.3) as

 z  H 0

e 1 e

(6.4)

We are going to use ρ pc to denote primary consolidation settlement rather than  z , so ρ pc  H 0

e 1 e

(6.5)

The void ratio at the end of the consolidation under load P is e  e0  e  e0  5.3.8

 z H 0

(1  e0 )

(6.6)

Effects of Vertical Stresses on Primary Consolidation

Figure 6.2: Three plots of settlement data from soil consolidation. We can apply additional loads to the soil and for each load i ncrement we can calculate the final void ratio from Eq. (6.6) and plot the results as shown by segment AB in Fig. 6.2. Three types of graphs are shown in Fig. 6.2 to illustrate three different ways of plotting the data from our test. Figure 6.2a is an arithmetic plot of the void ratio versus vertical effective stress. Figure 6.2b is a similar plot except the vertical stress is plotted on a logarithmic scale. Figure 4.3c is an arithmetic plot of the vertical strain (  z ) versus vertical effective stress. The segment AB in Figs. 6.2a,c is not linear because the settlement that occurs for each increment of loading brings the soil to a denser state from its initial state and the soil’s permeability decreases. Therefore, doubling the load from a previous increment, for example, would not cause a twofold increase in settlement. The segment AB (Figs. 6.2a–c) is called the 87


v i r g i n c on s o l i d at i o n li n e or norm al consolidation line (NCL). In a plot of log  z' versus e, the NCL is approximately a straight line. '

At some point of the vertical effective stress, say  zc , let us unload the soil incrementally. Each increment of unloading is carried out only after the soil reaches equilibrium under the previous loading step. When an increment of load is removed, the soil will start to swell by absorbing water from the burette. The void ratio increases but the increase is much less than the decrease in void ratio from the same magnitude of loading that was previously applied. ' Let us reload the soil after unloading it to, say  zu . The reloading path

CD is convex compared with the concave unloading path, BC . The average slopes of the unloading path and reloading path are not equal but the difference is generally small for many soils. We will represent the unloading-reloading path by an average slope BC and refer to it as the

r e c o m p r e s s i o n li n e or the u n l oa d i n g - r e lo ad i n g li n e (URL). A comparison of the soil’s response with typical material responses to loads as shown in Figs. 4.4 and 4.5 reveals that soils can be considered to be an elastoplastic material (see Fig. 4.5). The path BC represents the elastic response while the path AB represents the elstoplastic response of the soil. Loads that cause the soil to follow path BC will produce elastic settlement (recoverable settlement of small magnitude). Loads that cause the soil to follow path AB will produce settlements that have both elastic and plastic (permanent) components. Once the past maximum vertical effective stress,  zc' , is exceeded the slope of the path followed by the soil, DE , is the same as that of the initial loading path AB. Unloading and reloading the soil at any subsequent vertical effective stress would result in a soil’s response similar to paths BCDE . 5.3.9

Primary Consolidation Parameters The primary consolidation settlement of the soil (settlement that

occurs along path AB in Fig. 6.2) can be expressed through the slopes of the curves in Fig. 6.2. We are going to define two slopes for primary consolidation. One is called the coefficient of compression or c om p r e s s i o n i n d e x , C c, and is obtained from the plot of e versus log  z' (Fig. 6.2b). C c  

e2  e1

log

' ( z ) 2

( z' )1

e

 log

' ( z ) 2

(no units)

(6.7)

( z' )1 88


where subscripts 1 and 2 denote two arbitrarily selected points on the NCL. The other is called the m odulus of volume compres si bility , m v, and '

is obtained from the plot of  z versus  z (Fig. 6.2c).

 z  mv   ' ( z ) 2  ( z' )1 ( z' ) 2  ( z' ) 1 ( z ) 2  ( z ) 1

 m 2     kN 

(6.8)

where subscripts 1 and 2 denote two arbitrarily selected points on the NCL. Similarly,

we

can

define

the

slope

BC in

Fig.6.2b

as

the

r e c o m p r e s s i o n i n d e x , C r, which we can express as C r  

e 2  e1

log

' z 2 ' z 1

( )

e zr



log

( )

( z' ) 2

(6.9)

( z' )1

where subscripts 1 and 2 denote two arbitrarily selected points on the URL. The slope BC in Fig.6.2c is called the modulus

of volume

r e c o m p r e s s i b i l i t y , m vr, and is expressed as mvr  

( z ) 2  ( z )1 ( z' ) 2  ( z' )1



 zr log

( z' ) 2

 m 2     kN 

(6.10)

( z' )1

where subscripts 1 and 2 denote two arbitrarily selected points on the URL. From Hooke’s law, we know that Young’s modulus of elasticity is

 z' E   z ' c

(6.11)

where the subscript c denotes constrained because we are constraining the soil to settle only in one direction (one dimensional consolidation). We can rewrite Eq. (6.11) as ' E c 

1 mvr

(6.12)

The slopes C c, C r , mv, and mvr are taken as positive values to satisfy our sign convention of compression or recompression as po sitive. 5.3.10 Effects of Loading History In our experiment, we found that during reloading the soil follows the normal consolidation line once the past maximum vertical effective stress is exceeded. The history of loading of a soil is locked in its fabric and the soil maintains a memory of the past maximum effective stress. To understand how the soil will respond to loads, we have to unlock its memory. If a soil were to 89


be consolidated to stresses below its past maximum vertical effective stress, then settlement would be small because the soil fabric was permanently changed by a higher stress in the past. However, if the soil were to be consolidated beyond its past maximum effective stress, settlement would be large for stresses beyond its past maximum effective stress because the soil fabric would now undergo further change from a current loading that is higher than its past maximum effective stress. The practical significance of this soil behavior is that if the loading imposed on the soil by a structure is such that the vertical effective stress in the soil does not exceed its past maximum vertical effective stress, the settlement of the of the structure would be small, otherwise significant permanent settlement would occur. The preconsolidation stress defines the limit of elastic behavior. For stresses that are lower than the preconsolidation stress, the soil will follow the URL and we can reasonably assume that the soil will behave like an elastic material. For stresses greater than the preconsolidation stress the soil would behave like an elastoplastic material. 5.3.11 Overconsolidation Ratio We will create a demarcation for soils based on their consolidation history. We will label a soil whose current vertical effective stress or '

overburden effective stress,  z 0 , is less than its past maximum vertical effective stress or p r e c on s o li d a t i on s t r e s s ,  zc' , as an overconsolidated

s oi l . An overconsolidated soil will follow a void ratio versus vertical effective stress path similar to CDE (Fig. 6.2) during loading. The degree of overconsolidation called, overcons olidation ratio , OCR, is defined as

OCR 

'  zc

 z' 0

(6.13)

If OCR = 1, the soil is normally consolidated soil. Nor mally consolidated

s oi ls follow paths similar to ABE (Fig. 6.2). 5.3.12 Possible and Impossible Consolidation Soil States The normal consolidation line delineates possible from impossible soil states. Unloading of a soil or reloading it cannot bring it to soil states right of the NCL, which we will call impossible soil states (Fig. 6.2). Possible soil states only occur on or to the left of the NCL.

90


6.2

Calculation of Primary Consolidation Settlement

6.2.1

Effects of Unloading/Reloading of a Sample Taken from Field Let us consider a soil sample that we wish to take from the field at a

depth z (Fig.6.3a). We will assume that the groundwater level is at the surface. The current vertical effective stress or overburden effective stress is '  z' 0  ( sat   w ) z   z

and the current void ratio can be found from  sat (chapter 2).

On a plot of e

versus log  z' , the current vertical effective stress can be represented as A as depicted in Fig. 6.3b. To obtain a sample, we would have to make a borehole and remove the soil above it. The act of removing a soil and extracting the sample reduces the total stress to zero; that is, we have fully unloaded the soil. F rom the principle ' ' of effective stress (Eq.4.39),   u . Since  cannot be negative – that is,

soil can not sustain tension – the pore water pressure must be negative. As the pore water pressure dissipates with time, volume changes (swelling) occur.

Figure 6.3: (a) Soil sample at a depth z below ground surface. (b) Expected one-dimensional response. Using the basic concepts of consolidation described in section 6.2, the sample will follow an unloading path AB (Fig.6.3b). The point B does not correspond to zero effective stress because we cannot represent zero on a logarithmic scale. However, the effective stress level at the start of the logarithmic scale is assumed to be small (  0 ). If we were to reload our soil sample, the reloading path followed depends on the OCR. If OCR=1 (normally consolidated soil), the path followed during reloading would be BCD (Fig.6.3b). '

The average slope ABC is C r. Once  z 0 is exceeded, the soil will follow the normal consolidation line, CD, of slope C c. If the soil were overconsolidated, 91


OCR>1, the reloading path followed would be BEF because we have to reload the soil beyond its preconsolidation stress before it behaves like a normally consolidated line. The average slope of ABE is C r and the slope of EF is C c. The point E, marks the preconsolidation stress. Later in this chapter, we will determine the position of E from laboratory tests (section 6.5). 6.2.2

Primary Consolidation Settlement of Normally Consolidated Fine-grained Soils Let us consider a site consisting of a normally consolidated soil on

which we wish to construct a building. We will assume that the increase in vertical stress due to the building at depth z , where we took our soil sample, is  z . (Recall that you can find  z using the methods described in section 4.10.) The final vertical stress is '  fin   z' 0   z

The increase in vertical stress will cause the soil to settle following the NCL and the primary consolidation settlement is ρ pc  H 0

e  ' H 0  ; OCR  1 C c log fin ' 1  e0 1  e0  z0

(6.14)

' ' where e  C c log( fin  z0 ).

6.2.3

Primary Consolidation Settlement of Overconsolidated Fine-grained Soils If the soil is overconsolidated, we have to consider two cases

depending on the magnitude of  z . We will approximate the curve in the

log  z' versus e space as two straight lines, as shown in Fig. 6.4. In case 1, the increase in  z is such that  fin   z' 0   z is less than  zc' (Fig.6.4a). In this case, consolidation occurs along the URL and ρ pc  H 0

e  ' H 0 '  ;  fin   zc' C r log fin ' 1  e0 1  e0  z0

(6.15)

92


Figure

6.4:

Two

cases

to

consider

for

calculating

settlement

of

overconsolidated fine-grained soils. In case 2, the increase in  z is such that  fin   z' 0   z is greater ' than  zc (Fig.6.4b). In this case, we have to consider two equations, one

along the URL and the other along the NCL. The equation to use i n case 2 is ρ pc 

 '  '  H 0  '  C r log zc'  C c log fin ;  fin   zc' '  1  e0   z 0  zc 

(6.16)

or ρ pc 6.2.4

'  '  fin H 0   C r log( OCR )  C c log ' ;  fin    zc' 1  e0   zc 

(6.17)

Procedure to Calculate Primary Consolidation Settlement The procedure to calculate primary consolidation settlement is as

follows '

1. Calculate the current vertical effective stress (  z 0 ) and the current void ratio (e0) at the center of the soil layer for which settlement is required. 2. Calculate the applied vertical stress increase (  z ) at the center of the soil layer using the appropriate method in section 4.10. 3. Calculate the final effective vertical stress  fin   z' 0   z . 4. Calculate the primary consolidation settlement. a. If the soil is normally consolidated (OCR=1), use Eq. (6.14). b. If the soil is overconsolidated (OCR>1) and  fin   zc' , use Eq. (6.15). ' c. If the soil i s overconsolidated (OCR>1) and  fin   zc , use Eq. (6.16 or

6.17). In which, H 0 in the equations represents the thickness of the soil layer. You can also calculate the primary consolidation settlement using mv. 93


However, unlike C c, which is constant, mv varies with stress levels. You should compute an average value of mv over the stress ranges  z' 0 to  fin . The primary consolidation settlement, using mv, is ρ pc  H 0 mv  z

(6.18)

The advantage of using Eq. (6.18) is that mv is readily determined from displacement data in consolidations tests: you do not have to calculate void ratio changes from the test data as required to determine C c. 6.2.5

Thick Soil Layers For better accuracy, when dealing with thick layers (H 0 > 2 m), you

should divide the soil layer into sublayers (about two to five sublayers) and find the settlement for each sublayer. Add up the settlement of each sublayer to find the total primary consolidation settlement. You must remember that the value of H0 in the primary consolidation equations is the thickness of the sublayer. An alternative method is to use a harmonic mean value of the vertical stress increase for the sublayers in the equations for primary consolidation settlement. The harmonic mean stress increase is

 z 

n(  z )1  ( n  1)( z ) 2  ( n  2)( z ) 3      (  z ) n n  ( n  1)  (n  2)      1

(6.19)

where n is the number of sublayers and the subscripts 1, 2, etc., mean the first (top) layer, the second layer, and so on. The advantage of using the harmonic mean is that the settlement is skewed in favor of the upper part of the soil layer. You should recall from chapter 4 that the increase in vertical stress decreases with depth. Therefore, the primary consolidation settlement of the upper portion of the soil can be expected to be more than the lower portion because the upper portion of the soil layer is subjected to higher vertical stress increases. EXAMPLE 6.1

The soil profile at a site for a proposed office building consists of a layer of fine sand 10.4 m thick above a layer of soft normally consolidated clay 2 m thi ck. Below the soft clay is a deposit of soft sand. The groundwater table was observed at 3 m below ground level. The void ratio of the sand is 0.76 and the water content of the clay is 43%. The building will impose a vertical stress increase of 140 kPa at the middle of the clay layer. Estimate the primary consolidation settlement of the clay. Assume the soil above the water table to be saturated.

S t r at eg y You should write down what is given or known and draw a diagram 94


of the soil profile (see Fig. E6.1). In this problem, you are given the stratigraphy, groundwater level, vertical stres s increase, and the following soil parameters and soil condition: e0 (for sand) = 0.76; w (for clay) = 43% H0 = 2 m,  z  140 kPa, Cc = 0.3, Gs = 2.7 Since you are given a normally consolidated clay, the primary consolidation settlement response of the soil will follow path ABE (Fig.6.2). The appropriate equation to use is Eq. (6.14).

FIGURE E6.1 EXAMPLE 6.2

Assume the same soil stratigraphy as in Example 6.1. But now the clay is overconsolidated with an OCR=2.5, w =38%, and C r=0.05. All other soil values given in Example 6.1 remain unchanged. Determine the primary consolidation settlement of the clay.

S t r at eg y Since the soil is overconsolidated, you will have to check whether the preconsolidation stress is less than or greater than the sum of the current vertical effective stress and the applied vertical stress at the center of the clay. This check will determine the appropriate equation to use. In this problem, the unit weight of the sand is unchanged but the clay has changed. EXAMPLE 6.3

A vertical section through a building foundation at a site is shown in Fig.E6.3. The average modulus of volume compressibility of the clay is mv=5×10-5 m2 /kN. Determine the primary consolidation settlement.

S t r at eg y To find the primary consolidation settlement, you need to know the vertical stress increase in the clay l ayer from the building load. Since the clay layer is finite, we will have to use the vertical stress influence values in Appendix B. If we assume a rough base, we can use the influence values specified by Milovic and Tournier (1971) or if we assume a smooth base we can use the values specified by Sovinc (1961). The clay layer is 10 m thick, so i t is best to subdivide the clay layer into sublayers  2 m thick. 95


Fig.E6.3 EXAMPLE 6.4

A laboratory test on a saturated clay taken at a depth of 10 m below the ground surface gave the following results: C c=0.3, C r=0.08, OCR=5, w =23%, and Gs=2.7. The groundwater level is at the surface. Determine and plot the variation of water content and overconsolidation ratio with depth up to 50 m.

S t r at eg y The overconsolidation state lies on the unloading/reloading line, so you need to find an equation for this line using the data given. Identify what given data is relevant to finding the equation for unloading/reloading line. Here you are given the slope, C r, so you need to use the other data to find the complete question. You can find the coordinate of one point on the unloading/reloading line from the water content. 6.3

Terzaghi One-dimensional Consolidation Theory

6.3.1

Derivation of Governing Equation We now return to our experiment described in Section 6.1 to derive the

theory for time rate of settlement using an element of the soil sample of thickness dz and cross-sectional area dA=dxdy (Fig. 6.5). We will assume the following: 1. The soil is saturated, isotropic and homogeneous. 2. Darcy’s law is valid. 3. Flow only occurs vertically. 4. The strains are small. We will use the following observations made in section 6.1: 1. The change in volume of soil ( V ) is equal to the change of pore water expelled ( V w ), which is equal to the change in the volume of the voids ( V v ). Since the area of the soil is constant (the soil is laterally constrained), the change in volume is directly proportional to the change in height. 2. At any depth, the change in vertical effective stress is equal to the change 96


in excess pore water pressure at that depth. That is,  z  u . '

Figure 6.5: One-dimensional flow through a two-dimensional soil element. For our soil element in Fig. 6.5, the inflow of water is qv dA and the outflow over the elemental thickness dz is (q v  (q v  z )dz )dA . The change in flow is then (q v  z ) dzdA . The rate of change in volume of the soil, must eq ual the change in flow. That is,

V q v  dzdA t  z Recall Eq. (6.2) that the volumetric strain

(6.20)

 p  V V  e /(1  e0 ) , and

therefore

V 

e ' dz dA  mv  z dz dA  mv u dz dA 1  e0

(6.21)

Note that Eqs. (6.5 and 6.18) are also used to obtain Eq. (6.21). Substituting Eq. (6.21) into Eq.(6.20) and simplifying, we obtain

q v u  mv  z t

(6.22)

The one-dimensional flow of water from Darcy’s law is q v  Ak z i  Ak z

h  z

(6.23)

where k z is the coefficient of permeability in the z direction. Partial differentiation of Eq. (6.23) with respect to z gives

qv  2h  k z 2  z  z

(6.24)

The pore water pressure at any time from our experiment in section 6.1 is u  h w

(6.25)

where h is the height of water in the burette. Partial differentiation of Eq.(6.25) with respect to z gives

 2 h 1  2u   z 2  w  z 2

(6.26) 97


Substituting Eq.(6.26) into Eq.(6.24), we get

qv k z  2 u   z  w  z 2

(6.27)

Equating Eq.(4.22) and Eq.(4.27), we obtain 2 u k z  u  t mv  w  z 2

(6.28)

We can replace k z mv  w

by a coefficient C v called the coefficient of consolidation

The units of C v are length2 /time, for example, cm2 /min. Rewriting Eq. (6.28) by substituting C v, we get the general equation for one-dimensional consolidation as

u  2u  C v 2 t  z

(6.29)

Eq.(4.29) is often called the Terzaghi one-dimensional consolidation

equation because Terzaghi (1925) developed it. 6.3.2

Solution of Governing Consolidation Equation Using Fourier Series The solution of any differential equation requires a knowledge of the

boundary conditions. By specification of the initial distribution of excess pore water pressures at the boundaries, we can obtain solutions for the spatial variation of excess pore water pressure with time and depth. Various distributions of pore water pressures within a soil layer are possible. Two of these are shown in Fig. 6.6. One of these is a uniform distribution of initial excess pore water pressure with depth (Fig. 6.6a). This may occur in thin layers of fine-grained soils. The other (Fig. 6.6b) is a triangular distribution. This may occur in a thick layer of fine-grained soils. The boundary conditions for a uniform distribution of initial excess pore water pressure in which double drainage occurs are When t = 0, u  u 0  u z . At the top boundary, z = 0, u  0 . At the bottom boundary, z  2 H dr , u  0, where H dr is drainage path length.

98


Figure 6.6: Two types of excess pore water pressure distribution with depth. A solution for the governing consolidation equation (6.29), which satisfies these boundary conditions, is obtained using the Fourier series, 

u( z, t )   m0

 Mz   exp( M 2T v ) sin M  H dr 

2 u 0

(6.30)

where M  ( 2)( 2m  1) and m is a positive integer with values from 0 to  and, T v 

C v t 2

H dr

(6.31)

where T v is known as the tim e factor , it is a dimensionless term. A plot of Eq. (6.30) gives the variation of excess pore water pressure with depth at different times. Let us examine Eq.(6.30) for an arbitrarily selected isochrone at any time t or time factor T v as shown in Fig. 6.7. At time t =0 (T v=0), the initial excess pore water pressure, u 0 , is equal to the applied vertical stress throughout the soil layer. As soon as drainage occurs, the initial excess pore water pressure will immediately fall to zero at the permeable boundaries. The maximum pore water pressure occurs at the center of the soil layer because the drainage path there is the longest. At time t > 0, the total applied vertical stress increment  z at depth z is equal to  z'  u z . After considerable time ( t  0 ), u z  0 and

 z'   z . We now define a parameter, U z, called t h e d e g r e e o f consolidation or consolidation ratio , which gives us the amount of consolidation completed at a particular time and depth. This parameter can be expressed mathematically as

99


U z  1 

m  Mz  u z 2  1   sin   exp( M 2T v ) u 0 m  0 M  H dr 

(6.32)

Figure 6.7: An isochrone illustrating the theoretical excess pore water pressure distribution with depth. The consolidation ratio is equal to zero everywhere at the beginning of the consolidation ( u z  u 0 ) but increases to unity as the initial excess pore water pressure dissipates. A geotechnical engineer is often concerned with the average degree of consolidation, U , of a whole layer at a particular time rather than the consolidation at a particular depth. The shaded area in Fig. 6.7 represents the amount of consolidation of a soil layer at any given time. The average degree of consolidation can be expressed mathematically from the solution of the one-dimensional consolidation equation as U  1 

m

2

 M

m 0

2

exp( M 2T v )

(6.33)

Figure 6.8: Relationship between time factor and average degree of consolidation. 100


Figure 6.8 shows the variation of the average degree of consolidation with time factor T v for a uniform and triangular distribution of excess pore water pressure. A convenient set of equations for double drainage, found by curve fitting Fig. 6.8, is T v 

  U 



2

 for U  60%

4  100 

(6.34)

T v  1.781  0.933 log(100  U ) U  60%

and

(6.35) The time factor corresponding to every 10% of average degree of consolidation for double drainage conditions is shown in the inset table in Fig. 6.8. The time factor corresponding to 50% and 90% consolidation are often used in interpreting consolidation test results. T v = 0.848 for 90% and T v = 0.197 for 50% consolidation. 6.4

Secondary Compression Settlement You will recall that consolidation settlement consisted of two parts. The

first part is primary consolidation, which occurs at early times. The second part is secondary compression, or creep, which takes place under a constant effective stress. The physical reasons for secondary compression in soils are not fully understood. One plausible explanation is the expulsion of water from micropores; another is viscous deformation of the soil structure. Primary consolidation is assumed to end at the intersection of the projection of the two straight parts of the curve as shown in Fig. 6.9.The secondary compression index is C   

(e t  e p ) log(t / t p )



e log(t / t p )

; t  t p

(6.36)

Overconsolidated soils do not creep significantly but creep settlements in normally consolidated soils can be very significant. (SOME PART MISSING).

101


Figure 6.9: Secondary compression. EXAMPLE 6.5

A soft clay layer 1.5 m thick is sandwiched between layers of sand. The initial vertical total stress at the center of the clay layer is 200 kPa and the pore water pressure is 100 kPa. The increase in vertical stress at the center of the clay layer from a building foundation is 100 kPa. What is the vertical e ffective stress and excess pore water pressure at the center of the clay layer when 60% of consolidation occurs?

S t r at eg y You are given increment in applied stress and the degree of consolidation. You know that the initial change in excess pore water pressure is equal to the change in applied vertical stress. From the data given decide on the appropriate equation, which in this case is Eq. (6.32). SOLUTION 6.5

u 0  100 kPa , Vertical total stress :

u z  u 0 (1  U z )  100(1  0.6)  40 kPa ,

 z  200  100  300 kPa , Total pore water pressure : 100  40  140 kPa

Current vertical effective stress :  z'   z  u z  300  140  160 kPa

Or alternatively , Initial vertical effective stress : 200 - 100  100 kPa , u z  u 0 (1  U z )  100(1  0.6)  40 kPa , Increase in vertica l effective stress at 60% consolidat ion :  z'  100 - 40  60 kPa Current vertical effective stress :  z'  100  60  160 kPa

6.5

One-dimensional Consolidation Laboratory Test

6.5.1 Oedometer Test The one-dimensional consolidation test, called the oedometer test, is ' used to find C c, C r, C α, mv, and  zc . The coefficient of permeability k z, can also

be calculated from the test data. A disk of soil is enclosed in a stiff metal ring and placed between two porous stones in a cylindrical container filled with water, as shown in Fig. 6.10. A metal load platen mounted on top of the upper 102


porous stone transmits the applied vertical stress (vertical total stress) to the soil sample. Both the metal platen and the upper porous stone can move vertically inside the ring as the soil settles under the applied vertical stress. The ring containing the soil sample can be fixed to a container by a collar (fixed ring cell, Fig. 6.10b) or is unrestrained (floating ring cell, Fig.6.10c). Incremental loads, including unloading sequences, are applied to the platen and the settlement of the soil at various fixed times, under each load increment, is measured by a displacement gauge. Each loading increment is allowed to remain on the soil until the change in settlement is negligible and the excess pore water pressure developed under the current load i ncrement has dissipated. For many soils, this usually occurs with in 24 hours, longer monitoring times may be required for exceptional soil types, for example, montmorillonite. Each loading increment is doubled. The ratio of the load increment to the previous load is called the load increment ratio (LIR); conventionally, LIR = 1. To determine C r, the soil sample is unloaded using a load increment ratio – load decrement divided by current load – of 2. At the end of the oedometer test, the apparatus is dismantled and the water content of the sample is determined. It is best to unload the soil sample to a small pressure before dismantling the apparatus, because if you remove the final consolidation load completely, a negative excess pore water pressure that equals the final consolidation pressure would develop. This negative excess pore water pressure can cause water to flow into the soil and increase the soil’s water content. Consequently, the final void ratio calculated from the final water content would be erroneous. The data obtained from the one-dimensional consolidation test are as follows: 1. Initial height of the soil, H 0, which is fixed by the height of the ring. 2. Current height of the soil at various time intervals under each load (time-settlement data). 3. Water content at the beginning and at the end of the test, and the dry weight of the soil at the end of the test. You now have to use these data to determ ine Cc, Cr, Cα, Cv, mv, and  zc' . We will start with finding C v. 6.5.2 Determination of the Coefficient of Consolidation There are two popular methods to calculate C v. Taylor (1942) proposed one method called the r o ot t i m e m e t h o d . Casagrande and Fadum (1940) proposed the other method called the l o g t i m e m e t h o d . The root time 103


method utilizes the early time response, which theoretically should appear a straight line in a plot of square root of time versus displacement gauge reading. 6 . 5 . 2 . 1 R o o t t i m e m e t h o d Let us arbitrarily choose a point, C , on the

displacement versus square root of time factor gauge reading as shown in Fig. 6.11. We will assume that this point corresponds to 90% consolidation (U =90%) for which T v = 0.848 (Fig. 6.8). If point C were to lie on a straight line, the theoretical relationship between U and T v would be U =0.98√ T v; that is if you substitute T v=0.848, you will get U =90%. At early times, the theoretical relationship between U and T v is given by Eq.(6.34); that is, U 

4 

T v  1.13 T v ; U  0.6

The laboratory early time response is represented by a straight line OA in Fig.6.10. You should note that O is below the initial displacement gauge reading because there is an initial compression of the soil before consolidation begins. The ratio of the gradient of OA and the gradient of the theoretical early time response, line OCB, is

1.13 T v 0.98 T v

 1.15

We can use this ratio to establish the time when 90% consolidation is achieved in the one-dimensional consolidation test. The procedure, with reference to Fig. 6.10a, is as follows: 1. Plot the displacement gauge reading versus square root of time. 2. Draw the best straight line through the initial part of the curve intersecting the ordinate (displacement reading) at O and the abscissa ( Time ) at A. 3. Note the time at point A; let us say

t A .

4. Locate a point B, 1.15 t A , on the abscissa. 5. Join OB.

104


Figure 6.10: (a) Root time method and (b) Log time method to determine C v. 6. The intersection of the line OB with the curve, point C, gives the displacement gauge reading and the time for 90% consolidation (t 90). You should note that the value read off the abscissa is

t 90 . Now when U=90%,

T v=0.848 (Fig.6.9) and from Eq. (6.31) we obtain: C v 

2 0.848 H dr

t 90

Where H dr is the length of the drainage path. 6 . 5 . 2 . 2 L o g T i m e M e t h od In the log time method, the displacment gauge

readings are plotted against the logarthim of times. A typical curve obtained is shown in Fig. 6.10b. The theoretical early time settlement response in a plot of logarithm of times versus displacement gauge readings is a parabola. The experimental early time curve is not normally a parabola and a correction is often required. The procedure, with reference to Fig. 6.10b, is as follows: 1. Project the straight portions of the primary consolidation and secondary compression to interesect at A. The ordinate of A, d 100, is the displacement gauge for 100% primary consolidation. 2. Correct the initial portion of the curve to make it a parabola. Select a time t 1, point B, near the head of the initial portion of the curve (U <60%) and then another time t 2, point C , such that t 2=4t 1. 105


3. Calculate the difference in displacement readings, d  d 2  d 1 , between t 2 and t 1. Plot a point D at a vertical distance d from B. The ordinate of point D is the corrected initial displacement gauge reading, d 0, at the beginning of primary consolidation. 4. Calculate the ordinate for 50% consolidation as d 50=(d 100+d 0)/2. Draw a horizonatl line through this point to intersect the curve at E . The abscissa of point E is the time for 50% consolidation, t 50. 5. You will recall (Fig. 6.9) that the time factor for 50% consolidation is 0.197 and from Eq. (6.31) we obtain C v 

2 0.197 H dr

t 50

(6.37)

The log time method mkes use of the early (primary consolidation) and later time response (secondary compression) while the root time method only utilizes the early time response, which is expected to be a straight line. In theory, the root time method should give good results except when non-linearities arising from secondary compression cause substantial deviations from the expected straight line. These deviations are most pronounced in fine-grained soils with organic materials. 6.5.3 Determination of Void Ratio at the End of a Loading Step To determine  zc' , C c, C r , and C  , we need to know the void ratio for each loading step. Recall that at the end of the consolidation test, we determine the water content (w ) of the soil sample. Using this data, initial height (H 0), and the specific gravity (Gs) of the soil sample, you can calculate the void ratio for each loading step as follows: 1. Calculate the final void ratio, efin=wGs. 2. Calculate the total consolidation settlement of the soil sample during the test, ( z )fin = d fin - d i, where d fin is the final displacement gauge reading and d i is the displacement gauge reading at the start of the test. 3. Back-calculate the initial void ratio, e0=efin+ e , where e is found from Eq.(6.4) as

e 

( z ) fin H 0

(1  e0 )

efin  Therefore,

e0 

1

(  z ) fin

H 0 (  z ) fin H 0

4. Calculate e for each loading step using Eq. (6.6). 106


6.5.4 Determination of the preconsolidation stress Now that we have calculated e for each loading step, we can plot a graph of the void ratio versus the logarithm of vertical effective stress as '

shown in Fig. 6.12. We will call Fig. 6.12 the e versus log  zc curve. You will now determine the preconsolidation stress using a method proposed by Casagrande (1936). The procedure, with reference to Fig. 6.11a, is as follows: 1. Identify the point of maximum curvature, point D, on the initial part of the curve. 2. Draw a horizontal line through D. 3. Draw a tangent to the curve at D. 4. Bisect the angle formed by the tangent and the horizontal line at D. 5. Extend backward the straight portion of the curve (the normal consolidation line), BA, to intersect the bisector line at F . 6. The abscissa of F is the preconsolidation effective stress,  zc' . A simpler method that is also used in practice is to project the straight portion of the initial recompression curve to intersect the b ackward projection of the normal consolidation line at F as shown in Fig. 6.11b. The abscissa of F is  zc' .

Figure 6.11: Determination of the preconsolidation stress using (a) Casagrande’s method and (B) Simplified method. 6.5.5 Determination of the Compression and Recompression Indices 107


The slope of the normal consolidation line, BA, gives the compression index, C c. To determine the recompression index, C r, draw a line (BC ) approximately mid-way between the unloading and reloading curves (Fig. 6.11a). The slope of this line is the recompression index (C r). Field observations indicate that, in many instances, the predictions of the total settlement and the rate of settlement using oedometer test results do not much record settlement from actual structures. Disturbances from sampling and sample preparation tend to decrease C c and C v. Schmertmann (1953) suggested a correction to the laboratory curve to obtain a more representative in situ value of C c. His method is as follows: Locate a point A at ' , e 0 ) and a point B at ordinate 0.42e0 on the laboratory e versus coordinate (  zo

log  z' curve, as shown in Fig. 6.12. The slope of the line AB is the corrected

value for C c.

Figure 6.12: Schmertmann’s method to correct C c for soil disturbances. 6.5.6 Determination of the Secondary Compression Index The secondary compression index, C  , can be found by making a plot similar to Fig. 6.9. You should note that Fig. 6.9 is for a single load. The value of C  usually varies with the magnitude of the applied loads and other factors such as the LIR. EXAMPLE 6.6

At a vertical stress of 200 kPa, the void ratio of a saturated soil sample tested in an oedometer is 1.52 and lies on the normal consolidation line. An increment of vertical stress of 150 kPa compresses the sample to a void ratio of 1.43. 1. Determine the compression index C c of the soil. 2. The sample was unloaded to a vertical stress of 200 kPa and the void ratio increased to 1.45. Determine the slope of the recompression index, C r. 108


3. What is the overconsolidation ratio of the soil at stage (2)? 4. If the soil were reloaded to a vertical stress of 500 kPa, what void ratio would be attained?

S t r at eg y Draw a sketch of the soil response on an e versus log  z' curve. Use this sketch to answer the various question. SOLUTION 6.6

Step 1:

C c is the slope of AB shown in Fig. E6.5. C c 

Step 2:

C r is the slope of BC in Fig. E6.5. C r  350

Step 3： OCR 

200

(1.43  1.52) log( 350 / 200)

 (1.43  1.45) log(350 / 200)

 0.37

 0.08

 1.75  500    1.43  0.37 log 1.43  1.37  350 

e D  e B  C c log

Step 4:

EXAMPLE 6.7

During a one-dimensional consolidation test, the height of a normally consolidated soil sample at a vertical effective stress of 200 kPa was 18 mm. The vertical effective stress was reduced to 50 kPa, causing the soil to swell by 0.5 mm. Determine mvr and E ’ c.

S t r at eg y From the data, you can find the vertical strain. You know the increase in vertical effective stress, so the appropriate equations to use to calculate mvr and E ’ c are Eqs. (6.10) and (6.11). SOLUTION 6.6  z 

 z H 0



0.5 18

 0.028

m vr 

 z 0.028   1.9  10 4 m 2 / kN ' 150  z

E c' 

1 mvr

 5236 kPa

EXAMPLE 6.8

The following readings were taken for an increment of vertical stress of 20 kPa in an oedometer test on a saturated clay sample, 75 mm in diameter and 20 mm thick. Drainage was permitted from the top and bottom boundaries. Time（min）

0.25

1

2.25

4

9

16

25

36

ΔH (min)

0.12

0.23

0.33

0.43

0.59

0.68

0.74

0.7 109


6 Determine the coeffiient of consolidation using the root time method.

S t r at eg y Plot the data in a graph of displacement reading versus time and follow the procedures in Section 6.5.2.1. SOLUTION 6.8 Step 1:

t 90  3.22 min

Step 2:

H dr 

1/ 2

; t 90 = 10.4 m.

20  (20  0.89) 4

 9.8 mm

C v 

2 0.848 H dr

t 90

 7.8 mm 2 / min

t (min 1 / 2 )

6.6

Evaluation of Total Soil Settlement In general, the total settlement  total of a foundation can be given as:

 total   e   pc   sc where,  e

(6.38)

= immediate settlement.

 pc = primary consolidation settlement.  sc

= secondary compression settlement.

The immediate settlement is sometimes referred to as the elatic settlement. In granular soils this is the predominant part of the settlement, whereas in saturated inorganic silts and clays the primary consolidation settlement predominates. The secondary compression settlement forms the major part of the total settlement in highly organic soils and peats. The procedures for calculating primary consolidation and scondary compression settlements are covered in the previous sections in detail. Equtaions. (4.49 and 4.50) in chapter 4 are used to calculate the immediate or elatic settlement of a foundation subjected to a uniform circular load and Eq. (4.60) is used to calculate the immediate or elatic settlement of a foundation subjected to a uniform rectangular load. The details of elatic or immediate settlement calculations for different types of loading conditions that are based on the 110


elatic-half space theory can be found in different books such as “Advanced soil mechanics” by B.M. Das and will not b e covered in this course.

111


112


Figure 6.10: a) A typical consolidation apparatus, b) a fixed ring cell and c) a floating ring cell. 6.7

Relationship Between Laboratory and Field Consolidation The time factor (TV) provides a useful expression to estimate the

settlement in the field from the results of a laboratory consolidation test. If two layers of the same clay have the same degree of consolidation, then their time factors and coefficients of consolidation are the same. Hence, (C V t ) field (C V t ) lab  T V  2 2 ( H dr ) lab ( H dr ) field

(6.39)

EXAMPLE 6.8

A sample, 75 mm in diameter and 20 mm high, taken from a clay layer 10 m thick, was tested in an oedometer. It took the laboratory sampl e 15 minutes to reach 50% consolidation. Note that in an oedometer test a sample has two drainage boundaries. (1) If the clay layer in the field has the same drainage condition as the laboratory sample, calculate how long it will take the 10 m clay layer to achieve 50% consolidation. (2) How much more time would take the 10 m clay layer to achieve 50% consoliadtion if drainage existed only on one boundary.

S t r at eg y You are given all the date to directly use Eq. (6.39). For part (1) there is double drainge in the field and the lab, so the drainage path is one-half the soil thickness. For part (2), there is single drainge in the field, so the 113


Soil Mechanics I.handout

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