Soil Mechanics I CE317

Transcription

1 Soil Mechanics I CE317 Civil Engineering Department/ Third Stage Semester (Fall ) 1

2 Course Name: Soil Mechanics I Code: CE317 Name of Lecturer Office No.: Room # 217, 1 st floor Classroom and Meeting Time: Room #308, Sunday/ Group A (9:00-10:50 a.m.), Room #308, Sunday/ Group B (11:00-12:50 a.m.), Room #124 (soil Lab.), two hours per week for each group divided according to the time table. Number of Hours: Two hours per week for theory and another two per week for laboratory. Course Description: This class presents the principles of soil mechanics. It considers the following topics: the origin and nature of soils; volume-weight relations, soil consistency, soil classification, the total and effective stresses, hydraulic conductivity and seepage, compaction, and laboratory methods used to evaluate soil properties. Course Outcomes: Students who successfully complete this course will be able to: 1- Apply basic mathematics, science and engineering principles to solve engineering problems. 2- Classify soil type according to its engineering properties. 3- Understand the basic phase relations of soil. 4- A close understanding of the special behavior of fine grained soil through studying its states limits ( liquid limit, Plastic limit and Shrinkage limit ), and/or its indices (Plasticity index, liquidity index ) 5- Can calculate the earth overburden pressure. 6- Perform common soil tests to identify physical and mechanical properties of soils. 7- Be familiar with soil mechanics tests and determines which test is needed in designing civil engineering projects and/or solving engineering problems. 8- Demonstrate the ability to write clear technical laboratory reports. 9- Demonstrate the ability to work in groups and understand and apply ethical issues associated with decision making and professional conduct in the lab and field environment. Coarse Contents: 1- Introduction. 2- Soil formation and composition, description of individual soil particles, clay minerals. 3- The physical state of a soil sample (Volume-Weight Relations) 4- Soil consistency 5- Soil classification 6- Definitions of (Clay thixotropy, Types of samples, Soil sensitivity, Over consolidation ratio) 7- Stresses within soil mass and capillarity. 8- Fluid flow in soils: one dimensional flow. 9- Compaction. 2

3 References: 1- Das, B.M., Principle of Geotechnical Engineering, 5th edition. 2- Arora, K. R., Soil Mechanics and Foundation Engineering, 6 th edition. 3- Craig, R.F., Craig s Soil Mechanics, 7 th edition. Course Evaluation Criteria: Method Percentage Attendance 5 Quizzes 15 Laboratory Reports 10 Midterm Examination 30 Final Examination 40 Regulations and Requirements: 1- The passing percent of attendance is 80 from the total coarse hours; if the absence exceeded 20% then the student will not be able to perform the final examination. The student must attend in the time specified for the course, any delay will be considered as absence. 2- The student required to bring to the class: Soil Mechanics lecture notes, scientific calculator (mobile phones are not allowed to use). 3- Homework will be given at the end of the lecture. Submitting the homework must be done at the start of the next lecture of this course. The lecturer will not accept any delay in submitting the homework for any reason. 4- The student must be ready for a quiz about the subjects of previous lectures at any time during the course hours. 5- Participation in the class and paying attention during the lecture are important for collecting extra credits and the reverse is also true, disturbing the lecture will have its consequences on the student. The engineering profession is highly regarded by the public because those who practice it do so with ethical and social consciousness. The same is expected of students in this course. 6- Students are required to keep and assemble in a suitable folder the following : Quiz papers Tests question sheets Laboratory Experiment Reports Lecture notes and class notes 7- Keep all the lecture notes after the end of the semester because it is related to the course (Soil Mechanics II CE322) in the second semester. 3

4 1.1: Definition of Soil Mechanics Chapter One: Introduction Soil mechanics is the science that applies laws and principles of physics especially continuous and particulate mechanics and hydraulics of porous media to solve problems encountered in civil engineering related to force-deformation behavior of soil and the flow of fluids through soil. Or: Soil mechanics is the science dealing with the behavior of soil in engineering practice or how does a soil mass behave when acted on by external forces. Geotechnical Engineering can be defined The branch of Civil engineering that deals with the properties of soils and rocks and their capability of supporting structures placed on or under them. 1.2: Historical Background The properties and behavior of soils have exercised the minds of builders from time immoral. The very little attempt appears to have been made to deal with soil problems on a scientific basis until towards the end of the eighteenth century, when Coulomb published his theory of earth pressure and suggested his well-known law of failure of soils. Rankine s theory followed about a century later, and these two classical theories still form the basis underlying modern methods of estimating earth pressure. However, it was Karl Terzaghi ( ) who first brought theory and empiricism together. By his researches and consulting work, Terzaghi has, with justification, earned the title of The Father of Soil Mechanics 1.3: What Makes Soil Mechanics Interesting and Challenging? Soil stands amongst most variable and difficult of all materials to understand and model. Difficulties arise when one attempts to predict the behavior of soil based on a mathematical model such as: 1-The material we are dealing with may be as weak as soft clay or as strong as hard clay or clay shale. 2- Soil properties vary in both the horizontal and vertical directions and it can be change with time, stress, and environment. 4

5 3- Every site has different soil conditions. We cannot specify in advance the material properties that must be used for construction. 4- Soil "hidden" underground and data on small fraction of deposit. 1.4: Areas Involving the Application of Soil Mechanics 1- Foundation design and selection. - Foundations "shallow" e.g., spread footings for buildings - "deep" e.g., piles for offshore platform. 2- Slopes and Excavations. -Cut slopes for highway -Excavation for subway 3- Retaining Structures. - Slurry wall with tieback anchors -Gravity retaining wall 4- Earth Structures and Embankments -Compacted earth fill for dam -Landfill for waste storage 5- Erosion Control. 6- Seepage under, through and into structures. 7- Waste Disposal System. Chapter Two: Soil Formation and Composition 2.1: Formation and Nature of Soils Soil is any uncemented or weakly cemented accumulation of mineral particles formed by weathering of rocks, the void spaces between particles contains water and/or air. The weak cementation can be due to due to carbonates or oxides precipitated between the particles or due to organic matters. 5

6 A. Physical or Mechanical weathering causes disintegration of the rocks into smaller particle sizes, the processes that cause physical weathering are: 1- Freezing and thawing 2- Temperature changes 3- Erosion (Abrasion) 4- Activity of plants and animals including man Particles formed by in this type of weathering will retain the same properties as that of parent rock. The shape of soil particles will be approximately equidimensional: angular, sub-angular, and rounded. Each particles being in direct contact with adjoining particles. The structure of the resulting soil might be loose, medium, or dense depending on the way in which the particles are packed together. 6

7 B. Chemical weathering causes decomposition in rocks by: 1- Oxidation : union of oxygen with minerals in rocks forming another minerals 2- Hydration water will enter the crystalline structure of minerals forming another group of minerals 3- Hydrolysis the release Hydrogen from water will union with minerals forming another minerals 4- Carbonation when Co2 is available with the existence of water the minerals changed to Carbonates. Chemical weathering leads to the formation of groups of crystalline particles of collided size (less than 0.002mm) known as the clay minerals. There are two types of soil caused by the effect of erosion: 1- Sedimentary Soils: a- Residual Soils: Material formed by disintegration of underlying parent rock or partially indurated material. It will form Sand (which considered to be favorable foundation conditions), and Clay. Variable properties requiring detailed investigation. Deposits present favorable foundation conditions except in humid and tropical climates. b- Organic Soils: Accumulation of highly organic material formed in place by the growth and subsequent decay of plant life. It is very compressible, entirely unsuitable for supporting building foundations. organic soils can be: i- Peat: A somewhat fibrous aggregate of decayed and decaying vegetation matter having a dark color and odor of decay. ii- Muck: Peat deposits which have advanced in stage of decomposition to such extent that the botanical character is no longer evident. 2- Transported Soils: It constitutes when the products of weathering don t remain at their original location. During the transportation the size and shape of particles can undergo changes and sorted into certain size range. The agents of transportation are: 1- Gravity. 2- Wind. 3- Water. 4- Glaciers. Types of transported soil: A- Alluvial Soils: Material transported and deposited by running water. B- Aeolian Soils: Material transported and deposited by wind. C- Glacial soils: Material transported and deposited by glaciers, or by melt water from the glacier. 7

8 D- Marine Soils: Material transported and deposited by ocean waves and currents in shore and offshore areas. 2.2: Special Soils (problematic soil) 1- Expansive Soils: Expansive soils are distinguished by their potential for great volume increase upon access to moisture. Soils exhibiting such behavior are mostly Montmorillonite clays and clay shales. Expansive soils can be identified by either their plasticity limit or a swell test. 2- Collapsing Soils: Collapsing soils are distinguished by their potential to undergo large decrease in volume upon increase in moisture content even without increase in external loads. Examples: Loess (sugar clay), weakly cemented sands and silts where cementing agent is soluble (e.g., soluble gypsum, halite, etc.), certain granite residual soils, deposits of collapsible soils are usually associated with regions of moisture deficiency. 3- Permafrost and Frost Penetration: Volume Increase from underground ice formation leads to heave of structure. Silts are the most susceptible to frost heave. 4- Man-made and Hydraulic Fills: Found in coastal facilities, levees, dikes and tailings dams, it has high void ratio, subject to large amount of settlement, uniform gradation but variable grain size within same fills, high liquefaction potential, lateral spreading, and can be easily eroded. 5- Limestone and Related Soils: such as Karst Topography and Calcareous Soils. Limestone is very soluble, and the expansion of underground voids can lead to sinkholes. It can originate from biological processes such as sedimentation of skeletal debris and coral reef formation. Limestone can be found in north of Iraq area. 6- Quick Clays: Quick clays are characterized by their great sensitivity or strength reduction upon disturbance. All quick clays are of marine origin. Because of their brittle nature, collapse occurs at relatively small strains. Slopes in quick clays can fail without large movements. 7- Dispersive Clays: Easily eroded by low water velocities, When placed into embankments, tunnels and gullies easily form (piping). Can be dealt with chemical treatment of the soil, use of geotextiles or blockage using different types of walls. 8- Submarine Soils Found in continental shelf deposits at water depths up to several hundred feet. 8

9 2.3: Description of Individual Soil Particles A sample of soil consists of an assemblage of many individual soil particles with air and/or water filling the voids among the particles. The physical properties of the soil can be described by using the following categories: 1- Particle Size (Grain Size): It Can be determined by sieve analysis and hydrometer test. Boulder > 300mm Cobble 150mm to 300mm Gravel 2.0mm to 150mm Sand 0.06mm to 2.0mm Silt 0.002mm to 0.06mm Clay < mm 2-Particle Shape: Equidimentional, disk, sphere, plate, and rod 3- Degree of Roundness or Angularity The degree of roundness refers to the sharpness of the edges and corners of particles. Angular particle soils generally exhibit better engineering properties. 1-Angular 2-Sub-Angular 3-Rounded 4-Well rounded 5-Flanky particles 6-Needle like particles 9

10 4- Surface Texture: -dull or polished -smooth or rough 5-Color: Color is a useful particle character to the geologists, those how works in mining. There are standard cards that specify each type of soil. Soil Cohesion: Cohesionless Soils Generally are granular or coarse grained Particles do not naturally adhere to each other Have higher permeability Cohesive Soils Generally are fine grained Particles have natural adhesion to each other due to presence of clay minerals Have low permeability Coarse-grained, Granular or Cohesionless Soils: Excellent foundation material for supporting structures and roads. The best embankment material. The best backfill material for retaining walls. Might settle under vibratory loads or blasts. Dewatering can be difficult due to high permeability. If free draining not frost susceptible Fine-Grained or Cohesive Soils: Very often, possess low shear strength. Plastic and compressible. Loses part of shear strength upon wetting. Loses part of shear strength upon disturbance. Shrinks upon drying and expands upon wetting. Very poor material for backfill. Poor material for embankments. Practically impervious. 10

11 Clay slopes are prone to landslides. Silts: Characteristics: - Relatively low shear strength - High Capillarity and frost susceptibility - Relatively low permeability -Difficult to compact Compared to Clays: -Better load sustaining qualities - Less compressible - More permeable - Exhibit less volume change 2.4: Clay Minerals Particles Clay minerals are very tiny crystalline substances evolved primarily from chemical weathering of certain rock forming minerals; they are complex alumina silicates plus other metallic ions. All clay minerals are very small with colloidal sized (D < 1 m). Because of their small size and flat shape, they have very large specific surfaces (high surface area to mass ratio). There is usually a negative electric charge on the crystal surfaces and electro chemical forces on these surfaces are therefore predominant in determining their engineering properties. In order to understand why these materials behave as they do, it will be necessary to examine their crystal structure in some detail. The basic structure units of the most clay minerals consist of silica tetrahedron, alumina octahedron Silica tetrahedron 11

12 Alumina octahedron The various clay minerals are formed by the stacking of combination of the basic sheet structures with different forms of bonding between the combined sheets. Clay minerals tend to form flat, plate like shape. Types of Clay Minerals a) Kaolinite: The combined silica-alumina sheets are hold together fairly tightly by Hydrogen bonding. b) Illite: The combined sheets are linked together by fairly weak bonding due to Potassium Ions held between them. 12

13 c) Montmorilonite: The space between the combined sheets is occupied by water ( H 2 O). Properties of montmorilonite: 1-weak bonding. 2- Considerable swelling due to additional water being adsorbed between the combined sheets. 13

14 2.5: A preview of Soil Behavior The sandy soil composed of deserted particles; the particles are not strongly bounded together so it is free to move. Thus soil is considered as a particulate system, and soil is a multiphase nature (dry (solid + air), partially saturated (solid + air + water, fully saturated (solid + water)). Soil Deformation Nature: The deformation that occur in soil mass, results from an accumulation of several forces acting on the soil particles which causes a compression pressure on the particles that will move above each other to decreasing the voids, which eventually will lead to deform the contact surface of the soil particles, and changing the particle shape as well. The load applied on a soil mass has two components: (N) Normal force and (T) tangential force. 14

15 1- Deformations in Dry Soil: a. Deformation due to contact b. Bending of plate-like particles. c. Inter-particle sliding *the stress-strain behavior of soil will be strongly nonlinear and irreversible. 2- Chemical Interaction: Since the soil is a multiphase system consisting of a mineral phase called the mineral skeleton, plus a fluid phase called the pore fluid, so the quantity of fluid effect on the shear resistance between the soil particles. 15

16 3- Physical Interaction: In saturated soil, the pores are filled with water therefore sustain a pressure on the particles called (pore pressure) Hydrostatic pressure: The pressure in the pore water, at any point equals to the unit weight of water times the depth of the point below the water surface and there is no flow of water. Since the soil is permeable thus the water can flow through soil and interact with mineral skeleton, altering the magnitude of force at the contacts between particles and influencing the compression and shear resistance of the soil. Sharing the load: When load is applied to a soil, it ll have sudden changes, these changes are carried jointly by pore fluid and by the mineral skeleton, the change in pore pressure will cause water to move through the soil, hence the properties of the soil will change with time (Consolidation Settlement) 16

17 Soil Structure: the way the soil particles arrange themselves has an effect on the shear strength, compressibility and permeability of soils. Factors that affect the soil structure are: 1- The shape, size, and mineralogical composition of soil particles, 2- The nature and composition of soil water. Structures in Cohesionless Soil The structures generally encountered in cohesionless soils can be divided into two major categories:- 1. Single grained structure 2. Honeycombed structure In this structure, relatively fine sand and silt form small arches with chains of particles as shown in the figure below. Soils exhibiting honeycombed structure have large void ratios and they can carry ordinary static load. However, under heavy load or when subjected to shock loading, the structure breaks down, resulting in large settlement. 17

18 Structures in Cohesive Soils 1. Dispersed structure 2. Flocculated structure 18

19 Chapter Three: The physical state of a soil sample (Volume-Weight Relations) Since the soil is a multiphase system, the physical and engineering properties of soil spends on the percentage of each element (solid, gas and water). 3.1: Phase Diagram Assumptions and Definitions: - Weight of air W a = 0 -In a dry Soil: Water weight and volume = 0, W w =0, V w =0 - Volume of voids V v include all non-soil volume, both air and water. - In a Saturated Soil: Air volume V a = 0 - Only water and solids appear in completely saturated soil 3.2: Basic Formulas 1) Total volume (V) = V a + V w +V s 2) Total weight (W) = W w + W s 3) Unit weight of water (γ w ) = γ w =9.81 kn/m 3 γ w =1 g /cm 3 γ w = 1000 kg/m 3 γ w = 62.4 lb/ft 3 19

20 4) Total unit weight (γ t ) = units (kn/m 3, kg/m 3, g/cm 3, etc) 5) Unit weight of soil solids (γ s ) = units (kn/m3, kg/m3, g/cm3, etc) 6) Dry unit weight (γ d ) = units (kn/m 3, kg/m 3, g/cm 3, etc) 7) Saturated Unit Weight (γ sat ) = = 8) Submerged unit weight (buoyant unit weight) (effective unit weight) (γ b ) or ( γ ) = γ sat - γ w 9) Specific gravity of soil solids (G s ): is the ratio of soil solids unit weight to the unit weight of distilled water. (unit less) G s = Typical Specific Gravities for Soil Solids Soil Type G s Quartz Sand Silt Clay Chalk Loess Peat Except for organic soils, range is fairly narrow 10) Total specific gravity (G t ) = 11) Porosity (n): the percent of void volume to total volume. n= x 100% 12) Void Ratio (e): the ratio of voids volume to solids volume e = 13) Degree of Saturation (S): The percentage of void volume which is filled with water. S= x 100% When S=100% (Saturated Soil) S=0 (Dry Soil) 100 > S > 0 (Partially Saturated Soil) 14) Moisture Content (Water content) (ω): the weight percentage of water to soil solids 20

21 ω= x 100% 3.3: Mathematical Relations Note: in these rules, every percent should be changed to ratio (divided by 100) before using it. e. S = G s. ω γ t = γ w = G s γ w γ d = = γ b = γ w e = *Try to prove all of the aforementioned mathematical relations. Example (1): A container of soil has a weight of 168 g. On drying the weight became 142 g, the specific gravity of the sand is 2.65, and the volume of the container is 90 cm 3. Find the water content, dry unit weight, total unit weight, voids ratio and porosity. 21

22 Example (2): A sample of soil has, total unit weight of 1.94 g/cm 3, specific gravity of soil solids = 2.65 and void ratio of 0.6. a) Find the moisture content and degree of saturation b) If the degree of saturation became 95% at the same void ratio, then find the total unit weight of soil. Example (3): For a submerged soil sample, the submerged unit weight was 1.08 g/cm 3 and the specific gravity of solids were If the soil sample shrink during drying with 2/3 of its volume. Find the weight of 1.5 m 3 of the dry soil. 22

23 Example (4): prove that ɣ= (in such proves you must start from the left hand side, and with the original rule) 3.4: Density of Cohesionless Soil (granular soils) The natural density of cohesionless soil is conveniently compared with its loose and dense state in terms of voids ratio, this comparison or relative compactness is expressed by the form density index. Density Index (Relative Density) (R D) or (I D ) = x100%..equ. (1) Or by unit weight, R D = x x 100%..equ.(2) Prove that equ.(1)=equ.(2) Where: e max = voids ratio in the loosest state e min = Voids ratio in the densest state. e field = natural void ratio or any given voids ratio. γ d max = dry unit weight of soil in the densest state. γ d min = dry unit weight of soil in the loosest state. γ d field = natural dry unit weight. 23

24 State of Packing Relative Density % Very loose 0-15 loose Medium or compacted Dense Very Dense Example (5): A soil sample with total specific gravity of 1.91, specific gravity of soil solids = 2.69 and the moisture content was 29%. Find the porosity, void ratio, and degree of saturation. Example (6) Given: - Sand Backfill Unit Weight = 109 pcf - Water Content = 8.6% - Specific Gravity of Solids = e max = e min = Find: - Void Ratio - Relative Density 24

25 Example (7) Given: -Total Mass = kg -Total Volume = m3 -Dry Mass = kg -Specific Gravity of Solids = 2.7 Find: - Wet Density -Dry Unit Weight - Void Ratio - Water Content 25

26 Example (8): A saturated sample with voids ratio of 0.6 and soil solids specific gravity of 2.65, find the porosity and water content. 3.5 Question Bank of Chapter Three: 1) Prove: n.ɣ s = e. Ɣ d 2) Prove: Ɣ sat =n( )Ɣ w 3) Prove: γsat= γdry + n γw 4) Prove: γ d = ( ) γ w 5) A natural sand deposit has a water content of 32%, it was found that the maximum and minimum void ratios at the same water content were 0.97 and 0.33 respectively. Determine the natural unit weight for this deposit. Specific gravity=2.65, relative density=80%. 6) A sample of silty clay has a void ratio of 0.8. The soil is allowed to absorb water and its saturated density was found to be 1.92 gm/cm3. Determine the water content of the saturated sample. 7) A given soil mass has a moisture content of 10.5% and a void ratio of 0.67.The specific gravity of the soil solids is It is required to construct three cylindrical test specimens of diameter 3.75 cm and height 7.5 cm from this soil mass. Each specimen should have a moisture content of 15% and a dry density of 1.6 gm/cm3. Determine the weight of given soil to be used for this purpose, and the weight of water that s need to be added. 8) Show that at saturation the moisture (water) content is: 26

27 9) A soil sample has a unit weight of pcf and a saturation of 50%, when its saturation increased to 75%, its unit weight rises to pcf. Determine the voids ratio and the specific gravity G s of this soil. 10) 11) 12) 13) 14) What will be the volume of excavated soil which needed to produce 270,000ft3 of compacted fill? 15) 16) 17) 27

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29 4.1: Definitions Chapter Four: Soil Consistency and Atterberg s Limits The term consistency refers to the relative state with which a soil can be formed. While Atterberg s Limits are water contents at certain limiting or critical stages in soil behavior (especially, fine- grained soils). They, along with the natural water content (ω n ) are the most important items in the description of fine- grained soils and they are correlated with the engineering properties & behavior of fine- grained soils. Depending upon the water contents the following four state of consistency are used: 1- The liquid state 2- The plastic state 3- The semi-solid state 4- The solid state The water contents at which the soil passes from one state to the next are known as consistency limits or Atterberg s limits. They are: 1- Liquid Limit W L (L.L): It is the water content at which the soil first shows a small but definite shearing resistance if the water content is reduced. Casagrande defined the liquid limit as a water content at which a standard groove cut in the remolded soil sample by a grooving tool will close over a distance of 13 mm (1/2 ) at 25 blows of the L.L cup falling 10 mm on a hard rubber base. (See the figure below) 29

30 In practice, it is difficult to mix the soil so that the groove closure occurs at exactly 25 blows, so Casagrande did the following: * Sometimes one point liquid limit test can be used because, for soils of similar geologic origin, the slopes of the flow curves are similar. Where: tan β = slope of flow curve =0.121 (not equal for all soils) n (no. of drops) must be between ( 20 30) for best results. 2-The plastic Limit W P (P.L.): Is defined as the minimum water content at which the soil can be rolled into a thread approximately 3mm in diameter without breaking. 30

31 It should break up into segments about 3 10 mm (1/8 3/8 inch) long. If the thread crumbles at diameter smaller than 3 mm, the soil is too wet. If the thread crumbles at diameter greater than 3 mm, the soil past the P.L 3- The Shrinkage Limit (W S ) (S.L.): The maximum water content at which if decreased, will not cause a decrease in volume of soil mass. Plasticity Index (P.I.): is the average consistency at which the soil exhibits plastic properties. P.I. = L.L - P.L P.I Soil class 0 Non plastic 1-5 Semi-plastic 5-10 Low plastic Medium plastic High plastic More than 50 Very high plastic Consistency Index (I c ) I c = C r = W n : natural water content or water content in the field. 31

32 Liquidity Index (L.I.): indicates in what part of its plastic range a given soil mass lies. L.I = L.I < the soil is in Brittle state L.I between (0 1) ---- the soil is in plastic state L.I >1 --- the soil is in viscous liquid state Flow index (F.I) = Activity of the clay (A): Activity= ( ) Give an indication to the amount of water attractive to the clay. This term used for identifying the swelling potential of clay soils and for certain classification properties. Activity Less than 0.75 Class Non active Normally active Active Activity Type of clay minerals Kaolinite Illite Montmorillonite 4.2: Factors Affecting on Atterberg s Limits 1. Shape and size of particles: The plasticity increase as the specific surface area increases. As the grains size get smaller the plasticity increases while grains with flaky shape had more plasticity characteristics than other shapes. 2. Amount of clay minerals: The plasticity increase as the amount of clay minerals increases. 3. Type of clay minerals: Montmorilonite high plasticity 32

33 Illite Kaolinite medium plasticity Low plasticity 4. Type of positive ions: The type of absorbed ions will affect the plasticity characteristics such as Na, Mg will give high plasticity while Ca will give low plasticity. Sodium or Magnesium high plasticity Calcium low plasticity 5. Amount of organic materials: Increasing the organic materials leads to increase in water absorption and thus increasing the plasticity. 4.3: Applications of Atterberg s Limits: It gives an indication on the following characteristics in soil: 1- Shear Strength 2- Capillarity (Water-holding capacity) 3- Compressibility 4- Permeability :Soil that has high P.I. has low permeability 5- Rate of volume change: decrease with increasing P.I. Example (1): Laboratory test on a soil sample yielded the following results: Liquid limit=54% Plastic limit=25% Natural moisture content =29% The percent finer than0.002mm =18% (a) Determine the liquidity index of the soil. (b) Find out the activity number and comment on the nature of the soil. 33

34 Example (2): The following data were obtained from the liquid & plastic limits tests for a soil with ωn = 15 % Required: a- Draw the flow curve & find the liquid limit. b- Find the plasticity index of the soil c- Find L.I, C.I, F.I. 4.4: Question Bank of Chapter Four 1) Explain with the aid of sketches Atterberg s limits. 2) A saturated sample of clay soil with shrinkage limit of 22%, has a natural water content of 35%.What would its dry volume be as a percentage of its original (natural) volume if G s =2.7? 3) The following results were obtained by performing the liquid limit test: Weight of Weight of can Weight of can+ Number of can+ wet soil (gm) dry soil (gm) blows (gm) Determine the liquid limit by using the equation. 4) Define the liquid limit and shrinkage limit. What are the factors that effect on Atterberg s limits? 34

35 Chapter Five: Soil Classification The objective of classifying soil is to arrange them into groups according to their properties and engineering behavior, thus, if the group to which a soil belongs is determined, it would be possible to predict its behavior. A soil classification system is meant to provide an accepted and systematic method of describing the various types of soil and eliminating the personal factors. Most of the soils classification systems that have been developed for engineering purposes are based on simple index properties such as particle size distribution & plasticity. Although there are several classification systems now in use, none is totally definitive of any soil for all possible applications, because of the wide diversity of soil properties. 5.1: Particle size classification: In this system the soil is arranged into different groups or fractions according to the particles size distribution. A number of arbitrary systems have been suggested differing mainly in the size specified as limits of the different fractions. a- MIT Scale: Suggested by the Massachusetts Institute of Technology. b. US Bureau of soil and PRA (Public Roads Administration) 5.2: Textural Classification Chart: Texture is a term indicating the coarseness or fineness of a soil and the amount or quality of each particle-size group of particles that constitute the soil. Depending on the various percentage of sand, silt-size and clay-size fraction soil is given a textural name on a triangular chart. 35

36 In general classification systems divided soils into the following categories on the basis of particle size. Gravel; Sand; Silt; and Clay, but the nature of soils are mixtures of particles from several size groups, so if we know the principle components of the soils, we can named the soils such as Sandy Clay, Silty Clay ; and so forth. One of these systems is the system developed by AASHTO (American Association of State Highway and Transportation Official).the following chart is used to classify the soil, It is based on the particle size limits: Sand size mm in diameter Silt size mm in diameter Clay size smaller than mm in diameter. The textural classification system is useful for classifying soils consisting of different constituents. The system assumes that the soil does not contain gravel, however if the soil contains a certain percentage of soil particles larger than 2mm size, a correlation is required in which the sum of the percentages of sand, silt and clay is increased to 100% and the textural classification would be done based on these corrected percentages. In this system the term loam is used to describe a mixture of sand, silt and clay partciles in various proportions. The term loam originated in agricultural engineering. This term is not used in soil engineering. In order to eliminate the term loam, the Mississipi River Commission (USA) proposed a modified triangular diagram. The term loam is replaced by soil engieerng terms such as silty clay. The principal 36

37 component of the soil is taken as a noun and the less prominent component as an adjective. Example(1): Classify the soil with, Sand 43%, Silt 44%, and clay 13%. Example(2): Classify the soil with Sand 67%, Silt 27%, and Clay 6%. 37

38 Example (3): classify the following soil samples by using the textural soil chart. Particle-size distribution (%) Soil Gravel Sand Silt Clay A B : Particle Size Distribution: The results of mechanical analysis are usually plotted as a particle-size distribution curve representing particle size on a logarithmic scale and the percent finer on an ordinary scale as shown in figure. The position and general shape and slope of the curve indicate respectively the type and grading of the soil. 38

39 The Unified Soil Classification System USCS: The original form of this system was proposed by Casagrande in 1942 during World War 2, it was revised in At present it widely used among engineers. This system classifies soils under two broad categories: 1- Coarse grained soils that are gravelly and sandy in nature with less than 50% passing through the no.200 sieve. The group symbols start with prefixes of either G or S. 2- Fine grained soils with 50% or more passing through the no. 200 sieve. The group symbols start with prefixes M; C; O & Pt. (see the following The group symbols start with prefixes M; C; O & Pt. 39

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41 Step 1- determine the percent of soil passing no. 200 sieve (F). If F < 50%, the soil will classify as Coarse grained soil gravelly or sandy soil, then go to step 2. If F > = 50%, the soil will classify as Fine grained soil silty or clayey soil, then go to step 3 Step 2 Determine the percent of soil passing no. 4 & retained on no 200 sieve (F1). If F 1 < the soil will take the symbol G (gravel or gravelly soil). If F 1 the soil will take the symbol S (sand or sandy soil) To state the degree of gradation whether to be well (W) or poor (P) the following criteria shall be meet together and the soil will be well graded otherwise the soil will be poorly graded. C u greater than 4 for gravel & greater than 6 for sand C c between 1 and 3 41

42 Then if F < 5% examine GSD & find Cu & Cc and the soil will take one of the following symbol GW, SW, GP, SP according to the above criteria. If F is between 5% - 12% besides the GSD characteristics (Cu & Cc) we shall use the plasticity characteristics such as (L.L & P.I ) with the plasticity chart to define the dual symbol such as GW GM, SW SM, GP GM, SP SM, GW GC, SW SC, GP GC, SP SC. If F > 12% we use the plasticity characteristics (L.L & P.I) with the plasticity chart to state the soil symbol such as GM, GC, SM, SC, GM GC or SM SC. Step 3 For fine grained we use the plasticity characteristics (L.L & P.I) with the plasticity chart to state the soil symbol such as OL or ML, CL ML, CL when L.L <50% but if L.L >50% the symbol will be OH or MH, CH. To state whether the soil is inorganic (M or C) or organic (O) we shall examine the color and changes in L.L & P.I after drying for the soil such test will not describe here. After we classify the soil and give it a symbol, knowing its significant properties we can state the engineering use of it. Very Important Note: For Fine Soils, if the coarse fraction less than 15%, the name will stay the same. While if the coarse fraction between (15-29) % add With Sand or With Gravel at the end of the name depending on the one that is more dominated. If the coarse fraction was more than or equal to 30%, then add "Sandy" or "gravely " to the start of the name depending on which one is more dominated. Example (1) Following are the results of a sieve analysis and L.L & P.L tests for two soils Sieve Size Soil 1 % passing Soil 2% passing No (4.75mm) No.10 (2mm) No.40 ( mm) No No (0.075mm) L.L 20 - P.L 15 - P.I 5 NP (not plastic) 42

43 Example (2): Classify the soil according to the USCS Sieve Size % passing 25.4mm L.L= 40%, P.L.=30%. 43

44 Example (3): Use the USCS to classify the soil with the following data: - 70% of the materials are retained on the No.200 sieve. - More than 50% of the percent above is retained on the 4.75mm sieve. - L.L. = 39%, P.L. = 20% Example (4): Determine the value of the liquid limit of a soil from the following test data. If the plastic limit of this soil is 20%, classify the soil according to plasticity chart (A-line chart). Number of blows Water content %

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46 5.4: Question Bank of Chapter Five 1) Classify the following soils by using the textural classification system. Particle Size Distribution (%) Soil Sample Gravel Sand Silt Clay ) The results of a sieve analysis test for soil sample are shown below; (a) Fill the empty columns and draw the grain-size distribution curve. (b) Classify the soil by using the Unified Soil Classification System. Sieve No. Sieve opening Mass of Soil retained on each sieve (g) Pan Percent of mass retained on each sieve (%) Cumulative Percent retained (%) Percent Finer (%) 3) Define the liquid limit and liquidity index and determine the value of the liquid limit of a soil from the following test data. If the plastic limit of this soil is 20%, classify the soil according to plasticity chart (A-line chart). Number of blows Water content % ) A soil sample was tested in the laboratory with the following results: Liquid limit=non plastic, Plastic limit=non plastic Sieve size 25mm 19mm 12.5mm 9.5mm 4.75mm 2mm 0.425mm 0.150mm 0.075mm % passing Classify the soil sample by unified soil classification system. 5) The following results were obtained from sieve analysis on a soil sample: 46

47 1- Draw the particle size distribution curve. 2- Classify the soil by using USCS, L.L=40, P.L.=30 Sieve Number Pan Diameter(mm) Percentage of Passing ) Use the USCS to classify the soil with the following data: Sieve No % passing L.L. =50%, P.L. =40%, D 10 =0.18mm, D 30 =0.34, D 60 =0.71mm 7) 8) 9) 47

48 6.1: Types of Samples Chapter Six: Special Clay Characteristics In order to understand the clay sensitivity, an introduction on the types of soil samples must be presented. There are two types of soil samples according to the degree of disturbance (disturbance means: deformations, stress relief, change in water content and other properties) that occurs during the extraction: 1-Disturbed Samples: samples which contain all of contents of in-situ materials, but it suffer from a degree of disturbance in structure. These samples are used for visual classification, grain size analysis, Attarberg s limits and chemical tests. 2-Undisturbed Samples: samples that obtained from sampler which designed to preserve as closely as possible the natural structure of material used for shear, consolidation and permeability also can be used for all tests for which disturbed samples used for. 6.2: Clay Sensitivity The degree of disturbance or undisturbance of clay sample due to remolding is expressed by sensitivity (St). S t = ( ) ( ) (without any change in water content) Where: qu (undisturbed): unconfined compression strength in the natural or undisturbed state. qu (remolded): unconfined compression strength in remolded state). Sensitivity Classification 1 Insensitive 2 to 4 Normal or low sensitive 4 to 8 Sensitive 8 to 16 Extra sensitive >16 Quick 48

49 6.3: Thixotropy of Clays Thixotropy is the phenomenon of strength loose strength gain with no change in volume and water content. 6.4: Stress History The stresses to which a soil has been subjected during its formation up to the present are referred to as its stress history. During deposition, effective stresses will increase as more soil particles are placed, and during erosion effective stresses will decrease as soil removed. There are two types of clay according to the stress history: 1- Normally Consolidated Clay: It is clay that has undergone deposition only (only increase in stresses during its history) The stress at the present time= the maximum stresses that the soil was ever subjected to. 49

50 2- Over consolidated Clay: It is a clay that has been subjected to an effective stress in its past stress history ( the effective stress existing at the present time (p o). ) larger than The over consolidated Ratio (OCR): OCR= OCROCR Soil Type 1 Normally consolidated Lightly over consolidated >4 (might reach up to 50) Heavily over consolidated 50

51 6.5: Question Bank of Chapter Six 1) Explain with the aid of sketches the following: a) Sensitivity of clay b) Normally consolidated clay and over consolidated clay. c) Clay Thixotropy 2) What is the difference between normally consolidated clay and over consolidated clay? Explain with sketches. 3) Define the clay sensitivity, list the classification of soils due to sensitivity ranges 4) List the types of soil samples and explain them. 5) What is the difference between clay sensitivity and clay activity? Chapter Seven: Stresses in Soil Mass 7.1: Shear Resistance between Soil Particles The resistance of soil to deformation is influenced strongly by shear resistance at contacts between particles. Shear Resistance: Defined to be the force that must applied to cause movement between particles. f = = coefficient of friction = tanø 51

52 Where: Ø in soil is the angle of internal friction and it varies according to the soil type: Ø for sand about (30-45 ) Ø for clay about (8-20 ) 7.2: Stresses within a soil mass: Refer to the soil profile shown in the figure below; N v and N h represent the normal forces in the vertical and horizontal directions. T v and T h represents the shear forces in the vertical and horizontal directions. 52

53 7.3: Vertical Geostatic Stresses σ v = z for constant σ v = dz if varies with depth Vertical Stress: is the weight of soil above the point under consideration without changing in the state of soil. No shear stress in this case. (Only principal stresses) Overburden pressure: (in case of stratified soil) σ v = 1 z 1 + 2z 2 + 3z 3 + Or σ v = 7.4: Total, Effective and Neutral Pressure in Soil - There are three types of stresses exerted by the soil weight. - The pressure at certain level that arises from the soil and the materials filling the pore spaces is called Total Stress (σ) - For saturated or submerged soils, the total stress consists from two parts, effective stress and neutral pressure (pore water pressure) - Effective pressure (σ ): the measure of the stress existing within the mineral grain structure (without the material filling the pores) - Neutral Pressure (pore water pressure) (u) : is the pressure in voids spaces or pores which exist between and around the mineral grains. Total Pressure σ: σ= σ + u Neutral Pressure= pore water pressure= u 53

54 For Calculations: it is necessary to locate the ground water level. 1- Above water table (ground water level): σ= γ t * z = σ u= zero 2-Under the ground water table: The soil will be submerged in water under the ground water level, therefore the pore water pressure will be greater than zero. u= γ w h w (h w : the vertical distance from the point considered to the water level.) σ = σ-u Example (1): for the ground conditions given, determine the total and effective stresses with depth. 54

55 Example (2): for the ground conditions given, determine the total and effective stresses and plot showing the variation in stresses with the depth. 55

56 7.5: Horizontal Stresses The relation between vertical stresses and horizontal stresses is equal to (K) (coefficient of lateral stresses) K o : coefficient of lateral stresses at rest. K a : active coefficient of lateral stresses K p : passive coefficient of lateral stresses. σ h=σ v * k o K o =1- sin Ø (for sand and normally consolidated clay) K o < 1.0 K o = (1-sinØ ) OCR sinø (for over consolidated clay) K o > 0 Example (3): Draw the profile for total and effective stresses in the vertical and horizontal and pore water pressure. ɣ t = 2.1 ton/m 3 ɣ w =1.0 ton/m 3 Ko=0.4 56

57 7.6: Capillarity Water Capillary: water that is drawn up above the water table. Intermediate Water: water which from a certain amount of rain water is hold by the action of surface tension. Soil Water: the zone effected by evaporation. 57

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59 Example (4): find total stress, vertical stress and pore water pressure for the profile shown below 59

60 7.7: Question Bank of Chapter Seven 1) A sand deposit is 13m thick and overlies a bed of soft clay. The ground water table is 5m below the ground surface. If the sand above the ground water table has a degree of saturation of 45%, plot the diagram showing the variation of vertical total stress, pore water pressure and effective stress in the sand. The void ratio of the sand is 0.7. Take Gs=2.65 If the water table increased till it reached the ground surface, how much will be the difference in stresses. 2) The depth of water in a well is 3 m, below the bottom of the well lays a layer of sand 5 meters thick overlying a clay deposit. The specific gravity of the solids of sand and clay are respectively 2.64 and Their water contents are respectively 25 and 20 percent. Compute the total, effective and pore water pressures at points A and B shown in Figure below. If water in the well is pumped out up to the bottom of the well, estimate the change in the pressures at the points. 3) Calculate the horizontal stresses (total, effective and pore water pressure at point A, the angle of internal friction and unit weight of the soil are shown on the figure below. 60

61 4) 5) 61

62 Chapter Eight: One Dimensional Fluid Flow 8.1: The Aim of Studying the Fluid Flow in Soil 1- To determine rate of flow (quantity of water) beneath and through the earth dam. 2- To determine the strength of soil. 3- To determine the settlement of soil. 8.2: Definitions Any given mass of soil consists of solid particles of various sizes with interconnected void spaces. The continuous void spaces in a soil permit water to flow from a point of high energy to a point of low energy. Permeability is defined as the property of a soil that allows the seepage of fluids through its interconnected void spaces. Heads: The elevation head (he): is the Distance from the datum line to the point considered. Pressure head (hp): is the height to which the water raises in the piezometer. u= hp * γw Total head (ht): ht Piezometer: = hp + he The tube by which the water pressure can be m easured at any point in the soil. 62

63 There are two important points to be considered: 1- The flow between any two points depends on the difference of total head. 2-Any elevation head can be selected for a datum line. Darcy's Law : Darcy demonstrated experimentally that for laminar flow condition in saturated soil, the rate of flow or discharge per unit time is proportional to the hydraulic gradient. 63

64 Coefficient of Permeability (K): is the average velocity of flow that will occur through the total cross-sectional area of soil under unit hydraulic gradient. The unit is usually (m/day), (ft/day). If the soil is having (k) greater than 10-3mm/sec, it is classified as previous. If the soil is having (k) between (10-3 mm/sec- 10-5mm/sec), it is classified as semiprevious. If the soil is having (k) less than 10-5 mm/sec (imprevious) 8.3: Determination of Coefficient of Permeability 1- Laboratory Methods 2- Field Methods 3- Indirect Methods 4- Capillarity-Permeability Test 64

65 1- Laboratory Methods: a- Constant Head Method: (Used for relatively more permeable soils) Where L= length of the specimen h=the head that causing flow The discharge q= the volume of the water collected divided by the time 65

66 b- Variable-Head Method: (used for relatively less permeable soils because the quantity of water collected in the graduated jar of the constant head permeability test is very small and cannot be measured accurately.) Consider the instant when the head is h. For the small time dt, the head falls by dh. Then; 2- Field Methods: The coefficient of permeability of a soil deposit in-situ conditions can be determined by the following field methods: a- Pumping-out tests. b- Pumping-in tests. The pumping-out tests influence a large area around the pumping well and give an overall value of the coefficient of permeability of the soil deposit. The pumping-in tests influence a small area around the hole and therefore gives a value of the coefficient of permeability of the soil surrounding the hole. 3- Indirect Methods: 66

67 The coefficient of permeability of soil can also be determined indirectly from the soil parameters by (i) Computation from the particle size or its specific surface. Hazen conducted a formula for permeability for sand when the particle sizes between (0.1-3) mm (ii) Computation from the consolidation test data. 3- Capillarity-Permeability Test. 8.4: Factors affecting on the coefficient of permeability 1. Shape and size of the soil particles. 2. Void ratio: Permeability increases with increase in void ratio. 3. Degree of saturation: Permeability increases with increase in degree of saturation. 4. Composition of soil particles: For sands and silts this is not important; however, for soils with clay minerals this is one of the most important factors because permeability depends on the thickness of water held to the soil particles. 5. Soil structure: Fine-grained soils with a flocculated structure have a higher coefficient of permeability than those with a dispersed structure. 6. Viscosity of the permeant. 7. Density and concentration of the permeant. 67

68 Example (1): What is the value of permeability in mm/sec for a soil tested by using the head permeameter? The results were: t=4min Volume of water collected=300ml Difference in head=50mm Distance between points=100mm Dia. of sample=100mm Example (2): Find the value of permeability in mm/sec for a soil tested on the falling head permeability, Initial height of water=1500mm, Final height of water=605mm, Time of test=281sec, Length of sample=150mm, Diameter of sample=100mm, Diameter of the stand pipe=5mm 68

69 Example (3): Draw the elevation, pressure, and total heads for the system shown and find the velocity through the system. 69

70 Example (4): Draw the elevation, pressure and total heads for the system shown and find the velocity through the system. 70

71 Example (5): Draw the elevation, pressure and total heads for the system shown and find the velocity through the system 71

72 Example (6): Draw the elevation, pressure and total heads for the system shown and find the velocity through the system 72

73 8.5: Effective Stress in Soil with Fluid Flow For the system shown, in order to see the effect of fluid flow through the soil on the effective stresses, let s consider two cases and compare the stresses: 1) Assuming that there is no flow and the pressure is hydrostatic. 2) Taking the real case, that is having flow in upward direction inside the soil mass he hp ht Elev. σv us u σ v(s) σ v

74 Seepage Force: *The seepage force acts on the direction of flow. *A force per unit volume of soil: J= = = * γw Thus, J= i * γw The total downward stress=z γw + γt * L The total upward stress= γw * (L+z+h) For equilibrium; Downward force= upward force (z γw + γt * L) A = γw * (L+z+h) A = ic= : critical hydraulic gradient. If ic= 1 then Quick Condition (Piping or Boiling will happen). At ic= 1 σ= u σ =0 74

75 Shear Strength =0 When a cohesion less soil (sand or gravel ) is subjected to water flow condition that result in a zero effective stress, the strength of soil becomes zero and quick condition thus exists. Because cohesive soil (clay) has strength even at zero effective stress then it does not necessarily become quick. Example (1) : A 125m layer of soil is subjected to an upward seepage head of 1.85m; what depth of coarse sand would be required above soil to provide a factor of safety against piping equals to 2; let γsat for both soils be 20.4 kn/m³ 75

76 8.6: Question Bank for Chapter Eight 1) 2) 76

77 3) 4) 77

78 9.1: Definitions Chapter Nine: Compaction In construction of highway embankments, earth dams and many other engineering structures, loose soils must be compacted to improve their strength by increasing their unit weight. Compaction is: Densification of soil by removing air voids using mechanical equipment; the degree of compaction is measured in terms of its dry unit weight. In general, soil densification includes: compaction and consolidation. Compaction: is one kind of densification that is realized by rearrangement of soil particles without outflow of water. It is realized by application of mechanic energy. It does not involve fluid flow, but with moisture changing altering. Consolidation: is another kind of densification with fluid flow away. Consolidation is primarily for saturated clayey soils. Water is squeezed out from its pores under load. 9.2: Objectives for Compaction 1- Increasing the bearing capacity of foundations; 2- Decreasing the undesirable settlement of structures; 3- Control undesirable volume changes; 4- Reduction in hydraulic conductivity; 5- Increasing the stability of slopes. 9.3: There are 4 control factors affecting the extent of compaction: 1. Compaction Effort: With the development of heavy rollers and their uses in field compaction, the Standard Proctor Test was modified to better represent field compaction As the compaction effort increases, the maximum dry unit weight of compaction increase The optimum moisture content decreases to some extend 78

79 2. Soil type and Gradation; Fine grain soil needs more water to reach optimum; and coarse grain soil needs less water to reach optimum. 3. Moisture Content; Water added to the soil during compaction acts as a softening agent on the soil particles, The dry unit weight (γ d ) increases as the moisture content increases TO A POINT Beyond a certain moisture content, any increase in moisture content tends to reduce the dry unit weight. 4. Dry unit weight (dry density). 9.4: Zero-air-void unit weight At certain water content, that is the unit weight to let no air in the voids, It is clear that in the above equation, specific gravity of the solid and the water density are constant, the zero-air-void density is inversely proportional to water content w. 79

80 For a given soil and water content the best possible compaction is represented by the zero-airvoids curve. The actual compaction curve will always be below. For dry soils the unit weight increases as water is added to the soil because the water lubricates the particles making compaction easier. As more water is added and the water content is larger than the optimum value, the void spaces become filled with water so further compaction is not possible because water is a kind like incompressible fluid. This is illustrated by the shape of the zero-air-voids curve which decreases as water content increases. General Notes: * According to the procedure of standard Proctor test and modified Proctor test, the soil must be sieved on sieve no.4 (4.75mm) before performing the test, but if the percentage of soil retained on a 4.75 mm sieve is more than20%, the larger mold of internal diameter, effective height of 127.3mm and capacity 2250 ml is used. In this case, 56 blows required for each layer. The rest of the procedure is similar to that in the standard Proctor test. *The placement water content when the sheep-foot rollers, smooth wheel rollers and vibratory rollers are used is of the order of the optimum water content obtained in modified Proctor test. *For important works, a full-scale test is conducted in the field to determine the placement water content, the thickness of the layer, mass and speed of the roller and the number of passes. Sometimes, in case of unimportant works, the placement water content is taken equal to the optimum water content of the Standard Proctor test for light compaction and equal to the modified Proctor test for heavy compaction. However, the field water content is sometimes kept intentionally different from the optimum water content in order to achieve or improve a specific engineering property of the soil. 9.5: Relative Compaction This term is occasionally used to describe the degree of compaction. Where: Relative Compaction R c = x 100% In highways degree of compaction of embankments usually varies from 90% to 100% of the Standard Proctor s Density. In earth dam construction the degree of compaction varies from 85 to 100% of Modified Proctor s Density. V.I. Note: There is a difference between the relative compaction (R c ) and the relative density (D r ) described in chapter three. 80

81 Example (1): 81

82 Example (2): 82

83 Example (3): 83

84 9.6: Question Bank for Chapter Nine 1) 2) 3) 4) 5) 84

85 List of Experiments: 1- Determination of moisture content 2- Specific gravity of soil solids 3- Sieve analysis 4- Hydrometer analysis 5- Liquid limit test 6- Plastic limit test 7- Shrinkage limit test 8- Constant head permeability test 9- Falling head permeability test 10- Standard Proctor test 11- Modified Procotr test Text Book: Soil Mechanics Laboratory I Soil Mechanics Laboratory Manual, Sixth Edition, Braja B. Das, New York, OXFORD UNIVERSITY PRESS,

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