Third Year. Soil Mechanics Lectures. Students

Transcription

1 Soil Mechanics Lectures Third Year Students Includes: Soil Formation, Basic Phase Relationship, soil classification, compaction, Consistency of soil, one dimensional fluid flow, two dimensional fluid flows, stresses within the soil, consolidation theory, settlement and degree of consolidation, shear strength of soil, earth pressure on retaining structure.

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4 Soil Chapter One Soil Formation and Basic-Relation ships Is any uncemented or weakly cemented accumulation of mineral particles formed by weathering of rocks, the void between the particles containing water/ or air. Weak cementation can be due to carbonates or oxides precipitated between the particles or due to organic carbonates or oxides precipitated between the particles or due to organic matter. Depending on the method of deposition, soils can be grouped into two categories: 1- Residual soils: The soils which remain at the place of disintegration of parent rock. 2- Transported soils : The soils, which carried away from their place of disintegration to some other place by transporting agencies. The transporting agencies may be classified as: i) Water ii) wind iii)gravity iv) Ice So in general soil is formed from disintegration of rocks over laying the earth crust. rain wind Weathering Which are usually results from atmospheric processes action on the rock at or near the earth surface. 4

5 1- Mechanical (Physical weathering): All type of actions that cause a disintegration of the parent rocks by physical means such as, gravity, wind and water. The product of this type is rounded, sub rounded or granular, its products called coarse grained soil e.g. (gravel and sand ) they present in nature in a single grain structure. Coarse grained soil Sand & Gravel Cohesion less soil It properties are the same as parent rock. 2- Chemical weathering All types of chemical reactions that occur between the minerals of the rock and the environment (air, water ---et.) and will end up by disintegration of parent rock into fine grain particles; these products have different properties from the parent rock. They present in nature as a lumps of number of plate like particles. The physical property of this product does not reflect the same properties of the parent rocks. Fine grained soil Silt and clay Cohesive material Its properties do not reflect the same properties of the parent rocks. Soil Gravel, Dia > 2 Sand Dia: equivalents diameter (mm) Silt Clay 5

6 Clay minerals: There are two basic structure units that form types of the minerals in the clay: a) Tetrahedral Unit : Consists of four oxygen atoms (or hydroxyls, if needed to balance the structure) and one silicon a tom. Elevation Tetrahedral sheet ھيدرات الرباعية تحتوي على جزيئة واحدة من 4 جزيئات من ايون االوكسجين. السليكون + b) Octahedral Unit (consist of six hydroxyl ion at apices of an octahedral enclosing an aluminum ion at the center). 6

7 Formation of Minerals The combination of two sheets of silica and gibbsite in different arrangements and condition lead to the formation of different clay minerals such as : 1- Kaolinite Mineral : This is the most common mineral is the kaolin. The structure is composed of a single tetrahedral sheet and a single alumina octahedral sheet as shown in figure below: 1- Strong Hydrogen Bond So not affected by water 2- And its also called China clay 3-2- Illite has a basic structure consisting of two silica sheets with a central alumina sheet. There is a potassium bond between the layers. 3- Montmorillonite unit: The basic structural unit is similar to that of Illite. Highly affected by water ويتكون عادة من سحق ال االيت Highly affected by water with high shrinkage and Swell and it is called expansive soil. وعادة تظھر ھذه الحالة في المناطق الصحراوية وشبه الصحراوية 7

8 Clay Particle water relations: In nature every soil particle is surrounded by water. Since the centers of positive and negative charges of water molecules do not coincide, the molecules behave like dipoles. The negative charge on the surface of the soil particle therefore attracts the positive (hydrogen) end of the water molecules. More than one layer of water molecules sticks on surface with considerable force decrease with increase in the distance of the water molecule from the surface. The electrically attracted water surrounds the clay particle is known as the diffused double-layer of water. The water located within the zone of influence is known as the adsorbed layer as shown in figure: Clay Particle Diffuse double layer Adsorbed water layer surrounding a soil particle Clay structures: 1) - Dispersed structure 2) - flocculated structure 8

9 Distinguish between flocculated and dispersed structures Flocculated Dispersed More strength Permeability is higher Low compressibility Lower strength permeability is less higher compressibility Basic Relationships: Wt = Ww + Ws Where : total weight of soil Weight of water : Weight of solid Weight of air 0 Volume Vt = Vv + Vs = Va + Vw + Vs Total Volume : Volume of Void : Volume of air Volume of water : Volume of Soild 9

10 1- Unit Weight Density = = 2- Water content % % = 3- Void ratio, e e = v v 4- Porosity (n%) % = 5- Air content A% % = = 6- Bulk Density (total density), ρ = 7- Dry density, = ( ) ( ) 8- Dry unit weight ( ) = 9- Specific gravity, ( ) 10

11 = = = = = = ( its value range between ) 10- Solid Density, =, = Some Useful Correlation: 1- S.e =. 2- = 3- = 4- = ( ) 5- = 6- = ( ) 7- = 8- = = ( ) = = 9- = = 10-. = = 11- = = Some typical values of void ratio, moisture content in a saturated condition, and dry unit weight for soils in a natural state are given in the following table: Table 1- Void ratio, Moisture Content, and Dry Unit Weight for some Typical Soils in a Natural State. Type of Soil Void ratio Natural moisture content in a saturated state (%) Dry unit weight, ( ) Loose uniform sand Dense uniform sand

12 Loose angulargrained silty sand Dense angulargrained silty sand Stiff clay Soft clay Note: the weight of one kilogram mass is Newton 1 kg = N Example- 1: In its condition a soil sample has a mass of 2290 g and a volume of 1.15*10-3 m 3. After being completely dried in an oven the mass of the sample is 2035g. The value of G s for the soil is Determine the bulk density, unit weight, water content, void ratio, porosity, degree of saturation and air content. Solution: = =.. Unit weight, = = 1990 = 1.99 Water content, = = e = = = = = = % = (1 + ) ( ) Porosity, n= =.. = ~0.35. =. Degree of saturation, S=... = % Air content, A = n (1- S) = 0.35( 1-.62)=

13 Example 2: a moist soil has these values : = , m = kg, = 9.8 %, = Determine:,,,, ( %), by soild? Solution: = = = / = =.. = =. = e = 0.48 = 1 + = = S.e =. S = 2.66 * S = 54.3% = =. = 12.7 = = = =. = = = 1.25 = = Example 3:In the natural state, a moist soil has a volume of and weighs N. The oven dry weight of the soil is N. If = Calculate the moisture content, moist unit weight, dry unit weight, void ratio, porosity and degree of saturation. Solution: = =... = 15.6 % = = = = 19.1 = = = ~

14 e=, = =.. = = = e =.. = 0.6 = =.. = =. S. 0.6= 2.71 * S = 70.46% Example 4: A soil specimen has a volume of 0.05 m 3 and a mass of 87.5 kg. If the water content is 15% and specific gravity is Determine 1) void ratio 2) porosity 3) dry unit weight 4) saturated unit weight 5) degree of saturation. Solution: = = 0.15 =. = = = 1750 / = 76 = = = e = =.. = 0.77, n = =. = = 1 + = = = = / 10 = / S.e =. S*0.77 =2.68* 0.15 S= 52.2% Example 5: Show that = ( ) Solution: take the right hand side : 14

15 Example 6: Given mass of wet sample = 254 gm, void ratio = , volume of air = 1.9 cm 3, mass of solid =210 gm. Determine degree of saturation, air content and dry unit weight. Solution: mt = 254 gm, ms = 210 g = = 44 = = 44 1 = 44 = + = = =. = 74 = = = 95.8% = (1 ) = (1 0.95) = = = + + = = 120 = = 1.75 = 17.5 / Example 7: A soil specimen is 38 mm in diameter and 76 mm long and its natural condition weighs 168 gm when dried completely in an oven the specimen weighs gm. The value of = what is the degree of saturation of the specimen? Solution: Dia = 38 mm = 3.8 cm L= 76 mm = 7.6 cm = ( ) 7.6 = = = 37.5 = = = = =

16 = ( ) = = + = = = =.. = 97.6% Example 8: Given: mass of wet sample =254.1gm, void ratio = , volume of air = 1.9 cm 3, mass of solids = 210 gm. Determine: Degree of saturation, Air content, dry unit weight. Solution: Mass of water = = 44.1 gm Volume of water = = = = =. = 95.8% = 44.1 = = = 46 = 75. A=. (1 ) =. ( ) = = + = = 121 = = = 2.8 = 1 + = = Or = = = / = = = / Example 9: A soil specimen have void ratio of 0.7, = Calculate the dry unit weight, unit weight and water content at degree of saturation of 75%. Solution : 16

17 = = =. 10 = 16, = / = = = = = / % = = 19.1 / 0.75* 0.70 = 2.72 * = 19.3% Example 10 : Prove that S. e =. Take the right hand side;. = = = = Example 11: Show that = = = = = Example 12 : Prove that = = = = = References : 1- Soil mechanics R.F. Craig 17

18 2- Soil mechanics T.W.Lamb, R.V.Whitman 3- Soil Mechanics Basic Concepts and Engineering Applications A.Aysen 18

19 Chapter Two Plasticity of Fine Grained Soils Plasticity is the ability of a soil to undergo unrecoverable deformation at constant volume without cracking or crumbling. It is due to the presence of clay minerals or organic material. Consistency limits (Atterberg limits): Atterberg, a Swedish scientist developed a method for describing the limit consistency of fine grained soils on the basis of moisture content. These limits are liquid limit, plastic limit and shrinkage limit. Liquid limit (L.L): is defined as the moisture content in percent at which the soil changes from liquid to plastic state. 19

20 Plastic Limit (P.L.): The moisture contents in % at which the soil changes from plastic to semi solid state. Shrinkage Limit (S.L.): The moisture contents in % at which the soil changes from semi solid to solid state. Plasticity Index (P.I.): it is the range in moisture content when the soil exhibited its plastic behavior:.. =... Liquidity Index (L.I. or IL) : a relation between the natural moisture contents ( ) and (L.L.) and (P.L.) in form: If LI > 1 Then the soil at Liquid state If LI = 1 then the soil at L.L. If < 1 then the soil below L.L. 20

21 Activity: is the degree of plasticity of the clay size fraction of the soil and is expressed as: =. % كلما زادت الفعالية كلما دلت على لدونة التربة عالية Plasticity Chart: based on Atterberg limits, the plasticity chart was developed by Casagrande to classify the fine grained soil. Some useful notes: Constant at all stages Degree of saturation (S %) at S.L. and up to =100% 21

22 Degree of Saturation in the region from S.L. and below < 100% = =.. =.. Relative Density: is the ration of the actual density to the maximum possible density of the soil it is expressed in terms of void ratio. (%) = 100 Or (%) = 100 : The void ratio of the soil in its loosest condition : The void ratio of the soil in its densest condition : The void ratio of the soil in its natural condition : Maximum dry unit weight (at ) : Minimum dry unit weight (at ) : Natural dry unit weight (at ) RD Description 0 - loose medium Dense 22

23 Example 1: for a granular soil, given, = 17.3, relative density = 82%, = 8% and = If = what would be? what would be the dry unit weight in the loosest state? Solution: = =. 10 = 0.53 = =.. = 0.94 ( ) = 1 + = = Example 2: a granular soil is compacted to moist unit weight of at moisture content of 18%. What is relative density of the compacted soil? Given, = 0.85, = 0.42 = 2.65? Solution: = ( ) =. (. ) 10 = 0.52 = = = = 76.74% Example 3: A dry sample of soil having the following properties, L.L. = 52%, P.L. = 30%, = 2.7, e= Find: Shrinkage limit, density, dry unit weight, and air content at dry state. Solution 23

24 Dry sample = = =.. 1*0.53 =2.7 *S.L S.L.= 19.6% = =.. 1 = = = = Case is dry s=0 = = = 34.6% 1 + = = Example 4: A saturated soil sample has a volume of 20 at its L.L Given L.L= 42%, P.L.= 30%, S.L.= 17%, = Find the min. volume the soil can attain. The minimum volume occurs at S.L. or at dry state. = + : is constant along all state. At L.L.. =. 24

25 1*e= 2.74 * = = = 20 = = 9.3. =..... = =.. =.. = = =10.7 1*.. = = = 4.33 = Example 5: A sample of saturated clay had a volume of 97 and a mass of (0.202 kg). When completely dried at the volume of the sample was (87 ) and it's mass (0.167 kg). Find a) - initial water content. b)- shrinkage limit c)- specific gravity 25

26 Solution: ρ = = = 2.08 / 2.08 = (1) = = = 21% 167. =. 1*e = 0.21 (2) Solving (1) and (2) = = 2.69 But =.. = = = W= kg = W= kg At dry state : = = 1.96 / 1.92 = = = 15% 26

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28 Chapter three Soil Compaction Soil compaction is one of the most critical components in the construction of roads, airfield, embankments and foundations. The durability and stability of a structure are related to the achievement of proper soil compaction. Structural failure of roads, airfield and the damage caused by foundation settlement can often be traced back to the failure to achieve proper soil compaction. Compaction of soil: Compaction is the process of increasing the density of a soil by packing the particles closer together with a reduction in the volume of air only. Compaction increases the dry density and decreases the void ratio. Purpose of compaction: 1- Increase shear strength of soil 2- Reduce void ratio thus reduce permeability 3- Controlling the swell-shrinkage movement 4- Reduce settlement under working load 5- Prevent the buildup of large water pressure Factors affecting compaction: Water content Type of soil Compaction energy or effort All these factors are shown in the following figures: 28

29 The effect of types of soil on the dry density using the same compaction Energy. 29

30 Theory of compaction: Different in compaction energy and types of soil Compaction is the process of reducing the air content by the application of energy to the moist soil. From compaction test we can find: 1- There is a unique relationship between the water content and the dry density for specific compaction energy. 2- There is one water content (O.M.C.) (Optimum moisture content) at which the max dry density is achieved The two above points can be clearly shown through the following Figure: 30

31 Compaction Test Compaction curve The compaction test is performed to determine the relationship between the moisture content and the dry density of a soil for specific compactive effort. The compactive effort is the amount of mechanical energy that is applied to the soil mass. Several different methods are used to compact soil in field, and some examples include tamping, kneading, vibration, and static load compaction. This test will be carried out by using impact compaction method using the type of equipment and methodology developed by R.R.Proctor in 1933, therefore, the test is also known as the proctor test. Two types of compaction tests are routinely performed: (1) The standard Proctor and (2) The modified Proctor test. Type test of Standard Proctor Modified Proctor No. of No. of Volume Weight of Height of layer blows per of mold hammer drops layer ( ) (kg) cm Compaction Effort =.. Test Procedure : 1- a sufficient quantity of air-dried soil in large mixing pan (say 3 kg) 2- Determine the weight of the compaction mold with its base (without the collar). 3- Start with initial water such (3% of Soil weight) 4- Add the water to the soil and mix it thoroughly into the soil until the soil gets uniform color (see figure B and C). 5- Assemble the compaction mold to the base, place soil in the mold and compact the soil in the number of equal layers specified by the type of compaction method (see photo D and E). The number of drops per layer is dependent upon the type of compaction. The drops should be applied at a uniform rate not exceeding around

32 seconds per drops, and the rammer should provide uniform coverage of the specimen surface. 6- The soil should completely fill the cylinder and the last compacted layer must extend slightly above the collar joint. If the soil below the collar joint at the completion of the drops, the test point must be repeated. 7- Carefully remove the collar and trim off the compacted soil so that it is completely even with the top of the mold.(see photo F). 8- Weigh the compacted soil while it's in the mold and to the base, and record the mass (see Photo G). Determine the wet mass of the soil by subtracting the weight of the mold and base. 9- Remove the soil from the mold using a mechanical extruder (see Photo H) and take the soil moisture content samples from the top and bottom of the specimen (see Photo i). Determine the water content. 10- Place the soil specimen in the large tray and break up the soil until it appears visually as if it will pass through the #4 sieve, add 3% more water on the soil and remix as in step 4. Repeat step 5 through 9 until a peak value is reached followed by two slightly lesser compacted soil masses. Analysis: 1- Calculate the moisture content of each compacted soil specimen. 2- Compute the wet density in grams per cm 3 of the compacted soil by dividing the wet mass by the volume of the mold used. 3- Compute the dry density using the weight density and the water content determined in step 1. Use the following formula: = Plot the dry density values on the y-axis and the moisture contents on the x- axis. Draw a smooth curve connecting the plotted points. 5- On the same graph draw a curve of Saturation line ( Zero air void line ) using the following Equation : =. =. For S= 1 = Assume values of water content and find dry density, then plot the zero air void line which must be parallel to the moist side of compaction curve and never intersect it, If so that mean there is some error. To plot Air content line (A%) use the following equation: 32

33 = (1 ) 1 + The following Figures give the steps used in the test: 33

34 Compaction Equipments: 1- Sheep's foot roller (for cohesive soil) 2- Pneumatic roller (many different soils) 3- Vibratory rollers (mainly for granular material) 4- Grid rollers 5- Power Rammer 6- Vibratory plates. Compaction Of cohesion less soil: Moisture content has little or no influence on the granular soils (except when the soil is fully saturated). 34

35 Soil Mechanics Lectures Third year Student Their state of compaction can be obtained by relating dry density to the minimum and maximum dry densities and as in the following equation: = = Where RD : Relative density Compaction in the field: The results of laboratory tests are not directly applicable to the field compaction, because 1- The Laboratory tests are carried out on material smaller than 20 mm size. 2- Compactive efforts are different and apply in different method. Relative compaction: Or Degree of compaction is a means of comparing the field density with Laboratory results and is defined as the ratio of the dry density in the field to the maximum dry density in the Laboratory and in most construction works, the degree of compaction is specified as 95 % or more.. = %. So by using sand replacement method, find dry density at field then check the R.C The Optimum moisture content can be useful in field as follows : If < then add water and compact the soil 35

36 Example 1 : If = then compact the soil directly If > then either postponed the compaction to other time or add some additive (such as cement or lime) to accelerate evaporation of extra water. Measurement of field Density 1- Core cutter 2- Sand Replacement method 3- Air-Ball on method 4- Penetrating Needle 5- Radiation Technique. The following results were obtained from a standard compaction test. Determine the Optimum moisture content and maximum dry density. Plot the curves of 0%, 5% and 10% air content and gives the value of air content at the maximum dry density. Given the volume of standard mold is 1000 cm 3 and =.. Mass (gm) Water content (%) Solution : Calculate dry density for each test and tabulate the results. (%) (%) A %

37 Comaction Curve Comaction Curve From Figure: The =., and the Optimum Moisture content = 10% Comaction Curve Zero air void line % 5A= % 10A = Figure show the Zero air void line and a line of 5 and 10% air content 37

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39 Chapter Four Soil Classification Classification of soil is the separation of soil into classes or groups each having similar characteristics and potentially similar behaviour. A classification for engineering purposes should be based mainly on mechanical properties: permeability, stiffness, strength. The class to which a soil belongs can be used in its description. The aim of a classification system is to establish a set of conditions which will allow useful comparisons to be made between different soils. The system must be simple. The relevant criteria for classifying soils are the size distribution of particles and the plasticity of the soil. Particle Size Distribution for measuring the distribution of particle sizes in a soil sample, it is necessary to conduct different particle-size tests. Wet sieving is carried out for separating fine grains from coarse grains by washing the soil specimen on a 75 micron sieve mesh. 1- Dry sieve analysis is carried out on particles coarser than 75 micron. Samples (with fines removed) are dried and shaken through a set of sieves of descending size. The weight retained in each sieve is measured. The cumulative percentage quantities finer than the sieve sizes (passing each given sieve size) are then determined. The resulting data is presented as a distribution curve with grain size along x-axis (log scale) and percentage passing along y-axis (arithmetic scale). U.S. Standard sieve sizes A set of sieves for a test in the laboratory 39

40 Hydrometer (Sedimentation) analysis is based on the principle of sedimentation of soil grains in water. When a soil specimen is dispersed in water, the particles settle at different velocities, depending on their shape, size, and weight. For simplicity, it is assumed that all the soil particles are spheres, and the velocity of soil particles can be expressed by Stokes law, according to which: 40

41 In this method, the soil is placed as a suspension in a jar filled with distilled water to which a deflocculating agent is added. The soil particles are then allowed to settle down. The concentration of particles remaining in the suspension at a particular level can be determined by using a hydrometer. Specific gravity readings of the solution at that same level at different time intervals provide information about the size of particles that have settled down and the mass of soil remaining in solution. The results are then plotted between % finer (passing) and log size. Grain-Size Distribution Curve The size The results are then plotted between % finer (passing) and log size. Grain-Size Distribution Curve the size distribution curves, as obtained from coarse and fine grained portions, can be combined to form one complete grain-size distribution curve (also known as grading curve). A typical grading curve is shown. Grain-size distribution curve From the complete grain-size distribution curve, useful information can be obtained such as: 1. Grading characteristics, which indicate the uniformity and range in grain-size distribution. 2. Percentages (or fractions) of gravel, sand, silt and claysize. Grading Characteristics A grading curve is a useful aid to soil description. The geometric properties of a grading curve are called grading characteristics. 41

42 To obtain the grading characteristics, three points are located first on the grading curve. D60 = size at 60% finer by weight D30 = size at 30% finer by weight D10 = size at 10% finer by weight The grading characteristics are then determined as follows: 1. Effective size = D10 2. Uniformity coefficient, Cu, = 3. Curvature coefficient, Cc, = ( ) If Cu > 4 for gravel and Cu > 6 for sand and Cc between 1 and 3 indicates a well-graded soil (GW for gravel and SW for sand ). i.e. a soil which has a distribution of particles over a wide size range The consistency of a fine-grained soil refers to its firmness, and it varies with the water content of the soil. A gradual increase in water content causes the soil to change from solid to semisolid to plastic to liquid states. The water contents at which the consistency changes from one state to the other are called consistency limits (or Atterberg limits). The three limits are known as the shrinkage limit (WS), plastic limit (WP), and liquid limit (WL) as shown. The values of these limits can be obtained from laboratory tests. (as explained in chapter 3) 42

43 Classification Based on Grain Size The range of particle sizes encountered in soils is very large: from boulders with dimension of over 300 mm down to clay particles that are less than mm. Some clay contains particles less than mm in size which behave as colloids, i.e. do not settle in water. According to grain Size analysis: Gravel, sand, silt, and clay are represented by group symbols G, S, M, and C respectively. Physical weathering produces very coarse and coarse soils. Chemical weathering produces generally fine soils. Coarse-grained soils are those for which more than 50% of the soil material by weight has particle sizes greater than mm. They are basically divided into either gravels (G) or sands (S). According to gradation, they are further grouped as well-graded (W) or poorly graded (P). If fine soils are present, they are grouped as containing silt fines (M) or as containing clay fines (C). For example, the combined symbol SW refers to well-graded sand with no fines. Both the position 43

44 and the shape of the grading curve for a soil can aid in establishing its identity and description. Some typical grading curves are shown. Curve A- poorly-graded SAND Curve B - a well-graded GRAVEL-SAND (i.e. having equal amounts of gravel and sand) Fine-grained soils are those for which more than 50% of the material has particle sizes less than mm. Clay particles have a flaky shape to which water adheres, thus imparting the property of plasticity. A plasticity chart, based on the values of liquid limit (WL) and plasticity index (IP), The 'A' line in this chart is expressed as IP = 0.73 (WL - 20). 44

45 Depending on the point in the chart, fine soils are divided into clays (C), silts (M), or organic soils (O). The organic content is expressed as a percentage of the mass of organic matter in a given mass of soil to the mass of the dry soil solids. Soil classification using group symbols is as follows: 45

46 Activity "Clayey soils" necessarily do not consist of 100% clay size particles. The proportion of clay mineral flakes (< mm size) in a fine soil increases its tendency to swell and shrink with changes in water content. This is called the activity of the clayey soil, and it represents the degree of plasticity related to the clay content. = % ( h ) Where PI is plasticity index = Liquidity Index In fine soils, especially with clay size content, the existing state is dependent on the current water content (w) with respect to the consistency limits (or Atterberg limits). The liquidity index (LI) provides a quantitative measure of the present state. Example 1: = 46

47 Following are the results of a sieve analysis. Make the necessary calculations and draw a particle size distribution curve. U.S.sieve size Mass of soil retained on each sieve (g) Pan 12 Solution: Solution: 47

48 Find, = = = 1.8 = ( ) = (. ).. = 0.71 % passing # 200 less than 50% so the soil is coarse, and since % passing # 4= 100 so the soil is sand and since Cu less than 6, so the soil is SP. Home work :1- Following are the results of a sieve analysis: U.S. Sieve No. Mass of soil retained on each sieve (g) pan 31.2 I- Plote the grain size distribution curve. II- Calculate the uniformity coefficient,, and cofficient of gradation, A)-For a soil given: D10 = 0.1 mm D30 = 0.41 mm D60 = 0.62 mm B)-for a soil given: D10 = mm D30 = 0.29 mm D mm 48

49 Home work 1: 2- Classify the following soil according to USCS U.S. Sieve No. Mass of soil retained on the sieve in g Pan 23.5 Pl: 25, L.L. = 40 49

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51 SOIL PERMEABILITY Chapter Five Soil Permeability and Flow A material is permeable if it contains continuous voids. All materials such as rocks, concrete, soils etc. are permeable. The flow of water through all of them obeys approximately the same laws. Hence, the difference between the flow of water through rock or concrete is one of degree. The permeability of soils has a decisive effect on the stability of foundations, seepage loss through embankments of reservoirs, drainage of sub grades, excavation of open cuts in water bearing sand, rate of flow of water into wells and many others. Hydraulic Gradient When water flows through a saturated soil mass there is certain resistance for the flow because of the presence of solid matter. However, the laws of fluid mechanics which are applicable for the flow of fluids through pipes are also applicable to flow of water through soils. As per Bernoulli's equation, the total head at any point in water under steady flow condition may be expressed as Total head = pressure head + velocity head + elevation head Hydraulic Gradient When water flows through a saturated soil mass there is certain resistance for the flow because of the presence of solid matter. The laws of fluid mechanics which are applicable for the flow of fluid through pipes are also applicable to flow of water through soils. The total head at any point in water under steady flow condition may be expressed as: Total head = pressure head + velocity head + elevation head 51

52 The flow of water through a sample of soil of length L and crosssectional area A as shown in figure1: = = Figure (1) flow of water through a soil sample For all practical purposes the velocity head is a small quantity and may be neglected. The water flows from the higher total head to lower total head. So the water will flow from point B to C. = ( + ) - ( + ) Where, and = h, and = Pressure Head. The loss of head per unit length of flow may be expresses as : Where i is the hydraulic gradient. = h 52

53 Hydraulic gradient: The potential drop between two adjacent equipotentials divided by the distance between them is known as the hydraulic gradient. DARCY'S LAW Darcy in 1856 derived an empirical formula for the behavior of flow through saturated soils. He found that the quantity of water q per sec flowing through a cross-sectional area of soil under hydraulic gradient / can be expressed by the formula q = kia or the velocity of flow can be written as = Where k is termed the hydraulic conductivity (or coefficient of permeability) with units of velocity. The coefficient of permeability is inversely proportional to the viscosity of water which decreases with increasing temperature; therefore, permeability measurement at laboratory temperatures should be corrected to the values at standard temperature of 20 0 C using the following equation. Where : Coefficient of permeability at 20 0 C : Cofficient of permeability at Lab. Temperture 0 C Viscosity of water at lab. Temperature Viscosity of water at 20 0 C 53

54 Table (1) :The of at different temperature. DISCHARGE AND SEEPAGE VELOCITIES: Figure below shows a soil sample of length L and cross-sectional area A. The sample is placed in a cylindrical horizontal tube between screens. The tube is connected to two reservoirs R1 and R2 in which the water levels are maintained constant. The difference in head between R1 and R2 is h. This difference in head is responsible for the flow of water. Since Darcy's law assumes no change in the volume of voids and the soil is saturated, the quantity of flow past sections AA, BB and CC should remain the same for steady flow conditions. We may express the equation of continuity as follows qaa = qbb = qcc If the soil be represented as divided into solid matter and void space, then the area available for the passage of water is only Av. If vs. is the velocity of flow in the voids, and v, the average velocity across the section then, we have 54

55 Where is the area of the void, is the seepage velocity, is the approach velocity A: is the cross sectional area of the sample = = = Where n : is the porosity of the soil METHODS OF DETERMINATION OF HYDRAULIC CONDUCTIVITY OF SOILS (Coefficient of permeability). Coefficient of permeability Laboratory methods: 1- Constant head permeability method 2- Falling head permeability method 3- Indirect determination from consolidation test Field methods: 1- Pumping test 55

56 2- Bore hole tests Laboratory methods: 1- Constant head permeability test: The coefficient of permeability for coarse soils can be determined by means of the constant-head permeability test (Figure below): A steady vertical flow of water, under a constant total head, is maintained through the soil and the volume of water flowing per unit time (q): A series of tests should be run, each at different rate of flow. Prior to running the test a vacuum is applied to the specimen to ensure that the degree of saturation under flow will be close to 100%. 2- Falling head permeability test: For fine soils the falling-head test (Figure below) should be used. In the case of fine soils, undisturbed specimens are normally tested. The length of the specimen is l and the cross-sectional area A. the standpipe is filled with water and a measurement is made of the time ( ) for water level (relative to the water level in the reservoir) to fall from h h. At any intermediate time t the water level in the standpipe is given by h and its rate of change by. At time t the difference in total head between the top and bottom of the specimen is h. then applying Darcy's law: h = h h h = 56

57 = h h = 2.3 log h h Ensure that the degree of saturation remains close to 100%. A series of tests should be run using different values of h h Figure: Laboratory Test (a) Constant Head (b) Falling Head Example 1: A constant head permeability test was carried out on a cylindrical of sand 4 in. in diameter and 6 in. in height. 10 Of water was collected in 1.75 min, under a head of 12 in. Compute the hydraulic conductivity in ft/year and the velocity of flow in ft/sec. Solution: = = 10, = =

58 = h = 12 6 = 2, = 105 Therefore = = = / Velocity of flow = = = / Example 2 A sand sample of 35 cross sectional area and 20 cm long was tested in a constant head permeameter. Under a head of 60 cm, the discharge was 120 ml in 6 min. The dry weight of sand used for the test was 1120 g. and = Determine (a) the hydraulic conductivity in cm/sec. (b) the discharge velocity, and (c) the seepage velocity. Solution: = QL hat Q = 120 ml, t =6 min, = 35, = 20, h = 60 = = / Discharge velocity, v= ki = = / Seepage velocity = = = 1.6 = 1 + = 1 = = = 1 + =

59 = =.. = DIRECT DETERMINATION OF K OF SOILS IN FIELD: 1- Field test in unconfined aquifer Pumping Test in an unconfined aquifer 2- Field test in Confined aquifer There are two cases : Case 1- when h > Case 2-when h < 59

60 Empirical Relations for Hydraulic Conductivity Several empirical equations for estimating hydraulic conductivity have been proposed over the years. Granular Soil For fairly uniform sand (that is, a small uniformity coefficient), Hazen (1930) proposed an empirical relationship for hydraulic conductivity in the form: C= a constant that varies from 1.0 to 1.5 D10 = the effective size (mm) ( ) = The above equation is based primarily on Hazen s observations of loose, clean, filter sands. A small quantity of silts and clays, when present in a sandy soil, may change the hydraulic conductivity substantially. The accuracy of the values of k determined in the laboratory depends on the folloeing factors: 1- Temperture of the fluid 2- Viscosity of fluid 60

61 3- Trapped air bubbles present in the specimen 4- Degree of saturation 5- Migration of fines during testing 6- Duplication of field conditions in the laboratory. The coefficient of consolidation of saturated cohesive soils can be determined by laboratory consolidation tests. This will be listed in details in "consolidation of soil". Table -1 Typical Values of Hydraulic Conductivity of Saturated Soils Exercise:

62 HEADS AND ONE-DIMENSIONAL FLOW There are three heads which must be considered in problem involving fluid flow in soil (Figure 4): 1- Pressure head (h ) : is the pizometer reading = pore water pressure /unit weight of water 2- Elevation head at any point (h ): is the vertical distance above or below some reference elevation or datum plane. 3- Total head, h = h + h Figure (4) illustration of types of head(after Taylor, 1948). 62

63 Example 1: For the Setup shown (Figure 5a), plot, ht, he, hp and the velocity of flow? Datum 5 6 Figure(5a) Points ht(ft) (Figure 5b) he(ft) (Figure 5b) hp=ht-he(ft) (Figure 5b) V(ft/min) * =Ki (Figure5c) = = Figure(5b) Figure(5c) * = = = 2 Approch velocity = ki= 1*2= 2 ft/min Seepage velocity = =. = 6 / 63

64 Example 2 For the setup shown(figure 6a), Draw, ht, he, hp and velocity of flow? Datum 2 Figure ( 6) a b c 1- Direction of flow is upward flow (look to the water'symbole usaually water flow from higher one to lower one) 2- List all point with direction of flow 3- Construct a table to solve the problem Points ht(ft) (Figure 6b) he(ft) (Figure 6b) = * = = = hp=hthe(ft) (Figure 6b) V(ft/min) * =Ki (Figure6c) 64

65 Approch velocity = ki= 1*0.667= ft/min Seepage velocity = =.. = 2 / Example 3: For the setup shown(figure below), Draw, ht, he, hp and velocity of flow? Datum 1 2 Figure (7a) 3 Ft/min) 4 5 Figure heads vs horizontal distance Solution: Figure (7c) 1- Assume any arbitrary line representing the datum and let it at elevation =0(Figure 7a). 2- The flow will be in horizontal direction (elevation head is constant) 3- Construct the table Since pressure =3.4 psi=3.4*144= lb/ft 2 h = =. = 7.84 ft. Points ht(ft) he(ft) hp(ft)

66 Approch velocity = ki= 1*1.3= 1.3 ft/min = h h = = Seepage velocity = =.. = 3.9 / Example 4 For the setup shown in figure 8:a) - Calculate the pressure head, elevation head, total head and head loss at points B, C,D and F in centimeter of water. b)-plot the heads versus the elevation. Figure shows the Set up of example 4 Solution (1): points ht(cm) he(cm) hp(cm) Head loss B

67 C D F Figure (8)Example 4 Example (5): for the setup shown Calculate and plot total head, elevation head and pressure head. 67

68 Example 5 Solution example 5: Points Ht(cm) He(cm) Hp(cm) Head loss A B C D

69 Solution Example 5 Example -6 : For the set up shown, draw the variation of total head, pressure head and elevation head along points A,B,C,D and E. 69

70 Example 6 Example 6 Solution of Solution of Example 6 point Total Elevation Pressure Head Loss 70

71 Head(cm) Head(cm) Head(cm) (cm) A B C D E Example 7: For the setup shown, Find total head (ht), Elevation head (he) and Pressure head(hp) for the soil the setup shown. 5 m Datum 5 m 5 m = 10 A B C D E Example 7 setup 4 m 4 m 6 m 3 m 71

72 72

73 Example 8:For Setup shown: Soil I, A= 0.37 m 2, n=0.5 and k= 1 cm/sec, Soil II, A= m 2, n=0.5 and k=0.5 cm/sec m 2 Soil I 3 Soil II m 1.2 m 0.6 m 5 0 Datum Points ht(m) he(m) hp(m) Solution : 1- qi=qii = / * (.. ) = 0.37 =. / 0.1 (1) (.. ) 2- h + h = 3.6 (2) From Equation( 1 ) h = h Substitute in Equation h + h = 3.6 Approach Velocity= ki h = 2.4 h = 1.2 Approach velocity for soil I= = 1. (.. ) = 1 / 73

74 Seepage velocity = =. = 2 / For soil II Approach velocity= = 0.5. (.. ) = 2 / Seepage velocity (II) = =. = 6 / Velocity (cm/sec) 74

75 Example 9: For the setup shown draw ht, he, hp and find the seepage force. A 0.6 m B n=0.33, k=0.5 cm/sec Soil = 20.9 C 0.3 m 0.6 m Datum Supporting Screen Poi nts ht(c m) he(c m) hp(c m) A B C (ht (cm (he(cm (hp(cm E Elevation ( /) (cm) ( ) ( ) *10=3kN/m *10= *20.9 = *10=

76 3 3 = + 6 kn/m Water pressure on soil sample (a) Boundary water pressure (b) Buoyancy water pressure(static) (c) Pressure lost in seepage. 76

77 Water Force on Soil: = = h = Seepage forces usually act with direction of flow. Quick Condition: The shear strength of cohesionless soil is directly proportional to the effective stress. When a cohesionless soil is subjected to a water condition that results in zero effective stress, the strength of the soil becomes zero and quick condition exists. Quick condition: occurs in upward flow( for cohesionless soil) and when the total stress equals to pore water pressure. = 0 = h = 0 h = = = : The gradient required to cause a quick condition, termed critical gradient. Example 10: Excavation is been carried out as shown in the figure. Find: 1- the depth Z that could caused boiling at the bottom of clay layer. 2-The depth ( Z) for the factor of safety against boiling equal to 2 at the bottom of the clay layer. 3- What is the thickness of the raft foundation that should be used before boiling occurs. If an uplift pressure of 60 kn/m 2 at the bottom of excavation exist( = 25 / ). 4- Find the seepage force at an element of 0.2 m cube located at the center of silt layer. 77

78 Solution: h = = = =. h = 14 h = h h = = 4 To find Z F. down = F. upward A(8-z) *20 = 4.0 *10 *A z= 6 m 2-. = 2 = 3- F down ward = F upward (8 ) = 4 78

79 t*5*10*25 = 60*5*10 = = 2.4 m (thickness of concrete) 4- Seepage force = i = *10* (0.2)3 = 0.06 kn Summary of Main Points: 1- In soils = 2- There are three heads of importance to flow through porous media: elevation head ( h ), pressure head (h )and total head (h ). 3- Flow depends on difference in total head. 4- The seepage force per a volume of soil is i* and acts in the direction of flow. 5- "Quick", refers to a condition where in a cohesion less soil loses its strength because the upward flow of water makes the effective stress become zero. Exercise 1: For the setup shown. Plot to scale elevation head, pressure head, total head and seepage velocity versus distance along the sample axis. Exercise 2: For the setup shown, compute the vertical force exerted by the soil on screen A and that on screen B. Neglect friction between the soil and tube. G=

80 Exercise 3: For the set up shown: Steady vertical seepage is occurring. Make scaled plot of elevation versus pressure head, pore pressure, seepage velocity, and vertical effective stress. Determine the seepage force on a 1 ft cube whose center is at elevation -15 ft. G for all soils = Exercise 4: An 8 m thick layer of silty clay is overlaying a gravel stratum containing water under artesian pressure. A stand pipe was inserted into the gravel and water rose up the pipe to reach a level 2m above the top of the clay. The clay has a particle specific gravity of 2.7 and a natural moisture content of 30 percent. The permeability of the silty clay is 3*10-8 m/sec. It proposed to excavate 2 m into the soil in order to insert a wide foundation which, when constructed, will exert a uniform pressure of 100 kn/m 2 on to its supporting soil. Determine (a) The unit rate of flow of water through the silty clay in m 3 per year before the work 80

81 commences. (b) the factor of safety against heaving: i) at the end of excavation ii) after construction of the foundation? 81

82 Two Dimensional fluids Flow Most problems of flow are two dimensional flows, e.g. are shown in Fig. below: Impervious layer Impervious layer Impervious layer Impervious layer The purpose of studying the flow in two Dimension are : 1- To find the amount of seepage per meter length (i.e. rate of flow). 2- Pressure distribution (pore water pressure) 3- Stability against piping or boiling. 4- Pizometer levels of selected point required. 82

83 Seepage Theory: The general case of seepage in two dimensions will now be considered. In same principle used in one dimensional problem applied (Darcy's law & Continuity flow state). Consider the two dimensional steady state flows in the fig. Concrete Dam +. Element A +. Impervious X Take element A with dimension dx, dy and dz Rate of flow ( ) the flow entering the element ( + ) + ( + ) Since the flow is steady so = dy dz + dx dz =( + By simplification ) + ( + ) + = (1) Darcy's law = = = h 83

84 = h = h Sub. In equation (1) +( ) = 0 h + h = 0 Consider a function (, ) so that = = = = = = (, ) = h(, ) + Where c is a constant Thus if the function (, ) is given a constant value equal to & it will represent a curve a long which the value of total head (h ) is constant. If the function (, ) is given a series of constant value,, etc a family of curves, such curves are called equipotentials and this will corresponding to total head h, h, h h from the total differ nation. 84

85 =. + 0 = dx + dx = The second function ψ(x,y) called the flow line = ψ = ψ = ψ = ψ + = 0 ψ ψ = 0 ψ, Satisfy the Laplace equation A gain a series of ψ using ψ, ψ, ψ ψ Is selected and this function ψ = dψ + dψ 0 = + = 85

86 = Flow net: The graphical representation of the Laplace equation is represented by the two families of curve: 1- Equipotential lines: A series of lines of equal total head e.g. h, h, h h 2- Flow lines: A family of the rate of flow between any two adjacent flow lines is constant. For isotropic soil: The flow net is formed by a mesh of the intersection of two lines with the following limitation 1- Each element is a curvilinear square ψ ψ ψ ψ 9 Flow channel 1 Summary of the main points: 1- Laplace equation govern's the steady state flow in two dimensions 2- The solution is represented by two families of curve a- Equipotential lines 86

87 b- Flow lines 3- The intersection of the two lines is represented by a flow net of square elements 4- Each element is a curvilinear square with dimension 1 5- The rate of flow is expressed /m length by = h Q: rate of flow/m Where k: Coefficient of permeability lines H: Total head difference between the first and last equipotential :. h :. ( ) Steps in drawing a flow net:- The first step is to draw in one flow line, upon the accuracy of which the final correctness of the flow net depends. There are various boundary conditions that help to position the first flow line, including: 1- Buried surface (e.g. the base of the dam, sheet pile) which are flow lines as water cannot penetrate into such surface. 2- The junction between a permeable and impermeable material which is also a flow line : for flow net purpose a soil that has a permeability of one-tenth or less the permeability of the other may regard as impermeable. 3- The horizontal ground surface on each side of the dam which are equipotential lines. The procedure is as follows: a- Draw the first flow line hence establish the first flow channel b- Divide the first flow line into squares ( ) 87

88 c- Project the equipotentials beyond the first flow channel, which give an indication of the size of the square in the next flow channel. d- With compasses determine the position of the next flow line; draw this line as a smooth curve and complete the squares in the flow channel formed. e- Project the equipotentials and repeat the procedure until the flow net is completed. Example: Figure example for flow net construction Rate of flow = = 1 h h 88

89 h = = *k * ( ) = this is for one channel Assume No. of channel = Nf = = Where H= difference in water level (upstream and downstream. Note: Example 1: 89

90 Example 2 90

91 Example 3 91

92 Example 4 92

93 Example 6 93

94 Example 7 94

95 95

96 96

97 Example 8: A flow net for around a single row of sheet piles in a permeable soil layer is shown in figure below. Given that k= 5* 10-3 cm /sec. a) How high (above the ground surface )will the water rise if pizometers are placed at points a,b, c and d? b) What is the rate of seepage under sheet pile? Solution: From flow net = 3, = 6 h = = h = = h = = h = h = = So pizometer reading at point a = = m above surface So pizometer reading at point b = = 3.89 m above surface So pizometer reading at point c = =2.25 m above surface 97

98 q =kh Example 9 = / (5 1.67) =. / A deposit of cohesion less soil with a permeability of 3*10-2 cm/sec has a depth of 10 m with an impervious ledge below. A sheet pile wall is driven into deposit to a depth of 7.5 m. The wall extends above the surface of the soil and 2.5 m depth of water acts on one side. Determine the seepage quantity per meter length of the wall. Example 10- For the flow net shown below includes sheet-pile cutoff wall located at the head water side of the dam in order to reduce the seepage loss. The dam is half kilometer in width and the permeability of the silty sand stratum is 3.5 *10-4 cm /sec. Find (a) the total seepage loss under the dam in liters per year, and (b) would the dam be more stable if the cutoff wall was placed under its tail-water side? 98

99 Solution: a) Notice that h = 6.0, the number of flow channels = 3 = 10 by using = h q = (3.5*10-4 ) (6.0 ) = m 3 /sec/m Since the dam is 500 meters wide, the total Q under the dam is Q= Lq= 500m (6.3*10-4 m 3 /sec)( ) = 100 b) - No: Placing the cutoff wall at the toe would allow higher uplift hydrostatic pressure to develop beneath the dam. Home work: 1. Two lines of sheet piles were driven in a river bed as shown in figure. The depth of water over the river bed is 8.20 ft. The trench level within the sheet piles is 6.6 ft. below the river bed. The water level within the sheet piles is kept at trench level by resorting to pumping. If a quantity of water flowing into the trench from outside is 3.23 ft 3 /hour per foot length of sheet pile, what is the hydraulic conductivity of the sand? What is the hydraulic gradient immediately below the trench bed? (Ans 1 x 10-4 ft/sec, 0.50). 99

100 100

101 101

102 Stresses within the soil: Stresses within the soil Types of stresses: 1- Geostatic stress: Sub Surface Stresses cause by mass of soil a- Vertical stress = h b- Horizontal Stress = Note : Geostatic stresses increased lineraly with depth. 2- Stresses due to surface loading : a- Infintly loaded area (filling) b- Point load(concentrated load) c- Circular loaded area. d- Rectangular loaded area. Introduction: At a point within a soil mass, stresses will be developed as a result of the soil lying above the point (Geostatic stress) and by any structure or other loading imposed into that soil mass. stresses due Geostatic soil mass 1- = h (Geostatic stress) =, where : is the coefficient of earth pressure at rest. 102

103 Soil Mechanics Lectures Third year Student EFFECTIVESTRESS CONCEPT: In saturated soils, the normal stress (σ) at any point within the soil mass is shared by the soil grains and the water held within the pores. The component of the normal stress acting on the soil grains, is called effective stressor intergranular stress, and is generally denoted by σ'. The remainder, the normal stress acting on the pore water, is knows as pore water pressure or neutral stress, and is denoted by u. Thus, the total stress at any point within the soil mass can be written as: = + u This applies to normal stresses in all directions at any point within the soil mass. In a dry soil, there is no pore water pressure and the total stress is the same as effective stress. Water cannot carry any shear stress, and therefore the shear stress in a soil element is carried by the soil grains only. = = 1- W.T.L very deep. h 2- When W.T.L. at the ground surface:. h u= h* = = = = 103

104 2- Stresses due to filling: = h 3-Stresses due to external loading: a- Stresses due to point load : Boussinesq (1885)(French mathematician) developed the following equation to calculate vertical stress increment, in soils due to point load on the surface: Type of point load: the load under the wheel of car. Then, = Where factor, = Example 1: In road pavement design, the standard vehicle axel is defined as an axel with two single wheels as shown below. For a particular vehicle group, the standard axel load (P) is given as 80 kn and the distance between two wheels (L) is 1.8 m. what is the vertical stress increment in the sub-grade at 4 m depth directly under a wheel if this axel is running (consider wheel load as a point load ): 104

105 Solution: =, / = ( ) Therefore, To find = + / ( ) = ( ) = ( ) / ( ) = ( ) = ( 0 4 ) / / = = = () + () = = b- Circular footing =, F: factor, can be find from the following figure, where R: is the radius of the circular area X: is the distance from the center of the circle to the point Z : is the depth of the point. 105

106 c- Rectangular loaded area. = (, ) F(n,m) : is the factor (use figure below) 106

107 Approximate method to find at Z using 2:1 method: 107

108 = ( + ) ( + ) Where : Is the net surface load. Example: Plot the variation of total and effective vertical stresses, and pore water pressure with depth for the soil profile shown below: Solution: Within a soil layer, the unit weight is constant, and therefore the stresses vary linearly. Therefore, it is adequate if we compute the values at the layer interfaces and water table location, and join them by straight lines. At the ground level, σv= 0 ; = 0; and u=0 At 4 m depth, σv= (4)(17.8) = 71.2 kpa; u = 0 = 71.2 kpa At 6 m depth, σv= (4)(17.8) + (2)(18.5) = kpa u = (2)(9.81) = 19.6 kpa = = 88.6 kpa At 10 m depth, σv= (4)(17.8) + (2)(18.5) + (4)(19.5) = kpa u = (6)(9.81) = 58.9 kpa = = kpa At 15 m depth, σv= (4)(17.8) + (2)(18.5) + (4)(19.5) + (5)(19.0) = kpa u = (11)(9.81) = kpa 108

109 = = kpa The values of σv, u and Computed above are summarized in Table 6.1. Variation of, with depth Negative pore pressure (suction): Below the water table, pore pressure are positive. In dry soil, the pore pressure are positive. In dry soil, the pore pressure is zero. Above the water table, when the soil is saturated, pore pressure will be negative. u= h The height above the water table to which the soil is saturated is called the capillary rise, and this depends on the grain size and type(and thus the size of pores): -In coarse soils capillary rise is very small -In silts it may be up to 2m _In clays it can be over 20m 109

110 Soil Mechanics Lectures Third year Student Example : For the soil profile shown find the total, effective and pore water pressure:? Total Stress Pore water pressure Effective stress 4 5 Given: = 16, = 20 6 Solution: At depth z=0, = 0, = 0, = 0 At depth z= 2m = 2 16 = 32, = 0, At depth Z= 5 m = = 92 = 32, = 3 10 = 30, =

111 Example: For the soil profile shown: Determine, total, effective and pore water pressure for the following conditions:1- water table 3 m above the ground level, 2- water table at the ground level m below the surface,4-2 m below the surface? Stresses (kpa) Case /effective1 effective2case = 16, = 20 depth (m) case /effective Depth (m) WT Total stress(kpa) u(kpa) ) 0 Case *20=90 6*10= Case *20=60 3*10= Case *16= *2= Case *16= *20=52 1*10=

112 Example : A building 20 m * 20m results in a uniform surface contact pressure of 150 kpa. Determine the increase in vertical pressure at depth of 10 m below a) the center of the building b) the corner of the building. Estimate the additional pressure at both locations of a tower 5m *5m placed at the center of the building imposing 300 kpa uniform additional pressures. At the corner: nz= 10 n*10=20, n=2 mz=10 m*10=20, m=2 F(1,1)= h = = m * 20m Top Veiew Front Veiew At the center : N*10=10 n=m=1 From figure (rectangular ) f(1, 1)= at the center =4*0.176*150= kpa = at the center: n*10=2.5 n=m= 0.25 f(0.25,0.25)= m 5 m h = 4*0.0261*300=31.32 kpa at the corner due to tower for e1: 12.5 m n*10=12.5 n=m= 1.25 f(1.25,1.25)= For e2=e3 n*10=7.5 n=0.75 m*10=12.5 m=1.25 f(0.75,1.25)= for e3: n*10=7.5, 12.5 m 12.5 m 12.5 m +e1 7.5 m 12.5 m-e2 +e3 5 m n=m=0.75 f(0.75,0.75)=0.136 =(0.211-(0.165*2)-0.136)*300=5.1 kpa 112

113 Example: A distributed load of 50 kn/ is acting on the flexible rectangular area 6*3 m, as shown in figure. Determine the vertical stress at point A which is located at depth of 3 m below the ground surface. ( = 18.5 /. Solution: e1=e3, n*3= n=1.5, m*3= 1.5, m= 0.5 f(1.5,0.5)=0.131 e2=e4, n*3= 1.5, n=0.5, m*3=1.5,m=0.5 f(0.5,0.5)=0.085 = ) = 21.6 / Total stresses = Geostatic stress+ Total Stresses= 18*3+21.6=75.6 kpa Example : For the same example.if the foundation at 0.5 m below ground surface: q net= *18= 41 kn/ e1=e3, n*2.5=4.5, n=1.8, m*2.5= 1.5, m=0.6 f ( 1.8,0.6) =0.155 e2=e4, n*2.5=1.5, n= 0.6, m*2.5=1.5, m=0.6 f (0.6,0.6)= = ) = / Total stresses= 18* = kn/ 113

114 Principal Stresses and Mohr Circle Principal Stresses: There are exist at any stressed point in three orthogonal (i.e. naturally perpendicular) planes on which there are zero shear stresses. These planes are called the principal stress planes. The normal stresses that acts on these three planes are principal stresses. The largest of these three stresses is called the major principal stress, the smallest is called the minor principal stress, and the third is called the intermediate stress. When the stresses in the ground are geostatic, the horizontal plane through a point is a principal plane and so too are all vertical planes through the point. When ( = < 1, = and = =. when > 1, the situation is reversed. The shear stresses on any orthogonal planes (planes meeting at right angles) must be numerically equal ( =. Mohr circle: it is concerned only with the stresses existing in two dimensions, the state of stress in plane that contains the major and minor principal stresses,. The stresses will be considered positive when compressive. Equations for state of stress at a point. is positive when counter clock wise is measured counter clockwise from the direction of. Mohr diagram for state of stress at a point 114

115 = = ( 2 2 sin 2 cos 2 Where and are the stresses acts on any planes, the direction and magnitude of principal stresses can be found. Origin of planes: is a point on the Mohr circle, denoted by Op, with following property: 1- A line through Op and any points (A) of the Mohr circle will be parallel to the plane on which the stresses given by point A act. 2- If the plane and Op are known then the line parallel to the plane passes through Op and intersect Mohr circle, the intersect point represent the point which acts on that plane. To find the stresses, there are two method: 1- By drawing (prefer) 2- By using equations 1- By drawing To determine the normal and shear stresses on any plane, we have to do the following: 1-Draw the stress- state on Mohr circle (compression (+) and ( ). 2-Find the point denoted by Op origin of planes by either select and draws a line parallel to the plane on which is acting until it intersects Mohr Circle & the point of intersection is OP, or use (the same principal is apply to ). 3-From OP draw a line parallel to the plane you want to find stress on it. The point of intersection with Mohr Circle represents the stress & shear stress you need. 115

116 ملخص : طريقة الرسم : ھناك ) 3) عوامل رئيسية مرتبطة مع بعضھا ھي OP stress االجھاد Plane مستوي نقطة اصل المستوي : OP ھي نقطة وھمية وحيدة لكل دائرة ووحيدة فقط وتعين من اجھاد ومستوي. اي خط يمر في OP ويقطع دائرة مور فان ھذا الخط يمثل المستوي الذي تعمل عليه ذالك االجھاد. Principle Stress: Stresses acting normal on mutually orthogonal planes with no shear stresses. Principle Planes: The planes on which there is zero shear stresses. Case one: Given required Example 1: Find stresses on plane B-B? 116

117 Answer: = 2 = = 40 / = 20 / = = = 30 / 2 = 10 / By drawing: 1- Locate points (40,0) and (20,0). 2- Draw circle, using these points to define diameter. 3- Draw line through point (20,0) and parallel to plane on which stress (20,0)acts. 4- Intersection of with Mohr circle at point (40,0) is the origin Of planes. 5- Draw Line through OP and parallel to B-B 6- Read coordinates of X where intersect Mohr circle. So on B-B = 25 = 8.7 Method 2 using Equations : = + = cos 2 cos( 2 120) = 30 + ( 5) =

118 = = ( ) sin 2 2 ( 40 20) sin(2 120) = 8.66 / 2 Case 2 given, required Using Equation: = = ( + ) 2 + We usually take the largest couple. = = = 2 2 = 2 Example 2 : Find the stresses on Horizontal plane D-D? 1-Draw point 1(40,0), point (20,0) 118

119 2- From point (20,0) draw a line through point (20,0),parallel to the plane that the force (20,0) acts. 3- The point of intersection with Mohr circle represent the OP. 4- From the op draw a horizontal line (line parallel to horizontal plane) the point of intersection with Mohr circle represent the stresses on horizontal plane. ( 35, 8.7). Example 3 : Find the Magnitude and direction of priciple stresses. 119

120 1- Locate points (40,-10) and (20,10) 2- Erect diameter and draw Mohr circle. 3- Draw through (40,-10) parallel to BB 4- Intersection of with circle gives Op. 5- Read from graph. 6- Line through Op and give plane on which acts. Solution by Equations : 1- Make use of that the sum of normal stresses is a constant: = = = Use the following equation: 2 = ( + ) + 2 ( ) = = 200 = = + = = = Use stress pair in which ; (40, 10) 2 2 = = =

121 2 = 45 = Angle from horizontal to major principle stress direction = 52.5 Examples : to be added Stress Path (p-q diagram): 121

122 122

123 Chapter Seven Consolidation of soil A stress increase caused by the construction of foundations or other loads compresses the soil layers. The compression is caused by (a) deformation of soil particles, (b) relocations of soil particles, and (c) expulsion of water or air from the void spaces. In general, the soil settlement caused by load may be divided into three broad categories: 1. Immediate settlement, which is caused by the elastic deformation of dry soil and of moist and saturated soils without any change in the moisture content. Immediate settlement calculations are generally based on equations derived from the theory of elasticity. 2. Primary consolidation settlement, which is the result of a volume change in saturated cohesive soils because of the expulsion of water that occupies the void spaces. 3. Secondary consolidation settlement, which is observed in saturated cohesive soils and is the result of the plastic adjustment of soil fabrics. It follows the primary consolidation settlement under a constant effective stress. This chapter presents the fundamental principles for estimating the consolidation settlement: Consolidation : is the gradual reduction in volume of a fully saturated soil of low permeability due to drainage of some of the pore water, the process continuing until the excess pore water pressure set up by an increase in total stress has completely dissipated. The simplest case is that of one dimensional consolidation in which condition of zero lateral strain. Swelling: is the reverse of consolidation, is the gradual increase in volume of a soil under negative excess pore water pressure. 123

124 Consolidation settlement: is the vertical displacement of the surface corresponding to the volume change at any stage of the consolidation process. Consolidation settlement will result, for example, if a structure is built over a layer of saturated clay or if the water table is lowered permanently in a stratum overlying a clay layer on the other hand, if an excavation is made in a saturated clay, heaving (the reverse of settlement) will result in the bottom of the excavation due to swelling of the clay. Fundamentals of Consolidation: When a saturated clayey soil layer is subjected to a stress increase, the pore water pressure suddenly increases. So at time =0 = while =0 (Figure 1-b), but after a time t < while > 0 (Figure 1-c) and after very long time ( ) = 0 while = (Figure 1-d). 124

125 . Figure 125

126 Spring Analogy: the following figures illustrate the soil-spring analogy: The Oedometer (consolidation)test : Figure (7-2) Dial gage Test Procedure: Figure (7-2) Consolidation Cell 1- Determine All data for soil sample such as diameter, weight and height. Also determine Gs, initial water content, initial void ratio by using S*e= Gs. W(s=100%). 2- Set the sample in the consolidation test and apply initial stress p1= 25 kpa and record the dial reading for a period of (24 hrs) at the times 0,0.25, 0.5, 1,2,4,8,15, 30, 60, 120,240,480,1440 min. from the load application. 3- At the end of (24 hrs), double the applied stress (p2= 2p1=50 kpa 126

127 And record the dial gage reading for 24 hrs at similar times to those of step Repeat step 3 by doubling the applied stress and recording the dial gage reading. This process is repeated till we reach a stress of 1600 kpa (some times we reach 3200 kpa). This process last 7 days and it called loading stage. 5- Unload (3/4) the applied stress (i.e. remove 1200 kpa and leave 400 kpa) and record the dial gage reading for (24 hrs). 6- After 24 hrs, unload (3/4) the remaining stress (remove 300 kpa and keep 100 kpa)and record dial reading for 24 hrs. then remove all the applied stress and record dial readings. 7- Determine the final water content of the soil sample. Example : The results of an odometer test is given below:- Time Dial gage readings (min) 25 kpa 50 kpa 100 kpa 200 kpa 400 kpa 800 kpa 1600 kpa 400 kpa 100 kpa Mass of empty ring = gm Mass or ring +wet soil = gm Mass of ring + dry soil = gm Dia. Of ring = 75 mm Ht of ring = 19 mm 127

128 Dial gage Coeff. = " Initial water content = 31.14% Specific Gravity of soil solids (Gs) = 2.76 Initial Degree of saturation = 100% Required: void ratio at each load increment. Solution : 1- find using S.e = Gs * = = Voids Solids 2- Since the volume of solid particles not change (incompressible) So Assume = 1. So = So the Total volume ( = 1 + ) So any change in the volume due to apply load is due to change in void so = = 1 + Since the consolidation in one dimension, and there is no lateral strain so the axial strain ( = ) will be equal to volumetric strain ( ). = 128

129 Applied Pressure (kpa) Final Dial gage reading Dial change *0.0001*25.4 mm h( ) Thickness of sample at the end of (24 hrs) mm Change in void ratio = h (1 + ) From consolidation test the following result we can get : Void ratio at the end of 24 hrs e=

130 Compression Recompression or Reloading The slop of this curve at any stress range can be defined as : = : Coeff. Of compressibility change in effective stress corresponding change in void ratio Unloading Figure e- (results from oedometer test) = 1 + : Coeff. Of volume change : Coeff. Of Compressibility : Initial void ratio Plot the ( ) * the slop of first portion of the loading curve is defined = Log Which is equal to the slope of the unloading curve, and the first portion of the reloading curve = = : Reloading index 130

131 : Expansion index The slope of the last portion of the loading curves is defined as : = : Compression index Note : can be obtained using empirical correlation from Liquid Limit (L.L) = (. 10) An empirical correlation between is given by: 0.1 Log Figure e- log (find ) Pre-Consolidation Pressure ( ) : The max. Effective stress that has been experienced by the soil in the past or at the present. How to find 1- Produce back the straight line (BC) of the curve 2- Draw the tangent to the curve at D and bisect the angle between the tangent and the horizontal through D 3- The vertical through the point of the bisector and CB gives the approximate value of the pre consolidation 131

132 Normally and Over consolidation Clays: pressure. A cording to the stress history, the clay can be: 1- Normally Consolidated clay(n.c.c.) Which represent the clay at which the existing effective overburden pressure ( = ) is the largest stress experienced by the soil at the present time and in the past thus : For N.C.C. = 2- Over Consolidated Clay(O.C.C.) Represent the clay which has experienced a stress in the past larger than the existing effective overburden pressure ( ) acting at the present time. For O.C.C. > Over Consolidation Ratio: (O.C.R.) O.C.R.= If O.C.R. =1 = If O.C.R. >1 > Consolidation Settlement: To calculate the final consolidation settlement ( at t= ) use one of the following : 1- = 2- = 3- = log 4- For O.C.C. we have to Check: a- If + then use : = b- If + > log then use : 132

133 = log log Where Scf : Final consolidation settlement : Initial void ratio : Change in void ratio = H : thickness of compressed clay layer : Coefficient of volume compressibility for the stress range : change in effective stress between initial and finial conditions Compression index : reloading index : initial effective overburden pressure. : Pre consolidation pressure Note About settlement Calculation: 1- Scf : usually find at the middle of compressed clay layer (represent the worse case). 2- Calculate the increase in vertical stress at the middle of clay layer by either using 2:1 method or using Chart (figures 8.6 and 8.4). 3- Check whether the clay is N.C.C. or O.C.C., and use proper equation mention above. 4- For more accurate result you can divide the clay layer into sub layers (i.e not more than 2 m). Then find the total settlement by summing the settlement for each layer. 133

134 Example: For the soil Profile shown find the final consolidation settlement? 1- = = =.. ( ) = = = h = = 99 / 3- = log = = at the middle of the clay layer before fill application = ( 7.3 2) = U= ( ) + (.. ) = = = =... ( ) log. =

135 Degree of Consolidation (U% ) Degree of consolidation can be finding by Figure: Assumed linear e- relation 135

136 Terzaghi's theory of one-dimensional consolidation: The assumptions made in the theory are: 1- The soil is homogenous. 2- The soil is fully saturated. 3- The solid particles and water are incompressible. 4- Compression and flow is one-dimensional (vertical). 5- Strains are small. 6- Darcy's law is valid at all hydraulic GRADIENTS. 7- The coefficient of permeability and the coefficient of volume compressibility remain constant throughout the process. 8- There is a unique relationship, independent of time, between void ratio and effective stress. There are three variable in the consolidation equation: = (Consolidation equation) 1- Depth of the soil element in the layer (z). 2- The excess pore water pressure (U) 3- The time elapsed since application of the loading (t) Where u: is the excess pore water pressure. : is the coefficient of consolidation. = Where is the coefficient of volume changes ( ). Where Time factor = h 136

137 t : the time h : is the drainage path h =H/2 ( clay layer Between two permeable layers). h = H (for clay layer between one permeable and one impermeable layer). h = H h = H : Degree of Consolidation There is a relation between the and Degree of consolidation 1- Degree of consolidation at specific depth So the graphical solution of Terzaghi's one-dimensional consolidation equation using non-dimensional parameters is as shown in following figure: h Figure Variation of degree of consolidation with time factor and h. 137

138 The variation of total consolidation with time is most conveniently plotted in the form of the average degree of consolidation ( U ) for the entire stratum versus dimensionless time, and the this is illustrated below: Figure : Average Degree of consolidation and Time factor There are useful approximation relating the degree of consolidation and the time factor For < 0.60 = For > 0.60 = (100- U) Determination of coefficient of consolidation: There are two methods: 1- The log time method (casagrande method): a)-by plotting the dial reading versus log time Then choice any time (small time ) then find the dial reading at 4t then move above the dial reading a distance equal to then you can find the dial reading at t=

139 b)-take the tangent to the curves at the end of the curve so you can find the dial reading at 100% consolidation. c)-now dial reading at t= 09 and dial reading at the end of consolidation so you can find the dial reading for 50% consolidation. e)-so the coefficient of consolidation can be find by the following equation = 2- The root time method (Taylor) 139

140 =. Consolidation Settlement Time: To estimate the amount of consolidation which occur and the time, it is necessary to know: 1- The boundary drainage conditions 2- The loading condition 3- The soil parameters including initial void ratio, coefficient of compressibility, compressibility index and coefficient of consolidation. All these are obtained from consolidation test on representative sample. So to plot the variation of settlement versus time curve: 1- Find the time required to complete the consolidation ( = 1). 2- Assume different value of time start from (t=0) to time (t) find in step 1 3- For each value of time t, find the using ( = ). 4- For each value of (from step 3) find ( ). 5- Find the final consolidation ( ) using any proper equations. 6- Find the settlement at time t =. 7- Plot the settlement-time curve. Time (t) = = h 140

141 Figure Variation of settlement with time Example: For the soil Profile shown find the final consolidation settlement? 1- = = =.. ( ) = = = h = = 99 / 3- = log 141

142 = = at the middle of the clay layer before fill application = ( 7.3 2) = U= ( ) + (.. ) = = = Example : =... ( ) log. = Astrata of consolidated clay of thickness 10 ft drained on one side only. The hydraluic conductivity of k = 1.863*10 in /sec. and cofficient of volume compresebility = Determine the ultimate value of the compression of the stratum by assuming a uniformly distributed load of 5250 and also determine the time required for 20 percent and 80 percent consolidation. Solution : Total compression = = = in. For determining the relationship between U% and T for 20% consolidation use the equation = 4 % 100 For 80% consolidation use the equation = = = log(100 80) Therefore = log(100 80) =0.567 Example 142

143 Example : The loading period for a new building extended from may 1995 to may In May 2000, the average measured settlement was found to be cm. it is known that the ultimate settlement will be about cm. Estimate the settlement in may Assume double drainage to occur. Solution: For the majority of practical cases in which loading is applied over a period, acceptable accuracy is obtained when calculating time-settlement relationships by assuming the time datum to be midway through the loading or construction period. = h = 4 = The settlement is requried for t=9 years, that is up to may Assuming as a starting point that at t = 9 years, the degree of consolidation will be = under these conditions. = Is =, = Settlement at time = = = = Where is a constant. Therefore. = = Therefore at t=9 years, =. = 0.48 and since the value of U is less than 0.60 the assumption is valid. So the Settlement is cm. 143

144 Example : An oedometer test is performed on a 2 cm thick clay sample. After 5 minutes 50% consolidation is reached. Aftrer how long a time would the same degree of consolidation be achieved in the field where the clay layer is 3.70 m thick? Assume the sample and the clay layer have the same drainage boundary conditions (double drainage). Solution : The time factor T is defined as = Where = half the thickness of the clay for double drainage. Here, the time factore T and coefficient of consolidation are the same for both the sample and the field clay layer. The parameter that changes is the time t. Let and be the times required to reach 50% consolidation both in the oedometer and field respectively. = 5. = = ( ) =

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151 Example Example : 151

152 Example : 152

153 Example: 153

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155 Chapter Eight Shear Strength The strength of a material is the greatest stress it can sustain; So that the unit of strength is the same as stress (Pa in SI unit system). Shear Failure in Soils: Significant of Shear Strength: The safety of any geotechnical structure is dependent on the strength of the soil; If the soil fails, the structure founded on it can collapse. Understanding shear strength is the basis to analyze soil stability problems like lateral pressure on earth retaining structures, slope stability, and bearing capacity. 155

156 Slope Failure in soils: 156

157 West side of foundation sank 24 ft. 157

158 158

159 Shear Strength in Soils The shear strength of a soil is its resistance to shearing stresses. It is a measure of the soil resistance to deformation by continuous displacement of its individual soil particles Shear strength in soils depends primarily on interactions between particles Shear failure occurs when the stresses between the particles are such that they slide or roll past each other Shear Strength in Soils: Soil derives its shear strength from two sources: Cohesion between particles (stress independent component) Cementation between sand grains Electrostatic attraction between clay particles Frictional resistance between particles (stress dependent component) Shear strength of soils: Cohesion (C): is a measure of the forces that cement particles of soils. 159

160 Internal Friction angle (φ), is the measure of the shear strength of soils due to friction Mohr-Coulomb Failure Criteria This theory states that a material fails because of a critical combination of normal stress and shear stress, and not from their either maximum normal or shear stress alone. Mohr Columb Failure criteria = + tan 160

161 = + tan Where : is the shear strength of soil (kpa) C: cohesion (kpa) : is the angle of internal friction (degree) : is the cohesion in term of effective stress : angle of internal friction in term of effective stress The above equations are the shear strength of soils in terms total stress and in term of effective stress. c = 0 for pure sand = 0 for normally consolidated clay : Also some time called the drained angle of internal friction. Mohr-Columb shear Failure criteria =

162 = + + = + + ( + ) Also in term of total stress = + + ( + ) Determination of Shear strength Parameters (C,, and ) The shear strength parameters of a soil are determined in the lab primarily with two types of tests: 1) Direct Shear Test; and 2) Triaxial Shear Test. -Direct Shear Test: The test equipment consists of a metal box in which the soil specimen is placed The box is split horizontally into two halves Vertical force (normal stress) is applied through a metal platen Shear force is applied by moving one half of the box relative to the other to cause failure in the soil specimen. 162

163 Direct Shear Test Procedure: 1- Measure inner side of shear box and find the area(6 *6 ). 2- Make sure top and bottom halves of shear box are in contact and fixed together. 3- Place the soil (sand) in three layers in the mold using funnel. 4- Place cover on top of sand. 5- Place the box in machine. 6- Apply normal force. (usually the student divided in to three groups, each group use run with one weight, 2 kg, 4 kg and 6 kg the load is applied through a level arm : Note lever arm loading ratio 1:10, 2kg weight =20 kg). 7- Start the motor with selected speed (0.1 in/min) so that the rate of sharing is at a selected constant rate. 8- Take the horizontal dial reading, vertical dial reading and proving ring reading (shear load). Record the readings on the data sheet. 9- Continue taking readings until the horizontal shear load peaks and then falls, or the dial reading gage stopped or at 10% of the the length of the 163

164 sample. 10- Do the calculation : Find the = ( ) 11- Plot the results (shear stress Vs horizontal displacement) the result will be as in figure, take 3 peak stresses. Then plot the relation between shear strength and normal stresses. Find the slop of this curve, find, the intersection of the line with y-axis gives the value of C. 164

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166 The disadvantages of direct shear test are: Advantages: 1- The test's simplicity and, in the case of sands, the ease of specimen preparation 2- The travel of the machine can be reversed to determine the residual shear strength values, which is shear strength parameters at large displacements. 3- Shear box represents a cheaper method in determining the drained shear strength parameters for coarse-grained soil. Disadvantages 1) The main one: drainage conditions cannot be controlled. 2) As pore water pressure cannot be measured, only the total normal stress can be determined, although this is equal to the effective normal stress if the pore water pressure is zero. 3) Only an approximation to the state of pure shear is produced in the specimen and shear stress on the failure plane is not uniform, failure occurring progressively from the edges towards the center of the specimen. 4) The area under the shear and vertical loads does not remain constant throughout the test. Triaxial shear test : This experiment over come all the disadvantages of direct shear test: 1- Can be used for all types of soils 2- Pore water pressure can be measured 3- The area corrected be used 4- The failure plane Triaxial shear test is the most widely used and is suitable for all types of soils. A cylindrical specimen, generally having a length to diameter ratio of 2, is used in the test and is stressed under 166

167 conditions of axial symmetry in the manner shown in figure. Soil Sample in triaxial apparatus A sample of soil in a sealed triaxial test cell. Confining pressure is applied using water surrounding sample as shown in following figure: 167

168 Valve A Figure ( ) triaxial compression testing device 1- Triaxial test is more reliable because we can measure both drained and undrained shear strength. 2- Generally 1.4 diameter (3 tall) or 2.8 diameter (6 tall) specimen is used. 3- Specimen is encased by a thin rubber membrane and set into a plastic cylindrical chamber. 4- Cell pressure is applied in the chamber (which represents σ3 by pressurizing the cell fluid (generally water). 5- Vertical stress is increased by loading the specimen (by raising the platen in 168

169 strain controlled test and by adding loads directly in stress controlled test, but strain controlled test is more common) until shear failure occurs. Total vertical stress, which is σ1 is equal to the sum of σ3 and deviator stress (σd). Measurement of σd, axial deformation, pore pressure, and sample volume change are recorded. Depending on the nature of loading and drainage condition, triaxial tests are conducted in three different ways. UU Triaxial test (unconsolidated undrained triaxial test) CU triaxial test (consolidated undrained triaxial test) CD triaxial test ( consolidated drained triaxial test) Unconsolidated Undrained Triaxial test: As drainage is not permitted and consolidation is not necessary, this test is very quick and also called a quick test. In this test as drainage is not permitted the pore water pressure increases after application of as well as after the application of so = and = so the total excess pore water pressure will be as follows: = + But = This test is common in clayey soils. Application: UU triaxial test gives shear strength of soil at different confining stresses. Shear strength is important in all types of geotechnical designs and analyses. Test procedure: 1- Measure diameter, length, and initial mass of the specimen. 2- Set a soil specimen in a triaxial cell. 3- Increase the cell pressure to a desired value (100 kpa for the first case and 169

170 200 kpa in the second case and 300 kpa in the third case). 4- Shear the specimen by strain control record, and record the 5- Continue the test until the proving ring stop reading or use 10% After completion of test measure of the soil specimen again and put it into the oven then find the moisture content of the sample. Take the sample out the cell and Sketch the mode of failure. Repeat the test for the second specimen too (200 kpa of cell pressure and third specimen 300 kpa of cell pressure). calculation : 1- = ( = h ). 2- Calculate the corrected area ( ): = 3- Calculate the stress = 4- Plot versus axial strain separately for three tests. 5- Plot versus for three tests in the same plot 6- Plot Mohr circle using and at failure (note they should give the same value) 7- Plot the tangent to the three Mohr circles you plotted in above stage 8- Make a straight line, which is tangent to all three Mohr's circles. The slop give the angle of internal friction =0, the point of the intersection of the tangent and y-axis give the cohesion ( ) = 2 170

171 Cu triaxial test : consolidated undrained test In this test consolidation is permitted by opening drainage valve through applying, or cell pressure, or all around pressure. Then the drainage valve closed during applying the deviator stress loading. = + Because valve A open during applying cell pressure, therefore B=0 So the increase in pore water pressure is due to increase in deviator stress only. = CD triaxial test : consolidated drained triaxial test In this test valve A is open all time and the pore water pressure parameters A and B = 0. So this test is called drained test also called long term test and slow test. There is volumetric change: So = where : is the volumetric strain = In this case: = and = usually occurs in sandy soils The drained and undrained condition: Drained condition occurs when there is no change in pore water pressure due to external loading. In a drained condition, the pore water can drain out of the soil easily, causing volumetric strains in the soil. Undrained condition 171

172 Occurs when the pore water is unable to drain out of the soil. In an undrained condition, the rate of loading is much quicker than the rate at which the pore water is able to drain out of the soil. As a result, most of the external loading is taken by the pore water, resulting in an increase in the pore water pressure. The tendency of soil to change volume is suppressed during undrained loading. The existence of either a drained or an undrained condition in a soil depends on: The soil type (e.g. fine-grained or coarse-grained) Geological formation (fissures, sand layers in clays, etc.) Rate of loading For a rate of loading associated with a normal construction activity, saturated coarse-grained soils (e.g. sands and gravels) experience drained conditions and saturated fine-grained soils (e.g. silts and clays) experience undrained conditions If the rate of loading is fast enough (e.g. during an earthquake), even coarse-grained soils can experience undrained loading, often resulting in liquefaction. Unconfined compression test: This test subjects the soil to an axial compressive load between two platens as shown in the Figure. There is no confinement of the sample in the radial direction. The load is recorded using a proving ring or a load cell and the axial deformation of the soil sample is recorded using a dial gauge. 172

173 Unconfined compression test = 0, only suitable for cohesive soil. To find the stress = 1 = ( ) 1 Significance of unconfined compression test A quick test to obtain the shear strength parameters of cohesive (fine grained) soils either in undisturbed or remolded state. The test is not applicable to cohesion less or coarse grained soils The test is strain controlled and when the soil sample is loaded rapidly, the pore 173