Soil Mechanics & Engineering: Understanding Soil's Natural Opponents, Exams of Engineering

Download Soil Mechanics & Engineering: Understanding Soil's Natural Opponents and more Exams Engineering in PDF only on Docsity! 1 CHAPTER: 1 SOIL MECHANICS( B.Tech. Hons.Civil) Question: 1. Describe in Detail that the soil is used as construction material? Ans: The soil can be considered as the oldest and the most complex of the construction materials used by engineers. Unless if the structure is built on hard rock, every structure, whether it is a building of any kind, a bridge, a dam, a railroad, an airport, or a hydraulic structure, rest on soil therefore, the choice of an adequate foundation on the soil is must for construction of project. The stability and function of a structure depend upon the behavior of the soil upon which it is built. Because of ignorance of the principles of soil mechanics, Engineers have been faced, in the past, in various parts of the world with an increasing number of failures due to following problems. 1) Unanticipated action of water. 2) Frost action in soils. 3) Unexpected settlements of soils. 4) Lateral displacement of soil. 5) Other unexpected performance of the soil. Many earth dams have collapsed because engineers were unable to find the accurate 2 performance of compacted soil. The collapse of tunnels, bridge piers and abutments, and the failure of many earth retaining structures and various structures have been occurred because Engineers were unable to evaluate satisfactory how much pressure this soil would exert on these structures under various conditions. Differential settlements of the foundations under large engineering structures were directly responsible for serious structural damage. Q : 2 Explains that the Problems/Properties of soils and foundation are very difficult and Complex as compared with the superstructure. Ans. The mechanical properties of soils are more complex and difficult to determine than those of steel, concrete, or wood. No material has greater variation of properties than soil, because it is not a manufactured standard product like steel. This is because the soil with which the engineer must work was placed by nature in a great variety of kinds and conditions. Thus it can be understood that soils produce the most serious problems of design because it is not homogeneous. Soil investigations and knowledge of the physical and mechanical properties of soil can help to protect the owner of a building and the engineer in charge from unforeseeable conditions and the troubles that usually occurred to them. In the course of time engineers have learned that the problems dealing with soils and foundations are 5 theoretically and practically soils, by means of which and upon which engineers build their structures. Soil mechanics sometimes also called Geo- technique. The soil mechanics treats soil as a construction material. Behavior of soil under static and dynamic load, as well as under the influence of water and temperature, is studied theoretically and experimentally by using knowledge of soil mechanics, it also studies the mutual interaction of structure and soil. Soil mechanics is the application of the laws of mechanics and hydraulics to engineering problems dealing with sediments and other unconsolidated accumulations of solid particles produced by the mechanical and chemical disintegration of rocks regardless of whether or not they contain an admixture of organic. “Soil Mechanics” is that discipline of engineering science which deals with the properties, behavior & performance of soil as a structural material. The practice of engineering which applies the principles of soil mechanics to the design of engineering structures is called soil engineering. Purpose: The purpose of soil mechanics is to replace by scientific methods the design applied in foundation engineering in the past. Q: 2. Describe the objective of soil mechanics in detail. 6 Ans: Following are the objectives of soil mechanics. 1. To perform engineering soil surveys. 2. To develop rational soil sampling devices and soil sampling method. 3. To develop suitable soil testing devices and soil testing methods. 4. To collect and classify information on soils and their physical properties in the light or fundamental knowledge of soil mechanics, earthwork, and foundation engineering. 5. To investigate the physical properties of soils and to determine coefficients to characterize these soil properties. 6. To evaluate the soil test results and their application as a construction material. 7. To endeavor to gain more light in understanding the physical processes which actually take place in soils subjected to various factors such as static and dynamic loads, water and temperature. 8. To apply the knowledge of soil mechanics for the solution of practical engineering problems. 9. To replace by scientific methods, the design used in foundation and earthwork engineering in the past. 7 Question: 3. Explain that what problems related to soil should be considered before execution of any project? Ans: Following are some soil mechanics problems: A. How deep borings for soil exploration should be made. B. What is the bearing capacity of a soil on its surface and at carious depths to carry various loads? C. What is the load to be applied on a particular soil? D. What is the intensity and what is the stress distribution in a soil induced by various kinds of loading. E. How thick should be a layer of a good soil over a poor one in order to prevent other foundation form punching. F. Does a soil possess properties (friction and cohesion) which will assure satisfactory stability for foundations? G. How much counterweight should be placed as a remedial measure against lateral motion of soil, masses in order to maintain the stability of a structure? H. When does settlement cease. I. What is the mutual interaction between soil and foundation, and what kinds and magnitude stresses are induced in the foundation and soil of a rigid highway or runway pavement due to schemes under consideration? 10 In soil engineering a sharp demarcation between rock and soil is no longer made. In this sense the term soil is adopted and used also in soil mechanics, and in this book only the engineer’s definition of the terms soil is to be understood from as engineer’s viewpoint( soil is a material by means of which and upon which he builds his structures) c) Formation of soil: d) Soil which is a complex mixture of inorganic matter that may or not contain decomposed organic residues and other substances and which blankets the earth’s crust, is formed by the process f weathering, (disintegration and decomposition) of rocks and minerals through the action of many natural physical or mechanical and chemical agents into smaller and smaller particles The factors of weathering in the process of soil formation may be atmospheric, such as the work of temperature wind and water erosion and transportation by the water and glaciers; plant and animal life; 11 Question: 2. Distinguish between Glacial Soil & Alluvial Soil: Ans. A). Glacial Soils: The soil, which have been transported & deposited by glaciers. The glacial ice sheet filled up river valleys with the so-called glacial drift. Glacial drift is the glacial deposit formal types of the superficial material of rock debris of any sort, handled in any way, by the glacier-for example by erosion, transportation, deposition from ice, or running melt waters emanating from the ice. Glacial deposits consist of boulders, rock fragment, gravel, sand, silt, and clay in various proportions. The position of the southern boundary of the continental glacier is distinctively recognized by the geologic feature called the terminal moraine. The landform of the terminal moraine consisting of glacial debris material is characterized by irregular, hills. Their width varies from one to two miles, and their height can be about 30m. One of the engineering aspects of terminal moraine soil materials containing salty clay and clay is the disposal of domestic wastes by means of domestic septic tanks. Septic tanks built in clayey soils, soils which by nature are of low permeability, may function improperly, thus creating unpleasant odors another engineering aspect of glacial soils relative to foundation engineering is the thickness of the glacial drift. One ridges of bedrock the glacial drift may be 12 thin, whereas other glacial drift may be thick. The thickness of the glacial drift may be said to be from about 6 to 12m. The dam site originally selected had to be avoided because of a loose drift material; it could not properly support the weight of the dam, nor could it meet satisfactory permeability requirements. The thickness (or) depth of the glacial drift in a valley has its economic aspect, since this factor governs the cost of a dam regardless of whether it is of earth or concrete or a highway, or railway. It also governs the cost of foundations of bridges, lock and other engineering and hydraulic structures. Excavation operations in glacial state require a power shovel or explosives. Ordinarily, however, excavations in glacial till present no problem. b. Alluvial “Soils”: Alluvial soils occur in former and present flood plains and deltas often forming very thick deposits. Question: 3. Distinguish between Residual Soil & Wind Borne Soil also explain glacial clay in details. Ans. a) Glacial Clay (Carved Clay): During the course of geologic times the lakes were filled with stratified silt and clay, the so-called varve clay. The Swedish term varve means distinctly marked annual deposit of sediment. Excavations in glacial clay reveal the neat annual layers (or) varve. The varve indicate that the clay had once been the sediment at the bottom of a lake. The varve consist of a sequence of a thicker layers, light in color which is the summer 15 4. 5. Calcareous: Calcareous Soil denotes a soil containing calcium carbonate. Such a soil effervesces when tested with weak hydrochloric acid. 6. Caliches: It is a soil conglomerate of gravel, sand, and clay, the particles of which are cemented by calcium carbonate. 7. Cobbles: Cobbles Are rock fragments ranging in size from about 75 to 300mm in diameter. It is a diatomaceous earth which is a fine, light grey, soft sedimentary deposit of the siliceous remains. Diatoms are minute unicellular marine organisms. Such a soil is very porous, and of fine structure. 8. Gumbo: It is a very fine-particles clay deposit. This clay is dark; plastic, and very sticks. When moist it is sticky and spongy. It is one of the most difficult soils to handle either in excavation or as a road material. 9. Humus: It is the term used to denote a dark brown organic amorphous earth of the topsoil. It consists of partly decomposed vegetal matter. As a foundation or road material, humus is undesirable because it continues to decay, shrinks, and also holds water. 16 10. Loam: It is a mixture of sand, silt, and clay, sometimes containing some organic matter, such as humus. The terms sand, silt, and clay here refer to the particle size. 11. Loess: Denotes a uniform, and usually range between 0.01mm to 0.05mm. The color of loess is yellowish light brown. Loess is encountered in dry continental regions. Slopes of cuts made in it are able to stand nearly vertically. Question: 5. a ) Differentiate between Cohesive & Non-Cohesive Soil ‘b’ Define the following marl, mud, peat, pebbles, quicksand. Ans: a) Cohesive Soils: The soils possessing cohesion are called cohesive soils.there are two types of cohesions (1) True cohesion the intermolecular attraction of the soil particles for each other throughout the soil mass in called true cohesion. 2 Apparent Cohesion: Binding of the soil mass together by the action of surface tension farces of the soil moisture is called apparent. Non-Cohesive Soils: The soils which have been formed by the uncemented weathered rock particles is called Non-Cohesive Soils for example gravel sand. B). 17 1. Marl: Is a very loose term denoting deposits which consist of mixtures of sands, or clays or loam. The proportion of carbonate lime, however, may not fall below about 15% nor dos the quantity of clay rise above 75%. 2. Mud: Designates a mixture of silt or clay with water. The constancy of mud is an almost fluid mass. 3. Peat: Is an organic soil formed of vegetal matter under conditions of excessive moisture, as found in swamps. Peat is very compressible, and therefore unsuitable for supporting even the lightest foundations. 4. Pebbles: Are a constituent part of gravels with diameters about 50mm to 75mm. 5. Quicksand: Is not a special type of soil, but a condition. Any granular material through which an upward flow of water takes place may become “Quick” under proper hydraulic conditions. Q -6 Write a note on the Following (1) Soil Profile (2) Lime Soil (3) Hard Pan (4) Till Ans (1) Soil Profile: It is a vertical cross section of the actual soil strata at a given site showing a natural sequence and thickness of soil layers below the ground surface 20 a. Granular or single –grained structures b. Honeycomb or cellular structures and c. Flocculent structures. 4. Specific Gravity The specific gravity G of dry solids of a soil is defined as the ration of the density Ys of a given volume of the soil solids to the greatest density Yw (at +4°c) of an equal volume of pure water. G = Weight of soil particles in grams = Ys Ws (1) (Volume of soil particles) (1.000) ______ Vs Ys yw Where Ys = unit weight of soil solids = Ws = _Weight of Solid (Soil) Vs Vol. of Soil Yw = unit weight of water Ws = VsYs also Vs = Ws ________________________ (2) G.W The quantity G is dimensionless and it shows how many times heavier the substance of the solid 21 particles of soil is than an equivalent amount of water. Q-2 Define and Explain (1) porosity (2) void ratio Ans. 1- Porosity. The total volume of all voids within a unit volume of soil regardless of whether fully or partially filled with liquid is called the soil porosity .Let there be a soil sample some portion of which is consist of voids Vv and remaining will be solid as show in fig. Vv 1 1 1- n n 1 Solid Volum Explanation: Mathematically it is written as. Porosity = n = Volume of all voids in soil × 100% ……….. (1) Total volume of soil Or n = Vv ×100 = Vv × 100 ……………. (2) V Vs+Vv V = Vv +Vs 22 Where Vs = volume of solids and Vv volume of voids V= total volume of soil sample ………… (3) Vv = V-V Also n can be expressed as Porosity n= V - Vs = V _ Vs = 1 _ Vs = 1 - Ws ………. (4) V V V V G. yw.V Where Ws = weight of solids of soil and Vs = Ws G.yw G = specific gravity of the sol and Yw = unit weight of water 2 Void Ratio The quantity expressing the ratio of the volume of voids to the volume of solids is termed the relative porosity or void ration. e = Volume of Voids in soil Volume of Solids in soil 25 Or n + ne =e n (1+e) =e n = e ……………………… (3) 1+e Also we know that e= Vv Vs Vv = e.Vs as e= n 1-n Vv = Vs (n ) ………………………… (4) 1-n Total volume = V= Vv + Vs = Vs = V- Vv Vv =nv Vs = V-nv Vs = v (1-n) ………………………… (5) Similarly Vv = V-Vs Vs = (1-n) V =V- (1-n) V =V –V+ nv So Vv =nv …………………………. (6) Now 26 Void ratio = e= Vv Vs By putting Values of Vv and Vs we have e = n v = n (1-n) V (1-n) e = n ………………… (7) 1-n Total volume = V = Vv Vs , again putting values of Vv and Vs V= nv + (1-n) V V= nv + V-nv V=V ………………………… (8) If volume of air voids is considered then total volume V = Vs + Vv We know that Vv = Vw + Va V = Vs + Vw + Va …………………… (9) Where Va = volume of air Vw = volume of water Q-4 A )Define the soil moisture content and explain it by giving suitable example. B) Define and explain the soil phases. Ans. Definition The Soil moisture content is defined as that amount of water which is contained in the solids of the soil. The amount of water contained in 27 the void of soil natural state is termed the natural moisture moisture content of the soil. Explation: The knowledge of moisture content in soil is necessary in soil compaction control; the soil moisture content is determined by the so – called gravimetric method by drying the soil sample in drying oven at 105 oC is attained the moisture content w is expressed as a percentage of the dry weight of the soil and called the absolute moisture content of the soil: W = weight of water ×100 % = Ww *100 % Weight of solids Wd For example let Ww = 11.0 g and Wd = 67.0 g. then the moisture content based on dry weight is calculated as W= 11.0 ×100 % = 16.4% 67.0 Q-5 What do you mean (a) soil texture (b) degree of saturation C) Percent Air Voids Ans. a) Soil texture. The term soil texture is used to express the percentage of the three main fractions (sand site and 30 Description of details for performing particle size analysis of soils given by various agencies in their manuals or specification of standard test method of soils. The material retained on sieves Nos.4,10,20,40,60,100,140,and 200 are designated as “plus4” material (+4),+10,+20,+40,+60,+100,+140,+and +200 material, respectively. This means soil particles which are larger than the openings in a corresponding mesh screen. Conversely, minus4” material (-4),-10,-20,-40,-60,- 100,-140,and 200 soil material are those passing the corresponding mesh screen. Q-7 Define and write down the equation of following (1) unit weight of soil (2) Dry unit weight (3) submerged unit weight of soil. Ans. 1) Unit weight of soil. The unit weight of soil is the weight of soil per unit of it s volume. This weight comprises the. Whole soil mass, i.e., the solid particles plus all voids; the latter may not cantina soil moisture. The weight of air considered negligible and therefore ignored. The soil unit weight Y is expressed as the ratio of the weight W, of volume V, of soil mass to that volume Y= W/V where Y is the density. The unit weight can be expressed in g/cm3 or kg/m3 the unit weight of water Yw is its weight Ww per unit of its volume (Vw), Yw=Ww/Vw. 31 (2) Dry unit Weight of Soil. Definition. This is the weight of a soil per unit volume after lowering the ground Water table: (degree of saturation: S=O) Yd= (1-n). G.yw Where Yd = dry unit weight of soil, n = porosity, G = specific gravity of soil particles, and Yw = unit weight of water (3) Submerged unit weight of soil. Definition. This is the weight of a soil per its unit volume before lowering the ground water table: mathematically it can be written as YSub = (1-n) (G-1) Yw ………………………… (a) Ratio: = Yd = (1-n) G.yw = G > 1, Ysub (1-n) (G_1) Yw G-1 Or, assuming for quartz sand G = 2.65, the ration is (Yd) = 1.606, Y.sub 32 Which means that the dry unit weight of the sand soil is approximately 1.6 times grates grater than the submerged unit weight. Q-8 Derive the “stokes law, for the sedimentation of a soil particle suspended in a Liquid. Ans. Derivation of stocks Law. D = 6p r.m.r B r o W - B D B r o W Let a spherical soil particle hating a mass = m weight = w , radius = r falling with a terminal velocity “V” through a homogeneous liquid that is at rest and of infinitely large volume as compared with the soil particle is subjected to the following forces 1) its weighty “w” (2) boy ant force of a liquid ‘B” (3) effective weight of soil particle = W- B (4) viscous retarding (or) frictional force on a sphere is called drage force “D” the terminal velocity “V” can be calculated from the equation . W= D+B ………………………… (1) 35 2 V= 2/9 g * (Ys-Yw) r2 η = 2/9.g × d 2 4 (Ys- Yw) η V= g/18. (Ys-Yw) d2 ………………………….. (12) η Where. V = h/t h = distance of fall (or) sedimentation height of particle during a time period “t” (or) the sedimentation time “V” Varies W.R.T r2 From (11) r2 = V 2. g. (Ys-rw) 9 η taking under root from both sides r = V* 9/2 * η/g . (Ys-rw) Q-3A. Write a note on following 1. Soil. 2. Formation 3. Particle sizes of soil ? 36 Ans. 1. The term soil according to engineering point of view is defined as the material by means of which and upon which engineers build a structure. The term soil includes the entire thickness of the earth crust i.e. from ground surface to bed rock which is accessible and physible for practical utilization in solving engineering problems. It is also composed loosely bound mineral grains of various sizes cs - Soas eck. age og Tar en Rp eS ey ne Le Ipos ih weal whe latter contains, besides solids, the-liquid phase or s St aaa nt) and the gaseous phase (e.g., air). pies cus TABLE 5-1. SPECIFIC GRAVITIES OF SOME SoILs Nos. Soil Tipes meee Gravity 1 z 3 I 5 2 ae clay 2.34 3 Clay 2.63-2,73-2.81 4 Huras 2.44-2.53-2.92 5 Kaolin 137 : 6 seas 2.47-2.50-2,58 ‘ E 2.65-2.75 1 Line 270 8 Peat 1.26-1.50-1.80 9 Peat, sphagnum, 25% decomposed 0.50-0.70-0.80 10 Quartz sand 2.64-2.65 u , Quarizite 2.65 12 Silt 2.68-2.72 13° Silt with organic admixtures 2,40-2.50 The specific gravity G of dry solids of a soil is defined as the ratio of the density y: of a given volume of the soil solids to the greatest density “yw (att " o of an equal volume of pure water. The density of the substance of the soil particles is "their mass per unit volume of the particles. The density of water at +4°C is 1.000. Therefore, specific gravity is calculated as follows: - ctighesni crams We Is aus ‘srave- number shown itech el i. = Fic. 5-4. Sieving process. (The classification is in accordance with ASTM (1964). . . Reapproved 1972. pp. 69-79.) ete sere oe Soe RANT Colloids Kd rast <0.001 — Sl. Soil Particle Size Analysis. The soil physical make-up, or textural mosaters tn determined hu meane af ite Taste 5-5. SPECIFIC SURFACE AREA OF SOME NEW JERSEY SOILS Agronomic name TRB Soil Specific surface Predominant of soil Classification area, m’/g minerals 1 EG eee tan 4 Merrimac A-I-b za Quartz, feldspar Dunellen A-24 5.7 Quartz, feldspar, 3 : muscovite Montalto + * . A-24 ~ 22.6 Quartz, feldspar . muscovite Penn A-24 22.4 Quartz, feldspar, : muscovite Lansdale ‘ A-7-4 “ATS Quartz,feldspar ‘The specific surface area was determined by the SOR-BET method 37  40 CHAPTER: 7 SOIL MECHANICS Question1: Describe the Standard Proctors Test by the graph for the Compaction of a Soil Sample Ans : Definition: The standard proctor soil compaction test is performed as follows A 6lb soil sample of the soil material passing the No.4 sieve is air-dried, thoroughly mixed, and compacted n a standard compaction cylinder, 4.6 in. high, 4in. in diameter, and 1/30 cu ft. in volume. The soil compaction in the 41 mold is performed in three layers by means of metal rammer, 2inch. in diameter, weighing 5.5lb. Each of the three layers of soil should receive 25 blows from the rammer falling freely from a height of 12in. The net dry weight of the compacted soil is then determined, as well as the compacted moisture content. These two quantities, moisture content and dry density, form a pair of coordinates or a point “I” on a dry density graph, as shown in Fig (A). Compaction is repeated at various moisture contents. Fig A The calculation of the dry density of the compacted sol in terms of its wet density and moisture content is calculated for each test as follows: 42 W = A – B 100% (1) B – C Wd = Ww (2) 1 + (w/100)3 Where w = Percent of moisture by oven-dry weight in the specimen of the soil, A = Weight of dish and wet soil, B =Weight of dish and oven-dried soil, C =Weight of dish, Wd =Dry weight per cubic foot, Ww =Wet weight per cubic foot, The soil moisture contents are plotted on the graph A, as abscissas, and the dry unit weights as ordinates. When the plotted points are connected with a smooth line, a curve is obtained the shape of which is, generally, of a hyperbolic form. In order to determine better the course of the curve, it is advisable to obtain the test points before, at, and past the peak of the curve. The moisture content corresponding to the peak, or maximum ordinate of the curve, is termed the “optimum moisture content” at this stage 100% compaction can be achieved. Question: 2. Define the following Soil compaction, Soil consolidation, Density of soil ,Relative Density ,Densification & Optimum Moisture Ans 45 1 + (Gw/100(1)) 1+Gw/100 l+e = (1+Gw x 1) 100 Where e = void ratio = G.w /100 Question:4 Derive the dry unit weight equation for any constant degree of saturation as a function of moisture content, specific gravity of the soil, and unit weight of the water. Ans. i. Degree of saturation “s” of any soil sample is a sunder: S = Vw = Vv – Va = 1 - Va = 1-na ----- -----@ Vv Vv Vv How we will calculate the value Vw and Vu. ii. Volume of water (VW) in soil: Vw = W.Wd ----------- (b) w iii. Total volume the of voids occupied by air and water in each unit volume of soil: Vv = 1-Vs 1= Wd ---------- -(c) G.w Where G = specific gravity of a soil, Vv = volume of voids wd = dry unit weight of a soil. iv. Putting Equation: C  b in @: S = Vw Vv 46 W .Wd ---------- -(d) w [1–Wd/G.w] v. Rearrangement of equation (d): S = w.Wd ----------- (d) w [G.w – wd] G.w or, = W.Wd Gw - Wd G G S = w.Wd w – _Wd_ G By Cross Multiplication we have: S x (w – Wd) = w.Wdx1 G S.w – S.Wd = w.Wd G Where s = degree of saturation, w = moisture content, w = unit weight of a water Wd= dry unit weight of a soil Subtracting w.wd from both sides we have: S .w – SWd – w.Wd = w.Wd - w.wd G S .w – S.Wd – w.Wd = 0 G 47 S.w - Wd (S + w) =0 (7) G vi. Dry unit of a soil Wd at any moisture content “w” and at any degree of saturation (s) can be obtained from equation:7; Wd = w.S S/G + w Wd = w.S S + w.G G G = Gw. = w.G S + w.G S + w.G S S S W(d) = w.G (e) 1 + w.G S CHAPTER: 8 Question: 1. define the following terms by the line diagram. 1. Consistency 2. Plasticity 50 as plasticity index = P.I = WP.L (plasticity index) = liquid limit – plastic limit. P.I. = wL.L. – w P.L. (%) Explanation: The plasticity index, P.I., indicates the moisture range through which a cohesive soil has the properties of a plastic material. According to Atterberg when P.I. <7, the soil is said to be of low plasticity. When 7 < P.I. <17, the soil is of medium plasticity. When “P.I. > 17” the soil is of high plasticity. All of the Cohesive Soils can be at three different consistencies, depending upon the Moisture Content present . When the moisture content of a soil is greater than the liquid limit i.e when w> wl.l the Soil is in liquid state .When w= wl.l the Moisture Content is at the Liquid Limit of the Soil. When Wp.l < w < Wl.l the Soil is in a Plastic State of Consistency. When Ws.l < w < Wp.l the Soil is in a Semi Solid state of Consistency. When w < Ws.l the Soil is in the solid state of Consistency. Maany Construction Companies in High Way Engineering use the Plasticity Index not to exceed 6, whereas others spicify P.I < 3 There is a very pronounced difference in the Plasticity of some soil materials which are composed of very fine particles., a clay composed of equally fine material will exhibit a marked plasticity . Non Cohesive Soils do not possess plasticity , the limit of which therefore can not be determined by test. 51 Reactive Stress in Soil Question: 3. Describe shrinkage limit of a soil by the diagram as well as by the equation, also explain that why is it necessary to study for Dams and earth work? Ans: When a moist, cohesive soil is subjected to drying out, it loses moisture and shrinks. During the drying process the compressive forces produced by the pore water compress the particles of the skeleton of the cohesive soil together into a compact, coherent mass and thus density it as shown in Fig. Hence, the void ratio of the soil decreases. The moisture loss continues down to certain moisture content. When this moisture content is attained, any further decrease in moisture content ceases, and no further decrease in volume will take place. Meniscus At this point it can be noted that the soil changes its color from dark to light and the soil ceases to shrink. The moisture content at this condition is termed the shrinkage limit. At the shrinkage limit the soil passes from the semisolid to the solid state. The soil shrinkage limits vary from about 10 to 15 percent by dry weight of soil. The shrinkage limit is expressed as follows: wi Wd – w S.L. Wd = (Vi – Vf ) yw3 (1) Surface tension Moisture Film Soil Particle 52 Where: wi Wd = initial amount of soil moisture, in grams, before drying out; wi = initial moisture content, in percent; Wd = dry weight of a soil sample, in grams; w S.L. Wd = amount of moisture, in grams, at shrinkage limit; yw = 1=unit weight of water =1 g/cm 3; Vi = initial volume of a soil sample before drying; Vf = final volume of a soil sample after drying. The final volume Vf is the dry volume Vs. Equation (1), thus, represents an expression of loss in moisture in a shrinking process. From this equation the shrinkage limit, WS.L, is expressed as under. wS.L. = wi – w = wi – (Vi – Vs) yw Wd Where w = moisture loss in percent by dry weight upon drying out the soil sample from its initial moisture content with to the moisture content W.S.L. Which corresponds to the shrinkage limit of the soil? The concept shrinkage limit of cohesive soil is useful in evaluating the behavior of slopes of dams and cuts, particularly relative to the possibility of development of cracks in earthworks. In rainy seasons shrinkage cracks may be filled with water in its turn may slide down. The effects of shrinkage are more pronounced in cohesive soils which have high liquid limit values. 55 classification system would be a system which satisfies best the engineering profession. One of the reasons for the existence of the many soil classification systems is, perhaps, the fact that the geologist, the agronomist, and the engineer have slightly different concepts about the matter of soil. In engineering, soil is defined as the material by means of which & upon which engineers build their structures, as to their performance under load, water, and temperature. Due to different properties of the materials the soil engineers have to work with very complex indeed. Question: 2. What is the “AASTHO”, Soil Classification System? Ans: The “AASTHO” Soil Classification System was presented by the highway representatives. From which it can be noted that all soils are classed in seven groups, from A-1 to A-7. This system classifies soils based on their texture, physical properties (such as liquid limit and plasticity index), and no their expected field performance as sub grade materials for supporting pavements. The highway representatives' system has been adopted as one of the “AASHTO” standards for sub grade soil classification. The “AASHTO” system was appeared in the 1974 and 1976 specifications for a highway material divides the A-1, A-2 and A-7 groups into subgroups. Also, the AASHTO system permits making a relative evaluation of the effect of both coarse and fine soil particles within a given group. 56 A-2 groups are borderline materials between soil groups A-1 and A-3, and silt clay materials of Groups A-4, A-5, A-6, and A-7. It includes all materials containing 35% or less passing the 0.075mm sieve which cannot be classified as A-1 or A-3, due to fines content or plasticity or both, in excess of the limitations for these groups. The two subgroups A-2-4 and A-2-5 include various granular materials containing 35% or less passing the 0.075 mm sieve having the characteristics of the A-4 and A-5 groups. These groups include such materials as gravel and coarse sand with silt contents or plasticity indexes in excess of the limitations of group A-1 and fine sand with nonplastic silt content in excess of the limitation of group A-3. Subgroups A-2-6 and A-2-7 include materials similar to those described under subgroups A-2-4 and A-2-5 except that the fine portion contains plastic clay having the characteristics of the A-6 and A-7 group. The subgroup A-7-5 indicates elastic soil with moderate plasticity indexes, which is subject to considerable volume change. The A-7-6 subgroup comprehends soils with high plasticity indexes and subject to large volume changes. Question: 3. Explain “Group Index” in detail by giving suitable example. Ans: The group index is calculated from the following formula; Group index (G.I) = (F-35) [0.2 + 0.005 (LL -40)] + 0.01 (F-15) (Pl-10) where F = 57 percentage passing through No.200 sieve expressed as a; LL=Liquid limit; Pl= plasticity index. The group index should be reported to the nearest whole number. According to these specifications, when the calculated group index is negative, the group index shall be reported as zero (0). When calculating the group index of A-2-6 and A-2-7 subgroups, only the Pl portion of the formula shall be used. Under average field and construction conditions, drainage and compaction, the good qualities of a sub grade material are rated as being in inverse ratio to its group index. For example, a group index the value of which in zero generally indicates a good sub grade material. A group index the value of which is 20 indicates a poor sub grade material. The group index is used for determining empirically the combined thickness of flexible bituminous or Portland cement concrete pavement, base and sub base courses to be placed on a given sub grade. Question: 4. Write a note on Unconfined Soil Classification system. Ans: This soil classification system is in use by the corps of engineers and by many consultants and agencies. The unified soil classification system was developed jointly by the bureau of reclamation and 60 No project which need excavation below ground surface should be started before getting accurate information about the opponent water. In foundation and highway engineering the opponent water is termed as Public Enemy No 1 (b) Hygroscopic soil moisture content. Soil which appears quite dry must contain an amount of water the so called hygroscopic moisture. In technical literature hygroscopic moisture is also termed as adsorbed moisture, contact moisture or surface moisture. This form of soil moisture in a dense state surrounds the surface of an individual soil particle as a very thin film of water. Hydroscopic moisture is not concern with the ground water therefore it does not take part in the fluctuation on the ground water table nor it does transmit hydrostatic pressure. Hydroscopic moisture film is attached rigidly to the soil particles with some physical force due to these forces the hydroscopic soil moisture film become dense. Hydroscopic moisture can be removed by drying the soil partials at + 105C The difference between the weight of air dried soil sample and its weight after oven dried at +105C determines the amount of Hydroscopic moisture present in a soil. 61 Hydroscopic moisture has very definite effect on the cohesion and plasticity of a clayey. One can see that hydroscopic moisture is present in considerable amount in fine particle soil, this is because of the large internal surface area is such soils. Q NO 2 Describes the following in the detail (1) Moisture film (2) Free surface water (3) Rain fall or Runoff Ans (1) Moisture film. Moisture film forms in the soil due to condensation of water vapor. The moisture film is attached to the surface of the soil particles as film . The moisture film is held by molecular forces of considerable intensity but not as large as in the case of the hygroscopic moisture film. Moisture film is connected to the ground water table but is not affected by the gravity. The moisture film is set in motion or migration by the application to the soil system of an external energy, the moisture in soil then will migrate from places of higher temperature to the places of lower temperatures. Water film water freezes below 0 C depending upon the thickness of the film, the degree of stressed condition and the intensity and duration of freezing. 62 ii) Free surface water. Free surface water condition (precipitation, runoff, melting snow and flood water) should be investigated when a structure comes in direct or indirect contact, or when it influences the ground water in one way or another. For the construction work it is important to determine the amount and the fluctuation of the free water. iii)Rainfall and run off Rainfall and run off are erosive agents that are very destructive because they detach soil and transport it away. The tremendous force of the uncontrolled down slope running water creates channels in unprotected soil. Hence uncontrolled water can wash out road railroad and canal embankment dams and bridge piers foundation of structures. It can destroy coffer dams for foundation work and equipment and construction material or the eroded and transported material may be deposited in reservoirs, thus filling them up with silt and reducing the volume needed to store water for irrigation flood control domestic water supply power and recreation. The freezing point ,the boiling point, the surface tension and the viscosity of the frees surface water correspond to those of ordinary water. 65 Capillary water is that soil moisture which is located within the voids of capillary size of the soil. Capillarity in general is a phenomenon of the rise or depression of liquids in tubes having a bore so fine as to be comparable in diameter with a hair ,. the rise takes place and the liquid is held by means of a force called the surface tension force at the top of the water column in a capillary tube. Capillary movement in soil is the movement of the soil moisture through the minute pores between the soil particles. The minute pores serve as capillary tubes through which the soil moisture rises above the ground water table. Movement results whenever the soil moisture surface tension pulls is increased by loss of water through evaporation from the ground surface. Thus capillary water is hydraulically and continuously connected to the groundwater table or to a perched groundwater table and can be raised against the force of gravity. For capillary rise in soil to exist all of the voids should be completely filled with capillary water. The capillary saturated zone between the groundwater table and the plane of the menisci is called the closed capillary fringe. This contains no air the thickness of this closed capillary fringe depends mainly on the fineness of the soil particles. The larger the pore size (diameter of the capillary) the less the height of rise or the less the capillary fringe. 66 Above the closed capillary fringe there is the so called open capillary fringe i.e. the air containing capillary zone which reaches to the height of the menisci in the finest pores of the soil, here the larger pores are not filled with capillary moisture. Capillary water cannot be drained by means of drainage systems installed within the capillary fringe but it can be controlled by lowering the groundwater table. The drainage system must be installed in the groundwater to pull it down together with the capillary fringe thus controlling the capillary height to which the capillary water can rise. Capillary water can also be removed by heating or by its evaporation at ordinary temperatures. It freezes at about –1C. Capillary mo8isture may translocate in soil in any direction not just vertically upwards. The presence of capillary moisture in cohesive soil decreases considerably its cohesion and stability which transforms the soil into a sticky plastic condition which makes it difficult to work with to compact it or even to mix. The freezing of a capillary saturated soil result in an accumulation of a considerable amount of moisture which causing heaves and other damages to roads. 67 Q NO 5 Describe the capillary rise in a soil ,also explain that why does the inclination of capillary tube not effect on capillary rise? Ans. Capillary Rise The capillary rise takes place by means of a force called the surface tension force of the menisci acting at the top of the water column in a capillary tube between water and the wall surface of the tube . The force which brings the capillarity is also called the capillary force. The curved upper crescent shaped surface of the liquid column in the capillary tube is termed meniscus. This meniscus is concave relative to its top when the liquid wets the wall of the capillary as in the case of water. It is convex relative to its top as in the case of mercury. When the capillary tube is placed in a liquid that wets the walls of its surface the liquid will rise in the capillary above of the free water surface of the liquid into which the capillary is dipped . The height of the capillary rise in the tube depends upon the magnitude of the force of the surface tension pulling the liquid upward, as compared to the force of gravity pulling the liquid column downward. When the two forces are balanced further rise is prevented and a condition of 70 Q NO 6 EXPLAIN THE FOLLOWING (a) Effect of temperature on surface tension (b) Suspended water in a soil. Ans Effect of Temperature on Surface Tension. The surface tension of liquid at room temperatures are of the order of 0.03g/cm. For water the surface tension is more than double this figure and is equal to about 0.08g/cm When we increase the temperature of the liquid the moment of the molecules will increase due to this the force of Cohesion will decrease which becomes a cause for reduction of Surface Tension. Suspended Water in Soil Suspended water in soil is observed in nature in the rapid draw down of a water table for example on emptying a reservoir quickly or on lowering the groundwater table , upon such a lowering the continuity of the capillary fringe is torn off .. Also suspended water in soil can be encounter under the following condition. If the upper layer of silt or peat possesses capillary properties and its capillaries are saturated, then the soil water in the said layer may be hanging from the menisci of the peat. This phenomenon is analogous to the one in which a glass bell B , with a 71 capillary tube C at its top, is immersed into water and pulled out meniscus m-m. the suspension of water from the upper layer of soil possessing capillary properties can take place during lowering of the gravitational water after high tides or floods. In foundation work or highway construction, upon the removal of thee upper capillary silt or peat layer removal of upper menisci water will drain out from the sand layer into the aquifer below the ground water table. The hydrostatic pressure condition for suspended and free water are also shown in Fig. 10/6. The surface tension phenomenon and the ascent of soil moisture are most pronounced in soil composed mainly of fine sands silts or silt clays.The particle size is enough smaller than that of sands of afford considerable height of moisture above the groundwater table and at the same time large enough to provide voids of such a size as to permit rapid translocation of soil moisture through them. This tendency of silt results in the rapid loss of the stability of such a soil under certain conditions QNO 7 WHAT WILL BE THE EFFECT OF SURFACE TENSION ON SOIL MASS ALSO EXPLAIN THE EFFECT OF SURFACE ON LOWERING OF CAPPILARY FRINGE. ANS. Effect of Surface Tension on Soil Mass. 72 At all points where moisture menisci touch soil particles surface tension forces act causing a grain to grain pressure within the soil similar to that of the capillary tube in compression under the surface tension forces. This grain to grain pressure is recognized as the capillary pressure and is also called the intergranular or contact or effective pressure. This intergranuloar pressure tends to force the solid particles together with a pressure equal and opposite to the tension through the water. These compressive stresses on the soil contribute to the strength and stability of the soil mass. This surface tension induced strength of soil under proper condition of particle size, temperature, salinity and other factors. The surface tension induced strength Pc of soil due to temporary and may be destroyed entirely upon the full saturation of soil for example high tides or during flood conditions eliminate the interface menisci and the contact pressure Pc reduces to zero. Lowering of Capillary Fringe. The effect of surface tension forces on the performance of soils is thus very significant. This significance can be cited by another example capillary moisture water cannot be drained out of the soil particularly from silt and clay by any system of drainage installed within the capillary fringe. This is because capillary moisture is held within the soil by 75 constant amount t Pc of capillary pressure (Fid A) for example the vertical stress of a soil (self weight) and the capillary moisture acting on a unit area can be expressed as. If the capillary watering in a soil causes an additional load upon the solid particles, then the effect of this phenomenon on soil would be its tendency to decreases the swelling pressure of a soil. The whole thickness of the soil layer in question is affected by the additional height and weight of the closed capillary fringe by the additional load because it can be imagined that the capillary moisture might be attached and suspended to the surfaces of the menisci. Apparent Cohesion. The capillary or effective pressure as can now be under stood endows the soil with an additional bond which is called the apparent cohesion C. The solid phase of the of the system become more compressed and densified by this pressure , the effect of which is the consolidation of the soil mass identified by this pressure the effect of which is consolidation of the soil material. Apparent cohesion permits the excavation of deep cuts in soil without shoring up the excavation walls. However with an increase in moisture in the 76 voids of the soil the surface tension forces may become destroyed and after the amount of excess water in the soil has reached a certain magnitude, the walls of the excavation slump out and collapse. QNO 9 DISCUSS THE POLARITY OF WATER MOLECULES IN A SOIL ALSO DESCRIBE THE ORIONTATION OF WATER MOLECULE,s Ans Polarity According to Debby a molecule of water can be represented as a system of electrical charges arranged in a polar structure .One would expect that in the polar water molecules electrical poles would exist. Hence water is designated as a polar substance. Fig “B” shows a simplefield diagrammatic representation of the geometrical arrangement of the itoms in a polar water molecules. The plus and the minus poles form a dipole. Therefore generally polar molecules are called dipoles. The polarity of water is very important in that it gives water the ability to support ionization. Ionization is a process which results in the formation of ions in a certain medium in this case water. The polarity of water can thus be considered a link in the mechanism of the soil moisture transfer in the moisture film phase. 77 Orientation of Water Molecules Because of the presence of Electrical charges, the polar water molecules have among themselves a mutual electrostatic interaction and orientation , namely the positive pole of one molecule is attracted to the negative pole of another molecule and vice versa . As a result of mutual interaction, polar molecules of one and the same substance combine among themselves and form complex with ions. Fig A illustrate the concept of dipole water molecules by a positive ion called cation .In the case of Negative ion the attraction and orientation of dipole water molecules is thge converse of that shown in Fig A. 80 Called the potential Energy of that particle . 6. Kinetic Energy: Energy produced due to the velocity of flow of the particle is called kinetic energy. 7. Aquifer: The pervious soil layer is called aquifer. 8. Artesian Water: If the ground water flowing under pressure through a pervious layer of a soil (fully confined from its top & bottom) between impermeable geologic formations, such a ground water is called artesian water. 9. Turbulent Flow A flow in which each liquid particles do not have a definite paths of individual particle also cross each other is called turbulent flow. Question: 3. Define & explain the Darcy’s Law. Ans: definition : The filter velocity “v” of a laminar flow of water is directly proportional to the hydraulic gradient “i” Explaination The flow of seepage water is calculated by means of Darcy’s law of artificial filtration through a uniform, un-stratified soil. The quantity expressing flow velocity in centimeters per second is defined as: V = ki ……………….. (i) 81 FIG:_A It can be seen that the coefficient of permeability k is the filter velocity of laminar flow of water at a hydraulic gradient of i=l. In other words Darcy's law of permeability of saturated soil (S=l) to water is directly proportional to the hydraulic gradient i. This velocity is the ratio of flow with respect to the unit gross cross-sectional area of the soil. The proportionality in Darcy’s law is valid for laminar flow up to a certain critical gradient. i(C) at which the flow velocity is critical, as shown in diagram A. Beyond point L, where i > icr the filtration is turbulent, with a seepage velocity, v> vcr in the turbulent region the seepage velocity can be approximately expressed as v = C i, where C is a coefficient characterizing the seepage medium, the soil, and the turbulence of flow. a water particle possesses energy in the following major forms: 82 1) Potential energy owing to its position or height; 2) pressure energy because of its weight and also the kinetic energy due to its velocity of flow. Question: 4. By Using the Darcy’s Law Prove That Q=Ki At. Ans: A water particle possesses energy in the following major forms: Potential energy owing to its position or height; pressure energy because of its weight and also the kinetic energy due to its velocity of flow. If the velocity of the ground water flow were zero, then all of the water levels in the piezometric tubes would be at the same elevation; With flowing groundwater, the water level in each of the consecutive piezometric tubes is at a lower elevation than in the preceding one. The difference in these pressure heads, h = h1 - h2, between both ends of a streamline represents the loss in potential energy along the length. L’ used up in overcoming frictional resistance to flow in the voids as shown in Fig: “B”. 85 water as shown in Fig (A). The pervious soil layer is termed the aquifer. If a well is drilled into the soil with groundwater in the aquifer, then, depending upon the magnitude of the pressure p at the well point, artesian water will flow up through the well. The height of rise, hf = p/w, of artesian water may be higher than the ground surface. Just like the fountain of a water. When the height of rise of artesian water is less than the elevation of the ground surface, then it is called the semi artesian water. If excavation in soil is carried down so deep that the bottom of the excavation pit is broken through by the pressure of the artesian water, and then usually there is very little one can do to correct the conditions in the spoiled pit. Usually such sites are abandoned and a new site is sought if costly foundation methods are to be avoided. 86 FIG: A Fig B The danger of the presence of artesian water at a construction site relative to excavation and stability of the bottom of the pit is illustrated in Fig. To prevent the bottom of the pit from being broken through by water, the minimum thickness t of the bottom at pressures in equilibrium should be: w h = s t, (1) t = (w / s).h (cm) At an uplift coefficient of  =1. That is, the bottom of the pit, is then considered as absolutely impermeable, so that the bottom surface b-b receives the full, undiminished pressure with an uplift 87 coefficient ( <1.0, “t” then comes out less than (w / s).h If the porosity of the soil is n=30%, then the uplift coefficient ’ is theoretically calculated as.  =1.21.n2/3 This calculation shows the thickness “t” to be less than before namely:  = (1.21) (0.3)2/3 = 0.54 And t =  w/s h = (0.54) (t). Question: 6. Explain Artesian ground water flow and comments that what problems can occur in driving of piles under foundation in the area sites of Aquifer Ground Water. Ans If the groundwater flowing under pressure through a pervious layer of soil is fully confined from its top and bottom between impermeable geologic formations, such a groundwater is termed artesian water as shown in Fig (A). The pervious soil layer is termed the aquifer. If a well is drilled into the soil with groundwater in the aquifer, then, depending upon the magnitude of the pressure p at the well point, artesian water will flow up through the well. The height of rise, hf = p/w, of artesian water may be higher than the ground surface. Just like the fountain of a water. When the height of rise of artesian water is less than the elevation of the ground surface, then it is called the semi artesian water. If excavation in soil is carried down so deep that the bottom of the 90 FIG: A If the groundwater flows through a compound soil layer consisting of several different types of soil, and if each layer is of different permeability perpendicular to stratification of the layered soil system, then an average system's coefficient of permeability can be calculated if the permeability’s k1, k2, k3, ....... . , kn are known. With the symbols as shown in Fig: (A) the average coefficient of permeability kv perpendicular to the bedding planes is calculated on the principle of continuity of flow. Kv= L (cm/s) …… (1) 91 L1 / k1 + L2 / k2 + L3 / k3 +......+ Ln /kn The discharge through perpendicular to stratification in unit cross –sectional area is: q=v=Kv h/L = h x L (cm3/s) L1 / k1 + L2 / k2 + L3 / k3 +......+ Ln /kn x L = h (cm3/s) …… (2) L1 / k1 + L2 /k2 + L3/k3 +......+Ln /kn Question: 8. Write a note on the permeability parallel to the stratification. Ans: The problem, in soil, foundation, and highway engineering compound-layer systems in several strata, rather than single are often encountered as shown in Fig (A) in such a case the permeability to water of each of the particular single component soil layers of the stratified soil system may be, and usually is, different because of the various types of soil. Besides, the permeability perpendicular to the stratified soil layers is different from that parallel to the stratification. it therefore appears necessary to determine a weighted coefficient of permeability for the compound, 92 stratified soil system. + If the groundwater flows through a compound soil layer of different permeability’s parallel to the stratification then all flow lines are also hydraulically parallel to the bound arise of the stratification. Besides, the hydraulic gradient i is the same at every point of each layer, because in this case i does not depend upon the permeability of the soil layer. Therefore, the average flow velocity vh for an L-unit- thick compound layer of soil is calculated as: 95 FIG B Discharge Q=V/t (cm3/s) …… (1) Seepage velocity v=Q/A (cm/s) …..… (2) Hydraulic gradient i=h/L (cm/cm) …… (3) The coefficient of permeability is evaluated by means of Darcy's law per unit cross-sectional area, unit time, and unit gradient as: k=v/i = Q/A.L (cm/s) ………… (4) Also K = Q = Q.L ………… (5) A.h/L A.h PREPARED BY CHECKED BY NAZAR AHMED BHUTTA GHAZANFAR ALI Class: B-Tech (Honor) Civil Subject: Soil Mechanics 96 CHAPTER: 15 SOIL MECHANICS Question: 1. Define & explain the Settlement? Definition: If a clayey, saturated soil is subjected to a structural compressive load, then the volume change in the soil is attributed mainly to the expulsion of water out of the voids of the cohesive soil. Volume changes in the vertical direction bring a settlement of the soil. Explanation: Settlement of soil may be defined as the vertical, downward displacement of soil, brought about by a volume change in the soil due to a decrease in the volume of the voids in the soil, occurring after the beginning of construction. As the soil settles, so does the foundation. Thus the magnitude of the settlement of the foundation is the same as for the soil upon which the foundation rests. In other words, settlement is the sinking of a structure due to a compressive deformation of the underlying soil. If the contact pressure at the base of the footing on the soil is uniform, and uniformly distributed, settlements may be tolerable. In contradiction, non- uniform, or differential settlement of a structure may be disastrous, leading to cracking of the structural member and ultimately the collapse of the structure. This is particularly true with statically indeterminate structures such as continuous beams on more than two supports, frames, arches, and others. In these structures, settlement of a support induces 97 supplemental moments, and of these additional bending moments are not taken into account in proportioning the structural members, the structure may turn out to be too weak to resist the additional moments, and may start to crack. A tall building may lean because of unequal or differential settlements of the soil, and so would two adjacent structures as for example, two oil storage tanks or two towers built next to each other: these tanks would lean toward each other because of the pressure distribution overlap in soil from these two tanks. However, a uniform, tolerable settlement should not be construed as a failure because settlement of a soil as a function of load is a normal physical phenomenon. Uniform settlement is settlement which is brought about when the entire structure, under uniform pressure distribution on a uniform, homogeneous soil material, settles evenly without causing additional stresses in the structure. The term “differential settlement” is used then when some parts of a structure settle more than others. Question: 2. Discuss the various causes of the soil settlement. Ans: Following are the causes of the settlement: 1. Static loads on soil. 2. Dynamic forces from vibrations produced by machinery, traffic, pile driving operations, explosions, earthquakes, and various impacts on soil due to collapse of structures . These factors