Geo-Technical Engineering-I Part-A Unit-01 Introduction

Home , Void ratio

Compiled by: Prof.B.S.Chawhan M.Tech(Geo-Tech Engg), Asst.Professor,CED,Government.Engineering College,Haveri-581110(12/4/2011-Till date) 1 GEO-TECHNICAL ENGINEERING-I PART-A UNIT-01 INTRODUCTION History of Soil Mechanics: The weathered rocky materials found on the earth’s surface, such as sand, clay, silt, gravel, peat, etc., are broadly called as soils; and the science which analysises and defines their actions, properties and behaviors, is called Soil-Mechanics, or Soil- Engineering or Geotechnical Engineering. The science of soil-mechanics is of a recent origin, say as far back as 1925 or so. Prior to that, engineering structures were generally designed and constructed on different kinds of soils, or by different kinds of soils, without considering the peculiarities of the given foundation soil or construction soil, respectively. But, however, nowadays, this science has developed to a large extent. Almost all the engineering structures, founded on the earth (or soil), or structures constructed with the soil (such as earth dams, etc), are thus nowadays designed quite rationally, considering the properties and possible behavior of the existing available soil. Only a few years back, we get these days due to their rational scientific and assured designs. The failures of structures have, thus become very rare. We, therefore, owe a great deal of thanks to the pioneer research workers of this science-the important of whom include W.Fellinius and A.Atterberg-two Swedish scientists, Dr.Karl Terzaghis (who published the very first book on this science in 1925), Dr.Arthur Casagrande, Dr.Glennon Gilboy, Late Donald W.Taylor, Ralph B.Peck, A.W.Skempton, etc., etc. Dr.Terzaghi among them is known as the father of soil mechanics. Karl Terzaghi writing in 1951, on The Influence of Modern Soil Studies on the Design and Construction of Foundations, commented on foundations as follows; Foundation can appropriately be described as a necessary evil. If a building is to be constructed on an outcrop of sound rock, no foundation is required. Hence, in contrast to the building itself which satisfies specific needs, appeals to the aesthetic sense, and fills its matters with pride, the foundations merely serve as a remedy for the deficiencies of whatever whimsical nature has provided for the support of the structure at the site which has been selected. On account of the fact that there is no glory attached to the foundations, and that the sources of success or failure are hidden deep in the ground, building foundations have always been treated as step children; and their acts of revenge for the lack of attention can be very embarrassing. Definition of Soil: The term “soil” has different meanings in different scientific fields. It has originated from the Latin word Solum. According to agricultural scientist “the loose material on the earth’s crust consisting of disintegrated rock with an admixture of organic matter, which supports plant life”. According to geologist “the disintegrated rock material which has not been transported from the place of origin”. According to Civil Engineer “the loose unconsolidated inorganic material on the earth’s crust produced by the disintegration of rocks, overlying hard rock with or without organic matter”. Origin of soil: Soil may remain at the place of its origin or it may be transported by various natural agencies. It is said to be residual in the earlier situation and transported in the latter. Rocks or soils are generally classified into three groups according to their mode of origin- 1.Igneous rocks 2.Sedimentary rocks 3.Metamorphic rocks All rocks are derived from the material of igneous rocks. Infact, 95% of the volume of the outermost 10 miles of the globe is composed of rocks of igneous origin. Sedimentary rocks are formed of hardened, loose sediments and are composed of plant and animal remains and of end product of chemical and physical disintegration of igneous, sedimentary or metamorphic rocks. About 75% of the rocks exposed on the earth’s surface are sedimentary rocks or metamorphic rocks derived from them. Metamorphic rocks are produced when a rock changes from its primary form to a new form by internal process like pressure, heat and chemically active fluids, with the limitation that during the transportation the rock remained essentially solid. Formation of soils: Soils is formed by the process of ‘Weathering’ of rocks, that is, disintegration and decomposition of rocks and minerals at or near the earth’s surface through the actions of natural or mechanical and chemical agents into smaller and smaller grains. The factors of weathering may be atmospheric, such as changes in temperature and pressure; erosion and transportation by wind, water and glaciers; chemical action such as crystal growth, oxidation, hydration, carbonation and leaching by water, especially rainwater with time. Obviously, soils formed by mechanical weathering (that is, disintegration of rocks by the action of wind, water and glaciers) bear a similarity in certain properties to the minerals in the parent rock, since chemical changes which could destroy their identity do not take place.

Compiled by: Prof.B.S.Chawhan M.Tech(Geo-Tech Engg), Asst.Professor,CED,Government.Engineering College,Haveri-581110(12/4/2011-Till date) 2 It is to be noted that 95% of the earth’s crust consists of igneous rocks, and only the remaining 5% consist of sedimentary and metamorphic rocks. However, sedimentary rocks are present on 80% of the earth’s surface area. Felsphers are the minerals, abundantly present (60%) in igneous rocks. Amphiboles and pyroxenes, quartz and micas come next in that order. ‘Leaching’ is the process where by water-soluble parts in the soil such as Calcium Carbonates are dissolved and washed out from the soil by rainfall or percolating subsurface water. ‘Laterite’ soil, in which certain areas of kerala abound, is formed by leaching. Harder minerals will be resistant to weathering action, for example, Quartz presenting igneous rocks. But, prolonged chemical action may affect even such relatively stable minerals, resulting in the formation of secondary products of weathering, such as clay minerals-illite, kaolinite and montmorillonite.

The following figure shows a typical soil profile.

A-horizon 30 to 50cm

B-horizon 60 to 100cm

C1-horizon 3 to 4m

C2-hirizon below 4 to 5m.

The above figure shows a typical soil profile. Fig.1. A typical soil profile Where, A-Light brown loam; leached, B-Dark brown clay, leached C1-Light brown silty clay oxidized and unleached. C2-Light brown silty clay unoxidized and unleached. SOME COMMONLY USED SOIL DESIGNATIONS: The following are some commonly used soil designations, their definition and basic properties; 1. Bentonite: Decomposed volcanic ash containing a high percentage of clay mineral montmorillonite. It exhibits high degree of shrinkage and swelling. 2. Boulder clay: Glacial clay containing all sizes of rock fragments from boulders down to finely pulverized clay minerals. It is also known as ‘Glacial till’. 3. Black cotton soil: Black cotton soil containing a high percentage of montmorillonite and colloidal material; exhibits high degree of shrinkage and swelling. The name is derived from the fact that cotton grows well in the black soil. 4. Caliche: Soil conglomerate of gravel, sand and clay cemented by calcium carbonate. 5. Hard pan: Densely cemented soil which remains hard when wet. Boulder clays or glacial tills may also be called hard pan- very difficult to penetrate or excavate. 6. Laterite: deep brown soil of cellular structure, easy to excavate but gets hardened on exposure to air owing to the formation of hydrated iron oxides. 7. Loam: Mixture of sand, silt and clay size particles approximately in equal proportions; sometimes contains organic matter. 8. Loess: Uniform wind-blown yellowish brown silt clay; exhibits cohesion in the dry condition, which is lost on wetting. Near vertical cuts can be made in the dry condition. 9. Marl: Mixtures of calcareous sands or clays or loam; clay content not more than 75% and lime content not less than 15%. 10. Moorum: Gravel mixed with red clay. 11. Top-soil: Surface material which supports plant life. 12. Verved Clay: Clay and silt of glacial origin, essentially a lacustrine deposit; varve is a term of Swedish origin meaning thin layer. Thicker silt varies of summer alternate with thin clay varves of winter.

Compiled by: Prof.B.S.Chawhan M.Tech(Geo-Tech Engg), Asst.Professor,CED,Government.Engineering College,Haveri-581110(12/4/2011-Till date) 3 Sizes of soils: Gravel 80mm to 4.75mm Sand 4.75mm to 0.075mm Silt 0.075mm to 0.002mm Clay less than 0.002mm.

PHASE DIAGRAM:

Va Air Wa

Vv V Water Wv Vw Ww

Ws Solid 1=Vs (Soil)

Fig:(2) Soil- three phase diagram representing a soil mass (Partially saturated),(Volumes and weights of phases).

Va=volume of air Wa=Weight of air Vw=Volume of water Ww=Weight of water Vv=Volume of voids Wv=Weight of material occupying void space Vs=Volume of solids Ws=Weight of solids V = total volume of soil mass W =Total weight of soil mass ,Wv=Ww. Porosity (n): Porosity of a soil mass is the ratio of the volume of voids to the volume of the soil mass. i.e., n=Vv X100 (1) V Where,Vv =Va+Vw; V =Va+Vw+Vs Void ratio (e): Void ratio of a soil mass is defined as the ratio of the volume of voids to the volume of solids in the soil mass. It is denoting by the letter symbol e and is generally expressed as a decimal fraction; i.e., e = Vv (2) Vs Where, Vv= Va+Vw Void ratio is used more than porosity in soil mechanics to characterize the natural state of soil. Degree of saturation (Sr): Degree of saturation of a soil mass is defined as the ratio of the volume of water in the voids to the volume of voids. It is designated by the letter Sr and is commonly expressed as a percentage. i.e., Sr = Vw X100 (3) Vv Where,Vv = Va+Vw for a fully saturated soil mass, Vw=Vv. Therefore, for a saturated soil mass Sr=100%. For a dry soil mass, Vw is zero. Therefore, for a perfectly dry soil sample Sr is zero. In both these conditions, the soil is considered to be a two-phase system. The degree of saturation is between zero and 100%, the soil mass being said to be ‘perfectly’ saturated-the most common condition in nature. Percentage Air Voids (ηa): ‘A Percentage air void of a soil mass is defined as the ratio of the volume of air voids to the total volume of the soil mass. It is denoted by the letter symbol ηa and is commonly expressed as a percentage. i.e., ηa= Va X100 (4) V

Compiled by: Prof.B.S.Chawhan M.Tech(Geo-Tech Engg), Asst.Professor,CED,Government.Engineering College,Haveri-581110(12/4/2011-Till date) 4 Air Content (ac): ‘Air content’ of a soil mass is defined as the ratio of the volume of air voids to the total volume of voids. It is designated by the letter symbol ac and is commonly expressed as a percentage. i.e., ac= VaX100 (5) Vv Water(Moisture) content(w): ‘Water content’ or ‘Moisture content’ of a soil mass is defined as the ratio of the weight of water to the weight of solids(dry weight) of the soil mass. It is denoted by the letter symbol w and is commonly expressed as a percentage. i.e., w= Ww X100 (6) Ws(or Wd) =(W-Wd) X100 (6a) Wd Unit weight of Solids (γs): ‘Unit weight of solids’ is the weight of soil solids per unit weight volume of solids alone. It is also 3 sometimes called the absolute unit weight of a soil. It is denoted by the letter symbol γs and it is expressed interms of gm/cc or t/m or KN/m3. i.e., γs = Ws (7) Vs Bulk(mass) Unit weight(γ): ‘Bulk unit weight’ or ‘Mass unit weight’ of a soil mass is defined as the weight per unit volume of the soil mass. It is denoted by the letter symbol γ and it is expressed interms of gm/cc or t/m3 or KN/m3. i.e., γ = W (8) V W=Ww+Ws V=Va+Vw+Vs The term’ density’ is loosely used ‘unit weight’ in soil mechanics, although, strictly speaking, density means the mass per unit volume and not weight. Unit weight of water (γw): ‘Unit weight of water’ is the weight per unit volume of water. It is denoted by the letter symbol (γw) and it is expressed interms of gm/cc or t/m3 or KN/m3. i.e., γw = Ww (9) Vv 0 It has a convenient value @ 4 C, which is the standard temperature for this purpose. γo is the symbol used to denote the unit 0 weight of water @ 4 C. 3 3 3 The value of γo is 1gm/cm or 1000kg/m or 9.81KN/m .

Saturated Unit Weight (γsat): The saturated unit weight is defined as the bulk unit weight of the soil mass in the saturated condition. 3 3 This is denoted by the letter symbol γsat and it is expressed interms of gm/cc or t/m or KN/m . Submerged (Buoyant) Unit weight (γ' or γsub): The submerged unit weight or Buoyant unit weight is defined as it is the submerged weight of soil solids (Ws)sub per unit of total volume of the soil. It is denoted by the letter symbol γ' or γsub. γ' or γsub.= (Ws)sub (10) V (Ws)sub is equal to the weight of solids in air minus the weight of water displaced by the solids. This leads to; (Ws)sub=Ws-Vs.γw (11)

(Ws)sub =(W/V)- γw……….(dividing by V on both sides) V γ'= γsat- γw (12) It may be noted that a submerged soil is invariably saturated, while a saturated soil need not be submerged. The equation (12) may be written as a direct consequence of ‘Archimedes’ principle which states that the apparent loss of weight of a substance when weighed in water is equal to the weight of water displaced by it. Thus, γ'= γsat- γw, since these are weights of unit volumes. Dry Unit weight (γd): The dry unit weight is defined as the weight of soil solids per unit of total volume. The dry unit eight is denoted by the letter symbol γd and is given by; γd=Ws(orWd (13) V Specific Gravity of Solids (G): The specific gravity of soil solids is defined as the ratio of the weight of solids to the unit weight of water at the standard temperature(4oC). This is denoted by the letter symbol G and is given by G= γs/ γw (14)

Compiled by: Prof.B.S.Chawhan M.Tech(Geo-Tech Engg), Asst.Professor,CED,Government.Engineering College,Haveri-581110(12/4/2011-Till date) 5 CERTAIN IMPORTANT RELATIONSHIPS:

Va Air Wa

Vv V Water Wv Ww Vw

Ws Solid (Soil) 1=Vs

Fig: (3) Soil- three phase diagram (Volumes and weights of phases) showing additional equivalents on the weight side.

Jan/Feb/2006 1. Derive relationship between void ratio, water content, specific gravity and degree of Saturation. Solution: Procedure. 1. Draw the three phase soil system diagram (As shown in figure above (3). 2. By definition; w =Ww/Ws, as fraction…….(a) Sr =Vw/VV, as fraction……. (b) e =Vv/Vs……------(c) 3. Sr.e =Vw/VvXVv/Vs=Vw/Vs… (1) 4. w =Ww/Ws=Vw.γw/ Vs.γs = Vw.γw/ VsG.γw------G=γs/γw =Vw/Vs.G=e.Sr/G------(From eqn.1)

w.G=e.S r (1)

(Note: This is valid even if both and Sr are expressed as percentage). For saturated condition, Sr=1. Wsat=e/G or e=Wsat.G (2) July/2005 and Jan/2007 2. Considering soil as a three phase system, derive an expression for the density of the soil in the form γd=(1- ηa)G γw/1+wG with usual notations. Solution: Procedure. 1. Draw the three phase soil system diagram (As shown in figure above (3)). 2. By definition; V=Va+Vw+Vd……….(a) 3. Replace volume of water by weight ,of equation(a) V=Va+Ww/ γw+Wd/ γs 4. Dividing by V, we get V/V=Va/V+Ww/V.γw+Wd/V.γs 1= Va/V+w.Wd/V.γw+Wd/V.γs 1= ηaw.Wd/V.γw+Wd/V.γs………… (ηa=Va/V). 1- ηa= w.Wd/V.γw+Wd/V.γs =w. γd/ γw+ γd/ γs = w. γd/ γw+ γd/ G.γw…………….. (G= γs/ γw) = γd/ γw(w+1/G) γd =(1- ηa) γw/(w+1/G) Multiplying G on right side top & bottom

γ d =(1- η a )G γ w /1+wG (1)

Compiled by: Prof.B.S.Chawhan M.Tech(Geo-Tech Engg), Asst.Professor,CED,Government.Engineering College,Haveri-581110(12/4/2011-Till date) 6 Jan/2005 3.With usual notation, show that γb= (G+Se)/(1+e) γw. Solution: Procedure. 1. Draw the three phase soil system diagram (As shown in figure above (3). 2. By definition; γb=Ws+Ww/V=(Vsγs+w.Ws)/Vs(1+e) where,(V=Vs+e=Vs+Vv/Vs=1+e=Vs(1+e) by L.C.M………..Vs=1 = Vs.G.γw+w.Ws/Vs (1+e)------G =γw/γs =Vs.G.γw +w.Ws =Gγw + w.γs =Gγw +w.G.γw =Gγw(1+w) =(G+wG)γw=(G+e.Sr)γw Vs(1+e) Vs(1+e) (1+e) (1+e) 1+e 1+e 1+e 1+e 1+e γ =(G+eS) γ /1+e (1) b w July/2004 4.With the usual notations, derive from first principles an expression for Sr=w/ [γw/ γb(1+w)-1/G. Solution: Procedure. 1. Draw the three phase soil system diagram(As shown in figure above(3)). 2. We know that; γb=G.γw(1+w)/1+e (1) 3. e=wG/Sr (2) 4.γb=Gγw (1+w) /1+wG/Sr 5. Riversing both sides, γw(1+w)=1+wG/Sr γb G =1+ w ------(dividing by G on top & bottom of RHS) G Sr Or w= γw(1+w)- 1 Sr γb G

S r= w (3) γw(1+w)- 1 γb G Jan/2003 1. Derive an expression for buoyant unit eight of soil in the form γ'=(G-1) γw/1+e Solution: Procedure. 1. Draw the three phase soil system diagram (As shown in figure above(3)). 2. The submerged unit weight γ' may be written as; γ'= γsat- γw = [(G+e) γw/1+e]- γw = γw[G+e -1] 1+e

γ '=(G-1) γ w (1) 1+e Jan/2006 1. How many cubic meters of soil can be formed with void ratio of 0.5 from 100m3 of soil having void ratio 0.7? Solution: Given data a. V=? for e=0.5 b. V=100m3 of e=0.7 To find: V=? Procedure: 1. Draw a three phase soil diagram.

Compiled by: Prof.B.S.Chawhan M.Tech(Geo-Tech Engg), Asst.Professor,CED,Government.Engineering College,Haveri-581110(12/4/2011-Till date) 7

Va Air Wa

Vv V Water Wv Ww Vw

Ws Solid (Soil) 1=Vs

Fig: (4) Soil three phase diagram.

2. By considering above figure; Vs=1 3 V=Vs+e=1+0.7=1.7m ….(Assume block S/A=1X1units) 3. When the soil is put to formed, Vs=1, V=1+0.5=1.5m3 4. For every 1.7m3 of soil excavated to form of 1.5 m3. e.Therefore, total quantity of soil can be formed is =1.5/1.7X100………..(1.7m3---100m3 1.5m3----? ) =88.235m3. July/2005 and Jan/2007. 2. The maximum and minimum dry density weights of sand determined in the laboratory are 20KN/m3 and 15KN/m3 respectively. If the relative density of the sand is 74%, determine the in-situ porosity of the sand deposit. Take G=2.6. Solution: Given data 3 γdmax= 20KN/m 3 γdmin= 15KN/m ID= 74% G= 2.6 To find: n=? Procedure: 1. We have; γd= Gγw or 1+e γdmax= G γw 1+emax emax= G γw -1 γdmax =2.6X9.81 -1=0.2753 20

2.γdmin=G γw 1+emin emin= G γw -1 γdmin =2.6X9.81 -1=0.7004 15 3. We have; ID = emax-e emax-emin

Compiled by: Prof.B.S.Chawhan M.Tech(Geo-Tech Engg), Asst.Professor,CED,Government.Engineering College,Haveri-581110(12/4/2011-Till date) 8 -e=IDX(emax-emin)-emax

-e=0.74X(0.2753-0.7004)-0.2753 -e=-0.589874 e=0.589874 4. In-situ porosity n; n= e =Vv 1+e V n= 0.589874/1+0.589874 n=0.371 or 37.10% Jan/2005 3. A fully saturated soil sample has a water content of 27% and bulk unit weight of 19.5KN/m3. Compute: i) Dry unit weight, ii) Specific gravity of soil solids, iii) Void ratio and iv) Porosity. Solution: Given data 3 a.w=27%, b. γb= 19.5KN/m . c. Sr= 1…..(fully saturated) To find : a.γd=?, b. G=?, c. e=?, d. n=? 1. Dry density: 3 γd= γb/1+w=19.5/1+0.27=15.354KN/m . 2. Specific gravity: G= γs/γw=15.354/9.81=1.565 3. Void ratio: e =wG/Sr=0.27X1.565/1=0.422 4. Porosity: n= e/1+e=0.422/1+0.422=0.297 or 29.71% July/2004 4. A partially saturated soil from an earth fill has a natural water content of 22% and a bulk unit weight of 18.5KN/m3. Assuming that the specific gravity of soil solids is 2.67, compute the degree of saturation and void ratio. If subsequently the soil gets saturated, determine the dry unit weight, buoyant unit weight and saturated unit weight. Solution: Given data 3 a. w=22%, b. γb=18.5KN/m , c. G=2.67. To find; i.a. Sr=?, b. e=? ii.a.Sr=1. To find; a. γd=?, b. γ'=?, c. γsat=? Procedure: 1st case; 1. Dry density; 3 γd= γb/1+w=18.5/1+0.22=15.163KN/m . 2. We have; γd= Gγw/1+e 1+e= Gγw/ γd e= (Gγw/ γd)-1 e =(2.67X9.81/15.163)-1 e=0.727 3. Degree of saturation; Sr=wG/e =0.22X2.67/0.727 =0.8076 Or Sr= 80.76%. 2nd Case; 1. Dry density; 3 γd= γb/1+w=18.5/1+0.22=15.163KN/m . 2.γsat= (G+e) γw 1+e 3 γsat =(2.67+0.727)X9.81= 19.296KN/m . 1+0.727 3. Buoyant unit weight; 3 γ'= γsat- γw=19.296-9.81=9.486KN/m .

Compiled by: Prof.B.S.Chawhan M.Tech(Geo-Tech Engg), Asst.Professor,CED,Government.Engineering College,Haveri-581110(12/4/2011-Till date) 9 Jan/2004 5. A soil sample weighing 16.5KN/m3 has a water content of 28.5%. The specific gravity of soil particle is 2.7. Determine i). dry unit weight, ii) void ratio, iii) porosity & iv) degree of saturation. Solution: Given data: 3 a. γb=16.5KN/m , b. w=28%, c. G=2.7 To find; a. γd=?, b. e=?, c.n=? and d.Sr=? Procedure: 1. Dry density; 3 γd= γb/1+w=16.5/1+0.28=12.89KN/m . 2. Void ratio: γd= Gγw 1+e 1+e= Gγw/γd e= (Gγw/γd)-1 e=(2.7X9.81/12.89)-1 e=0.054 3. Porosity; n=e/1+e=1.054/1+1.054=0.513 or 51.35% 4. Degree of saturation; Sr=wg/e Sr=0.28X2.7/1.054=0.7172 or 71.72% Jan/2003 6. A sample of clay soil of volume 1x10-3m3 and weight 0.0174KN, after being dried out in an oven had a weight of 0.01368KN. If the specific gravity of the particles was 2.69, find for the sample in its original condition. i) the water content, ii) the void ratio, iii) the saturated unit weight and iv) the dry unit weight. Solution: Given data; -3 3 a.V=1X10 m , b. Wt= 0.017KN(wet), c. Wd= 0.01368KN, d. G=2.69 To find: a. w=?, b.e=?, c. ηa=? , d. γsat=? and e.γd=?. Procedure: 1. Water content(w); w= Weight of water X100 Weight of dry soil w=(0.01762-0.01368) X100 =28.80% 0.01368 2. Bulk unit weight (γb); -3 3 γb= Wet weight= 0.01762/1X10 =17.62KN/m Total volume 3. Dry unit weight(γd); γd= γb/1+w 3 γd=17.62/1+0.2880=13.68KN/m . 4. Void ratio(e); γd= Gγw 1+e 1+e= Gγw/γd e= (Gγw/γd)-1=(269X9.81/13.68)-1=0.929 5. The air voids(ηa); ηa=n.ac where, ac=1-Sr Sr=wG/e=0.288X2.69/0.929=0.8339 or 83.39% ac=1-0.8339=0.1601 n=e/1+e=0.929/1+0.929=0.4815 or 48.15% ηa=0.4815X0.1601=0.0771 or 7.71%. 6.The saturated unit weight(γsat); 3 γsat= (G+e) γw= (2.69+0.929)X9.81 = 18.40KN/m 1+e 1+0.929 7. An earth embankment is to be compacted to a density of 19KN/m3 at a moisture content of 14percent. The in-situ density and water content of the borrow pit are 18KN/m3 and 8% respectively. How much excavation should be carried out from the borrow pit for each m3 of the embankment?. Solution: Given data

Compiled by: Prof.B.S.Chawhan M.Tech(Geo-Tech Engg), Asst.Professor,CED,Government.Engineering College,Haveri-581110(12/4/2011-Till date) 10 a. Borrow pit a. Borrow pit 3 γb=18KN γb=19KN/m w=8 w=14%. To find; a. Quantity of soil (m3)=? From borrow pit Procedure: 1.γd= γb/1+w 1. γd= γb/1+w 3 3 γd=18/1+0.08=16.67KN/m . γd=19/1+0.14=16.67KN/m . 2. The excavation from borrow pit per m3 of embankment is γd(embankment) =16.67 =1.0 γd(embankment) 16.67 3.γd (embankment) = γd(borrow pit) 1m3 (emb)=1m3 (borrow pit) Dec/2010 8. The maximum and minimum dry unit weights of sand, determined in the laboratory are 21kN/m3 and 16kN/m3 respectively. If the relative density of sand is 60%, determine insitu porosity of sand deposit. Take G=2.65. Ans: a.emax=0.625, b.emin=0.238, c.e=0.3928 and d.n=28.20% May/June-2010 9. A fully saturated sample has a water content of 25% and unit weight of 20kN/m3. Calculate: i. Dry unit weight ii. Specific gravity iii. Void ratio iv. Porosity and 3 v. Its unit weight, when it has S=50%. Use γw=10kN/m . 3 3 Ans: a.γd=16kN/m , b.G=2.67, c.e=0.67, d.n=40% and,γ=18kN/m . Dec.09/Jan.10 10. A soil sample weighing 19kN/m3 has a water content of 30%. The specific gravity of soil particles is 2.68. Determine void ratio, porosity and degree of saturation. 3 Ans: a.γd=14.415kN/m , b.e=0.7988, c.Sr=1, and d.n=44.40%. Dec.08/Jan.09 11. In an earth embankment under construction the bulk unit is 16.50kN/m3 at water content 11%. If the water content has to be increased to 15%, compute the quantity of water to be added per cubic meter of soil. Assuming no change in void ration, determine the degree of saturation at this water content. Take Gs=2.70. Ans. Given data 3 a.γ=16.50kN/m , b.w=11%, c.wincreased=15%, d.Gs=2.70 3 To find: a.Quantity of water=? In weight and m , b.Sr=? Procedure: 3 1. γd=γb/1+w=14.865kN/m 3 2. ww/ws= γd------(ws=1m ) ww=14.865kN 3. ww/ws=11%. 4. ww=0.11x14.865=1.635kN 5. When water content=15%, then 6. ww/ws=0.15, then ww=0.15x14.865=2.30kN 7. Weight of water required to be added to increase water content from 11% to 15%, then =(2.230-1.635=0.595kN 3 8. Volume of water to be added=ww/γw=0.595/9.81=0.0606m 9. γd=G.γw/1+e 14.865=2.70x9.81/1+e, then e=0.782 10. Sr=wG/e=0.15x2.7/0.782x100=51.81% 2002Scheme-May/June-2010 12. A soil sample had a porosity of 40%. The specific gravity of solids is 2.70. calculate i. Void ratio ii. Dry density iii. Unit weight, if the soil is 50% saturated iv. Unit weight, if the soil is completely saturated. 3 3 3 Ans. a. e=0.667, b. γd=15.89kN/m ,c.w=0.124,d.γ=17.85kN/m ,e.wsat=0.247, and f. γsat=19.81kN/m . 13. A sample of soil has volume of 1x106mm3 and weights 19N. The water content of the soil 15% and G=2.7. Find i) Dry unit 3 weight ii) Void ratio iii) degree of saturation iv) water content at saturation. (γw=9.81kN/m ). 6 -3 -3 3 3 Ans.a.γbulk=W/V=19N/1x10 =19x10 /1x10 =19kN/m , b.γd=16.52kN/m , e.=0.603,Sr=67.15%,w=22.33%.

Compiled by: Prof.B.S.Chawhan M.Tech(Geo-Tech Engg), Asst.Professor,CED,Government.Engineering College,Haveri-581110(12/4/2011-Till date) 11

14. The mass of a moist soil sample having a volume of 0.0057m3 is 10.5kg. The moisture content (w) and the specific gravity of soil solids (Gs) were determined to be 13% and 2.68 respectively. Determine a. Moist density, ρ in 3 3 kg/m , b.Dry density, ρd in kg/m .,c.Void ratio (e), d.Porosity (n) and f.Degree of saturation.

3 3 15. For a saturated soil, given w=40% and Gs=2.71, determine the saturated and dry unit weight in lb/ft and KN/m .

16. The mass of a moist soil sample collected from the field is 465gms, and its oven dry mass is 405.76gms, the specific gravity of the soil solids was determined In the laboratory to be 2.68. If the void ratio of the soil in the natural state is 0.83, find 3, 3 the following; a .The moist density of the soil in the field in kg/m b. The dry density of the soil in the field in kg/m , c.The mass of water in kilograms, to be added per cubic meter of soil in the field for saturation.

17. The maximum and minimum dry density weights of sand determined in the laboratory are 22KN/m3 and 17KN/m3 respectively. If the relative density of the sand is 76%, determine the in-situ porosity of the sand deposit. Take G=2.67.

18. There are two borrow areas A and B which have soils with void ratios of 0.80 and 0.70, respectively. The inplace water content is 20%, and 155, respectively. The fill at the end of construction will have a total volume of 10,000m3, bulk density of 2Mg/m3 and a placement water content of 22%. Determine the volume of the soil required to be excavated from both areas. G=2.65. 19. a compacted cylindrical specimen, 50 mm dia and 100 mm length, is to be prepared from oven-dry soil. If the specimen is required to have a water content of 155 and the percentage air voids of 20%, calculate the mass of the soil and water required for the preparation of the sample. Take G=2.69.

20. A sample of clay was coated with paraffin wax and its mass, including the mass of wax, was found to be 697.5gm. The sample was immersed in water and the volume of the water displaced was found to be 355ml. The mass of the sample without wax was 690.0gm, and the water content of the representative specimen was 18%.

21. A moist soil sample weighs 3.52N. After drying in an oven, its weight is reduced to 2.9N. The specific gravity of solids and the mass specific gravity are, respectively, 2.65 and 1.85. Determine the water content, void ratio, porosity and the degree 3 of saturation. Take γw=10kN/m .

22. A soil has a porosity of 40%, the specific gravity of solids of 2.65 and a water content of 12%. Determine the mass of water required to be added to 100m3 of this soil for full saturation.

23. The dry unit weights of a sand in the loosest and densest states are respectively 1.36 and 2.18gm/cc. Assuming the specific gravity of solids as 2.67, determine the relative density of sand with porosity of 30%.

24. An earthen dam containing 60million cubic metre of soil at 0.8void ratio is to be constructed. There are four available borrow pits, which are designated as 1,2,3 and 4. The void ratio of the soil in each pit, and estimated cost of moving the pit soil to the dam site are given below in table below.

Pit (1) Void ratio (2) Cost of moving per cubic metre of soil Rs. (3) 1 0.90 2.00 2 1.20 1.80 3 1.60 1.70 4 2.00 1.60

What will be the least earth-moving cost?.

25. 3.3lakh cubic metre of soil is removed from a site. This dry soil has an insitu void ratio of 1.20. (a0 How many cubic metre of a fill having a void ratio of 0.70 could be constructed from this? (b) With 2.70 as the specific gravity of the soil particles, what will be the weight of the soil moved?.

GOOD-LUCK