[ [GEOTECHNICAL ENGG-1] SOLVED QUESTION PAPER]

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1 1 a) With the help of three phase diagram, define the terms bulk density, dry density, void ratio and water content. Bulk Density-It is defined as the ratio of total weight to total volume of soil mass. Dry density -It is defined as the ratio of weight of soil solids to total volume of soil mass. Void Ratio-It is defined as the ratio of volume of voids to volume of solids. Water Content-It is defined as ratio of weight of water to weight of solids. 1 b) with usual notation prove that Se=WG Useful weight-volume relations can be developed by considering a soil mass is which the volume of soil solids is unity, as shown in Figure. Since from the definition of void ratio given in Eq. the volume of voids is equal to the void ratio, e. the weight of soil solids can be given by Dept of Civil Engg Page 1

2 i. c) A soil sample weighing 19 kn/m 3 has a water content of 30%. The specific gravity of soil particles is Determine void ratio, porosity and degree of saturation. Dept of Civil Engg Page 2

3 2) a) What is consistency of soil? List and define consistency limit. Consistency of soil-consistency is the relative ease with which soil can be deformed. It is Applicable to fine grained soils whose consistency depends on water content. List of consistency limit. Liquid limit Plastic limit Shrinkage limit Definition Liquid limit- It is the water content corresponding to an arbitrary limit between liquid and plastic states of consistency of a soil. It is minimum water content at which soil is still in liquid state, but possessing small shear strength and exhibiting some resistance to flow. Plastic limit - It is the water content between plastic and semi solid state. It is a minimum water content at which soil will just begin to crumble when rolled in to a thread of approximately 3 mm diameter Shrinkage Limit it is the water content between semi solid and solid state. It is the lowest water content at which soil is in fully saturated condition. It is the maximum water content at which any reduction in water content will not reduce the volume of soil mass. 2 b) Briefly explain the correction to be applied to a hydrometer reading. The recorded hydrometer reading may require the following corrections depending in the situation. Correction for meniscus- The solution containing soil will be opaque. Hence, it is difficult to obtain reading corresponding to lower meniscus. Rh is taken to upper meniscus and correction is applied. The correction is always positive. Correction for temperature- Hydrometer is calibrated at 27 o C. If the laboratory temperature is higher, the liquid will be less viscous and hydrometer sinks more. Rh will be less than actual needing positive correction. Similarly, low laboratory condition requires negative correction. Dept of Civil Engg Page 3

4 Correction for dispersing agent-a dispersing agent is added to water to disperse the soil particles and allow them to settle individually. Dispersing agent increases the density of water and hydrometer floats higher. Therefore the correction is always negative. 2 c) Following results were obtained from a liquid limit test on a clay sample, whose plastic limit is 20% Number if Blows Water content (%) Plot the curve and obtain: I. liquid limit II. III. plasticity index flow index From the graph liquid limit is % Plasticity index = liquid limit plastic limit = = 28.51% Dept of Civil Engg Page 4

5 3 a) Explain soil classification as per IS classification system. Particle Size classification (IS) Gravel-Retain on 4.75mm sieve Coarse sand-passing 4.75mm and retain on 2mm Medium sand passing 2mm and retain on 425micron Fine sand passing 425micron and retain on 75micron Silt-passing 75micron and retain on 2micron Clay-particle size less than 2micron. Indian Standard Classification (ISC) System Broadly classified into 3 division Coarse grained soil-50% or more of the total material by weight is retained on 75micron sieve. Fined grained soil-more than 50% of soil passing 75micron sieve Organic soil-consist of decomposed vegetation. Classification of coarse grained soil Coarse grained soils are sub divided into gravel (G) and sand (S) The soil is termed as gravel when more than 50% of coarse fraction is retained on 4.75mm sieve Dept of Civil Engg Page 5

6 The soil is termed as sand when more than 50% of the coarse fraction is passing 4.75mm sieve. Case-1-Fines less than 5% GW-well graded gravel GP-poorly graded gravel SW-well graded sand SP-poorly graded sand Cu > 4, Cc -1 to 3, gravel > sand Cu, Cc not in range, gravel > sand Cu > 6, Cc -1 to 3, sand > gravel Cu, Cc not in range, sand > gravel Case-2-Fines between 5% - 12% GW-GM-well graded gravel, contain silt Cu > 4, Cc -1 to 3, gravel > sand, silt > clay GW-GC-well graded gravel, contain clay Cu > 4, Cc -1 to 3, gravel > sand, clay > silt GP-GM-poorly graded gravel, contain silt Cu, Cc not in range, gravel > sand, silt > clay GP-GC-poorly graded gravel, contain clay Cu, Cc not in range, gravel > sand, clay > silt SW-SM-well graded sand, contain silt Cu > 6, Cc -1 to 3, sand > gravel, silt > clay SW-SC- well graded sand, contain clay Cu > 6, Cc -1 to 3, sand > gravel, clay > silt SP-SM-poorly graded sand, contain silt Cu, Cc not in range, sand > gravel, silt > clay SP-SC-poorly graded sand, contain clay Cu, Cc not in range, sand > gravel, clay > silt Case-3-Fines >12% GC-clayey gravel gravel > sand, clay > silt, Ip > 7% GM-silty gravel gravel > sand, clay < silt, Ip < 7% SC-clayey sand sand > gravel, silt <clay, Ip >7% SM-silty sand sand > gravel, silt > clay, Ip < 7% Classification of Fine Grained Soil Based On Plasticity Chart Case-1 Liquid Limit (> 50%) CH-high plastic clay MH-high plastic silt OH-high plastic organic soil Dept of Civil Engg Page 6

7 Case-2 Liquid Limit (35% - 50%) CI-medium plastic clay MI- medium plastic silt OI- medium plastic organic soil Case-3 Liquid Limit (0-35%) CL-low plastic clay ML- low plastic silt OL- low plastic organic soil 3 b) Explain any two fields test to identify silt from clay. Dilatancy test: It consists of placing a pat of moist soil in the palm of the hand, and then shaking the hand. If a shiny, moist surface appears on the soil after shaking it in the open hand and then becomes dull and dry when the pat is squeezed by closing the hand, a non- plastic soil (e.g. silt) is indicated. If it is clay, the water cannot move easily and hence, it continues to look dark. If it is a mixture of silt and clay, the relative speed with which the shine appears may give a rough indication of the amount of silt present. This test is also known as shaking test or water mobility test. Rolling test: A thread is attempted to be made out of a moist soil sample with a diameter of about 3-mm. If the material is silt, it is not possible to make such a thread without disintegration and crumbling. If it is clay, such a thread can be made even to a length of about 30-cm and supported by its own weight when held at the end. This is also called the Toughness test. Dept of Civil Engg Page 7

8 3 c) With neat figure explain the structure of three clay minerals. Kaolinite Structural sheet consist of an alumina sheet (gibbsite) combined with silica sheet. Thickness of one structure unit-7a 0 This mineral is formed by stacking one over other, several such basic units Structural units are bonded with hydrogen bond Shape-hexagonal, thickness-0.05micron. illite Structural sheet consist of an alumina sheet is sandwiched between two silica sheet Thickness of one structure unit-10 A 0 Structural units are bonded with potassium ion Bond is stronger than montmorillonite Dept of Civil Engg Page 8

9 Montmorillonite- Montmorillonite is the most common mineral of the montmorillonite group of minerals. The basic structural unit consists of an alumina sheet sandwiched between two silica sheets; Successive structural units are stacked one over another, like leaves of a book. Fig shows two such structural units. The thickness of each structural unit is about 10Å. The two successive structural units are joined together by a link between oxygen ions of the two silica sheets. The link is due to natural attraction for the cat ions in the intervening space and due to Vander Waal force. The negatively charged surfaces of the silica sheet attract water in the space between structural units. This results in expansion of the mineral. It may also cause dissociation of the mineral into individual structural units of thickness 10Å. The soil containing a large amount of mineral montmorillonite exhibits high shrinkage and high swelling characteristics. The water in the intervening space can be removed by heating to C. 5 a) Explain Mohr coulomb theory. The theory can be expressed by the equation f=f (ϭ) f = shear stress at the failure plane, F (ϭ) =function of normal stress. If the normal stress and shear stress corresponding to failure are plotted then a curve is obtained. The curve is called as strength envelope. Coulomb defined the function F(ϭ) as a lines function of ϭ and gave the following strength equation. =c + ϭ tan Where, c = cohesion (intercept of shear axis) Dept of Civil Engg Page 9

10 ϭ = normal stress = angle of internal friction. (Slope of the straight line) Mohr theory also stated that the shear strength depends on normal stress, but the relation is not liner as shown in fig-b. 5 b) Explain sensitivity and thixotrophy of clay The degree of disturbance of undisturbed clay sample due to remoulding is expressed by Sensitivity Sensitivity = [qu undisturbed] / [qu remoulded] The loss of strength due to remoulding is partly due to: Dept of Civil Engg Page 10

11 Permananent destruction of the structure due to in-situ layers and Reorientation of the molecules in the adsorbed layers. Classification of soils based on Sensitivity Thixotropy When sensitive clays are used in construction, they loose strength due to remoulding during contruction operations.however, with passage of time the strength again increases, though not to the same original level. This phenomenon of Strength loss- strength gain with no change in volume or water content is called Thixotropy. Reasons for the gain in strength with time: Rehabilitation of the molecular structure of the soil Due to thixotropic property 5 c) The results of shear box test are as follows: Normal stress (kn/m2) Shear stress (kn/m2) Determine shear parameter. Would the failure occur on the plane within the soil mass when the shear stress is 122 kn/m2 and normal stress is 246 kn/m2? Dept of Civil Engg Page 11

12 Cohesion = 12.5 kn/m 2 and Angle of Internal Friction = The point lies within failure envelop as shown in graph hence the soil is safe. 6 a) Write the difference between modified and standard proctor test. Sl.no Description Standard Proctor Test Modified Proctor Test 1 Soil For Lab Test passing 4.75mm sieve passing 4.75mm sieve 2 Volume of Mould 945 cm cm 3 3 Number of Layers 3 No 5 No 4 Number of Blow Per Layer 25 No 25 No 5 Weight of Hammer 2.6 kg 4.9 kg 6 Energy 60 kg-mm/cm mm/cm 3 Dept of Civil Engg Page 12

13 6 b) Briefly explain factors affecting compaction Factors Affecting Compaction. Increase in dry density is mainly depends upon following factors Water content Compactive Effort Method of Compaction Types of Soil Admixture Water content-as water content increases dry density increases till OMC is reached, at this stage, air voids attains constant volume with further increase in water content, the air voids do not decrease, but the total voids increases and the dry density decreases. Compactive Effort -Different curves are obtained for different compactive efforts. A greater compactive effort reduces the optimum moisture content and increases the maximum dry density. Dept of Civil Engg Page 13

14 Method of Compaction-For the same amount of compaction effort, dry density will be depend upon method of compaction such as dynamic or static etc. Types of Soil-fine grained soil have more air voids. These attain lower dry density compared with the coarse grained soil. Fine grained soil has high plasticity, very low dry density and very high optimum moisture content as shown in fig. Admixture-the properties of soil can be modified by adding materials called admixture which accelerates the process of densification 6 c) The following data were obtained from standard proctor compaction test: Water content (%) Bulk unit weight kn/m I. Plot the compaction curve and determine MDD and OMC II. Draw zero-g.r. void line. III. Also determine saturation at MDD. Take G = 2.6. Water content w % Bulk density kn/m 3 Dry density kn/m 3 Zero air voids kn/m Dept of Civil Engg Page 14

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16 7 a) Explain mass-spring analogy Spring Analog for Primary Consolidation Cylinder fitted with a piston having valve. The cylinders filled with water and contain a spring if specified stiffness.let the initial length of spring is 100mm and the stuffiness of spring is 10 mm/n. Case-1 When the load P (1N) is applied to the piston, with its valve closed, entire load is shared by water and spring does not take any load. t = 0. Pw + Ps = P where Pw is load taken by water, Ps is load taken by spring, P = total load For case-1, P=1N, when the valve is closed Ps = 0 Hence Pw = P. At any time t = 0 Case-2 Dept of Civil Engg Page 16

17 If the valve is now gradually opened, water starts escaping from the cylinder, spring starts sharing some load and decrease in its length occurs. Pw +Ps = 1 at any time t Case-3 As more and more water escapes, the load carried by spring increases, when steady state reaches, the water stops escaping, and finally entire load is taken by spring. Ps = 1 and Pw = 0 t = tf This load causes a decrease in length of the spring by 10mm. the final length is 90 mm as shown in fig. Dept of Civil Engg Page 17

18 7 b) With neat sketch, explain how pre consolidation pressure is determined by casagrande method. Conduct an odometers test on the undisturbed soil sample obtained from the field. Plot e - log σ plot as shown. The equilibrium void ratio at the end of each of the pressure increments are used in obtaining e - log σ plot. Select the point of maximum curvature (Point A) on the e - log σ curve Draw a tangent at the point of maximum curvature (Point A) Draw a horizontal line AC Draw the bisector line AD between the tangent and horizontal line Extend the normally consolidated line to intersect the bisector line at P The vertical effective stress corresponding to point of intersection P is the preconsolidation pressure (σ pc) 7 c) A soil sample 20mm thick takes 20 min to reach 20% consolidation. Find the time taken by a clay layer 6mm thick to reach 40% consolidation. Assume double drainage in both cases. Dept of Civil Engg Page 18

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20 Merits 8 a) What are the advantages and limitations of vane shear test? How do you conduct the test in laboratory? The test is simple and quick It is ideally suited for determination of the in-situ un-drained shear strength of nonfissured, fully saturated clay The test can be conveniently used to determine the sensitivity of the soil Demerits The test cannot be conducted on the fissured clay or the clay containing silt or sand laminations The test does not give accurate results when the failure envelope is not horizontal For conducting test in a laboratory, a specimen of diameter 38mm and height 75mm is prepared and fixed to the base of the apparatus. The vane is slowly lowered in to the specimen till the top of the vane is at a depth of 10 to 20 mm below the top of the specimen. The readings of the strain indicator and torque indicator are taken Shear strength S Where T =Torque applied D = Diameter of vane H1= Height of vane Dept of Civil Engg Page 20

21 8 b) Explain square root of time fitting method for determination of coefficient of consolidation. Square-root time curve fitting method for determination of coefficient of consolidation. Theoretical curve between square root of time factor ( T V ) and degree of consolidation (U) is obtained in the figure. The curve is straight up to U=60% and the x-axis corresponding to U=90% point on curve is 1.15 times the x-axis of point of intersection of straight line portion of the curve produced with the horizontal line at U=90%. ` Theoretical Curve The laboratory consolidation curve is obtained by plotting dial gauge reading (R) as y-axis and square root of time ( t) on x-axis. Dept of Civil Engg Page 21

22 Lab Graph The initial dial gauge reading at t=0 is denoted by R0 and it corresponds to U=0 as shown in fig. Line A is drawn coinciding with the straight portion of the lab curve. It intersect at y-axis at Rc then Rc represents the corrected zero reading. From point Rc another line B is drawn such that every point on it has abscissa 1.15 times that of corresponding point on A-line. Let the curve B intersect at point P the point P corresponding to U=90% and its coordinates denoted as t 90 and R90.The coefficient of consolidation is calculated from the following equation. Where (Tv)90 = time factor corresponding to U=90% d=average drainage path. Dept of Civil Engg Page 22

23 8 c) With neat sketch, explain vane shear test. Vane Shear Test The un-drained strength of soft clays can be determined in a laboratory by vane shear test. The test can also be conducted in the field on the soil at the bottom of bore hole. The apparatus consists of a vertical steel rod having four thin stainless steel blades or vanes fixed at its bottom end. Height of the vane should be equal to twice the diameter. Dept of Civil Engg Page 23