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GEOTECHNICAL ENGINEERING- 5th semester, (18SCV531), , MODULE 1, CLASSIFICATION OF SOILS:, For Engineering purposes, soils may classified by the following system., , 1. Particle size classification, 2. Textural classification, 3. Highway Research board classification, 4. Unified soil classification &, 5. IS classification, Particle size classification:, •, , In this system, soils are arranged according to grain size., , •, , The following terms are used to indicate grain size., ,  – gravel,  – Sand,  – Silt and,  – Clay, These terms are used only as designation of particles size, and do not signify the naturally, occurring types. The most commonly used Particle size soil classification systems are,, 1. U.S.Bureau of soil and public road administration system of U.S., 2. International soil classification system., 3. M.I.T classification, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 1

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4. I.S classification based on M.I.T classification, , Soil Classification:, It is necessary to adopt a formal system of soil description and classification in order to describe, the various materials found in ground investigation. Such a system must be meaningful and concise, in an engineering context, so that engineers will be able to understand and interpret., , It is important to distinguish between description and classification:, Description of soil is a statement that describes the physical nature and state of the soil. It can be a, description of a sample, or a soil in situ. It is arrived at by using visual examination, simple tests,, observation of site conditions, geological history, etc., , Classification of soil is the separation of soil into classes or groups each having similar, characteristics and potentially similar behaviour. A classification for engineering purposes should, be based mainly on mechanical properties: permeability, stiffness, strength. The class to which a, soil belongs can be used in its description., , The aim of a classification system is to establish a set of conditions which will allow useful, comparisons to be made between different soils. The system must be simple. The relevant criteria, for classifying soils are the size distribution of particles and the plasticity of the soil., , For measuring the distribution of particle sizes in a soil sample, it is necessary to conduct different, particle-size tests., Wet sieving is carried out for separating fine grains from coarse grains by washing the soil, specimen on a 75 micron sieve mesh., Dry sieve analysis is carried out on particles coarser than 75 micron. Samples (with fines removed), are dried and shaken through a set of sieves of descending size. The weight retained in each sieve is, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 2

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measured. The cumulative percentage quantities finer than the sieve sizes (passing each given sieve, size) are then determined., The resulting data is presented as a distribution curve with grain size along x-axis (log scale) and, percentage passing along y-axis (arithmetic scale)., Sedimentation analysis is used only for the soil fraction finer than 75 microns. Soil particles are, allowed to settle from a suspension. The decreasing density of the suspension is measured at, various time intervals. The procedure is based on the principle that in a suspension, the terminal, velocity of a spherical particle is governed by the diameter of the particle and the properties of the, suspension., In this method, the soil is placed as a suspension in a jar filled with distilled water to which a, deflocculating agent is added. The soil particles are then allowed to settle down. The concentration, of particles remaining in the suspension at a particular level can be determined by using a, hydrometer. Specific gravity readings of the solution at that same, level at different time intervals provide information about the size of particles that have settled, down and the mass of soil remaining in solution., The results are then plotted between % finer (passing) and log size, , .Grain-Size Distribution Curve:, The size distribution curves, as obtained from coarse and fine grained portions, can be, combined to form one complete grain-size distribution curve (also known as grading curve)., A typical grading curve is shown., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 3

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From the complete grain-size distribution curve, useful information can be obtained such as:, , 1. Grading characteristics, which indicate the uniformity and range in, , grain-size distribution., 2. Percentages (or fractions) of gravel, sand, silt and clay-size., , Grading Characteristics, , A grading curve is a useful aid to soil description. The geometric properties of a grading, curve are called grading characteristics, , ., To obtain the grading characteristics, three points are located first on the grading curve. D60 =, size at 60% finer by weight, D30 = size at 30% finer by weight, D10 = size at 10% finer by weight, , The grading characteristics are then determined as follows:, , 1. Effective size = D10, , 2. Uniformity coefficient,, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 4

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3. Curvature coefficient,, Both Cu and Cc will be 1 for a single-sized soil., Cu > 5 indicates a well-graded soil, i.e. a soil which has a distribution of particles over a, wide size range. Cc between 1 and 3 also indicates a well-graded soil., Cu < 3 indicates a uniform soil, i.e. a soil which has a very narrow particle size range., , Indian Standard Soil Classification System:, , Fine-grained soils are those for which more than 50% of the material has particle sizes less, than 0.075 mm. Clay particles have a flaky shape to which water adheres, thus imparting the, property of plasticity., , A plasticity chart , based on the values of liquid limit (WL) and plasticity index (IP), is, provided in ISSCS to aid classification. The 'A' line in this chart is expressed as IP = 0.73, (WL - 20)., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 5

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Depending on the point in the chart, fine soils are divided into clays (C), silts (M), or, organic soils (O). The organic content is expressed as a percentage of the mass of organic, matter in a given mass of soil to the mass of the dry soil solids.Three divisions of plasticity, are also defined as follows., , Low plasticity, , WL< 35%, 35% < WL<, , Intermediate plasticity, 50%, High plasticity, , WL> 50%, , The 'A' line and vertical lines at WL equal to 35% and 50% separate the soils into various, classes., For example, the combined symbol CH refers to clay of high plasticity., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 6

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Soil classification using group symbols is as follows:, Group Symbol Classification, Coarse soils, GW, , Well-graded GRAVEL, , GP, , Poorly-graded GRAVEL, , GM, , Silty GRAVEL, , GC, , Clayey GRAVEL, , SW, , Well-graded SAND, , SP, , Poorly-graded SAND, , SM, , Silty SAND, , SC, , Clayey SAND, , Fine soils, ML, , SILT of low plasticity, , MI, , SILT of intermediate plasticity, , MH, , SILT of high plasticity, , CL, , CLAY of low plasticity, , CI, , CLAY of intermediate plasticity, , CH, , CLAY of high plasticity, , OL, , Organic soil of low plasticity, , OI, , Organic soil of intermediate plasticity, , OH, , Organic soil of high plasticity, , Pt, , Peat, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 7

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PERMEABILITY OF SOIL:, Pressure, Elevation and Total Heads, In soils, the interconnected pores provide passage for water. A large number of, such flow paths act together, and the average rate of flow is termed the coefficient, of permeability, or just permeability. It is a measure of the ease that the soil, provides to the flow of water through its pores., , At point A, the pore water pressure (u) can be measured from the height of water, in a standpipe located at that point. The height of the water column is the, pressure head (hw)., w, , hw = u/, To identify any difference in pore water pressure at different points, it is, necessary to eliminate the effect of the points of measurement. With this in view,, a datum is required from which locations are measured., The elevation head (hz) of any point is its height above the datum line. The, height of water level in the standpipe above the datum is the piezometric head, (h)., h = hz + hw, Total head consists of three components: elevation head, pressure head, and, velocity head. As seepage velocity in soils is normally low, velocity head is, ignored, and total head becomes equal to the piezometric head. Due to the low, seepage velocity and small size of pores, the flow of water in the pores is steady, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 8

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and laminar in most cases. Water flow takes place between two points in soil due, to the difference in total heads., , Darcy's law states that there is a linear relationship between flow velocity (v) and, hydraulic gradient (i) for any given saturated soil under steady laminar flow, conditions., , If the rate of flow is q (volume/time) through cross-sectional area (A) of the soil, mass, Darcy's Law can be expressed as, v = q/A = k.i, where k = permeability of the soil, i = ∆h/L, ∆h = difference in total heads, L = length of the soil mass, The flow velocity (v) is also called the Darcian velocity or the superficial, velocity. It is different from the actual velocity inside the soil pores, which is, known as the seepage velocity, vS. At the particulate level, the water follows a, tortuous path through the pores. Seepage velocity is always greater than the, superficial velocity, and it is expressed as:, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 9

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where AV = Area of voids on a cross section normal to the direction of flow, n = porosity of the soil, , Permeability of Different soils:, Permeability (k) is an engineering property of soils and is a function of the soil, type. Its value depends on the average size of the pores and is related to the, distribution of particle sizes, particle shape and soil structure. The ratio of, permeabilities of typical sands/gravels to those of typical clays is of the order of, 106. A small proportion of fine material in a coarse-grained soil can lead to a, significant reduction in permeability. For different soil types as per grain size, the, orders of magnitude for permeability are as follows:, , Soil, Gravel, Coarse sand, Medium sand, Fine sand, Silty sand, Silt, Clay, , k (cm/sec), 100, 100 to 10-1, 10-1 to 10-2, 10-2 to 10-3, 10-3 to 10-4, 1 x 10-5, 10-7 to 10-9, , In soils, the permeant or pore fluid is mostly water whose variation in property is, generally very less. Permeability of all soils is strongly influenced by the density, of packing of the soil particles, which can be represented by void ratio, (e) or porosity (n)., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 10

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For Sands, In sands, permeability can be empirically related to the square of some, representative grain size from its grain-size distribution. For filter sands, Allen, Hazen in 1911 found that k, , 100, , (D10)2, , cm/s, , where D10= effective grain size in cm., , Different relationships have been attempted relating void ratio and permeability,, such as k= e3/(1+e), and, They have been obtained from the Kozeny-Carman equation for laminar flow in, saturated soils., , where ko and kT are factors depending on the shape and tortuosity of the pores, respectively, SS is the surface area ofsolid particles per volume of solid material, The equation can be reduced to a simpler form as, , For Silts and Clays, For silts and clays, the Kozeny-Carman equation does not work well, and, log k versus e plot has been found to indicate a linear relationship., , For clays, it is typically found that, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 11, , 2, , .

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where Ckis the permeability change index and ek is a reference void ratio., , Laboratory Measurement of Permeability:, Constant Head Flow, Constant head permeameter is recommended for coarse-grained soils only since for, such soils, flow rate is measurable with adequate precision. As water flows through a, sample of cross-section area A, steady total head drop h is measured across length L., , Permeability k is obtained from:, Falling Head Flow:, Falling head permeameter is recommended for fine-grained soils., Total head h in standpipe of area a is allowed to fall. Hydraulic gradient varies with, time. Heads h1 and h2 are measured at times t1 and t2. At any time t, flow through the, soil sample of cross-sectional area A is, , (1), , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 12

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T, , Flow in unit time through the standpipe of cross-sectional area a is, , =, , ----------------- (2), , Equating (1) and (2) ,, , or, Integrating between the limits,, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 13

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Module 2:, SHEAR STRENGTH OF SOIL, Necessity of studying Shear Strength of soils :, • Soil failure usually occurs in the form of “shearing”, along internal surface within the soil., Shear Strength:, • Thus, structural strength is primarily a function of shear strength., • The strength of a material is the greatest stress it can sustain, • The safety of any geotechnical structure is dependent on the strength of the soil, • If the soil fails, the structure founded on it can collapse, Thus shear strength is, “The capacity of a material to resist the internal, and external forces which slide past each other”, Significance of Shear Strength :, - Engineers must understand the nature of shearing resistance in order to analyze soil, stability problems such as;, - Bearing capacity, - Slope stability, - Lateral earth pressure on earth-retaining structure, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 14

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Shear Failure under Foundation, Load, , Slope Stability Failure as an Example of Shearing Along Internal Surface, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 15

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At failure, shear stress along the failure surface reaches the shear, , Thus shear strength of soil is, “The capacity of a soil to resist the internal and, external forces which slide past each other”, Shear Strength in Soils :, - The shear strength of a soil is its resistance to shearing stresses., - It is a measure of the soil resistance to deformation by continuous displacement of its, individual soil particles., - Shear strength in soils depends primarily on interactions between particles., -Shear failure occurs when the stresses between the particles are such that they slide or, roll past each other, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 16

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Components of shear strength of soils, Soil derives its shear strength from two sources:, – Cohesion between particles (stress independent component), • Cementation between sand grains, • Electrostatic attraction between clay particles, – Frictional resistance and interlocking between particles (stress dependent component), Cohesion :, Cohesion (C), is a measure of the forces that cement particles of soils, , Internal Friction :, Internal Friction angle (f), is the measure of the shear strength of soils due to, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 17

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friction, , ANGLE OF REPOSE, Angle of Repose determined by:, , , Particle size (higher for large particles), , , , Particle shape (higher for angular shapes), , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 18

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, , Shear strength (higher for higher shear strength), , Stresses:, Gravity generates stresses (force per unit area) in the ground at different points. Stress on, a plane at a given point is viewed in terms of two components:, Normal stress (ζ): acts normal to the plane and tends to compress soil grains towards, each other (volume change), Shear stress (η): acts tangential to the plane and tends to slide grains relative to each, other (distortion and ultimately sliding failure)., Factors Influencing Shear Strength:, The shearing strength, is affected by:, – soil composition: mineralogy, grain size and grain size distribution, shape of particles,, pore fluid type and content, ions on grain and in pore fluid., – Initial state: State can be describe by terms such as: loose, dense, overconsolidated,, normally consolidated, stiff, soft, etc., – Structure: Refers to the arrangement of particles within the soil mass; the manner in, which the particles are packed or distributed. Features such as layers, voids, pockets,, cementation, etc, are part of the structure., Mohr-Coulomb Failure Criteria:, This theory states that a material fails because of a critical combination of normal stress, and shear stress, and not from their either maximum normal or shear stress alone., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 19

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Direct Shear Test:, Dry sand can be conveniently tested by direct shear tests. The sand is placed in a, shear box that is split into two halves . A normal load is first applied to the specimen., Then a shear force is applied to the top half of the shear box to cause failure in the, sand. The normal and shear stresses at failure are, , Direct shear test in sand: (a) schematic diagram of test equipment; (b) plot of, test results to obtain, , the friction angle, ϕ, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 23

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Range of relative density and corresponding range of angle of friction for, coarse-grained soil, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 24

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Triaxial Tests, Triaxial compression tests can be conducted on sands and clays shows a schematic, diagram of the Triaxial test arrangement. Essentially, it consists of placing a soil, specimen confined by a rubber membrane in a Lucite chamber. An all-round confining, pressure (ζ3) is applied to the specimen by means of the chamber fluid (generally, water or glycerin). An added stress (Δζ) can also be applied to the specimen in the, axial direction to cause failure (Δζ=Δζf at failure). Drainage from the specimen can be, allowed or stopped, depending on the test condition. For clays, three main types of, tests can be conducted with Triaxial equipment:, Triaxial test:, 1. Consolidated-drained test (CD test), 2. Consolidated-undrianed test (CU test), 3. Unconsolidatedundrained test (UU, test), Major, effective, , Principal, stress, , =ζ3=Δζf=ζ1=ζ′1, Minor, effective, , Principal, stress, , =ζ3=Δζ′3, Changing ζ3 allows several tests of this type to be conducted on various clay, specimens. The shear strength parameters (c and ϕ) can now be determined by plotting, Mohr‟s circle at failure, as shown in figure and drawing a common tangent to the, Mohr‟s circles. This is the Mohr-Coulomb failure envelope. (Note: For normally, consolidated clay, c≈0). At failure, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 26

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For consolidated-undrained tests, at failure, Major Principal total stress =ζ3=Δζf=ζ1, Minor principal total stress =ζ3, Major principal effective stress =(ζ3+Δζf)−uf=ζ′1, , Minor principal effective stress =ζ3− uf=ζ′3, Changing ζ3 permits multiple tests of this type to be conducted on several soil, specimens. The total stress Mohr‟s circles at failure can now be plotted, as shown in, figure , and then a common tangent can be drawn to define the failure envelope. This, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 27

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total stress failure envelope is defined by the equation, s=ccu+ζtanϕcu, Where ccu and ϕcu are the consolidated-undrained cohesion and angle of friction, respectively (Note: ccu≈0 for normally consolidated clays), Similarly, effective stress Mohr‟s circles at failure can be drawn to determine the, effective stress failure envelopes. They follow the relation expressed in equation ., For unconsolidated-undrained triaxial tests Major principal total stress=ζ3=Δζf=ζ1, Minor principal total stress =ζ3, The total stress Mohr‟s circle at failure can now be drawn, as shown in figure. For, saturated clays, the value of ζ1−ζ3=Δζf is a constant, irrespective of the chamber, confining pressure, ζ3.The tangent to these Mohr‟s circles will be a horizontal line,, called the ϕ=0 condition. The shear stress for this condition is, , Where, B=Skempton′s pore pressure parameter, Similarly,the pore pressure ud is the result of added axial stress, Δζ, so ud=A Δζ, Where, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 28

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A=Skempton′s pore pressure parameter However,, Δζ=ζ1−ζ3, Combining equations gives u=ua+ud=Bζ3+Aζ1−ζ3, The pore water pressure parameter B in soft saturated soils is 1, so u=ζ3+A(ζ1−ζ3), The value of the pore water pressure parameter A at failure will vary with the type of soil., Following is a general range of the values of A at failure for various types of clayey soil, encountered in nature., , Triaxial test equipment, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 29

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UNCONFINED COMPRESSION TEST, The unconfined compression test is a special type of unconsolidated-undrained, Triaxial test in which the confining pressure ζ3=0, as shown in figure. In this test an, axial stress, Δζ, is applied to the specimen to cause failure (that is, Δζ=Δζf). The, corresponding Mohr‟s circle is shown in figure . Note that, for this case, u, , Unconfined compression test: (a) soil specimen; (b) Mohr‟s circle for the test; (c), variation of qu with the degree of saturation, Major principal total stress =Δζf=qu Minor principal total stress = 0, , The axial stress at failure, Δζf=qu is generally referred to as the, unconfined compression strength. The shear strength of saturated clays under this, condition (ϕ=0,, The unconfined compression strength can be used as an indicator for the consistency, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 30

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of clays. Unconfined compression tests are sometimes conducted on unsaturated soils., With the void ratio of a soil specimen remaining constant, the unconfined compression, strength rapidly decreases with the degree of saturation shows an unconfined, compression test., , Unconfined compression test in progress (courtesy of Soiltest, Inc., Lake Bluff, Illinois), Vane Shear Test:, Fairly reliable results for the undrained shear strength, c,, (S : 0 concept), of very soft, to medium cohesive soils may be obtained directly from vane shear tests. The shear, vane usually consists of four thin, equal-sized steel plates welded to a steel torque rod., First, the vane is pushed into the soil. Then torque is applied at the top of the torque, rod to rotate the vane at a uniform speed. A cylinder of soil of height ft and diameter r/, will resist the torque until the soil fails. The undrained shear strength of the soil can be, calculated as follows. If I is the maximum torque applied at the head of the torque rod, to cause failure, it should be equal to the sum of the resisting moment of the shear, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 31

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force along the side surface of the soil cylinder (M.) and the resisting moment of the, shear force at each end (M,,), , where d : diameter of the shear van c/z : height of the shear vane, For the calculation of M., investigators have a, , several types of distributing of, , shear strength mobilization at the ends of the soil cylinder:, l. Triangular. Shear strength mobilization is c,, at the periphery of the soil cylinder and, decreases linearly to zero at the center., 2, Uniform. Shears strength mobilization is constant ( that is, c)from the periphery to the, center of the soil cylinder., 3. Parabolic. Shear strength mobilization is c,, at the periphery of the soil cylinder and, decreases parabolically to zero at the center., These variations in shear strength mobilization are shown in Figure .In general, the, torque,I at failure can be expressed as, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 32

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(a) resisting moment of shear force; (b) variations in shear strength, mobilization, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 34

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MODULE 3, CONSOLIDATION, Civil Engineers build structures and the soil beneath these structures is loaded. This results in, increase of stresses resulting in strain leading to settlement of stratum. The settlement is due, to decrease in volume of soil mass. When water in the voids and soil particles are assumed as, incompressible in a completely saturated soil system then - reduction in volume takes place, due to expulsion of water from the voids. There will be rearrangement of soil particles in air, voids created by the outflow of water from the voids. This rearrangement reflects as a volume, change leading to compression of saturated fine grained soil resulting in settlement. The rate, of volume change is related to the rate at which pore water moves out which in turn depends, on the permeability of soil. Therefore the deformation due to increase of stress depends on, the “Compressibility of soils”, As Civil Engineers we need to provide answers for, 1. Total settlement (volume change), 2. Time required for the settlement of compressible layer The total settlement consists of, three components, 1. Immediate settlement., 2. Primary consolidation settlement, 3. Secondary consolidation settlement (Creep settlement) St = Si + Sc + Ssc, Elastic Settlement or Immediate Settlement, This settlement occurs immediately after the load is applied. This is due to distortion, (change in shape) at constant volume. There is negligible flow of water in less, pervious soils. In case of pervious soils the flow of water is quick at constant volume., This is determined by elastic theory., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 35

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Primary Consolidation Settlement, , Figure Settlement versus Time, It occurs due to expulsion of pore water from the voids of a saturated soil. In case of, saturated fine grained soils, the deformation is due to squeezing of water from the, pores leading to rearrangement of soil particles. The movement of pore water depends, on the permeability and dissipation of pore water pressure. With the passage of time, the pore water pressure dissipates, the rate of flow decreases and finally the flow of, water ceases. During this process there is gradual dissipation of pore water pressure, and a simultaneous increase of effective stress as shown in the above Figure. The, consolidation settlement occurs from the time water begins move out from the pores to, the time at which flow ceases from the voids. This is also the time from which the, excess pore water pressure starts, , reducing (effective stress increase) to the time at, , which complete dissipation of excess pore water pressure (total stress equal to, effective stress). This time dependent compression is called “Consolidation, settlement”., Primary consolidation is a major component of settlement of fine grained saturated, soils and this can be estimated from the theory of consolidation., In case of saturated soil mass the applied stress is borne by pore water alone in the initial, stages, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 36

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With passage of time water starts flowing out from the voids as a result the excess pore, water pressure decreases and simultaneous increase in effective stress will takes place., The volume change is basically due to the change in effective stress After considerable, amount of time (t =0) flow from the voids ceases the effective stress stabilizes and will, be is equal to external applied total stress and this stage signifies the end of primary, consolidation., , Secondary Consolidation Settlement:This is also called Secondary compression (Creep). “It is the change in volume of a fine, grained soil due to rearrangement of soil particles (fabric) at constant effective stress”., The rate of secondary consolidation is very slow when compared with primary, consolidation., , Figure Effective Stress versus, Time, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 37

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Figure Mechanism of volume change in saturated fine grained soil under external, loading, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 38

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When saturated soil mass is subjected to external load decrease in volume takes place, due to rearrangement of soil particles. Reduction in volume is due to expulsion of, water from the voids. The volume change depends on the rate at which water is, expelled and it is a function of permeability., The total vertical deformation (Consolidation settlement) depends on, 1. Magnitude of applied pressure, 2. Thickness of the saturated deposit We are concerned with, _ Measurement of volume change, _ The time duration required for the volume change, , Spring Analogy, The consolidation process is often explained with an idealized system composed of a, spring, a container with a hole in its cover, and water. In this system, the spring, represents the compressibility or the structure itself of the soil, and the water which, fills the container represents the pore water in the soil., On figure , the tube on the left of the container shows the water pressure in the container., , Figure: process of consolidation, 1. The container is completely filled with water, and the hole is closed. (Fully saturated, soil), 2. A load is applied onto the cover, while the hole is still unopened. At this stage, only the, water resists the applied load. (Development of excessive pore water pressure), , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 39

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3. As soon as the hole is opened, water starts to drain out through the hole and the spring, shortens. (Drainage of excessive pore water), 4. After some time, the drainage of water no longer occurs. Now, the spring alone resists, the applied load. (Full dissipation of excessive pore water pressure. End of consolidation), , Terzaghi‟s Spring Mass AnalogyTerzaghi‟s model consists of a cylindrical vessel with a series of piston separated by, springs. The space between springs is filled with water the pistons are perforated to allow, for passage of water. Piezometers are inserted at the centers of different compartment to, measure the pressure head due to excess pore water pressure., Terzaghi has correlated the spring mass compression process with the consolidation of, saturated clay subjected to external load. The springs and the surrounding water represent, the saturated soil. The springs represent the soil skeleton networks of soil grains and, water in the vessels represents the water in the voids. In this arrangement the compression, is one dimensional and flow will be in the vertical direction. When pressure is applied this, will be borne by water surrounding the spring, , There will be no volume change. After some time „t‟ there will be flow of water, through perforation beginning from upper compartment. In the lower compartment the, volume of water remains constant since the flow is in upward direction., Due to flow of water in the upper segment there will be reduction in volume due to this, spring‟s get compressed and they being to carry a portion of the applied load. This, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 40

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signifies a reduction in excess hydrostatic pressure or pore water pressure and increase, in effective stress in the upper segments. Whereas there will be no dissipation of, excess hydrostatic pressure in lower compartments. The isochrones indicate that with, passage of time there is flow of water from the lower compartments leading to gradual, dissipation of excess hydrostatic pressure. At time t = 0 when no more pore water, flows out the excess hydrostatic pressure will be zero in all compartments and the, entire load is carried by springs., , Figure Compression of Spring mass, The compression of a spring mass system is analogous to the consolidation of a saturated, fine grained soil deposit subjected to external pressure., Soil Compressibility, Sand deposit compresses immediately on load application. Loose sand compresses more, than dense sand. Loose and dense sand deposits tend towards the same void ratio., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 41

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Figure Void ratio-effective stress and compression-time plots for sand, Compression of fine grained soil (clay), , Figure Void ratio-effective stress and compression-time plots for clay, Time dependent compression takes longer time compared to sand. The magnitude of, compression is also large., Compression of Fine Grained Soil, The compressibility of fine grained soils can be described in terms of voids ratio versus, effective stress, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 42

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A laboratory soil specimen of dia 60mm and height 20mm is extracted from the, undisturbed soil sample obtained from the field. This sample is subjected to 1D, consolidation in the lad under various pressure increments. Each pressure increment is, maintained for 24 hrs and equilibrium void ratio is recorded before the application of the, next pressure increment. Then a plot of void ratio versus effective stress is made as shown, in above figure. When the sample is recompressed from point D it follows DE and, beyond C it merges along BCF and it compresses as it moves along BCF, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 43

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Figure Void ratio versus effective stress(on arithmetic plot), , Figure Void ratio versus logarithm of effective stress (semi-log plot), During the initial stages (at low effective stress) sample follows recompression path, (portion AB) and undergoes less compression. Beyond this is the virgin compression, line (portion BC) also called the normal compression line and the sample undergoes, large compression., 1. BC – Virgin compression curve also called normal consolidation line, 2. From „C‟ when the sample is unloaded, sample expands and traces path CD (expansion, curve unloading), , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 44

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3. Sample undergoes Permanent strain due to irreversible soil structure and there is a, small elastic recovery., 4. The deformation recovered is due to elastic rebound, 5. When the sample is reloaded-reloading curve lies above the rebound curve and makes, an hysteresis loop between expansion and reloading curves., 6. The reloaded soils shows less compression., 7. Loading beyond „C‟ makes the curve to merge smoothly into portion EF as if the soil is, not unloaded., , Terzaghi‟s 1D Consolidation Equation, , Figure Saturated soil Strata, Assumptions:,  The soil medium is completely saturated,  The soil medium is isotropic and homogeneous,  Darcy‟s law is valid for flow of water,  Flow is one dimensional in the vertical direction,  The coefficient of permeability is constant,  The coefficient of volume compressibility is constant,  The increase in stress on the compressible soil deposit is constant, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 45

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 Soil particles and water are incompressible, One dimensional theory is based on the following hypothesis, 1. The change in volume of soil is equal to volume of pore water expelled., 2. The volume of pore water expelled is equal to change in volume of voids., 3. Since compression is in one direction the change in volume is equal to change in height., The increase in vertical stress at any depth is equal to the decrease in excess pore water, pressure at the depth, , Limitation of 1D consolidation, 1. In the deviation of 1D equation the permeability (Kz) and coefficient of volume, compressibility (mv) are assumed constant, but as consolidation progresses void, spaces decrease and this results in decrease of permeability and therefore, permeability is not constant. The coefficient of volume compressibility also changes, with stress level. Therefore Cv is not constant, 2. The flow is assumed to be 1D but in reality flow is three dimensional, 3. The application of external load is assumed to produce excess pore water pressure, over the entire soil stratum but in some cases the excess pore water pressure does not, develop over the entire clay stratum., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 46

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The solution of variation of excess pore water pressure with depth and time can be, obtained for various initial conditions., Uniform excess pore water pressure with depth, 1. Single Drainage (Drainage at top and bottom impervious), 2. Double Drainage (Drainage at top and bottom), , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 47

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1. The depth z, 2. The excess pore water pressure _u, 3. The time (t) after application of loading, The above variables are expressed in the form of the following non-dimensional terms, as, , Terzaghi‟s solution for onedimentional consolidation, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 49

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This indicates the progress of consolidation with time and depth for a, given set of boundary conditions., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 50

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Figure: Selection of clay ayer and Excess pore water pressure distribution, , uz= Degree of consolidation at a particular depth at any given time, From practical point of view, the average degree of consolidation over the entire depth, at any given time is desirable. At any given time uz varies with location and hence the, degree of consolidation also varies., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 51

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The average degree of consolidation for the whole soil deposit at any time is given by, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 52

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Preconsolidation Pressure, It is the maximum effective stress experienced by a soil in its stress history (past, existence), For the soil loaded along the recompression curve AB the effective stress close to, point B will be the preconsolidation pressure., If the soil is compressed along BC and unloaded along CD and then reloaded along, DC the effective stress close to point C will be the new preconsolidation pressure., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 56

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Figure: Void ratio versus effective stress (log scale), , Effect of Stress History, Based on the stress history (preconsolidation pressure) soils are classified as, 1. Normally Consolidated Soils, 2. Over Consolidated Soils, 3. Under Consolidated Soils, Normally Consolidated Soils, It is a soil deposit that has never subjected to a vertical effective stress greater than the, present vertical stress., , Void ratio versus effective stress (log scale), , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 57

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Under Consolidated Soils, A soil deposit that has not consolidated under the present overburden pressure, (effective stress) is called Under Consolidated Soil. These soils are susceptible to, larger deformation and cause distress in buildings built on these deposits., Over Consolidated Soils, It is a soil deposit that has been subjected to vertical effective stress greater than the, present vertical effective stress., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 58

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Determination of coefficient of consolidation (Cv) from laboratory data-, , The coefficient of three graphical procedure are used, 1. Logarithm of time method, 2. Square root of time method, 3. Hyperbola method, Log – time curve fitting method, The basis for this method is the theoretical (Uz) versus log Tv curve and experimental dial, gauge reading and log t curves are similar., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 59

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Log-time curve fitting method, Steps, 1., , Plot the dial reading of compression for a given pressure increment versus time to log, , scale, 2., , Plot two points P and Q on the upper portion of the consolidation curve (say, , compression line) corresponding to time t1 and t2 such that t2=4t1, 3., , Let x be the difference in dial reading between P and Q. locate R at a vertical distance x, , above point P, 4., , Draw a horizontal line RS the dial reading corresponding to this line is d0 which, , corresponds with 0% consolidation., 5., , Project the straight line portion of primary and secondary consolidation to intersect at, , point T. The dial reading corresponding, T, , is d 100, , and, , this, , to, , corresponds, , to, , 100%, , consolidation., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 60

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1. Plot the dial reading and square root of time i.e T for a pressure increment as shown in, figure., 2. Draw a tangent PQ to the initial portion of the plot as shown in fig., 3. Draw a line PR such that OR=1.15OQ., 4. The intersection of the line PR with the second portion of the curve i.e point S is, marked., , Time Rate of consolidation-, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 62

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Therefore the time required for a given degree of consolidation is proportional to the, length of the drainage path, If the time required to reach a certain degree of consolidation is measured in the, laboratory on a sample obtained from the field, The, , time, , taken, , by, , the, , can, , be, , predicted, , field, , deposit, , by, , using, , of, , known, , thickness, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 63

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FOUNDATION SETTLEMENT, , Foundation settlement:, , , The main criteria for designing a building are that the settlement should not exceed, the permissible value., , , , The foundation settlements are of three types., 1., , Immediate or elastic settlement., , 2., , Primary consolidation settlement., , 3., , Secondary compression settlement., , i. Immediate settlement:, The elastic settlement takes place immediately or in a shorter time after the load has, been placed. In clay it is generally termed as „distortion‟ settlement. It is mainly due, to change in shape of soils with change in volume or water content. The immediate, settlement of a foundation of width „B‟ on a surface of semi infinite, isotropic soil is, given by:, , Table 7.1:If values for footings, Shape, , Flexible, , Rigid, , Circle, , 1.00, , 0.86, , Square, , 1.12, , 0.82, , Rectangle L/B=1.5, , 1.36, , 1.06, , Rectangle L/B=2, , 1.52, , 1.20, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 66

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ii. Primary Settlement:, It is due to gradual expulsion of pore water from voids of the soil resulting in, dissipation of excess pore pressure and therefore increase in effective stress. Primary, consolidation is computed using Terzaghi‟s theory of consolidation:, , iii. Secondary compression settlement:, This type of settlement will be occurring at constant effective stress with volume, change occurring due to re arrangement of soil particles., Mode of settlement:, , , Uniform settlement: A structure with a rigid raft or mat will experience a uniform, settlement where settlement at all the sides of the footing will be same . Here the, entire footing settles through out due to its own weight., , , , Differential settlement: it is the difference between the total settlement between two, points., , , , Tilt: where the entire structure rotates as consequences of no uniform settlement., , , , Note: the angular distortion is the ratio of differential settlement between two columns, to the spacing between them., , Maximum and differential settlement of a building: (IS1904-1978):, For steel ISOLATED FOOTING, structures, , Sand, , and, , RAFT FOOTING, , hard Soft clay, , Sand, , clay, , and, , hard Soft clay, , clay, , Max, , Diff, , Max, , Diff, , Max, , Diff, , Max, , Diff, , 50mm, , 0.0033L, , 50mm, , 0.0033L, , 75mm, , 0.0033L, , 100mm, , 0.0033L, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 67

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MODULE 4:, STRESSES IN SOIL, The Knowledge of distribution of stresses within a soil mass induced by loads applied on the, surface is necessary for a proper foundation design. In this chapter we will be studying on the, stresses and displacements in soil mass, due to various surface loadings., At a certain point within a soil mass the stresses are caused due to both surface loadings as, well as due to self weight of soil above it. The stress components due to both these loadings, can be found separately, and then added algebraically to get then final stresses at the points., Boussinesq‟s theory:, In 1885 Boussinesq‟s published an equation to determine the state of stress in a sub grade, material. The stresses induced in a soil mass due to surface loading can be calculated from, the theory of elasticity if the following assumptions are made:, 1. Soil mass is an elastic medium, for which the modulus of elasticity E in constant., 2. The soil mass is homogeneous., 3. The soil mass is isotropic, that is it has identical elastic properties in all directions, through any point of it., 4. The soil mass is infinite, that is it extends infinitely in all directions below a level, surface., , Q, , ζz, Z, r, , ., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 68

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., ζz= (Q. Ib)/Z2, Where Ib= (3/2π). [1/ {1+(r/Z2)}] 5/2., Ib = Boussinesq‟s influence factor., Vertical stress distribution diagram (Pressure distribution diagram):, With the help of Boussinesq‟s equation the following diagrams can be presented graphically, after doing necessary calculation., i., , The vertical stress distribution diagram with depth at a distance „r‟ away from line of, action of force., , ii., , The vertical stress distribution on horizontal plane „Z‟ units below the ground surface., , iii., , Vertical stress‟ isobar‟ diagram., , Isobars: An isobar is a curve or contour connecting all points below the ground surface of, equal vertical pressure. An isobar is a spatial curved surface of the shape of a bulb, because, the vertical pressure on a given horizontal plane is the same in all directions at points located, at equal radial distances around the axis of loading., Pressure bulb: The zone in a loaded soil mass bounded by an isobar of a given vertical, pressure intensity is called pressure bulb., Westergaard‟s equation:, , Boussinesq‟s stress distribution theory assumes among other, , things isotropy for the soil mass. Natural deposits of soil are seldom isotropic. Water, deposited sedimentary soils, which are quite common in occurrence are formed by, deposition, alternately of horizontal layers of silts and clays. For this condition,, Westergaard‟s solution is better suited. Westergaard‟s (1938) assumed an elastic soil to be, laterally reinforced by numerous, closely spaced, horizontal sheets of negligible thickness but, of infinite rigidity that allow only vertical movement but prevent the mass as whole from, undergoing any lateral strain., Wetergaard‟s equation ,, , Westergaard‟s equation for the vertical stress for a point load for µ=0 is of the form., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 69

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ζz= (Q/z2) {1/ [π( 1+ 2(r/ z)2)3/2]}, ζz= (Q/z2). Iw., Where, Iw= {1/ [π (1+ 2(r/ z) 2)3/2]} is called as Westergaard‟s factor, which is also a function, of (r/z) ratio, similar to the Boussinesq‟s influence factor Kb., Comparison of Boussinesq‟s and Westergaard‟s analysis:, Boussinesq‟s theory is based on a number of simplifying assumptions, that the soil is elastic,, isotropic, homogeneous, and infinite. Soils are seldom homogeneous and soil density may, vary abruptly from point to point. Moreover the contact surface between soil and footing, neither is hardly an ideal plane, nor are stresses on the plane of contact are uniform. The, theory does not provide any estimate of safe bearing capacity. Despite these shortcomings,, Boussinesq‟s theory is a good approximation and is very popular., For layered soils conditions, on the other hand Westergaard‟s theory is better idealization of, field conditions. It assumes that elastic material is sandwiched between numerous, horizontal,, thin sheets off inelastic material which permit downward deformation but prevent any lateral, deformation. An example of such a layered soil is a deposit made up of alternating thin strata, of compressible clay and relatively incompressible sand. The choice of which analysis to use, will depend on how closely the field conditions match the basic assumptions. For layered, soils Westergaard‟s analysis is clearly more suitable., For r/z ratios less than 11.5, Westergaard‟s equations predict value of ζz smaller than, Boussinesq‟s equation and for r/z ratios, , equal to or greater than 1.5 there is no much, , difference., , Contact pressure:, The pressure transmitted from the base of a foundation to the soil is termed as contact, pressure. This depends on the rigidity of the foundation structure and the nature of the soil., The presence of thick compressible layer like soft clay beneath a flexible foundation presents, a bowl shaped settlement profile with more settlement at the centre and almost zero at the, edge. But pressure distribution is uniform. This is the conventional distribution pattern used, in the calculation of stresses and settlements., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 70

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An extremely rigid footing on the same clay will settle a uniform amount across its breadth,, thus a rigid footing has to transmit high contact pressure near the edges than at the centre so, as to maintain a uniform settlement., For flexible foundation resting on a non cohesive soil the distribution of contact pressure is, uniform but the edge of the foundation experience a large settlement because of lack of, confining pressure., The settlement of a rigid footing on sand layer is uniform and the contact pressure increases, from zero at the edges to a maximum at the centre., , Line loads:, ,  By applying the principle of the above theory, the stresses at any point in the mass, due to a line load of infinite extent acting at the surface may be obtained.,  The state of stress encountered in this case is that of a plane strain condition. The, strain at any point P in the Y-direction parallel to the line load is assumed equal to, zero. The stress ζy normal to the XZ-plane is the same at all sections and the shear, stresses on these sections are zero.,  The vertical ζz stress at point P may be written in rectangular coordinates as,, , where, I z is the influence factor equal to 0.637 at x/z =0., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 71

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Strip loads:, , •, , Fig. shows a load q per unit area acting on a strip of infinite length and of constant, width, , •, , B., , If the point P lies below the centre of strip then,, , Vertical Stress below the Center of a Uniformly Loaded Circular Area:, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 72

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•, , Uniform load intensity q per unit area acting on a circular area of radius „R‟. The, vertical stress ζz at a point „A‟, „z‟ units below the centre of loaded is given by,, , Vertical Stress Caused by a Rectangular Loaded Area:, , •, , The intensity of a vertical stress under rectangular loaded area can be determined by,, , a. Equivalent point load method, b. Newmark‟s influence chart., , 1., , Equivalent point load method:, , L, a, , b, Q1, , x1, , 1, , r1, , Q2, y1, , r2 2, , r3, , Q3, , B, Q4, , r4, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 73

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•, , It is an approximate method for calculating vertical stress at any point due to any, loaded area., , •, , The entire will be divided into numerous small area units., , •, , The total distributed load over a unit area is replaced by a point load of same, magnitude acting at the centroid of the area unit., , •, , The distributed loads over the whole area are replaced by a number of points loads, situated at the centroids of the various area units., ζz= 1[Q1IB1+Q2IB2+……….Qn IBn]/z2, Newmark‟s influence chart:, , 2., •, , The vertical pressure ζ z at depth z below a corner of uniformly loaded rectangular, area of length L and width B has been obtained by Newmark on integrating the, equation,, ζz= (Q. Ib)/Z2, Where Ib= (3/2π). [1/ {1+(r/Z2)}] 5/2., After integrating the equation obtained as, ζz= (q. Kb), , Where q= uniform load intensity, Kb = influence factor given the equation,, , Where,, m= B, z, n= L, z, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 74

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The values of Kb for different values of m and n are determined by the influence chart., Newmark‟s influence chart:, Based on the equation of vertical stress under a circular loaded area, Newmark developed an, influence chart to compute the vertical stress due to load of any shape below any point inside, or outside the loaded area., Equation of circular loaded area,, , The construction of Newmark‟s chart is outlined as below:, , Let a uniformly loaded area of radius „r‟ be divided into 20 area units or sectors. If „q‟ is the, intensity of the load and ‟ζz‟ is the vertical stress at a depth „z‟ below the centre of the, loaded area then each area unit such as OBC produces a stress equal to (ζz/20) at the point, under consideration., , If if be made equal to an arbitrary fixed value, say 0.005, we have, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 75

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Let a concentric circle of radius „r2‟ be drawn and is divided into another 20 area units. Each, area unit such as BCB‟C‟ produces a vertical stress of 0.005q at a depth of „z‟ below the, center of loaded area. Thus stress due to the loaded area OB‟C‟ produces a vertical stress of, 2x 0.005q., , circle, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , 0.27, , 0.4, , 0.52, , 0.64, , 0.77, , 0.92, , 1.11, , 1.39, , 1.91, , ∞, , no, r/z, , From the above equation r10= ∞ and hence tenth circle cannot be drawn, The influence chart is drawn on the basis of data given in the table for „z‟ equal to the, distance AB shown in the chart., If it is required to compute vertical stress ζz at a depth „z‟ direct below any point p on the, loaded area a scale drawing of loaded area is drawn preferably on a tracing sheet using a, scale that corresponds to line AB on the chart (AB= Z m). The outline of the loaded area is, then superimposed on the chart such that the point P coincides with the centre of the chart., The no of area units „N‟ enclosed by the outline is counted. Then the required vertical stress, ζz is calculated., ζz= N.q. influence value., , i., , The vertical stress distribution on horizontal plane „z‟ units below the ground, surface:, The vertical stress on a horizontal plane at a depth „z‟ is given by,, ζz= (Q. Ib), , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 76

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z2, „z‟ being a specified depth for several assumed value of r, r/z is calculated and Ib is, found for each, the value of ζz is then computed., For r=0, ζz will be maximum of 0.4775 Q ., z2, For r=2z, it‟s only about 1.8% of max., For r=3z, it‟s only about 0.3% of max., If Q is taken as unity then the diagram becomes what is known as the influence, diagram for a vertical stress @ A., , Table1: Distance at a specified depth (z=1) say,, r, , r/z, , Ib, , ζz, , 0, , 0, , 0.4775, , 0.4775Q, , 0.25, , 0.25, , 0.4103, , 0.4103Q, , 0.5, , 0.5, , 0.2733, , 0.2733Q, , 0.75, , 0.75, , 0.1565, , 0.1565Q, , 1, , 1, , 0.0844, , 0.0844Q, , 1.25, , 1.25, , 0.0454, , 0.0454Q, , 1.5, , 1.5, , 0.0251, , 0.0251Q, , 1.75, , 1.75, , 0.0144, , 0.0144Q, , 2, , 2, , 0.0085, , 0.0085Q, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 77

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ii., , The vertical stress distribution diagram with depth at a distance „r‟ away from, line of action of force:, , The variation of vertical stress with the depth at a constant radial distance from the, axis of the load may be shown by horizontal ordinate as in the fig below., As „z‟ increases r/z value decreases for a constant value of r., As r/z decreases Ib value in the equation for ζz increases, but since z2 is involved in, the denominator, ζz value first increases with depth and attains a maximum value and, then decreases with further increase in the depth., Table 2: variation of vertical stress with depth for a constant value of r (say r=1, unit)., Ib, z, , r/z, , Ib, z, , 2, , ζz, , 0.5, , 2, , 0.0085, , 0.00340, , 0.0034Q, , 1, , 1, , 0.0844, , 0.0844, , 0.0844Q, , 2, , 0.5, , 0.2733, , 0.0683, , 0.0683Q, , 5, , 0.2, , 0.4329, , 0.0173, , 0.0173Q, , 10, , 0.1, , 0.4657, , 0.0047, , 0.0047Q, , LATERAL EARTH PRESSURE, The structures which are used to hold back a soil mass are called as retaining structures., Retaining walls, Sheet piles, Basement walls, are some of the retaining structures. Soil that is, retained at a slope steeper than it can retain by virtue of its shear strength exerts a pressure on, the retaining wall and this pressure is called as earth pressure. The material that retained is, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 78

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called as backfill. The portion of the back fill lying above a horizontal plane at the elevation, of the top of a wall is called as “Surcharge”. The inclination of surcharge surface to the, horizontal is known as the “surcharge angle β”. The side of the wall in contact with the, back fill is called as “Back Face”. Back of the face is said to have batter which is represented, by the batter angle ϴ, also called as angle of inclination of the back., , Effect of wall movement on earth pressure:, When the wall is rigid and unyielding the soil mass will be in state of rest and there, will be no deformations and displacements. The earth pressure corresponding to this state is, called as” Earth Pressure at Rest”., , Relationship between wall movement and earth pressure., , If the wall rotates about its toe, thus moving away from the backfill, the soil mass expands, resulting in decrease in earth pressure. A portion of the backfill breaks away located next to, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 79

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the retaining wall tends to break away from the rest of the back fill and move downwards and, outwards. The force acting on the retaining wall is called “Active Earth Pressure”., On the other hand, if the wall is pushed towards the backfill the soil is compressed and offers, resistance to movement and earth pressure gradually increases. If the force reaches a value, which a backfill cannot with stand, failure occurs and slip surface develops and the pressure, value reaches the maximum value and maintained constant with further wall movement, this, pressure is called as the “Passive Earth Pressure”., , The actual pressure and the passive earth pressure developed correspondingly to 2 limiting, states of equilibrium and the soil mass is said to be in Plastic equilibrium at these states., , Earth Pressure at Rest:, , H, Po, , po= koγH., Where ko = co-efficient of earth pressure at rest., , Where µ= Poisson‟s ratio., Soil, , ko, , Dense solid, , 0.4- 0.45, , Loose sand, , 0.45- 0.5, , Consolidated soil, , 1.0-4.0, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 80

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For sands and OC clays, ko can be expressed as, ko= 1- sinΦ‟,, Consider an element of soil at depth z, being acted upon by vertical stress ζv and horizontal, stress ζh. There will be no shear stress. The lateral strain εh = (ζh-μ(ζh+ ζv))/E The earth, pressure at rest corresponding to lateral strain = zero, ζh =μ(ζh+ ζv), , Earth pressure theories:, The 2 classical theories of earth pressure are those due to Coulomb (1776) and Rankine, (1857)., , Rankine theory of earth pressure:, Assumptions:, 1., , The soil mass is homogeneous, isotropic and cohesion less., , 2., , The soil is in plastic equilibrium during active and passive pressure conditions., , 3., , The rupture (slip) surface is planar surface which is obtained by considering the, , plastic equilibrium of soil., 4., , The backfill surface is horizontal., , 5., , Back of the wall is vertical., , 6., , Back of the wall is smooth., , We know that, ζ1= ζ3 tan2 (45+ Φ/2) + 2c tan (45+ Φ/2)., As per assumptions, soil is cohesion less i.e. c=0, Then the equation will be, ζ1= ζ3 tan2 (45+ Φ/2)…. (1), In active case (active Rankine case) ζx goes on decreasing and maintained as pa., Here ζ1= ζz = γz= major principal stress., ζ3= pa., Therefore, the equation 1 will be, γz= pa tan2 (45+ Φ/2)., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 81

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b) Submerged backfill:, , H1, H1/3, , P1, , H, γ', P2, (H-H1)/2, , P3, (H-H1)/3, kaγH1, , kaγ‟ (H-H1) γw (H-H1), , c) The effect of surcharge: When a back fill carries a uniformly distributed load of, intensity q per unit area, the effective vertical pressure ζz at any depth within the back, fill increases by q and therefore the increase in the active earth pressure at a depth of, H is given by, kaγH +ka q., q, , H, , ka q, , ka Hγ, , d) Sloping ground surface:, Consider a sloping surface behind the wall be inclines at an angle β with the, horizontal. The Rankine theory assumes that the vertical pressure on soil element, within the back fill and the lateral pressure on the vertical plane of the element are, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 83

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conjugate stresses i.e. the direction of one pressure is parallel to the plane on which, the other pressure acts. Both being conjugate the lateral and vertical pressures have, the same angle β., Consider a soil element at a depth z within a back fill with sloping surface as shown, in fig. below., , stresses due to obliquity β., , is given by,, , β, , A, , H, , Their interrelationship, , Pa, , kaγH.cosβ, , H/3, B, , The top plane of the element is parallel to the ground surface. The intensity of, vertical pressure is given by, ζz‟= γ z cosβ., The pressure ζz‟ and ζx‟ are not the principal stresses due to obliquity β. Their, interrelationship is obtained by means of a Mohr‟s circle, Co efficient of active earth for sloping back fill is given by,, ka=, e) Inclined back of the wall:, , The total active earth pressure Pa for the wall of height H is given by,, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 84

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2c√Ka, -, , zo= 2c /(γ√Ka), , 2zo, , H, , γ H Ka, It can be seen that up to a depth of zo, a –ve pressure exists. In the zone (z=0 to z= zo) the soil, will be in state of tension in active case. In practice since the soil cannot take tension in the, active case, the tension cracks tends to develop and the pressure cannot act on the wall., Hence while calculating the total active thrust the tension zone is usually ignored and only, area of the pressure distribution diagram below zo is considered., The total net pressure is given by,, Pa= 0.5 γ H2 cot2α- 2cH cotα., Where α= (45+ Φ/2), It‟s clear from the pressure diagram that from active thrust is zero for a depth equal to 2zo., Thus it is implied that in cohesive soil a vertical cut can be made up to a depth of 2zo without, having to provide any lateral support. This depth is called critical depth of vertical cut., The critical height Hc is given by, Hc= 4c/ (γ √ (Ka)., Passive pressure of cohesion less soil on retaining walls:, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 86

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H, Pp= Total Passive thrust= 0.5, 2, , kaγH ., , H/3, pp= kpγH., γ= kp. γ H, , pp, , Total passive resistance,, Pp= 0.5 γH2kp., Passive pressure of cohesive soil on retaining walls:, q, , H, , kpq 2c√kp, , γ H kp, , LIMITATIONS OF RANKINE‟S THEORY:, i., , Rankine‟s theory assumes that back of the wall to be smooth, no frictional forces, assumed to exist between soil and the wall, thus lateral pressure is taken to act, parallel to back fill surface. But, in practice with the movement of wall considerable, friction may develop between the soil and the wall and as a consequence the earth, pressure will be inclined at certain angle to the normal of the wall., , ii., , The Rankine‟s assumption of smooth wall surface results in over estimation of, active pressure and under estimation of passive pressure., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 87

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iii., , It is difficult to use Rankine‟s theory for the cases of irregular back fill surfaces or, where surcharge loads are complex., , Coulomb‟s Theory of Earth Pressure:, Coulomb‟s theory of earth pressure involves the consideration of sliding wedge which tends, to break away from the rest of the back fill upon wall movement. When the wall moves, outwards the sliding wedge moves downwards and outwards; and when the wall is pushed, towards the back fills the sliding wedge moves upwards and inwards. The Coulomb theory, takes into account the friction b/w back of the wall and soil by introducing the angle of, friction δ., Assumptions made in Coulomb‟s theory:, 1., , The back fill is dry, cohesion less, homogeneous and isotropic in nature., , 2., , The soil possesses both friction and cohesion can also be treated with coulomb‟s, , theory. The backfill surface is planar and can be inclined., 3., , The back of the wall can be inclined to the vertical., , 4., , The failure surface is a plane surface which passes through the heel of the wall., , 5., , The sliding wedge is considered to be rigid body and the earth pressure is obtained by, , considering the limiting equilibrium of the sliding wedge as a whole., FAILURE UNDER ACTIVE CONDITIONS, For cohesion less back fill the three forces are keeping the wedge in equilibrium are:, i., , Weight of the wedge, , ii., , Soil reaction force (R) on the reaction plane., , iii., , Reaction to the force (Pa) between the wall and the soil, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 88

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3., , A second line AG known as Earth Pressure Line is drawn at an angle ψ (90-θ-δ) with, , W., 4., , Lines are drawn through points D1, D2, D3, D4…..parallel to the earth pressure line to, , intersect the correspondingly slip, , planes at E1, E2, E3…., AD1E1,, , AD2E2,, , AD3E3…..represent the Force Triangles corresponding to trial wedges ABC1, ABC2,, ABC3…… and lengths D1E1, D2E2….represent the corresponding active earth pressure., 5., , The points E1, E2, E3…… are joined to get the Earth Pressure Locus., , 6., , To obtain the maximum active thrust (Pamax), a tangent to AF is drawn to the Cullman, , curve. Then the line DE is drawn parallel to the earth pressure line AG through the point of, tangent E. then DE represents max active thrust when converted to a scale. The critical slip, line will be AEC., , Pa=, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 90

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= 0. 5H1.BC.1. γ. DE/AD, Pa= 0.5H1. DE. γ., Rebhan‟s Graphical Method:, Based on the principle postulated by Rebhan‟s Poncelet suggested a graphical method to, determine the earth pressure on a rough wall for a non cohesive, homogeneous, and inclined, back fill., Step by step procedure is as follows:, 1. Choose a suitable scale and draw the retaining wall along with the back fill., 2. Draw a line AD from a point A making an angle Φ with the horizontal., 3. Draw another line AE at an angle ψ= 90- θ-δ., 4. Draw a semi circle with AD as diameter., 5. Through B draw a line parallel to pressure line to intersect AD at F., 6. Erect a perpendicular at F to intersect the semi circle at G., 7. With A as centre AG as radius draw a semi circle to meet line AD at J., 8. Through J draw a line parallel to pressure line to intersect the back-fill surface at C., 9. With J as centre, JC as radius, draw an arc to meet AD at K., 10. Join KC and AC., 11. Find the area of the triangle JKC, then total active thrust, Pa= (area of Δ JKC). γ., STABILITY OF SLOPES, Slope of earth are of two types: 1. Natural slope., 2. Man made slope., The natural slopes are those that exist in nature and are formed by natural causes. Such, slopes exist in hilly areas., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 91

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The sides of cutting, the slopes of embankments constructed for taking roads, railway lines,, canals etc and slopes of earthen dams constructed for storing water are some examples for, manmade slopes., , Stability of slopes:, Earth slopes whether natural or manmade can be broadly classified into two groups:, a) Infinite slopes., b) Finite slopes., , The term infinite slope is used to designate a constant slope of infinite content. The long, slope of a mountain is an example of this type, whereas finite slopes are limited in extent., The slopes of earthen embankment dams are example of finite slopes., , Slope stability:, Slope stability is one of the important considerations in the design of construction of earth, dams. The result of slope failure can be often catastrophic involving the loss of considerable, properties and many lives., , Causes of failure of slopes:, The important factors that cause instability in a slope and lead to failure are:, 1. Gravitational force., 2. Force due to seepage water., 3. Erosion of surface of slopes due to the flowing water., 4. The sudden lowering of water adjacent to a slope., 5. Force due to earth quake., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 92

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Types of failure of finite slopes:, Two basic types failure of a finite slope may occur:, Face failure., a. Slope failure., Toe failure., , b. Base failure., , Df<1, , D f=1, , Df>1, , H. Df, , H.Df, , Df. H, , H, , Face failure, , Toe failure, , Base failure, , If failure occurs along a surface of sliding that intersects the slope at or above the toe, the, slide is known as slope failure., , The slope failure is called as Face failure if the arc passes above the toe., If the arc passes through the toe then it is called as Toe failure., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 93

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On the other hand the soil beneath the toe of the soil is weak, the failure occurs along a, surface that passes at some distance below the toe of the slope. Such type of failure is called, as Base failure., , If Df H represents the depth of lowest point of the failure surface from the top of the slope,, then Df is called as depth factor and H is the height of the slope., , For face failure Df <1,, For toe failure Df =1,, For base failure Df >1., , Assumptions made in stability analysis of slope:, , 1., , The soil is homogeneous so that the properties such as unit weight and shear strength, , parameters (c, Φ) can be assumed to remain constant at every point in the entire soil mass., 2., , Slip surface is assumed to be along a circular arc., , 3., , Shear resistance developed along a soil slip surface is assumed to be uniform., , 4., , The Coulomb‟s equation for shear strength, η= c + tanΦ is assumed to be valid., , Methods of stability analysis of finite slopes:, There are several methods of stability analysis, important being are:, Φ=0 analysis, a., , Swedish method (Slip Circle method)., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 94

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c-Φ analysis (method of slices)., , b. Friction circle method., c. Taylor‟s stability number., , A. Swedish method:, i., , Purely cohesive slopes (Φ=0 analysis):, Let AB be the trial slip surface and R be the radius of the slip circle. θ be the, , angle subtended by the slip circle at the centre., , „W‟ is the weight of soil mass and „c‟ is the unit cohesion., Weight „W‟ provides the driving force and CL represents the resisting force. Taking moment, about the centre of rotation „O‟,, Driving moment MD= wx., Resisting moment MR= CLR= CR2θ., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 95

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Therefore the factor of safety against sliding,, F= MR, MD, , ii., , ., , = CR2θ ., w. x, , Slope made up of C-Φ soil or C-Φ analysis:, , In this method the soil mass contained b/w slip circle and slope is divided into number of, slices each of width „b‟., , The weight „W‟ of each slice is assumed to act at its centre. Let N and T represent normal and, tangential components. Consider a typical slice 4, its weight W4. Its weight is resolved into, normal component N4 and tangential component T4. As the normal component passes through, the centre of the rotation, it will not cause any driving moment. However the tangential, component T4 causes a driving moment T4R., If „C‟ is the unit cohesion. ΔL is the curved length of each slice then the resisting force from, Coulomb‟s equation,, , C ΔL+ N3 tanΦ., For the entire slip surface AB, the driving moment, MD= (∑T) R., Resisting moment, MR= (C∑ΔL + (∑N) tanΦ) R, = (CL+ ∑N tanΦ) R, Where ∑T is the algebraic sum of the entire tangential component., ∑N is the algebraic sum of the entire normal component., , Factor of safety, F= MR/MD., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 96

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Calculation of ∑N and ∑T:, First resolve the mid ordinate of each slice into normal and tangential components. Let N1,, N2, and N3……. represent normal component and T1, T2, T3….represent tangential, component. The N and T rectangles are drawn as below., , ∑N= (Area of N rectangle) γ., ∑T= (Area of T rectangle) γ., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 97

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MODULE 5:, BEARING CAPACITY, The structural foundations can be broadly classified into 2 categories:, 1. Shallow Foundation, 2. Deep Foundation., General Requirements of Foundation:, 1., , Location and Depth criteria., , 2. Shear Failure/ Bearing Capacity Criteria., 3., , The Settlement Criteria., , Definitions:, 1. Footing: A footing is a portion of the foundation of a structure that transmits loads, directly to the soil., 2. Foundation: A foundation is that part of the structure which is in direct contact with, and transmits load to the ground., 3. Foundation soil: It is the upper part of the earth mass carrying the load of the, structure., 4. Bearing Capacity: The supporting power of the soil or rock is referred to as Bearing, Capacity., 5. Gross Pressure intensity (q): The Gross Pressure intensity is the total pressure at the, base of the footing due to the weight of the superstructure, self-weight of the footing, and, the weight of the earth fill., 6. Net Pressure Intensity (qn): It is defined as the excess pressure or the difference in, intensities of the gross pressure after the construction of the structure and the original, overburden pressure. If D is the depth of the footing, then, qn = q - ζ= q - γD., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 98

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7. Ultimate bearing capacity (qf): It is defined as the minimum gross pressure intensity at, the base of the foundation at which the soil fails at shear., 8., , Net Ultimate Bearing Capacity (qnf): It is the minimum net pressure intensity, , causing shear failure of the soil., qnf= qf - ζ= qf - γD, 9., , Net Safe Bearing Capacity (qns): The Net safe bearing capacity is the net, , ultimate bearing capacity divided by a factor of safety F, qns = qnf, F, 10., , Safe Bearing Capacity (qs): the maximum pressure the soil can carry safely, , without risk of shear failure is called as Safe bearing capacity., qs = qnf + γD., F, 11., , Safe Bearing Pressure:, , It is the intensity of loading that will cause the, , permissible settlement or specified settlement for the structure., , Minimum Depth of Foundation : Rankine‟s Analysis:, Rankine‟ considered equilibrium of 2 soil elements:, 1. Immediately below the foundation (element1)., 2. Just beyond the edge of the footing (element2)., When load on footing is increased and approaches a value of qf, a state of equilibrium, state is reached under the footing., For shear failure of element 1, element 2 also fails by lateral thrust from element1., During the state of shear failure, the following stress relationship exists., ζ1=ζ3tan2α+ 2c tanα., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 99

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ζ1=ζ3tan2α, , For cohesion less soil,, For the element2,, , ζ3 = ζv = γ D., , Therefore,, , ζ1 = ζh = γ D tan2α., ζ3 = ζh = ζ1 of element 2, , For the element1,, , = γ D tan2α, Therefore,, , ζ1 = γ D tan4α., , But,, , ζ1 = qf, qf = γ D tan4α, , -------------(1), Eq 1 gives the bearing capacity of cohesion less soil as zero at the ground surface., ------------(2), Types of Bearing Capacity Failures:, 1. General Shear Failure., 2. Local Shear Failure., 3. Punching shear Failure., , 1), , General Shear Failure:, , When the pressure approaches a valve qf, the state of plastic equilibrium is reached, initially in the soil around the edges of the footing. Then it gradually spreads, downwards and outwards. Ultimately the State of Equilibrium is attained fully, throughout the soil above the failure surface. The failure is accompanied by bulging of, sheared mass of soil. Final slip moment would occur only at one side, accompanied by, tilting of footing. Such a failure occurs in soils of low compressibility., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 100

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2), , Local Shear Failure:, , In this type, only a partial plastic equilibrium will be developed. Due to this the failure, surfaces do not reach the ground surface and only slightly heaving occurs. In this type, of failure, the tilting of foundation does not occur. This type of failure is associated, with soils of high compressibility and in sands having relative density lying between, 35 and 70 percent. Ultimate bearing capacity in such a failure is not well defined., 3), , Punching shear Failure:, , This type of failure occurs only when there is relatively high compression of soil under, the footing. This may occur in relatively loose sand with relative density less than, 35%. This may also occur in soil of low compressibility if the foundation is located at, considerable depth. In this type of failure there is no heaving of soil mass or tilting of, the footing. Relatively large settlement occurs in this mode. Ultimate bearing capacity, is not well defined., , Assumptions Made in Terzaghi‟s Analysis:, 1), , The soil is homogeneous and isotropic and its shear strength is represented by, , Coulomb's equation., 2), , The strip footing has a rough base and the problem is essentially 2 dimensional., , 3), , Failure zone do not extend above the horizontal plane through the base of the, , footing., 4), , The elastic zone has straight boundaries inclined at ψ=ϕ, to the horizontal and, , the plastic zone fully develop., , Terzaghi‟s Bearing Capacity Theories for different footings:, Strip footing:, qf= CNc+ γDNq+0.5 γ BNγ., Square Footing:, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 101

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qf=1.3CNc+ γDNq+0.4 γ BNγ., Circular Footing:, qf=1.3CNc+ γDNq+0.3 γ BNγ., Where, Nc, Nq, Nγ are dimensionless bearing capacity factor depending upon angle of, internal friction., ϕ, , General Shear Failure, , Local Shear Failure, , Nc, , Nq, , Nγ, , Nc, , Nq, , Nγ, , 0, , 5.7, , 1, , 0, , 5.7, , 1, , 0, , 5, , 7.3, , 1.6, , 0.5, , 6.7, , 1.4, , 0.2, , 10, , 9.6, , 2.7, , 1.2, , 8.0, , 1.9, , 0.5, , 15, , 12.9, , 4.4, , 2.5, , 9.7, , 2.7, , 0.9, , 20, , 17.7, , 7.4, , 5.0, , 11.8, , 3.9, , 1.7, , 25, , 25.1, , 12.7, , 9.7, , 14.8, , 5.6, , 3.2, , 30, , 37.2, , 22.5, , 19.7, , 19.0, , 8.3, , 5.7, , 34, , 52.6, , 36.5, , 35.0, , 23.7, , 11.7, , 9.0, , 35, , 57.8, , 41.4, , 42.4, , 25.2, , 12.6, , 10.1, , 40, , 95.7, , 81.3, , 100.4, , 34.9, , 20.5, , 18.8, , 45, , 172.3, , 173.3, , 297.5, , 51.2, , 35.1, , 37.7, , 48, , 258.3, , 287.9, , 780.1, , 66.8, , 50.5, , 60.4, , 50, , 347.5, , 415.1, , 1153.2, , 81.3, , 65.6, , 87.1, , Limitations in Terzaghi‟s Analysis.:, 1., , As the soil compresses, ϕ changes: slight downward movement of footing, , may not develop fully the plastic zones., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 102

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2., , Error due to assumptions no. 3 increases with depth of foundation and hence, , theory is suitable for shallow foundation only., 3., Bearing Capacity Based on Brinch Hansen‟s Analysis:, Brinch Hansen has Proposed a general equation for Ultimate Bearing Capacity of soil:, qf= c. Nc. sc. dc. ic. gc. bc+ζo. Nq. sq. dq.iq. gq. bq+ 0.5γ. B. Nγ. sγ. dγ. iγ. gγ. bγ., Where,, , ,, , ,, , and,, , where, ζo = effective over burden pressure at foundation level, s= shape factor, d= depth, factor, i= load inclination factor, g= ground inclination factor, b=base inclination factor., For vertical loads i=1,, For horizontal ground surface g=1,, For horizontal base b=1,, For strip footing S=1, , Effect of Ground water table on bearing capacity:, The Terzaghi‟s equation for bearing capacity is given by:, qf= CNc+ γDNq+0.5 γ BNγ., Water is known to effect on its unit weight and also the shear parameters. When the soil is, submerged under water the effective unit weight is to be used in the computation of bearing, capacity. However, the effect of water table on the shear parameter of the foundation soil is, usually considered small it is ignored. But the effective unit weight is roughly half of the, saturated unit weight consequently there will be 50% reduction in the value of corresponding, term in the bearing capacity formula. For locations of ground water table within a depth of, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 103

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the width of the foundation below the base and the GL. The equation for the ultimate bearing, capacity may be modified as follow:, qf= C*Nc+ γ1*D*Nq *Rw2+0.5* γ2*B*Nγ.*Rw1, , where Rw1 and Rw2 are the reduction factors for the water table at Zw1 and Zw2, respectively, Rw1= 0.5[1+Zw1], B, , Rw2= 0.5[1+Zw2], D, , Zw2, D, , B, , Zw1, GWT, , Effect of eccentricity of loading:, If total load „V‟ on the footing acts eccentrically (i.e. line of action of V not passing through, the center of gravity of footing are), the width B and length L should be reduced as under:, B‟= B-2ex., L‟= l-2ey., A‟=B‟* L‟., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 104

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i., , Static Cone Penetration Test: (SCPT), , In this method continuous record of resistance of soil is obtained by penetrating steadily, under static pressure a cone with a base of 10cm2 and an angle of 60o at the vertex. The cone, is carried at the lower end of a steel driving rod which passes through a steel tube with, external diameter equal to the base of the cone. Initially the cone alone is driven into the soil, mass up to 8cm. the maximum value of resistance is recorded. The steel tube is then pushed, down up to the cone. And then both together are penetrated through a depth of 20cm to give, the total of the cone resistance and the frictional resistance along the tube. Among the, sounding tests the SCPT is the best and can give more reliable values in locations below the, water table and where SPT fails., , Meyerh (1950) suggested allowable bearing soil pressure based on cone penetration test, values restricting the settlement not to exceed 2.5mm, 𝑞𝑎=𝑞𝐶𝑆/30, for B≤1.22m, 𝑞𝑎=𝑞𝐶𝑆/50∗{(0.3+𝐵)/𝐵}2 , for B>1.2m, Where, N is the SPT value, and B is the width of the footing in „m‟,, , ii., , Standard penetration test: (SPT), , The test is performed in a clean hole, 55 to 150mm in diameter. A casing or drilling mud is, used to support the sides of the hole. A split spoon sampler (OD 50.8mm and ID 35mm), resting on the bottom of the bore hole is allowed to sink under its own weight. It is then, seated 15cm with the blows of the hammer falling through a height of 75 cm. Next the, sampler is driven by 30 cm or 50 blows. The number of blows required to affect each 15cm, penetration is recorded. The first 15 cm drive may be considered as seating drive. The total, blows required for the second and third 15cm of penetration is termed as penetration, resistance N. If the split spoon sampler is driven less than 45cm then penetration resistance, shall be for the last 30cm of penetration. The entire sampler may sometimes sink under its, own weight when very soft sub soil stratum is encountered. Under such conditions it may not, be necessary to give any blows to sampler and SPT value to be indicated as the observed, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 105

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value of N is corrected for overburden and dilatancy/ submergence. Correction for over, burden: the N value for cohesion less soil can be corrected for overburden. No= C n. N,, Where Cn is the normalizing correction factor. Correction due to dilatancy/ submergence: The, value No obtained after applying over burden correction is corrected further for dilatancy if, the stratum consists of fine sand and silt below the water table, for values of N greater than, 15 using following expression:, Ne= 15+ (No-15), Teng(1928) proposed an equation based on SPT value ‟N‟ to determine net allowable bearing, pressure for a settlement of 25mm: 𝑞𝑛𝑎=34.5 (𝑁−3) ∗{(0.305𝐵+1)/0.7𝐵}, Meyerh(1950) suggested that, 𝑞𝑛𝑎=12𝑁, for B≤1.22m and 𝑞𝑛𝑎=8𝑁∗{(0.305𝐵+1)/0.305𝐵},, for B>1.2m, Where, N is the SPT value, and B is the width of the footing in „m‟,, , Plate load test:, Plate load test is a field test to determine the ultimate bearing capacity of the soil and, the probable settlement under the given loading. This test essentially consists in, loading a rigid plate at the foundation level and determining the settlement, corresponding to each load increment. The ultimate bearing capacity is taken at the, load at which the plate starts sinking at a rapid rate., Bearing plate: the bearing plate is either circular or square made up of mild steel of, not less than 25mm in thickness and varying in size from 300 to 750mm with, chequered or grooved bottom., Test pit: the test pit normally at the foundation level, having generally a width equal, to five times the test plate width(Bp)., Loading arrangements: the loading of the test plate may be applied with the help of, hydraulic jack. The reaction of the hydraulic jack may be borne by either of the, following methods:, a) Gravity loading frame method., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 106

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b) Reaction truss method., , i., , Gravity loading method:, , In this case a platform is constructed over a vertical column resting on a test plate. Loading is, done with the help of sand bags, stones, or concrete blocks. When the load is applied to the, plate it sinks or settles. The settlement of the plate is measured with the help of sensitive dial, gauges. For square plate 2 dial gauges are used. The dial gauges are mounted on, independently supported datum bar. As the plate settles the ram of the dial gauges moves, down and settlement is recorded. The load is indicated on the load gauge of the hydraulic, jack., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 107

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ii., , Reaction truss method:, , The truss is held to the ground through the soil anchors. These anchors are firmly driven in, the soil with the help of hammers. The reaction truss is usually made of mild steel sections., Guy ropes are used for the lateral stability of the truss., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 108

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Setting of the plate:, The test plate is placed over a fine sand layer of maximum thickness of 5mm, so that the, center of plate coincides with the center of the girder with the help of plumb and bob. It is, should be horizontally levelled to avoid eccentric loading. The hydraulic jack is centrally, placed over the plate with the loading column in between the jack and reaction beam., , Load increments:, A minimum seating pressure of (0.7t/m2) is applied and removed before the test. The load is, applied without any impact, eccentricity etc. The load is measures over the pressure gauge, attached to the pumping unit kept over the pit, away from the testing plate, through the, extending the pressure pipes., Settlement and observations:, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 109

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Settlement should be observed for each load increments after the interval of 1, 2.25,4, 6.25, 9,, 16, 25 min and thereafter hourly intervals to the nearest 0.02mm. In case of clayey soils, the, time-settlement curve shall be plotted at each load stage and load shall be increased to next, stage either when the curve indicates that the settlement has exceeded 70 to 80 % of probable, ultimate settlement at that stage., Effect of size of bearing plate:, The bearing capacity of sands and gravels increases with the size of the footing. The relation, ship can be expressed as under:, , The settlement of the footing varies with its size., , Limitations of plate load test:, i., , The test result reflects only the character of the soil located within the depth less than, twice the width of the bearing plate., , ii., , It is essentially a short duration test and hence the test does not give the ultimate, settlement, particularly in the case of cohesive soil., , iii., , For the clayey soils the ultimate pressure for a large foundation is the same as that for, the test plate. But in dense sandy soils the bearing capacity increases with the size of, the foundation and the test on the smaller size bearing plates tend to given, conservative values., , Design of rectangular combined footing:, , The design of rigid rectangular combined footing consists in determining the location of, center of gravity of the column loads and using length and width dimensions such that, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 110

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centroid of the footing and the center of gravity of column loads coincide. The resulting, pressure distribution will be rectangular with the pressure intensity, 𝑞=(𝑃1+𝑃2)/𝐵., The column loads may be considered to be concentrated loads and the resulting shear force, and bending moment diagrams can be plotted. The maximum bending moment should be, adopted as the design value for the reinforced concrete footing, which should also be checked, for maximum shear etc., , Design of trapezoidal combined footing:, When the two column loads are unequal, with the outer column carrying heavier load, and, when there is space limitation beyond the outer column trapezoidal footing is provided. In, order to determine widths a and b, the following relations are used:, , Where,, , = allowable soil pressure, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 111

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By taking moments about the property line or left edge and on simplifying,, , By above 2 equations B1 and B2 can be found, thus the solution leads to:, , The pressure intensities q1 and q2 are calculated, once B1 and B2 are obtained:, , and, The bending moments and shear force diagrams are drawn, the maximum values are used for, the purpose of design., B1=0 where,, For a rectangular shape,, Thus, a trapezoidal combined footing solution exists when, , is such that:, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 112

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MODULE1, 1. What are the methods used for general exploration?, a) Subsurface penetration, b) Ground water exploration, c) Rock Cuttings, d) All of the mentioned, 2. For pile foundations, the depth of exploration at the start of the work is, _________, a) 10 meters, b) 40 meters, c) 70 meters, d) 200 meters, 3. The feature that helps to estimate the relative density of coarse-grained soil is, _______, a) Soundings, b) Shallow test pits, c) Exploratory borings, d) Geophysical method, 4. The number and disposition of bore holes are varied, depending upon _______, a) Surroundings, b) Strata, c) Subsoil condition, d) Ground water, 5. The various method of site exploration can be grouped under, which of the, following?, a) Open excavations and Borings, b) Soil strata, c) None of the mentioned, d) All of the mentioned, 6. Hand auger can be used for depths up to ________, a) 7 m, b) 6 m, c) 2 m, d) 10 m, 7. Wash boring cannot be used for _________ type of soil strata., a) Cohesive soil, b) Cohesion less soil, c) Boulder, d) All of the mentioned, 8. The commonly used geophysical method for site exploration is ________, a) Gravitational method, b) Electrical resistivity, c) Magnetic method, d) All of the mentioned, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , 01, , 01, , 01, , 01, , 01, , 01, , 01, , 01, , Page 115

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9. Electrical resistivity method is based on measurement of _____________, a) Specific resistance, b) Voltage, c) Potential drop, d) Current, 10. Exploratory borings in general exploration is carried out by using, ___________, a) Auger, b) Bore equipment, c) Well curb, d) All of the mentioned, , 01, , 11. Hand auger can be used for depths up to ________, a) 7 m, b) 6 m, c) 2 m, d) 10 m, , 01, , 12. Wash boring cannot be used for _________ type of soil strata., a) Cohesive soil, b) Cohesion less soil, c) Boulder, d) All of the mentioned, , 01, , 13. The commonly used geophysical method for site exploration is ________, a) Gravitational method, b) Electrical resistivity, c) Magnetic method, d) All of the mentioned, , 01, , 14. Electrical resistivity method is based on measurement of _____________, a) Specific resistance, b) Voltage, c) Potential drop, d) Current, , 01, , 15. The commonly used penetration test are _______, a) IS penetration test, b) Cone penetration test, c) Dutch standard test, d) All of the mentioned, , 01, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , 01, , Page 116

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16. In the seismic refraction method, the waves sent along the ground surface is, picked by _________, a) Geo satellite instrument, b) Geophone, c) Wave detector, d) All of the mentioned, , 01, , 17. The methods of site investigation are dependent upon ________, , 01, , (A) Climatic condition, B) Nature of engineering project, (C) Local topography, (D) All of the mentioned, , 18. Depending upon the details, the site exploration may be classified as _______, , 01, , (A) General and Detailed, (B) Complex, (C) None of the mentioned, (D) All of the mentioned, , 19. The type of boring method that can be used for both rock and soils are, ________, , 01, , (A) Shell boring, (B) Wash boring, (C) Auger boring, (D) Rotary boring, , 20. Which of the following method is adopted for fast boring?, , 01, , (A) Cylindrical augers, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 117

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(B) Percussion boring, (C) Rotary boring, (D) Wash boring, , MODULE 2, 21. The vertical stress at point P1 due to the point load Q on the ground surface as, shown in figure is σ2. According to Boussinesq‟s equation, the vertical stress at, point P2 shown in figure will be, , 01, , (A) ζz/2, (B) ζz, (C) ζ2z, (D) 4ζz, 22. According to the assumptions in Boussinesq‟s, the soil is _______________, a) Homogeneous and Isotropic, b) Non Homogeneous, c) None of the mentioned, d) All of the mentioned, , 01, , 23. A curve or contour connecting all points below the ground surface of equal, vertical pressure., a) pressure bulb, b) contact pressure, c) isobar, d) none of the above, , 01, , 24. The pressure transmitted from the base of a foundation to the soil is termed as., a) pressure bulb, , 01, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 118

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b) contact pressure, c) isobar, d) none of the above, 25. The zone in a loaded soil mass bounded by an isobar of a given vertical, pressure intensity is, a) pressure bulb, b) contact pressure, c) isobar, d) none of the above, , 01, , 26. The Boussinesq‟s influence factor is denoted by, a) KB, b) IB, c) BI, d) none of the above, , 01, , 27. The Westergaard‟s influence factor is denoted by, a) KB, b) IB, c) BI, d) none of the above, , 01, , 28. When z=0, the Boussinesq‟s influence factor value is equal to, a) 0.5, b) 0.445, c) 0.637, d) 0, , 01, , 01, 29. The following equation relates to:, a) Vertical stress below the center of uniformly loaded circular area, b) Vertical stress below the Rectangular loaded area, c) vertical stress due to line loads, d) vertical stresses due to strip loads, 01, 30. The following equation relates to:, a) Vertical stress below the center of uniformly loaded circular area, b) Vertical stress below the Rectangular loaded area, c) vertical stress due to line loads, d) vertical stresses due to strip loads, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 119

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31. The Westergaard‟s influence factor is denoted by, a) KB, b) IB, c) BI, d) none of the above, 32. When z=0, the Boussinesq‟s influence factor value is equal to, a) 0.5, b) 0.445, c) 0.637, d) 0, , 33. The following equation relates to:, , 𝑞, , [, , ( ), , ], , a) Vertical stress below the center of uniformly loaded circular area, b) Vertical stress below the Rectangular loaded area, c) vertical stress due to line loads, d) vertical stresses due to strip loads, , 34. The following equation relates to:, , ∗ [, , ( ), , ], , a) Vertical stress below the center of uniformly loaded circular area, b) Vertical stress below the Rectangular loaded area, c) vertical stress due to line loads, d) vertical stresses due to strip loads, , 35. The following equation relates to:, , 𝑞, , ( ⁄ ), , ⁄, , ), , a) Vertical stress below the center of uniformly loaded circular area, b) Vertical stress below the Rectangular loaded area, c) vertical stress due to line loads, d) vertical stresses due to strip loads, , 36. The following equation relates to:, , 𝑛, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 120

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a) Vertical stress below the center of uniformly loaded circular area, b) Vertical stress below the Rectangular loaded area, c) vertical stress due to line loads, d) vertical stresses due to strip loads, , 37. The following equation relates to:, , ∗, , ⁄, , a) Vertical stress below the center of uniformly loaded circular area, b) Vertical stress below the Rectangular loaded area, c) vertical stress due to point load- Boussinesq‟s Equation, d) vertical stress due to point load- Westergaard‟s Equation, , 38. The following equation relates to:, , ∗, , ∗, , ⁄, , a) Vertical stress below the center of uniformly loaded circular area, b) Vertical stress below the Rectangular loaded area, c) vertical stress due to point load- Boussinesq‟s Equation, d) vertical stress due to point load- Westergaard‟s Equation, , 39. When r/z=0,, a) Value of IW is less than that of IB by33%, b) Value of IW is less than that of IB by 67%, c) Value of IB is less than that of IW by33%, d) Value of IW is less than that of IB by67%, 40. Contact pressure mainly depends upon, a) allowable bearing pressure of soil, b) type of footing, c) nature of foundation structure and nature of soil, d) none of the above, , MODULE 3, 41. On designing retaining walls it is necessary to take care of __________, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , 01, , Page 121

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exerted by soil mass., a) Erosion, b) Lateral pressure, c) Surcharge, d) Lateral stress, 42. The material retained or supported by the retaining structure is called, __________, a) Surcharge, b) Support wall, c) Back fill, d) All of the mentioned, , 01, , 43. The coefficient of earth pressure when the soil is at equilibrium is, ___________, a) ζv /ζh, b) ζh /ζv, c) ζv × ζh, d) ζ1 / ζ3, , 01, , 44. In active stress, the major principal stress ζ1 acting on the wall will be in, __________ plane., a) Vertical, b) Horizontal, c) Inclined, d) Zero, , 01, , 01, 45. In passive stress, the major principal stress ζ1 acting on the wall will be in, __________ plane., a) Vertical, b) Horizontal, c) Inclined, d) Zero, 46. The position of the backfill lying above the horizontal plane at the top of wall, is called _________, a) Active state, b) Plasticity, c) Surcharge, d) Slip lines, 47. Based on the assumptions of Rankine‟s theory, the soil mass is __________, a) Stratified, b) Submerged, c) Homogeneous, d) All of the mentioned, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , 01, , 01, , Page 122

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48. The value of K0(coefficient of earth pressure at rest) for loose sand is, __________, a) 0.6-0.65, b) 0.5-0.55, c) 0.45-0.50, d) 0.8-1, , 01, , 49. The value of K0(coefficient of earth pressure at rest) for dense sand is, __________, a) 0.6-0.65, b) 0.5-0.55, c) 0.45-0.50, d) 0.4- 0.45, , 01, , 50. The expression for K0 as given by Jacky is ___________, a) K0 = 1 – sin θ, b) K0 = sin θ, c) K0 = 1 – cos θ, d) K0 = 1 + sin θ, , 01, , 51. Slopes is classified into ____________ types., a) 2, b) 3, c) 4, d) 5, 52. Which of the following is an example of slopes extending to infinity?, a) Inclined face of Earth dams, b) Embankments, c) Cuts, d) None of the mentioned, , 53. In Swedish circle method ___________ analysis cases are considered., a) θC = 0, b) C – θ and θ = 0, c) None of the mentioned, d) All of the mentioned, 54. Analysis of stability of slopes is used for determining ___________, a) Shearing strength and Stressed internal surface, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 123

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b) Properties of the soil, c) None of the mentioned, d) All of the mentioned, 55. Factor of safety with respect to height is given by ______________ equation., a) FC = HC / H, b) FC = C / Cm, c) FC = ηf / η, d) FC = ηC / η, MODULE 4, 56. Local shear failure is associated with soils having _________, a) High compressibility, b) High pore pressure, c) Low porosity, d) Low compressibility, , 01, , 57. Which of the following is not one of the characteristics of a local shear failure?, a) Failure is defined by large settlements, b) Failure surface do not reach the ground surface, c) Failure is sudden, d) Ultimate bearing capacity is not well defined, , 01, , 58. State of equilibrium is fully developed in which of the following bearing, capacity failures?, a) Local shear failure, b) General shear failure, c) Punching shear failure, d) All of the mentioned, , 01, , 59. Which of the following N factors has the widest range of values?, a) Nc, b) Nq, c) Nγ, d) All of the mentioned, , 01, , 60. Which of the following equation has been recommended by Indian standard, for finding reduction factor in water table?, a) R w = 0.5 (1 + zw2/B), b) R w = (1 + z w2/B), c) R w = c Nc + ζ. Nq + 0.5γBNγ, d) R w = z w2 / B, , 01, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 124

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61. The plate load test is essentially a ___________, a) Laboratory test, b) Field test, c) Graphical method analysis, d) None of the mentioned, , 01, , 62. The loading to the test plate is applied with __________, a) Fluid tube, b) Hydraulic jack, c) Sand bags, d) Cross-joists, , 01, , 63. A seating pressure of ___________ is applied on the plate before starting the, load test., a) 70 g/cm2, b) 30 g/cm2, c) 50 g/cm2, d) 100 g/cm2, , 01, , 64. The two commonly used penetration tests are ____________, a) Standard penetration test, b) Cone penetration test, c) All of the mentioned, d) None of the mentioned, , 01, , 65. The split tube used in static cone penetration test, is commonly known as, ______________, a) Split spoon sampler, , 01, , b) Split tube sampler, c) Tube sampler, d) All of the mentioned, 66. The loading to the test plate is applied with __________, a) Fluid tube, b) Hydraulic jack, c) Sand bags, d) Cross-joists, 67. A seating pressure of ___________ is applied on the plate before starting the, load test., a) 70 g/cm2, b) 30 g/cm2, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 125

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c) 50 g/cm2, d) 100 g/cm2, 68. The two commonly used penetration tests are ____________, a) Standard penetration test, b) Cone penetration test, c) All of the mentioned, d) None of the mentioned, 69. The split tube used in static cone penetration test, is commonly known as, ______________, a) Split spoon sampler, b) Split tube sampler, c) Tube sampler, d) All of the mentioned, , 70. The total blow required for the second and third 15 cm of penetration in, standard penetration test is taken as ____________, a) Seating drive, b) Penetration resistance, c) Overburden pressure, d) Dilatancy/submergence, 71. The parameters Nc, Nq, Nγ in the equations of bearing capacity failure are, known as _________, a) Constant head, b) Bearing capacity factors, c) Effective pressure, d) Load intensity, 72. According to the assumptions in Terzaghi‟s analysis, the soil is, _______________, a) Homogeneous and Isotropic, b) Non Homogeneous, c) None of the mentioned, d) All of the mentioned, MODULE 5, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 126

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73., , For simple spread footing on clayey soil( for steel structure), the, differential settlement should not exceed __________, a) 1/ 400, b) 1/300, c) 1/100, d) 1/3, 74. For simple spread footing on clayey soil(for RCC structure), the differential, settlement should not exceed __________, a) 1/ 400, b) 1/300, c) 1/660, d) 1/3, , 01, , 75. For raft footing on clayey soil(for RCC structure), the differential settlement, should not exceed __________, a) 1/ 500, b) 1/300, c) 1/660, d) 1/3, , 01, , 76. Foundations can be broadly classified under __________, a) Shallow foundation and Deep foundation, b) Pile foundation, c) None of the mentioned, d) All of the mentioned, , 01, , 77. A foundation is said to be shallow if its depth is ________ than its width., a) Equal to and Less than, b) Greater than, c) None of the mentioned, d) All of the mentioned, , 01, , 78. A foundation is said to be deep if its depth is ________ than its width., a) Equal to and Less than, b) Greater than, c) None of the mentioned, d) All of the mentioned, , 01, , 79. which of the following, is a type of shallow footing?, a) Spread footing, b) Pile foundation, c) Pier foundation, d) Well foundation, , 01, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , 01, , Page 127

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80. When two column loads are unequal, which of the possible footing can be, provided?, a) Strap footing, b) Raft footing, c) Trapezoidal combined footing, d) Mat footing, , 01, , 81. Based on the function, piles can be classified into ___________ types., a) 4, b) 6, c) 9, d) 3, 82. Based on the method of installation, piles can be classified into ___________, types., a) 4, b) 6, c) 9, d) 3, , 01, , 01, , Module I:, Theory questions:, What is the importance of soil exploration? On what basis the number and depth of, , 1.1, , borings is decided for the following civil engineering work?, i., , Building foundation., , ii., , Road cuts and fills., , iii., , Dam site., , 1.2, , Write a note on geo physical exploration stating its limitations., , 1.3, , Explain the terms with a aid of a neat sketch of the sampling tube:, i., , Inside clearance, , ii., , Outside clearance, , iii., , Area ratio, , iv., , Recovery ratio, , Determine the area ratio for the following details and state the type of sampler. Outer dia of, cutting edge is 75mm, wall thickness is 1.7mm., 1.4, , Explain with neat sketches the different methods of dewatering in construction site., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 128

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1.5, , Explain typical bore hole log., , 1.6, , Enumerate the objectives of subsurface exploration., , 1.7, , Explain with a neat sketch the seismic refraction method of exploration., , 1.8, , List the different methods used to control ground water during excavation. Explain, , any two methods., 1.9, , Distinguish between undisturbed, disturbed and representative sample of soil. What, , are the tests conducted on these samples in the laboratory?, 1.10, , Give the procedure of standard penetration test., , 1.11, , Explain Hvorselev‟s method of determining the level of ground water table., , 1.12, , Explain vacuum method of dewatering., , 1.13, , Explain briefly stabilization of bore holes., , 1.14, , Explain electrical resistivity method of soil exploration., , 1.15, , Explain the uses of seismic refraction method., , Problematic questions:, 1.a, , A sampling tube has inner diameter 70mm, with a cutting edge having 68mm, , diameter. Its outside diameters are 72 and 74mm respectively. Determine the area ratio,, inside clearance and outside clearance of the sampler. This tube is pushed at the bottom of a, bore hole to a distance of 550mm, with length of sample recovered being 530mm. find the, recovery ratio., 1.b, , Estimate the ground water level given the following data:, , Depth upto which water is bailed out=15m., Water rise on the first day= 0.8m., Water rise on the second day= 0.7m., Water rise on the third day=0.6m., 1.c, , Establish the location of ground water in a clayey stratum, water in box hole was, , bailed out to a depth of 10.67m below ground surface and rise of water was recorded at 24, hours interval. H1=64.0cm, H2= 57.9cm, H3= 51.8cm., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 129

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To establish the location of ground water table in a clayey strata water in the bore hole, , 1.d, , is bailed out to a depth of 12m below ground surface. Rise in water was recorded at 24 hours, interval as:, First day= 0.6m, second day= 0.55m, third day= 0.5m, Estimate the ground water table, Determine the area ratios of the samplers of the following data and also comment on, , 1.e, , the values obtained in regard to sample disturbance:, i., , Split spoon sampler, Do=50mm, Di=35mm., , ii., , Drive tube, Do=100mm, Di=90mm., , iii., , Shelby tube, Do=50mm, Di=47mm., , 1.f, , Water in bore hole was bailed out to a depth of 10.67m and the level of water in the, , bore hole was observed to rise by 0.65m on the next day and 0.6m on the second day and, 0.5m on the third day. Estimate the position of the ground water table., 1.g, , Estimate the position of the ground water table from the following data obtained from, , the field. Depth up to which water is bailed out is 30m. Raise in the water levels: on 1st day is, 2.2m, 2nd day is 1.8m, 3rd day is 1.5m., 1.h, , Given the following data from seismic refraction test plot the time versus distance and, , compute wave velocities V1, V2, V3 and thickness of I and II layers., D1= 30m T1= 0.10sec, D2= 90m T2=0.15sec, D3=210m T3=0.20 sec, , MODULE II:, Theory questions:, 1. What are the assumptions and limitations of Boussinesq‟s and Westergaard‟s theory, of stresses in soils?, 2. Explain the use of Newmark‟s chart for finding vertical pressure at a point., 3. Explain the construction & use of Newmark‟s chart., 4. Distinguish between Boussinesq‟s and Westergaard‟s theory of stress distribution., 5. Derive an expression to find vertical pressure under a uniformly loaded circular area., 6. Explain the term pressure bulb and isobar., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 130

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Problematic questions:, 1. A circular area 8m in dia carries a uniformly distributed load of 5kN/m2. Determine, the vertical stresses at the depths of 3m, 5m, 8m and 12m. Plot the variation with, depths., 2. Construct an isobar for a vertical stress of 40kN/m2 when ground surface is subjected, to a concentrated load of 1000kN., 3. Using Newmark‟s chart determine the vertical pressure at the depth of 5m below any, one corner of the rectangular area 2mx4m which is loaded with an intensity of, 15t/m2. Take influence factor as 0.005., 4. A rectangular area 4mx2m is uniformly loaded with a load intensity of 80kN/m2 at, the ground surface. Calculate the vertical pressure at a point 3m below one of its, corner:, i., , By equivalent area method, , ii., , By Newmark‟s table given., Newmark‟s influence factor „K‟, N, m, 0.5, , 1.0, , 1.0, , 0.1202, , 0.1752, , .0, , 0.135, , 0.1999, , 5. A point load of 500kN due to monument acts on the ground surface. Calculate the, vertical pressures at point 5m directly below the load and at a distance of 4m from, axis of loading. Assume µ = 0. Using, i., , Boussinesq‟s analysis, , ii., , Westergaard‟s analysis., , 6. A concentrated load of 250kN is applied at the ground surface. Determine the vertical, stress along the axis of load at a depth of 10m and a radial distance of 5m at the same, depth., 7. Find the intensity of vertical pressure and horizontal shear stress at a point 4m directly, below 20kN point load acting on a horizontal ground surface., 8. A point load of 1000kN acts at the ground surface. Compute the vertical stresses at, 6m depth., i., , On the axis of load, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 131

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ii., , 1.5m away from the axis., Use Westergaard‟s analysis. Take µ= 0.3., , 9. A water tank is supported by a ring foundation having outer diameter of 10m and, inner diameter of 7.5m. The ring foundation transmits uniform load intensity of, 160kN/m2. Compute the vertical stress induced at a depth of 4m below the centre of, ring foundation, using, i., , Boussinesq‟s analysis, , ii., , Westergaard‟s analysis. Taking µ=0., , 10. A point load of 800kN acts at the ground surface. Compute the vertical stresses at 8m, depth., i., , On the axis of load, , ii., , 2m away from the axis. Use Westergaard‟s analysis. Take µ= 0.25., , 11. Find intensity of vertical pressure at appoint 3m directly below 25kN point load, acting on a horizontal ground surface. What will be the vertical pressure at a point 2m, horizontally away from the axis of loading and at same depth of 3m? Use, Boussinesq‟s equation., 12. For a concentrated load of 500kN acting on the ground surface, draw an isobar for a, vertical stress of 20kN/m2., , Module 3:, Theory questions:, 1., , Distinguish between the active and passive earth pressure., , 2., , Explain the step by step procedure to determine lateral earth pressure by Cullman‟s, , graphical method., 3., , Differentiate between Rankine theory and Coulomb theory., , 4., , Explain with aid of a sketch the step by step procedure to determine the magnitude, , and position of active earth pressure by Rebhan‟s graphical method when the backfill surface, is equal to the angle of internal friction., 5., , Describe the Rebhann‟s graphical method of finding active earth pressure on a, , retaining wall., 6., , State the assumptions made in Rankine‟s theory for active earth pressure. Derive the, , expression for ka considering back fill with horizontal surface., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 132

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7., , Derive the equations for the earth pressure co- efficient ka, kp, k0., , Problematic questions:, 4.a, , A retaining wall 9.0 m high retains a Cohesionless backfill. The top 3.0m of fill has a, , unit weight of 18kN/m3 with ϕ= 320. The rest has unit weight of 22kN/m3 and ϕ= 220., Determine active earth pressure on the wall and its position., 4.b, , A retaining wall 6m height has its back fill inclined at positive batter angle of 10 o., , The surface of the backfill rises from top of the wall at a slope of 20o up to a horizontal, distance of 3m away from top of the wall and thereafter surface of backfill becomes, horizontal. Determine the magnitude and position of active earth pressure by Cullman‟s, graphical method. Take wall friction= 20o, angle of internal friction= 30o, bulk density of, back fill as 18kN/m3., 4.c, , Vertical wall 6m high, backfill horizontal carrying uniform distributed surcharge of, , 45kN/m2. Φ1= 36o in top 3m and ϕ2= 32o in bottom 3m, γ1= 19.8kN/m3 in top 3m and γ2=, 19kN/m3 in bottom 3m. Find the total active earth pressure force and point of application., 4.d, , For the retaining wall shown in fig, draw the earth pressure diagram for the active, , state and final total active earth pressure per unit length of wall, by Rankine‟s theory. Assume, sand above water table is dry., , q= 14 kN/m2, 3m, 10m, 7m, , 4.e, , sand, gwt, G=2.65, sand e= 0.65, = 30o, , A vertical wall 6m high with a vertical smooth back has sand behind it. The level of, , sand is horizontal with a uniform surcharge of 12kN/m2 on the surface of sand. The water, table behind the wall is at 2m below the surface of sand. The unit weight of sand above water, table is 18 kN/m3 and saturated unit weight of sand is 21kN/m3,, , = 30o. Sketch the active, , earth pressure distribution. Calculate the active earth pressure and point of its application on, the wall per meter length of the wall., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 133

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4.f, , Compute the intensities of active and passive earth pressure at a depth of 8m in a dry, , cohesion less sand with an angle of internal friction 30o and unit weight of 18kN/m3. What, will be the intensities of active and passive earth pressures, if the water table rises to the, ground level? Take saturated unit weight of sand as 22kN/m3., 4.g, , A vertical retaining wall 10m high supports a back fill having unit weight of, , 18kN/m3. Angle of shearing resistance of 20o and unit cohesion of 10kN/m2. The back fill has, a horizontal surface flushed with the top of wall and carries a uniform surcharge load of, 20kN/m2. Determine active earth pressure and its point of application per unit length of wall., Sketch the active earth pressure distribution., 4.h, , A retaining wall of height 10m supports cohesion less soil with following properties:, , G= 2.65, e= 0.65,, , = 30o. Water table lies at 3m depth. Surface of back fill is horizontal and, , carries a surcharge of intensity 14kN/m2. Draw lateral active earth pressure distribution, diagram. Determine total active earth pressure and its point of application., 4.i, , A retaining wall 8m high supports sandy back fill with l= 0.6, G= 2.65 and, , = 30o., , Water table is at a depth of 2m from ground surface. Draw active earth pressure diagram and, find magnitude and point of application of total active earth force. Assume soil above water, table has degree of saturation of 50%., 4.j, , A retaining wall of height 6m retains two layers of soil. Top 2m is cohesion less soil, , with γ= 18kN/m3,, , = 30o and bottom 4m is cohesive soil with, , = 0o, c=20kN/m2 and γ, , =16kN/m3. Calculate the total active earth pressure on the retaining wall and its position, above the base of retaining wall., 4.k, , Excavation was been done for a foundation trench in plastic clay having unit weight, , of 23kN/m3. When the depth of excavation reached 8m, the trench failed. On the assumption, that, , = 0o., , Calculate the value of cohesion of the clay., , MODULE 4:, Theory questions:, 4.1, , What are the causes of failures of slopes?, , 4.2, , What is stability number? Discuss the use of stability chart., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 134

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4.3, , Explain the friction circle method of stability analysis for slopes., , 4.4, , Mention the types of slopes failure and various causes of slope failure., , 4.5, , Explain the method of slices for c-, , soils to determine the factor of safety against, , failure., 4.6, , Explain the Fellinious method for stability analysis of slopes., , 4.7, , Explain the Swedish method of slices stating clearly the assumptions made., , 4.8, , What are the different types of failure of finite slopes? Explain with neat sketches., , When do you expect each of these failures?, Problematic questions:, 1.i, , A finite slope of height 10m is inclined at 40o with respect to horizontal. If the, , Fellinious directional angles are α=35o and β=26o. Calculate the factor of safety against toe, failure of the slope given γ= 18kN/m3, C= 10kN/m2 and ϕ= 15o., 1.j, , Fig shows the detail of an embankment made of pure cohesive soil having c=, , 30kN/m3. Unit weight of soil is 20kN/m3. Determine the factor of safety against sliding along, the trial circle shown., , o, , R= 10m, , 1:1, , 1.k, , 6m, , A new canal is excavated to a depth of 5m below ground level through a soil having, , following properties:, C= 14kN/m2, ϕ= 20o, e= 0.67, g= 2.70, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 135

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Calculate FOS with respect to cohesion when canal is running full. Slope of the canal is 45 o., What will be the change in FOS if slope is changed to 40o to vertical?, i, , 30, , 45, , 60, , 75, , 90, , Ns, , 0.0525, , 0.062, , 0.097, , 0.134, , 0.182, , A laboratory test gave the following soil parameters, c=40kN/m2, ϕ= 30o. The, , 1.l, , expected parameters of the mobilized shearing resistance are, c= 25kN/m2, ϕm= 22o. The, average effective pressure on the failure plane is 150kN/m2. Find the factor of safety with, respect to average shearing strength, cohesion and internal friction., An embankment is made of a soil having c= 25kN/m2 and ϕ= 20o and γ= 19kN/m3., , 1.m, , The slope is 1.5 horizontal to 1 vertical and of 9m in height. Determine factor of safety along, a slip circle passing through the toe. The centre of slip circle is located by directional angles, α= 26o and β= 35o. Use method of slices analysis., Determine the safe depth of cutting at a slope angle 40o which is to be made in a soil, , 1.n, , with c=30kN/m2, ϕ= 15o and γ= 20kN/m3. The required factor of safety is 1.5. Take the, stability number Sn= 0.098., An embankment is inclined at an angle of 35o and its height is 15m. The angle of, , 1.o, , shearing resistance is 15o and the cohesion intercept is 200kN/m2 the unit weight of soil is, 18kN/m3. If the Taylor‟s stability number is 0.06. Determine the factor of safety with respect, to cohesion., A long natural slope in a c- ϕ soil is inclined at12o to horizontal. The water table is at, , 1.p, , the surface and seepage is parallel to the slope. If a plane slip has developed at a depth of 4m,, determine the factor of safety. Take c= 8kN/m2, ϕ = 22o and γsat= 19kN/m3., A 5m deep canal has side slope of 1:1. The properties of soil are C=20kN/m 2, ϕ= 10o,, , 1.q, , e= 0.8 and G=2.8. If Taylor‟s stability number is 0.108 determine the factor of safety with, respect to cohesion when the canal runs full. Also find the same in case of sudden draw, down, if Taylor‟s stability number for this condition is 0.137., An embankment is to be constructed with C= 20kN/m2, ϕ= 20o, γ= 18kN/m3, FOS =, , 1.r, , 1.25 and height = 10m. Estimate required side slope. Taylor‟s stability numbers are as, follows for slope angle., i, , 10, , 20, , 30, , 45, , 60, , 75, , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , 90, , Page 136

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Ns, , 0, , 0.005, , 0.025, , 0.062, , 0.097, , 0.134, , 0.182, , Module 5: BEARING CAPACITY., Theory questions:, 6.1, , Define:, , i., , Ultimate bearing capacity., , ii., , Safe bearing capacity., , iii., , Allowable bearing capacity., , iv., , Net ultimate bearing capacity, , 6.2, Explain The ISI procedure of conducting the plate load test. Discuss the validity of, the results in the design of foundations., 6.3, , List the assumptions made in Terzaghi‟s analysis., , 6.4, What is the effect of size and shape of the footing on bearing capacity when the, footing is resting on:, i., , Cohesion less soil., , ii., , Cohesive soil., , 6.5, , Discuss the effect of ground water table on the bearing capacity of soils., , 6.6, , What is standard penetration test? Explain., , 6.7, , Distinguish between general shear failure and local shear failure., , 6.8, , List the factors affecting bearing capacity of soil., , 6.9, , Explain the three types of failures that can occur below a footing., , 6.10 Explain the Terzaghi‟s theory for bearing capacity of soil. Give the Terzaghi‟s, equation for the bearing capacity of strip, circular, square and rectangular footings., 6.11, , Explain, , i., , Plate load test., , ii., , Standard penetration test, , 6.12 Write down the Brinch Hansen‟s bearing capacity equation and explain the terms, involved., 6.13, , Explain the terms allowable soil pressure and ultimate bearing capacity., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 137

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=456+111.38B+(19*1.5), =484.5+111.38B---------------------------------------(i), , Actual load intensity= load, area, =1500/(B2)-----------------------------(ii), Equating (i) and (ii), AFTER CALCULATION B=1.51M, , 6.e, A strip footing 1m wide and a square footing 1m wide are placed at a depth of 1m, below the ground surface. The foundation soil has cohesion of 10kPa, angle of friction of 26o, and unit weight of 18kN/m3. Taking bearing capacity from the following table calculate the, safe bearing capacity of soil using Terzaghi‟s theory. Use factor of safety of 3., Φ, , Nc, , Nq, , Nγ, , 15o, , 12.9, , 4.4, , 2.5, , 20o, , 17.7, , 7.0, , 5.0, , 25o, , 25.1, , 12.7, , 9.7, , 6.f, A square footing placed at a depth of 1m is required to carry a load of 1000kN. Find, the required size of footing given the following data: c=10kN/m2. ϕ= 38o, = 19kN/m3 NC=, 61.35, Nq= 48.93 and Nγ= 74.03. Assume water table is at base of footing., 6.g, A square footing located at a depth of 1.3m below ground has to carry a safe load of, 800kN. Find size of footing, if the desired factor of safety is 3. Assume e=0.55, degree of, saturation =50%, G=2.67, c= 8kN/m2. Use Terzaghi‟s analysis for general shear failure., Assume NC= 37.2, Nq= 22.5 and Nγ= 19.7., 6.h, Design a square footing to carry a safe load of 1200kN on a sandy soil at a depth of, 1.5m below GL, given = 16kN/m3, N=25, permissible settlement is 40mm. water table is at, the base of footing. Use factor of safety F=2., , PRASANNA P RAO, ASST PROF, DEPT., OF CIVIL ENGINEERING, SUCET MUKKA, , Page 141