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So, shear strength is a very important concept in soil mechanics or geotechnical engineering because, to support foundations in soils or to support any structures on the soil through foundations, the shear strength of the soil should be sufficient to withstand any loading conditions. Shear strength has interesting historical development. So, let us briefly understand those historical developments and then go into the shear strength of unsaturated soils. So, shear strength the concept of shear strength has been derived or brought into existence in the modern soil mechanics by Coulomb through his work inspired by Leonardo Da Vinci of the 16th century. So, Coulomb in the 18th century has given friction laws that made the understanding of shear strength of soils. So, it was Coulomb in the 18th century gave the friction laws based on the shear strength the soil has been derived and he looks at soil as single-phase, he does not differentiate soil and rock. So, generally, if you take a soil mass and thefailure of the soil mass along a slip plane, he is observed it could be there for rock or soil it should act in the same manner. So, a single-phase system, similar to the single-phase system he considers, the soil is like an entire soil is one particular he has considered, just like a rock mass and he proposes the shear stress is proportional to the total stress of the soil. So, based on this the direct shear apparatus have been developed, so in the direct shear apparatus, which is a very popular shear stress apparatus which we use after in the soil
mechanics laboratory. So, in direct shear apparatus, soil mass is compacted and it can be it is like two half system, and one half can be moved against another half. Generally, the one half can be fixed, supplying the shear stress the other half can be moved related to the other half. So, a normal load of N can be applied. So, divide by the cross-sectional you should get the
total stress which is acting on soil mass. So, this is total stress that is acting on a soil mass this is the total stress. A simply the stress which is acting on a soil mass and shear is applied and soil moves relative to the other half and dial gauge can be loaded and which can be connected again to a fixed
system. So, this dial gauge can record what is the amount of strain that is taking place. Similarly, one more dial gauge can be loaded, kept on the system, kept on the top of the surface. So, therefore, vertical movement during shear operation and horizontal movement or the shear strain can be recorded at any given instance. So, this led to a shear stress versus
shear strain diagram for different soils. For OC soils, for normally consolidated soils they exhibit this kind of behavior. They normally consolidate clays or loosely compacted soils, soils or sands; they exhibit this kind of behavior and overconsolidated clays or densely compacted. Sands exhibit this kind of behavior and there they reach a critical state at one particular strain. So, this is τ critical and this is the τ peak, this is τ peak. So, this is exhibited in the system, but initially, the direct shear apparatus meant for only dry sands, this is meant for dry sands and however, we using this for measuring the shear strength of different soils, the clay soils and (Refer Time: 06:08) soils and also including some moisture. So, generally, this is not meant for moist soils, there is dry soil, dry sands soils are generallyused in these particular apparatus. So, this leads to shear stress versus shear strain and this is under one particular normal stress applied. Similarly, the test can be repeated with different normal stress; therefore, one can obtain the relationship between normal stress and shear. So, you get as the normal stress increases, the shear stress peak stress increases. So, based on that either you get this kind of a linear relationship between τ and σ. However, there are there is a lot of debate about why this does not follow through the origin and there are some schools of thought they believe that it should a non-linear behavior something like this and it should not be linear. But let us assume that it has cohesion and an angle of internal friction ϕ, but
generally, for sands, the line passes through the origin. In the case of clays, you have a cohesion term that is coming into the picture and in the unsaturated shear test also you get a cohesion term and we will discuss that little later. So, generally, this c cohesion term is absent in the case of sands. So, essentially this leads to τ = σ tan ϕ, this is the relationship you obtain which is attributed to Coulomb's friction loss this is on the failure plane. So, this ϕ is a material constant angle of internal friction and σ is normal stress or total stress, which is a stress state variable, and τ, is shear stress and this is also a state variable. So, here one can add cohesion to make it more generalized, some schools of thought they do not mind adding cohesion term along with this. So, this is essential to attribute whatever the non-linearity or to attribute the intercept which is coming into the picture while plotting τ versus σ data. So, this is what we get.
So, however, Terzaghi called Terzaghi 1920, he realizes that the total stress concept should be revised because he looks at the soil as a two-phase system; soil solids and water. When you have such a system, the strength of the soil which is exhibited during the application of the loading can differ, depending on the boundary conditions. So, for example, if I have a soil system are solids + water system, your consolidation test also we have seen that when you apply a load. So, initially, no settlement because the hydraulic conductivity of the soil is very less that water does not come out. So, depending on the drainage condition soil exhibits different strengths. That is because, when you apply normal load the pore water pressure can be built up to the same value as you applied to the applied normal stress and soil does not fail and soil behaves differently altogether. And if the water is allowed to come out through the boundaries then, it exhibits different strengths. So, then shear strain effects started coming into the picture and so two-phase system so then, the modification to the total stress concept has come into the picture. The plain normal stress is taken up by the soil solids that are σ ` effects on stress + pore water pressure. So, when the drainage is absent, there are no drainages then what is the loading condition that is applied, there is normal stress which should be = u and σ ` is absent. So, in the triaxial test, the concept of the triaxial test comes out where, we can control the drainage very well, in the direction apparatus we cannot control the drainage.So, therefore, generally, we do not use soils with a certain good amount of moisture in it for shear strength testing. So, we take soils in triaxial test apparatus, and here sorry the triaxial test apparatus is designed for testing soils construing drainage conditions. So, essentially the soil is seen as a two-phase system, so fully saturated conditions. So, the soil sample is prepared for standard sample 38 mm by 76 mm length 38 mm diameter and 76 mm length. And the samples are fully saturated through the backpressure valve ok. So, the sample is placed on a porous disk and the sample is surrounded by the flexible membrane so that it does not directly interact with the water, which is available all around. So, here the water
is allowed around the sample and now by applying pressure in this water, we can maintain all-round pressure. So, σ 3 all-round pressure can be maintained around the soil sample and the backpressure valve is connected and through which we can saturate the soil sample. So, here once the soil is completely saturated, so whatever the amount of σ 3 you apply, here, the porous pressure within the soil sample would be the same. So, whatever the σ 3 you apply, Δ σ 3 whatever the change you make all around the sample that would be = Δu, in the beginning in the test. And to may, to apply the shear stresses on the soil loading ram is placed and load is applied vertically. So, this deviates the stress from the all-round pressure or consolidation pressure and which shears the soil sample. So, as you can control the drainage valve, this the back pressure valve, it is the same. So, in this
particular case as you can control the drainage and we can conduct different tests altogether. So, here as of σ `, the effect to stress concept has been brought in by Terzaghi, which is σ ` = σ mines u. Σ ` affects to stress is a one which controls the share of the soil. So, therefore, Coulomb's equation is modified to be torque equals to σ ` tan ϕ ` + c `. So, this is effective stress and this ϕ ` and c ` are the effective metal constants are ϕ ` is an effective angle of internal friction and effective cohesion. So, τ we do not use τ ` because water does not contribute to share. So, this is τ ` = τ only.
So, this is the equation that is modified based on Terzaghis inputs, and therefore now, depending on the drainage conditions the soil exhibits different strengths, and the consolidated generally different test can be conducted in this particular set up. So, youcan apply all-round pressure σ 3 and you can consolidate the soil to whatever the extent possible or whatever the extent you want to consolidate and after that shearing is started,
so that the shear stress is applied on the soil and so on the soil sample you can apply σ 1 on one direction and σ 3 on the other direction. This is in a radial direction. So, if the soil sample is shown like this, this is in a radial
direction which is acting. So, this is σ 1, this is σ 3. So, the σ 1 is nothing but σ 3 + Δ σ. So, this is deviator stress. So, soil fails under this combination of σ 1 and σ 3. So, essentially you get with deviatoric stress, the actual strain can be measured. So, here also you can obtain for OC soil and NC soil are loosely compacted and densely compacted sands are coarse-grained, soils loosely compacted sands this is a behavior that is executed. So, you have Δσ at the peak, but we consider the critical state's critical values. So, this is a Δ σ at failure, if you consider this is a Δ σ at failure under certain strain values then,
here as the failure plain direction is not known. So, we cannot directly use a Coulomb's principle we have to utilize Mohr theory as well. So, this is a Mohr-Coulomb concept. So, the Mohr-Coulomb concept you have τ 1 vexes and σ or σ ` on the x access. So, then one test with σ 3 has initially the soil state is this, initially, the soil is consolidated to say σ 3. Now, keeping the σ the 3 constant Δ σ is increased, that means, this is σ 1. And which is continuously increased and increased to one particular value where it fails. So, finally, this is the value of failure. So, we can conduct many more tests so that we can obtain the failure plain, this could be somewhat like this. So, here this is the angle of internal friction, ϕ ` on this access, this is cut at say one particular point here and this is c` and if you join the center here and pool for planes is this because the on the soil sample, σ one is acting on horizontal plain and σ 3 is acting on vertical plane. So, σ 1 is acting on the horizontal, and σ 3 is acting on the vertical plane, this is the pole for planes. So, p so front pole for plains, if you draw a line joining the tangent, so, this is a failure plain, we can call θ. So, there is a point at which this is a failure plain direction. So, this is a failure plain direction, it should be the same as θ. So, now, as this side is the
same as this side, this radius of the more circle and this side opposite angle is θ. So, this, therefore, should be θ 2 and this should be 180 - 2 θ and this is 2 θ. And this is the c ` cot ϕ, this from this value is c`cot ϕ.And if you take sin ϕ `, sin ϕ ` and this is σ 1, - σ 3 ` by 2 divided by so, this is again c ` cot ϕ + this is σ 3 f these are all f at failure, σ 3 ` this is the effect of stress we are plotting. So, σ 3 ` + σ 1 ` σ 1 - σ 3 by 2, therefore, it is σ 1 +, σ 1, σ 3 by 2 So, if you simplify this, this comes out to be 2 c `, cos ϕ ` + σ 3 ` + σ 1 ` = σ 1 ` - σ 3 `. We simplify this sorry this is sin ϕ, σ 3 ` + σ 1 ` times sin ϕ ` = σ 1 ` - σ 3 `. If you simplify this we get σ 1 ` = σ 3 ` times 1 + sin ϕ by 1 - sin ϕ ` + 2 c ` cos ϕ ` divided by 1 - sin ϕ ` So, this is σ 1 ` into σ 3 ` tan square 45 + ϕ ` by 2 + 2 c ` tan 45 + ϕ ` by two. So, essentially if we have two sets of σ 1 and σ 3 at failure, this is from one test. If you have two sets of the data we can determine c and ϕ, if you do not have c cohesion it is for only pure sands then if you have one set of data you can determine what is the angle of
internal friction. This is using formula or graphically you can plot and then you can determine C ` and ϕ ` it is are the effects of stress parameters. Similarly, here these effective stress parameters you can obtain when the drainage is allowed. So, when the drainage is open. So, then pressure dissipates at any given instants then you will get C ` and ϕ`. So, if the drainage valve is closed during the shearing operation. So, then the shear
strength that is exhibited is different and it exhibits s u undrained shear strength data undrained shear strength values. So, either you get total stress parameters or affective stress parameters based on your analyses. You get either shears undrained shear strength or affect to stress parameters
like this in a given test. You can get both total stress parameters and affective stress parameters if you conduct a consolidated undrained test, but measuring porous pressure during the shearing then you get both combinations of affective stress parameters and total stress parameters, that we will use for their analysis for long term analysis and short term analysis. The effective stress parameters are used for the long term analysis and undrained parameters are used in the short term analysis in case of clays. So, the Terzaghis concept of a two-phase system that is the introduction of the effective stress concept helped in understanding the soil behavior during shearing and the influence of drainage conditions very well.
In 1959, Bishop the gentleman British gentleman called Bishop came up with, now it comes to unsaturated soils how the soil behavior can be very well predicted are can be understood very well using the concept of negative for water pressure. So, if we draw a profile, if we have a soil mass this is a ground surface and this is the groundwater table. So, far we have assumed that the groundwater table is either at the ground surface or no groundwater table exists. So, essentially we consider only a twophase system, which consists of whether soil solids and pure air or soil solids and pure water. So, we then in a real situation we see that the soil consists of three phases. In that scenario, let us assume that the pure this is the groundwater table and this is the ground surface and how the in our earlier assumption where we considered that the if the water table is up to the ok, in the early assumption the what we did is we considered total stress total stress distribution. So, even in the soils, it may very differently. So, this is what we consider and affective stress the pure water pressure does not exists. So, therefore this is
σ or σ ` profile, but if you correctly analyze the pure water pressure varies in this particular manner total stress. Positive in the downward and negative in upward at the hydrostatic condition, this is what you have seen when you have a capillary inserted in the water table. In a beaker of water, a small capillary is inserted into a beaker of water. So, then there is a reason water
table the pure water pressure distribution if required this is 0 here and sorry this is how itgoes. And this is positive and this is negative. This downward positive because water is water level is here and this is positive and this is negative. Similarly, at the hydrostatic condition, this should be negative this the negative pore water pressure and this is positive pure water pressure so above this is the condition. And if you see your total stress variation this is how the total stress varies. So, this is σ plot and this is a negative in this direction So, according to the effective stress principle the variation should be the effective stress principle σ ` = σ - u, u means pure water pressure. So, as we so far dealt with the two-phase system in basic soil mechanics we generally, use u to indicate pore water pressure. So, here we have two
pure pressures, one is the water pressure and air pressure to distinguish that we use a suffix w here this is water pressure. So, if σ ` = σ - u w then σ - here uw is negative. So, then they should be σ + uw. So, that means, affective stress should be somewhat like this and it should be plotted from here onwards. So, here whatever the uw you have that will be here. So, this is this
value uw. Because a σ is 0 there you have uw and this is how it should vary σ `. But after we have seen that generally if you take dry clays or dry soils the magnet should of uw is very high. So, negative pure water pressure or suction of the clays theoretical it can go to 10 power 6 kilo Pascal’s nearly dry state or residual state. If this goes that much then the effective stress should be nearly infinity and soil should never fail, but then we see
failures very often. So, therefore, the contribution of uw, the contribution of negative pure water pressure to the total stress for effective stress calculations, is not 100 percent. So, then what is the question is how much percentage of negative pure water pressure contributes to your effective stress calculation is a question. So, there are several proposals and there are
several theories put forward, and finally, it is agreed that there is a certain percentage that is less than 100 percent will contribute to the effective stress. So, this is less than 100 percent. So, now, the Bishop in 1959 has come up with he has come up with 2 stress variables 2 independent stress variables, but in a lamped manner. He has proposed an effective stress equation which is σ ` = σ - u a + χ u a - uw. So, here σ - u a is normal stress, net normal stress and this is matric suction.And this χ parameter is the effective stress parameter, which where is between 0 and 1. If χ becomes 0, χ becomes 0 the equation becomes σ ` = σ. So, this indicates the soil is
completely dry. So, if u a is used as gauge pressure, then this is σ = σ ` = σ. Say if χ = 1, then this equation becomes σ ` = σ - uw, which indicates completely saturated soil. This is completely dry. So, χ = 0 indicates completely dry soil and χ = 1 indicates fully saturated soil. So, generally for unsaturated soils, this χ parameter varies between 0 and 1. And this led to controlling matrix suctions in the shearing test. So, net normal stress is
applied to the soil and how to control the matrix suction is another question, that is how the research began during that time.
So, then a modified shear box test has been developed. So, here in the modified share box tests, the important inclusion is that the HAE disk, High Air Entry Disc is added. So, this a shear box and you have 2 halves here and the soil is specimen is here and here one coarse porous stone is placed and here HAE disk is placed High Air Entry Disc is placed. And this whole setup is confined in a one closed volume airtight volume and which is connected to one air pressure valve and air pressure can be controlled in this particular
manner. So, here the air pressure can be controlled in the system. And the air can go through this coarse porous stone, but then this High Air Entry Disc
which is saturated with a water chamber here and air cannot enter into it. This is closedand this is completely closed, air cannot enter into it. So, the soil specimen you can take initially saturated soil which is in direct contact with the HAE disk and should have hydraulic equipment in the beginning. Then so through the coarse porous stone, the air can enter, but then the air exert some pressure on the air can enter into it and then it can desaturate the soil sample and it cannot air cannot enter into the HAE disk until the air
pressure - water pressure u a - uw = the air entry of the HAE disk. So, this is exactly similar to what we use in the access translation technique. So, in the
access translation technique, you have an HAE disk, High Air Entry Disc. And you have a water chamber; you can control the water pressure. So, generally, we keep it we keep this value close to the atmospheric pressure and this is connected to the air pressure valve, and u a is applied here. So, here u a - uw is applied on the soil system here. If you place a soil here this porous disk is completely saturated and this is the soil mass as a specimen, the specimen is placed here. So, initially, if you keep a saturated soil specimen here, under the application of u a - uw that is a load or suction load is applied then, the soil starts bleeding and water comes out and so by measuring what is our amount of water that is come out from the soil mass we can determine the water content and here (Refer Time: 35:15) you are controlling the suction. So, that is how we established the soil-water characteristic curve. So, in this particular case, in the direct shear apparatus, the same technique is applied.
Here the High Air Entry Disc is placed above this the soil sample is placed. So, then you have a water chamber, here a coarse porous disk is placed so that air can enter out and. So, by controlling u a - uw you can maintain one particular matric suction within a soil sample. So, by maintaining on particular matric suction under this control matric suction you are controlling u a - uw, similarly the σ - u a is also controlled. So, under particular net vertical stress σ - ua you can apply in the soil system and under this particular controlling two stress variables u a - σ - u a and u a - uw you can conduct the shearing test right. So, this is how the shear test direct shear test is modified by in cooperating the suction control technique there is an access translation technique into this one, and shear stress at different control net normal stress and matric suction values can be obtained.We will discuss the results a little later, we will only discuss the; now we will discuss thehow the test is modified So, this particular setup is also modified for
exerting different stress, this is one particular way of doing it, this is by controlling using the access translation technique or this also can be done by, osmotically controlling the matric suction within a soil mass. So, earlier I have discussed this particular system, this particular methodology, generally,
if you look at the principal if you have a U shape tube you have a water level same on both sides initially and if you keep one semipermeable membrane, which is like a filter which you keep and then add some salts and the salt molecule size is such that, it cannot pass through the semipermeable membrane. The semipermeable membrane is placed here, then salt is added on this side, so then if the semipermeable membrane is not present the salt must have diffused the salt solution here high concentration C = C naught
and here concentration is 0, had it been that the semipermeable membrane is not present. The salt must have diffused from the left hand to the left hand side to the right hand side, salt concentration should have become the same. However, as semipermeable membrane separates these two chambers and the salt corrupt migrate through this semipermeable membrane. So, water has to move from the right hand right side to the left hand side and dilute this solution. So, therefore, at equilibrium, if you see, so at equilibrium oilier level
was somewhere here and now the level is here and the level on the right-hand side decreases. So this is the condition there is an osmatic head of h o will be available here. So, that means, to make these two-level same, you need to apply had cosmetic head of h naught to bring it to this particular level. So, this concept is utilized in modifying the direct shear test to conduct unsaturated shear stress shear strength stress. So, essentially you will have
a soil mass. And initially, it is saturated and then now this is connected to porous stone, and then which is connected to a reservoir of test solution polyethylene glycol, the molecular size of this polyethylene glycol PEG is much larger than the semipermeable membrane. A semipermeable membrane will be placed somewhere here then, this is loaded in the same
manner as it was earlier. So, this is the load that is applied or σ; σ - ua.So, here the PEG solution is placed in contact with soil, so as the molecules of the PEG cannot enter into the entry through the semipermeable membrane, the water will be drawn out from the soil mass and it moves down and then what the essential PEG solution gets dilute diluted. So, PEG concentration gets changed, and water content in the soil system decreases, which causes by which causes particular matrix suction maintained within the soil mass.
So, this can be obtained because the knowing PEG concentration and what is the size of these molecules and what is the semipermeable membrane size etcetera this can be obtained and standard using the standard techniques this can be obtained and the matric suction within the soil sample can you maintained. After that, this is sheared in the same manner. So, either using access translation technique or osmotic technique osmotically controlled control direct shear test. So, an osmotic control test also can be conducted. Either way, the matric suction, and normal stress are two independently maintained, and the strength of the shear stress can be conducted. Similarly, the triaxial test is also conducted in the same manner where, you have a triaxial test is also conducted in the same manner where we control the matrix suction within a soil sample by applying, so air pressure and we place High Air Entry Disc here. So, this the soil sample and this is a High Air Entry Disc and the coarse porous disk is placed here and you have air pressure control and so there is an airpressure line, so through which air pressure is control within the soil mass and air cannot enter into the High Air Entry Disc until u - u w is = the air entry of the High Air Entry Disc. So, this is how the soil starts bleeding and it attains particular it decreases what? The water content of the soil sample decreases when the air pressure is increased. And at equilibrium it attains one particular suction value that is u a - u w and net normal stressing can be applied by applying the pneumatic stress, shear stress can be by applying one you can anyways control the σ 3 all-round pressure of the in the soil sample σ 3 and by controlling these 2 we can shear the soil sample by applying pneumatic stress on a soil sample. So, this is similar to the direct shear apparatus and whatever the test setup.
Essentially the access translation or osmotic concepts are brought into the triaxial and direct shear apparatus for modification or for controlling the matrix suction within the soil sample. So, under this control matrix suction environment u a - u w and σ - ua, so the affective stress can apply. So under this controlled environment, the τ now will become the equation, so essentially the Bishops effective stress per have effective stress is σ ` = σ - u
a + χ * u a - uw. So if you write in terms of τ, this is σ - u a + χ* u a - uw * tan ϕ ` + C `. So, here you have another effect (Refer Time: 44:50) to estimate and C ` and ϕ ` also you have. So these are material constants. So, χ is the function of now moisture content θ or matrix suction. So, these 2 stress variables are controlled independently in this all this stress, and the shear strength of the soil sample is the shear strength of soil sample can be determined by understanding the by estimating the material constant like C ` ϕ ` and χ variation with θ. Then you can estimate how the shear strength varies with moisture content. Thank you for this lecture and will continue for discussing how the saturated soil can be obtained from modifying the direct shear apparatus and modify the triaxial stress apparatus using the Bishop’s concept of effective stress. Thank you.
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