Welcome to lecture series on advanced geotechnical engineering and we are discussing module 4 shear strength of soils. In the previous lecture we introduced ourselves to methods for determining shear strength of the soils in the laboratory and then we said that we have discussed about the direct shear test and what are the stress states in direct shear test and demerits of the direct shear test. Then we said that the other type of very popular testing which is triaxial compression test and which is the physical model test which actually stimulates the stress conditions and drainage conditions as in the soil and it actually has got versatile applications. So in this particular module 4 and lecture 8 we are going to discuss about the different types of triaxial test and pertinent stress parts. So this is the module 4 in the lecture 8 on the stress strain relationship and the shear strength of soils and we are as informed we are going to concentrate on the triaxial behavior particularly with reference to unconsolidated unconsolidated undrained test, unconfined compression test, consolidated undrained and consolidated drain triaxial test and the stress parts for the in the triaxial test. So this as said most widely used shear strength test and is suitable for all types of soil and a cylindrical specimen generally having L by D is equal to 2 because this is required to maintain the principal planes to remain as principal planes during all stages of the triaxial test. Suppose the failure plane passes through the upper and bottom portions then there can be possibility of the shear stress has generated and that in such situations those planes cannot be called as principal planes. So in triaxial test we have this radial plane which is vertical plane as well as these horizontal planes these planes actually remain as principal planes throughout the test and the typical specimen diameters has been mentioned 38 mm, 100 mm widely used otherwise we also have possibilities of 300 mm diameter and even the larger sizes triaxial tests are actually becoming in vogue now. So in order to simulate the confining stresses it is actually done through by filling the chamber with water and the water is actually pressurized by all round pressure and which is actually called as the chamber pressure or cell pressure. So this is the typical triaxial cell which is actually shown and the sample which is having L by D is equal to 2 that is 38 mm diameter or 100 mm diameter and 200 mm height or 75 mm diameter and 150 mm height is actually placed and in order to prevent interaction of sample with the water there is a membrane which is actually placed and so that that membrane ensures that water tightness and there are different ports will be there one port is to the supply water into the soil sample and one port is to supply to take to measure the volume changes and another port is to measure any changes in the pressure within the sample and the loading is actually applied from the top that is actually called as deviator load. So when we have the sample which is confined with all round pressures then the pressures all round for the samples are sigma 3 and then when in order to induce the loading let us say the example that we consider an example that we have got a soil strata and we have got a sample at a certain depth let us say about 5 meters depth and above that suppose we are constructing a foundation and the building loading is actually increasing gradually. So that is actually increase in the incremental load is actually applied by this you know this York and this loading arm and this loading arm induces that sigma 1. So sigma 1 is equal to sigma 3 plus P by A so sigma 1 minus sigma 3 is actually called as the deviated load. So in the mechanism basically is that intermediate principle stress sigma 2 must be equal to major sigma 1 or minor sigma 3 it is so as to facilitate representation of the stress state in two dimensional Mohr circle and the cylindrical specimen is placed inside the perspex cell filled with water. So that the acrylic you know cell will be filled with water the specimen is covered with a latex sheet so as to avoid the direct contact with water and the specimen is initially loaded by the cell pressure and then basically during that stage if the consolidation has to happen then the consolidation will be allowed and then thereafter once the consolidation is completed a deviated stress is then applied gradually on the sample with the help of ram actually. A duct at the bottom of the sample allows the water to pass to the sample which is further monitored and which actually can have a volume increase or you know which can actually measure the volume changes and the pore water pressure transducer actually measures positive or negative pore water pressures depending upon the stress history of the soil being tested. So fine grained soil can stand the mould without any support but the coarse grained soil samples have to be kept in some supporting mould until the application of negative pore water pressure to the sample through the drainage duct. So the fine grained soil samples are actually prepared you know on the without any fine grained soil can actually support but you know coarse grained soils or sandy soil samples have to be kept with some suction and because of that you know then afterwards the suction is actually released gradually. If the cell pressure is increased to say sigma Cp then the isotropic pressure is taken entirely by the pore water and the pore water you know the pressure increases but no change occurs in the effective stress because if there is no volume change then the pore water pressure will you know no change in the volume, no change in the effective stress. Now we have different drainage conditions like as we mentioned that actually makes you know the combinations in triaxial test or different types of triaxial test. One is that you know we have two steps, one step one is that you know during that consolidation of the sample and then shearing stage. So in this basically the two stages are there you know stage one where the consolidation can happen and the stage two is the shearing can happen. Under all all round pressures sigma C drainage valve condition will be open if you are having if you are allowing consolidation. So the sample the water in the sample drains out once the consolidation is actually completed then you know it is actually the shear loading. In this also the shear loading actually it actually has got the drainage valve condition can be open and close it can be open if you are having consolidated drained triaxial test and it can be closed if you are having thinking of consolidated undrained triaxial test and it can be closed in both consolidation and you know drainage shearing stages and that is actually unconsolidated undrained triaxial test. So unconsolidated undrained triaxial test is basically a quick test where the sample is as prepared will be tested without any consolidation by applying three different cell pressures and if the cell pressure is actually is if the sample is actually saturated and the even at different cell pressures the deviated load you know will be same. So with that what will happen is that we get the undrained shear strength of a soil and the failure you know failure plane or the failure envelope will be horizontal. So we have certain practical considerations in which we will be actually doing this consolidated drained and consolidated undrained and unconsolidated undrained triaxial test. For example for unconsolidated undrained triaxial test where actually when you are constructing you know an embankment on the soft soil without any you know waiting period that means that you do not allow the consolidation to take place and you do not allow the even the consolidation during the shear then you know a rapid construction on a particular soil deposit on a let us say a soft clay deposit actually simulates the unconsolidated undrained condition or you know let us say the rapid loading of a founding on a foundation on a unsaturated undrained clay is also an indication of you know the example of unconsolidated undrained test or let us assume that a catastrophic you know failure of a vertical cut under of a saturated soil under undrained condition is also to some extent indicates an example of unconsolidated undrained test in such situations one need to you know do the unconsolidated undrained triaxial test but we have some practical examples in which we can actually say that you know the different types of you know test can be done depending upon the practical situation the one what we have here is some examples for the CD analysis for clays this is after holes and coax 1981 and consider an embankment constructed very slowly in the sense that embankment constructed in stages and each stage the consolidation has been allowed and between each stage there is an adequate waiting period then embankment was actually constructed very slowly in stages or a soft clay deposit. So this example in the suppose if you are having a situation like embankment construction you know very slowly in stages for a soft clay deposit then you know one has to do the consolidated drained triaxial test and this also indicates this also called the drained parameters and where the pore water pressure dissipation actually completely occurs. So with that what we can we call is that these effective shear strength parameters like effective cohesion and effective friction angle can be obtained and these are actually also indicate for example when we are doing a slope stability analysis particularly under effective stress conditions we have to use for the long term stability of a particular you know slope or embankment we have to use the we need the effective shear strength parameters like C dash and phi dash. We are actually discussing about the example for you know some field examples for the you know where we actually adopt the consolidated drained triaxial test parameters. The other example what we are discussing in this for the CD test particularly where we can actually get the drained parameters is the earth dam with steady state seepage conditions wherein we actually have an earthen embankment or earthen dam constructed with a core and a casing. The casing is actually having you know material which is medium the low having low plasticity materials fine grained soil but with low plasticity materials and this is the upstream water level and this is the periodic surface and the water flows and then through a drain it actually went out. Now this is the failure surface typical failure surface so under steady state seepage conditions under long term conditions and you know then when we wanted to evaluate the stability of an earthen dam subjected to steady state seepage condition one need to adopt the effective shear strength parameters that is C dash and phi dash especially obtained from by allowing the consolidation that is the consolidation drainage during the consolidation as well as you know during shearing stages. So both the stages actually happen very very slow so in a way they give you know the both the stages what will actually happen is that particularly in the drainage stage the dissipation of pore water pressure is almost close to zero because you know the raise of the pore water pressure will not be there at all because the drainage is actually happening continuously. The another set of examples like for example similar situation we are having an embankment on soft clay but what we have done is that you know embankment is constructed after you know waiting for a period of you know for consolidation for one that is stage one and the stage two was embankment is raised subsequent to consolidation of a soil under the stage one loading of an embankment. So in this case what actually happen is that you actually allowed the consolidation to happen and during that stage actually drainage is allowed then volume changes in the soil were actually recorded but when stage two happened and the no drainage actually happened all of a sudden there is a you know the failure which actually could have reason. So this actually situation you know arise and for this type of situation that you know we can actually say that consolidated undrained parameters are very useful. And the other example like the earthen dam example when you take we have a core and we have the casing and this is the water level one and water level two assume that the water level one to two drops you know suddenly that is called rapid drawdown case and in this case rapid drawdown if it occurs behind an earthen dam and no drainage of the core when it is take place that is the during steady state seepage conditions the consolidation and all these things actually happened though the shear tau f is nothing but the shear strength under steady state seepage conditions prior to the drawdown but what is actually happened is that the drawdown is so sudden there is no drainage actually during the you know the period of the drawdown. So in such situations the shearing actually happens you know under undrained conditions so this is an example for a consolidated you know undrained triaxial test parameters. So here when you have this consolidated undrained case then in such situations what we can use that we use these you know depending upon the typical situations we actually select the type of the test and then we use this parameters for conducting the different special cases of the you know the triaxial test. So we have you know in principle consolidated undrained unconsolidated undrained and consolidated drained triaxial test and there is a special case of you know a triaxial test where unconfined compression test what we call and which is you know which is you know without any cell pressure the sample is actually taken to failure so the space one is actually taken to failure with no confinement that is unconfined compressive strength test and unconsolidated undrained test is actually specimen is taken to failure with no drainage permitted neither in consolidation time nor in you know shearing stages and consolidated undrained test actually here the drainage is actually initially open to allow the consolidation to take place and so that the pore water pressure dissipates to zero and then it is closed so that the failure can actually happen without any drainage that is the condition where the shearing stage you know the drainage will be close. So then third type of test is that the consolidated drained triaxial test the drainage valve is initially open to allow the pore water pressure to dissipate to the zero so that is as usual in the previous stage previous type of test consolidated undrained test the sample is allowed to consolidate and the pore water pressure is you know UI is allowed to dissipate to zero and it is kept open while the specimen is actually taken to the failure at a sufficiently slow rate. So the strain rate at which the samples are actually you know tested for example unconfined compression test and unconsolidated undrained test they are actually tested at 1.25 mm per minute and this is the you know the standard rate you know the strain rate at which the sample this particular 1.25 mm per minute is actually selected keep in view that you know the pore water no volume changes actually take place and the pore water pressure dissipation also will not actually takes place particularly in unconfined compression strength test actually because it is not actually having you know exposure these application of cell pressure the sample is actually not allowed to undergo the changes due to in the temperatures. So we have you know in case of consolidated drained and consolidated undrained triaxial test the strain rates are actually calculated based on the permeability of the soil. Suppose if we are having a merengue clay with clay of high compressibility let us say CH then in situations what will happen is that the permeability of the soil is so low the even the test is actually conducted very very slow rate. Suppose if you are having a silty type of soil the stress the stress the test is actually conducted this the rate strain rate is selected as the permeability of the you know silty soil is actually more than clay soil so the test can actually can be done at a slightly rapid rate compared to you know the rate at which actually it was done for a clay. So this particular you know the stresses and strains in a sample in the triaxial compression test basically it is an axisymmetric condition where sigma 2 dash is equal to sigma 3 dash or sigma 2 is equal to sigma 3 and epsilon 2 is equal to epsilon 3 epsilon 2 and 3 are nothing but epsilon 2 in the intermediate in intermediate direction that is intermediate principle stress direction and so in case of triaxial test because of the cylindrical nature of the sample sigma 2 is equal to sigma 3 and epsilon 2 is equal to epsilon 3.   So thus q dash is equal to q shear is actually unaffected by the pore water pressure so here when you are actually having a sample which is confined with the membrane and then cell pressure sigma 3 and the axial total stress is sigma 1 is equal to sigma 3 plus p by a sigma 3 will be acting in all directions sigma 3 will be acting in all directions and the sigma 1 is actually applied so sigma 1 is equal to sigma 3 plus p by a. So the d weight stress is nothing but sigma 1 minus sigma 3 is equal to p by a so it is also called as sigma 1 minus sigma 3 is equal to sigma d by b by a and the axial strain in the major principle stress direction that is epsilon 1 is equal to delta z by h naught is the original height of the sample and delta z is the vertical compression of the sample or vertical strain actually experienced by the sample so epsilon 1 is equal to delta z by h naught and radial strain that is the bulging if it actually happens for example for a loose sand or for a normally consolidated soil then there is epsilon r is equal to delta r by r naught delta r is the change in the radius of the sample to the original radius of the sample and the d weight strain is actually given by epsilon d is equal to 2 third of epsilon 1 minus epsilon 3 and volumetric strain is nothing but epsilon v is equal to epsilon 1 plus 2 epsilon 3. So with p dash is equal to sigma 1 plus 2 sigma 3 dash by 3 and p is equal to 2 sigma 1 plus 2 sigma 3 by 3 and q dash is equal to q is equal to sigma 1 minus sigma 3 we will use these deliberations for plotting the stress paths for the different types of triaxial test. So as we have discussed the types of triaxial test and we said that consolidated and drained triaxial test that is the CD test which is very slow test and if you look into the different stress states in the during the sample testing in the step 1 is that at the end of consolidation what will happen is that the sample is allowed to consolidate during by allowing the valve to open. So the pore water pressure dissipation will be 0 so then we actually have sigma v dash is equal to sigma v dash is equal to sigma v and sigma h dash sigma dash hc is equal to sigma hc and during axial stress increase that with the sigma v and then delta sigma increase and here as we are not allowing the pore water the stress state the shearing is actually done at such a slow rate that the pore water pressure dissipation will be more or less close to 0. In that situation then we have got sigma v dash is equal to sigma v plus delta sigma is equal to sigma 1 dash. So sigma dash h is equal to sigma dash h is equal to sigma 3 dash at failure then these things turn out to be that is delta sigma turn out to be delta sigma f then sigma hc remains constant and the pore water pressure again still remains constant at failure then sigma dash vf is equal to sigma v plus delta sigma f is equal to sigma dash you know 1 sigma 1f that is the failure and sigma dash h hf is equal to sigma dash h is equal to sigma dash 3f. So this is the principle stress in the radial direction at failure and sigma 1 dash up is the principle stress in the direction the major principle stress direction. So sigma 1 is equal to sigma vc vc plus delta sigma and sigma 3 is equal to sigma hc the d weight stress or d weight is given by q or delta sigma d is equal to sigma 1 minus sigma 3. So during the consolidated drained triaxial test or if we are having a cu test the volume change actually takes place. So this particular volume change of the sample during the consolidation so till the consolidation is completed the sample is actually allowed to consolidate and thereafter once this consolidation is completed once the consolidation stage is completed typically for a clay it may last for about more than 4 to 5 days or sometimes about a week for completion of consolidation. And in this particular slide the stress strain relationship during shearing is actually given and this is for a case of consolidated undrained triaxial test wherein here what we have done is that we actually have you know not allowed the drainage to take place during the shear and then you know what you can actually happen is that we have a situation that here the d weight stress actually variation is like this and then there is a decrease and decrease actually takes place. So this is for the valued for dense sand and Oc clay and this is for the loose sand or normally conserved clay you can see that dense sand and Oc clay there is a hardening takes place and after attaining a peak a distinct peak is actually after exhibiting the distinct peak there is a you know a softening actually takes place. So here in this case what will actually happen is that you know here this is actually for CD test that is for consolidated drain test and where in actually here the sample if you can see that loose sand and normally conserved clay there is a compression actually takes place but in case of dense sand and Oc clay initially there will be some compression thereafter there is an increase in volume that is volume actually increases. So this is actually because you know we have discussed earlier the dilatancy behavior of the soil so in case of triaxial test also particularly when you are actually allowing the sample volume to change during the shearing by allowing the drainage to happen that is in the case of consolidated drain test. So what will happen is that you actually have got you know the volume changes actually takes place you know in the way which is actually shown but during all stages the pore water pressure remains to be same the pore water pressure remains to be same. And the CD test when you actually have done say if you wanted to see how actually we can determine a different the parameter the shear strength parameters. So what one need to do is that we have to do let us say minimum 3 samples or 4 samples and each sample is actually tested for different cell pressures or different confining stresses. So confining stress 1 that is sigma 3A and confining stress B say sigma 3B and confining stress C is sigma 3C then you know we actually have got different types of you know the stress deviator stress versus axial strain curves. So once you know the maximum or peak values and for those peak values that is this is nothing but the delta sigma D. So we can actually with that as this as diameter one if you draw the Mohr circle you actually can draw different Mohr circles and depending upon let us say that if it is you know normally consolidated soil and which are loose sand then actually we have a situation like you know C dash is equal to 0 that passes through the origin and tau sigma envelope will be like this with where pi is equal to pi dash here which is you know the angle of inclination of the Mohr Coulomb failure envelope. See when this actually interacts with this thing so this is the failure plane which is nothing but 45 plus 5 by 2 inclination which actually happens and this is the angle of inclination of the failure. So this is the typical so you need to have these you know the parameters basically to determine different types of sets of Mohr circles can be drawn with different cell pressures and different deviated loads and so in case of you know when you consolidated drain test as the cell pressure is increased the Mohr circle diameter keeps on increasing the deviated load keeps on increasing. So the strength parameters C and phi obtained from CD test since U is equal to 0 in consolidated drain triaxial test sigma is equal to sigma dash and therefore you know the C is equal to C dash and phi is equal to phi dash and what we call these are CD and pi D this is nothing but drained cohesion and drained angle of internal friction which is called and these parameters basically represent the long term conditions in a soil where consolidation is actually allowed and then drainage actually taking place during the application of the loading period. So in such situations the consolidated drain test the parameters are actually are vital. For strictly speaking for example if you are having a normally consolidated clay if that is actually identified as normally consolidated clay and as CD is equal to 0 and Mohr circle is actually passing through the horizon the failure envelope actually passing through the horizon therefore one CD test would be sufficient to determine the pi D of the sand or loose sand or normally consolidated clay. So one CD test would be sufficient to determine the pi D the drain friction angle of a loose sand or normally consolidated clay. For more consolidated clays for the soils which are actually have been subjected to certain pre-consolidation pressure the CD of a more consolidated soil is not equal to 0 or when you are actually having a very dense sand or dense sand deposits and where in that case also the CD may not equal to 0. So more consolidated clay and normally consolidated clay when we look into this here initially up to pre-consolidation pressure the envelope actually the Mohr column failure envelope runs like this then beyond that it actually changes into the normally consolidated this is valid like it joins with the normally consolidated failure envelope. So the point of transition where it actually changes again the behavior is actually like the pre-consolidation pressure that is sigma C and where in up to that pre-consolidation pressure stage the sample actually exhibits a drained cohesion and then thereafter once it crosses sigma C and there is a possibility that it actually merges with the normally consolidated Mohr column failure envelope. Now let us discuss about the stress paths during the CD test. So what we have drawn is that we said that there are two stages so we have actually represented here Q sigma 1 minus sigma 3 and P is equal to sigma 1 plus 2 sigma 3 by 3 or P is equal to sigma 1 P dash is equal to sigma 1 dash plus 2 sigma 3 dash by 3 because P and P dash are indicated here. So during the initial stage that is consolidation phase you know what it has been indicated that the stress path actually follows from here to here and thereafter it is subjected to shear so this is during the shearing stage. So the shearing stage actually starts from here and this inclination of this is actually indicated as 3 to 1 3 vertical 1 horizontal. So if you look into this here this delta sigma 1 is equal to delta sigma 1 dash because no load is actually applied and delta sigma 3 is equal to delta sigma 3 dash. So delta sigma 3 is equal to delta sigma 1 dash and delta u is equal to 0 so the consolidation phase actually is indicated here. So delta sigma 1 is equal to delta sigma 1 dash is equal to delta sigma 3 is equal to delta sigma 3 dash and delta sigma 1 greater than 0 delta u is 0 at the end of consolidation. So during the consolidation once it is self-pressure is applied the pore water pressure increases to the excess pore water pressure and then you know it dissipates. So that is at the end of the consolidation the delta u is equal to 0 with that what actually happens is that delta p dash is equal to delta p is equal to delta sigma 1 plus 2 delta sigma 1 by 3. So it gets simplified to delta p is equal to delta sigma 1 and delta q is equal to delta sigma 1 minus delta sigma 3 as delta sigma 1 is equal to delta sigma 3 delta q is equal to 0. So delta q by delta p dash is equal to delta q by delta p is equal to 0 so because the slope of this is actually 0 it actually passes along the p dash line that is q is equal to 0 line it actually passes and subsequently what actually happens is that the sample is actually subjected to you know the subjected to shearing. So here during the shearing phase sigma 3 is equal to sigma 3 dash delta sigma 3 is equal to 0 and delta u is equal to 0 because no pore water pressure you know changes are actually allowed because the sample is not subjected to very slow shearing rate and sigma 1 dash is equal to sigma 3 dash plus p by a so with this you know this also indicated as the effective shear stress path and total stress path both are actually identical in this case. So for the case here stage 2 delta sigma 1 is equal to delta sigma 1 dash greater than 0 and delta sigma 3 is equal to delta so we are not actually changing the cell pressure so delta sigma 3 dash is equal to 0 and delta u is equal to 0 during the shear stage. So delta p dash is equal to delta p with delta sigma 1 and remaining other things are 0 delta sigma 2 and delta sigma 3 is 0. So with that what will actually happen is that delta sigma 1 by 3 is the what we get. So this is actually delta sigma 1 by 3 and delta q is equal to delta sigma 1 and delta sigma 3 so this is equal to you know delta sigma 3 is equal to 0 what you have is that we have only delta sigma 1. So delta q by delta p dash is equal to delta q by delta p is equal to because of delta q is equal to delta sigma 1 and delta p dash is equal to delta sigma 1 by 3 which is nothing but delta Q by delta P is equal to 3. So because of this reason what you can see is that delta P is equal to delta sigma 1 by 3 or delta Q is equal to delta sigma 1. So delta Q by delta P is equal to is given as 3 here that is the 3 vertical one horizontal. So next we introduce ourselves to the consolidated and drained triaxial test and wherein we said that during consolidation the drainage is actually allowed, but during the shearing the drainage valve is actually closed. So shearing is done to represent a particular place where the practical cases are examples have been discussed here. So this is for total and neutral and effective stress. So here as we are actually not allowing the drainage in the shearing stage, so depending upon the stress history of the soil the pore water pressure changes occur accordingly. Suppose if you are having a normally consolidated or loose sand sample their pore water pressure will be positive and in case we are having you know a work consolidated soil are very dense sand sample or dense are very dense sand sample the pore water pressure is actually negative after initially positive and then subsequently changes to negative. So at the end of the consolidation the pore water pressure is actually 0, so then the effective stress is sigma dash vc is equal to sigma vc sigma dash hc is equal to sigma hc. Then during the axial stress increase the pore water pressure is can be plus or minus delta u it can be positive or it can be negative depending upon that the changes sigma dash v is equal to sigma vc plus delta sigma plus or minus delta u is equal to sigma 1 dash and sigma dash h is equal to sigma hc plus or minus delta u is equal to sigma 3 dash. So at failure so you can see that during the stress increase the drainage is actually closed here so because of that you know the sample is not allowed to drain then because of that there is a changes in the pore water pressure and at failure again no drainage so with that you know again the failure shear stresses are the pore water pressure changes are plus or minus delta uf. So sigma dash f is equal to sigma vc plus delta sigma f plus or minus delta uf is equal to sigma dash 1f and similarly in the horizontal direction sigma dash hf is equal to sigma hc plus or minus delta uf is equal to sigma dash 3f. So the volume changes in the sample which is actually same as you know the consolidated undrained triaxial test which is similar and in this case Cu test the stress strain relationship during shear which is again you know very similar to that but only thing is that here the pore water pressure aeration which actually takes place like this is positive and the pore water pressure is negative if you are having a dense sand and water consolidated clay. So in this case when you have Cu test basically to determine the shear strength parameters here we can actually get total strength parameters as well as effective strength parameters and for example for the first case which is actually shown here the total strength parameters how to obtain actually shown here and we are having a case where you know two confining pressures which are actually shown sigma 3a and sigma 3b and two more circles are shown and when the envelope is actually is drawn this is the Mohr-Colem envelope for the total stresses and this is the Mohr circle but depending upon the pore water pressure you know depending upon the you know variation of the pore water pressure we can actually get the Mohr circles accordingly positive or negative so with that what will happen is that we will get the effective strength parameters that is the Mohr-Colem failure envelope shifts accordingly and so this is for the strength parameters for C and phi in case of so with the consolidated under-interaction test we actually we get both drain parameters as well as undrained parameters. So the shear strength parameters in terms of total stresses are actually called as C consolidated undrained and phi Cu in case of shear strength parameters in terms of effective stresses are called as C dash and phi dash so C dash is equal to Cd and phi dash is equal to phi d they are called the drain friction angle and drain coefficient. So the stress parts for the consolidated drain triaxial test so we have seen consolidated drain triaxial test and in the case of consolidated drain triaxial test we notice that the stress parts for the you know effective stress part and total stress part both look alike but in case of you know the consolidated undrained test where the shearing changes actually happens so because of that you know during the shearing the pore water pressure changes actually happens so the effective stress part actually runs like this schematically which is shown here. So we have isotropic consolidation phase delta u is equal to 0 and delta p dash is equal to delta p is equal to delta sigma 1 plus 2 delta sigma 1 by 3 which is nothing but delta sigma 1 so delta q is equal to delta sigma 1 and delta sigma 3 with delta sigma 3 is equal to 0 and delta sigma 1 is equal to 0 here delta q by delta p dash is equal to 0. So with that what will actually happen is that delta q is equal to delta sigma 1 and so with with this actually what will happen with the delta Q is equal to 0 and delta P is equal to delta sigma 1 delta Q by delta P dash is equal to delta Q by delta P is equal to 0. So in this case also during the consolidation phase the stress path actually runs like this then depending upon effective stress path means it runs like this, total stress path means it runs like this. So here in this case again delta P is equal to delta sigma 1 by 3 and delta Q is equal to delta sigma 1. So delta Q by delta P is equal to 3 so this actually runs like this. So this is for the shearing stages which are shown and this is for the initial consolidation phase which are actually shown. So for effective stress parameters and effective stress path and total stress path the shearing phases is actually indicated here delta sigma 1 greater than 0 where in delta sigma 3 is equal to 0. So delta sigma 1 dash is equal to delta sigma 1 minus delta U which is equal to 0 and delta sigma 3 dash is equal to minus delta U and delta P is equal to delta sigma 1 by 3 is equal to delta sigma 1. So delta P is equal to delta sigma 1 by 3 and delta Q is equal to delta sigma 1. So delta Q by delta P is equal to which is nothing but delta sigma 1 by 3 delta sigma 1 by delta sigma 1 by 3 becomes 3 that is for total stress path. But in this case you know delta P dash is equal to delta P minus delta U which is nothing but delta sigma 1 by 3 minus U that is you know what we have done is that delta U the forward pressure change actually has been subtracted and delta Q is equal to delta sigma 1 where delta Q by delta P dash is equal to delta sigma 1 divided by delta sigma 1 by 3 minus delta U. So that is when you simplify that one because this is you know different. So we have 3 divided by 1 minus 3 to the ratio of delta U by delta sigma 1. So because of this particular variation the effective stress path actually runs like this. Now as we said that we have another class of you know the triaxial test which is a quick test which is actually called unconcerned undrained triaxial test. The purpose of the unconcerned undrained triaxial test is to determine the undrained shear strength of a soil of a saturated soil. Predominantly if it is saturated then we actually have you know the undrained coefficient of a or undrained shear strength of a soil. So this is a quick test so neither during consolidation nor shearing stage excess pore water pressure is allowed to drain but there is a possibility that the pore water pressure develops but is not allowed to drain. So this indicates you know these are you know sometimes you actually get for catastrophic loading or shuttle loading the shear strength parameters this is parameters which are actually obtained from this can be used for the design. So this is basically for short term considerations this can be applied. So different stress states or conditions are actually shown here. So this is actually the sample as prepared 0 the effective total stresses on the sample on outside 0 the pore water pressure inside the sample is actually negative and then because of that you know sigma dash v is equal to positive uf will be there. So because of that negative pore water pressure the sample actually stands vertical so the effective stress here is that sigma dash v0 is equal to uf sigma dash h0 is equal to uf. So all round actually it maintains an effective stress of equivalent to the negative pore water pressure which is actually there in the soil. Then after application of the you know hydrostatic cell pressure then you know what we have done is that we applied sigma c the total stress on the sample changes to sigma c and sigma c here. Now the pore water pressure is nothing but minus uf that is negative plus delta uc because we applied sigma c because of that there is an increase in soil sample pressure and 100% saturation means the water actually attracts the load and then you know sigma dash v c is equal to sigma c plus uf minus uc. So this is another you know set of this thing will come and the sigma dash hc is equal to uf will be there. During the application of the actual load these are the stress states which is nothing but you know because of the shear there is an increase in the pore water pressure that gets added here. So these stress states actually different stages actually the stages are given one is just application of the just placement of the sample before application of cell pressure then other one is actually after the application of cell pressure and then during the application of the shear load and then at failure. So the more circles are you know for the 100% saturated clay are indicated like this. So you can see that this is independent of the you know whatever the cell pressure we applied for example you know this is a case where you know 0 cell pressure but when you actually have a cell pressure of say 50 kilo Pascal 100 kilo Pascal and 200 kilo Pascal and the soil is completely saturated and unconsolidated and drying test is performed then you actually have a case where you know you get the horizontal failure envelope that is more column failure envelope the pi t is equal to 0 or pi u is equal to 0 and so the tau f is equal to shear strength is equal to c u and that is the more column failure envelope in the horizontal only. So this indicates that you know the stress is actually independent of the stress. So whatever may be the confining pressure you apply that much you know the deviator load is generated in such a way that diameter of the more circle remains constant. Say for example when you are having a partially saturated clay and initially when you do unconsolidated and drying test using a with partially saturated samples then actually you have got you know when degree of saturation is less than 100%. So with high cell pressures when you have high confining pressures and because of the cavitation actually what happens is that the water is actually the air which is within the sample is shunted out with water. So what will happen is that the 100% saturation is actually ensured so and at very high confining pressures the more column failure envelope tend to become horizontal so then the sample actually becomes you know 100% saturated. So in that case again it maintains the horizontal plateau for the more column failure envelope. Otherwise initially for partially saturated samples actually you have got curvilinear more column failure envelopes can actually happen and you know sometimes people mislead by using these values per Cu and Phi U as strength parameters and undrained conditions. So the total stress parts during unconsolid and undrained tri-shell tests are actually given here. So in this case because neither consolidation nor the shearing stage there is a drainage. So what is actually happening is that the stress part actually starts directly here this is a total stress part only and where you have sigma 1 this is the sigma 1 is equal to sigma 3 plus P by A and sigma 3 and delta U is not equal to 0. So initial phase in initial stage is that delta sigma 1 is equal to delta sigma 3 and delta U is not equal to 0. So delta P is equal to delta sigma 1 delta Q is equal to 0. So delta Q by delta P is equal to 0 that is at this point and the shearing phase delta sigma 1 is greater than 0 and delta sigma 3 is equal to 0. So delta P is equal to delta sigma 1 by 3 delta Q is equal to delta sigma 1. So with this again the inclination of this is delta sigma 1 by you know is again 3 so delta p is equal to delta sigma 1 by 3 delta q by delta p is equal to 3. So as we have been discussing that there is also unconfined compressive strength test this is one thing basically to determine the undrained shear strength of a saturated clay very quickly and in this case is a special case of triaxial test which is called with sigma 3 is equal to 0 the cell pressure is actually 0. So in this particular slide where a sample actually loaded which is actually shown with sigma 1 that is the deviated load because sigma 3 is equal to 0 and the sample is actually not confined with you know the chamber pressure. So in this case basically this is a quick test where it can actually give undrained shear strength of a saturated sample. Sometimes this is also used for partially saturated soils basically to get the you know unconfined compressive strength of a sample and then there afterwards you can actually get the undrained cohesion the sample need not be completely you know saturated but truly speaking it should be to determine the undrained shear strength of a saturated clay quickly and the deviated load is actually increased rapidly until the soil sample fails pore water cannot drain from the soil. So the sample is sheared of at the constant volume the sample is actually sheared at constant volume without any changes in the you know volumes and the without changes in the pore water pressure. So the stress states are actually given here as the sample is prepared this is same as unconsolidated undrained traction compression test and this is neutral stress or pore water pressure is minus uf. So initial effective stress is that sigma dash v naught is equal to uf. Similarly during the application of load you can see that there is a pore water pressure change occurs but not allowed to drain. So the test is actually done such a way that no dissipation of pore water pressure takes place no volume change under no volume change conditions. So sigma dash v is equal to the delta u plus uf that is the uc that is during the initial conditions. So this is at failure conditions. So the total stress path during the unconfined compression test again it actually resembles to this it actually passes through you know the origin and the if delta u is measured it would have been negative since it would be negative since sigma 3 is equal to 0 that the sigma dash 3 is equal to sigma 3 minus delta u is equal to minus delta u. So delta u must be negative because as sigma dash 3 cannot be negative so soils cannot actually sustain tension so sigma 3 must be positive. So the effective stress path is unknown in case of unconfined compression test because the pore water pressures or changes are normally not normally measured they are not measured normally. If delta u is measured it would be negative since sigma 3 is equal to 0 and this delta u must be negative because as sigma dash 3 cannot be negative the soils cannot sustain the tension so the sigma dash 3 must be actually positive. So here the delta u if at all you measure during unconfined compression test the pore water pressure would be negative and the resultant Mohr circles for the unconfined compression test are actually gone here. So this is the total stress circles because this starts from origin because sigma 3 is equal to 0 and this is the Mohr Coulomb failure envelope and this is the total stress Mohr circle and if at all means the effective stress will be like this the effective stress Mohr circle towards the right side because delta u is negative and if it is positive it comes this side. So the effective stress circle cannot be determined in the UC test this is only just indicated here and this is you know the as the phi u is equal to 0 the failure plane is actually only 45 plus phi by 2 it is only 45 degrees so the failure plane is actually indicated here and this point where it actually intersects that is actually the horizontal envelope of a Mohr Coulomb variation envelope so the Cu is actually shown here. So basically the results from the UC test are actually you can lead to usage of the estimate of the short term bearing capacity of a fine-grained soils for foundations and basically for unconsolidated rain test also we can use this logic and the estimate the short term stability of the slopes or ethan dams and determine the stress strain characteristics under fast undrained loading conditions. So if wanted to see this stress strain characteristics of a material under fast loading conditions and this actually particular test is actually used. So this is the typical variation of sigma 1 with UC1 so you can see that the sample where the failure surface is actually about you know is inclined like this and so you can see that this is sigma 1 and this is epsilon 1 and so the average of these two values which actually gives this is the deviated load sigma 1 by 2 gives the you know 136 by 2 which actually gives the undrained cohesion of a soil sample. So in this case this is a partially soil saturated soil which is actually having you can say silty clay type of sample where you know you can say that you know the undrained cohesion is about 65 to 8 kilopascals. So we also have some you know results which are actually typical results of consolidated undrained triaxial test and silty sand samples are actually shown here. So you can see that the sample is actually tested with three different cell pressures 50 kilopascals and 100 kilopascals and 150 kilopascals. So this is actually 50 kilopascals cell pressure with tested at sigma 1 – sigma 3 and this is at 50 kilopascals, 100 kilopascals and 150 kilopascals and this is the excess pore water pressure measured because this is you know mostly you can actually see that the pore water pressure is positive because this is you know close to normally consolidated sample state. So the sample at the end of the shear you can see that the sample actually has undergone major portion of bulging and you know then the subsequent shear failure actually happened this is actually within the membrane and once the membrane is actually taken out this is the case and then the stress parts actually are drawn here with the P is equal to sigma 1 plus 2 sigma 3 by 3 from the test data which is actually obtained. So these are the total stress parts that is for 50 kilopascals cell pressure 100 kilopascals and 150 kilopascals and then these are you know the stress parts for you know the effective stress for 50, 100 and 150 and these are the more circles for the silty sand sample which is actually tested. So the sample found to have the undrained parameters of the undrained parameters are see 7 kilopascals and 532 degrees and the drained parameters which is you know 2 kilopascals and 35 degrees so this is the drained parameters. And this is for example of a consolidated undrained triaxial test and a fine sand having maximum void ratio 0.778 and 0.54 to minimum void ratio and the void ratio after the consolidation stage is very dense nature and deviator stress versus axial strain is shown for two different cell pressures 50 and 100 kilopascals and this is the pore water pressure here you can see that because of the dense sand the pore water pressure variation is actually is negative then these are the typical stress parts. In this particular slide you know we have seen we are seeing a status of sample particularly very dense sand sample after termination of consolidated undrained triaxial test. So you can see that the sample bulging as well as you know the shearing actually which has taken place with pi by 4 plus 36 by 2, 45 plus you know 45 plus 36 by 2 that is the angle of inclination of the failure plane about 63 degrees is the failure plane and wherein you can see that you know the dense sand sample actually exhibits you know the failure plane which is distinct failure plane and when you actually have loose sand and normally consolidated soils we have you know predominantly bulging actually takes place. So in this particular lecture what we have done is that you know we try to understand about the different types of triaxial test particularly unconsolidated and drained triaxial test and consolidated undrained and consolidated drain triaxial test and unconfined compression test is special case of so all these tests which actually we have done for as a compression case and we know that you know by maintaining different combinations of cell pressure and axial pressures we can also do extension test by using the triaxial compression test. But in this particular case mostly we are covered about the unconfined compression test or UU and CU and CD triaxial compression test pertinent details and then we also have discussed about the stress parts pertinent to that with the Q and P and P dash and then we also have discussed about how we can actually determine the drain parameters and undrained parameters depending upon the situation like for example undrained bearing capacity or you know when you are actually have you know short term stability of a slope then we can actually think of using undrained parameters undrained strength parameters when you wanted to have a long term stability of a slope or long term stability of an embankment then long term bearing capacity then we actually have to use and the effective strength parameters like drain parameters C dash and pi dash.


The race to Laugh Tale is heating up, because the recent chapter showed us another Yonko officially punching in his ticket to join the chase for the One Piece, with Buggy declaring his desire to become the Pirate King. Luffy, Shanks, Blackbeard and now Buggy are all after the same goal and the road to witnessing this highly anticipated clash of styles and personalities begins. We also learned that Crossgill's notoriety is increasing, with the bounty system they placed on the Marines, resulting in the death of a beloved Vice Admiral.  And in this video, we're going to dive deeper into all these new events that are guaranteed to have massive implications for the future of the story. The chapter begins in a shocking way, announcing the death of Vice Admiral T-Bone and it's even more shocking to find out who was responsible. We see Sengoku and Suru discussing his death and find out that one of the very same civilians they swore to protect is the culprit, an old man from Pepe Kingdom where every year over a thousand people die of starvation. And this is an important detail surrounding the death of T-Bone because of what we know of him. He was portrayed to us as one of the good Marines who will do anything and sacrifice everything for his subordinates. We saw this when he was first introduced back in Walter 7, T-Bone had an over-dramatic approach when it came to showing his kindness, which contrasts his scary appearance. He has a hobby of saving people and a motto of a hundred good deeds per day. It's also worth noting that the man who supposedly killed T-Bone seems feeble and someone not used to violence, only acting out for the sake of providing for his starving family. The man just doesn't scream fighter, let alone someone who could beat T-Bone who was shown capable of cutting a sea king in half. After all these details put together makes T-Bone's death somewhat suspicious and if we were to speculate on the reality of it, we can assume that T-Bone's death played out in an over-dramatic way. Say maybe T-Bone went to Pepe Kingdom and saw the people starving and as part of his 100 good deeds per day told the people to kill him and claim the bounty on his head to feed the culprit's family. Or it could even be a setup to another classic one piece joke where the man did actually try to kill T-Bone, but whatever the man did wasn't enough and T-Bone actually survived but is now thought to be dead because, well, he does look like he is, yohoho.  But this is also the thing I love about One Piece. Here we have a very minor character who was given an obvious personality quirk and a dedicated moment to showcase his ability. Even if he's not a significant character in the grand scheme of things, we still feel the impact of his supposed death because of our familiarity with him and the way we were geared towards liking this scary yet adorable man as opposed to just using a random throwaway character to move the plot forward. By the way, I just want to acknowledge how Sengoku seems to be more relaxed now that he's not a fleet admiral. Seeing him ask for Garp, not to scold him, but just to have him try a new rice cracker flavor is so adorable. But of course, Garp continues to be a headache for him even after no longer being his subordinate. Also, I wonder if he and Suru will take action and also become a part of the upcoming story since Suru's niece might be in danger as well now. And of course, barring any flashbacks, it opens a way for these two legends to showcase their abilities or what's left of it. While there may be some uncertainty surrounding T-Bone's death, the news of it guarantees one thing and that is the ever-increasing threat of Cross-Gild who are becoming one of the Marines' biggest headaches. Tapping into people's desire for wealth increases the seriousness and danger that this faction poses to the world government who are now faced with the difficult conflict of how to protect themselves against the people they swore to protect and serve. And it's scary to think that Cross-Gild is just barely scratching the surface because we also learn that Crocodile and Newhawk are planning to increase their influence to an unparalleled level to form their utopia and become a military state. But the most intriguing part is their plan to build such overwhelming power. Cross-Gild's plan is working, but they still need to increase their military forces. And this could suggest that they're going to recruit other forces from all over the place. So don't be surprised if you see the return of a familiar face or two to join this faction. And I could throw in as many notable names in there as I want, but why don't I leave that up to you? Leave a comment below on who you would like to see eventually join Cross-Gild. Now onto my favorite EVERYTHING about this chapter, the man himself, Buggy the Clown. As with every other time we see Buggy, his appearance in this chapter is filled with comedy. We learn that Buggy took responsibility for building the Cross-Gild ship, but his ship rights overdid it with Buggy-centric features, angering both Crocodile and Newhawk. We also get a look at Cross-Gild's meeting room, but after having a closer look inside, it seems to act more of a torture room for Buggy. And I'm not even sure if he's welcome in this room outside of his punishment, because apart from the ones that Crocodile and Newhawk are sitting on, the only other chair available is a spiky iron chair. It's also a cute detail that the island where Buggy is staying is fittingly named Empty Bluffs. But in this case, Buggy showed more than just force bravado. He showed real guts. As of right now, out of the three of them, Buggy is the one who is showing the most pirate-like qualities. Crocodile and Newhawk's approach is that of businessmen. We're arguably considering risks and opportunities, whereas Buggy is the one fearlessly going for the big prize. Buggy now confirmed to also have the dream of becoming the pirate king. In revealing this, the chapter also extended the scene from chapter 434, which showed Buggy refusing Shanks' request to join his crew. We learned that Buggy always believed in Shanks and was even willing to support Shanks, helping him achieve this goal of becoming the pirate king of their generation. And it's a real shock to find out that the actual cause of their now-distant relationship is because Buggy was disappointed that Shanks changed his mind about going after the One Piece. From his point of view, it seemed like Shanks was giving up and not following their beloved captain's footsteps, which raises further intrigue about what Shanks has really been up to. If he wasn't ready then, why is he ready now? It begs the question of whether his final conversation with Roger had something to do with his decision to postpone his chase for the One Piece or whether it's something else, such as it simply being Shanks' way of remembering his captain, ensuring Roger's name lives on a little longer as the pirate king instead of quickly being replaced by a new king. This reveal also provides an interesting what-if scenario as to what Shanks would have been able to achieve compared to Luffy at a similar age had he been more active in his pursuit. In Buggy's case, going after Captain John's treasure might have been more of a consolation prize in his mind since he thought he couldn't compete with someone like Shanks, which led to my favorite Buggy character moment, him realizing that fate has brought him to equal footing with Shanks again and so reclaiming his dream to go after the most important treasure to become the pirate king. It's just such a good scene. It's a very inspiring moment to witness someone with a forgotten dream at one of his weakest points, under threat, his head disembodied and still an overwhelming urge to yell out his true desire. It is time for Buggy to claim the One Piece. Chills. It reminded me of the scene at Onigashima where Kaito and Big Mom declared they're going after the One Piece and their subordinates were all cheering like Buggy's crew. And ever the pirate that he is, Buggy knows the way, he knows the objective and he finds other means within his ability to achieve it. The most hilarious part about this is that Crocodile and Mihoek are now reluctant passengers in this race. Their dynamic with Buggy was already becoming a favorite, but now we've added another layer with Buggy dragging the meticulous Crocodile and Mihoek who just want to be left alone to his fall to the top, cementing my full investment into this crew. I would honestly love nothing more than for Mihoek and Crocodile to get to that point when they actually end up genuinely liking Buggy and actively help him to get to the One Piece. It's one of those ideas that makes your nose tingle from happiness when you think about it. Honestly, seeing those two save Buggy from danger and delivering dialogue along the lines of, Hey, hands off our captain. It's probably something very unlikely to happen, but boy would that be peak. Wealth, fame, power, right? Also, speaking of things that make you happy, Luffy being mentioned in the same breath as Shang's by another powerful character like Mihoek is just plain awesome. The Yonko threat is for real. I'm giddy just thinking about their inevitable clash because this Yonko battle royale is setting up to be the most epic clash to date. Just think of all the big names involved so far and the potential matchups we could be witnessing. You can write down names, shuffle them, pick any two up, and it's guaranteed to be a big, exciting matchup. It doesn't matter who runs into who out of these four powerhouse groups, it's a classic in the making. This chapter also just casually revealed that Sabo is alive. As we get an update on the Revolutionary Army, some members we saw back in chapter 905 have officially been introduced as the vice-captains to the army commanders, and while they all seem interesting and have their own quirks, as is the standard with one piece characters, I think we can all agree that the most notable one is Ahiru, the girl with robotic arms whose speech patterns includes making mechanical noises, adding to the list of waifus in this arc. Moda, the milk maiden who helped Ace in his cover story and was last seen in chapter 904, also made an appearance and may even find her way to prominence as she's now joining the Revolutionary Army, along with the other people from Lelusia. Adding to the parallel that Sabo has now saved the woman who saved Ace, I love how Sabo's character has become more than just Ace's replacement. The recent events involving Sabo really turned him into one of the most crucial parts of piecing together some of the biggest mysteries. We're finally about to see what really happened at the reverie from Sabo's perspective, and the excitement I have cannot be contained. With the scenes rapidly changing from one to another, Sabo's story is one that I'm really interested in, and if I'm being honest, even more than any of the fights happening right now, just because this will be the reveal of something we've been wondering about for a long time. And even if you're someone who would rather witness a fight in a chapter, apart from the excitement in finding out answers to questions like who really assassinated King Cobra and why was it made out to be Sabo? And the more intriguing what or who did Sabo see on the empty throne, Sabo's flashback may also show how exactly the battle between the Revolutionary officers against Fujitora and Ryokugu went down. So there really is just a lot to be excited about in Sabo's upcoming story. And no matter what we get in the next chapter, it is guaranteed to be exciting. Back to Egghead with the Straw Hats and Vegapunk, Resuming Garp versus Kuzan, another step towards the man marked in flames, or back here where we left off, Sabo telling his story. Seriously, can you think of any other form of media that has this effect? Every week we're getting fed a hype chapter after hype chapter, and I'm here to eat up whatever one piece is cooking. So as always, thank you for listening to another one of my rambles. Let me know your thoughts, please do subscribe. This is Joygirl and I'll see you again soon.


The Ministry of Economic Affairs created a small policy-forming nature summit, all natuartop. We were tasked with promoting the summit and driving engagement among Dutch citizens. Over the course of a single day and night, events and activities by nature organizations took place across the country. All activities appeared on an interactive map that the policy-forming group could monitor and be inspired by during their meeting. Yes, as it turned out, a lot of people care about preserving Dutch nature. They just needed a fun original platform for which to voice their thoughts and ideas. Output from the day has contributed to the vision for a new natural environment policy, which will be launched by the Secretary of State any day now.


Hi everybody, Joe for jasmiescasebreaks.com with the Revolution 2 and 3 filler. You're guaranteed a team or a box of revolution. So we only sold 19 spots outright, I want to make sure I got 19 here, right? We only sold 19 spots outright and then we're just randomizing your names and matching you up with something here. Top name gets a box of revolution basketball, second name gets the magic and pick your team 2, third name gets the magic and pick your team 3, so on and so forth. All right, so and there's the whole list right here. Now without further ado, let's do it. Let's roll the dice, randomize that list three and a four, seven times. Big thanks everybody here for making this happen. Lucky seven times, everyone gets something. One, two, three, four, five, six and once more, seventh and final time. After seven, there you go. Francis with the box. Let's gather everybody's names after seven and here's how it shakes out for you. And we'll put a little rooftop next to your name so you know you won those spots in the filler. So Francis, you win the box. We'll send you a sealed box of revolution basketball, it's current year. Justin with the magic and two, Serenity with the magic and three, Steve with the bowls and two and three, both in our cases. Brandon with the Celtics in three, Sixers in two, Steve with the Sixers in three, T-Wolves in two, Brandon with the T-Wolves in three, Cody with the Nets in two, Serenity with the Nets in three, Evan with the Knicks in two, Bryan with the Hawks in three, Chris with the Knicks in three, Craig with the Spurs in three, Woon with the Heat in three, Josh with the Wizards in picker team three and Brandon with the Blazers in picker team three. There you go, ladies and gentlemen. Those two breaks coming up in separate videos tonight. Jaspyscasebrace.com. Bye-bye.


we now turn to looking at some of the different RNN architectures that are used, particularly for natural language processing. So one choice is whether to use a simple RNN or something that is a gated recurrent network such as an LSTM. We're going to not look at that right now, but instead we'll look at a variety of architectures that depend upon what the structure of the input is and what the output is. Language models, which are the basis upon which almost everything is built, takes in a sequence of words, a subsequence, I went to the and predicts the next word, store. I went to the store yesterday. Sequence labeling takes in a sequence of words, I went to the store and puts a label shopping or travel or study or not safe for work or hate speech. We'll look at later. So a sequence to a label. It's also not uncommon in NLP to label each individual token in the sequence. So I goes to a pronoun, I went, went goes to a verb and so forth. We label each item in it. And finally, sequence to sequence. Taking a whole sequence of words, I went to the store and output a different sequence, a translation into German or Chinese or a summary or some sort of other transformation of the sequence. So we'll explore each of these in sequence as it were. So the core one language model is of the structure we've seen before. Take in the first word, take its embedding, put it into a neural net along with the null hidden state output in the state that hidden state goes the next copy of the neural net. This is unrolled in time. The embedding, the next word goes in so forth and so forth. And at the end of the subsequence, whatever it is, we then predict what the next word is going to be. So that's a language model. It's great because it doesn't require any labels.  So this is great because it's unsupervised or self-supervised, does not need extra labels. Given a model that you've trained with that language model, you can then take that exact same model you've already trained, run it forward over a sequence of words, get the hidden state at the edge and put that hidden state into a very simple neural net and predict a label. Is this sentence about travel? Is this sentence about shopping? Is this sentence about friendship? So it's a way to map a sequence to a hidden state and the hidden state then to a label. You could also skip the self-supervised training and go straight from sequence to label, but that's usually harder because you don't have lots of labels.


So the the kind of argument that's usually given is that if we don't have something like social security and Medicare The the elderly will starve in the streets now. There's an interesting test case for this What happens when people don't have old-age welfare programs and it's called America for every year until 1935 which by the way if you can do the math is longer than we've had social security since 1935 Now did the elderly starve in the streets and it turns out that amazingly no Amazingly people figured out a way to live without handouts for over a hundred years and they It's pretty fantastic when you think about it They were very entrepreneurial in their whole approach to life And so I'm just gonna go through some of the mechanisms that they took in order to not as they say starve in the streets So first of all people worked And it's kind of funny today where we think of like ah works a real burden But actually the people pushing for a welfare state complained for many years because the elderly those darn people refused to retire The fact is that they found their lives more meaning for more meaningful more valuable more enjoyable with work than without it and they saved I Who has a savings of more than 0% if you do your you are the 1% because most Americans don't People at that era saved between 1 8th and 1 9th of their income Which is I mean that's impressive Credit obviously is you know if you if you couldn't support yourself various forms of credit were available Turning to friends family community Informal ways of helping people the number one does anybody know what the number one thing people did so first of all before industrialization before People were really dealing with retirement. What was the number one retirement plan anybody know Close that comes later. The first retirement plan was actually death We worked and then you died and then you didn't have a retirement plan But now we have industrialization people are living longer and yeah, that's exactly what they did the number one thing People did as they lived with their kids and When you think about it social security is in effect like an indirect way of us all taking care of other people's grandparents Like it's in effect the government wheels over somebody else's grandma shoves her into our lawn and says here you watch her for a while But no what most people did is that they took care of their own parents if they needed it There were also things called mutual aid societies now Anybody heard of like an elk's lodge or a moose lodge or something? I'd love to see these re-emerged They're kind of cool but these are in effect the descendants of mutual aid societies and these were Community groups in effect where you would pay your monthly membership due and you would be entitled to certain kinds of then support from the club and It might be things like unemployment insurance It might be things like private health care you'd have a lodge doctor you could see or it could be a retirement thing They had often very they had retirement homes where if you couldn't support yourself You would go and you'd be able to stay at the the Society's retirement lodge and finally there was charity private charity which in this era was abundant again, we're talking about pre-1935 and People per capita were giving more to charity then than they did today And so what were the results of this was it was it people starving in the streets? No, the fact is that What was happening with immigration where people fleeing the shores of America trying to get away from this wretched area? Where people were starving the streets and going to those countries that had? welfare programs Anybody know what was going on with immigration at this time? This is the biggest wave of immigration to America Indeed the people pushing for welfare state programs, which they started doing really in the 1880s model after Bismarck's Germany. They said well, we don't have to worry about old people in America They're doing just fine. We have to worry about poor people So they spent 30 years trying to create welfare programs for poor people and then they realized oh Americans Don't care about poor people. Let's focus on old people because most of us one day aspire to being old So let's let's do that and then they eventually succeed in 1935 as we talked about but the the best statistics we have in terms of what's going on then and we don't have great statistics So take these with a bit grain of salt But the best ones we have tell us that by 1935 Only 8% of the elderly need any form of formal charity whether it's private or whether it's the kind of miniscule government-level programs going on and what I'm showing here is I want us to Remember what's going on historically So I'm talking about 8% of the elderly need any kind of formal help And you can extend this line back to infinity But basically for hundreds and hundreds and thousands and thousands of years human beings have been living just above subsistence We're dealing with a time frame right here where Capitalism economic freedom has only started to enrich us and already people are productive and smart enough So that only 8% need any sort of formal Assistance in order to achieve economic security So for thinking about economic security then the bottom line is that Clearly a welfare state is not necessary and Indeed it undermines our economic security by giving the government the power to take away as much of our Economic resources as it decides is in the public good But now let's turn to the biggie Watkins I want to help people You're being a jerk who's just talking about the cost. What about the benefit to these people who need help? What I want to stress and the major point I want to make about this is that whether or not you want to help people is Totally irrelevant to whether or not there should be a welfare state Totally irrelevant First of all we've seen people are helping people immensely during this time Most people don't need help the vast majority of us are able to support ourselves throughout our lifetime including old age But those who did the vast majority are being helped privately voluntarily. You're able to help the people that you care about Yeah Sure So the the question is how do we compare the help available today versus before the welfare state? if we just compare the amount of giving As a percent of GDP it was today It's about 1.5 percent that number might not be exactly right But it's in the ballpark 1.5 percent of GDP goes to charity then it was over 2 percent Something close to two and a half percent. So a vast great a vastly greater percentage is going in that way It's more difficult once you start comparing. Well, how do we count the welfare state? Does that count as some kind of percent of giving or something like that? But the bottom line is that it increases significantly. It basically was double as a percent of what people are actually taking home the so the major fact though is then If we're talking about is it good to help people do I want to help people it's irrelevant the way I like to think about is this So imagine that you Are in the situation not where you want to help somebody, but you're in need of help Let's say you need an operation that you can't afford So you go over to your neighbor's house and you knock on the door and you say One of two things one is hey, I need an operation. I can't afford it Would you please help me fair enough? The other one is hey, I need an operation. You owe me What the welfare state is about and this is kind of the key point I want to stress The welfare state is not an issue of I want to help people that welfare state is an issue of people are in need and that Imposes a debt on everybody else. It's an entitlement Under the welfare state logic now, I've asked thousands of people How would you behave if you were in that situation and I've never met a person's yeah I would regard myself as entitled to somebody's help, you know, he might want to send his kid to school He might want to pay off his mortgage, but darn he owes me first and foremost But that's what the entitlement state says that's what the whole welfare state is built around It's the idea of you are entitled if you're in need regardless of its source Regardless of other people's hopes dreams priorities and rights They owe you because you're in need so the Often what will happen When anybody challenges the welfare state is that you're the jerk who wants to throw Grandma off a cliff right you want to take away her handouts You're gonna throw her off a cliff and I think the bottom line here is that The exact opposite is the truth So first of all the idea that not giving somebody handouts is the equivalent of throwing them off a cliff I mean, that's like saying if a thief came to your house and was like hey give me your wallet And you said no you're throwing him off a cliff because how is he gonna buy his Doritos? No, the truth is the exact opposite. It's young people today by the welfare state who are being thrown off a cliff 205 trillion dollars Already you're working a month and a half without pay and once it gets to two three four months That is what it means to be thrown off a cliff That is what it means to be detracted from actually achieving the things that you want to achieve in this world Who knows who Russell Brand is? Often I speak to older audiences and they have no clue and this story just goes poop. So I guess over a year ago. I went and I was supposed to do Russell show brand X and We were supposed to debate the issue of self-interest But after the show is talking to him and I was telling him about this whole problem and I saying look He's kind of a socialist so I was saying look the welfare state is treating young people as servants and These debt numbers are gonna turn people into actual servants and he looked at me thoughtfully which was quite an achievement and he said Isn't that the point and he's right the whole point of the welfare state is to force you to serve other people And so there's a reason that nobody blinks the numbers I gave you this is not some secret theory that I've unearthed from the archives of like Don Watkins conspiracy theory everybody in Washington knows this Everybody in Washington will admit it except for people like Elizabeth Warren who are basically debt deniers who say oh Let's increase Social Security, but setting that aside This is this is a well-known problem problem But they don't blink an eye because the whole idea is that you should be serving others You shouldn't be so focused on your happiness You guys are in this room to what learn to be healthier learn to be more fit learn to get better relationships learned That is selfish you guys should not be focused on that You should be focused on serving others and the welfare state it's gonna make you and So that's the deeper meaning of the debt draft is It's making you do exactly what the welfare state says you should do which is be a servant I mean basically what the welfare state does is it divides us into a cast of burdens and resources We're burdens when we receive handouts right and then we're resources were in a position to pay for other people's handouts Now that is a really really awful way to view human beings in my view In my view and I think this was the founders view which is that you have a right to exist for your own sake You have a right to come into this room and try to make the best of your life Even if those guys out there are not making the best of their lives The whole idea that you have to serve others and subjugate yourself to them is a really ignoble view of human relationships Each of us should be free to make the most of our own lives So how do we get this so wrong if what I'm saying is right Then how is it common sense? How is it conventional wisdom that the welfare state is a great Achievement and so I want to say a few words about this before we wrap up We're taking a different look of something that we've been diving into in the last couple of years now We have Don Watkins right here How social security is sabotaging the land of self-reliance bring it on home By the way, if you can do the math is longer than we've had social security since 1935 Now did the elderly starve in the streets and it turns out that amazingly no amazingly people figured out a way to live without handouts for over a hundred years and they It's pretty fantastic when you think about it They were very entrepreneurial in their whole approach to life And so I'm just gonna go through some of the mechanisms that they took in order to not as they say starve in the streets So first of all people worked It's kind of funny today where we think of like ah works a real burden But actually the people pushing for a welfare state complained for many years because the elderly those darn people refused to retire The fact is that they found their lives more meaningful more meaningful more valuable more enjoyable with work Than without it. So the first reason I think is what I call the collectivist premise and This is looking at the welfare of the group rather than the well-being of the individual and says what's good for quote? society as a whole How should we distribute society's resources in a fair way? I think the better way The more rational way to think about political issues is to focus on the individual And I think if you think about what's good for the individual you come to a very different conclusion So the two essentials in my judgment of what makes us happy and successful life And I think this came through in all the talks today is Thought and effort. This is Isaac Newton and Steve Jobs. It's You