 Welcome to lecture 2 of module 1 for the course advanced geotechnical engineering. In the previous lecture we have understood about origin of soils and their type of soils. In this lecture we would like to understand about phase properties, phase relationships using phase diagrams and we will try to solve some couple of problems. So as we discussed soil is a three phase material where you have a soil composition which consists of solids which consists of air, water and solids. So if you separate the soil composition into three predominant phases they are air, water and solids and if you put weights on the left hand side as on the weights on the right hand side where weight of air is equal to 0 and weight of water and weight of solids and if you put volume on left hand side the volume occupied by voids which comprises of volume of air and volume of water and volume of solids. So volume of voids plus volume of solids is total volume. So V is equal to Va plus Vs plus Vw plus Va and weight total weight is equal to weight of solids plus weight of water. So this is a case for a partially saturated soil. Now based on the phase diagram whatever we have discussed we can deduce number of volumetric ratios and these are used in soil mechanics and geotechnical engineering particularly they are known as void ratio E porosity N degree of saturation SR air content AC air void ratio or percentage void percentage air voids NA. So from the subsequent slides we will try to understand about these definitions of these and interrelations among these properties. So void ratio is defined as the ratio of the volume of voids to the volume of solids which can be written as E is equal to Vv by Vs as it can be seen here E is equal to Vv by Vs. So volume of voids V suffix V refers to that portion of the volume of the soil not occupied by the solid grains. Since the relationship between volume of air and volume of voids volume of water usually changes with ground water conditions as well as imposed loads. Suppose if the ground water table changes or fluctuates or because of the imposed loads if the soil undergoes some compression or some expulsion of water there is a possibility that volume of air and volume of water can change. It is convenient to designate all the volume not occupied by the solid grains as a void space that is volume of voids. So the volumes which are not occupied by the solid grains is indicated as the volume of voids. So void ratio is defined as Vv by Vs. Now if you look into for any solid material E is equal to 0 that means this absence of voids and E greater than 1 that means that the volume of voids is many times more than the volume of solids. For a typical soil the void ratio ranges from say 0.82 or 0.4 to around 5.2. In this case in the table here the typical soils are given say for example you have got say uniform sand and which is in the loose state which we are going to discuss in subsequent lectures. So if you have got a uniformly graded sand particles the void ratio is about 0.85. If you have got mixed grain sand that means that and it is in dense condition in that case the void ratio can be as low as 0.43. If you have got a glacial clay the void ratio can be up to 1.2. And soft highly organic clay that means the soil which actually has got organic matter the clay which actually has got organic matter can have void ratio up to 3.0. In the case of soft Mentonite the void ratio can be as high as 5.2. So the void ratio for sandy or sandy soils is relatively low and clay soils is relatively high. So if you make this observation here the void ratio of sandy soils is on the lower side compared to void ratio of the clay soil. So in subsequent lectures we will understand why this particular observation. So in nature even though the individual void spaces are larger in coarse granular soils that means that in nature if you look into this even though the individual void spaces are larger in coarse grain soils the void ratios of fine-grained soils are generally higher than those of coarse grain soils. So in the previous slide we observed that the void ratio of sandy soils are relatively lower compared to clay soils. That means that if you put as a coarse grain soil the void ratios are of the individual grains the pore spaces in nature even though the individual void spaces are pore spaces are larger in coarse grain soils the void ratio of fine-grained soils are generally higher than those of coarse grain soils. Then we define one more parameter called porosity. This is nothing but is a ratio of volume of voids to total volume. This is generally expressed in a percentage. So if you take Vv plus Vs is equal to total volume. If you write V is equal to 1 plus Vv by Vs into Vs so you can write that as 1 plus Vv into Vs that means that you can actually get an interrelation between n porosity and void ratio which is nothing but n is equal to E by 1 plus E. If you look into this the porosity provides a measurement a measure of the permeability of a soil. If the soil is more porous the permeability is high that means that the porosity provides a measure of the permeability of a soil. Porosity is a property of a soil which is defined as it is ease with which the water can flow through the soil. The porosity indicates here a large porosity provides a measure of the permeability of a soil. So this porosity n of soil cannot exceed 100% that means that it ranges between 0 and 100. Porosity n of a natural deposit can be a function of the shape of grains that means that the type shape of the grain, uniformity of grain size whether it is same size or different sizes and the conditions of sedimentation. So for natural sands the porosity is in the range of n is equal to 25 to 50% and for a soft natural clays the porosity can be in the range of 30 to 60%. So we have defined porosity and porosity we said that which is ratio of vv by v that is volume of voids to the total volume and the porosity n of a natural deposit is a function of shape of the grain, uniformity of grains and the conditions of sedimentation. Now if you look into this in soil mechanics or in geotechnical engineering we use void ratio. The reason is out of the void ratio e and porosity void ratio is used frequently in soil engineering because of the particular region if you see the definition here e is equal to vv by vs n is equal to vv by capital V. So any change in a volume is a direct consequence of a similar change in volume of voids. While vs remains constant. So because of this if you see the numerator and denominator vs which is actually remains constant in case of void ratio. But in case of porosity both numerator and denominator undergoes a change. So hence void ratio e is frequently used in soil engineering. Now another definition which is actually called as water content which is very much important. The water content of a soil is defined as weight of water to weight of the solids expressed as a percentage. So ww that is w suffix w is nothing but weight of water, ws is nothing but the weight of the solids in dry state. So natural water content of fine grained soils is greater than coarse grained soils and there is no upper limit for the water content. So this water content is also called as moisture content or in case if it is in a natural state it is called natural moisture content or in situ moisture content. So the water content w which is defined as weight of water to weight of solids and which is actually not having any upper limit and for some soft soil deposits the water content can be up to 500%. And we also said that the natural water content of fine grained soils is high greater than coarse grained soils. Then there is another one important term which indicates the degree of saturation of a soil. So which is defined as volume of water to volume of whites. That means that volume of water within volume of whites. So this how it can be deduced let us see through this slide. For a fully saturated soil water system since all the whites will be completely filled with water we can write v suffix v gamma w as weight of water that means that weight of water is nothing but gamma w that is the unit weight of water which is in SI units can be taken as 9.81 kilo Newton per meter cube. So w w is equal to v v gamma w. So for partial saturation we can write v v minus v a gamma w is equal to weight of water that means that what we did is that we have taken that weight of that volume of whites which is completely saturated soil volume of whites is equal to volume of water. Now for partial saturation we can write v v minus v a into gamma w. Now the relationship between SR the degree of saturation can be obtained as SR is equal to v v minus v a by gamma w divided by v v gamma w. So if you simplify this you will get v suffix w by v suffix v. So SR is the ratio of volume of water to the volume of white spaces this generally expressed as percentage if this it ranges from 0 to 100. If SR degree of saturation is equal to 100 that means that the soil is completely saturated. If you have a soil with 80% saturation then we can say that the soil is partially saturated. So for fully saturated soil SR that s suffix r or degree of saturation is equal to 100%. So soil can be partially saturated and if you look into this diagram if this is the ground surface what you see is the ground surface and this is the water table location the hydrostatic ground water table location. The soil above this remains in partially saturated this is because of the temperature fluctuations as well as the partial the water evaporation which takes place in this zone. The soil below the water table is said to be completely saturated except some minor occurrence of air bubbles otherwise you can say that the soil is completely saturated it means that all the voids which are actually filled with water. So at SR is equal to 100% all voids are completely filled with the water. So below the water table we can say that the soil is completely saturated above the water table unless the type of soil is different you actually have a partially saturated soil occurs. So this particular slide shows a typical variation of degree of saturation for sands. So this is strictly varied for sands based on the condition of sand if it is dry the degree of saturation is equal to 0 humid 1 to 25% and damp 26 to 50% moist 51 to 75% wet 76 to 99 saturated 100 that means that this is degree of saturation if it is 100 then we can say that the sands of sand deposit is completely saturated. So degree of saturation SR fine or silty sands are moist wet or saturated clays are always completely or nearly saturated except in the layer of soil subjected to seasonal variation of temperature and moisture fluctuations. So what we discussed in the previous slide is that clays are always completely or nearly saturated except in layer of soil subjected to seasonal variation of temperature and moisture. Now we define one more parameter which is called air content. Air content is called A suffix C is indicated by A suffix C which is nothing but volume of air in volume of whites that means that which is ratio of volume of air to volume of whites degree of saturation we defined as volume of water to volume of whites. So here the air content is nothing but volume of air in volume of whites. So we can write here VA plus V suffix W minus V suffix W by VV. So if you write VA plus VW as VV minus VW we can write this as VW by V is SR that is degree of saturation so we can write VV by V is equal to 1 so the air content is equal to AC is equal to 1 minus SR. So if SR is equal to 1 that means that for a saturated soil air content AC is equal to 0 that means that for a saturated soil air content A suffix C is equal to 0. For a dry soil air content is 1 because all the voids are filled with air or air within the voids. Then another parameter which is actually called as air void ratio or percentage void percentage air voids. So another parameter which is actually called as air void ratio or percentage air voids N suffix A which is defined as volume of air to total volume. So by writing VA into VV divided by V into VV and using N is equal to VV by V which is nothing but porosity we have defined earlier and A suffix C air content is equal to VA by VV we can write air void ratio or percentage air voids as NAC that means that if you substitute AC is equal to 1 minus SR we can write percentage air voids NA is equal to porosity into 1 minus SR. So again if you see when SR is equal to 1 percentage air voids NA is equal to 0 that means that air void ratio percentage air voids NA is equal to N into 1 minus SR. The another important parameter as far as soil engineering or geotechnical engineering is concerned is unit weight of the soil and this actually has got number of terminologies depending upon the state of saturation or whether it is in the dry state, whether it is in saturated state or whether it is in moist or partially saturated state. So the unit weight general definition is nothing but weight of the soil to the total volume. Weight of the soil mass to the total volume is one of the most important physical properties of the soil. The unit weight must be expressed with due regard to the state of soil. So as I said earlier the unit weight must be expressed with due regard to the state of soil that means that whether it is a dry or moist or saturated. So unit weight is a function of the unit weight of the solid constituents, the type of the grains which are composed of and porosity whether it is closely packed or loosely packed and degree of saturation. So the application of why do we want the unit weight of the soil? If you have got a retaining wall and if the soil is placed behind the wall the soil exerts the pressure. So if you wanted to calculate the lateral pressures you need to calculate what is the earth pressure. So in this case in order to compute earth pressure you need the unit weight of the soil. For example if you wanted to determine a vertical stress or a total stress at a particular depth because of the certain depth of the soil or certain stata of the soil then you need to know the unit weight of the different layers which are actually above that particular point of interest. So if you wanted to determine the stresses in the soil or say earth pressures behind the walls you need to know what is the unit weight of the particular material. For a partially saturated soil bulk unit weight or gamma bulk which is also called as moist unit weight which is nothing but total weight of soil mass to the ratio of the total weight of soil mass to total volume. So gamma bulk is equal to we can write as weight of water plus weight of solids divided by volume of water plus volume of solids plus volume of air. Weight of the air is equal to 0 that means you are here having only weight of water plus weight of solids divided by total volume. For a saturated soil gamma bulk is equal to gamma saturated. Gamma suffix B is equal to gamma saturated so because and volume of air is equal to 0. In that situation for a gamma sat it is defined as weight of water plus weight of solids divided by volume of water plus volume of solids where gamma sat is nothing but the saturated unit weight of the soil. For example the saturated unit weight of the soil can occur if you find a soil stata below the ground water table. Another case for example dry unit weight so dry unit weight which is indicated as gamma suffix D. For a dry soil gamma D is equal to Ws divided by Va plus Vs. Here because the soil is in dry volume of water occupied in the voids is 0 that means that volume of water is equal to 0. In that situation it is written as for a dry soil Ws divided by total volume which is nothing but volume of air plus volume of solids. So gamma D is equal to Ws by V we can write this as W minus W suffix W so W minus W by V we can write as W by V minus W Ws. So if you write Ws by V as gamma D we can write gamma D as gamma bulk minus water content into gamma D. So if you rearrange these terms we can write gamma D is equal to gamma bulk by 1 plus W. So this particular relationship is used widely which is nothing but gamma D is equal to gamma bulk divided by 1 plus water content. So here the unit weights of gamma D are in kilo Newton per meter cube, gamma bulk in kilo Newton per meter cube in water content expressed in percentage but here is indicated in decimal form W in terms of decimal form. For example if the soil actually has got say 20% water content here in this expression you need to use W is equal to 0.2. So typical weights typical values of unit weight for soils are given in this particular slide. So if you see here typical values of unit weight for soils is given for a soil type gravel the saturated unit weight can be 20 to 22, the dry unit weight can be 15 to 17, sand it can be 18 to 20 and the dry unit weight can be 13 to 16, silt it can be 18 to 20 and 14 to 18 in dry state. For clay the saturated unit weight can be 16 to 22 and the dry unit weight can be 14 to 21. For some non-soil like materials like say municipal solid waste they are even lighter than water. Some the unit weight ranges from in the loose state for a fresh municipal solid waste it can be as low as 9 kilo Newton per meter cube. For industrial waste like coal ash the dry unit weight can be in the range of 12 kilo Newton per meter cube that means that these light materials can be used in soil engineering or geotechnical engineering for some construction purposes if you wanted to do on construction on soft ground. The another parameter which is called specific gravity the specific gravity is defined as the ratio of the unit weight of the substance to the unit weight of water gamma w at 4 degree centigrade. So in soil mechanics or geotechnical engineering the specific gravity generally refers to the specific gravity of the solid particles. So we use specific gravity of the solid particles that is G suffix s and is defined as the unit weight of the solid particles to that of water that is Gs is indicated as gamma s by gamma w which is nothing but gamma s if you write it as Ws divided by capital Vs that is Vs nothing but volume of the solids into gamma w. So Gs we can write it as Ws by Vs gamma w or weight of solids is equal to Gs Vs gamma w that means that if you know the specific gravity of the solids in a given soil mass if you know the volume of solids and with the unit weight of water you can determine the weight of the solids. How to determine the specific gravity say in the laboratory if you see this simple illustration which is shown in this slide. Let us take you have got a particular material under consideration is filled in a specific gravity bottle and if you take weight of empty specific gravity bottle and if you indicate that weight as W1 and if you take the desired soil in dry state air cooled and dry state and if you take that weight that is W2 is equal to W1 empty bottle plus the dry soil and W3 what you do is that W2 plus water without any entrapped air. So in order to do that you have to continuously shake and subjected to say some sort of boiling wherein the whites or entrapped air in the soil particles can be removed. So if you determine that that weight is indicated as W3 is W2 plus water. Then once after doing that entire specific gravity bottle is filled with water and that weight is indicated as W4 that is W1 plus water. Now so we have got W1, W2, W3, W4 now we need to find out the specific gravity of the soil under consideration means. So you need to take weight of the soil solids. How do you get that is nothing but W2 minus W1, W2 is the weight of the soil minus W1 is the empty weight of the bottle. Then weight of the water volume equivalent to that of water. So in order to get this what you need to do is that you need to subtract weight of water minus weight of water occupied that is W3 minus W2. So what we are doing is that we are actually taking the water volume equivalent to that of soil solids. So in order to get that what you need to do is that W4 minus W1 minus W3 minus W2 you need to do. If you do that so you will get the expression Gs is equal to W2 minus W1 within parenthesis divided by W4 minus W1 minus W3 minus W2. For most soils the specific gravity of the solids ranges from 2.5 to 2.9. Here I have given for a typical minerals which we are going to discuss in the next lecture kaolinite, illite and martybalite these are the three predominant minerals in fine grain soils. So if you see here for a kaolinite based soil the specific gravity is around the range of 2.62 to 2.66 for a illite based soil 2.6 to 2.86 martybalite based soil the specific gravity can be in the range of 2.75 to 2.78. So normally for sands the specific gravity of the solids Gs will be equivalent to 2.65. And for say non-soil like materials say pondash has specific gravity in the range of 2.2 to 2.4 in some cases for very light coal ash or a pondash it can be as low as 2. And for a iron ore which is based on hematite type mineral where 4.4 to 5.2 it can have you know very high specific gravity. So for a partially saturated soil we have discussed that it is gamma bulk that is the gamma bulk is nothing but the bulk unit weight or most unit weight of the soil. So for a partially saturated soil mass specific gravity is defined G suffix m as gamma B by gamma W. In case of say GM dry that is mass specific gravity in dry state which is nothing but a gamma D by gamma W for dry soil, GM saturated which is nothing but a gamma sat by gamma W for saturated soil. So GM dry is nothing but a mass specific gravity in dry state which is nothing but a ratio of gamma D by gamma W. GM sat which is nothing but a mass specific gravity in a saturated state. Then we knew that the soils when they are actually below the water table they are on the bionint state. So we defined in order to calculate stresses at certain points below the water table and all you need to know the bionint unit weight or submerged unit weight of the soil. So submerged unit weight or bionint unit weight is indicated by gamma sub or gamma dash. So gamma dash is nothing but submerged unit weight of the soil. So how to get relation between saturated unit weight of the soil and unit weight of water. So further if you see the simple illustration here we have taken a ground surface this is the ground surface and this is the ground water table and this is the soil mass which is completely saturated weight of water and weight of solids which is nothing but weight of soil mass which is submerged below the water table and volume of soil and volume of solids and volume of water on the left hand side here. So we can write the weight of soil inside the water divided by total volume. So this can be written as weight of solids plus weight of water minus V gamma w. So that much volume is displaced treating whole soil mass as one unit. We can write as Ws plus W, Ws plus W minus V gamma w by V. So this can be written as Ws by W suffix W by V minus gamma w. Which is nothing but V and V will get cancelled. So Ws plus W by V is nothing but weight of saturated soil divided by volume which is nothing but a saturated unit weight of soil. So we get a relationship here gamma submerged is equal to gamma sat minus gamma w. So if you wanted to say compute the submerged unit weight of soil you need to take the unit weight of water, subtract the unit weight of water from the saturated unit weight of the soil. So gamma dash is equal to gamma sat minus gamma w. So now if you wanted to determine the phase relationships you can actually use different approaches like you can assume a volume which is nothing but volume of voids, volume of solids and volume of water, volume of solids, volume of water and volume of air in case of partially saturated soil and weight of water, weight of solids and weight of air which is equal to zero. So there are two approaches which are there, one is called specific volume approach. In these the volume of the solids is set as one. Volume of solids is set as actually one, unit volume approach which is sometimes is also very convenient is set assumed as V is equal to one. So this is assumed and these particular approaches are used if the volume is not known. So using specific volume approach Vs is put as equivalent to unit volume. So in the specific volume, so here this is another which is actually used in which is nothing but total volume per unit volume of solids. So V is equal to one plus e which is nothing but Vv by Vs that is nothing but V by Vs. So total volume per unit volume of solids is nothing but V is equal to V by Vs. So here the specific volume is defined as one plus e. So for a dry soil in a two phase system what you have here is weight of air is equal to zero. This is called two phase system, weight of solids is nothing but here we have set based on the unit volume approach. Here specific volume approach, here Vs is equal to one we have set here and volume of voids is equal to volume of air because the soil is dry. So Ws is equal to Gs Vs gamma w, Vs being one it is nothing but a Gs gamma w and Wa is equal to zero. So this is total weight of soil mass in case of a dry soil state two phase system is nothing but W is equal to Ws. So for the soil is dry from the definition of void ratio we can write Vv by Vs. So Vv is nothing but e, Vs is nothing but one. So Gs is equal to gamma s by gamma w based on that Ws by Vs gamma w. So we can write Ws is equal to Gs gamma w and gamma d or dry unit weight of the soil mass we can write as Ws by V which is nothing but a Gs gamma w by one plus e. So here for a dry soil we can write a relation like gamma d is equal to Gs gamma w by one plus e. So this is very useful if you wanted to if you know the dry unit weight of soil and the specific gravity of soil what will be the void ratio of the soil mass. That means that e is equal to Gs gamma w by gamma d by minus one. So if you rearrange the terms gamma d is equal to Gs gamma w divided by one plus e we can use this relationship. In case of basic relationship here for a saturated soil in this case weight of air is equal to nothing but e gamma w because weight of air is zero here, weight of air is zero and weight of water is equal to e gamma w and weight of solids is equal to Gs gamma w. So total weight of the soil mass is equal to Gs gamma w plus e gamma w. So from this if you look into the volume scale total volume is nothing but one plus e. So the saturated unit weight of soil mass gamma sat is nothing but W that is saturated unit weight of the entire soil mass divided by total volume. So it is nothing but Gs gamma w plus e gamma w divided by one plus e. So like this by using this phase diagrams one can deduce the interrelation between the soil properties. So for the soil when the soil is fully saturated from the definition of water content we can write as W is equal to weight of water to weight of solids. So weight of water in the previous slide we have seen as e gamma w divided by weight of solids is nothing but Gs gamma w because Vs is equal to one we have set. So with that if you look into this the relationship is that e is equal to W Gs. So e the void ratio is nothing but water content times the specificity of the soil solids for a fully saturated case. So for a fully saturated case gamma sat the saturated unit weight of the soil mass is nothing but W by V which is nothing but Gs gamma w plus e gamma w by one plus e. So the relationship with what we can write is that gamma sat is equal to Gs plus e into gamma w divided by one plus e. Similarly now we have seen the dry state of the soil and saturated state of the soil but if you have got a partially saturated or we define this as a three phase system. So in this case you have weight of water, weight of solids and weight of air that is nothing but zero. So but here as a volume we have got volume of solids which we have set as one because of the specific volume approach and volume of voids which is nothing but e but which is the summation of volume of air plus volume of water. So from the definition of degree of saturation of soil we can write Sr is equal to volume of water to volume of voids. So volume of water which is nothing but here if you see weight of water is nothing but W Gs by gamma w, Ws is equal to Gs gamma w from the water content definition we have written here weight of water is equal to W Gs by gamma w. So here by writing this what we get is that a relationship between for a partially saturated soil between void ratio, water content, specific gravity of the solids and degree of saturation where e is equal to W Gs by Sr. For Sr is equal to one e is equal to W Gs that we have deduced previously for a saturated soil. So here gamma d that is nothing but Ws that is dry state that is Gs gamma w divided by v that is nothing but a volume which is nothing but 1 plus e but e is nothing but W Gs by Sr. So for a partially saturated soil if you see the relationship is that gamma d is equal to Gs gamma w plus 1 plus W Gs by Sr. So this is valid for a partially saturated soil. Now let us try to derive some interrelationship between gamma d the dry unit weight of the soil specifically at the solid Gs, water content and Na. So we knew that how to derive that let us say that we have total volume v is equal to Vs plus Vw plus Va. So what we are doing is that by dividing both left hand side and right hand side by volume we can get 1 is equal to Vs by v plus Vw by v plus n suffix a which is nothing but Va by v we have written as Na. Using Vs is equal to Ws by Gs gamma w and Vw is equal to W suffix w by gamma w and by writing Ww is equal to W times Ws we can get that Na if you take it to the left hand side we get 1 minus Na Vs by v plus Vw by V and if you substitute these expressions here and simplify you will get gamma d by gamma w into water content plus 1 by Gs. So if you write this expression in terms of gamma d we can write gamma d is equal to 1 minus Na Gs gamma w plus Wgs. So when the soil becomes completely saturated we knew that Na is equal to 0 so in this case what will happen is that gamma d is equal to Gs gamma w by 1 plus Wgs this is actually the expression what we derived for saturated soil. So this is for a partially saturated soil if you have the relationship between gamma d Gs w and Na can be given by gamma d into 1 minus Na Gs gamma w divided by 1 plus Wgs. Now having seen the specific volume approach let us see the unit volume approach also is sometimes very easy to deduce interrelationships or phase relationships in the different phases of the soil. For a partially saturated soil let us consider and we say that it is a three phase system and here what we did is that total volume is set as 1. So if you set the total volume as 1 then the portion which is the volume of voids is nothing but n because porosity is nothing but volume of voids divided by total volume. So Vv by 1 is equal to n nothing but this becomes small n that is porosity and this portion becomes 1 minus n so the weight of the solids is given as Gs into 1 minus n into gamma w Gs into 1 minus n into gamma w. Based on the water content definition we can write W into Gs 1 minus n gamma w by taking gamma w out we can write here the volume of water as Wgs into 1 minus n the rest is volume of the air and WA is equal to 0 here. So for a partially saturated soil and if you want total weight of the soil mass which is nothing but Gs into 1 minus n gamma w plus Wgs into 1 minus n gamma w by total volume 1. So for a partially saturated soil three phase system we have n is equal to Vv by v with V is equal to 1 n is equal to volume of voids so we can write E is equal to Vv by vs that is nothing but n by 1 minus n we have deduced earlier. So we can write gamma d is equal to Ws by v that is nothing but 1 minus n into Gs gamma w that is we are expressing in terms of say porosity that is 1 minus n into Gs gamma w. Gamma bulk from the previous slide we can say that W by v which is nothing but 1 minus n into Gs gamma w plus Wgs into 1 minus n gamma w. Now percentage air voids NA can be defined as VA by v so which is nothing but Vv minus Vw by v which is which can be very simply it can be written as n minus Wgs into 1 minus n because total volume is equal to 1. For a completely saturated soil similarly we can actually set and use this approach and we can write gamma sat is equal to Gs into 1 minus n gamma w plus n gamma w because for in case of a completely saturated soil you have got n gamma w that is the weight of water and weight of solid is nothing but Gs into volume is nothing but 1 minus n gamma w. So here the total volume set is equal to set as 1 and n is the volume of water. So W is equal to we can write it as n gamma w divided by Gs into 1 minus n gamma w which is nothing but E by Gs. So E is equal to Wgs is what actually we have got for a completely saturated soil. For a dry soil similarly so we have seen for partially saturated soil and seen for completely saturated soil and for a dry soil with the same approach we can write gamma d is equal to Ws by v which is nothing but 1 minus n into Gs gamma w divided by total volume is equal to 1 because of this we can write this as 1 minus n into Gs gamma w and let us also discuss some additional phase properties and relations particularly which are used in unsaturated soil mechanics and porosity can be defined with respect to each of the phase of the soil. So we have defined normally the porosity as volume of voids in a total volume of soil but we have got different phases of soil particularly you have got a solid phase that is solids are there and water is there air or gas is there. So soil particle porosity n suffix S is defined as volume of solids in total volume that is vs by v this is expressed in percentage and water porosity which is nothing but n suffix w which is volume of water in total volume. So this is also referred as volumetric water content theta w so this is particularly used in soil science or in nowadays it is widely used in unsaturated soil mechanics. Air porosity n suffix a is equal to va by V so air porosity na is equal to va by V and water porosity nw is equal to vw divided by V and this is also referred as volumetric water content theta w. Now the water and air porosities represent their volumetric percentages in the soil. So water and air porosities they represent their volumetric percentages in the soil that is volume of water in the total volume, volume of air in total volume. The soil particle velocity can be visualized as the percentage of the total volume comprised of soil particles. So the soil particle porosity is can be visualized as the percentage of total volume comprised of soil particles. So the soil particle porosity can be visualized as the percentage of total volume comprised of soil particles. But if you look into this ns is nothing but volume of soil solids to volume of total volume plus n is nothing but volume of whites in a total volume. So the summation of ns plus n is equal to ns plus na plus nw should be equal to 100%. That means that here na is water, the air porosity, nw that is nothing but water porosity, ns is nothing but the soil particle porosity. The summation of ns plus n is equal to ns plus na plus nw has to be equal to 100%. Some additional phase properties and relationships if you derive volumetric water content or you know we have defined in the previous slide as theta w as volume of water to capital V. So using degree of saturation as the vw by vv and n is equal to vv by v we can write and using n is equal to n is equal to e by 1 plus e. If you substitute here for degree of saturation we can write it in terms of theta w in terms of Sr vv by v, how we have written is nothing but vw is equal to Sr into vv we have written divided by capital V, vv by v is nothing but porosity. So we can write the volumetric water content is n times porosity times degree of saturation. So here by using n is equal to e plus 1e we can write e Sr divided by 1 plus e. So if you wanted to get say the relationship between the gravimetric water content which is what we have defined conventionally the water content and theta w the volumetric water content we can actually obtain for a partially saturated soil as by using e is equal to substituting e is equal to wgs by Sr if you substitute here then what we get is that theta w is equal to wgs Sr by Sr plus wgs. So if you have this one you will get a relationship between volumetric water content theta w and the gravimetric water content w. So theta w is equal to wgs Sr plus divided by Sr plus wgs. So having seen the phase relationships among the soil properties let us take some example problems and try to solve them. So in this example problem in this slide it is given as a 0.8 meter cube soil specimen weighs 17 kilo Newton and has a moisture content of 9%. The specificity of the soil solids is 2.7 using the phase relationships. So we need to calculate gamma, gamma d, e, porosity, volume of water and degree of saturation. So the solution runs like this. What is given is that volume is given 0.9 meter cube and weight is given as 17 kilo Newton and water content is given as 9% and specific gravity of the solids is given as 2.7. So from the definition of unit weight we can write gamma is equal to w by v. So the gamma, the bulk weight it can be defined as 17 divided by 0.9. We can write it as 18.89 kilo Newton per meter cube. Using gamma d is equal to gamma by 1 plus w which is nothing but gamma is nothing but gamma bulk divided by 1 plus w we can calculate what is the dry unit weight of the soil. So from this if you look into this w s is equal to gamma d v which is nothing but 17.33 into 0.9 which is nothing but 15.59 kilo Newton and w w is nothing but w minus w s nothing but weight of water in the soil mass is nothing but 1.41 kilo Newton. Then using specific gravity is equal to gamma s by gamma w, w s by v s gamma w we can determine volume of solids as w s by g s gamma w which is nothing but 0.5889 meter cube. Total volume is equal to v s plus v v and which is nothing but v v is obtained as 0.311 meter cube. So void ratio is nothing but the volume of voids to volume of solids. So in this case the soil mass has void ratio of 0.528, porosity is nothing but volume of voids to the total volume and total volume of the soil mass is given as 0.9. So the porosity which is obtained as 0.346 is expressed as 34.6 percent. And the volume of water is nothing but weight of water divided by gamma w and gamma w is equal to taken as 9.81 kilo Newton per meter cube. So with that volume of water as 0.143 meter cube. So the degree of saturation is given as volume of water to total volume of voids. So based on this the problem for a given problem the degree of saturation is obtained as about 45.9 percent. And if you express this in what we calculated in the soil composition for a partially saturated soil total volume is 0.9 if you apportion this volume of solids is 0.589 meter cube and volume of water is 0.143, volume of air is 0.168. So total put to there you will get 0.9 meter cube, volume of voids is nothing but volume of water plus volume of air which is nothing but 0.311 meter cube. And weight of water which is 1.41 and weight of solids is 15.59 total weight of soil mass is Ws plus W 17 kilo Newton per meter cube and V is equal to Vs plus Vw plus Va which is equal to 0.9 meter cube. So by using these phase diagrams one can actually deduce the properties of soil and interrelationship between weight and volume ratios. So in this lecture what we understood is that some phase relationships among soil properties and we have defined number of soil properties particularly like volumetric ratios and water content and we also discussed about the relationship between gravimetric water content and theta W volumetric water content theta W. So in the next lecture we are going to discuss about soil particle sizes and their arrangement and particularly soil particle sizes and arrangement, their particular arrangement. And then we will discuss about mineralogy, clay mineralogy and then that leads to the discussion about the type of clay minerals present in the soil.