Skip to main content
Loading Explore guides…
Soil Phase Relationships · Acedra
Home
Explore
Engineering
Soil Phase Relationships
Sign in to save
Chat
If YouTube asks you to sign in, open the video on YouTube.
Watch on YouTube
Reload player
Transcript
28 segments
Copy
Hide video
0:00
this video is the first of three videos that I have created to talk about the phase relationships for soils. In this video, I first introduce the phase diagram, then I will explain the basic parameters in the soil mechanics that help us understand the soil and their behavior. Let's first see what is a phase diagram. Let's imagine this is a bucket full of soil, or it is a soil sample from a certain depth under the ground. This soil sample has the total volume of Vt and the total mass of
0:41
Mt. This soil sample includes solid particles which are these orange particles in this sample and also some voids that are filling the gaps between these particles. These voids typically consist of water and air and typically the more words a soil has the more problematic that soul could be... So in order to understand voids and their effect in soils we use phase diagrams. A phase diagram is basically a way to divide and present the three components of soils. The phase diagram is an
1:38
imaginary way of showing the soil sample that is in this bucket with this difference that it is like the solid particles here are melted down in the sample, then the water sat on it, then the air sat on top of the water. The importance of using a phase diagram is that we can show how the total volume and the total mass are divided between the three components of the soil, that are air water and solids... so if you look at this left side, the mass side, we can see the total mass is equal to the
2:28
mass of solids plus the mass of water, and obviously the mass of air is equal to zero... and then... if you look at the right side the volume side, you can see the total volume of a soil sample is equal to the volume of solids plus the volume of voids...and then... we can also see here that the volume of voids in a soil sample is equal to the volume of water plus the volume of it air. Now... you may ask which one of these volumes and masses we can measure in the lab?
3:08
The answer is that in a soil lab we can measure the total volume, the mass of solids, and also the mass of water, and then... once we have these three values we can calculate all the other masses and volumes within a soil sample. Now, we hopefully better understand the phase diagram and are ready to learn about the basic parameters in soil mechanics that are based on phase relationships. These parameters can be divided in three categories the first one is the one that
3:47
is only based on the volumetric ratios. The second one is the one that is only based on mass ratios, and the third one is the one that is based on the ratio between masses and volumes. The first parameter that we want to talk about is based on the volumetric ratiosm and it is the void ratio. Void ratio is shown by "e", and is equal to the ratio between the total volume and the volume of solids that are shown here. The void ratio is typically expressed as a decimal,
4:28
and mathematically, the values of void ratio can be anywhere between zero to infinity. Let's imagine you have a solid rock sample in which the volume of voids are equal to zero. In this case, the porosity of that rock sample is also equal to zero. On the other hand, let's imagine you have a soil sample that has a very large volume of voids. In that case, the word ratio of that soil sample will be infinity...but practically the typical values of void ratio in SANDs is
5:14
somewhere between 0.0 to 1.0, and in CLAYs it is somewhere between 0.3 to 1.5. The next parameter here is the porosity. The porosity is shown by "n" and is equal to the ratio between the volume of voids and the total volume, which you can see here. The porosity numerically shows how much voids we have in our soil sample within the total volume. Porosity is typically shown as a percentage, and mathematically, the values of porosity can be anywhere between 0.0 to
6:11
100%. the 0.0 porosity will be when we have our rock example in which the volume of voids is equal to zero, and anr hundred percent porosity is something that doesn't happen in practice but is where the volume of voids equals the total volume, basically when the volume of soil particles is equal to zero. In this case the porosity will be 100%. Now, that we are familiar with void ratio and the porosity, let's see how we can derive an equation that relates these two... so let's start with
6:56
the equation for the porosity, which is "n" equal is equal to Vv divided by the total volume. Now... in this equationm we can replace Vt by the Vs plus the VV, then if we rewrite this equation, we will have "n" is equal to VV divided by Vs Plus Vv. Now... the next thing that we can do here is that we can divide both numerator and the denominator by Vs, to be able to find the void ratio equation here... So as you can see in the numerator, we derived the equation for "e" as well as in the
7:50
denominator, and then Vs divided by Vs is equal to 1... So we can finally write the equation that relates these two, which is "n" is equal to "e" divided by 1 plus "e"... So we could also start from the equation of the void ratio which was "e" is equal to VV divided by the Vs. If we had taken similar steps but with just a little bit of difference, we could find the other equation that relates "n" and "e" which is... "e" is equal to and divided by 1 minus "n". These two are
8:39
very important equations that we can use a lot in of phase relationship problems. There is another equation that is really helpful to use in phase relationships problems that is the equation that relates "e" Vt and Vs... so to derive this equation, let's start with the equation of the void ratio... then what we can do is that we can add 1 to the both sides of the equation... so we can write "e" plus 1 is equal to VV divided by Vs plus 1... Now, the 1 on the right side of
9:26
the equation can be replaced with Vs divided by Vs which is equal to 1. So we can rewrite this equation as "e" plus 1 is equal to Vv devided by Vs plus Vs divided by Vs so we can get a common denominator between these two fractions and write "e" plus 1 is equal to Vs plus Vv or VV Plus Vs divided by Vs... so we can see that in the numerator we have the total volume... so finally we were able to derive the equation that we wanted that relates "e", the total volume, and the volume of solids,
10:26
which is "e" plus 1 is equal to the total volume divided by the volume of solids The last parameter to talk about, in this category, is the degree of saturation which is shown by Sr or sometimes S, and is equal to the ratio between the volume of water to the volume of voids. This parameter simply quantifies how much water exists between the volume of voids in our sample. This parameter is usually expressed as a percentage, and both mathematical and practical values
11:13
of the degree of saturation are anywhere from 0 to 100%. When the degree of saturation is zero the sample is called a dry sample. When it is 100%, the sample is called a saturated sample. In a dry sample, as you can see from the equation, the volume of water is equal to 0. In a saturated sample, the volume of water is equal to the volume of voids, so all the voids in this sample are filled with water; therefore. the volume of air is equal to zero, and all the other samples that have
12:02
saturated degrees or degree of saturations between hundred percent to zero are called partially saturated samples.
12:18
All right... now, we are done with the first category and we can move on to the next one that is only based on mass rartios. In this category, we only have one parameter that is the moisture content. Moisture content is defined as the ratio between the mass of water to the mass of solids in a soil sample. This parameter is normally expressed as a percentage, and the typical values of a moisture content can be from zero which is in a dry sample, which has a mass of water of zero,
13:02
to a very very large number, which will happen in a sample that has, as far as mass, has a mass of water that is much larger than mass of solids in it. The last category that we want to talk about here today is the category that is only based on volumes and masses of a soil sample, and the only parameter here is the density. We are all familiar with densities from physics and chemistry, and we all know that "density is mass per unit volume", and it is shown by Rho and
13:58
the equation based on the definition is mass divided by volume... So in the geotech engineering, when we talk about the densities, we can talk about three different densities for a soil sample. The first one is the density of the soil as a whole. The second one is the density of the solid particles,
14:33
and the third one is the density of water. We all know that the density of water is known, and it is equal to a thousand kilograms per cubic meter or one gram per cubic centimeter. The density of this soul is shown by Rho usually, or it's sometimes shown by Rho sub "t". and "t" stands for the total, and is equal to the total mass divided by the total volume. On the other hand the density of the solid particles is usually shown my Rho sub "s", and "s" stands
15:24
for solid particles, and it is equal to the mass of solids divided by the volume of solids. As far as the values of Rho sub T and Rho sub s we can say that Rho sub T or the total density of soil can be anywhere between a thousand to twenty four hundred kilogram per cubic meter. The density of the solid particles, on the other hand, can be anywhere between 2,500 to 2,800 kilogram per cubic meter. The last thing to mention here is that the unit of the density is kilogram per cubic meter in the
16:26
SI units, as we were saying here, and then in the English unit, the usit is pounds per cubic feet which is also shown as PCF. In talking about densities of soils, there are a few densities that are very important to know and use in phase relationship problems. The first one is the saturated density. The saturated density is the density of the soil sample when all the voids in the sample are filled with water, or in another word, the volume of air is equal to zero.
17:20
In this case, the equation is shown by Rho sub SAT, and it will be mass of solids plus mass of water divided by the total volume. The other important density that a soil has is the dry density. The dry density is pretty much used a lot in compaction problems.
17:52
The dry density of a soil is when there is no water in the soil, so mass of water is equal to zero. Looking at the equation above you can say that the dry density is equal to only mass of solids, because it will be the only mass in the soil, divided by the total volume. Then, the last one is the submerged density. I'm going to make a separate video to explain the buoyancy effect and the submerged density, but for now, let's just say the submerged density is shown by Rho Prime and is equal to
18:49
the saturated density minus the density of water. There are two more parameters that I did not include in the list of parameters on top, that are kind of related to the density. We are all again familiar with them from physics and chemistry. The first one is the Specific Gravity. The specific gravity by definition is equal to the ratio of the density of an object to the density of water. In geotechnical engineering, when we talk about the specific gravity, we are
19:42
interested to know what is the specific gravity of the solid particles, so we use Rho sub s here, and then we call it G sub s, and s stands for the solid particles. The second parameter, and the last parameter in this video is the unit weight. We all know what the unit weights are. Unit weights are density times gravitational acceleration, and as density was mass divided by volume, when we multiplied mass by gravitational acceleration, we know we have weight divided by volume.
20:28
So whatever densities that we had in the previous parts of this video, if we multiply it by g we will have a unit weight. For example, we had the density of solid particles. If we multiply it by g, we will have the unit weight of solid particles. We had the dry density a sample. If we multiply it by g, we will have the unit weight of dry sample. We had the density of water. If you multiply it by g, we will have a unit weight of water. This concludes this video that was about the
21:14
phase diagram and some of the most fundamental parameters in soil mechanics. I hope you liked. it please let me know in the questions if you have any question, and I hope you will see there are other videos that complement this video
Here to help you learn
Sign in to keep a private conversation with your course tutor.
Ask me anything about the material…
Sign in to chat