Hello, this is Professor Kitch.
Welcome to this webcast from my introduction to geotechnical engineering course.
This webcast is from chapter 4 of the textbook and covers sections 4.1 – 4.3 on the particulate
nature of soils and 3-phase diagrams When you’ve completed this webcast, you
should be able to Describe the three-phased nature of soil
Draw 3-phase diagrams representing soil Complete a 3-phase diagram by computing weights
and volumes of soil phases and using a completed 3-phase diagram, you should be able to compute:
Unit weight, gamma Dry unit weight, gamma-sub-d
Porosity, n Saturation, S
Void ratio, e Moisture content or sometimes call water content,
w.
So let’s get into solving weight volume problems using 3-Phase diagrams
When we examine soil, we see that it not a continuous mass like steel or other engineering
materials.
It is a particulate material consisting of three distinct parts of phases: Solid mineral
particles, Air of other gases, and water or other liquids
We use a three phase diagram to represent the particulate nature of soil.
The lower box in the diagram represents the soil particles.
The middle box represents the water or other fluids
And the top box represents the air or gas portion of the soil
The left side of the diagram will represent the weights of each phase using units of either
pounds or kilonewtons The right side of the diagram will represent
the volumes of each phase using the units of either cubic feet or cubic meters
On weight side, we use W-sub-s to represent the weight of the solids, W-sub-w to represent
the weight of the water, and W-sub-t to represent the total weight.
The weight of the air, or gas, is zero.
On the volume side, we use V-sub-s to represent the volume of the solids, V-sub-w to represent
the volume of the water, V-sub-a to represent the volume of the air, and V-sub-t to represent
the total volume We also define the variable V-sub-v, to represent
the volume of all the voids.
The volume of the voids is equal to the volume of the water plus the volume of the air.
It can be convenient to draw the 3-phase diagram using mass instead of weight.
When we using this method, we almost always use SI units.
The mass units are expressed in grams and the volume in cubic centimeters.
In this form the left side of the diagram uses M-sub-s to represent the mass of the
solids, M-sub-w to represent the mass of the water, and M-sub-t to represent the total
mass.
The variables for the volumes on the right side of the figure are the same as before.
We’ll explain later when it’s useful to use this formulation.
There are two different types of problems that appear when working with phase diagrams.
The first type of problem appears when we are given exact quantities for the soil component
phases.
Here we are given weights in pounds or kilonewtons and volumes in cubic feet or cubic meters.
The second type of problem appears when we are given relative quantities of the soil
component phases.
Here we are given quantities such as water content, degree of saturation, or porosity.
This webcast will cover the first type of problems.
The second type will be covered in a separate webcast
So here’s a type one example problem We’re given a cylindrical specimen of soil,
2.4 inches in diameter and 6 inches tall.
We weigh the specimen and determine it has total weight of 1.95 pounds.
We then put the specimen in an oven over night to dry it out and weigh it again the next
day.
It now weighs 1.65 pounds.
Since all the water has been removed by oven drying, this represents W-sub-s, the weight
of the solids.
We will assume the specific gravity of the solids is 2.66.
We’re asked to find the total unit weight, dry unit weight, void ratio, porosity, and
water content.
We’ll start our solution process by creating a 3-phase diagram.
On the left side of the diagram will record the weight in pounds, and on the right we’ll
record the volume in cubic feet.
Starting on the left side, we’ll enter the information we already know.
The total weight is 1.95 pounds and the weight of the solids is 1.65 pounds.
It’s easy then to determine the weight of the water as the total weight minus the weight
of the solids, or 1.95 minus 1.65, which gives us the weight of the water as 0.30 pounds.
We record this on the diagram.
We now move to the volume side of the diagram.
Since we know the original dimension of the specimen, we first compute the total volume
as pi r-squared times the height of the specimen, or pi times 2.4 inches divide by 2, quantity
squared time 6 inches.
However, this will give use the volume in cubic inches and we want it in cubic feet.
To make the conversion from cubic inches to cubic feet, we multiply by 1 foot over 12
inches, quantity cubed.
This will give us the proper units of cubic feet.
Completing this computation we determine a total volume of 0.0157 cubic feet.
We record that on the diagram.
Now we must use the unit weight of the water and solid phases to convert the weights on
the left hand side to volumes on the right.
I usually start with water, because, like every good civil engineer, I long ago memorized
that the unit weight of water is 62.4 pounds per cubic foot or 9.81 kilonewtons per cubic
meter.
So the volume of water is 0.30 pounds divided by 62.4 pounds per cubic foot, or 0.0048 cubic
feet, which we record on the diagram.
The volume of solids will be equal to the weight of the solids divided by the unit weight
of the solids.
However, we will have to use the specific gravity of the solids and the unit weight
of water to get this quantity.
So, the volume of the solids will be equal to 1.65 pounds divided by 2.66 (the specific
gravity of the solids) divided by 62.4 pounds per cubic foot.
This gives us a value of 0.0099 cubic feet, which we record on the diagram.
Next, we compute the volume of air by subtracting the volume of solids and volume of water from
the total volume.
This computation is 0.0157 minus 0.0048 minus 0.0099 which equals 0.001 cubic feet of air.
Finally, we compute the volume of the voids by summing the volume of the water and air.
So the void volume is 0.0010 plus 0.0048 or 0.0058.
We now have a complete three phase diagram with the weights and volumes of all phases.
The last step of the problem is to solve for the quantities we’re interested in using
the information on the phase diagram.
First we compute the total unit weight as the total weight divided by the total volume
or 1.95 pounds divided by 0.0157 cubic feet.
This gives us 124.2 pounds per cubic feet.
The dry unit weight is computed as the weight of the solids divide by the total volume,
or 1.65 pounds divided by 01.0157 cubic feet.
This gives us 105.1 pounds per cubic feet.
Next, we compute the void ratio.
Remember the void ratio is one of those weird soils properties where it’s based on the
amount of solid material rather than the total amount of material.
The void ratio is the volume of voids divided by the volume of the solids (not the total
volume).
This is 0.0058 cubic feet divided by 0.0099 cubic feet or 0.586.
By convention the void ratio is presented in decimal form rather than as a percentage.
The porosity is similar to the void ratio, except it is based on the total volume of
material.
The porosity is computed as the volume of the voids divided by the total volume, or
0.0058 divided by 0.0157 time 100 percent.
The porosity is most commonly presented as a percentage.
This helps to keep it from being confused with the void ratio.
Four our specimen, the porosity is 38.9 percent.
The degree of saturation is computed as the volume of water divided by the volume of the
voids, or 0.0048 divide by 0.0058 times 100 percent.
This gives us a saturation of 82.8 percent.
Finally, we compute the moisture content (or water content) as the mass of water divide
by the mass of solids.
The moisture content, like the void ratio, is based on the quantity of solid material
rather than the total quantity of material.
For our specimen this will be 0.30 pounds divide by 1.65 pounds times 100 percent
of 18.3 percent.