 So, let's go through a few examples of using this one-dimensional consolidation theory to calculate settlement time, or consolidation time, within soils. Let's say we had an example like this where we had a layer of clay, and we knew, we found out the CV value from tests, and let's say the clay was 10 meters thick. And over that clay layer, we had a sand which is incompressible, but was highly permeable, so we knew that water could flow through the sand. But underneath the clay, we have an impermeable bedrock, so an incompressible material that's also impermeable. And if we subjected that sand or that soil strata to an increase in stress, the question we might want to answer is, how long would it take to reach 90% consolidation? So the first step we need to do is relate this degree of consolidation, U, to a TV value. Now you can use an equation or a table to derive that, but if you look at the table and I've put a link to the table on my website, you can see that the relationship between U and TV for U equals 90% gives us a TV value of 0.848. So for U 90%, TV value equals 0.848. So if we go back to the consolidation equation, a TV value is equal to the CV value multiplied by time all over D squared. So we're giving the CV value in the equation, and the TV value is derived from the amount of consolidation we're interested in. What we're interested to find out, or the question is asking us, is what is T within this? So what we need to do is rearrange this equation in terms of T. So T is equal to your TV value multiplied by D squared all over CV. So this one is, so your TV value is derived from the degree of consolidation, your CV value is an inherent property of the material. What is the D value here? What's the drainage pathway? Well, you can see that the impermeable bedrock, or the bedrock underneath the material is impermeable. So there's only one way for the water to fly out of that material, and that's through the sand at the top. So the drainage pathway, D, is equivalent to the thickness of the layer. So we also know D now. So we can work out the time, and that's equal to the TV value, which is 0.848, multiplied by the drainage pathway, which is squared, which is the thickness of the layer squared, in this case 100, divided by the CV value, which is 2. Now it's important, when you're doing this problem, to make sure that the length unit within your CV value, in this case it's meters, is the same unit of length in the thickness of your layer, or your drainage pathway. If this was in, let's say, centimeters squared per year, or per month, then you'd need to convert the units to be the same as your drainage pathway. So when you square them here, you can see that they cancel out within this equation. So what you're left with is a value, in terms of units, your TV value is dimensionless or unitless, your D value is in meters, and you're squaring it, so you've got meters squared, and you're dividing it by meters squared per year, so you have meters squared on the bottom, and you multiply that by a year, and you can see that the two meters squared, meters squared cancel out, or square meters cancel out, and you're left with a value of years, so your time value here is in the years. So if you do that equation, what you end up with is an answer in 42 years. So this is quite interesting because we found out a time it would take for your soil to reach 90% consolidation, but we didn't say anything about the stresses, we just said that there was a stress applied, we didn't give it a value. So in the next example, I'll go through when you might include a stress within this calculation. Okay, let's suppose we had an example that looks a bit like this. So we've replaced the bedrock now with incompressible sand, so we know that this material is now permeable. And we've expressed that on the top of the soil layers, we were sticking 100 kilonewton per meter squared stress. Now, we haven't got the CV value for all of clay, but we've found out our MV value, the coefficient of volume compressibility. And we've also found out the permeability of the clay. We might want to ask the question now, well, after one year, what is the total settlement? So we need to, in this case, employ both sets of consolidation calculations. We need to work out what the total settlement is, or total maximum settlement possible. And then what the degree of consolidation is after one year. So the first step is to say, well, what is the maximum settlement, given that new load? So we ask, well, what is delta H? The maximum settlement for 100 kilonewton per meter squared in this clay layer. So if we multiply by MV, middle sample thickness, and the change in stress, we take our change in stress to be 100, or initial soil thickness to be 10 meters. And the MV value here to be 0.3, and remember we have to convert from meter squared per meganewton to meter squared per kilonewton to keep it in the same relative units of stress. So we have 0.3 divided by 1,000 to convert the unit, multiplied by 10, multiplied by 100. That equals 0.3 meters, or 30 centimeters. And that seems like a reasonable value for 10 meters of clay. I wouldn't expect that to be 30 meters or 300 meters. So it seems like a reasonable value. Okay, so the next step is then to say, well, what is our degree of consolidation? What is U after one year? So what is U after one year? So we need to work backwards this time from the TV value. So we need to find out what our TV value is. And we do that using this equation, CV multiplied by T over D squared. Now, we know that T now is, or in our question, it's one year. We need now a D squared value. So the drainage pathway is now the thickness of the layer over 2 because water can flow both through the sand at the top and the permeable sand at the bottom. So our D is equal to 10 over 2 now, the thickness of the layer over 2. But what is our CV value? Well, the CV value, we need to return to the equation that we're allowed CV to permeability is K over the unit weight of water times by the MV value. Now, the unit weight of water, or there's a different color pen here. The unit weight of water is equal to the density of water, which is one megagram per meter cubed, multiplied by the acceleration due to gravity, which is 9.8, or we'll just say 10. So the unit weight of water is 10 kilonewtons per meter cubed. The permeability is given in the question, and the MV value is given in the question. Again, we need to be careful that the units of MV and meter squared per meganewtons are the same as the units of unit weight. So we don't have meganewtons, but we convert that to kilonewtons. So we do the same as what we've done up here. So your CV value equals the permeability, which is 10 to the minus 9 meters per second, divided by your unit weight of water, which is 10, multiplied by the MV value, which is 0.3. And we divide that by 1,000 to convert it into a meter squared per kilonewton. So we can sort of putting the 1,000 down here. We can just multiply the top by 1,000. So our CV value comes out at 10.7 meters squared per. Okay, so we now have our T value. That was what the question is asking for. We have a d value and we have a cv value. So we should be able to work out what our tv value is, the time factor. So if we put everything into this equation, our cv value is 10.7 meter squared per year. Our t value is one year. So we have to be careful that all time the t value is the same as the units of the cv value, which is pa. If the cv value was meters squared per second or centimeters squared per second or per month or something, we'd need to convert. But we can just put it straight in here. So 10.7 times 1. And the d value is 10 over 2, which is 5. 5 squared is 25. So the time factor value comes out at 0.428. Now if we take the tv value, the time factor value, and we use either equations or the table to derive the degree of consolidation, we'll see that our degree of consolidation is equal to 72%. So what that's saying really is that after one year, this soil, 72% of the total settlement has been reached. So the total settlement is this value here, 0.3 meters. So we know that 72% of that value has been reached after one year. So the total settlement after one year in terms of meters is equal to 70% of 0.3 meters. So delta H at one year is equal to 0.3 multiplied by 0.72. 72% of that, which comes out at 22 centimeters or 0.22 meters.