Hello this is Professor Kitch.
Welcome to this webcast on section 11.1.
This is the second of two webcasts on Terzaghi's consolidation theory.
In this webcast we'll solve the differential equation we derived in part 1 and do some
computations To solve the differential equation we derived,
we'll need to apply certain boundary conditions and it's critical that you know these boundary
conditions You're also expected to distinguish between
a doubly and singly drained system and find the maximum drainage path for a given problem
And finally, you should be able to apply the solution and compute excess pore pressure
as a function of space and time during consolidation.
Here's the differential equation we derived in the previous webcast
First we'll notice that the solution to this differential equation must be a function of
both z and time, since it contains partials of the excess pore pressure with both z and
t.
As you'll recall from you differential equations class, a specific solution to any differential
equation requires applications of boundary conditions so let's define our boundaries.
To do that let's consider a compressible layer that has a free draining material on the top
and the bottom.
In the field, this would be a saturated clay with sand layers on the top and bottom.
The hydraulic conductivity of the sand is so much higher than that of the clay that
it's essentially freely draining.
We need to define a few new variables.
First we'll define the depth into the compressible layer from the top drainage boundary as z-dr.
Next we need to introduce the concept of the maximum drainage distance.
That is the longest path that any molecule of water has to travel to get out of the compressible
soil to a free boundary.
In this case, it should be obvious that the water above the mid-height of the layer will
travel to the top boundary and the water below mid-height will travel to the lower boundary.
So the greatest distance any water molecule has to travel is from the middle out, and
the maximum drainage distance, H-dr, is half the thickness of the layer
Or the layer had a thickness of 2H-dr.
So with z-dr and H-dr defined, let determine our boundary conditions.
We have three degrees of freedom in our differential equation so we'll need three boundary conditions.
We'll use the top drainage boundary first.
If this is a free draining boundary, then the excess pore pressure must be zero at this
location for all times.
That is u-e at z-dr equal zero for all time t, is equal to zero.
Our second boundary condition is similar and comes from the lower drainage boundary.
In this case u-e for z equal 2 H-dr is zero for all t.
For our third boundary condition, we'll consider our initial loading conditions.
Recall that we assumed that delta-sigma-z was applied instantaneously and this caused
an immediate increase in the excess pore pressure, equal to delta-sigma-z, throughout the compressible
layer.
We can write that as u-e for all z at time zero is equal to the constant u-i and u-i
is equal to the applied stress delta-sigma-z.
So now we have our boundary conditions, and we find the solution to our differential equation.
As you know, the best way to solve a differential equation is to find a mathematician who's
already solved it for you.
Seriously, we've got much more important things to do rather than solve differential equations:
hang out with friends at the pub, like a friend's new Facebook post,
send a few tweets, get to level 250 in candy crush,
enjoy a cold one.
So to leave you time for the finer things in life
here's the solution to our differential equation, which is equation 11.17 in your text
This might look a bit complicated at first, but it's really not that bad.
It's a series solution and these terms are just counting constants that are a function
of n.
For example this first term is equal to 3 over pi when n is zero and 4 over 3 pi when
n is 1, etcetera.
The three key variables are here U-e over delta-sigma-z
Z-dr over H-dr And T-v
All three of these are formulated as dimensionless variable
T-v is called the time factor and is our dimensionless measure of time.
It's equal to c-v times time, over H-dr squared.
You should check the dimensions of T-v and satisfy yourself that it is indeed dimensionless.
Our dimensionless Depth is z-dr over H-dr And our dimensionless measure of excess pore
pressure is u-e over delta-sigma-z Since our solution is formulated in terms
of dimensionless variables, we only have to solve it once.
We can use the dimensionless solution for any case we're interested in by converting
the dimensioned variables in our particular problem into these three dimensionless ones.
Here's our solution to equation 11.17.
Using this figure we can determine the excess pore pressure in a consolidating soil layer
as function of both location and time.
The vertical axis represents our dimensionless depth into our compressible layer, z-dr over
H-dr.
Using our double drained example, z-dr over H-dr equal to 1 would be the mid-point of
our layer and z-dr over H-dr of 0.5 would a fourth of the way into the layer
The horizontal axis represents our dimensionless excess pore pressure, that is, the current
excess pore pressure divided by the initial excess pore pressure, which was generated
immediately after loading, or u-e over delta-sigma-z.
So at u-e over delta-sigma-z equal to 0.5, half of the initial excess pore pressure has
been dissipated and half of it remains The curves in the figure are called isochrones
because each represents a specific point in time.
For example this isochrones represents the time where the time factor, T-v, is equal
to 0.2 The isochrones represent increasing time from
right to left So if we wanted to know what the excess pore
pressure was one quarter of the way down from the top of our consolidating layer at time
when the time factor was 0.2, We'd start at a z-dr over H-dr of 0.5 and
draw a line to the isochrones where T-v is equal to 0.2
Then we'd draw a line down to the x-axis and find out that the ratio of the current excess
pore pressure, u-e, to the initial excess pore pressure, delta-sigma-z was 0.55.
We'll do an example later to clarify this computation
I think it's helpful to do a little illustration to explain how the excess pore pressure is
changing within the clay layer during consolidation.
Let's consider this clay layer.
Since it's doubly drained, the height of the layer is twice the maximum drainage distance
or 2 H-dr.
We're going to consider cross-section A-A' And plot the excess pore pressure along this
cross-section at various points in time We'll start immediately after the fill is
placed on top of the soil.
At this instant in time the excess pore pressure jumps up to the initial condition and it is
equal to the applied stress delta-sigma across the entire height of the clay layer
At some time later, t-1, some of the excess pore pressure will have dissipated.
At the two boundaries it will be zero, because the boundaries are freely draining, and force
the excess pore pressure to zero.
It will be greatest in the center of the layer which is furthest from the drainage boundaries.
At some time later, t-2, there will be additional dissipation of the excess pore pressure and
our isochrone might look like this.
The pore pressure continues to dissipate Eventually all the excess pore pressure will
dissipate.
Theoretically, this takes an infinite amount of time.
From a practical standpoint, it takes a finite amount of time.
So the isochrone in our dimensionless solution represent the actual dissipation of excess
pore pressure in the field.
Let's solve part of example 11.2 from your text as an illustration
Here we have a compressible clay layer 10 m thick, sandwiched between a sandy silt and
a silty sand layer.
We propose to construct a 3.5 m thick fill on top of the sand layer
Let's compute both the total and excess pore pressures at points A and B in the clay layer
at a time 10 years after the fill has been placed
First we must determine the drainage conditions.
Since both the sandy silt below the clay and silty sand above it will have hydraulic conductivities
much greater than the clay, it's safe to assume the clay layer is doubly drained.
In that case the maximum drainage distance will the half the layer thickness or 5 m
Next, let's compute the change in vertical effective stress, delta-sigma-z
This will be the unit weight of the fill times the height of fill
or 19.7 kilonewtons per meter cubed times 3.5 m,
which is 68.9 kilopascals We'll be using Figure 11.4 to solve the problem
so we'll need dimensionless values for all our variables
First we'll need to know the dimensionless time, T-v, which is c-v time t over H-dr squared.
We're given c-v as 3 time 10 to the minus 3 meters squared per day.
The time is 10 years which we need to covert to days and we divide by H-dr squared or 5
squared The resulting time factor is 0.44
At point A we'll compute the static pore pressure before
consolidation simply as gamma-water times the height of groundwater table above A which
will come out to be 69.6 kPa To determine the excess pore pressure, we'll
need our dimensionless depth, z-f over H-dr or 5 over 5 which is one.
We then go to Figure 11.4 and draw a line at
z-f over H-dr equal to 1.
We want to know the answer when T-v is 0.44 so we'll have to interpolate between the 0.4
and 0.5 isochrones.
We'll sketch in our own isochrone at T-v equal to 0.44.
Where this isochrone intersects the blue line we draw a vertical line down to the x-axis
and see that it corresponds to a value of u-e over delta-sigma-z of 0.43
So our dimensionless excess pore pressure at point A is 0.43
and we compute the true pore pressure as 0.43 times delta-sigma-z (which is 68.9)
and get 29.6 kilopascals And the total pore pressure at point A 10
years after the fill is placed is simply the static pore pressure plus the excess pore
pressure or 69.9 plus 29.6 which is 99.2 kPa.
We repeat the process at point B Here the static pore pressure is 9.8 time
2.5 plus 2.1 or 45.1 kPa Our dimensionless depth at B is 2.5 over 5
or 0.5 So now we enter Figure 11.4 at z-dr over H-dr
of 0.5 We find where that line intersects our interpolated
isochrone of T-v equal to 0.44 and find that u-e over delta-sigma-z is 0.31
With this value of u-e over delta-sigma-z ,
we compute the actual excess pore pressure as 0.31 times 68.9,
which is 21.4 kilopascals.
And our total pore pressure at B is 45.1 plus 21.4 or 66.5 kPa
We can also answer the question, �When will the excess pore pressure at point A be 10
kilopascals?
To do this, we first compute the dimensionless excess pore pressure, which will be 10 over
68.9 or 0.15 We then go to Figure 11.4
And enter at u-e over delta-sigma-z equal to 0.15.
And find the point where this line intersects z-dr over H-dr equal to 1, which is the dimensionless
depth of point A At the intersection,
we interpolate the time factor to be equal to 0.87
So we know that T-v is 0.87 And we convert that back to true time with
the equation t is equal to T-v times H-dr squared over c-v.
Which will be 0.87 times 5 squared divided by 3 times 10 to the minus 3 ,
or 7.250 days which is 19.8 years.
So all we need to do to solve a consolidation problem is change the actual variables into
dimensionless values Go to Figure 11.4, input two dimensionless
variables, and get back the third dimensionless value.
We then convert that dimensionless output back to the true value we want.
It's really pretty simple.
Let's look at one more case before we conclude this webcast.
Here we have a singly drained compressible layer.
On the bottom it's underlain by an impervious material, perhaps something like a dense shale,
and on the top it's overlain by a freely draining sand.
The longest path any water molecule will have to travel to exit the clay layer now is the
full height of the clay layer The solution to this problem will exactly
the same as the doubly drained case, except that H-dr is now the full height of the layer
rather than one half the height of the layer.
Let's look at the distribution of excess pore pressure along A-A' for this case
When the fill is placed, we'll again have a uniform initial excess pore pressure equal
to delta-sigma-z As time goes on
the excess pore pressure will dissipate.
But this time the maximum excess pore pressure will be at the impervious boundary at the
bottom of the layer rather than in the middle of the layer.
The isochrones in this case are the same as those in Figure 11.4
except we need to consider just the upper half of the figure.
All the equations remain the same.
We just have to use the correct H-dr in our calculations.
Let's take a moment now and review the original assumptions we used in Terzaghi's consolidation
theory First we assumed the soil was homogenous and
isotropic.
This may be a reasonable assumption depending upon the site.
If we have a site with only one major compressible soil this might work.
If we have a site like the Kansai airport site, this won't work
Our assumptions about soil saturation, incompressibility of solids and water, and Darcy's law are all
reasonable for soils below the groundwater table.
The assumption that the load was applied instantaneously will depend upon the actual construction sequence.
Sometime projects take years to complete and this assumption is not valid.
The assumption that all the total stress change is initially transferred to excess pore pressure,
is usually ok.
However our assumption that the soil properties c-v and k don't change isn't so good.
We saw that c-v is really a function of the effective stress which will be increasing
as the soil consolidates.
And, obviously, the hydraulic conductivity will decrease as the soil consolidates.
Finally or assumption of 1-D consolidation and flow generally seems to be satisfactory
So the biggest uncertainties in determining the time rate of settlement are the homogeneity
of the site, the rate at which loads are applied, and how soil properties change during consolidation.
So let's summarize The solution we developed to the consolidation
differential equation is appropriate for either a doubly or singly draining system.
It assumes the initial excess pore pressure is uniform and equal to delta-sigma-z, which
corresponds to a quickly applied load.
If we have different drainage conditions or if the load is applied gradually, this solution
is not appropriate Our solution is in the form of dimensionless
variables for time, depth, and excess pore pressure
And we can use figure 11.4 to determine the excess pore pressure for any location within
the consolidating layer at any time.