Posted in Academic Issues, Geotechnical Engineering

## The Equivalent Thickness Method for Estimating Elastic Settlements

In our very popular post Analytical Boussinesq Solutions for Strip, Square and Rectangular Loads we discuss the use of the theory of elasticity (as originally formulated by Boussinesq) to estimate the stresses and settlements under foundations. We start by giving methods of estimating the stresses under various configurations of rectangular foundations (the circular ones are discussed in the post Going Around in Circles for Rigid and Flexible Foundations.) We then show the use of superposition to expand the use of these results for complex foundations (with further discussion in our post Superposition, and Using Point Loads in Place of Distributed Ones.) We then show the estimation of deflections for simple rigid and flexible foundations. But when it comes to deflections for more complex situations…crickets.

This post is an attempt to solve the “crickets” problem through the use of a method shown in Tsytovich (1976). It’s doubtless useful for preliminary calculations and to enhance our understanding of how settlements of foundations in one place can affect adjacent structures. It also uses some of the linkage between elastic and consolidation settlement theory which is discussed in From Elasticity to Consolidation Settlement: Resolving the Issue of Jean-Louis Briaud’s “Pet Peeve”.

$\epsilon_{{x}}={\frac {p\beta}{E}}$ (1)

where we swap p for $\sigma_x$ as the vertical pressure.

Now let us define an equivalent height heq. Keep in mind that we are assuming that the soil’s reaction to vertical pressure is that of a laterally confined specimen; the equivalent height is the height of that equivalent specimen. Multiplying both sides of Equation (1) by this equivalent height,

$h_{{{\it eq}}}\epsilon_{{x}}={\frac {h_{{{\it eq}}}p\beta}{E}}$ (2)

Since by definition

$\epsilon_x = {\frac {s}{h_{{{\it eq}}}}}$ (3)

where s is the settlement, Equation (1) becomes

$s={\frac {h_{{{\it eq}}}p\beta}{E}}$ (4)

Equation (7) of the last linked post tells us that

$E={\frac {\beta}{m_{{v}}}}$ (5)

where

$\beta = 1-2\,{\frac {{\nu}^{2}}{1-\nu}}$ (7)

Combining Equations (4) and (5) yields

$s=h_{{{\it eq}}}pm_{{v}}$ (8)

To compute the equivalent height, we turn to our post Analytical Boussinesq Solutions for Strip, Square and Rectangular Loads, where we modify the equation presented for the deflection of rectangles and squares with Equation (5) above to obtain

$s={\frac {\omega\,pb\left (1-{\nu}^{2}\right )m_{{v}}}{\beta}}$ (9)

The value of $\omega$ is discussed in that post for squares and rectangles and for circles in Going Around in Circles for Rigid and Flexible Foundations. These are very complex (and in the case of circles have no closed form solution.) For convenience the table for these is reproduced below.

If we equate the right hand sides of Equations (8) and (9) and solve for heq, we have at last

$h_{eq} = {\frac {\omega\,b\left (1-{\nu}^{2}\right )}{\beta}}$ (10a)

We can also substitute Equation (7) into Equation (10a) and obtain

$h_{eq} = {\frac {\left (-1+\nu\right )^{2}b\omega}{1-2\,\nu}}$ (10b)

If we define

$A = {\frac {\left (1-\nu\right )^{2}}{1-2\,\nu}}$ (10*)

we can also write the equation thus

$h_{eq} = A\omega b$ (10c)

It should be evident that there are several computational routes to obtain the equivalent height, which is then substituted into Equation (8) to obtain the settlement. Let us consider these options:

• We could tabulate values of $A \omega$ for various foundation configurations and then use these to compute the equivalent height using Equation (10c). This is given in Tsytovich (1976).
• We could determine values for $A$ (it is simply a function of Poisson’s Ratio $\nu$), obtain $\omega$ using the table above and then compute the equivalent height using the width of the foundation $b$. Values for both $A$ and $\beta$ are shown in graphical form as a check for computations.
• We could perform direct substitution into Equations (10a) or (10b.) Equation (10a) is probably the best as it will be necessary to compute $\beta$ using Equation (7).

## Worked Example

As an example, let us consider the same example from the post Analytical Boussinesq Solutions for Strip, Square and Rectangular Loads. The foundation diagram is shown below; we are interested in the settlement at Point A.

The complete solution is given in the same spreadsheet as the solution for the stress problem, which you can access here. The geometry is the same and the loading is the same: 5 kPa on the yellow (2 m x 10 m foundation) and 15 kPa on the brown foundation (2 m x 6 m.) Keep in mind that, for this method, the value b is always the smaller of the two, and that goes for the “void” foundations as well.

We do not need the depth from the surface; we are only interested in surface deflections. We do need the elastic modulus and Poisson’s Ratio of the soil, which are E = 10,000 kPa and ν = 0.25.

The superposition is exactly the same as before, using the following diagram as before:

The superposition scheme is as follows:

• Yellow Foundation Positive, corners ABFG, pressure +5 kPa
• Yellow Foundation Negative, corners ABJH, pressure – 5kPa
• Brown Foundation Positive, corners ACEG, pressure +15 kPa
• Brown Foundation Negative, corners ABFG, pressure – 15 kPa

We will only go through the calculations for the first one; you can view the spreadsheet for the rest. We proceed as follows:

• We compute A and β as follows:
• From Equation (10*), A = (1-0.25)2/(1-(2)(0.25)) = 1.125
• From Equation (7), β = 1-(2)(0.25)2/(1-0.25) = 0.833
• You can verify these using the plot of these parameters.
• We compute the coefficient of volume compressibility by using Equation (5), mv = 0.833/10000 = 0.0000833 1/kPa
• We compute the value of α = l/b = 10/6 = 1.6667
• We compute the corner value for ω (since we are dealing with corners as was the case with stresses.) We can use the table for ω or we can compute it using the formulae from Analytical Boussinesq Solutions for Strip, Square and Rectangular Loads, but it is ω = 0.71.
• We then use Equation (10c) to compute the equivalent height, thus heq = (1.125)(0.71)(6) = 4.8 m. This value will be different for each corner point considered.
• Using Equation (8), the settlement for this portion of the analysis is s = (4.8)(5)(0.000083333) = 0.002 m = 1.998 mm.

You repeat this process for all four “foundations.” Keep in mind that the negative foundations will result in negative settlements. A summary of the results is as follows:

Posted in Deep Foundations, Pile Driving Equipment

## About that “Warrington Method” For Vibratory Pile Drivability — vulcanhammer.info

Every now and then something comes up that you really didn’t expect. That took place with a paper published this year cited “W.J. Lu, B. Li, J.F. Hou, X.W. Xu, H.F. Zou, L.M. Zhang, “Drivability of large diameter steel cylinders during hammer-group vibratory installation for the hong kong–zhuhai–macao bridge,” Engineering (2022), doi: https://doi.org/10.1016/j.eng.2021.07.028.” (You can […]

About that “Warrington Method” For Vibratory Pile Drivability — vulcanhammer.info
Posted in Civil Engineering, Deep Foundations

## The Paper “Vibratory and Impact-Vibration Pile Driving Equipment” Cited — vulcanhammer.info

It’s happened again: the paper “Vibratory and Impact-Vibration Pile Driving Equipment” has been cited by Mohammed Al-Amrani and M Ikhsan Setiawan in their paper “Prefabricated and Prestressed Bio-Concrete Piles: Case Study in North Jakarta.” The abstract of their paper is here: In this research, we will talk about Prefabricated and Prestressed Concrete piles in general and […]

The Paper “Vibratory and Impact-Vibration Pile Driving Equipment” Cited — vulcanhammer.info

## Determining the Degree of Consolidation

This is the last (hopefully) post in a series on consolidation settlement. We need to start by a brief summary of what has gone before. Note: the material for this derivation and those that preceded it have come from Tsytovich with some assistance from Verruijt.

## Review

In the post From Elasticity to Consolidation Settlement: Resolving the Issue of Jean-Louis Briaud’s “Pet Peeve”, we discussed the issue of how much soils (especially cohesive ones) settle through the rearrangement of particles. We were able to start with the theory of elasticity and, considering the effects of lateral confinement, define the coefficient of volume compression $m_v$ by

$m_v = \frac{\beta}{E}$ (1)

where E is the modulus of elasticity and $\beta$ is a factor based on Poisson’s Ratio and includes the effects of confinement, be that in an odeometer or in a semi-infinite soil mass. We also showed that, for a homogeneous layer,

$\delta_p = m_v H_o \sigma_x$ (2)

where $\delta_p$ is the settlement of the layer, $H_o$ is the thickness of the layer and $\sigma_x$ is the uniaxial stress on the layer. The problem is that $m_v$ is not constant, and the settlement more accurately obeys the law

$\delta_p = \frac{C_c H_o}{1+e_o} \log{\frac{\Delta p + \sigma_o}{\sigma_o}}$ (3)

where $C_c$ is the compression index, $e_o$ is the initial void ratio of the layer, $\Delta p$ is the change in pressure induced from the surface, and $\sigma_o$ is the average effective stress in the layer.

Turning to the post Deriving and Solving the Equations of Consolidation, we first determined that the change in porosity $\Delta n$ could, for small deflections, be equated to the change in strain $\epsilon$. From this we could say that

$\Delta n = m_v \Delta \sigma_x$ (4)

The change in porosity, for a saturated soil whose voids are filled with an incompressible fluid (hopefully water) induces water flow,

${\frac {\partial }{\partial x}}q(x,t)=-{\frac {\partial }{\partial t}} {\it n}(x,t)$ (5)

where $q(x,t)$ is the flow of water out of the pores and $n(x,t)$ is the porosity as a function of position and time. The flow of water is regulated by the overall permeability of the soil, and all of this can be combined to yield

${\frac {k{\frac {\partial ^{2}}{\partial {x}^{2}}}u(x,t)}{{\it \gamma_w }}}=m_{{v}}{\frac {\partial }{\partial t}}\sigma_{{x}}(x,t)$ (6)

where $k$ is the permeability of the soil and $\gamma_w$ is the unit weight of water. Defining

$c_v = \frac{k}{m_v \gamma_w}$ (7)

and making some assumptions about the physics, we can determine the equation for consolidation as

$c_{{v}}{\frac {\partial ^{2}}{\partial {x}^{2}}}u(x,t)={\frac {\partial }{\partial t}}u(x,t)$ (8)

where \$latex u(x,t) is the pore water pressure. If we invoke the effective stress equation and solve this for the boundary and initial conditions described, we have a solution

$\sigma_{x}(x,t)=p\left(1-\frac{4}{\pi}\left(\sin(1/2\,{\frac{\pi\,x}{h}}){e^{-1/4\,{\frac{{\it c_v}\,{\pi}^{2}t}{{h}^{2}}}}}+1/3\,\sin(3/2\,{\frac{\pi\,x}{h}}){e^{-9/4\,{\frac{{\it c_v}\,{\pi}^{2}t}{{h}^{2}}}}}+1/5\,\sin(5/2\,{\frac{\pi\,x}{h}}){e^{-{\frac{25}{4}}\,{\frac{{\it c_v}\,{\pi}^{2}t}{{h}^{2}}}}}\cdots\right)\right)$ (9)

## The Degree of Consolidation

One thing that our theory presentation demonstrated was the interrelationship between pore pressure, stress and deflection. We know what the ultimate deflection will be based on Equation (3) above (or more complicated equations when preconsolidation is taken into consideration.) But how does the settlement progress in time?

We start by defining the degree of consolidation thus:

$U = \frac{\delta(t)}{\delta_p}$ (10)

where $\delta(t)$ is the settlement at any time before complete settlement. For the specific case (governing equations, initial equations and boundary conditions) at hand, the degree of consolidation–the ratio of settlement at a given point in time to total settlement–can be determined as follows:

$U_{o}=\intop_{0}^{h}\frac{\sigma_{x}(x,t)}{ph}dx$ (11)

In this case the result is divided by the uniform pressure p and the height h. Let us further define the dimensionless time constant

$T_{v}=\frac{c_{v}t}{h^{2}}$ (12)

That being the case, if we integration Equation (9) with Equation (11), we obtain

$U_{o}=1-\sum_{n=1}^{\infty}4\,{\frac {{e^{-1/4\,{\it Tv}\,{n}^{2}{\pi }^{2}}}\left (\cos(n\pi )\cos(1/2\,n\pi )-\cos(n\pi )-\cos(1/2\,n\pi )+1\right )}{{n}^{2}{\pi }^{2}}}$ (13)

otherwise put

$U_{o}=1-8\,{\frac {{e^{-1/4\,{\it Tv}\,{\pi }^{2}}}}{{\pi }^{2}}}-{\frac {8}{9}}\,{\frac {{e^{-9/4\,{\it Tv}\,{\pi }^{2}}}}{{\pi }^{2}}}-{\frac {8}{25}}\,{e^{-{\frac {25}{4}}\,{\it Tv}\,{\pi }^{2}}}{\pi }^{-2}\cdots$ (14)

As was the case with Equation (9), only the odd values of n are considered; the even ones result in zero terms.

It is regrettable that, in defining $T_v$, the value $\frac{\pi^2}{4}$ was not included, as using Equation (14) would be much simpler. For certain cases, it is possible to use the first two or three terms. In any case the usual method for determining $T_v$–and by extension the degree of consolidation–is generally done either using a graph or a table, as is shown in the graph at the start of the post (repeated below:)

The notation is a little different. We use the variable $U_o$ to emphasise that we are dealing with the “standard” case. The above graph also gives approximating equations; it is easy to see that, for $T_v > 0.2$, the equation given is simply the first two terms of Equation (14). The distinction between the drainage length h ($H_{dr}$ in the graph above) and the layer thickness H is clear.

## Conclusion

We have covered the basic, classic case of consolidation settlement in this post and its predecessors From Elasticity to Consolidation Settlement: Resolving the Issue of Jean-Louis Briaud’s “Pet Peeve” and Deriving and Solving the Equations of Consolidation. We trust that this presentation has been enlightening and informative.

## Deriving and Solving the Equations of Consolidation

In an earlier post we discussed consolidation settlement. For situations where soil a) decreases its volume due to rearrangement of the particles and b) does so over a relatively long period of time due to difficulties in expelling pore water pressures, we need to know how long it takes to reach maximum settlement, in addition to know what that settlement is and what deflections we might achieve along the way.

This is generally a two part process: a) determining the dissipation of pore water pressure and b) determining the amount of settlement at a given time. This post will focus on the first part of the process; we will deal with the second in a later post. This is a well-worn path in geotechnical engineering but hopefully this derivation is a little simpler than others.

## Developing the Differential Equation

### Relating Porosity to Strain

Let us begin by considering the system above (from Tsytovich (1976)), a uniformly loaded soil layer which is saturated (always) and clay (usually.) The first thing we need to note is that, while the diagram uses z as the variable of length in the vertical direction, from this post we will use the variable x (sorry for the confusion.) The second thing is that we assume the water and the solids to be incompressible. The third thing is that all the changes that take place do so because of changes in the voids where the water is resident. We first note the definition of porosity as

$n={\frac {V_{{v}}}{V_{{t}}}}$ (1)

where

• n = porosity
• Vv = volume of voids
• Vt = total volume

That being the case, the relationship between the porosity of the soil (due to changes in the volume of the voids) and the flow rate of the water can be expressed as

${\frac {\partial }{\partial x}}q(x,t)=-{\frac {\partial }{\partial t}} {\it n}(x,t)$ (2)

We can envision a differential volume having a height x and an area A. Since the problem is one-dimensional, the areas cancel out and the ratio of the void height to the total height is

$n={\frac {x_{{v}}}{x_{{t}}}}$ (3)

where

• xv = height of the voids
• xt = height of the solids

We want to determine the change in porosity from some state 0 to some state 1, just as we did with void ratio in this post. That change can be expressed as follows:

${\frac {x_{{{\it v0}}}}{x_{{{\it t0}}}}}-{\frac {x_{{{\it v1}}}}{x_{{{\it t0}}}-x_{{{\it v0}}}+x_{{{\it v1}}}}}$ (4)

We can assume that the change in void volume xv0 – xv1 << xt0, in which case Equation (4) can be simplified to

$\Delta{{{\it n}}}={\frac {x_{{{\it v0}}}-x_{{{\it v1}}}}{x_{{{\it t0}}}}} = \frac{\Delta x}{x_t}$ (5)

Now we can say, for the small increments we are dealing with here, that the change in strain is equal to the change in porosity,

$\Delta n = \Delta \epsilon$ (6)

From this and our previous post, we can thus use the change in strain to come to the following:

$\Delta n = m_v \Delta \sigma_x$ (7)

Stating this differentially,

${\frac {\partial }{\partial t}}{\it n}(x,t)=m_{{v}}{\frac {\partial }{\partial t}}\sigma_{{x}}(x,t)$ (8)

In this way we emphasise that both porosity and uniaxial stress are functions of depth and time. We now combine Equations (2) and (8) to yield

${\frac {\partial }{\partial x}}q(x,t)=-m_{{v}}{\frac {\partial }{\partial t}}\sigma_{{x}}(x,t)$ (9)

### Including Permeability

Darcy’s Law (or more properly d’Arcy’s Law) states that

$q(x,t) = -k{\frac {\partial }{\partial x}}H(x,t)$ (10)

where k is the coefficient of permeability and H is the hydraulic head. We then combine Equations (9) and (10) to obtain

$k{\frac {\partial ^{2}}{\partial {x}^{2}}}H(x,t)=m_{{v}}{\frac {\partial }{\partial t}}\sigma_{{x}}(x,t)$ (11)

Noting that

$H(x,t) = \frac{u(x,t)}{\gamma_w}$ (12)

where $\gamma_w$ is the unit weight of water and $u(x,t)$ is the excess pore water pressure generated by the decrease in porosity, we substitute this with the result of

${\frac {k{\frac {\partial ^{2}}{\partial {x}^{2}}}u(x,t)}{{\it \gamma_w }}}=m_{{v}}{\frac {\partial }{\partial t}}\sigma_{{x}}(x,t)$ (13)

### Putting It All Together

We now define the parameter

$c_v = \frac{k}{m_v \gamma_w}$ (14)

At this point we need to stop and make an observation: this whole process wouldn’t be worth too much if $c_v$ wasn’t constant (or reasonably so.) The reason it is is that the coefficient of permeability $k$ and the coefficient of volume compressibility $m_v$ both decrease as the void ratio/porosity decrease, and do so at roughly the same rate. That being the case, we substitute Equation (14) into Equation (13) to have at last

$c_{{v}}{\frac {\partial ^{2}}{\partial {x}^{2}}}u(x,t)={\frac {\partial }{\partial t}}\sigma_{{x}}(x,t)$ (15)

## Tidying up the Physics, Governing Equation, Boundary and Initial Conditions, and the Solution

It should be evident that getting to Equation (15) was a major triumph for geotechnical theory. But it’s also evident that there’s one glaring problem: the dependent variable on the left hand side is not the same as the one on the right. It is here that we need to clarify some assumptions behind our equations.

A basic assumption in consolidation theory is that, when the load at the surface is applied, all of this additional load is initially borne by the pore water pressure. Because of the aforementioned permeability, the water will want to “head for the exits,” i.e., the permeable boundaries of the layer being compressed. When the particles have rearranged themselves and the excess pore water has been squeezed out, the settlement should stop (until secondary compression kicks in.) During this process the load on the pore water is being progressively transferred to the soil particles until consolidation has stopped (which, in theory, it never does, as we will see) and the load is completely handed off to the particles.

That being the case, the consolidation equation (15) should be rewritten as

$c_{{v}}{\frac {\partial ^{2}}{\partial {x}^{2}}}u(x,t)={\frac {\partial }{\partial t}}u(x,t)$ (16)

At this point we should invoke the effective stress equation

$\sigma_x(x,t) = p - u(x,t)$ (17)

where p is the applied pressure, and do the following to determine the time and distance history of the vertical stress $\sigma_x(x,t)$:

• Determine the solution of Equation (16), which will be our governing equation.
• Determine the initial and boundary conditions for the problem.
• Solve the problem for $\sigma_x(x,t)$ using Equation (17).

There are several ways to solve the governing equation for this problem. Verruijt employed Laplace Transforms to accomplish this; these are very useful, as was demonstrated by Warrington (1997) for the wave equation. For this analysis, we will use separation of variables, a method which is not as fundamental as one might like (this is a demonstration of a method that is) but which is easier to follow than most of the others.

The separation of variables begins by assuming the solution is as follows:

$u(x,t) = X(x)T(t)$ (18)

This means that the solution is a product of two functions, one of time and the other of distance, and that those functions are separate one from another. Substituting this into Equation (16) and rearranging a bit yields

${\frac {{\frac {d^{2}}{d{x}^{2}}}X(x)}{X(x)}}={\frac {{\frac {d}{dt}}T(t)}{{\it c_v}\,T(t)}}$ (19)

Since the left and right hand sides are equal to each other, they are equal to a third variable, which we will designate as $\beta^2$. In reality they are equivalent to the eigenvalue $\lambda = \beta^2$, but the utility of the squared term will become evident. In any case,

${\frac {{\frac {d^{2}}{d{x}^{2}}}X(x)}{X(x)}}=-{\beta}^{2}$ (20a)

${\frac {{\frac {d}{dt}}T(t)}{{\it c_v}\,T(t)}}=-{\beta}^{2}$ (20b)

The solutions to these equations are, respectively,

$X(x)={\it C_1}\,\cos(\beta\,x)+{\it C_2}\,\sin(\beta\,x)$ (21a)

$T(t)={e^{-{\it c_v}\,{\beta}^{2}t}}{\it C_3}$ (21b)

We need to pause and consider the boundary conditions. As the problem is shown at the top of the article, the boundary conditions are as follows:

$X(0) = 0$ (22a)

$X'(h) = 0$ (22b)

Equation (22a) represents a Dirichelet boundary condition and Equation (22b) represents a Neumann boundary condition. While the equation for $X(x)$ can certainly be solved for this boundary condition, a simpler way would be to do the following:

• Mirror the problem shown above at x = h so that you have two permeable boundaries.
• You now have a layer of 2h thickness with Dirichelet boundary conditions on both sides. At the centre of this new layer, there is no flow; the water above it flows upward, and the water below flows downward.
• Problems in the field can have either one or two permeable boundaries. The distance h is NOT the thickness of the layer but the longest distance the trapped pore water must travel to escape. Confusing h with the thickness of the layer is a common mistake and should be avoided at all costs.

That said, the second boundary condition is now

$X(2h) = 0$ (22c)

Returning to Equation (21a), because of Equation (22a) $C_1 = 0$ as the cosine is by definition unity at this point. If we then set x = 2h and X(2h) = 0, for real, non-zero values of $C_2$ the boundary condition can be satisfied if and only if

$\beta = 1/2\,{\frac {n\pi }{h}}$ (23)

This is the square root of the eigenvalues. Any integer value of n > 0 is valid for this, and we will have recourse to them all to produce a complete orthogonal set (Fourier Series) to solve the problem.

This change will affect both Equations (21a) and (21b). Combining constants into the coefficient $B_n$, substituting the results into Equation (18) and making it an infinite sum for the Fourier series yields

$u(x,t) = {\it B_{n}}\,\sin(1/2\,{\frac{n\pi\,x}{h}}){e^{-1/4\,{\frac{{\it c_v}\,{n}^{2}{\pi}^{2}t}{{h}^{2}}}}}$ (24)

At this point we need to consider our initial conditions $f(x)$. Although many different distributions of initial pore pressure are possible, the simplest one–and the one most commonly used–is a uniform pressure p, which is the same as the surface pressure compressing the layer. For a complete orthogonal set, the coefficients $B_n$ can be computed by (Kreyszig (1988)) as

${\it B_n}=\int _{0}^{2\,h}f(x)\sin(1/2\,{\frac {n\pi \,x}{h}}){dx}{h}^{-1}$ (25)

Substituting $f(x) = p$ and performing the integration,

$B_n = 2\,{\frac{p\left(1-\cos(n\pi)\right)}{n\pi}}$ (26)

Substituting this into Equation (24) and taking the complete sum, we have at last

$u(x,t)=\sum_{n=1}^{\infty}2\,p\left(1-\cos(n\pi)\right)\sin(1/2\,{\frac{n\pi\,x}{h}}){e^{-1/4\,{\frac{{\it c_v}\,{n}^{2}{\pi}^{2}t}{{h}^{2}}}}}{n}^{-1}{\pi}^{-1}$ (27)

More simply we can say that

$u(x,t)=\frac{4p}{\pi}\left(\sin(1/2\,{\frac{\pi\,x}{h}}){e^{-1/4\,{\frac{{\it c_v}\,{\pi}^{2}t}{{h}^{2}}}}}+1/3\,\sin(3/2\,{\frac{\pi\,x}{h}}){e^{-9/4\,{\frac{{\it c_v}\,{\pi}^{2}t}{{h}^{2}}}}}+1/5\,\sin(5/2\,{\frac{\pi\,x}{h}}){e^{-{\frac{25}{4}}\,{\frac{{\it c_v}\,{\pi}^{2}t}{{h}^{2}}}}}\cdots\right)$ (28)

The vertical pressure on the soil skeleton is determined by combining Equations (17) and (28) to yield

$\sigma_{x}(x,t)=p\left(1-\frac{4}{\pi}\left(\sin(1/2\,{\frac{\pi\,x}{h}}){e^{-1/4\,{\frac{{\it c_v}\,{\pi}^{2}t}{{h}^{2}}}}}+1/3\,\sin(3/2\,{\frac{\pi\,x}{h}}){e^{-9/4\,{\frac{{\it c_v}\,{\pi}^{2}t}{{h}^{2}}}}}+1/5\,\sin(5/2\,{\frac{\pi\,x}{h}}){e^{-{\frac{25}{4}}\,{\frac{{\it c_v}\,{\pi}^{2}t}{{h}^{2}}}}}\cdots\right)\right)$ (29)

It is interesting to note that, although Equation (27) includes all positive non-zero values of n, only the odd ones end up in Equations (28) and (29). For even values of n, $B_n = 0$.

## Conclusion

At this point we have derived the equations of pressure dissapation for consolidation settlement. In a future post we will deal with the second part of the problem, namely how much of the total anticipated settlement has taken place at any given time from initial loading.

## Reference

Kreyszig, E. (1988) Advanced Engineering Mathematics. Sixth Edition. New York: John Wiley and Sons