Soil mechanics - Chapter 16 pot

Chapter 16

ANALYTICAL SOLUTION

In this chapter an analytical solution of the one dimensional consolidation problem is given. In soil mechanics this solution was first given by

Terzaghi, in 1923. In mathematics the solution had been known since the beginning of the 19th century. Fourier developed the solution to

determine the heating and cooling of a metal strip, which is governed by the same differential equation.

16.1 The problem

The mathematical problem of one dimensional consolidation has been established in the previous chapter. The differential equation is

∂p

∂t = cv

∂

2p

∂z2

, (16.1)

with the initial condition

t = 0 : p = p0 =

q

1 + nβ/mv

, (16.2)

in which q the load applied at time t = 0. It is assumed that the load remains constant for t > 0.

The boundary conditions are, for the case of a sample of height h, drained at its top and impermeable at the bottom,

z = 0 :

∂p

∂z = 0, (16.3)

z = h : p = 0. (16.4)

These equations describe the consolidation of a soil sample in an oedometer test, or a confined compression test, with a constant load, and drained

............................................................................................................................................................................................................

.

.................................................................................................................................................................................................................................................................................................................................

.........................................................................................................................................................................................................

.........................................................................................................................................................................................................

.........................................

......................................... ........................................

......................................... ........................................

......................................... ........................................

......................................... ........................................

......................................... ........................................

......................................... ........................................

......................................... ........................................

......................................... ........................................

......................................... ........................................

......................................... ........................................

......................................... ........................................

......................................... ........................................

......................................... ........................................

......................................... ........................................

......................................... ........................................

.............................................................................................................................................................................................................................................................................................................

............................................................................................................................................................................................................................................................................................................

......................................................................................

...................................................................................... ......................................................................................

...................................................................................... ......................................................................................

...................................................................................... ......................................................................................

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.............................................................................................................................................................................................................................................................................................................

............................................................................................................................................................................................................................................................................................................

Figure 16.1: Consolidation.

only at the top of the sample. The equations also apply to a sample of thickness 2h, drained both

at its top and bottom ends. The top half of such a sample drains to the upper boundary, and the

lower half drains to the lower boundary. The center line acts as an impermeable boundary. The

same problem occurs in case of a layer of clay between two very permeable layers, when the soil is

loaded, in a very short time and over a very large area, by a constant load. If the area is very large

it can be assumed that there will be no lateral deformations, and vertical flow only. The load can

be a surcharge by an additional sand layer, applied in a very short time.

96