Back to Volume
Change
Consolidation 
Based on part of the GeotechniCAL reference package
by Prof. John Atkinson, City University, London 
The amount of settlement which occurs in a given time depends on the
If soil is unloaded (e.g. by excavation) the excess pore pressures may be negative.
Back to The process of consolidation and settlement
Consider a site on clay soil with initial steadystate groundwater conditions.
An embankment is built, the loading is undrained: the pore pressure in the soil increases,
seepage flow and therefore volume changes commences. As consolidation takes place,
settlement occurs, and continues at a decreasing rate until steadystate conditions are regained.
Click on the buttons to see the sequence of loading and pore pressure changes. 
Back to The process of consolidation and settlement
A general theory for consolidation, incorporating threedimensional flow vectors is complicated and only applicable to a very limited range of problems in geotechnical engineering. For the vast majority of practical settlement problems, it is sufficient to consider that both seepage and strains take place in one direction only; this usually being vertical.
Onedimensional consolidation specifically occurs when there is no lateral strain, e.g. in theoedometer test
Onedimensional consolidation can be assumed to be occurring under wide foundations.
Back to Consolidation
A simple onedimensional consolidation model consists of rectilinear element of soil subject to vertical changes in loading and through which vertical (only) seepage flow is taking place.
There are three variables:
Back to Onedimensional consolidation theory
Consider the element of consolidating soil. In time dt:
· the seepage flow is dq
(q = A k i = A k dh/dz)
· the change in excess pressure is
· the thickness changes by
dH = m_{v} dzds´
It can be shown that the basic equation for onedimensional consolidation
is:
By defining the coefficient of consolidation as
this can be written:
Back to Onedimensional consolidation theory
Solutions to the onedimensional consolidation equation can be obtained by plotting the variation of with the depth in the layer at given elapsed times. The resulting curves are called isochrones. (Gk. iso = equal; kronos = time)
The figure shows a set of supposed standpipes inserted into a consolidating
layer.
Before loading, the pore pressure in the drain is zero. At the base of
each standpipe there is some initial pore pressure u= u_{o}, the
excess pore pressure = 0.
Immediately after the loading is applied the standpipes will each show an initial excess pore pressure of _{i}, thereafter the excess pore pressure will dissipate.
Click on the following time intervals to observe the changes in across the thickness of the layer with time.

Back to Isochrones
At the drainage surface, isochrones are steepest and
= 0.
At the impermeable (k = 0) base the seepage velocity is zero since V =
ki; the isochrones will therefore be at 90° to the impermeable boundary.
Between two isochrones the change in thickness in time dt,
i.e. (t_{2}  t_{1}), is dH
= m_{v} dz d,
where dz.d
is the shaded area.
Thus, the settlement at the surface of the layer is given by:
r = DH = m´_{v}
area OAB
Back to Onedimensional consolidation theory
(z,t) is excess pore pressure at depth z after
time t.
The solution depends on the boundary conditions:
The general solution is obtained for an overall (average) degree of consolidation
using nondimensional factors.
Back to Terzaghi's solution
· Degree of consolidation at depth z
· Time factor
· Drainage path ratio
The differential equation can now be written as:
If the excess pore pressure is uniform with depth, the solution is:
Putting U_{t} = r_{t}/r¥ = average degree of consolidation in the layer at time t:
Back to Terzaghi's solution
Back to Onedimensional consolidation theory
Back to Solution using parabolic isochrones
and average degree of consolidation,
the general solution is
This is valid for 0 < t < tc
At t = t_{c}, n = H =
Giving
and
U_{t} = 0.3333
Back to Solution using parabolic isochrones
and average degree of consolidation,
the general solution is
This is valid for t_{c} < t < t¥
At t = t_{c}, n = H =
Giving
and
U_{t} = 0.3333
Back to Consolidation
Back to The oedometer test
Vertical static load increments are applied at regular time intervals (e.g. 12, 24, 48 hr.). The load is doubled with each increment up to the required maximum (e.g. 25, 50, 100, 200, 400, 800 kPa). During each load stage thickness changes are recorded against time.
After full consolidation is reached under the final load, the loads are removed (in one or several stages  to a low nominal value close to zero) and the specimen allowed to swell, after which the specimen is removed and its thickness and water content determined. With a porous stone both above and below the soil specimen the drainage will be twoway (i.e. an open layer in which the drainage path length, d = H/2)
Back to Consolidation
Back to Determination of c_{v} from test results
where d = drainage path length
[d = H for oneway drainage, d = H/2 for twoway drainage]
Other appropriate timeinterval values could be used:
e.g. U_{50}, ÖT_{50}, Öt_{50} , etc.
Back to The RootTime method
On the straight line:
ÖT_{90} = AB = 0.9 x Ö(p/4)
= 0.7976
On the curved portion:
ÖT_{90} = AC = Ö0.848 = 0.9209
Thus, a line drawn through points O and C has abscissae 1.15 times greater
than those of the straight line (OB). [0.9209/0.7976 = 1.15]
After the laboratory results curve has been plotted, line OB is drawn,
followed by line OC: this crosses the laboratory curve at point (ÖT_{90},U_{90})
and locates Öt_{90}
The coefficient of consolidation is therefore:
Back to The RootTime method
On the straight line:
ÖT_{90} = AB = 0.9 x Ö(3/4) = 0.7794
On the curved portion:
ÖT_{90} = AC = Ö0.716 = 0.8462
Thus, a line drawn through points O and C has abscissae 1.086 times greater
than those of the straight line (OB). [0.8462/0.7794 = 1.086]
After the laboratory results curve has been plotted, line OB is drawn,
followed by line OC: this crosses the laboratory curve at point (ÖT_{90},U_{90})
and locates Öt_{90}
The coefficient of consolidation is therefore:
Back to Determination of c_{v} from test results
Back to Consolidation
where d = drainage path length
[d = H for oneway drainage, d = H/2 for twoway drainage]
T_{v} and t are coupled to a given degree of consolidation
Back to Calculation of settlement times
The final consolidation settlement of a layer of clay 5.0 m thick is calculated to be 280mm. The coefficient of consolidation for the loading range is 0.955 mm²/min. There is twoway drainage, upward and downward. Calculate the time required for (a) 90% consolidation settlement, (b) a settlement of 100 mm.
(a) Drainage path length, d = 5.0/2 = 2.50 m = 2500 mm
For U_{90}, T_{90} = 0.848. Then
(b) For 100 mm settlement, U_{t} = 100/280 = 0.357
and since U_{t} < 0.6, T_{v} = 0.357² x p/4
= 0.100
Then time for 100mm settlement
Back to Calculation of settlement times
A layer of clay has a thickness of 4.0 m and drains both upward and downward. A laboratory test has yielded a coefficient of consolidation for the appropriate loading range of 0.675 mm²/min. The final consolidation settlement has been calculated to be 120mm. Provide estimates of the consolidation settlement that may be expected 1yr, 2yr, 5yr and 10yr after construction.
Drainage path length, d = 2000 mm
When U_{t} < 0.6, use U_{t} = Ö(4T_{v}/p)
When U_{t} > 0.6,
c_{v} = 0.645 mm²/min = 928.8 mm²/day
time t (years) 
time t (days) 
T_{v} = c_{v}t/d² 
Ut (<0.6) 
Ut (>0.6) 
r_{c} (mm) at time t 

1  365  0.0848  0.328  39  
2  730  0.1965  0.465  56  
5  1825  0.4238  0.735  0.715  86 
10  3650  0.8475  0.900  108  
23.6  8613  2.0  0.994  119 
Back to Consolidation
Back to Consolidation
Back to Secondary compression or creep
Back to Secondary compression or creep