Correction for construction Settlement

1. Boundary Condition for
Solution
& Settlement Correction of
1D-Consolidation
By
Sanchari Halder
Submitted to
Professor Dr. Myoung Soo
Won

2. Contents
 Boundary Condition for Solution of 1D-
Consolidation.
 Correction of Settlement for Construction
Period:

3. We can analyze the strain of a saturated clay layer subjected to a stress increase .Considering the case where a layer of saturated clay of thickness H
that is confined between two layers of sand is being subjected to an instantaneous increase of total stress of Δσ. This incremental total stress will be
transmitted to the pore water and the soil solids. This means that the total stress, Δσ, will be divided in some proportion between effective stress and
pore water pressure.
It follows this equation:
Because clay has a very low hydraulic conductivity and water is incompressible as compared with the
soil skeleton, at time t = 0, the entire incremental stress, Δσ, will be carried by water (Δσ= Δ u) at all
depths (Figure b). None will be carried by the soil skeleton—that is, incremental effective stress (Δσ´) =
0. After the application of incremental stress, Δσ, to the clay layer, the water in the void spaces will start
to be squeezed out and will drain in both directions into the sand layers. By this process, the excess
pore water pressure at any depth in the clay layer gradually will decrease, and the stress carried by the
soil solids (effective stress) will increase. Thus, at time 0 < t < ∞,
However, the magnitudes of Δσ´ and Δu at various depths will change (Figure c), depending on the
minimum distance of the drainage path to either the top or bottom sand layer.
Theoretically, at time t= ∞, the entire excess pore water pressure would be dissipated by drainage from
all points of the clay layer; thus, Δu=0. Now the total stress increase, Δσ, will be carried by the soil
structure (Figure d). Hence, Δσ= Δσ´
This gradual process of drainage under an additional load application and the associated transfer of
excess pore water pressure to effective stress cause the time-dependent settlement in the clay soil
layer.

4. Figure 1(a) shows a layer of clay of thickness 2Hdr (Note: Hdr
length of maximum drainage path) that is located between two
highly permeable sand layers.
If the clay layer is subjected to an increased pressure of Δσ,
the pore water pressure at any point A in the clay layer will
increase. For one-dimensional consolidation, water will be
squeezed out in the vertical direction toward the sand layer.
Figure 1(a): Clay layer Undergoing consolidation;

5. Figure 1(b) shows the flow of water through a prismatic element at A. For the soil element shown,
Rate of outflow of water – Rate of inflow of water = Rate of volume change
……….(1)
……….(2)
1 & 2
……….(3)
;
According to Laplace’s Equation:
;

6. During consolidation, the rate of change in the volume of the soil element is equal to the rate of
change in the volume of voids. Thus,
But (assuming that soil solids are incompressible)
and
…….(4)
Substitution for and Vs in Eq. (4) yields, ……… (5)
Where e0 initial void ratio.
Combining Eqs. (3) and (5)
gives,
……… (6)

7. The change in the void ratio is caused by the increase of effective stress (i.e., a decrease of
excess pore water pressure). Assuming that they are related linearly, we have
………(7)
Combining Eqs. (6) and (7)
gives,
Where, mv = coefficient of volume compressibility =
Or, Where, cv = coefficient of consolidation
=
Thus,
…(8)

8. Eq. (8)
is the basic differential equation of Terzaghi’s consolidation
theory and can be solved with the following boundary
conditions:
Boundary Condition for Solving 1D-Consolidation
Equation

9. 1. Initial Condition,
at time t = 0 ; u = Δσ
2. Boundary Conditions
at any time
where z = 0 ; u = 0
For Double Drainage,
z = 2H ; u = 0
The Initial & Boundary Conditions:

10. The solution yields, …..(9)
The time factor is a non dimensional number. Because consolidation progresses by the
dissipation of excess pore water pressure, the degree of consolidation at a distance z at
any time t is
where, uz= excess pore water pressure at time
t.
…….. (10)

11. Equations (9) and (10) can be
combined to obtain the degree of consolidation at any depth z.
This is shown in Figure 2.
The average degree of consolidation for the entire depth
of the clay layer at any time t can be written from Eq. (10)
as
Figure2: Variation of Uz withSubstitution of the expression for excess pore water pressure uz
given in Eq. (9) into Eq. (11) gives
……. (11)

12. The variation in the average degree of
consolidation with the non dimensional time
factor Tv , is given in Figure 3, which represents
the case where u0 is the same for the entire
depth of the consolidating layer.
The values of the time factor and their
corresponding average degrees of consolidation
for the case presented in Figure 2 may also be
approximated by the following simple
relationship:
Figure
3:
Sivaram and Swamee (1977) gave the following
equation for U varying from 0 to 100%:

13. Example Problem 1:

14. Solution:

16. Correction of Settlement for Construction
Period:
 In practice, structural loads are
applied to the soil not
instantaneously but over a period
of time.
 Terzaghi proposed an empirical
method of correcting the
instantaneous time–settlement
curve to allow for the construction
period.
Load
Settlement
Loading curve
Instanous
Curve
Effective Construction
Time
Time

17. S
R
Q
P
½t1 ½ tc tc
Correction for construction period:
 The net load(P′) is the gross load less the weight of
soil excavated, and the effective construction
period(tc) is measured from the time when P′ is zero.
 It is assumed that the net load is applied uniformly
over the time tc and that the degree of consolidation
at time tc is the same as if the load P′ had been acting
as a constant load for the period tc/2.
 Thus the settlement at any time during the
construction period is equal to that occurring for
instantaneous loading at half that time.
Load
O
Settlement
Loading curve
Instanous
Curve
Currected Curve
½ tc
t1
T
Effective Construction
Time
P1
P´
Time
S
c
Sc
1

18. Example Problem 2:
20 kn/m3
γsd = 17 kn/m3
γsw = 19 kn/m3
γcw=20 kn/m3
γw =9.8 kn/m3Cv ,=1.26 m2/year
Cc =.32

19. Example Problem:
Required
Equation:
Settlement
,
σ1
20 kn/m3
γsd = 17 kn/m3
γsw = 19 kn/m3
γcw=20 kn/m3
γw =9.8 kn/m3
Cv ,=1.26 m2/year
Cc =.32

20. Solution:
20 kn/m3
γsd = 17 kn/m3
γsw = 19 kn/m3
γcw=20 kn/m3
γw =9.8 kn/m3
3
m
Cv ,=1.26 m2/year
Cc =.32

21. Solution:
Cv ,=1.26 m2/year
Cc =.32

22. Correction for Settlement:
Settlement for Instantaneous
Loading,
Scf =182 mm
Correction of Settlement for construction
period,
Sc =61 mm
182
mm
61
mm

23. Thank you all….