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1.
GEOTECHNICAL ENGINEERING - II
Engr. Nauman Ijaz
SETTLEMENT ANALYSIS
Chapter # 03
UNIVERSITY OF SOUTH ASIA
2.
SETTLEMENT
When a soil deposit is loaded, deformation will
occur due to change in stress. The total vertical
downward deformation at the surface resulting
from the load is called Settlement.
Similarly when load is decreased (e.g during
excavation) the deformation may vertically
upward and is known as swelling.
Estimate of settlement and swelling are made
using identical procedures.
3.
TYPES OF SETTLEMENT
a)
b)
Types with respect to Permanency.
Permanent settlement
Temporary Settlement.
Types with respect to Mode of Occurrence.
a)
b)
c)
a)
b)
Primary consolidation settlement (Sc)
Secondary consolidation settlement (Ss)
Immediate settlement (Si)
Types with respect to Uniformity.
Uniform settlement
Differential settlement
4.
Types with respect to Permanency
Permanent Settlement (Irreversible
settlement):
•
•
•
•
•
This type of settlement is caused due to distortion brought
about by sliding and rolling of particles under the action of
applied stresses.
The sliding and rolling will reduce the voids resulting in
reduction of volume of soil deposit.
The increased in stresses may also crush the soil particles
while alter the material and produce some settlement.
This type of settlement is permanent and undergoes
insignificant recovery upon removal of load.
Settlement due to consolidation (both primary and Secondary)
generally falls under this category.
5.
TEMPORARY SETTLEMENT
Settlement due to elastic compression
of soil are usually reversible and
recover a major part upon load
release.
Immediate settlement falls under this
category.
This settlement is generally small in
soils.
6.
TYPES WITH RESPECT TO MODE OF
OCCURRENCE:
PRIMARY CONSOLIDATION SETTLEMENT (Sc):
These settlements are time dependent or
long term settlements and completion time
varies from 1- 5 years or more.
This is also known as Primary consolidation
(i.e the settlement caused due to expulsion
of water from the pores of saturated fine
grained soils (clays).
This type of settlement is predominant in
saturated inorganic fine grained soil (clays).
7.
SECONDARY CONSOLIDATION
This is the consolidation under constant
effective stress causing no drainage.
This is very predominant in certain Organic
soils, but insignificant for inorganic soil.
This is similar to creep in concrete.
8.
IMMEDIATE SETTLMENT
This type of settlement is predominant in
coarse grained soils of high permeability
and in unsaturated fine grained soils of low
permeability.
The completion time is usually few days
(say about 7 days).
Usually this type of settlement is
completed during construction period and
is called build in settlement.
Also known as short term settlement.
9.
Settlement produced due to inadequate
shear strength of the soil mass is caused
due to bearing capacity failure of soil.
Settlement due to lateral expulsion of soils
from underneath the foundation is an
example of this category.
This settlement can not be estimated using
present knowledge of soil mechanics but
can be controlled easily by controlling
bearing capacity.
So the total settlement;
St = Si + Sc + Ss
10.
TYPES WITH RESPECT TO UNIFORMITY
UNIFORM SETTLEMENT:
When all the points settle with an equal amount,
the settlement is known as uniform settlement.
This type of settlement is possibly only under
relatively rigid foundation loaded with uniform
pressure and resting on uniform soil deposit,
which is a very rare possibility.
This type of settlement may not endanger the
structure stability but generally affects the utility of
the structure by jamming doors/windows,
damaging the utility lines ( sewer, water supply
mains etc)
11.
DIFFERENTIAL SETTLEMENT
When different parts of a structure settle by
different magnitude, the settlement is called
differential settlement.
This is very important as it may endanger
the structural stability and may cause
catastrophic failure.
If soil is granular, then differential settlement
will be 2/3 of the total maximum settlement.
In case of cohesive soil, possible differential
settlement is about 1/3 of the maximum
settlement.
12.
SETTLEMENTS OF FOUNDATIONS
NO SETTLEMENT * TOTAL SETTLEMENT * DIFFERENTIAL SETTLEMENT
Uniform settlement is usually of little consequence in a building, but
differential settlement can cause severe structural damage
16.
PLASTIC AND ELASTIC
DEFORMATION
All materials deform when subjected to an
applied load.
If all this deformation is retained when load is
released, it is said to have experienced Plastic
deformation.
Conversely if the material returns to its original
size and shape when the load is released, it is
said to have experienced elastic deformation.
Soil exhibits both plastic and elastic
deformation.
17.
ANGULAR DISTORTION
Angular distortion between two points
under a structure is equal to the
differential settlement between the
points divided by the distance between
them.
Angular distortion is also known as
Relative Rotation.
Differential settlement = ∆S = Smax – Smin
Angular Distortioin = ∆S /L
18.
LIMITS OF ALLOWABLE AND TOTAL
DIFFERENTIAL SETLLEMENT
Tolerable differential settlement of buildings, in inches,
recommended maximum values in parentheses.
CRITERION
ISOLATED FOUNDATION
ANGULAR
DISTORTION
RAFT
1/300
GREATEST
DIFFERENTIAL
SETTLEMENT
1³/4 (11/2) (CLAYS)
1³/4
(1) (SANDS)
MAXIMUM
SETTLEMENT
CLAYS
3 (21/2)
3- 5 (21/2 - 4)
SANDS
2 (11/2)
2-3 (11/2 - 21/2)
19.
For normal structures with isolated foundation,
total settlements up to 50mm and differential
settlement between adjacent columns up to
20mm is acceptable.
According to Euro Code 7, the maximum
acceptable relative rotation for open frames
and load bearing or continuous brick wall are
likely to range from about 1/2000 to about
1/300 to prevent the occurrence of a
serviceability limit state in the structure.
According to Bjerrum safe limit to avoid
cracking in the pannel walls of frame structure
( partition wall) = 1/500
20.
Large total and differential settlement may be
acceptable provided the relative rotations remains within
acceptable limits and provided total settlement do not
cause problems with the services entering the structure
or cause titling.
Average maximum settlement of Structures on Perma
frost. (a layer of soil beneath the surface that remains
frozen through out the year)
Structure
Average Max Settlement (mm)
Reinforced Concrete
150 at 40mm/year
Masonry ,Precast Concrete
200 at 60mm/year
Steel frames
250 at 80mm/year
Timber
400 at 129mm/year
21.
CAUSES OF SETTLEMENT
Following are the major causes of
settlement;
Changes in Stress due:
a) Applied structural load or excavation
.
b) Movement of ground water table
.
c) Vibrations due to Machines, Earth quake.
Desiccation due to surface drying and/or
plants life. (Desiccation = Removal of water from soil
Loss (evaporation) of water / effective stress(inter-granular stresses)
increase /Mass shrinkage will start ) ( Reason= High fines content,
Volume of water is the direct function of shrinkage)
22.
Changes due to structure of soil.
Adjacent excavation.
Mining Subsidence.
Swelling and Shrinkage.
Lateral expulsion of soil.
Land slides.
23.
Remedial Measures
a)
b)
Philosophy of remedial measures is to;
Reduce or eliminate settlement.
Design structure to with stand the Settlement.
To reduce or eliminate stresses following
considerations can be followed;
Reduce Contact pressure.
Reduce Compressibility of soil deposits using
various ground improvement techniques (
Stabilization, pre-compression, vibro-flotation etc).
Remove soft compressible material such as peat,
muck. etc
24.
Built slowly on cohesive soils to avoid
lateral expulsion of soil mass, and to give
time for pore pressure dissipation.
Consider using deep foundation (piles or
piers).
Provide lateral restraint against lateral
expulsion.
To achieve uniform settlement one can
resolve to;
Design of footing for uniform pressure.
Use artificial cushion underneath the less
settling foundation parts of the structure.
25.
Built different parts of foundation of
different weight on different soil at
different depths.
Built the heavier parts of structure first
(such as towers) and lighter parts later.
26.
CONSOLIDATION
(OEDOMETER) TEST
This test is performed to determine the magnitude and
rate of change in volume of a laterally confined soil
specimen undergoes when subjected to different vertical
pressure
To compute the consolidation settlement in a soil we
need to know stress- strain properties. (i.e relationship
between( σZ & εz ).
This normally involves bringing the soil sample to the
laboratory, subjecting it to a series of loads and
measuring corresponding settlements.
27.
This test is known as consolidation test or
Oedometer test.
We mostly interested in engineering properties
of natural soils as they exist in the field, so
consolidation tests are usually performed on
high quality Undisturbed samples.
It is also important for samples that were
saturated in the field to remain so during
storage and testing.
If the sample is allowed to dry, a process we
call Desiccation, negative pore pressure will
develop and may cause irreversible changes in
the in the soil.
29.
The test begins by applying vertical normal load P.
It produces a vertical effective stress of;
σ’Z = P/A - U
Where;
σ’Z
= Vertical effective stress.
P
A
U
= Applied Load.
= Cross sectional area of soil specimen.
= pore water pressure inside the soil specimen.
The water bath barely covers the specimen , so the pore
water pressure is very small as compared to the vertical
stress and thus may be ignored;
σ’Z = P/A
30.
The vertical strain εz is noted by monitoring the
dial gage, for each corresponding increase in load.
εz =
Change in Dial Gage Reading
Initial height of the sample
Increase the load to some higher value and allow the soil to
consolidate again, thus obtaining a second value of (σZ ,εz).
This process will continues until we have reached the
desired peak vertical stress; from this loading sequence we
obtain the loading curve ABC.
We then incrementally unload the sample and allow it to
rebound thus producing unloading curve CD. Shown in the
figure presented in the next slide.
Data is plotted on logarithmic scale.
31.
AB representing the Recompression Curve
BC representing the Virgin Curve ,CD representing the Rebound Curve
32.
SIGNIFICANCE
The consolidation properties determine from
the consolidation test are used to estimate
the magnitude of both primary and
secondary settlement.
The consolidation parameters we find out;
1. Compression Index (Cc)
.
2. Recompression Index (Cr)
.
3. Coefficient of Volume change (mv) .
4. Pre-consolidation pressure
.
5.
6.
e – field Density
Coefficient of Consolidation (Cv)
.
.
33.
The first 4 parameters can be determined
from e ~ σ’Z graph.
Field density is determined by;
γd = Gs γw
1+e
34.
Cc = Compression Index
σ’Z0 = Initial vertical effective stress
σ’Zf = Final vertical effective stress
Cc = ∆e/ log(σ’Zf / σ’Z0)
35.
Cr =
σ’C =
Recompression Index
Pre- consolidation
pressure
Pre-consolidation pressure is
the maximum pressure that
the soil has been subjected in
the past. It varies with depth
and is used to identify over
consolidation of specimen.
36.
NORMALLY CONSOLIDATED (NCC)
A soil is said to be normally
consolidated when;
σ’C = σ’
0Z
σ’C =
σ’
0Z
Pre-consolidation Pressure.
= Present effective overburden pressure.
37.
OVER CONSOLIDATED (OCC)
The soil is said to be over consolidated
when;
σ’C > > > σ’0Z
1.
2.
3.
4.
This shows that soil has been subjected to some over burden
pressure in the past which has been removed.
This over burden pressure may be due to;
Snow loading
Past Structure which now has been removed.
Level of ground is lowered.
Lowering of the ground water table.(In the past ground water
table is high)
38.
OVER CONSOLIDATED RATIO (OCR)
Over consolidated ratio is defined
as ratio of pre-consolidation
pressure to present overburden
pressure.
OCR = σ’C
/ σ’
0Z
Greater the value of OCR the more
the soil is consolidated
40.
εf
As;
we have;
=
εL
∆H/H =
∆h/h
∆V/Vo =
∆e/ (1+eo)
∆H/H = ∆e/ (1+eo)
∆H
= H × ∆e/ (1+eo)
Cc = ∆e / log(σ’Zf / σ’Z0)
∆e = Cc H log(σ’Zf / σ’Z0)
∆H = Cc H log(σ’Zf / σ’Z0)…….(A)
(This is the equation for NCC soil)
1+eo
σ’Z0 = Initial Pressure or present Over burden pressure.
σ’Zf = Final Pressure.
41.
FOR OCC
For over consolidated soil
we have two cases;
CASE # 01
When ;
σ’Z0
<
∆H = Cr
σ’zf < σ’C
H log(σ’Zf / σ’Z0)…(B)
1 + eo
Cr = Coefficient of
Recompression
42.
Cr = Coefficient of
Recompression
CASE # 02
When;
σ’Z0 < σ’C < σ’zf
∆H
=
Cr H log(σ’c / σ’Z0 ) + Cc H log(σ’Zf / σ’Z0)
1+eo
1+eo
43.
OVER CONSOLIDATION MARGIN
The σ’C values from the laboratory represent
the pre-consolidation stress at the sample
depth.
However we have to compute σ’C at other
depths. To do so compute the overconsolidation
margin σ’m, using σ’z0 at the sample depth and
the following equation;
σ’m
= σ’C - σ’z0
44.
EXAMPLE
A 3m deep compacted fill is to be placed
over the profile shown in the figure. A
consolidated test on a sample from point A
produced the following results;
Cc = 0.40
Cr = 0.08
eo = 1.10
σ’ = 70.0 Kpa
Compute the ultimate consolidation
settlement due to weight of this fill.
C
46.
σ’Zf
σ’Zf
σ’Zf
=
σ’Z0
+
γfillHfill
= σ’Z0 + (19.2KN/m³) (3.0m)
=
σ’Z0
+
57.6Kpa
Compute the initial vertical stress at sample location.
σ’Z0 = Σ γH - U
σ’Z0 = (18.5KN/m³) (1.5m) + (19.5KN/m³) (2.0m)
+ (16.0KN/m³) (4.0m) – (9.8KN/m³) (6.0m)
σ’Z0 = 72.0 Kpa.
At the sample σ’C ͌ σ’Z0
Clay is Normally consolidated
Cc/ (1+e0) = 0.40/(1+1.10) = 0.190
For sand ; Cc/ (1+e0) = 0.008 (this value is taken from table
corresponding to the value of Dr)
47.
At Mid Of layer
Layer
H(m)
σ’Z0
σ’Zf
(Kpa)
Cc/
(1+e0)
δc (mm)
(Kpa)
1
1.5
13.9
71.5
0.008
8
2
2.0
37.4
95.0
0.008
6
3
3.0
56.4
114.0
0.190
174
4
3.0
75.0
132.6
0.190
141
5
4.0
96.7
154.3
0.190
154
δc, ult
483mm
Using the equation for NCC
δc = 480mm Round off
∆H = Cc
H log(σ’Zf / σ’Z0)…(B)
1 + eo
48.
EXAMPLE #02
An 8.5m deep compacted fill is to be
placed over the soil profile shown in
the figure. Consolidation test on
samples from point A and B produced
the following results
Sample A
Sample B
Cc
0.25
0.20
Cr
0.08
0.06
e0
0.66
0.45
σ’C
101kpa
510kpa
49.
σ’Zf
σ’Zf
σ’Zf
=
σ’Z0
+
γfillHfill
= σ’Z0 + (20.3KN/m³) (8.5m)
=
σ’Z0
+
172.6 Kpa
Compute the initial vertical stress at sample location.
σ’Z0 = Σ γH - U
σ’Z0 = (18.3KN/m³) (2.0m) + (19.0KN/m³) (2.0m)
– (9.8KN/m³) (2.0m)
σ’Z0 = 55.0 Kpa.
σ’Zf = σ’Z0 + 172.6 Kpa
σ’Zf = 55.0 Kpa + 172.6 Kpa = 227.6Kpa
σ’Z0 < σ’C < σ’zf
55kpa < 101Kpa < 227.6 Kpa
(Over consolidated Case#2)
We will use this equation
∆H = Cr H log(σ’c / σ’Z0 ) +
1+eo
Cc H log(σ’Zf / σ’Z0)}
1+eo
50.
σ’m
σ’m
σ’c - σ’Z0
= 101 – 55 = 46Kpa
=
Therefore σ’c at any depth in the stiff silty clay stratum is equal to
σ’z0 + 46Kpa.
For sample B:
σ’Z0 = Σ γH - U
σ’Z0 = (18.3KN/m³) (2.0m) + (19.0KN/m³) (7.0m) + (19.5KN/m³)
(10m) – (9.8KN/m³) (17m)
σ’Z0 = 198.0 Kpa.
σ’Zf = σ’Z0 + 172.6 Kpa
σ’Zf = 198.0Kpa + 172.6 Kpa
σ’Zf = 370.6 Kpa
σ’Z0
< σ’zf
< σ’
C
(Over consolidated Case#1)
198kpa < 370Kpa < 510 Kpa
∆H =
Cr H log(σ’Zf / σ’Z0)
1 + eo
52.
SETTLEMENT DUE TO SECONDARY
CONSOLIDATION
Secondary consolidation is also known as creep settlement and it is
actually a continuation of the volume change that started during
primary consolidation.
This settlement takes place at constant effective stress (i.e after all the
pore pressure has been dissipated).
Co-efficient of secondary consolidation is defined as the vertical strain
which occurs during one log cycle of time following completion of
primary consolidation and is given by;
Cα = ∆H/Ho = (D1 – D2)/Ho
where;
∆H = the change in sample height during consolidation following the
completion of primary consolidation.
Ho = Consolidation sample height under a given pressure.
D1 & D2 = dial gage reading along secondary compression curve
against any time t1 and t2 where t2 = 10t1
53.
Settlement due to secondary
consolidation is now computed as
follow;
Sc = (Cα) (Ho) (log tsc/tp)
Where;
Ho = thickness of compressible stratum.
tp & tsc = the time for completion of primary
consolidation and time for which secondary
consolidation is to be calculated.
In sand, settlement caused by secondary
compression is negligible, but in peat, it is
very significant.
54.
FOOTINGS ON SANDS AND GRAVEL
In theory the method used to predict
settlement of spread footings on clays and
silts also could be used for sands and
gravels but to use this method we need to
compute Cc and Cr in these soils, which is
very difficult or impossible because of the
difficulties in obtaining undisturbed
samples.
Because of this limitation we will use a
different approach in computing
settlement in sands and gravels.
55.
Settlement in sands and gravel are
generally much smaller than those in
soft or medium clays.
The most common method used is that
developed by Schmertmann.
56.
SCHMERTMANN'S METHOD
For Square Footing
Σ
Σ
For Strip Footing
q = P/A + γCDf - U
q’o =γsoilDf
57.
SCHMERTMANN'S METHOD
C1 = Correction factor for footing depth. .
C2 = Correction factor for Creep.
t = Time in years
for which
settlement is
required
Modulus of Elasticity Es
58.
Square
Strip
Modified triangular vertical strain influence factor distribution
diagram
The strain influence factor accounts for both the distribution of stresses
below the footing and nonlinear soil behavior immediately below the
footing. The peak value is shown in the graph.
59.
For Square footing (L/B =1)
Iz = 0.1 + (Zc/0.5B) ( Ipeak – 0.1) (For Z < 0.5B)
Iz = (2/3) Ipeak (2 – Zc/B) (For Z> 0.5B)
For a strip Footing (L/B > 10)
Iz = 0.2 + (Zc/B) ( Ipeak – 0.2) (For Z < B)
Iz = (1/3) Ipeak (4 – Zc/B) (For Z > B)
60.
IMPORTANT CONSIDERATIONS
Primarily used to estimate immediate settlement
of foundations in sand.
Specially useful when CPT data are available.
Results are compatible with field
measurements.
Based on analysis of vertical strain distribution
with a linear elastic half space subjected to a
uniform pressure
61.
Example
A strip footing 2.0 × 23 m is subjected to load
450KN/m. The depth of footing is 2m. There is a
deep deposit of sand of unit weight 16KN/m³, the
water table is deep below the surface.
The variation of cone penetration
resistance (qc) with depth (z) is shown in
the next figure.
62.
SOLUTION
As it is a strip footing;
For strip footing;
Σ
Net foundation base pressure = qn = q – q’o
Total foundation pressure = q = P/A + γCDf – U = (450/2) + 2×23.5 – 0 = 272KN/m²
Effective overburden pressure at foundation level = q’o = γsoilDf = 16 × 2 = 32KN/m²
Net foundation base pressure = qn = q – q’o = 272 – 32 = 240 KN/m²
63.
Now we calculate;
C1 = 1 – 0.5 (32/240) = 0.0933
Now we calculate C2;
C2 = 1 + 0.2 log(5/0.1) = 1.339
64.
Now we calculate Ipeak;
= 0.5 + 0.1 undr Root(240/64)
=0.7015
P’o = 2× 16 + 2× 16 = 64KN/m²
As we know that for a strip Footing (L/B > 10).
Iz = 0.2 + (Zc/B) ( Ipeak – 0.2) (For Z < B)
Iz = (1/3) Ipeak (4 – Zc/B)
(For Z > B)
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