ce5101 lecture 5 - settlements and elastic stress distribution (sep 2013)
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CE5101 Lecture 5 - Settlements and Elastic Stress Distribution (SEP 2013)TRANSCRIPT
CE 5101 Lecture 5 CE 5101 Lecture 5 CE 5101 Lecture 5 CE 5101 Lecture 5 –––– SETTLEMENTS SETTLEMENTS SETTLEMENTS SETTLEMENTS and Stress Distributionand Stress Distributionand Stress Distributionand Stress Distribution
Sep 2013
Prof Harry Tan
9/11/2013 1
Outline
9/11/2013 2
• Foundation Requirements
• Elastic Stress Distribution
• Concept of Effective Stress
• Settlements of Soils - Immediate, Delayed, and Creep Compression
• Hand Calculations
• SPREADSHEET Calculations (UNISETTLE)
• Finite Element Analysis (PLAXIS)
Requirements for Foundation Design
• Adequate Safety (degree of utilisation of soil strength)
• Acceptable Deformations(Movements and Settlements limits)
9/11/2013 3
What is Adequate Safety(Lambe and Whitman Pg 196)
AA
-2.000 -1.000 0.000 1.000 2.000 3.000 4.000 5.000 6.000 7.000 8.000 9.000 10.000 11.000 12.000
0.000
1.000
2.000
3.000
4.000
5.000
6.000
7.000
8.000
9.000
Plastic Points
Plastic Mohr-Coulomb point Tension cut-off point
E
Stress
Strain
2c
Failure Load = 866 kPa
E = 34.48 MN/m2
c=167.6 kN/m2
νννν = 0.3
Yielded
Zone
9/11/2013 4
0 200 400 600 800 1.00E+03
-0.35
-0.30
-0.25
-0.20
-0.15
-0.10
-0.05
0.00
Footing Load (kPa)
Footing CL Settlement [m]
Chart 1
Point A
Load vs Settlement Behaviour of Flexible Footing
First
Yield Local Shear
Failure
General Shear
Failure
Settlement (m)
Load (kPa)
9/11/2013 5
0 0.05 0.10 0.15 0.20 0.25 0.30 0.35
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
Settlement (m)
Factor of Strength Reduction
Chart 2
Point A
Factor of Safety as Measure of Degree of Utilisation of Soil Strength
FS = 4.338; q = 200 kPa; qult = 868 kPa
FS = 2.164; q = 400 kPa; qult = 866 kPa
FS = 1.291; q = 670 kPa; qult = 865 kPa
9/11/2013 6
What is Allowable Settlements
Effects on:
• Appearance of Structure
• Utility of Structure
• Damage to Structure
9/11/2013 7
Types of Settlements
ρUniform settlement
Non-uniform settlement
Angular distortion = δδδδ/L
δ
L
L
δ
Angular distortion =
δδδδ/(L/2)
9/11/2013 8
Allowable Settlements
Controlled by type of settlements (Sowers 1962)
• Total settlements
• Tilting
• Differential settlements
9/11/2013 9
Limiting Angular Distortions (Bjerrum, 1963)
1/150 - Considerable cracking in panel and brick walls, safe limit
for flexible brick wall h/l<1/4, limit where structural damage of
general buildings is to be feared
1/250 - Limit where tilting of high rigid buildings become visible
1/300 - Limit where first cracking in wall panel expected,
difficulties with overhead cranes expected
9/11/2013 10
1/500 - Safe limit for buildings where cracking is not
permissible
1/600 - Limit of danger for frames with diagonals
1/750 - Limits where difficulties with machinery sensitive to
settlements are to be feared (high tech plants)
Limiting Angular Distortions (Bjerrum ,1963)
Max Distortion (δδδδ/L) vs Max Differential Settlement :
1/500 vs 25 mm
1/300 vs 45 mm
1/200 vs 70 mm
1/100 vs 150 mm
For Stiff Footing,
Max Diff Sett = 1/4 Max Sett
For Flexible Footing,
Max Diff Sett = 1/2 Max Sett
Typical Foundation Design,
Max Diff Sett < 25 mm
9/11/2013 11
9/11/2013 12
Limiting Angular Distortions (Bjerrum ,1963)
200 mm
350 mm
100 mm
• Stiff Footing
max diff sett =1/4
max sett
• Flexible Footing
max diff sett=1/2
max sett
9/11/2013 13
• ER2010 – Review Paper by
Ed Cording et al –
Assessment of excavation-
induced building damage
• Damage is due to combined
effects of:
• Angular distortion; β• Lateral strain
• Bending strain
Basic Soil Phase Relationships
Air
Water
Solids
MASSVOLUME
0
Mw Mt
Ms
Mw
Va
Vw
Vs
Vv
Vt
Densities (kg/m3)
Total, ρρρρt = Mt/Vt
Dry, ρρρρd = Ms/Vt
Solids, ρρρρs = Ms/Vs
Water, ρρρρt = Mw/Vw
Saturated, ρρρρsat
= (Ms+Mw+ ρρρρw Va)/Vt
Ratios
Water content, w = Mw/Ms
Void ratio, e = Vv/Vs
Porosity, n = Vv/Vt
Degree of saturation = Vw/Vv9/11/2013 14
Basic Soil Phase RelationshipsDefine Vs=1, Vv=e
eS ρρρρw = w ρρρρs
Air
Water
Solids
MASSVOLUME
0ρρρρt
ρρρρs
Mw
(1-e)S
eS
1
e
1+e
Densities (kg/m3)
Total, ρρρρt = ρρρρs (1+w)/(1+e)
= ρρρρd (1+w)
Dry, ρρρρd = ρρρρs /(1+e)
Saturated, ρρρρsat
= (Ms+Mw+ ρρρρw Va)/VtRatios
Water content, w = ρρρρt/ ρρρρd - 1
Void ratio, e = ρρρρs/ρρρρd - 1
Porosity, n = e/(1+e)
Degree of saturation = S = wρρρρs / eρρρρw 9/11/2013 15
Common soil mineral densities
Mineral Type Solid Density, kg/m3
Calcite 2800
Quartz 2670
Mica 2800
Pyrite 5000
Kaolinite 2650
Montmorillonite 2750
Illite 27009/11/2013 16
Typical range of saturated densities
Soil Type Saturated Density, kg/m3
Sands;gravels 1900-2300
Silts 1500-1900
Soft Clays 1300-1800
Firm Clays 1600-2100
Peat 1000-1200
Organic Silt 1200-1900
Granular Fill 1800-22009/11/2013 17
The Textbooks on Foundations — They come no better
This is one of the few showing
more than one soil layer
9/11/2013 18
The Reality — With a bit of needed “W” add-on
9/11/2013 19
The Reality — Getting closer, at least
9/11/2013 20
So, with this food for thought,
on to the Fundamental
Principles
9/11/2013 21
Determining the effective stress is the key to geotechnical analysis
• The not-so-good method:
)1(' iwt −−= γγγ
9/11/2013 22
h∆=∆ '' γσ
)'(' hz ∆∑= γσ
γ’ = buoyant
unit weight
It is much better to determine, separately, the total stress and the pore pressure. The effective stress is then the total stress minus the pore pressure.
uz −= σσ '
)( hz ∆∑= γσ
9/11/2013 23
Determining pore pressure
u = γw h
The height of the column of water (h; the head representing the water pressure) is
usually not the distance to the ground surface nor, even, the distance to the
groundwater table. For this reason, the height is usually referred to as the “phreatic
height” or the “piezometric height” to separate it from the depth below the groundwater
table or depth below the ground surface.
The pore pressure distribution is determined by applying the facts that
(1) in stationary conditions, the pore pressure distribution can be assumed to be linear
in each individual soil layer
(2) in pervious soil layers that are “sandwiched” between less pervious layers, the pore
pressure is hydrostatic (that is, the vertical gradient is unity)
SAND
Hydrostatic distribution
CLAY
Non-hydrostatic distribution,
but linear
SAND
Hydrostatic distribution
Artesian phreatic head
GW
DEPTH
PRESSURE
9/11/2013 24
Distribution of stress below a a small load area
9/11/2013 25
)()(0
zLzB
LBqqz
+×+
××=
The 2:1 method
The 2:1-method can only be used for distributions directly under the center
of the footprint of the loaded area. It cannot be used to combine (add)
stresses from adjacent load areas unless they all have the same center. it is
then only applicable under the area with the smallest footprint.
Boussinesq Method for stress from a point load
9/11/2013 26
2/522
3
)(2
3
zr
zQqz
+=
π
Newmark’s method for stress from a loaded area
Newmark (1935) integrated the Boussinesq equation over a finite area and
obtained a relation for the stress under the corner of a uniformly loaded
rectangular area, for example, a footing
9/11/2013 27
π40
CBAIqqz
+×=×=
2222
22
1
12
nmnm
nmmnA
+++
++=
1
222
22
++
++=
nm
nmB
−++
++=
2222
22
1
12arctan
nmnm
nmmnC
m = x/z
n = y/z
x = length of the loaded area
y = width of the loaded area
z = depth to the point under the corner
where the stress is calculated
(1)
• Eq. 1 does not result in correct stress values near the ground surface. To represent the stress near the ground surface, Newmark’s integration applies an additional equation:
9/11/2013 28
π
π
40
CBAIqqz
−+×=×=
For where: m2 + n2 + 1 ≤≤≤≤ m2 n2
(2)
Stress distribution below the center of a square 3 m wide footing
0 20 40 60 80 100
-6
-4
-2
0
STRESS (KPa)
DE
PTH
(m
)
0.01 0.10 1.00 10.00
0.00
0.05
0.10
0.15
0.20
0.25
m and n (m = n)
INF
LU
EN
CE
FA
CTO
R,
I
9/11/2013 29
Eq. (1)
Eq. (2)
Eq. (1)
Eq. (2)
Comparison between Boussinesq, Westergaard, and 2:1 distributions
0
1
2
3
4
5
0 25 50 75 100
STRESS (%)D
EP
TH
(d
iam
ete
rs)
Boussinesq
Westergaard
2:1
0
1
2
3
4
5
0 25 50 75 100
SETTLEMENT (%)
DE
PT
H (d
iam
ete
rs)
Boussinesq
Westergaard
2:1
9/11/2013 30
0
1
2
3
4
5
0 25 50 75 100
STRESS (%)D
EP
TH
(d
iam
ete
rs)
2:1
Westergaard
Boussinesq
0
1
2
3
4
5
0 25 50 75 100
SETTLEMENT (%)
DE
PT
H (d
iam
ete
rs)
Boussinesq
Westergaard
2:1
9/11/2013 31
0
1
2
3
4
5
0 25 50 75 100
STRESS (%)D
EP
TH
(d
iam
ete
rs)
2:1
Westergaard
Boussinesq
Characteristic
Point; 0.37b
from center
0
1
2
3
4
5
0 25 50 75 100
SETTLEMENT (%)
DE
PT
H (d
iam
ete
rs)
Boussinesq
Westergaard
2:1Characteristic
Point; 0.37b
from center
Below the characteristic point, stresses for flexible and stiff footings are equal9/11/2013 32
Now, if the settlement distributions are so
similar, why do we persist in using
Boussinesq stress distribution instead of
the much simpler 2:1 distribution?
Because a footing is not alone in this world;
near by, there are other footings, and fills,
and excavation, etc., for example:
9/11/2013 33
The settlement imposed
outside the loaded
foundation is often critical
0
1
2
3
4
5
0 25 50 75 100
SETTLEMENT (%)
DE
PT
H (d
iam
ete
rs)
Boussinesq
Outside Point Boussinesq
Center Point
Loaded
area
9/11/2013 34
Calculations using Boussinesq distribution can be used to determine how stress
applied to the soil from one building may affect an adjacent existing building.
EXISTING
ADJACENT
BUILDING
NEW
BUILDING
WITH LARGE
LOAD OVER
FOOTPRINT
AREA
2 m2 m 4 m
0
5
10
15
20
25
30
0 20 40 60 80 100
STRESS (%)
DE
PT
H (m
)STRESSES ADDED
TO THOSE UNDER
THE FOOTPRINT OF
THE ADJACENT
BUILDING
STRESSES
UNDER THE
FOOTPRINT
OT THE
LOADED
BUILDING
STRESSES
UNDER AREA
BETWEEN THE
TWO BUILDINGS
9/11/2013 35
Calculation of Stress Distribution
Depth σσσσ0 u0 σσσσ’0 σσσσ1 u1 σσσσ’1
(m) (KPa) (KPa) (KPa (KPa) (KPa) (KPa)
Layer 1 Sandy silt ρρρρ = 2,000 kg/m3
0.00 0.0 0.0 0.0 30.0 0.0 30.0
GWT 1.00 20.0 0.0 20.0 48.4 0.0 48.4
2.00 40.0 10.0 30.0 66.9 10.0 56.9
3.00 60.0 20.0 40.0 85.6 20.0 65.6
4.00 80.0 30.0 50.0 104.3 30.0 74.3
Layer 2 Soft Clay ρρρρ = 1,700 kg/m3
4.00 80.0 30.0 50.0 104.3 30.0 74.3
5.00 97.0 40.0 57.0 120.1 43.5 76.6
6.00 114.0 50.0 64.0 136.0 57.1 79.0
7.00 131.0 60.0 71.0 152.0 70.6 81.4
8.00 148.0 70.0 78.0 168.1 84.1 84.0
9.00 165.0 80.0 85.0 184.2 97.6 86.6
10.00 182.0 90.0 92.0 200.4 111.2 89.2
11.00 199.0 100.0 99.0 216.6 124.7 91.9
12.00 216.0 110.0 106.0 232.9 138.2 94.6
13.00 233.0 120.0 113.0 249.2 151.8 97.4
14.00 250.0 130.0 120.0 265.6 165.3 100.3
15.00 267.0 140.0 127.0 281.9 178.8 103.1
16.00 284.0 150.0 134.0 298.4 192.4 106.0
17.00 301.0 160.0 141.0 314.8 205.9 109.0
18.00 318.0 170.0 148.0 331.3 219.4 111.9
19.00 335.0 180.0 155.0 347.9 232.9 114.9
20.00 352.0 190.0 162.0 364.4 246.5 117.9
21.00 369.0 200.0 169.0 381.0 260.0 121.0
Layer 3 Silty Sand ρρρρ = 2,100 kg/m3
21.00 369.0 200.0 169.0 381.0 260.0 121.0
22.00 390.0 210.0 180.0 401.6 270.0 131.6
23.00 411.0 220.0 191.0 422.2 280.0 142.2
24.00 432.0 230.0 202.0 442.8 290.0 152.8
25.00 453.0 240.0 213.0 463.4 300.0 163.4
0
5
10
15
20
25
0 100 200 300 400 500
STRESS (KPa)
DE
PT
H (
m)
SAND
CLAY
SAND
9/11/2013 36
HYDROSTATIC PORE PRESSURE DISTRIBUTION
Calculation of Stress DistributionSTRESS DISTRIBUTION (2:1 METHOD) BELOW CENTER OF EARTH FILL (Calculations by means of UniSettle)
ORIGINAL CONDITION (no earth fill) FINAL CONDITION (with earth fill and artesian pore pressure in sand)
Depth σσσσ0 u0 σσσσ’0 σσσσ1 u1 σσσσ’1
(m) (KPa) (KPa) (KPa (KPa) (KPa) (KPa)
Layer 1 Sandy silt ρρρρ = 2,000 kg/m3
0.00 0.0 0.0 0.0 30.0 0.0 30.0
GWT 1.00 20.0 0.0 20.0 48.4 0.0 48.4
2.00 40.0 10.0 30.0 66.9 10.0 56.9
3.00 60.0 20.0 40.0 85.6 20.0 65.6
4.00 80.0 30.0 50.0 104.3 30.0 74.3
Layer 2 Soft Clay ρρρρ = 1,700 kg/m3
4.00 80.0 30.0 50.0 104.3 30.0 74.3
5.00 97.0 40.0 57.0 120.1 43.5 76.6
6.00 114.0 50.0 64.0 136.0 57.1 79.0
7.00 131.0 60.0 71.0 152.0 70.6 81.4
8.00 148.0 70.0 78.0 168.1 84.1 84.0
9.00 165.0 80.0 85.0 184.2 97.6 86.6
10.00 182.0 90.0 92.0 200.4 111.2 89.2
11.00 199.0 100.0 99.0 216.6 124.7 91.9
12.00 216.0 110.0 106.0 232.9 138.2 94.6
13.00 233.0 120.0 113.0 249.2 151.8 97.4
14.00 250.0 130.0 120.0 265.6 165.3 100.3
15.00 267.0 140.0 127.0 281.9 178.8 103.1
16.00 284.0 150.0 134.0 298.4 192.4 106.0
17.00 301.0 160.0 141.0 314.8 205.9 109.0
18.00 318.0 170.0 148.0 331.3 219.4 111.9
19.00 335.0 180.0 155.0 347.9 232.9 114.9
20.00 352.0 190.0 162.0 364.4 246.5 117.9
21.00 369.0 200.0 169.0 381.0 260.0 121.0
Layer 3 Silty Sand ρρρρ = 2,100 kg/m3
21.00 369.0 200.0 169.0 381.0 260.0 121.0
22.00 390.0 210.0 180.0 401.6 270.0 131.6
23.00 411.0 220.0 191.0 422.2 280.0 142.2
24.00 432.0 230.0 202.0 442.8 290.0 152.8
25.00 453.0 240.0 213.0 463.4 300.0 163.4
9/11/2013 37
Aquifer with
artesian head
Stress Distribution
0
5
10
15
20
25
0 100 200 300 400 500
STRESS (KPa)
DE
PT
H (
m)
0
5
10
15
20
25
0 100 200 300 400 500
STRESS (KPa)
DE
PT
H (
m)
SAND
CLAY
SAND
Stress from Fill
Artesian Pore Pressure Head
9/11/2013 38
The distribution for the hydrostatic case
Effective Stress Concept
Effective stress = Total stress - Pore pressure
σ σ σ σ’ = σ σ σ σ - u
Unit weight γγγγ = ρρρρg (kN/m3)
g assumed to be 10 m2/s
Total vertical stress or overburden stress,
σσσσz = γγγγ t.z
9/11/2013 39
Pore Water Pressure (PWP)
Pore water pressure u = uw = uss + uexc
uss = Steady state condition, hydrostatic or steady seepage
uexc = Excess pwp due to soil loading
Buoyant unit weight
γγγγ ‘ = γγγγ t - γγγγ w for hydrostatic condition
γγγγ ‘ = γγγγ t - γγγγ w + i γγγγ w
i = hydraulic gradient (head diff/ distance)
i is negative for upward artesian flow9/11/2013 40
Elastic Stress Distribution with Depth
LB
1(H):2(V)zL+z
B+z
Q
qo
qz
BL
Qqo =
))(( zLzB
BLqq oz
++=
• Only can be used for stress at centre of
loaded area
• Cannot use for combined effects of two or
more loaded areas, unless they have same
centres
9/11/2013 41
Boussinesq Distribution (1885)
Q
r
z
qz
Assumes: isotropic linear elastic halfspace,
Poisson ratio = 0.5
For Point load Q kN
2/522
3
)(2
)3(
zr
zQqz
+=
π
Integrate for Line Load, P kN/m
222
3
)(
2
rz
zPqz
+=
π
Integrate for Rectangular Area, get Fadum’s Influence Chart
Fig.1; for Circular Area, get Foster and Alvin Chart Fig.2
9/11/2013 42
Weastergaard Distribution (1938)
Q
r
z
qz
Assumes: isotropic linear elastic halfspace,
Poisson ratio = 0, rigid horizontal layers
For Point load Q kN
2/322 ))/(21( zrz
Qqz
+=
π
Differences with Boussinesq is small
For wide flexible loaded areas Westergaard method is preferred
9/11/2013 43
UniSettle 2.4 21 Jan 2000
EX02.STL page 1
Effective Stress Comparison, ( 4.11 , 8.11 )
-----------------------------------------------------------------
BOUSSINESQ WESTERGAARD 2:1
Depth Ini. Fin. Ini. Fin. Ini. Fin.
Stress Stress Stress Stress Stress Stress
(m) (kPa) (kPa) (kPa) (kPa) (kPa) (kPa)
-----------------------------------------------------------------
Layer 1 Any Soil 0. kg/m^3
0.00 100.0 0.0 100.0 0.0 100.0 0.0
0.10 97.9 0.0 89.3 0.0 93.7 0.0
0.20 94.3 0.0 79.4 0.0 87.9 0.0
0.30 89.4 0.0 71.4 0.0 82.6 0.0
0.40 83.2 0.0 64.4 0.0 77.9 0.0
0.50 76.8 0.0 58.4 0.0 73.5 0.0
0.60 70.9 0.0 53.2 0.0 69.4 0.0
0.70 65.6 0.0 48.9 0.0 65.7 0.0
0.80 61.1 0.0 45.1 0.0 62.3 0.0
0.90 57.2 0.0 41.9 0.0 59.2 0.0
1.00 53.7 0.0 39.1 0.0 56.3 0.0
1.10 50.8 0.0 36.7 0.0 53.5 0.0
1.20 48.1 0.0 34.5 0.0 51.0 0.0
1.30 45.8 0.0 32.5 0.0 48.7 0.0
1.40 43.7 0.0 30.8 0.0 46.5 0.0
1.50 41.7 0.0 29.2 0.0 44.4 0.0
1.60 40.0 0.0 27.8 0.0 42.5 0.0
1.70 38.3 0.0 26.4 0.0 40.7 0.0
1.80 36.8 0.0 25.2 0.0 39.1 0.0
1.90 35.4 0.0 24.1 0.0 37.5 0.0
2.00 34.0 0.0 23.0 0.0 36.0 0.0
Comparison of stresses under a
3m square footing below its
characteristic point
Characteristic point is point
where vertical stress is equal for
both rigid and flexible footing,
this point is located at 0.37B
and 0.37L from centre of
rectangular footing, or 0.37R of
circular footing
Results show that under
characteristic point 2:1 method
is similar to Boussinesq result
9/11/2013 44
AA
BCDEFGHIJKLMNOPQRS
-1.000 0.000 1.000 2.000 3.000 4.000 5.000 6. 000
-0.000
1.000
2.000
3.000
4.000
5.000
E ffective mean stresses
Extrem e eff ective mean s tress -882.98*10-3
kN/m2
[ kN/m2]
A : -0.900
B : -0.850
C : -0.800
D : -0.750
E : -0.700
F : -0.650
G : -0.600
H : -0.550
I : -0.500
J : -0.450
K : -0.400
L : -0.350
M : -0.300
N : -0.250
O : -0.200
P : -0.150
Q : -0.100
R : -0.050
S : 0.000
T : 0.050
Elastic Stress Bulb for Circular Footing
9/11/2013 45
Elastic Settlement for Circular Flexible Load
q
R
D = ∞∞∞∞
Surface settlements is given by Terzaghi,1943 as:
3.),( FigseerfisI
IE
qR
µ
ρ
ρ
ρ=
Edge settlement = 0.7 centre settlement
Centre settlement is :
footingFlexibleforE
Rqz )1(2 2µρ −=
footingRigidforE
Rqz )1(
2
2µπ
ρ −=
9/11/2013 46
Flexible and Rigid Footing
AA*
0.08 0.16 0.24 0.32 0.40 0.48 0.56 0.64 0.72 0.80 0.88 0.96 1.04
9.76
9.84
9.92
10.00
10.08
10.16
10.24
10.32
10.40
10.48
A
A
-0.300 -0.200 -0.100 0.000 0.100 0.200 0.300 0.400 0.500 0.600 0.700 0.800 0.900 1.000 1.100
9.300
9.400
9.500
9.600
9.700
9.800
9.900
10.000
10.100
10.200
Flexible Footing
E=1000 kPa, µµµµ=0, q=10 kPa, sett= 20 mm
-0.300 -0.200 -0.100 0.000 0. 100 0.200 0.300 0.400 0.500 0.600 0.700 0.800 0.900 1.000 1.100
9.200
9.300
9.400
9.500
9.600
9.700
9.800
9.900
10.000
10.100
10.200
AA*
0.08 0.16 0.24 0.32 0.40 0. 48 0.56 0.64 0.72 0.80 0.88 0.96 1.04
9.76
9.84
9.92
10.00
10.08
10.16
10.24
10.32
10.40
10.48
Rigid Footing
E=1000 kPa, µµµµ=0, q=10 kPa, sett= 16 mm
9/11/2013 47
Approximate Ratio at corner, centre and edge to average settlement
Flexible Load Area Rigid
Footing
Foundation
Depth
Corner/
Ave
Edge/
Ave
Centre/
Ave
Rigid/
Ave
H/L=∞∞∞∞ 0.6 0.9 1.2 0.9
H/L=1 0.5 0.7 1.3 0.8
H/L=1/4 0.4 0.7 1.3 0.8
9/11/2013 48
Elastic Settlement for Other Flexible Load
B
D = ∞∞∞∞
Corner settlements is given by Terzaghi,1943 as:
4.)/(
)1(2
FigseeLBfisI
IE
qB
ρ
ρ
µρ
−=
Points other than corner for any combination of
rectangles can be obtained by superposition
For centre of square loaded area:
)1(12.1 2µρ −=E
Bqz
qL
9/11/2013 49
Superpostion Principle for Rectangles
= + + +zσ
- - +=
zσ
9/11/2013 50
Elastic Settlement for Other Flexible Load
For Flexible Rectangular Loaded area, Corner settlement ρρρρ,
with finite depth of elastic layer, use Steinbrenner chart
5.
)21()1(
)1(
21
2
2
1
2
2
FigingivenFandF
FFI
IE
qB
µµµ
µρ
ρ
ρ
−−+−=
−=
B
Finite
D
qL
9/11/2013 51
Equivalent Footings for Pile Groups Settlements
L
2/3L
2
1Homogeneous
Clay
Equivalent Ftg
Ground level
L
2/3L
2
1
Soft Clay
Firm Layer
Equivalent Ftg
Ground level
9/11/2013 52
design analysis design analysis design analysis design analysis
isisisis
SettlementSettlementSettlementSettlement
9/11/2013 53
Movement, Settlement, and Creep
Movement occurs as a result of an increase of stress, but the term should be reservedto deformation due to increase of total stress. Movement is the result of a transfer ofstress to the soil (the movement occurs as necessary to build up the resistance to theload), and when the involved, or influenced, soil volume successively increases as thestress increases. For example, when adding load increments to a pile or to a plate in astatic loading test (where, erroneously, the term "settlement" is often used). As a term,movement is used when the involved, or influenced, soil volume increases as the loadincreases.
Settlement is volume reduction of the subsoil as a consequence of an increase in
effective stress. It consists of the sum of "elastic" compression and deformation due toconsolidation. The elastic compression is the compression of the soil grains (soilskeleton) and of any free gas present in the voids, The elastic compression is oftencalled "immediate settlement”. It occurs quickly and is normally small (theelastic compression is not associated with expulsion of water). The deformation due toconsolidation, on the other hand, is volume change due to the compression of the soilstructure associated with an expulsion of water—consolidation. In the process, theimposed stress, initially carried by the pore water, is transferred to the soil structure.Consolidation occurs quickly in coarse-grained soils, but slowly in fine-grained soils. Asa term, settlement is used when the involved, or influenced, soil volume stays constantas the effective stress increases.
9/11/2013 54
Movement, Settlement, and Creep
The term "settlement" is normally used for the deformation resulting from the
combined effect of load transfer, increase of effective stress, and creep during
long-term conditions. It is incorrect to use the term “settlement” to mean
movement due to increase of load such as in a loading test.
Creep is compression occurring without an increase of effective stress.
Creep is usually small, but may in some soils add significantly to the
compression of the soil skeleton and, thus, to the total deformation of the
soil. It is then acceptable to talk in terms of creep settlement.
9/11/2013 55
Strain
Linear Elastic Deformation (Hooke’s Law)
ε = induced strain in a soil layer
= imposed change of effective stress in the soil layer
E = elastic modulus of the soil layer (Young’s Modulus)
Young’s modulus is the modulus for when lateral expansion is allowed, which may be the case for soil loaded by a small footing, but not when the load is applied over a large area. In the latter case, the lateral expansion is constrained (or confined). The constrained modulus, D, is larger than the E-modulus. The constrained modulus is also called the “oedometer modulus”. For ideally elastic soils, the ratio between D and E is:
9/11/2013 56
)21()1(
)1(
νν
ν
−+
−=
E
D
'σ∆
E
'σε
∆=
ν = Poisson’s ratio
Stress-Strain
9/11/2013 57
Stress-strain behavior is non-linear for most soils. The
non-linearity cannot be disregarded when analyzing
compressible soils, such as silts and clays, that is, the
elastic modulus approach is not appropriate for these soils.
Non-linear stress-strain behavior of compressible soils, is
conventionally modeled as follows.
where ε = strain induced by increase of effective stress from σ‘0 to σ‘1Cc = compression index
e0 = void ratio
σ‘0 = original (or initial) effective stress
σ‘1 = final effective stress
CR = Compression Ratio = (MIT)
0
1
0
1
0 '
'lg
'
'lg
1 σ
σ
σ
σε CR
e
Cc =+
=
01 e
CCR c
+=
9/11/2013 58
In overconsolidated soils (most soils are)
9/11/2013 59
)'
'lg
'
'lg(
1
1 1
00 p
c
p
cr CCe σ
σ
σ
σε +
+=
where σ‘p = preconsolidation stress
Ccr = re-compression index
The Janbu Method
])'
'()
'
'[(
1 01 j
r
j
rmj σ
σ
σ
σε −=
9/11/2013 60
The Janbu tangent modulus approach, proposed by Janbu (1963; 1965; 1967; 1998),
and referenced by the Canadian Foundation Engineering Manual, CFEM (1985; 1992),
applies the same basic principles of linear and non-linear stress-strain behavior. The
method applies to all soils, clays as well as sand. By this method, the relation between
stress and strain is a function of two non-dimensional parameters which are unique for a
soil: a stress exponent, j, and a modulus number, m.
Janbu’s general relation is
where σ‘r is a “reference stress = 100 KPa
j > 0
The Janbu Method
9/11/2013 61
Dense Coarse-Grained Soil j = 1
Cohesive Soil j = 0
Sandy or Silty Soils j = 0.5
0
1
'
'ln
1
σ
σε
m=
'1
)''(1
01 σσσε ∆=−=mm
'2
1)''(
2
101 σσσε ∆=−=
mm
KPa
ksf
)''(5
101 σσε −=
m
pm
''(2
1 σσε −=
KPa
ksf
There are direct mathematical conversions
between m and the E and Cc-e0
22
00 69.02lg10ln13.2
110ln
εε==
+=
+=
cc C
e
C
em
9/11/2013 62
For E given in units of KPa (and ksf), the relation between the
modulus number and the E-modulus is
m = E/100 (KPa)
m = E/2 (ksf)
For Cc-e0, the relation to the modulus number is
Typical and Normally Conservative Virgin Modulus Numbers
SOIL TYPE MODULUS NUMBER STRESS EXP.
Till, very dense to dense 1,000 — 300 (j = 1)
Gravel 400 — 40 (j = 0.5)
Sand dense 400 — 250 (j = 0.5)
compact 250 — 150 - " -
loose 150 — 100 - " -
Silt dense 200 — 80 (j = 0.5)
compact 80 — 60 - " -
loose 60 — 40 - " -
Silty clay hard, stiff 60 — 20 (j = 0)
and stiff, firm 20 — 10 - “ -
Clayey silt soft 10 — 5 - “ -
Soft marine clays
and organic clays 20 — 5 (j = 0)
Peat 5 — 1 ( j= 0)
For clays and silts, the recompression modulus, mr, is often five to tentimes greater than the virgin modulus, m, listed in the table
9/11/2013 63
0.40
0.60
0.80
1.00
1.20
10 100 1,000 10,000
Stress (KPa) log scale
Void
Ratio
(- -
)
m = 12
(CR = 0.18)
p'c
0
5
10
15
20
25
10 100 1,000 10,000
Stress (KPa) log scale
Str
ain
(%
)
p 10p
Cc
Cc = 0.37
e0 = 1.01 p'c
p 2.718p
1/m
Slope = m = 12
Evaluation of compressibility from oedometer results
Void-Ratio vs. Stress and Strain vs. Stress — Same test data
9/11/2013 64
Comparison between the Cc/e0 approach
and the Janbu method
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40 0.60 0.80 1.00 1.20
VOID RATIO, e0
CO
MP
RE
SS
ION
IN
DE
X, C
c
0
5
10
15
20
25
30
35
0.400.600.801.001.20
VOID RATIO, e0
VIR
GIN
MO
DU
LU
S N
UM
BE
R,
m
9/11/2013 65
Data from a 20 m thick sedimentary deposit of medium compressibility.
Do these values
indicate a
compressible soil, a
medium compressible
soil, or a non-
compressible soil?
The Cc-e0 approach implies that the the compressibility varies by 30± %.
However, the Janbu methods shows it to vary only by 10± %. The
modulus number, m, ranges from 18 through 22; It would be unusual to
find a clay with less variation.
What about “Immediate Settlement”
and Consolidation?
50
75
100
125
150
0.40 0.60 0.80 1.00 1.20
VOID RATIO, e0
NO
RM
AL
IZE
D
Cc
(%)
50
75
100
125
150
0.400.600.801.001.20
VOID RATIO, e0
NO
RM
AL
IZE
D
m
(%)
9/11/2013 66
CE 5101 - SETTLEMENTS AND 1D Compression Theory
Settlements of Soils - Immediate, Delayed, and Creep Compression
Delayed Consolidation Compression•mv method•e-logP method•Janbu method
Terzaghi’s Theory of 1D ConsolidationEffects of Drainage and Initial Stress Distribution
SPREADSHEET Calculations (UNISETTLE)Finite Element Analysis (PLAXIS)
9/11/2013 67
Foundation Settlement Issues
How Much settlements will occur?
Interested in Ultimate settlements in fully drained state, as well as long-term creep settlements
How Fast and how long will it take for most of settlements to occur?
Involved Consolidation and Secondary Compression theories to estimate rate of settlements, and
Methods to accelerate settlements and minimise long-term settlements
9/11/2013 68
Types of Ground Movements and Causes of Settlements
•Compaction - due to vibrations, pile driving, earthquake
•Elastic Volumetric Settlement in OC Clay in recompression,
use E’ and νννν’ or recompression index, Cr or κκκκ
•Immediate or Undrained Settlement - Distortion without
volume change, use Eu and ννννu
•Moisture changes - Expansive soils, high LL and PI, high
swelling and shrinkage
Swell Potential (%) = 0.1(PI-10) log (s/p’)
•Effects of vegetation - related to moisture changes by root
system
•Effects of GWT lowering - shrinkage ad consolidation
9/11/2013 69
Types of Ground Movements and Causes of Settlements
•Effects of temperatures - Frost heaving,
drying by furnace and boilers
•Effects of seepage and scouring - Erosion by
piping, scouring and wind action, mineral
cement dissolved by GW eg limestone, rock
salts and chalk areas
•Loss of lateral support - Footings beside
unsupported excavation, movement of natural
slopes and cuttings
•Effects of mining subsidence - collapse of
ground cavities
•Filled ground - settlements of the fill soils,
compaction, consolidation, and creep
9/11/2013 70
9/11/2013 71
Shrink/ swell potential is function of Clay Activity = %clay fraction/ PI
Swell Potential (%) = 0.1(PI-10) log (s/p’)
Where s=suction before construction and p’=final bearing pressure
1D Settlements
Soil deformations are of two types:• Distortion (change of shape 2D effects)•Compression (change of volume)
Components of Settlements:
ncompressiosecondarys
ncompressiosettlementionconsolidats
distortionsettlementimmediateswhere
ssss
s
c
i
scit
=
=
=
++=
,
,
9/11/2013 72
Undrained Immediate Settlements
Distortion of clay layer:•Calculate by elastic theory eg Janbu Chart
layerclay of ratio sPoisson'
layerclay of modulus Undrained E
B dthfootong wi on loadAverage q
by JanbufactorsessdimensionlIandI
arealoadedflexibleforsettlementimmediateswhere
E
qBIIs
10
i
i
=
=
=
=
=
−=
υυυυ
νννν )1( 2
10
9/11/2013 73
Immediate Settlements in Clays by Janbu
9/11/2013 74
Example on Janbu Chart: Foundation 4x2m with q=150kPa,
located at 1m in Clay layer 5m thick with Eu=40MN/m2. Below
is second clay layer of 8m thickness and Eu=75 MN/m2.What
is average settlement under foundation? Assume νννν=0.5
H
D
B(((( )))) mm5.35.01
40
2*150*7.0*9.0s
7.0 2,4/2L/B and 24/2H/B
:MN/m40E withlayer,clay upper Consider (1)
9.0 0.5, 1/2D/B 2;4/2L/B Now
2
1i
1
2
u
0
====−−−−====
====⇒⇒⇒⇒================
====
====⇒⇒⇒⇒================
µµµµ
µµµµ
(((( )))) mm3.25.0175
2*150*85.0*9.0s
85.0 2,4/2L/B and 612/2H/B
:MN/m75E withcombined, layers two Consider (2)
2
2i
1
2
u
====−−−−====
====⇒⇒⇒⇒================
====
µµµµ
(((( )))) mm9.15.0175
2*150*7.0*9.0s
7.0 2,4/2L/B and 24/2H/B
:MN/m75E withlayer, upper Consider (1)
2
3i
1
2
u
====−−−−====
====⇒⇒⇒⇒================
====
µµµµ
mm 3.9 1.9 -2.3 5.3s
ssss
;inciplePrionSuperposit By
i
3i2i1ii
====++++====
−−−−++++====
9/11/2013 75
1D Primary Consolidation Settlements
Time delayed Primary Consolidation Compression:
water
solids
0e
1
e∆
0H
H∆
)()(
∆
00
0
volumeverticalc
0
volumevertical
HHHs
e1
e
H
H
εεεεεεεε
εεεεεεεε
==∆=
+==
∆=
'vv P where P), vs(e methodm σσσσ=
e
0e
fe
fP0P P
P
e -ility compressib of coeffnav
∆
∆==
0
v
0
ve1
a
P
1
)e(1
e -change volume of coeffnm
++++====
++++========
∆∆∆∆
∆∆∆∆
0
0
vc
0v0
0
c
HPe1
a s
HPm He1
e sH
∆∆∆∆
∆∆∆∆∆∆∆∆
∆∆∆∆
++++====
====++++
========
9/11/2013 76
mv and Constrained Modulus D
For wide loaded area, get 1D compression under Ko condition, elastic modulus to apply is called the Constrained Modulus D defined by:
(((( ))))(((( ))))(((( ))))
0.35)( ratio Poisson drained Soil
elasticity of modulus drained Soil E
changevolume of tCoefficien mwhere
211
1E
m
1D
v
v
<<<<====
====
====
−−−−++++
−−−−========
υυυυ
υυυυυυυυ
υυυυ
9/11/2013 77
Sc by (e vs logP) Method
Normally Consolidated Clays - non-linear stress strain for soil,
but linear in logP
e
0e
fe
fP0P Plog
indexncompressioCc =
(((( ))))0
f
c0fcP
PlogCPlogPlogCe ====−−−−====∆∆∆∆
0
f
0
c0c
P
Plog
e1
CHHs
+== ∆∆∆∆
0
cc
e1
CC CRRatio, nCompressio
++++======== εεεε
9/11/2013 78
Sc by (e vs logP) Method
Over-Consolidated Clays (Pf < Pc), Preconsolidation Pressure
e
0e
fe
fP0P Plog
indexncompressioCc =
(((( ))))0
f
r0frP
PlogCPlogPlogCe ====−−−−====∆∆∆∆
cr
0
f
0
r0c
C than smaller times10 to5 is C as
ssettlementsmall
P
Plog
e1
CHHs
++++======== ∆∆∆∆
0
rr
e1
CC RR, Ratio ncompressioRe
++++======== εεεε
cP
indexswellingC
indexionrecompressC
s
r
=
=
9/11/2013 79
Sc by (e vs logP) Method
Over-Consolidated Clays (Pf > Pc), Preconsolidation Pressure
e
0e
ce
fP0P Plog
cC
(((( ))))0
cr0cr1
P
PlogCPlogPlogCe ====−−−−====∆∆∆∆
ssettlementlarge mean willThis
P
PlogC
P
PlogC
e1
H
e1
eeHHs
c
f
c
0
cr
0
0
0
210c
++
=
+
+==
∆∆∆∆∆∆∆∆∆∆∆∆
cP
sr CorC
fe(((( ))))
c
f
ccfc2P
PlogCPlogPlogCe ====−−−−====∆∆∆∆
9/11/2013 80
Sc by Janbu Tangent Modulus MethodCanadian Foundation Manual 1985
Janbu (1963, 1965,1967,1998); For Cohesionless Soils; j>0
Theory is used in Program UniSettle (by UNISOFT, Canada)
(((( ))))
kPa100 stresseffective verticalReference
test field or lab from obtained number, modulus Janbu m
exponent stress Janbuj
P stresseffective vertical final
P stresseffective vertical initial
stresseffective vertical inchange by induced strain vertical where
mj
1 get to Integrate m
'M
,
r
f
,
f
0
,
0
v
j
,
r
,
0
j
,
r
,
f
v
j1
'
r
''
rt
========
====
====
========
========
====
−−−−
====
====
∂∂∂∂
∂∂∂∂====
−−−−
σσσσ
σσσσ
σσσσ
σσσσσσσσ
σσσσσσσσ
εεεε
εεεεσσσσ
σσσσσσσσ
εεεε
σσσσ
9/11/2013 81
Cohesionless Sands and Silts; j=0.5
Normally Consolidated Cohesionless Soils
−−−−====
========
σσσσσσσσεεεε,
0
,
fv
,
r
5m
1
kPa100 σ0.5jSOIL TYPE m
SAND DENSE 400-250
MEDIUM 250-150
LOOSE 150-100
SILT DENSE 200-80
MEDIUM 80-60
LOOSE 60-40
9/11/2013 82
Cohesionless Sands and Silts; j=0.5
Over-Consolidated Cohesionless Soils
m than larger times10 to5 usually is m number, modulus OC
kPa inpressure dationpreconsoli
where
5m
1
5m
1
kPa100 'P
'f
r
,
P
,
P
,
f
,
0
,
Pr
v
,
r0.5j; for CASE
====
−−−−++++
−−−−====
========>>>>
σσσσ
σσσσσσσσσσσσσσσσ
σσσσσσσσ
εεεε
σ
−−−−====
====≤≤≤≤ ====
σσσσσσσσεεεε
σσσσσσσσ
,
0
,
fr
v
,
r
'
p
'
f
5m
1
kPa,100 ; for CASE ,0.5j σ
9/11/2013 83
Cohesive Soils; j=0
Normally Consolidated Clays
++++====
++++====
++++====
====
========
c
0
c
0
,
0
,
f
0
c
,
0
,
f
v
,
r
C
e13.2
C
e110lnm
loge1
Cln
m
1
kPa100 0j
σσσσσσσσ
σσσσσσσσ
εεεε
σ
SOIL TYPE m
SILTYCLAYS
HARD,STIFF
60-20
AND STIFF,FIRM
20-10
CLAYEYSILT
FIRM TOSOFT
10-5
SOFTMARINECLAYS
20-5
ORGANICCLAYS
10-3
PEAT 5-1
9/11/2013 84
Over-consolidated Clays; j=0
Over- Consolidated Clays
++++====
++++====
++++====
++++====
++++++++
++++====
++++
====
r
0
r
0r
c
0
c
0
,
P
,
f
0
c
,
0
,
P
0
rv
,
P
,
f
,
0
,
P
r
v
C
e13.2
C
e110lnmand
C
e13.2
C
e110lnm
loge1
Clog
e1
C
lnm
1ln
m
1
σσσσσσσσ
σσσσσσσσ
σσσσσσσσ
σσσσσσσσ
εεεε
εεεε
9/11/2013 85
Linear Elastic Soil; j=1
(((( ))))
100
Em
m100E,therefore
m100
1
mj
1
,
0
,
f
j
,
r
,
0
j
,
r
,
f
v
====
====
−−−−====
−−−−
==== σσσσσσσσ
σσσσσσσσ
σσσσσσσσ
εεεε
9/11/2013 86
Janbu compared to e-logP
9/11/2013 87
Oedometer Results in linear scale by Janbu
9/11/2013 88
Calculation of Settlement
•Determine soil profile to get initial effective stresses
•Determine soil compressibility parameters, e-logP, Cc,
Cr and Pc OR Janbu modulus number, m and mr
•Determine final effective stresses due to imposed loads,
excavations, fills, GWT changes etc
•Divide each soil layer into sublayers, calculate strain
caused by change from initial to final effective stresses
in each sublayer
•Calculate the settlement for each sublayer and the
accumulated settlement
9/11/2013 89
UNISETTLE Calculation of Settlement
9/11/2013 90
Hand Calculation
9/11/2013 91
UniSettle Input
9/11/2013 92
InputLoading and Excavation
9/11/2013 93
Results
9/11/2013 94
Settlements Distribution
9/11/2013 95
Alternative Conditions
9/11/2013 96