Topic #5
Effective Stress Concept
CE3348 Spring 2019
Exam #1 Grade Distribution
100908070605040302010
60
50
40
30
20
10
0
Grades
Frequency
CE
33
48
-Geo
tech
nic
al
Eng
inee
rin
g
Ph
ysic
al P
rop
erti
es o
f So
ils
Weight-Volume Relations Phase Diagrams
Soil Texture Atterberg Limits
Aggregate Geometry (AIMS)
Soil Classification USCS Method
AASHTO Method
Soil Compaction
Mec
ha
nic
al A
na
lysi
s o
f So
ils
Effective Stress Pore Water Pressure Concept
Geostatic Stresses Effective Stress
Calculations
Flow of Water in Soils
One Dimensional Flow Theory Darcy’s Law
Two Dimensional Flow Theory Flow Nets
External Stresses Boussinesq Theory Newmark Method
Westergaard Theory
Shear Strength of Soils
Mohr-Coulomb Theory
Direct Shear Test
Triaxial Tests
Consolidated Drained (CD)
Consolidated Undrained (CU)
Unconsolidated Undrained (UU)
Mohr Circle
Settlement Analysis
Immediate Settlement
Primary Consolidation
Secondary Compression
Time Rate of Settlement
Decomposition of Total Stress to
Effective Stress and Pore Water Pressure
This slide shows a saturated soil element carrying a total vertical stress of σz. This
total stress is carried by both the pore water pressure, u, and the effective stress,
σ’z. The effective stress is the portion carried by the interparticle contact forces.
4
Effective Stress and Pore Water Pressure
Case 1: Dry Sand under Load (Q)
The load applied at the surface of the soil is transferred
to the soil grains in the mold through their points of
contact.
If the load is quite considerable, it would result in the
compression of the soil mass in the mold.
If the cross sectional area of the cylinder is A, the
average stress at any level X-Y may be written as:
Since this stress is responsible for the deformation of the
soil mass, it is termed the intergranular or effective stress.
We may therefore write:
A
Qa
' a
5
Effective Stress and Pore Water Pressure
Case 2: Fully Saturated Sand under Load (Q)
Closed Valve
If the same load Q is placed on the piston, this load will
not be transmitted to the soil grains in the same way as
in the case 1.
If we assume that water is incompressible, the external
load will be transmitted to the water in the pores.
The pressure that is developed in the water is called the
pore water pressure or neutral stress (u). This pore
water pressure prevents the compression of the soil
mass. For this specific case, pore pressure can be
calculated as:
A
Qu
6
Effective Stress and Pore Water Pressure
Case 3: Fully Saturated Sand under Load (Q)
Open Valve
If the valve provided in the piston is opened, immediately there
will be expulsion of water through the hole in the piston. The
flow of water continues for some time and then stops.
The expulsion of water from the pores decreases the pore water
pressure and correspondingly increases the intergranular
pressure. At any stage the total pressure Q/A is divided between
water and the points of contact of grains. A new equation may
therefore be written as:
Total Pressure = Intergranular Pressure + Pore Water Pressure
Final equilibrium will be reached when there is no
expulsion of water. At this stage, the pore water pressure
u=0. All the pressure will be carried by the soil grains.
Therefore: '
u '
7
Effective Stress Concept
As in other materials, stresses may act in soils as a result of an external load and the volumetric weight of the material itself.
The pressure transmitted through grain to grain at the contact points through a soil mass is termed as intergranular or effective pressure (’). It is known as effective pressure since this pressure is responsible for the decrease in the void ratio or increase in the frictional resistance of a soil mass.
If the pores of a soil mass are filled with water (saturated state) and if the pressure induced into the pore water tries to separate the grains, this pressure is termed as positive pore water pressure (u). The effect of this pressure is to decrease the frictional resistance of the soil mass.
8
P
N
P Interparticle force
P
T
= + u
uA
ΣP'
A
P
c
N
9
10
Particles in Solid Contact
= + u
11
Summary of Important Points
Effective stress represents the actual intergranular pressure that
occurs between soil particles. This effective stress is the stress
that influences shear strength of the soil and volume changes or
settlements.
Total water pressure at a point is termed the neutral stress (u),
for it acts equally in all directions (hydrostatic stress state).
It is important to recognize that the neutral stress in saturated
state acts to reduce the intergranular stress that develops
between soil particles. This condition frequently has an adverse
effect on the strength of a soil.
12
Capillary Rise and
the Effective Stresses in Soils
13
Definitions of Zones
Notice that the soil can
be saturated above the
phreatic surface in the
zone of capillary rise.
The zone of capillary
rise may be only a
fraction of an inch in
sands, but can be several
feet in clay or silty soils.
The phreatic zone is
that zone below the
ground water table.
14
Menisci Formation in Glass Tubes
The attraction between the water and the
capillary tube affects the shape of the air–
water interface at the top of the column of
water.
For water and glass, the shape is concave
downward; that is, the water surface is
lower at the center of the column than at
the walls of the tube. The resulting curved
liquid surface is termed the meniscus.
15
Capillary Rise
Notice that the pore water pressure above the
free water surface is negative.
w
c
cw
y
d
CosTh
CosTdhd
F
4
4
0
2
Notice that the height of
capillary rise is inversely
related to the tube diameter.
Surface Tension acting around the
circumference of the tube should be
in equilibrium with the weight of the
water rose in the glass tube.
16
Capillary Rise and Particle Size
Fines Silts and Clays
Sands Gravels
Fines Silts and Clays
Sands Gravels
Reminder: The inter-connected voids in particulate soils
resemble nano-tubes that acts as channels of water.
Menisci Radius and
Inclination of Capillary Tube
In the capillary tube, at
the level equal to the free
surface of the water in
the supply pan, the
hydrostatic pressure of
the water is zero.
Hydrostatic pressures
increase below that free
water surface (u= w . Z).
19
Expected Height of Capillary Rise
At normal room temperature (ambient
temperature), the magnitude of the surface
tension is close to 0.005 lb/ft or 0.064 N/m.
– Therefore:
dh
d
CosTh
c
w
c
31
4
In this equation both the diameter of the capillary
tube and the height of capillary rise are in mm.
20
Vadose Zone (Summary)
Ground Water Table (GWT):
Equilibrium elevation at which
the pore water pressure is equal to
atmospheric pressure.
Vadose Zone: By definition, vadose
zone is the zone above the ground
water table (GWT).
Vadose zone can be partially
saturated or fully saturated due to
capillary action.
Observation wells (bore holes) are
used to determine the depth of the
vadose zone.
Notice that the pore water
pressure is negative in
vadose zone. (+)
(-)
U=w.Z
21
Within the zone of capillary rise, a plot of
pore water pressure versus elevation is
simply an extension of the plot below the
groundwater table.
Above the zone of capillary rise, the pore
water accumulates near the particle contact
points and surface tension forces develop
between these pockets of pore water and the
adjacent solid particles. These forces produce
tensile stresses (negative pore water
pressures) in the water. The magnitude of the
negative pore water pressures in this zone
depends on the soil type and other factors,
and can be much greater (i.e., more negative)
than those in the capillary zone.
22
Apparent Cohesion
In partially saturated granular soils, the
surface tension phenomenon contributes to
the strength of the soil mass.
The available water collects between where
soil particles touch, forming a wedge of
moisture but leaving the center portion of
the void filled with air. Thus, an air–water
interface, or meniscus, is formed.
The surface tension in the meniscus imposes
a compressive force onto the soil particles,
increasing the friction between particles and
improves the shear strength of the soil.
The strength gain in partially saturated granular soils due to surface
tension phenomenon is termed apparent cohesion.
23
Pore Water Pressure and Strength
Negative pore water pressure tends to pull the particles
together therefore it will enhance the shear strength of soils.
24
Sand Castle
Sand castle is an example of
how the surface tensional
forces pulls the sand grains
together.
Keep in mind that you need
to wet the sand to be able to
build a sand castle. Dry sand
will just simply flow.
25
geometry
Surface Tension and Water Bug
27
Geostatic Stresses in Soils
28
Imaginary Column of Soil to Compute
the Geostatic Vertical Stress (σz)
In a soil mass, the vertical stress caused
by the soil at a point below the surface is
equal to the weight of the soil lying
directly above the point. Vertical stress
thus increases as the depth of the soil
overburden increases.
The vertical stress can be calculated as
the weight of a “column” of soil
extending above a unit area.
σv1 = γ1H1
σv2 = γ1H1 + γ2H2
σv3 = γ1H1 + γ2H2 + γ3H3
Point 1
Point 2
Point 3
29
Point 1
Point 2
Point 3
(a) Saturated soil due to capillary action (use sat) (b) Submerged soil (use ’)
Where the soil surface is below water (such as in oceans and lakes), the
effective stress should be computed by using the submerged or effective soil
unit weight multiplied by the depth measured from the soil surface.
30
Forces Acting on an Element of Soil When the Seepage Forces
are Acting Vertical Upward
Buoyancy Effect
When a soil exists below the groundwater table,
the submerged soil particles are subject to a
buoyant force resulting from the hydrostatic water
pressure, the same phenomenon that acts
on any submerged solid. Reminder
’ = sat - w Submerged unit weight of soil
31
Variation of Stresses with Depth
Plots of the variations of the total stress, pore water pressure, and effective stress, respectively, with depth for a submerged layer of soil placed in a tank with no seepage.
- u = 32
A piezometer is a device used to
measure static liquid pressure in
a system by measuring the height
to which a column of the liquid
rises against gravity.
Piezometers are widely used to
measure the pressure head (or
more precisely, the piezometric
head) of groundwater at a
specific point.
33
Pore Water Pressure Measurement Piezometers
Types of Piezometers Standpipe Piezometers
The standpipe piezometer is the most basic of piezometer types. It consists of a filter tip joined to a riser pipe. Water
flow from the surrounding soil into the standpipe. Readings are obtained with a water level indicator.
Advantages: Simple, reliable, not electrical, no calibrated components.
Limitations: Accuracy depends on skill of operator; reading requires a man on site; remote reading not possible;
slower to show changes in pore-water pressure.
Vibrating Wire Piezometers
The vibrating wire piezometer is the most commonly used piezometer and is suitable for almost all applications. It
consists of a vibrating wire pressure transducer and signal cable. It can be installed in a borehole, embedded in fill, or
suspended in a standpipe. Readings are obtained with a portable readout or a data logger.
Advantages: Easy to read, very accurate; good response time in all soils; easy to automate; reliable remote readings.
Limitations: Must be protected from electrical transients.
Pneumatic Piezometers
The pneumatic piezometer operates by gas pressure. It consists of a pneumatic pressure transducer and pneumatic
tubing. It can be installed in a borehole, embedded in fill, or suspended in a standpipe. Readings are obtained with a
pneumatic indicator.
Advantages: Reliable, remote reading possible, not electrical, indicator can be calibrated at any time.
Limitations: Accuracy depends on skill of operator; difficult and expensive to automate, so reading requires man on
site; reading time increases with length of tubing; pneumatic tubing can be blocked by condensation if not frequently
charged with dry nitrogen gas.
34
Effect of Ground Water Table (GWT)
A- Water Table Level at Ground Surface
The submerged weight of the soil (’), is
termed the effective soil weight, and the
subsurface stress that results is termed the
effective stress.
Effective stress represents the actual
intergranular pressure that occurs between
soil particles.
If the total weight of soil as it exists above
the water table is t (the wet weight of soil
before its submerged), and the unit weight
of soil at saturated state is sat , the effective
stress can be calculated as: 𝜎′𝑧 = 𝛾
𝑆𝑎𝑡𝑧 -𝛾
𝑤𝑧
𝜎′𝑧 = (𝛾𝑆𝑎𝑡
-𝛾𝑤) 𝑧 = 𝛾′ 𝑧
𝜎′𝑧 = 𝜎𝑧- 𝑢
35
Effect of ground Water Table (GWT)
Example #1
A- Water Table Level at Ground Surface
36
Effect of Ground Water Table (GWT)
B- Water Table Level above Ground
Surface
In general, to calculate the effective stress,
simply calculate the total soil pressure
(disregarding buoyancy effects) and then
subtract the hydrostatic pressure (the neutral
pressure) at the point being analyzed.
The pore water pressure (u) is the unit
weight of water w multiplied by the depth
below the water table.
𝜎′𝑧 = [𝛾𝑆𝑎𝑡𝑧 + 𝛾
𝑤ℎ] − 𝛾𝑤(𝑧 + ℎ)
= 𝛾𝑆𝑎𝑡𝑧 -𝛾
𝑤𝑧 = (𝛾
𝑆𝑎𝑡-𝛾𝑤)𝑧
𝜎′𝑧 = 𝜎𝑧- 𝑢
𝜎′𝑧 = 𝛾′𝑧
37
Effect of ground Water Table (GWT)
Example #2
B- Water Table Level above Ground Surface
38
Effect of Ground Water Table (GWT)
C- Water Table Level Below Ground Level
Above the water table, the effective soil
weight is the total soil weight, including the
pore water; below the water table, the
effective soil weight is the submerged or
buoyant weight.
The pore water pressure u is the unit weight
of water w multiplied by the depth below
the water table.
𝜎′𝑧 = 𝛾𝑡𝑧𝑎+ 𝛾𝑆𝑎𝑡𝑧𝑏- 𝛾𝑤𝑧𝑏
= 𝛾𝑡𝑧𝑎 + (𝛾𝑠𝑎𝑡-𝛾𝑤)𝑧𝑏 𝜎′𝑧 = 𝛾𝑡𝑧𝑎 + 𝛾′𝑧𝑏
A
za
zb 𝜎′𝑧 = 𝜎𝑧- 𝑢
39
Effect of ground Water Table (GWT)
Example #3
C- Water Table Level below Grand Level
40
Horizontal Stress (’h)
Horizontal stresses can be calculated from the vertical stresses.
The ratio of the lateral effective stress (horizontal stress) to
vertical effective stress is called coefficient of lateral earth
pressure (Ko).
'
'
v
ho
pressuresoilVertical
pressuresoilHorizontalK
’z
Z
’h
41
Coefficient of Lateral Earth Pressure (Ko)
In a horizontal, uniform soil mass, the lateral movement of the soil at any
depth is not possible, because the confining pressure is equal in all
horizontal directions, therefore, a state of static equilibrium exists and
the soil is in the at-rest condition.
The magnitude of K0 for a given soil mass is affected by the soil
deposit’s stress history.
Soils that have been subjected to heavy loading at some time in their
history, such as now-dense granular (sand or gravel) deposits would have
had to develop resistance to high lateral stress in order to remain stable.
Deposits that have not been exposed to heavy loading, such as loose
granular soils and soft clays, would not have developed high lateral
strength.
Typically, dense granular soils end up having lower values of K0
compared to loose granular soils and soft clays.
42
How to Calculate Horizontal Stresses
I. Calculate vertical effective stress (’z )
II. Calculate horizontal effective stress (’h )
’h = ’z x Ko
III. Remember that pore water pressure is the same in horizontal and vertical directions (hydrostatic stress state), therefore:
uh = uz = u
IV. Calculate horizontal total stress (h)
h = ’h + u
43
Example #4
The water table in an 8m thick silty sand deposit lies at a depth 3m below the ground level. The entire soil above the water table is saturated through capillary rise. The saturated unit weight is 18.8 KN/m3. Plot the variation of total and effective vertical stresses and pore water pressure with depth.
5 m
3 m
44
5 m
3 m
0
1
2
3
4
5
6
7
8
-50 0 50 100 150 200
De
pth
(m
)
Stress (kPa)
Total Stress (kPa)
Pore Water Pressure (kPa)
Effective Stress (kPa)
Note the negative pore water
pressure above the ground level
due to capillary action.
sat 18.8 KN/m3
w 9.81 KN/m3
z (m) v u 'v
0 0 -29.43 29.43
1 18.8 -19.62 38.42
2 37.6 -9.81 47.41
3 56.4 0 56.4
4 75.2 9.81 65.39
5 94 19.62 74.38
6 112.8 29.43 83.37
7 131.6 39.24 92.36
8 150.4 49.05 101.35
45
𝜎𝑇𝑜𝑡𝑎𝑙 = 𝛾𝑖𝑍𝑖
𝑛
𝑖=1
Example #5
Calculate and Plot the variation of total, effective
vertical stresses, and pore water pressure with
depth for the following soil profile.
4 m
4 m
2 m
5 m
sat=19.8 KN/m3
sat=21.3 KN/m3
t=17.3 KN/m3
sat=17.8 KN/m3 Soil Layer #1
Soil Layer #2
Soil Layer #3
46
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0 50 100 150 200 250 300 350
De
pth
(m
)
Stress (kPa)
Total Stress (kPa)
Pore Water Pressure (kPa)
Effective Stress (kPa)
The slope of the pore water pressure plot
stays constant with depth.
z (m) v u 'v
0 0 0 0
1 17.3 0 17.3
2 34.6 0 34.6
3 51.9 0 51.9
4 69.2 0 69.2
5 87 9.81 77.19
6 104.8 19.62 85.18
7 124.6 29.43 95.17
8 144.4 39.24 105.16
9 164.2 49.05 115.15
10 184 58.86 125.14
11 205.3 68.67 136.63
12 226.6 78.48 148.12
13 247.9 88.29 159.61
14 269.2 98.1 171.1
15 290.5 107.91 182.59
17.3
17.8
19.8
21.3
Notice the change in the slope
of the total stresses and
effective stresses due to
different layer properties.
47
Example #5-Solution
Example #6
Sensitivity analysis for the variation of effective
stress for different values of saturated unit
weight in the second layer.
4 m
4 m
2 m
5 m
sat= Variable
sat=21.7 KN/m3
t=17.3 KN/m3
sat=17.8 KN/m3 Soil Layer #1
Soil Layer #2
Soil Layer #3
48
Keep in mind that some of the assumed values of the saturated unit weights in this
example are not realistic and are exaggerated. The purpose of this example is to
demonstrate the impact of varying saturated unit weight on the distribution of the
effective stresses in the soil layer.
10 15 20 25 30 35
A B C D E F
z (m)
0 0 0 0 0 0 0
1 17.3 17.3 17.3 17.3 17.3 17.3
2 34.6 34.6 34.6 34.6 34.6 34.6
3 51.9 51.9 51.9 51.9 51.9 51.9
4 69.2 69.2 69.2 69.2 69.2 69.2
5 79.19 79.19 79.19 79.19 79.19 79.19
6 87.18 87.18 87.18 87.18 87.18 87.18
7 87.37 92.37 97.37 102.37 107.37 112.37
8 97.56 112.56 127.56 142.56 157.56 172.56
9 117.75 147.75 177.75 207.75 237.75 267.75
10 147.94 197.94 247.94 297.94 347.94 397.94
11 159.83 209.83 259.83 309.83 359.83 409.83
12 193.42 243.42 293.42 343.42 393.42 443.42
13 248.71 298.71 348.71 398.71 448.71 498.71
14 325.7 375.7 425.7 475.7 525.7 575.7
15 424.39 474.39 524.39 574.39 624.39 674.39
3 (kN/m3)
'v (kPa)
49
Example #6-Solution
Notice the change in the magnitude and the slope of the effective stress distribution lines
with increasing values of saturated unit weight in the second layer.
Soil Layer #2
7
7.5
8
8.5
9
9.5
10
0 50 100 150 200 250 300 350 400 450 500
De
pth
(m
)
Stress (kPa)
A
B
C
D
E
F
50
Example #6-Solution