Chapter 5

Example Problems Involving

In Situ Stresses Under
Hydrostatic Conditions

5.0 GENERAL COMMENTS

The fluid in soils and porous rocks can be present in two forms, namely, as
follows:
l Free water occupying part or all of the voids between the particles, and
l Adsorbed water films surrounding clay particles.
If the voids are completely filled with water, the material is saturated (i.e.,
S ¼ 100%) and the moisture is said to be continuous. If, on the other hand, the
voids are only partially filled with water, the soil is unsaturated. The moisture
is discontinuous and forms “wedges” of water between adjacent particles and
moisture films around them.
Water is an important factor in most geotechnical engineering design and
construction activities. The presence of water strongly affects the engineering
behavior of most soils, especially fine-grained ones (e.g., silts and claysd
recall the discussion of Atterberg limits in Chapter 2).

5.1 SURFACE TENSION

The boundary or interface between air and fluid in the voids is of particular
importance. At liquideair interfaces, the greater attraction of water molecules
to each other (due to cohesion) rather than to molecules in air (due to adhe-
sion) creates an unbalanced molecular attraction of the water. This in turn
gives rise to surface tension, a force that acts parallel to the surface of the
water in all directions1 and causes water to behave as if its surface was covered
with a stretched elastic membrane.

1. Sowers, G.B., Sowers, G.F., 1970. Introductory Soil Mechanics and Foundations. Macmillan
Publishing Co., Inc., New York, NY.

Soil Mechanics. http://dx.doi.org/10.1016/B978-0-12-804491-9.00005-7

Copyright © 2017 Elsevier Inc. All rights reserved. 205
206 Soil Mechanics

Surface tension is a contractive tendency of the surface of a liquid that

allows it to resist an external force. Surface tension is evident, for example,
any time an object that is denser than water is able to float or run along the
water surface.
Because of the relatively high attraction of water molecules for each other,
water has a high surface tension compared to that of most other liquids.
Surface tension can be visualized as a tensile force per unit length (Ts) along
the interface between air and water, acting parallel to the water surface.
Surface tension thus has the dimension of force per unit length (FL1) or of
energy per unit area. The magnitude of this force is approximately
Ts ¼ 72.8 mN/m ¼ 7.426  102 g/cm at 20 C ¼ 4.988  103 lb/ft
(recall that in soil mechanics tensile forces and stresses are denoted by a
negative sign).

5.1.1 Surface Tension Phenomena

Surface tension manifests itself in several aspects of soil behavior, namely,
l In a hole dug in the ground, soil is found to be saturated long before the
groundwater table is reached. This results from the capillary rise of water
in the voids, a phenomenon that is discussed in Section 5.2.
l If a sample of saturated clay is dried, it decreases in volume in the process.
Surface tension acting in the soil voids serves to compress the soil
microfabric and decreases the volume of the sample.
l Dry sand cannot be molded into a ball. However, if the sand moistened, it
can be packed and easily shaped. This moist strength is attributed to the
tension in the interparticle moisture films. If the moist sand is immersed in
water, the moisture films disappear and the sand will again lose its ability
to be molded.

5.2 CAPILLARY PHENOMENA IN TUBES

Capillary rise or capillarity is a phenomenon in which liquid spontaneously
rises or falls in a narrow space such as a thin tube or in the voids of a porous
material. Surface tension is an important factor in the phenomenon of capil-
larity. The surface adhesion forces or internal cohesion present at the interface
between a liquid and a solid stretch the liquid and form a curved surface called
a meniscus (Figure 5.1A). The meniscus is the curve in the upper surface of a
liquid close to the surface of the container or another object, caused by surface
tension. It can be either concave or convex, depending on the liquid and the
surface. Adhesion forces between water and a solid form a concave meniscus.
Internal cohesion in mercury, on the other hand, pulls down the liquid to form
a convex meniscus. Menisci are thus a manifestation of capillary action.
The stress associated with menisci is known as the capillary tension. It is
computed for a cylindrical tube of diameter d (Figure 5.1A) by considering the
 In Situ Stresses Under Hydrostatic Conditions Chapter j 5 207

FIGURE 5.1 (A) Schematic illustration of a meniscus and capillary tension. (B) Geometry
associated with the meniscus.

force developed by the stretched meniscus. If a is the angle of contact between

the meniscus and the solid material (such as glass), then the total unbalanced
upward force (Fu) developed along the perimeter of the meniscus is
Fu ¼ ðTs cos aÞðpdÞ (5.1)
where a zero air pressure has been assumed. For an interface consisting of
water and air-dried glass, a ¼ 0 degree. For an interface consisting of water
and oven-dried glass, a ¼ 45 degrees.
The downward force (Fd) is equal to the weight of the water, which is
computed from the product of the unit weight of water and the volume of the
water that has risen to a capillary height of hc in the tube (Figure 5.2A). Thus,

2
pd
Fd ¼ gw ðhc Þ (5.2)
4
For force equilibrium, Fu ¼ Fd, which leads to the following result:
4Ts cos a
hc ¼ (5.3)
gw d
For example, in the case of an interface consisting of water in contact with
air-dried glass, the surface tension a ¼ 0, giving,
4Ts 1
hc ¼ 0 hc f (5.4)
gw d d
208 Soil Mechanics

FIGURE 5.2 (A) Schematic illustration of capillary tension. (B) Water Hanging on a meniscus.

Furthermore, since Ts ¼ 72.8 mN/m, the magnitude of the capillary rise

as given by Eq. (5.3) is thus




mN N kN
4 72:8
m 1000 mN 1000 N 2:968  105 0:00003
hc ¼ 3
¼ z
ð9:81 kN=m Þd d d
(5.5)
where d and hc have units of meters.
If h < hc, the angle a adjusts so as to satisfy the following equation:
gw dh
cos a ¼ (5.6)
4Ts
The capillary stress is obtained by dividing Fu from Eq. (5.1) by the cross-
sectional area (A) of the tube, giving
Fu ðTs cos aÞðpdÞ 4Ts cos a
sc ¼ ¼ ¼ ¼ gw hc (5.7)
A pd 2 d
4
The capillary tension can be related to the radius of the meniscus (rm) by
considering the geometry of the meniscus (Figure 5.1B). In particular,
d/2 ¼ rm$cosa. Substituting for d into Eq. (5.7) gives
2Ts
sc ¼ (5.8)
rm
Thus for water in contact with air, the capillary tension stress is dependent
only on rm and varies inversely with it.
The maximum capillary tension occurs when the meniscus radius is smallest,
which corresponds to the case where the meniscus is tangent to the tube,
implying that a ¼ 0 degrees and thus rm ¼ d/2. The maximum capillary tension
will thus be scmax ¼ 4Ts =d, where it is understood that scmax will be negative.
 In Situ Stresses Under Hydrostatic Conditions Chapter j 5 209

5.3 CAPILLARY PHENOMENA IN SOILS

If pore water were subject only to the force of gravity, the soil above the
groundwater table would be perfectly dry. In reality, however, every soil in the
field is completely saturated for a certain distance above the groundwater
table2; above this level, it is only partially saturated. Figure 5.3 shows
the relation between a hypothetical saturated aquifer3, the capillary zone, and
the zone of varying degree of saturation.
If a soil is saturated, the airewater interfaces disappear and the capillary
tension becomes zero. When a saturated soil is exposed to open air, capillary
tension develops as soon as evaporation creates menisci at the surface4. Since
the moisture in a saturated soil is continuous, the water tension stress devel-
oped at the airewater interfaces is felt throughout the mass.

FIGURE 5.3 Schematic illustration of a saturated aquifer, capillary zone, and zone of varying
saturation.

2. The water table or phreatic surface refers to the locus of the levels to which water would rise in
observation wells, i.e., where the water pressure head is equal to the atmospheric pressure
(where gauge pressure ¼ 0). It is commonly visualized as the “surface” of the geomaterials that
are saturated with groundwater in a given vicinity.
3. An aquifer is an underground layer of water-bearing permeable rock or unconsolidated materials
(gravel, sand, or silt) from which groundwater can be extracted using a water well.
4. Capillarity is the reason that soils shrink as they dry out. In particular, capillary menisci pull the
particles together.
210 Soil Mechanics

FIGURE 5.4 Schematic illustration of pore pressure distribution with depth in a soil deposit with
capillary rise above the groundwater table.

The water obeys the law of hydrostatics; thus, u ¼ gw z, where z is

measured positive downward. The capillary rise of water in a soil above the
groundwater table illustrates the combined effect of capillary tension and
hydrostatic pressure. Referring to Figure 5.2A, at the groundwater elevation
(free surface) the water pressure is zero. Below the free surface the pressure
increases according to the aforesaid expression for u. In the capillary zone
above the free surface, the water pressure decreases linearly, again in accor-
dance with this expression (only with z being negative). Figure 5.4 shows the
distribution of both positive (compressive) and negative (tensile) pore pressure
with depth for a hypothetical soil with capillary rise above the groundwater
table. Above the capillary zone the pore pressure will be a nonlinear function
of the degree of saturation.
In contrast to the capillary phenomena discussed in Section 5.2, the
continuous voids in soils have a variable width and are by no means straight.
Indeed, the interconnected voids in a soil form a collection of irregular but
definite capillary tubes. The maximum capillary tension that can develop will
vary from point to point, depending on the pore diameter and degree of
saturation5. Consequently, the capillary tube analogy is not directly applicable.
The thickness of the capillary zone in a soil thus depends on the size and
shape of the pores. This, in turn, is a function of the particles sizes and shapes,

5. Sowers, G.B., Sowers, G.F., 1970. Introductory Soil Mechanics and Foundations. Macmillan
Publishing Co., Inc., New York, NY.
 In Situ Stresses Under Hydrostatic Conditions Chapter j 5 211

as well as the void ratio. As the particle size decreases, the size of the voids
likewise decreases, and the height of capillary rise in the soil (hc) increases.
Thus, in general,
1 c
hc f 0 hc ¼ (5.9)
deff deff
where deff is some effective pore size and c is a constant of proportionality.
When hc is expressed in centimeters, c is typically taken to equal 0.3.
Along these same lines, Terzaghi and Peck6 proposed the following
empirical expression:
C
hc ¼ (5.10)
eD10
where hc is in units of centimeters, e is the void ratio, D10 is the effective grain
size (in centimeters) as determined from a sieve analysis (recall Chapter 2),
and C (units of cm2) is an empirical constant that depends on the shape of
particles and on surface impurities. It varies between 0.1 and 0.5 cm2 for loose
and dense sands, respectively.

5.4 IN SITU STRESSES IN SOILS UNDER HYDROSTATIC

CONDITIONS
This section briefly reviews key aspects related to total stress, pore fluid
pressure, and effective stress concept in soils. In all cases, the pore fluid is
assumed to be at rest; i.e., under hydrostatic7 conditions.

5.4.1 Total Stress

Consider the saturated soil deposit without seepage shown in Figure 5.5. The
vertical total stress (sv) is obtained by summing up the densities of the solid
and fluid phase above some point, multiplied by the gravitational acceleration
(g). Mathematically this is written as
Z h
sv ¼ rg dz (5.11)
0

If rg remains constant throughout the soil, then

sv ¼ rgh ¼ gh (5.12)
where rg ¼ g is the moist unit weight of the soil and h is the depth below the
ground surface (the origin of the coordinate system used).

6. Terzaghi, K., Peck, R.B., 1967. Soil Mechanics in Engineering Practice, second ed. John Wiley
and Sons, New York, NY.
7. Hydrostatics or fluid statics is the branch of fluid mechanics that studies incompressible fluids at
rest.
212 Soil Mechanics

FIGURE 5.5 Saturated soil deposit under hydrostatic conditions.

5.4.2 Pore Fluid Pressure

From fluid mechanics, it is known that under hydrostatic (no seepage) con-
ditions the pore fluid pressure (u) at some depth h is simply
u ¼ rw gh ¼ gw h (5.13)

Remark: The pore fluid pressure is also called the “neutral stress” because it has
no shear stress components.8

5.4.3 Effective (Intergranular) Stress

The effective stress is defined as follows:
s0 ¼ s  u (5.14)
where s and u are again the total stress and pore fluid pressure, respectively.
The vertical effective stress is thus
s0v ¼ sv  u (5.15)
The effective stress is approximately the force per unit area carried by the solid
phase (i.e., the soil skeleton); it controls a soil’s volume change and strength. For
example, increases in s0 lead to a denser state of packing in cohesionless soils.

8. By definition, a liquid cannot support static shear stresses; it only has normal stress components
that act equally in all directions.
 In Situ Stresses Under Hydrostatic Conditions Chapter j 5 213

Thus, the vertical total stress for the soil element shown in Figure 5.5 is
sv ¼ gsat h (5.16)
where gsat is the saturated unit weight of the soil. The pore fluid pressure at
this point is
u ¼ rw gh ¼ gw h (5.17)
Finally, the vertical effective stress at this point is thus
s0v ¼ sv  u ¼ gsat h  gw h ¼ ðgsat  gw Þh ¼ gb h (5.18)
where gb is the buoyant or submerged unit weight of the soil.

5.5 RELATIONSHIP BETWEEN HORIZONTAL AND

VERTICAL STRESSES
From hydrostatics the pressure in a liquid is the same in all directions. This is
not, however, true for soils, as the state of stress in situ is not necessarily
hydrostatic.
The determination of the magnitude of the horizontal stress is not
as straightforward as the vertical total stress (recall the discussion of
Section 5.1.1). As such, the general relationship between horizontal and ver-
tical total stresses is
sh ¼ Ksv (5.19)
where K is a positive earth pressure coefficient.

Remark: Since the groundwater table can fluctuate, the total stress will change
with such fluctuations. Thus, K is not a constant.

To remove the effect of a variable groundwater table on the determination

of the horizontal stress, it is expedient to work in terms of effective stresses, i.e.,
s0h ¼ K0 s0v (5.20)
where K0 is the coefficient of lateral earth pressure at rest; it is independent of
the location of the groundwater table. Even if the groundwater table fluctuates,
K0 will remain unchanged so long as the same soil layer is considered and its
density remains unchanged.
The magnitude of K0 is very sensitive to the geologic and engineering
stress history that a soil has been subjected to in the past. In natural soil
deposits,
l K0 ¼ 0.4e0.5 for sedimentary soils.
l K0 may be as large as 3.0 for very heavily preloaded soils.
214 Soil Mechanics

EXAMPLE PROBLEM 5.1

General Remarks
This example problem illustrates the manner in which to compute the capillary
rise in a glass tube.

Problem Statement
Compute (a) the capillary tension (in g/cm) in a 0.002 mm diameter oven-
dried glass tube and (b) the height of capillary rise (in feet) in the tube.

Solution

a) The capillary tension is computed from Eq. (5.7), i.e.,

4Ts cos a
sc ¼ (5.1.1)
d
Since a ¼ 45 degrees for an interface consisting of water and oven-
dried glass tube, Eq. (5.1.1) gives
 
2 g 
4Ts cos a 4 7:426  10 cm ðcos 45 Þ
sc ¼ ¼  cm  ¼ 1.050  103 g=cm2
d ð0:002 mmÞ
10 mm
(5.1.2)

b) The height of capillary rise in the tube is given by Eq. (5.7), i.e.,
sc ð1:050  103 g=cm3 Þ
sc ¼ gw hc 0 hc ¼ ¼ ¼ 1:050  103 cm (5.1.3)
gw 1:0 g=cm3
Converting to units of feet gives



  in ft
hc ¼ 1:050  10 cm3
¼ 34.5 ft (5.1.4)
2:54 cm 12 in
EXAMPLE PROBLEM 5.2
General Remarks
This example problem illustrates the manner in which to estimate capillary rise
above the groundwater table in a sandy soil.

Problem Statement
The effective grain size (D10) of a medium sand is 0.15 mm. The void ratio of
the sand in a dense configuration is 0.45; in a loose configuration it is 0.81.
What is the estimated capillary rise for this sand?
 In Situ Stresses Under Hydrostatic Conditions Chapter j 5 215

Solution
In the loose configuration, C ¼ 0.1 cm2. Thus,
0:1 cm2
hc ¼  cm  ¼ 8.2 cm (5.2.1)
ð0:81Þ 0:15 mm 
10 mm
In the loose configuration, C ¼ 0.5 cm2. Thus,
0:5 cm2
hc ¼  cm  ¼ 74 cm (5.2.2)
ð0:45Þ 0:15 mm 
10 mm

EXAMPLE PROBLEM 5.3

General Remarks
This example problem relates the effective pore size deff to the effective grain
size D10.

Problem Statement
Compute (a) the maximum capillary tension and (b) the theoretical height of
capillary rise in a soil whose effective grain size (D10) is 0.016 mm if the
effective pore size (deff) is estimated to be (D10)/5.

Solution

a) Recalling that the capillary tension in a glass tube is given by Eq. (5.7), i.e.,
4Ts cos a
sc ¼ (5.3.1)
d
The maximum value will be realized for a ¼ 0 degree. Replacing d by
deff, the estimated maximum capillary tension in the sand is thus



mN N
4 72:8
4Ts m 1000 mN 2
scmax ¼ ¼   ¼ 9:100  10 N=m
4
deff 1 m (5.3.2)
ð0:016 mmÞ
5 1000 mm
¼ 9.100  101 kN=m2
where deff is some effective pore size.
216 Soil Mechanics

b) The capillary rise in a glass tube is given by the second part of Eq. (5.7), i.e.,
4Ts cos a sc
sc ¼ ¼ gw hc 0 hc ¼ (5.3.3)
d gw
The magnitude of the estimated maximum capillary rise in the sand is thus
scmax 9:100  101 kN=m2
hc ¼ ¼ ¼ 9.3 m (5.3.4)
gw 9:81 kN=m3

EXAMPLE PROBLEM 5.4

General Remarks
This example problem illustrates how total stress, pore pressure, and effective
stress are computed in the case where the groundwater table lies above the
ground surface.

Problem Statement
Consider a case where the groundwater table is located above the surface of a
saturated soil deposit (Figure Ex. 5.4A). Such conditions are typical of soils in
lakes and in oceans. Determine the variation with depth below the groundwater
table of the total stress, pore fluid pressure, and effective stress.

FIGURE EX. 5.4A Soil deposit with groundwater table above ground surface under hydrostatic
conditions.
 In Situ Stresses Under Hydrostatic Conditions Chapter j 5 217

FIGURE EX. 5.4B Schematic illustration of variation with depth of vertical total stress.

The vertical total stress distribution with depth has two contributions. The
first (gw d) is from the layer of water that is located over the soil layer; the second
(gsat$H ) is due to the saturated unit weight of the soil. Figure Ex. 5.4B shows the
variation of vertical total stress with depth below the groundwater table.
The pore fluid pressure varies with depth in the usual linear fashion, i.e.,
u ¼ gwz. Figure Ex. 5.4C shows the variation with depth of the pore fluid pressure.
Finally, the vertical effective stress is the difference between the total stress
and the pore pressure. The maximum value is thus
s0v ¼ sv  u ¼ ðgw d þ gsat HÞ gw ðd þ HÞ ¼ ðgsat  gw ÞH ¼ gb H (5.4.1)
where gb is the buoyant unit weight9 of the soil. Figure Ex. 5.4D shows the
variation of the effective stress with depth below the groundwater table.

EXAMPLE PROBLEM 5.5

General Remarks
This example problem illustrates how pore pressures, and thus effective stresses,
are computed when capillary rise is present above the groundwater table.

Problem Statement
Given the soil profiles shown in Figures Ex. 5.5A and Ex. 5.5B, compute the
total stress, pore fluid pressure, and effective stress at (a) points A and B in
Figure Ex. 5.5A and (b) points C and D in Figure Ex. 5.5B.

9. Recall the discussion of saturated and buoyant (submerged) unit weights given in Chapter 1.
218 Soil Mechanics

FIGURE EX. 5.4C Schematic illustration of variation with depth of pore fluid pressure.

FIGURE EX. 5.4D Schematic illustration of variation with depth of vertical effective stress.
 In Situ Stresses Under Hydrostatic Conditions Chapter j 5 219

FIGURE EX. 5.5A Soil deposit with capillary zone extending to ground surface.

FIGURE EX. 5.5B Soil deposit with capillary zone not extending to ground surface.
220 Soil Mechanics

Solution

a) In the case of the soil deposit shown in Figure Ex. 5.5A, the capillary rise
above the groundwater table extends to the ground surface. In this figure,
g1 and g2 are the saturated unit weights of the upper and lower soil layer,
respectively.
At point A:
The vertical total stress is
sv ¼ g1 ðH  hA Þ (5.5.1)
Since point A lies above the groundwater table, the pore fluid pressure
is negative, i.e.,
u ¼ gw hA (5.5.2)
The vertical effective stress is thus
s0v ¼ sv  u ¼ g1 ðH  hA Þ þ gw hA ¼ g1 H ðg1  gw ÞhA ¼ g1 H  gb1 hA
(5.5.3)
where gb1 is the buoyant unit weight of the upper soil layer.
At point B:
The vertical total stress is
sv ¼ g1 H þ g2 hB (5.5.4)
Since point B lies below the groundwater table, the pore fluid pressure
is positive, i.e.,
u ¼ gw hB (5.5.5)
The vertical effective stress is thus
s0v ¼ sv  u ¼ g1 H þ g2 hB gw hB ¼ g1 H þ ðg2  gw ÞhB ¼ g1 H  gb2 hB
(5.5.6)
where gb2 is the buoyant unit weight of the lower soil layer.

b) In the case of the soil deposit shown in Figure Ex. 5.5B, the capillary rise
above the groundwater table does not extend to the ground surface. In this
figure, g is the moist unit weight of the soil in the zone of varying degree of
saturation, and g1 and g2 are again the saturated unit weights of the upper
and lower soil layer, respectively.
At point C:
The vertical total stress is
sv ¼ gðH  hcap Þ þ g1 ðhcap  hC Þ (5.5.7)
 In Situ Stresses Under Hydrostatic Conditions Chapter j 5 221

Since point C lies above the groundwater table, the pore fluid pressure
is negative, i.e.,
u ¼ gw hC (5.5.8)
The vertical effective stress is thus
s0v ¼ sv  u ¼ gðH  hcap Þ þ g1 ðhcap  hC Þ þ gw hC
¼ gðH  hcap Þ þ g1 hcap ðg1  gw ÞhC ¼ gðH  hcap Þ þ g1 hcap  gb1 hC
(5.5.9)
where gb1 is again the buoyant unit weight of the upper soil layer.
At point D:
The vertical total stress is
sv ¼ gðH  hcap Þ þ g1 hcap þ g2 hD (5.5.10)
Since point D lies below the groundwater table, the pore fluid pressure
is positive, i.e.,
u ¼ g w hD (5.5.11)
The vertical effective stress is thus
s0v ¼ sv  u ¼ gðH  hcap Þ þ g1 hcap þ g2 hD  gw hD
¼ gðH  hcap Þ þ g1 hcap þ ðg2  gw ÞhD ¼ gðH  hcap Þ þ g1 hcap þ gb2 hD
(5.5.12)
where gb2 is again the buoyant unit weight of the lower soil layer.

EXAMPLE PROBLEM 5.6

General Remarks
This example problem illustrates the manner in which in situ stresses are
computed under hydrostatic conditions in the presence of capillary rise above
the groundwater table.

Problem Statement
A 10 m thick soil deposit overlies a layer of soft rock. The groundwater table is
approximately 5 m above the surface of the rock and the height of capillary rise is
approximately 3.5 m. The soil has an average void ratio (e) of 0.36 and a specific
gravity of solids (Gs) equal to 2.68. No seepage is present at the site. Determine
the variation with depth of the vertical total stress, the pore fluid pressure, and the
vertical effective stress in the deposit at depths of 2.5, 5.0, 7.5, and 10.0 m below
the ground surface assuming (a) capillary rise above the groundwater table as
stated above and (b) no capillary rise. Where not saturated, the soil has a moisture
content (w) of 6% and a degree of saturation (S) equal to 45%.
222 Soil Mechanics

Solution
Figure Ex. 5.6A shows the single soil profile and the extent of the capillary
rise.
First, all of the necessary unit weights are determined. For the unsaturated
portion of the soil deposit,
gw ðGs þ SeÞ ð9:81 kN=m3 Þ½2:68 þ ð0:45Þð0:36Þ
g¼ ¼ ¼ 20:50 kN=m3
1þe 1 þ 0:36
(5.6.1)
Similarly, for the unsaturated portion of the soil deposit,
gw ðGs þ eÞ ð9:81 kN=m3 Þ½2:68 þ 0:36
g¼ ¼ ¼ 21:93 kN=m3 (5.6.2)
1þe 1 þ 0:36

a) For capillary rise above the groundwater table as shown in Figure Ex.
5.6A, the vertical total stress, pore fluid pressure, and vertical effective
stresses are next computed at the requested depths.
At a depth of 2.5 m:
The vertical total stress is
   
sv ¼ 20:50 kN=m3 ð1:5 mÞ þ 21:93 kN=m3 ð1:0 mÞ ¼ 52.68 kN=m2
(5.6.3)

Extent of capillary rise

5.0 m

3.5 m

Soil deposit
5.0 m

Rock layer
FIGURE EX. 5.6A Profile consisting of a single soil layer.
 In Situ Stresses Under Hydrostatic Conditions Chapter j 5 223

The pore fluid pressure is

 
u ¼  9:81 kN=m3 ð1:0 mÞ ¼ 9.81 kN=m2 (5.6.4)
The vertical effective stress is thus
s0v ¼ sv u ¼ 52:68 ð9:81Þ ¼ 62.49 kN=m2 (5.6.5)
At a depth of 5.0 m:
The vertical total stress is
   
sv ¼ 20:50 kN=m3 ð1:5 mÞ þ 21:93 kN=m3 ð3:5 mÞ ¼ 107.5 kN=m2
(5.6.6)
The pore fluid pressure is
u ¼ 0.0 (5.6.7)
The vertical effective stress is thus
s0v ¼ sv  u ¼ 107:5  0 ¼ 107.5 kN=m2 (5.6.8)
At a depth of 7.5 m:
The vertical total stress is
   
sv ¼ 20:50 kN=m3 ð1:5 mÞ þ 21:93 kN=m3 ð6:0 mÞ ¼ 162.3 kN=m2
(5.6.9)
The pore fluid pressure is
 
u ¼ 9:81 kN=m3 ð2:5 mÞ ¼ 24.53 kN=m2 (5.6.10)
The vertical effective stress is thus
s0v ¼ sv  u ¼ 162:3  24:53 ¼ 137.8 kN=m2 (5.6.11)
At a depth of 10.0 m:
The vertical total stress is
   
sv ¼ 20:50 kN=m3 ð1:5 mÞ þ 21:93 kN=m3 ð8:5 mÞ ¼ 217.2 kN=m2
(5.6.12)
The pore fluid pressure is
 
u ¼ 9:81 kN=m3 ð5:0 mÞ ¼ 49.05 kN=m2 (5.6.13)
The vertical effective stress is thus
s0v ¼ sv  u ¼ 217:2  49:05 ¼ 168.1 kN=m2 (5.6.14)
Figure Ex. 5.6B shows the variation with depth of the vertical total
stress, pore pressure, and vertical effective stress.
224 Soil Mechanics

0.0

Vertical total stress

Depth below ground surface (meters)

Pore pressure
2.0 2.50 m Vertical effective stress

4.0
5.00 m

6.0

7.50 m

8.0

10.00 m
10.0
0.0 50.0 100.0 150.0 200.0
Stress or pore pressure (kilonewtons per square meter)
FIGURE EX. 5.6B Variation with depth of vertical total stress, pore fluid pressure, and vertical
effective stress when considering capillary rise.

b) If capillary rise above the groundwater table is ignored, the vertical total
stress, pore fluid pressure, and vertical effective stresses are next computed
at the requested depths. The degree of saturation above the groundwater
table is still 45%.
At a depth of 2.5 m:
The vertical total stress is
 
sv ¼ 20:50 kN=m3 ð2:5 mÞ ¼ 51.25 kN=m2 (5.6.15)
The pore fluid pressure is
u ¼ 0.0 (5.6.16)
The vertical effective stress is thus
s0v ¼ sv  u ¼ 51:25  0 ¼ 51.25 kN=m2 (5.6.17)
At a depth of 5.0 m:
The vertical total stress is
 
sv ¼ 20:50 kN=m3 ð5:0 mÞ ¼ 102.5 kN=m2 (5.6.18)
The pore fluid pressure is
u ¼ 0.0 (5.6.19)
 In Situ Stresses Under Hydrostatic Conditions Chapter j 5 225

The vertical effective stress is thus

s0v ¼ sv  u ¼ 102:5  0 ¼ 102.5 kN=m2 (5.6.20)
At a depth of 7.5 m:
The vertical total stress is
   
sv ¼ 20:50 kN=m3 ð5:0 mÞ þ 21:93 kN=m3 ð2:5 mÞ ¼ 157.3 kN=m2
(5.6.21)
The pore fluid pressure is
 
u ¼ 9:81 kN=m3 ð2:5 mÞ ¼ 24.53 kN=m2 (5.6.22)
The vertical effective stress is thus
s0v ¼ sv  u ¼ 157:3  24:53 ¼ 132.8 kN=m2 (5.6.23)
At a depth of 10.0 m:
The vertical total stress is
   
sv ¼ 20:50 kN=m3 ð5:0 mÞ þ 21:93 kN=m3 ð5:0 mÞ ¼ 212.2 kN=m2
(5.6.24)
The pore fluid pressure is
 
u ¼ 9:81 kN=m3 ð5:0 mÞ ¼ 49.05 kN=m2 (5.6.25)
The vertical effective stress is thus

s0v ¼ sv  u ¼ 212:2  49:05 ¼ 163.1 kN=m2 (5.6.26)

Figure Ex. 5.6C shows the variation with depth of the vertical total
stress, pore fluid pressure, and vertical effective stress.

EXAMPLE PROBLEM 5.7

General Remarks
This example problem illustrates the manner in which in situ stresses are
computed in the presence of capillary rise above the groundwater table.

Problem Statement
Figure Ex. 5.7A shows the soil profile at a specific site. No seepage is present
at the site. The following properties are known for the respective soil layers:
l Sand layer: Gs ¼ 2.70; moisture content of 30%.
l Silt layer: saturated unit weight of 127 lb/ft3.
l Weald clay layer: buoyant unit weight 45 lb/ft3.
226 Soil Mechanics

0.0

Vertical total stress

Depth below ground surface (meters)

Pore pressure
2.0 2.50 m Vertical effective stress

4.0
5.00 m

6.0

7.50 m

8.0

10.00 m
10.0
0.0 50.0 100.0 150.0 200.0
Stress or pore pressure (kiloNewtons per square meter)
FIGURE EX. 5.6C Variation with depth of vertical total stress, pore fluid pressure, and vertical
effective stress when ignoring capillary rise.

Assume the groundwater table to be 8 ft below the ground surface with

capillary rise in the sand layer that extends to the ground surface. Determine
the vertical total stress, the pore fluid pressure, and the vertical effective stress
at depths of 0, 8, 20, 25, and 45 ft.

Solution
The correct unit weights to use for the respective soil layers are first deter-
mined. It is important to note that in this problem, all of the layers are saturated.
Using the expression developed in Case 1.3 of Chapter 1, for the sand layer
Gs w ð2:70Þð0:30Þ
e¼ ¼ ¼ 0:810 (5.7.1)
S 1:0
Thus, from Case 1.8 of Chapter 1,
ðGs þ eÞgw ð2:70 þ 0:810Þð62:4 lb=ft3 Þ
gsat ¼ ¼ ¼ 121:0 lb=ft3 (5.7.2)
sand
1þe 1 þ 0:810
For the silt  layer the saturated unit weight is given, i.e.,
gsat silt ¼ 127lb ft3 .
Finally, for the Weald clay,
gsat clay ¼ g0 þ gw ¼ 45:0 þ 62:4 ¼ 107:4 lb=ft3 (5.7.3)
 In Situ Stresses Under Hydrostatic Conditions Chapter j 5 227

FIGURE EX. 5.7A Soil profile consisting of a sand, silt, and clay layer (not to scale).

At a depth of 0 ft:
The vertical total stress is
sv ¼ 0.0 (5.7.4)
The pore fluid pressure is
 
u ¼  62:4 lb=ft3 ð8 ftÞ ¼ 499.2 lb=ft2 (5.7.5)
The vertical effective stress is thus
s0v ¼ 0 ð 499:2Þ ¼ 499.2 lb=ft2 (5.7.6)
228 Soil Mechanics

At a depth of 8 ft:
The vertical total stress is
 
sv ¼ 121:0 lb=ft3 ð8 ftÞ ¼ 968.0 lb=ft2 (5.7.7)
The pore fluid pressure is
u ¼ 0.0 (5.7.8)
The vertical effective stress is thus
s0v ¼ 968:0  0 ¼ 968.0 lb=ft2 (5.7.9)
At a depth of 20 ft:
The vertical total stress is
 
sv ¼ 121:0 lb=ft3 ð20 ftÞ ¼ 2420.0 lb=ft2 (5.7.10)
The pore fluid pressure is
 
u ¼ 62:4 lb=ft3 ð12 ftÞ ¼ 748.8 lb=ft2 (5.7.11)
The vertical effective stress is thus
s0v ¼ 2420:0  748:8 ¼ 1671.2 lb=ft2 (5.7.12)
At a depth of 25 ft:
The vertical total stress is
 
sv ¼ 2420.0 lb=ft2 þ 127:0 lb=ft3 ð5 ftÞ ¼ 3055.0 lb=ft2 (5.7.13)
The pore fluid pressure is
 
u ¼ 62:4 lb=ft3 ð17 ftÞ ¼ 1060.8 lb=ft2 (5.7.14)
The vertical effective stress is thus
s0v ¼ 3055:0  1060:8 ¼ 1994.2 lb=ft2 (5.7.15)
At a depth of 45 ft:
The vertical total stress is
   
sv ¼ 3055.0 lb=ft2 þ 127:0 lb=ft3 ð5 ftÞ þ 107:4 lb=ft3 ð15 ftÞ
(5.7.16)
¼ 5301.0 lb=ft2
The pore fluid pressure is
 
u ¼ 62:4 lb=ft3 ð12 þ 10 þ 15 ftÞ ¼ 2308.8 lb=ft2 (5.7.17)
The vertical effective stress is thus
s0v ¼ 5301:0  2308:8 ¼ 2992.2 lb=ft2 (5.7.18)
Figure Ex. 5.7B shows the variation with depth of the vertical total stress, pore
fluid pressure, and vertical effective stress.
 In Situ Stresses Under Hydrostatic Conditions Chapter j 5 229

0.0 0.0 ft
Vertical total stress
8.0 ft Pore pressure
Depth below ground surface (feet)

10.0 Vertical effective stress

20.0 20.0 ft

25.0 ft

30.0

40.0

45.0 ft

50.0
-1000.0 0.0 1000.0 2000.0 3000.0 4000.0 5000.0 6000

Stress or pore pressure (pounds per square foot)

FIGURE EX. 5.7B Variation with depth of vertical total stress, pore fluid pressure, and effective
stress in soil profile consisting of a sand, silt, and clay layer.

EXAMPLE PROBLEM 5.8

General Remarks
This example problem illustrates the manner in which in situ stresses are
computed under hydrostatic conditions. The second part of the problem
considers the rapid rise of the groundwater table and its affect on the pore fluid
pressure and effective stress state.

Problem Statement
Borehole data at a site reveals the soil profile shown in Figure Ex. 5.8A. The
groundwater table is found at a depth of 3.6 m. No seepage is present at the
site.
Some details pertaining to the soil profile are given as follows:
l The top 2.0 m consists of very fine, wet sand with silt. Laboratory tests
indicate that for this soil the moisture content (w) is 5%, the degree of
saturation (S) is 40%, and the specific gravity of solids (Gs) equals 2.69.
l The next 3.4 m consists of fine sand. Laboratory tests indicate that for this
soil, Gs ¼ 2.68. Above the groundwater table, w ¼ 8% and S ¼ 78%.
Below the groundwater table, w ¼ 12%.
l The final 15.2 m consists of soft blue clay. Laboratory tests indicate that for
this soil, w ¼ 32% and Gs ¼ 2.71.
230 Soil Mechanics

Depth

0.0 m
Sand with silt w = 5%, S = 40%, Gs = 2.69 2.0 m
2.0 m
w = 8%, S = 78%, Gs = 2.68 1.6 m
3.6 m

w = 12% 1.8 m
Fine sand

5.4 m

w = 32%, Gs = 2.71

Soft blue clay

15.2 m

20.6 m

Rock layer
FIGURE EX. 5.8A Soil profile based on borehole data (not to scale).

a) Determine the vertical total stress, pore pressure, and vertical effective
stress at depths of 0.0, 2.0, 3.6, 5.4, and 20.6 m. If the coefficient of lateral
earth pressure at rest (K0) is equal to 0.55, determine the lateral effective
stress at the same depths.
b) If the groundwater table were to rise rapidly to the ground surface,
determine the vertical total stress, pore pressure, and vertical effective
stress at depths of 2.0, 5.4, and 20.6 m. Assume that the void ratio in the
sand layers remains unchanged during the rise in groundwater.

Solution
First, all of the necessary unit weights are determined. Since the moisture
content, degree of saturation, and the specific gravity of solids are known for
 In Situ Stresses Under Hydrostatic Conditions Chapter j 5 231

each of the soils, it is timely to substitute the relation e ¼ Gs w=S into the
general expression for moist unit weight to give the following relation:
gw Gs ð1 þ wÞ gw Gs ð1 þ wÞ
g¼ ¼
 (5.8.1)
1þe Gs w
1þ
S
For the very fine wet sand with silt, w ¼ 5%, S ¼ 40%, and Gs ¼ 2.69. The
moist unit weight is thus
ð9:81 kN=m3 Þð2:69Þð1 þ 0:05Þ
g¼
¼ 20:74 kN=m3 (5.8.2)
ð2:69Þð0:05Þ
1þ
0:40
For the fine sand above the groundwater table, w ¼ 8%, S ¼ 78%, and
Gs ¼ 2.68. The moist unit weight is thus
ð9:81 kN=m3 Þð2:68Þð1 þ 0:08Þ
g¼
¼ 22:27 kN=m3 (5.8.3)
ð2:68Þð0:08Þ
1þ
0:78
For the fine sand below the groundwater table, w ¼ 12%, S ¼ 100%, and
Gs ¼ 2.68. The moist unit weight of the fine sand is thus
ð9:81 kN=m3 Þð2:68Þð1 þ 0:12Þ
g¼
¼ 22:28 kN=m3 (5.8.4)
ð2:68Þð0:12Þ
1þ
1:00
Finally, for the soft blue clay, w ¼ 32%, S ¼ 100%, and Gs ¼ 2.71. The
moist unit weight of this soil is thus
ð9:81 kN=m3 Þð2:71Þð1 þ 0:32Þ
g¼
¼ 18:79 kN=m3 (5.8.5)
ð2:71Þð0:32Þ
1þ
1:00

a) The vertical total stress, pore fluid pressure, and vertical and horizontal
(lateral) effective stresses are next computed at the requested depths.
At a depth of 0.0 m:
sv ¼ 0.0 u ¼ 0.0; s0v ¼ s0h ¼ 0.0 (5.8.6)
At a depth of 2.0 m:
The vertical total stress is
 
sv ¼ 20:74 kN=m3 ð2:0 mÞ ¼ 41.48 kN=m2 (5.8.7)
232 Soil Mechanics

Since the soil is not saturated, capillary rise in the very fine wet sand
with silt is ignored. The pore fluid pressure is thus
u ¼ 0.0 (5.8.8)
The vertical effective stress is thus equal to the vertical total stress, i.e.,
s0v ¼ sv  u ¼ 41.48 kN=m2 (5.8.9)
The horizontal (lateral) effective stress is thus
 
s0h ¼ K0 s0v ¼ 0:55 41:48 kN=m2 ¼ 22.81 kN=m2 (5.8.10)
At a depth of 3.6 m:
The vertical total stress is
 
sv ¼ 41:48 kN=m2 þ 22:27 kN=m3 ð1:6 mÞ ¼ 77.11 kN=m2 (5.8.11)
Since the soil is not saturated, capillary rise in the fine sand is ignored.
The pore fluid pressure is thus
u ¼ 0.0 (5.8.12)
The vertical effective stress is thus again equal to the vertical total
stress, i.e.,
s0v ¼ sv  u ¼ 77.11 kN=m2 (5.8.13)
The horizontal (lateral) effective stress is thus
 
s0h ¼ K0 s0v ¼ 0:55 77:11 kN=m2 ¼ 42.41 kN=m2 (5.8.14)
At a depth of 5.4 m:
The vertical total stress is
 
sv ¼ 77:11 kN=m2 þ 22:28 kN=m3 ð1:8 mÞ ¼ 117.2 kN=m2 (5.8.15)
The pore fluid pressure is
 
u ¼ 9:81 kN=m3 ð1:8 mÞ ¼ 17.66 kN=m2 (5.8.16)
The vertical effective stress is thus
s0v ¼ sv  u ¼ 117:2  17:66 ¼ 99.54 kN=m2 (5.8.17)
Finally, the horizontal (lateral) effective stress is
 
s0h ¼ K0 s0v ¼ 0:55 99:54 kN=m2 ¼ 54.75 kN=m2 (5.8.18)
At a depth of 20.6 m:
The vertical total stress is
 
sv ¼ 117:2 kN=m2 þ 18:79 kN=m3 ð15:2 mÞ ¼ 402.8 kN=m2 (5.8.19)
 In Situ Stresses Under Hydrostatic Conditions Chapter j 5 233

0.0
2.0 m Vertical total stress
Pore pressure
Depth below ground surface (meters)

3.6 m
Vertical effective stress
5.0 5.4 m Horizontal effective stress

10.0

15.0

20.0 20.6 m

0.0 100.0 200.0 300.0 400.0

Stress or pore pressure (kilonewtons per square meter)
FIGURE EX. 5.8B Variation with depth of vertical total stress, pore fluid pressure, and vertical
and horizontal effective stress.

The pore fluid pressure is

 
u ¼ 9:81 kN=m3 ð1:8 þ 15:2 mÞ ¼ 166.8 kN=m2 (5.8.20)
The vertical effective stress is thus
s0v ¼ sv  u ¼ 402:8  166:8 ¼ 236.0 kN=m2 (5.8.21)
Finally, the horizontal (lateral) effective stress is
 
s0h ¼ K0 s0v ¼ 0:55 236:0 kN=m2 ¼ 129.8 kN=m2 (5.8.22)
Figure Ex. 5.8B shows the variation with depth of the vertical total
stress, pore fluid pressure, and vertical and horizontal effective stress.

b) If the groundwater table rises rapidly to the ground surface, the very fine,
wet sand with silt, as well as the fine sand will be saturated. For both soils,
the rapid rise in groundwater is assumed to take place without change
in the void ratio. The initial void ratio in the wet sand with silt layer
before the rise in groundwater table is
Gs w ð2:69Þð0:05Þ
e¼ ¼ ¼ 0:336 (5.8.23)
S 0:40
234 Soil Mechanics

Since this void ratio is assumed to be unchanged, the saturated unit

weight in the wet sand with silt layer is thus
gw ðGs þ eÞ ð9:81 kN=m3 Þð2:69 þ 0:336Þ
gsat ¼ ¼ ¼ 22:22 kN=m3 (5.8.24)
1þe 1 þ 0:336
For the entire fine sand layer,
ð9:81 kN=m3 Þð2:68Þð1 þ 0:12Þ
gsat ¼ ¼ 22:28 kN=m3 (5.8.25)
1 þ ð2:68Þð0:12Þ
At a depth of 0.0 m:
sv ¼ 0.0; u ¼ 0.0; s0v ¼ s0h ¼ 0.0 (5.8.26)
At a depth of 2.0 m:
The vertical total stress is
 
sv ¼ 22:22 kN=m3 ð2:0 mÞ ¼ 44.44 kN=m2 (5.8.27)
The pore fluid pressure is
 
u ¼ 9:81 kN=m3 ð2:0 mÞ ¼ 19.62 kN=m2 (5.8.28)
The vertical effective stress is thus
s0v ¼ sv  u ¼ 44:44  19:62 ¼ 24.82 kN=m2 (5.8.29)
Finally, the horizontal (lateral) effective stress is
 
s0h ¼ K0 s0v ¼ 0:55 24:82 kN=m2 ¼ 13.65 kN=m2 (5.8.30)
At a depth of 3.6 m:
The vertical total stress is
 
sv ¼ 44:44 kN=m2 þ 22:28 kN=m3 ð1:6 mÞ ¼ 80.09 kN=m2 (5.8.31)
The pore fluid pressure is
 
u ¼ 9:81 kN=m3 ð3:6 mÞ ¼ 35.32 kN=m2 (5.8.32)
The vertical effective stress is thus
s0v ¼ sv  u ¼ 80:09  35:32 ¼ 44.77 kN=m2 (5.8.33)
The horizontal (lateral) effective stress is thus
 
s0h ¼ K0 s0v ¼ 0:55 44:77 kN=m2 ¼ 24.63 kN=m2 (5.8.34)
At a depth of 5.4 m:
The vertical total stress is
 
sv ¼ 80:09 kN=m2 þ 22:28 kN=m3 ð1:8 mÞ ¼ 120.2 kN=m2 (5.8.35)
 In Situ Stresses Under Hydrostatic Conditions Chapter j 5 235

The pore fluid pressure is

 
u ¼ 9:81 kN=m3 ð5:4 mÞ ¼ 52.97 kN=m2 (5.8.36)
The vertical effective stress is thus
s0v ¼ sv  u ¼ 120:2  52:97 ¼ 67.23 kN=m2 (5.8.37)
The horizontal (lateral) effective stress is thus
 
s0h ¼ K0 s0v ¼ 0:55 67:23 kN=m2 ¼ 37.00 kN=m2 (5.8.38)
At a depth of 20.6 m:
The vertical total stress is
 
sv ¼ 120:2 kN=m2 þ 18:80 kN=m3 ð15:2 mÞ ¼ 406.0 kN=m2 (5.8.39)
The pore fluid pressure is
 
u ¼ 9:81 kN=m3 ð20:6 mÞ ¼ 202.1 kN=m2 (5.8.40)
The vertical effective stress is thus
s0v ¼ sv  u ¼ 406:0  202:1 ¼ 203.9 kN=m2 (5.8.41)
The horizontal (lateral) effective stress is thus
 
s0h ¼ K0 s0v ¼ 0:55 203:9 kN=m2 ¼ 112.1 kN=m2 (5.8.42)
Figure Ex. 5.8C shows the variation with depth of the vertical total stress, pore
fluid pressure, and vertical and horizontal effective stress.

EXAMPLE PROBLEM 5.9

General Remarks
This example problem illustrates the manner in which in situ stresses are
computed under hydrostatic conditions. The second part of the problem con-
siders a lowering of the groundwater table (e.g., by the process of dewatering)
and its effect on the pore fluid pressure and effective stress state.

Problem Statement
Figure Ex. 5.9A shows a soil profile consisting of a 3 m thick gravel fill, an
8 m thick layer of sand, a 10 m thick layer of soft silty clay, and an 8 m thick
stiff clay layer. The 2 m portion of the sand layer above the ground water layer
is saturated by capillary rise, while the gravel fill is unaffected by the capil-
larity. No seepage is present at the site. The following properties are known for
the respective soil layers:
l Gravel fill: Gs ¼ 2.75, w ¼ 12.5%, S ¼ 67%, gd ¼ 17.8 kN/m3, K0 ¼ 1.20.
l Sand layer: Gs ¼ 2.69, w ¼ 16.0%, K0 ¼ 0.470.
 236 Soil Mechanics

0.0
2.0 m
Vertical total stress
Depth below ground surface (meters)

3.6 m Pore pressure

5.0 Vertical effective stress
5.4 m
Horizontal effective stress

10.0

15.0

20.0 20.6 m

0.0 100.0 200.0 300.0 400.0

Stress or pore pressure (kilonewtons per square meter)
FIGURE EX. 5.8C Variation with depth of vertical total stress, pore fluid pressure, and vertical
and horizontal effective stress.

l Soft silty clay: Gs ¼ 2.65, w ¼ 65.0%, K0 ¼ 0.658.

l Stiff clay layer: Gs ¼ 2.68, w ¼ 20.0%, K0 ¼ 1.00.
where K0 is the coefficient of lateral earth pressure at rest.

a) Determine the vertical total stress (sv), pore pressure (u), vertical effective
stress (s0v ), and horizontal effective stress (s0h ) at depths of 1.5, 4, 8, 16, and
25 m.
b) If the groundwater table is next lowered by 6 m, what is the vertical total
stress, the pore pressure, vertical effective stress, and horizontal effective
stress at a depth of 16 m? Assume that the sand layer remains saturated.

Solution
The correct unit weights to use for the respective soil layers are first deter-
mined. For the gravel fill:
 
g ¼ gd ð1 þ wÞ ¼ 17:8 kN=m3 ð1 þ 0:125Þ ¼ 20:03 kN=m3 (5.9.1)
 In Situ Stresses Under Hydrostatic Conditions Chapter j 5 237

Depth

0.0 m
Gravel fill
3.0 m

5.0 m

Sand

11.0 m

Soft silty clay

21.0 m

Stiff clay

29.0 m

Bedrock
FIGURE EX. 5.9A Soil profile consisting of gravel fill, and sand, silty clay, and stiff clay layers.

For the sand layer:

g Gs ð1 þ wÞ ð9:81 kN=m3 Þð2:69Þð1 þ 0:16Þ
gsat ¼ w ¼ ¼ 21:40 kN=m3
1 þ Gs w 1 þ ð2:69Þð0:16Þ
(5.9.2)
For the soft silty clay layer:
g Gs ð1 þ wÞ ð9:81 kN=m3 Þð2:65Þð1 þ 0:65Þ
gsat ¼ w ¼ ¼ 15:76 kN=m3
1 þ Gs w 1 þ ð2:65Þð0:65Þ
(5.9.3)
238 Soil Mechanics

For the stiff clay layer:

gw Gs ð1 þ wÞ ð9:81 kN=m3 Þð2:68Þð1 þ 0:20Þ
gsat ¼ ¼ ¼ 20:54 kN=m3
1 þ Gs w 1 þ ð2:68Þð0:20Þ
(5.9.4)
a) The hydrostatic stresses are determined for the initial location of the
groundwater table as shown in Figure Ex. 5.4A.
At a depth of 1.5 m:
The vertical total stress is
 
sv ¼ 20:03 kN=m3 ð1:5 mÞ ¼ 30.04 kN=m2 (5.9.5)
The pore fluid pressure is
u ¼ 0.0 (5.9.6)
The vertical effective stress is thus
s0v ¼ sv  u ¼ 30.04 kN=m2 (5.9.7)
The horizontal effective stress is thus
 
s0h ¼ K0 s0v ¼ ð1:20Þ 30:04 kN=m2 ¼ 36.05 kN=m2 (5.9.8)
At a depth of 4.0 m:
The vertical total stress is
   
sv ¼ 20:03 kN=m3 ð3:0 mÞ þ 21:40 kN=m3 ð1:0 mÞ ¼ 81.49 kN=m2
(5.9.9)
The pore fluid pressure is
 
u ¼  9:81 kN=m3 ð1:0 mÞ ¼ 9.81 kN=m2 (5.9.10)
The vertical effective stress is thus
s0v ¼ 81:49 ð9:81Þ ¼ 91.30 kN=m2 (5.9.11)
The horizontal effective stress is thus
 
s0h ¼ K0 s0v ¼ ð0:470Þ 91:30 kN=m2 ¼ 42.91 kN=m2 (5.9.12)
At a depth of 8.0 m:
The vertical total stress is
   
sv ¼ 20:03 kN=m3 ð3:0 mÞ þ 21:40 kN=m3 ð5:0 mÞ ¼ 167.1 kN=m2
(5.9.13)
 In Situ Stresses Under Hydrostatic Conditions Chapter j 5 239

The pore fluid pressure is

 
u ¼ 9:81 kN=m3 ð3:0 mÞ ¼ 29.43 kN=m2 (5.9.14)
The vertical effective stress is thus
s0v ¼ 167:1  29:43 ¼ 137.7 kN=m2 (5.9.15)
The horizontal effective stress is thus
 
s0h ¼ K0 s0v ¼ ð0:470Þ 137:7 kN=m2 ¼ 64.70 kN=m2 (5.9.16)
At a depth of 16.0 m:
The vertical total stress is
   
sv ¼ 20:03 kN=m3 ð3:0 mÞ þ 21:40 kN=m3 ð8:0 mÞ
  (5.9.17)
þ 15:76 kN=m3 ð5:0 mÞ ¼ 310.1 kN=m2
The pore fluid pressure is
 
u ¼ 9:81 kN=m3 ð11:0 mÞ ¼ 107.9 kN=m2 (5.9.18)
The vertical effective stress is thus
s0v ¼ 310:1  107:9 ¼ 202.2 kN=m2 (5.9.19)
The horizontal effective stress is thus
 
s0h ¼ K0 s0v ¼ ð0:658Þ 202:2 kN=m2 ¼ 133.0 kN=m2 (5.9.20)
At a depth of 25.0 m:
The vertical total stress is
   
sv ¼ 20:03 kN=m3 ð3:0 mÞ þ 21:40 kN=m3 ð8:0 mÞ
   
þ 15:76 kN=m3 ð10:0 mÞ þ 20:54 kN=m3 ð4:0 mÞ ¼ 471.1 kN=m2
(5.9.21)
The pore fluid pressure is
 
u ¼ 9:81 kN=m3 ð20:0 mÞ ¼ 196.2 kN=m2 (5.9.22)
The vertical effective stress is thus
s0v ¼ 471:1  196:2 ¼ 274.9 kN=m2 (5.9.23)
The horizontal effective stress is thus
 
s0h ¼ K0 s0v ¼ ð1:0Þ 274:9 kN=m2 ¼ 274.9 kN=m2 (5.9.24)
Figure Ex. 5.9B shows the variation with depth of the vertical total
stress, pore fluid pressure, and vertical effective stress.
240 Soil Mechanics

0.0
1.50 m
Depth below ground surface (meters)

4.00 m Vertical total stress

5.0 Pore pressure
Vertical effective stress
8.00 m Horizontal effective stress

10.0

15.0
16.00 m

20.0

25.00 m
25.0

0.0 100.0 200.0 300.0 400.0 500.0

Stress or pore pressure (kilonewtons per square meter)
FIGURE EX. 5.9B Variation with depth of vertical total stress, pore fluid pressure, and vertical
and horizontal effective stress when ignoring capillary rise.

b) The hydrostatic stresses are next determined for the case where the
groundwater table is lowered by 6 m.
At a depth of 16.0 m:
The vertical total stress is unchanged, i.e.,
   
sv ¼ 20:03 kN=m3 ð3:0 mÞ þ 21:40 kN=m3 ð8:0 mÞ
  (5.9.25)
þ 15:76 kN=m3 ð5:0 mÞ ¼ 310.1 kN=m2
The pore fluid pressure is
 
u ¼ 9:81 kN=m3 ð5:0 mÞ ¼ 49.05 kN=m2 (5.9.26)
The vertical effective stress is thus
s0v ¼ 310:1  49:05 ¼ 261.1 kN=m2 (5.9.27)
Finally, the horizontal effective stress is thus
 
s0h ¼ K0 s0v ¼ ð0:658Þ 261:1 kN=m2 ¼ 171.8kN=m2 (5.9.28)

EXAMPLE PROBLEM 5.10

General Remarks
This example problem illustrates how information is obtained from in situ
stresses computed under hydrostatic conditions.
 In Situ Stresses Under Hydrostatic Conditions Chapter j 5 241

Problem Statement
At a given site the soil stratum consists of a thick clay layer. The groundwater
table is located 1.5 m below the ground surface. Above the groundwater table
the degree of saturation (S) is 92.5%. If the specific gravity of solids (Gs) is
equal to 2.71 and the void ratio in the clay is 1.21, determine (a) at which
depth (d) below the ground surface the vertical effective stress will be equal to
120 kPa and (b) the vertical effective stress at this depth immediately after the
groundwater table is lowered by 2.0 m.

Solution

a) Above the groundwater table, the moist unit weight is computed using the
expression determined in Case 1.7 of Chapter 1, i.e.,
gw ðGs þ SeÞ ð9:81 kN=m3 Þ½2:71 þ ð0:925Þð1:21Þ
g¼ ¼ ¼ 17:0 kN=m3
1þe 1 þ 1:21
(5.10.1)
Below the groundwater table, the moist unit weight is computed using
the expression determined in Case 1.8 of Chapter 1, i.e.,
gw ðGs þ eÞ ð9:81 kN=m3 Þ½2:71 þ 1:21
gsat ¼ ¼ ¼ 17:4 kN=m3 (5.10.2)
1þe 1 þ 1:21
At a depth (d) below the ground surface the vertical total stress is
sv ¼ gð1:5 mÞ þ gsat ðd  1:5 mÞ (5.10.3)
The pore pressure at the same depth is
u ¼ gw ðd  1:5 mÞ (5.10.4)
Finally, the vertical effective stress at a depth (d) below the ground
surface is
s0v ¼ sv  u ¼ gð1:5 mÞ þ ðgsat  gw Þðd  1:5 mÞ (5.10.5)
Solving Eq. (5.10.5) for d gives
s0v  gð1:5 mÞ
d¼ þ 1:5 m (5.10.6)
ðgsat  gw Þ
Substituting all known values into Eq. (5.10.6) gives the desired result
ð120 kPaÞ ð17:0 kN=m3 Þð1:5 mÞ
d¼ þ 1:5 m ¼ 14.0 m (5.10.7)
ð17:4  9:81Þ kN=m3
242 Soil Mechanics

b) Since the vertical effective stress is computed immediately following the

lowering of the groundwater table, the unit weights of the partially and
fully saturated portions of the clay layer remain unchanged. The vertical
total stress at a depth d ¼ 14.0 m below the ground surface is
sv ¼ gð1:5 mÞ þ gsat ð14:0  1:5 mÞ
   
¼ 17:0 kN=m3 ð1:5 mÞ þ 17:4 kN=m3 ð14:0  1:5 mÞ ¼ 243:0 kPa
(5.10.8)
The pore pressure at the same depth is now
 
u ¼ gw ð14:0  1:5  2:0 mÞ ¼ 9:81 kN=m3 ð10:5 mÞ ¼ 103:0 kPa
(5.10.9)
The vertical effective stress at a depth d ¼ 14.0 m below the ground
surface is thus
s0v ¼ sv  u ¼ 243:0  103:0 ¼ 140.0 kPa (5.10.10)