@JozveBartarOfficial

12.6 General Comments on Direct Shear Test 379

Solution 1A2 ϭ 1p/4 2 a 50 2
Area of the specimen
1000 b ϭ 0.0019634 m2. Now the following

table can be prepared.

Test Normal Normal Peak shear Speak Residual Tr ‫؍‬ Sresidual
no. force, N Tf ‫ ؍‬A A
stress, S؅ force, Speak (kN/m2) shear
(N) (kN/m2) (N) (kN/m2)
force,
Sresidual

(N)

1 150 76.4 157.5 80.2 44.2 22.5

2 250 127.3 199.9 101.8 56.6 28.8

3 350 178.3 257.6 131.2 102.9 52.4

4 550 280.1 363.4 185.1 144.5 73.6

The variations of tf and tr with sЈ are plotted in Figure 12.18. From the plots, we
find that

Peak strength: tf1kN/m2 2 ϭ 40 ؉ S؅ tan 27
Residual strength: tr 1kN/m2 2 ϭ S؅ tan 14.6

(Note: For all overconsolidated clays, the residual shear strength can be expressed as
tr ϭ s¿ tan frœ

where frœ ϭ effective residual friction angle.)

300

250

Shear stress, t (kN/m2) 200

150 tf versus sЈ
100

50 27Њ ϭ fЈ tr versus sЈ

0 cЈ ϭ 40 kN/m2 Figure 12.18
0 frЈ ϭ 14.6Њ

50 100 150 200 250 300 Variations ■
Effective normal stress, sЈ (kN/m2) 350 of tf and tr

with sЈ





































































414 Chapter 12: Shear Strength of Soil

12.18 Stress Path

Results of triaxial tests can be represented by diagrams called stress paths. A stress path is
a line that connects a series of points, each of which represents a successive stress state
experienced by a soil specimen during the progress of a test. There are several ways in
which a stress path can be drawn. This section covers one of them.

Lambe (1964) suggested a type of stress path representation that plots qЈ against pЈ
(where pЈ and qЈ are the coordinates of the top of the Mohr’s circle). Thus, relationships
for pЈ and qЈ are as follows:

p¿ ϭ s1œ ϩ s3œ (12.51)
2 (12.52)

q¿ ϭ s1œ Ϫ s3œ
2

This type of stress path plot can be explained with the aid of Figure 12.53. Let us con-

sider a normally consolidated clay specimen subjected to an isotropically

consolidated-drained triaxial test. At the beginning of the application of deviator
stress, s1œ ϭ s3œ ϭ s3, so

p¿ ϭ s3œ ϩ s3œ ϭ s3œ ϭ s3 (12.53)
2

and

q¿ ϭ s3œ Ϫ s3œ ϭ 0 (12.54)
2

Shear stress, or qЈ F
fЈ F'

a

D

DЉ B
DЈ

s3 ϭ s3Ј 45Њ A s1 ϭ s1'

O I ⌬sd s, sЈ, or pЈ

(⌬sd)f

Figure 12.53 Stress path—plot of qЈ against pЈ for a consolidated-drained triaxial test on a
normally consolidated clay

12.18 Stress Path 415

For this condition, pЈ and qЈ will plot as a point (that is, I in Figure 12.53). At some other
time during deviator stress application, s1œ ϭ s3œ ϩ ⌬sd ϭ s3 ϩ ⌬sd; s3œ ϭ s3. The
Mohr’s circle marked A in Figure 12.53 corresponds to this state of stress on the soil spec-

imen. The values of pЈ and qЈ for this stress condition are

p¿ ϭ s1œ ϩ s3œ ϭ 1s3œ ϩ ¢sd 2 ϩ s3œ ϭ s3œ ϩ ¢sd ϭ s3 ϩ ¢sd (12.55)
2 2 2 2

and

q¿ ϭ 1s3œ ϩ ¢sd 2 Ϫ s3œ ϭ ¢sd (12.56)
22

If these values of pЈ and qЈ were plotted in Figure 12.53, they would be represented by
point DЈ at the top of the Mohr’s circle. So, if the values of pЈ and qЈ at various stages of
the deviator stress application are plotted and these points are joined, a straight line like ID
will result. The straight line ID is referred to as the stress path in a qЈ-pЈ plot for a con-
solidated-drained triaxial test. Note that the line ID makes an angle of 45Њ with the hori-
zontal. Point D represents the failure condition of the soil specimen in the test. Also, we

can see that Mohr’s circle B represents the failure stress condition.
For normally consolidated clays, the failure envelope can be given by tf ϭ sЈ tan fЈ.

This is the line OF in Figure 12.53. (See also Figure 12.22.) A modified failure envelope
now can be defined by line OFЈ. This modified line commonly is called the Kf line. The
equation of the Kf line can be expressed as

q¿ ϭ p¿ tan a (12.57)

where a ϭ the angle that the modified failure envelope makes with the horizontal.
The relationship between the angles fЈ and a can be determined by referring to

Figure 12.54, in which, for clarity, the Mohr’s circle at failure (that is, circle B) and lines

Shear stress, or qЈ F
fЈ FЈ

a

C
D

O s3 ϭ ␴3Ј O' s1 ϭ s1Ј s, sЈ, or pЈ

(⌬sd)f

Figure 12.54 Relationship between fЈ and a

416 Chapter 12: Shear Strength of Soil

OF and OFЈ, as shown in Figure 12.53, have been redrawn. Note that OЈ is the center of
the Mohr’s circle at failure. Now,

DO¿ ϭ tan a
OO¿

and thus, we obtain

s1œ Ϫ s3œ

tan a ϭ s1œ 2 s3œ ϭ s1œ Ϫ s3œ (12.58)
ϩ s1œ ϩ s3œ

2

Again,

CO¿ ϭ sin f¿
OO¿

or

s1œ Ϫ s3œ

sin f¿ ϭ s1œ 2 s3œ ϭ s1œ Ϫ s3œ (12.59)
ϩ s1œ ϩ s3œ

2

Comparing Eqs. (12.58) and (12.59), we see that

sin f¿ ϭ tan a (12.60)

or

f¿ ϭ sinϪ11tan a 2 (12.61)

Figure 12.55 shows a qЈ– pЈ plot for a normally consolidated clay specimen

subjected to an isotropically consolidated-undrained triaxial test. At the beginning of the
application of deviator stress, s1œ ϭ s3œ ϭ s3. Hence, pЈ ϭ s3œ and qЈ ϭ 0. This relationship
is represented by point I. At some other stage of the deviator stress application,

s1œ ϭ s3 ϩ ¢sd Ϫ ¢ud
and

s3œ ϭ s3 Ϫ ¢ud
So,

p¿ ϭ s1œ ϩ s3œ ϭ s3 ϩ ¢sd Ϫ ¢ud (12.62)
2 2

12.18 Stress Path 417

FЈ

Shear stress, or qЈ Effective stress Mohr’s circle
U Total stress Mohr’s circle

a UЉ ␴1
O UЈ s, sЈ, or pЈ

s3
s3Ј I s1Ј

⌬ud

Figure 12.55 Stress path—plot of qЈ against pЈ for a consolidated-undrained triaxial
test on a normally consolidated clay

and

q¿ ϭ s1œ Ϫ s3œ ϭ ¢sd (12.63)
22

The preceding values of pЈ and qЈ will plot as point UЈ in Figure 12.55. Points such as UЉ
represent values of pЈ and qЈ as the test progresses. At failure of the soil specimen,

1 ¢sd 2 f (12.64)
p¿ ϭ s3 ϩ 2 Ϫ 1 ¢ud 2 f

and

1 ¢sd 2 f (12.65)
q¿ ϭ

2

The values of pЈ and qЈ given by Eqs. (12.64) and (12.65) will plot as point U.
Hence, the effective stress path for a consolidated-undrained test can be given by the curve
IUЈU. Note that point U will fall on the modified failure envelope, OFЈ (see Figure 12.54),
which is inclined at an angle a to the horizontal. Lambe (1964) proposed a technique to
evaluate the elastic and consolidation settlements of foundations on clay soils by using the
stress paths determined in this manner.

Example 12.9

For a normally consolidated clay, the failure envelope is given by the equation tf ϭ sЈ
tan fЈ. The corresponding modified failure envelope (qЈ-pЈ plot) is given by Eq. (12.57)
as qЈ ϭ pЈ tan a. In a similar manner, if the failure envelope is tf ϭ cЈ ϩ sЈ tan fЈ, the
corresponding modified failure envelope is a qЈ-pЈ plot that can be expressed as qЈ ϭ
m ϩ pЈ tan a. Express a as a function of fЈ, and give m as a function of cЈ and fЈ.

418 Chapter 12: Shear Strength of Soil

Solution

From Figure 12.56,

AB AB a s1œ Ϫ s3œ b
AC CO ϩ OA 2
sin f¿ ϭ ϭ ϭ ϩa s1œ ϩ s3œ

c¿ cot f¿ 2 b

So, (a)
s1œ Ϫ s3œ ϭ c¿ cos f¿ ϩ a s1œ ϩ s3œ b sin f¿ (b)
22

or
q¿ ϭ m ϩ p¿ tan a

Comparing Eqs. (a) and (b), we find that

m ϭ c؅ cos F؅

and
tan a ϭ sin f¿

or
a ϭ tanϪ11sin F؅ 2

fЈ

Shear stress tf ϭ cЈ ϩ s' tan fЈ

B

C cЈ s1' Ϫ s3'
2
O s3Ј Normal stress
cЈ cot fЈ s1Ј ϩ s3Ј A s1Ј

2

Figure 12.56 Derivation of a as a function of fЈ and m as a function of cЈ and fЈ ■

12.19 Summary and General Comments

In this chapter, the shear strengths of granular and cohesive soils were examined. Laboratory
procedures for determining the shear strength parameters were described.

In textbooks, determination of the shear strength parameters of cohesive soils
appears to be fairly simple. However, in practice, the proper choice of these parameters for
design and stability checks of various earth, earth-retaining, and earth-supported structures

Problems 419

is very difficult and requires experience and an appropriate theoretical background in geot-
echnical engineering. In this chapter, three types of strength parameters (consolidated-
drained, consolidated-undrained, and unconsolidated-undrained) were introduced. Their
use depends on drainage conditions.

Consolidated-drained strength parameters can be used to determine the long-term
stability of structures such as earth embankments and cut slopes. Consolidated-undrained
shear strength parameters can be used to study stability problems relating to cases where
the soil initially is fully consolidated and then there is rapid loading. An excellent example
of this is the stability of slopes of earth dams after rapid drawdown. The unconsolidated-
undrained shear strength of clays can be used to evaluate the end-of-construction stability
of saturated cohesive soils with the assumption that the load caused by construction has
been applied rapidly and there has been little time for drainage to take place. The bearing
capacity of foundations on soft saturated clays and the stability of the base of embank-
ments on soft clays are examples of this condition.

The unconsolidated-undrained shear strength of some saturated clays can vary
depending on the direction of load application; this is referred to as anisotropy with respect
to strength. Anisotropy is caused primarily by the nature of the deposition of the cohesive
soils, and subsequent consolidation makes the clay particles orient perpendicular to the
direction of the major principal stress. Parallel orientation of the clay particles can cause
the strength of clay to vary with direction. The anisotropy with respect to strength for clays
can have an important effect on the load-bearing capacity of foundations and the stability
of earth embankments because the direction of the major principal stress along the poten-
tial failure surfaces changes.

The sensitivity of clays was discussed in Section 12.13. It is imperative that sensi-
tive clay deposits are properly identified. For instance, when machine foundations (which
are subjected to vibratory loading) are constructed over sensitive clays, the clay may lose
its load-bearing capacity substantially, and failure may occur.

Problems

12.1 For a direct shear test on a dry sand, the following are given:
• Specimen size: 75 mm ϫ 75 mm ϫ 30 mm (height)
• Normal stress: 200 kN/m2
• Shear stress at failure: 175 kN/m2
a. Determine the angle of friction, fЈ
b. For a normal stress of 150 kN/m2, what shear force is required to cause failure
in the specimen?

12.2 For a dry sand specimen in a direct shear test box, the following are given:
• Angle of friction: 38Њ
• Size of specimen: 2 in. ϫ 2 in. ϫ 1.2 in. (height)
• Normal stress: 20 lb/in.2
Determine the shear force required to cause failure.

12.3 The following are the results of four drained, direct shear tests on a normally
consolidated clay. Given:
• Size of specimen ϭ 60 mm ϫ 60 mm
• Height of specimen ϭ 30 mm

420 Chapter 12: Shear Strength of Soil

Test Normal Shear
no. force force at
(N) failure (N)

1 200 155
2 300 230
3 400 310
4 500 385

Draw a graph for the shear stress at failure against the normal stress, and
determine the drained angle of friction from the graph.
12.4 Repeat Problem 12.3 with the following data. Given specimen size:
• Diameter ϭ 2 in.
• Height ϭ 1 in.

Test Normal Shear
no. force force at
(lb) failure (lb)

1 60 37.5
2 90 55
3 110 70
4 125 80

12.5 The equation of the effective stress failure envelope for a loose, sandy soil was
obtained from a direct shear test at tf ϭ sЈ tan 30Њ. A drained triaxial test was
conducted with the same soil at a chamber confining pressure of 10 lb/in.2.

Calculate the deviator stress at failure.

12.6 For the triaxial test described in Problem 12.5:

a. Estimate the angle that the failure plane makes with the major principal

plane.

b. Determine the normal stress and shear stress (when the specimen failed) on a
plane that makes an angle of 30Њ with the major principal plane. Also, explain

why the specimen did not fail along the plane during the test.
12.7 The relationship between the relative density, Dr , and the angle of friction, fЈ, of

a sand can be given as fЈ ϭ 25 ϩ 0.18Dr (Dr is in %). A drained triaxial test on
the same sand was conducted with a chamber-confining pressure of 18 lb/in.2. The

relative density of compaction was 60%. Calculate the major principal stress at

failure.

12.8 For a normally consolidated clay, the results of a drained triaxial test are as

follows.
• Chamber confining pressure: 15 lb/in.2
• Deviator stress at failure: 34 lb/in.2
Determine the soil friction angle, fЈ.
12.9 For a normally consolidated clay, fЈ ϭ 24Њ. In a drained triaxial test, the
specimen failed at a deviator stress of 175 kN/m2. What was the chamber
confining pressure, s3œ ?

12.10 For a normally consolidated clay, fЈ ϭ 28Њ. In a drained triaxial test, the
specimen failed at a deviator stress of 30 lb/in.2. What was the chamber confining
pressure, s3œ ?

Problems 421

12.11 A consolidated-drained triaxial test was conducted on a normally consolidated

clay. The results were as follows:
s3 ϭ 250 kN/m2

1 ¢sd 2 f ϭ 275 kN/m2
Determine:
a. Angle of friction, fЈ
b. Angle u that the failure plane makes with the major principal plane
c. Normal stress, sЈ, and shear stress, tf , on the failure plane
12.12 The results of two drained triaxial tests on a saturated clay are given next:
Specimen I: Chamber confining pressure ϭ 15 lb/in.2

Deviator stress at failure ϭ 31.4 lb/in.2
Specimen II: Chamber-confining pressure ϭ 25 lb/in.2

Deviator stress at failure ϭ 47 lb/in.2

Calculate the shear strength parameters of the soil.

12.13 If the clay specimen described in Problem 12.12 is tested in a triaxial apparatus with a
chamber-confining pressure of 25 lb/in.2, what is the major principal stress at failure?

12.14 A sandy soil has a drained angle of friction of 38Њ. In a drained triaxial test on the
same soil, the deviator stress at failure is 175 kN/m2. What is the chamber-

confining pressure?

12.15 A consolidated-undrained test on a normally consolidated clay yielded the follow-

ing results:
• s3 ϭ 15 lb/in.2
• Deviator stress: (⌬sd)f ϭ 11 lb/in.2

Pore pressure: (⌬ud)f ϭ 7.2 lb/in.2
Calculate the consolidated-undrained friction angle and the drained friction angle.

12.16 Repeat Problem 12.15 with the following:
s3 ϭ 140 kN/m2

1 ¢sd 2 f ϭ 125 kN/m2
1 ¢ud 2 f ϭ 75 kN/m2
12.17 The shear strength of a normally consolidated clay can be given by the equation
tf ϭ sЈ tan 31Њ. A consolidated-undrained triaxial test was conducted on the
clay. Following are the results of the test:
• Chamber confining pressure ϭ 112 kN/m2
• Deviator stress at failure ϭ 100 kN/m2

Determine:

a. Consolidated-undrained friction angle

b. Pore water pressure developed in the clay specimen at failure

12.18 For the clay specimen described in Problem 12.17, what would have been the

deviator stress at failure if a drained test had been conducted with the same
chamber-confining pressure (that is, s3 ϭ 112 kN/m2)?
12.19 For a normally consolidated clay soil, fЈ ϭ 32Њ and f ϭ 22Њ. A consolidated-

undrained triaxial test was conducted on this clay soil with a chamber-confining pres-
sure of 15 lb/in.2. Determine the deviator stress and the pore water pressure at failure.
12.20 The friction angle, fЈ, of a normally consolidated clay specimen collected during
field exploration was determined from drained triaxial tests to be 25Њ. The

unconfined compression strength, qu, of a similar specimen was found to be
100 kN/m2. Determine the pore water pressure at failure for the unconfined

compression test.

422 Chapter 12: Shear Strength of Soil

12.21 Repeat Problem 12.20 using the following values.

f¿ ϭ 23°
qu ϭ 120 kN/m2
12.22 The results of two consolidated-drained triaxial tests on a clayey soil are as follows.

Test no. S3œ S1œ 1failure2
(lb /in.2) (lb /in.2)

1 27 73
2 12 48

Use the failure envelope equation given in Example 12.9—that is, qЈ ϭ m ϩ pЈ
tan a. (Do not plot the graph.)
a. Find m and a
b. Find cЈ and fЈ
12.23 A 15-m thick normally consolidated clay layer is shown in Figure 12.57. The
plasticity index of the clay is 18. Estimate the undrained cohesion as would be
determined from a vane shear test at a depth of 8 m below the ground surface.
Use Eq. (12.35).

3 m g ϭ 16 kN/m3
Groundwater table

15 m

gsat ϭ 18.6 kN/m3

Dry sand Clay Rock Figure 12.57

References

ACAR, Y. B., DURGUNOGLU, H. T., and TUMAY, M. T. (1982). “Interface Properties of Sand,”
Journal of the Geotechnical Engineering Division, ASCE, Vol. 108, No. GT4, 648–654.

ARMAN, A., POPLIN, J. K., and AHMAD, N. (1975). “Study of Vane Shear,” Proceedings,
Conference on In Situ Measurement and Soil Properties, ASCE, Vol. 1, 93–120.

AMERICAN SOCIETY FOR TESTING AND MATERIALS (2004). Annual Book of ASTM Standards,
Vol. 04.08, Philadelphia, Pa.

BISHOP, A. W., and BJERRUM, L. (1960). “The Relevance of the Triaxial Test to the Solution of
Stability Problems,” Proceedings, Research Conference on Shear Strength of Cohesive Soils,
ASCE, 437–501.

BJERRUM, L. (1974). “Problems of Soil Mechanics and Construction on Soft Clays,” Norwegian
Geotechnical Institute, Publication No. 110, Oslo.

References 423

BLACK, D. K., and LEE, K. L. (1973). “Saturating Laboratory Samples by Back Pressure,” Journal
of the Soil Mechanics and Foundations Division, ASCE, Vol. 99, No. SM1, 75–93.

CASAGRANDE, A., and CARRILLO, N. (1944). “Shear Failure of Anisotropic Materials,” in
Contribution to Soil Mechanics 1941–1953, Boston Society of Civil Engineers, Boston.

CASAGRANDE, A., and HIRSCHFELD, R. C. (1960). “Stress Deformation and Strength
Characteristics of a Clay Compacted to a Constant Dry Unit Weight,” Proceedings, Research
Conference on Shear Strength of Cohesive Soils, ASCE, 359–417.

CHANDLER, R. J. (1988). “The in situ Measurement of the Undrained Shear Strength of Clays
Using the Field Vane,” STP 1014, Vane Shear Strength Testing in Soils: Field and Laboratory
Studies, ASTM, 13–44.

COULOMB, C. A. (1776). “Essai sur une application des regles de Maximums et Minimis á
quelques Problèmes de Statique, relatifs ál’Architecture,” Memoires de Mathematique et de
Physique, Prséentsé, ál’Academie Royale des Sciences, Paris, Vol. 3, 38.

KENNEY, T. C. (1959). “Discussion,” Proceedings, ASCE, Vol. 85, No. SM3, 67–79.
LADD, C. C., FOOTE, R., ISHIHARA, K., SCHLOSSER, F., and POULOS, H. G. (1977). “Stress

Deformation and Strength Characteristics,” Proceedings, 9th International Conference on Soil
Mechanics and Foundation Engineering, Tokyo, Vol. 2, 421–494.
LAMBE, T. W. (1964). “Methods of Estimating Settlement,” Journal of the Soil Mechanics and
Foundations Division, ASCE, Vol. 90, No. SM5, 47–74.
LOH, A. K., and HOLT, R. T. (1974). “Directional Variation in Undrained Shear Strength and Fabric
of Winnipeg Upper Brown Clay,” Canadian Geotechnical Journal, Vol. 11, No. 3, 430–437.
MOHR, O. (1900). “Welche Umstnäde Bedingen die Elastizittäsgrenze und den Bruch eines
Materiales”? Zeitschrift des Vereines Deutscher Ingenieure, Vol. 44, 1524–1530, 1572–
1577.
ROSENVQIST , I. TH. (1953). “Considerations on the Sensitivity of Norwegian uQick Clays,”
Geotechnique, Vol. 3, No. 5, 195–200.
SEED, H. B., and CHAN, C. K. (1959). “Thixotropic Characteristics of Compacted Clays,”
Transactions, ASCE, Vol. 124, 894–916.
SIMONS, N. E. (1960). “The Effect of Overconsolidation on the Shear Strength Characteristics of
an Undisturbed Oslo Clay,” Proceedings, Research Conference on Shear Strength of Cohesive
Soils, ASCE, 747–763.
SKEMPTON, A. W. (1954). “The Pore Water Coefficients A and B,” Geotechnique, Vol. 4, 143–147.
SKEMPTON, A. W. (1957). “Discussion: The Planning and Design of New Hong Kong Airport,”
Proceedings, Institute of Civil Engineers, London, Vol. 7, 305–307.
SKEMPTON, A. W. (1964). “Long-Term Stability of Clay Slopes,” Geotechnique, Vol. 14, 77.

13 Lateral Earth Pressure:
At-Rest, Rankine, and Coulomb

Retaining structures such as retaining walls, basement walls, and bulkheads commonly are
encountered in foundation engineering as they support slopes of earth masses. Proper
design and construction of these structures require a thorough knowledge of the lateral
forces that act between the retaining structures and the soil masses being retained. These
lateral forces are caused by lateral earth pressure. This chapter is devoted to the study of
the various earth pressure theories.

13.1 At-Rest, Active, and Passive Pressures

424 Consider a mass of soil shown in Figure. 13.1a. The mass is bounded by a frictionless wall

of height AB. A soil element located at a depth z is subjected to a vertical effective pressure,
soœ , and a horizontal effective pressure, shœ . There are no shear stresses on the vertical and
horizontal planes of the soil element. Let us define the ratio of shœ to soœ as a nondimen-
sional quantity K, or

K ϭ shœ (13.1)
soœ

Now, three possible cases may arise concerning the retaining wall: and they are
described

Case 1 If the wall AB is static—that is, if it does not move either to the right or to the left
of its initial position—the soil mass will be in a state of static equilibrium. In that case, shœ

is referred to as the at-rest earth pressure, or

K ϭ Ko ϭ shœ (13.2)
soœ

where Ko ϭ at-rest earth pressure coefficient.

Case 2 If the frictionless wall rotates sufficiently about its bottom to a position of AЈB
(Figure 13.1b), then a triangular soil mass ABCЈ adjacent to the wall will reach a state of

13.1 At-Rest, Active, and Passive Pressures 425

A At-rest pressure AЈ A Active pressure CЈ
soЈ z ⌬La

soЈ z

H KosoЈ ϭ shЈ KasoЈ ϭ shЈ
tf ϭ cЈ ϩ sЈ tan fЈ tf ϭ cЈ ϩ sЈ tan fЈ

B B (b)
(a) Passive pressure CЉ

A AЉ
⌬Lp

z

KpsoЈ ϭ shЈ
soЈ

tf ϭ cЈ ϩ sЈ tan fЈ

B
(c)

Figure 13.1 Definition of at-rest, active, and passive pressures (Note: Wall AB is frictionless)

plastic equilibrium and will fail sliding down the plane BCЈ. At this time, the horizontal
effective stress, shœ ϭ saœ , will be referred to as active pressure. Now,

K ϭ Ka ϭ shœ ϭ saœ (13.3)
soœ soœ

where Ka ϭ active earth pressure coefficient.

Case 3 If the frictionless wall rotates sufficiently about its bottom to a position
AЉB (Figure 13.1c), then a triangular soil mass ABCЉ will reach a state of plastic

426 Chapter 13: Lateral Earth Pressure: At-Rest, Rankine, and Coulomb

Earth pressure, shЈ

Passive pressure, spЈ

At-rest pressure, shЈ

Active pressure, saЈ

Wall tilt ⌬La ⌬Lp Wall tilt
H H

Figure 13.2 Variation of the magnitude of lateral earth pressure with wall tilt

Table 13.1 Typical Values of ⌬La /H and ⌬Lp/H

Soil type ⌬La /H ⌬Lp /H

Loose sand 0.001–0.002 0.01
Dense sand 0.0005–0.001 0.005
Soft clay 0.02 0.04
Stiff clay 0.01 0.02

equilibrium and will fail sliding upward along the plane BCЉ. The horizontal effective
stress at this time will be shœ ϭ spœ , the so-called passive pressure. In this case,

K ϭ Kp ϭ shœ ϭ spœ (13.4)
soœ soœ

where Kp ϭ passive earth pressure coefficient
Figure 13.2 shows the nature of variation of lateral earth pressure with the wall

tilt. Typical values of ⌬La/H (⌬La ϭ AЈA in Figure 13.1b) and ⌬Lp /H (⌬Lp ϭ AЉA in
Figure 13.1c) for attaining the active and passive states in various soils are given in

Table 13.1.

13.2 AT-REST LATERAL EARTH PRESSURE

Earth Pressure At-Rest

The fundamental concept of earth pressure at rest was discussed in the preceding sec-
tion. In order to define the earth pressure coefficient Ko at rest, we refer to Figure 13.3,

13.2 Earth Pressure At-Rest 427

A z
soЈ ϭ gz

H shЈ ϭ Kogz

tf ϭ cЈ ϩ sЈ tan fЈ

Figure 13.3
B Earth pressure at rest

which shows a wall AB retaining a dry soil with a unit weight of g. The wall is
static. At a depth z,

Vertical effective stress ϭ soœ ϭ gz
Horizontal effective stress ϭ shœ ϭ Kogz

So,

Ko ϭ shœ ϭ at-rest earth pressure coefficient
soœ

For coarse-grained soils, the coefficient of earth pressure at rest can be estimated by
using the empirical relationship (Jaky, 1944)

Ko ϭ 1 Ϫ sin f¿ (13.5)

where fЈ ϭ drained friction angle.
While designing a wall that may be subjected to lateral earth pressure at rest, one

must take care in evaluating the value of Ko. Sherif, Fang, and Sherif (1984), on the basis
of their laboratory tests, showed that Jaky’s equation for Ko [Eq. (13.5)] gives good results
when the backfill is loose sand. However, for a dense, compacted sand backfill, Eq. (13.5)
may grossly underestimate the lateral earth pressure at rest. This underestimation results
because of the process of compaction of backfill. For this reason, they recommended the
design relationship

Ko ϭ 11 Ϫ sin f2 ϩ c gd Ϫ 1 d 5.5 (13.6)
gd1min2

where gd ϭ actual compacted dry unit weight of the sand behind the wall
gd1min2 ϭ dry unit weight of the sand in the loosest state 1Chapter 3 2

428 Chapter 13: Lateral Earth Pressure: At-Rest, Rankine, and Coulomb

The increase of Ko observed from Eq. (13.6) compared to Eq. (13.5) is due to over-
consolidation. For that reason, Mayne and Kulhawy (1982), after evaluating 171 soils, rec-
ommended a modification to Eq. (13.5). Or

Ko ϭ 11 Ϫ sin f¿ 2 1OCR 2 sinf¿ (13.7)

where

OCR ϭ overconsolidation ratio

ϭ preconsolidation pressure, scœ soœ
present effective overburden pressure,

Equation (13.7) is valid for soils ranging from clay to gravel.
For fine-grained, normally consolidated soils, Massarsch (1979) suggested the

following equation for Ko:

PI 1% 2 (13.8)
Ko ϭ 0.44 ϩ 0.42 c 100 d

For overconsolidated clays, the coefficient of earth pressure at rest can be approxi-
mated as

Ko 1overconsolidated2 ϭ Ko 1normally consolidated2 1OCR (13.9)

Figure 13.4 shows the distribution of lateral earth pressure at rest on a wall of
height H retaining a dry soil having a unit weight of g. The total force per unit length
of the wall, Po, is equal to the area of the pressure diagram, so

Po ϭ 1 KogH2 (13.10)
2

tf ϭ cЈ ϩ sЈ tan fЈ
Unit weight ϭ g

H

H
3

KogH

Figure 13.4 Distribution of lateral earth pressure at-rest on a wall