1

SOIL MECHANICS: PART B

B1. SHEAR STRENGTH AND FAILURE CRITERIA

Strength is an essential soil parameter used in the design of geotechnical structures. It is the

ultimate shear stress that the soil can sustain before it fails. Obviously we do not design for

geotechnical structures to be on the brink of failure, but it must be proven that the structure is

safe. We start this section with a reminder of Mohr circles applied to soils, as this will be

useful to demonstrate the application of strength to design in section B3.

B1.1 Mohr circle applied to soils

Mohr circle is the representation of the state of stress (, ) on any plane at a point in a two-dimensional field (assuming plane strain conditions: no movement in the direction

perpendicular to the plane). A point in a stress field is conventionally represented by a small

element to help us visualize (small element A in Figure 1).

A

Figure 1 Small soil element (A) behind a retaining wall.

On each face of the element are acting a normal stress () and a stress stress (). In soil mechanics the sign convention is: for normal stresses, compression is positive, and for shear

stresses, anti-clockwise is positive.

Lets consider an element inclined at degrees from the horizontal, and call the stresses

acting on the bottom plane (, ) (Figure 1). The stresses acting on the perpendicular plane

are (, ), with = (otherwise the element would spin) and is positive if it is

on the element anti-clockwise. The 2-D plane contains the principal stresses and which can be resolved as follows (Figure 2).

a

b

c

ab=1

unit l

ength

Figure 2 Principal stresses in the ground.

2

cos + sin = 1cos (1)

sin cos = 3 sin (2)

We obtain:

22sin 31

(3)

2cos22

3131

(4)

Equations 3 and 4 describe a circle of centre [

2

31 ; 0] and radius equal to

2

31

(see Figure 3).

2

31

2

31

s

Figure 3 Mohr circle of soil stresses.

2

31 s is the mean stress and 2

31 t is the maximum shear stress.

Note that the angle was measured from the major principal plane, i.e. the plane on which

acts. Lets now assume that the principal stress plane is at an angle from the horizontal, with axes (x,y) (x and y are the horizontal and vertical axes on the page; z is the horizontal

axis perpendicular to the page; Figure 4):

3

yx

xy

y

x

1

Horizontal plane on

which y acts.

Figure 4 Stresses acting on small soil element.

The origin of planes, or pole P, gives a real physical orientation to any stress state. We can

determine the origin of planes on Mohr circle by drawing the physical plane (which is

horizontal) from point (c) which represents the stress pair which acts on the plane ------

(Figure 5). That plane intersects the opposite side of Mohr circle at point P, the origin of

planes. From trigonometry, the planes from point P to any other point on the circle (i.e. 1) will be at half the angle subtended at the centre. For example, the physical plane on which the

major principal stress 1 acts which passes through P is at to the horizontal and 2 at the centre.

yyx

a

c

b o

d

P

xxy

Figure 5 Determination of origin of planes or pole P.

Figure 6 below shows how to use Mohr circle and the origin of planes to determine the stress

pair (, ) on a plane inclined at from the horizontal in the same soil element as in Figure 4. The pole P is found in the same way as in Figure 5. In the Mohr circle (Fig. 6) the

inclined plane crosses the circle in i. the coordinates in i are:

2cos22

3131

and

22sin 31

4

yyx

i

c

d

P

Reference plane

x

Figure 6 Determination of stresses acting on plane at from the horizontal.

B1.2 Loading tests

Soil loading tests are mainly carried out in order to determine the strength, stiffness and

compressibility characteristics of the soil. These tests also inform us on the full stress-strain

response of the soil under different conditions of loading and drainage.

The compressibility of soil is determined by carrying out one-dimensional compression tests

in an oedometer apparatus. The change in void ratio with stress during loading and unloading

is recorded so we can plot the compression and swelling curves of the soil (refer to Dr

Kwoks lecture notes and laboratory class). By timing the evolution of volume change with time at every stage of the loading we can also measure the consolidation characteristics of the

soil, which inform us on how long a fine-grained soil would take to reach full settlement for

example.

The strength of the soil is determined by different ways, and there is not one strength but

several that we can measure in the laboratory. The first type may be the unconfined

compressive strength (UCS) obtained by loading vertically an unconfined (typically

cylindrical) specimen of soil and measuring the maximum axial load it can sustain (the UCS

is that maximum load divided by the area of the specimen). This is only possible for soil

specimens with enough cohesion to hold together without confinement. Granular soils such

as sands are typically subjected to shear box tests, where they are sheared along a given

horizontal plane and under a constant normal load. In granular soils like sands, they are

usually drained. The ultimate shear stress that the soil reaches is unique for a given normal

stress, and independent of the density of the soil. We determine the angle of friction ' from

the ratio of shear stress to normal stress (which is equal to tan'). The main disadvantages of the shear box test are (1) that there is no drainage control (2) each test only gives one point in

- plane, so we cannot plot the Mohr circle for that test; see later.

Triaxial tests are common in commercial laboratories and research laboratories. They are

performed on cylindrical specimens confined by a surrounding cell pressure. There are two

stresses acting on the specimen, the axial stress (a) and the radial stress (r), and because of axisymmetry they provide a three-dimensional picture of the stress state (Figure 7). The axial

and radial stresses are principal stresses (no shear stress on boundaries) so that we know the

Mohr circle at every stage of the test. Corresponding strains are the axial strain (a) and the

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radial strain (r). Isotropic compression tests (a=r) and shear tests can be performed in the triaxial apparatus, and when they are more advanced both cell pressure and axial pressures

can be controlled. Logging of enough data allows determining the evolution of stiffness with

strain level a vital information in design as different geotechnical structures may be required to withstand different levels of strain to be serviceable.

a

ru

Figure 7 Schematic diagram of triaxial specimen.

In research laboratories more advanced tests are performed e.g. hollow cylinder or true

triaxial tests (allow three independent stresses to be applied; good to determine the anisotropy

of the soil), cyclic tests, dynamic tests (for offshore wave loading or earthquake loading).

These are very rarely commissioned for site investigations, which usually consist of the

simplest (and cheapest) tests and are carried out in commercial laboratories.

B1.3 Strength and failure criteria

The strength of a soil is the ultimate state of stress that it can sustain before it fails. We can

talk of tensile strength, compressive strength, shear strength, and although they seem to be

different they all refer to the same fundamental measurement of the soil strength. We can

understand this if we try to determine these strengths using Mohr circles. This is illustrated in

Figure 8. In Figure 8(a), the element is subjected to axial tension with no confining pressure.

The tensile strength is the maximum value of shear stress (f) that the element can withstand. These types of tests are very difficult to carry out on soils. Here the Mohr circle is drawn

using the values of horizontal stress (zero) and axial stress at failure (f), which is negative

because of being tension. The strength f is equal to the radius of the Mohr circle. In Figure 8(b), a soil element is subjected to axial compression with no confining pressure. The Mohr

circle is determined from the lateral stress (zero) and the axial stress at failure (f), which is

positive. The strength f is again equal to the radius of the circle. Finally, lets consider that a block of soil has failed and is sliding along a failure plane (Figure 8(c)). The effective

stresses on the failure plane ('f, f) determine the limiting size of the Mohr circle at failure. The strength of the material is the maximum shear stress, and is equal to the radius of the

Mohr circle. All three strengths are thus equal to the largest Mohr circle of stresses than can

be sustained.

6

f f f

0 00 0

f

f

f

f

f f f

ff

(a) (c)(b)

Figure 8 (a) Tensile strength (b) compressive strength (c) mobilized strength on shear plane.

But this does not mean that for a given soil there is a unique value of strength. We must

consider other factors that may affect the value and nature of the strength measured, for

example is the pore water pressure, or the stress-volume state of the soil, important? Lets consider one of the classic laboratory tests to measure strength in soil: the shear box test. If

we carry out a test under a normal stress , we may measure a shearing resistance , but if

we perform the test under a normal stress , we will find a shear strength . The shear strength measured depends on the externally applied normal pressure. However

whether the soil is fully saturated or dry (for sand) has no impact on the measured strength:

the water has no shearing resistance per se. This means that if the shear strength is , then

=. If the soil is only partially saturated (there is air as well as water in the voids), then the water tension can act as glue between particles and give an apparent strength to the soil. This is called suction. The behaviour of partially saturated soils however is difficult and

beyond the scope of undergraduate courses, so in this course we shall deal with soils that are

either dry or fully saturated.

i. Friction

Lets come back to the shear box experiment, and consider three tests performed on saturated

sand under three different normal effective pressures 1

7

data curve towards the origin. Fitting a straight line to curved data is often done for

convenience. Equation (5) is usually referred to as Mohr-Coulomb equation. It will be seen

later that this equation can be generalized to include peak states and dilatancy.

c

1 32

1

2

3

Figure 9 Dependence of shear strength on normal stress and definition of cohesion and angle

of friction.

ii. Critical state

If we carry out a shear test in a shear box or in a triaxial apparatus, we will always reach

failure, when the stresses and volume of the specimen become constant under continued

shearing. This is called critical state. Failure occurs when the effective stress Mohr circle reaches the Mohr-Coulomb failure envelope (Figure 10; here c=0).

, a

rpole

r a

u

Figure 10 Total and effective Mohr circles at failure during triaxial testing.

Lets consider standard triaxial tests where the cell pressure (r) is kept constant while the

axial stress (a) increases. If the specimen was consolidated isotropically prior to shearing,

initially, a=r. Lets first consider a drained test. The pore water pressure (u) in the soil

remains constant during the test (Figure 11). The Mohr circle increases in size as a is increasing, until failure is reached.

8

,

Figure 11 Total and effective Mohr circles during drained triaxial compression.

Lets now consider an undrained test. The pore water pressure (u) is not constant (Figure 12).

The Mohr circle increases in size as a is increasing, until failure is reached.

,

Figure 12 Total and effective Mohr circles during undrained triaxial compression.

Although there are no shear stresses on the boundaries of the specimen, the test is a shearing

test in which the soil is taken to failure, and a failure plane develops. The stresses on any

plane can be derived from construction from the pole. The failure plane is represented on the

Mohr circle by the plane passing through the pole and the failure point at the intersection

between the circle and the failure line. The maximum shear stress is at 45 from the

horizontal, but the failure plane is inclined at an angle of 45+/2, being the angle of friction of the soil (see exercise).

Most soils have little or no strength at zero effective confining stress. The true failure

envelope must pass through the origin and therefore may be curved (e.g. Figure 13). Equation

(5) is a straight line just to fit the data. In order to determine the shape of the failure envelope

one needs to carry out tests at different confining pressures r.

9

c

Figure 13 Real failure envelope.

Mohr circles are fine to see the conditions at failure, but it is often useful to see the course of

the test with stress paths in terms of invariants. The triaxial invariant stresses are:

3

2 rap

(mean stress) (6)

3

'2'' raupp

(mean effective stress) (7)

raraqq ''' (deviatoric stress) (8)

[Note that q and q' are interchangeable.]

The triaxial strains used are:

rav 2 (volumetric strain) (9)

ras 3

2

(shear strain) (10)

Lets now compare shearing data from drained tests performed in the shear box apparatus and in the triaxial apparatus (Figure 14). The data show similar trends, the behaviour depending

on the initial state of the soil. For example, a dense specimen will show peak strength (point

P on the figure) before softening to reach the stable state of critical state (point C on the

figure). A loose specimen will harden all the way to critical state, its peak strength

corresponding to its critical state strength. On clay specimens, dense-type behaviour is found

for over-consolidated specimens, while loose-type behaviour is found for normally or lightly

consolidated specimens.

10

n

u=0

h

v

r

a

u=0

a

v

1 10 100

/n

h (mm)

v

h (mm)

dil.

comp.

p

ctanc

p

c

r

c

1 10 100

q/p

a

(%)

v

a

(%)

dil.

comp.

c

r

p

M r

p

c

c

(a) (b)

Figure 14 Stress-strain and volumetric responses during shearing (a) shear box test (b)

triaxial test.

The peak state (P), defined as max. '/n' or q'/p', is reached at small strains (1%). It

corresponds to the maximum rate of dilation (max. v/a or v/h).

Critical state (C), where the soil deforms in a turbulent flow, is reached at

medium strains (10%).

In the shear box apparatus, the strength is determined in terms of peak ('p)

and critical state ('c or 'cs) angle of friction. In the triaxial apparatus, critical state is defined by the gradient of the critical state line M (=q'/p' at failure).

For dilative tests, at minimum volume, q'/p'=M.

Figure 15 illustrates typical stress paths for shear box test and standard drained

triaxial test. Specimens denser (dry) than critical state will dilate before they reach critical state while specimens looser (wet) than critical state will contract before they reach critical state. For a given soil and a given initial stress the specimens reach

the same critical state point. If we carry out a series of tests at different normal stress

(shear box) or different confining pressures (triaxial test) we will define the location

of the critical state points. They define a line, the critical state line (CSL). There is a

unique CSL for a given soil, defined in space e-- (shear box test) or v-q'-p' (triaxial test).

11

n

n

e

q

p

v

p

cM

3

1

p

c

Critical state line

(CSL)

Normal

consolidation line

CSL

Swellingline

c

p

= n tanc

c

start

(a) (b)

Figure 15 Location of the critical state line (a) shear box test (b) triaxial test.

A typical value for the critical state angle of friction of clays is 22-28, and for sands

30-35 if they have rounded particles and around 40 if they are more angular (hence

more frictional). The angle of friction 'c depends on the mineralogy and grain shapes. It can be linked to the triaxial stress ratio q/p at critical state, M, by:

Mc = 6sin'c / (3-'c) for compressive shearing (11)

Note that equation (11) becomes Me=6sin'e / (3+'e) for extension, so that if the angle

of friction is constant ('e='c='cs), the M-value in compression, Mc, is different to the M-value in extension, Me. There is no intercept on the q-axis for critical states i.e. the

ultimate strength is entirely frictional. If we plot the CSL in v-lnp', we find a straight

line, parallel to the Normal Consolidation or Compression Line NCL (Figure 16):

For NCL: 'ln pNv (12)

For CSL: 'ln pv (13)

12

ln p

v

CSL

iso-NCL

p=1kPa

Figure 16 Schematic diagram of NCL and CSL (note slope of swelling line, )

13

iii. Peak states

We have noticed however that when a specimen is initially dense, the stress-strain

response displays peak strength before softening to the critical state. The peak

strength is a result from the dilation of the dense soil. We have seen earlier that in

clays, peak envelopes are fitted to the soil with an intercept c and a peak angle of

friction p. In sands, it is preferred to have the peak envelope passing through the

origin (no cohesion). The peak angle of friction p is then equal to the sum of the

critical state angle of friction (cs) and the dilation angle (p) (see Figure 17). We thus have to be careful which definition of peak angle of friction we mean.

CSL

p?

c

cs

peak

failure

p

Figure 17 Definition(s) of peak strength and peak angle of friction.

Unlike the critical state envelope, the peak envelope is influenced by the initial stress

and volume (state or overconsolidation ratio (OCR or R) refer to Part A by Dr Kwok).

iv. Undrained shear strength

The undrained shear strength is the strength of the soil when it is loaded very rapidly

so there is no time for drainage. Unlike in drained loading, where the stress path can

vary (e.g. constant p test, or constant r test), the undrained stress path is unique and depends on the development of pore water pressure. Thus while we can determine

different drained strengths for the soil depending on the stress path chosen, the

undrained shear strength is unique. It is equal to: su = cu= qmax/2 (i.e. radius of Mohr

circle at peak or critical state as appropriate). su is often used in design for short term

conditions, which can be most critical for some structures e.g. foundations. The use of Su can be convenient, but predicting the direction of the undrained stress path and

hence the change of pore pressure is difficult - if p at failure cannot be estimated then the shear strength q cannot be either.

The undrained shear strength depends on the soil stress state, thus it is important when

trying to measure su in the laboratory to start the test from the correct in-situ stress. As

a soil sample is extracted from the ground with little disturbance, its mean effective

stress, which is unknown (p'0), remains constant. If the specimen is installed in a

triaxial apparatus and subjected to undrained isotropic compression, the rigidity of the

water will mean that any increase in total stress will be entirely taken by the pore

water (u=p), so that the effective stress remains unchanged (p'=0) we are

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assuming the soil is isotropic. Thus whatever the isotropic total stress imposed to the

specimen, the pore water response during isotropic undrained compression will be

such that the mean effective stress remains constant, equal to p'0. If the axial total

stress is then increased, the specimen will be sheared and will eventually fail. The

value of qmax obtained in the undrained test is equal to 2su. It is not necessary to

measure the pore water pressure generated during the shear test only qmax is needed, equal to the size of the Mohr circle tangent to the failure line. The same Mohr circle

will be found whatever the confining pressure on the specimen (Figure 18).

,

cseffective

stresses alternative

total stresses

ar ar ar a

r

uu

Su

Figure 18 Mohr circles during undrained shearing and determination of Su.

From the critical state theory, su is only a function of the specific volume, v, which

means it is a function of the water content only (Figure 19).

v

lnp'

CSL: v= lnp'

pcs

A C B

Figure 19 Volumetric stress paths during undrained shearing.

For an undrained test, the volume is constant, thus the stress path in a v-lnp' plot is

horizontal. If A and B are two specimens with different stresses but same volume,

they will reach the critical state line (CSL) at the same point C. The mean effective

stress at failure can be determined form the equation of the CSL, and the undrained

shear strength can be derived:

vp cs exp' (14)

cscs pMq ' (15)

15

vMqs csu exp

22 (16)

For a saturated soil, v=Gs.w, thus the undrained shear strength is only dependent on

the water content. Because of localisation of strains in tested specimens (we never

really have homogeneous deformations throughout the soil specimen), the strength

predicted during site investigation is often underestimated hence it is not correct in

practice.

The undrained shear strength is also indirectly measured when performing a fall cone

liquid limit test. For the majority of clays, the liquid limit corresponds to a strength of

about 2kPa, while the plastic limit corresponds corresponds to about 200kPa.

Assuming a semi-logarithmic relationship between Su and the water content (Figure

20), we can derive an expression for Su in function of the liquidity index LI

(PLLL

PLwLI

):

)210200 LIus (17)

w

logSu

200kPa

2kPa

PL LL

Figure 20 Dependence of undrained shear strength on water content.

v. Residual strength

The alignment of platy shaped particles in high plasticity soils at large displacements

allows the strength to reduce below the Critical State to the Residual Strength (Figure

21). Because of localisation of strains we must plot displacements.

q

a (%)

peak

critical state

1-5 10-30 shear displacement (mm)

high plasticity soil

low plasticity soil

q

a (%)

peak

critical state

1-5 10-30 shear displacement (mm)

high plasticity soil

low plasticity soil

Figure 21 Stress-strain response during shearing to large deformation definition of

residual strength.

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Summary

Critical state is a state of constant q and constant v or u with continued shear strain: there is turbulent shearing of the soil particles.

Critical state must not be confused with the residual strength of a plastic clay mobilised at large displacements when platy clay particles align: in this case

we have laminar shearing.

The critical state angle of shearing resistance, 'cs, which corresponds to the angle of repose of an infinite slope, is related to the CSL gradient, M, by:

sin 'cs = 3M/(6+M) for compression.

(Note that MeMc if 'cs= constant - Mohr-Coulomb)

Drained and undrained tests from normally consolidated or overconsolidated initial states all define one unique critical state line

at large strains, typically 10-30% s.

Peak states are reached at small-intermediate strains (typically 0.5-5%). They result solely from work done in dilation.The peak corresponds to the

maximum rate of dilation and hence depends on the current combination

of stress & volume (overconsolidation ratio).

'cs is dependent on the nature of the soil particles (shape, size, mineralogy).

A residual state (r) can be reached at very large strains for plastic clays only: the clay deforms in a laminar flow, with deformations concentrated

on a plane (e.g. landslides).

17

B2. APPLICATIONS OF SOIL MECHANICS IN ENGINEERING PRACTICE

Typical problems in geotechnical engineering are the design of foundations, of slopes

and of retaining structures. The geotechnical engineer must find solutions (i) that are

safe from collapse (limit state) and (ii) ensure that the ground movements caused by

construction do not damage the appearance or reduce the useful life of a structure

(service state). In this course, we will focus on (i), in which we will find the ultimate

load (or condition) when geotechnical structures fail (the ultimate condition). As long

as the actual loading condition is smaller than the ultimate loading condition, the

geotechnical structures are considered to be safe.

There are various ways to design geotechnical structures. The two extremes are:

Complex method: model the soil behaviour as accurately as possible and incorporate the stress-strain model into a numerical method such as the finite

element method to design geotechnical structures. The use of numerical

methods allows us to solve problems with complex boundary/ initial

conditions observed in field. However, this can become a time consuming

effort and requires expertise knowledge of the conditions and of constitutive

models.

Simplified method: identify the important soil parameters that are relevant to the design objective, and use relatively simple mathematics to solve a

boundary value problem. The boundary/ initial conditions may also need to be

simplified to obtain the solution. The simplified method leads to formulae

available in the design codes. The solution can be obtained easily, but

engineering judgement is required to assess the potential errors associated with

the simplification.

The solutions must satisfy:

Equilibrium (the forces must balance i.e. the sum of forces in any direction or sum of moments about any point is equal to zero).

Compatibility (the displacements must balance i.e. no holes appear and no soil is eaten up).

Material properties (stresses and strains in the soil are correctly related by strength and stiffness).

Usually it is not possible to find simplified solutions which satisfy all three criteria,

due to (1) complex material behaviour (2) complex geometry of problem (boundary

conditions) (3) self weight effects (gravity). The drawback of using simplified

analyses is that we cannot find easily exact solutions, so we calculate upper and lower

bound solutions as close to each other as possible. For example, Figure 29 shows the

load-settlement curve for a foundation; a simplified solution assuming that the soil has

failed (upper bound) will be on the unsafe side (the load or pressure determined is

higher than the load or pressure that the soil can bear) use a factor of safety; a simplified solution assuming that the soil is in equilibrium (lower bound) will be on

the safe side. The exact solution is in between the two.

18

qqf

safe lower bound unsafe upper bound (a)

(b)

Figure 22 Illustration of upper and lower bound.

The upper bound solution is found by calculating the load or stress that causes the soil

to fail. By trying different mechanisms of failure different solutions are found. The

lower the upper bound solution, the closer it is to the exact solution (Figure 22b). The

lower bound solution is found by calculating the load or stress under which the soil is

in equilibrium. By trying different scenarii of stress discontinuities different solutions

are found. The higher the lower bound solution, the closer it is to the exact solution.

When upper bound and lower bound solutions are equal, the solution is thus exact.

B2.1 Limit equilibrium method

The upper and lower bound plasticity methods give us a range of acceptable values

for design; it can be the maximum load that a foundation can carry for example, or the

maximum height than a cut can sustain without failure. Simplified solutions that do

not even obey two of the criteria cited are often used in geotechnical engineering.

They are known as limit equilibrium analyses, and were introduced by Coulomb in

1776 for analyzing rigid retaining walls. They typically assume a failure mechanism

but with the failing soil mass in equilibrium. A simple analysis of the equilibrium of

the forces at the boundaries of the system is carried out, neglecting internal stresses.

19

Solutions are not unique need to find the worse case scenario i.e. the largest active load on a retaining wall.

B2.2 Plasticity solutions

The mathematical theory of plasticity can be used to find the ultimate load or

dimension for a problem. There are two approaches:

Upper bound approach we will find the kinematically admissible state, i.e. only mechanisms of failure are considered and compatibility of stress is

ignored. The failure mechanism that gives the smallest ultimate solution

should be closer to the correct solution.

Lower bound approach we will find the statically admissible state, i.e. failure mechanisms are ignored and compatibility of stresses is ensured. The

stress state that gives the largest ultimate solution should be closer to the

correct solution.

i. Stress discontinuity line and stress fan for clay

Stresses near geotechnical structures are not all equal, for example the stress directly

beneath a foundation is different from the stress at the edge of the foundation. There

are thus discontinuities in the soil stress field. We can find the rotation and magnitude

of stresses either side of a discontinuity by assuming that the stress equilibrium is

maintained within each block and across each discontinuity line. Lets consider Figure 23, with two failing soil elements adjacent to a discontinuity line. Here, the stresses in

the elements satisfy: (i) na = nb and na = nb and (ii) ta tb

Figure 23 Stress equilibrium across the discontinuity line

The stress equilibrium on the discontinuity is maintained by having the stress states

acting on the discontinuity line the same on both sides of the discontinuity line.

For a general stress state condition (1(A)

20

Figure 24 Rotation of major principal stress direction by

The soil has an undrained shear strength cu, which is the same in A and B (Mohr

circles have the same radius cu. The circles intersect at the common stress state on the

discontinuity line (Figure 24). This stress state is acting on the horizontal planes on

elements A and B: we can draw the physical planes where it is acting and find the

poles for circles A and B, P(A) and P(B). The change of total stress across the

discontinuity line is related to the rotation of the direction of the major principal

stress. [angle between planes where A and B act: = ( ) () = =

90 - ]

s = 2cu cos = 2cusin= 1(B) 1(A) (18)

Example (Figure 25): na = nb = 0. Two Mohr circles join at the tip. The major principal stress of one circle is equal to the minor principal stress of the

other circle. 1(B) = 1(A) + 2 cu. The rotation of the major principal stress

direction = /2 = 90.

Figure 25 90 degrees rotation of major principal stress direction

Example (Figure 26): Two Mohr circles join with 2cu separation. 1(B) =

1(A) + 2cu Rotation of the major principal stress direction = 45.

21

(a)

(b)

Figure 26 45 degrees rotation of major principal stress direction

Stress fan: Using the stress discontinuity concept, it is possible to have a change of stress from one region to another across a fan of discontinuities as

shown in Figure 27. The fan angle f is equal to the rotation of the direction

of the major principal stress across the fan. s = 2cu sin~ 2cu when is very small. Integration gives:

s = 2cu = 2cu f (19)

Figure 27 Stress fan

22

ii. Application No. 1: open trench in clay (vertical cut)

Lets consider a vertical cut in clay (Figure 28).

Upper bound solution: if the soil finds a mechanism to fail, it will. Lets now assume that the collapse mechanism is the simple planar failure illustrated in the figure.

45

HW

Figure 28 Schematic diagram of a vertical cut with planar failure mechanism.

We can calculate the work done by the soil during the failure. Because the soil is

assumed to have failed, the equilibrium criterion is not satisfied, so our upper bound

analysis will only satisfy compatibility (the block slides along the slip/failure plane)

and material properties (undrained shear strength cu). We can draw a displacement

diagram for the different forces involved: compatibility imposes that the diagram

closes (Figure 29).

v v

Figure 29 Displacement diagram for vertical cut.

We use as unit displacement v the displacement of the weight W per unit length. By

trigonometry in the right angle triangle, we can calculate that W=H2/2, with the unit weight of the soil. The shear force per unit length on the slip plane is: T=

(failure)x area, with (failure) = cu and area=H/cos(45)= H2; we obtain T= cu H2 (note that it is in opposite direction to the weight). From the displacement diagram it

has moved by 2 v. The sum of the external work (by the weight) and the internal work (on slip plane) for the failing block of soil is therefore:

H2/2 x v - cuH2 x 2 v = 0 (20)

We thus derive the height at which this collapse mechanism would occur:

H = 4cu/ (21)

Lower bound solution: lets now take another approach: lets assume that the soil is in equilibrium (lower bound; Figure 30). There are no displacements so the two

criteria satisfied in this solution are the equilibrium and the material properties.

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Hv

h= 0 Figure 30 Schematic diagram of a vertical cut in equilibrium.

Figure 31 Schematic diagram of a vertical cut in equilibrium.

We consider the equilibrium of a small element just next to the cut. The Mohr circle

of total stresses for this element is drawn in Figure 31. The horizontal total stress is

equal to zero. The vertical stress is equal to the weight per unit area: v = H. Because this a total stress (undrained) analysis, the radius of the Mohr circle is equal to cu, and

we have:

H = 2cu (22)

The maximum height for the cut to be in equilibrium is therefore:

H=2cu/ (23)

We have determined a lower bound solution (H=2Su/) and an upper bound solution

(H=4Su/) for the vertical cut problem. By refining our choice of failure mechanism we could approach a better approximation.

iii. Application No.2: shallow strip foundation

Lets finally consider the case of a strip shallow foundation of width L subjected to a load F. The short-term solution is undrained.

Upper bound solution: one of the simplest mechanisms of failure we can imagine is

a two-wedge mechanism (Figure 32). Here we take the two wedges as isosceles

triangles (with two sides of same length L; the third side is of length L/cos(45) =

L2). We represent the shallow foundation as a surface foundation, as it is easier to model, but we have to take account of the fact that it is embedded below the surface.

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To do so we consider the weight of the soil at the depth of the foundation (called S in

the figure, for surcharge). For this simple example though we will assume no effect of the soil weight. The wedge directly underneath the foundation, which moves

downward, is called A and the corresponding wedge which moves upward is called B.

The soil is immobile (referred to as O).

F

BA

O

S

Figure 32 Schematic diagram of a foundation with wedge failure mechanism.

Lets draw the displacement diagram for the different forces acting internally (movements along shear planes between O and A, and between O and B; relative

movement between wedges A and B; Figure 33) and externally (weight of the

structure and foundation, F, and surcharge, S). Note that here because there is no self-

weight effect the term in S will be zero. Compatibility dictates that F and S move by

the same amount in opposite directions, that the movements of F and A relative to O

are the same, and similarly B and S move in the same amount relative to O.

O(fixed)

F

S B

A

v

2v

Figure 33 Displacement diagram for foundation.

We can now equate the work done by the external forces and the internal dissipated

energy:

F x v - 0 = cu x L2 x v2 + cu x L2 x v2+ cu x L x 2v (24)

The maximum load that the foundation can carry before failure is:

F = 6cuL (25)

Lower bound solution: lets now assume that the foundation is in equilibrium. There is a discontinuity in the stresses directly beneath the foundation and those beside it

(Figure 34).

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F

A BI II

Figure 34 Schematic diagram of a foundation with vertical discontinuities.

Lets call qF the bearing capacity of the foundation (qF = F/L). Lets consider two small elements underneath (A) and beside the foundation (B), at a depth z. The

vertical and horizontal total stresses acting on element A are A,v = z + qF and A,h

respectively. The stresses acting on element B are B,v = z and B,h = A,h. Lets draw the Mohr circles for these elements (Figure 35):

Figure 35 Mohr circles for elements A and B.

We assume that the foundation does not create any shear stresses on the horizontal

and vertical planes. In element A, the major principal stress is the vertical stress A,v,

and the minor principal stress is A,h. By continuity, in element B the major principal

stress is B,h and the minor principal stress is B,v. By using the geometry of the circles and the undrained shear strength property of the soil (which defines the size of

the total stress Mohr circles at failure), we obtain:

A,v B,v = 4cu = (z + qF) z (26)

qF = 4cu (27)

By choosing more complex discontinuity patterns (e.g. fan) we would obtain a better,

higher, lower bound value. See examples for more upper and lower bound solutions.

B3. CONCLUSIONS

Even though this part of the course used the ultimate state as a design criterion, it is

really the service state that should be considered (ground deformations will affect the

structures). We are not designing for failure but for structures to be serviceable. So

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both strength and stiffness are important! You will come across serviceability when

learning about retaining earth systems, foundations.

REFERENCES

Coulomb, C.A. (1776). Essai sur une application des rgles des maximi et minimi

quelques problmes de statique relatifs larchitecture. Mmoire acadmique royal prsentant divers savants. Vol. 7. Paris.

GEO manual No.2, published by the Geotechnical Engineering Office, Civil

Engineering and Development Department, Hong Kong SAR Government

BIBLIOGRAPHY

Powrie, W. (1997). Soil mechanics: concepts and applications. Spon Press.

Atkinson, J.H. (2007). An introduction to the mechanics of soils and foundations

through critical state soil mechanics. Spon Press.