Concept in Geotechnical and Foundation Engineering

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NPTEL Syllabus

e-Book on Concepts and

Techniques in Geotechnical and

Foundation Engineering - Webcourse

COURSE OUTLINE

This is in the nature of a ‘covering’ course for students who are

undergoing or have undergone courses in GeotechnicalEngineering and Foundation Engineering, at theundergraduate or postgraduate level. It is prepared as alearning aid in the hands of the student and a teaching aid inthe hands of the teacher. Geotechnical and foundationengineering professionals will also find the material useful inreinforcing their understanding of the subject they are dealing

with. The course material consists of text, figures (withANIMATION), audio and video clippings, the latter wherevernecessary and possible. Text and figures will also adopt acolour scheme to differentiate and highlight the material in the

order of importance. In short, the course is designed to createan environment of effective learning of the subject matter on ane-platform.

COURSE DETAIL

Sl.No.

Topic

1 Void ratio – porosity: general relationshipTotal and effective stress – a theoretical buildingblock

2 Shear strength – Mohr-Coulomb failure criterion

3 Earth pressure – active and passive – similarity with‘arching’ - active earth pressure onstem of cantilever retaining wall – impliedassumption

NPTELhttp://nptel.iitm.ac.in

CivilEngineering

Pre-requisites:

Basic courses in geotechnical

engineering and foundationengineering

Additional Reading:

1. Kurian, N. P. AnIntroduction to ModernTechniques inGeotechnical andFoundation Engineering,

Narosa PublishingHouse, New Delhi, AlphaScienceInternational, U. K., 2013.

2. Kurian, N. P., ShellFoundations –Geometry,Analysis, Design andConstruction,Narosa PublishingHouse, New Delhi, AlphaScience International,

U.K.,2006.

Coordinators:

Dr. Nainan P. Kurian


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Submerged unit weight – combined earth and waterpressures

4 Bearing capacity – relevance of shear failure –‘skirted footings’

5 Permeability: water table, hydraulic gradient, quicksand, filters

6 Consolidation – short term and long termperformanceCompaction – wet and dry densities

7 Foundation design phases – geotechnical design –bearing capacity and settlement factors‘net loading intensity’ – influence of water table ongeotechnical design

8 Compensated rafts

9 Special piles – inclined pile, tapered pile,

underreamed pile, screw pile‘Thermal analogy’ for analysing expansive soils andfoundations interacting with expansive soils

10 Negative skin frictionPile group actionPiled rafts

11 Soil pressure for structural design’ – in normal andswelling soils.Spring bed analogy for soilsColumn action – soil reaction

12 Soil-structure interaction – continuous elastic andWinkler modelsNonliner Winkler model, continuous Winkler modelInfluence of rigidity on differential settlements

13 Conical, spherical and hypar shell foundationsInstallation of precast shell foundation by ‘centrifugalblast compaction’

Department of CivilEngineeringIIT Madras


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14 Plate bearing testStandard penetration testPile load tests

15 Cantilever footing – constructionSimplex pile – constructionUnderreamed pile construction, half bulbCut support by ‘prestressing’ struts

16 Pile driving – by hammer impact, vibrationDriving steel, R.C. sheet pilesWell foundation – sinking

17 Drainage by well points – lowering of ground watertableFoundation dewatering

18 Stabilisation of boreholes and trenches by drillingmud

19 Reinforced earth – principle – Telescope and Hitexmethods of constructionBack-to-back construction of reinforced earth vs.continuous stripsReduction of settlement by reinforced earthSoil nailing

20 Diaphragm walls – construction, trench cutterGround anchors – construction, uses

21 Bored piles - constructionBored pile walls – secant piles, tangent piles,intermittent pilesMetro lines – construction by the ‘cut and cover’method

22 ‘Gabions’ for retaining structuresTerramesh and Green Terramesh for slopestabilisation

23 Retainin wall with relievin shelves


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Controlled yielding technique to reduce lateral earthpressure

24 Vibroflot – rotation of eccentric massVibrocompactionVibroreplacement, stone columns

25 SoilcreteSoilfrac

26 Dynamic compaction

27 Sand drains

Vacuum consolidation

28 Pile dynamic testing

29 Pressuremeter testingCentrifugal testing of geotechnical models

30 Dilatometer testingPiezocone testing

31 V-piles – static installationBox jacking

32 Sanitary Landfill construction

Bamboo-reinforced soil-cement for rural construction

33 Petronas and Burj Khalifa Towers – piled-raftfoundationConstruction of the Suez and Panama Canals

34 Geotechnical intervention in the restoration of theLeaning Tower of Pisa

35 Prestressed concrete piles – splicing


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36 Granular anchor piles in expansive soils

37 Multistoreyed structures with basement – Top-downconstruction

38 R.C. pavement construction

39 Statnamic, Osterberg tests

40 Dilatometer testingPiezocone testing

41 Cathodic protection of marine structures

42 Beach Management SystemGeneral

43 Functions and Scales

44 SI Units

References:

1. Gulhati, S. K. And Datta, M. J. Geotechnical Engineering,Tata McGraw-Hill

Publ. Co. Ltd., New Delhi, 2005.2. Venkatramaiah, C. Geotechnical Engineering, (3rd edn.)

New Age InternationalPublishers, New Delhi, 2006.

3. Kurian, N. P. Design of Foundation Systems –Principlesand Practices (3rd edn.)Narosa Publishing House, New Delhi, Alpha ScienceInternational, U.K.,2005.

A joint venture by IISc and IITs, funded by MHRD, Govt of India http://nptel.iitm.ac.in

http://nptel.iitm.ac.in/


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Module 1 / Topic 1

VOID RATIO (e) – POROSITY (n ) RELATIONSHIP

1.1 Definitions (Fig 1.1)

e =V V

V s

n =V V

V

Note: n is always expressed as a

percentage unlike e which is

expressed as a number.

1.2 Relationships

1.2.1 n vs e (Fig 1.2)

n =e

1 + e (1.1)

1.2.2 e vs n (Fig 1.3)

e = n1 - n

(1.2)

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Notes

1) Eq. (1.1) is of the same general form as y = x

a

+

bx

. The features of this relationship

are illustrated in detail in Sec. 51.10; also see Kurian (2005: App.E – Sec.12).

2) It may be noted that Fig.1.3 can be obtained by rotating Fig.1.2 anticlockwise by

900 and viewing from the reverse side.

3) Whereas e can exceed the value of 1 (unity), n cannot exceed 100 % which is its

upper limit.

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Module 1 / Topic 2

THE “EFFECTIVE STRESS PRINCIPLE” – A THEORETICAL BUILDING

BLOCK

The ‘effective stress principle’ was enunciated by Terzaghi (1925) in his celebratedbook ‘Erdbaumechanik’ which was the first seminal publication heralding the birth of

modern Soil Mechanics.

To the extent effective stress controls the mechanics of saturated soils (soils

whose pore space is filled with water), it can be called a determinant of the engineering

behaviour of soils (Gulhati and Datta, 2005).

However, what is interesting is the fact that effective stress is not a real quantity –

in the sense of being a quantity that can be physically measured – but a rather fictitious

quantity, dwelling in the realm of concepts.

2.1 Definition

Effective stress or (pressure) p is defined as the difference between two quantities

which can be measured or determined, namely total stress (or pressure) and pore water

pressure, or simply pore pressure, which is the pressure of water existing in the pore

space.

Referring to Fig. 2.1 which depict a simple static condition, at level A-A the total

pressure due to overburden =

x h, where

is the saturated unit weight of the soil.

The soil being saturated, there is a continuous body of water running through the pore

space in the soil. Hence the pore water pressure at level A-A due to the head of water

above = x h. (h can be determined by a piezometer if it is different from the static

head.) If we call effective pressure, p, as per the above definition we can state,

p = p – u, where u is the pore pressure due to the head h.

Hence, p = x h - x h

= ( - ) h

= x h, (2.1)

where is the submerged unit weight of the soil. (Please see Sec.4.5.7 explaining

how the submerged unit weight is obtained as the difference between the saturated unit

weight of the soil and unit weight of water).

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Just as water is a continuous body running through the pore space, the soil

skeleton is also continuous thanks to the mechanical contact between the grains which

constitute the solid phase of the soil. Hence the total pressure at any depth is sustained

together by the soil grains and the pore water. Therefore the effective pressure (or

stress), in a physical sense, can be looked upon as the stress transmitted from grain to

grain at their points of contact, and in tha t sense it is called the ‘intergranular pressure.’

But a closer examination, which follows, will reveal that, in a real sense it is not the same

as the physical quantity described above, but only a conceptual quantity, defined as the

difference between two real quantities, viz. the total pressure and the pore water

pressure.

2.2 Examination of the nature of the effective stress

Let us assume the solid phase of the soil medium as consisting of small spherical

balls or beads of identical size place one above the other as shown in Fig.2.2. Let us

also leave some distance between the columns of spheres so that the pore space is filled

with water and forms a continuous body.

At level A-A running through the points of contact between the spherical balls, if the total

pressure is p, the total force.

P = p x A, where A is the total area over which p acts.

If the balls are perfectly rigid, the contact between the balls is a point, which theoretically

has no area (Fig. 2.2a). However, if the balls are of some softer material, a small area

can be assumed over which contact exists. (Note that if the area of contact is 0, thecontact stress would be ∞; even if one has a small positive value for the area of contact,

the contact stress would be less than ∞, but will still have a very high value.)

Let us call this small area A’ .

A = A’ + Aw (Fig.2.2b) where, Aw is the area occupied by

water.

Hence we can state,

P = P’ + u Aw , where P’ is the part of P transmitted

through the solid phase.

Dividing by A,P

A =

P '

A +

u.Aw

A

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Since ≃ A, we can state, p =P'

A + u

= p + u, from which

p = p – u (2.2)

From the above one notes that p is not P ' divided by A' , but the full area A over which p

acts. It is this fact which gives p its fictitious attribute.

Let us now look at Fig. 2.3 which depicts the actual situation in a saturated soil.

The plane C-C passing through the actual points of contact between the soil particles is

wavy , and its area is slightly higher than A which is actually its projected area on a

horizontal plane. If this area is treated as equal to A' , the same situation as in Eq. (2.2)

will repeat here also.

Thus p is not the actual ‘intergranular’ pressure in so far as it relates not to the

small area A' which is the actual area of contact, but A the total area. (It may still be called

‘intergranular’ pressure in a literary sense, but not in a quantitative sense.)

It is indeed amazing to note that a whole body of knowledge in the field of modern

geotechnical engineering has been built on such a seemingly innocuous concept!

P.S.: The above picture has a parallel in ‘permeability’ (Topic 6 ), where k the coefficientof permeability is defined in relation to the total cross sectional area of the soil and not

the actual cross sectional area of the pore space through which water flows.

References

1. Gulhati, S. K. and Datta, M. (2005), Geotechnical Engineering , New Delhi, Tata

McGraw-Hill, xxviii + 738 pp.

2. Terzaghi, K. (1925), Erdbaumechanik auf bodenphysicalischer Grunlage, Franz

Deuticke, Leipzig und Wien.

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Modules 2,3 / Topic 3

SHEAR STRENGTH: THE CHARACTERISTIC STRENGTH OF SOIL

3.1 Introduction

The strength of a material in any mode is the highest or ultimate value of stress it can

sustain or resist in that mode. The basic modes occurring are tension, compression, shear

and torsion. The characteristic strength of concrete is its compressive strength, and that

of steel its tensile strength. The characteristic strength of soil , however, is its shear

strength. Concrete also has tensile strength – even though very minor – and shear

strength. In the same way steel has compressive strength of comparable magnitude as

the tensile strength, and a shear strength of nearly half that value (Kurian, 2005: Sec.

7.4.3). In the case of soil also small cylindrical samples can be extracted and tested in

compression under an all round pressure (as in the triaxial test) or without it (as in the

unconfined compression test). The strengths so obtained are known to be functions of

the shear strength of the soil. Soil, however, has negligible strength in tension.

The fact that a body of soil can stay in a slope (Fig.3.1a) is because it possesses shear

strength.Since water has no shear strength, the surface of a still body of water must

always remain horizontal (Fig.3.1b).

The pressure exerted by soil on a retaining wall (see Topic 4) is a function of its shear

strength. Since water has no shear strength, the pressure exerted by water on a weir or

dam is higher, notwithstanding its lesser unit weight – typically double the earth pressure.

In fact herein lies the essence of the stabilizing action of ‘drilling mud’ on the sides of cutsas in boreholes and trenches (Topic 44).

The bearing capacity of a foundation, such as a footing, which transmits loads from

the superstructure on to the soil below, is also a function of the shear strength of the soil.

The soil, in wedge form, fails in shear (Topic 5 ) and we use the term ‘shear failure’ also

for failure in bearing capacity. Since water has no shear strength, it has also no bearing

capacity.

All the above go to prove that shear strength is a fundamental property of a soil on

which depends the pressure exerted by the soil and the pressure borne or resisted by the

soil. In fact, the entire body of Soil Mechanics is built on the basic fact that the

characteristic strength of soil is its shear strength.

3.2 Shear strength parameters

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The general case of a soil having c > 0, > 0 is called a cohesive soil, because

the term ‘cohesive’ per se does not rule out the presence of friction. Hence the term

‘ideally’ for describing a purely c > 0, = 0 soil. On the other hand the term ‘cohesionless

for the second type does rule out the presence of cohesion.

One may also call the above soils c –soil, -soil and c - soil, using the respectivesymbols.

Fig.3.3 shows all the cases mentioned above. (In the c –case, since the shear strength

line is parallel to the x -axis, it is independent of . Since the shear strength line in the

- case starts at the origin, c is absent in s.)

3.3. Determination of c and from the direct shear test

The shear box carrying the soil sample is in two halves (Fig.3.4). Under a given normalstress , the bottom half of the box is moved sideways until the soil fails by shear at the

interface of the two halves. The corresponding failure load is noted and plotted in the

figure (Fig.3.4). The test is repeated at different values of and a straight line is fitted

through the points so obtained. c and are obtained as the y -intercept and the inclinationof the fitted line, as shown.

3.4. The Mohr’s circle

The Mohr’s circle provides a graphical means of determining the normal stress and

shear stress on any plane in a 2D biaxial stress situation.

Fig.3.5 shows the cross section of a long rectangular prism of any material subjected

to stresses (vertical) and 3(lateral). Being a 2D case, the prism is, theoretically,

infinitely long and whatever happens on the cross section shown in the figure is identical

at all parallel cross sections, i.e. in the length direction perpendicular to the plane of the

paper (in this case the monitor screen). and 3 are ‘principal’ (normal) stresses since

they are unaccompanied by shear stresses on the respective planes. There is no

principal stress 2 on the cross sectional face. i.e. in the length direction, which makes it

a purely biaxial stress situation.

The Mohr’s circle is drawn on a plot (Fig. 3.5) with (the normal stress) on the x -axis

and s (the shear stress) on the y -axis. The principal stresses and 3 are plotted on

the x -axis as OB and OA and a circle is drawn (only half the circle is shown) with AB as

diameter. This circle is called the ‘Mohr’s circle’.

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Our effort is to determine the normal stress n and the shear (tangential) stress t on a

plane inclined at to the horizontal. This is accomplished by drawing a line inclined at

2 from the centre of the Mohr’s circle C. The point of intersection of this line with the

circumference of the circle, D gives n and t on the inclined plane as shown. Note that the

same point of intersection D can be obtained by drawing a line from A inclined at . (This

follows from the result that the same arc such as BD subtends at any point on the

circumference half the angle which it subtends at the centre.) One can draw the complete

circle and get the stress picture on all planes with varying from 0 to 180. In this respect

point A is called the ‘origin of planes’ or the ‘pole’.

3.4.1 The Mohr-Coulomb failure theory for soils

The Mohr-Coulomb failure theory integrates the Mohr’s circle at failure with the shear

strength (Coulomb) line as shown in Fig.3.6. In this figure the is the value of at

failure against the given 3. It is found that the shear strength line is tangential to thiscircle with the point of contact A representing failure. This means, shear failure occurs

on a plane inclined at (45 +

2) (Fig. 3.6). The normal and shear stresses acting on the

failure plane are marked in the figure. The ratio ( s/ ) has the highest value in this plane,

which earns it the name ‘the plane of maximum obliquity’ (Gulhati and Datta, 2005, Sec

11.8). It is called so in the sense that the angle will attain the maximum value when

(s / ) is the highest. (see figure which shows increasing by increasing s at the same ,

and decreasing at the same s.)

What is, however, interesting is the fact that failure does not occur on the plane of

maximum shear (Point B) which is inclined at 45. It is easily verified that in this plane:

s =−

2 , which is the radius of the Mohr’s circle, and

=+

2

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3.5 Determination of c, from the triaxial shear test

The triaxial test provides facility for failing a small cylindrical soil sample by increasing

to failure against a given constant 3. (In the actual test the sample is subjected to an

allround pressure 3 and the deviator stress is increased till failure occurs (Fig3.7a).) So = 3+ deviator stress at failure. Since the shear strength line is tangential to the

Mohr’s circle, if two tests are conducted on two different but identical samples at two

different values of 3, two Mohr’s circles can be drawn. Since the shear strength line

must be tangential to both the circles, a common tangent – called ‘Mohr’s envelope’ – is

drawn from which c and follow as shown in Fig. 3.7b.

It is necessary for one to clearly appreciate that this is a biaxial case resulting from

axisymmetry. Being axisymmetric, what happens on any diametric plane is the same as

what happens on any other diametric plane, which makes it a purely 2D or biaxial case.(The corresponding picture in the rectangular prism case (Fig.3.5) was that what happens

on any cross sectional plane is the same as what happens on any other cross sectional

plane all of which are parallel, and perpendicular to the longitudinal axis.)

If it is a -soil, since only one shear strength parameter, viz. is to be determined,

one test would suffice. Since c = 0, the Mohr’s envelope starts at the origin and mustbe tangential to the Mohr’s circle (Fig. 3.8).

If it is a c-soil, again one test would be sufficient, and since =0, the tangent must

be horizontal and pass through B (Fig. 3.9) giving c as shown. It is noted in this case that

failure occurs at the 45 plane.

If it is an unconfined compression test (3 = 0) on a c -soil or a predominantly c -soil,

= qu (Fig. 3.10) and c = qu

2 , where q

u is the unconfined compressive strength.

In the limit, if it is water for which c = = 0, the x -axis itself is the Mohr’s envelope

which means the Mohr’s circle is a point lying on the x-axis (Fig. 3.11). Since and 3

coincide at this point, = 3, or pv

= ph

.

3.6 The effective stress parameters c ' and ϕ'

Our discussion so far veered round to total shear strength parameters c and . It is

relevant in respect of saturated soils to investigate the effective stress parameters c’ and

′ , taking into account the influence of pore water pressure on the results. At the failure

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plane in a saturated soil the presence of pore water obviously does not contribute to shear

strength simply because water has no shear strength. Therefore frictional failure (slip)

can only occur along the points of grain contact at the failure plane produced by the

effective normal stress ′ and the effective angle of internal friction ′. It is therefore

reasonable and necessary to rewrite the shar strength equation in terms of the effective

stress parameters as:

s’ = c’ + σ' tanϕ' (3.2)

If the soil in the field is in a saturated state, and if it has facility to drain under load

(consolidation – Topic 7 ), it would be more relevant to relate the long term behaviour of

the soil to its effective stress parameters. (Note that even when water is slowly but

continuously draining under consolidation, the soil remains saturated at all levels of

consolidation. The pore pressure, however, will be negligible at advanced stages of

consolidation (Murthy, 1974: Ch.13).)

3.6.1 Determination of the effective stress parameters

Determination of the effective stress parameters c ' and ′ can be achieved by the

same triaxial test if we can either measure pore pressure in the sample during test under

and 3 and , or else, allow the sample to drain under load and then conduct the test.

The former approach is indeed faster where it is possible to conduct the triaxial test

with facility for pore pressure measurement. The total and effective Mohr’s circles can be

drawn as shown in Fig. 3.12 from which one can get c , and ′, ′ .

One cannot predict for certain how different ′ and ′ would be compared to c and in quantitative terms since it depends on several interacting parameters. A typical result

could be, c’ < c and ′ > .

Analysis using c and is called total stress analysis and that using c’ and ′ is calledeffective stress analysis. Total stress analysis is more relevant in the ‘short term’ and

effective stress analysis, in the ‘long term’. The short/long term differentiation is on

account of consolidation. While the theory of consolidation can predict quantitatively the

time it takes for a given percentage of consolidation to occur, short/long term behavioursare not expressed quantitatively in relation to time.

Total and effective stress parameters can be based on ‘undrained (subscript U) or

‘quick’ tests and ‘drained (subscript D) tests respectively. The unconfined compression

test is always an undrained or quick test.

Summarising the above, we can state the following terms in mutual association.

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short term behaviour – undrained (quick) test – total stress analysis

long term behaviour – drained (slow) test – effective stress analysis

Between total stress (short term) and effective stress (long term) analysis, design must

cater to the more critical of the two states.

The above applies to cohesive soils because of the time-dependent nature of its

behaviour thanks to consolidation, It is not relevant in the case of a cohesionless soil like

sand where, in the field, drainage and therefore consolidation takes place instantaneously

on the application of the load.

3.7 Undraind/drained triaxial tests

The triaxial test provides facility for the drainage of the sample in two phases, 1) after

the application of the all round pressure (3), and 2) during the stagewise increase of the

deviator stress until failure occurs. Accordingly we have three types of tests, which are:

1) Undrained test or ‘quick’ test (UU) in which drainage is not allowed in both the above

phases,

2) Consolidated-undrained test or consolidated-quick test (CU) where the sample is

allowed to consolidate or drain under the all round pressure, but not allowed to drain

during the increase of the deviator stress until failure, and

3) Drained or ‘slow’ test (DD) in which the sample is allowed to drain both under the all

round pressure and under all the stages of the increase in deviator stress until failure.

The purpose of carrying out a particular test is to simulate the field conditions to the

extent possible. Hence the choice regarding which triaxial test to conduct depends upon

how far the test conforms to the actual drainage conditions of the soil existing in the field

(Murthy, 1974: Ch. 13).

For foundations in clayey soils, UU tests are preferred because, thanks to their very

low permeability, the soil will be in an undrained state during the initial phase of the

application of load,such as from a footing, which can therefore be more critical.

CU tests are used where the soil has undergone consolidation before the applicationof any fresh loading. This condition is typical of very fine sand, silt or silty sand with

relatively low permeability.

DD tests are generally conducted in sand where, because of their relatively high

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completed at the end of the loading process. DD tests are therefore ‘slow’ only in the

case of clayey soils.

In drained tests the sample undergoes reduction in volume due to the exit of pore

water by drainage. On the other hand, in undrained tests, where pore pressures (u) are

measured for the determination of the effective stress parameters, there is no change ofvolume accompanying the test (Gulhati and Datta, 2005: Sec.11.4).

P.S.: Referring to Fig. 3.1a, the angle which a natural slope makes with the horizontal

may be called the angle of repose. For a perfectly clean and dry sand or gravel, this angle

is approximately equal to the angle of internal friction of the sand in the loosest state.

This is, however, not the case with other soils.

Reference

Murthy, V. N. S. (1974), Soil Mechanics and Foundation Engineering , Dhanpat Rai &

Sons, Delhi.

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Modules 4, 5, 6 / Topic 4

EARTH PRESSURES

4.1 Introduction

A retaining structure, such s a retaining wall, retains (holds in place) soil on one side

(Fig. 4.1). The lateral pressure exerted by the retained soil on the wall is called earth

pressure. It is necessary for us to quantitatively determine these pressures as they

constitute the loading on the wall for which it must be designed both geotechnically and

structurally, the former ensuring the various aspects of stability of the wall (stability

analysis) and the latter, catering to the structural action induced in the wall by the forces

(Kurian, 2005: Sec.1.1.2). Since we deal with the limiting values of these pressures, earth

pressures are ultimate problems in Soil Mechanics. This means, at this stage the soil is

no longer in a state of elastic equilibrium, but has reached the stage of plastic equilibrium.

In situations such as the one shown in Fig. 4.2, which involves grading (removal) and

filling, the in-situ soil itself may be used as the fill. The soil which thus stays in contact

with the wall is called backfill in the sense of being the fill at the back of the wall or a fill

which is put back. However, where we have the choice for a fresh backfill material, we

would go in for cohesionless soils of high internal friction (ϕ), and permeability (k ) to aid

fast drainage (Kurian, 2005: Sec.6.1).

A retaining wall permits a backfill with a vertical face. The alternative to a retaining wall

to secure the sides is to provide a wide slope (Fig. 4.3) as, for example, in road

embankments, but this needs the availability of adequate land to ensure the desired levelof stability of slopes, which may not be available in all instances. It may be noted in this

connection that sometimes backfills may themselves be laid in slopes to reduce the

heights of the wall (Fig. 4.4).

It is ideal that the water table (free water level in soil) stays below the base of the wall

without allowing it to rise into the backfill, adding to the load on the wall – with water

pressure adding to the earth pressure from the submerged soil below the water table –

for which it may not have been designed, rendering the design inadequate in such

eventualities. However, in situations such as water-front structures (Fig. 4.5), we have to

reckon with water, since water table will eventually rise in the backfill and attain the samelevel as the free water in the front.

We shall first consider earth pressure due to dry backfills.

4.2 The limiting values of earth pressures

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Earth pressures attain their ultimate or ‘limiting’ values depending on the relative

movement of the wall with respect to the backfill.

Thus from a stationary position, if the wall starts moving away from the soil, the

pressure exerted by the soil on the wall starts decreasing until a stage is reached when

the pressure reaches its lowest value (Fig.4.6). This means that there will be no furtherreduction in the pressure, if the wall moves further away from the soil. This limiting

pressure is called active earth pressure.

On the other hand, if the wall is made to move towards the soil, i.e. the wall pushes

the soil, the pressure exerted by the soil on the wall starts increasing until a stage is

reached when the pressure attains its maximum value (Fig.4.6) and as before, there will

be no further increase in the pressure if the wall moves further towards the soil. This

limiting pressure is called the passive earth pressure. In the initial at-rest state, the soil

is in a state of elastic equilibrium. From this state it reaches the states of plastic

equilibrium at the limiting active and passive states. The initial value of the earth pressuremay be called neutral earth pressure or earth pressure at-rest .

Fig. 4.7 shows quantitatively typical at-rest, active and passive earth pressure

distributions on the retaining wall. While the active pressure is about 2/3 of the at-rest

value, the passive pressure is nearly 6 times the at-rest value, or 9 times the active value

in a cohesionless soil with ϕ = 30. Further, in a similar manner, the passive state ismobilised at a much higher value of wall movement than the active state. Quantitatively,

the lateral wall movements are typically 0.25 % and 3.5 % of the wall height for the active

and passive pressure conditions to get fully mobilised, respectively (Venkatramaiah,

2006: Sec. 13.3).

A word of explanation is due with regard to the names active and passive. In the active

case, soil is the actuating element the movement of which leading to the active condition.

In the passive case, the actuating element is the wall leading the soil to a passive state

of resistance against the approaching wall (Venkatramaiah, 2006: Sec. 13.3).

4.3 Determination of earth pressures by earth pressure theories

Earth pressures are determined by earth pressure theories. The two basic theories

available for this purpose are the Rankine’s t heory and the Coulomb’s theory . Of the two,the Coulomb’s theory is the older one; we shall, however, take up the Rankine’s theory

first because of its theoretical form.

However, before setting out on the above theories of limiting earth pressures, it is

necessary for us to look at earth pressure at-rest which should be treated as a starting

case. The soil being in elastic equilibrium at this stage, we should be able to proceed

with it based on theory of elasticity considerations.

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4.4 Earth pressure at rest

Fig. 4.8 shows an element of soil at a depth z in a semi-infinite soil mass. (Semi-

infinite means the mass extends in the +x, -x, +y, -y directions, but only in the +z

(downward) directions, all to infinity. If it extends equally also in the –z direction (upwards)

it would have made a fully infinite space.) The vertical and horizontal stresses in the

element are shown. The element can deform (undergo strain) in the vertical direction

only since the soil extends to infinity in the horizontal directions. Let the modulus of

elasticity and Poisson’s ratio of the soil be and μ respectively. Setting the lateral strain,obtained from theory of elasticity to 0,

=

- μ (

+

) = 0 (4.1)

Multiplying by - μ ( ) = 0 (1- μ) = μ =

− (4.2)

Referring to Fig 4.8

= . Therefore = − .

If − is denoted as and named coefficient of earth pressure at-rest , we canset

= . (4.3) being a constant, it is noted that also increases linearly with depth as itself,starting with 0 at the surface (z = 0)

If we now revert to Topic 1, it is seen that the - μ relationship is of the same form asthe e-n relationship. Hence will plot against as in Fig. 1.3.

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We note that = 0 when μ = 0 , a condition giving rise to 0 horizontal pressure.Further setting = 1,

− = 1

μ = 1- μ

2μ = 1

μ =

At this value of μ, = = . or in other words, and will plot identically with depth. When μ varies from 0 to 0.5, will vary as increasing fractions of , as can be noted from Fig. 4.9.

Because of the difficulty in determining μ of a soil reliably, various empirical formulae

have been suggested among which the one attributed to Jaky (1944) is an early favourite.

It states: = 1 (4.4)Fig 4.10 plots against . It bears comparison with Fig. 16 (Kurian, 2005: Sec. 6.4.1)for which it is plotted till = 90. It is noted from Figs. 4.9 and 4.10 that increases with

μ , but decreases with . At = 0 , applying to water , = 1, following which = .On the other hand, at = 90, applying to rock, = = 0 4.5 Rankine’s theory for active and passive earth pressures (1857)

Before we take up Rankine’s theory of earth pressure, we shall try to establish

analytically the relationship between , the principal stresses, based on theMohr-Coulomb failure theory (Fig.4.11).

In the figure,

CA = CD =−

OC = OA+AC = = + EO = c cot ϕ

CD = EC sin ϕ

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i.e.,

= + Multiplying by 2

− = 2 = ( 2 1 = 1 2 = +− 2 −

Similarly, = −+ 2 + Trignometrically −+ =45 )

+− =45

+ =45

− =45

Hence we can state

= 45 2 45 (4.5) = 45 2 45 (4.6)

Referring to Fig. 4.11, one may look upon the c-ϕ case as the ϕ -case with the origin

shifting from E to O.

4.5.1 Rankine’s expressions for active and passive earth pressures

In Fig. 4.12 let OA represent the vertical (principal) stress. Mohr’s circles I and II ar e

drawn on either side of A without gap. In case I the soil is laterally relieved leading to

reduction in reaching the limiting active value at failure. In case II the soil ispushed into itself and reaches the limiting passive value of at failure.

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Using Eqs. (4.5) and (4.6) we can now state:

= 45 2 45 (4.7)

= 45 2 45 (4.8)Calling 45 = ,’the coefficient of active earth pressure’, and

45 = ,’the coefficient of passive earth pressure’,and further = , and, therefore, = ,

and

being reciprocals of each other,

we can further state,

= . √ (4.9) = . 2√ (4.10)

In the case of a cohesionless soil, with c = 0,

= (4.11) = (4.12)

On the other hand, in the case of an ideally cohesive soil for which is 0, = √ = 1, we have = 2 (4.13) = 2 (4.14)

It may be noted that if c also = 0 (case of water ) we get = = = .


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Note further that, since the plane of failure is inclined at = (45 ), it followsthat = , = 2 The above results pertaining to

and c -soils can be directly obtained from the respective

Mohr ’s circles as shown in Figs. 4.13 and 4.14.

4.5.2 Failure planes

The failure plane in the active state is inclined at = ( 45 ) to the horizontal. If thefull Mohr’s circle is drawn, the potential failure planes are as shown in Fig. 4.15

(Venkatramaiah, 2006: Sec.13.6.1). In the passive case, like in the active case, the failure

plane should be reckoned from the point . It can be identified that the failure planes atpassive failure are inclined at (

45 ) to the horizontal. (The arcs of the Mohr’s circles

subtending these angles are highlighted in Fig. 4.12.) The picture is the same for -soil.In the case of the c -soil, these planes are inclined at 45 to the horizontal.4.5.3 Variation of active and passive earth pressure coefficients

It follows from the Rankine’s theory that the higher the , the higher the shear strength,the lesser the active pressure and the higher the passive pressure.

It is interesting to note that the Rankine’s theory for earth pressure developed for soil

can be extended to water (ϕ = 0 ) on the one hand and rock ( ϕ =90), on the other.When ϕ = 0, = = = 1

Therefore, = = = = = ℎ This is the hydrostatic pressure condition, applying to water.

If ϕ increases, decreases and increases. The latter increases much faster than theformer decreases, until we reach ϕ = 90 at which = 0 and = ∞. As a result, =ℎ, = 0 and = ∞.The variations of and , and also their square roots, with areshown in Fig. 4.16. 4.5.4 Plots of and c-ϕ case

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Since c and ϕ are constants, the first part of and , as per Eqs. (4.9) and (4.10)plot linearly like , but the second parts are constants. Fig 4.17 shows the sum of theseeffects. (Note that when two plots are to be added they should be drawn on opposite

sides, whereas if one is to be subtracted from the other they should be drawn on the same

side.)

It is observed from Eqs. (4.9) and (4.10) that is decreased and is increased onaccount of the contribution of c . As a result of the subtraction, Fig 4.16a shows a tensile

zone to a depth z which can be determined by setting,

=

. Therefore = √ (4.15)Since soil cannot exist in a state of tension, it is likely that it breaks contact with the

support over this depth (Kurian, 2005: Sec. 8.8).

c-case

Fig. 4.18 shows the active and passive pressure variations in the c -case.

To obtain the depth z at which the net pressure is 0,

Setting z = 2c , from which z = 4.5.5 Effect of surcharge on the backfill

There are instances such as in port and harbour structures where the backfill is

subjected to heavy surcharges such as due to supporting roads, railway tracks and heavy

stationary equipments. Like any other vertical load such as the self weight of the backfill,

these surcharges add to the lateral pressure on the wall the effect of which must be taken

into account in its design.

In order to consider the influence of the surcharge, its effect is reduced to an equivalent

downward pressure, q per unit area, Fig. 4.19.

The lateral active pressure due to surcharge is q

which is uniform with depth, q and

being constants. To this will be added the active earth pressure as shown in the figure.The same figure can be obtained by converting the surcharge pressure q as an

equivalent additional height (h’) of the backfill which is obtained by setting

γ h’ = q, from which h’ = ( ) (4.16)

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The pressure diagram on the wall alone for the full height of the backfill including the

additional height is the same as the earlier pressure diagram as shown in the figure.

4.5.6 Earth pressure due to layered backfills

Rankine’s theory can easily accommodate layered backfills, if the layers concernedare horizontal.

Let us consider the example shown in Fig. 4.20.

B̅ = ℎ (4.17) at B+ = ℎ

√ (4.18)

at C = ( ℎ ℎ √ (4.19)The above means that there is an immediate transition at B thanks to the difference in the

shear strength parameters of layers I and II. As a result it is seen that at + undergoesa sudden decrease thanks to the presence of c, being the same in the present case.

Theoretically speaking, at B̅ applies to a point in layer I lying infinitesimally abovepoint B, whereas at point + applies to a point in layer II lying infinitesimally below pointB. If one asks what is its value exactly at point B, the theoretical answer is, it is not the

average of B̅ and at +, but, simply, it is not defined at B.4.5.7 Earth pressure due to submerged backfills

If the backfill is submerged fully or partially, i.e. to full height or partial height, there is

a continuous body of water running through the pore space in the soil below the water

table. The water over this height will exert full hydrostatic pressure on the wall. To this

will be added the pressure due to submerged soil over this depth and the dry soil above

(Fig. 4.21).

While submergence causes a reduction in the unit weight of the soil, the shear strength parameters c and remain unchanged. Submerged unit weight (Kurian, 2005: Sec.2.7.1)

The submerged weight of a continuous (i.e. non-porous) body is its weight in air

subtracted by the weight of water displaced by the body. In other words,

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submerged weight of an object = weight in air of the object - weight of a body of water

having the same volume as the object.

That is to say, w s = w – v. (4.20)Unit weight is the weight per unit volume. In submerged unit weight sub of the soil weare concerned with is the weight of the solid particles in the soil in a unit volume whichare in a state of submergence.

Fig. 4.22 represents a unit volume in which

sub = weight of solids – weight of an equal volume of water.In order to simplify calculation, we add to both the parts on the R.H.S. a constant which

is the water to fill the pore space. The constant being the same, the result is, we have

saturated unit weight as the first term and unit weight of water as the second term on the

R.H.S. The final result is the familiar result,

= - . (4.21)Interestingly enough, this follows Eq.(4.20) with in place of w , as it should,representing the whole body.

4.5.8 Combined pressures (Kurian, 2005: Sec.6.4.2)

What follows is an important matter which every student/geotechnical engineer should

clearly understand, appreciate and assimilate.

If we take the unit weight of dry soil as 15 kN/m and = 30 (c = 0), Ka = andtherefore the active earth pressure at any depth h m = 5h kN/m. being 10 kN/m,the water pressure at the same depth = 10 h kN/m, which is twice the value of the activeearth pressure. (This is important since many, at least among the lay public, may tend to

assume that water being thinner, the corresponding pressure is also lower!)

When the backfill is saturated,

= ℎ ℎ = ℎ = ℎ . but ≠ ℎ . ,

but = ( ℎ . ℎ

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The importance of the above result is illustrated in Fig. 4.23. For a case where = and = 30, it is seen that, while the active pressure intensity is only 1/3 of the waterpressure, the passive pressure intensity is 3 times the water pressure or 9 times the active

soil pressure. (The first of the above statements means that water pressure is three times

the active pressure due to submerged soil, which we noted above as twice the active

pressure due to dry soil. Further if = , it follows that the active pressure due to drysoil is twice the same due to submerged soil.) It is obvious from the figure that walls

designed for active soil pressure are unsafe if the soil is allowed to get saturated!!

4.5.9 Need for retention (Kurian, 2005: Sec.6.4.1)

Fig. 4.24 draws attention to the need for retention in water, soil and intact rock.

Since water has no shear strength, its surface must always remain horizontal;

therefore water must be fully retained.

On the other extreme, if we treat intact rock as a medium with = 90, its sides canremain vertical, calling for no support since K a = 0.

Because of its shear strength, soil can remain in a slope. This means that only the fill

placed over this slope, which is needed to maintain a horizontal surface, requires support,

which therefore may be described as partial. This is, however, a qualitative statement as

the next section will show that the active pressure on the retaining wall is not exactly due

to such a wedge.

4.6 The Coulomb’s theory of earth pressure (1776)

The earth pressure theory propounded by Coulomb involves the consideration of a

critical wedge in the backfill adjoining the retaining wall the failure of which by shear at

the interface with the intact backfill and the wall gives rise to the active and passive failure

conditions. It involves the mechanical analysis of trial wedges for equilibrium at the stage

of ‘incipient’ or imminent failure by shear in the above manner (Fig4.25). It involves a

geometrical trial and error approach and therefore more tedious than the theoretical

approach followed by Rankine.

4.6.1 Coulomb’s method for the determination of active earth pressure

Let us take the general case of a retaining structure with an inclined back face,

supporting and inclined backfill in a c - soil (Fig. 4.26a). At the wall-soil interface weassume an angle of wall friction . The analysis is per unit length of the wall which makesit a purely 2D_case.

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action of the wedge on the wall and this is equal in magnitude and opposite in direction

to P obtained as above.

Now for different values of , we have to repeat the above work and determine thecorresponding values of P . We now make a plot of the values of P so obtained against

(Fig. 4.26c ). Join the points so obtained by a smooth curve and by drawing a horizontalline (i.e. parallel to the - axis) touching the curve tangentially we determine the highest value of P which is the active thrust . We can note the corresponding value of whichgives us the critical wedge causing the active thrust . (Note that we cannot go by thehighest value of P from among the individual results obtained. A curve must necessarily

be drawn because the peak may generally lie between two values and need not coincide

with any single value.). The P a that we have determined is the reaction of the wall on the

wedge. The action of the wedge on the wall which we are investigating is a force of the

same magnitude of P a, acting exactly in the opposite direction. It is this action on the wall

that we need for the design of the wall.

Reverting to the trial wedges, we can look upon the picture as the weight of the wedge

acting downwards, which we have noted as the force with which earth attracts the mass

of the wedge, being held back by the forces C, R and P.

If we want to make the picture more general by adding an adhesion component a at

the wall-backfill interface, a total tangential force A is generated in the direction AB, which

must be entered at point c at the end of which is to be drawn the line parallel to R . (Note

that adhesion at the wall-soil interface is similar to c at the backfill-backfill interface, i.e.

within the soil. In other words, a and δ at the wall-soil interface correspond to c and

within the soil. And just like in the case of c and , shear strength at the interface can bewritten as,

s’ = a + tanδ (4.22)In the case of the -soil (c = 0) the only difference is that C (and A) do not appear. In

the c -case ( = 0), on the other hand, we have to deal with only W , C (and A), N and P . As regards the influence of the parameters, the higher the values of c , , a and δ,

the lesser the value of P a as can be identified from the force polygon. This picture will

reverse sign when we come to the passive case.

4.6.2 Coulomb’s method for the determination of passive earth pr essure

The point of departure in the passive case is, since the wall pushes the soil, the wedge

moves upwards causing shear failure along AB and AC. This causes reversal in the

direction of the forces C, N tan and P tan δ (Fig. 4.27a).

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The polygon of forces (Fig. 4.27b) starts with W marked as ab. bc represents C . At c

a line is drawn parallel to R and at a, a line parallel to P . They intersect at d; ad gives the

value of P corresponding to this trial wedge. The P values so obtained from several trial

wedges are plotted against (Fig 4.27c ) and the minimum value so obtained by drawinga horizontal line (i.e. parallel to the

- axis) tangential to the curve, gives Pp.

It is important to note that Ka which gives the minimum value of earth pressure is

obtained as a maximum in Fig. 4.26c , and Kp which gives the maximum value of earth

pressure is obtained as the minimum value in Fig.4.27c , both being optimum values.

4.7 Comparison between Rankine’s and Coulomb’s theories of earth

pressure

A fundamental difference between the two theories is that, while Rankine’s theory

gives pressure distribution, Coulomb’s theory gives only total thrust. One can of course

obtain distribution from the latter, by assuming the nature of variation, such as linear-

triangular.

Rankine’s theory, though theoretically elegant, has several limitations as it goes by the

concept of principal stresses, without even recognising the presence of the wall. Hence

it cannot take adhesion and wall friction into account, leading to conservative values of

the active earth pressure. It can be extended to backfills with single slopes, but an

inclined wall-backfill interface is difficult to accommodate.

Coulomb’s theory, on the other hand, is more versatile as it can accommodate wall

with inclined interface, sloped backfills with even more slopes than one, with practicallythe same ease as vertical wall with horizontal backfill. It can also account for wall friction

and adhesion leading to more realistic results. However, being a geometrical trial and

error approach, it is certainly more tedious and time consuming unlike Rankine’s theory,

the expressions from which can be programmed and put on computer to yield fast results.

(While on this issue, it may be added that, trial wedge approach can also be programmed

on the computer for obtaining faster results.)

It is now time to show that both Rankine’s theory and Coulomb’s theory give the same

results for the basic case, as shown below.

Let us consider the basic active case of a retaining structure with a vertical face and

no wall friction, supporting a -soil with a horizontal surface. Consider a trial wedge whichmakes an angle with the horizontal (Fig.4.28). Angle BAC is therefore (90- =, say.The force polygon (triangle of forces) consists of P, W and R .

From the triangle of forces,

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= 9 0

Therefore, = + W = .ℎ .= ℎ

Rewriting, P =+ (4.23)

For P to a maximum, = 0

i.e., tan(

x

.

= 0

i.e.,+

+ .+ = 0 i.e.,

+ +−+ = 0

2 = 0 sin2 =2

2 = 90 = 4 5

Therefore = 4 5 , (4.24)which defines the critical failure plane. It is noted that the result is the same as obtained

from the Rankine’s theory.

Substituting so obtained in the expression for P (Eq.4.23), we get, = ℎ

= ℎ −

+


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= ℎ −

−+

= ℎ 45 x (90 45 ) =

ℎ 45 x 45

= ℎ 45

= ℎ−+ (4.25)

which is the same result as obtained from Rankines theory, as per Eq. (4.11).

It is indeed interesting to observe how both the theories converge to the same result

in this case which establishes the soundness of both the approaches.

4.8 Conclusion

We may close with a significant question in relation to active earth pressure which has

a bearing on design. The question pertains to the ‘assumption’ of active pressure which

has the potential of making the design based on it unsafe!

Active pressure being the lowest value of pressure, it is logical to ask, have we taken

any steps in the design to ensure active conditions to develop? The significance of the

question is that, if active conditions are not developing, the pressures are higher, and the

retaining wall designed for active pressure will be inadequate, and in the limit, even

unsafe!

The answer to this question is in fact simple. The design of the wall implies an

assumption that if the wall is designed for active pressure, the same being the lowest

pressure, the design of the wall will be the thinnest , and since it is thin, it will deflect

enough (Fig. 4.29) resulting in active conditions and the corresponding active pressures

to develop, considering especially the low value of deformation needed to mobilise the

active condition (Kurian, 2005: Sec.12.1).

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Modules 7, 8 / Topic 5

BEARING CAPACITY – THE LOAD BEARING ROLE OF SOIL

Every civil engineering structure, from the point of view of structural design, is

a load-bearing system. This is true whether it is a building, bridge, highway pavement

or railway track. The loads from the superstructure are to be transmitted to the soil or

rock below in a satisfactory manner as per the requirements of design.

Let us take the case of a building supported on soil. (Note that in the majority

of cases civil engineering structures are founded on soil.) If it is a multi-storeyed

framed structure, as for a residential or commercial complex, the loads are transmitted

by the columns to the soil, say through footings. The geotechnical design of these

footings, by which we arrive at their plan dimensions, must ensure that there is

adequate factor of safety against the failure of the soil in bearing, and that the structure

should not undergo excessive settlement.

Our concern in this section is the supporting or load-bearing role of the soil.

The ‘bearing capacity’ of the soil is the maximum load the soil can bear at which failure

occurs in the bearing mode. This makes it an ultimate problem in Soil Mechanics like

the earth pressure problem (Topic 4). At this stage the soil fails in shear and its

inherent property of shear strength (Topic 3) comes into play. The load at which the

soil fails in bearing is called its bearing capacity , and what give soil its bearing capacity

is its shear strength. Considering a body of water with a free surface, since water has

no shear strength it cannot bear loads, which soil in a similar state can do thanks toits shear strength (Fig. 5.1).

Like shear strength, its functions ‘earth pressure’ and ‘bearing capacity’ are

expressed in pressure or intensity terms (F/L2, e.g. kN/m2). The designer’s job is to

make sure that the pressure transmitted to the soil through the footing (Fig. 5.2)

matches the bearing capacity of the soil to an adequate factor of safety, such as 3.

Being an ultimate problem, the soil is in a state of plastic equilibrium at its bearing

capacity. However, if the load coming through the footing is of the order of 1/3 of the

load causing failure in bearing capacity – i.e. the factor of safety is 3 – the soil can be

deemed to be in a state of elastic equilibrium at working loads.

5.1 Reckoning bearing capacity

Let us consider, say a square footing, of width B installed at a depth Df (Fig.

5.3a). (The footing is shown as a rigid block to indicate that it will not fail (structurally)

before the soil fails in bearing.) We apply a central load of P on the footing and

measure the corresponding settlement S. Let us increase the load in stages and

measure the corresponding S at each stage. We cannot increase the load indefinitely

and a stage will be reached when the soil cannot take any further load. If we plot P vs.

S (load-settlement diagram) we will get a curve marked A in the figure. The curve ends

in a vertical part which represents a stage when S increases without any gain in P.

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Documents

Concept in Geotechnical and Foundation Engineering