Course Code Course Name CECE 4131 Geotechnical Engineering - II

Mr. Jayaram.D.K

MARK DISTRIBUTION

TOTAL MARKS – 100

THEORY -100 MARKS – (2/3)- 66.67 MARKS

PRACTICAL – 100 MARKS – (1/3) – 33.33 MARKS

Assessment Plan

(The course is a combination of Theoretical and Practical hours (For 2T + 2P contact hours)2/3 x Theoretical part marks + 1/3 x Practical part marks= Total marks out of 100

THEORY MARKS DISTRIBUTION

1. Assignments – 10 marks2.Quizzes – 20 marks

3.Mid term exam – 20 marks4.Final exam -50 marks

All assignments are to be submitted to the lecturer “ with in class time only”

Penalities (deduction of marks) for late submissions

Course DetailsCourse code Course Title(2T+2p) Group No Course Lecturer Academic Year Semester Credit

points Passing Marks

Student Details Theory Practical

Sl.No ID Name Gender Course Work MidTerm Exam

FinalExam

TotalMarks

(2/3) TotalMarks

Part – I Reports

Part- III Assessment

Total Marks

(1/3)Total Marks

Total Course marks (TT+TP)

Letter Grade

Grade point

Remarks

Quiz mark(Total)

Assignment Mark (Total)

20 10 20 50 100 TT 60 40 100 TP 100

Attendance Policy

• 10 Minutes late absent and not allowed in the class

• Health & Safety:-Lab Coat & Safety Shoes are compulsory in practical classes.

CECE 4131 Geotechnical Engineering II 3 Credit Hours

Prerequisites: CECE 3230

Goal To introduce the student to the principles of soil mechanics and to enable him/her

to apply such principles to civil engineering problemsObjectives Outcomes

The course should enable the student to:1. Understand the applications ofGeotechnical Engineering in the design offootings, retaining walls and in the assessment ofstability of slopes2. Understand types of deep foundations and theirdesign principles.

1. Apply basic concept to solve lateral earth pressureproblems and identify how they affect structures

2. Solve bearing capacity problems

3. Design retaining walls and footings

4. Apply basic concept to solve slope stability problems

5. Understand pile foundations and their design principles.

6. Work in a team

UNIT-1EARTH PRESSURE THEORIES

Introduction: In the design of retaining wall, sheet piles

or other earth retaining structures, it is necessary to compute

the lateral pressure exerted by the retained mass of the soil.

A retaining wall is used to for maintaining the ground surfaces

at different elevation on either side of it. The material retained

by the structure is called backfill which may have its top

surface horizontal or inclined. The position of the backfill

lying above a horizontal surface at the elevation of the top of

the wall is called the surcharge and its inclination to the

horizontal is called surcharge angle β

FIGURE SHOWS THE RETAINING WALL

Active and Passive earth pressure

Active earth pressure and Passive earth pressure

During the active state, the wall moves away from backfill and a

certain portion of backfill located immediately behind the wall

breaks away from the rest of the soil mass. This wedge shape

portion of soil is called failure wedge. The resisting force due to

shear strength of soil is developed in an upward direction along the

failure plane (or slip lines) as shown in Fig.

In passive case the wall moves towards the fill, due to

some thrust etc.The magnitude of the lateral earth

pressure depends upon the movement of the wall

relative to the backfill and upon the nature of soil or fill.

When the soil moves away from the backfill, mobilization of the

internal resistance of the soil, which builds up in directions away

from the wall takes place and hence earth pressure on the wall

decreases. The decrease of earth pressure continues upto a point

until the full

resistance has been mobilized. The earth pressure does not

decrease beyond this point with further movement of the wall and

is called as active earth pressure.

If on the other hand, the wall moves towards the fill, the earth pressure increases, because the shearing resistance builds up in direction towards wall. The pressure reaches at the point when the shearing resistance of the soil has been fully mobilized. Any further movement of the wall does not increase the pressure. The maximum pressure is called the passive earth pressure.

Failure surfaces for active and passive states

Rankine’s Theory: As originally proposed,Rankine’s theory of lateral earth pressure isapplied to uniform cohisionless soil only. Later it was extended to cohesive soils by Resaland Bell. The theory has also been extended to stratified, partially immersed and submergedsoil. This theory is valid when:

The back of the wall is vertical and smoothThe level of the soil behind the wall is horizontalSoil mass is semi-infinite, homogeneous, dry and cohesionlessThe length of the wall is long in comparison to the height so that it can be treated two dimensional situations.

Coulomb’s theory: He highlighted that if the wall is not

friction less unlike assumed in Rankine’s case the lateral

earth pressure applied by the soil is not normal to the wall

but at an angle or in other word has both horizontal and a

vertical components.Solutions from the coulomb theory can be obtained from the situations

when the back of the wall is not vertical but is inclined at an angle α to the

horizontal and the soil behind the wall is not horizontal but inclined at

an angle β to the horizontal

Assumptions:

The backfill is dry and cohesionless, homogeneous,

istropic and elastically undeformable but breakable.

The slip surface is the plane which passes through the heel

of the wall.

The sliding wedge itself acts as a rigid body and the

value of earth pressure is obtained by considering the

limiting equilibrium of the sliding wedge as a whole.

Bearing Capacity Of Shallow Foundation

Bearing Capacity Of Shallow Foundation

* A foundation is required for distributing the loads of the superstructure on a large area. * The foundation should be designed such that a) The soil below does not fail in shear & b) Settlement is within the safe limits.

Basic Definitions :

1) Ultimate Bearing Capacity (qu) : The ultimate bearing capacity is the gross pressure at the base of the foundation at which soil fails in shear.

2) Net ultimate Bearing Capacity (qnu) : It is the net increase in pressure at the base of foundation that cause shear failure of the soil.

Thus, qnu = qu – γDf (overbruden pressure)

3) Net Safe Bearing Capacity (qns) : It is the net soil pressure which can be safely applied to the soil considering only shear failure. Thus, qns = qnu /FOS

FOS - Factor of safety usually taken as 2.00 -3.00

4) Gross Safe Bearing Capacity (qs) : It is the maximum pressure which the soil can carry safely without shear failure. qs = qnu / FOS + γ Df

5)Net Safe Settlement Pressure (qnp) :It is the net pressure which the soil can carry without exceeding allowable settlement.

6)  Net Allowable Bearing Pressure (qna ): It  is the net bearing pressure which can be used for design of foundation. Thus,          qna = qns        ; if qnp > qns

 qna = qnp       ; if qns > qnp It is also known as Allowable Soil Pressure (ASP).

Modes of shear Failure : Vesic (1973) classified shear failure of soil under a foundation base into three categories depending on the type of soil & location of foundation. 1) General Shear failure.2) Local Shear failure.3) Punching Shear failure

General Shear failure –

Strip footing resting on surface Load –settlement curve

of dense sand or stiff clay

* The load - Settlement curve in case of footing resting on surface of dense sand or stiff clays shows pronounced peak & failure occurs at very small stain.

* A loaded base on such soils sinks or tilts suddenly in to the ground showing a surface heave of adjoining soil

* The shearing strength is fully mobilized all along the slip surface & hence failure planes are well defined.

* The failure occurs at very small vertical strains accompanied by large lateral strains.

* ID > 65 ,N>35, Φ > 360, e < 0.55

2) Local Shear failure -

* When load is equal to a certain value qu(1), * The foundation movement is accompanied by sudden jerks. * The failure surface gradually extend out wards from the foundation. * The failure starts at localized spot beneath the foundation & migrates out ward part by part gradually leading to ultimate failure. * The shear strength of soil is not fully mobilized along planes & hence failure planes are not defined clearly. * The failure occurs at large vertical strain & very small lateral strains. * ID = 15 to 65 , N=10 to 30 , Φ <30, e>0.75

Strip footing resting on surface Load –settlement curve

Of Medium sand or Medium clay

3) Punching Share failure -

* The loaded base sinks into soil like a punch.

* The failure surface do not extend up to the ground surface.

* No heave is observed.

* Large vertical strains are involved with practically no lateral

deformation.

* Failure planes are difficult to locate 222

Terzaghi’s Bearing Capacity Analysis –Terzaghi (1943) analysed a shallow continuous footing by making some assumptions –

* The failure zones do not extend above the horizontal plane passing through base of footing

* The failure occurs when the down ward pressure exerted by loads on the soil adjoining the inclined surfaces on soil wedge is equal to upward pressure.

* Downward forces are due to the load (=qu× B) & the weight of soil wedge (1/4 γB2 tanØ)

* Upward forces are the vertical components of resultant passive pressure (Pp) & the cohesion (c’) acting along the inclined surfaces.

 

Effect of water table on Bearing Capacity :* The equation for ultimate bearing capacity by Terzaghi has been developed based on assumption that water table is located at a great depth .* If the water table is located close to foundation ; the equation needs modification.

i) When water table is located above the base of footing -

* The effective surcharge is reduced as the effective weight below water table is equal to submerged unit weight. q = Dw.r +x.γsub

put x = Df-Dw q = γsub Df +( γ- γsub)Dw

ii) When water table is located at depth y below base :

* Surcharge term is not affected.

* Unit weight in term is γ = γsub + y ( γ – γsub) B

Thus,

qu = c’Nc + γ Df Nq + 0.5B γ Nr

When y = B ; W.T. at B below base of footing.

qu = c’Nc + γ Df Nq + 0.5 B γ Nr

Hence when ground water table is at b ≥ B, the equation is not affected.

Settlement of foundation :a) Settlement under loadsSettlement of foundation can be classified as-1. Elastic settlement (Si): Elastic or immediate settlement takes place during or immediately after the construction of the structure. It is also known as the distortion settlement as it is due to distortions within foundation soil.

2. Consolidation settlement (Sc): Consolidation settlement occurs due to gradual expulsion of water from the voids at the soil. It is determined using Terzaghi's theory of consolidation.

3. Secondary consolidation settlement (Ss): The settlement occurs after completion of the primary consolidation. The secondary consolidation is non-significant for inorganic soils.

DESIGN OF RETAINING WALL AND FOUNDATIONS

Design of simple column square footingDesign of simple column rectangular footing

Design of combined footingDesign of pile foundationDesign of retaining wall

JAYARAM D K

FOUNDATION

• The foundation of a structure is the part of the structure which transfers the load to the soil on which it rests.

• The ground surface in contact with the lower surface of the foundation is called the base of the foundation

• The ground on which the foundation rest is called the subgrade or foundation soil.

SUBSTRUCTURE AND SUPER STRUCTURE

• SUBSTRUCTURE:The structure below the ground level it is called

sub structureSUPER STRUCTURE:The structure above the ground level it is called

as super structure

Types of Foundations

• Shallow FoundationsIf the depth of the foundation is equal to or less

than its width the foundation is classified as shallow foundation

(i) Wall Footing(ii)Column or Isolated Footing(iii)Combined Footing(iv)Mat Footing

• Deep foundationIf the depth of the foundation is greater than its width

it is called as deep foundation.(i)Well foundation(ii)Pile foundation

Bearing Capacity of soil:Ability of the soil to resists the load with out failure.Causes of failure of foundations:(i) Unequal settlement of subsoil(ii)Shinkage of soil below the foundation due to

withdrawal of moisture

Safe Bearing capacity of the different soils

Types of Soil Safe Bearing Capacity of soil ( KN/m2)

1.Hard Dry Clay2.Sand and clay mixed3.Firm clay4.Fine confined wet sand5.Fine dry sand6.Coarse sand7.Soft rock8.Hard rock (mixture of gravel and clay)9.Medium hard rocks

35020020020035045065090011002750

Formula for finding the depth of the foundation:

2

D- depth of the foundation in mp- Safe bearing capacity of the soilr-Specific weight of the soil0-Angle of repose

Design 1: Find the area and the depth of foundation required for a column carrying an axial load of 1250 KN. The safe bearing capacity of the soil is 120 KN/m2 . The density of the soil is 18 KN/m3 and has an angle of repose of 30 degree.

Solution:Load on the column = 1250 KNApproximate weight of foundation = 125 KN( take 10 % of total weight)Total load = load on the column + approximate weight

of the column.= 1250 + 125 = 1375 KN

Area of the foundation = total load / safe bearing capacity of soil

= 1375/120= 11.46 m2

Provide a foundation area of 12 m2Determination of depth of the foundation:

Minimum depth of 2the foundation = (p/γ){(1-sinΦ)/(1+sin Φ)}= (120/18) {(1-sin30)/(1+sin30)}2= 0.75 m.

FRAMED STRUCTURE SHOWING FOOTING

WALL FOOTING

ISOLATED FOOTING OR COLUMN FOOTING

COMBINED FOOTING

STRAP FOOTING

CONTINUOUS FOOTING

RAFT FOOTING

WELL FOOTING

PILE FOOTING

Foundation…• Pile :

• A slender, structural member

consisting steel or concrete or timber.• It is installed in the

ground to transfer the structural loads to soils at some significant depth below the base of the structure.

Foundation…Pile caps are thick slabs used to tie a group of piles together to support and transmit column loads to the piles.

Pile Foundations The term ‘Pile Foundation’ denotes a construction for the

foundation of a wall or pier which is supported on piles.

Where Used : stratum of required bearing capacity is at greater depth steep slopes are encountered Compressible soil or water-logged soil or soil of made-up type

Examples: Piles are used for foundation for buildings, trestle-bridges and water front installations (piers, docks etc ).

Advantages: Provides a common solution to all difficult foundation site

problems Can be used for any type of structure and in any type of soil

SOFT STRATA

HARD STRATA

PILE FOUNDATION

Foundation…• Deep foundation :• Deep foundation consists of

pile and pier foundations.• This consists in carrying

down through the soil a huge masonry cylinder which may be supported by the sides of soil or may be supported on solid rock (hard stratum).

• Pile foundation :• Pile is an element of

construction used as foundation. It may be driven in the ground vertically or with some inclination to transfer the load safely.

Foundation…• Pile foundation…• Loads are supported in two

ways.• If the load is supported by

the effect of friction between the soil and the pile skin, it is called friction pile.

• Friction piles may be made of cast iron, cement concrete, timber, steel, wrought iron and composite materials.

• If the load is supported by resting the pile on a very hard stratum, it is called load bearing pile.

• Load bearing piles are steel sheet piles, concrete piles and timber piles.

• Piles may be cast-in-situ or precast.• They may be cased or

uncased.

Load Bearing

Pile

Friction Piles

Load bearing pile

Friction Pile

TYPES OF PILES• Concrete Piles

• i) Cast-In-Situ Concrete Piles• a) Cased cast-in-situ b) Uncased cast-in-situ• ii) Precast Concrete Piles

• Steel Piles• i) H-Piles ii) Cylindrical piles iii) Tapered piles

• Timber Piles

• Composite Piles

TYPES OF PILE CONSTRUCTION1. Displacement Piles

• It cause the soil to be displaced radially as well as vertically as pile shaft is driven or jacked into the ground.

• b) Non Displacement Piles (Replacement piles)• It cause the soil to be removed and the resulting hole filled with

concrete or a pre cast concrete pile is dropped into the hole and grouted in.

• Displacement Pile Non Displacement pile

METHOD OF INSTALLATION

• Dropping Weight or Drop Hammers• - commonly used method of insertion of displacement piles

• Diesel Hammers• -Most suitable to drive pile in non cohesive granular soil

• Vibratory Hammers or vibratory method of pile driving • -very effective in driving piles through non cohesive granular soil

• Jacking Method Of Insertion

Pile installation using Drop Hammer

• Pile Driving Rig - temporarily support the pile that being

• driven and to support the pile hammer.

Design of column footing:Design 2: A square column 500mm X 500mm carries an axial load of 1500 KN . Design the column and the square footing for the column. The safe bearing capacity of the column is 225 KN/m2. Use M20 and Fe 415 steel.

Design of Column:

Load on the column W = 1500 KN

Factored load Pu= 1.5 x 1500 = 2250 KNOver all area of the column section Ag= 500 x 500 = 250000 mm2Area of the steel = Asc

Area of the concrete = Ac = Ag- Asc= 250000-Asc

Ultimate load Pu=0.4 fck Ac + 0.67 fy Asc2250000 = 0.4 X 20 X ( 250000 – Asc) + 0.67 X 415 x AscAsc= 925.75 mm2Assume 29 mm dia bars Provide 4 bars of 20 mm diaLateral ties :•¼ diameter of the longitudinal bar•5mmFrom the above two take the greater one so provide the diameter of 6 mm dia bar.Pitch of lateral ties:(i)Least lateral dimension of the column = 500mm(ii)16 times the diameter of the longitudinal bars =16 X 20 = 320 mm

(iii)48 times the diameter of the ties = 48 X6Provide 6 mm dia ties at 250 mm c/c.

Design of the foundation:Load on the column = 1500 KNApproximate weight of the footing at 10 % of the column load

= 150 KNTotal load = 1650 KNSafe bearing capacity of the soil = 225 KN/m2Area of foundation = 1650/225 = 7.333 m2

BX B = 7.333

Breadth of foundation B = Γ(7.333) = 2.71 say 2.75 m

So the area of the foundation is 2.75 X 2.75 m

Net upward pressure = load on the column / area of the footing= 1500000/(2.75 X 2.75) = 198347.11 N/m2.

Depth of the foundation = Minimum depth of the foundation = (p/r){(1-sinΦ)/(1+sin Φ)}2

= (225/18) {(1-sin30)/(1+sin30)}2

= 1.4 m

Determination of the depth of the concrete slab below the footing.Critical section for bending moment is = (2750-500)/2 = 1125 mm = 1.125 m

Maximum bending moment = M = 198347.11 x 2.75 X 1.125 x( 1.125/2) = 345170 Nm

Factored moment Mu = 1.5 X M= 1.5 X 345170 = 517755 Nm.

To find the depth of the slab in the foundation.Mu= 0.138 fck bd2

517755000 = 0.138 X 20 X 500 (width of column) X d2

d= 613 mm

D= 613 + 12/2 +60 = 691 mm

(12- dia of bar , 60 – clear cover for footing)

The depth of slab of the foundation is increased by 30 % D= 691 X ( 0.3 X 691) = 900 mmd=900 – (12/2)- 60 =822 mm.

Determination of quantity of steel required:Ast = Pt X b x d

Determination of area of main reinforcement:

• Ast= Percentage of steel X b X d• Percentage of steel =

2

Factored moment Mu = 1.5 X M= 1.5 X 345170 = 517755 NmFck = 20 N/mm2And Fe =415 N/mm2

b=500 mmAndd=822mmSubstitute all the values in the above formula we getPt= 0.47 % Ast = Pt X b x d= (0.47/100) x 500 x 822 = 1932 mm2 2

No of bar =Total area/ area of one bar= 1932/ (3.14/4)x12Assume 12 mm dia bars so provide 18 bars of 12 mm diameter.•Here the column is square so provide the same reinforcement on both the directions.

REINFORCEMENT DETAILS OF COLUMN AND FOOTING:

2. A rectangular column footing 600 mm X 400 mm carries an axial load of 800 KN . Design a rectangular footing to support the column . The safe bearing capacity of the soil is 200 KN/m2 . Use M20 concrete and Fe415 steel.Load on the column = 800000N

Approximate weight of the foundation take 10 % of the weight of the column = 80000N

Total load = 880000N

Safe bearing capacity of the soil is given as 200 KN/m2 = 200000N/m2

DESIGN OF RECTANGULAR FOOTING

Area of the foundation = Total load / Safe bearing capacity of the soil.= 880000/200000 = 4.40 m2To find the length and breadth of the foundation “ in case of square footing its easy because by taking square root we get all the values “

Area = 4.4 BL = 4.4B= 4.4 /L

Equating the projections on both sides beyond the footing½ ( B-0.4) = ½ (L-0.6)Sub B Value½ ( {4.4/L}-0.4) = ½ (L-0.6)

Solving the above equation we get L = 2.2 m

Sub this is B value we getB= 4.4/L = 4.4/ 2.2 = 2 m

Now find the projections on both the axis= 0.8 m

Net upward pressure = column load / Area of the footing= 800000/4.4=181820 N/m2

Determination of reinforcement in section xx axis and YY axis.

Bending moment Myy = 181820 X 2.2X 0.8 X (0.8/2).= 128000 Nm

Factored Moment Muy = 128000 X 1.5 = 192000 Nm

Determine the depth Muy = 0.138 Fck bd2

192000 = 0.138 x 20 x 600 x d2

d= 341 mm.Bending moment Mxx = 181820 X 2 X 0.8 X (0.8/2).

= 116364.8 NmFactored Moment Mux = 116364.8 X 1.5

= 174547.2NmDetermine the depth Mux = 0.138 Fck bd2

174547.2 = 0.138 x 20 x 400 x d2

d= 398 mm.From the above two depth take the greater one

d= 398 mm

Providing 10 mm dia bars at a clear cover of 70 mmD = 398 + (10/2) + 70 = 473 mm

The overall depth may increased by 30%= 473 + (0.3 X473)= 614.9 mm

D=620 mm.

Effective Depth of the footing

d= 620 – (10/2) – 70= 545 mm.

Take Muy and find Longer direction steel

Factored Moment Muy = 128000 X 1.5 = 192000 Nm

Determination of quantity of steel required:Ast = Pt X b x dSub b= 600mmd=545mmFck=20N/mm2Fe=415 N/mm2In the Pt formula and find Pt Pt=0.32 %Ast =( 0.32/100) X600 x545=1046.4mm2 2

No of bar =Total area/ area of one bar= 1046.4/ (3.14/4)x12Assume 12 mm dia bar .Provide 12 mm dia bars of 10 numbers.

Determination of area of main reinforcement:

• Ast= Percentage of steel X b X d• Percentage of steel =

2

Take Mux and find Shorter direction steel

Factored Moment Factored Moment Mux = 116364.8 X 1.5 = 174547.2Nm

Determination of quantity of steel required:Ast = Pt X b x dSub b= 400mmd=545mmFck=20N/mm2Fe=415 N/mm2In the Pt formula and find Pt Pt=0.45 %Ast =( 0.45/100) X400 x545=981mm2 2

No of bar =Total area/ area of one bar= 981/ (3.14/4)x12Assume 12 mm dia bar .Provide 12 mm dia bars of 9 numbers.

Reinforcement details of rectangular footing:

Combined footingWhenever two or more columns in a straight line are carried on a single spread footing, it is called a combined footing. Isolated footings for each column are generally the economical.

Combined footings are provided only when it is absolutely necessary, as1.When two columns are close together, causing overlap of adjacent isolated footings2.Where soil bearing capacity is low, causing overlap of adjacent isolated footings3.Proximity of building line or existing building or sewer, adjacent to a building column

53

2. Slab and beam type

3. Strap type

Types of combined footing

1. Slab type

89

Design Steps

• Locate the point of application of the column loads on the footing.

• Proportion the footing such that the resultant of loads passes through the center of footing.

• Compute the area of footing such that the allowable soil pressure is not exceeded.

• Calculate the shear forces and bending moments at the salient points and hence draw SFD and BMD.

• Fix the depth of footing from the maximum bending moment.• Calculate the transverse bending moment and design the

transverse section for depth and reinforcement. Check for anchorage and shear.

90

Design of combined footing – Slab and Beam type

1. Two interior columns A and B carry 700 kN and 1000 kN loads respectively. Column A is 350 mm x 350 mm and column B is 400 mm X 400 mm in section. The centre to centre spacing between columns is 4.6 m. The soil on which the footing rests is capable of providing resistance of 130 kN/m2. Design a combined footing by providing a central beam joining the two columns. Use concrete grade M25 and mild steel reinforcement.

91

Solution: Datafck = 25 Nlmm2, fy= 250 N/mm2, fb = l30 kN/m2 (SBC), Column A = 350 mm x 350 mm, Column B = 400 mm x 400 mm, c/c spacing of columns = 4.6 m, PA = 700 kN and PB = 1000 kNRequired: To design combined footing with central

beam joining the two columns.Ultimate loadsPuA= 1.5 x 700 = 1050 kN, PuB = 1.5 x 1000 = 1500 kN

92

Proportioning of base size

Working load carried by column A = PA = 700 kNWorking load carried by column B = PB = 1000 kNSelf weight of footing 10 % x (PA + PB) = 170 kN Total working load = 1870 kN Required area of footing = Af = Total load /SBC

=1870/130 = 14.38 m2

Let the width of the footing = Bf = 2m Required length of footing = Lf = Af /Bf = 14.38/2 = 7.19m Provide footing of size 7.2m X 2m,Af = 7.2 x 2 = 14.4 m2

93

Then x = (PB x 4.6)/(PA + PB) = (1000 x 4.6)/(1000 +700) = 2.7 m from column A.If the cantilever projection of footing beyond column A is ‘a’ then, a + 2.7 = Lf /2 = 7.2/2, Therefore a = 0.9 m

Similarly if the cantilever projection of footing beyond B is 'b' then, b + (4.6-2.7) = Lf /2 = 3.6 m,

Therefore b = 3.6 - 1.9 = 1.7 m The details are shown in Figure

For uniform pressure distribution the C.G. of the footing should coincide with the C.G. of column loads. Let x be the distance of C.G. from the centre line of column A

94

C

700 kN 1000 kN

4600 mm b=1700 a=900

x R

Combined footing with loads

A B D

pu=177 kN/m2

wu=354 kN/m

95

Learning Outcomes:

• After this class students will be able to do the After this class students will be able to do the complete design and detailing of different types of complete design and detailing of different types of retaining walls.retaining walls.

DESIGN AND DETAILING OF RETAINING WALLS

96

Gravity retaining wall

GL1

GL2

Retaining walls are usually built to hold back soil mass.  However, retaining walls can also be constructed for aesthetic landscaping purposes.

RETAINING WALL

BACK SOIL

97

Batter

Drainage HoleToe

Cantilever Retaining wall with shear key

98

Photos of Retaining walls

99

Classification of Retaining walls

• Gravity wall-Masonry or Plain concrete

• Cantilever retaining wall-RCC(Inverted T and L)

• Counterfort retaining wall-RCC• Buttress wall-RCC

100

Counterfort

Gravity RWT-Shaped RW

L-Shaped RW

BackfillBackfill

Counterfort RW

ButtressBackfill

Buttress RW

Tile drain

Weep hole

Classification of Retaining walls

101

Earth Pressure (P)

• Earth pressure is the pressure exerted by the retaining material on the retaining wall. This pressure tends to deflect the wall outward.

• Types of earth pressure :

• Active earth pressure or earth pressure (Pa) and

• Passive earth pressure (Pp).

• Active earth pressure tends to deflect the wall away from the backfill.

Pa

GL

Variation of Earth pressure

102

Factors affecting earth pressure• Earth pressure depends on type of backfill,

the height of wall and the soil conditions

Soil conditions: The different soil conditions are

• Dry leveled back fill• Moist leveled backfill• Submerged leveled backfill• Leveled backfill with uniform surcharge• Backfill with sloping surface

103

Analysis for dry back fills

Maximum pressure at any height, p=kah Total pressure at any height from top,

pa=1/2[kah]h = [kah2]/2

Bending moment at any height M=paxh/3= [kah3]/6

Total pressure, Pa= [kaH2]/2 Total Bending moment at bottom, M = [kaH3]/6

Pa

Hh

kaH

M

GL

GL

H=stem height

104

• Where, ka= Coefficient of active earth pressure• = (1-sin)/(1+sin)=tan2• = 1/kp, coefficient of passive earth pressure• = Angle of internal friction or angle of repose• =Unit weigh or density of backfill

• If = 30, ka=1/3 and kp=3. Thus ka is 9 times kp

105

• pa= ka H at the bottom and is parallel to inclined surface of backfill

• ka=

• Where =Angle of surcharge Total pressure at bottom

=Pa= ka H2/2

22

22

coscoscoscoscoscoscos

Backfill with sloping surface

GL

106

Stability requirements of RW

• It should not overturn• It should not slide• It should not subside, i.e Max. pressure at

the toe should not exceed the safe bearing capacity of the soil under working condition

107

Check against overturning

Factor of safety against overturning = MR / MO 1.55 (=1.4/0.9)Where,

MR =Stabilising moment or restoring moment

MO =overturning moment

MR>1.2 MO, ch. DL + 1.4 MO, ch. IL0.9 MR 1.4 MO, ch IL

108

Check against Sliding

• FOS against sliding • = Resisting force to

sliding/• Horizontal force

causing• sliding• = W/Pa 1.55

(=1.4/0.9)

• 1.4 = ( 0.9W)/Pa

Friction W

SLIDING OF WALL

109

Maximum pressure at the toe

Pressure below the Retaining Wall

T

x1

x2

W1

W2

W3

W4

b/2b/6e

xb

H/3

Pa

W

Hh

Pmax

Pmin.

R

110

be

bW 61Pmin

be

bW 61Pmax

• Let the resultant R due to W and Pa • lie at a distance x from the toe.• X = M/W, • M = sum of all moments about toe.

• Eccentricity of the load = e = (b/2-x) b/6

• Minimum pressure at heel= >Zero.

• For zero pressure, e=b/6, resultant should cut the base within the middle third.

• Maximum pressure at toe=• SBC of soil.

111

Depth of foundation

• Rankine’s formula:

• Df =

•• =•

2

sin1sin1

SBC

2ak

γSBC Df

112

Preliminary Proportioning (T shaped wall)

• Stem: Top width 200 mm to 400 mm

• Base slab width b= 0.4H to 0.6H, 0.6H to 0.75H for surcharged wall

• Base slab thickness= H/10 to H/14• Toe projection= (1/3-1/4) Base

width

H

200

b= 0.4H to 0.6H

tp= (1/3-1/4)bH/10 –H/14

113

Design of Cantilever RW

• Stem, toe and heel acts as cantilever slabs

• Stem design: Mu=psf (ka H3/6)• Determine the depth d from Mu = Mu, lim=Qbd2

• Design as balanced section or URS and find steel

• Mu=0.87 fy Ast[d-fyAst/(fckb)]

114

Curtailment of bars

Ast Provided

Ast/2

Ast

Dist.fromtop

h2

Every alternate bar cut

Ast

Ast/2 h2

Ldt

h1c

h1

Cross section Curtailment curve

Effective depth (d) is Proportional to h

Bending moment is proportional to h3

Ast is αl to (BM/d) and is αl to h2

22

21

2

1..hh

AAei

st

st

115

Design of Heel and Toe

1. Heel slab and toe slab should also be designed as cantilever. For this stability analysis should be performed as explained and determine the maximum bending moments at the junction.

2. Determine the reinforcement. 3. Also check for shear at the junction. 4. Provide enough development length.5. Provide the distribution steel

116

Design a cantilever retaining wall (T type) to retain earth for a height of 4m. The backfill is horizontal. The density of soil is 18kN/m3. Safe bearing capacity of soil is 200 kN/m2. Take the co-efficient of friction between concrete and soil as 0.6. The angle of repose is 30°. Use M20 concrete and Fe415 steel.

Solution Data: h' = 4m, SBC= 200 kN/m2, = 18 kN/m3, μ=0.6, φ=30°

Cantilever RW design

117

Depth of foundation

• To fix the height of retaining wall [H]• H= h' +Df

• Depth of foundation

• • Df =

• = 1.23m say 1.2m , • Therefore H= 5.2m

2

sin1sin1

SBC

H

200

b

Df

h1 h

118

Proportioning of wall• Thickness of base slab=(1/10 to1/14)H• 0.52m to 0.43m, say 450 mm

• Width of base slab=b = (0.5 to 0.6) H• 2.6m to 3.12m say 3m

• Toe projection= pj= (1/3 to ¼)H• 1m to 0.75m say 0.75m

• Provide 450 mm thickness for the stem at the base and 200 mm at the top

H=5200 mm

200

b= 3000 mm

tp= 750 mm

450

119

• Ph= ½ x 1/3 x 18 x 4.752=67.68 kN• M = Ph h/3 = 0.333 x 18 x 4.753/6 • = 107.1 kN-m• Mu= 1.5 x M = 160.6 kN-m

• Taking 1m length of wall,• Mu/bd2= 1.004 < 2.76, URS • (Here d=450- eff. Cover=450-50=400 mm)• To find steel• Pt=0.295% <0.96%• Ast= 0.295x1000x400/100 = 1180 mm2

• #12 @ 90 < 300 mm and 3d ok• Ast provided= 1266 mm2 [0.32%]

Design of stem

Or Mu = [kaH3]/6

Pa

h

kah

M

Df

120

• Curtail 50% steel from top• (h1/h2)2 = 50%/100%=½

• (h1/4.75)2 = ½, h1 = 3.36m

• Actual point of cutoff• = 3.36-Ld=3.36-47 φbar = 3.36-

0.564 = 2.74m from top.• Spacing of bars = 180 mm c/c

< 300 mm and 3d ok

Curtailment of bars-Stem

Ast Provided

Ast/2

Ast

Dist.fromtop

h2

Every alternate bar cut

Ast

Ast/2 h2

Ldt

h1c

h1

121

• Development length (Stem steel)• Ld=47 φbar =47 x 12 = 564 mm

• Secondary steel for stem at front• 0.12% GA • = 0.12x450 x 1000/100 = 540 mm2

• #10 @ 140 < 450 mm and 5d ok

• Distribution steel • = 0.12% GA = 0.12x450 x 1000/100 =

540 mm2

• #10 @ 140 < 450 mm and 5d ok

H=5200 mm

200

b= 3000 mm

tp= 750 mm

450

Design of stem-Contd.,

L/S ELEVATION OF WALL

#16 @ 190

#12 @ 180

#12 @ 90

#10 @ 140

#10 @ 140 C/S OF WALL

Drawing and detailing

PILE FOUNDATION AND THEIR DESIGN PRINCIPLES

M.Jayaram

Pile FoundationsPile Foundations Based on its function pile foundations are classified as

1. End Bearing Pile 4. Tension piles (or) Uplift piles2. Friction (Shaft friction) Pile 5. Anchor pile 7.Sheet piles3. Compaction piles 6. Fender and Dolphins pile 8. Batter piles

Some of the important classification of piles Based on Materials: Timber, Concrete, Steel, or combination of any of them. Types of Soil condition: Gravel, Clay, Sand, Rock Based on Loads: Vertical load, Horizontal load ( water, soil, earthquake) Based on construction: Cast in-situ, Pre-cast, Bored piles, Driven piles

BASED ON CONSTRUCTION

Driven PilesCast in Place Cast in Place (permanent casing/shell) Pre-cast Steel Timber

Bored PilesLarge Diameter (>600mm) Small Diameter (300-600mm) Tubed Augured Piles Continuous Flight Auger (CFA) PilesBattered pilesMini-Piles Augured Displacement

Pile MaterialsPile Materials Steel: H- piles, Steel pipeSteel: H- piles, Steel pipe Concrete: Cast in-situ (cast at site) or PrecastConcrete: Cast in-situ (cast at site) or Precast Wood : (Timber Piles)Wood : (Timber Piles) CompositeComposite

END BEARING PILESEND BEARING PILES

End bearing piles are those which terminate in relatively hard, impenetrable stratum such as rock or very dense sand and gravel. This pile act as a laterally transmitted column. The load being transmitted to the toe and resisted by the hard soil or rock.

ROCK

SOFT SOILPILES

End Bearing PileEnd Bearing Pile

Fig: End bearing file

ROCKROCK

SOFT SOILSOFT SOILPILESPILES

FRICTION PILESFRICTION PILES

Friction piles: In this pile the load is transmitted to the soil through the adhesion or skin frictional resistance along the shaft of the piles. In cohesionless soils like sands of medium to low relative density friction piles are often used to increase the density and thus the shear strength.

SOFT SOILPILES

Site Cast Concrete PilesSite Cast Concrete Piles

Cased Piles Uncased Piles

Advantages and Disadvantages of Wood Advantages and Disadvantages of Wood pilespiles

Advantages:Advantages: +The piles are easy to handle+ Relatively inexpensive where timber availability is more.+ Sections can be joined together and excess length can be

easily removed.Disadvantages:Disadvantages:-- The piles will rot above the ground water level. Have a

limited bearing capacity.-- Can easily be damaged during driving by stones and boulders.-- The piles are difficult to splice and are attacked by marine

borers in salt water.

Advantages and Disadvantages of Bored and cast in-situ (Non-displacement Piles)

Piles

• + Length can be readily varied to suit varying ground conditions.• + Soil removed in boring can be inspected and if necessary sampled or in- situ test made.• + Can be installed in very large diameters.• + End enlargement up to two or three diameters are possible in clays.• + Material of piles is not dependent on handling or driving conditions.• + Can be installed in very long lengths.• + Can be installed with out appreciable noise or vibrations.• + Can be installed in conditions of very low headroom.• + No risk of ground heave.• -- Susceptible to "waisting" or "necking" in squeezing ground.• -- Concrete is not placed under ideal conditions and cannot be subsequently inspected.• -- Water under artesian pressure may pipe up pile shaft washing out cement.• -- Enlarged ends cannot be formed in cohesionless materials without special techniques.• -- Cannot be readily extended above ground level especially in river and marine structures.• -- Boring methods may loosen sandy or gravely soils requiring base grouting to achieve

economical base resistance.

Precast Concrete PliesPrecast Concrete Plies

Types of PileTypes of Pile

• The pile installation procedure varies considerably, and has an important influence on the subsequent response

• Three categories of piles are classified by method of installation as below:– Large displacement piles

• They encompass all solid driven piles including precast concrete piles, steel or concrete tubes closed at the lower end

– Small displacement piles• They include rolled steel sections such as H-pile and open-end

tubular piles– Replacement piles

• They are formed by machine boring, grabbing or hand-digging.

Advantages and Disadvantages of Displacement Pile Advantages and Disadvantages of Displacement Pile

Advantages Disadvantages Pile material can be inspected

for quality before driving May break during driving

Construction operation affect by ground water

Noise and vibration problems

Can driven in very long lengths Cannot be driven in condition of low headroom

Construction operation not affected by ground water

Noise may prove unacceptable. Noise permit may be required

Soil disposal is not necessary Vibration may prove unacceptable due to presence of sensitive structures, utility installation or machinery

AdvantagesAdvantages and and DisadvantagesDisadvantages of Replacement Pile (A/D) of Replacement Pile (A/D)

Advantages Disadvantages Less noise or vibration problem Concrete cannot be inspected after

installationEquipment can break up practically all kinds of obstructions

Liable to squeezing or necking

Can be installed in conditions of low headroom

Raking bored pile are difficult to construct

No ground heave Drilling a number of pile groups may cause ground loss and settlement of adjacent structures

Depth and diameter can varied easily

Cannot be extended above ground level without special adaptation

Combinations of vertical, horizontal and moment loading may be applied at the soil surface from the overlying structure

For the majority of foundations the loads applied to the piles are primarily vertical

For piles in jetties, foundations for bridge piers, tall chimneys, and offshore piled foundations the lateral resistance is an important consideration

The analysis of piles subjected to lateral and moment loading is more complex than simple vertical loading because of the soil-structure interaction.

Pile installation will always cause change of adjacent soil properties, sometimes good, sometimes bad.

Loads applied to PilesLoads applied to PilesVV

MH

LOAD CARRYING CAPACITY OF PILESLOAD CARRYING CAPACITY OF PILES

Load Carrying Capacity Of Piles: The ultimate load carrying capacity or Ultimate bearing capacity or Ultimate

bearing resistance (Qup ) of a pile is defined as the maximum load which can be carried by a pile, and at which the pile continues to sink without further increase of load.

The allowable load (Qa) is the safe load which pile can carry safely and is determined on the basis of (1) Ultimate bearing capacity divided by suitable factor of safety.(2) The permissible settlement and (3) the Overall Stability of the pile foundation

The load carrying capacity of the pile can be determined by the following methods: Dynamic Formulae (1. Engineering News Formula, and 2. Hileys Formulas) Static formulae Pile Load Tests Penetration Test

Engineering News formulaEngineering News formula• This formula was proposed by A.M.Wellington (1818) to determine the allowable

bearing capacity as:• Qa = WH/F(S+C) ………………….(1)

• Where, Qa = Allowable Load, • W = Weight of Hammer in kg, H = Height of fall of a Hammer (cm)• S = Final Set (Penetration) per blow, usually taken as average penetration in cm per

blow for the last 5 blows of a Drop Hammer and 20 Blows for a Steam Hammer• C= Empirical Constant (2.5 for Drop Hammer, 0.25 for Single (or) Double acting

steam hammer • (1) for Drop Hammers, Qa = WH/6(S+2.5) ………………….(2)

• (2) for Single acting Steam Hammers, Qa = WH/6(S+0.25) ……….(3)

• (3) for Double acting Steam Hammers Qa = {(W+ap)H}/6(S+0.25) …….(4)• Where a = effective area of piston and p = mean effective steam pressure (kg/cm2)

Testing ProcedureThe pile load test can be performed either on a working pile or on a test pile. A rigid steel plate (circular or rectangular steel plate) is placed on the top of pile projection.A calibrated jack plate is mounted on the plate to measure the applied load. A reaction jack is

borne by the Truss or a Platform. The truss can be anchored to the ground with the help of anchor piles.

The load is applied in equal increments of about one-fifth of its estimated allowable load. The settlements were recorded with the help of three dial gauges of sensitive to 0.02mm, arranged symmetrically over the test plate. For each load increments, the rate settlement becomes less than 0.02mm per hour. The test piles are loaded until ultimate load is reached.

In general, the test load is increased to a value of 2.5times the values of estimated allowable load or to a load which cause a settlement of equal al to 1/1o of the pile diameter, whichever occurs earlier. The ultimate load can be obtained by plotting the graph for Load Vs Settlement curve. If the ultimate load can not obtained from the plot, then the allowable load can be obtained as follows:

1. One-half to one-third of the final load which causes settlement equal to 10% of pile diameter.2. Two–third (2/3) of final load which cause a settlement of 12 mm, or 3. Two-third of final load which causes a net settlement (residual settlement after removal of

load) of 6mm

Pile Load TestPile Load Test

145

Test of 6’ diameter Type-I Shaft at UCLA

Modes of failure

• The soil is always failure by punching shear.• The failure mode of pile is always in buckling

failure mode.

147

Test of 6’ diameter Type-I Shaft at UCLA

SETTLEMENT REDUCING PILESSETTLEMENT REDUCING PILES

Settlement reducing piles are usually incorporated beneath the central part of a raft foundation in order to reduce differential settlement to an acceptable level. Such piles act to reinforce the soil beneath the raft and help to prevent dishing of the raft in the centre.

FAILURE OF PILE FOUNDATION & REMEDIESFAILURE OF PILE FOUNDATION & REMEDIES

Pile foundation It is widely used deep foundation

for complex geologic conditions with kinds of load conditions, especially for soft soil foundation.

Pile foundation has large bearing capacity, well stability and small differential settlement compared to other foundation types.

But pile foundations may also get damaged and fail specially during earthquakes.

The failure of the pile foundation may result from any of The failure of the pile foundation may result from any of the following causes:the following causes:

Remedies to prevent failure of pile Remedies to prevent failure of pile foundation:foundation:Early repair such as Encasement or Replacement of piles Removal of partial loadUnderpinning

Problems: Problems: Exercise:1A wooden pile is being driven with a drop hammer weighing 20kN and having a free fall of 1.0m. The penetration in the last blow is 5 mm. Determine the load carrying capacity of the pile according to the Engineer’s news formula. Solution:1Given Data: 1.A Wooden Pile is Diriven 2.Weight of Drop Hammer = W=20kN3.Height of fall H = 1.0m4.Penetration in the last blow S = 05mm

Solution:1To find: 1.Load carrying capacity of the pile2. using Engineering News formula

Solution:1Load carrying capacity Qa = WH / F(S+C) = 20*100) / 6(0.5+2.5) = 111.10kN

Exercise:2A reinforced concrete pile weighing 30kN (inclusive of helmet and dolly) is driven by a drop hammer weighing 40kN and having an effective fall of 0.8m. The average set per blow is 1.4cm. The total temporary elastic compression is 1.8cm. Assuming the coefficient of restitution as 0.25 and factor of safety of 2. Determine the ultimate bearing capacity and allowable load for the pile.

Exercise:2Given Data: 1.A Reinforced concrete Pile weighing P=30kN is Driven 2.Weight of Drop Hammer = W=40kN3.Height of fall H = 0.80m4.Penetration in the last blow S = 1.40cm5.Total elastic compression =C = 1.80 cm6.Co-efficient of restituion =e = o.257.Factor of safety = f = 28.W > e*P

Exercise:2To find: 1.Load carrying capacity of the pile2. Allowable Load

Solution:2ƞb = (W+P*e2) / (W+P) = (40 +30*0.252) / (40+30) = 0.597Ultimate Bearing Capacity Qf = {(ƞh *H)Wƞb }/ (S+C/2) = {80*40)*0.597} / (1.4+1.8/2) = 830 kN

Solution:2Allowable Load (Qa) = Qf / F = 830 / 2 = 415 kN

Exercise:3Design a friction pile group to carry a load of 3000kN including the weight of thepile cap at a site where the soil is uniform clay to depth of 20m, underlain by a rock.Average unconfined compressive strength of the clay is 70kN/m2 .The clay may beassumed of normal sensitivity and normally loaded, with liquid limit of 60%. A factorof safety of 3 is required against shear failure.

Exercise: 3 (1) : Given Data: 1.A Reinforced concrete Pile weighing P=3000kN is Driven 2.Depth of clay layer = 20 m3.Average unconfined compressive strength qu = 70kN/cm2 ; (c = qu/2; c=70/2 =35)4.Length of the pile L = 10 m5.Diameter of the pile D = 0.50 m6.Spacing of pile S =3*d =3 * 0.50 = 1.50m7.Number piles = n8.Factor of safety f = 29.Nc = 910.Ap = Area of Piles (B*B)

Exercise:3 (2) : To find: 1.Design the group piles

Solution:4 (3) General formula Qup = As rf

Pile acting as a Single pile Qug = n c π DL 3000 = n (35/5) π *0.5 *10 n = {3*3000}/{35* π *0.5 *10} = 16.37 (Adopt 16 Nos.)Number of piles n =16 PilesModify the Length of pile L have to be increased by the ratio of (16.37/16) * 10 = 10.23m (Adopt 11 m)

Pile acting as a group Qug= Asg *rf

where, B = 3 * 1.5 + 0.50 = 5.0mLoad taken by group action, As g= {4 * B* L*c}+Ap *cNc = {4*5 *11*35/3}+{25*35/3*9 = 5191.7 which is more the 3000kN (Hence safe)

4. In a 16 pile group, the pile diameter is 45 cm and centre to centre spacing of the square group is 1.5 m. If c = 50 kN/m2, determine whether the failure would occur with the pile acting individually, or as a group? Neglect bearing at the tip of the pile. All the piles are 10 m long. Take m = 0.7 for shear mobilization around each pile.

•

Exercise:4 (1)Given Data: 1.16 nos. of Group Pile 2.Diameter D = 600 mm 3.Spacing = 1.20m4.Length of Pile L = 10m on Soft clay5.Cohesion c + 30 kN/m2

6.weighing P=30kN is Driven 7.Adhession factor m = 0.608.Neglecting the bearing resistance

Exercise:4 (2) :To find: 1.Ultimate Load carrying capacity of the Group pile

Solution:4 (3) General formula Qup = As rf

Pile acting as a Single pile Qun = n Qup

= n As rf

Area of Shaft (As)= π D *L; = π *0.60*10 = 18.85m2

Unit Skin friction, (rf)= m *c = 0.60 * 30 = 5428.70 kN

Pile acting as a group Qug= Asg *rf

As g= {4 * B* L} where, B = 3 * 1.2 + 0.60 = 4.20m = {4*4.2 *10} = 168 m2 and (rf)= c = 30 kN/m2

Solution:4 (4)Ultimate Load carrying capacity of the pile (Qug) = Asg*rf = 168 * 30 = 5040 kN Ultimate Bearing Capacity of Single pile = Ultimate Bearing Capacity group pile 5428.70 = 5040 (consider lesser of the Two values)

END Pile foundations• END Pile foundations

• END Pile foundations

W

Q u

Q b

Q s

Basic ConceptBasic ConceptThe ultimate axial load of (Qu ) of a single pile may be considered to be the sum of its skin friction and end-bearing resistances , that are mobilized by the applied load. i.e

Qu =Qb+Qs-W ………………….(1) =Ab*q’b +ΣAs*q’s -W where Ab, and As = Area of base and Shaft respectivelyq’b =Ultimate net bearing capacity of soil at the end of the pile q’s = Mobilized adhesion OR frictional resistance along the shaft of the pile and W=weight of the pile – (minus) weight of soil replacedW = 0.25 πd2 L (ϒp –ϒ) ; D is diameter and L is length of the pileϒp =Average density of pileQu = total pile resistance, Qb is the end bearing resistance (Qu = Cu *Nc* Ab) and Qs Qs is Shaft resistance capacity (Qs = Σ(αCu *Nc* ΔAs) Nc = 9 for intact clay and 6.75 for fissured claysAlpha = adhesion factor =0.45 (soft clay =1, Overconsolidated clay = 0.30General behaviourShaft resistance fully mobilized at small pile movement (<0.01D)Base resistance mobilized at large movement (0.1D)

Loading

Settlement

Behaviour of Frictional Pile

Loading

SettlementBehaviour of End Bearing

Pile

Qu

QS

QB

Qu

QB

QS

Piles founded on dense soils Important to adopt good

construction practice to enhance shaft friction and base resistance

Shaft and base grouting useful in enhancing pile capacity

Piles founded on strong stratumNot much benefit in enhancing base resistanceImportant to adopt good construction practice to enhance shaft frictionShaft grouting useful in enhancing pile capacity

W

Qs

QB

QT

ho

D

QQDESDES = Q = QBB/F/FBB + Q + Qs s /F/Fss ––WW…………(2)(2)d

Ultimate Limit State Design Ultimate Limit State Design

Where FB and FS is the factor of safety of components of end bearing strength and shaft friction strength

Qb=Ab[cbNc+Po(Nq-1)+d/2N+Po] -WpWhere Ab = area of the base , cb = the cohesion at the base of the pile, Po = the overburden stress at the base of the pile, and d = the width (diameter) of the pile.

QQUU = Q = QBB + Q + Qss––WW…………(3)(3)

End Bearing ResistanceAssumptions :Assumptions :

1. The weight of the pile is similar to the weight of the soil displaced of the pile

=> Wp=AbPo

2. The length (L) of the pile is much greater than its width d

=> Wp=AbPo+ AbdN/2

3. Similarly for Nq approximately equal to Nq-1

Qb=Ab[cbNc+Po(Nq-1)+d/2N+Po] –Wp

=> Qb=Ab[cbNc+PoNq]

End Bearing resistance for Bore pile in granular soils

Due to the natural of granular soil, the c’ can be assumed equation to zero. The ultimate end bearing resistance for bored pile in granular soils may be express in terms of vertical effective stress, ’v and the bearing capacity factors Nq as :

QB=AB Nq v’

Nq is generally related to the angle of shearing resistance ’. For general design purposed, it is suggested that the Nq value proposed by Berezantze et al (1961) as presented in Figure ?? are used. However, the calculated ultimate base stress should conservatively be limited to 10Mpa, unless higher values have been justified by load tests.

Shaft Friction ResistanceShaft Friction Resistance

The ultimate shaft friction stress qs for piles may be expressed in terms of mean vertical effective stress as :

qs =c’+Ksv’tans

qs =v’ (when c’=0) Where Ks= coefficient of horizontal pressure which depends on the relative density and state

of soil, method of pile installation, and material length and shape of pile. Ks may be related to the coefficient of earth pressure at rest,

K0=1-sin as shown in Table 1. Qv’ = mean vertical effective stress s’ = angle of friction along pile/soil interface (see table2) = shafte friction coefficient (see Table 3)

Qs = pLqs

Where p is the perimeter of the pile and L is the total length of the pile

Total and Effective Stress Analysis

• To determine drained or undrained condition, we may need to consider the following factors:– Drainage condition in the various soil strata– Permeability of soils– Rate of application of loads– Duration after the application of load

• A rough indicator will be the Time Factor (Tv=cvt/d2)

GROUND IMPROVEMENT TECHNIQUES

GROUTING

GROUTING• Grouting is a process of ground improvement attained by

injecting fluid like • material into subsurface soil or rock.• Grouting is the injection specially formulated cement of stable

suspensions • or liquid into pores, fissures or voids, or the jetting of cement

mixtures at • high flow rate and pressure into the soil to create soil-

cement to increase • the strength.

• Producing mass concrete structures and piles• Fixing ground anchors for sheet pile walls, concrete pile walls,

retaining walls tunnels etc• Repairing a ground underneath a formation or cracks and

structural • Defects on building masonry or pavement.• Fixing the tendons in prestressed post tensioned concrete• Filling the void between the lining and rock face in tunnel works

APPLICATIONS

(a) Suspension grouts: These are multi-phase systems capable of forming sub systems after being subjected to natural sieving processes, with chemical properties which must be carefully scrutinized so as to ensure that they do not militate against controlled properties of setting and strength. Water in association with cement, lime, soil, etc., constitute suspensions. Emulsion (asphalt or bitumen) with water is a two-phase system which is also included under suspension.

(b) Solution grouts: These are intimate one-phase system retaining an originally designed chemical balance until completion of the relevant reactions. Solutions in which the solute is present in the colloidal state are known as colloidal solutions. Chemical grouts fall into this category.

GROUTING MATERIAL TYPES

MATERIALS USED FOR GROUTING

• Cement and water

• Cement, rock flour and water

• Cement, clay and water

• Cement clay, sand and water

• Asphalt• Clay and water• Chemicals

MATERIALS USED FOR GROUTINGCommon admixtures used with cement grouts:

1. Calcium chloride ]2. Sodium hydroxide ]-----for accelerating setting

time 3. Sodium silicate ]4. Gypsum ]5. Lime sugar ]-----for retarding setting

time.6. Sodium tannate ]7. Fine bentonite ]8. Clay ]9. Ground shale ]--for reducing cost of grout strength of and

reduces grout

10. Rock flour ]

MATERIALS USED FOR GROUTING

GROUTINGMATERIALS• Cement and water• Cement, rock flour and

water• Cement, clay and water• Cement clay, sand and

water• Asphalt• Clay and water• Chemicals

COMMON ADMIXTURES USED WITH CEMENT 1. Calcium chloride ]2. Sodium hydroxide ]-----for accelerating

setting time 3. Sodium silicate ]4. Gypsum ]5. Lime sugar ]---for retarding setting time.

6. Sodium tannate ]7. Fine bentonite ]8. Clay ]9. Ground shale ]--for reducing cost of grout and reduces the

strength of grout

10. Rock flour ]

Grout is injected into the soil at low pressure and fills the voids without significantly changing the soil’s structure and volume. Variety of binders are used with this technique, the choice of which is dictated mainly by the permeability of the soil.

When the coefficient of permeability is greater than 10-2cm/sec, water-cement mixes are used and for permeability as low as 10-5 cm/sec, the more expensive resin based grouts are used. Soils with K values lower than 10-6 cm/sec are normally not groutable by permeation.

PERMEATION

Disadvantages•Grouting adjacent to unsupported slopes may be ineffective.•Not suitable in decomposable materials.•Danger of filling underground pipes with grout.•Effectiveness questionable in saturated clays

COMPACTION PERMEATION

Thick slurries can not penetrate fine cracks and higher injection pressures would cause fracturing of ground foundations. Because of the higher water requirements of micro fine cement, the slurry remains fluid enough to flow into and penetrate fine sands and small cracks in rock.

These cements can treat finer grained sands not possible to treat with Portland cement alone. They are also used to stabilize waste plumes.

MICROFINE CEMENT

A key advantage of chemical grouting is the ability to introduce grout into soil pores without any essential change in the original soil volume and structure, thus changing the support capability of granular soils without disturbing them. Another advantage is the ability to be less disruptive and enable tunneling to proceed without over-excavation. A possible drawback of chemical grouting is that only certain soil types are amenable. Another barrier to the use of chemical grouting techniques in the recent is increasing concern regarding potential pollution by chemical grouting in urban areas. Two trends have addressed this issue: 1. Improvement of grouts through the development of new formulae that enhance the penetrability of particulate suspensions and meet the strictest specifications for environmental safety2. Development of alternative techniques which by-pass the penetrability restraints, such as jet grouting which allows the treatment of most types of soil, independent of its grain size and permeability, using simple cement grouts

CHEMICAL GROUTING

COMPACTION PERMEATION

1.Compensation (hydrofracture) grouting uses high-mobility grout to split the ground and thereby create lifting or densification under structures or other facilities.

2.The ground is deliberately split by injecting stable fluid cement-based grouts at high pressures in order to increase total stress by the wedging action of successive thin grout lenses, to fill unconnected voids, and possibly to consolidate the soil locally under injection.

3.This process is often undertaken as a reaction to movements while tunnel excavation is in progress.

4.It is important to keep in mind that the effects of compensation (hydrofracture) grouting are difficult to control and the potential danger of damaging adjacent structures by the use of high pressure may prove prohibitive

COMPENSATION GROUTING

1. It is a technology in which high- pressure jets of cement grout are discharged sideways into the borehole wall to simultaneously excavate and then mix with the soil.

2. The outstanding feature of jet grouting is the ability to treat a whole range of soils, from silty sands to cohesive deposits, by means of simple cement grouts.

3. Jet grouting can be performed in soils with a wide range of granulometries and permeabilites.

JET GROUTING

Disadvantages•Grouting adjacent to unsupported slopes may be ineffective.•Not suitable in decomposable materials.•Danger of filling underground pipes with grout.•Effectiveness questionable in saturated clays

COMPACTION PERMEATION

These are special cements used to treat fine grained soil. It is not possible to treat with Portland cement alone.

Thick slurries can not penetrate fine cracks and higher injection pressures would cause fracturing of ground foundations. Because of the higher water requirements of micro fine cement, the slurry remains fluid enough to flow into and penetrate fine sands and small cracks in rock.

MICROFINE CEMENT

A key advantage of chemical grouting is the ability to introduce grout into soil pores without any essential change in the original soil volume and structure, thus changing the support capability of granular soils without disturbing them. Another advantage is the ability to be less disruptive and enable tunneling to proceed without over-excavation. A possible drawback of chemical grouting is that only certain soil types are amenable. Another barrier to the use of chemical grouting techniques in the recent is increasing concern regarding potential pollution by chemical grouting in urban areas. Two trends have addressed this issue: 1. Improvement of grouts through the development of new formulae that enhance the penetrability of particulate suspensions and meet the strictest specifications for environmental safety2. Development of alternative techniques which by-pass the penetrability restraints, such as jet grouting which allows the treatment of most types of soil, independent of its grain size and permeability, using simple cement grouts

CHEMICAL GROUTING

1.Compensation (hydrofracture) grouting uses high-mobility grout to split the ground and thereby create lifting or densification under structures or other facilities.

2.The ground is deliberately split by injecting stable fluid cement-based grouts at high pressures in order to increase total stress by the wedging action of successive thin grout lenses, to fill unconnected voids, and possibly to consolidate the soil locally under injection.

3.This process is often undertaken as a reaction to movements while tunnel excavation is in progress.

4.It is important to keep in mind that the effects of compensation (hydrofracture) grouting are difficult to control and the potential danger of damaging adjacent structures by the use of high pressure may prove prohibitive

COMPENSATION GROUTING

1. In this method, high- pressure jets of cement grout are discharged sideways into the borehole wall during excavation to strengthen the earth wall.

2. The jet grouting is to treat a all range of soils, (silty sands to cohesive deposits).

3. Jet grouting can be performed in wide range to permeability problem in soils.

Advantages:1. the ability to use very small drilling tools (90mm diameter) to create

large elements (1.2m to 2.4m diameter) using pressure and flow; 2. the ability to drill underneath obstacles and solidify zones which are

hard to access; 3. the use of technically sophisticated techniques such as high-powered

pumps and monitoring devices with continuous measurement of all operational parameters.

JET GROUTINGJET GROUTING

JET GROUTING

A grouting plant includes a mixer, an agitator, a pump, and piping connected to grout holes. Two systems: single line type and circulating type. In the circulation type, the unused grout is returned to the agitator and in the single-line type the grout refused is wasted. The basic items required for a grouting plant and their functions are:(a) Measuring tank-to control the volume of grout injected.(b) Mixer-to mix the grout ingredients (c) Agitator-to keep the solid particles in suspension until pumped (d) Pump-to draw the grout from the agitator to deliver to the pumping line.(e) Control fittings-to control the injection rate and pressure so that the hole can be regularly blend with water and thin grout.

GROUTING PLANT AND EQUIPMENT

SCHEMATIC REPRESENTATION

PRECAUTIONS

The following are the precautions while mixing a grout:Water is placed first in the mixer. Mixer is run at the maximum speed before adding the cement.Grout is mixed in batches.Ingredients have to be measured in volumeEnough water should be maintained to cover the rotor while it

is functioning.Mixer should not be allowed to run for more than a few minutes

between batches.Mixers should be cleaned thoroughly after completion of work.

Stone & Perforated Pipe Drainage Mat & Perforated Pipe

Drainage MethodsDrainage Methods

DampproofingTypically, a liquid asphalt

applied with a roller or sprayer

Not an effective barrier for water under pressure.BUT, will prevent ground

‘moisture’ from migrating through a wall.

Typically used in conjunction will drainage pipe.

Earth Slope Stability Analysis

CECE 4131 – Geotechnical Engineering – II

JAYARAM D K

STABILITY OF SLOPESINTRODUCTION:

Earth embankments are commonly required for Railways, Roadways, Earth Dams, Levees and River training works. The stability of those embankments or slopes, should be thoroughly analyzed, since their failure my lead to loss of human life, as well as economic loss.

I. The failure of a mass soil located beneath a slope is called slide. It involves the movement of soil mass either downward or outward from the position.Types of slopes: 1) Infinite slope 2) Finite slope

1) Infinite slope: If a slopes represents the boundary surface of a semi-infinite soil mass, and the soil properties for all depth below the surface are constant, it is called infinite slope.

2) Finite slope: If the slope is of limited extent of it s boundary, it is called finite slope

Failure of SLOPE • Modes of slope failure: An exposed ground surface that stands at an angle with the

horizontal is called unrestrained slope. – The slope can be natural or man-made.– It can fails in various modes– The failures are classified in to five major categories:– 1. Fall 2. Topple 3. Slide 4. Spread 5. FlowFall: this is the detachment of Soil and or Rock fragments that fall down a slope, and

large amount of soil mass has slide down a slop.Topple: this is a forward rotation of Soil/Rock mass about an axis below the centre of

gravity of mass being displaced.Slide: It is the downward movement of soil mass occurring on a surface of ruptureSpread: this is a form of slide by rotation. It occurs by sudden movement of water

bearing seams of sand silts overlain by clays or loaded by fills.Flow: this is a downward movement of soil mass similar to a viscous fluid

II. Causes of Mass Movements

Seismic forces, Weaken the soil layer from the continuous exposure of groundwater or Chemical waste leachate, High Groundwater water level, Excavation for the construction,

III. Types of Slope Movements

FACTOR OF SAFETY• The task of the engineer charged with analysing slope is to determine the factor of safety. It is

defined as • Factor of safety with respect to strength (Fs =τf / τd ); • (τf = Average shear strength of the soil / • τd = Ave. Shear Stress developed along the potential failure surface)

• Shear strength of soil consists of Two components: Cohesion and Friction, it can be writen as τf = c’ + σ’ tan ϕ’ (σ’ = normal stress on the potential failure surface)

Methods of finite slope analysis1. Culmann’s method of planar failure surface - Suitable for very steep slope2. The Swedish (Slip circle) Circle method3. The Friction circle methods4. Bishop’s method

• Culmann’s method : Planar failure surface:• Culmann (1866) considered, a simple failure mechanism of a slope of homogeneous soil with plane

failure surface passing through the toe of the slope.• Let AB be any probable slip plane.• The wedge ADB is in equilibrium under the action of three forces

– (1) Weight of the wedge W = ½(AB)*h* ϓ; = ½(L)*h* ϓ– (2) The cohesive force C along the surface AB, resisting motion Cm*L– (3) The reaction R, inclined at an angle (ϕm) to the normal

IV. Infinite Slope Stability Analysis

Free Body Diagram of a Representative Slice of Infinite Slope with Cohesionless Soil

Stability of infinite slopes

Considering the problem of slope stability, of an infinite slope shown in fig. 15.7 To get the strength of the soil mass τf= c’ + σ’ tanϕ Assume Porewater pressure is zero To Evaluate the factor of safety against a possible slope failure along a plane AB

located at a depth H, below the ground surface The slope failure can occur by the movement of soil above the plane AB from right

to left. Fs = c’/(γH cos2βtanβ) + (tanϕ’/ tanβ) For COHESSIONLESS (granular) soils, c’=0, and Fs becomes equal to (tanϕ’/

tanβ), this indicates that in an infinte slope in sand. Factor of safety,Fs is independent, and the slope is stable as long as β < ϕ’ For soil possesses Cohesion and friction, the depth of the plane along which

critical equilibirium occurs may be determined by substituting Fs=1, and H=Hcr. Hcr = (c’/γ )*(1/ γ cos2β(tanβ - tanϕ’)

Types of stability Analysis Procedure• The slope stability Analysis is divided in to two major classes, namely • 1.Mass procedure, 2. Method of SlicesMass Procedure: • The mass of the soil above the surface of sliding is taken as unit.• The soils that forms the slope is assumed to be homogeneous• This procedure is not considered for the case in most natural slopes

Methods of Slices:The soil above the surface of sliding is divided into a number of vertical parallel slices.The stability of each slice is calculated separatelyThis method the soil and porewater pressure can be taken into considerationThe variation of the normal stress along the potential failure surface also taken into account.

Landslide Overview Map of the Conterminous United States

Debris flows that blocked Interstate-70 during Labor Day weekend, 1994

Earth flow in Cincinnati, Ohio-This slide shows material being removed by highway crew along the Columbia Parkway, Cincinnati, Ohio.

Case Studies

Government Hill School in Anchorage 27 March1964 Alaska Earthquake - 1964

This building hangs over the head scarp of a landslide in decomposed bedrock that was triggered by the 1995 Kobe earthquake. Several homes were buried and over 30 people killed by the landslide.

http://cee.engr.ucdavis.edu/faculty/boulanger/geo_photo_album/Earthquake%20hazards/Landslides/EQ%20landslides%20-%20Main.html

In 1970, an earthquake induced rock and snow avalanche on Mt. Huascaran, Peru.

Cross-Sections Before and After the 1971 Earthquake

•San Fernando EQ 1971•Magnitude: 6.7•Deaths: 48•Injuries: 2000•Damage: $511 million•Crest dropped from 142 ft to 112 ft

http://quake.usgs.gov/prepare/factsheets/LADamStory/

Limit of landslides triggered by the Northridge Earthquakeand area of greatest landslide concentration (shaded)

“Failure rates correlated with: (1) shaking severity; (2) slope steepness; (3) strength and engineering properties of geologic materials; (4) water saturation; (5) existing landslide areas; and (6) vegetative cover.”

17 January 1994 Northridge Earthquake (M = 6.7)

U.S. Hwy 95 – Bonners Ferry Landslide

http://www.landslidetechnology.com/landslides/bonnersferry.htm

Landslides in Idaho

Clearwater County Project