lecture-1 shear strength of soils

7/30/2019 Lecture-1 Shear Strength of Soils

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CHAPTER ONE

Shear Strength of Soils

By Yada T.Nov/2012



Introduction

In this chapter we will define, describe, anddetermine the shear strength of soils. When you

complete this chapter, you should be able to; Determine the shear strength of soils.

Understand the difference between drained and undrainedshear strength. Determine the type of shear test that best simulates field

conditions. Interpret laboratory and field test results to obtain shear

strength parameters.



1.0 Introduction

The safety of any geotechnical structure is

dependent on the strength of the soil.

If the soil fails, a structure founded on it can collapse,

endangering lives and causing economic damages.

Soils fail either in tension or in shear.However, in the majority of soil mechanics problems

(such as bearing capacity, lateral pressure against

retaining walls, slope stability, etc.), only failure in

shear requires consideration.



Strength of different

materials

Steel

Tensile

strength

Concrete

Compressive

strength

Soil

Shear

strength

Presence of pore waterComplex

behavior



Cont.…

Expansive Soil



Cont.…



Cont…



Typical "roller-

coaster" roadcaused by

expansive soils

Cracks in exterior walls, as aresult of upward soil

expansion



Embankment

Strip footing

Shear failure of soils

Soils generally fail in shear

At failure, shear stress along the failure surface(mobilized shear resistance) reaches the shear strength.

Failure surface

Mobilized shear

resistance



Retaining

wall




Retaining

wall

At failure, shear stress along the failure surface(mobilized shear resistance) reaches the shear strength.

Failure

surface

Mobilized shear

resistance




Shear failure mechanism

The soil grains slideover each other alongthe failure surface.

No crushing of individual grains.

failure surface



Shear failure mechanism

At failure, shear stress along the failure surface ()reaches the shear strength (f ).



Cont’d…

The shear strength of soils is, therefore, paramount

importance to geotechnical engineers.

The shear strength along any plane is mobilized by

cohesion and frictional resistance to sliding between

soil particles. The cohesion(c) and angle of friction (ϕ) of a soil are

collectively known as shear strength parameters.



Coulomb’s Frictional Law

You may recall Coulomb’s frictional law from your

courses in statics and physics. If a block of weight (W)

is pushed horizontally on a plane (Fig. 1.1a), the

horizontal force ( H ) required to initiating movement

is:

Where:- µ=coefficient of static friction between the

block and the horizontal plane.

NB coefficient of static friction is independent of

the area of contact.

W H



Cont’d…

• It is, however, strongly dependent on;

the nature of the surface in contact the type of material

the condition of the surface and so on.

The angle between the resultant force R and thenormal force N (Fig. 1.1) is called the friction angle

1

tan



Figure 1.1: (a) Slip plane of a block. (b) A slip plane in a

soil mass

In terms of stresses, Coulomb’s law is

expressed as: tann f



Cont’d…

Where:-

( = T/A) the shear stress when slip is initiated

( =N/A) the normal stress on the plane on which slip is

initiated

Coulomb’s law requires the existence or the

development of a critical sliding plane, also

called slip plane or failure plane. In the case of

the block the slip plane is at the interfacebetween the block and the horizontal plane.

f

n



Mohr’s Circle for Stress

The stress states at a point within a soil mass can berepresented graphically by a very useful and widely

used devise known as Mohr’s circle for stress.

The stress state at a point is the set of stress vectors

corresponding to all planes passing through that point.

For simplicity, we will consider a two-dimensional

element with stresses as shown in Fig. 1.2a.

Let’s draw Mohr’s circle. First, we have to choose asign convention. In soil mechanics, compressive

stresses and clockwise shear are generally assumed to

be positive.



Cont’d…

We will also assume that

x z

1 1



Mohr’s Circle for Stress

The two coordinates of the circle are ( ) and

Recall from your strength of materials course that, for

equilibrium ..

Plot these two coordinates on a graph of shear stress(ordinate) and normal stress (abscissa) as shown by A

and B in Fig. above.

Draw a circle with AB as the diameter. The circle

crosses the normal stress axis at 1 and 3, where shear

stresses are equal to zero.

The stresses at these points are the major principal

stress, , and the minor principal stress, .

zx z ,

zx x ,

zx xz

11

1 3

zx ,



Cont’d…

The principal stresses are related to the stresses

and by the following relations:

z x ,

z x ,

zx

2

2

1 22 zx

x z x z

2

2

322

zx x z x z

The angle between the major principal stressplane and the horizontal plane( ).



Cont’d…

.

The stresses on a plane oriented at an angle

to the major principal stress plane are:

The maximum shear stress is at the top of the

circle with magnitude:

x

zx

1tan

2cos22

3131

2sin2

31



Cont’d…

231

max



Cont’d…



Summary of today Session

Safety of geotechnical structures concerning bearing capacity,

slope stability, lateral earth pressure against retaining wall

problems are depend on shear strength of soils.

The cohesion(c) and angle of friction (ϕ) of a soil are known as

shear strength parameters.

Define cohesion and angle of friction of soil.

Derive major principal , minor principal stress and maximum

shear stress from Mohr's circle?

What is failure plane?

What is the fails happen in soil, concrete, and steel?

Search structure surround your environments failed

due to shear failure.Free charge

assignment



.



Next class:

Mohr-Coulomb Failure Criteria



Mohr-Coulomb Failure Criteria

In 1900, Mohr presented a theory for rupture in

materials. According to this theory, failure along aplane in a material occurs by a critical combination of

normal and shear stresses, and not by normal or shear

stress alone.

The functional relation between normal and shear

stress on the failure plane can be given by

s=f( )

where:- S=shear stress at failure

= Normal stress on failure plane.



Cont’d..

Coulomb (1776) suggested that the shear strength of a soil

along a failure plane could be described by:

Where:- c is cohesion and ϕ is the angle of friction of the soil.

This Equation is generally referred to as the Mohr –Coulomb

failure criteria.

tann f c



Mohr-Coulomb Failure Criterion

(in terms of total stresses)

f is the maximum shear stress the soil can take without failure,

under normal stress of .

tan c f

c

CohesionFriction angle

f




(in terms of effective stresses)

u '

f is the maximum shear stress the soil can take without failure,

under normal effective stress of ’.

’

'tan'' c f

c’

’ Effectivecohesion

Effectivefriction anglef

’

u = pore waterpressure




'tan'' f f c

Shear strength consists of two components:

cohesive and frictional.

’f

f

’

'

c’ c’

’f tan ’ frictionalcomponent



c and are measures of shear strength.

Higher the values, higher the shear strength.



Soil elements at different locations

Failure surface

Mohr Circles & Failure Envelope

X X

X ~ failure

YY

Y ~ stable

’

'tan'' c f




Y

c

c

c

Initially, Mohr circle is a point

c+

The soil element does not fail if

the Mohr circle is contained withinthe envelope

GL




Y

c

c

c

GL

As loading progresses, Mohr

circle becomes larger…

.. and finally failure occurs when

Mohr circle touches the

envelope



Mohr circles in terms of total & effective stresses

= X

v’

h’ X

u

u

+

v’ h’

effective stresses

uv h

X

v

h

total stresses

or ’



Failure envelopes in terms of total & effective stresses

= X

v’

h’ X

u

u

+

v’ h’

effective stresses

uv h

X

v

h

total stresses

or ’

If X is on

failure

c

Failure envelope in

terms of total stresses

’

c’

Failure envelope in terms

of effective stresses

Cont



Cont.… • Mohr circle for effective and normal stresses

• Cohesion is the minimum shear strength at zero

normal stress, since tensile stress is ignored in the

soil or minimum.

'''

tan n f c

Cont



Cont…

• From the above graph the equation becomes

• This equation can be written in terms of the

principal stresses

• Considering OB = c’ and OC = , we have

245

2

90 '0

'

''2

2''

sincos2

OC OBCF R zx x z

''

2'3

'1 sincos

2

OC OB

''3

'1

'''3

'1 sin)(cos2 c

2)( '3

'1

'

''

'

''3

'1'

'''

3'1

sin1

sin12

sin1

sin1

sin1

cos2

sin1

sin1

cc

Cont



Cont…

or

or

'

''

'

''

1

'

3 sin1

sin1

2sin1

sin1

c

)2

45tan(2)2

45(tan'

''

2'3

'1

c

)245tan(2)245(tan

''

'2'

1

'

3

c

'

3'1

'

3

'

1'sin

'

'

'

1

'3

sin1

sin1

'

'

'

3

'1

sin1

sin1

)2

45(tan'

2

'

1

'3

)

245(tan

'2

'

3

'

1



Drained and undrained Shear strength

Drained condition occurs when the excess pore water pressure

developed during loading of a soil dissipates, i.e.resulting in volume changes in the soil.

Loose sands, normally consolidated clays and lightly over

consolidated clays tend to compress or contract, whilst dense

sands and heavily over consolidated (OCR > 2) clays tend toexpand during drained condition.

Undrained condition occurs when the excess pore water

pressure cannot drain, at least quickly from the soil, i.e.

During undrained shearing, the volume of the soil remainsconstant. Consequently, the tendency towards volume change

induces a pressure in the pore water.

If the specimen tends to compress or contract during shear, then

the induced pore water pressure is positive.

0u

0u



Cont’d..

Positive pore water pressures occur in loose sands, normally

consolidated clays and lightly over consolidated clays. If the specimen tends to expand and swell during shear, the

induced pore water pressure is negative. It wants to expand and

draw water into the pores, but it can not. Negative pore water

pressures occur in dense sands and heavily over consolidated(OCR > 2) clays.

The shear strength of a fine-grained soil under undrained

condition is called the undrained shear strength, Su. The

undrained shear strength Su is the radius of Mohr’s total stress

circle; that is

22

)()(

2

'

3

'

1

'

3

'

131

uuS u



1, A sample of clay has a cohesive strength of 80kN/m2,and

an angle of shearing resistance of 100.calculate the

shearing strength of clay at a normal stress of 100kN/m2

Solution - 1

Given c = 80kN/m

= 100 and

= 100kN/m2

The shearing strength is

tan

c20

/ 63.9710tan10080 mkN



• A cube sample of soil 100 mm×100 mm is subjected to

the forces shown in Fig. below. Determine

(a) The principal stress 1 & 3.

(b) The maximum shear stress, and

(c) The stresses on a plane oriented at Ѳ=300 clockwise

to the major principal stress plane.

SOLUTION 2



SOLUTION -2

• Thus

• The principal stresses are:

• Max shear stress

500

1.0*1.0

5

A

F z z

3001.0*1.0

3

A

F x x

100

1.0*1.0

1

A

T xz

1001.0*1.0

1

A

T zx

kpa42.541)100(2

300500

2

300500 2

2

1

kpa58.258)100(2

300500

2

300500 2

2

3

42.1412

58.25842.541

2

31max

SOLUTION 2



SOLUTION - 2

The stress on a plane(Ѳ=300 cw = 150 ccw) is given by:-

The angle between major principal plane and horizontal

plane :-

01

1

15.22]

30042.541

100[tan][tan

x

zx

kpa71.470150*2cos2

58.25842.541

2

58.25842.541

2cos22

3131

KPa47.122150*2sin2

58.25842.5412sin

2

31

D t i ti f h t th t f



Determination of shear strength parameters of

soils (c, or c’, ’)

Laboratory tests on specimens

taken from representative

undisturbed samples

Field tests

Most common laboratory tests

to determine the shear strength

parameters are,

1.Direct shear test2.Triaxial shear test

3. unconfined compression test

1.Vane shear test

2. Cone penetration test

3.Standard penetration test

4. Pocket penetrometer

5. Fall cone6. Pressuremeter



Preconsolidation Stress or past maximum effective stress, is themaximum vertical effective stress that a soil was subjected to in

the past σ’zc.

• Normally Consolidated Soil is one that has never experienced

vertical effective stresses greater than its current verticaleffective stress (σ’zc = σ’zo ).

• Over consolidated Soil is one that has experienced vertical

effective stresses greater than its current vertical effective stress

(σ’zo<σ’zc ).• Over consolidation ratio, OCR , is the ratio by which the current

vertical effective stress in the soil was exceeded in the past

(OCR = σ’zc / σ’zo ).

Remember those key terms

L b t t t



Laboratory tests

Field conditions

z

vc

vc

hchc

Before construction

A representative

soil samplez

vc +

hchc

After and during

construction

vc +

Laboratory tests

vc +



Laboratory tests

Simulating field conditions in

the laboratory

Step 1Set the specimen in

the apparatus and

apply the initial

stress condition

vc

vc

hchc

Representative soilsample taken

from the site

0

00

0

Step 2

Apply the

corresponding field

stress conditions

hchc

vc +

vc

vc

1 Di h



1.Direct shear test

Schematic diagram of the direct shear apparatus

Direct shear test



Direct shear test

Preparation of a sand specimen

Components of the shear box Preparation of a sand specimen

Porous

plates

Direct shear test is most suitable for consolidated drained tests

specially on granular soils (e.g.: sand) or stiff clays

Di t h t t



Direct shear test

Leveling the top surface of

specimen

Preparation of a sand specimen

Specimen preparation

completed

Pressure plate

Direct shear test



Direct shear test

Test procedure

Porous

plates

Pressure plate

Steel ball

Step 1: Apply a vertical load to the specimen and wait for consolidation

P

Proving ring to

measure shear

force

S

Direct shear test



Direct shear test

Step 2: Lower box is subjected to a horizontal displacement at a constant rate

Step 1: Apply a vertical load to the specimen and wait for consolidation

PTest procedure

Pressure plate

Steel ball

Proving ring to

measure shear

force

S

Porous

plates

Interface tests on direct shear apparatus



Interface tests on direct shear apparatus

In many foundation design problems and retaining wall problems, it is required

to determine the angle of internal friction between soil and the structural

material (concrete, steel or wood)

tan' a f cWhere,

ca = adhesion,

= angle of internal friction

Foundation material

Soil

P

S

Foundation material

Soil

P

S

Th di h b i l l li bl h



The direct shear box test particularly applicable to those

foundation design problems where it is necessary to determine

the angle of friction between the soil and material of which the

foundation is constructed.The friction between the base of the concrete footing

underneath soil, in such case ,the lower box with soil and the

upper box contain the foundation material.

The soil is used for test are either undisturbed sample or

remolded. If undisturbed ,the specimen has to be carefully

trimmed and filled in to the box.

If remolded ,the soil is placed in to the box in layer at the

required initial water content and tamped to the required dry

density .Porous stones may be placed on the top and bottom part of the

sample to facilitate drainage.

Direct shear test



Direct shear test

Shear box

Loading frame toapply vertical load

Dial gauge to

measure vertical

displacement

Dial gauge to measurehorizontal displacement

Proving ring to

measure shear

force

Observation/ analysis test results



Observation/ analysis test results

1. For each specimen plot the shear stress Vs shear

displacement (horizontal displacement).

Plot the graph of shear strength Vs normal stress for the threespecimens and calculate the shear strength parameters for the

soil.

Type of Test : Direct shear test. Deter. of moisture content

Calibration factor = __ Moisture Content (%) =___

Normal Load Applied = 1,2,3 kg *9.81m/s²

Horizontal Dial Gauge Constant: 20 Division

Calibration factor=0.001mm/Division

Due to normal load_________ Normal stress_________Kpa

ARBA MINCH UNIVERSITY

CIVIL ENGINEERING DEPARTMENT



GEOTECHNICAL ENGINEERING LABORATORY

Direct Shear Test

Project:

LOCATION:

Sample No. TP 2 Sample Depth, m: 2.20Thickness of sample: 25 mm Ring Calib. Factor: 0.70 N/div Wet unit weight, kN/M3: 18.56

Length of sample : 60 mm Rate of strain : 0.1 mm/min Dry Unit Weight, kN/M3: 14.62

Width of sample: 60 mm Moisture content, % 27.0 Sample Condition: Disturbed

Applied Vertical Stress 100 kPa Applied Vertical Stress 200 kPa Applied Vertical Stress 300 kPa

Horizontal

Displacement [mm]

Corrected

Area[mm2]

Proving

Ring Reading

Shear load[N]

Shear

Stress[kPa]

Proving

Ring Reading

Shear load[N]

Shear

Stress[kPa]

Proving

Ring Reading

Shear load[N]

Shear

Stress[kPa]

0.0 3600 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

0.5 3570 50.00 35.00 9.80 90.00 63.00 17.65 110.00 77.00 21.57

1.0 3540 75.00 52.50 14.83 110.00 77.00 21.75 130.00 91.00 25.71

1.5 3510 105.00 73.50 20.94 140.00 98.00 27.92 170.00 119.00 33.90

2.0 3480 140.00 98.00 28.16 170.00 119.00 34.20 195.00 136.50 39.222.5 3450 175.00 122.50 35.51 200.00 140.00 40.58 230.00 161.00 46.67

3.0 3420 190.00 133.00 38.89 230.00 161.00 47.08 258.00 180.60 52.81

3.5 3390 210.00 147.00 43.36 260.00 182.00 53.69 300.00 210.00 61.95

4.0 3360 240.00 168.00 50.00 290.00 203.00 60.42 340.00 238.00 70.83

5.0 2880 - - - - - - 610.00 427.00 148.26

Max. shear stress,kPa 50.00 Max. shear stress,kPa 60.42 Max. shear stress,kPa 148.26



Analysis of test results

sampletheof sectioncrossof Area(P)forceNormal stressNormal

sampletheof sectioncrossof Area

(S)surfaceslidingat thedevelopedresistanceShearstressShear

Note: Cross-sectional area of the sample changes with the horizontal

displacement

Shear load = Proving ring reading x Ring Calibration Factor

Corrected Cross-sectional =Ao[1+[ΔL⁄3]]

Direct shear tests on sands



f1

Normal stress = 1

How to determine strength parameters c and

S h e a r s t r e s s ,

Shear displacement

f2

Normal stress = 2

f3

Normal stress = 3

S h e a r s t r e s s a t f a

i l u r e , f

Normal stress,

Mohr – Coulomb failure envelope

Direct shear tests on sands



Some important facts on strength parameters c and of sand

Sand is cohesionless hence

c = 0

Direct shear tests are drainedand pore water pressures are

dissipated, hence u = 0

Therefore,

’ = and c’ = c = 0

Direct shear tests on clays



Direct shear tests on clays

Failure envelopes for clay from drained direct shear tests

S h e a r s t r e

s s a t f a i l u r e , f

Normal force,

’

Normally consolidated clay (c’ = 0)

In case of clay, horizontal displacement should be applied at a very

slow rate to allow dissipation of pore water pressure (therefore, one

test would take several days to finish)

Overconsolidated clay (c’ ≠ 0)

Advantages of direct shear apparatus



Advantages of direct shear apparatus

Due to the smaller thickness of the sample, rapid drainage can

be achieved

Can be used to determine strength parameters c and

Clay samples can be oriented along the plane of weakness or

an identified failure plane

Disadvantages of direct shear apparatus

Failure occurs along a predetermined failure plane

Area of the sliding surface changes as the test progresses

Non-uniform distribution of shear stress along the failure surface

2.Triaxial Shear Test



2.Triaxial Shear Test

Soil sample atfailure

Failure plane

Porous

stone

impervious

membrane

Piston (to apply deviatoric stress)

O-ring

pedestal

Perspex

cell

Cell pressure

Back pressure Pore pressure or

volume change

Water

Soilsample

Triaxial Shear Test



Triaxial Shear Test

Specimen preparation (undisturbed sample)

Sampling tubes

Triaxial Shear Test



Triaxial Shear Test


Edges of the sample are

carefully trimmed

Setting up the sample in the

triaxial cell

Triaxial Shear Test



Triaxial Shear Test

Sample is covered with a

rubber membrane and sealed Cell is completely filled

with water


Triaxial Shear Test



Triaxial Shear Test


Proving ring to

measure the

deviator load

Dial gauge to

measure verticaldisplacement

In some tests

Types of Triaxial Tests deviatoric stress



Is the drainage valve open?

yes no

Consolidated

sampleUnconsolidated

sample

Is the drainage valve open?

yes no

Drained

loading

Undrained

loading

Under all-around cell pressure c

cc

c

cStep 1

( = q)

Shearing (loading)

Step 2

c c

c+ q

Consolidated- drained test (CD Test)



Deviator stress (q or d) = 1 – 3

( )

1 = VC +

3 = hC

CD tests How to determine strength parameters c and



D e v i a t o r s t r e s s ,

d

Axial strain

S h e a r

s t r e s s ,

or ’

Mohr – Coulomb

failure envelope

(d)fa

Confining stress = 3a(d)fb

Confining stress = 3b

(d)fc

Confining stress = 3c

3c 1c3a 1a

(d

)fa

3b 1b

(d)fb

1 = 3 + (d)f

3

CD tests



Strength parameters c and obtained from CD tests

Since u = 0 in CD

tests, = ’

Therefore, c = c’

and = ’

cd and d are used todenote them

CD tests Failure envelopes



S h e a r s t r

e s s ,

or ’

d

Mohr – Coulomb

failure envelope

3a 1a

(d)fa

For sand and NC Clay, cd = 0

Therefore, one CD test would be sufficient to determine d of sand

or NC clay

CD tests Failure envelopes



For OC Clay, cd ≠ 0

NC clay, cd =0

or ’

3 1

(d)f

c

c

OC NC

3.Unconfined Compression Test (UC Test)



p ( )

1 = VC +

3 = 0

Confining pressure is zero in the UC test

Unconfined Compression Test (UC Test)



p ( )

1 = VC + f

3

= 0

S h e a r s t r e s s ,

Normal stress,

qu

Note: Theoritically qu = cu , However in the actual case qu < cu due

to premature failure of the sample

A i t #1



Assignment #1

Due date:- @ Next Class



In-situ/ Field Tests/ shear tests

Vane shear test Cone penetration test

Standard penetration test

1. Vane shear test

In soft and saturated clays, where undisturbedspecimen is difficult to obtain, the undrained shearstrength is measured using a shear vane test.

A diagrammatic view of the shear vane apparatus is

shown in Fig. below. It consists of four thin metal blades welded

orthogonally (900) to a rod where the height H is twicethe diameter D. Commonly used diameters are 38, 50and 75 mm.

1.Vane shear test



PLAN VIEW

This is one of the most versatile and widely used devices used for investigating

undrained shear strength (Su) and sensitivity of soft clays

Bore hole (diameter

= DB)

h > 3DB)

Vane

D

H

AppliedTorque, T

Vane T

Rupture

surfaceDisturbed

soil

Rate of rotation : 60 – 120 per minute

Test can be conducted at 0.5 m vertical

intervals



Vane shear testAfter the initial test, vane can be rapidly



Since the test is very fast,

Unconsolidated Undrained (UU)

can be expected

Su

Su

h

, p y

rotated through several revolutions until the

clay become remoulded

peakultimate

Shear displacement

StengthUltimate

StengthPeak ySensitivit

Some important facts on vane shear test



Insertion of vane into soft clays

and silts disrupts the natural soilstructure around the vane causing

reduction of shear strength

The above reduction is

partially regained aftersome time

Cu as determined by vane

shear test may be a

function of the rate of

angular rotation of the

vane

Correction for the strength parameters obtained from



vane shear test

Bjerrum (1974) has shown that as the plasticity of soilsincreases, su obtained by vane shear tests may give unsafe

results for foundation design. Therefore, he proposed the

following correction.

Su(design) = l Su(vane shear)

Where, l = correction factor = 1.7 – 0.54 log (PI)PI = Plasticity Index

Cont’d…



• The maximum torque is measured by a

suitable instrument and equals to the momentof the mobilized shear stress about the central

axis of the apparatus. The undrained shear

strength is calculated from

)6 / 2 / (2 D H D

T S u

2 Static Cone Penetrometer test



2. Static Cone Penetrometer test

Static Cone Penetrometer test



Force required for the inner rod to push the tip (Fc) and the total

force required to push both the tip and the sleeve (Fc + Fs) will bemeasured

Point resistance (qc) = Fc/ area of the tip

Sleeve resistance (qs) = Fs/ area of the sleeve in contact with soil

Friction Ratio ( f r ) = qs / qc ×100 (%)

Various correlations have been developed to determine soil

strength parameters (c, , ect) from f r

Standard Penetration Test, SPT



SPT is the most widely used test procedure to determine the

properties of in-situ soils

63.5 kg

0.76 m

Drill rod

0.15 m

0.15 m0.15 m

Number of blows = N1



Standard penetration resistance (SPT N) = N2 + N3

Number of blows for the first 150 mm

penetration is disregarded due to the

disturbance likely to exist at the bottom of the

drill hole

The test can be conducted at every 1m vertical

intervals

Various correlations have been developed to determine soil

strength parameters (c, , ect) from N



Standard Penetration Test, SPT



SPT (Manual operation)



Fig. Common In-Situ Tests for Geotechnical Site Characterization of Soils.

Summary of chapter one



Discuss shear strength parameters of soils.

What is the reason in case of soil thatgeotechnical structures such as shallow

foundation, pile foundation, retaining walls,

embankment dam and etc are failed?What is difference between positive pore water

pressure and negative pore water pressure?

Describe the methods used to determine shearstrength parameters of soils.

#2 assignment



#2 assignment

1. Write short note on difference of laboratorytest and in-situ test for shear strength of soils,

and give your reason which is important/

better/ from both of tests method.

lecture-1 shear strength of soils

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