ce 371 soil mechanics - civil cmu · 2017-08-09 · ce 371 soil mechanics suriyah thongmunee, ph.d....

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CE 371 Soil Mechanics

Suriyah Thongmunee, Ph.D.

Compressibility and Consolidation

Textbook: Braja M. Das, "Principles of Geotechnical Engineering", 5th E.


2

Lecture Outline:

1. Elastic Settlement

2. Consolidation Settlement

3. One-Dimensional Consolidation

4. Consolidometer and Standard Test

5. Pressure-Void Ratio Curves

6. Determination of Preconsolidation Pressure

7. Computation of Consolidation Settlement

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Soil engineering problems are of two types. The first type includes all cases wherein there is no possibility of the stress being sufficiently large to exceed the shear strength of the soil, but the settlement due to the imposed stress is greater than allowable settlement. The second type includes cases in which there is danger of shearing stresses exceeding the shear strength of the soil

3



When structures are built on soils, they transfer loads to the subsoil through the foundations. The effect of the loads is felt by the soil normally up to a depth of about two to three times the width of the foundation.

As a result, the soil within this depth gets compressed due to the imposed stresses. The compression of the soil mass leads to the decrease in the volume of the mass which results in the settlement of the structure.

4

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The settlement of structure is caused by (a) deformation of soil particles (b) relocations of soil particles (c) expulsion of water or air from the void spaces.

The settlement of the soil mass due to the imposed stresses may be almost immediate (elastic) or time dependent according to the permeability (consolidation) characteristics of the soil

In general, the soil settlement caused by loads may be divided into three categories: (1) Elastic settlement (2) Primary consolidation settlementand (3) Secondary consolidation settlement. The total settlement can be determined by summing up these settlement

5



(1) Elastic settlement or immediate, which is caused by the elastic deformation of dry soil and of moist and saturated soils without any change in the moisture content. Elastic settlement calculations generally are based on equations derived from the theory of elasticity.

(2) Primary consolidation settlement, which is the result of a volume change in saturated cohesive soils because of expulsion of the water that occupies the void spaces.

(3) Secondary consolidation settlement, which is observed in saturated cohesive soils and is the result of the plastic adjustment of soil fabrics. It is an additional form of compression that occurs at constant effective stress.

6

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This chapter presents the fundamental principles for estimating the elastic and consolidation settlements of soil layers under superimposed loadings.

The total settlement of a foundation can then be given as

where

ST = Total settlement

Sc = Primary consolidation settlement

Ss = Secondary consolidation settlement

Se = Elastic settlement

7

When foundations are constructed on very compressible clays, the

consolidation settlement can be several times greater than the elastic

settlement.


Elastic Settlement

Elastic, or immediate, settlement of

foundations (Se) occurs directly after the application of a load without a change in the moisture content of the soil.

The elastic settlement of a “flexible foundation” may be expressed as

8

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Elastic Settlement

9

The elastic settlement of a “rigid foundation” can be estimated as

Textbook: Braja M. Das, "Principles of Geotechnical Engineering", 5th E. 10

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Elastic Settlement

Due to the nonhomogeneousnature of soil deposits, the magnitude of Es may vary with depth. For that reason, Bowles (1987) recommended using a weighted average value of Es.

13

Typical value of Es

Typical value of ms


Example 1

14

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Example 1

15


Consolidation Settlement

“It is quite reasonable and rational to assume that the solid matter and the pore water are relatively incompressible under the loads usually encountered in soil mass”

16

The spring represents the mineral skeleton in actual soil mass, while the water below the piston is the pore water under saturated conditions in the soil mass

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17

This process is called “consolidation”.

The process opposite to consolidation is called

“swelling”.



Consolidation may be due to one or more of the following factors:

1. External static loads from structure

2. Self-weight of the soil such as recently placed fills

3. Lowering of the ground water table

4. Desiccation

5. Permeability of soil

6. Skeleton of soil

18

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One-Dimensional Consolidation

19

Based on the consolidation process, we can analyze the strain of a saturated clay layer subjected to a stress increase.


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Consolidometer and Standard Test

The compressibility of a saturated clay is determined by means of the apparatus as shown in below figure. This apparatus is well known as oedometer.

21

2-5 cm in height and 6-10 cm in dia.



The main purpose of the consolidation test is to determine the magnitudeand rate of settlement.

• Loads are applied in steps in such a way that successive load intensity is twice the preceding load.

• Commonly used load intensities are 25, 50, 100, 400, 800 and 1600 kN/m2.

• Each load is allowed to stand until compression has practically ceased (≤24hr.)

• The dial reading are taken at elapsed times of ¼, ½, 1, 2, 4, 8, 15, 30, 60, 120, 240, 480, 1440 minutes from the time of new increment of load is put on the sample.

• After completion of greatest load, the load is removed in the steps to provide data for plotting the expansion curve of the soil.

• Required data of soil sample are %wbefore, %wafter, Gs, T˚, A and H.

22

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The general shape of deformation of the specimen against time for a given load increment is shown in Figure.

23

Initial compression, which is caused mostly by preloading.

Primary consolidation,

Secondary consolidation, which occurs after complete dissipation of the excess pore water pressure, when some deformation of the specimen takes place because of the plastic readjustment of soil fabric.



Moreover, we could obtain 2 relationships

1. The relationship between void ratio (settlement) and effective stress for estimating preconsolidation pressure (Pc), compression index (Cc) and swelling index (Cs).

2. The relationship between deformation (settlement) and time for estimating coefficient of consolidation (Cv).

24

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Pressure-Void Ratio Curves

The pressure and void curve can be obtained if the void ratio of the sample at the end of each increment of load is determined.

Next is step-by-step procedure for obtaining the pressure-void curve.

25

1Calculate the height of solids, Hs, in soil specimen using the equation



26

2Calculate the initial height of void, Hv, in soil specimen as

3Calculate the initial void ratio, e, in soil specimen as

4After completion of each increment load, calculate the change of void and new void ratio if the deformation is ΔH1.

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If all of effective stress and corresponding void ratio are determined. Then, we can plot void ratio in normal scale and effective stress in log scale.

27


Example 2

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Example 2

29


Pressure-Void Ratio Curves for sand

it has no consolidation tests of sand because the compression of sand is instantaneous. 90% of compression has taken place within a period of less than 2 minutes.

30

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Pressure-Void Ratio Curves for clay

Consolidation tests are usually conducted on samples of clay. Because the compression of clay under external load is time-dependent compression due to the discharge rate of pore water in the soil sample.

The compressibility of characteristics of clay depend on many factors in previous topic.

The most important factors are

1. whether the clay is normally consolidated or overconsolidated.

2. Whether the clay is sensitive or insensitive.

31


?


A soil in the field at some depth has been

subjected to a certain maximum effective past pressure in its geologic history. This maximum effective past pressure may be equal to or lessthan the existing effective overburden pressure at the time of sampling.

During the soil sampling, the existing effective overburden pressure is also released, which results in some expansion.

32

Steeper slope

Flat slope

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When this specimen is subjected to a

consolidation test, a small amount of settlement will occur when the applied effective pressure is less than the maximum effective past pressure.

When the effective pressure on the specimen becomes greater than the maximum effective past pressure, the change in the void ratio is much larger, and the e–log s ’ relationship is practically linear with a steeper slope.

33

Steeper slope

Flat slope


Determination of Preconsolidation Pressure

1. By visual observation, establish point a, at which

the e–log s ’ plot has a minimum radius of curvature.

2. Draw a horizontal line ab.

3. Draw the line ac tangent at a.

4. Draw the line ad, which is the bisector of the angle bac.

5. Project the straight-line portion gh of the e–log s plot back to intersect line ad at f. The abscissa of point f is the

preconsolidation pressure, sc.

34

The maximum effective past pressure or well known as preconsolidationpressure can be determined from the laboratory e–log s ’ plot

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Empirical relationship



Empirical relationship

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Normally Consolidated and Overconsolidated

The ratio of the preconsolidation pressure,sC, to the present effective

overburden pressure, s’, is called the overconsolidation ratio (OCR).

37

The OCR leads us to the two basic definition of clay based on stress history.

- Normally consolidated clay, whose present that the present effective overburden pressure is the maximum past pressure (OCR = 1).

- Overconsolidated clay, whose present that the present effective overburden pressure is less than that which the soil experienced in the past (OCR > 1).


Overconsolidation of a clay may have been caused due to some of the following factors.

1. Weight of an overburden of soil which has eroded.

2. Weight of a continental ice sheet that melted

3. Desiccation of layers close to the surface

Experience (Murthy) indicates that the natural moisture content is commonly close to the liquid limit for normally consolidated clay whereas for overconsolidated clay the natural moisture content is close to the plastic limit.

38


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Pressure-Void Ratio Curves : av and mv

40

Coefficient of Compressibility

Coefficient of Volume Compressibility

σ

e

ee

am

oo

vv Δ

Δ•

+1

1=

+1=

o

ov σσ

ee

σ

ea

-

-=

Δ

Δ=

1

1

De

Dp

eo

e1

σo σ1

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Pressure-Void Ratio Curves : Cc and Cr

41

Δ+

Δ=

σ

σσlog

eCc

Δ+

Δ=

σ

σσlog

eCr

Normally consolidation

recompression

Over consolidation

De

s s+Ds

Preconsolidation

Pressure, sc

sc

Recompression

Index, Crrebound

Compression

Index, Cc


Calculation of settlement: ODPC

42

So,

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Calculation of settlement: e-p

• From coefficient of volume compressibility

* mv is not constant value

43

σ

e

ee

am

oo

vv Δ

Δ•

+1

1=

+1=

σme

ev

o

Δ=+1

Δ

HσmS vc Δ=∴

De

Dp

eo

e1

σo σ1


Example 3.1

44

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Calculation of settlement: e-log p

• For normally consolidated clay (after s’o)

• For overconsolidation clay (before s’o)

46

s’o

Cc

Cr

Cr

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Cc and Cr by empirical formulas

47

Skempton (1944)

Rendon-Herrero (1983)

Nagaraj and Murty (1985)

Cc

Cs

or

Cr Nagaraj and Murty (1985)

Kulhawy and Mayne (1990)


Example 3.2

48

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Example 3.2 : Part a

49


Example 3.2 : Part b

50

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Example 3.2 : Part c

51


Example 4

52

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Example 4 : part a

53


Example 4 : part b

54

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Example 4 : part c

55


Secondary Consolidation Settlement

Secondary consolidation settlement is the settlement of soil due to the plastic adjustment of soil fabrics.

The secondary compression index can be defined as

56

The magnitude of the secondary consolidation can be calculated as

p

α'α

'αs e

CCwhere

t

tlogHCS

+1==

1

2

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Secondary Consolidation Settlement

57


Example 5

58

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Time Rate of Consolidation

The total settlement caused by primary consolidation resulting from an increase in the stress on a soil layer can be calculated by the use of the equations as follows.

However, they do not provide any information regarding the rate of primary consolidation.

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Terzaghi (1925) proposed the first theory to consider the rate of

one-dimensional consolidation for saturated clay soils. The mathematical

derivations are based on the following six assumptions (also see Taylor, 1948):

1. The clay–water system is homogeneous.

2. Saturation is complete.

3. Compressibility of water is negligible.

4. Compressibility of soil grains is negligible (but soil grains rearrange).

5. The flow of water is in one direction only (that is, in the direction of

compression).

6. Darcy’s law is valid.

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62

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63

Using Darcy’s law, we have

The rate of change in the volume of the soil element is equal to the rate of change in the volume of voids.



64

The change in the void ratio is caused by the increase of effective stress (i.e., a decrease of excess pore water pressure). Assuming that they are related linearly, we have

av = coefficient of compressibility

mv = coefficient of volume compressibility

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65

Unit : L2/t



is the basic differential equation of Terzaghi’s consolidation theory and can be solved with the following boundary conditions:

66

where m = an integer

M = (π/2)(2m +1)uO = initial excess pore water pressureTv = time factor = cvt / H2

dr

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67

Because consolidation progresses by the dissipation of excess pore water pressure, the degree of consolidation at a distance z at any time t is

∑∞

0=

-

2

32-1=

m

TM

drz

veH

Mzsin

MU ( )



68

However, for analysis of settlement in general case, the average settlement of the clay layer are mainly considered. The degree of consolidation at time t can be written as

100×=c

)t(c

S

SU

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Example 6

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Example 7

71


Example 7

72

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Example 7

73


Determination of Coefficient of Consolidation

Logarithm of Time Method

74

50

2

v

50

t

2

H196.0

C

196.0T

Final Compression

=

=

t 4t t50

d50

d100

=

=

Initial

Compression

Primary

Consolidation

Secondary

ConsolidationFinal Compression

U = 0.5¯

ds

U = 1.0¯

U = 0¯

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Determination of Coefficient of Consolidation

Square Root of Time Method

75

Initial

Compression

Primary

Consolidation

100 หนวย 15 หนวย

Secondary

Consolidation

U = 0¯

U = 0.9¯

U = 1.0¯

t90

90

2

v

90

t

2

H848.0

C

848.0T

ds

d90

d100

100

90

dd

dd

s100

s90


Example 7

76

Using the logarithm-of-time method and square-root-time, determine cv. The average height of the specimen during consolidation was 2.24 cm, and it was drained at the top and bottom.

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Example 7.1 using log-of time

77

0.01 0.1 1 10 100 1000 10000

0.54

0.52

0.50

0.48

0.46

0.44

0.42

0.40

0.38

sett

lem

ent

(cm

)

time (min)

=

=

t 4t

t50

d50

d100

U = 0.5¯

ds

U = 1.0¯

U = 0¯

19 min

( )

sec

cm.c

min

cm.c

/..c

t

H.c

.T

v

v

v

drv

24

2

2

50

2

50

10×1702=

0130=

19

2242×1960=

1960=

1960=


Example 7.2 using square-root-of time

78

0 10 20 30 40 50

0.54

0.52

0.50

0.48

0.46

0.44

0.42

0.40

0.38

sett

lem

ent

(cm

)

square root time (min)

100 หนวย15 หนวย

U = 0¯

U = 0.9¯

U = 1.0¯

t90ds

d90

8.5 min

( )

sec

cm.c

min

cm.c

..

/..c

t

H.c

.T

v

v

v

drv

24

2

2

90

2

90

10×4532=

01470=

58×58

2242×8480=

8480=

8480=

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Quiz 1

79

Calculate the total settlement at center of the footing as shown below by using the data of clay layer from the example 7.1.1. After 5 year2. After 15 years of construction.

Very stiff layer

Loose to dense layerEs = 30000 kN/m2

Normal consolidation clay900=020= 0 .e,.cα

Es = 4000 kN/m2

μs = 0.30

“Education is the best provision for the journey to old age.”

Aristotle

อางอง

• Braja M. Das :"Principles of Geotechnical Engineering 5th Edition

• V.N.S. Murthy : Geotechnical Engineering

• เอกสารการสอน อ.อนรทธ ธงไชย

• รปภาพจาก Google images

3/14/2015

Soil Mechanics .......... Suriyah 1

Soil Mechanics


Shear Strength of Soil

Textbook: Braja M. Das, "Principles of Geotechnical Engineering", 5th E. ………. Suriyah Thongmunee, Ph.D.


2

Lecture Outline:

1. Mohr-Coulomb Failure Criterion

2. Direct Shear Test

3. Triaxial Test

4. Unconfined Compression Test

5. Vane Shear Test

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The “shear strength” of a soil mass is the internal resistance per

unit area that the soil mass can offer to resist failure and sliding along any plane inside it. One must understand the nature of shearing resistance in order to

analyze soil stability problems, such as bearing capacity, slope stability, and lateral pressure on earth-retaining structures.

3


Mohr-Coulomb Failure Criterion

Mohr (1900) presented a theory for rupture in materials that contended that a material fails because of a critical combination of

normal stress and shearing stress and not from either maximum normal or shear stress alone.

For most soil mechanics problems, Coulomb (1776) approximated the

shear stress on the failure plane as a linear function of the normal stress.

4

or

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The equations are expressions of shear strength based on total stress and effective stress.

c’ = 0 (sand and inorganic silt)

≈ 0 (normally consolidated clays)

> 0 (Overconsolidated clays)

The angle of friction, φ’, is sometimes referred to as the drained angle of friction.

5

or



6

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Inclination of the Failure Plane

7


Determination of Shear Strength Parameters

There are several methods now available to determine the shear strength parameters (i.e., c, φ, c’, φ’) of various soil specimens in the laboratory. They are as follows:

• Direct shear test

• Triaxial test

• Unconfined Compression Test

• Van shear test (field test)

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Direct Shear Test

9

Direct shear test can be either stress

controlled or strain controlled.In stress-controlled tests, the shear force

is applied in equal increments until the specimen fails. The failure occurs along the plane of split of the shear box. After the application of each incremental load, the shear displacement of the top half of the box and The change in the height of the specimen are measured by a horizontal dial gauge and vertical dial gauge, respectively.


Direct Shear Test

In strain-controlled tests, a constant rate of shear displacement is applied to one-half of the box by a motor that acts through gears. The constant rate of shear displacement is measured by a horizontal dial gauge. The resisting shear force of the soil corresponding to any shear displacement and The volume change of the specimen during the test can be measured by proving ring and dial gauge.

The advantage of the strain-controlled tests is that in the case of dense sand, peak shear resistance (that is, at failure) as well as lesser shear resistance (that is, at a point after failure called ultimate strength) can be observed and plotted.

10

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Direct Shear Test : each normal stress

11

At large shear displacement, the void ratios of loose and dense sands become practically the same, and this is termed the “critical void ratio”.


Direct Shear Test : various normal stresses

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Drained Direct Shear Test on Saturated Sand and Clay

13

Similar to the ultimate shear strength in the case of sand, at large shearing displacements, we can obtain the residual shear strength of clay (τr) in a drained test.


Direct Shear Test

14

Skempton (1964)

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Ex. of DSTs

15

Test no. Normal weight

(kg)

Shear force

(kg)

Cross section area

(cm ×××× cm.)

Normal Stress

(kN/m2)

Shear Stress

(kN/m2)

1 4 5.80 5.5 ×××× 5.5 13 17.5

2 8 6.94 5.5 ×××× 5.5 26 22.4

3 12 8.10 5.5 ×××× 5.5 39 26.2

4 16 9.60 5.5 ×××× 5.5 52 31.0

Determine shear strength of soil from direct shear test results as shown in below table and find principle stress of Test No.2

222

11

222

11

kN/m 17.5 )01.0((5.5)

000)(9.81)(1/1 5.8

A

T

kN/m 13 m/cm) 01.0(cm) (5.5

kN/1000N) N/kg)(1 kg)(9.81 (4

A

P

===

===

τ

σ


Ex. of DSTs

100

40 50 60 70 803020

10

20

30

40

50

From Figure, it was found thatc = 13 ton/m2ϕ = 19.2 deg

Principle stress can be determined by drawing graph as shown or calculating using below equations

where

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General Comment on DST

• Simple and economical for sandy soil but not fail along the weakest plane

• Shear stress distribution over shear plane is not uniform

• Interface friction angle between soil and foundation material can be easily determined by DST --- Similar equation with shear strength of soil

17


General Comment on DST

18

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Triaxial Shear Test

• What is Triaxial Shear Test?Shear test that confining pressure is considered

and applied by compression of fluid in camber. Axial load is applied via vertical loading ram.

• It has 2 ways for applying the vertical stress until soil specimen is failed.

o Stress Control – measure deformationo Strain Control – measure vertical stress

19

Most reliable methods for

determining shear strength


Triaxial Shear Test

• Standard types of triaxial tests1. Consolidated-drained test or drained test (CD test)

2. Consolidated-undrained test (CU test)

3. Unconsolidated-undrained test (UU test)

20

∆∆∆∆u= 0ττττ

σσσσn

∆∆∆∆u= 0

σσσσn

∆∆∆∆u= 0 ∆∆∆∆u>

0

ττττσσσσn σσσσn

∆∆∆∆u> 0 ∆∆∆∆u> 0ττττ

σσσσnσσσσn

C,φ

Ccu, φcu

Cuu, φuu

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Triaxial Shear Test

Triaxial shear test is the test of saturated soil (%Sr =100)

21

How to saturate and

check %Sr

When the saturated soil specimen is firstly subjected the confining pressure, the pore pressure increases suddenly. The ratio of pore pressure and confining pressure is defined as B parameter. It is used to check the degree of soil saturation. Theoretical values of B at complete saturation are shown in below table.

B = Skempton’s pore pressure parameter


Consolidated-drained test (CD test)

22

CD test procedure

• Saturates the specimen

• Checks %Sr

• Applies confining pressure

• Open drainage gage

• Waits until uc≈ 0 and record volume change

• Slowly applies deviator stress and record volume change

Typical test results

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23

CD test results of normally consolidated soil



24

CD test results of overconsolidated soil

For overconsolidated soil, the strength parameter will consist of friction and cohesion parameters. These parameters can be determined by conducting at least 2 tests with different confining pressure.

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Ex. of CD test

25

From the CD-Test results of sand sample, please calculate 1. Internal friction angle, ϕ, 2. Angle between failure plane and principle plane, θ,3. Normal stress, σn, and shear stress, τn, on failure plane

σ3= 276 kN/m2

(Δσd)f = 276 kN/m2

s3

s3

s3

Deviator Stress kN/m2

Axial Strain

Cell pressure = 276 kN/m2

276


Ex. of CD test

26

2

d311

233

kN/m 552 276 276

)(

kN/m 276

=+=

∆+====

fσσσσσσ

φστ tan n=

01-

31

31

5.19)333.0(sin '

0.333 276552

276 - 552

2''

2''

OA

AB ' sin

==

=+

=

+

−

==

φ

σσ

σσ

φ

R

OA

B

o

o

o

o

7.542

5.1945

2

'45

=+=

+= φθ

kN/m 130 54.7) (2sin 2

276 - 552 '

kN/m 368 57.4) (2 cos 2

276 - 552

2

276 552

2 cos 2

''

2

'' '

2f

2

3131

=×=

=×++=

−++=

τ

θσσσσσ n

φφφφ

++++= =

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Consolidated-undrained test (CU test)

27

CU test procedure


• Checks %Sr

• Applies confining pressure

• Waits until uc≈ 0 and record volume change

• Closed drainage gage

• Applies deviator stress and record pore pressure

A = Skempton’s pore pressure parameter

/



28

CD test results

Typical test results

Loose sand

Or NC soil

Dense sand

Or OCR soil Decreases due to dilation of soil

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29

CD test results of normally consolidated soil



31

CD test results of overconsolidated soil

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Unconsolidated-undrained test (UU test)

32

//

CU test procedure


• Checks %Sr

• Closed drainage gage

• Applies confining pressure and record pore pressure

• Applies deviator stress and record pore pressure

UU test method is appropriate for fully saturated

cohesive soil.


CD & CU & UU tests

33

CD : Q

CU : P

UU : R

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Unconfined Compression Test (UC test)

34

• Special test of UU test with confining pressure is zero

• Value of Cu slightly lower Cu obtained from UU test

• Fully saturated cohesive soil with completely undrained

• Rapid loading

• Useful in practice : test preparation and testing is fast


Ex. of CU and UC test

35

From CU test result of clay sample, please calculate

1. Total stress strength envelope, ϕ,2. Effective stress strength envelope, ϕ’,3. Angle of failure plane of sample 1, θ,4. Unconfined compressive strength of

sample 2 , qu,

Test No.

σ3

(kPa)(∆σd)f(kPa)

∆uf

(kPa)σ1

(kPa)σ’3(kPa)

σ’1(kPa)

1 200 150 140 350 60 210

2 400 300 280 700 120 420

0

01-

31

31

1-

01-

31

31

1-

8.15

8.15 400700

400 - 700 sin

2

2 sin

8.15 200350

200 - 350 sin

2

2 sin

=

=+

=

+

−

=

=+

=

+

−

=

avgφ

σσ

σσ

φ

σσ

σσ

φ

3/14/2015




36



sample 2 , qu,

Test No.

σ3

(kPa)(∆σd)f(kPa)

∆uf

(kPa)σ1

(kPa)σ’3(kPa)

σ’1(kPa)

1 200 150 140 350 60 210

2 400 300 280 700 120 420

01-

31

31

7.33)556.0(sin '

0.556 60210

60 - 210

2''

2''

' sin

==

=+

=

+

−

=

φ

σσ

σσ

φ



37



sample 2 , qu,

Test No.

σ3

(kPa)(∆σd)f(kPa)

∆uf

(kPa)σ1

(kPa)σ’3(kPa)

σ’1(kPa)

1 200 150 140 350 60 210

2 400 300 280 700 120 420

θθθθ

σ'1

σ'3

62°°°°

o61.8 2

33.7 45

2

' 45

oo

o

=+=

+= φθ

3/14/2015




38



sample 2 , qu,

Test No.

σ3

(kPa)(∆σd)f(kPa)

∆uf

(kPa)σ1

(kPa)σ’3(kPa)

σ’1(kPa)

1 200 150 140 350 60 210

2 400 300 280 700 120 420

kPaq

q

u

fdu

30031

=∆=−= σσσ


Sensitivity and Thixotropy of Clay

After UC test, the collapsed soil is remolded and tasted again. The UC strength will reduce greatly. This property is called “Sensitivity”

Degree of sensitivity is ratio of undisturbed and remolded UC strength. Some clay turn to viscos fluid (high sensitivity) is called “Quick clay”

39

Mos

t cl

ayH

ighl

y floc

cule

nt

mar

ine

clay

Rosenqvist (1953)

3/14/2015



Sensitivity and Thixotropy of Clay

Why the UC strength of soil reduces ?

because the clay particle structure was destructed during remolding.

After remolding, the soil is kept for a while. Then, the UC test of the soil is again carried out. The UC strength of soil will be greater then that of the remolded soil. This phenomenon is called “Thixotropy”.

40

Most soil is partially

thixotropic material. Its

behavior is shown here


Related researches on Thixotropy

Seed and Chan (1959) carried out the shear strength tests of three compacted soil with water content near the plastic limit to study the thixotropic strength regain.

it was founded that the tendency of thixotropic strength regain is exponential

growth. Moreover, the strength of all compacted soil developed more than 10%

within one week.

41

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Strength Anisotropy in Clay

So, UU shear strength of saturated soil can vary, depending direction of applied load. It can be called

“strength anisotropy”. It is caused by the nature of the deposition of the cohesive soils, and subsequent consolidation.

42

For construction of embankment, Lateritic soil is filled and compacted in construction area. Due to that process, it may change the direction of major principle stress acting on soil element beneath the embankment as shown in figure.

วรยา ฉมอRอย, 2553


Strength Anisotropy in Clay

some soil particles can orient perpendicular or parallel to major principle stress that can be cause the soil strength variation with direction.

Casagrande and Carrillo (1944) proposed the relationship between strength and direction.

43Based on Loh and Holt, 1974

3/14/2015



Vane Shear Test

Vane shear test is commonly used to determine the undrained shear strength of soft cohesive soils. The test results is fairly reliable.

The vane is pushed into the soil and torque is, then, applied at head of vane rod with uniform speed. The maximum torque to cause soil failure is used to estimate Cu by using below equation.

44

Soil covers top blade3u R 28

3T Cor

πτ =

Soil is not cover top blade 3u R 26

3T Cor

πτ =


HW.

45

Test No.σσσσ3

(kPa)

(∆σ∆σ∆σ∆σd)f

(kPa)

1 25 85

2 60 101

3 105 113

4 154 163

From the CD-Test results of sand sample, please calculate 1. Internal friction angle, ϕ, and cohesion, c, for test 1 & 22. Internal friction angle, ϕ, and cohesion, c, for test 2 & 33. Internal friction angle, ϕ, and cohesion, c, for test 3 & 44. Draw Mohr-Coulomb of each test and each failure envelope in the same graph

3/14/2015



Aristotle

อ<างอง

• Braja M. Das :"Principles of Geotechnical Engineering 5th Edition• V.N.S. Murthy : Geotechnical Engineering• วรยา ฉมนRอย (2553) ผลกระทบของความไมaสมนยตaอพฤตกรรมการรบแรงเฉอนของดนเหนยวกรงเทพ,

วารสารวทยาศาสตรnและเทคโนโลย ปqท 18 ฉบบท 4• เอกสารการสอน อ.อนรทธn ธงไชย• รปภาพจาก Google images

10/18/2015

Aj.Suriyah 1

Soil Mechanics


Lateral Earth Pressure I



2

Lecture Outline:

1. Introduction

2. At rest, active and passive pressure

3. Earth pressure at rest

4. Rankine’s theory of active pressure

5. Rankine’s theory of passive pressure

10/18/2015

Aj.Suriyah 2


Introduction

3

Knowledgeof

lateral forcesthat act

betweenSoiland

Structuresis

Needed


At rest, active and passive pressure

4

Consider a soil element located at a depth zclosed to frictionless wall in below figure. The soil element is subjected to a effective overburden pressure, σ’o, and a effective horizontal

pressure, σ’h.

The ratio of σ’h to σ’o is defined as earth pressure coefficient, K.

3 possible cases may arise depending on the movement of soil mass or retaining wall

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Aj.Suriyah 3



5

If the wall does not move to the right or to the left compared with its initial position, the soil element will be in a state of static equilibrium. In the case, the σ’h is referred to as the at-rest earth pressure and the K is referred to as the at-rest earth pressure coefficient.



6

If the wall moves to the left compared with its initial position, the soil mass BC will reach a state of plastic equilibrium. In the case, the σ’h is referred to as the active earth pressure and the K is referred to as the active earth pressure coefficient.

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7

If the wall moves to the right compared with its initial position, the soil mass BC will also reach a state of plastic equilibrium. In the case, the σ’h is referred to as the passive earth pressure and the K is referred to as the passive earth pressure coefficient.



8

hp = Kpv

ho = Kov

ha = Kav

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9

Change of horizontal stress due to the wall

movement

v is constant

h decreases when soil mass expands (wall move away)

h increases when soil mass is compressed


Earth pressure at rest

At rest earth pressure coefficient, Ka, and overburden effective stress,o, are necessary for computing the horizontal effective stress. Ka can be determined by below equation

10

-------------Loose sand

Dense sand

10/18/2015

Aj.Suriyah 6


Ex.1

11

if wall is static, calculate the lateral force, P0, per unit length of the wall. Also, determine the location of the resultant force from point c

Determine σ’o Determine σ’h Determine u


Ex.1if wall is static, calculate the lateral force, P0, per unit length of the wall.

Also, determine the location of the resultant force from point c

Z

10/18/2015

Aj.Suriyah 7


Rankine’s theory of active pressure

13

)/(K

σK)/(

σ σ

c

)/(

σ σ

)/(c)/(σσ

oa

vaov

ha

oov

ha

oohav

245tan

1

245tan

0c ; sscohesionle is soil if

)2/45tan(

2

245tan

245tan2245tan

2

2

2

2

ha= 3

v = 1 avaha Kc σ K σSo 2 ,


Rankine’s theory of active pressure

14

avaha Kc σ K σSo 2 , What happened ??

This area is subjected the tension force. However, soil is weak when subjected the tension force. consequently, tension crack will occur easily in this part.

So, engineer usually analyzes this kind of problem after tension crack occurs

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Aj.Suriyah 8


Ex.2A retaining wall with saturated clay backfill is shown in below figure.

Determine 1. depth of tension crack 2. Pa before tension crack occurs 3. Pa

after tension crack occurs


Rankine’s theory of passive pressure

16

hp= 1

v = 3

)/(K

Kσ)/( σ σ

)/(c)/(σσ

op

pvo

vhp

oovhp

245tan

245tan

0c ; sscohesionle is soil if

245tan2245tan

2

2

2

pa

pvphp

KKMoreover

Kc σ K σSo

1 ,

2 ,

10/18/2015

Aj.Suriyah 9


Ex.3

17

Determine all lateral forces acting on sheet pile wall as shown in figure

SOIL 1

SOIL 2

= 18 kN/m3

c = 0 , = 38o

= 20 kN/m3

c = 10 kN/m2, = 28o

50kN/m2


aa K2.c. - .K vha

SOIL 1

SOIL 2

= 18 kN/m3

c = 0 , Ka= 0.24

= 20 kN/m3

c = 10 kN/m2, Ka=0.36

50kN/m2

18

Ex.3 Determine all lateral forces acting on sheet pile wall as shown in figure

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Aj.Suriyah 10


SOIL 1

SOIL 2

= 18 kN/m3

c = 0 , Kp= 4.17

= 20 kN/m3

c = 10 kN/m2, Kp = 2.78

pvhp Kc..2.K p

19

Ex.3 Determine all lateral forces acting on sheet pile wall as shown in figure


HW.

20

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Aj.Suriyah 11


HW.

21


Aristotle

อางอง





10/18/2015

Aj.Suriyah 1

Soil Mechanics


Bearing Capacity for shallow foundation


Bearing Capacity

2

Lecture Outline:

1. Introduction

2. Bearing capacity failure

3. Modes of bearing capacity failure

4. Bearing capacity analysis

5. Bearing capacity equation

6. Factor of safety

10/18/2015

Aj.Suriyah 2


Introduction


Technical words• หนวยแรงแบกทาน (Bearing Pressure)• หนวยแรงแบกทานสงสด (Ultimate Bearing Capacity)• หนวยแรงแบกทานทยอมให (Allowable Bearing Capacity)• คาความปลอดภย (Safety Factor)

What is foundation ?? A lowest part of an architectural structure which transfers the loads from superstructure to soil layers beneath the superstructure

Introduction

10/18/2015

Aj.Suriyah 3


Introduction

ขนตอน วตถประสงค ทฤษฎ ขอมล

ออกแบบขนาด/รปราง

(Proportional Design)

ใหฐานรากสามารถสงถายแรงกระทาจากโครงสรางใหดนรบไดอยางปลอดภย

• การวเคราะหหนวยแรงในมวลดน

• การวเคราะหการทรดตวชนดน

• การวเคราะหกาลง แบกทานชนดนฐานราก

• นาหนกบรรทกฐาน

ราก

• คณสมบตชนดนฐานราก

การออกแบบโครงสราง (Structural

design)

ใหฐานรากมความแขงแรง สามารถรบนาหนกบรรทกและแรงตานจากดนได

• การวเคราะหหนวยแรงภายในโครงสราง

• การวเคราะหกาลงรบแรงวสดโครงสราง (e.g. RC design)

• คณสมบตกาลงรบแรงวสดโครงสราง(e.g. คอนกรต/เหลก)

Design of foundation


Introduction

6

Types of foundation• Shallow foundation

o Spread foundationo Wall foundationo Mat foundationo etc.

• Deep foundationo Bore pileo Driven pileo etc.

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Aj.Suriyah 4


หนวยแรงแบกทานสงสดททาใหมวลดนเกดการวบต

= กาลงแบกทาน

= Q3

Bearing capacity failure

Q1 < Q2 < Q3


Modes of bearing capacity failure

Local Shear Failure

ดนฐานรากหลวม

Punching Shear Failure

ดนฐานรากหลวมมาก

General Shear Failure

ดนฐานรากแนน - แนนมาก

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Aj.Suriyah 5


Modes of bearing capacity failure

o Density of soil

o Depth of foundation

Influencing factors


Bearing capacity analysis

o Shear strength of soil, c, o Unit weight of soil,

o Depth of foundation, q = D

o Size and shape of foundation, B

Controlling factors of bearing capacity

สถานะสมดลพลาสตก:

/2) (45 tan 2 /2)(45 tan 231 c

10/18/2015

Aj.Suriyah 6

Ka = 1 / Kp

22/12/32/12/5

22/12/112/3

)(2)(4

1

)(2)(4

1

pppppult

pppppppult

qKKKcKKBq

qKKKcKKKBKq

γqcult NB qN cN q 2

1

22/12/32/12/5

22/12/112/3

)(2)(4

1

)(2)(4

1

pppppult

pppppppult

qKKKcKKBq

qKKKcKKKBKq

10/18/2015

Aj.Suriyah 7


Bearing capacity analysis

Nc, Nq, N = Bearing Capacity Factors

= ขนอยกบคาสมประสทธมมเสยดทาน,

พจนท 1 = กาลงแบกทานจากคาสมประสทธกาลงรบแรงเฉอน, c

พจนท 2 = กาลงแบกทานจากอทธพลของนาหนกดนกดทบ, q

พจนท 3 = กาลงแบกทานจากอทธพลของนาหนกดนใตฐานราก, γ

Bearing capacity :

γqcult NB qN cN q 2

1


Bearing capacity equation by other researchers

10/18/2015

Aj.Suriyah 8


Bearing capacity equation by other researchers

ผวจย เงอนใข รปสมการ

Terzaghi -

Meyerhof แรงกระทาใน แนวดง แรงกระทาใน แนวเอยง

Hansen ทวไป

= 0

bgidsBNbgids Nq bgids cNq qqqqqqccccccult .5 0

qdsSq cuult )(1 5.14 c'

sBNNqscNq qccult 0.5

dsBNdsNqds cNq qqqcccult 0.5

idBNidNqid cNq qqqcccult 0.5


ปจจยทมอทธพล ตวประกอบ สมการทใช

สมประสทธมมเสยดทาน, Bearing Capacity Factors: Nc , Nq , N

Terzaghi, Meyerhof, Hansen

รปรางฐานราก Shape Factors: sc , sq , s Terzaghi, Meyerhof, Hansen

ความลกฐานราก, D Depth Factors: dc , dq , d Meyerhof, Hansen

มมเอยงของแนวแรงกระทา, i Inclination Factors: ic , iq , i Meyerhof, Hansen

มมเอยงของระนาบผวดน, Ground Factors: gc , gq , q Hansen

มมเอยงของระนาบฐานราก, Base Factors: bc , bq , b Hansen

Component factors in bearing capacity equation

10/18/2015

Aj.Suriyah 9


Bearing Capacity Factors by Terzaghi


Bearing Capacity Factors by Meyerhof

10/18/2015

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Bearing Capacity Factors by Hansen


Factor Strip Circular Square

sc 1.0 1.3 1.3

sγ 1.0 0.6 0.8

Other Factors by Terzaghi

10/18/2015

Aj.Suriyah 11


Bearing Capacity Factors by Meyerhof


Bearing Capacity Factors by Hansen

H = Qh = horizontal component of the inclined loadV = Qu = vertical component of the inclined load

10/18/2015

Aj.Suriyah 12


How to choose the bearing capacity equation

Researcher Conditions

Terzaghi (1943) ระดบฐานรากอยไมลกจากผวดนมาก (D/B <1)

ดนรองรบฐานรากเปนดนเหนยว (= 0)

ฐานรากแบบแถบ จตรส กลม

แรงกระทาฐานรากอยในแนวดง

Meyerhof (1963) ระดบฐานรากอยลกจากผวดน (D/B >1)

ดนรองรบฐานรากมคาสมประสทธความตานทานแรงเฉอน c และ ฐานรากสเหลยมผนผา กวาง B ยาว L

แรงกระทาฐานรากอาจเอยงทามมกบแนวดง

Hansen (1970) ใชเหมอนสมการ Meyerhof ในทกกรณ และใชไดเพมในกรณตอไปน

ระดบผวดนอยในแนวเอยง หรอ ระดบฐานรากอยในแนวเอยง


Factor of safety

A factor of safety, Fs, of about 3 or more is generally used to calculate the value of the allowable bearing capacity. An Fs of 3 or more is not considered too conservative because soils are neither homogeneous nor isotropic in nature.

There are two basic definitions of the allowable bearing capacity of shallow foundations. They are gross allowable bearing capacity, qall, and net allowable bearing capacity, qnet.

24where

10/18/2015

Aj.Suriyah 13


Ex.1

25


Ex.2

26

10/18/2015

Aj.Suriyah 14


Aristotle

อางอง






HW.

28

10/18/2015

Aj.Suriyah 15

Kp =tan2 (45+ϕ/2)

Qh = horizontal component of the inclined loadQu = vertical component of the inclined load

10/18/2015

Aj.Suriyah 1

Soil Mechanics


Slope stability I



2

Lecture Outline:

1. Introduction

2. Modes of slope failures

3. Factor of safety

4. Stability of infinite slopes

10/18/2015

Aj.Suriyah 2


Introduction

What is earth slope? ……………………. is a ground surface that stands at an angle with the horizontal plane which can be natural or man-made.

3

Earth slope

Earth retaining structure


Introduction

What is slope failure?

……………………. Is the downslope movement of soil mass in response to a gravitational stresses.

4

10/18/2015

Aj.Suriyah 3


Modes of slope failures

It can be classified the slope failures into six major categories. They are

1. Fall. masses are detached from steep slope/cliff along surfaces with little or no shear displacement.

2.Topple. This is a forward rotation of soil and/or rock mass about an axis below the center of gravity of

mass being displaced.

5



3. Slide. Rotational slides (slumps): masses slide outwards and downwards on one or more concave-

upward failure surfaces that impart a backward tilt to the slipping mass, which sinks at the rear and heaves at the toe. Translational (planar) slides: movements occur along planar failure surfaces that may run more-or less parallel to the slope.

4. Spread. involve the fracturing and lateral extension of coherent rock or soil masses due to plastic flow or

liquefaction of subjacent material.

6

10/18/2015

Aj.Suriyah 4



5. Flow. slow to rapid movements of saturated or dry materials which advance by flowing like a viscous

fluid, usually following an initial sliding movement.

6. Complex. A complex slide involves one of the main types of movement followed by two or more of the

other main types of movement.

7


Factor of safety

For analyzing slope stability, engineer has to determine the factor of safety. The factor of safety, or safety factor, FS, is generally defined as

8

When FS 1 ; slope failsWhen FS > 1 ; slope is stable

Generally, the safety factor of 1.5 with respect to strength is acceptable for design of slope stability

10/18/2015

Aj.Suriyah 5


Factor of safety

=

In case of cohesive soilϕ = 0

In case of cohesionless soilc = 0

So,


Stability of infinite slopes without seepage

10

10/18/2015

Aj.Suriyah 6


Stability of infinite slopes with seepage

11

sincosH satu

HH

H

wsat

22

2,

coscos

cos


Ex.1

10/18/2015

Aj.Suriyah 7


Ex.1


Ex.2

14

10/18/2015

Aj.Suriyah 8


Ex.2

15


HW.

16

10/18/2015

Aj.Suriyah 9


Aristotle

อางอง




• http://www.bgs.ac.uk/landslides/how_does_BGS_classify_landslides.html


ce 371 soil mechanics - civil cmu · 2017-08-09 · ce 371 soil mechanics suriyah thongmunee, ph.d....

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