Analyse de Sol

8/11/2019 Analyse de Sol

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University of Tennessee atChattanooga

ENCE 3610 Soil Mechanics

Soil Composition:

Size Gradation of Soil ParticlesPhase States of Soil


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Gradation of Particle Size


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Sieve Analysis

Primarily applied to

granular (cohesionless)

soils

Passes soil sample

through a series ofsieves of varying mesh

fineness

Different portions of soil

with different grain sizepass through each mesh

Distribution of grain sizes

constructed and plotted


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Sieve Analysis for

Cohesionless Soils


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Results of

Sieve Analysis

Note semi-logarithmic scale


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Grain Size and Sieve Opening

Comparisons


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Particle Size Designation and

Coefficients D

xdesignates particle size

for which x percent of sample

has passed

D10

effective sizeparticle

size at which 10% of the

sample has passed

Useful to determine

permeability

Coefficient of Curvature Cc

Another quantity useful in

determining the gradationof a soil

Uniformity Coefficient Cu

Well gradedeven

distribution of different

particle sizesCu > 10

Poorly gradedmost

particles in a narrow sizerangeC

u < 5

Gap Graded some

particle size ranges are

missing

10

60

D

D=C

u

6010

2

30

DD

D=C

c


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Soil Gradation


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53.18

13.0

41.2

10

60

D

D=C

u 29.0)41.2)(13.0(

30.0 2

6010

2

30

DD

D=C

c


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Passing #4 and #200 Sieve

Portion Passing #200 (0.074 mm) Sieve

Measure of whether soil is cohesive orcohesionless (50%)

In this case, portion is approximately 4% ofsample, so soil is definitely cohesionless

Portion Remaining on #4 Sieve

Measure of whether a soil is a gravel or a sand(50%)

Usually taken as a percentage of soil notpassing #200 sieve

For this sample, percentage is (10078)/0.96= 23%, so soil is sand


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Fine Grained Soils: Hydrometer Test

Soil grains ending up on the pan can also

be graded using the hydrometer test

Soil is placed in suspension in water, then the

progress of its sedimentation is used as anindicator of the distribution of particle size

Not used very often in geotechnical analysis;

particle size distribution not as critical with

fine particles and clay minerals


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Angularity


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Angularity

Angularity

AngularSharp Edges

SubangularEdges distinct but well rounded Subrounded

Rounded

Very Round

Angular particled soils generally exhibit better

engineering properties; also can frequently

pass larger particles through a given sieve

size


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Soil Composition

Weight-VolumeRelationships

Water or MoistureContent

Unit Weight or Mass Specific Gravity

Relative Density Particle Size and Shape

Grain Size Tests Sieve Tests (Coarse-

Grained Soils) Hydrometer Tests (Fine-

Grained Soils)

Plasticity and theAtterberg Tests


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Basic Concepts

Soil is a collection of particles that do not form

an totally solid substance

Soil is a combination of: Soil material in particles

Air

Water

The relationship between this combination

defines much of what any particular soil can

do to support foundations


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Phase Diagram

Assumptions and

Definitions:

Weight of air = 0

Dry Soil: Waterweight and

volume = 0

Volume of voids

include all non-soil volume, both

air and water


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Saturated Soil

Saturated Soil: Air

volume = 0

Only water and

solids appear incompletely saturated

soil


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Basic Formulas

solidswaterairtotal V+V+V=V

solidswatertotal W+W=W

solidswatertotal M+M=M

Or

xxx V=W Or xxx V=M


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Systems of Units Used in Phase

Calculations U.S. Units

Most common within

the U.S., generally not

used elsewhere

Basic units of weight: 1pound (lb.) or kip (1000

lbs.)

Basic unit of length:

feet

Will frequently write psf

and pcf instead of lb/ft2

and lb/ft3

SI Units

Technically the official

system everywhere

else

Basic units of weight: Nor kN

Basic unit of length: m

MKS Units

Commonly used inmetric countries

Basic unit of mass: kg

Basic unit of length: m


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Weight and Volume Relationships

xwxx VG=W

xwxx VG=M

In most cases, calculations in soil mechanics aredone on a weight basis.

Exceptions include wave propagation problems

(earthquakes, pile dynamics, etc.)


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Soil Composition

Constants Void Ratio: e = Vv/Vs Porosity: n = Vv/V 100%

Water or MoistureContent: w = Ww/Ws100%

Unit Weight = W/V

kN/m3or lb/ft3 Dry Unit Weight d= Ws/V

kN/m3or lb/ft3

Unit Weight of Solids, s=Ws/Vs, kN/m

3or lb/ft3

Specific Gravity of SolidGs= s/w

Degree of Saturation S =Vw/Vv100%


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Specific Gravity and Density

Unit Weight of Water (w) 62.4 pcf = 0.624 kcf

9.81 kN/m3

Density of Water

1.95 slugs/ft3 1000 kg/m3

1 g/cm3= 1 Mg/m3= 1Metric Ton/m3

Typical Values for SoilParticles

Quartz Sand: 2.642.66

Silt: 2.672.73 Clay: 2.702.9

Chalk: 2.602.75

Loess: 2.652.73

Peat: 1.301.9

Except for organicsoils, range is fairlynarrow

ww

x

M=

W=G


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Example 1

Given:

Total Volume = 1 cu. ft.

Total Weight = 140 lb.

Dry Weight = 125 lb.

Find

Water Content

Wet Unit Weight

Dry Unit Weight

By Definition:

Dry Unit Weight = Dry

Weight = 125 lb/ft3

Wet Unit Weight = Total

Weight = 140 lb/ft3

Solve for Weight of Water

WT= W

s+ W

w

140 = 125+Ww

Ww= 15 lb/ft3

Solve for Water Content

w=Ww/W

s= W

w/125 =

15/125 = 0.12 = 12%

W+W=W


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Example 2

Given:

Total Mass = 18.18 kg

Total Volume = 0.009 m3

Dry Mass = 16.13 kg

Specific Gravity of Solids =

2.7

Find

Wet Density

Dry Unit Weight

Void Ratio

Water Content

Degree of Saturation

Convert masses to weights

Wt= (18.18)(9.8) = 178.2 N

= 0.178 kN

Ws= (16.13)(9.8) = 158.1 N

= 0.158 kN

Compute Weight of Water

Wt= W

s+W

w

.178 = .158 + Ww

Ww= .02 N

Compute Water Content

w = Ww/W

s

w = .02/.158 = .127 = 12.7%

solidswaterairtotal V+V+V=V

solidswatertotal W+WW


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Example 2

Compute Volumes Volume of Water

Vw= W

w/

w

Vw= .02 kN/9.8 kN/m3=

0.00205 m3

Volume of Solids

Vs= W

s/ s= ws/(Gs w)

Vs= 0.158/((9.8)(2.7)) =

0.00597 m3

Volume of Air

Va= V

tV

wV

s

Va= 0.009-0.00205-

0.00597 = .00098 m3

Compute Wet Unit Weight

wet= W

T/V

T= 0.178/0.009 =

19.78 kN/m3

Compute Dry Unit Weight

dry

= Ws/V

T= 0.158/0.009 =

17.58 kN/m3

Void Ratio e = V

v/V

s= (V

w+V

a)/V

s=

(.00205+0.000976)/.00597 =.507

Compute Degree of

Saturation S = V

w/(V

w+V

a)

S = .00205/(.000976+.00205) =.677 = 67.7%


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Example 3

Given Saturated Soil

Void Ratio = 0.45

Specific Gravity of Solids = 2.65

Find

Wet Unit Weight

Water Content

Assumptions

Va= 0

Vt= 1

Vs+ V

w= 1

water= 62.4 lb/ft3

Solve for Volumes

e = Vw/V

s= 0.45

Vw= 0.31 ft3

Vs= 0.69 ft3

Compute Wet Unit Weight Weight of Soils =

wV

sG

s=

(62.4)(0.69)(2.7) = 114 lb

Weight of Water = wV

w=

(62.4)(0.31) = 19.4 lb

Total Weight = 114 + 19.4 =

133.4 lb Since volume is unity, total

weight is also net unit weight =133.4 pcf

Compute Water Content

w = Ww/W

s= 19.4/114 = 0.17 =

17%


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Example 4

Given Well Graded Sand

Specific Gravity of Solids = 2.65

Void Ratio = 0.57

Porosity = 36.5%

Saturated Soil

Find

Wet and Dry Unit Weight of Soil

Solution

Set sample volume = 1 ft3

Total Volume = 1= Vw+ V

a+ V

s

Use porosity to compute volumeof voids

n = Vv/Vt Vv= nVt= (.365)(1) = .365 ft

3

Compute volume of solids Vs= VtVv= 1 - ,365 = .635 ft

3

Vv= Vw+ Va Since soil is saturated, Va= 0

and Vw= .365 ft3

Dry Unit Weight W

s=

wG

sV

s=

(62.4)(2.65)(.635) = 105 lb/ft3

Weight of Water

Ww= wVw= (62.4)(.362) =

22.6 lb/ft3

Wet Unit Weight W

t= W

w+ W

v= 127.6 lb/ft3


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Phase Relationships


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Questions?

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