§1.Soil Physical Characteristics and Classification ◆ Soil Formation ◆ Composition and Structure of Soil ◆ Soil Model Expressions----Phase Relationships

§1.Soil Physical Characteristics and Classification

◆ 　 Soil Formation

◆ 　 Composition and Structure of Soil

◆ 　 Soil Model Expressions----Phase Relationships 　　　

◆ 　 Relative density of the granular soils

◆ 　 Consistency of cohesive soils

◆ 　 Soil Compaction

◆ 　 Classification

Civil Engineering Department of Shanghai University Soil Mechanics Chapter 1Civil Engineering Department of Shanghai University Soil Mechanics Chapter 1

§1.1 Soil Formation

To the civil engineer, Soil is any uncemented or weakly accumulation of mineral particles formed by the weathering of rocks, the void space between the particles containing water and /or air. Weak cementation can be due to carbonates or oxides precipitated between the particles or due to organic matter.

♦ Residual soil If the products of weathering remain at their original location they constit

ute a residual soil.

♦ Transported soil If the products are transported and deposited in a different location they

constitute a transported soil, the agents of transportation being gravity, wind, water and glacier.



Chemical Process(Crystalline particles)

Physical Process(Single Grain)

Weathering

Temperature

Water

Ice/glacier

Wind

Disintegration (freezing/thawing)

Air

Water

CO2

Liquid (acid, oxygen, etc)

§ 1.2 Composition and Structure of Soil

1.2.1 Particle Size Analysis ♦ Particle sizes vary

considerably from those measured in microns (clays) to those measured in meters (boulders)---a difference of one million! Most of natural soils are composite soils and the distribution of these sizes gives very useful information about the engineering behavior of the soil.

♦ The distribution is determined by separated the particles using two processes—sieving and sedimentation

1.2.2 Sieving

♦ Sieves separate particles in the range between 75mm and 60μm(gravel and sand) and sedimentation separates particles less than 60μm(silt and clay). Particles greater than 75mm are cobble size and not usually include in the tests. They are removed before testing and an estimate made or their proportion.


1.2.3 Sedimentation Test

♦ This test procedure is based on Stockes’ Law, which governs the velocity at which spherical particles settle in a suspension: the larger the particles the greater is the settling velocity and vice versa. The law does not apply to particles smaller than 0.0002mm.The size of a particles is given as the diameter of a sphere which would settle at the same velocity as the particle. The soil sample is pretreated with hydrogen peroxide to remove any organic material. The sample is then made up as a suspension in distilled water to which a deflocculating agent has been added to ensure that all particles settle individually.

♦ The suspension is placed in a sedimentation tube. From Stoke’ Law it is possible to calculate the time ,t,for particles of a certain size D(the equivalent settling diameter) ,to settle a specified depth in the suspension. If, after a the calculated time t, a sample of the suspension is drawn off with a pipette at the specified depth below the surface, the sample will contain only particles smaller than the size D at a concentration unchanged from that at the start of


sedimentation. If pipette samples are taken at the specified depth at times corresponding to other chosen particle sizes the particle size distribution can be determined from the weights of the residues.

Stokes’ Law

d- 颗粒 Gs- 土粒比重 G w- 温度为 T ℃ 时水的比重 ρ-4 时水的密度1.2.4 Particle size distribution curves ♦ The particle size distribution of a soil is presented as a curve on a semi

logarithmic plot, the ordinates being the percentage by weight of particles smaller than the size given by the abscissa. The flatter the distribution curves the larger of the range of particle sizes in the soil; the steeper the curve the smaller the size range.

♦ D10 The size such that 10% of the particles are smaller than that size is denoted by D10 . D10 is defined as effective size.

Other sizes such as D30 and D60 (Limited Size)can be defined in a similar way.

♦ The general slope and shape of the distribution curve can be described by means of the coefficient of uniformity (Cu) and the coefficient of curvature (Cc), defined as follow:

)(.)(

1800

0 T

L

gGGd

wWTS

Particle size distribution curves


♦ The higher the value of the coefficient of coefficient of uniformity the larger the range of particles sizes in the soil. A well graded soil has a coefficient of curvature between 1 and 3.

1.2.5 Structure of Soil♦ The nature of each individual particle in a soil is derived from the minerals it

contains, its size and its shape. These are affected by the original rocks from which the particle was eroded, the degree of abrasion and comminution during erosion and transportation and decomposition and transportation and disintegration due to chemical and mechanical weathering. The various clay minerals are formed by the stacking of combinations of the basic sheet structures with different forms of bonding between the combined sheets.

♦ Structure of clay minerals ♦ Illite( 伊里石 ) has a basic structure consisting of a sheet of alumina

octahedrons between and combined with two sheets of by magnesium and iron and in the tetrahedral sheet there is partial substitution of silicon by aluminum. The combined sheets are linked together by fairly weak bonding due to potassium ions held between them.


♠ Kaolinite （高岭石） consists of a structure based on a single sheet of silica

tetrahedrons combined with a single sheet of alumina octahedrons. There is very limited isomorphous substitution. The combined silica-alumina sheets are held together fairly tightly by hydrogen bonding: a kaolinite particle may consist of over one hundred stacks.

Montmorillonite （蒙脱石） has the same basic structure as illite. In the octahedral sheets is occupied by water molecules and cations other than potassium. There is a very weak bond between the combined sheets due to these ions. Considerable swelling of montmorillonite can occur due to additional water being adsorbed between the combined sheets.


§1.3 Soil Model Expressions----Phase Relationships

♦ Soils can be of either two-phase or three-phase composition. In a completely dry soil there are two phases, namely solid soil particles and pore air. A fully saturated soil is also two–phase, being composed of solid particles and pore water. A partially saturated soil is three-phase, being composed of solid soil particles, pore water and pore air .The components of a soil can be

represented by a phase diagram as show in Fig.1.1a.

(a) Fig.1.1a. (b)


♦ The specific gravity of the solid soil particle (G s) is given by

The bulk density (ρ) of a soil is the ratio of the total mass to the total volume

The unit weight(γ) of a soil is the ratio of the total weight(a force) to the total volume, i.e:

The water content (w ，％ ), or moisture content (m) is the ration of the mass of water to the mass of solid in the soil

S

w

M

Mw


Ws

ss V

MG

V

M

V

Mg

V

W

TEST: The water content is determined by weighing a sample of the soil then drying the sample in an oven at a temperature of 105-110 and reweighing. Drying should continue until the differences bet℃ween successive weightings at four-hourly intervals are not greater than 0.1% of the original mass of the sample. A drying period of 24h is normally adequate for most soils.

The degree of saturation (Sr ，％ ) is the ration of the volume of water to the total volume of void space.

The degree of saturation can range between the limits of zero for a completely dry soil and 1(or 100%) for a fully saturated soil.

The void ratio (e) is the ratio of the volume of voids to the volume

of solids

v

wr V

VS

S

V

V

Ve


♦ The porosity (n) in the ratio of the volume of voids to the total volume of the soil

♦ The void ratio and the porosity are inter-related as follows

V

Vn V

n

ne

1 e

en

1


♦ From the definition of the void ratio, if the volume of solids is 1 unit then the volume of voids is e units. The mass of solids is then Gsρw and, from the definition of water content, the mass of water is w Gsρw. The volume of water is thus w Gs. these volumes and massed are represented in Fig.1.1b

♦ The following relationships can now obtained. ♦ The degree of saturation can be expressed as

♦ In the case of a fully saturated soil, S r=1,hence

e

wGS sr

swGe


♦ The bulk density of a soil can be expressed as:

(1.11)

Or from Equation :

(1.12)

♦ For a fully saturated soil (S r=1)

(1.13)

♦ For a completely dry soil (S r=0)

(1.14)

♦ Equations similar to (1.11) to (1.14) apply in the case of unit weights, for example:

ws

e

wG

1

)1(

e

wGS sr

wrs

e

eSG

1

ws

sat e

eG

1

ws

d e

G

1

ws

e

wG

1

)1(

wrs

e

eSG

1


♦ where γw is the unit weight of water. Convenient units are KN/m3, the unit weight of water being 9.8KN/m3.

♦ When a soil in-situ is fully saturated the solid soil particles(volume1 unit weight Gsγw)are subjected to upthrust(γw) . Hence the buoyant unit weight (γ‘)is given by :

(1.18)

i .e: (1.19)

wswws

e

G

e

G

1

1

1'

wsat '


Example 1 A sample of soil in natural state , given V=210㎝ 3 ， W=0.0035KN ， After oven Drying the soil had a weight of 0.0031KN. If the particle specific gravity was 2.67, compute itsγ,ω, e, n, Sr,γd ,γsat

337.16

1031.0

0035.0m

KN

%9.12

1031.0

101.35.33

3

s

w

W

W

343

10161.11067.2

101.3m

WV

s

ss

•Method 1

V’= 210 ㎝ 3 =0.21×10-3m3, W=3.5N=0.0035KN


3410939.0 mVVV sv

3

3

3

4

4

343

2.9'

2.19

8.14

%6.4210939.0

104.0

104.010

101.35.3

%7.44

81.0

mKN

V

VW

mKN

V

VW

mKN

V

W

S

mW

V

n

V

Ve

sws

wvssat

sd

r

w

ww

s

v


3

3

3

3.9'

3.191

8.141

%5.42

%8.441

81.017.16

7.26%9.1211

1

mKN

mKN

e

e

mKN

w

e

wGS

e

en

we

wsat

wssat

d

r

s

Method 2


For a sample of soil withγs=27kN/m3, Given the volume was 12Cm3 ， the weight was 0.2N, After oven drying the soil had a weight of 0.165N. Determine the e and Sr in natural state.

Method 1

Example 2

%4.59964.0

7.2165.0

165.02.0

,

%4.59

964.0

1012102.0

165.0165.02.0

1271

1

6

3

e

wGSor

V

VS

we

sr

v

wr

s


73.5

27

wW

W

sw

ss

573.0

10

73.5 wV

%4.59%100964.0

73.5%100

964.011

1

v

wr

sv

V

VS

wVe

• Method 2


In its natural condition a soil sample has a mass of 2290g and a volume of 1.15×10-3m3.After being completely dried in an oven the mass of the sample is 2035g. The value of Gs for the soil is 2.68. Determine the bulk density, unit weight, water content, void ratio, porosity, degree of saturation .

Exercise 1

Bulk density,

Unit weight,

Water content,

333

99.119911015.1

290.2m

Mgm

kgV

M

33 5.19500,198.91990m

KNm

NV

Mg

%5.122035

20352290

s

w

M

Mw


52.0152.1199.1

1125.168.211

w

s wGe

%3452.1

52.0

1

e

en

%5.6452.0

68.2125.0

e

wGS sr


For a soil in natural state, given e=0.8, and Gs=2.68.

(a) Determine the moist unit weight, dry unit weight, and degree of saturation.

( b) If the soil is made completely saturated by adding water,what would its moisture content be at that time? Also find the saturated unit weight.

Solution Part(a): the moist unit weight is e

wG ws

1

1

311.18

8.01

24.0181.968.2m

KN

Example P.5Civil Engineering Department of Shanghai University Soil Mechanics Chapter 1Civil Engineering Department of Shanghai University Soil Mechanics Chapter 1

1-4. A saturated soil sample weighs 0.4N and its volume is 21.5 cm3. The weight and the volume are 0.33N and 15.7 cm3after being non-completely (partly) dried in an oven for a period. The corresponding degree of saturation is 75%.Determine the water content w, void ration e and dry unit weight γd before drying.

1-4. 某饱和土样重 0.4N ，体积为 21.5 cm3 。放入烘箱内烘一段时间后取出，称得其重量为 0.33N ，体积减小至 15.7 cm3 ，饱和度为 75% 。试求该土样烘烤前的含水量 w 、孔隙比 e 及干容重 γd 。

解：设烘一段时间后，孔隙体积为 Vv2 ，孔隙水所占体积为 Vw2 ，则：在烘后状态：

在烘前状态：

联立求解得： Vv2=4.8cm3 ， Vw2 =3.6cm3

%752

2 v

wr V

VS

%100)7.155.21(

1010/10)33.04.0(

2

632

v

wr V

VS

3631 3.41010/10)33.04.0(6.3 cmVw

31 6.10)7.155.21(8.4 cmVv

%3610106.10104.0

10106.1063

6

972.06.105.21

6.10

e

3

63

68.13%361

)105.21/(104.0

1 mkN

d

§1.4 Relative density of the granular soils

♦ The relative density is a term generally used to describe the degree of the granular soils. Relative density is defined as

minmax

max

ee

eeDr

Where e max=maximum possible void ratio

e min=minimum possible void ratio

e =void ratio in natural state of soil


♦ According to the equation, the relative density of a soil in it’s densest state (e= e min ) is 100%and the relative density of in it’s densest state (e= e max) is 0. The state of packing can be classified by the Dr :

Loose 0< Dr ≤1/3 Medium dense 1/3< Dr ≤2/3 Dense 2/3< Dr ≤1

A sample of sand, given γ=14.7 kN/m3, w=13%, γdmin=12kN/m3 , γdmax=16.6kN/m3 . Estimate its compaction state.

Example


1,1

1,1

maxmin

minmax

d

s

d

s

d

sd

ee

ew

11

11

maxmin

min

minmax

max

d

s

d

s

d

s

d

s

ree

eeD

Solution:

3

1278.0

13

6.16

126.16

1213max

minmax

min

d

dr

d

dd

dD

So, it is loose.


♦ The equation can also be expressed as

d

dr

d

dd

dD

max

minmax

min

Where the γ dmax ,γdmin ,γd are the maximum, minimum, and natural-state dry unit weights of the soil.

♦ The γdmax ,γdmin ,γd can be acquired by the experiments. Then the relative density can be determined.


§1.5 Consistency of cohesive soils1.5.1 Atterberg limits

♦ Atterberg limits are developed for describing the limit consistency of fine-grained soils on the basis of moisture content. These limits are the liquid limit, the plastic limit, and the shrinkage limit.

♦ The liquid limit LL is defined as the moisture content, in percent, at which the soil changes from a liquid state to a plastic state. The moisture contents(in percent)at which the soil changes from a plastic to a semisolid state and from a semisolid state to a solid state are defined, respectively, as the plastic limit Lpand the shrinkage limit Ls .

Liquid

State

Plastic

State

Semisolid

State

Solid

StateLiquid plastic shrinkage

limit LL limit Lp limit Ls

Moisture contents


TEST——♦ The Atterberg limits of cohesive soil depend on several factors, such as several factors, such as the amount and type of clay minerals and type of absorbed cation.

♦ The difference between the liquid limit and the plastic limit of a soil is defined as the plasticity index PI: PI=LL-PL Where LL is the liquid limit and PL is the plastic limit


粘性土的分类《建筑地基基础设计规范》 GB50007-2002

塑性指数 Ip 土的名称

Ip>17 粘土

10<Ip≤17 粉质粘土

注 : 塑性指数由相应于 76g 圆锥体沉入土样中深度为 10mm 时测定的液限计算而得。

♦ The relative consistency of a cohesive soil can be defined by a ratio called the liquidity index LI. It is defined as

1.5.2 Liquidity Index

PI

PLw

PLLL

PLwLI nn

♦ Where w n is the natural moisture content.

If w n =LL, then LI=1.

If w n =PL,then LI= 0. thus for a natural deposit which is in a plastic state, the value of LI varies between 1 and 0. A natural soil deposit with w n > LL will have a LI greater than 1.In an undisturbed state, these soils may be stable; however, a sudden shock may transform them into a liquid state. Such soils are called sensitive clays.


粘性土的状态 , 可按下表分为坚硬、硬塑、可塑、软塑、流塑。

粘性土的状态《建筑地基基础设计规范》 GB50007-2002

液性指数 IL 状态液性指数 IL 状态

IL≤0 坚硬 0.75< IL≤1 软塑

0< IL≤0.25 硬塑 IL>1 流塑

0.25< IL≤0.75 可塑

1.5.3 Activity

♦ Based on laboratory test results for several soils, Skempton made the observation that, for a given soil, the plasticity index is directly proportional to the percent of clay-size fraction (i.e. percent by weight finer than 0.002mm in size). With this observation , Skempton defined a parameter A called activity as

Where C is the percent of clay-size fraction, by weight.

♦ Activity has been used as an index property to determine the swelling potential of expansive clays.

C

PIA


§1.6 Soil Compaction♦ The degree of compaction of a soil is measured in terms of dry density, i.e. the mass of solids only per unit

volume of soil. It is apparent that the dry density is given by

wd

1

Where ρis the bulk density of the soil ,w is the water content. ♦ The dry density of a given soil after compaction depends on the water content and the energy supplied by the compaction equipment. ♦ The compaction characteristics of a soil can be assessed by means of standard laboratory tests. The soil is compacted in a cylindrical mould using a standard compactive effort. After compaction, the bulk density and water content of the soil are determined and the dry density calculated. For a given soil the compaction process is repeated at least five times, the water


content of the sample being increased each time. Dry density is plotted against water content and a curve of the form shown in Fig below is obtained:

♦ The curve shows that for a particular method of compaction there is a particular value of water content, known as the optimum water content (w opt) at which a maximum value of dry density is obtained. At low values of water content most soils tend to be stiff and are difficult to compact. As the water content is increased the soil becomes more workable, facilitating compaction and result in high dry densities. At high water contents, however, the dry density decreases with increasing water content, an increasing proportion of the soil volume being occupied by water.


§1.7 Classification

1.7.1 The British Soil Classification System♦ The British Soil Classification System is shown in detail in Table

1.4. Reference should also be made to the plasticity chart(Fig.1.6).The soil groups in the classification are denoted by group symbols composed of main and qualifying descriptive letters having the meanings given in Table1.5.


1.7.2The Unified Soil Classification System

♦ In the Unified Soil Classification System, developed in the United States, the group symbols consist of a primary and a secondary descriptive letter. The letters and their meaning are given in Table 1.6.The Unified Soil Classification System, including the laboratory classification criteria, is detailed in Table 1.7 and the associated plasticity chart is shown in Fig1.7.Classification may be based on either laboratory or field test procedures. Soil exhibiting the characteristics of two groups should be given a boundary classification denoted by dual symbol connected by a hyphen.


1.7.3 Chinese classificationCivil Engineering Department of Shanghai University Soil Mechanics Chapter 1Civil Engineering Department of Shanghai University Soil Mechanics Chapter 1

《建筑地基基础设计规范》◈ GB50007-2002

《公路土工试验规程》◈ JTJ 051-93

1.7.3 Chinese classification

A Gravels ♦ The gravels are those which more than 50% of coarse material is of

gravel size(coarser than 2mm). ♦ The classification of the gravels

土的名称颗粒形状颗粒级配

漂石圆形及亚圆形为主粒径大于 200mm 的颗粒超过全重块石棱角形为主 50 ％卵石圆形及亚圆形为主粒径大于 20mm 地颗粒超过全重碎石棱角形为主 50 ％圆砾圆形及亚圆形为主粒径大于 2mm 的颗粒超过全重角砾棱角形为主 50 ％


B Sands

♦ The sands are those which more than 50% of coarse material is of sand size(finer than 2mm and coarser than 0.075mm).

The classification of the sands

土的名称颗粒级配砾砂粒径大于 2 mm 的颗粒占全重的 25 ～ 50 ％粗砂粒径大于 0 。 5mm 的颗粒超过全重 50 ％中砂粒径大于 0 。 25mm 的颗粒超过全重 50 ％细砂粒径大于 0 。 075mm 的颗粒超过全重 85 ％粉砂粒径大于 0 。 075mm 的颗粒超过全重 50 ％


C Clays

♦ The clays are often classified by the the soils’ age of sediment or the plasticity index PI.

The classification of the clays according to the age of sediment 土的名称沉积年代工程性质老粘性土第四纪更新世及较高的强度和较低的压缩性其以前沉积的粘性土一般粘性土第四纪全新世（文化期以前）工程性质变化大新近沉积的粘性土文化期以来沉积的粘性土欠固结，强度较低

The classification of the clays according to the PI

土的名称粘质粉土粉质粘土粘土塑性指数 PI≤10 10 ＜ PI≤17 PI ＞ 17


D Granular soils♦ The granular soils are t

hose which more than 50% of the

material is finer than 0.075mm. The granular soils are classified by the Plasticity chart. The Plasticity chart was advanced by A.Casagrande in 1948:

Fig-The line A,B,C divided the chart into six sectors .We can know

which class the soil belong to by the value of the PI and LL

which can get from the experiment.

lineA:PI=0.73(LL-20) lineB:LL=50 lineC:LL=28SLM ：粉质低液限砂土 CL: 低液限粘土 CLM: 粉质

低液限粘土ML,MI: 粉土 CIM: 粉质中液限粘 CI: 中液限粘土CH: 高液限粘土 CV: 很高液限粘土


Documents

§1.Soil Physical Characteristics and Classification ◆ Soil Formation ◆ Composition and Structure of Soil ◆ Soil Model Expressions----Phase Relationships