Geotechnics - Solved Subjects 1-13 (1)

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Subject 1 – soil structure. Structure of clay

Three main phases : solid, liquid and air (vapourous)

The structure of the soil represents one the important properties of the mineral grains. Soilstructure is responsible for the integrity of the system and for the response to externally applied

and internally induced sets of the forces and fluids.

Soil structure can be defined as the property of soil, which provides the geometric arrangement

of the particles or mineral grains and the antiparticle forces which may act upon them.

Four types of structures : grained, honeycomb, floccules and mixed

Subject 2- soil structure

Soil texture may be defined as the visual appearance of a soil based on a qualitative

composition of soil grain sizes in a given soil mass.

The relative sizes and shapes of particles, along with their distribution, define the soil

texture.

Soil texture is used for the classification of soils based on a visual grain description and

connection of the particles, which compose them (cohesionless or cohesive).

Soils are divided into coarse – grained and fine – grained soils on the basis of their texture,

and the dividing reference size is that which is visible to the naked eye (about 0.05 mm).Sands and gravels, in this respect, appear to be coarse textured.

Cohesionless soils can be : homogeneous and inhomogeneous

Cohesive soils can be : homogeneous, layered and in lens form

Subject 3 – grain size distribution. Sieve analysis

The solid phase of soil is an inhomogeneous material consisting of three different phases. To

properly classify a soil one must know the grain – size distribution on it.

To obtain the grain size distribution of a soil in laboratory such methods are used:

The sieve analysis of coarse – grained soil:

The hydrometer analysis for fine – grained soil.

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For the sieve analysis we will use:

-

A set of standardized sieves of sizes varying from 80mm to 75 micro meters

-

A balance or a scale capable to measure 0.1% of the weight sample

-

A sample of soil which is dried at constant weight

-

Sample should have 500g-

We will weigh the material left on each sieve and if that material is greater than 20% from the

original sample we will have to use the Hydrometer Analysis method.

Subject 4 – Hydrometer analysis (Stoke’s law)

This laboratory method is based on the principle of sedimentation of soil particles in water.The use of an immersion hydrometer to measure the specific weight of the liquid is well known.

The principle can be extended to the measurement of the varying specific weight of a soil

suspension as the grains settle, thereby determining the grain size distribution diagram.

The diameter of the soil particle still in suspension at time “t” can be determined by Stoke`s law.

The ternary diagram is useful to identify the soil, giving a name to each of them and to classify

the soils.

The finer percentage in each sieve determined by a sieve analysis is plotted on semilogarithmic

graph paper.

The grain diameter, d, is plotted on the logarithmic scale, and the finer percentage is plotted on

the arithmetic scale.

Subject 5 – Uniformity coefficient. Soil separate size limits

From the grain – size distribution curves of coarse – grained soil two parameters can be

determined:

The uniformity coefficient (Un or Cu)

%10

%60

d

d U n

The coefficient of gradation (Cv or Cz)

%10%60

2

v%30

d d

d C








Subject 6 – importance of the mineral composition of clay

The clay minerals, commonly found in soils, belong to the larger mineral family termed

PHYLLOSILICATES , which also contains other layer silicates. The clay minerals usually occur

in small particle size.

The two basic units in the clay minerals structures are the SILICA TETRAHEDRON , with atetrahedral silicon ion coordinated with four oxygen atoms, and the ALUMINIUM or

MAGNESIUM OCTAHEDRON , wherein aluminum or a magnesium ion is octahedral

coordinated with six oxygen atoms or hydroxyls.

KAOLINITE is composed of alternating silica and octahedral sheets. The tips of silica

tetrahedral and one of the planes of atoms in the octahedral sheet are common.

HALLOSYTE is a particularly interesting member of the kaolinite subgroup.

Two distinct forms of this mineral exist: one a no hydrated form having the same structural

composition as kaolinite and the other a hydrated form consisting of unit kaolinite layersseparated from each other by a single layer of water molecules.

ILLITE perhaps the most commonly occurring clay mineral found in the soil encountered in

engineering practice has a structural similar to that of mica and is termed “illite” or “hydrous

mica”. The basic structural unit for illite is the three-layer silica-gibbsite-silica sandwich that

forms pyrophyllite.

MONTMORILLONITE has a structural consisting of an octahedral sheet sandwiched

between two silica sheets

Subject 7 – Types of water in soil

The kinds and properties of water in soils may vary depending on its content and on the forces of

interaction of water with mineral particles, which are mainly determined by hydrophility of these

particles.

The types are :

The first criterion refers to the state of aggregation of water .Using this criterion we can

distinguish the following forms of water:

-

fluid water (liquid)

- solid water (ice)

- gaseous water (water vapors)

Another criterion is given by the forces which on the water molecules namely the nature

of the force fields which act on these molecules.








The structural form of water molecule is:

- chemical bound water in two forms: hydratation water (constitution water,

crystallization water) and zeolitic water;

- physical bound water in two forms: hydrocolloid water and gravitational water; The

physical bound water is due to electro molecular forces that develop between the soils

mineral grains and the water molecules called also hydrocoloidal forces.

These forces are the result of the interaction between the solid grains and the water

molecules and give peculiar properties to the soil.

- free water in two forms: - capillary water(surface tension) and gravitational water. The

forces which act on the capillary water are the gravity and the surface tension, forces that

in a certain moment can balance each other.

Subject 8 – Indices for soil characterization

In nature, soils are three – phase systems consisting of solid soil particles, water and air (or

gas). In order to develop the weight – relationships for a soil, the three phases can be separated as

shown in Figure. 2.9.

Based on this, the volume relationships can be defined as follows:

To determine the physical properties of soils we have to know three simplest characteristics:

specific weight, s , of solid particles of the soil;

unit weight, , of the soil of natural structure;

natural moisture content or water content w, of the soil.

the specific weight, s , can be defined as follows:

S

S s

V

W (2.3.)

where:

WS – the weight of the solid particles;VS – specific volume or volume of soil solids.

unit weight, , of the soil of natural structure , can be defined as follows:

V

W (2.4)

where:

W = Ws + Ww + Wa total weight of a soil specimen;








V = Vs + Vv – total volume of soil.

the moisture content, w, can be defined as follows:

100s

w

W

W w (2.5)

where:wW - weight of water;

sW - weight of the soil solids.

The weight relationships are moisture content, moist unit weight, dry unit weight and

saturated unit weight.

More useful relations can now be developed by considering a representative soil specimen in

which the volume of soil solids is equal to unity.

Porosity, n, is ratio of the volume of voids to the volume of the soil specimen, or

V

Vn v [%] (2.6)

where:

V = total volume of soil.

vV = volume of voids;

sV = volume of solids soil.

Void Ratio, e, is the ratio of the volume of voids to the volume of solids soil in a given soil

mass and can be written as,

v

s

Ve

V (2.7)

where:

It can also be seen that:

e1

e

V

V

V

V

V

V

VV

V

V

Vn

s

v

s

s

s

v

vs

vv

(2.8)

Degree of Saturation, S, is the ratio of the volume of water in the void spaces to the

volume of voids, and it is generally expressed as a percentage. So:

v

w

r V

VS (2.9)

where: wV = volume of water and vV - total volume








Subject 9 – Loose and dense states for cohensionless soils

Physical state of cohesionless soils may be appreciated function of porosity amount expresses by

the void ratio, which also defined the extreme states: the loosest state and the dense state.

In granular soils, can be measured relative density Dr , or density index ID, by the relation:

minmax

max

ee

ee I D Dr

(2.20)

where: maxe = void ratio of the soil in the loosest state;

mine = void ratio in the densest state;

e = in situ void ratio.

The relative density of soil

Consolidation state Dr

Loosened 0.33

Average consolidation0.34…0.66

Compacted0.67…1.00

The relative density can also be expressed in terms of dry unit weight, or

d

d

d d

d d

r D

max

minmax

min

(2.21)

where: d in situ dry unit weight;

maxd dry unit weight in the densest state – that is, when the void ratio is mine ;

mind dry unit weight in the loosest state – that is, when the void ratio is maxe

.

Also, the consolidation capacity iC determined with the relation 2.22, allows the following

classification of soils:

min

minmax

e

eeC

i

(2.22)

The foundation on sands with D I > 0.5 can be …. without a supplementary consolidate.








The consolidation capacity of soil

Consolidation capacity Ci

Small 0.40

Average 0.40 … 0.60

Large 0.60

The density of sand

Subject 10 – Plasticity of soils. Atterberg’s limit

To characterize a clayey soil from this physical point of view, we use following property indices:

Liquid limit wL, plastic limit w p and shrinkage limit ws, called also Atterberg’s limits or

characteristic humidities.

The liquid limit (wL) is the water content about which the soil behaves as a viscous liquid (a soil,

water mixture with no measurable shear strength).

The plastic limit (w p) is the water content below which the soil no longer behaves as a plastic

material and when it is worked crumples.

Shrinkage limit (ws) is the water content defined as the humidity below which no further soil

volume change occurs with ………. drying.

The behavior of cohesion soils depends on its mineral composition the water content, the degree

of saturation and its structure.

Kinds of Sand Dense Medium Dense Loose

The void ratio (e)

Gravelly, coarse-grain and

medium-grain sands<0,55 0,55 – 0,70 >0,70

Fine-grain sands <0,60 0,60 – 0,75 >0,75

Silt sands <0,60 0,60 – 0, 80 >0,85








Using this method the upper plastic limit can be defined as being the humidity at which a

slot made in the soil paste in the bowl of the device is closed on a length of 12 mm after 25 drops

of the cup, from a height of 10 mm with a frequency of 120 drops per min.

The plasticity index expresses quantitatively the soil plasticity and is calculated with the

relation:I P =w L -w P

The plastic limit is the water content which a soil element will start to crumble when rolledinto a pencil shape of 3.0mm diameter.

The consistency index expresses the physical state of a cohesive soil, which depends onhumidity and the value of this coefficient is determined with the relation:

I C =P L

L

ww

ww

The difference between the liquid limit and the plastic limit of a soil is defined as the

plasticity index (P.I.), or (I.P.):

() = − = − (2.25)

The current state, in terms of Atterberg limits, is defined by the liquidity index LI.

LI =

(2.26)

The amount of water that is bound to a clay surface dependents on the type of mineral, and

this phenomenon is accounted for by the activity defined as (Skempton 1953):

A

0.002

PII =

A

where: 0.002 2A A

activity index (2.27)

PI = 0.73(LL- 20)

PI – plasticity index; LL-liquidity index.

Figure 2.15 represents the relation between the volume and weight of on soil sample at its

different stages of plasticity, beginning from the dry state and ending at liquid state.

It shows clearly the different ranges of plasticity.








Definition of Atterberg’s limit

Subject 11 – Capillarity and capillary effects

We suppose that the capillary menisci are an elastic membrane. If on its two faces act the

same forces, its shape its plane. If the external pressing, e p is bigger than the internal pressing,

i p with u , the menisci tensions S T appear.

The capillary action is attributed to electro – mechanical forces existing between the water

molecules. If the adhesion forces between a liquid and any other material are larger than the

intermolecular attraction of the liquid, the surface of the dissimilar material will be “wetted” byliquid.

Any quantity of liquid will behave as the surface of a tightly stretched skin due to the

intermolecular attractive forces in the interior. This phenomenon is termed surface tension. Since

surface tension is a material property of liquids and depends on intermolecular attraction it will

be temperature dependent.

In the capillarity rise in a tube is considered there is an atmospheric pressure. It is well-

known fact that when capillary tube is placed in water, the water level in the tube rises. This is

caused by the surface tension effect.

When in a hollow open ended tube is inserted into a container of liquid and if the liquid wets

the contact surface it will climb the inside walls of the tube because of surface tension. The water

in the capillary tube is submitted to stresses.








r

T

r

T uS S

u

cos**2*2

1

The weight of the capillary water is transmitted to the skeleton so that its weight increases

the weight of the skeleton with the weight of the water comprised in the capillary rise. Thus

results the capillary pressure.

Since surface tension is a material property of liquids and depends on intermolecular

attraction it will be temperature dependent.

So for equilibrium 0h

F and 0vF

The behavior of cohesive soils depends on its mineral composition, the water content, the

degree of saturation and its structure.

If the equilibrium of the water column in the capillary tube is now considered, the downward

acting force is the weight of the water, and the upward acting force is the vertical component of

the reaction of the meniscus along the circumference (figure 2.18).

cos4

2

scw T d hd

(2.29)

and for a glass tube it follows that:

mm

md d

T h

w

s

c

03.04

(2.30)

The pore pressure in the capillary tube above the actual outside level is negative and itsvalue is obtained by considering that for any elevation z must be: (figure 2. 19)

0 zu w (2.31)

from which:

zu w z (2.32)

for z = 0 to ch zu w (2.33)

and z = 0 u=0

2.9.3 Vapor pressure

In a given soil mass, the interconnected void spaces can behave like a number of capillary

tubes with varying diameters. The surface – tension force may cause water in the soil to rise

above the ground water table as shown in the height of the capillary rise will depend on the

diameter of the capillary tubes.










v - velocity (cm/sec)

k – coefficient of permeability of soil (cm/sec)

The hydraulic gradient (i) may be defined as:

i =∆

(2.36)

where: h - piezometric head difference between the sections at AA and BB

L - distance between the sections AA and BB;

Figure 2.20. Geometrical significance of the Darcy’s law

These sections are perpendicular to the direction of flow.

Subject 13 – Compression and consolidation. Phenomenon definition

The mechanisms will be evoked the deformations of soils are:

- volume changes due to the extrusion of pore air and pour water;

- shear distortions producing particle and fabric unit displacement or yield phenomena with

or without measurable pore water extrusion and with development of slip planes.

Compressibility describes the volumetric response behaviour of the soil mass, and recent

and common usage of the term has restricted it to describing behaviour characteristics under

compression.

The change in volume of a soil mass with time due to the extrusion of pore water is said to

be a process of consolidation

The overall load-volume change performance is identified as a stress - strain -time phenomenon

and can be called rheological behaviour.








Subiect 14. Establishment of compressibility “in situ” with plate

In situ test

The establishing of compressibility "in situ" with plate

Figure 3.6 –Loading plate

The test results are plotted in a diagram which contains:

- The variation of clear pressure ( pn) functions of time;

- The variation of settlement (s) functions of time;

- The variation of settlement (s) functions of clear pressure.

= = / = = /

Figure 3.7 - The establishing of the limit pressure of proportionality.

With this diagram, we establish the limit pressure of proportionality (p1) until there is

proportionality between p and s.

n ef f p p D (3.16)

- rigidity factor

lS

l

pk tg

s (3.17)








iS

i

pk tg

s (3.18)

1 11.5( )i i i is s s s (3.19)

- linear deformation module2

2

s

(1 ) DE=k D(1 ) l

l

p

s

(3.20)

= form factor

Subiect 15. Lab tests and and mechanical values for compression indices

Laboratory test’s

A one-dimensional consolidation process can be simulated in laboratory by compressing a

soil specimen in a special testing apparatus called oedometer or consolidometer.

Figure 3.2 Oedometer

This apparatus models the behavior of a soil volume at a certain depth beyond the axis of afoundation.

The load is applied step by step and after the application of a certain load one waits until the

deformation due to this load stops.

The result of the oedometer test is plotted in the compression - settlement curve.

Compressibility tests of soils are carried out in devices with rigid walls (oedometers) in order to

ensure that the soil is compacted in one direction only (to prevent lateral expansion).

The relationship between moisture content and pressure can be represented in the diagram which

is called the compression curve








Subiect 16. The compression settlement curve, geotechnical indices

The mechanism will be evoked the deformations of soils are:

- volume changes due to the extrusion of pore air pour water

- shear distortions

2 1

p2 p1

pM=

p p

h

h

[kPa] (3.1)

The compression curve can be easily reconstructed into “void ratio – pressure” coordinates

(figure 3.4).

Result the compression – void ratio curves (e – p and e – log p).








Subiect 17. The compression void ratio curve, geotechnical indices

Figure 3.4 Compression - void ratio curve

Coefficient of compressibility

0

v v 0

1(1 )

100a m (1 )

ee

e p p

(3.2)

where: - Initial void ratio

Figure 3.5

– Mechanical model

The correlation between pressure and void ratio can be deduced in the following way.

For the sample having the transversal section equal to A and the volume V, we can write:

V A h (3.3)Loading this sample which the pressure p it’s volume will decrease.

(1 ) (1 ) V

S V S S

S

V V V V V V e

V

(3.4)

where: V = total volume [cm]








A=/ 4 total surface [cm]

Vs = volume of the skeleton [cm]

Vv = volume of voids [cm]

For the pressure p1, the volume results:

1 1 1 1

(1 ) S S

V A h V V V e (3.5)

1(1 ) (1 )1 S S S V=V-V V e V e V e

(3.6)

1 11

1 1 1 1 1

- (1 ) - (1 )-

(1 ) 1

h h s s

h s

A A V e V eV V V e

V V A V e e (3.7)

1 1 1

1 1 1

- 1 -1- -

1 1

h h e e e e

h e e (3.8)

1

1 1

h

h 1

e e

e

or

11

1

h h1

e e

e

(3.9)

where:1 2 1

1 2 1

h

h 1

e e p p

e p p

(3.10)

1

1 2 1

2 1

h 1

1h

e e

e p p

p p

(3.11)

Result:

v1v

2 1 1 1

a1 1m

1 1 M

e e

p p e e

2cm

daN

(3.12)

Subiect 20. Ultimate shear resistance of soils at a direct shear test

The Direct Shear Test

This test is simple on principle. The soil specimen is encased in a shear box (Figure 3.18).

This has been divided into two halves in order to allow relative movements. After a normal load

is first applied to the specimen, one half of the box is pushed horizontally, and the other half

remains fixed. This way the vertical and the horizontal displacements can be measured.








Figure 3.18 – Schematic diagram of the direct shear test

The major disadvantage of this test is that it is impossible to control the drainage. The state

of stress cannot be completely determined because the only stresses that are known are the

normal and the shear stresses on the horizontal plane.

Subiect 18. The yield or failure criteria. Mohr-Coulomn failure criteria

Yield and Field Criterion

General Introduction

By applying the Mohr – Coulomb failure criteria one can determine the shear strength of a

soil (s).

The effective stress is calculated with the equation:

' tans c (3. 36)

where:σ - effective normal stress on plane of shearing;

c - cohesion (or apparent cohesion);

- internal friction angle.

For the most day – to – day work, the shear strength parameters of a soil are determined by

standard laboratory tests: the direct shear test, the triaxial test,the unconfined compression test.

The normal and shear stresses at failure can be determined as:

σ =

and τ = s=

(3. 37)

where:

A = area of the failure plane soil

For sand f or s tg (3. 38)

clay f or s tg c








Figure 3.13 – Stress state in the soil mass

The locus of all (σ,τ) referring to all plans passing through L point is Mohr’s circle.

Several tests of this type can be conducted by varying the normal load. The angle of friction

of the sand can be determined by plotting a graph of s vs. sigma.

Subiect 19. General terms of clay status in soil. Mohr’s Circle

If we have to know the main directions knowing the efforts, we plot the Mohr’s circle: Aand B points have these efforts as co – ordinates. Then we plot the point by drawing parallels to

the co – ordinate.

Figure 3.15 – The establishing of the principal stresses








This theory may be applied also to the case spatial state of stresses.

*

Figure 3.16 Mohr’s representation of stresses in three dimensional systems

For sands, the angle of friction usually ranges from 260 to 45

0. It increases with relative

density of compaction .The approximate range of relative density of compaction and the

corresponding range of the angle of friction for various sands are given in Table 2.4, chapter two.

Mohr envelopes are often curves.

Figure 3.17 – Mohr’s representation of stresses in the state of rupture

The analytical form of the failure criterion can be expressed in several manners (forms).

- Skempton, Bishop(gives straight line);

- Rendulic, Henkel(gives straight line);

Coulomb expressed in its most general form the failure surface:

- The total stress failure envelope can be defined by the equation:

s= c +tanϕ

Subiect 21. Unconfined compression test of cohesive soils

The Unconfined Compression test

The unconfined compression test is a special type of unconsolidated-undrained test. In this test,

, the major principal total stress is applied vertical to cause failure (Figure 3.21 ,a ) and stresses

= = 0.

The corresponding Mohr’s circle is shown in (Figure 3.21, c)








Figure

3.21 – Unconfined compression test

a) soil specimen b)Variation of qu with the degree of saturation

c) Mohr’s circle for the test;

The shear strength of saturated clays can be given as:

s= c =

for ϕ = 0

where: q = the axial stress at failure

s = shear strength

c = cohesion force

The unconfined compression strength can be used with the specifications on the consistency of

clays from the table below

Subiect 22. Triaxial compression test of soils

The principle of triaxial test:

By referring to a triaxial test, the strength parameters can be obtained by means of:

Consolidated – Drained test (C.D.)

Consolidated – Undrained test(C.U.)

Unconsolidated – Undrained test(U.U.)

For clays, three main types of test can be conducted by means of triaxial equipment:

Consolidated drained tests:








Figure 3.20– a) Consolidated

– drained test

The shear strength parameters ( ) can now be determined by plotting Mohr’s circle

at failure as shown in Figure 3.20 (a).

σ +Δσ = σ = σ

′

(3. 40)

σ = σ′ - major principal effective stress

σ = σ′ - minor principal effective stress

Consolidated Undrained Testes:

The total stress Mohr’s circles of test this type and effective stress can be presented in Figure

3.20, (b).

The Unconsolidated Undrained Test:

The total stress Mohr’s circle at failure can be determined with the relation:

+∆ - the major principal total stress - the minor principal total stress

when :

∆ = − = constant

=0 and Coulomb’s line is horizontal

The shear stress for this condition can be given as:

= =

(3. 42)

Subiect 24. Stress distribution in half-space action of a concentrated force

State of stresses in half space

Boussinesq, in 1885, developed the mathematical relationships for the determination of the

normal and shear stresses at any point. It is considered that the medium is homogeneous elastic,

and isotropic and due to concentrated point load located at the surface.

In figure 4.1 are according to his analysis for a vertical punctual load (at point M).








Figure 4.1 - State of stresses in

half-space vertical punctual load

Subiect 25. Distribution of stresses in horizontal and vertical planes (punctual load)

4.1.1 Vertical punctual load

The stresses in a point of the half – space are:

22

25 33

22

25 33

23 1 2 1

2 3

23 1 2 1

2 3

x

y

R z xP x y z

R R R z R R z R

R z yP y z z

R R R z R R z R

(4.1)

3

55 22 2

3 3 1

2 2

1

z

P z P

R zr

z

2

5

2

5

3

2

3

2

xz zx

yz zy

P xz

R

P yz

R








25 3

23 1 2

2 3 xy yx

R z xyP xyz

R R z R

It can be determined by coordinates z, and r:

2 2r x y

and2 2 2 2 2 2

R x y z r z

where: P = vertical punctual load and

x, y, z = coordinates of the point M.(figure 4.1,c)

Figure 4.1 – (c) State of stresses in half – space vertical punctual load

Subiect 26. Determination of compressive stresses by the corner point method

The method of corner points

This method is used for the determination of compressive stresses, when the surface of

loading can be divided into such rectangles that the point considered becomes a corner point.

Let us explain this by discussing three main cases Figure 4.11.

- Point A is on the contour of the rectangle of the external pressures (a);- Point A is inside the pressures rectangle (b);

- Point A is outside the pressures rectangle (c) and (d).

Case (a) 1 2( )

Z c cK K p (4.20)

1 11

1

,c

a bK f

b z

and 2 2

2

2

,c

a bK f

b z








a) b)

c)

d)

Figure 4.11 Angle Method

Case (b) 1 2 3 4( ) Z c c c cK K K K p

(4.21)

1 11

1

2 22

2

3 33

3

4 44

4

,

,

,

,

c

c

c

c

a bK f

b z

a bK f

b z

a bK f

b za b

K f b z

1 2 3 4( )

Z c c c ck k k k p (4.22)

1 2 3 4( )

Z c c c ck k k k p (4.23)








2cosd

P

d

0

0d

2cos y

P

d

(4.24)

Subiect 28. Determination of the active compression zone by the equivalent layer method

Equivalent layer method

This method combines the solution based on the theory of elasticity and the solution for the

settlement obtained for a confined specimen of soil of constant thickness in which the

distribution of stresses is uniform like in the oedometer.

In Figure 5.6 are shown the two situations for a partially uniform distributed load.

221

1e e

ps h

E

and

21r s pb

E

21

1 2e

h b A b

1i

hh m

n (5.15)

ii zi

i

hs pm (5.16)

Result:1

i n i zii

i

hS p

M

(5.17)

The condition is that for a both foundations the settlements should be equal:

1e

s m A p h p b A M

Subiect 29. Method of elementary layer wise summation

Method of summation of the settlements of the elementary strips (figure 5.4)

We consider a rectangular foundation embedded at a depth Df . The foundation soil is

divided in strips taking into account the stratification and the different values of the modulus of

deformation for each layer, termed elementary layer.

On the vertical of the center of the foundation are computed the stresses in the middle of

each layer. The settlement is equal to the sum of the settlements of each layer.








1

100m

n ziii

i

s h E

(cm) (5.10)

Where:

zi

- the stress in the middle of layer i at depth z;

ℎ - thickness of the layer i;

– modulus of linear deformation at the layer i;

=0.7

STAS 3300 Method’s

When the surface of the foundation has a width larger then 10m we can use following

relation to compute the settlement (after the Romanian Standard a soil is considered

incompressible when E > 1000kPa),Figure 5.5.

21

11

n i in i

i

k k s m p B

E

(5.13)

where:

- net pressure;

– width of the foundation;

– Poisson’s ratio;

– modulus of deformation;

= (

,

) - adimensional coefficient (Table 5.4);

- depth of the point;

- coefficient related to the conditions of work.

Subiect 32. Rankine’s active earth pressure theory

6.3 Rankine’s Method

The lateral earth pressure condition described involves walls that do not yield at all,

however, if a wall tends to move away from the soil a distance x as shown in figure 6.3.

However, with x >0 h will be less than

vk 0 .

The failure condition in the soil mass permit determination of the horizontal stress at this time is

equal to .h

The horizontal stress h is referred to as the Rankine active pressure.

6.3.1 The Rankine’s Active Earth Pressure

The Mohr’s circle corresponding to wall displacements of 0 x and x >0 are show as

circles 1 and 2 respectively in figure 6.3.[b]








Referring to the limit equilibrium equation, it can be calculated the principal stresses for

Mohr’s circle that touches the Mohr-Coulomb failure envelope, which can be given by the

Rankine’s Active Earth Pressure.

Figure

6.3 Rankine’s Active Earth Pressure theory

Subiect 33. Rankine’s passive earth pressure theory

6.3.2 The Rankine Passive Earth Pressure

If

the wall is pushed in to the soil mass by an amount x < 0 as shown in fig.6.6, the horizontal

stresses at a depth z, can be defined as the Rankine passive pressure, or ph p 3 .

Figure.6.6 Rankine’s passive earth pressure theory








Using the preceding equation shows the passive pressure diagram for the wall shown in

figure 6.7.

Figure 6.7 Rankine’s passive earth diagram’s

b1) - q > c : 2 A

p p p p q k c k ; 2 B

p p p p p h k q k c k ;

b2) - q < c : 2 A

p p p p c k q k ; 2 B

p p p p p h k q k c k ; (6.14)

If we consider the soil without cohesion, the relations are:

p p p p p p K cK q zK cK qK z p )( (6.15)

where:

2tan 45

2

pk

= Rankine passive earth pressure coefficient (6.16)

The passive force per unit length of the wall can be determined from the area of the pressure

diagram, or

212

2 p p p pP h K q h K c h K

Subiect 34. Pressure distribution from soil’s own weight (coulomb’s theory)

6.4.1 Coulomb’s Active Earth Pressure

The basic assumptions for the earth–pressure theory, proposed by C.A. Coulomb in 1776 are

as follows:

1.

The soil is isotropic and homogeneous and possesses both internal friction and cohesion;

2. The rupture surface is a plane surface;

3. The friction forces are distributed uniformly along the plane rupture surface and f = tan

(f = friction coefficient); 4. The failure wedge is a rigid body;

5. There is wall friction, i.e., the failure wedge moves along the back of the wall,developing frictions forces along the wall boundary;








6. Failure is a two-dimensional problem considered a unit length of infinitely long body.

Deficiencies Coulomb’s Equations

The principal deficiencies in the Coulomb theory are in the assumption of an ideal soil and

that a plane plan defines the rupture surface.

The calculus based on the Coulomb theory for cohesionless soil is derived from as follows:fig. 6.8.

Figure.6.8 Coulomb’s earth theory

The area of wedge ABC1 of figure 6.8 is the active forces P a and can be determined

applying the law of sines to (fig. 6.8).

)180sin()sin(

W p a

(6.18)

or

)180sin(

)sin(

W Pa

(6.19)

From relation 6.19 it can be seen that value of )( f Pa , that is, all other terms for a given

problem are constant, and the value of aP of primary interact is the largest possible value.

Subiect 36. Culmann’s graphical method for active earth pressure

The graphical solution for determination the earth pressure of soil

Culmann’s solution








Cullman consider wall friction, irregularity of the backfill (either concentrated or distributed

loads) and the angle of internal friction of the soil.

The solution is applicable only to cohesion less soils (modified it can be used for soils with

cohesion).

In this discussion the solution is applicable only to cohesionless soils, although with

modifications it can be used for soils with cohesion.

This method can be adapted to stratified deposits of varying densities, but the angle of

internal friction must be the same throughout the soil mass.

A rigid plane rupture surface is assumed. Essentially, the solution is a graphical

determination of the maximum value of soil pressure, and a given problem may have several

graphical maximum points, of which the largest value is chosen as the design value.

A solution can be made for both active and passive pressure.

Steps in the Cullman solution for active pressure are as follows:

Figure

6.12

Figure 6.13

1. Draw the retaining wall to any convenient scale, together with the ground line, location of

source irregularities, point loads, surcharges, and the base of the wall when the retaining wall is

cantilever type.

2. From the point A lay off the angle ∅ with the horizontal plane, locating the line AC.

3. Lay of the line AD at an angle of with line AC. The angle Ψ is computed as :

where: - angle back of wall makes with the horizontal ;

- angle of wall friction.

4.Draw assumed failure wedges as ABC1, ABC2…ABCn.








These should be made utilizing the backfill surface as a guide, so that geometrical shapes such as

triangles and rectangles are formed.

5.Find the weight Wn of each of the wedges by treating as triangles, trapezoids or rectangle,

depending on the soil stratification, water in soil, and other conditions of geometry.

6.Along the line AC, plot to a convenient weight scale, the wedge locating the points W1, W2

...Wn.

7.Through the points just established (steps 6) draw lines parallel to AD to intersect the

corresponding side of the triangle as W1 to side AC1, W2 to side AC2…Wn to side ACn.

8.Through the locus of points established on the assumed failure wedges, draw a smooth curve

(the Culmann line). Tangent to this curve and parallel to the line AC draw a tangent line. It may

be possible to draw tangents to the curve at several points, if so, draw all possible tangents.

9.Through the tangent point established in step 8, project a line back to the AC line, which is also

parallel to AD.

10.The value of this to the weight scale is Pa, and a line through the tangent point from A, is the

failure surface.11.When several tangents are drawn, choose the largest value Pa.

Subiect 37. Poncelet’s graphical method for active earth pressure

Poncelet Graphical Process

Poncelet has given a graphic method to compute the active and passive earth pressure

based on the rule of Rebhann. The surface of the fill in this case is plane.

1. It is built the natural , , , , , , H slope line BC, angle (internal friction angle).

2. Through the superior edge A is drawn the orientation line which makes with the direction AB

angle , obtaining at the intersection with BC the point D.

3. On the natural slope line is being built a semicircle with the diameter BC.

4. From the point D is rise a perpendicular BC, till meeting the semicircle in point P.

5. Planning the point P on BC in F having as planning center the point B.

6. From F is banded A parallel to the orientation line which meets the free plane surface of the soil

in point G.

A) Without over charge

B) With overcharge

7. Its planning G on BC, in H, with the center in F:

I.1

2a ef P

II.1

2

ef

a pP

8. Are united G with H, is descending perpendicular from G on HT, resulting point I.








Subiect 38. Types of retaining walls

A retaining wall is a wall that provides lateral support for a vertical or near-vertical slope of

soil.

It is common in many construction projects and the most common types of retaining wallmay be classified as follows:

- Gravity retaining walls;

- Semi gravity retaining walls;

- Cantilever retaining walls;

- Counter fort retaining walls.

Gravity retaining walls depend on their owned weight and any soil resting on the masonry

for their stability (Figure 7.1)








Figure 7.1

Gravity Retaining wall types

In many cases, a small amount of steel may be used for the construction of gravity walls,

these walls are referred to as semigravity walls.

Cantilever retaining walls are economical up to a height of about 8 m and are made of

reinforced concrete that consists of a thin stem and a base slab (Figure 7.2 a)

Counter fort retaining walls are similar to cantilever walls except for the fact that, at regular

intervals, they have thin vertical concrete slabs known as counter forts that tie the wall and the

base slab together (Figure 7.2 b).

Figure 7.2

Cantilever retaining wall types

7.2 The design of retaining walls

Consider a vertical wall of height H, as shown in Figure 7.2, retaining a soil having a unit

weight of γ.

Figure 7.3

Approximately dimensions for various components of retaining wall








Subiect 39. The study of the retaining wall stability

7.3 Stability Checks

To check the stability of a retaining wall, the following steps are necessary:1. Check for overturning about its toe ;

2. Check for sliding failure along its base;

3. Check for bearing capacity failure of the base;

4. Check for settlement;

5. Check for overall stability.

7.3.1 Check for overturning

In Figure 7.4 shows the forces acting on a gravity and cantilever retaining wall with the

assumption that the Rankine active pressure is acting along a vertical plane AB trawl through the

heel. Pp is the Rankine passive pressure; its magnitude can be given as.

Figure 7.4 Check for overturning

The factor of safely against overturning about the toe – that is, about point C in Figure 7.4

can be expressed as:

= ∑

∑ (7.1)

where :

∑

= sum of the moments of forces tending to overturn about point C;

∑ = sum of the moments of forces tending to resist overturning about point C.








Subiect 41. Slope stability for the homogeneous soil mass limited by a slipping plane

The factor of safety against sliding may be expressed by the equation:

1,3 R

l

d

F F f

F

(7.2)

Where:

RF = sum of the horizontal resisting forces;

DF = sum of the vertical forces.

Figure 7.5 Check for sliding along the base The thus, the maximum resisting force that can be derived from the soil per unit length of the

wall along the bottom of the base is area of cross section A =B 1.

Subiect 45. Correlation between pressure and settlement (bearing capacity)

This chapter discusses in detail the evaluation of the safe load – bearing capacity and

settlement of foundations.

The soil must be capable of carrying the loads from any engineered structure placed on it.

The shallow foundations must have two main characteristics:1. The foundation should be safe against shear failure in the soil that supports it;2. The foundation should not undergo excessive settlement.

The variation of the load per unit area on the foundation )(nq with the foundation settlement

– represents the limiting shear resistance, or the ultimate bearing capacity of the foundation.








Figure 8.1 Stages of soil

settlements

Model tests have shown that, by increasing the load, the soil beneath the foundation

behaves differently.

In the first phase a) almost only the compression of the soil takes place, the shear stresses

being neglected.

a) Stresses in elastic

state b) Apparition of plastic zones

Figure 8.2

a) Plastic zone

at shallow b) Plastic zones at deep

foundation foundationFigure 8.3 Bearing capacity failure in soil under a rough rigid continue foundation

Increasing the load, the magnitude of these shear stresses at boundaries of the foundations

develops and slides take place. As far as the magnitude of these shear stresses is not too large

and their development in space is restrained, danger does not exist.








Subiect 47. Soil calculus of the limit state of deformation

8.4.2 Computation of deformations limit state DLS

Is made for loads from fundamental groups, corresponding to an ultimate limit state

(ULS), (if the deformations of foundation soil can result in displacements and deformations of

the building, compatible with it structure) or corresponding to a normal serviceability limit state(NSLS), (if the soil deformations hinder the normal exploitation of construction).

Computation of deformation limit state in the foundation soil done by observing the double

condition:

s s in the case of check DLS

t t in the case of check NSLS

where: ,s t are possible displacements or deformations of the construction, due to

displacement of deformations of the foundation soil, computed with fundamental loads for ULS,

respectively NSLS;

-

,s t displacements or deformations permissible for the structure, respectively for the

technological process, established by the designer of the structure or by technological designer.

Computation of possible deformations (possible settlements) is made by applying the

models (examples) of computation given. In this chapter it is provided that this computation is

made without considering the secondary settlement. This settlement will be taken into account if

some important settlement from secondary consolidation can occur. In this case, it will apply the

computation relations known from specialty literature, as well as the experimental values of

consolidation coefficient C .

Subiect 48. Soil calculus to the limit state of bearing capacity

8.4.3 Computation of the limits state of carrying capacity

The calculus of the limit state of carrying capacity (LSCC) is made in the following

situations:

for all constructions founded on difficult soils;

for all constructions founded on saturated soils, subjected to loads, fast applied;

for special constructions, founded on rocky soil;

for all constructions to which the foundations transmit important horizontal loads (H >0,1V), where H and V are respectively, the horizontal and vertical component of the load

on the foundation plate, Figure 8.7;

for all types of constructions whose foundations are placed on slopes or near them;

for existing constructions in which the loading regime is to be modified.The limit state of carrying capacity (LSCC) of foundation soil corresponds to zone II from

Figure 8.1, where the effective tangential stress is equal to the shear strength of the soil that








results in its failure, and implicitly, in the loss of stability of foundation soil and of construction.

This limit state (LSCC) is always an ultimate limit state (ULS).

The calculus of foundation soil for the limit state of carrying capacity is done by observing

the condition:

Q < mR (8.39)

where:

Q – the design load applied on foundation soil, produced from special group actions, that

can be either effective pressure, slipping force, or overturning moment, etc.;

R – design carrying capacity of foundation soil that can be either critical pressure, shear

strength, or stability moment, etc.;

m – working conditions coefficient.

The carrying capacity in the case of direct foundations with horizontal foundation plate is

verified with relation:

' p m pc cr (8.40)

where: pef = V I L’B’ is the medium vertical pressure on the foundation plate resulting from

design loads of special group;

V is vertical compound of the same load;

L’ = L – 2l and B’ = B – 2l are reduced length and respectively reduced width of

foundation plate;

e and e L B are the transverse axes, about longitudinal axis of foundation plate respectively;

mc is the working conditions coefficient equal to 0,9 and cr p is critical pressure.

In the case of stony grounds , pcr r strength of the rock in saturated state. For non –

stony grounds, the critical pressure is determined with relation:' p B N i qN i c N icr q q q c c c

(8.41)

This relation is used when the slope of design loads resultant about vertical ,5 the

foundation plate is horizontal, and the significance is approximately horizontal.

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