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26. Determination of compressive stresseses by the corner points method

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

be divided into such rectangles that the point considered is 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 22

2

,ca b

K fb z

=

p p

I. I .

II.

II .

III . IV .A

A

b

a

b

a

b b

a

a

b a

b

a

1 2

43

1

1 1 2

2

2

3 4

a)

b)

I. I.II . II.

III. III. IV .IV .A

Ab

a

a a b

a

b b b b

b

a

a ab

a

1 2

3

4

1 1

1 2

22

3

3

3

4

4

4

c) d)

Figure 4.11 Angle Method


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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 fb z

a bK f

b z

a bK 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)

27. Stress distribution in the case of planar problem

Using the diagrams of stress distribution over horizontal and vertical sections of soil, for the case

the planar pressure problem it is easy to construct curves of equal stresses isobars.

11z zP = (4.25)

1kg/cm

2

3

4

5

76

89

10

1520

50

P=100t

z=0.5m

z=1m

z=1.5m

z=2.0m

Figure 4.13 Isobars in half space. Punctual load


3/20

Note that the planar problem has a very important property consisting in that al component

stress ,z y

andz in the given plane zOx are independent of the deformation characteristics of

the linearly deformable half-space. The calculus relation for stresses in cartesian coordinate is

being:32 c o s

z

P

R

=

22 sin cosy

P

R

=

22 c o s s i nP

R

=

where: P = the load for a length equal to unity;

= the angle between R and the vertical axis;

R = the distance between the origin and the point considered.

28. Determination of the active compressive zone by the quivalent layer method

Method of equivalent layer

Figure 5.6. Equivalent layer method

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

=

( )

( )

21

1 2e

h b A b

= =

1ih

h mn

= (5.15)

ii zi

i

hs p

m= (5.16)

Result:1

i n izii

i

hS p

M

=

== (5.17)


4/20

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

1es m A p h p b A

M = = 5.18)

29.

30.+ 31.

for cohesionless soil

2

1 3

2

1 3

tan 452

tan 45 2 tan 452 2

c

= +

= + + +

for cohesion soil (6.7)

s=c+tan

s

1

01

T

0

s=c+

tan

c

s

13

01

T

03 02 c*ctg

a) cohesionless soil b) cohesive soil

Figure 6.4 The limit equilibrium state on soil

where: =1 major principal stress = v

3 = minor principal stress = h

Thus:

a) Cohesionless soil

213 1 1

2

tan (45 )2tan (45 )

2

aK

= = =

+

(6.8)


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b) Cohesive soil

213 1 1

2

2tan (45 ) 2 tan(45 ) 2

2 2tan ( 45 ) tan(45 )

2 2

a a

cc k c K

= = =

+ +

(6.9)

where:2tan 45

2ak

=

Rankine active earth pressure coefficient . (6.10)

Result:

a ap h k for cohesionless soil=

; (6.11)

and

2a a ap h k c k for cohesive soil= .

32.

The Rankines Active Earth Pressure

The Mohrs 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, can be calculated the principal stresses for

Mohrs circle that touches the Mohr-Coulomb failure envelope can be given by the Rankines

Active Earth Pressure.

h

v

H

z

45+/245+/2x

Wallmovement

to left

z

Rotation of wallabout this point

h

3 h kv v 1

Shearstress

Normalstress

1

2

3

s=c+

tan

(s)

a) b) Figure

6.3 Rankines Active Earth Pressure theory


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If the displacement of the wall, x > 0, continue to increase there will be a time when the

corresponding Mohrs circle will just touch the Mohr-Coulomb envelope defined by the equation:

ctg += (6.6)

The three circles marked the failure condition in the cohesive soil mass.

The horizontal stress 3 is referred to as the Rankine active pressure. The failure planes in the

soil mass at this time will make angles of

+

245

o with the horizontal.

33.

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 depthz, can be defined as the Rankine passive pressure, or

ph p== 3 .

Figure.6.6 Rankines passive earth pressure theory

h

vH

z

45-O/245-O/2

z

Rotation aboutthis point

x

Direction ofwall movement

O

s=tg+

c

ab

0

c

Normal stress

vh v p =h0 = kh

Shearstress


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36.

The graphical solution for determination the earth pressure of soilCulmanns 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 graphicaldetermination 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:

h

A

D

C

W1

W2

W3

Pamax

Culmann's lineTa

ngent

W2=W1+W1'

=

W

Pa

R

=

B

C1

C2

C3

W1

W1

=

Figure 6.12


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W1W1

W

B

C1

C2C3

Culmann'sline

Tangent

W1

W2W3Pa

max

C

=-=90-

=90W2=W1+W1'

D

A

Figure 6.13

1. Draw the retaining wall so any convenient scale, together with the ground line, location ofsource 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 is computed as :

= +

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

- angle of wall friction.4.Draw assumed failure wedges as ABC1, ABC2ABCn.

These should be made utilizing the backfill surface as a guide, so that geometrical shapes such astriangles 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 AC2Wn 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.


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37.

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 overcharge7. Its planning G on BC, in H, with the center in F:

I.1

2a efP =

II.1

2

ef

a pP =

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

q[KH/m2]

A

C

PB

G

F

A

4

D

O

H nPavPah

Pa

ORIZONTALA

PLANULD

EALUNECARE

DREAPTADEORIENTARE

z

R=OC=OB

ORIZONTALA

R'=BE

R"=FG

5

1

3

6

7

2


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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 a common in many construction projects and the most common types of retaining wall

may 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)

a b c

d e f g Figure 7.1 Gravity Retaining wall types

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

this walls find are referred to as semigravity walls.

Cantilever retaining walls is 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).

b)

wall wall

counterfort

plate

foundation platefoundation plate

a)

wall

spurfoundation plate

Figure 7.2 Cantilever retaining wall types


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39.RETAINING WALLS 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 witch 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.

Ba)gravity wall

A

C B

PaPav

Pah

Pp

qheelqtoe

11c1=0

22c2

H

Df

b)cantilever wall

b'

B

C

Pp

qtoe

A

PaPav

Pah

H'

qheel22c2

B

11c1=0

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.

7.3.2 Check for sliding along the base

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

1,3R

l

d

FF f

F=

(7.2)

Where:


12/20

RF = sum of the horizontal resisting forces;

DF = sum of the vertical forces.

B

Pp

1(kN/m ) 1( C)c 1=0 (kPa)

R'

Pah

Ws 1

D

D'

3o

Ws 2

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.

7.3.3 Check for the concrete section

Check for section 1 1

, =

and (7.2)

where: A = B 1,0 and W =

Pah

Df

W1

Ws2 1o1B'

1.0m

W2

Ws1

W3

PavPa

1o1

Pp

B'

1 1

section1-1

a

Figure 7.6 Stress calculation in the most stressed 1-1 for a weight retaining wall.


13/20

7.3.4 Check for bearing capacity failure

The vertical pressure as transmitted to the soil by the base slab of the retaining wall should

be checked against the ultimate bearing capacity of the soil (figure 7.7).

The pressure distribution under the base slab can be determined by using the simple principles of

mechanics of materials:

12

V M yp

A I

=

(7.3)

61

V e

A B

1max admp p ; 2min 0p (7.4)

where: netM =moment=( )V e of all the forces, function of the central point O2I=moment of inertia per unit length of the base section.

31

12

BI

=

2

By =

= bearing capacity of soil ( , , )

The relationships for the ultimate bearing capacity of a shallow foundation were discussed in

chapter 8.

B/2

1

1

22

Pah

Ws''

Df

Cpmin=pheel=p2

p1=pmax=ptoe

E

W

Ws'

a' a''

1 13

32

2

p4

p3

Bo2

02

B

1.0m

p1>0p2>0

p2


14/20

11

M 1 - 1 = Pa *1 a1h'

Pah'

Section1-1

q1q 3

a' 2

2

M 2 - 2 =(2q1+ q 3)* A a226a'

Section2-2

a ''3

3

W s ''

q = W s '' /a ' '

M 3 - 3 =q * a ' '

2

2

A a 3

Section 3-3

where:

P

- horizontal resulting force of the earth;WW - the soil weight;

W the wall weight.


15/20

40.SLOPE STABILITY

Slopes may be man made and slopes may also be naturally formed. The soil moves from

the high points to a lower level.The several forces produce shear stresses through the soil mass and a movement will occur

unless the shearing resistance on every possible failure surface throughout the mass is sufficiently

larger than the shearing stress. The shearing resistance depends on the shear strength of the soil

and other natural factors.

A stability analysis involves making an estimate of both the failure model and the shear

strength.

In general slope stability is a plane strain problem.

It is usual to investigate a typical cross section which is one unit thick with plane strain

ignoring the perpendicular strains ( and stresses).

The main difficulty in the stability analysis slopes lies in the necessity of relation

calculations for the determination of the most slip surface, respectively of the safely factor.

In order to eliminate this difficulty, this has also repercussions on the accuracy by admitting

certain.

41.

Methods of computing stability homogeneous soil mass limited by a slipping plane.

The factor of safety will be:

tg

tg

h

c

tg

tg

W

LcF

WLctgW

TRF

s

s

+

=+

=

+==

)sin(

sin2

sin

sincos

(9.12)

for 0=c

tg

tgFs = (9.13)

for 0= )sin(

sin2

=

h

cFs (9.14)

A slope is steady if 50,1sF


16/20

43.Fellenius method

The investigation carried out in Sweden (1928), confirmed the surface of failure of earth

slopes resemble the shape of a circular arc.

The method of analysis is as follows.

The soil mass above assumed slip circle is divided into a number of vertical slices ofequal width. The forces used in the analysis acting on the slices are shown inFigure. 9.11

i

bi B C

1:m

bi

hE

E'

T'

Wi

NiTi

ci

Wi

(i)

i

li

i

o'

r

A

Wi Ni

Ti

Clay(i)

1(-)

1(-)

i(+)

i(+)

Ti=WisiniT'=Ntg=Wcos1tg

Ci=cili

0

Active

xi

Passive

r

x

Figure 9.11 Fellenius Method

The forces are:

1.The weight W of the slice;

1= iiii bhW (9.25)

1= iiii bhW

where:

i - total unit weight of soil

ih - average height of slice

ib - width of slice.

2.The normal component of the weight N;

ii

n

n

i WNN cos==

(9.26)


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3.The tangential components of the weight T;

ii

n

i

i WTT sin0

== =

(9.27)

4.The effective frictional ( F ) and cohesive ( C ) resistances.

The frictional force F acting on the base of any slice resisting the tendency of the slice to move

downward is :,tan=NF

or (9.28)

The cohesive force C opposing the movement of the slice and

acting at the base of the slice:

lcC = ' (9.29),

c - the effective unit cohesion;

' - the effective angle of friction.

These forces acting on the base of the slice which is designated asi.

The effective normal pressure ,N acting on the base of any slice is

UNN =' (9.30)

luU =

The moments of the actuating and resisting forces about the point of rotation may be written

as a ratio between the stability moment and actuating moment.

The factor safety sF may now be written as:

iR

TRCFR

M

MF

Tn

i

i

ni

ii

n

ni

r

s

s

=

==

++

==

0

0

)(

(9.31)

where:

M s - the resisting moment;

rM - the actuating moment.

Several trial circles must be investigated in order to locate the critical circle, which is the one

having the minimum value of sF .

'tan)(' = UNF


18/20

44.Friction Circle Method

The principle of the method:

A trial circle with centre of rotation O and radiusR sin, whereR is the radius of the trial

circle, a circle is drawn.

Any line tangent to the inner circle must intersect the trial circle at an angle,

with R.Therefore, any vector representing an intergranular pressure at obliquity to an element of

rupture it must be tangent to the inner circle. This inner circle THOSE is called the friction

circle or the - circle.

The friction circle method of slope analysis is a convenient approach for both graphical and

mathematical solutions.

It was given this name because the characteristic assumption of the method refers to the

circle.

The forces considered in the analysis are:

-The total weight W of the mass above the trial circle. This the force acting through the centreof the mass;-The resultant boundary neutral force U;-The resultant intergranular force,R, acting on the boundary;-The resultant cohesive force C.

To a homogeneous slope, the factor of safety is determined, considering the soil mass as an entire

stiff mass.

We consider two particular cases:

a) For 1=sF and o0= (Figure 9.13 a)We establish necc with the expression

AD

Ccnec = (9.35)

Figure.9.13 .a Friction circle when 00 =

b) For 1=sF and [ ]kPAc 0= We establish nec with the relation :

R

rarcnec sin= (9.36)

Determination of factor of safety with respect to strength Figure.9.13.b is a section of a dam.

AD is the trial arc. The force rW is drawn as explained earlier.


19/20

A

B D

1:m

HW r

R

R

or=Rsinnec

Figure 9.13.b Friction circle when [ ]kPac 0=

The factor of safety is:

MO

LOFs = (9.37)

The o45 line, representing FFc intersects the curve to give the factor of safety sF for this

trial circle.

cnec

nec0

M

L(,c)

Figure 9.13 c Wedge block analysis

9.9 Block method for stability of slope

If the soil mass has some slipping surface, we can obtain the blocks by cutting the mass with

vertical plans in order to obtain for each block an homogeneous slipping surface, with one value

of c and ' .


20/20

li

i1

i1

i+1

i

i

i

i1i1

i-1

i

i+1

Eic i R i

Si

Ti

W

N i

ui

(R i ci)tg i

E+E -

Figure 9.14 Blocks method for the stability of slopes

( ) ( ) ( ) 0coscos'tan' 11 =++ iiiiiiiiiiii EEURlcT

( ) ( ) ( ) 0sinsin 11 =++ iiiiiiiii EEURN (9.38)

( )fiiiiiiii TTfEE +=

,

11 ,,,

where: =iE active pressure for (i+1) block or passive earth pressure of (i+1) block

=iT the active earth pressure for i block

=fiT the shearing force opposing

If:

iE0 i block is instable.

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