2014 -2015 Soil Third - الجامعة التكنولوجية · Soil Mechanics Lectures 2014 -2015...
Transcript of 2014 -2015 Soil Third - الجامعة التكنولوجية · Soil Mechanics Lectures 2014 -2015...
Soil
Mechanics
Lectures
2014-2015
Third Year
Students Includes: Soil Formation, Basic Phase Relationship, soil classification,
compaction, Consistency of soil, one dimensional fluid flow, two
dimensional fluid flows, stresses within the soil, consolidation theory,
settlement and degree of consolidation, shear strength of soil, earth
pressure on retaining structure.
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Chapter One
Soil Formation and Basic-Relation ships
Soil
Is any uncemented or weakly cemented accumulation of mineral particles
formed by weathering of rocks, the void between the particles containing
water/ or air. Weak cementation can be due to carbonates or oxides
precipitated between the particles or due to organic carbonates or oxides
precipitated between the particles or due to organic matter.
Depending on the method of deposition, soils can be grouped into two
categories:
1- Residual soils:
The soils which remain at the place of disintegration of parent rock.
2- Transported soils :
The soils, which carried away from their place of disintegration to
some other place by transporting agencies.
The transporting agencies may be classified as:
i) Water ii) wind iii)gravity iv) Ice
So in general soil is formed from disintegration of rocks over laying the
earth crust.
Weathering
Which are usually results from atmospheric processes action
on the rock at or near the earth surface.
rain wind
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1- Mechanical (Physical weathering):
All type of actions that cause a disintegration of the parent rocks by
physical means such as, gravity, wind and water. The product of this type is
rounded, sub rounded or granular, its products called coarse grained soil e.g.
(gravel and sand ) they present in nature in a single grain structure .
Coarse grained soil
• Sand & Gravel
• Cohesion less soil
• It properties are the same as parent rock.
2- Chemical weathering
All types of chemical reactions that occur between the minerals of the
rock and the environment (air, water ---et.) and will end up by disintegration
of parent rock into fine grain particles; these products have different
properties from the parent rock. They present in nature as a lumps of
number of plate like particles.
The physical property of this product does not reflect the same properties of
the parent rocks.
Fine grained soil
• Silt and clay
• Cohesive material
• Its properties do not reflect the same properties of the parent rocks.
Dia: equivalents diameter (mm)
Soil
Sand Silt
���≤
Clay Gravel , Dia
> 2��
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Clay minerals: There are two basic structure units that form types of
the minerals in the clay:
a) Tetrahedral Unit : Consists of four oxygen atoms (or hydroxyls, if
needed to balance the structure) and one silicon a tom.
Elevation Tetrahedral sheet
.جزيئات من ايون ا#وكسجين 4+ السليكون ھيدرات الرباعية تحتوي على جزيئة واحدة من
b) Octahedral Unit (consist of six hydroxyl ion at apices of an
octahedral enclosing an aluminum ion at the center).
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Formation of Minerals
The combination of two sheets of silica and gibbsite in different arrangements
and condition lead to the formation of different clay minerals such as :
1- Kaolinite Mineral :
This is the most common mineral is the kaolin. The structure is composed of a
single tetrahedral sheet and a single alumina octahedral sheet as shown in figure
below:
1- Strong Hydrogen Bond So not affected
by water
2- And its also called China clay
3-
2- Illite has a basic structure consisting of two silica sheets with a central alumina sheet.
There is a potassium bond between the layers.
3- Montmorillonite unit: The basic structural unit is similar to that of Illite.
Highly affected by water
ويتكون عادة من سحق ال ا#يت
Highly affected by water with high shrinkage and Swell and it is called expansive soil.
وعادة تظھر ھذه الحالة في المناطق الصحراوية وشبه الصحراوية
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Clay Particle –water relations:
In nature every soil particle is surrounded by water. Since the centers of positive and
negative charges of water molecules do not coincide, the molecules behave like dipoles.
The negative charge on the surface of the soil particle therefore attracts the positive
(hydrogen) end of the water molecules. More than one layer of water molecules sticks on
surface with considerable force decrease with increase in the distance of the water
molecule from the surface. The electrically attracted water surrounds the clay particle is
known as the diffused double-layer of water. The water located within the zone of
influence is known as the adsorbed layer as shown in figure:
Diffuse double layer
Adsorbed water layer surrounding a soil particle
Clay structures:
1) - Dispersed structure
2) - flocculated structure
Clay Particle
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Distinguish between flocculated and dispersed structures
Flocculated Dispersed
More strength Lower strength
Permeability is higher permeability is less
Low compressibility higher compressibility
Basic Relationships:
Weight Wt = Ww + Ws
Where: total weight of soil
∶ Weight of water
� : Weight of solid
∶ Weight of air ≈0
Volume Vt = Vv + Vs = Va + Vw
+ Vs
�� ∶ Total Volume
��: Volume of Void
�� : Volume of air
�� ∶ Volume of water
�� : Volume of Soild
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1- Unit Weight – Density
�����= �� ������� ������� =
!
2- Water content %
"# % = $$$% ∗ '(()*+# =
,$,% ∗ '((
3- Void ratio , e
e = v/v0 4- Porosity (n%)
1% =3334 ∗ '((
5- Air content A%
5% =3634 ∗ '((
6- Bulk Density (total density), 74 ρ8 =9:/:
7- Dry density ,
7;*< =,%34 (=, #,>⁄ ))*(B=,>) 8- Dry unit weight (C;*<) C;*< =$%34 (BD ,>⁄ )
9- Specific gravity , E%
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E% = 7%7$ = ,% 3%⁄7$ = ,%
3%∗7$
E% = C%C$ = $% 3%⁄C$ = $%
3%∗C$ ( its value range between 2.6- 2.85)
10- Solid Density, 7% 7% =,%3% , C% =$%3%
Some Useful Correlation:
1- S.e = E%. +# 2- 1 = G'HG 3- G = 1'I1
4- 5 = 1(' − %) 5- 5 = GI+∗E%'HG
6- 74 =E%('H+)'HG 7$)*C4 =E%('H+)'HG C$
7- 74 =E%H%∗G'HG 7$)*C4 =E%H%∗G'HG C$ 8- 7% =E%HG'HG 7$)*C% =
E%HG'HG C$
9- 7;*< = E%'HG7$)*C; =E%'HGC$
10- 7GKK. = 7 = 7%64 −7$ 11- CGKK = C = E%I''HG C$
Some typical values of void ratio, moisture content in a saturated condition, and dry unit
weight for soils in a natural state are given in the following table:
Table 1- Void ratio, Moisture Content, and Dry Unit Weight for some Typical Soils in a
Natural State.
Type of Soil Void ratio Natural moisture
content in a
saturated state (%)
Dry unit weight ,C;
( BD ,>⁄ ) Loose uniform
sand
0.8 30 14.5
Dense uniform
sand
0.45 16 18
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Loose angular-
grained silty sand
0.65 25 16
Dense angular-
grained silty sand
0.4 15 19
Stiff clay 0.6 21 17
Soft clay 0.9-1.4 30-50 11.5-14.5
Note: the weight of one kilogram mass is 9.806 Newton
1 kg = 9.806 N
Example- 1: In its condition a soil sample has a mass of 2290 g and a volume of 1.15*10-3
m3. 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 and air content.
Solution:
M� =NO = P.PQR
S.ST∗SRUV = 1990 Z[ �\⁄ = 1.99N]^V
Unit weight ,_ = N]O = 1990 ∗ 9.8 = 19500b �\⁄ = 19.5 Zb �\⁄
Water content , c =N�N� =PPQRIPR\T
PR\T = 0.125de12.5%
_� =f�(1 + ch)1 + i _�
19.5 = 2.68(1 + .125)1 + i ∗ 10
e = 0.538
Porosity, n= kSHk =
R.T\lS.T\l = 0.3490~0.35
p. i = f�. ch Degree of saturation , S=
R.SPT∗P.qlR.T\l = 62.267%
Air content, A = n (1- S) = 0.35( 1-.62)= 0.132
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Example 2: a moist soil has these values : � = 7.08 ∗ 10I\�\, m = 13.95 kg , c =9.8%,f� = 2.66.Determine:
M, Mt , i, u, p(%), vdwx�idyyxz�i{|}~��ie�u{vdwx�idyyxz�i�{ by soild?
Solution:
M = �� =13.95
7.08 ∗10I\ = 1970.3Z[/�\
Mt = ����SH� =SQ�R.\SHR.RQl = 1794.4
�]^V
Mt = f�1 + i M�
1794.4 = P.qqSHk ∗ 1000 e = 0.48
u = i1 + i =
0.481.48 = 0.324
S.e = f�. c S. 0.48 = 2.66 * 0.098 S = 54.3%
Mt =^�O� 1794.4 = ^�
�.Rl∗SRUV �� = 12.7Z[
�� = � −�� �� = 13.95 − 12.7 = 1.25Z[
∴ v� =^��� = S.PTSRRR = 0.00125�\
v� =v� −v� v� = 7.08 ∗ 10I\ − 0.00125 = 0.00583�\
Example 3:In the natural state, a moist soil has a volume of 0.0093 �\ and weighs 177.6
N. The oven dry weight of the soil is 153.6 N. If f� = 2.71 . Calculate the moisture
content, moist unit weight, dry unit weight, void ratio, porosity and degree of saturation.
Solution: ch =���� = S��.qIST\.q
ST\.q = 15.6%
_� =�� =177.60.0093 = 19096
b�\ = 19.1 Zb �\⁄
_t =��� = 153.60.0093 = 16516
b�\~16.52 Zb �\⁄
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e= O�O� ,�� =
������ =
R.ST\qP.�S∗SR = 0.0058�\
∴ �� = 0.0093 − 0.0058 = 0.0035�\e = R.RR\TR.RRTl = 0.6 u = kSHk =
R.qSHR.q = 0.375
p. i = f�. c S. 0.6= 2.71 * 0.156 S = 70.46%
Example 4: A soil specimen has a volume of 0.05 m3 and a mass of 87.5 kg. If the water
content is 15% and specific gravity is 2.68. Determine 1) void ratio 2) porosity 3) dry unit
weight 4) saturated unit weight 5) degree of saturation.
Solution:
M� =��v� = 87.50.05 = 1750Z[/�\
~h =^�^� = 0.15 = l�.TI^�^� �� = 76Z[
v� = ��f�M� = 76
2.68 ∗ 1000 = 0.028�\
e = /�/� =R.RPSqR.RPl = 0.77,n =
�SH� =
R.��SHR.�� = 0.43
_t�� = f�1 + i_� =2.68
1 + 0.77 10 = 15.14Zb/�\
_��� =f� + i1 + i _� =2.68 + 0.771 + 0.77 10 = 19.49Zb/�\
S.e = f�.ch S*0.77 =2.68* 0.15 S= 52.2%
Example 5: Show that _��� =_tH( ����∗��)
Solution: take the right hand side :
_tH� kSHk∗����
��SHk∗���kSHk∗���
��HkSHk ∗�������
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Example 6: Given mass of wet sample = 254 gm, void ratio = 0.6133, volume of air = 1.9
cm3, mass of solid =210 gm. Determine degree of saturation, air content and dry unit
weight.
Solution: mt = 254 gm, ms = 210 g �� = 254 − 210 = 44[�
v� =��M� = 441 = 44y�\
v� =v� + v� = 44 + 1.9 = 45.9 0.6133 = �T.Q�� →∴ v� = 74y�\
p = v�v� = 4445.9 = 95.8% → = u(1 − �) =
0.61331 + 0.6133 (1 − 0.95)
= 0.019
Mt�� =��v�
v� =v� +v�¡� +v� = 44 + 1.9 + 74.84 = 120y�\
∴ Mt�� =210120 = 1.75[�y�\ → _t�� = 17.5Zb/�\
Example 7: A soil specimen is 38 mm in diameter and 76 mm long and its natural
condition weighs 168 gm when dried completely in an oven the specimen weighs 130.5
gm. The value of f� = 2.73. what is the degree of saturation of the specimen?
Solution: Dia = 38 mm = 3.8 cm
L= 76 mm = 7.6 cm
v� = (3.82 )P¢ ∗ 7.6 = 86.192y�\
�� = 168 − 130.5 = 37.5[�
v� =37.51 = 37.5y�\
v� = ~�f� ∗ _� =
130.52.73 ∗ 1 = 47.8y�\
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v� = 86.192 − (37.5 + 47.80) = 0.889y�\
v� =v� +v� = 37.5 + 0.889 = 38.389y�\ ∴ p = ���� =
\�.T\l.\lQ = 97.6%
Example 8: Given: mass of wet sample =254.1gm, void ratio = 0.6133, volume of air =
1.9 cm3, mass of solids = 210 gm. Determine: Degree of saturation, Air content, dry unit
weight.
Solution:
Mass of water = 254.1 -210= 44.1 gm
Volume of water = ���� = 44.1y�\
i = ���� → 0.6133 =
���� =
��H���� = ��.SHS.Q��
0.6133 = 46v� → v� = 75. y�\
p = ���� =��.S�q = 95.8%
A= kSHk (1 − �) =
R.qS\\SHR.qS\\ (1 − 0.958) = 0.0157
v� =v� + v� = 46 + 75 = 121y�\
f� = ~�v�_� =21075 ∗ 1 = 2.8
_t�� = f�1 + i_� → _t�� =2.8
1 + 0.6133 ∗ 10 = 17.355Zb �\⁄
Or Mt�� =^��� = PSRSPS = 1.7355[�/y�\
∴ _t�� =Mt�� ∗ [ = 1.7355 ∗ 10 = 17.355Zb/�\
Example 9: A soil specimen have void ratio of 0.7 , f� = 2.72. Calculate the dry unit
weight, unit weight and water content at degree of saturation of 75%.
Solution :
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_t�� = ��SHk _� =P.�PSHR.� ∗ 10 = 16
�£^V,
_��� =2.72 + 0.71 + 0.7 ∗ 10 = 20.11Zb/�\
_¤ = _ = _k¥¥ =_��� −_� = 20.11 − 10 = 10.11Zb/�\
_�����T% =2.72 + 0.75 ∗ 0.71 + 0.7 ∗ 10 = 19.1Zb/�\
0.75* 0.70 = 2.72 *c → c = 19.3%
Example 10 : Prove that S. e = f�. ch Take the right hand side;
f�. ch = ���� ∗
������
= �������� ∗
���� =
���� ∗
���� = p ∗ i
Example 11: Show that _t�� = ��SH�
��SH� =
�� ��⁄SH�� ��⁄ = �� ��⁄�����
��=
��������= ����
Example 12 : Prove that u = kSHk kSHk =
�� ��⁄SH�� ��⁄ = �� ��⁄
�������
= ���� = i
References :
1- Soil mechanics
R.F. Craig
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2- Soil mechanics
T.W.Lamb, R.V.Whitman
3- Soil Mechanics
Basic Concepts and Engineering Applications
A.Aysen
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Chapter Two
Plasticity of Fine Grained Soils
Plasticity is the ability of a soil to undergo unrecoverable deformation at
constant volume without cracking or crumbling. It is due to the presence of clay
minerals or organic material.
Consistency limits (Atterberg limits):
Atterberg, a Swedish scientist developed a method for describing the limit
consistency of fine grained soils on the basis of moisture content. These limits
are liquid limit, plastic limit and shrinkage limit.
Liquid limit (L.L): is defined as the moisture content in percent at which the
soil changes from liquid to plastic state.
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Plastic Limit (P.L.): The moisture contents in % at which the soil changes from
plastic to semi solid state.
Shrinkage Limit (S.L.): The moisture contents in % at which the soil changes
from semi solid to solid state.
Plasticity Index (P.I.): it is the range in moisture content when the soil exhibited
its plastic behavior:
¦. §. = ¨. ¨– ¦. ¨.
Liquidity Index (L.I. or IL) : a relation between the natural moisture contents
(cª) and (L.L.) and (P.L.) in form:
If LI > 1 Then the soil at Liquid state
If LI = 1 then the soil at L.L.
If ¨§ < 1 then the soil below L.L.
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Activity: is the degree of plasticity of the clay size fraction of the soil and is
expressed as:
y��v��} = ¦. §%d¬yw�}��iz�e��ywi�
كلما زادت الفعالية كلما دلت على لدونة التربة عالية
Plasticity Chart: based on Atterberg limits, the plasticity chart was developed
by Casagrande to classify the fine grained soil.
Some useful notes:
v� ∶ Constant at all stages
Degree of saturation (S %) at S.L. and up to =100%
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Degree of Saturation in the region from S.L. and below < 100%
v�t�� =v���®.¯.-------- v�t�� =v���®.¯. it�� =i®.¯.
Relative Density: is the ration of the actual density to the maximum possible
density of the soil it is expressed in terms of void ratio.
°�(%) = i^�±Iiªi^�± −i^¡ª ∗ 100
Or °�(%) = �²³�´�²µ ∗ �²µU�²³¶µ�²³�´I�²³¶µ ∗ 100
i^�±: The void ratio of the soil in its loosest condition
i^¡ª: The void ratio of the soil in its densest condition
iª: The void ratio of the soil in its natural condition
_t^�±: Maximum dry unit weight (ati^¡ª)
_t^¡ª: Minimum dry unit weight (ati^�±)
_tª: Natural dry unit weight (atiª)
RD Description
0 - S\ loose
13 −
23
medium
23 − 1
Dense
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Example 1: for a granular soil, given,_t�� = 17.3 �£^V, relative density = 82%,
c = 8% and f� = 2.65. Ifi^¡ª = 0.44. what would be i^�±? what would
be the dry unit weight in the loosest state?
Solution:
_t�� = ��SHkµ ∗ 10 17.3 = P.qTSHkµ ∗ 10
∴ iª = 0.53 °� = k³�´Ikµk³�´Ik³¶µ ∗ 100
0.82 = k³�´IR.T\k³�´IR.�� ∴ i^�± = 0.94
∴ _t��(��wdd�i��) = f�1 + i^�± _� =
2.651 + 0.94 ∗ 10
= 13.65 Zb �\⁄
Example 2: a granular soil is compacted to moist unit weight of 20.45 Zb �\⁄
at moisture content of 18% . What is relative density of the compacted soil?
Given, i^�± = 0.85, i^¡ª = 0.42�u{f� = 2.65? Solution:
_ = ��(SH�¸)SHkµ _� 20.45 = P.qT(SHR.Sl)SHk ∗ 10
∴ iª = 0.52 °� = k³�´Ikµk³�´Ik³¶µ =
°� = 0.85 − 0.520.85 − 0.42 ∗ 100 = 76.74%
Example 3: A dry sample of soil having the following properties, L.L. = 52%,
P.L. = 30%, f� = 2.7, e= 0.53. Find: Shrinkage limit, {��density, dry unit
weight, and air content at dry state.
Solution
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Dry sample it�� =i�¹�¡ª��]k = 0.53
∴ p. i�.º =f�. ch�.» 1*0.53 =2.7 *S.L
S.L.= 19.6%
Mt�� = ��SHk M� Mt�� = P.�
SHR.T\ 1 = 1.764]^h^V
∴ _t�� =Mt�� ∗ [ = 1.764 ∗ 10 = 17.64 Zb �\⁄ Case is dry s=0
∴ = u = i1 + i =
0.531 + 0.53 = 0.346
∴ = 34.6%
Example 4: A saturated soil sample has a volume of 20 y�\ at its L.L Given
L.L= 42% , P.L.= 30% , S.L.= 17% , f� = 2.74. Find the min. volume the soil
can attain.
The minimum volume
occurs at S.L. or at dry
state.
v� = v� +v� v�: is constant along all
state.
At L.L.
p. i = f�. ch
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1*e= 2.74 *0.42
i¯.¯.= 1.1508
i = v�v� = 20 − v�v�= 1.1508
∴ v� = 9.3y�\
∴ p. i = f�. c®.¯. i�.¯. = 0.4658
i = ���� v�®.¯. =v�®.¯.= 4.33 + 9.3
∴ v�¯.¯. = 20- 9.3 =10.7y�\
1* i�.¯. = 2.74 ∗ 0.17
v�®.¯. = 0.465 ∗ 9.3 = 4.33y�\
= 13.63y�\
Example 5: A sample of saturated clay had a volume of 97y�\ and a mass of
(0.202 kg). When completely dried at the volume of the sample was (87y�\) and it's mass (0.167 kg). Find
a) - initial water content. b)- shrinkage limit c)- specific gravity
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Solution:
ρ� =��v� =0.20297
= 2.08[�/y�\
2.08 = f� + i1 + i ∗ 1 − − − −− −(1) ch =���� =
202.167167 = 21%
p. i = f�.ch 1*e = f� ∗ 0.21 − − − (2)
Solving (1) and (2)
i = 0.565�u{f� = 2.69
But Mt�� = ��SHk M� ∴ i®.¯. = 0.4
p. i®.¯ = 2.69 ∗ c®.¯.
At dry state : Mt�� = ^��� =1.96[�/y�\
1.92 = P.qQSHk½.¾ ∗ 1
1 ∗ 0.4 = 2.69 ∗ c®.¯. c®.¯. = 15%
3 = ¿À#,>
W= 0.167 kg
3 = ÁÀ#,>
W= 0.202 kg
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Chapter three
Soil Compaction
Soil compaction is one of the most critical components in the construction of roads,
airfield, embankments and foundations. The durability and stability of a structure are
related to the achievement of proper soil compaction. Structural failure of roads, airfield
and the damage caused by foundation settlement can often be traced back to the failure to
achieve proper soil compaction.
Compaction of soil:
Compaction is the process of increasing the density of a soil by packing the particles
closer together with a reduction in the volume of air only. Compaction increases the dry
density and decreases the void ratio.
Purpose of compaction:
1- Increase shear strength of soil
2- Reduce void ratio thus reduce permeability
3- Controlling the swell-shrinkage movement
4- Reduce settlement under working load
5- Prevent the buildup of large water pressure
Factors affecting compaction:
• Water content
• Type of soil
• Compaction energy or effort
All these factors are shown in the following figures:
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The effect of types of soil on the dry density using the same
compaction Energy.
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Different in compaction energy and types of soil
Theory of compaction:
Compaction is the process of reducing the air content by the application of energy
to the moist soil. From compaction test we can find:
1- There is a unique relationship between the water content and the dry density
for specific compaction energy.
2- There is one water content (O.M.C.) (Optimum moisture content) at which
the max dry density is achieved
The two above points can be clearly shown through the following Figure:
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Compaction curve
Compaction Test
The compaction test is performed to determine the relationship between the
moisture content and the dry density of a soil for specific compactive effort. The
compactive effort is the amount of mechanical energy that is applied to the soil
mass. Several different methods are used to compact soil in field, and some
examples include tamping, kneading, vibration, and static load compaction. This
test will be carried out by using impact compaction method using the type of
equipment and methodology developed by R.R.Proctor in 1933, therefore , the test
is also known as the proctor test.
Two types of compaction tests are routinely performed: (1) The standard Proctor
and (2) The modified Proctor test.
Type of
test
No. of
layer
No. of
blows per
layer
Volume
of mold
(#,>) Weight of
hammer
(kg)
Height of
drops
cm
Standard
Proctor
3 25 1000 2.5 30
Modified
Proctor
5 25 1000 4.5 45
Compaction Effort = $4)KÂ6,,G*∗;*)Ã%ÂGÄ=Â4∗D).)KÅÆ)$%∗D).)KÆ6<G*
Ç)ÆÈ,G)K,)Æ;
Test Procedure :
1- a sufficient quantity of air-dried soil in large mixing pan (say 3 kg)
2- Determine the weight of the compaction mold with its base (without the
collar).
3- Start with initial water such (3% of Soil weight)
4- Add the water to the soil and mix it thoroughly into the soil until the soil gets
uniform color (see figure B and C).
5- Assemble the compaction mold to the base, place soil in the mold and
compact the soil in the number of equal layers specified by the type of
compaction method (see photo D and E).
The number of drops per layer is dependent upon the type of compaction.
The drops should be applied at a uniform rate not exceeding around 1.5
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seconds per drops, and the rammer should provide uniform coverage of the
specimen surface.
6- The soil should completely fill the cylinder and the last compacted layer
must extend slightly above the collar joint. If the soil below the collar joint at
the completion of the drops, the test point must be repeated.
7- Carefully remove the collar and trim off the compacted soil so that it is
completely even with the top of the mold.(see photo F).
8- Weigh the compacted soil while it's in the mold and to the base, and record
the mass (see Photo G). Determine the wet mass of the soil by subtracting
the weight of the mold and base.
9- Remove the soil from the mold using a mechanical extruder (see Photo H)
and take the soil moisture content samples from the top and bottom of the
specimen (see Photo i). Determine the water content.
10- Place the soil specimen in the large tray and break up
the soil until it appears visually as if it will pass through the #4 sieve, add 3%
more water on the soil and remix as in step 4. Repeat step 5 through 9 until a
peak value is reached followed by two slightly lesser compacted soil masses.
Analysis:
1- Calculate the moisture content of each compacted soil specimen.
2- Compute the wet density in grams per cm3 of the compacted soil by dividing
the wet mass by the volume of the mold used.
3- Compute the dry density using the weight density and the water content
determined in step 1. Use the following formula:
Mt�� = M�k�1 + ch 4- Plot the dry density values on the y-axis and the moisture contents on the x-
axis. Draw a smooth curve connecting the plotted points.
5- On the same graph draw a curve of Saturation line ( Zero air void line ) using
the following Equation :
Mt�� = ��SHkM�
p. i = f�. ch For S= 1 ∴ i = f�ch Assume values of water content and find dry density, then plot the zero air
void line which must be parallel to the moist side of compaction curve and
never intersect it , If so that mean there is some error.
To plot Air content line (A%) use the following equation:
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Mt�� =f�(1 − )1 + cÉf� M�
The following Figures give the steps used in the test:
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Compaction Equipments:
1- Sheep's foot roller (for cohesive soil)
2- Pneumatic roller (many different soils)
3- Vibratory rollers (mainly for granular material)
4- Grid rollers
5- Power Rammer
6- Vibratory plates.
Compaction Of cohesion less soil:
Moisture content has little or no influence on the granular soils (except when the soil is fully
saturated) .
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Their state of compaction can be obtained by relating dry density to the minimum and
maximum dry densities and as in the following equation:
°� = i^�± −iª��Ê��ºi^�± −i^¡ª = Ë_t^�±_t Ì _t −_^¡ª_ �± −_^¡ª
Where RD : Relative density
Compaction in the field:
The results of laboratory tests are not directly applicable to the field compaction, because
1- The Laboratory tests are carried out on material smaller than 20 mm size.
2- Compactive efforts are different and apply in different method.
Relative compaction:
Or Degree of compaction is a means of comparing the field density with Laboratory results and
is defined as the ratio of the dry density in the field to the maximum dry density in the
Laboratory and in most construction works, the degree of compaction is specified as 95 % or
more.
ÍGÆ64Ä3GÎ),Ã6#4Ä)1Í. Î = C;*<KÄGÆ;C,6Ï64Æ6Å. ≥ ÁÑ%)*6%%ÃG#ÄK<Ä14ÂG$)*B%
So by using sand replacement method, find dry density at field then check the R.C
The Optimum moisture content can be useful in field as follows :
If ch¥¡kºt < cÒÓ� then add water and compact the soil
1.5
1.6
1.7
1.8
1.9
2
2.1
2.2
2.3
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If ch¥¡kºt =cÒÓ� then compact the soil directly
If ch¥¡kºt > cÒÓ� then either postponed the compaction to other time or add some additive
(such as cement or lime) to accelerate evaporation of extra water.
Measurement of field Density
1- Core cutter
2- Sand Replacement method
3- Air-Ball on method
4- Penetrating Needle
5- Radiation Technique.
Example 1 :
The following results were obtained from a standard compaction test. Determine the Optimum
moisture content and maximum dry density. Plot the curves of 0%, 5% and 10% air content and gives
the value of air content at the maximum dry density. Given the volume of standard mold is 1000 cm3
and E% = Ô. À. Mass (gm) 1768 1929 2074 2178 2106 2052 2007
Water content (%) 4 6 8 10 12 14 16
Solution :
Calculate dry density for each test and tabulate the results.
+(%) 4 6 8 10 12 14 16
7$G4 1.768 1.929 2.074 2.178 2.106 2.052 2.007
7;*< =, #,>⁄ 1.7 1.82 1.92 1.98 1.88 1.8 1.73
+(%) A % 4 6 8 10 12 14 16
7;*< =, #,>⁄ 0 2.44 2.32 2.22 2.13 2.04 1.96 1.88
7;*< =, #,>⁄ 5 2.32 2.2 2.11 2.02 1.94 1.86 1.79
7;*< =, #,>⁄ 10 2.20 2.09 2.00 1.92 1.84 1.76 1.69
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From Figure: The C;*<,6Ï = '. Á¿ BD ,>⁄ ,
and the Optimum Moisture content = 10%.
Figure show the Zero air void line and a line of 5 and 10% air content
1.4
1.5
1.6
1.7
1.8
1.9
2
2.1
0 5 10 15 20
Comaction Curve
Comaction Curve
1.4
1.6
1.8
2
2.2
2.4
2.6
0 5 10 15 20
Comaction Curve
Zero air void line
A=5 %
A = 10 %
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Chapter Four
Soil Classification
Classification of soil is the separation of soil into classes or groups each having similar
characteristics and potentially similar behaviour. A classification for engineering
purposes should be based mainly on mechanical properties: permeability, stiffness,
strength. The class to which a soil belongs can be used in its description.
The aim of a classification system is to establish a set of conditions which will allow
useful comparisons to be made between different soils. The system must be simple. The
relevant criteria for classifying soils are the size distribution of particles and the plasticity
of the soil. Particle Size Distribution for measuring the distribution of particle sizes in a
soil sample, it is necessary to conduct different particle-size tests. Wet sieving is carried
out for separating fine grains from coarse grains by washing the soil specimen on a 75
micron sieve mesh.
1- Dry sieve analysis is carried out on particles coarser than 75 micron. Samples (with
fines removed) are dried and shaken through a set of sieves of descending size. The
weight retained in each sieve is measured. The cumulative percentage quantities
finer than the sieve sizes (passing each given sieve size) are then determined.
The resulting data is presented as a distribution curve with grain size along x-axis
(log scale) and percentage passing along y-axis (arithmetic scale).
U.S. Standard sieve sizes
A set of sieves for a test in the laboratory
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Hydrometer (Sedimentation) analysis is based on the principle of sedimentation
of soil grains in water. When a soil specimen is dispersed in water, the particles
settle at different velocities, depending on their shape, size, and weight. For
simplicity, it is assumed that all the soil particles are spheres, and the velocity of
soil particles can be expressed by Stokes’ law, according to which:
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In this method, the soil is placed as a suspension in a jar filled with distilled water to
which a deflocculating agent is added. The soil particles are then allowed to settle down.
The concentration of particles remaining in the suspension at a particular level can be
determined by using a hydrometer. Specific gravity readings of the solution at that same
level at different time intervals provide information about the size of particles that have
settled down and the mass of soil remaining in solution. The results are then plotted
between % finer (passing) and log size. Grain-Size Distribution Curve The size
The results are then plotted between % finer (passing) and log size. Grain-Size
Distribution Curve the size distribution curves, as obtained from coarse and fine grained
portions, can be combined to form one complete grain-size distribution curve (also
known as grading curve). A typical grading curve is shown.
Grain-size distribution curve
From the complete grain-size distribution curve, useful information can be obtained
such as: 1. Grading characteristics, which indicate the uniformity and range in
grain-size distribution. 2. Percentages (or fractions) of gravel, sand, silt and clay-
size. Grading Characteristics A grading curve is a useful aid to soil description. The
geometric properties of a grading curve are called grading characteristics.
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To obtain the grading characteristics, three points are located first on the grading
curve. D60 = size at 60% finer by weight
D30 = size at 30% finer by weight
D10 = size at 10% finer by weight
The grading characteristics are then determined as follows:
1. Effective size = D10
2. Uniformity coefficient, Cu, ÕÊ =Ö×ØÖ�Ø
3. Curvature coefficient, Cc , Õh = (ÖVØ)ÙÖ×ØÖ�Ø
If Cu > 4 for gravel and Cu > 6 for sand and Cc between 1 and 3 indicates a
well-graded soil (GW for gravel and SW for sand ). i.e. a soil which has a
distribution of particles over a wide size range
The consistency of a fine-grained soil refers to its firmness, and it varies with the
water content of the soil.
A gradual increase in water content causes the soil to change from solid to semi-
solid to plastic to liquid states. The water contents at which the consistency
changes from one state to the other are called consistency limits (or Atterberg
limits). The three limits are known as the shrinkage limit (WS), plastic limit (WP),
and liquid limit (WL) as shown. The values of these limits can be obtained from
laboratory tests. (as explained in chapter 3)
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Classification Based on Grain Size The range of particle sizes encountered in
soils is very large: from boulders with dimension of over 300 mm down to clay
particles that are less than 0.002 mm. Some clay contains particles less than 0.001
mm in size which behave as colloids, i.e. do not settle in water.
According to grain Size analysis:
Gravel, sand, silt, and clay are represented by group symbols G, S, M, and C
respectively. Physical weathering produces very coarse and coarse soils. Chemical
weathering produces generally fine soils.
Coarse-grained soils are those for which more than 50% of the soil material by
weight has particle sizes greater than 0.075 mm. They are basically divided into
either gravels (G) or sands (S). According to gradation, they are further grouped as
well-graded (W) or poorly graded (P). If fine soils are present, they are grouped as
containing silt fines (M) or as containing clay fines (C). For example, the
combined symbol SW refers to well-graded sand with no fines. Both the position
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and the shape of the grading curve for a soil can aid in establishing its identity and
description. Some typical grading curves are shown.
Curve A- poorly-graded SAND
Curve B - a well-graded GRAVEL-SAND (i.e. having equal amounts of gravel
and sand)
Fine-grained soils are those for which more than 50% of the material has particle
sizes less than 0.075 mm. Clay particles have a flaky shape to which water adheres,
thus imparting the property of plasticity.
A plasticity chart , based on the values of liquid limit (WL) and plasticity index
(IP),
The 'A' line in this chart is expressed as IP = 0.73 (WL - 20).
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Depending on the point in the chart, fine soils are divided into clays (C), silts (M),
or organic soils (O). The organic content is expressed as a percentage of the mass
of organic matter in a given mass of soil to the mass of the dry soil solids.
Soil classification using group symbols is as follows:
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Activity "Clayey soils" necessarily do not consist of 100% clay size particles.
The proportion of clay mineral flakes (< 0.002 mm size) in a fine soil increases its
tendency to swell and shrink with changes in water content. This is called the
activity of the clayey soil, and it represents the degree of plasticity related to the
clay content.
= ¦§%yw�}¬e�y��du(~i�[ℎ�)
Where PI is plasticity index = ¨¯ −¦
Liquidity Index In fine soils, especially with clay size content, the existing state is
dependent on the current water content (w) with respect to the consistency limits
(or Atterberg limits). The liquidity index (LI) provides a quantitative measure of
the present
state.
ÛÜ = + − ÝÛÛÛ − ÝÛ
Example 1:
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Following are the results of a sieve analysis. Make the necessary calculations
and draw a particle –size distribution curve.
U.S.sieve size Mass of soil retained on each
sieve (g)
4 0
10 40
20 60
40 89
60 140
80 122
100 210
200 56
Pan 12
Solution:
Solution:
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Find ÕÞ,ÕÉ
ÕÊ =�qR�SR =0.270.15 = 1.8
Õh = (ÖVØ)ÙÖ×ØÖ�Ø =(R.S�)ÙR.P�∗R.ST = 0.71
% passing # 200 less than 50% so the soil is coarse , and since % passing # 4= 100
so the soil is sand and since Cu less than 6 , so the soil is SP.
Home work :1- Following are the results of a sieve analysis:
U.S. Sieve No. Mass of soil retained on each sieve
(g)
4 0
10 21.6
20 49.5
40 102.6
60 89.1
100 95.6
200 60.4
pan 31.2
I- Plote the grain –size distribution curve.
II- Calculate the uniformity coefficient ,ÕÞ, and cofficient of gradation ,ÕÉ
A)-For a soil given:
D10 = 0.1 mm
D30 = 0.41 mm
D60 = 0.62 mm
B)-for a soil given:
D10 = 0.082 mm
D30 = 0.29 mm
D60 0.51 mm
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Home work 1:
2- Classify the following soil according to USCS
U.S. Sieve
No.
Mass of soil
retained on the
sieve in g
4 0
6 0
10 0
20 9.1
40 249.4
60 179.8
100 22.7
200 15.5
Pan 23.5
Pl: 25, L.L. = 40
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Chapter Five
Soil Permeability and Flow
SOIL PERMEABILITY
A material is permeable if it contains continuous voids. All materials such as
rocks, concrete, soils etc. are permeable. The flow of water through all of them obeys
approximately the same laws. Hence, the difference between the flow of water through
rock or concrete is one of degree. The permeability of soils has a decisive effect on the
stability of foundations, seepage loss through embankments of reservoirs, drainage of
sub grades, excavation of open cuts in water bearing sand, rate of flow of water into
wells and many others.
Hydraulic Gradient
When water flows through a saturated soil mass there is certain resistance for the flow
because of the presence of solid matter. However, the laws of fluid mechanics which
are applicable for the flow of fluids through pipes are also applicable to flow of water
through soils. As per Bernoulli's
equation, the total head at any point in water under steady flow condition may be
expressed as
Total head = pressure head + velocity head + elevation head
Hydraulic Gradient
When water flows through a saturated soil mass there is certain resistance for the flow
because of the presence of solid matter. The laws of fluid mechanics which are
applicable for the flow of fluid through pipes are also applicable to flow of water
through soils. The total head at any point in water under steady flow condition may
be expressed as:
Total head = pressure head + velocity head + elevation head
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The flow of water through a sample of soil of length L and cross-
sectional area A as shown in figure1:
ßà = áà +¦à_� +vàP2[
ßÉ = áÉ + ¦h_� +vhP2[
Figure
(1) flow of water through a soil sample
For all practical purposes the velocity head is a small quantity and may
be neglected.
The water flows from the higher total head to lower total head. So the
water will flow from point B to C.
ßà −ßÉ = (áà +âã��)- (áÉ +â��)
Where,áà and áÉ = äwiv���duℎi�{, ¦à and ¦É = Pressure Head.
The loss of head per unit length of flow may be expresses as :
� = ℎ¨
Where i is the hydraulic gradient.
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Hydraulic gradient:
The potential drop between two adjacent equipotentials divided by the
distance between them is known as the hydraulic gradient.
DARCY'S LAW
Darcy in 1856 derived an empirical formula for the behavior of flow through
saturated soils. He found that the quantity of water q per sec flowing through a
cross-sectional area of soil under hydraulic gradient / can be expressed by the
formula
q = kiA
or the velocity of flow can be written as
� = å
Where k is termed the hydraulic conductivity (or coefficient of permeability) with
units of velocity. The coefficient of permeability is inversely proportional to the
viscosity of water which decreases with increasing temperature; therefore,
permeability measurement at laboratory temperatures should be corrected to the
values at standard temperature of 200C using the following equation.
Where ZPR : Coefficient of permeability at 200 C
Zæ : Cofficient of permeability at Lab. Temperture0 C
çæ Viscosity of water at lab. Temperature
çPR Viscosity of water at 200C
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Table (1) :The of èéèÙØ at different temperature.
DISCHARGE AND SEEPAGE VELOCITIES:
Figure below shows a soil sample of length L and cross-sectional area A. The
sample is placed in a cylindrical horizontal tube between screens. The tube is
connected to two reservoirs R1 and R2 in which the water levels are maintained
constant. The difference in head between R1 and R2 is h. This difference in head is
responsible for the flow of water. Since Darcy's law assumes no change in the
volume of voids and the soil is saturated, the quantity of flow past sections AA, BB
and CC should remain the same for steady flow conditions. We may express the
equation of continuity as follows
qaa = qbb = qcc
If the soil be represented as divided into solid matter and void space, then the area
available for the passage of water is only Av. If vs. is the velocity of flow in the
voids, and v, the average velocity across the section then, we have
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Where � is the area of the void,
v� is the seepage velocity,
v is the approach velocity
A: is the cross sectional area of the sample
v� = ∗ ¨ � ∗ ¨ v = v�v� v =
vu
Where n : is the porosity of the soil
METHODS OF DETERMINATION OF HYDRAULIC
CONDUCTIVITY OF SOILS (Coefficient of permeability).
Coefficient of permeability
Laboratory methods:
1- Constant head permeability method
2- Falling head permeability method
3- Indirect determination from consolidation test
Field methods:
1- Pumping test
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2- Bore hole tests
Laboratory methods:
1- Constant head permeability test:
The coefficient of permeability for coarse soils can be determined by means of the
constant-head permeability test (Figure below): A steady vertical flow of water,
under a constant total head, is maintained through the soil and the volume of water
flowing per unit time (q):
A series of tests should be run, each at different rate of flow. Prior to running the
test a vacuum is applied to the specimen to ensure that the degree of saturation
under flow will be close to 100%.
2- Falling head permeability test:
For fine soils the falling-head test (Figure below) should be used. In the case of
fine soils, undisturbed specimens are normally tested. The length of the specimen
is l and the cross-sectional area A. the standpipe is filled with water and a
measurement is made of the time (�S)for water level (relative to the water level in
the reservoir) to fall from ℎR�dℎS. At any intermediate time t the water level in
the standpipe is given by h and its rate of change by − t¹t� . At time t the
difference in total head between the top and bottom of the specimen is h. then
applying Darcy's law:
−� {ℎ{� = êℎw
−�ë {ℎℎ = êw
¹�
¹Øë {�¹
R
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∴ ê = �w �S wuℎRℎS = 2.3
�w �S log
ℎRℎS
Ensure that the degree of saturation remains close to 100%. A series of tests should
be run using different values of ℎR�u{ℎS
Figure: Laboratory Test (a) Constant Head (b) Falling Head
Example 1:
A constant head permeability test was carried out on a cylindrical of
sand 4 in. in diameter and 6 in. in height . 10 �u\
Of water was collected in 1.75 min, under a head of 12 in. Compute the
hydraulic conductivity in ft/year and the velocity of flow in ft/sec.
Solution:
Z = ï ��
ï = 10�u\, = 3.14 ∗ 4P4 = 12.56�uP
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� = ℎ¨ = 126 = 2, � = 105�iy
Therefore Z = SRSP.Tq∗P∗SRT = 3.79 ∗
SRUV¡ª�kh = 31.58 ∗ 10IT¬�/�iy
Velocity of flow =Z� = 31.58 ∗ 10IT ∗ 2 = 6.316 ∗ 10I�¬�/�iy
Example 2
A sand sample of 35 y�P cross sectional area and 20 cm long was tested
in a constant head permeameter. Under a head of 60 cm, the discharge
was 120 ml in 6 min. The dry weight of sand used for the test was 1120
g. and f� = 2.68. Determine (a) the hydraulic conductivity in cm/sec. (b)
the discharge velocity, and (c) the seepage velocity.
Solution:
Z = QL∆hAt Q = 120 ml, t =6 min, = 35y�P, ¨ = 20y�, �u{ℎ = 60y�
Z = SPR∗PRqR∗\T∗q∗qR= 3.174 ∗ 10I\y�/�iy
Discharge velocity, v= ki =31.74 ∗ 10I\ ∗ qRPR = 9.52 ∗ 10I\y�/�iy
Seepage velocity v� _t =~�v =
112035 ∗ 20 = 1.6 [� y�\⁄
_t = _�f�1 + i dei = f�_t − 1
i = 2.681.6 − 1 = 0.675
u = i1 + i = 0.403
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v� = �ª = Q.TP∗SRUVR.�R\ = 2.36 ∗ 10IP y� �iy⁄
DIRECT DETERMINATION OF K OF SOILS IN FIELD:
1- Field test in unconfined aquifer
Pumping Test in an unconfined aquifer
2- Field test in Confined aquifer
There are two cases :
Case 1- when ℎR > ßR
Case 2-when ℎR < ßR
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Empirical Relations for Hydraulic Conductivity
Several empirical equations for estimating hydraulic conductivity have been
proposed over the years.
Granular Soil
For fairly uniform sand (that is, a small uniformity coefficient), Hazen (1930)
proposed an empirical relationship for hydraulic conductivity in the form:
Z(y� �iy⁄ ) = Õ�SRP
C= a constant that varies from 1.0 to 1.5
D10 = the effective size (mm)
The above equation is based primarily on Hazen’s observations of loose,
clean, filter sands. A small quantity of silts and clays, when present in a
sandy soil, may change the hydraulic conductivity substantially.
The accuracy of the values of k determined in the laboratory depends on the folloeing factors:
1- Temperture of the fluid
2- Viscosity of fluid
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3- Trapped air bubbles present in the specimen
4- Degree of saturation
5- Migration of fines during testing
6- Duplication of field conditions in the laboratory.
The coefficient of consolidation of saturated cohesive soils can be determined by laboratory
consolidation tests. This will be listed in details in "consolidation of soil".
Table -1 Typical Values of Hydraulic Conductivity of Saturated Soils
Exercise:
1-
2-
3-
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HEADS AND ONE-DIMENSIONAL FLOW
There are three heads which must be considered in problem involving fluid flow in
soil (Figure 4):
1- Pressure head (ℎÓ): is the pizometer reading = pore water pressure /unit
weight of water
2- Elevation head at any point (ℎk): is the vertical distance above or below
some reference elevation or datum plane.
3- Total head, ℎ = ℎÓ + ℎk
Figure (4) illustration of types of head(after Taylor, 1948).
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Example 1: For the Setup shown (Figure 5a), plot, ht, he, hp and the velocity of
flow?
Points ht(ft)
(Figure 5b)
he(ft)
(Figure 5b)
hp=ht-he(ft)
(Figure 5b)
V(ft/min)*=Ki
(Figure5c)
1 12 12 0 2
2 12 10 2 2
3 12 8 4 6
4 =SPHRP = 6 5 1 6
5 0 2 -2 6
6 0 0 0 2
*� = ¹���VI¹���ö¯VUö = SPIRq = 2
Approch velocity = ki= 1*2= 2 ft/min
Seepage velocity =�ª =
PR.\\\ = 6¬�/��u
1
2
3
4
5
6
Figure(5a)
Datum
Figure(5b) Figure(5c)
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Example 2
For the setup shown(Figure 6a) , Draw, ht, he , hp and velocity of flow ?
1- Direction of flow is upward flow (look to the water'symbole usaually water
flow from higher one to lower one)
2- List all point with direction of flow
3- Construct a table to solve the problem
Points ht(ft)
(Figure 6b)
he(ft)
(Figure 6b)
hp=ht-
he(ft)
(Figure 6b)
V(ft/min)*=Ki
(Figure6c)
1 16 16 0
2 16 2 14
3 16 + 122 = 14
5 9
4 12 8 4
5 12 12 0 *� = ¹���ÙI¹���÷¯ÙU÷ = SqISPq = 0.667
1
2
3
4
Figure ( 6) a b c
5
Datum
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Approch velocity = ki= 1*0.667= 0.667 ft/min
Seepage velocity =�ª =
R.qq�R.\\\ = 2¬�/��u
Example 3: For the setup shown(Figure below) , Draw, ht, he , hp and velocity of
flow ?
Solution:
1- Assume any arbitrary line representing the datum and let it at elevation
=0(Figure 7a).
2- The flow will be in horizontal direction (elevation head is constant)
3- Construct the table
Since pressure =3.4 psi=3.4*144= 489.6 lb/ft2
ℎz = Ó�k��Ê�kʪ¡��k¡]¹�Ò¥���k� =
�lQ.qqP.�
= 7.84 ft
Points ht(ft) he(ft) hp(ft)
1 7.84 0 8
Figure heads vs horizontal distance
Figure (7a)
Figure (7c)
Datum 1
2 3
4
5
Ft/min)
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2 7.84 3 5.84
3 3.92 3 0.92
4 0 3 -3
5 0 0 0
� = ℎ���P −ℎ����¨PI� =7.84 − 06 = 1.3
Approch velocity = ki= 1*1.3= 1.3 ft/min
Seepage velocity =�ª =
S.\R.\\\ = 3.9¬�/�iy
Example 4
For the setup shown in figure 8:a) - Calculate the pressure head, elevation head,
total head and head loss at points B, C,D and F in centimeter of water. b)-Plot the
heads versus the elevation.
Figure shows the Set up of example 4
Solution (1):
points ht(cm) he(cm) hp(cm) Head loss
B 40 35 5 0
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C 40 20 20 0
D 20 7.5 12.5 20
F 0 -5 5 40
Figure (8)Example 4
Example (5): for the setup shown Calculate and plot total head, elevation head and
pressure head.
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Example 5
Solution example 5:
Points Ht(cm) He(cm) Hp(cm) Head loss
A 5 -5 10 0
B 5 -5 10 0
C 2.5 -5 7.5 2.5
D 0 -5 5 5
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Solution Example 5
Example -6 : For the set up shown , draw the variation of total head, pressure head
and elevation head along points A,B,C,D and E.
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Example 6
Solution of
Example 6
Solution of Example 6
point Total Elevation Pressure Head Loss
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Head(cm) Head(cm) Head(cm) (cm)
A 18.5 8.5 10 0
B 18.5 6.5 12 0
C 16 4 12 2.5
D 13.5 1.5 12 5
E 13.5 0 13.5 5
Example 7: For the setup shown, Find total head (ht) , Elevation head (he) and
Pressure head(hp) for the soil the setup shown.
Example 7 setup
4 m 4 m 6 m 3 m
5 m
5 m
A C B D E
ZS = 10ZP 5 m
ZS ZP
5 m Datum
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Example 8:For Setup shown: Soil I , A= 0.37 m2 , n=0.5 and k= 1 cm/sec,
Soil II, A= 0.186 m2 , n=0.5 and k=0.5 cm/sec.
Points ht(m) he(m) hp(m)
1 3.6 3.6 0
2 3.6 2.4 1.2
3 2.4 1.2 1.2
4 0 0.6 -0.6
5 0 0 0.0
Solution :
1- qI=qII
Zø�ø ø =Zøø�øø øø = Sh^/�khSRR *
∆¹ù(P.�IS.P) ∗ 0.37 =
R.Tú9/0�úSRR ∗ ∆¹ùù
(S.PIR.q) ∗ 0.1 − − − −(1) 2- ∆ℎø +∆ℎøø = 3.6 − − −−−−−−−− − − −−−−(2)
From Equation( 1 ) -----------∆ℎø = 0.502∆ℎøø Substitute in Equation 2--------- 0.502∆ℎø +∆ℎøø = 3.6
∴ ∆ℎøø = 2.4� ∆ℎø = 1.2�
Approach Velocity= ki
Approach velocity for soil I=Zø�ø = 1 ∗ S.P(P.�IS.P) = 1y�/�iy
Soil I
Soil II
0
0.6 m
1.2 m
2.4 m
3.6 m 1
2
3
4
5 Datum
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Seepage velocity = �ª =
SR.T = 2y�/�iy
For soil II
Approach velocity=Zøø�øø = 0.5 ∗ P.�(S.PIR.q) = 2y�/�iy
Seepage velocity (II) = �ª =
PR.\\\ = 6y�/�iy
0 2 6
Velocity (cm/sec)
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Example 9: For the setup shown draw ht, he, hp and find the seepage force .
Poi
nts
ht(c
m)
he(c
m)
hp(c
m)
A 0.9 0.9 0
B 0.9 0.6 0.3
C 1.5 0.0 1.5
Elevation
(cm) ∆û�(�£/^Ù) û�(Zb�P) x(Zb�P)
0.9
0 0 0
0.3
*10=3kN/m2
0.6 3 0.3*10=3 0
0.6*20.9 =12.54
0 15.54 1.5*10=15 0.54
0
0.3
0.6
0.9
-1.55 E-1 0.3 0.6 0.9 1.2 1.5
ht (cm(
he(cm(
hp(cm(
Soil
0.3 m
0.6 m
0.6 m
Supporting
Screen
_� = 20.9 Zb �\⁄
n=0.33, k=0.5 cm/sec
A
B
C Datum
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3
= +
6 kN/m2
9
15
Water pressure on soil sample (a) Boundary water pressure (b) Buoyancy water
pressure(static) (c) Pressure lost in seepage.
3
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Water Force on Soil:
ü = piiz�[i¬deyi�dwx�id¬�d�w = ℎ _�¨ = �_�
Seepage forces usually act with direction of flow.
Quick Condition:
The shear strength of cohesionless soil is directly proportional to the
effective stress. When a cohesionless soil is subjected to a water condition that
results in zero effective stress, the strength of the soil becomes zero and quick
condition exists.
Quick condition: occurs in upward flow( for cohesionless soil) and when the
total stress equals to pore water pressure .
ûk¥¥kh� = 0 = ¨ _� − ℎ _� = 0
ℎ¨ = � =
_¤_� = �h
�h: The gradient required to cause a quick condition, termed critical gradient.
Example 10: Excavation is been carried out as shown in the figure. Find: 1-
the depth Z that could caused boiling at the bottom of clay layer.
2-The depth ( Z) for the factor of safety against boiling equal to 2 at the
bottom of the clay layer.
3- What is the thickness of the raft foundation that should be used before
boiling occurs. If an uplift pressure of 60 kN/m2 at the bottom of
excavation exist(_hÒªh�k�k = 25Zb/�\). 4- Find the seepage force at an element of 0.2 m cube located at the center of
silt layer.
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Solution:
ℎ�S = 21.5
1- � = \� = ∆¹�UUÙSR^ = PS.TI¹�ÙSR −−−−−∴ ℎ�P = 14�
∴ ℎzP =ℎ�P −ℎiP = 14 − 10 = 4�
To find Z
F. down = F. upward
A(8-z) *20 = 4.0 *10 *A z= 6 m
2- þ. p = tÒ�ª���t¥Ò�hkÊÓ���t¥Ò�hk
2 = (8 − ) ∗ 20 ∗ 4 ∗ 10 ∗
∴ á = 4�
3- F down ward = F upward
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t*5*10*25 = 60*5*10
∴ � = qRPT= 2.4 m (thickness of concrete)
4- Seepage force = i_�vdwx�i
= \� *10* (0.2)
3 = 0.06 kN
Summary of Main Points:
1- In soils v = Z�
2- There are three heads of importance to flow through porous media: elevation
head ( ℎk), pressure head (ℎÓ)and total head (ℎ�). 3- Flow depends on difference in total head.
4- The seepage force per a volume of soil is i*_�and acts in the direction of
flow.
5- "Quick", refers to a condition where in a cohesion less soil loses its strength
because the upward flow of water makes the effective stress become zero.
Exercise 1: For the setup shown. Plot to scale elevation head, pressure head, total
head and seepage velocity versus distance along the sample axis.
Exercise 2: For the setup shown, compute the vertical force exerted by the soil on
screen A and that on screen B. Neglect friction between the soil and tube. G= 2.75.
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Exercise 3: For the set up shown: Steady vertical seepage is occurring. Make scaled
plot of elevation versus pressure head, pore pressure, seepage velocity, and vertical
effective stress. Determine the seepage force on a 1 ft cube whose center is at
elevation -15 ft. G for all soils = 2.75.
Exercise 4: An 8 m thick layer of silty clay is overlaying a gravel stratum
containing water under artesian pressure. A stand –pipe was inserted into the
gravel and water rose up the pipe to reach a level 2m above the top of the clay. The
clay has a particle specific gravity of 2.7 and a natural moisture content of 30
percent. The permeability of the silty clay is 3*10-8
m/sec. It proposed to excavate
2 m into the soil in order to insert a wide foundation which, when constructed, will
exert a uniform pressure of 100 kN/m2
on to its supporting soil. Determine (a) The
unit rate of flow of water through the silty clay in m3 per year before the work
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commences. (b) the factor of safety against heaving: i) at the end of excavation ii)
after construction of the foundation?
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Two Dimensional fluids Flow
Most problems of flow are two dimensional flows, e.g. are shown in Fig.
below:
The purpose of studying the flow in two Dimension are :
1- To find the amount of seepage per meter length (i.e. rate of flow).
2- Pressure distribution (pore water pressure)
3- Stability against piping or boiling.
4- Pizometer levels of selected point required.
Impervious layer
Impervious layer
Impervious layer
Impervious layer
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Seepage Theory:
The general case of seepage in two dimensions will now be considered.
In same principle used in one dimensional problem applied (Darcy's law
& Continuity flow state).
Consider the two dimensional steady state flows in the fig.
Take element A with dimension dx , dy and dz
Rate of flow ( 塪)the flow entering the element
(v± +t�´t´ {�){}{ + (v� +t��t� {}){�{
Since the flow is steady so 塪 =åÒÊ� v±dy dz +v�dx dz =(v± +t�´t´ {�){}{ + (v� +
t��t� {}){�{
By simplification
{v± {� +
{v�{} = − − − − − − −−−− (1)
Darcy's law = v± =−Z t¹t±
∴ {v±{� = −Z{Pℎ{�P
v� + {��{� . {�
v± + {�±{± . {±
X
v±
v�
Element A
Concrete Dam
Impervious
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v� =−Z {ℎ{}
∴ {v�{} = −Z{Pℎ{}P
Sub. In equation (1)
−Z tÙ¹t±Ù +(−Z tÙ¹t�Ù) = 0
{Pℎ{�P +
{Pℎ{}P = 0¨�zw�yiiåx���du
Consider a function ∅(�, }) so that
v± ={∅{�
v� ={∅{}
v± = t∅t± = −Z t¹t±
v� = t∅t� = −Z t¹t�
∅(�, }) = −Zℎ(�, }) + y Where c is a constant
Thus if the function ∅(�, }) is given a constant value equal to ∅S & it
will represent a curve a long which the value of total head (ℎS) is
constant. If the function ∅(�, }) is given a series of constant value ∅S ,∅P , ∅\ etc a family of curves, such curves are called equipotentials
and this will corresponding to total head ℎS ,ℎP , ℎ\ ---------ℎª from
the total differ nation.
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{∅ = t∅t± . {� +t∅t� {}
0 = v± dx + v�{} −v±dx = v�{}
The second function ψ(x,y) called the flow line
v± =−{ψ
{}
v� = −{ψ
{�
{v±
{�= −
{Pψ
{�{}
{v�
{}= −
{Pψ
{}{�
{v±
{�+{v�
}= 0
{Pψ
{�{}−{Pψ
{}{�= 0
∴ ψA�, }@Satisfy the Laplace equation
A gain a series of ψ using ψS, ψP, ψ\ − −−−ψª
Is selected and this function
ψ =dψ
{�{� +
dψ
{}{}
0 = v�{� +A−v±@{}
v�{�= v±{}
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{}
{�= v�
v±
Flow net:
The graphical representation of the Laplace equation is represented by the
two families of curve:
1- Equipotential lines: A series of lines of equal total head e.g.
ℎS, ℎP, ℎ\ − −−−− −ℎª
2- Flow lines: A family of the rate of flow between any two adjacent
flow lines is constant.
For isotropic soil:
The flow net is formed by a mesh of the intersection of two lines with
the following limitation
1- Each element is a curvilinear square ψS
¤
º ≅ 1
Summary of the main points:
1- Laplace equation govern's the steady state flow in two dimensions
2- The solution is represented by two families of curve
a- Equipotential lines
ψψ
ψ
∅∅
∅
Flow channel
90
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b- Flow lines
3- The intersection of the two lines is represented by a flow net of
square elements
4- Each element is a curvilinear square with dimension ¤
º≅ 1
5- The rate of flow is expressed /m length by
å = Zℎ£¥
£t
Q: rate of flow/m
Where k: Coefficient of permeability
H: Total head difference between the first and last equipotential
lines
b¬:bd. d¬¬wd~yℎ�uuiw� b{:bd. d¬{edz�(iåx�zd�iu���w{edz�) Steps in drawing a flow net:-
The first step is to draw in one flow line, upon the accuracy of which the
final correctness of the flow net depends. There are various boundary
conditions that help to position the first flow line, including:
1- Buried surface (e.g. the base of the dam, sheet pile) which are flow
lines as water cannot penetrate into such surface.
2- The junction between a permeable and impermeable material
which is also a flow line : for flow net purpose a soil that has a
permeability of one-tenth or less the permeability of the other may
regard as impermeable.
3- The horizontal ground surface on each side of the dam which are
equipotential lines.
The procedure is as follows:
a- Draw the first flow line hence establish the first flow channel
b- Divide the first flow line into squares (| ≅ ¨)
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c- Project the equipotentials beyond the first flow channel, which
give an indication of the size of the square in the next flow
channel.
d- With compasses determine the position of the next flow line;
draw this line as a smooth curve and complete the squares in the
flow channel formed.
e- Project the equipotentials and repeat the procedure until the
flow net is completed.
Example:
Figure example for flow net construction
Rate of flow
å = Z�
∆å = | ∗ 1 ∗ Z ∗ ∆¹�¨ �ℎ����¬dedui¬wd~yℎ�uuiw
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∆ℎ� = ßu{
∆å = ¤¯ *k * �£t (| ≅ ¨)
∆å = Z �£t this is for one channel
Assume No. of channel = Nf
∴ å = ∆åb¬ = êßb¬b{
Where H= difference in water level (upstream and downstream.
Note:
Example 1:
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Example 2
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Example 3
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Example 4
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Example 6
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Example 7
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Example 8:
A flow net for around a single row of sheet piles in a permeable soil layer is
shown in figure below. Given that k= 5* 10 -3
cm /sec.
a) How high (above the ground surface )will the water rise if pizometers
are placed at points a,b , c and d?
b) What is the rate of seepage under sheet pile?
Solution:
From flow net b¥ = 3,bt = 6
∆ℎ = 5 − 1.676 = 0.555 ℎ���zd�u�� = 15 − 0.555 = 14.445�
ℎ���zd�u�| = 15 − 2 ∗ 0.555 = 13.89�
ℎ���zd�u�y = ℎ���zd�u�{ = 15 − 5 ∗ 0.555 = 12.225�
So pizometer reading at point a = 14.445-10= 4.445 m above surface
So pizometer reading at point b = 13.89-10= 3.89 m above surface
So pizometer reading at point c = 12.225-10=2.25 m above surface
1
2
3 4
5
6 1
3
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q =kH £�£² = Ñ ∗ '(IÑ�/�iy ∗ (5 − 1.67) ∗
\q = ¿. >ÔÑ ∗ '(IÑ ^V
�kh /�
Example 9
A deposit of cohesion less soil with a permeability of 3*10-2
cm/sec has a
depth of 10 m with an impervious ledge below. A sheet pile wall is driven into
deposit to a depth of 7.5 m. The wall extends above the surface of the soil and 2.5
m depth of water acts on one side. Determine the seepage quantity per meter length
of the wall.
Example 10-
For the flow net shown below includes sheet-pile cutoff wall located at
the head water side of the dam in order to reduce the seepage loss. The
dam is half kilometer in width and the permeability of the silty sand
stratum is 3.5 *10-4
cm /sec. Find (a) the total seepage loss under the
dam in liters per year , and (b) would the dam be more stable if the cutoff
wall was placed under its tail-water side?
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Solution:
a) Notice that ∆ℎ = 6.0�, the number of flow channels b¥ =3�u{bt = 10 by using å = Z∆ℎ £�£²
q = (3.5*10-4h^�kh) �
^SRRh^� (6.0�) �
\SR�
= 6.3 ∗ 10Iqm3/sec/m
Since the dam is 500 meters wide, the total Q under the dam is
Q= Lq= 500m (6.3*10-4
m3/sec)(
SRVº¡�k��S^V ) �31.5 ∗ 10q �kh�k��� =
100^¡ºº¡Òªº¡�k���k��
b) - No: Placing the cutoff wall at the toe would allow higher uplift
hydrostatic pressure to develop beneath the dam.
Home work:
1. Two lines of sheet piles were driven in a river bed as shown in figure. The depth of water
over the river bed is 8.20 ft. The trench level within the sheet piles is 6.6 ft. below the river
bed. The water level within the sheet piles is kept at trench level by resorting to pumping.
If a quantity of water flowing into the trench from outside is 3.23 ft3/hour per foot length of
sheet pile, what is the hydraulic conductivity of the sand? What is the hydraulic gradient
immediately below the trench bed? (Ans – 1 x 10-4 ft/sec, 0.50).
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Stresses within the soil
Stresses within the soil:
Types of stresses:
1- Geostatic stress: Sub Surface Stresses cause by mass of soil
a- Vertical stress û� = ∑ _ℎ
b- Horizontal Stress û� =êRû�
Note : Geostatic stresses increased lineraly with depth.
2- Stresses due to surface loading :
a- Infintly loaded area (filling)
b- Point load(concentrated load)
c- Circular loaded area.
d- Rectangular loaded area.
Introduction:
At a point within a soil mass, stresses will be developed as a result of the soil
lying above the point (Geostatic stress) and by any structure or other loading
imposed into that soil mass.
1-
stresses due Geostatic soil mass
û� = _ℎ (Geostatic stress)
û¹ =ZRû� , where ZR : is the coefficient of earth pressure at
rest.
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EFFECTIVESTRESS CONCEPT: In saturated soils, the normal stress (σ) at any
point within the soil mass is shared by the soil grains and the water held within the
pores. The component of the normal stress acting on the soil grains, is called
effective stressor intergranular stress, and is generally denoted by σ'. The
remainder, the normal stress acting on the pore water, is knows as pore water
pressure or neutral stress, and is denoted by u. Thus, the total stress at any point
within the soil mass can be written as: û = û�+ u
This applies to normal stresses in all directions at any point within the soil mass. In
a dry soil, there is no pore water pressure and the total stress is the same as
effective stress. Water cannot carry any shear stress, and therefore the shear stress
in a soil element is carried by the soil grains only.
1- W.T.L very deep
û� = _¡t��ℎ¡
¡�ªÒ.Ò¥º��k�
¡�S
2- When W.T.L. at the ground surface:
û� = _¡���ℎ¡
¡�ªÒ.Ò¥º��k�
¡�S
u= h*_�
û� =û� − x=
_�ʤ = _k¥¥ = _��� −_�
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2- Stresses due to filling: ∆û¥¡ºº =ℎ¥¡ºº ∗ _¥¡ºº 3-Stresses due to external loading:
a- Stresses due to point load :
Boussinesq (1885)(French mathematician) developed the following
equation to calculate vertical stress increment ∆û, in soils due to point
load on the surface:
Type of point load: the load under
the wheel of car.
Then, ∆û = ��Ù §z
Where factor, §Ó
§Ó =3
2¢ 11 + �e�
P�TP
Example 1:
In road pavement design, the standard vehicle axel is defined as an axel with two
single wheels as shown below. For a particular vehicle group, the standard axel load
(P) is given as 80 kN and the distance between two wheels (L) is 1.8 m. what is the
vertical stress increment in the sub-grade at 4 m depth directly under a wheel if this
axel is running (consider wheel load as a point load ):
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Solution:
∆û = ��Ù §z,
§Ó = 32¢ 11 + (e)P
�T/P
To find ∆û = ∆ûà +∆û�
(§â)à = 32¢ 11 + (e)P
�T/P= 32¢ 1
1 + (1.84 )P�T/P= 0.301
(§â)� = 32¢ 11 + (e)P
�T/P= 32¢ 1
1 + (04)P�T/P= 0.477
Therefore,
∆û = ï�áP (§â)� +ïàáP (§â)à =
404P ∗ 0.477 +
404P ∗ 0.301 = 1.945Z¦�
b- Circular footing
∆û¥ = ¬ ��° , °� ∗ ∆å�ui��
F: factor, can be find from the following figure �� , where R: is the radius of the circular area
X: is the distance from the center of the circle to the point
Z : is the depth of the point.
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c- Rectangular loaded area.
∆ûk±�k�ª�ººÒ�t = þ(u,�) ∗ å�ªk�� F(n,m) : is the factor (use figure below)
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Approximate method to find ∆û at Z using 2:1 method:
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∆û = å�ªk�∗à∗¯(� + ) ∗ (¨ + á)
Where�k�: Is the net surface load.
Example: Plot the variation of total and effective vertical stresses, and pore water
pressure with depth for the soil profile shown below:
Solution:
Within a soil layer, the unit weight is constant, and therefore the stresses vary
linearly. Therefore, it is adequate if we compute the values at the layer
interfaces and water table location, and join them by straight lines. At the
ground level,
σv= 0 ; û� = 0; and u=0 At 4 m depth,
σv= (4)(17.8) = 71.2 kPa; u = 0
∴û� = 71.2 kPa
At 6 m depth, σv= (4)(17.8) + (2)(18.5) = 108.2 kPa u = (2)(9.81) = 19.6 kPa
∴ û� = 108.2 – 19.6 = 88.6 kPa At 10 m depth, σv= (4)(17.8) + (2)(18.5) + (4)(19.5) = 186.2 kPa u = (6)(9.81) = 58.9 kPa
∴û� = 186.2 – 58.9 = 127.3 kPa
At 15 m depth,
σv= (4)(17.8) + (2)(18.5) + (4)(19.5) + (5)(19.0) = 281.2 kPa u = (11)(9.81) = 107.9 kPa
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∴û�= 281.2 – 107.9 = 173.3 kPa The values of σv, u and
û� Computed above are summarized in Table 6.1.
Variation of û�,x�u{û� with depth Negative pore pressure (suction):
Below the water table, pore pressure are positive. In dry soil, the pore
pressure are positive. In dry soil, the pore pressure is zero. Above the
water table, when the soil is saturated, pore pressure will be negative.
u=−ℎ�_�
The height above the water table to which the soil is saturated is called
the capillary rise, and this depends on the grain size and type(and thus
the size of pores):
-In coarse soils capillary rise is very small
-In silts it may be up to 2m
_In clays it can be over 20m
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Example :
For the soil profile shown find the total, effective and pore water pressure:?
Given:
_t�� = 16Zb
�\, _��� = 20
Zb
�\
Solution:
At depth
z=0, û�Ò��º = 0, x = 0, ûk¥¥ = 0
At depth
z= 2m û�Ò��º = 2 ∗ 16 = 32�£
^Ù, x = 0, ûk¥¥ = 32
�£
^Ù
At depth
Z= 5 m û�Ò��º = 2 ∗ 16 + 3 ∗ 20 = 92�£
^Ù, x = 3 ∗ 10 = 30
�£
^Ù , ûk¥¥ = 32
�£
^Ù
0
1
2
3
4
5
6
0 20 40 60 80 100
Total Stress
Pore water
pressure
Effective stress
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Example:
For the soil profile shown: Determine , total, effective and pore water pressure for the following
conditions:1- water table 3 m above the ground level, 2- water table at the ground level 3- 1.0 m
below the surface,4- 2 m below the surface?
_t�� = 16Zb
�\,
_��� = 20Zb
�\
Depth
(m)
WT Total stress(kPa) u(kPa) ûk¥¥ AZ¦�)
0 Case 1 30 30 0
3 30+3*20=90 6*10=60 30
0 Case 2 0 0 0
3 3*20=60 3*10=30 30
0 Case 3 0 0
1 1*16=16 0 16
3 16+20*2=56 20 36
0 Case 4 0 0 0
2 2*16=32 0 32
3 32+1*20=52 1*10=10 42
0
0.5
1
1.5
2
2.5
3
3.5
0 10 20 30 40 50
Case
1/effectivecase 2effective
case
3/effective
Stresses (kPa)
de
pth
(m
)
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Example :
A building 20 m * 20m results in a uniform surface contact pressure of 150 kPa.
Determine the increase in vertical pressure at depth of 10 m below a) the center of
the building b) the corner of the building. Estimate the additional pressure at both
locations of a tower 5m *5m placed at the center of the building imposing 300 kPa
uniform additional pressures.
At the corner:
nz= 10 n*10=20 , n=2
mz=10 m*10=20, m=2
F(1,1)= 0.2325
∆û���ℎiydeuie{xi�d�xe¬�yiwd�{ =
0.2325 ∗ 150 = 34.875Z¦�
At the center :
N*10=10
n=m=1
From figure (rectangular ) f(1, 1)= 0.085
∆û at the center =4*0.176*150= 105.6 kPa
∆û{xi�d�d~ie =
at the center:
n*10=2.5
n=m= 0.25
f(0.25,0.25)= 0.0261
∆û{xi�d�d~ie|iwd~�ℎiyiu�ie ∆û = 4*0.0261*300=31.32 kPa
∆û at the corner due to tower
for e1:
n*10=12.5
n=m= 1.25
f(1.25,1.25)=
0.211
For e2=e3
n*10=7.5
n=0.75
m*10=12.5
m=1.25
f(0.75,1.25)=
0.165
for e3:
n*10=7.5,
n=m=0.75
f(0.75,0.75)=0.136
=(0.211-(0.165*2)-0.136)*300=5.1
kPa
Top Veiew
Front Veiew
20 m * 20m
5 m
5 m
5 m
12.5 m
12.5 m
12.5 m
12.5 m +e1
-e2
7.5 m
12.5 m
+e3
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Example: A distributed load of 50 kN/�P is acting on the flexible rectangular area
6*3 m, as shown in figure. Determine the vertical stress at point A which is located
at depth of 3 m below the ground surface. (_ = 18.5Zb/�P@.
Solution:
e1=e3, n*3=4.5 -------n=1.5, m*3= 1.5, m= 0.5
f(1.5,0.5)=0.131
e2=e4 , n*3= 1.5 , n=0.5, m*3=1.5,m=0.5
f(0.5,0.5)=0.085
∆û = 50 ∗ A0.131 ∗ 2 + 2 ∗ 0.085) = 21.6Zb/�P
Total stresses = Geostatic stress+∆û
Total Stresses= 18*3+21.6=75.6 kPa
Example : For the same example .If the foundation at 0.5 m below ground surface:
q net= 50- 0.5*18= 41 kN/�P
e1=e3, n*2.5=4.5, n=1.8, m*2.5= 1.5, m=0.6
f ( 1.8,0.6) =0.155
e2=e4, n*2.5=1.5, n= 0.6, m*2.5=1.5, m=0.6
f (0.6,0.6)= 0.109
∆û = 41 ∗ A0.155 ∗ 2 + 2 ∗ 0.109) = 21.648Zb/�P
Total stresses= 18*2.5+21.648=66.648 kN/�P
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Principal Stresses and Mohr Circle
Principal Stresses:
There are exist at any stressed point in three orthogonal (i.e. naturally
perpendicular) planes on which there are zero shear stresses. These planes are
called the principal stress planes. The normal stresses that acts on these three planes
are principal stresses. The largest of these three stresses is called the major
principal stress ûS , the smallest is called the minor principal stressû\, and the third
is called the intermediate stressûP. When the stresses in the ground are geostatic,
the horizontal plane through a point is a principal plane and so too are all vertical
planes through the point. When ( Z = ��
��@ Z < 1, û�= ûS and û¹= û\ = ûP .
when Z > 1, the situation is reversed. The shear stresses on any orthogonal
planes (planes meeting at right angles) must be numerically equal (�¹ =��@. Mohr circle: it is concerned only with the stresses existing in two dimensions, the
state of stress in plane that contains the major and minor principal stresses ûS ,û\.
The stresses will be considered positive when compressive.
Equations for state of stress at a point. � is positive when counter clock wise � is measured counter clockwise from the
direction of ûS.
Mohr diagram for state of stress at a point
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û� =ûS +û\2+
ûS − û\
2cos2�
� =(ûS −û\@
2sin2�
Where û� and � are the stresses acts on any planes, the direction and magnitude
of principal stresses can be found.
Origin of planes: is a point on the Mohr circle, denoted by Op, with following
property:
1- A line through Op and any points (A) of the Mohr circle will be parallel to
the plane on which the stresses given by point A act.
2- If the plane and Op are known then the line parallel to the plane passes
through Op and intersect Mohr circle , the intersect point represent the point
which acts on that plane.
To find the stresses , there are two method:
1- By drawing (prefer)
2- By using equations
1- By drawing
To determine the normal and shear stresses on any plane , we have to do
the following:
1-Draw the stress- state on Mohr circle (compression (+) and ( ).
2-Find the point denoted by Op origin of planes by either select ûS and draws a line
parallel to the plane on which ûSis acting until it intersects Mohr Circle & the
point of intersection is OP, or use û\ (the same principal is apply to û\) .
3-From OP draw a line parallel to the plane you want to find stress on it. The point
of intersection with Mohr Circle represents the stress & shear stress you need.
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: ملخص
:طريقة الرسم
عوامل رئيسية مرتبطة مع بعضھا ھي ) 3( ھناك
stressا#جھاد -1
Planeمستوي -2
OPنقطة اصل المستوي -3
OP :ھي نقطة وھمية وحيدة لكل دائرة ووحيدة فقط وتعين من اجھاد ومستوي.
.ويقطع دائرة مور فان ھذا الخط يمثل المستوي الذي تعمل عليه ذالك ا#جھاد OPاي خط يمر في
Principle Stress:
Stresses acting normal on mutually orthogonal planes with no shear stresses.
Principle Planes:
The planes on which there is zero shear stresses.
Case one: Given �'61;�> required ��61;��
Example 1:
Find stresses on plane B-B?
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Answer:
ûS = 40Zb/�
P
û\ = 20Zb/�P
Õ = ûSHû\
2=
40 + 20
2= 30Zb/�P
° = ��I�V
P=�RIPR
P= 10Zb/�P
By drawing:
1- Locate points (40,0) and (20,0).
2- Draw circle, using these points to define diameter.
3- Draw line � � through point (20,0) and parallel to plane on which stress
(20,0)acts.
4- Intersection of � � with Mohr circle at point (40,0) is the origin
Of planes.
5- Draw Line �� �� through OP and parallel to B-B
6- Read coordinates of X where �� �� intersect Mohr circle.
So on B-B �û = 25z��� = −8.7z��
Method 2 using Equations :
û� =ûS +û\2 +ûS − û\2 cos 2�
û� = �RHPRP +�RIPRP cos( 2 ∗ 120) = 30 + (−5) = 25
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� =(ûS −û\)2 sin 2�
� =(40 − 20)2 sin(2 ∗ 120) = −8.66Zb/�P
Case 2 given ��61;�� , required �'61;�>
Using Equation:
û� = ∑ ûS +û\2
ûS − û\2 = !(û� −ûS + û\2 )P + ��P We usually take the largest couple.
ûS = ûS + û\2 +ûS − û\2
û\ = ûS + û\2 −ûS − û\2
��u2� =��
ûS − û\
2
= 2��ûS − û\
Example 2 : Find the stresses on Horizontal plane D-D?
1-Draw point 1(40,0), point (20,0)
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2- From point (20,0) draw a line through point (20,0) ,parallel to the plane
that the force (20,0) acts .
3- The point of intersection with Mohr –circle represent the OP.
4- From the op draw a horizontal line (line parallel to horizontal plane) the
point of intersection with Mohr circle represent the stresses on horizontal
plane. ( 35, 8.7).
Example 3 :
Find the Magnitude and direction of priciple stresses.
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1- Locate points (40,-10) and (20,10)
2- Erect diameter and draw Mohr circle.
3- Draw �� �� through (40,-10) parallel to BB
4- Intersection of �� �� with circle gives Op.
5- Read ûS�u{û\ from graph.
6- Line through Op and ûS give plane on which ûS acts.
Solution by Equations :
1- Make use of that the sum of normal stresses is a constant: ûS + û\2 = ∑ û�2 = 40 + 202 = 30z�� 2- Use the following equation:
�ûS −û\2 � = !"û� − (ûS + û\
2)#P
+ $��%P
(��I�VP ) = &$20 − 30%P + $10%P = √200 = 14.14z�� 3- ûS = ���H�VP � + ���I�VP � = 44.14z��
û\ = ���H�VP � − ���I�VP � = 15.86z�� 4- Use stress pair in which û���w�e[i��; (40,−10)
��u2� = 2��ûS − û\ = −2028.28 = −0.707
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2� = −45° � = −22.5° 5- Angle from horizontal to major principle stress direction = 52.5°
Examples : to be added
Stress Path (p-q diagram):
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Chapter Seven
Consolidation of soil
A stress increase caused by the construction of foundations or other loads
compresses the soil layers. The compression is caused by (a) deformation of soil
particles, (b) relocations of soil particles, and (c) expulsion of water or air from the
void spaces. In general, the soil settlement caused by load may be divided into
three broad categories:
1. Immediate settlement, which is caused by the elastic deformation of dry soil and
of moist and saturated soils without any change in the moisture content. Immediate
settlement calculations are generally based on equations derived from the theory of
elasticity.
2. Primary consolidation settlement, which is the result of a volume change in
saturated cohesive soils because of the expulsion of water that occupies the void
spaces.
3. Secondary consolidation settlement, which is observed in saturated cohesive
soils and is the result of the plastic adjustment of soil fabrics. It follows the primary
consolidation settlement under a constant effective stress.
This chapter presents the fundamental principles for estimating the consolidation
settlement:
Consolidation : is the gradual reduction in volume of a fully saturated
soil of low permeability due to drainage of some of the pore water, the
process continuing until the excess pore water pressure set up by an
increase in total stress has completely dissipated.
The simplest case is that of one – dimensional consolidation in which
condition of zero lateral strain.
Swelling: is the reverse of consolidation, is the gradual increase in
volume of a soil under negative excess pore water pressure.
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Consolidation settlement: is the vertical displacement of the surface
corresponding to the volume change at any stage of the consolidation
process.
Consolidation settlement will result, for example, if a structure is built
over a layer of saturated clay or if the water table is lowered permanently
in a stratum overlying a clay layer on the other hand, if an excavation is
made in a saturated clay, heaving (the reverse of settlement) will result in
the bottom of the excavation due to swelling of the clay.
Fundamentals of Consolidation: When a saturated clayey soil layer is
subjected to a stress increase, the pore water pressure suddenly increases.
So at time =0 ∆x = ∆û while ∆û=0 (Figure 1-b), but after a time t ∆x < ∆û while ∆û > 0 (Figure 1-c) and after very long time (� ≈∞)∆x = 0 while∆û = ∆û (Figure 1-d).
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.
Figure
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Spring Analogy: the following figures illustrate the soil-spring analogy:
Figure (7-2)
The Oedometer (consolidation)Test :
Figure (7-2) Consolidation Cell
Test Procedure:
1- Determine All data for soil sample such as diameter, weight and height. Also
determine Gs, initial water content , initial void ratio by using S*e= Gs .
W(s=100%).
2- Set the sample in the consolidation test and apply initial stress p1= 25 kPa
and record the dial reading for a period of (24 hrs) at the times 0,0.25, 0.5,
1,2,4,8,15, 30, 60, 120,240,480,1440 min. from the load application.
3- At the end of (24 hrs) , double the applied stress (p2= 2p1=50 kPa
Dial gage
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And record the dial gage reading for 24 hrs at similar times to those of step
2.
4- Repeat step 3 by doubling the applied stress and recording the dial gage
reading. This process is repeated till we reach a stress of 1600 kPa (some
times we reach 3200 kPa). This process last 7 days and it called loading
stage.
5- Unload (3/4) the applied stress (i.e. remove 1200 kPa and leave 400 kPa) and
record the dial gage reading for (24 hrs).
6- After 24 hrs , unload (3/4) the remaining stress (remove 300 kPa and keep
100 kPa)and record dial reading for 24 hrs. then remove all the applied stress
and record dial readings.
7- Determine the final water content of the soil sample.
Example : The results of an odometer test is given below:-
Time
(min)
Dial gage readings
25
kPa
50
kPa
100
kPa
200
kPa
400
kPa
800
kPa
1600
kPa
400
kPa
100
kPa
0
0 0 127.5 218 320 443 598 841 1038 1028 597
0.25 95 164 250 366 510 695 904 1038 937
0.5 101 168 255 373 515 704 910 1038 924
1 104 172 262 381 528 725 920 1036 904
2 110 179 268 391 544 739 932 1035 880
4 115 185 278 402 560 763 946 1033 835
8 115 189 284 413 570 783 963 1031 800
15 121 195 291 421 582 802 977 1030 765
30 124 198 300 428 591 805 979 1029 680
60 125.5 204 303 430 596 822 1000 1028 655
120 126 205 306 435 597 831 1012 1028 622
1440 127.5 218 320 443 598 841 1038 1028 597 420
Mass of empty ring = 99.18 gm
Mass or ring +wet soil = 266.4 gm
Mass of ring + dry soil = 226.68 gm
Dia. Of ring = 75 mm
Ht of ring = 19 mm
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Dial gage Coeff. = 0.0001"
Initial water content = 31.14%
Specific Gravity of soil solids (Gs) = 2.76
Initial Degree of saturation = 100%
Required: void ratio at each load increment.
Solution :
1- find iR using S.e = Gs * c
iR = 2.76 ∗ 0.3114
iR = 0.8595
2- Since the volume of solid particles not change (incompressible)
So Assume v� = 1.
So iR = v�
So the Total volume (v� = 1 + iR)
So any change in the volume due to apply load is due to change in void so
∆i = iR −i¥
∈�= ∆i1 + iR
Since the consolidation in one dimension, and there is no lateral strain so the axial
strain (∈�= ∆¹� ) will be equal to volumetric strain (∈�). ∆¹� = ∆kSHkØ
Voids
Solids
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Applied
Pressure
(kPa)
Final
Dial gage
reading
Dial change
*0.0001*25.4
mm ∆ℎ(��) Thickness
of sample
at the end
of (24 hrs)
mm
Change in
void ratio ∆i= ∆ℎß (1+ iR)
Void
ratio at
the end of
24 hrs
e=iR −∆k
0
25
50
100
200
400
800
1600
400
100
0
0
127.5
218
320
443
598
841
1038
1028
597
420
0.32385
0.22987
0.25908
0.31242
0.3937
0.61722
0.5
0.0254
-1.09474
-0.4496
19
18.676
18.446
18.187
17.875
17.481
16.864
16.364
16.389
17.484
17.934
0.0317
0.0225
0.0254
0.0306
0.0385
0.0604
0.0489
-0.0025
-0.10714
-0.044
0.8595
0.8278
0.8053
0.7799
0.7493
0.07108
0.6504
0.6015
0.604
0.7111
0.755
From consolidation test the following result we can get :
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¦�
The slop of this curve at any stress
range can be defined as :
�� = ∆i∆¦� �� : Coeff. Of compressibility ∆¦ ∶� change in effective stress ∆i ∶corresponding change in void
ratio
Figure e- ¦� (results from oedometer test)
Log¦�
�� = ��1 + iR
��: Coeff. Of volume change �� : Coeff. Of Compressibility iR: Initial void ratio
Plot the ¨d[¦� vie�x�vd�{e���d(i)* the slop of first portion of the
loading curve is defined
Õ� = ∆i∆wd[¦� Which is equal to the slope of
the unloading curve, and the
first portion of the reloading
curve
∴ Õ� = Õk = ∆i∆wd[¦� Õ�: Reloading index
Compression
Recompression or
Reloading
Unloading
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Õk: Expansion index
The slope of the last portion of the loading curves is defined as :
Õh = ∆i∆wd[¦� Õh: Compression index
Note : ∗ Õh can be obtained using empirical correlation from Liquid Limit (L.L)
Õh = 0.009(¨. ¨ − 10) An empirical correlation between Õh�u{ Õ� is given by:
Õ� ≅ 0.1Õh
Log¦�
Figure e- log z (find ¦h� )
Pre-Consolidation
Pressure (¦h� ) : The max. Effective stress
that has been experienced
by the soil in the past or at
the present.
How to find ¦h�
1- Produce back the
straight line (BC) of
the curve
2- Draw the tangent to
the curve at D and
bisect the angle
between the tangent
and the horizontal
through D
3- The vertical through
the point of the
bisector and CB
gives the
approximate value
of the pre
consolidation
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pressure.
Normally and Over consolidation Clays:
A cording to the stress history, the clay can be:
1- Normally Consolidated clay(N.C.C.)
Which represent the clay at which the existing effective overburden pressure
(¦R� = û�) is the largest stress experienced by the soil at the present time
and in the past thus :
For N.C.C. ¦h� = ¦R� 2- Over Consolidated Clay(O.C.C.)
Represent the clay which has experienced a stress in the past larger than the
existing effective overburden pressure (¦R� ) acting at the present time.
For O.C.C. ¦h� > ¦R� Over Consolidation Ratio: (O.C.R.)
O.C.R.= â�âØ�
If O.C.R. =1 ∴ ¦h� = ¦R� ∴ b. Õ. Õ. If O.C.R. >1 ∴ ¦h� > ¦R� ∴ ,. Õ. Õ. Consolidation Settlement:
To calculate the final consolidation settlement ( at t= ∞) use one of the following
:
1- py¬ = ∆kSHkØ ß
2- py¬ = ��∆ûß
3- py¬ = ɸSHkØ ß log âØ� H∆â�âØ�
4- For O.C.C. we have to Check:
a- If ¦R� + ∆¦� ≤ ¦h� then use :
py¬ = É-SHkØ ß log âØ� H∆â�âØ�
b- If ¦R� + ∆¦� > ¦h� then use :
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py¬ = Õ�1 + iR ß log ¦h�¦R� + Õh1 + iR ß log¦R� + ∆¦�¦h�
Where Scf : Final consolidation settlement
iR: Initial void ratio
∆i: Change in void ratio = iR −i¥
H : thickness of compressed clay layer
�� : Coefficient of volume compressibility for the stress range
∆¦� : change in effective stress between initial and finial conditions
Õh ∶Compression index
Õ�: reloading index
¦R� : initial effective overburden pressure.
¦h� : Pre consolidation pressure
Note About settlement Calculation:
1- Scf : usually find at the middle of compressed clay layer (represent the worse
case).
2- Calculate the increase in vertical stress at the middle of clay layer by either
using 2:1 method or using Chart (figures 8.6 and 8.4).
3- Check whether the clay is N.C.C. or O.C.C., and use proper equation
mention above.
4- For more accurate result you can divide the clay layer into sub layers (i.e not
more than 2 m). Then find the total settlement by summing the settlement for
each layer.
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Example:
For the soil Profile shown find the final consolidation settlement?
1- py¬ = ∆kSHkØ ß = kØIk�SHkØ ß = S.l\IS.�SHS.l\ (11.6 − 7.3) = 0.653�
2- py¬=��∆¦� ß---------∆¦� = ℎ¥¡ºº ∗ _¥¡ºº = 22 ∗ 4.5 = 99Zb/�P
3- py¬ = ɸSHkØ ß log âØ� H∆â�âØ�
¦R� = û = û − x at the middle of the clay layer before fill application
∴ û = (7.3 − 2) ∗ 18.22 ∗ Ë11.6 − 7.32 Ì ∗ 16.34 = 131.7 Zb�P
U= .(7.3 − 2.9) +(SS.qI�.\P )/ ∗ 9.807 = 64.236 �£Ù ¦R� = û131.7 − 64.236 = 67.46 �£Ù py¬ = S.RQTTSHS.l\ (11.6 − 7.3) log q�.�qHQQq�.�q = 0.653�
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Degree of Consolidation (U% )
Degree of consolidation can be finding by
Figure: Assumed linear e-û relation
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Terzaghi's theory of one-dimensional consolidation:
The assumptions made in the theory are:
1- The soil is homogenous.
2- The soil is fully saturated.
3- The solid particles and water are incompressible.
4- Compression and flow is one-dimensional (vertical).
5- Strains are small.
6- Darcy's law is valid at all hydraulic GRADIENTS.
7- The coefficient of permeability and the coefficient of volume compressibility
remain constant throughout the process.
8- There is a unique relationship, independent of time, between void ratio and
effective stress.
There are three variable in the consolidation equation:
0Ê0� = Õ� 0ÙÊ0�Ù (Consolidation equation)
1- Depth of the soil element in the layer (z).
2- The excess pore water pressure (U)
3- The time elapsed since application of the loading (t)
Where u: is the excess pore water pressure. Õ�: is the coefficient of consolidation.
� = �_�
Where �� ∶ is the coefficient of volume changes (�P }i�e⁄ ).
1� = Õ��ℎt�P
Where 1� ∶ Time factor
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t : the time ℎt�: is the drainage path
ℎt�=H/2 ( clay layer Between two permeable layers).
ℎt�= H (for clay layer between one permeable and one impermeable layer).
2�: Degree of Consolidation
There is a relation between the 1� and Degree of consolidation
1- Degree of consolidation at specific depth
So the graphical solution of Terzaghi's one-dimensional consolidation
equation using non-dimensional parameters is as shown in following figure:
Figure Variation of degree of consolidation with time factor and
á ℎt�⁄ .
ℎt�= H
ℎt� ℎt�= H
áℎt�
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The variation of total consolidation with time is most conveniently plotted in the
form of the average degree of consolidation ( U ) for the entire stratum versus
dimensionless time 1�, and the this is illustrated below:
Figure : Average Degree of consolidation and Time factor
There are useful approximation relating the degree of consolidation and the time
factor
For 2 < 0.601� = 3� 2P
For 2 > 0.601v = 1.781 − 0.933wd[10(100- U)
Determination of coefficient of consolidation:
There are two methods:
1- The log time method (casagrande method):
a)-By plotting the dial reading versus log time
Then choice any time (small time ) then find the dial reading at 4t then move
above the dial reading a distance equal to ∆� then you can find the dial
reading at t= 0.
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b)-Take the tangent to the curves at the end of the curve so you can find the
dial reading at 100% consolidation.
c)-Now dial reading at t= 09 and dial reading at the end of consolidation so
you can find the dial reading for 50% consolidation.
e)-So the coefficient of consolidation y� can be find by the following
equation.
y� = 0.196�P�TR
2- The root time method (Taylor)
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Õ�= R.l�l¹²-Ù�4Ø
Consolidation Settlement Time:
To estimate the amount of consolidation which occur and the time, it is necessary
to know:
1- The boundary drainage conditions
2- The loading condition
3- The soil parameters including initial void ratio, coefficient of
compressibility, compressibility index and coefficient of consolidation. All
these are obtained from consolidation test on representative sample.
So to plot the variation of settlement versus time curve:
1- Find the time required to complete the consolidation ( 1� = 1).
2- Assume different value of time start from (t=0) to time (t) find in step 1
3- For each value of time t , find the 1� using ( 1� = É��¹²-Ù ). 4- For each value of 1� (from step 3) find ( 2��).
5- Find the final consolidation ( ph¥)using any proper equations.
6- Find the settlement at time t 5ph� = 2�� ∗ ph¥6. 7- Plot the settlement-Time curve.
Time (t) 1� = Õ��ℎt�P 2�� ph¥ ph� = 2�� ∗ ph¥
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Figure Variation of settlement with time
Example: For the soil Profile shown find the final consolidation settlement?
1- py¬ = ∆kSHkØ ß = kØIk�SHkØ ß = S.l\IS.�SHS.l\ (11.6 − 7.3) = 0.653�
2- py¬=��∆¦� ß---------∆¦� = ℎ¥¡ºº ∗ _¥¡ºº = 22 ∗ 4.5 = 99Zb/�P
3- py¬ = ɸSHkØ ß log âØ� H∆â�âØ�
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¦R� = û = û − x at the middle of the clay layer before fill application
∴ û = (7.3 − 2) ∗ 18.22 ∗ Ë11.6 − 7.32 Ì ∗ 16.34 = 131.7 Zb�P
U= .(7.3 − 2.9) +(SS.qI�.\P )/ ∗ 9.807 = 64.236 �£Ù ¦R� = û131.7 − 64.236 = 67.46 �£Ù py¬ = S.RQTTSHS.l\ (11.6 − 7.3) log q�.�qHQQq�.�q = 0.653�
Example :
Astrata of consolidated clay of thickness 10 ft drained on one side only.
The hydraluic conductivity of k = 1.863*10Il in /sec. and cofficient of volume
compresebility �� = 8.6 ∗ 10I� ¡ªÙº¤ . Determine the ultimate value of the
compression of the stratum by assuming a uniformly distributed load of 5250 w| ¬�P⁄ and also determine the time required for 20 percent and 80 percent
consolidation.
Solution : Total compression
p� = ��ß∆¦ = 8.6 ∗ 10I� ∗ 10 ∗ 12 ∗ 5250 ∗ SS�� = 3.763 in.
For determining the relationship between U% and T for 20%
consolidation use the equation
1 =¢
42%100 de1 = 3.144 ∗ 20P100 = 0.0314
For 80% consolidation use the equation
1 = 1.781 − 0.933 log(100 − 80)
Therefore 1 = 1.781 − 0.933 log(100 − 80) =0.567
Example
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Example : The loading period for a new building extended from may 1995
to may 1997. In May 2000, the average measured settlement was found to be 11.43
cm. it is known that the ultimate settlement will be about 35.56 cm. Estimate the
settlement in may 2005. Assume double drainage to occur.
Solution:
For the majority of practical cases in which loading is applied over a
period, acceptable accuracy is obtained when calculating time-settlement
relationships by assuming the time datum to be midway through the loading or
construction period.
�� = 11.43y�~ℎiu� = 4}i�e��u{p = 35.56y�
The settlement is requried for t=9 years, that is up to may 2005. Assuming as
a starting point that at t = 9 years, the degree of consolidation will be = 0.60. under
these conditions. 2 = 1.13√1.
Is p�S = pi��wi�iu������i�S , p�P = Settlement at time �P
p�Sp�P = 2S2P = !1S1P = !�S�P p�uyi1 = Õ��ßt�P
Where É��²-Ù is a constant. Therefore
SS.�\®�Ù = 7�Q de��P = 17.15y�
Therefore at t=9 years , 2 = S�T\T.Tq = 0.48 and since the value of U is less
than 0.60 the assumption is valid. So the Settlement is 17.15 cm.
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Example :
An oedometer test is performed on a 2 cm thick clay sample. After 5 minutes
50% consolidation is reached. Aftrer how long a time would the same degree of
consolidation be achieved in the field where the clay layer is 3.70 m thick? Assume
the sample and the clay layer have the same drainage boundary conditions (double
drainage).
Solution :
The time factor T is defined as 1 = É���²-Ù
Where ßt� = half the thickness of the clay for double drainage.
Here , the time factore T and coefficient of consolidation are the same for
both the sample and the field clay layer. The parameter that changes is the time t.
Let �S and �P be the times required to reach 50% consolidation both in the
oedometer and field respectively. �S = 5��u. Õ��Sßt�SP = Õ��Pßt�PP
�P = (�²-Ù�²-�
)P�S = \�RP
P ∗ 5 ∗ SqR ∗ SP� {�}� ≈ 119{�}�
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Example
Example :
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Example :
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Example:
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Chapter Eight
Shear Strength
•The strength of a material is the greatest stress it can sustain;
•So that the unit of strength is the same as stress (Pa in SI unit system).
Shear Failure in Soils:
Significant of Shear Strength:
• The safety of any geotechnical structure is dependent on the strength of the
soil;
• If the soil fails, the structure founded on it can collapse.
• Understanding shear strength is the basis to analyze soil stability problems
like lateral pressure on earth retaining structures, slope stability, and –
bearing capacity.
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Slope Failure in soils:
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West side of foundation sank 24 ft.
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Shear Strength in Soils •The shear strength of a soil is its resistance to shearing stresses.
•It is a measure of the soil resistance to deformation by continuous displacement of
its individual soil particles
•Shear strength in soils depends primarily on interactions between particles
•Shear failure occurs when the stresses between the particles are such that they
slide or roll past each other
Shear Strength in Soils: Soil derives its shear strength from two sources:
–Cohesion between particles (stress independent component)
• Cementation between sand grains
• Electrostatic attraction between clay particles
–Frictional resistance between particles (stress dependent component)
Shear strength of soils:
Cohesion (C): is a measure of the forces that cement particles of soils.
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Internal Friction angle (φ), is the measure of the shear strength of soils due to
friction
Mohr-Coulomb Failure Criteria
•This theory states that a material fails because of a critical combination of normal
stress and shear stress, and not from their either maximum normal or shear stress
alone.
Mohr Columb Failure criteria
� = y + ûª tan9
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� = y + ûª tan 9
Where � : is the shear strength of soil (kPa)
C: cohesion (kPa)
9: is the angle of internal friction (degree)
y :is the cohesion in term of effective stress
9: angle of internal friction in term of effective stress
The above equations are the shear strength of soils in terms total stress and in term
of effective stress.
c = 0 for pure sand
y = 0 for normally consolidated clay
9: Also some time called the drained angle of internal friction.
Mohr-Columb shear Failure criteria
p�u9 = ûS − û\2Õyd�9� + ûS + û\2
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�' = �> ' + #)%:' − %Ä1: + Ô#
#)%:' − %Ä1:
�' = �>461Ô Ë;Ñ + :ÔÌ + Ô# 461(;Ñ + :Ô) Also in term of total stress
�' = �>461Ô �;Ñ +:Ô� + Ô#461(;Ñ + :Ô)
Determination of Shear strength Parameters (C , � , : and :)
The shear strength parameters of a soil are determined in the lab primarily with two
types of tests: 1) Direct Shear Test; and 2) Triaxial Shear Test.
-Direct Shear Test:
• The test equipment consists of a metal box in which the soil specimen is placed
• The box is split horizontally into two halves
• Vertical force (normal stress) is applied through a metal platen
• Shear force is applied by moving one half of the box relative to the other to cause
failure in the soil specimen.
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Direct Shear Test Procedure:
1- Measure inner side of shear box and find the area(6 *6 ).
2- Make sure top and bottom halves of shear box are in contact and fixed
together.
3- Place the soil (sand) in three layers in the mold using funnel.
4- Place cover on top of sand.
5- Place the box in machine.
6- Apply normal force. (usually the student divided in to three groups ,
each group use run with one weight, 2 kg, 4 kg and 6 kg the load is
applied through a level arm : Note lever arm loading ratio 1:10 , 2kg
weight =20 kg).
7- Start the motor with selected speed (0.1 in/min) so that the rate of
sharing is at a selected constant rate.
8- Take the horizontal dial reading, vertical dial reading and proving ring
reading (shear load). Record the readings on the data sheet.
9- Continue taking readings until the horizontal shear load peaks and then
falls, or the dial reading gage stopped or at 10% of the the length of the
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sample.
10- Do the calculation :
Find the %ÂG6*%4*G%% = %ÂG6*Æ)6;5*G6(><#,Ô)
11- Plot the results (shear stress Vs horizontal displacement) the
result will be as in figure, take 3 peak stresses. Then plot the relation
between shear strength and normal stresses. Find the slop of this curve,
find , the intersection of the line with y-axis gives the value of C.
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The disadvantages of direct shear test are:
Advantages:
1- The test's simplicity and , in the case of sands, the ease of specimen
preparation
2- The travel of the machine can be reversed to determine the residual shear
strength values, which is shear strength parameters at large displacements.
3- Shear box represents a cheaper method in determining the drained shear
strength parameters for coarse-grained soil.
Disadvantages
1) The main one: drainage conditions cannot be controlled.
2) As pore water pressure cannot be measured, only the total normal stress
can be determined, although this is equal to the effective normal stress if
the pore water pressure is zero.
3) Only an approximation to the state of pure shear is produced in the
specimen
and shear stress on the failure plane is not uniform, failure occurring
progressively from the edges towards the center of the specimen.
4) The area under the shear and vertical loads does not remain constant
throughout the test.
Triaxial shear test :
This experiment over come all the disadvantages of direct shear test:
1- Can be used for all types of soils
2- Pore water pressure can be measured
3- The area corrected be used
4- The failure plane
Triaxial shear test is the most widely used and is suitable for all types of soils. A cylindrical
specimen, generally having a length to diameter ratio of 2, is used in the test and is stressed under
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conditions of axial symmetry in the manner shown in figure.
Soil Sample in triaxial apparatus
A sample of soil in a sealed triaxial test cell. Confining pressure is applied using
water surrounding sample as shown in following figure:
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Figure ( ) triaxial compression testing device
1- Triaxial test is more reliable because we can measure both drained and
undrained shear strength.
2- Generally 1.4” diameter (3” tall) or 2.8” diameter (6” tall) specimen is used.
3- Specimen is encased by a thin rubber membrane and set into a plastic
cylindrical chamber.
4- Cell pressure is applied in the chamber (which represents σ3 by pressurizing
the cell fluid (generally water).
5- Vertical stress is increased by loading the specimen (by raising the platen in
Valve A
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strain controlled test and by adding loads directly in stress controlled test,
but strain controlled test is more common) until shear failure occurs. Total
vertical stress, which is σ1 is equal to the sum of σ3’ and deviator stress
(σd). Measurement of σd, axial deformation, pore pressure, and sample
volume change are recorded.
Depending on the nature of loading and drainage condition, triaxial tests
are conducted in three different ways.
UU Triaxial test (unconsolidated undrained triaxial test)
CU triaxial test (consolidated undrained triaxial test)
CD triaxial test ( consolidated drained triaxial test)
Unconsolidated Undrained Triaxial test:
As drainage is not permitted and consolidation is not necessary, this test is very
quick and also called a quick test. In this test as drainage is not permitted the pore
water pressure increases after application of û\ as well as after the application of
ût so ∆x{xi�dû\ = �∆û\ and∆x{xi�dût = ∆ût so the total excess pore
water pressure will be as follows:
∆x = �∆û\ + ∆ût
But ût = ûS −û\
This test is common in clayey soils.
Application: UU triaxial test gives shear strength of soil at different confining
stresses. Shear strength is important in all types of geotechnical designs and
analyses.
Test procedure:
1- Measure diameter, length, and initial mass of the specimen.
2- Set a soil specimen in a triaxial cell.
3- Increase the cell pressure to a desired value (100 kPa for the first case and
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200 kPa in the second case and 300 kPa in the third case).
4- Shear the specimen by strain control record ��, and record the ût
5- Continue the test until the proving ring stop reading or use 10%
After completion of test measure of the soil specimen again and put it into the
oven then find the moisture content of the sample.
Take the sample out the cell and Sketch the mode of failure.
Repeat the test for the second specimen too (200 kPa of cell pressure and third
specimen 300 kPa of cell pressure).
calculation :
1- =� = ∆¯ (∆¨ = vie��y�w{i¬de����dud¬�ℎi�ziy��iu).
2- Calculate the corrected area ( h): h = �ØSI>
3- Calculate the stress û = ¯Ò�t
�¸
4- Plot ût versus axial strain separately for three tests.
5- Plot ût versus =� for three tests in the same plot
6- Plot Mohr circle using ûS and û\ at failure (note they should give the same ût value)
7- Plot the tangent to the three Mohr circles you plotted in above stage
8- Make a straight line, which is tangent to all three Mohr's circles. The slop give the angle
of internal friction 9 =0 , the point of the intersection of the tangent and y-axis give the
cohesion (yx)
yx = ût2
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Cu triaxial test : consolidated undrained test
In this test consolidation is permitted by opening drainage valve through applying û\,or cell pressure, or all around pressure. Then the drainage valve closed during
applying the deviator stress ût loading.
∆x = �∆û\ + ∆ût
Because valve A open during applying cell pressure, therefore B=0
So the increase in pore water pressure is due to increase in deviator stress only.
∆x = ∆ût
CD triaxial test : consolidated drained triaxial test
In this test valve A is open all time and the pore water pressure parameters
A and B = 0.
So this test is called drained test also called long term test and slow test.
There is volumetric change:
So h = R SI>�SI>� where=�: is the volumetric strain =� =∆�
�Ø
In this case: y = y and 9 = 9 usually occurs in sandy soils
The drained and undrained condition:
Drained condition
• occurs when there is no change in pore water pressure due to external
loading.
• In a drained condition, the pore water can drain out of the soil easily, causing
volumetric strains in the soil.
Undrained condition
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• Occurs when the pore water is unable to drain out of the soil.
• In an undrained condition, the rate of loading is much quicker than the rate
at which the pore water is able to drain out of the soil.
• As a result, most of the external loading is taken by the pore water, resulting
in an increase in the pore water pressure.
• The tendency of soil to change volume is suppressed during undrained
loading.
The existence of either a drained or an undrained condition in a soil
depends on:
• The soil type (e.g. fine-grained or coarse-grained)
• Geological formation (fissures, sand layers in clays, etc.)
• Rate of loading
• For a rate of loading associated with a normal construction activity,
saturated coarse-grained soils (e.g. sands and gravels) experience
drained conditions and saturated fine-grained soils (e.g. silts and clays)
experience undrained conditions
• If the rate of loading is fast enough (e.g. during an earthquake),
even coarse-grained soils can experience undrained loading, often
resulting
in liquefaction.
Unconfined compression test: This test subjects the soil to an axial
compressive load between two platens as shown in the Figure. There is no
confinement of the sample in the radial direction.
The load is recorded using a proving ring or a load cell and the axial deformation of
the soil sample is recorded using a dial gauge.
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Unconfined compression testû\ = 0, only suitable for cohesive soil.
To find the stress
û = ¦ h
h = R( 11 − =�
)
Significance of unconfined compression test
• A quick test to obtain the shear strength parameters of cohesive (fine
grained) soils either in undisturbed or remolded state .
• The test is not applicable to cohesion less or coarse grained soils
• The test is strain controlled and when the soil sample is loaded rapidly, the
pore
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pressures (water within the soil) undergo changes that do not have enough
time to dissipate Hence the test is representative of soils in construction sites
where the rate of construction is very fast and the pore waters do not have
enough time to dissipate.
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Chapter Nine
Lateral Earth Pressure
Lateral earth pressure
Is the pressure that soil exerts in the horizontal direction. The lateral earth
pressure is important because it affect the consolidation behavior and strength of
the soil and because it is considered in the design of engineering structures such as
retaining walls, basements, tunnel, deep foundations.
Coefficient of lateral earth pressure, k is defined as the ratio of the horizontal stress
to the vertical stress.
There are three coefficients:
1- Coefficient of earth
pressure at rest
(êR) = 1 − ��u9
2- Coefficient of active earth
pressure (ê�) = SI�¡ª?SH0@A?
3- Coefficient of passive
earth pressure (êÓ) =SH0@A?SI�¡ª? = S��
êR : When the wall is at rest and the material is in its natural stable then the
pressure applied by material is known as earth pressure at rest
ê� : when the wall moves away from the backfill , there is a decrease in the
pressure on the wall this decrement continues until a minimum value is reach after
which there is no reduction in the pressure and the value become constant this kind
of pressure is known as active earth pressure.
êÓ : when the wall moves toward the backfill, there is an increase in the pressure
on the wall and this increase continues until maximum value is reach after which
there is no increase in the pressure and the value will become constant ,this kind of
pressure is known as passive earth pressure.
Active
pressure
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Consider the wall shown in the figure
1- If no movement of the wall takes place, the soil is at rest and the vertical and horizontal
stresses acting on element A and B are: