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THEORETICAL SUBJECTS
for the graduating examination - Civil Engineering specialization (ICE)
Discipline: SOIL MECHANICS
1. Soils components – solid phase, chemical and mineralogical composition.
Answer 1:
From physical point of view, soils are
dispersed, three-phase mediums, generally
composed from the following phases: solid phase
(solid particles composing the mineral skeleton of
soil); liquid phase (water in the spaces between the
solid particles, named voids); gaseous phase (air or
gases in voids unoccupied by water).
Fig. 1. Soils components:
1 – solid particle; 2 - water; 3 – air
The mineral skeleton of soils was formed by physical weathering and chemical
weathering of the minerals contained in the pre-existing rocks (primary minerals resulted
from physical weathering and secondary minerals resulted from chemical weathering of the
primary minerals, resulting new minerals). Most frequent primary minerals parts of sandy and
silty soils are: quartz, feldspar, calcite, mica, etc. Characteristic to clays are secondary
minerals as: montmorillonite, kaolinite, etc.
2. Physical characteristics of the soils – solid particles density and the density of the soils
(ρs, γs, ρ, γ).
Answer 2:
Solid particles density represents the ratio between the solid particles mass Ms from a
soil sample and their volume Vs; is expressed by the relationship:
s =
s
s
V
M [g/cm
3]
The solid particles density varies between relatively restricted limits (2,6-2,8 g/cm3).
In laboratory the solid particles density is determined using the picnometer.
Unit weight of the solid particles is the ratio between the weight of the solid
particles, Gs from a soil sample and their volume, Vs. The calculus relationship is: s =
gV
gM
V
Gs
s
s
s
s
[kN/m3] in which g is the gravitational acceleration.
Bulk density represents the ratio between the mass of a soil sample M and its total
volume V, including the volume of the voids (spaces between the solid particles). It’s
expressed by the relationship: = V
M [g/cm
3]
Bulk density varies widely, for the same soil, with the same porosity (voids volume),
function of the water content.
Bulk unit weight is the ratio between the weight of a soil sample G and its volume V:
= gV
gM
V
G
[kN/m
3]
3. Moisture content and degree of saturation of the soils (w, Sr).
Answer 3:
Moisture content of a soil represents the ratio between the water mass Mw from the
voids of a soil quantity and the mass of the solid particles Ms from the same soil quantity. It’s
quoted with w and is expressed by the relationships:
w
s
w
M
M or: in percent w 100
M
M
s
w [%]
In laboratory, moisture content is determined by soil samples drying (in the oven at a
temperature of 105°C) till a constant mass.
Degree of saturation is defined as the ratio between the volume of water from a soil
sample and the voids total volume of the same soil sample:
p
wr
V
VS
Function of the degree of saturation value the soils are classified into:
- Dry soil, Sr ≤ 0,40;
- Wet soil, 0,40 < Sr ≤ 0,80;
- Very wet soil, 0,80 < Sr ≤ 0,90;
- Practically saturated soil, 0,90 < Sr ≤ 1,00.
4. Void ratio, porosity and the density index of cohesionless soils (e, n%, ID).
Answer 4:
Void ratio (e) represents the ratio between the voids volume Vp and the volume of the
solid particles for the considered soil sample: e = Vs
Vp
Porosity (n) expresses the ratio between the voids volume and the total volume of the
considered soil quantity:
n =V
Vp or in percent: n = 100V
Vp [ % ] where: Vp – volume of the voids
from the considered soil sample; V – total volume of the considered soil sample.
Density index, ID is defined by the following relationship: ID =
minmax
max
ee
ee
where: emax – void ratio corresponding to the loose state of the soil; emin – void ratio
corresponding to the compacted state of the soil; e – void ratio corresponding to the natural
state of the soil.
5. Plasticity limits, plasticity index and consistency index (wL, wP, IP, IC).
Answer 5:
Moisture contents that define the plastic cohesive soils behaviour are named plasticity
limits.
Plastic limit wp, represents the minimum moisture content from which a clayey soil
behaves as a plastic body, the soil passing from a hard (semisolid) state into a plastic state.
Liquid limit wL represents the maximum moisture content till which a clayey soil has
a plastic behaviour, the soil passing from a plastic state into a yielding state. For moisture
contents greater than wL the soil yields under self-weight.
Property of the cohesive soils to behave, in a certain moisture content limits, as a
plastic body, is named plasticity. Quantitatively, plasticity is expressed by plasticity index Ip,
that represents the moisture content interval in which cohesive soils are in plastic state, being
defined by the relationship: Ip = wL - wp.
Consistency index Ic expresses quantitatively the consistency state of the cohesive
soils, being defined by the following relationship: Ic =
p
L
pL
L
I
ww
ww
ww
.
6. Study of the soils compressibility in the laboratory. Oedometric test.
Answer 6:
In laboratory, for the compressibility study is used an apparatus named oedometer
(fig. 1). During this test, upon the soil sample is applied, through a piston, a vertical
compression load in steps. For water drainage from soil sample voids, the soil sample is
placed between two porous plates.
Fig. 1. Oedometer Fig. 2. Load-settlement curve
The main characteristic of the compressibility test consists is the fact that the lateral
deformation of the soil sample is completely hindered.
On the basis of the compressibility test can be drawn the load-settlement curve (fig. 2).
From the load-settlement curve is determined: specific deformation:
0
ii
h
h100
[%] and
oedometric modulus of deformation:
ppptgM
12
12
M value is computed for the pressures: p1 = 200 kPa and p2 = 300 kPa; this value is
quoted M2-3.
7. Shear strength of soils, definition, Coulomb’s Law.
Answer 7:
The application of an exterior load upon a soil mass (fig. 1) and its own weight
develops in its mass normal and tangential stresses. Normal stresses produce reduction in
voids volume and tangential stresses tend to displace soil particles laterally one towards the
other. The shear strength of soils f opposes to displacements produced by tangential stresses,
being generated by the bound forces between its particles.
Fig. 1. Shear strength emphasizing Fig. 2. Coulomb’s line:
a – cohesionless soil; b – cohesive soil
By the shear strength of a soil we understand the resistance opposed by it to
shearing of the bonds between the particles, being equal as value with the tangential
stress that produces shearing.
Coulomb’s Law: - for cohesionless soils: f = tg,
- for cohesive soils: f = tg + c
The equation of the lines presented above is called the Coulomb’s line, defined by
two parameters: inclination to the horizontal, representing the angle of interior friction, and
the ordinate to the coordinate system origin, representing the cohesion of the soil c.
8. Lateral Earth pressure of soils. Earth pressure diagrams due to distributed overload.
Answer 8:
In the case when on the surface of the retained soil mass acts a uniform distributed
load q (fig. 1), it is replaced by a soil layer, with a height i and with the same unit weight as
one of the soil behind the retaining wall: qi
So, by this replacement, can be considered that on H+i height is found a homogeneous
soil layer with unit weight, for which corresponds a triangular pressure diagram (abc).
Fig. 1. Pressure diagram
Values of the pressure in B and A points are:
aaB KqKip
aaaA KiKHKiHp
aa KqKH
The pressure diagram corresponding to
lateral Earth pressure due to distributed overload q is
represented by afed rectangle and that corresponding
to the soil behind the retaining surface by fbe
triangle.
9. Retaining walls. Retaining walls classification and check of the pressure distribution
on the base of the retaining wall.
Answer 9:
Retaining structures are classified in gravity retaining walls, cantilever retaining walls,
precast retaining walls and reinforced soil retaining structures.
For pressure verification on the ground all exterior loads are reduced to the middle
of the wall foundation base, obtaining N, M and T. The ground pressure is:
Fig. 1. Calculus scheme for
gravity retaining walls
B
e61
B
N
6
B1
M
1B
N
W
M
A
Np
22,1
where B –
width of the wall base, e = M/N – eccentricity of N
load towards the middle of the wall base.
The retaining wall dimensions are properly
chosen if the condition is fulfilled:
p1 ≤ pall in which pall represents the
allowable pressure on foundation ground at the level of
contact surface with the foundation base.
If the above condition is not fulfilled, the
retaining wall dimensions must be changed.
10. Retaining walls. Retaining walls stability checks.
Answer 10:
Fig. 1. Calculus scheme for
gravity retaining walls
Stability check to overturning.
Under the action of the horizontal component Pah
of the Earth lateral pressure, the retaining wall can lose
its stability by overturning around the side in front of the
foundation. Stability check to overturning is assured if
the following condition is fulfilled:
Mr ≤ mrMs in which:
Mr – overturning bending moment;
Ms – stability bending moment;
mr – workability coefficient equal to 0,8.
In the case of calculus scheme from fig. 1, the
two bending moments can be written as: Mr = Pahh and
Ms = Gd + PavB
Stability check to sliding along the base. This verification consists in comparing the
friction force between the foundation base and the soil with the resultant of the horizontal
loads (T component), according to relationship: T < mhμN in which:
N – resultant of vertical loads (N = G + Pav);
mh - workability coefficient equal to 0,8;
μ – friction coefficient between the foundation base and soil: where, μ = tg and =
(1/3…1/2) is the angle of friction between the soil and the retaining wall base or μ = tg ,
where is the soil internal angle of friction. This relationship is used in the case of realizing a
plain concrete base key placed at the wall foundation base level.
V. CASE STUDY / PROBLEMS
Discipline: SOIL MECHANICS
Problem 1
Determine the physical characteristics of a sand that in natural state has the moisture
content w = 25%, unit weight γ = 17,5 kN/m3 and solid particles unit weight γs = 26,5 kN/m
3.
Solution 1:
The unit weight of the soil in dry state results from the relationship:
3kN/m 0,14100251
5,17
1
wd
The porosity is determined with the relationship:
%2,471005,26
0,145,26100%
s
dsn
The void ratio is given by the relationship:
894,01002,471
1002,47
1
n
ne
Problem 2
A saturated clay sample weights in natural state, m1 = 490,2 g, and after drying,
m2 =368,2 g. The solid particles unit weight, γs, was determined in the laboratory and is 27,2
kN/m3. Determine the following physical characteristics of the clay: moisture content, w, void
ratio, e, porosity, n.
Solution 2:
The moisture content is given by the relationship:
%1,331002,368
2,3682,490100
2
21
m
mmw
The void ratio is:
90,010
2,27
100
1,33
w
swe
The porosity is:
%4,4710090,01
90,0100
1
e
en
Problem 3
For a clayey soil was determined the moisture content, w = 40%, the plastic limit, wP
= 15% and the liquid limit, wL = 60%. Determine the plasticity index, IP and the consistency
index, IC.
Solution 3:
The plasticity index is given by the relationship:
%451560 PLP wwI
The consistency index is given by the relationship:
44,01560
4060
PL
LC
ww
wwI
Problem 4
Determine the oedometric modulus of deformation, M2-3 and the deformation modulus
of the soil, E for a clayey sand (with the consistency index, IC = 0,55 and void ratio, e = 0,47)
that has the following specific deformations: for a load of 50 kPa, ε0 = 1,20%, for 100 kPa
ε1 = 2,13%, for 200 kPa, ε2 = 3,95%, for 300 kPa, ε3 = 5,15% , for 500 kPa, ε4 = 7,49%, and
for 300 kPa, ε5 = 7,31%, for 100 kPa, ε6 = 6,70% and that are represented on a curve below:
Solution 4:
The oedometric modulus of deformation is given by the relationship:
kPappp
M 33,8395,315,5
200300
23
20030032
The soil deformation modulus is given by the relationship:
32 MME o
Because the analyzed soil is a clayey sand, with the consistency index, IC = 0,55 and
with the void ratio, e = 0,47, the value of the correction coefficient M0 = 1,6 according to
STAS 8942/1-84.
So, kPaE 33,13333,836,1
Problem 5
On soil samples with the cross-section of 36 cm2 were carried out direct shear tests,
being obtained the following results:
σ 100,00 kPa 200,00 kPa 300,00 kPa
τmax 107 kPa 122 kPa 137 kPa
Determine the shear strength parameters, the angle of internal friction Φ and the
cohesion c (using the method of the smallest rectangles) and draw the Coulomb’s line.
It is specified that the relationships for the shear strength parameters determination
using the method of the smallest rectangles are:
2
11
2
111
n
i
n
i
n
fi
n
i
n
fii
n
n
tg
2
11
2
1111
2
n
i
n
i
n
i
n
fii
n
fi
n
i
n
c
Solution 5:
Using the method of the smallest rectangles, the angle of internal friction is given by
the relationship:
15,03002001003002001003
1371221073002001001373001222001071003tan
2222
Φ = 8,530
Using the method of the smallest rectangles, the soil cohesion is given by the
relationship:
kPac 923002001003002001003
3002001001373001222001071001371221073002001002222
222
With the aid of the pairs σ and τmax is drawn the Coulomb’s line.