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.