Soil Mechanics “Lecture notes”

Prepared by

Dr. Eng. Asad Hafudh Al-Defae

University of Wasit College of Engineering

Civil eng. Division

Karl Terzaghi (1883–1963)

Grading

• Attendance, Class Participation and quizzes

10% for each term

• Midterm Exam (20%)

• Laboratory attendance, participation and experiments

reports (10%)

• Final Exam (60%)

Description of sections

• Origin of Soil, soil composition and index

properties

• Weight-Volume Relationships, Plasticity and

Structure of Soil

• Engineering Classification of Soil

• Permeability

• Fluid flow in soil

• In Situ Stresses (Effective Stress Concept)

• Stresses in a Soil Mass

• Compressibility of Soil

• Shear Strength of Soil

• Soil Compaction

Ref.

• Textbook- Das, B.M., Principle of Geotechnical

Engineering 5th

edition.

• Soil mechanics laboratory manual 6th ed. Braja M. Das

2000 • Soil mechanics, R.F. Craig

• The term Soil has various meanings, depending upon the general field in which

it is being considered.

• To a Pedologist ... Soil is the substance existing on the earth's surface, which

grows and develops plant life.

• To a Geologist ..... Soil is the material in the relative thin surface zone within

which roots occur, and all the rest of the crust is grouped under the term ROCK

irrespective of its hardness.

• To an Engineer .... Soil is the un-aggregated or un-cemented deposits of mineral

and/or organic particles or fragments covering large portion of the earth's

crust.

Soil Mechanics is one of the youngest disciplines of Civil Engineering involving the

study of soil, its behavior and application as an engineering material.

According to Terzaghi (1948): "Soil Mechanics is the application of laws of

mechanics and hydraulics to engineering problems dealing with sediments and

other unconsolidated accumulations of solid particles produced by the mechanical

and chemical disintegration of rocks regardless of whether or not they contain an admixture of organic constituent."

• Soil Mechanics is the branch of civil engineering that concerns the application of the principles of mechanics, hydraulics and to smaller extent, chemistry, to engineering problems related to soils.

• The study of the science of soil mechanics equips a civil engineer with the basis scientific tools needed to understand soil behavior.

Define: Soil

• Naturally occurring particulate material.

• Formed, directly or indirectly, from solid rocks (i.e.

weathering)

• Composition of soil particles depends on composition of

parent rock

• The void spaces between the particles contain water and/or

air.

Applications

• Engineer uses the SOIL to build:

• On it: e.g. buildings

• In it: e.g. tunnels

• With it; e.g. earth dams

Basic points:

Every civil engineering structure, whether it be a

building, a bridge, a tower, an embankment, a road

pavement, a railway line, a tunnel or a dam, has to

be founded on the soil (assuming that a rock

stratum is not available) and thus shall transmit

the dead and live loads to the soil stratum.

Example: Bridge foundations

Example: Tunnels

Example: Earth dams

Example: Failure

Soil formation A- Geoloical origin (by nature).

1- mechanical weathering process (disintegration of rocks).

• Physical forces.

• Impact of sand grain carried by high winds or water or glaciers.

2- Chemical weathering of rocks

• Due to water, acids or alkaline

• Oxidation – union of oxygen with minerals in rocks forming

another minerals

• Hydration – water will enter the crystalline structure of

minerals forming another group of minerals

• Hydrolysis – the release Hydrogen from water will union with

minerals forming another minerals

• Carbonation – when Co2 is available with the existence of

water the minerals changed to Carbonates

B- Fill (man made)

Residual soil: the products of rock weathering remain at their

original location.

Transported soil: the products are transported and deposited in

a different location. (gravity, wind, water).

Marians soil: formed by deposition in the SO4.

Soil-Particle Size or Grain Sizes

Between boulder (D>75 mm) to Ultra fine grain (D<0.001mm)\

Coarse-grained, Granular or Cohesionless Soils • Excellent foundation material for supporting structures and roads. • The best embankment material. • The best backfill material for retaining walls. • Might settle under vibratory loads or blasts. • Dewatering can be difficult due to high permeability. • If free draining not frost susceptible Fine-Grained or Cohesive Soils • Very often, possess low shear strength. • Plastic and compressible. • Loses part of shear strength upon wetting. • Loses part of shear strength upon disturbance. • Shrinks upon drying and expands upon wetting. • Very poor material for backfill. • Poor material for embankments. • Practically impervious. • Clay slopes are prone to landslides.

Silts Characteristics • Relatively low shear strength • High Capillarity and frost

susceptibility • Relatively low permeability • Difficult to compact Compared to Clays • Better load sustaining qualities • Less compressible • More permeable • Exhibit less volume change

Solutions of soil engineering problems

• Soil Mechanics

• Geological exploration

• Experience

• Economic

Engineering Judgment solution of soil engineering problem

Soil may be divided into three main classes: 1. Coarse – grained or non-cohesive soil 2. Fine – cohesive soil 3. Organic soil

Coarse grain soil Boulders > 300mm Cobble 150-300mm Gravel 2-150mm Sand 0.06-2mm Fine grain soil Silt 0.002-0.06mm Clay <0.002mm

Building block of clay mineral • Silicon – Oxygen tetrahydra (Sio4) Silica sheet • Aluminium or Magnisium sheet (Al2(OH)3, Mg2(OH)3- Gibbsite sheet

Common clay minerals of (two layers sheet) • Kaolinite and Halloysite

Common clay minerals of (Three layers sheet)

• Muscovite (Mica) and Illite (Hydrous mica)

Consistency: When clay mineral are present in fine grained soil, and soil can be remolded in presence of some moisture without rumbling, this cohesive nature is due to the absorbed water surrounding the clay minerals.

Atterberg limits: the limits are

based on the concept that the fine

grain soil can exist in any four state

depends on its water content.

L.L. = Liquid limit P.L. = Plastic Limit. S.L. =Shrinkage limit. P.I. = Plastic index

P.I.= L.L. –P.L.

𝑭𝒍𝒐𝒘 𝑰𝒏𝒅𝒆𝒙 = 𝑰. 𝑭. =𝝎𝟏 − 𝝎𝟐

𝐥𝐨𝐠 (𝑵𝟐𝑵𝟏

)

𝑻𝒐𝒖𝒈𝒉𝒏𝒆𝒔𝒔 𝑰𝒏𝒅𝒆𝒙 = 𝑰. 𝒕. =𝑷. 𝑰.

𝑰. 𝑭.

𝑳𝒊𝒒𝒖𝒊𝒅 𝑰𝒏𝒅𝒆𝒙 = 𝑳. 𝑰. =𝝎𝒄 − 𝑷. 𝑳.

𝑳. 𝑳. −𝑷. 𝑳.

If L.I. < 0 𝝎𝒄 < 𝑷. 𝑳. Soil in solid or semi-solid state

If L.I. = 0 𝝎𝒄= 𝑷. 𝑳 Soil plastic limit

If L.I. > 0 𝝎𝒄> 𝑷. 𝑳 Soil in plastic state

If L.I. = 1 𝝎𝒄= 𝑳. 𝑳 Soil liquid limit

If L.I. < 0 𝝎𝒄> 𝑳. 𝑳 Soil in liquid state

Activity of Clay: because of great increase of surface area per mass (specific surface) it

may be expected that the amount of attached water will be influenced by the amount of

clay minerals that is present in the soil.

Activity of Clay =𝑃. 𝐼.

% 𝑏𝑦 𝑤𝑒𝑖𝑔𝑕𝑡 𝑓𝑖𝑛𝑒𝑟 𝑡𝑕𝑎𝑛 0.002

Activity of Clay depends on : • Specific surface • Amount of clay minerals • Type of clay mineral

The physical state of a soil sample

Total Volume = Vt = Vs + Vw + Va Total Weight = Wt = Ws + Ww

Porosity (n): is the ratio of void volume. n = Vv/Vt

Void Ratio (e):is the ratio of void volume to solid volume. e = Vv/Vs

now n = Vv/Vt = Vv/ Vv+ Vs = Vv/Vs/ Vv/Vs+1

e=e+1

Note: • n < 1 and is expressed as % • e may be any value greater or smaller than unity.

Example: A soil has a total volume of 250ml and a void ratio of 0.872. What is the volume of solids of the sample?

Example: A soil has a porosity of 0.45. What is the value of its void ratio?

100Ws

Wwwc

Water Content, W: is the ratio of water weight in a soil sample to the solids weight

100Vv

VwS

Degree of Saturation, S: is the ratio of water volume to void volume.

Bulk unit weight:

Dry Unit Weight

Submerged Unit Weight

Unit weight Unit Weight of Water

1.0g/cc =1000kg/ m3

9.8 kN/m3

Specific gravity (GS): specific gravity of soil solids of a soil is defined as the ratio of the density of a given volume of the solids to the density of

any equal volume of water at 4 C.

Soil type G

Gravel 2.65-2.68

Sand 2.62-2.65

Silt 2.66-2.7

Clay 2.68-2.8

Organic soils may be less than 2.0

n

neor

e

enei

VV

VV

VV

V

V

Vn

sv

sv

vs

vv

11..

/1

/

S

GwGwc

S

G

W

W

V

V

W

W

V

V

V

V

V

V

V

Ve

scs

w

ws

s

w

w

v

ss

ww

w

v

s

w

w

v

s

v

..1

/

/

ws

t

s

c

cwst

wss

s

w

w

v

s

w

svs

sws

vs

wst

e

SeG

G

Sew

e

wG

GV

Wande

V

Vw

W

W

VVV

WWW

VV

WW

V

W

1

1

)1(

,,,

)/1(

)/1(Relationship between n and e

Relationship between w, e, S, Gs

In term of Gs, e and w

In term of Gs, e and w s

ws

sat

wws

vs

wssat

e

eGor

e

eG

VV

WW

V

W

1

1

d in terms of Gs, e and w

d in terms of Gs, w, S and w

e in terms of d, Gs and w

weightunitVoidAirZero

Gw

GSif

S

Gw

G

e

G

VV

W

V

W

V

W

sc

wsd

sc

wswsd

vs

ssd

1,1

11

S

Gw

G

e

G

VV

W

V

W

V

W

sc

wswsd

vs

ssd

11

ws

wws

wsat

d

ws

wsd

vs

ssd

e

Gor

e

eG

Ge

Ge

VV

W

V

W

V

W

1

1

1

1

)1(

In term of Gs, e and w

Relationship between dry and total Unit weight

Example 1: A moist chunk of 25kg soil had a volume of 12000 cm3. After it is dried in an oven, the weight of the dry soil became 23kg. The specific gravity of the soil material is 2.65. Using a phase diagram, determine: Water content (wc), Unit weight of moist soil, Void ratio (e), Porosity (n), Degree of Saturation (S)

Solution: Given: W=25kg, V=12000 cm3, Ws=23kg, Gs=2.65 w=1g/cm3

Calculate water content: Ww=W-Ws, Ww=25-23=2kg, so, wc=Ww/Ws=8.7%

Calculate unit weight of moist soil: =W/V=25000/12000=2.08g/cm3 Calculate void ratio: s=Gs w=(2.65)(1)=2.65g/cm3, Vs=Ws/s=8679.2cm3 Vw=Ww/w=2000/1=2000cm3, Vv=V-Vs=12000-8679.2=3320.8cm3 e=Vv/Vs=3320.8/8679.2=0.38 Calculate porosity: n=Vv/V=3320.8/12000=0.28 Calculate degree of saturation: S=Vw/Vv=2000/3320.8=0.60=60%

Example 2: A 0.01 m3 of saturated soil sample has a unit weight of 1.6g/cm3. After it is dried in the oven the weight of the soil sample reduced to 13.5kg. Use a phase diagram, determine: Weight of water for the saturated soil, Ww Water content, wc Void ratio, e Porosity, n

0.01 m3

0.01 m3

13.5kg

13.5kg

Solution: Given: Ws=13.5kg, V=0.01 m3, =1.6g/cm3, w=1g/cm3 Calculate weight of the saturated soil: W= V=(1.6)(0.01)(1000000)=16kg Calculate water content: Wc=Ww/Ws=(W-Ws)/Ws=[(16-13.5)/13.5](100)=19%

Calculate void ratio: Vw=Ww/w=2500/1=2500cm3=0.0025 m3 Vs=V-Vw=0.01-0.0025=0.0075 m3 For saturated soil: Va=0, e=Vw/Vs=0.0025/0.0075=0.33 Calculate porosity: n=Vv/V=0.0025/0.01=0.25

Example 3: A soil sample has a porosity of 0.30 and specific gravity of 2.50. Use a phase diagram, determine: Void ratio, e Dry unit weight, d Unit weight if the soil is 50% saturated, 50 Unit weight if the soil is completely saturated, sat

Solution:

Given: n=0.3, Gs=2.5, w=1g/cm3=1000kg/m3 Calculate void ratio: assume V=1 m3 Vv=nV=(0.3)(1)=0.3 m3 Vs=V-Vv=1-0.3=0.7 m3 e=Vv/Vs=0.3/0.7=0.43 Calculate dry unit weight: s=w Gs=(1000)(2.5)=2500kg/m3 Ws=sVs=(2500)(0.7)=1750kg d=Ws/V=1750/1=1750kg/m3 Calculate moist unit weight if the soil is 50% saturated: Vw=SVv=(0.5)(0.3)=0.15 m3 Ww= w Vw=(1000)(0.15)=150 kg W=Ww+Ws=150kg+1750kg=1900 kg 50=1900/1=1900 kg/m3 Calculate saturated unit weight: Vw=SVv=(1)(0.3)=0.3 m3 Ww= w Vw=(1000)(0.3)=300 kg W=Ww+Ws=300+1750=2050kg sat=2050/1=2050 kg/m3

Example 4: A soil sample has a unit weight of 1.5g/cm3. The moisture content of this soil is 20% when the degree of saturation is 50%. Determine Void Ratio, Specific Gravity of soil solid, and Saturated Unit Weight.

V= 1 cm3 W=1.5g

Solution: Given: =1.5 g/cm3, Wc=20%, S=50%, w=1g/cm3 Step1: Assume V=1 cm3 W=V=1.5g Wc=Ww/Ws, so, Ww=0.2 Ws W=0.2Ws+Ws, so, Ws=W/(1+0.2) Ws=1.25g Ww=W-Ws=0.25g Vw=Ww/w=0.25cm3 Vv=Vw/S=0.5 cm3 Vs=V-Vv=0.5 cm3 e=Vv/Vs=1

Gs=s/w, s=Ws/Vs=1.25/0.5=2.5 g/cm3 Gs=s/w=2.5 For the saturated unit weight, Vv=Vw, so, Ww=wVv=0.5g sat=(Ww+Ws)/V=1.75g/cm3