Dr Mohammad Rezania

Department of Civil Engineering

February 2014

Lecture 6

H21G11: Geotechnics 1

Basic soil mechanics

1. Introduction1.1. Review of Stress1.2.Review of Strain1.3.Stress-Strain Analysis

2. Strength Analysis in Geotechnics2.1. Shear Strength2.2.Coulomb’s Failure2.3.Mohr Stress Circle

3. Material Behaviour4. Summary

Outline

1. Introduction

� All civil engineering projects impart loadsonto the ground that supports them.

� Loads produce stresses� Stresses might cause problems (when above soil strength),

such as failure or excessive settlement(deformation).

Introduction

Nicoll Highway Collapse(Singapore)

Leaning Tower of Pisa (Italy)

1.1. Introduction: Stress

Stress - a medicalterm for a wide range of strong external stimuli,

both physiological and psychological.

Stress – a mechanicalmeasure of force intensity, either within or on

the bounding surface of a body subjected to loads

What is the definition of stress?

F

F

Area, A

F

vσ

Vertical Force

CrossSectional Areav

F

Aσ = = Unit: N/m2 or Pa

usually use kPa (1000 kPa)

Note: in soil mechanics we use a sign convention that compressivenormal stresses are positive.

In Mechanics:

The intensity of force over an area is called stressIf we are simply concerned about the normal stress:

What is the definition of stress?

�The concept ofStress = Force /Area was introduced into theory of elasticity by Cauchy in about 1822.

�Universally used as an expedientin engineering design and analysis.

�Cannot be measured directly

�Gives no indication how forces are transmitted through a stressed material

Review of stress

Stressesin soils occur in all directions and to do stability analysis it is often necessary to relate the stresses in one particular direction to those in other directions.

• normal stress

σσσσ = Fn / A

• shear stress

ττττ = Fs / A

Review of stress (contd.)

3-D 2-D

Shear stress

Normal stress

(+)

(+)

(–)

The state of stress in the ground is three dimensional.

Review of stress (contd.)

It will be discussed in more details in Mohr

circle section.

1.2. Introduction: Strain

�When materials are subjected to stresses, they respond by deforming.

�Strain is an intensity of deformation given by a displacement over a unit gauge length

ε = dL / L 0 (%)

�Normal stresses produce normal strains and shear stresses produce shear strains.

Review of strain

= −dZZ

ε

�Normal strain (ε) is the change in length divided by the initial length.

dZ

Z

Normal strain (εεεε)

dH

Z = −dHZ

γ

�Shear strain (γγγγ) is the angle of deformation expressed in radians.

Shear strain (γγγγ)

1.3. Introduction: Stress-Strain Analysis

• normal strain

ε = δz / zo

• shear strain

γ = δh / zo

Stress-strain analysis

Strains in soils occur in all directions and it is often necessary to relate the strains in one particular direction to those in other directions.

General case

Stress-strain analysis (contd.)

The state of stress in the ground is three dimensional. The algebra for 3-D states becomes quite complex.

In practice, the majority of geotechnical problems are simplified into 1-D or 2-Dconditions.

• One-dimensional

• Axisymmetric

• Plane-strain

Stress-strain analysis (contd.)

Stress-strain analysis (contd.)

One-dimensional (Oedometric)-applicable to:�Ground conditions below wide foundation, embankments and excavations

�Consolidation tests in Oedometer

Geotechnics 3

Before loading

After loading

Geotechnics 3

Before loading

After loading

Before loading

After loading

Assume 1D for ground settlement calculation

(a simplified test for solving consolidation problems).

σ’z

εh=0 εh=0

∆H

H0

Stress-strain analysis (contd.)

In one-dimensional condition:�Horizontal strain (εh) is assumed to be zero.

Stress-strain analysis (contd.)

In one-dimensional condition:�Consider an element of soil where initially V s=1.

e0 is the initial void ratio (before stress increment)

0 0

V H

V Hvε ∆ ∆= =1. Average volumetric strain:

∆e

1

eo

Time = 0+ Time = ∞

01v

e

eε ∆=

+2. Average volumetric strain:

0 0

H

H 1

e

e

∆ ∆=+

Therefore:

σ’z

εh=0 εh=0

∆H

H0

At the beginning of 1-D compression test, the void ratio and thickness of the specimen were 0.891 and 19mm, respectively.

Find the final void ratio ef if the soil consolidated by 3.52mm.

Task 1: One dimensional example

23

σr , εr σr , εr

σa, εa

σa ≠ σr εa ≠ εr

Stress-strain analysis (contd.)

Axisymmetric (triaxial)-applicable to:�Ground conditions below circular foundation.

�Shear tests in triaxial cell.

σσσσ1, εεεε1

σσσσ2

εεεε2 = 0

σσσσ3, εεεε3

Stress-strain analysis (contd.)

Plane-Strain-applicable to:�Slopes, dams, embankments, retaining walls, etc.

�Shear tests in plane-strain (biaxial) apparatus.

Tarbela Dam (Islamabad, Pakistan)

Stress-strain analysis (contd.)

Earth dam can be idealised into a plane-strain problem.

2. Strength Analysis in

Geotechnics

Steel

Tensile

strength

Concrete

Compressive

strength

Soil

Shear

strength

void ratio & effective stress

(presence of pore water)

Complex

Behavior!

Strength of different materials (contd.)

Soil strength needs to be evaluated in many problems.

Failure under foundations

Soil strength needs to be evaluated in many problems.

Failure under foundations (contd.)

Soil strength needs to be evaluated in many problems.

A failure surface can be identified in many cases along which shear stress reaches the shear strength of soil.

Failure under foundations (contd.)

σnτ

τ

At failure : shear resistance = shear strength

τ = τf

σn

Applied shear stress

Soil shearresistance

Soil strength needs to be evaluated in many problems.

Shear strength (contd.)

Failure under foundations

Soil strength needs to be evaluated in many problems.

Failure in slopes

Soil strength needs to be evaluated in many problems.

Failure in slopes (contd.)

Soil strength needs to be evaluated in many problems.

A failure surface can be identified in many cases along which shear stress reaches the shear strength of soil.

Failure in slopes (contd.)

Lower San Fernando Dam Failure, 1971

Failure in slopes

Soil strength needs to be evaluated in many problems.

Failure in retaining walls

Soil strength needs to be evaluated in many problems.

Failure in retaining walls (contd.)

Soil strength needs to be evaluated in many problems.

A failure surface can be identified in many cases along which shear stress reaches the shear strength of soil.

Failure in retaining walls (contd.)

Failure in retaining walls

2. Shear Strength

σnτ

τ

At failure : τ = τf

σn

Applied shear stress

Soil shearresistance

Soil strength needs to be evaluated in many problems.�Soil generally fails in shear.

Shear strength

Direct Shear Test apparatus

Direct Shear Test apparatus (contd.)

RollersPorous plates

Top platen

Measured

Shear force

Normal load

Motor

drive

Motor

drive

Direct Shear Test apparatus (contd.)

RollersPorous plates

Top platen

Measured

Shear force

Normal load

Motor

drive

Direct Shear Test results

Example:�Dimensions of shear box: 100×100×100mm.

Horizontal Dial,

Δh (mm)

Shear Force,

Fh (N)

Vertical Dial,

Δv (mm)

0 0 0

1 226 -0.2

2 290 0.1

4 338 1.2

6 322 2.2

10 296 3.9

15 275 5.9

20 262 7.1

25 253 7.8

30 251 8.1

35 250 8.2

Horizontal Dial,

Δh (mm)

Shear Force,

Fh (N)

Vertical Dial,

Δv (mm)

0 0 0

1 27 -0.3

2 55 -0.6

4 92 -0.9

6 121 -1.2

10 158 -1.7

15 189 -2.4

20 209 -2.7

25 227 -3.0

30 234 -3.1

35 238 -3.1

Test 1: Dense sand, M = 2124g Test 2: Loose sand, M = 1891g

Direct Shear Test results

Example: Granular soils

Mineral frictions

Interlocking

Mineral frictionsDense:

Loose:

Critical State (CS)

Important features of soil response can be summarised in two general groups of soils.

�Type I soils:� Loose sands, NC and lightly OC clays (OCR<2)

�Type II soils:� Dense sands, and heavily OC clays (OCR>2)

Notes: Preconsolidation stressor past max effective stress, σ’ zc, is the max vertical effective

stress that soil was subjected to in the past.

Normally consolidated soilis one that has never experienced vertical effective stresses greater than its current vertical effective stress (σ’z=σ’zc).

Overconsolidated soilis one that has experienced vertical effective stresses greater than its existing vertical effective stress (σ’zc>σ’z).

Overconsolidation ratio (OCR) is the ratio by which the current vertical effective stress in the soil was exceeded in the past (OCR=σ’zc/σ’z).

Soil response in shearing

Important features of soil response can be summarised in two general groups of soils.

�Type I soils:� Loose sands, NC and lightly OC clays (OCR<2)

�Type II soils:� Dense sands, and heavily OC clays (OCR>2)

Notes: Preconsolidation stressor past max effective stress, σ’ zc, is the max vertical effective

stress that soil was subjected to in the past.

Normally consolidated soilis one that has never experienced vertical effective stresses greater than its current vertical effective stress (σ’z=σ’zc).

Overconsolidated soilis one that has experienced vertical effective stresses greater than its existing vertical effective stress (σ’zc>σ’z).

Overconsolidation ratio (OCR) is the ratio by which the current vertical effective stress in the soil was exceeded in the past (OCR=σ’zc/σ’z).

Soil response in shearing

Effective stress concept will be explained in coming

lectures.

Soil’s simple shear deformation

Type I soils

Type II soils

Soil’s simple shear deformation

Effect of the normal stress level:

Low normal force

Dilates

High normal force

Shears off

Soil’s simple shear deformation

Effect of the normal stress level:

Friction angle: will be discussed in more details, later!

Soil’s simple shear deformation

Effect of OCR:

2.2. Coulomb’s Failure

� In terms of stresses

Shear stress required to initiate slip: τf = tanφ’ (σn’) f

Coulomb’s failure criterion

Coulomb’s frictional law� In terms of forces

Horizontal force required to initiate movement: Hf = µW

W

N

H

H ≤ μΝ

τ

τ

σn'

σn'

µ: coefficient of static sliding friction

φ’: friction angle φ’= tan -1µ

Coulomb’s failure criterion (contd.)

The slip plane does not have to be horizontal!

τf vs. (σn’) f � straight line OA if φ’ = φ’ cs

Coulomb’s law: applicable at or near the CS!

What about the peak behavior?

( )W

tanH

f

n f

τϕ

σ′= =

′

HW

=µ + tanα

1 − µ tanα=

tanφ + tanα1 − tanφ tanα

Coulomb’s failure criterion (contd.)

Shear failure may be modeled using Coulomb’s frictional law.

⇒

ΣFx = 0⇒ H = N (sinα + µ cosα) (

ΣFz = 0⇒W = N(cosα− µsinα) (

⇒

( ) ( )tanf n fτ σ ϕ α′ ′= +

( ) ( ) ( )tan tantan

1 tan tanf n nf f

ϕ ατ σ σ ϕ αϕ α

′ +′ ′ ′= = +′−

Coulomb’s failure criterion (contd.)

p cs pϕ ϕ α′ ′= +

( ) ( )tancs n csfτ σ ϕ′ ′=

( ) ( )tanp n pfτ σ ϕ′ ′=

The effect of dilation:� Increase of shear strength; curved failure envelope

α>0 � soil expands during shearing

Coulomb’s law:� Gives information about the soil shear strength when slip is initiated

� No information on the strains at which soil failure occurs

Friction angle:� @ critical state: related to sliding friction

� @ peak state: depends on the soil’s capacity to dilate

The effects of cohesion, soil tension and cementation:

Coulomb’s failure criterion (contd.)

2.3. Mohr Stress Circle

We can calculate σV’ andσH’ but sometimes, we need to calculate stressesacting on other planes, e.g.the slip surface along a slope.

σv’

σH’

σσσσnf’

ττττnf’

Introduction

Definitions

Forces acting on soil:

Definitions (contd.)

Forces acting on soil:

Let’s zoom in:

y

x

z

Definitions (contd.)

Forces acting on soil:

Let’s zoom in:

Definitions (contd.)

Forces acting on a plane:

y

x

z

∆∆∆∆Fyx

∆∆∆∆Fxx

∆∆∆∆Fzx

Definitions (contd.)

Stresses:∆Ax =∆y.∆z

0Ax

xxxx

xAF

itlim→∆

∆∆=σ

0Ax

y xyx

xA

Fitlim

→∆

∆∆

=σ

0Ax

zxzx

xAF

itlim→∆

∆∆=σ

σxx σxy σxz

σyx σyy σyz

σzx σzy σzz

σσσσ =

Stresses in different planes

Stresses in different planes (contd.)

Stresses in different planes (contd.)

3-D Stresses:

y

x

z

σσσσxx σσσσxy σσσσxz

σσσσyx σσσσyy σσσσyz

σσσσzx σσσσzy σσσσzz

σ σ σ σ =

σσσσxz

σσσσzz

σσσσyz

σσσσxy

σσσσyy

σσσσzy

σσσσyx

σσσσxx

σσσσzx

Stresses in different planes (contd.)

2-D Stresses:

y

x

z

σσσσxz

σσσσzz

σσσσxx

σσσσzx

σσσσxx σσσσxz

σσσσzx σσσσzz

σ σ σ σ =

x

z

σσσσxz

σσσσzz

σσσσxx

σσσσzx

σσσσzz

σσσσxx

σσσσzx

σσσσxz

Stresses in different planes (contd.)

Sign conventions (reminder):�Normal stresses: compression positive�Shear stresses: counter-clockwise positive

ττττxz

ττττxz

σσσσz

σσσσx

ττττxz

σσσσz

σσσσx

ττττxzx

z +ive

-iveττττxz

ττττxz

σσσσz

σσσσx

ττττxz

σσσσz

σσσσx

ττττxzx

z -ive

+ive

Stresses in different planes (contd.)

ττττxz

ττττxz

σσσσz

σσσσx

ττττxz

σσσσz

σσσσx

ττττxzx

z

Stresses in different planes (contd.)

ττττxz

ττττxz

σσσσz

σσσσx

ττττxz

σσσσz

σσσσx

ττττxzx

z

ττττn

σσσσn

θθθθ

A

B

O

|AO| = |AB| cos θθθθ|BO| = |AB| sin θθθθ

Stresses in different planes (contd.)

Equilibrium of forces:

ττττxz

ττττxz

σσσσz

σσσσx

ττττxz

σσσσz

σσσσx

ττττxzx

z

ττττn

σσσσn

θθθθ

A

B

O

|AO| = |AB| cos θθθθ|BO| = |AB| sin θθθθ

σσσσn|AB|= σσσσx|BO|sin θθθθ+σσσσz|AO|cos θθθθ+ττττxz|BO|cos θθθθ+ττττxz|AO|sin θθθθ

σσσσn|AB|= σσσσx|AB|sin 2θθθθ+σσσσz|AB|cos 2θθθθ+ττττxz|AB|sin θ θ θ θ cos θθθθ+ττττxz|AB|sin θθθθ cos θθθθσσσσn=σσσσx sin 2θθθθ+σσσσz cos 2θ θ θ θ + 2ττττxz sin θθθθ cos θθθθ

θτ+θσ−σ+σ+σ=σ 2sin2cos22 xz

xzxzn

ττττn|AB|= - σσσσx|BO|cos θθθθ+σσσσz|AO|sin θθθθ

+ττττxz|BO|sin θθθθ - ττττxz|AO|cos θθθθ

ττττn= -σσσσxsin θθθθ cos θθθθ +σσσσzcos θθθθ sin θθθθ

+ττττxzsin 2θθθθ – ττττxzcos 2θθθθ

θτ−θσ−σ=τ 2cos2sin2 xz

xzn

(1)

(2)

Defined as σn that act on the surfaces where there is no shear stress, i.e. τn = 0:

Principal stresses

ττττn

A

B

O

ττττxzσσσσx

σσσσz

ττττxzx

z σσσσn

θθθθ

θτ−θσ−σ=τ 2cos2sin2 xz

xzn

Principal stresses (contd.)

xz

xz22tan

σ−στ=θ

A

B

O

ττττxzσσσσx

σσσσz

ττττxzx

z σσσσ1,3

θθθθ

θτ−θσ−σ=τ 2cos2sin2 xz

xzn 0=

Defined as σn that act on the surfaces where there is no shear stress, i.e. τn = 0:

(3)

xz2

2xzxz

1 22τ+

σ−σ+σ+σ=σ

Major principal stress:

xz2

2xzxz

3 22τ+

σ−σ−σ+σ=σ

Minor principal stress:

substituting Eq.(3) into Eq.(1):

Principal stresses (contd.)

xz2

2xzxz

1 22τ+

σ−σ+σ+σ=σ

Major principal stress:

xz2

2xzxz

3 22τ+

σ−σ−σ+σ=σ

Minor principal stress:

There are always 2 planes on which there is no shear stress. These planes are called principal planesand the normal stresses applied on them are principal stresses.

3

1

σσσσ1

σσσσ3

σσσσ1

σσσσ3

It is now possible to calculate the normal stress σσσσn and shear stress ττττn onany plane, as long as we know the principal stresses.

Principal stresses (contd.)

σn =σ1 +σ3

2+

σ1 −σ3

2cos2θ

τn =σ1 −σ 3

2sin2θ

Stresses on other planes based on the principal planes:�Assuming σσσσy=σσσσ1 and σσσσx=σσσσ3 it is possible to write Eqs. (1)

and (2) in terms of principal stresses.

θ is measured counter-clockwise

w.r.t. the major principal plane.

σσσσ3

σσσσ1

3

1

ττττn

σσσσn

θθθθ

�These equations are known as the double angle equationsand are awkward to use in the practice.

� It is easier to use a graphical procedure based on Mohr circle of stress (Mohr 1882) plotted in the τ-σspace.

�Mohr circle represents the state of stress at a point of equilibrium .

� It applies to any material, not just soil.

Mohr circle

(+)

(–)

Mohr circle (contd.)

σσσσ3

σσσσ1

3

1

ττττn

σσσσn

θθθθ

σσσσ

ττττ

(–)

(+)

Mohr circle (contd.)

σσσσ3

σσσσ1

3

1

ττττn

σσσσn

θθθθ

σσσσ

ττττ

σn

σ1σ3

τn

2θ2θ2θ2θ

σn =σ1 +σ3

2+

σ1 −σ3

2cos2θ

τn =σ1 −σ 3

2sin2θ

1 3

2

σ σ−radius =

1 3

2

σ σ+centre =

Mohr circle - Pole method

σσσσ3

σσσσ1

3

1

ττττn

σσσσn

θθθθ

Pole or Origin of Planes - a very useful property:�Any straight line drawn through the pole will intersect the

Mohr circle at a point which represents the state of stress on a plane inclined at the same orientation in space as the line.

σσσσ

ττττ

σσσσ1σσσσ3

σσσσ3

σσσσ1

3

1

ττττn

σσσσn

θθθθ

Pole or Origin of Planes - a very useful property:�Any straight line drawn through the pole will intersect the

Mohr circle at a point which represents the state of stress on a plane inclined at the same orientation in space as the line.

σσσσ

ττττ

σσσσ1

θθθθσσσσ3

OP

Pole orOrigin of Planes

Mohr circle - Pole method (contd.)

Task 2: Mohr circle example 1

Stresses on the 30o plane?

3

1σσσσ3=20

σσσσ1

σσσσ3

σσσσ1=40

30o

??

Task 3: Mohr circle example 2

Stresses on the 35o plane on the 20o inclined surface?

σσσσ1=52

σσσσ3

σσσσ1

σσσσ3=12

20o

�No simple methodexists for drawing Mohr circles to represent the general 3-D case, in which both normal and shear stresses act on all six faces of the cubical element.

�There is one simple case, however, which can be represented by three Mohr circles:� Cubical element which has only principal stresses acting

on the six faces.

Mohr circle for 3-D

Mohr circle for 3-D (contd.)

6. Material Behaviour

�The analyses developed for stresses and strains using Mohr circles are independent of the material and they are equally applicable to steel, concrete, soil or rock.

� In order to analyse any kind of structure or solid it is necessary to have relationships between stresses and strainsfor a specific material. These are called constitutive relationships.

�The link between stresses and strains is governed by the properties of the material.

Introduction

Strain, εεεε

Stress,

σσ σσ

Linear Elastic

Stress,

σσ σσStrain, εεεε

Nonlinear Elastic

Stress-strain relationships

Strain, εεεε

Stress,

σσ σσ

Elasto-plastic

Stress,

σσ σσ

Strain, εεεε

Elastic Perfectly Plastic

Stress-strain relationships (contd.)

Strain, εεεε

Stress,

σσ σσ

Elasto-PlasticStrain Hardening

Stress,

σσ σσ

Strain, εεεε

Elasto-PlasticStrain Softening

Stress-strain relationships (contd.)

� Elasticity: all of the work done by external stresses during an increment of deformation is recoveredon unloading.

� The response of an isotropic linear-elastic material is described using two independentelastic constant

� Young’s modulus (E)

� Poisson’s ratio (�)

uniaxial stress increment dE = =

uniaxial strain increment dσε

d-(lateral strain increment) = =

axial strain increment d⊥εν

ε�

Elasticity

σ

Volumetric strain, εv

Isotropic stress,

σ

dσ

dεv

σ

εv

τ

τ

Shear strain, γ

Shear stress,

τ

dτ

dγ

γ

Elasticity (contd.)

An alternative to the use of the practical parameters E and νννν is to use the more fundamental parameters G and K .

� Shear modulus (distortionaldeformation) G = dττττ/dγγγγ� Bulk modulus (volumetricdeformation) K = dσσσσ/dεεεεv

EG =

2(1+ ν)E

K =3(1- 2ν)

Elasticity (contd.)

� In soil mechanics G and K are preferable to E and ννννbecause it is important to consider shearing (change of shape) separately from compression (change of size).

�Therefore, it is usefulto be able to convert from Eand νννν into G and K .

�For isotropic elastic materials the elastic constants G and K can be expressed in terms of E and νννν in the following manner:

9KGE =

G + 3K3K -2G

=2(G + 3K)

ν

Elasticity (contd.)

�Knowing any two elastic quantities we can recover any other elastic parameters

�For example, if we know K andG; E andνννν can be obtained from:

Task 4: Elastic parameters example

In a triaxial compression test on an isotropic and elastic sample of soil, the radial stress is held constant at σσσσr = 200kPaand the axial stress is changed from σσσσa=350kPa to 360kPa.The strains for this increment were dεεεεa=0.05% and dεεεεr=-0.01%. Determine E, νννν, G and K .

Stiffness: the gradient of the stress-strain curve

Strain, ε

Stress,

σ

Gradient = stiffness

Stiffness

Stiffness (contd.)

7. Summary

Summary

�The state of stress in the ground is three dimensional.

�The majority of geotechnical problems are simplified into 1-D or 2-D conditions.

�Soil strength is the maximum shear stress that it can sustain.

�Principal stresses are normal stresses on planes with no shear stress.

�Knowing the principal stresses, the stresses on any other plane can be determined using the Mohr circle.

�Pole method is a practical way to calculate stresses on any plane from known principal stresses.

Tools for Mohr Circle

�Compass

�Protractor

�Ruler