ESS2222H Tectonics and Planetary Dynamics Lecture 2 Stress, … · 2020. 1. 21. · Tectonics and Planetary Dynamics Lecture 2 – Stress, Strain University of Toronto, Department

Tectonics and Planetary Dynamics

Lecture 2 – Stress, Strain

University of Toronto, Department of Earth Sciences,

Hosein Shahnas

1

ESS2222H

2

Review of Lecture 1

Evolution of the Solar System

Solar system formed about 4.6 billion year ago from low-density cloud of interstellar gas and dust (~99% Hydrogen & Helium).

Heavy elements in the nebula were cooked in the interior of large stars.

Left from other stars Several light years 200 AU

𝐹 = − 𝐺 𝑀𝑚

𝑟2+ 𝑚𝑣2

𝑟

Proto-Sun formed Gravitational collapse T 10 MK at he center

Static equilibrium in 50 Myr Sun was born from ~99% of the nebula mass (Main-sequence star)

Main-sequence star sub-giant Red-giant White dwarf

Hydrogen burning in core

Hydrogen is used up at the

star's core, nuclear fusion of

hydrogen begins In the outer

shells, more heat causes

expansion

Sun has exhausted it fuel, It

gets cooler and the core

collapses, gravitational energy

changes to heat

3

Review of Lecture 1

Contraction Theory (Suess, Dana)

Problems with this theory

a) The oceans and continents are permanent

b) Similarities of the plants and animals at different distant region

c) Mountain folds

d) Mountain roots

e) Radiogenic heating

Continental Drift (Wegener)

Evidence

a) The apparent fit of the costliness of the continents (jigsaw-puzzle)

b) Fossil correlation

c) Rock and mountain correlation

d) Similarities of the plants and animals at different distant region

e) Paleoclimate data (Glacial striations - Coal deposits)

f) Palaeomagnetism data

Wegener could not explain the responsible driving force

Plate Tectonics

Lithosphere – Mantle – Core

Lithosphere rigid, broken in parts, in motion wrt one another

Driving force: Mantle convection

Supercontinent theory (Pangaea)

Wilson cycle

4

Stress, Deformation, and Strain

Stress & Strain Tensors

Isobaric & Deviatoric Stress

Principal Axes & Principal Stress

Isotropic & Deviatoric Strain

5

Tensors

Scalar:

Scalar quantity is a tensor of rank zero, specified by a single

component; like: temperature (T).

Vector:

Vector quantity is a tensor of rank 1, specified by three components;

like: velocity (𝑉1, 𝑉2, 𝑉3).

Tensor of rank 2:

Tensor of rank 2 is an algebric object specified by nine components;

like: stress tensor, 𝜎11 𝜎12 𝜎13𝜎21 𝜎22 𝜎23𝜎31 𝜎32 𝜎33

Stress Tensor

𝑇𝑛 = 𝑙𝑖𝑚𝛿𝐹

𝛿𝑆 𝛿𝑆 → 0 force per unit area acting on surface with orientation n

𝑋 𝑏𝑜𝑑𝑦 𝑓𝑜𝑟𝑐𝑒 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑣𝑜𝑙𝑢𝑚𝑒

∆𝑆

∆𝑉

𝑋

𝑇 𝑛

𝛿𝑆

𝜎

𝜏

𝑇

Convention:

Positive outward (tension)

Negative inward (compression)

6

Stress Tensor

Consider a cubic material element. The tractions on the three

faces can be resolved into their Cartesian components, one

normal and two tangential to the face on which the traction acts:

𝑇𝑒1 = 𝜎11𝑒1 + 𝜎12𝑒2 + 𝜎13𝑒3 𝑇𝑒2 = 𝜎21𝑒1 + 𝜎22𝑒2 + 𝜎23𝑒3 𝑇𝑒3 = 𝜎31𝑒1 + 𝜎32𝑒2 + 𝜎33𝑒3

In summation notation: 𝑇𝑒𝑖 = 𝜎𝑖𝑗𝑛𝑗

𝑇𝑒𝑖 is the traction force acting on a surface with normal vector

along 𝑒𝑖.

𝑇𝑒1 = 𝜎11, 𝜎12, 𝜎13

𝑇𝑒2 = 𝜎21, 𝜎22, 𝜎23

𝑇𝑒3 = 𝜎31, 𝜎32, 𝜎33

𝑥1 ≡ 𝑥 𝑥2 ≡ 𝑦 𝑥3 ≡ 𝑧

s://commons.wikimedia.org/w/index.php?curid=5668647

𝜎𝑖𝑗 =

𝜎11 𝜎12 𝜎13𝜎21 𝜎22 𝜎23𝜎31 𝜎32 𝜎33

7

Equations of Equilibrium

The equations of equilibrium 𝐹 = 0

Consider the components of the surface force acting in the 𝑥1 − 𝑑𝑖𝑟.

𝜎11𝑑𝑥2𝑑𝑥3 and 𝜎11 +𝜕𝜎11

𝜕𝑥1𝑑𝑥1 𝑑𝑥2𝑑𝑥3





And the body force components:

𝜌𝑋1 𝑑𝑥1𝑑𝑥2𝑑𝑥3 𝜌𝑋2 𝑑𝑥1𝑑𝑥2𝑑𝑥3 𝑋𝑖: 𝑡ℎ𝑒 𝑏𝑜𝑑𝑦 𝑓𝑜𝑟𝑐𝑒 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑚𝑎𝑠𝑠 𝑖𝑛 𝑖 − 𝑑𝑖𝑟. 𝜌𝑋3 𝑑𝑥1𝑑𝑥2𝑑𝑥3

The condition of equilibrium of forces in 𝑥1 − 𝑑𝑖𝑟.

𝜎11 +𝜕𝜎11

𝜕𝑥1𝑑𝑥1 𝑑𝑥2𝑑𝑥3 − 𝜎11𝑑𝑥2𝑑𝑥3 + 𝜎21 +

𝜕𝜎21

𝜕𝑥2𝑑𝑥2 𝑑𝑥1𝑑𝑥3 − 𝜎21𝑑𝑥1𝑑𝑥3 +

𝜎31 +𝜕𝜎31

𝜕𝑥3𝑑𝑥3 𝑑𝑥1𝑑𝑥2 −𝜎31𝑑𝑥1𝑑𝑥2 + 𝜌𝑋1 𝑑𝑥1𝑑𝑥2𝑑𝑥3 = 0

https://commons.wikimedia.org/w/index.php?curid=5668647

8

Equations of Equilibrium

𝜕𝜎11

𝜕𝑥1+𝜕𝜎21

𝜕𝑥2+𝜕𝜎31

𝜕𝑥3+ 𝜌𝑋1 = 0

Similarly if we repeat this in 𝑥2 − 𝑑𝑖𝑟 and 𝑥3 − 𝑑𝑖𝑟 𝜕𝜎12

𝜕𝑥1+𝜕𝜎22

𝜕𝑥2+𝜕𝜎32

𝜕𝑥3+ 𝜌𝑋2 = 0,

𝜕𝜎13

𝜕𝑥1+𝜕𝜎23

𝜕𝑥2+𝜕𝜎33

𝜕𝑥3+ 𝜌𝑋3 = 0

Or 𝜕𝜎𝑗𝑖

𝜕𝑥𝑗+ 𝜌𝑋𝑖 = 0

Also 𝑁 = r × 𝐹 = 0 Torque (moment of force)

Equilibrium of moments about an axis parralel to 𝑥1:

𝜎23𝑑𝑥1𝑑𝑥3𝑑𝑥22+ 𝜎23 +

𝜕𝜎23𝜕𝑥2

𝑑𝑥2 𝑑𝑥1𝑑𝑥3𝑑𝑥22− 𝜎32𝑑𝑥1𝑑𝑥2

𝑑𝑥32− 𝜎32 +

𝜕𝜎32𝜕𝑥3

𝑑𝑥3 𝑑𝑥1𝑑𝑥12𝑑𝑥32= 0

Neglecting the higher orders 𝜎23 = 𝜎32 Similarly 𝜎13 = 𝜎31, 𝜎12 = 𝜎21

Or 𝜎𝑖𝑗 = 𝜎𝑗𝑖

𝐹

𝑟

𝜃

N = r × 𝐹 = 𝑟 𝐹 𝑆𝑖𝑛𝜃

9

Cauchy's Formula

Using this symmetry 𝜕𝜎𝑗𝑖


𝜕𝜎𝑖𝑗


Stress on a surface – Cauchy's formula

Consider a small tetrahedron

𝛿𝐴1 = 𝛿𝐴 𝑛1 𝛿𝐴2 = 𝛿A 𝑛2 𝛿𝐴3 = 𝛿A 𝑛3

𝛿V =1

3h 𝛿A

𝐹 = 0 Along 𝑥2 − 𝑑𝑖𝑟

𝑇2𝑛𝛿𝐴 − 𝜎12 𝑛1𝛿𝐴 − 𝜎22 𝑛2𝛿𝐴 − 𝜎32 𝑛3𝛿𝐴 +

1

3ρ ℎ 𝛿𝐴 𝑋2 = 0

For infinitesimally small volume ℎ → 0 𝑇2𝑛 = 𝜎12 𝑛1 + 𝜎22 𝑛2 + 𝜎32 𝑛3


y

x n1

n n2

𝛼 𝛽

𝑛1 = 𝐶𝑜𝑠 𝛽 𝑛2 = 𝐶𝑜𝑠 𝛼

10

Cauchy's Formula

Similarly 𝑇1𝑛 = 𝜎11 𝑛1 + 𝜎21 𝑛2 + 𝜎31 𝑛3 𝑇3𝑛 = 𝜎13 𝑛1 + 𝜎23 𝑛2 + 𝜎33 𝑛3

𝑇𝑖𝑛 = 𝜎𝑖𝑗 𝑛𝑗 Cauchy's formula

𝑇1𝑛

𝑇2𝑛

𝑇3𝑛=

𝜎11 𝜎12 𝜎13𝜎21 𝜎22 𝜎23𝜎31 𝜎32 𝜎33

𝑛1𝑛2𝑛3

Resolving the traction into two components

𝜎 = 𝑇𝑛 𝐶𝑜𝑠 𝛽 𝜏 = 𝑇𝑛 𝑆𝑖𝑛 𝛽

Making use of Cauchy’s formula

𝜎 = 𝑇𝑖𝑛𝑛𝑖 = 𝜎𝑖𝑗 𝑛𝑖 𝑛𝑗 normal stress

𝜏 = 𝑇𝑖𝑛𝑡𝑖 = 𝜎𝑖𝑗 𝑡𝑖 𝑛𝑗 shear stress

𝜎

𝜏

𝑇𝑛

𝑛

𝑡

𝛽 𝛿𝑆

11

Isotropic and Deviatoric Stress

Isotropic and deviatoric stress

In a continuous medium (like rocks, tectonic plates) complete specification of the state of system

requires knowledge of the stress tensor 𝜎𝑖𝑗 at each point as functions of the co-ordinates. It is

useful to break the stress into two parts, the isotropic and deviatoric parts:

The isotropic stress

𝜎𝑖𝑗0 =

1

3𝜎𝑘𝑘𝛿𝑖𝑗 = 𝜎0 𝛿𝑖𝑗 where 𝜎𝑘𝑘 = 𝜎11 + 𝜎22 + 𝜎33 𝜎0=

1

3𝜎𝑘𝑘 (mean normal stress)

𝜎𝑖𝑗0 =

𝜎0 0 0 0 𝜎0 0 0 0 𝜎0

The deviatoric stress

𝜎𝑖𝑗′ = 𝜎𝑖𝑗 − 𝜎𝑖𝑗

0 𝜎𝑖𝑗 = 𝜎0 𝛿𝑖𝑗+ 𝜎𝑖𝑗′

𝜎𝑖𝑗′ =

𝜎11 − 𝜎0 𝜎12 𝜎13𝜎21 𝜎22 −𝜎0 𝜎23𝜎31 𝜎32 𝜎33 − 𝜎0

(𝜎𝑘𝑘 = 0)

𝜎𝑖𝑗 =

𝜎0 0 0 0 𝜎0 0 0 0 𝜎0

+

𝜎11 − 𝜎0 𝜎12 𝜎13𝜎21 𝜎22 −𝜎0 𝜎23𝜎31 𝜎32 𝜎33 − 𝜎0

12

Principal Axes and Principal Stress

Principal axes and principal stress

In this coordinate system the only nonzero stress components are diagonal elements

𝜎𝑖𝑗𝑃 =

𝜎11𝑃 0 0

0 𝜎22𝑃 0

0 0 𝜎33𝑃

≡

𝜎1 0 0 0 𝜎2 0 0 0 𝜎3

The usefulness of the reasoning in terms of principal axes and principal stresses lies in the fact

that they give a clear picture of the state of stress at a point. We need to determine 6-

componets; three principal axes and three stress components.

This is a classic eigenvalue problem. Consider an arbitrary-oriented surface with unit normal n.

In general the direction of the traction and that of the normal don not coincide unless the later is

the principal axes.


13


The condition for n to be a principal direction is on the basis of Cauchy’s formula:

𝑇𝑖𝑛 = 𝜎𝑖𝑗 𝑛𝑗= 𝜎 𝑛𝑖 (𝜎11 𝑛1 + 𝜎12 𝑛2 + 𝜎13 𝑛3 = 𝜎11 𝑛1 ≡ 𝜎 𝑛1, 𝑓𝑜𝑟 𝑖 = 1)

Using 𝑛𝑖 = 𝛿𝑖𝑗𝑛𝑗

𝜎𝑖𝑗 − 𝜎 𝛿𝑖𝑗 𝑛𝑗 = 0

𝜎𝑖𝑗: stress tensor in the initial frame

𝜎: any component of the stress in principal axes

𝜎11 − 𝜎 𝑛1 + 𝜎12𝑛2 + 𝜎13𝑛3 = 0 𝜎21𝑛1 + 𝜎22 − 𝜎 𝑛2 + 𝜎23𝑛3 = 0 𝜎31𝑛1 + 𝜎32𝑛2 + 𝜎33 − 𝜎 𝑛3 = 0

14


Also Remark

𝑛12 +𝑛2

2 +𝑛32= 1

𝜎11 − 𝜎 𝜎12 𝜎13𝜎21 𝜎22− 𝜎 𝜎23𝜎31 𝜎32 𝜎33− 𝜎

𝑛1𝑛2𝑛3

= 0

It can be proved that nontrivial solution exits only if

the determinant of the coefficients vanishes.

𝜎𝑖𝑗 − 𝜎 𝛿𝑖𝑗 = 0

𝜎11 − 𝜎 𝜎12 𝜎13𝜎21 𝜎22− 𝜎 𝜎23𝜎31 𝜎32 𝜎33− 𝜎

= −𝜎3 + 𝐼1𝜎2 + 𝐼2𝜎 + 𝐼3 = 0

where 𝐼1 = 𝜎11 + 𝜎22 + 𝜎33 −𝐼2= 𝜎11 𝜎12𝜎21 𝜎22

+𝜎11 𝜎13𝜎31 𝜎33

+ 𝜎22 𝜎23𝜎32 𝜎33

𝐼3 =

𝜎11 𝜎12 𝜎13𝜎21 𝜎22 𝜎23𝜎31 𝜎32 𝜎33

15


It can be proved that a real-valued symmetric matrix always has always three real-valued

eigenvalues.

𝜎1, 𝜎2, 𝜎3 principal stresses

By convention: 𝜎1 > 𝜎2 > 𝜎3

Principal directions are obtained by solving 𝜎𝑖𝑗 − 𝜎 𝛿𝑖𝑗 𝑛𝑗 = 0 for 𝑛1, 𝑛2, 𝑛3 successively for the

case 𝜎 = 𝜎1, 𝜎 = 𝜎2 , 𝜎 = 𝜎3 𝜎11 − 𝜎𝑖 𝜎12 𝜎13𝜎21 𝜎22− 𝜎𝑖 𝜎23𝜎31 𝜎32 𝜎33− 𝜎𝑖

𝑛1𝑛2𝑛3

= 0 𝑖 = 1,2,3

𝜎11 − 𝜎𝑖 𝑛1 + 𝜎12𝑛2 + 𝜎13𝑛3 = 0 𝜎21𝑛1 + 𝜎22 − 𝜎𝑖 𝑛2 + 𝜎23𝑛3 = 0 𝜎31𝑛1 + 𝜎32𝑛2 + 𝜎33 − 𝜎𝑖 𝑛3 = 0

Quantities 𝐼1, 𝐼2, and 𝐼3 are the invariants of the stress tensor (first, second and third invariant).

16

Maximum Shear Stress

Maximum shear stress

We had

𝜎 = 𝑇𝑖𝑛𝑛𝑖 = 𝜎𝑖𝑗𝑛𝑖 𝑛𝑗 normal stress

𝜏 = 𝑇𝑖𝑛𝑡𝑖 = 𝜎𝑖𝑗𝑡𝑖 𝑛𝑗 shear stress

In principal coordinate system:

𝜎 = 𝜎1 𝑛12 + 𝜎2 𝑛2

2 + 𝜎3 𝑛32

𝜏 = 𝜎1 𝑛1𝑡1 + 𝜎2 𝑛2𝑡2 + 𝜎3 𝑛3𝑡3

where 𝜎1, 𝜎2, 𝜎3 are principal stresses and 𝑛𝑖 , 𝑡𝑖 are unit normal and unit tangent to the surface

element.

In matrix notation:

𝜎 =𝜎1 0 00 𝜎2 00 0 𝜎3

𝑛1𝑛2𝑛3

𝑛1𝑛2𝑛3= 𝜎1𝑛1 𝜎2𝑛2 𝜎3𝑛3

𝑛1𝑛2𝑛3= 𝜎1 𝑛1

2 + 𝜎2 𝑛22 + 𝜎3 𝑛3

2

𝜎

𝜏

𝑇𝑛

𝑛

𝑡

𝛽 𝛿𝑆

17


𝜏2 = 𝑇2 − 𝜎2

Using Cauchy’s formula 𝑇𝑖𝑛 = 𝜎𝑖𝑗 𝑛𝑗 𝑇𝑖

𝑛 2 = 𝜎12𝑛12 + 𝜎2

2𝑛22 + 𝜎3

2𝑛32

𝜏2 = 𝑇2 − 𝜎2 = 𝜎12𝑛12 + 𝜎2

2𝑛22 + 𝜎3

2𝑛32 − 𝜎1 𝑛1

2 + 𝜎2 𝑛22 + 𝜎3 𝑛3

2 2=

𝜎1 − 𝜎2 2 𝑛1𝑛2

2 + 𝜎2 − 𝜎3 2 𝑛2𝑛3

2 + 𝜎1 − 𝜎3 2 𝑛1𝑛3

2

Note that 𝑛1

2 1 − 𝑛12 = 𝑛1

2 𝑛22 + 𝑛3

2 and so on (𝑛12 +𝑛2

2 +𝑛32= 1)

From this equation we see that 𝜏 has a minimum (𝜏 = 0) for the following choices of n:

𝑛1 = ±1, 𝑛2= 𝑛3= 0 𝑛2 = ±1, 𝑛1= 𝑛3= 0 𝑛3 = ±1, 𝑛1= 𝑛2= 0

Which shows no shear stress act on three planes with normal in the co-ordinate directions

(principal planes) as one would expect from the choice of the principal axes as co-ordinate

system.

18


The planes on which the shear stress is maximum are obtained from the condition

𝜕𝜏

𝜕𝑛1=𝜕𝜏

𝜕𝑛2= 0

Since 𝑛1

2 +𝑛22 +𝑛3

2= 1, there is no need for the third derivative

Applying this to 𝜏2-equation, after some algebra, the extreme values of 𝜏 are

𝜏 =1

2𝜎1 − 𝜎2 for 𝑛1 = 𝑛2 =

1

2, 𝑛3= 0

𝜏 =1

2𝜎1 − 𝜎3 for 𝑛1 = 𝑛3 =

1

2, 𝑛2= 0 Since 𝜎1 > 𝜎2 > 𝜎3 𝜏𝑚𝑎𝑥 =

1

2𝜎1 − 𝜎3

𝜏 =1

2𝜎2 − 𝜎3 for 𝑛2 = 𝑛3 =

1

2, 𝑛1= 0

19

Examples

Ex. 1 – Find the principal axes and eigenvalues for the following stress tensor:

𝜎𝑖𝑗 =80 3030 40

Sol. 𝜎𝑖𝑗 − 𝜎 𝛿𝑖𝑗 𝑛𝑗 = 0

𝜎𝑖𝑗 − 𝜎 𝛿𝑖𝑗 = 0 80;𝜎 3030 40;𝜎

= 0

80 − 𝜎 40 − 𝜎 − 302 = 0 𝜎2 − 120𝜎 + 2300 = 0 𝜎1 = 96.05, 𝜎2 = 23.95 eigenvalues

For 𝜎1 = 96.05 𝑀𝑃𝑎:

80;96.05 3030 40;96.05

𝑛1𝑛2=00

80 − 96.05 𝑛1+ 30 𝑛2 = 0

𝑛12 + 𝑛2

2 = 1 n = 𝑛1𝑛2=0.880.47

y

x

30 MPa

40 MPa

80 MPa

20

Examples

Alternatively, if we assume 𝑛2= 1

80 − 96.05 𝑛1+ 30 = 0 𝑛1= 30

16.95 𝑛 =

1

30

16.95

2:1

30

16.95

1=0.880.47

Similarly for 𝜎2 = 23.95: 𝑛 = −0.470.88

Rename 𝑛 =0.880.47

as 𝑛1 =0.880.47

corresponding to the eigenvalue 𝜎1 = 96.05

and 𝑛 =−0.470.88

as 𝑛2 =−0.470.88

corresponding to the eigenvalue 𝜎2 = 23.95

y

x

30 MPa

40 MPa

80 MPa

y

x 𝒏𝟏 𝒏𝟐

𝜎′ =96.05 0

0 23.95

y'

x'

21

Examples

Ex. 2 – The state of stress at a point is given by:

𝜎𝑖𝑗 =2 1 31 2 − 23 − 2 1

Find:

a) the traction vector acting on a plane with unit normal of:

𝑛 = 1

3𝑒1 +

2

3𝑒2 −

2

3𝑒3

b) the shear and normal components of this traction vector.

Sol.

a) 𝑇𝑖

𝑛 = 𝜎𝑖𝑗 𝑛𝑗 Cauchy's formula

𝑇1𝑛

𝑇2𝑛

𝑇3𝑛=

𝜎11 𝜎12 𝜎13𝜎21 𝜎22 𝜎23𝜎31 𝜎32 𝜎33

𝑛1𝑛2𝑛3

=2 1 31 2 − 23 − 2 1

1/32/3−2/3

= 1

3

−29−3

𝑇𝑛 = −2

3𝑒1 + 3𝑒2 − 𝑒3

22

Examples

b) 𝜎 = 𝑇𝑛 ∙ 𝑛 = −2

3

1

3+ 3

2

3+ −1 −

2

3 ≈ 2.4

𝜏 = 𝑇𝑛 2 − 𝜎 2 = −2

3

2+ 3 2 + −1 2 − 2.4 2 ≈ 2.1

Ex. 3 – The state of stress at a point is given by:

𝜎𝑖𝑗 =5 0 0

0 − 6 − 120 − 12 1

Determine principal stress components and principal directions.

Sol.

𝜎𝑖𝑗 − 𝜎 𝛿𝑖𝑗 𝑛𝑗 = 0

𝜎𝑖𝑗 − 𝜎 𝛿𝑖𝑗 = 0

23

Examples

5;𝜎 0 00 6;𝜎 ;120 ;12 1;𝜎

= −10 + 𝜎 5 − 𝜎 15 + 𝜎 = 0

𝜎1 = 10, 𝜎2 = 5, 𝜎3 = −15

For 𝜎1 = 10 𝑀𝑃𝑎:

5;𝜎 0 00 6;𝜎 ;120 ;12 1;𝜎

𝑛1𝑛2𝑛3=000

(𝜎 = 𝜎1)

−5𝑛1 + 0𝑛2 + 0𝑛3 = 0

0𝑛1 − 16𝑛2 − 12𝑛3 = 0 𝑛1 = 0, 𝑛2= −3

5, 𝑛3=

4

5

0𝑛1 + 12𝑛2 + 9𝑛3 = 0 𝑛 = −3

5 𝑒2 +

4

5 𝑒3

𝑛12 +𝑛2

2 +𝑛32= 1

24

Examples

Similarly for 𝜎2 = 5 𝑀𝑃𝑎:

5;𝜎 0 00 6;𝜎 ;120 ;12 1;𝜎

𝑛1𝑛2𝑛3=000

(𝜎 = 𝜎2)

0𝑛1 + 0𝑛2 + 0𝑛3 = 0 0𝑛1 − 11𝑛2 − 12𝑛3 = 0 𝑛1 = 1, 𝑛2= 0, 𝑛3= 0 0𝑛1 − 12𝑛2 − 4𝑛3 = 0 𝑛 = 𝑒1 𝑛12 +𝑛2

2 +𝑛32= 1

And for 𝜎3 = −15 𝑀𝑃𝑎:

5;𝜎 0 00 6;𝜎 ;120 ;12 1;𝜎

𝑛1𝑛2𝑛3=000

(𝜎 = 𝜎3)

20𝑛1 + 0𝑛2 + 0𝑛3 = 0

0𝑛1 + 9𝑛2 − 12𝑛3 = 0 𝑛1 = 0, 𝑛2=4

5, 𝑛3=

3

5

0𝑛1 − 12𝑛2 + 16𝑛3 = 0 𝑛 =4

5 𝑒2 +

3

5 𝑒3

𝑛12 +𝑛2

2 +𝑛32= 1

25

Examples

𝑛 = −3

5 𝑒1 +

4

5 𝑒3 rename as 𝑛1 = −

3

5 𝑒2 +

4

5 𝑒3

𝑛 = −3

5 𝑒1 rename as 𝑛2 = 𝑒1

𝑛 =4

5 𝑒2 +

3

5 𝑒3 rename as 𝑛3 =

4

5 𝑒2 +

3

5 𝑒3

Note that the components of each principal direction, are the direction cosines.

e.g.,

𝑛1 = −3

5 𝑒2 +

4

5 𝑒3

𝐶𝑜𝑠𝛼 = 0

𝐶𝑜𝑠 𝛽 = −3

5

𝐶𝑜𝑠 𝛾 = 4

5

x3

x2

x1

𝑛3

𝑛2

𝑛1 𝜸

𝜷

𝜶

26

Stress Tensor in Principal axes

𝜎𝑖𝑗 =

𝜎1 0 00 𝜎2 00 0 𝜎3

Invariants

We obtained: 𝐼1 = 𝜎11 + 𝜎22 + 𝜎33

𝐼2 = −𝜎11 𝜎12𝜎21 𝜎22

−𝜎11 𝜎13𝜎31 𝜎33

− 𝜎22 𝜎23𝜎32 𝜎33

𝐼3 =

𝜎11 𝜎12 𝜎13𝜎21 𝜎22 𝜎23𝜎31 𝜎32 𝜎33

Invariants

In principal co-ordinate:

𝐼1 = 𝜎1 + 𝜎2 + 𝜎3 𝐼2 = 𝜎1𝜎2 + 𝜎2𝜎3 + 𝜎3𝜎1 𝐼2 = 𝜎1𝜎2𝜎3

27

Ex. 4 – The stress tensor at a point in two different co-ordinate systems are given by:

𝜎𝑖𝑗 =2 1 01 3 − 20 − 2 1

𝜎′𝑖𝑗 =

3

2

3

2

1

23

2 3 −

1

21

2 −

1

2 3

2

The first invariant:

𝐼1 = 2 + 3 + 1 =3

2+ 3 +

3

2= 6

Verify that the second and third invariants are:

𝐼2 = 6, and 𝐼3 = −3

for the matrix in the original coordinate and rotated frame..

Invariants

28

Axi-symmetric stress state

When two of the principal stresses are equal, only one of the principal directions will

be unique.

𝜎1, 𝜎2 = 𝜎3

Isotropic State of Stress

When all three components of the principal stresses are

equal, all directions are principal directions and the stress

tensor has the form of

𝜎𝑖𝑗 =𝜎 0 00 𝜎 00 0 𝜎

In all coordinate systems

𝜎1 = 𝜎2 = 𝜎3 ≡ 𝜎

Symmetries

𝑛1

𝑛2 𝑛3

𝑛1

𝑛2

𝑛3

29

Ex. 5 – The principal values (eigenvalues) for the following stress tensor

𝜎𝑖𝑗 =5 0 00 − 6 − 120 − 12 1

are 𝜎1 = 10, 𝜎2 = 5, 𝜎3 = −15 (𝜎1 > 𝜎2 > 𝜎3 ) So the maximum shear stress is given by:

𝜏𝑚𝑎𝑥 =1

2𝜎1 − 𝜎3 =

1

210 + 15 = 12.5

for 𝑛1 = 𝑛3 =1

2, 𝑛2= 0

Examples

x3

x2

x1

𝑛3

𝑛2

𝑛1 𝜸

𝜷

𝜶

𝑛2

𝑛3 𝑛1

𝝉𝒎𝒂𝒙

30

Ex. – What is wrong with this stress tensor?

𝜎𝑖𝑗 =5 34 7734 − 6 − 12−77 − 12 1

Examples

31

Two-Dimensional Problem

Two dimensional approximation

Geological problems involving stress can often be approximated to be approximately two-

dimensional.

𝜎𝑖𝑗 = 𝜎11 𝜎12𝜎21 𝜎22

And any surface is defined by its unit normal and unit tangent

𝑛𝑖 =𝑛1𝑛2=𝐶𝑜𝑠 𝜃𝑆𝑖𝑛 𝜃

𝑡𝑖 =𝑡1𝑡2=−𝑆𝑖𝑛 𝜃𝐶𝑜𝑠 𝜃

where 𝑛𝑖 is the angle between and the 𝑥𝑖 − 𝑎𝑥𝑖𝑠

From 𝜎 = 𝑇𝑖

𝑛𝑛𝑖 = 𝜎𝑖𝑗 𝑛𝑖 𝑛𝑗

𝜎 = 𝜎11 𝑛1 𝑛1 + 𝜎12 𝑛1 𝑛2 + 𝜎21 𝑛2 𝑛1 + 𝜎22 𝑛2 𝑛2

𝜎 = 𝜎11 𝐶𝑜𝑠 𝜃2+ 𝜎12 𝐶𝑜𝑠 𝜃 𝑆𝑖𝑛 𝜃 + 𝜎21𝑆𝑖𝑛 𝜃 𝐶𝑜𝑠 𝜃 + 𝜎22 𝑆𝑖𝑛 𝜃

2

𝜎

𝜏 x2

x1

ni

ti

𝜃

𝜎21

𝜎12

𝜎22

𝜎11

32


Since 𝐶𝑜𝑠 𝜃 𝑆𝑖𝑛 𝜃 = 1

2 𝑆𝑖𝑛 2𝜃

𝜎 = 𝜎11𝐶𝑜𝑠2𝜃 + 2 𝜎12 𝐶𝑜𝑠 𝜃 𝑆𝑖𝑛 𝜃 + 𝜎22𝑆𝑖𝑛

2𝜃 (𝜎12 = 𝜎21) 𝜎 = 𝜎11𝐶𝑜𝑠

2𝜃 + 𝜎12 𝑆𝑖𝑛 2𝜃 + 𝜎22𝑆𝑖𝑛2𝜃

From

𝜏 = 𝑇𝑖𝑛𝑡𝑖 = 𝜎𝑖𝑗 𝑡𝑖 𝑛𝑗

𝜏 = 𝜎11 𝑛1 𝑡1 + 𝜎12 𝑛1 𝑡2 + 𝜎21 𝑛2 𝑡1 + 𝜎22 𝑛2 𝑡2 𝜏 = − 𝜎11𝑆𝑖𝑛 𝜃 𝐶𝑜𝑠 𝜃 − 𝜎12𝑆𝑖𝑛

2𝜃 + 𝜎21𝐶𝑜𝑠2𝜃 + 𝜎22𝑆𝑖𝑛 𝜃 𝐶𝑜𝑠 𝜃

Using 𝐶𝑜𝑠2𝜃 =1

21 + 𝐶𝑜𝑠 2𝜃 and 𝑆𝑖𝑛2𝜃 =

1

21 − 𝐶𝑜𝑠 2𝜃

𝜎 =1

2𝜎11 + 𝜎22 +

1

2𝜎11 − 𝜎22 𝐶𝑜𝑠 2𝜃 + 𝜎12 𝑆𝑖𝑛 2𝜃

𝜏 = −1

2𝜎11 − 𝜎22 𝑆𝑖𝑛 2𝜃 + 𝜎12 𝐶𝑜𝑠 2𝜃

Principal co-ordinates

𝜎 = 𝜎1 =1

2𝜎11 + 𝜎22 +

1

2𝜎11 − 𝜎22 𝐶𝑜𝑠 2𝜃 + 𝜎12 𝑆𝑖𝑛 2𝜃

𝜏 = 0 =1

2𝜎22 − 𝜎11 𝑆𝑖𝑛 2𝜃 + 𝜎12 𝐶𝑜𝑠 2𝜃 𝜃 =

1

2arctan 2𝜎12/ 𝜎11 − 𝜎22

33


𝜃 +𝜋

2 (the other plane perpendicular to the first)

𝜎 = 𝜎11𝑆𝑖𝑛2𝜃 − 𝜎12 𝑆𝑖𝑛 2𝜃 + 𝜎22𝐶𝑜𝑠

2𝜃

𝜎 = 𝜎2 =1

2𝜎11 + 𝜎22 −

1

2𝜎11 − 𝜎22 𝐶𝑜𝑠 2𝜃 − 𝜎12 𝑆𝑖𝑛 2𝜃

𝜏 = 0 =1

2𝜎11 − 𝜎22 𝑆𝑖𝑛 2𝜃 − 𝜎12 𝐶𝑜𝑠 2𝜃

2𝜃 = arctan 2𝜎12/ 𝜎11 − 𝜎22 ,

𝑆𝑖𝑛 2𝜃 = 𝑆𝑖𝑛(arctan 2𝜎12/ 𝜎11 − 𝜎22 ) =

2𝜎12𝜎11−𝜎22

1:2𝜎12

𝜎11−𝜎22

2

𝐶𝑜𝑠 2𝜃 = 𝐶𝑜𝑠(arctan 2𝜎12/ 𝜎11 − 𝜎22 ) =1

1:2𝜎12

𝜎11−𝜎22

2

𝜎1 =1

2𝜎11 + 𝜎22 + 𝜎12

2 +1

4𝜎11 − 𝜎22

21/2

𝜎2 =1

2𝜎11 + 𝜎22 − 𝜎12

2 +1

4𝜎11 − 𝜎22

21/2

34

Mohr’s Circle

The state of stress using the Mohr’s Circle (Otto Mohr)

In this method the normal and shear stresses acting on a single plane are

represented by a single point on the Mohr circle.

The normal and shear stresses acting on two perpendicular planes are represented

by two points, one at each end of a diameter on the Mohr circle.

We obtained two parametric equations (𝜃 being parameter):

𝜎 =1

2𝜎11 + 𝜎22 +

1

2𝜎11 − 𝜎22 𝐶𝑜𝑠 2𝜃 + 𝜎12 𝑆𝑖𝑛 2𝜃

𝜏 = −1

2𝜎11 − 𝜎22 𝑆𝑖𝑛 2𝜃 + 𝜎12 𝐶𝑜𝑠 2𝜃

Eliminating 𝜃 from these equations will yield the non-parametric equation of the Mohr

circle:

𝜎 −1

2𝜎11 + 𝜎22

2=1

4𝜎11 − 𝜎22

2

𝐶𝑜𝑠22𝜃 + 𝜎12𝑆𝑖𝑛2𝜃

2 +1

2𝜎11 − 𝜎22 𝐶𝑜𝑠2𝜃 𝜎12𝑆𝑖𝑛2𝜃

𝜏2 =1

4𝜎11 − 𝜎22

2

𝑆𝑖𝑛22𝜃 + 𝜎12𝐶𝑜𝑠2𝜃

2 − 1


35

Mohr’s Circle

𝜎 −1

2𝜎11 + 𝜎22

2=1

4𝜎11 − 𝜎22

2

𝐶𝑜𝑠22𝜃 + 𝜎12𝑆𝑖𝑛2𝜃

2 +1


𝜏2 =1

4𝜎11 − 𝜎22

2

𝑆𝑖𝑛22𝜃 + 𝜎12𝐶𝑜𝑠2𝜃

2 − 1


𝜎 −1

2𝜎11 + 𝜎22

2+ 𝜏2 =

1

4𝜎11 − 𝜎22

2 + 𝜎122 Mohr’s circle

𝑋2 + 𝑌2 = 𝑅2

𝑿 ≡ 𝜎 −1

2𝜎11 + 𝜎22 , 𝒀 ≡ 𝜏, 𝑹 ≡

1

4𝜎11 − 𝜎22

2 + 𝜎122

𝜎𝑎𝑣𝑒 ≡1

2𝜎11 + 𝜎22 𝜎 − 𝜎𝑎𝑣𝑒

2 + 𝜏2 = 𝑅2

36

Mohr’s Circle

𝜎 − 𝜎𝑎𝑣𝑒2 + 𝜏2 = 𝑅2 Mohr’s circle centered at 𝜎, 𝜏 = (𝜎𝑎𝑣𝑒 , 0)

Stress Coordinates

a) Normal stress 𝜎: abscissae (horizontal)

b) Shear stress 𝜏: ordinates (vertical)

Note that

𝑹 ≡ 1

4𝜎11 − 𝜎22

2 + 𝜎122 =

1

2𝜎1 − 𝜎2

𝜏

𝜎

𝐴

𝐵

2𝜃

𝜎1 𝜎2 𝜎𝑎𝑣𝑒

𝜎 =1

2𝜎1 + 𝜎2 +

1

2𝜎1 − 𝜎2 𝐶𝑜𝑠 2𝜃

𝜎 =1

2𝜎1 + 𝜎2 −

1

2𝜎1 − 𝜎2 𝐶𝑜𝑠 2𝜃

𝜏 = −1

2𝜎1− 𝜎2 𝑆𝑖𝑛 2𝜃

1

2𝜎11+ 𝜎22 =

1

2𝜎1+ 𝜎2

𝜎1

𝜎2

𝜎11

𝜎22

𝜎12 𝜃

𝐴

𝐵

37

Mohr’s Circle in 3D

Mohr Circle Diagram

𝜎𝑖𝑗 =

𝜎11 𝜎12 𝜎13𝜎21 𝜎22 𝜎23𝜎31 𝜎32 𝜎33

In principal coordinate

𝜎𝑖𝑗 =

𝜎1 0 00 𝜎2 00 0 𝜎3

𝜶 = 𝑨𝒓𝒄 𝑪𝒐𝒔 𝒏𝟏 , 𝜷 = 𝑨𝒓𝒄 𝑪𝒐𝒔 𝒏𝟐 , 𝜸 = 𝑨𝒓𝒄 𝑪𝒐𝒔(𝒏𝟑)

a) Draw 𝐿1 parallel to 𝜏 passing through 𝜎1 b) Measure 𝛼 from this line and draw 𝐴𝐴′ c) Use center 𝐶3 (which doesn’t depend on 𝜎1 and draw arc 𝐴𝐴′ d) Repeat in similar way for the other angles

e) The normal and shear components for the plane with

normal 𝑛 are given by the coordinates of the intersection point P

https://www.rockmechs.com/mohr-circle-3d/

𝜎1 𝜎2

𝜎3

𝑛

38

Strain Tensor

The Infinitesimal Strain Tensor

Consider a continuous body occupying domain D and two neighbouring points 𝑃(𝑥𝑖) and 𝑄(𝑥𝑖 +𝑑𝑥𝑖) After displacement: 𝑃 𝑥𝑖 → 𝑃′(𝑥𝑖 + 𝑢𝑖) 𝑄 𝑥𝑖 + 𝑑𝑥𝑖 → 𝑄′(𝑥𝑖 + 𝑑𝑥𝑖 + 𝑢𝑖 + 𝑑𝑢𝑖) 𝑢𝑖 and 𝑢𝑖 + 𝑑𝑢𝑖 are displacements.

In general displacement of the particles results from: 1) Translation, 2) Rotation, 3) Deformation

Assume 𝑢𝑖 = 𝑢𝑖(𝑥, 𝑦, 𝑧) and 𝜕𝑢𝑖/𝜕𝑥𝑗 are small, then

d𝑢𝑖 ≈ 𝜕𝑢𝑖

𝜕𝑥𝑗d𝑥𝑗 + . . . (Taylor series)

Resolving the displacement into two parts:

𝜕𝑢𝑖

𝜕𝑥𝑗=1

2

𝜕𝑢𝑖

𝜕𝑥𝑗+𝜕𝑢𝑗

𝜕𝑥𝑖+1

2

𝜕𝑢𝑖

𝜕𝑥𝑗−𝜕𝑢𝑗

𝜕𝑥𝑖

where

휀𝑖𝑗 ≡1

2

𝜕𝑢𝑖

𝜕𝑥𝑗+𝜕𝑢𝑗

𝜕𝑥𝑖 infinitesimal strain tensor

39

Strain Tensor

Remark: Taylor series

𝑓 𝑥 ≈ 𝑓 𝑎 +𝑓′(𝑎)

1!𝑥 − 𝑎 +

𝑓′′(𝑎)

2!𝑥 − 𝑎

2+ . . =

𝑓 𝑘 (𝑎)

𝑘!𝑥 − 𝑎 𝑘∞

0

Example: Expansion of 𝑆𝑖𝑛(𝑥) around 𝑥 = 0

𝑆𝑖𝑛 𝑥 ≈ 𝑥 −𝑥3

3!+𝑥5

5!−𝑥7

7!

Rigid body rotation

𝜔𝑖𝑗 ≡1

2

𝜕𝑢𝑖

𝜕𝑥𝑗−𝜕𝑢𝑗

𝜕𝑥𝑗𝑖 rigid body rotation

Note that

휀𝑖𝑗 = 휀𝑗𝑖 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐 𝑡𝑒𝑛𝑠𝑜𝑟

𝜔𝑖𝑗 = −𝜔𝑗𝑖 anti − 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐 𝑡𝑒𝑛𝑠𝑜𝑟

Therefore

𝑢𝑖 + 𝑑𝑢𝑖 = 𝑢𝑖 +𝜕𝑢𝑖

𝜕𝑥𝑗𝑑𝑥𝑗 = 𝑢𝑖 + 휀𝑖𝑗𝑑𝑥𝑗 + 𝜔𝑖𝑗𝑑𝑥𝑗

𝑢𝑖: rigid body translation

휀𝑖𝑗: measure of deformation

𝜔𝑖𝑗: measure of rotation

40

Strain Tensor

We now show why 𝜔𝑖𝑗 represents a rotation. Consider a small rotation about 𝑥3 − axis, we can

write

𝛼 = tan𝛼 =𝑑𝑢2

𝑑𝑥1=

𝜕𝑢2𝜕𝑥1𝑑𝑥1

𝑑𝑥1=𝜕𝑢2

𝜕𝑥1 (APA’ triangle)

On the other hand

𝛼 = tan𝛼 = −𝑑𝑢1

𝑑𝑥2= −

𝜕𝑢1𝜕𝑥2𝑑𝑥2

𝑑𝑥2= −

𝜕𝑢1

𝜕𝑥2

Combining two results

𝛼 =1

2

𝜕𝑢2

𝜕𝑥1−𝜕𝑢1

𝜕𝑥2= 𝜔12

41

Strain Tensor

Strain tensor

Symmetric strain tensor with 6 independent components required to specify it at a point.

휀𝑖𝑗 =

휀11 휀12 휀13휀21 휀22 휀23휀31 휀32 휀33

=

𝜕𝑢1𝜕𝑥1

1

2

𝜕𝑢1𝜕𝑥2+𝜕𝑢2𝜕𝑥1

1

2

𝜕𝑢1𝜕𝑥3+𝜕𝑢3𝜕𝑥1

1

2

𝜕𝑢2𝜕𝑥1+𝜕𝑢1𝜕𝑥2

𝜕𝑢2𝜕𝑥2

1

2

𝜕𝑢2𝜕𝑥3+𝜕𝑢3𝜕𝑥2

1

2

𝜕𝑢3𝜕𝑥1+𝜕𝑢1𝜕𝑥3

1

2

𝜕𝑢3𝜕𝑥2+𝜕𝑢2𝜕𝑥3

𝜕𝑢3𝜕𝑥3

On-diagonal terms

Consider an on-diagonal term, say 휀22 =𝜕𝑢2

𝜕𝑥2. With reference to the figure suppose the volume is

stretched (or compressed) in the direction of 𝑥2, therefore:

𝑑𝑥2 𝑑𝑥2 + 𝑑𝑢2 Initial final

In this case 𝑢2 = 𝑢2(𝑥2 𝑜𝑛𝑙𝑦),

then 𝑑𝑢𝑖 =𝜕𝑢𝑖

𝜕𝑥𝑗𝑑𝑥𝑗 𝑑𝑢2 =

𝜕𝑢2

𝜕𝑥2𝑑𝑥2

42

Strain Tensor

𝑑𝑥2 + 𝑑𝑢2 = 1 +𝜕𝑢2

𝜕𝑥2𝑑𝑥2 = 1 + 휀22 𝑑𝑥2

Which shows that 휀22 represents “the change in the length per unit length” (or elongation in

𝑥2 − 𝑑𝑖𝑟.).

Similarly:

휀11: elongation in 𝑥1 − 𝑑𝑖𝑟. 휀33: elongation in 𝑥3 − 𝑑𝑖𝑟.

휀𝑖𝑖 > 0: expansion

휀𝑖𝑖 < 0: compression

Off-diagonal terms

Off-diagonal terms represent deformation.

Suppose shear stress acting on the faces

Normal to 𝑥2 and 𝑥3 axes. Dissolve the problem

In two steps as shown in the figure

𝜙′ ≈ tan𝜙′ =𝜕𝑢3𝜕𝑥2

𝜙′′ ≈ tan𝜙′′ =𝜕𝑢2𝜕𝑥3

43

Strain Tensor

Cubical dilatation Consider an element volume 𝑉 = 𝑑𝑥1𝑑𝑥2𝑑𝑥3, after deformation V + dV. The change in the

volume is:

dV = 𝑑𝑥1𝜕𝑢1𝜕𝑥1+ 𝑑𝑥1 𝑑𝑥2

𝜕𝑢2𝜕𝑥2+ 𝑑𝑥2 𝑑𝑥3

𝜕𝑢3𝜕𝑥3+ 𝑑𝑥3 − 𝑑𝑥1𝑑𝑥2𝑑𝑥3

Neglecting the higher order terms:

dV = 𝑑𝑥1𝑑𝑥2𝑑𝑥3 − 𝜕𝑢1𝜕𝑥1+𝜕𝑢2𝜕𝑥2+𝜕𝑢3𝜕𝑥3

𝑑𝑥1𝑑𝑥2𝑑𝑥3 − 𝑑𝑥1𝑑𝑥2𝑑𝑥3

𝑑𝑉 = 휀11 + 휀22 + 휀33 𝑑𝑥1𝑑𝑥2𝑑𝑥3

𝜃 =𝑑𝑉

𝑉= 휀11 + 휀22 + 휀33 = 휀𝑘𝑘 =

𝜕𝑢𝑘

𝜕𝑥𝑘 cubical dilatation

44

Strain Tensor

𝜙′ + 𝜙′′ =𝜕𝑢3

𝜕𝑥2+𝜕𝑢2

𝜕𝑥3= 2휀23 change in the angle of lines

Isotropic and deviatoric Strain

As in the case of stress, the strain tensor can be resolved into isotropic and deviatotic parts, i.e.,

The isotropic strain

휀𝑖𝑗0 =

1

3휀𝑘𝑘𝛿𝑖𝑗 = 휀0 𝛿𝑖𝑗 where 휀𝑘𝑘 = 휀11 + 휀22 + 휀33 휀0=

1

3휀𝑘𝑘 (mean normal stress)

휀𝑖𝑗0 =

휀0 0 0 0 휀0 0 0 0 휀0

≡ 휀0 𝛿𝑖𝑗 pure volume change

The deviatoric strain

휀𝑖𝑗′ = 휀𝑖𝑗 − 휀𝑖𝑗

0 휀𝑖𝑗 = 휀𝑖𝑗′ + 휀𝑖𝑗

0

휀𝑖𝑗′ =

휀11 − 휀0 휀12 휀13휀21 휀22 −휀0 휀23휀31 휀32 휀33 − 휀0

(휀𝑘𝑘 = 0) change of shape

휀′𝑖𝑗 = 휀𝑖𝑗 for 𝑖 ≠ 𝑗 the shear components of the strain deviator (angular deformation)

45

Strain Tensor

휀𝑖𝑗 =

휀0 0 0 0 휀0 0 0 0 휀0

+

휀11 − 휀0 휀12 휀13휀21 휀22 −휀0 휀23휀31 휀32 휀33 − 휀0

Note that

a) If 휀′𝑖𝑗 = 0 ∀ 𝑖, 𝑗

휀′𝑖𝑗 =

휀11 − 휀0 휀12 휀13휀21 휀22 −휀0 휀23휀31 휀32 휀33 − 휀0

휀11 = 휀0, 휀11 = 휀0, 휀11= 휀0

and off-diagonal elements = 0 purely volumetric deformation

b) If 휀𝑘𝑘 = 0 no volumetric deformation

Ex. 휀𝑖𝑗 =2 0 00 − 3 00 0 1

휀𝑘𝑘 = 0 no volumetric deformation but

Cube cuboid

46

Strain Tensor

Ex. 휀𝑖𝑗 =0 0 00 0 00 0 0

휀𝑘𝑘 = 0 no volumetric deformation

Ex. 휀𝑖𝑗 =5 0 00 5 00 0 5

휀𝑘𝑘 = 15

Ex. 휀𝑖𝑗 =−5 0 00 − 5 00 0 − 5

휀𝑘𝑘 = −15

47

Strain Tensor

c) But if 휀𝑖𝑗 = 0 for 𝑖 ≠ 𝑗 this is not sufficient condition for having purely volumetric

deformation

Ex. 휀𝑖𝑗 =2 0 00 3 00 0 1

if rotated, we can find non-zero off-diagonal elements

This means that the strain tensor (in a given co-ordinate) 𝑎 0 0 0 𝑏 0 0 0 𝑐

may result shape

deformation provided that the all three elongations are not equal.

Different representations of the stress tensor

Similar representations may be used for the strain tensor.

ESS2222H Tectonics and Planetary Dynamics Lecture 2 Stress, … · 2020. 1. 21. · Tectonics and Planetary Dynamics Lecture 2 – Stress, Strain University of Toronto, Department

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