VECTORS AND TENSORS - pioneer.netserv.chula.ac.thpioneer.netserv.chula.ac.th/~pwithit/Vectors.pdf2101202 – Mechanics of Materials I By Watanachai Smittakorn VECTORS AND TENSORS “A

2101202 – Mechanics of Materials I By Watanachai Smittakorn

VECTORS AND TENSORS

“A beautiful story needs a beautiful language to tell. Tensor is the language of mechanics.”

scalars

tensors

vectors

matrices

VECTORS

• defined as a line with magnitude and direction • denoted by

, , , , , ...AB PQ a u Fuuur r

• two vectors are equal if (both magnitude & direction are equal) • a unit vector is (a vector with a unit magnitude) • zero vector, denoted by 0, is (a vector with zero magnitude) • | AB|, |u|, v are magnitudes of AB, u, v

- 1 -


Vector addition:

• “parallelogram law” • commutative and associative

a + b = b + a

(a + b) + c = a + (b + c) Vector subtraction:

a – b = a + (-b)

Let e1, e2, e3 be the unit vectors in x1, x2, x3 directions,

x1

x2

x3

u

u1

u2

u3

1 2 3 1 2

2 2 21 2 3

( , , )u u u u u u

u u u u

= + + =

= = + +

1 2 3u e e e

u3

- 2 -


Scalar product (or dot product):

1 1 2 2 3 3

cos (0 )u v u v u v

θ θ π⋅ = + +

⋅ = ≤ ≤

u vu v u v

e.g., F · s Vector product (or cross product):

2 3 3 2 3 1 1 3 1 2 2 1

1 2 3

1 2 3

( ) ( ) (

sin (0 )

u v u v u v u v u v u v

u u uv v v

θ θ π

× = − + − + −

=

× = ≤ ≤

1 2

1 2 3

u v e e ee e e

u v u v

) 3

2

e.g., r × F Properties:

1 3 3

3 3 3 1

( )( )

( )k k k

× = − ×× + = × + ×× =× = × = × =

× = × = × =

× = × = ×

1 2 2

1 2 2 1

u v v uu v w u v u wu u 0e e e e e e 0e e e e e e e e eu v u v u v

HOMEWORK: 2.5, 2.7, 2.9

- 3 -


What is mechanics? Why is tensor important?

scalars

tensors

vectors

matrices

INDICIAL NOTATION

x1, x2,…, xn is denoted as xi, i = 1,…,n

i is an index with range of 1 to n

xi vs (xi) or X member vector/matrix

Summation convention:

3

1 1 2 2 3 31

i i i ii

a x a x a x a x a x=

+ + = ≡∑ Note: The repetition of an index in a term will denote a summation with respect to that index over its range. “dummy index” – one that is summed over “free index” – one that is not summed

- 4 -


Hence, aixi = ajxj = akxk

e.g., 1 2 3

1 2 32 2 2 2 2

1 2 3

1 1 2 2 3 3

i i

i i

u u uv v v

u u u u u uu v u v u v u v

= + +

= + +

= = + + =

⋅ = + + =

1 2 3

1 2 3

u e e ev e e e

uu v

(more examples: aijxj, aijxjk, aijxij, …)

Kronecker delta (δij):

1 if ,

0 if ij

ij

i j

i j

δ

δ

= =

= ≠

or

( )1 0 00 1 00 0 1

ijδ⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠

then 2

i i ij i j

ij i j

u u u u u

a a

δ

δ

= =

=

- 5 -


Matrices and Determinants Let A be an m×n matrix and B be an n×p matrix.

( ) 1,..., ; 1,...,

( ) 1,..., ; 1,...,ij

jk

a i m j n

b j n k p

= = =

= = =

A

B

the product of A and B is an m×p matrix defined as ( ) 1,..., ; 1,..., ; 1,...,ik ij jka b i m j n k p⋅ = = = =A B

11 12 13

21 22 23

31 32 33

11 22 33 12 23 31 13 21 32

11 23 32 12 21 33 13 22 31

det det( )ij

a a aa a a a

a a aa a a a a a a a aa a a a a a a a a

= =

= + +

− − −

A

- 6 -


Permutation symbol (εrst):

1 if , , permute as 1,2,3,1,2,3,...1 if , , permute as 3,2,1,3,2,1,...

0 otherwise

rst

rst

rst

r s tr s t

εεε

== −=

i.e.,

123 231 312

213 321 132

111 222 333 112 121 211

1,1,

... 0

ε ε εε ε εε ε ε ε ε ε

= = == = = −= = = = = = =

then,

1 2 3det( )ij rst r s ta a a aε=

rst s t ru vε× =u v e

( ) ( ) ijk i j ka b cε⋅ × = × ⋅ =a b c a b c

HOMEWORK: 2.16, 2.19

- 7 -


TRANSLATION AND ROTATION OF COORDINATES 2-D SPACE

• Translation

x

x’

y’y A

h

k

''

x x hy y k

= += +

or ''

x x hy y k

= −= −

- 8 -


• Rotation

x

y

x’

y’ A

θ θ

'cos 'sin'sin 'cos

x x yy x y

θ θθ θ

= −= +

or ' cos sin' sin cos

x x yy x y

θ θθ θ

= += − +

Using index notation,

' (i ij jx x i 1, 2)β= = where

cos( ', )ij i jx xβ =

11 12

21 22

cos sin( )

sin cosij

β β θ θβ

β β θ θ⎛ ⎞ ⎛ ⎞

= =⎜ ⎟ ⎜ ⎟−⎝ ⎠⎝ ⎠

- 9 -


The inverse transform is

' ( 1,2i ji jx x i )β= =

Note: (βij) or B is an orthogonal matrix:

1( ) ( ) ( )Tji ij ijβ β β −= =

1−⋅ = ⋅ =TB B B B I

Hence,

ik jk ijβ β δ=

i.e., Consider a unit vector along xi’-axis (βi1,βi2)

2 21 2( ) ( ) 1 ( 1,2i i iβ β+ = = )

)

Unit vector along xi’-axis (βi1,βi2) is perpendicular to unit vector along xj’-axis (βj1,βj2) if i≠j

1 1 2 2 0 (i j i j i jβ β β β+ = ≠

- 10 -


3-D SPACE

y

x

z

x’

y’

z’

A

In the same manner as 2-D (See proof in pp. 51-52)

'i ij jx xβ= 'i ji jx xβ=

where direction cosines

'ij i jβ ≡ ⋅e e which is a cosine of angle between xi’ and xj axes Law of transformation of any vector A:

'i ij jA Aβ= 'i ji jA Aβ= HOMEWORK: 2.34

- 11 -


ANALYTICAL DEFINITIONS OF SCALARS, VECTORS, AND CARTESIAN TENSORS

P

x1

x2

x3

x2’

x1’

x3’

Scalar (or tensor of rank 0)

• has only single component • equal in all frames of reference

e.g., temperature, moisture, mass, …

1 2 3 1 2 3( , , ) '( ', ', ')x x x x x xφ φ=

Vector (or tensor of rank 1) • has 3 components • obeys law of transformation

e.g., force, displacement, velocity, …

1 2 3 1 2 3

1 2 3 1 2 3

'( ', ', ') ( , , )( , , ) '( ', ', ')

i k

k i

u x x x u x x xu x x x u x x x

ik

ik

ββ

==

- 12 -


Tensor of rank 2

• has 9 components • obeys law of transformation

e.g., stress, strain, …

1 2 3 1 2 3

1 2 3 1 2 3

'( ', ', ') ( , , )

( , , ) '( ', ', ')ij mn im jn

mn ij im jn

x x x x x x

x x x x x x

σ σ β β

σ σ β β

=

=

Tensor of rank 3

ijke

Tensor of rank 4

ijklC

: Since based on rectangular Cartesian frame

called “Cartesian tensors”

- 13 -


PARTIAL DERIVATIVES

1 21 2

,

... nn

ii

i i

f f fdf dx dx dxx x xf dxx

f dx

∂ ∂ ∂= + +

∂ ∂ ∂∂

=∂

=

Comma notation:

,

,

,

ii

ii j

j

ijij k

k

xuux

x

φφ

σσ

∂≡

∂

∂≡

∂

∂≡

∂


- 14 -


EIGENVALUE PROBLEM For a symmetric matrix with real elements A=(aij)

11 12 13 1 1

21 22 23 2 2

31 32 33 3 3

a a a x xa a a x xa a a x x

λ⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟=⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

or ( )ij j i ij ij ja x x a x 0λ λδ= − =

e.g., stress, strain, buckling of column, vibration,…

• There are 3 real eigenvalues λ (λ1, λ2, λ3) obtained from

det( ) 0ij ija λδ− =

• There are 3 eigenvectors X corresponding to

the eigenvalues. They are mutually orthogonal (for distinct eigenvalues) and can be found from solving

ij j ia x xλ=

- 15 -


• If the coordinate system is rotated so as to coincide with the eigenvectors, then matrix A becomes diagonal matrix

1

2

3

0 0' 0 0

0 0

λλ

λ

⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟⎝ ⎠

A

• The eigenvalues are the extremum values of

aij. EXAMPLE: Find eigenvalues (and eigenvectors) for

4 2( )

2 1ijσ⎛ ⎞

= ⎜ ⎟−⎝ ⎠

Soln: 1 1

2 2

2

4 22 1

4 2det 0

2 1(4 )( 1 ) (2)(2) 0

3 8 04.70, 1.70

x xx x

λ

λλ

λ λ

λ λλ

⎧ ⎫ ⎧ ⎫⎡ ⎤=⎨ ⎬ ⎨ ⎬⎢ ⎥−⎣ ⎦ ⎩ ⎭ ⎩ ⎭

−⎡ ⎤=⎢ ⎥− −⎣ ⎦

− − − − =

− − == −

- 16 -


STRESS (Shames 2.2, 2.3, 2.4)

Consider a continuum (solid or fluid) of volume V with surface S under some external forcing conditions.

δA

δF

n

t

Stress vector (or traction) is defined as

20

forcelim (unit: )lengthA

dA dAδ

δδ→

= =F FT

which has components in normal and tangential directions as

0

0

lim

lim

n nn A

t tt A

F dFA dA

F dFA dA

δ

δ

δτδ

δτδ

→

→

= =

= =

- 17 -


STRESS NOTATION

P

Stress (or force intensity) at a point depends on

• position • orientation

x y

z τxx

τxy

τxz τyy

τyx

τyz

τzz

τzx

τzy

- 18 -


Stress tensor (a second-order tensor)

xx xy xz

ij yx yy yz

zx zy zz

τ τ ττ τ τ τ

τ τ τ

⎛ ⎞⎜ ⎟

= ⎜ ⎟⎜ ⎟⎝ ⎠

irst index identifies the plane (normal direction).

.g., τ is a stress acting on x-plane,

τ , τ , τ are called “normal stresses”.

s”.

ote: Stress tensor is symmetric, i.e., (see proof in

FSecond index identifies the direction of the stress. e xy

and pointing in y-direction.

xx yy zzτxy, τxz, τyx,… are called “shearing stresse

Nsection 2.4)

ij jiτ τ= or

xy yx

yz zy

xz zx

τ τ

τ τ

τ τ

=

=

=

- 19 -


Sign convention mponent acts

ection or

-ive: Therection or

XAMPLE:

(For normal stresses, positive means tension and

+ive: The coon +i face in +j diron –i face in –j direction component acts

on +i face in –j di on –i face in +j direction E

+τ −τ

negative means compression.) HOMEWORK: 2.36, 2.50

x y

z +τxx

+τxy +x face

−x face xx xx

−τxy

−τxx

- 20 -


STRESS TRANSFORMATION (Shames 15.2)

ΔS = area of ABC n = unit normal vector ΔS1, ΔS2, ΔS3 = areas of OBC, OCA, OAB h = distance between O and N

n

From Newton’s law (ΣFi = mai)

i

O

A

B

C

x1

x2

x3

n

N TΔS

γΔV

i iS SΔ = Δ

1 1 2 2 3 3i i i i iT S S S S V Vaτ τ τ γ ρΔ − Δ − Δ − Δ − Δ = Δ

- 21 -


- 22 -

3 3i ji j i iShSn aγ ρShΔT S τ Δ

Δ − =

as h→0,

Δ −

i ji jT nτ= “Cauchy’s formula”

Then,

nn i i ji j iT n n nτ τ= =

nt i i ji j iT t n tτ τ= =

XAMPLE: Find normal stress on plane whose nit normal vector

⎟= −⎜ ⎟⎜ ⎟⎝ ⎠

13 1 3

3

31 3 1 32 3 2 33 3 3 (2000)(0.6)(0.6) (2000)(0.6)(0.8) 2 2640 kPa

nn ji j in n n n n n n n

nn n n n n n

Eu

(0,0.60,0.80)=n 3000 1000 0−⎛ ⎞1000 2000 2000 kPa0 2000 0

ijτ ⎜

τ 11 1 1 12 1 2

21 2 1 22 2 2 23 2 n n n n n

τ τ τ τ

τ τ τ

= = + +

+ + +

= + × =

τ τ τ+ + +



- 23 -

Next, from stresses τij in xi-system, find stresses τij’ in xi’-system.

From

x1

x2

x3

xk’

x m’

'i ij jx xβ= where β is cosi eij n of angle between xi’ and xj and Cauchy’s formula

i ji jT nτ= where nj is unit normal vector to any plane If n is parallel to xk’ axis, then

jj kn β=

n

T’k

τkm’


and

'ki ji kjT τ β= The component T’k in xm’ direction which is stress (τkm’) on xk’ face in xm’ direction is

3

'

m

τ 1 1 2 2 3 3

1 1 2 2 3

' ( ' ' ' )

' ' '

'

k k kkm m

k k km m

ki mi

T T T

T T Tβ

T

β β

β

= + + ⋅

= + +

=

e e e e

Hence,

'km ji kj miτ τ β β=

x

y

z

x’

y’

z’

τ τkm’ ij

- 24 -


PRINCIPAL STRESSES & PRINCIPAL PLANES (Shames 15.3, 15.4)

rom Cauchy’s formula (for a state of stress at a

j

Fpoint)

T ni jiτ= where nj is unit normal vector to any plane, there exist three mutually perpendicular axes (called principal axes) where there is zero shear stress (i.e., Ti = τni). (See Figure 1.)

j i ij ij jnor ( ) 0ijn nτ = τ τ τδ− =

In matrix form:

or

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

11 12 13 1 1

21 22 23 2 2

31 32 33 3 3

n nn nn n

τ τ ττ τ τ ττ τ τ

⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎜ ⎟=⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

11 12 13 1

21 22 23 2

31 32 33 3

000

nnn

τ τ τ ττ τ τ ττ τ τ τ

−⎛ ⎞⎜ ⎟− =⎜ ⎟⎜ ⎟−⎝ ⎠

- 25 -


To solve this problem,

11 12 13

det 0τ τ τ τ⎜ ⎟21 22 23

31 32 33

τ τ τ τ

τ τ τ τ

−⎛ ⎞− =⎜ ⎟

⎜ ⎟−⎝ ⎠

Invariants (i.e., do not change with coordinate system):

gives

3 21 2 3 0I I Iτ τ τ− + − + =

yields 3 roots for τ: τ1, τ2, τ3

1 11 22 33

22 23 11 1311 122

32 33 31 3321 22

11 12 13

3 21 22 23

31 32 33

I

I

I

τ τ ττ τ τ ττ ττ τ τ ττ τ

τ τ ττ τ ττ τ τ

+

= + +

=

= +

OMEWORK: 15.14 H

- 26 -


Figure 1. Stress at a point in 3-D

- 27 -


MOHR’S CIRCLE (Shames 7.5)

is a graphical method of stress transformation (for a state of stress at a point)

• Plane stress (2-D) 0zz zx zyτ τ τ= = =

y τyy

τ τyyyxτ

Sign convention

• normal stress: + for tension −

• shear+ for clockwise − for counterclockwise

for compression stress:

L

K

z

x

yx τxy

τxy

ττ xx xx

L’ K’

τxx’ τxy’

τyx’ τyy’

θ

- 28 -


EXAMPLE:

L

K

1

2

4

aτ ⎜ ⎟

2

4 2 0⎛ ⎞2 1 0 kPj

0 0 0i = −

⎜ ⎟⎝ ⎠⎜ ⎟

τnn

2 2(2.5) (2) 3.2 kPaR = + =

:

3.2 1.7 kPa

principal stresses1

2 1.51.5 3.2 4.7 kPaτ

τ = −= + =

= −

maximum shear stress:

max ns R 3.2 kPaτ = = ±

K (4, -2)

L (-1, 2)

C

τns

τ1 τ2

R

- 29 -


• General state of stress (3-D) y, z coincide with principal axes

Let x,τ

10 20 30 40-10-20normal stress

(MPa)

shearing stress(MPa)

principal stresses: , 20 MPa1 2 340 MPa, 0τ τ τ= = = −

largest shear stress:

max 30 MPansτ =

x y

z τ1

τ2

3

τ1 τ2 τ3

- 30 -


EQUILIBRIUM EQUATIONS (Shames Consider a small element of a deformable body which is in equilibrium condition.

2.6)

x

y

z

dy

dz

dx

τzx

τ

Bx

τyx+(

0;

0

x

xxxx xx

yxyx yx

zxzx zx x

F

dx dydz dydzx

dy dxdz dxdzy

dz dxdy dxdy B dxdydzz

ττ τ

ττ τ

ττ τ

=

∂⎛ ⎞+ −⎜ ⎟∂⎝ ⎠∂⎛ ⎞

+ + −⎜ ⎟∂⎝ ⎠∂⎛ ⎞+ + − + =⎜ ⎟∂⎝ ⎠

∑

yx

τxx

∂τyx/ y)dy

τxx+(

∂

∂τxx/∂x)dx

τzx+( τzx/ z)dz ∂ ∂

- 31 -


Canceling terms and dividing by dxdydz, we get

0yxxx zxxB

x y zττ τ∂∂ ∂

+ + + =∂ ∂ ∂

Similarly for y and z directions, we have

0

0

xy yy zyy

yzxz zzz

Bx y z

Bx y z

τ τ τ

ττ τ

∂ ∂ ∂+ + + =

∂ ∂ ∂∂∂ ∂

+ + + =∂ ∂ ∂

as equilibrium equations (in differential form). However, using index notation, these equations can be written as

, 0ji j iBτ + =

- 32 -


- 33 -

EXAMPLE:

( )( )

( )

2 3

3

2 3

2 3 10 kPa

1000 2 10 kPa

2 10 kPa

zy

zx

zz

x y

x

xy x

τ

τ

τ

= + ×

= +

y

×

= + ×

Find resultant force.

1 12

0 0

(2000 3000 )

2167 kN

(1000 2000 )

917 kN

y zy

z zz

1 12

0 0

1 1

0 0

2000 kN

(1000 2000 )x zx

F dA x y dxdy

F dA x dxdy

F dA xy x dxdy

τ

τ

= = +

=

= = +

=

∫ ∫ ∫

∫ ∫ ∫

τ= = +∫ ∫ ∫=

0

1 m

1 m

x


- 34 -

STRAIN (Shames 3.1, 3.2, 3.3)

Displacement field u(x,y,z)

1. Translation 2. Rotation 3. Extension/Contraction 4. Distortion

EXAMPLE: Given the following displacement field,

u = [(x2+y) i + (3+z) j + (x2+2y) k] m what is the deformed position of a point originally at (3,1,-2)m?

C’ D’

x

y r

u

r’A B

C D A’

B’

2 2' (3,1, 2) ((3) 1,3 ( 2), (3) 2(1)) (13, 2,9)

= + = − + + + − +=

r r u

2101202 – Mechanics of Materials I By Watanachai Smittakorn 2101202 – Mechanics of Materials I By Watanachai Smittakorn

- 35 -

Translation

Rotation

r

A B

C D A’

A’

- 35 -


Extension/Contraction

☺ Normal strain

dy

xx

yy

zz

dxdxdy

dydz

dz

ε

ε

ε

Δ=

Δ=

Δ=

(+ for extension, − for contraction)

A’

dx dy Δdx

Δ

- 36 -


Distortion

A’

B’

C’

D’

εxy

εyx 90−γxy

☺ Shear strain

2

2

2

xyxy yx

γε ε

xzxz zx

yzyz zy

γε ε

γε ε

= =

= =

= =

is called “engineering shear strain”

+ for decrease in angle − for increase in angle

γ

- 37 -


STRAIN-DISPLACEMENT RELATIONS (Shames 3.4) Assume: infinitesimal deformation or displacement

( ), ,i x y zu u u u=

• Normal strain

xxx

yyy y

zzz

ux

u

u

z

ε

ε

ε

∂=

∂∂

=∂

∂=

∂

∂ux ∂x

∂uy

∂y

x

y

- 38 -


• Shear strain

∂ux

Since

yxxy

uuy x

γ∂∂

= +∂ ∂

then,

12

yxxy yx

uuy x

ε ε∂⎛ ⎞∂

= = +⎜ ⎟∂ ∂⎝ ⎠

Also,

12

12

x zxz zx

y zyz zy

u uz xu uz y

ε ε

ε ε

∂ ∂⎛ ⎞= = +⎜ ⎟∂ ∂⎝ ⎠∂⎛ ⎞∂

= = +⎜ ⎟∂ ∂⎝ ⎠

∂uy

∂x

∂y

x

y

0−γxy 9

- 39 -


Strain tensor

xx xy xz

ij yx yy yz

zx zy zz

ε ε εε ε ε ε

ε ε ε

⎛ ⎞⎜ ⎟

= ⎜ ⎟⎜ ⎟⎝ ⎠

By using index notation, εij may be written as

( ), ,12ij i j j iu uε = +

PLE: For the following displacement field, u = (sin xi + yzj + x2k) × 10-2 m,

what are the strain components at r = 2i + j − k m?

EXAM

cos cos 2100 100

1100 100

0

1 1 (0 0) 02 200

1 1 (0 2 )2 200

1 1 ( 0)2 200 200

xxx

yyy

zzz

yxxy yx

x zxz zx

y zyz zy

u xxu zy

uz

uuy x

u u xz xu u yz y

ε

ε

ε

ε ε

ε ε

ε ε

∂= = =

∂∂ −

= = =∂

∂= =

∂∂⎛ ⎞∂

= = + = + =⎜ ⎟∂ ∂⎝ ⎠∂ ∂⎛ ⎞= = + = + =⎜ ⎟∂ ∂⎝ ⎠∂⎛ ⎞∂

= = + = + =⎜ ⎟∂ ∂⎝ ⎠

4

200

1

- 40 -


COMPATIBILITY EQUATIONS Shames 3.5)

In solid mechanics, displacement field (ui) must be a single-valued & continuous function.

Also, strain field must satisfy certain requirements to be associated with the single-valued & continuous displacement field.

“compatibility equations”

(

discontinuity ≈ crack or fissure

HOMEWORK: 3.32

- 41 -


2

2

2

2 22

2 2

2 2 2

,

,

2 ,

2

yy xy yzzx

xy yz zxzz

xy yyxx

zx

z,yz xyxx zx

y z x x y

2 2

2 2

2 ,yz yy zz

2zz

z x y y z x

x y z z x y

x y y x

y z z y

z x x

ε ε

ε ε εε

ε ε εε

ε εε

ε ε ε

ε ε

∂ ∂

∂ ⎠∂ ∂ ∂⎛ ⎞∂∂

= − + +⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠∂ ∂⎛ ⎞∂∂ ∂

= − + +⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠∂ ∂∂

= +∂ ∂

∂ ∂ ∂

∂ ∂=

∂ ∂ ∂

ε ε⎛ ⎞∂ ∂∂= − + +⎜ ⎟∂ ∂ ∂ ∂ ∂⎝

∂ ∂

= +∂ ∂ ∂ ∂

2

2 .xx

zε∂

+∂

- 42 -


STRAIN TRANSFORMATION Shames 16.1, 16.2, 16.3)

The strain components at a point transform exactly as do the stress components at a point on a rotation of axes.

second-order Cartesian tensor

From strains εij in xi-system, find stresses εij’ in xi’-system.

(

x x’

y

y’

z

z’

'ij mn im jnε ε β β= where βij is cosine of angle between xi’ and xj

'i ij jx xβ=

- 43 -


EXAMPLE: The following displacement field a body under load:

x y xz z −⎡ ⎤= + + + − ×u i j k

s

describes the movement of

( ) ( )2 2 2 23 0.6 10 ft⎣ ⎦ Compute the normal strain at (0, 1, 3) in a direction s given a

0.6 0.8= +s i j

( )

2 0100

0

1.2 3.6100 100

1 1 22 2 100 100

1 1 0 0 02 2

1 1 0 02 2 100

xxx

yyy

zzz

yxxy

x zxz

y zyz

u xxuy

u zz

uu y zy x

u u

2.5

z y

ε

ε

ε

ε

ε

ε

∂= = =

∂∂

= =∂

∂ − −= = =

∂∂⎛ ⎞∂ +⎛ ⎞= + = =⎜ ⎟ ⎜ ⎟∂ ∂ ⎝ ⎠⎝ ⎠

∂ ∂⎛ ⎞= + = + =⎜ ⎟z xu u x

∂ ∂⎝ ⎠∂⎛ ⎞∂ +⎛ ⎞= + = =⎜ ⎟ ⎜ ⎟∂ ∂ ⎝ ⎠⎝ ⎠

'

2.5 3.6 (0.6)(0.8) 2 (0)(0)100 100

0.024

ij mn im jnε

ss xy sx sy yx sy sx zz sz sz

ε β β=

−⎛ ⎞ ⎛ ⎞= × +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

=

ε ε β β ε β β ε β β= + +

HOMEWORK: 16.1

- 44 -


Principal strains & principal axes:

Solving an eigenvalue problem

ij j in nε ε=

0xx xy xz

yx yy yz

zx zy zz

ε ε ε εε ε ε εε ε ε ε

−− =

−

yields

where A, B, and C are strain invariants:

3 2 0A B Cε ε ε− + − =

2 2 2

2 2 22

xx yy zz

xx yy xx zz yy zz xy yz xz

xx yy zz xy yz zx xy zz xz yy yz xx

A

B

C

ε ε ε

ε ε ε ε ε ε ε ε ε

ε ε ε ε ε ε ε ε ε ε ε ε

= + +

= + + − − −

= + − − −

See example 16.2) (

- 45 -


STRAIN MEASUREMENT AND STRAIN OSETTES R (Shames 8.6)

Strain gage – a device for measuring strain

Strain rosettes – used to measure the state of strain

at a point (2D)

- 46 -


INTRODUCTION TO ELASTICITY

(Hooke’s Law)

Solid Mechanics

force (F)

displacement (u)

stress (τ)

strain (ε)

equilibrium compatibility

constitutive relation

(material law)

e.g., constitutive relation for a bar in tension

τ τ

Eτ ε=

where E = modulus of elasticity or Young’s modulus

- 47 -


THREE-DIMENSIONAL HOOKE’S LAW FOR ISOTROPIC MATERIALS

) (Shames 6.2, 6.3

“isotropic materials” – have the same properties in all directions

nder τxx:

τxx τxx

τxx τxx

u

' xxxx E

τε = where E = modulus of elasticity

or Young’s modulus

and by Poisson effect

' ' xxyy xx

τ

' ' xxzz xx

E

E

ε νε ν

τε νε ν= − = −

= − = −

where ν = Poisson’s ratio

- 48 -


under τyy:

'' yyyy E

τε =

and

'' ''

'' ''

yyxx yy

yyzz yy

E

E

τε νε ν

τε νε ν

= − = −

= − = −

under τzz:

''' zzzz E

τε = and

''' '''

''' '''

zzxx zz

zzyy zz

Eτε νε ν

τE

ε νε ν= − = −

= − = −

- 49 -


τxx τxx

τyy

τyy

τzz

τzz

Hence, superposing the effects of the normal stresses

( )

( )

( )

1

1

1

xx xx yy zz

yy yy xx zz

zz zz xx yy

E

E

ε τ

E

ν τ τ

τ ν τ τ

ε τ ν τ τ

⎡ ⎤= − +⎣ ⎦

⎡ ⎤= − +⎣ ⎦

⎡ ⎤= − +⎣ ⎦

ε

- 50 -


For isotropic materials, shear stresses act independently of each other.

γxy

1

1

1

xyγ xτ yG

xγ z xz

yz yz

G

G

τ

γ τ

=

=

=

where G = shear modulus

ote: relation between the three material constants N

( )2 1EG

ν=

+

- 51 -


In matrix form: 1 0 0 0

1 0 0 0

1 0 0 0

10 0 0 0 0

10 0 0 0 0

10 0 0 0 0

xx xx

yy yy

zz zz

yz yz

xz xz

xy xy

E E E

E E E

E E E

G

G

G

ν ν

ν ν

ε τε τν νε τγ τγ τγ τ

⎤⎢ ⎥⎢ ⎥⎢ ⎥− −⎧ ⎫ ⎧ ⎫⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥− −⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥

⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭⎢ ⎥

⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

Inverse matrix:

44

66

0 0 00 0 0

0 0 0 0 00 0 0 0 0

⎡ − −

11 12 13

21 22 23

31 32 33 0 0 0

55

0 0 0 0 0

xx xx

yy yy

τ

zz zz

yz yz

xz xz

xy xy

C C CC C C

CC

ετ ετ εC C C

⎧ ⎫ ⎧ ⎫⎡ ⎤⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥τ

C

γτ γτ γ

⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪

⎪ ⎪ ⎪ ⎪⎥ ⎨ ⎬

⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪

⎣ ⎦⎩ ⎭ ⎩ ⎭

Here, Cij depend on only two independent variables (E, ν, or G)

= ⎢⎨ ⎬

- 52 -


ANISOTROPIC MATERIALS: GENERALIZED HOOKE’S LAW (Shames 6.5) In general for anisotropic materials:

11 12 13 14 15 16

21 22 23 24 25 26

31 32 33 34 35 36

41 42 43 44 45 46

51 52 53 54 55 56

61 62 63 64 65 66

xx x

yy yy

x

zz z

yz yz

z

xz xz

xy x

C C C C C CC C C C C CC C C C C CC C C C C CC C C C C CC C C C C C y

τ ετ ετ ετ γτ γτ γ

⎧ ⎫ ⎧⎡ ⎤⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪= ⎢ ⎥⎨ ⎬ ⎨

⎫

⎢ ⎥⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪⎣ ⎦⎩ ⎭ ⎩

⎪⎪⎪⎪

⎪⎪⎪⎭

Special case: for materials with symmetry about planes (e.g., wood)

called “orthotropic”

0 0 00 0 00 0 0

0 0 0 0 00 0 0 0 0

⎬⎪

three orthogonal

11 12 13

21 22 23

31 32 33

44

55

660 0 0 0 0

xx xx

yy yy

τ ε

zz zz

yz yz

xz xz

xy xyC

C C CC C CC C C

CC

τ ετ ετ γτ γτ γ⎢ ⎥⎪ ⎪ ⎪ ⎪⎣ ⎦⎩ ⎭ ⎩ ⎭

⎧ ⎫ ⎧ ⎫⎡ ⎤⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪= ⎢ ⎥⎨ ⎬ ⎨ ⎬

⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪

- 53 -


PLANE STRESS (Shames 7.1)

Consider a thin plate loaded in the midplane of

Hence,

the plate.

x

y y

z

0zz zx zyτ τ τ= = =

0 but 0)zx zy zz(for isotropic material: ε ε ε= = ≠

Transformation of stress and strain in 2-D can use Mohr’s circle.

small t

τyy

τxy

τxx

- 54 -


PLANE STRAIN (Shames 8.1)

A prismatic body constrained not to bend, stretch, or shorten is loaded in a direction normal to the centerline with no variation in the direction of the axis of the prism. Hence,

0zz zx zyε ε ε= = =

(for isotropic material: 0 but 0)zx zy zzτ = τ τ= ≠

Transformation of stress and strain in 2-D can use Mohr’s circle. e.g., a prismatic dam built between rigid rock supports

x

y

W H

- 55 -


EXAMPLE: A gas pipeline with 600 mm inside diameter hickness is carrying a pressure of 1.5 MPa

bove atmosphere. ) Fi in

the w

and 12 mm wall ta(a nd the principal stresses and maximum shearing stress

all. (b) If the pipe is made of steel with E=200 GPa and ν=0.28, find the principal strains in three dimensions.

σ

τ

σ1 2

(a)

hoop 1

1 2max

(1.5)(300) 37.5 MPa12

(1.5)(300)

18.75 9.375 MPa2 2

prt

praxial 218.75 MPa

2 2 12t

σ σ

σ σ= = = =×

σ στ

= = = =

−= = =

(b)

( )( ) ( ) ( )( ) ( )

( )( ) ( ) ( )( ) ( )

( )( ) ( ) ( )( ) ( )

36

61 1 2 3 9

66

2 2 1 3 9

66

3 3 1 2 9

0 (plane stress)

1 10 37.5 0.28 18.75 0 161 10200 10

1 10 18.75 0.28 37.5 0 41.3 10200 10

1 10 0 0.28 37.5 18.75 78.8 10200 10

E

E

E

σ

ε σ ν σ σ

ε σ ν σ σ

ε σ ν σ σ

−

−

−

=

= − + = − + =

= − + = − + =

= − + = − + = −

σhoop

σaxial

τmax

σ

- 56 -

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VECTORS AND TENSORS - pioneer.netserv.chula.ac.thpioneer.netserv.chula.ac.th/~pwithit/Vectors.pdf2101202 – Mechanics of Materials I By Watanachai Smittakorn VECTORS AND TENSORS “A