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Mechanical Behavior of Solids

Linear Elastic solids

E= (1D) klijklij C = or klijklij S = (3D)

where C is sometimes called the stiffness tensor and S is sometimes called the compliance

tensor. Both of them are 4th

order elastic moduli tensors.

Symmetry of elastic moduli tensors:

jiklijkl CC = , (minor symmetry)ijlkijkl CC =

The minor symmetries reduce the independent elastic constants from 81 to 36.

There is also a major symmetr in elastic moduli klijijkl CC = , which reduces the number of

independent elastic constants from 36 to 21.

We use the concept of energy & work to demonstrate the major symmetry of C and S :

uF,

Assume an increment of displacement at the bar end,

uuu +

The work done by the applied load should equal to the stored energy in the material,

( ) VAllAuFW ====

==V

Ww should be the stored elastic energy per unit volume, which is also called the

strain energy density.

( )ww = ,

=

w, ( ) =

0

dw

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Generalize to 3D

S

V

f

v

tv

( )

( )

VwV

VuVuf

VufVu

VufSun

VufSut

VufSutw

VVijij

Vjiiji

Vijij

iV

iV j

iij

iV

iiS

jij

iV

iiS

i

VS

dd

dd

dd

dd

dd

dd

,,

==

++=

+=

+=

+=

+=

vvvv

ijijw = is the strain energy density and

= ijijw must be integrable.

Knowing klijklij C = , we have

klijijklCw 2

1=

klij

klij

ijkl Cw

C =

=

2

Elastic moduli for isotropic materials:

Isotropic tensor: aaIvv

= ( ijij aa = )

The 4th

order isotropic tensors can be generally written as

jkiljlikklijijkl cccC 321 ++=

2

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Imposing the symmetry conditions indicate that the elastic moduli can be expressed in terms of

just two constants

jkiljlikklijijklC ++=

where and are called Lam constants.

Similarly, the can also be expressed in terms of just two constants asijklS

jkiljlikklijijkl ddS ++= 21

The constants , , and are to be determined experiments.1d 2d

Since there are only 2 independent elastic constants, it suffices to conduct experiments in 1D:

'

=E (Youngs modulus)

'

= (Poissons ratio)

:

We can specify the general stress-strain relation

ijijkkkljkiljlikklijklijklij ddddS 2121 2+=++==

to 1D loading 011 , 03322 == . In this case,

( )E

dddd 11112111211111 22

=+=+=

Ed 111111122

===

Therefore,

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Ed

=1

Ed

2

12

+=

and

( )jkiljlikklijijklEE

S

++

+=2

1

We thus obtain the so-called generalized Hookes law:

ijkkijijEE

+

=1

This equation can be inverted as follows.

qqkkqqqqEEE

2131

=+

=

The percentage change of volume,

( )p

EEV

Vkkkk

21321 =

==

Defines the so-called bulk modulu( )213

=

=E

VV

pK . (This modulus can also be

calculated using ab initio quantum mechanics techniques.)

Invert the generalized Hookes law:

( )( ) ijkkijij

ijkkijijkkijij

EE

EEEE

2111

211

++

+=

+=+=+

Comparing this to the Lam form of Hookes law

ijkkijkljkiljlikklijklijklij C 2+=++== , we identify

( )( )

211 +

=E

( )

+=

12

E

For shear deformation:

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== G

G

Let , in1=i 2=j ijkkijij 2+= ,

=== 1212 2

Therefore, the second Lam constant is just the shear modulus:

( )

+==

12

EG

Alternative derivation of the generalized Hookes law

I

II

III

Consider a linear superposition of strain in the principal stress directions

( )IIIIIIIIIIIII

IEEEEE

+++

==1

( )IIIIIIIIIIIIII

IIEEEEE

+++

==1

( )IIIIIIIIIIIIIII

IIIEEEEE

+++

==1

i.e.

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Therefore, we have

IEE

kk

+

=1

Through change of coordinates, this relation remains true for any other coordinate systems.

Orthotropic materials

An orthotropic material (e.g. wood, composites, fiber glass, etc) has three mutually perpendicular

symmetry planes.

1Sym

2Sym

3Sym

=

12

13

23

33

22

11

12

13

23

33

22

11

66C

C : (9 elastic constants)

66

55

44

33

2322

131211

0

00

000

000

000

C

CSym

C

C

CC

CCC

Cubic materials

Single crystal metals: FCC (Cu, Al, Ag, etc.); BCC (Fe, etc.)

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C : (3 elastic constants)

44

44

44

11

1211

121211

0

00000

000

000

C

CSym

CC

CC

CCC

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