Isotropic linear elastic response

1Isotropic linear elastic response

Range of Young’s modulus 2

0.2

8

0.6

1

Magnesium,

Aluminum

Platinum

Silver, Gold

Tantalum

Zinc, Ti

Steel, Ni

Molybdenum

Graphite

Si crystal

Glass-soda

Concrete

Si nitrideAl oxide

PC

Wood( grain)

AFRE( fibers)*

CFRE*

GFRE*

Glass fibers only

Carbon fibers only

Aramid fibers only

Epoxy only

0.4

0.8

2

4

6

10

20

40

6080

100

200

600800

10001200

400

Tin

Cu alloys

Tungsten

<100>

<111>

Si carbide

Diamond

PTFE

HDPE

LDPE

PP

Polyester

PSPET

CFRE( fibers)*

GFRE( fibers)*

GFRE(|| fibers)*

AFRE(|| fibers)*

CFRE(|| fibers)*

From Callister, Intro to Eng. Matls., 6Ed

E [G

Pa =

109

N/m

2 ]Metals Alloys

Graphite Ceramics

Semicond.

Composites Fibers

Based on data in Table B2, Callister 6Ed. Composite data based on reinforced epoxy with 60 vol% of aligned carbon (CFRE), aramid (AFRE), or glass (GFRE) fibers.

What is the source of both universality and range in

modulus?

Polymers

3

Materials are made of atoms, held together by atomic interactions • covalent and ionic bonding: ceramics, semiconductors (~200 N/m) • metallic bonding: metals (~ 20 N/m) • van der Waals interaction: polymers (~ 0.5 N/m)

Materials are made of many atoms, governed by thermodynamics • materials choose structures, phase variables (such as density) that

minimize free energy: A = U − TS • A: Helmholtz free energy • U: internal energy (bonding) • T: (absolute) temperature • S: entropy (disorder: kB log Ω)

Universality of linear elastic response

4Thermodynamic “equation of state”

E

Equilibrium bond length

Bond energy

P(V) =@A@V

��V

5

Materials are made of atoms, held together by atomic interactions • covalent and ionic bonding: ceramics, semiconductors (~200 N/m) • metallic bonding: metals (~ 20 N/m) • van der Waals interaction: polymers (~ 0.5 N/m)

Materials are made of many atoms, governed by thermodynamics • materials choose structures, phase variables (such as density) that

minimize free energy: A = U − TS • A: Helmholtz free energy • U: internal energy (bonding) • T: (absolute) temperature • S: entropy (disorder: kB log Ω)

Universality of linear elastic response

P(V) =@A@V

��V

P(�V + V0) =@A@V

��V0

+ �V@2A@V2

��V0

+12�V2 @

3A@V3

��V0

+ · · ·

= 0 +�VV0

0BBBB@V0@2A@V2

��V0

1CCCCA + · · ·

= ✏VK

✏k =1E�k ✏? = �

⌫E�k � =

1G⌧

• For small stresses the strains are linearly related to stresses:

• We can generalize these results by considering superposition 1.Each stress component (σx σy σz τxy τxz τyz) is considered individually 2.All of the strains from each stress component computed 3.Sum of all strains = material response to stress

Superposition principle 6

✏x

=1E

[�x

� ⌫(�y

+ �z

)]

✏y

=1E

[�y

� ⌫(�x

+ �z

)]

✏z

=1E

[�z

� ⌫(�x

+ �y

)]

�xy

=1G

⌧xy

�xz

=1G

⌧xz

�yz

=1G

⌧yz

• The superposition principle can relate our elastic moduli: • E: Young’s modulus (normal strain from uniaxial stress) • ν: Poisson’s ratio (perpendicular normal strain from uniaxial stress) • G: shear modulus (shear strain from shear stress) • K: bulk modulus (volume change from hydrostatic pressure)

Material property relationships 7

0 ⌧⌧ 0

!$

⌧ 00 �⌧

! 0 �/2�/2 0

!$

�/2 0

0 ��/2

!

� =1G⌧ �

2=

1E⌧ � ⌫

E(�⌧)

� =2(1 + ⌫)

E⌧

G =E

2(1 + ⌫)

• The superposition principle can relate our elastic moduli: • E: Young’s modulus (normal strain from uniaxial stress) • ν: Poisson’s ratio (perpendicular normal strain from uniaxial stress) • G: shear modulus (shear strain from shear stress) • K: bulk modulus (volume change from hydrostatic pressure)

Material property relationships 8

� =

0BBBBBB@

�p 0 00 �p 00 0 �p

1CCCCCCA ✏ =

0BBBBBBB@

� pE + 2⌫ p

E 0 00 � p

E + 2⌫ pE 0

0 0 � pE + 2⌫ p

E

1CCCCCCCA

�V = V(1 + ✏x

)(1 + ✏y

)(1 + ✏z

) � V

= V(1 + (✏x

+ ✏y

+ ✏z

) + · · · ) � V

�V

V

⇡ ✏x

+ ✏y

+ ✏z

= �3(1 � 2⌫)E

p = � p

K

K =E

3(1 � 2⌫)

9Anisotropic linear elastic response

10Isotropic stress/strain relations

✏x

=1E

[�x

� ⌫(�y

+ �z

)]

✏y

=1E

[�y

� ⌫(�x

+ �z

)]

✏z

=1E

[�z

� ⌫(�x

+ �y

)]

�xy

=1G

⌧xy

�xz

=1G

⌧xz

�yz

=1G

⌧yz

�x

=E

(1 + ⌫)(1 � 2⌫)

h(1 � ⌫)✏

x

+ ⌫(✏y

+ ✏z

)i

�y

=E

(1 + ⌫)(1 � 2⌫)

h(1 � ⌫)✏

y

+ ⌫(✏x

+ ✏z

)i

�z

=E

(1 + ⌫)(1 � 2⌫)

h(1 � ⌫)✏

z

+ ⌫(✏x

+ ✏y

)i

⌧xy

= G�xy

⌧xz

= G�xz

⌧yz

= G�yz

Is there a way to extend this to anisotropic response?

11General state of stress• Each point in a body has normal and shear stress components. • We can section a cubic volume of material that represents the state of

stress acting around the chosen point. • As the cube is at equilibrium the total forces and moments are zero:

• Infinitesimal cube = equal and opposite forces on opposite sides of cube • Note also: the values of all the components depend on how the cube is

oriented in the material (we’ll talk later about relating those values) • The combination of the state of stress for every

point in the domain is called the stress field.

• Stress × area = force Fi = Σj=xyz σij Aj

σxx=σx σxy=τxy σxz=τxz

σyx=τxy σyy=σy σyz=τyz

σzx=τxz σzy=τyz σzz=σz

Graphical stress tensor components 12

Defining strain 13

• We want to describe the dimension and shape change in a continuous cohesive body

• In a sufficiently small element, deformations of the element are all proportional to the size of the element

• length / length = unitless, % (10−2), mm/mm, µm/m (10−6), or in/in • Requires that we capture both the orientation of original vector and

change in that vector • Original relative position is a vector: one index = 3 numbers to describe • New relative position is a vector: one index = 3 numbers to describe • Strain is a tensor: two indices (coordinates) = 3×3 numbers to describe

• Normal strain describes a length change in a vector • Shear strain describes an orientation change in a vector

• Be aware: whether deformation changes length or changes orientation also depends on the original orientation

� = limAB,AC!0

✓⇡2� ✓0◆

Normal strain and shear strain 14

" = lim�s!0

�s0 � �s�s

• We can describe all the strains on an element in a body:

Normal strain and shear strain 15

normal strains

shear strains �xy

= (angle change in xy plane)= "

xy

+ "yx

deformation

"i j

=

0BBBBBBBBBBB@

✏x

12�xy

12�xz

12�xy

✏y

12�yz

12�xz

12�yz

✏z

1CCCCCCCCCCCA

✓xy

= (rotation in xy plane)

= "yx

� "xy

• Strain × length = length change

Graphical strain tensor components

εxx=ϵx εxy=½γxy εxz=½γxz

εyx=½γxy εyy=ϵy εyz=½γyz

εzx=½γxz εzy=½γyz εzz=ϵz

δℓi = Σj=xyz εij ℓj16

17Elastic constants: stiffnesses and compliances• Just as stress relates a vector (area) to another vector (force), and strain

relates a vector (position) to another vector (change in position), our elastic constants relate stresses to strains: 4th rank tensors

• 3×3×3×3 = 81 components! • But first two and last two are symmetric: xyzz = yxzz and zzxy = zzyx • And first pair and last pair can be swapped: xyzz = zzyx

• Stiffness is a second derivative of energy: Cijkl = d2U/dϵij dϵkl

• Results in 21 unique elastic constants. Better written with Voigt notation:

✏i j =X

kl

Sijkl�kl �i j =X

kl

Cijkl✏kl

compliance[GPa−1]

stiffness[GPa]

0BBBBBB@

�1 �6 �5�6 �2 �4�5 �4 �3

1CCCCCCA

0BBBBBBBBB@

e112 e6

12 e5

12 e6 e2

12 e4

12 e5

12 e4 e3

1CCCCCCCCCA

ei =6X

j=1

Sij� j

�i =6X

j=1

Cijej

1 2 3xx yy zz4 5 6yz xz xz

18Elastic constants: stiffnesses and compliances• The S and C matrices are inverses of each other • The 21 stiffness and compliance matrix entries have factors of 2 and 4 to

convert to tensor components: • Cab = Cijkl for a=1..6, b=1..6 • Sab = Sijkl for a=1..3 and b=1..3 • Sab = 2Sijkl for a=1..3 and b=1..6 or a=4..6 and b=1..3 or • Sab = 4Sijkl for a=4..6 and b=4..6

• Crystalline symmetry reduces the number of unique and nonzero entries

Symmetry operations

2-fold axis 3-fold axis 4-fold axis

mirror plane mirror plane

Rotating a cube around the body diagonal ⟨111⟩?

3-fold axis

21Elastic constants: stiffnesses and compliances• The S and C matrices are inverses of each other • The 21 stiffness and compliance matrix entries have factors of 2 and 4 to

convert to tensor components: • Cab = Cijkl for a=1..6, b=1..6 • Sab = Sijkl for a=1..3 and b=1..3 • Sab = 2Sijkl for a=1..3 and b=1..6 or a=4..6 and b=1..3 or • Sab = 4Sijkl for a=4..6 and b=4..6

• Crystalline symmetry reduces the number of unique and nonzero entries • Cubic symmetry is the most common for structural materials:

• C11 = C22 = C33 • C12 = C13 = C23 • C44 = C55 = C66 • all others zero

• Isotropic materials are cubic and C11−C12 = 2C44 (or S11−S12 = S44/2) • Hexagonal materials and aligned fiber composites have lower symmetry:

• C11 = C22 ≠ C33; C12 ≠ C13 = C23; C44 = C55 ≠ C66

• Isotropic in basal plane: 2C66 = C11−C12

• all others zero

Graphical compliance components

εij = Σk=xyz Σl=xyz Sijkl σkl

εxx

=

σxx1E

+−

σyy

-−νE

σzz

-−νE

Sxxxx = 1/E Sxxyy = -ν/E Sxxzz = -ν/ESxxkl = 0 for all other kl

Voigt and Reuss averages

C11 =13

(C11 + C22 + C33)

C12 =13

(C12 + C13 + C23)

C44 =13

(C44 + C55 + C66)

EVoigt

=

⇣C

11

� C12

+ 3C44

⌘ ⇣C

11

+ 2C12

⌘

2C11

+ 3C12

+ C44

Voigt average = isostrain

...Randomly oriented grains

Voigt and Reuss averagesReuss average = isostress

...Randomly oriented grains 1

EReuss=

15

⇣3S11 + 2S12 + S44

⌘

S11 =13

(S11 + S22 + S33)

S12 =13

(S12 + S13 + S23)

S44 =13

(S44 + S55 + S66)

EVoigt > Erandom polycrystal > EReuss

Grain structure and textureNi alloy grain structure

Each grain has a different orientation, and responds differently to applied stress

Polycrystalline response is an average of individual grain responses.

Texture is a preferential orientation of grains.

M. Groeber et al., AIP Conf. Proc. 712, 1712 (2004).

Ti / TiB metal-matrix compositeSEM backscatter: polish

SEM secondary e-: deep etch

Ci j =

0

BBBBBB@

419 92 113 0 0 092 523 63 0 0 0

113 63 418 0 0 00 0 0 196 0 00 0 0 0 179 00 0 0 0 0 220

1

CCCCCCAGPa

TiB: orthorhombic crystal

EVoigt = 442GPaEReuss = 435GPaETi = 110GPa

ETi+20%vol TiB = 153GPaS. Gorsse et al., Mat. Sci. Eng. A340, 80-87 (2003)

D. R. Trinkle, Scripta Mater. 56, 273-276 (2007)

Bovine femural bone: elastic constants

bone orthotropic stiffnesses:

along length: E3 = 21.7GPa

transverse: E1 = 11.6GPa

W. C. Buskirk et al., J. Biomech. Eng. 103, 67-72 (1981)

longitudinal (3)

transverse (1,2)

Ci j =

0

BBBBBB@

14 6.3 4.8 0 0 06.3 18.4 7.0 0 0 04.8 7.0 25 0 0 0

0 0 0 7.0 0 00 0 0 0 6.3 00 0 0 0 0 5.3

1

CCCCCCAGPa

Extracted from sound-speed measurements:C11 = r(v11)2 C22 = r(v22)2 C44 = r(v23)2

Composite behavior 28

Composite = matrix + reinforcementMatrix: continuous phase • transfers load to reinforcement • protects reinforcement from environment Types of matrix: • MMC metal matrix composite: designed for plastic strain

• better yield stress, tensile strength, creep resistance • CMC ceramic matrix composite: designed for fracture

• better toughness • PMC polymer matrix composite: designed for elastic and plastic

strain • better modulus, yield stress, tensile strength, creep • inexpensive, temperature range limited by polymer decomposition

Reinforcement: stronger, discontinuous phase • carries significant portion of load • classified by geometry

29

Spheroidite steel

cemented carbide

10µm

100µm

ferrite (bcc-Fe)

cementite (Fe3C)

Co matrix (Vm=10-15%)

WC particlesFrom Callister, Intro to Eng. Matls., 6Ed

Particle reinforcements 30

From Callister, Intro to Eng. Matls., 6Ed

100nm Tire rubber

rubber

carbon particles

S. Gorsse et al., Mat. Sci. Eng. A340, 80-87 (2003).

Ti/TiB MMC

alpha-Ti (hcp)

TiB needles

Particle reinforcements 31

fracturesurface

Mo + Ni3Al

alpha-Mo (bcc)Ni3Al (gamma’)

W. Funk et al., Met. Trans A19, 987-998 (1988).

From F.L. Matthews and R.L. Rawlings, Composite Materials; Engineering and Science, Reprint ed., CRC Press, Boca Raton, FL, 2000. (a) Fig. 4.22, p. 145 (photo by J. Davies); (b) Fig. 11.20, p. 349 (micrograph by H.S. Kim, P.S. Rodgers, and R.D. Rawlings).

Glass (E=76GPa) with SiC fibers (E=400GPa)

Fiber reinforcements: continuous, aligned 32

carbon-carbon composite

C matrix

C fibers

processed by laying down fibers in binder (pitch); high heat converts binder to C matrix

Randomly oriented fibers layered in 2D, not continuous

with composite

Fiber reinforcements: discontinuous, random 33

Determining mechanical behavior

isoload/isostress isostrain

equal load in phases equal strain in phases

34

Stress / strain response depends on • material properties (E, TS, σY) of matrix +

reinforcement • amount of matrix + reinforcement (Vm, Vr=1−Vm) • orientation of reinforcement relative to load • size and distribution of reinforcement • geometry (length of fibers, cross-sectional shape,

aspect ratio) Two limiting cases for analysis:

equal length/strain in phases

Similar to Voigt average

ℓreinforcement = ℓmatrix = ℓcomposite

εreinforcement = εmatrix = εcomposite

shared load: Fc = Fm + Fr

Fc

A=

Fm

A+

Fr

A

�c =Fm

Am

Am

A+

Fr

Ar

Ar

A�c = Vm�m + Vr�r ROM for stresses

Isostrain 35

equal load/stress in phases

Similar to Reuss average

Freinforcement = Fmatrix = Fcomposite

σreinforcement = σmatrix = σcomposite

`0c = `0m + `

0r

`c(1 + "c) = `m(1 + "m) + `r(1 + "r)1 + "c = Vm(1 + "m) + Vr(1 + "r)"c = Vm"m + Vr"r

shared length:

ROM for strains

Isoload / isostress 36

Cu 20% 40% 60% 80% Wvolume fraction (W)

100

200

300

E [G

Pa]

Elastic moduli of the composite are constrained by two limits: isostrain and isoload isostrain

isoload

Rule of mixtures: Cu particles + W matrix 37

Ec = VmEm + KcVpEp

(T.S.)c = Vm(T.S.)m + KsVp(T.S.)p

Empirical relations:

Kc ≠ Ks < 1

Orientation effects on tensile strength 38

Tensile stress not parallel to fibers has complex stress state: 3 limiting cases:

1.small misorientation: limited by fiber failure (σ‖ = σ cos2 θ)

2.large misorientation: limited by matrix tensile failure (σ⟂ = σ sin2 θ)

3.medium misorientation: limited by matrix shear failure (τ = σ cos θ sin θ)

(T.S.)

c

=�?k

cos

2 ✓

(T.S.)c =�??

sin2 ✓

(T.S.)

c

=⌧

m,y

cos✓ sin✓

� =

� cos

2 ✓ � cos✓ sin✓� cos✓ sin✓ � sin

2 ✓

!


Tensile stress not parallel to fibers has complex stress state: 3 limiting cases:

� =

� cos


2 ✓

!

(T.S.)

c

=�?k

cos

2 ✓

(T.S.)c =�??

sin2 ✓

(T.S.)

c

=⌧

m,y

cos✓ sin✓

0 30 60 90θ [degrees]

0

20

40

60

80

100

120Te

nsile

stre

ngth

[MPa

]

Tsai-Hill

smallmisorientation

intermediatemisorientation

largemisorientation


Tensile stress not parallel to fibers has complex stress state: Some limitations:

1.Predicts that tensile strength increases for small misorientation. 2.Predicts “cusps” in strength vs. misorientation angle. 3.Doesn’t account for multiaxial loading effects. Solution: Tsai-Hill failure criterion:

�k�?k

!2

� �k�?�? 2?

!+

�?�??

!2

+

⌧⌧m,y

!2

= 1

(T.S.)

c

=

266664

cos

4 ✓

�? 2

k+

sin

4 ✓

�? 2

?+ cos

2 ✓ sin

2 ✓

1

⌧2

m,y

� 1

�? 2

k

!377775�1/2

� =

� cos


2 ✓

!


Tensile stress not parallel to fibers has complex stress state:

� =

� cos


2 ✓

!

0 30 60 90θ [degrees]

0

20

40

60

80

100

120

Tens

ile st

reng

th [M

Pa]

Tsai-Hill

smallmisorientation

intermediatemisorientation

largemisorientation

(T.S.)

c

=

266664

cos

4 ✓

�? 2

k+

sin

4 ✓

�? 2

?+ cos

2 ✓ sin

2 ✓

1

⌧2

m,y

� 1

�? 2

k

!377775�1/2

Tsai-Hill smooths out cusps Never exceeds aligned T.S.

Isotropic linear elastic response

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