ME 7502 – Lecture 2 Effective Properties of Particulate and...

• Concepts from Elasticity Theory• Statistical Homogeneity, Representative Volume

Element, Composite Material “Effective” Stress-Strain Relations

• Particulate composite effective moduli• Unidirectional composite effective moduli• Lamina failure theories• Lamina constitutive relations

ME 7502 Lecture 2 1

ME 7502 – Lecture 2Effective Properties of Particulate and

Unidirectional Composites

ME 7502 Lecture 2 2

CONCEPTS FROM ELASTICITY THEORY

• STRESS AND EQUILIBRIUM– SURFACE TRACTION AND STRESS– STRESS EQUILIBRIUM

• DISPLACEMENT AND STRAIN– NORMAL AND SHEAR STRAIN, TENSOR STRAIN– STRAIN COMPATIBILITY

• LINEAR ELASTIC STRESS-STRAIN-TEMPERATURE RELATIONS

– ISOTROPIC MATERIALS– ORTHOTROPIC MATERIALS– TRANSVERSELY-ISOTROPIC MATERIALS

ME 7502 Lecture 2 3

STRESS AND EQUILIBRIUM

DEFINITION OF STRESSSURFACE TRACTION

INTERNAL DISTRIBUTED FORCE ON THE CUT SURFACE OF A STRUCTURE

ME 7502 Lecture 2 4

SURFACE TRACTION VECTOR ON A PLANE AT A POINT

kTjTiTAFT zyxA ++=

∆∆

= →∆ 0lim

T - EXPRESSED IN FIXED GLOBAL COORDINATES X, Y, Z

∆A - Infinitesimal area

T not perpendicular to surface, not parallel to n

ME 7502 Lecture 2 5

TRACTION AS “SURFACE STRESS” NORMAL AND PARALLEL TO THE SURFACE

nTn

•=σ

knjninn zyx

++=

),cos( inni = zyxi ,,=

222tnT σσ +=

22nt T σσ −=

tnT tn

σσ +=

t

nnTtσσ

−=

where,

since

Therefore, the tangential unit vector t is

ME 7502 Lecture 2 6

STRESS AT A POINT

POINT IS SURROUNDED BY A BOX WHOSE SIDES ARE PARALLEL TO COORDINATE AXES

+X FACE: OUTWARD NORMAL IS IN +X DIRECTION

-X FACE: OUTWARD NORMAL IS IN –X DIRECTION

…ETC.

ME 7502 Lecture 2 7

SURFACE TRACTION COMPONENTS ON RECTANGULAR VOLUME ELEMENT SURROUNDING POINT DEFINE STRESS

AND, SINCE , ETC.,

,xxxx σσ =−− yzzy −−= σσ

xyyx σσ = yzzy σσ = xzzx σσ =

xxxx σσ =

jiij σσ =

∴ INDEPENDENT STRESS COMPONENTS REDUCE TO 6

STRESS TENSOR IS SYMMETRIC

FROM FORCE EQUILIBRIUM:

FROM MOMENT EQUILIBRIUM:

ETC.

ME 7502 Lecture 2 8

COMPONENTS OF STRESS AT A POINT IN THREE-DIMENSIONAL RECTANGULAR CARTESIAN COORDINATES:

NOTE THAT ),,( zyxijij σσ =

• NORMAL STRESS COMPONENTS ALWAYS OCCUR ON AN ELEMENT IN PAIRS

• SHEAR STRESS COMPONENTS OCCUR IN QUADRUPLES

• IF THE MAGNITUDE AND SENSE OF A STRESS COMPONENT ON ONE FACE IS KNOWN, THE SAME STRESS COMPONENT WILL ALSO BE KNOWN ON OTHER FACES ON WHICH IT ACTS.

ME 7502 Lecture 2 9

MATRIX REPRESENTATION OF STRESS:

[ ]

=

=

=

zzyzxz

yzyyxy

xzxyxx

zzyzxz

yzyyxy

xzxyxx

zzzyzx

yzyyyx

xzxyxx

στττστττσ

σσσσσσσσσ

σσσσσσσσσ

σ

SIGN CONVENTION FOR STRESS COMPONENTS:

FACE OUTWARD DIRECTION OF STRESS SIGN OF STRESSNORMAL DIRECTION COMPONENT COMPONENTS

POSITIVE (+) POSITIVE (+) POSITIVE (+)NEGATIVE (-) NEGATIVE (-) POSITIVE (+)NEGATIVE (-) POSITIVE (+) NEGATIVE (-)POSITIVE (+) NEGATIVE (-) NEGATIVE (-)

SYMMETRY DIFFERENT NOTATION

ME 7502 Lecture 2 10

DIFFERENTAL EQUATIONS OF STRESS EQUILIBRIUM


IF NO BODY FORCES,

GET

0

0

0

3

33

2

23

1

13

3

23

2

22

1

12

3

13

2

12

1

11

=∂

∂+

∂∂

+∂

∂

=∂

∂+

∂∂

+∂

∂

=∂

∂+

∂∂

+∂

∂

xxx

xxx

xxx

σσσ

σσσ

σσσ

=∂

∂

j

ji

xσ

OR

0=∂

∂

j

ij

xσ

DIFFERENTAL EQUATIONS OF STRESS EQUILIBRIUM


SURFACE ELEMENT EQUILIBRIUM


DISPLACEMENT AND STRAIN

DEFINITION OF DISPLACEMENT

DISPLACEMENT

kwjviuu

u

++=

= DISPLACEMENT VECTOR u = x-component of displacement

v = y-component of displacement

w = z-component of displacement


STRAIN &STRAIN-DISPLACEMENT EQUATIONS


STRAIN-DISPLACEMENT EQUATIONS (cont’d)


STRAIN COMPATIBILITYDISPLACEMENTS MUST BE SINGLE-VALUED & CONTINUOUS (NO

HOLES OPEN IN THE STRUCTURE; 2 PARTICLES OF MATTER CANNOT OCCUPY SAME SPACE AT SAME TIME)

STRAINS MUST BE DERIVABLE FROM SINGLE-VALUED, CONTINUOUS DISPLACEMENTS (COMPATIBLE)

CONSIDERING EQUATIONS (3) TO (5), IT IS POSSIBLE TO HAVE VALUESOF εx, εy, AND γxy SUCH THAT WHEN (3) TO (5) ARE INTEGRATED TO OBTAIN u(x,y) and v(x,y), A UNIQUE SOLUTION CANNOT BE OBTAINED.

THEREFORE, THERE MUST BE SOME RELATIONSHIP BETWEEN εx, εy, AND γxy. CONSIDERING (5):

∴

IN TENSOR FORM, THE COMPATIBILITY EQUATIONS ARE:

EQUATION (9) REPRESENTS A TOTAL OF 81 (34) EQUATIONS; HOWEVER, ONLY SIX ARE INDEPENDENT.


6 INDEPENDENT STRAIN COMPATIBILITY EQUATIONS:

COMPATIBILITY EQUATIONS ARE EQUIVALENT TO STRAIN-DISPLACEMENT EQUATIONS, AND GIVE NO NEW INFORMATION

2

2

2

222

2

2

2

222

2

2

2

222

2

2

2

2

2

2

zxzxzx

yzzyzy

xyyxyx

yxzzyx

xzyyzx

zyxxzy

xxzzxzxz

zzyyyzyz

yyxxxyxy

xzyzxyzz

yzxyxzyy

xyxzyzxx

∂∂

+∂

∂=

∂∂∂

=∂∂

∂

∂∂

+∂

∂=

∂∂

∂=

∂∂

∂

∂

∂+

∂∂

=∂∂

∂=

∂∂

∂

∂

∂+

∂

∂+

∂

∂−

∂∂

=∂∂

∂

∂

∂+

∂

∂+

∂∂

−∂∂

=∂∂

∂

∂

∂+

∂∂

+∂

∂−

∂∂

=∂∂

∂

εεγε

εεγε

εεγε

εεεε

εεεε

εεεε


ELASTIC MODULI

LINEAR ELASTIC STRESS-STRAIN RELATIONS

+temp incr ∆T

Coefficient of linear thermal expansion αΑ = εΑ/∆T εΑ

∆T

αΑ

1


IMPORTANT POINTS FOR ISOTROPIC OR ORTHOTROPIC MATERIALS (PRINCIPAL MATERIAL AXES)

1. UNIAXIAL NORMAL STRESS σA CAUSES NORMAL STRAINS. FOREXAMPLE εA = σA/EA AND εT = -νATεA = -νATσA/EA

2. PURE SHEAR STRESS τAT CAUSES ONLY SHEAR STRAIN γAT (NO OTHER SHEAR STRESSES, NO OTHER SHEAR STRAINS)

3. UNIAXIAL NORMAL STRESS CAUSES ONLY NORMAL STRAINS (NO SHEAR STRAINS)

4. TEMPERATURE CHANGE ∆T CAUSES ONLY NORMAL STRAINS (NOSHEAR STRAINS)


ISOTROPIC MATERIALS (CAST MATERIALS, MOST METALS)PROPERTIES ARE DIRECTION-INDEPENDENT

ONLY 1-E, 1-G, 1-ν, 1-α∴

GG

GG

GG

TEEE

TEEE

TEEE

xyxy

xyxy

zxzx

zxzx

yzyz

yzyz

yyxxzzzz

zzxxyyyy

zzyyxxxx

2,

2,

2,

τε

τγ

τετγ

τε

τγ

ασ

νσνσε

ασνσνσ

ε

ασνσ

νσε

==

==

==

∆+−−=

∆+−−=

∆+−−=OR

ijijkkijij TEE

δαδσνσνε ∆+−+

=1

NOTE: CAN PROVE THAT

(NO RELATION BETWEEN

E AND ν ALONE)

3 INDEPENDENT THERMOELASTICCONSTANTS

{ E, ν OR E, G OR ν, G OR K,G}

AND α

∴

)1(2 ν+= GE

)3/(9 GKKGE +=


INVERTING PREVIOUS EQUATIONS YIELDS

xyxyxyxy

xzxzxzxz

yzyzyzyz

zzyyxxzzzz

zzyyxxyyyy

zzyyxxxxxx

GGGGGG

TE

TE

TE

ετγτετγτ

ετγτ

αννεεε

ννε

νσ

αννεεε

ννε

νσ

αννεεε

ννε

νσ

2,2,2,

211)(

211

211)(

211

211)(

211

====

==

∆

−+

−++−

++

=

∆

−+

−++−

++

=

∆

−+

−++−

++

=

OR

∆

−+

−−

++

= ijijkkijij TE δαννδε

ννε

νσ

211

211


ORTHOTROPIC MATERIAL σ−ε RELATIONS


TRANSVERSELY ISOTROPIC MATERIALS

TL ααααα === 321 ,

TEEE

TEEE

TEE

TTT

TT

L

LT

TT

TT

TL

LT

LT

TL

L

∆++−−=

∆+−+−=

∆+−−=

ασσνσνε

ασνσσνε

ασσνσε

3213

32

12

321

1 )(

LL

LL

TT

GG

GG

GG

2,

2,

2,

1212

1212

1313

1313

2323

2323

τετγ

τετγ

τετγ

==

==

==


ALSO, AS WITH GENERAL ORTHO,

T

TL

L

LT

EEνν

=

7 INDEPENDENT THERMOELASTIC CONSTANTS:

2 E’S, 1or2 G’S, 2or1 ν’s, 2 α’s

AN APPROXIMATION: MOST TRANSVERSELY-ISOTROPIC COMPOSITES HAVE GL~GT

SINCE T.I. PROPERTIES ARE NOTDIRECTION-DEPENDENT, )1(2 TTT GE ν+=


Anisotropic material properties: calculation using phase (fiber or

particulate) and matrix properties

EFFECTIVE COMPOSITE PROPERTIES

• Statistical Homogeneity: to calculate effective properties, it is first necessary to introduce a representative volume element (RVE), which must be large compared to typical phase region dimensions (i.e., reinforcement diameters and spacing)

RVE must be large enough so that average stress in RVE is unchanged as size increases:

28ME 7502 Lecture 2

• Effective properties of a composite material define relations between averages of field variables σ and ε

Cijkl* and Sijkl

* are reciprocals of one another

Overbars denote RVE averages

29ME 7502 Lecture 2

EFFECTIVE COMPOSITE PROPERTIES

PARTICULATE REINFORCED COMPOSITE MODULI

• Provided dispersion of particulate reinforcement is uniform, and provided orientation of non-spherical particulates is random, stress-strain relations of such composite materials will be effectively isotropic– Two independent elastic moduli– For convenience, these are selected to be the bulk

modulus (K) and the shear modulus (G)– All other elastic constants can be defined in terms of K

and G


• Effective elastic constants of particulate reinforced composites are obtained using multi-phase material solutions from elasticity theory

• Exact solutions are possible only in the case of spherical particles

• Approximate (bounding theory) results are used for other cases, such as non-spherical particles

• Example of a lower bound result is Arbitrary Phase Geometry (APG) lower bound on G*


)43(5)2(61

*

mmm

mmm

mi

im

GKGGKv

GG

vGG

++

+−

+≥

PARTICULATE REINFORCED COMPOSITE MODULI


RESULTS OF BOUNDING THEOREMS FOR PARTICULATE REINFORCED COMPOSITE MATLS

Best lower bound is from Arbitrary Phase Geometry (APG) boundBest upper bound is from Composite Spheres Assemblage (CSA) bound


• We are first interested in finding effective properties of a unidirectional composite (ply)

• Unidirectional composite is transversely isotropic

• x2x3 plane is plane of isotropy

UNIDIRECTIONAL COMPOSITE PROPERTIES

ME 7502 Lecture 2

• Material stiffness matrix for a transversely isotropic material is

There are 5 independent elastic constants and 2 independent CTE’s

34


ME 7502 Lecture 2

• The σ−ε relations of a transversely isotropic material are frequently written as

35


ME 7502 Lecture 2

• For the unidirectional composite subjected to a homogeneous stress state (e.g., uniform tension), states of stress and strain are not uniform but highly complex

• Variations of stress and strain on any transverse plane are random– No reason that one plane should be different from another, other

than statistical variation

• Such a material is statistically homogeneous, which allows the definition of effective elastic properties to be specified by relations between average stress and average strain– Unidirectional composite is statistically transversely isotropic due

to the random placement of fibers in the ply or lamina

36


ME 7502 Lecture 2

Effective σ−ε

relations of a

lamina:

37


ME 7502 Lecture 2

COMPOSITE CYLINDERS ASSEMBLAGE MODEL

• Micromechanical methods are used to compute the effective elastic properties of a unidirectional composite material in terms of fiber and matrix properties and these phase volume fractions– Composite Cylinders Assemblage (CCA) model is

used for axial properties– Generalized Self-Consistent Scheme (GSCS) is used

to calculate transverse properties

CCA model

GSCS model

38

ME 7502 Lecture 2

PHILOSOPHY OF CCA MODEL

• Unidirectional composite is considered as an assemblage of composite cylinders– Varying diameters but constant

df / dm ratio• Properties of homogeneous

cylinder which can be used to replace each heterogeneous cylinder provide the effective lamina properties– 5 elastic, 2 CTE’s

Schematic of CCA model

39

ME 7502 Lecture 2

CONCEPT OF GENERALIZED SELF-CONSISTENT SCHEME (GSCS)

• Single composite cylinder is embedded in the model effective material

• Properties of effective material which satisfy continuity of displacements and tractions across interface between matrix (a<r<b) and effective material are sought

• Motivation for GSCS: CCA can only bound transverse properties of ply Schematic of GSCS

model

40

ME 7502 Lecture 2

• Effective Elastic Properties are obtained by solving closed form solutions to Theory of Elasticity problems – Uniform stress states are applied to the two

(fiber and matrix) phase composite cylinder in cylindrical coordinates

– Details are provided in textbooks by Robert Jones, R.M. Christensen and others

41


ME 7502 Lecture 2

COMPLETE RESULTS OF CCA ANALYSES

1 = fiber; 2 = matrix42

ME 7502 Lecture 2

RESULTS OF CCA ANALYSES (cont’d)

1 = fiber; 2 = matrix43

ME 7502 Lecture 2

Composite material strength and failure

theories

44

ME 7502 Lecture 2

COMPOSITE STRENGTH & FAILURE THEORIES

• Unidirectionally reinforced composites are typically employed as lamina within laminates and thus are subjected to combined stress states– Stress state is at least plane stress– Therefore it is usually necessary to consider

strength of lamina under combined stress state

– Best approach is to consider analytical theory with limited number of experiments

45

ME 7502 Lecture 2

PRACTICAL LAMINA FAILURE CRITERION

• Fundamental assumption is that for plane stress state there exists a failure criteria of the form F(σ11, σ22, σ12) = 1

• Schematic representations are:

46

ME 7502 Lecture 2

LAMINA FAILURE CRITERION (cont’d)

Stress states inside curves (F<1) do not induce lamina failure

Stress states on or outside curves (F=1 or F>1) induce lamina failure

47

ME 7502 Lecture 2

• Two common types of failure criteria– Maximum Stress / Strain– Quadratic Interaction

• Maximum Stress or Strain– Assumption is that failure occurs when any single

stress (or strain) component reaches its ultimate value, regardless of values of other components

– Advantages:• Very simple to use and allows failure modes to be identified

– Disadvantages:• Not necessarily realistic since it ignores stress interaction• It is sometimes not conservative, depending on stress state

σ1

σ2

48


ME 7502 Lecture 2

• Quadratic Interaction Failure Criterion– General form is given by the equation

– Three popular forms of quadratic interaction:1) Tsai-Hill:

X = axial strength (tensile or compressive)Y = transverse strength (tensile or compressive)S = shear strength

F11σ112 + F22σ22

2 + F66σ122 + 2F12σ11σ12 + F1σ11 + F2σ22 = 1 (1)

(2)

49


ME 7502 Lecture 2

• Quadratic Interaction Failure Criterion2) Tsai-Wu Failure Criterion uses

F11σ112 + F22σ22

2 + F66σ122 + 2F12σ11σ12 + F1σ11 + F2σ22 = 1

Requires bi-axial testing

Tsai-Wu is basically a curve fit and does not specify failure mode

(1)

(3)

50


ME 7502 Lecture 2

(4)

51


ME 7502 Lecture 2

• Quadratic Interaction Failure Criterion3. Hashin Failure Criteria

52


ME 7502 Lecture 2

• Hashin Failure Criterion (cont’d)– Convenient method for examining failure criteria

experimentally is by means of off-axis specimens; here stress state is given by

– Testing in tension shows that for small off-axis angles, failure is in the fiber mode; for large angles, failure is by matrix mode

– Angle θo separating the two modes is found when (5) and (6) are satisfied simultaneously in terms of (7); result is

53


ME 7502 Lecture 2

Off-axis test specimen

Hashin Criteria Failure Modes

54


ME 7502 Lecture 2

Performance of Hashin

Interaction Failure

Criteria (for Gl/Ep

composites)

55



USUALLY USE LAMINATED COMPOSITES IN PLATE OR SHELL FORM WHERE PRIMARY STRESSES ARE IN THE PLANE OF THE LAMINA (1-2) (PLANE STRESS). LAYER STRESS-STRAIN BEHAVIOR THEN BECOMES:

∆

−

−−

−−

=

T

G

EE

EE

0200

011

011

2

1

12

22

11

12

2112

2

2112

212

1221

121

1221

1

12

22

11

αα

εεε

ννννν

ννν

νν

σσσ

∆

−

=

T

QQQQQ

00000

2

1

12

22

11

66

2212

1211

12

22

11

αα

εεε

σσσ

{ } [ ]{ } { }{ }TQ tt ∆−= 111 αεσ

OR

(D-2)

OR

DEFINE [Qt] MATRIX:

PLANE STRESS

[ ]

[ ]

1212

21112221221

222

12221111221

111

21

1

G

TE

TE

=

∆++−

=

∆++−

=

σ

αενενν

σ

αενενν

σ

LAMINA CONSTITUTIVE RELATIONS


{ } [ ]{ } { } TS ttt ∆+= ασε 111

THE INVERSE STRAIN-STRESS RELATIONSHIP IS GIVEN BY

T

G

EE

EE

∆

+

−

−

=

0

2100

01

01

2

1

12

22

11

12

21

12

2

21

1

12

22

11

αα

σσσ

ν

ν

εεε

TS

SSSS

∆

+

=

00000

2

1

12

22

11

66

2212

1211

12

22

11

αα

σσσ

εεε

(D-3)

LAMINA CONSTITUTIVE RELATIONS (cont’d)


CONSTITUTIVE RELATIONS IN OFF-AXIS (X,Y) LAMINATE COORDINATES HAVE

WANT

{ } [ ]{ } { }{ }TQ tt ∆−= 111 αεσ

{ } [ ]{ } { }{ }TQ txttx ∆−= αεσx

y

θx1x2

a. TRANSFORMATION OF STRESS AND STRAIN

STRESS

xy

τxy

σx

σyτx’y’

σx’σy’

θ



ENGINEERING STRAIN

TENSOR STRAIN

)(22 22

22

22

nmmnmn

mnmn

mnnm

xyyxyx

xyyxy

xyyxx

−++−=

−+=

++=

′′

′

′

γεεγ

γεεε

γεεε

)(

2

2

22

22

22

nmmnmn

mnmn

mnnm

xyyxyx

xyyxy

xyyxx

−++−=

−+=

++=

′′

′

′

εεεε

εεεε

εεεε

STRESS

)()sin(coscossin)(

2cossin2cossin

2cossin2sincos

2222

2222

2222

nmmnmn

mnmn

mnnm

xyyxxyxyyx

xyyxxyyxy

xyyxxyyxx

−++−=−+−=

−+=−+=

++=++=

′′

′

′

τσσθθτθθσστ

τσσθθτθσθσσ

τσσθθτθσθσσ

θθ sin,cos == nm



(D-4)

(D-5a, b) THEN

NOTE: (D-6)

THIS IS USED FOR STRESS AND TENSOR THERMOMECHANICAL STRAIN (WOULD NEED DIFFERENT IF ENGINEERING STRAIN!)[ ]e

CLT

[ ]

−−

−=−

22

22

22

1 22

nmmnmnmnmnmnnm

TCL

[ ]

−−−=

22

22

22

22

nmmnmnmnmn

mnnmTCL

{ } [ ]{ }xCL

x T σσ =′

{ } { } [ ] { } { }{ }TTT xtxtCL

txtx ∆−=∆− ′′ αεαε

DEFINE TENSOR 2-D TRANSFORMATION MATRIX

θθ sin,cos == nm



b. TRANSFORMED LAMINA STIFFNESS AND CTE MATRICES IF NO ∆T:

HAVE

WANT

KNOW

IF NO {σ}:

{ } [ ]{ }ttQ 11 εσ =

{ } [ ]{ }xttx Q εσ =

{ } [ ]{ }xCLT σσ =1

{ } { } Ttt ∆= 11 αε

{ } { } { } TT txtxt ∆=∆= ααε

{ } [ ]{ }xtCL

t T εε =1

SUB FOR 1 IN TERMS OF X:

[ ]{ } [ ][ ]{ }xtCL

txCL TQT εσ = [ ]{ } { } TT txt

CL ∆=∴ 1αε

MULT THRU BY [TCL]-1,

[ ] [ ]{ } [ ] { } TTTT tCL

xtCLCL ∆= −− 111 αε

{ } [ ] { } TT tCL

xt ∆=∴ − 11 αε

[ ] [ ]{ } [ ] [ ][ ]{ }xtCL

tCL

xCLCL TQTTT εσ 11 −− =

{ } [ ] [ ][ ]{ }xtCL

tCL

x TQT εσ 1−=∴

COMPARE WITH WANT

{ } [ ] { }tCL

t T 11 αα −=∴[ ] [ ] [ ][ ]CLt

CLt TQTQ 1−=∴ (D-7)



SUMMARY: IN OFF-AXIS (X,Y) LAMINATE COORDINATES

HAVE

WANT

{ } [ ]{ } { }{ }TQ tt ∆−= 111 αεσ

{ } [ ]{ } { }{ }TQ txttx ∆−= αεσx

y

θx1x2

TENSOR 2-D TRANSFORMATION MATRIX [ ]

−−−=

22

22

22

22

nmmnmnmnmn

mnnmTCL θθ sin,cos == nm

{ } [ ]{ }xCLT σσ =1

TRANSFORMATION EQUATIONS:

{ } [ ]{ }xtCL

t T εε =1

[ ] [ ] [ ][ ]CLt

CLt TQTQ 1−=∴

{ } [ ] { }tCL

t T 11 αα −=∴

STRESS AND STRAIN FROM LAMINATE TO LAYER COORDS

STIFFNESS AND CTE MATRICES FROM LAMINATE TO LAYER COORDS



T300/5208 Gr/Ep LAYER - FIND [0] LAYER PLANE STRESS [Q] AND [α] MATRICES ON-AXIS, AND OFF-AXIS FOR +45 AND -45 deg TO X-AXIS

EXAMPLE

MPaEMPaE3

2

31

103.1010181×=

×=MPaG 3

12

12

1017.728.0

×=

=ν16

2

161

1032106.0

−−

−−

°×=

°×−=

CC

α

α

LAYER THICKNESS t = 0.125mm ( ) 0159.028.0181

3.1012

1

221 === νν

EE

LAYER MATERIAL PROPERTIES:

X1

X2

X

Y

X

Y


LAYER MATERIAL AXIS STIFFNESS MATRIX, [Qt] AND CTE MATRIX, [α]

{ } 16

6

2

1

010.32106.0

0

−−

−

−=

= Cx

xot α

αα

[ ] MPaQt 31034.1400

0346.10897.20897.281.181

×

=

LAYER TRANSFORMATION MATRICES [TCL]k

2145sin,

2145cos ±=±==±= nm

[ ]

−−=

−−−=+

01

12

2

21

21

21

21

21

21

22

22

22

45

nmmnmnmnmn

mnnmTCL [ ]

−

−=

−−

−=−

+

011

22

21

21

21

21

21

21

22

22

22

1

45

nmmnmnmnmnmnnm

TCL

[ ]

−

−=−

011

21

21

21

21

21

21

45CLT[ ]

−−=−

−

01

1

21

21

21

21

21

21

1

45CLTSIMILARLY,

[ ]( ) ( )

( ) ( )

−−

−−

=

12

2112

2

1221

212

2112

121

2112

1

200

011

011

G

EE

EE

Qtνννν

ννν

ννν


LAYER OFF-AXIS STIFFNESS MATRICES [ ]tkQ

°+ 45 [ ] [ ] [ ][ ]

[ ] MPaQ

TQTQ

t

CLt

CLt

345

4514545

10

021

21

121

21

121

21

34.14000346.10897.20897.281.181

021

21

121

21

121

21

×

−

−

−

−

=

=

+

+−++

[ ] MPaQ t 345 10

18.9386.4286.4273.8566.5632.4273.8532.4266.56

×

=∴ +

SIMILARLY,

°− 45[ ] MPaQ t 3

45 1018.9386.4286.4273.8566.5632.4273.8532.4266.56

×

−−−−

=∴ −


LAYER OFF-AXIS THERMAL EXPANSION MATRICES { }kα

( ) ( ) Cnmx °×=×+×−=+= −−− /107.15103221106.0

21 666

22

12 ααα

( ) ( ) Cmny °×=×−+×=+= −−− /107.15106.0211032

21 666

22

12 ααα

( ) ( ) Cmnxy °×=×−−

±=−= −− /103.1610326.0

21

21 66

21 ααα

{ } ,/103.16

7.157.15

645 C°×

−=∴ −

+α { } C°×

= −

− /103.167.157.15

645α


IN MATHCAD,

α

6− 10 7−×

3.2 10 5−×

0

1K

=α

α1

α2

0

:=LAYER CTE MATRIX [α]

Q

181.811

2.897

0

2.897

10.346

0

0

0

14.34

GPa=Q

E1V

ν12E2V

⋅

0

ν21E1V

⋅

E2V

0

0

0

2 G12⋅

:=then

V 1 ν12 ν21⋅−:=let

LAYER MATERIAL AXIS STIFFNESS MATRIX [Qt] CTE MATRIX [α]

t 0.125 mm⋅:=LAYER THICKNESS:

α2 32 10 6−⋅ K 1−⋅:=ν21 0.01593=ν21E2E1

ν12⋅:=G12 7.17 GPa⋅:=E2 10.3 GPa⋅:=α1 0.6− 10 6−⋅ K 1−⋅:=

ν12 0.28:=E1 181 GPa⋅:=

GPa 1000 MPa⋅:=MPanewton

mm2:=LAYER MATERIAL PROPERTIES:


LAYER TRANSFORMATION MATRICES [TCL]k

angle direction cosines m θ( ) cos θ deg⋅( ):= n θ( ) sin θ deg⋅( ):=

m 45( ) 0.707= n 45−( ) 0.707−=

T θ( )m θ( )2

n θ( )2

m θ( )− n θ( )⋅

n θ( )2

m θ( )2

m θ( ) n θ( )⋅

2 m θ( )⋅ n θ( )⋅

2− m θ( )⋅ n θ( )⋅

m θ( )2 n θ( )2−

:=

T 0( )

1

0

0

0

1

0

0

0

1

= T 45( )

0.5

0.5

0.5−

0.5

0.5

0.5

1

1−

0

= T 45−( )

0.5

0.5

0.5

0.5

0.5

0.5−

1−

1

0

=

LAYER MATERIAL OFF-AXIS STIFFNESS MATRICES [Qbar]

Qbar θ( ) T θ( ) 1− Q⋅ T θ( )⋅:= Qbar 0( )

181.811

2.897

0

2.897

10.346

0

0

0

14.34

GPa=

Qbar 45( )

56.658

42.318

42.866

42.318

56.658

42.866

85.732

85.732

93.182

GPa= Qbar 45−( )

56.658

42.318

42.866−

42.318

56.658

42.866−

85.732−

85.732−

93.182

GPa=


LAYER CTE MATRICES [αbar]

αbar θ( ) T θ( ) 1−α⋅:=

αbar 45( )

1.57 10 5−×

1.57 10 5−×

1.63− 10 5−×

1K

= αbar 45−( )

1.57 10 5−×

1.57 10 5−×

1.63 10 5−×

1K

=αbar 0( )

6− 10 7−×

3.2 10 5−×

0

1K

=

ME 7502 – Lecture 2 Effective Properties of Particulate and...

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Transcript of ME 7502 – Lecture 2 Effective Properties of Particulate and...