apuntes compositos i-presentaciones

(1) Contain two or more distinct constituents.

(2) Are synthesized in a way that the form, distribution and amount of constituents are controlled in a predetermined way.

(3) Have unique, useful and superior performance that can be predicted from the properties, amounts and arrangements of constituents using principles of mechanics.

Engineering Composites - Defined

Material Two

Constituents Synthesized/ controlled

Properties Predictable

Engineering Composite

Fiberglass/Epoxy Yes Yes Yes Yes Cast Iron Yes No Yes No Steel/Polypropylene Yes Yes Yes Yes D.S. Eutectic Yes Yes Yes Yes Adobe Brick Yes Yes Yes Yes Wood Yes No No No Bone Yes No No No

Two-constituent materials

0 1 2 3 4 5 6 7

SPECIFIC STIFFNESS

KEVLAR/EPOXY

BORON/EPOXY

GR/EPOXY (HS)

GR/EPOXY(HM)

GR/EPOXY(UHM)

E-GLASS/EPOXY

ALUMINUM ALLOYS

Anisotropic - Material properties are different in all directions.

Orthotropic - Material properties are different in three mutually perpendicular directions.

Isotropic - Material properties are the same in all directions. Almost all engineering alloys such as aluminumAnisotropic and steel in the annealed condition are isotropic.

Preferred Orientation - A material that has some degree of anisotropy is said to have preferred orientation. Highly wrought alloys including aluminum and steel will have different properties in the direction of elongation. Drawn wire and extrusions a well known examples.

Random Orientation - A material that is said to have random orientation possesses some degree of isotropy. Engineering alloys are made up of an assembly of crystals, each of which may be orthotropic but on a macroscale exhibits apparent isotropy.

Homogeneous -Properties are the same from point to point in the material. This effect is very scale dependent. Even a material that consists of two or more phases or constituents

TERMS COMMONLY USED IN COMPOSITE SCIENCE AND TECHNOLOGY

Lamina - A single layer containing reinforcements in the plane of the layer. This is the usual building block for design and fabrication of composite structures.

Laminate - A laminate is a stack of lamina (usually with alternating or varying principal directions). The laminas are arranged in a particular way to achieve some desired effect.

Micromechanics - Predicts composite behavior from the interaction of constituentson the appropriate scale. In whisker composites the appropriate scale is a fraction of a micrometer. For reinforced concrete the appropriate scale is several centimeters.

Macromechanics - Predicts composite behavior by presuming homogeneous material. The effects of constituents are detected only as averaged apparent properties.

TERMS COMMONLY USED IN COMPOSITE SCIENCE AND TECHNOLOGY (Continued)

The reinforcement is the dimensionally controlled constituent (fixed geometry) that can either be particles, fiber or flakes.

Reinforcements have unique properties that contribute significantly to the properties of the composite.

REINFORCEMENTS

• Textile Fibers•Polymers - rayon, polyacrilonitrile, Kevlar®, Nomex®, nylon•Carbons - graphite, carbon, Celiox®

•Ceramics - glass, alumina, Nextel®, silica, silicon carbide

• Monofilaments (25 micron to 200 micron, continuous)•Boron on W, boron on carbon•SiC on W, SiC on C• Sapphire• W, Mo, steel, Be

•Whiskers (< 1 micron, discontinuous)•Silicon carbide• alumina• boron carbide• carbon

204 Twisted filaments

204 Untwisted filaments

Chopped for injectionmolding or SMC

Multiple Untwisted Strands

ROVING or TOW

Filament windingor pultrusion

Muliple Twisted Strands

UNTWISTED STRAND (END)

TWISTED STRAND (END)

Weaving

WOVEN ROVING CLOTH

Weaving

FILAMENT(10-25 micron)

1 10 100 10000.001 0.01 0.1 1.0 10

FIBER ASPECT RATIO

FIBER REINFORCED

SOLID SOLUTIONAND PRECIPI-TATION HARDEN-ING

DISPERSIONSTRENGTHENING

CERMETS

COMPOSITE STRENGTH AFFECTED BY REINFORCEMENT SIZEM

PARTICLE DIAMETER,MICROMETERS

PARTICLEREINFORCED

FIBER ORWHISKERREINFORCED

CONTINUOUSFIBER

PARTICULATE COMPOSITES FIBER COMPOSITES

3-0 3-0 3-0 3-1,3-2

fracture mechanics

multiple cracks

modified fracturemechanics

multiple cracks

DISPERISIONSTRENGTHENED

conventional fracture mechanics

single crack

The matrix is the continuous phase of the composite

Functions of the matrix

• Holding the fibers in place

• Protecting the fibers from reaction with the environment

• Transmitting load from fiber to fiber

• Protecting the fibers from mechanical abrasion

MATRIX

CLASSIFICATION OF COMPOSITE BY MATRIX TYPE

MatrixType

CommonDesignation

Matrix Properties Most EffectiveReinforcement

Stiffness Strength Ductility

PolymerFRPGRPCRPPMC

Low(0.2 TO

0.5)msi

Low(0.5 TO

Low(< 2%) Continuous Fiber

Metal MMCModerate(6 TO 16)

High(10 TO150) ksi

High(20%)

Continuous andDiscontinuousFiber

Ceramic CMCHigh

(20 TO80)msi

High(20 to 80)

Low(< 1%)

DiscontinuousFiber or Whiskerand Particulate

Matrix Materials for Engineering Composites

Matrix Type Positive Attribute Negative Attribute

Thermoset resins Low cost processing Brittle

Thermoplastic resins ToughFormable

Low thermal andsolvent resistantHigh cost to processfor filament

Carbon Very high temperatureapplications

Very high cost toprocess

Light metalsThermal resistantConductiveElectrical and thermal

Reacts with most fibers

Superalloys Oxidation resistant Heavy

RefractoryHigh temperaturestrength

HeavyNo oxidationResistant

Glass Corrosion resistantThermal expansion

Brittle

Glass/ceramics Corrosion andtemperature resistant

Brittle and expensive

Ceramic Very high temperatureOxidation Expensive

Glass Fibers

! M e c h a n ic a l s t r e n g t h

! O x id a t i o n r e s i s t a n t

! M o i s t u r e r e s i s t a n t

! H ig h t h e r m a l c o n d u c t iv i t y c o m p a r e d t o p o l y m e r

! L o w t h e r m a l c o n d u c t iv i t y c o m p a r e d t o m e t a l s

! L o w d e n s i t y ( 2 . 5 5 )

! L o w t h e r m a l e x p a n s io n

! E l e c t r i c i n s u l a t o r

! C o r r o s i o n r e s i s t a n t ( e x c e p t f o r H 3 P O 4 a n d H F )

! S o lv e n t , o i l a n d f u e l r e s i s t a n t

! Is o t r o p ic

! L o w m o d u lu s c o m p a r e d t o s t e e l

! S e l f - a b r a d i n g

! L o w f a t i g u e r e s i s t a n c e

! S i z i n g n e e d e d f o r a d h e s io n t o p o l y m e r s r e s u l t s in m o i s t u r ea b s o r p t i o n

Glass Fiber Characteristics

Package Former

Draw Roll

Sizing Applicator (Starch-oil Emulsion)

Platinum Bushing (1 mm holes)

Remelt Furnace

Glass Bead Hopper

Fiberglass Manufacturing

Glass Fiber Compositions

Designation Use Composition

E ElectricalInsulation

55%SiO2, 11%Al2O3, 6%B2O3,18%CaO, 5%MgO, 5%Other

S High StrengthComposites

65%SiO2, 25%Al2O3,10%MgO(Strength Expensive)

S-2 High StrengthComposites

Same as above but lessstringent QC

A ThermalInsulation

72%SiO2, 1%Al2O3, 10%CaO,3%MgO, 14%K2O

C ChemicalApplications

65%SiO2, 4%Al2O3, 6%B2O3,14%CaO, 3%MgO, 9%K2O

Glass Fiber Strength

FiberVirgin Forming Package

GPa ksi GPa ksi

S-glass 7.00 1000 5.00 725

E-glass 3.70 537 2.80 406

Pyrex 2.00 295 1.60 230

Fracture Strength

Surface Active Substance

WaterKerosene (Non-polar hydrocarbon)

Surface Chemistry Effects on Glass Fiber Strength

Diameter, µm

, % 25

3 25 50 75

Fiber Diameter Effects on Strength

FiberVirgin Strength LiqN2 Strength - 196BC

GPa ksi GPa ksi

S-glass 7.00 1000 7.00 1000

E-glass 3.70 540 4.50 780

Pyrex 2.00 295 2.60 375

Cryogenic Strength of Glass Fibers

CompositionE BulkGPa

EFiberGPa

E-glass (non-alkaline alumino-borosilicate)

86.1 73.5 14.6

S-glass (magnesia-alumino-silicate)

94.5 87.0 8.0

C-glass (soda-lime-alumino-silicate)

76.0 70.0 8.0

Pyrex (soda-borosilicate) 60.0 55.0 8.0

Elastic Properties of Bulk and Fiber Form of Glass

E Glass

Pyrex7

20 100

Relative Humidity, %

Elastic Post-Action

Time, min70

m E-Glass

Soda-lime-silicate300

Elastic Post-Action vs Composition

Time, min.

m 1.3 GPaKerosene

1.0 GPaWater

0.8 GPaSurfactant

E-Glass

Post-Action Effects with Environment

! Lower SiO2 content

! Increase beryllium and aluminum oxide content

! Increase heavy metal and rare earth oxides

! Non-silicate system -- lime-aluminate

! Density goes up

! Strength goes down

High Modulus Glass Fibers

High SpeedWinding

Heat Treatment

Continuous Strand

Forming Package

Glass ChoppingMachine

Roving Winder

ContinuousRoving

Woven Roving

Chopped Strand

Mat Machine

ChoppedStrand

WeavingMachine

Glass Fiber Products

Carbon Fibers

! Very high stiffness

! Low density

! High strength

! Thermally stable to (2000BC)

! Electrically conducting

! Low (negative) thermal expansion

! Chemically Inert

! High fatigue strength

! Oxidize at 1000BC

! Anisotropic

! Expensive

Characteristics of Carbon Fibers

0.669 nm

0.3345 nm

0.142 nm

Crystal Structure of Graphite

Property Diamond(Cubic)

Graphite (Hexagonal)

Bond Length (nm) 0.154 0.142 0.334

Conductivity (O-1m-1) <10-15 250 0.05

Thermal Cond. (Wm-1BC-1) 0.9 x 103 2 x 103 6

Thermal Exp. (BC-1) 0.8 x 10-6 -1.5 x 10-6 27 x 10-6

Young's Modulus (GPa) 1200 1060 36.5

Hardness (Moh) 10 .5-1

Density (Kg m -3) 3300 2265

"a" direction is parallel to the basal planes

"c" direction is perpendicular to the basal planes

Comparison of Graphite and Diamond Properties

Tensile axis1200

Anlge between basal planes and tensile axis, φ

0 45 90

Young’s Modulus of Graphite

Idealized Graphite Fiber Structure

Graphite Fiber Structure

Fiber axisSheath containingenclosed voids

Fiber surfaceconsisting ofbasal planes

Homogeneous main body with amean orientation of 12

Thin skin , 100 nm with large crystallite sizeand mean orientation of 7 o

Typical Graphite Fiber Structure

(1) Controlled slow heating from RT to 150EC

! Remove absorbed water

(2) Dehydration from structure 150E - 240EC

(3) Thermal scission 240E - 400EC

! Break C-O bonds

! Form H2O, CO2, and CO

(4) Carbonization 400E - 700EC

(5) Graphitization to 3000EC - Stretching to align basalplanes parallel fiber axis

Processing of Rayon Based Graphite Fibers

H Ethane

Ethylene

Polyethylene

Olefin Chemistry

V i n y l B e n z e n e (S ty rene )

P o l y v i n y l B e n z e n e (Po lys ty rene )

Acrylonitr i le Polyacrylonitr i le ( P A N )

Olefin Chemistry - PAN

P r e c u r s o rA c r l i c F i b e r

S T A B I L I Z A T I O N

C A R B O N I Z A T I O N

G R A P H I T I Z A T I O N

S U R F A C E T R E A T M E N T

n i t r i c a c i d e t c h i n g

p r e o x i d a t i o n a t 2 0 0 C - 3 0 0 C i n a i r f o r 1 - 2 h o u r s

h e a t t r e a t m e n t a t 1 2 0 0 C - 1 5 0 0 C i n n i r t o g e n f o r 3 0 - 6 0 s e c

h e a t t r e a t m e n t a t 2 0 0 0 C - 3 0 0 0 C i n n i t r o g e n / a r g o n f o r 1 5 - 2 0 s e c

H i g h m o d u l u s g r a p h i t e f i b e r s

PAN Processing of Carbon Fibers

+ H O2

I . O x i d a t i o n s t e p ( 2 0 0 - 2 5 0 C )o

F ibers a re p laced in t ens ion t o u n f o l d t h e c h a i n m o l e c u l e s

C o n s i d e r a p a i r o f u n f o l d e dP A N c h a i n m o l e c u l e s l y i n ga p p r o x i m a t e l y a d j a c e n t a n dpara l le l to one another

Ox id i z ing env i ronment i susual ly a i r

C r o s s - l i n k i n g b e t w e e na d j a c e n t c h a i n s o c c u r s b yi n c o r p o r a t i o n o f t h e o x y g e n

E x c e s s o x y g e n r e a c t sw i t h t h e h y d r o g e n t of o r m w a t e r

A commerc ia l ve rs ion o f th is c ross - l inked product i sm a r k e t e d u n d e r t h e t r a d e n a m e " C E L I O X "

O O O O

PAN Step I (Oxidation)

CH2 CH

CN C N

CH C H CH

+ H O2

PAN Conversion

C C C C C

H O HCN2 +

PAN Step II (Carbonization)

200 20

001200 1600 2000 2400 2800

Heat Treatment Temperature, C

siPAN Step III (Graphitization)

Manufacturer DesignationModulus

(Msi)Strength

(Ksi)Cost($/#)

Hercules AS2 33 400

AS4 33 520 21

Amoco T300 32 450 26

T650-35 35 645 28

Celion G30-500 34 550 24

G30-600 34 630 34

Graphil AP38-750 38 750

AP38-500 33 500 16

AP38-600 33 600 24

Hercules IM6 40 745 48

IM7 42 785 53

XIM8 45 750

XMS4 48 400

Amoco T650-42 42 720 53

T40 40 820 55

T1000 42 1002 326

Celion G40-600 43 620 45

G40-700 43 720 47

42-7A 42 725 59

Graphil AP43-600 43 650

Hercules HMU 52 400

Celion G50-300 52 360 58

Graphil AP50-400 50 400 55

AP53-650 53 650 100

AP53-750 53 750 110

Hercules UHMS 62 325 325

Celion GY-70 75 270 750

GY-80 83 270 850

Properties of PAN Based Carbon Fibers

Coal Tar Pitch

Mesophase

Heat 40 Hr400-500 C

Spun(Extrude) Precursor

(Raw) Fibers

Stabilization(Thermosetting)

GreenFibers

Carbonize andGraphitize1700 - 3000 CGraphite

Textile Fibers

Processing of Mesophase Pitch Fibers

PRESSURE

MESOPHASEDIE

Mesophase Pitch Spinning

Precursor Yield(%)

Density E(GPa)

Rayon 10 1.66 390 2.0

PAN I 40 1.83 350 1.5

PAN II 40 1.74 230 2.2

Mesophase(HS)

80 2.10 340 2.9

Mesophase(HM)

80 2.20 690 2.4

Comparison of Carbon Fibers

Fiber Cross-Sections

VSA16Pitch

HM3000PAN

T300PAN

GY70PAN

T50Rayon

Other Fibers

- C - N - C - C - C - C - C - C - N - C - C - C - C - C - C -H H H2 2 2 2 2

H H H H H H H2 2 2 2H2H

The amide linkage:

Aliphatic Amides (Straight-Chained-Satutated)

Nylon 6

Nylon 6/6

C - N - C - C - C - C- C - C - N - C - C - C - C - C - C - N -H H H H H H H H H H H2 2 2 2 2 22 222

- C - OH + - NH = - C - N + H O2

Hcarboxolic acid amine amide water

Aramide Fibers

NH . . . O = C

Hydrogen Bond

H HC C

Aromatic Amide

Benzene-Ring-Unsaturated

Advantages:

! Very high tensile strength (400-500 ksi)

! Very low density

! Low (negative) thermal expansion coefficient

! Resistant to acids,solvents, lubricants and oils

! Thermal stability

! High tensile elongation

Limitations:

! Degrades in ultraviolet light

! Moisture absorption

! Poor transverse properties

! Poor compressive properties

! Poor shear properties

Properties of Aramide Fibers

• High temperature strength

• Thermal stability

• Excellent in compression

• Oxidation resistant

• Corrosion resistant

• Low coefficient of thermal expansion

• High stiffness

• Expensive

• Difficult to machine composites containing ceramicfibers

• Relatively high density

• Brittle

Ceramic Fibers

(2) Spinning of polycarbosilane

! Distillation at 280BC to adjust molecular weightfor spinability

! Melt spinning 200-300BC

(3) Curing by oxidation 200BC

! Cross linking of polycarbosilane with oxygen

(1) Production of polycarbosilane

Processing of SiC Fibers

GRIND COKECARBON TUBE REACTOR

1600 C

RICE HULL STORAGE

WHISKER / CARBON SEPARATIONWHISKER /HULL RELIC SEPARATION

DRY CARBON OXIDATION ANALYSIS

DISPERSE

SiC Whisker Processing

Diameter (microns) 0.45 - 0.65

Length (microns) 10 - 80 (< 80%)

Surface area (M2/G) 3.0

Density (G/cm3) 3.2

Bulk density (G/cm3) 0.2 (Approximate)

Tensile strength ! GPa ! ksi

1.4 - 4.8200 - 700

Elastic modulus ! GPa ! Mpsi

420 - 690 60 - 100

SiC Whisker Properties

! Excellent in compression

! Ideal for metal matrix (permits some degree of fiberreaction)

Limitations

! Expensive

! Heavy (for W substrate)

! Brittle (large minimum bend radius)

Monofilaments (Thick Fibers)

1 curvature

bendradius

Minimum Bend Radius

Fiber s f

(GPa) D

(Microns)

E(GPa)

Carbon 2.1 11 520 1

Al2O3 FP 1.4 25 345 3

SiC Filaments 3.5 9 300 0.5

B, Sic or Borsic 2.8 200 400 14

Tungsten 1.1 500 400 91

Tungsten (FineDrawn)

4.1 75 400 4

Bend Radii Comparisons

Let-off spool

Gas inlet

Exhaust gas

Take-up spool

Mercury electrode

Tungsten filament

Reaction: 2BCl3+3H2→2B+6HCl

Boron Filament Processing

Tungsten Core

W B2 5

Bulk amorphous boron

Boron Fiber Structure

Residual Stresses in Boron Fibers

125 micron

25 micron

10 micron

1 micron

GLASS FIBER

CARBON FIBER

WHISKERMONOFILAMENT

TUNGSTEN OR CARBON CORE

Fiber Diameter Comparison

Manufacturer Designation

CompositionTensile

Strength,MPa

TensileModulus

, GPaDensity Diameter

NipponCarbon

Nicalon 50 Si, 31C, 10 O 2520-3290

182-210 2.55 10-20

Textron SCS-6 SiC on carbon core 3900 406 3.0 143

3M Nextel 312 62 Al22O33, 14 B22 O33, 15 SiO22 1750 154 2.7 11

DuPont FP >99 a -Al22O33 >1400 385 3.9 20

Sumitomo (spinel) 85 Al22O33, 15 SiO22 1800-2600

210-250 3.2 9-17

Ube Tyranno Si, Ti, C, O >2970 >200 2.4 8-10

Textron polymerpre

Si, C >2800 280-315 6-10

Dow Corning MPDZ 47 Si, 30 C, 15 N, 8 O 1750-2450

175-210 2.3 10-15

Dow Corning HPZ 59 Si, 10 C, 28 N, 3 O 2100-2450

140-175 2.35 10

Dow Corning MPS 69 Si, 30 C, 1 O 1050-1400

175-210 2.65 10-15

3M Nextel 440 70 Al22O33, 2 B22O33, 28 SiO22 2100 189 3.05 10-12

3M Nextel 480 70 Al22O33, 2 B22O33, 28 SiO22 2275 224 3.05 10-12

DuPont FP 166 Al22O33, 15-25 ZrO22 2100-2450

385 4.2 20

Continuous Ceramic Fibers

Fiber Dia.µ m

msisTU

GPasTU

µ/EFar

µ/EF?

E-Glass 10 2.54 72 11 3.45 500 4.8 2.8 2.8 0.20

S-Glass 10 2.49 87 13 4.30 625 5.0 1.6 1.6 0.22

PAN C T-300 7 1.76 228 34 3.20 470 1.4 -0.1

7.0 0.20

PAN C AS 7 1.77 220 32 3.10 450 1.2 -0.5

7.0 0.20

PAN C T-40 6 1.81 276 40 5.65 820 2.0 0.20

PAN C HMS 7 1.85 345 50 2.34 340 0.6 0.20

PAN C GY-70 8.4 1.96 483 70 1.52 220 0.4 0.20

Pitch C P-55 10 2.00 380 55 1.90 275 0.5 -0.5

Pich C P-100 10 2.15 690 100 2.20 325 0.3 -0.9

Kevlar 49 11.9

1.45 131 19 3.62 525 2.8 -1.1

33 0.36

Boron 140 2.70 393 57 3.10 450 0.8 2.8 2.8 0.20

SiC 133 3.08 400 58 3.44 485 0.8 0.8 0.8 0.20

Al22O33 FPfiber

20 3.95 379 55 1.90 285 0.4 4.6 4.6 0.20

Dia.µ m

Typical Fiber Properties

The Rule-of-Mixtures

What is it? –

Certain properties of materials with discrete phases can be predicted by the properties and volume amounts of their constituents.

M B BA AP P V P V= +

Volume Fraction of Fiber:

Volume fraction

In a two-component system consisting of one fiber and one matrix, the total volume of the composite is ,

hence c f mv v v= +

(1 )m fV V= −

Similarly the weight fractions Wf and Wmof the fiber and matrix respectively can be defined in terms of the fiber weight wf, the matrix weight, wm and the composite weight, wc.

Weight Fraction

c f mw w w= +

c c f f m mv v vρ ρ ρ= +

c f f m mV Vρ ρ ρ= +

Composite Density

c f mv v v= +

Composite Density in Terms of Weight Fraction

ww wρ ρ ρ

Convert between volume fraction and weight fraction

Use the definition of weight fraction and express the weights in terms of volumes and densities

f f ff f

ρ ρρ ρ

Voids in Composites

void actual theoreticalv v v= −

actual theoreticalvoid

actual

Divide by actual composite volume gives:

Express volumes in terms of weights and densities

theoretical actualvoid

theoretical

Vρ ρ

•The fibers are continuous, that is they extend for the length of the composite

•The fibers are aligned in one direction only

•The load is applied parallel to the direction of the fibers

•There is perfect bonding between the fiber and the matrix thereby preventing interfacial slip

Strength and stiffness parallel to fibers

MATRIX

Model of Unidirectional Composite

Pc, strains fibers , matrix, and composite the same amount

Pc is partitioned between the fiber and the matrix

Loads can be expressed in terms stress and cross-sectional area

Dividing by the cross section area of the composite

Areas Am, Af and Ac are equivalent to volumes Vm, Vf and Vc

ROM Strength

c f mε ε ε= =

c f mP P P= +

c c f f m mA A Aσ σ σ= +

f fm mc

σσσ = +

c m m f fV Vσ σ σ= +

Simple rule-of –mixtures can also be expressed in terms of strain and Young’s modulus

Strains in the composite and the constituents are equivalent

ROM Stiffness

c C m m m f f fE E V E Vε ε ε= +

c m m f fE E V E V= +

The Young’s moduli of fiber, Ef, the matrix, Em and the composite, Ec can be expressed in terms of stress and strain:

Because of strain equivalency the ratio of the stresses in the constituents are the same as the ratio of their Young’s moduli.

Similarly the ratio of the stress in one of the constituents, say the fiber, to the stress in the composite is given as

Consequences of ROM

f f fE σ ε= m m mE σ ε= c c cE σ ε=

Solving for the stress on the fiber

In terms of load

Consequences of ROM (Cont)

f f m m

E V E V

σσ =

P V EP V E

E VPE V

1 10 100

Vf=0.9

Vf=0.7

Vf=0.5

Vf=0.3

Vf=0.1

Load-Stiffness Relation

Stress-strain Curves For CompositesBrittle Fiber and Brittle Matrix With the Same Failure Strain

Strain

Fiber Fraction, Vf

0 1ε'fε'm

σmu σmu

matrix

Matrix contribution (1 )mu fVσ −

Fiber Contribution fu fVσ

Brittle Fiber Ductile Matrix With Different

Failure StrainsS

STRAIN

Vmin Vcrit0 1.0

FIBER FRACTION

' (1 )cu fu f m fV Vσ σ σ= + −

'm fE ε =

(1 )mu fVσ −

(1 )m fVσ ′ −

fu fVσ

'(1 ) (1 )mu f fu f m fV V Vσ σ σ− = + −

Minimum and Critical Volume Fractions

Minimum volume fraction is defined when

hence'

min 'mu m

fu mu m

V σ σσ σ σ

−=+ −

Critical volume fraction is defined when

' (1 )mu fu f m fV Vσ σ σ= + −

hence'

critfu m

Vσ σσ σ

A hybrid composite consists of two different types of fiber in an epoxy matrix. Using the properties of the constituents given below, draw the stress-strain diagram for the composite tested to failure in an elongation maintained test. Assume all constituents fail in a brittle manner.

Sample Problem

Constituent Weight %

Density, g/cc

Young’s Modulus, GPa

Tensile Strength, GPa

S-glass 40 2.49 87 4.3

Kevlar 49 25 1.45 131 3.6

Epoxy 25 1.16 0.75 0.031

The first step is to determine the order of failure of the constituents:

Solution (page1)

' 4.30.04943

87S glassf

S glassS glassE

σε −

−−

3.60.02748

131Kevlarf

KevlarKevlarE

σε = = =

' 0.0310.04133

0.75epoxyf

epoxyepoxyE

σε = = =

1 11.575

0.4 0.25 0.352.49 1.45 1.16

cS glass epoxyKevlar

S glass Kevlar epoxy

W WWρ

ρ ρ ρ−

= = =+ ++ +

Solution (page 2)Composite density from ROM

0.4(1.575)0.253

2.49S glass c

S glassS glass

−−

Volume fractions

0.25(1.575)0.272

1.45Kevlar c

KevlarKevlar

491 0.475epoxy S glass KevlarV V V−= − − =

Solution (page 3)The Young’s modulus of the composite before any constituent fails is designated E1

1 49 49S glass S glass Kevlar Kevlar epoxy epoxyE E V E V E V− −= + +

1 87(0.253) 131(0.272) 0.75(0.475) 57.67E = + + = GPa

The first constituent to fail is Kevlar49

2 S glass S glass epoxy epoxyE E V E V− −= +

2 87(0.253) 0.75(0.475) 22.37E = + =

When the epoxy fails

3 S glass S glassE E V− −=

3 87(0.253)E =

Solution (page 4)The stresses when the Kevlar fails

The stresses when the Epoxy fails

When the epoxy fails

'1 1 49 57.67(0.02748) 1.585KevlarEσ ε= = ='1 2 49 22.37(0.02748) 0.615KevlarEσ ε= = = GPa

'2 2 22.37(0.04133) 0.9246epoxyEσ ε= = =

'2 3 22.01(0.04133) 0.9097epoxyEσ ε= = =

'3 3 22.01(0.04911) 1.081S glassEσ ε −= = =

0 0.01 0.02 0.03 0.04 0.05 0.06

Strain

Solution (page 5)

Brittle matrix and ductile fiber with different failure strains

FIBER FRACTIONSTRAIN

Vtrans

matrix

stress on the fiber at matrix failure ' 'f f mEσ ε=

' (1 )cu f f mu fV Vσ σ σ= + −

At high fiber fraction the matrix can fail completely with the fibers holding the fractured matrix fragments in place

cu fu fVσ σ=

The transition from matrix-dominated to fiber-dominated failure

transfu mu f

σ σ σ=

Brittle Matrix Failure Transition

Brittle Matrix Failure Mechanism

For the matrix may split into a series of slabs held in place by the fibers. Slabs will range in thickness between X’ and 2X’.

FIBERMATRIX

The force on the matrix, is transferred from the fiber by shear, , hence

md Vσ2 rdxπ

Vd V rdx

rσ π τ

π= ⋅ ⋅

Model for calculating force on matrix

MATRIX

0 02muX

V rdx d

τ=∫ ∫

4x' 2x' <4x'

2x' x' <2x'

Slab Formation

2X’ is then the minimum distance to transfer sufficient force from the fiber to the matrix to cause matrix fracture, thus creating a slab. If a slab is 4X’ when it fracture it will crate two slabs exactly 2X’ thick. When they break there will be 4 slabs X’ thick. If a slab is between 2X’ and 4X’ it will fracture into a slab between X’ and 2X’, thus accounting for the range of observed slab thickness.

STRAIN

Slabing phenomenonin matrix

Slope = V f E

mu maxεε

Stress-strain diagram for brittle matrix-

ductile fiber composite

max 1 m mmu

V EV E

= − max 1

V EV E

2X’ slabs 4X’ slabs

Strain

0 Vf 1

( )minAbove : 1cu fu f m fV V Vσ σ σ ′= + −

( )minBelow : 1cu mu fV Vσ σ= −

minmu m

fu mu m

Vσ σ

σ σ σ′−

=′+ −

fu mu f

σ σ σ=

′+ −

fu mu f

σ σ σ=

′+ −

fu mu f

σ σ σ=

′+ −

matrix

minAbove : cu fu fV Vσ σ=

( )minBelow : 1cu mu f f fV V Vσ σ σ ′= − +

Stress-Strain Summary

FIBER FRACTIONSTRAIN

matrix

• For Vf < Vmin failure matrix failure results in failure of the composite• For Vf > Vmin matrix is split into thin slabs of thickness X< t < 2X

FIBERMATRIX

Ductile fiber-brittle matrix

4X 2X <4X

2X 2X X X <2X<2X

Slab Formation

• X is the minimum distance that the shear force is sufficient to break matrix and since the maximum stress will occur at the midplane of the slap the slab must be 2X thick to break. Smaller slabs would not allow the force to build up to the fracture stress of the matrix at the midplane of the slab.• Slabs exactly 4X thick can break into two slabs 2X thick which can then break into slabs X thick• Slabs 2X< t < 4X will split into two slabs X< t < 2X • Slabs < 2X cannot split any further

MATRIX

Force to Fracture Matrix

Force on matrix = total area for transfer x shear stress

Vd V rdx

rσ π τ

π= ⋅ ⋅

Solving for dx in terms of dσ and integrating

V rdx dV

V rX V

∫ ∫

Stress-Strain BehaviorCracking begins at εmu and ends at εmax when the last remaining slab splits

STRAIN

Slab splitting phenomenonin matrix

Slope = V f Ef

mu maxεε

max1 12

m m m mmu mu

f f f f

V E V EV E V E

ε ε ε

− ≤ ≤ −

The εmax limts are:

Transverse Properties

Every plane normal to the transverse direction will have a different fiber cross sectional area. This significantly complicates the stress analysis.

MATRIX

Slab Model

tftm/2 tm/2

matrix

reinforcement

Longitudinal

Transverse

Transverse Rule-of-Mixtures

c c m mf f

f mc mf

t t tt tt t

δ δ δε ε ε

ε ε ε

= += +

Elongations through the thickness are additive hence:

For a unit area thickness is the same as volume then:

c m mf fV Vε ε ε= +For serially connected constituents the loads on all constituents are the same

( )( )

fc mf f

V VE E E

VVE E E

σ σ σσσ σ

= == =

= + −

−= +

0 0.25 0.50 0.75 1.0

Fiber Volume Fraction , Vf

Longitudinal(Parallel Connected)

Transverse(Serial Connected)

ROM Plot

EEE V VE

=+ −

Rearranging for plotting

E V VE E

Comparison of slab model and realistic composites

Realistic Model Slab Model

Elasticity model

C= 0 C=1ISOLATED MATRIXFIBERS CONTIGUOUS

ISOLATED FIBERSMATRIX CONTIGUOUS

0<C<1PARTIAL CONTIGUITY

Transverse Young’s Modulus

( )( )

( ) ( )( ) ( )

2 22 1

f m m m f m m

m m f m m

T f f m m

f m f f m f m

m f m f m

K K G G K K VC

K G K K VE V

K K G G K K VC

K G K K V

ν ν ν

+ − −− +

+ + − = − + − + + −

+ − −

( )2 1f

− ( )2 1m

− ( )2 1f

+ ( )2 1m

The shear modulus

( ) ( )( )

( ) ( )( ) ( )

2f f m m f m f m m

L m fm f m m f m f m m

G G G V G G G G VG C G CG

G G G V G G G G V

− − + − −= − +

+ − + + −

Poisson’s ratio

( )( ) ( )

( ) ( )( ) ( )

( ) ( )2 2 2 2

f f m m m m m f m m m m f f m f f m f fLT

f m m m f m m f m m f m f m

K K G V K K G V K K G V K K G VC C

K K G G K K V K K G G K K V

ν ν ν νν

+ + + + + += − +

+ − − + + −

Halpin-Tsai Equations

transvers

Transverse Young’s Modulus

ξηη

+( )2 a

LOADLOAD

Cross Sectional Aspect Ratio

1.732ab

Rectangular Cross Section Reinforcement

STRESSSTRESS

For the case of then , hence and

STRESSSTRESS

0ξ = 2 0ab = 0a → b → ∞

Transverse Case 0ξ =

EE Vη

m f m m f

EEE V E V E

Results are identical to the serially connected constituent or slab model

Longitudinal Case

STRESSSTRESS b

ξ = ∞

For the case of and , then ξ = ∞a → ∞ 0b →

+ ∞=

−f m

T f f m mE E V E V= +then

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1Volume Fraction, V f

ROMHalpin-TsaiElastic C=0Elastic C=0.5Elastic C=1

Comparison of Transverse Modulus Models

Transverse Strength

• Transverse strength is always less than the matrix strength•Fibers act as stress concentrators•Defects in fiber-matrix bond can result in critical flaws

S Is the strength reduction factor

Assuming brittle fracture in the composite the transverse failure strain is

Estimates of Failure Strain

Analytical estimate:

ε επT mu

Empirical estimate:

( )ε εT mu fV* /= −1 1 3

Estimates of Strength

Analytical approach:

σ σπTu mu

Empirical approach:

( )1/31TTu mu f

Eσ σ= −

Limitations to volume fraction

Square Array Close Packing

Random Regular Open

Packing Geometry

square array packing

40.785f

π = =D

close packed array

28 0.90734

Weight of composite in air = 3.697g

Weight of composite in water = 1.636g

Weight of fiber after acid digestion = 2.595g

Density of fiber =2.5g/cc

Density of matrix=1.2g/cc

Find the theoretical and actual volume fraction and weight fraction of constituents

298.0697.3

595.2697.3701.0

697.3595.2 =−=====

)2.1(697.3107.1

)8.2(697.3595.2

79.10.1636.1697.3

−= water

watercc

cact ww

Acid Digestion Problem

Actual volumes

3.6972.065

2.595 1.0382.5

1.1020.918

cactual

actual

1.0380.503

0.9180.444

1 1 0.503 0.444 0.053

actual

= − − = − − =

Volume fractions

Acid Digestion Problem(page2)

Theoretical volume fractions

2.595 1.893.697 0.531

1 0.469

theoretical

f theoreticalf

= − =

Check on vV

1.89 1.79 0.0531.89

theoretical actualv

theoretical

V ρ ρρ

− −= = =

Acid Digestion Problem(page3)

Rule of Mixtures Strength Problem

Given the following constituent properties

Ffu=1.75 GPa σm=0.51 GPa Ef=230 GPa Em=157 GPa

a) What is the maximum fiber fraction at which matrix failure constitutes composite failure?

b) What is the strength of the composite at this fiber fraction?

c) What fiber fraction is necessary for a composite strength of 1.0Gpa?

d) What fiber fraction is required to achieve a composite strength of 0.54 GPa?

e) What is the composite strength at 20% fiber content?

f) What is the composite strength at 40% fiber content?

Strain Volume Fraction0 1

Rule of Mixtures Strength Problem(page2)

00325.0157

748.000761.0230

===′

=′=′===′

εσσ

σ’f

Rule of Mixtures Strength Problem(page3)

a) What is the maximum fiber fraction at which matrix failure constitutes composite failure?

337.0748.051.075.1

=′−+

=′+−

mutransitionf

ffufffmu

σσσσ

σσσ

b) What is the strength of the composite at this fiber fraction?

GPaGPaV

transitionfucu

59.0)337(.748.0)337.1(51.0)1(

59.0)337.0(75.1

cu =+−=′+−=

σσσ

Rule of Mixtures Strength Problem (page 4)

c) What fiber fraction is necessary for a composite strength of 1.0 GPa?

571.075.100.1

d) What fiber fraction is required to achieve a composite strength of 0.54 GPa?

126.051.0748.051.054.0

=−−

=−′−

′+−=

σσσσ

σσσ

Rule of Mixtures Strength Problem (page 5)

e) What is the composite strength at 20% fiber content?

VV fffmu

526.0)2.0(59.0)8.0(51.0

′+−=

σσσ

f) What is the composite strength at 40% fiber content?

GPaGPaV ffucu 70.0)40.0(75.1 === σσ

Transverse Stiffness Design Problem

You need to design a structure for operation at 750° F that must have a longitudinal modulus,EL of 35 mpsi and a transverse modulus, ET of 7 mpsi. The structure must also have a high but unspecified compressive strength.

Transverse Stiffness Design Problem (Page2)

Material Selection:

To meet temperature, stiffness and compression requirements I would select either boron or silicon carbide monofilaments. These fibers also are approximately isotropic.

psiEboron 000,000,57=

From organic material handbooks I found polyamide-imide meets the temperature requirement. It is also a low density material.

psiE imidepolyamide 000,720=−

Transverse Strength Problem

Estimate the transverse fracture strength of an S-glass/epoxy composite with 75% by weight fibers. The constituents have the following properties:

1.140.0723.1epoxy

2.484.3089S-glass

Density, g/ccσU, GPaE, GPa

1 11.917 /

(1 ) 0.75 0.252.48 1.14

g ccW W

= = =− ++

Composite density

Fiber volume fraction

0.75(1.917)0.58

2.48f c

ρ= = =

Transverse StrengthProblem (page 2)

η ξξ

−= =

13.298TE GPa=

( )1 31 0.051

TTU mu f

EV GPa

= − =

Adhesion between fiber and matrix is controlled byproperties of the interface.

High Degree of Adhesion:

! Provides efficient transfer of load between fiberand matrix.

! Controls properties of composite in directiontransverse to fibers.

! Influences shear properties.

! Reduces susceptibility to environmentaldegradation.

Poor Adhesion:

! Provides higher fracture toughness for cracknormal to reinforcement.

! Minimizes tri-axial state of stress in matrix betweenfibers (promotes ductility of matrix).

The Interface

Control Surface Tension - High degree of wettingbetween fiber and matrix promotes good bonding.

! Wetting can be controlled by:- Temperature- Composition of matrix- Composition of fiber surface

Interfacial Bonding & Wetting

cos0 complete wetting

180 unwetting

GS LS LGγ γ γ θθ

= +→

! Mechanical Locking- Whiskerizing by the CVD growth of SiCwhiskers on graphite

SiCl4 + CH3 º SiC + HCl

H2 + 2SiCl2(CH2) º 2SiC +4HCl

! Oxidation- Treating graphite fibers in concentratedHNO3 - Heating graphite fibers in air

Methods for Increasing Bonding

Matrix

Good Bond Poor Bond

Micro-indentation Test

SHORT BEAM GOOD BOND POOR BOND

IOSIPESCU SHEAR GOOD BOND POOR BOND

Shear Tests for Bond strength

Fibers unbonded

- Stiffness in bending is the moment of inertiaof each fiber x Ef x the number of fibers

Direct Effect of Interfacial Bonding

dD E mn=

Beam Stiffness with Bonded Fibers

14 4 12

B c nd

B f m nd

D E X dA

d dnm nm

D E E X dAmd nd md nd

D E E X dA

mn dD E E

= + − ⋅ ⋅ = + − = + −

For the graphite fiber (P100)/epoxy matrix system:

Stiffness Ratio for Bonded & Unbonded Fibers

100 10 (P100 graphite fiber)

E 0.2 10 (epoxy)

D 0.000434 12

mn dE E

mn dD E

E mn dDD E mn d

Unbonded fibers behave as missing fibers hence the Young'smodulus of a composite with some unbonded fibers are givenas:

neglecting the stiffness of the matrix

Effect of Unbonded fibers on Ec

Volume fraction of unbonded fibers

c f f m f f f

c f f m f

c c f f

E E V E V E V

E E V E V

E E E V

= + − −

= + −

0.050.150.200.40

0 0.2 0.4 0.6 0.8

Contribution of Unbonded Fibers

Interfacial Failure

Equilibrium forces on element of beam

yM MB A

Vdx M MMyIE

EE MyE I

Equilibrium Forces on Fiber

2fB fBP P rdxπ τ− = ⋅

Divide by cross section area of fiber

2fB fBP P rdxr r

π τπ π

− ⋅=

Gives stress on fiber

τσ =

Stress on fiber in terms of bending moment

E My dxE I r

usingM

Equilibrium Forces on Fiber(cont)

E My ME I r V

Solving for τ

VE yrE I

Dividing both sides byfσ

VE yrE I

E MyE I

Ratio of Interfacial to Fiber stress

M=PL/4

Failure Under Compressive Loads

! Microbuckling! Transverse splitting - Poisson tensile strain! Shear failure

Microbuckling

Extensional Mode Shear Mode

f m fLU f

V E EV

σ ′ =−

Transverse Splitting

SPLITTING

( )( )1/31

T L LT

LU LTT

L TULU

f f m m uLU

f f m m

E V E V VV V

ε =−ε ν′σ ν

ε =′

′ε′σ =ν

+ ε−′σ =ν + ν

Shear Failure

KINK BAND

Local rotation

10.000

100.000

0.0 0.2 0.4 0.6 0.8 1.0

Fiber Volume Fraction

Extensional BucklingShear BucklingTransverse Splitting

Compressive Strength of Composites

TRANSVERSE TENSILE FAILURE

! Matrix failure

! Interfacial failure (debonding)

! Fiber splitting

TRANSVERSE COMPRESSIVE FAILURE

! Matrix shear

! Debonding (interface shear)

! Fiber crushing

IN-PLANE SHEAR FAILURE

! Matrix shear

! Interface shear (debonding)

! Combination of above

Other Failure Modes

Short Fiber Composites

Model system: (Kelly-Tyson)• Fiber is perfectly elastic• Matrix is rigid/perfectly plastic

matrix

STRAIN

Elongation in Model System

All forces on fiber are transmitted from the matrix through fiber surface

LOAD LOAD

Stress on Fiber

CENTER-LINE

fσ σ+ d

Sum forces in element dx

Forces consist of •End forces σfAc•Shear forces τiAs

f i f f

r rdx r d r

dx d r

σ π τ π σ π σ π

=∫ ∫

Stress on Fiber (Cont.)

largest stress developed in fiber

stress due to loading on free cross-sectionf

Evaluating the integral

σ σ= +

Stress due to loading on free cross-section can be neglected

max2 i

Stress on Fiber ProfileIntegrating from fiber center to other end

f ld dx

τσ =∫ ∫

Gives stress profile in fiber

LENGTH ALONG FIBER

σmaxf

Ineffective LengthDiscontinuous but long fibers

Composite is loaded to cσ Where c cuσ σ<

f f cc

Eσ σ σ= =

LENGTH ALONG FIBER, X

maxfσ

lt is called the ineffective length

LENGTH ALONG FIBER

fσmaxfσ

If the composite contains fiber lengths , less than then the fibers cannot be loaded to their maximum potential hence

l tlmaxf fσ σ<

Eσ σ<

Fibers Less Than Ineffective Length

Critical LengthIf the composite is loaded to the ultimate or breaking strength, , the ineffective length becomes the critical length, and the discontinuous fibers in the composite can be loaded to failure hence,

and the critical length is given as

ffu cu

Eσ σ=

LENGTH ALONG FIBER

cfσ σc

Stress Transfer for Elastic Matrix and Elastic Fiber

Assume that both the fiber and the matrix are elastic

LOAD LOAD

• no yielding at the interface between fiber and matrix

• strain in the matrix and in the composites is considered equal

• strain in the fiber is clearly less than that of the compositem cε ε=

Matrix Displacement Around Fiber

DISTANCE FROM CENTER, zDIS

The matrix displacement w in the fiber direction is constant for any distance z from the fiber over all possible angles, θ , hence

Shear stress, τ, is also independent of θ , hence

The matrix displacement at the surface of the fiber is equal to the fiber displacement

( )w f z=

( )f zτ =

Multi-fiber Model

Summing the shear forces at the fiber surface and at a distance z form the center of the fiber gives

which reduces to

Expressing the shear stress in terms of shear strain and shear modulus of the matrix

2 2 ez dx r dxπ τ π τ=

τ τ=

rdudz zG

Solve for the Displacement

evaluates to

The ratio R/r depends upon the arrangement of the fibers which can be expressed as the packing factor, Pf.

For square array of fibers

For hexagonal array of fibers

r dzdu

=∫ ∫

lneR f

r Ru u

1ln ln

1 2ln ln

For a general arrangement

1ln ln

Using packing fraction and solving for

Taking the center of the fiber as the origin the stress transferred to the fiber can be written as

Combining above equations

(1 ) ln

m R fe

Calculating Stress Transfer

σ τ= −

(1 ) ln

f m R f

d E u udx P

−= −

Calculating Stress Transfer (Cont)

Differentiating

( )2 2

σσ ε= −

Fiber stress: Shear stress:

Shear Stress and Fiber Stress Distribution Along Fiber

DISTANCE ALONG FIBER

DISTANCE ALONG FIBER DISTANCE ALONG FIBER DISTANCE ALONG FIBER

STRAIN STRAIN STRAIN

RIGID PERFECTLYPLASTIC MATRIX

ELASTIC FIBER

ELASTIC MATRIX

ELASTIC FIBER

ELASTIC-PLASTIC MATRIX

ELASTIC FIBER

MATRIX

SHEAR STRESSFIBER STRESS

Comparison of Fiber and Shear Stress Profiles for Different Matrix Behavior

Average Stress in Short FibersThe rule-of-mixture for discontinuous (short) fiber reinforced composite is

c f f m mV Vσ σ σ= +

fσ Is the average stress on the fiber that depends on fiber length

For fiber length l<lt

σ τσ

Fiber Length Equal Ineffective Length

2 2f c f

σ σσ = =

S σcEE

Fiber Length > Ineffective Length

maxfσ

max max

tf f t

σ σσ

+ − =

∫∫

Fiber Length >> Ineffective Length

50 tl l=

max max

σ σ σ

= − = −

Composite Strength with Average Fiber Stress

cl l< ( )1icu f mu f

σ σ= + −

cl l= ( )11

2cu fu f m fV Vσ σ σ ′= + −

cl l> ( )1 12

ccu fu f m f

lσ σ σ ′= − + −

cl l>> ( )1cu fu f m fV Vσ σ σ ′= + −

cu f f m mV Vσ σ σ= +

Average Fiber Stress for Elastic Fiber and Elastic Matrix

tanh1f f

Comparison of Elastic –Perfectly Plastic and Elastic-Elastic Cases

DISTANCE ALONG FIBER

maxfσ

( )max 1 1 tf f

lσ σ = − −

q Is the ratio of the cross hatched area to the area given by max

2f tlσ

The cross-hatched area is then maxf tq lσ

ROM Strength Diagram

+ 1-( )VfV

1-( )VfV

Vmin Vcrit

minmu m

f mu m

Vσ σ

σ σ σ′−

=′+ −

Young’s Modulus of Short Fiber Composites

For transverse modulus

LOADLOAD ba

For a circular cross section fiber a b=

T T fT

ξ ηη

LOADLOAD d

Young’s Modulus of Short Fiber Composites

Longitudinal Modulus

L L fL

ξ ηη

Empirical Values for ξ

102 40 f

bξ = +

For EL102 40 f

dξ = +

For ET

1.7321040 f

bξ = +

For GLT

Correction for volume fraction becomes important for Vf>0.7

Random Orientation

5 38 8

2 18 8

random T L

randomrandom

random

STRAIN, %

COMPOSITE WITH MATRIX A

COMPOSITE WITH MATRIX B

MATRIX AMATRIX B

Stress-Strain Behavior of Short Fiber Composites

Typical Strength Behavior of Short Fiber Composites

00.2 0.60.4 0.8 1.00

ACTUAL MATRIX A

ACTUAL MATRIX B

ROM MATRIX AROM MATRIX B

FIBER VOLUME FRACTION

Fracture Behavior of Short Fiber Composites

CRACK OPENING DISPLACEMENT

D0.29 V

Depends on:

Fiber Properties

Interfacial Properties

Strain Rate

Temperature

Relatively insensitive to:

Matrix

Specimen Thickness

Notch Length

Fracture in Discontinuous Composites

Property

% Glass Fiber

0 10 20 30 40

B D B D B D B D B D

FlexuralStrength,(MPa)

58 21 _ 44 89 67 110 85 132 -

FlexuralModulus,(GPa)

3.0 0.9 - 2.4 5.4 4.0 6.5 5.5 7.5 -

Impact, (J/cm2) - 19 - 13 - 10 - 9 - -

Izod Impact,(J/cm)

0.1 - -- - 3.3 - 4.5 - 6.4 -

B - Brittle Matrix (Polystyrene)

D - Ductile Matrix (High Density Polyethylene)

Impact Behavior of Discontinuous Glass Fiber Composites

! Lower fatigue resistance than continuous fibercomposites

! Initiation by debonding at fiber-matrix interface

! Fiber ends are also initiation sites

! Contributes to thermal damage of matrix

! Results in susceptibility to moisture damage

! Ductile matrix can limit fatigue crack propagation

Fatigue of Short Fiber Composites

An orthotropic lamina is a sheet with unique and predictable propertiesand consists of an assemblage of fibers lying in the plane of the sheetand held in place by a matrix.

Continuous Discontinuous

L and T are the principal material directions sometimes referred to as 1and 2. The angle formed by the counter clockwise rotation from anarbitrary direction x to L is +?. The clockwise rotation produces -?.

Orthotropic Lamina

Deformation in Orthotropic Lamina

Orthotropic lamina loaded in principal directions or isotropic lamina loaded in any direction

a) Load in principal direction, L b) Shear load in LT direction

Deformation in Orthotropic Lamina

Orthotropic lamina loaded in an arbitrary direction

a) Extensional load in arbitray direction b) Shear load in arbitrary direction

Hookes's Law For AnisotropicMaterial

full tensor notation

ij ijkl klEσ ε=

Subscripts i,j,k,l = 1,2,3 ,hence there are a total of 34 or 81

Strain symmetry produces the result Eijkl = Eijlk

Stress symmetry produces the result Eijkl = Ejikl

Thermodynamic arguments produce the result Eijkl = Eklij

Leaving only 21 unique elastic constants

Expanded Elastic Constants in Full Tensor

1111 1122 1133 1123 1131 1112

1122 2222 2233 2223 2231 2212

1133 2233 3333 3323 3331 3312

1123 2223 3323 2323 2331 2312

1131 2231 3331 2331 3131 3112

1112 2212 3312 2312 3112 1212

E E E E E EE E E E E EE E E E E EE E E E E EE E E E E EE E E E E E

Contracted NotationReplace paired subscripts with a single subscript according to the following: 1 for 11, 2 for 22, 3 for 33, 4 for 23, 5 for 31, and 6 for 12.

i ij jCσ ε=

subscripts can have the values 1,2,3,4,5, and 6

11 12 13 14 15 16

12 22 23 24 25 26

13 23 33 34 35 36

14 24 34 44 45 46

15 25 35 45 55 56

16 26 36 46 56 66

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

Transformation Of Elastic Constants

To transform an elastic constant from the X axis to the X' axis use the following:

where aim, ajn, akr, als are direction cosines.

'mnrs im jn kr ls ijklE a a a a E=

Direction Cosines for Transformation Through Plane of Symmetry

X1 a11 = 1 a12 = 0 a13 = 0

X2 a21 = 0 a22 = 1 a23 = 0

X3 a31 = 0 a32 = 0 a33 = -1

X'1 X'2 X'3

'1111 1111

'1122 1122

'1131 1131

The result can only be true if . Examination of the stiffness matrix indicates that odd multiples of a33 will give negative and hence zero values for Eijkl. Applying this observation to all the coefficients then the following Eijkl are zero:

E1113, E1213, E2223, E1223, E1123, E1333, E2213, E2333

'1131 1131E E= − 1131 0E =

Elimination of Stiffness Coefficient with One Plane of Symmetry

Effect of Second Plane of SymmetryApply to a second plane, X2X3

X1 a11 = - 1 a12 = 0 a13 = 0

X2 a21 = 0 a22 = 1 a23 = 0

X3 a31 = 0 a32 = 0 a33 = 1

X'1 X'2 X'3

Elimination of Stiffness Coefficient with Second Plane of Symmetry

Odd multiples of a11 give negative (hence zero) Eijkl

T T T T

E1112 E2231 E2212 E3331 E3312 E2331 E2312

The T'ed coefficients are eliminated by the second plane of symmetry, unchecked coefficients fit the criteria but were already eliminated by the first plane of symmetry. Applying the third plane of symmetry does not eliminate any other coefficients.

Hooke's Law for the Orthotropic Material

11 12 131 1

12 22 232 2

13 23 333 3

4423 23

5531 31

6612 12

0 0 00 0 00 0 0

0 0 0 0 00 0 0 0 00 0 0 0 0

C C CC C CC C C

σ εσ εσ ετ γτ γτ γ

Reduced Stiffness MatrixFor a two dimensional lamina (sheet) the stresses through the thickness are so small that they can be neglected (plane stress), hence

3 23 31 0σ τ τ= = =

Hooke’s Law for plane stress

1 11 12 1

2 12 22 2

12 66 12

Q QQ Q

σ εσ ετ γ

11 1133

C= −

13 2312 12

C CQ C

C= −

22 2233

C= −

66 66Q C=.

For lamina that are transversely isotropic, 13 12C C=33 22C C=and

Hooke’s law in terms of compliance is

Compliance Matrix

1 11 12 1

2 12 22 2

12 66 12

S SS S

ε σε σγ τ

Sij's and Qij's are mutually inverse

[ ] [ ][ ]I S Q=

Inverting

2211 2

11 22 12

1122 2

11 22 12

1212 2

11 22 12

SQS S S

= −−

Inverting [S]

2211 2

11 22 12

1122 2

11 22 12

1212 2

11 22 12

Q Q QQ

SQ Q Q

= −−

Inverting [Q]

Experimental Determination of Compliance

1 11 12 1

2 12 22 2

12 66 12

1 11 1 12 2

2 12 1 22 2

12 66 12

S SS S

S SS SS

ε σε σγ τ

ε σ σε σ σγ τ

= += +=

expanded

Uniaxial Tensile Tests

1 11 1

νσε

2 22 2

νσε

Shear Test

126612

Compliance And Stiffness In Terms Of Engineering Constants

12 2112

= − = −

21 1 12 212

12 21 12 21

ν νν ν

ν ν ν ν

= =− −

Stress-strain Relation For Lamina in Any Direction

Tσ σσ στ τ

1 12 2

Tε εε ε

[ ]( )

cos sin 2cos sinsin cos 2cos sin

cos sin cos sin cos sin

Tθ θ θ θθ θ θ θ

θ θ θ θ θ θ

= − − −

Second Order Transformation Matrix

[ ]( )

cos sin 2cos sinsin cos 2cos sin

cos sin cos sin cos sin

Tθ θ θ θθ θ θ θ

θ θ θ θ θ θ

− −

Inverted transformation matrix

The inverted matrix can by found using 2 = -2 in th [T ] matrix

Transforming From Principal to Arbitrary Direction

Transforming stress from principal to arbitrary direction

Tσ σσ στ τ

Transforming strain from principal to arbitrary direction

1 12 2

Tε εε ε

Hookes Law For Stresses And Strains Applied In Arbitrary Direction

(1) Convert engineering strain in the arbitrary direction to tensorial strain in the arbitrary direction.

(2) Transform tensorial strain in the arbitrary direction to tensorial strain in the principal material direction.

(3) Convert tensorial strain in the principal material direction back to the engineering strain in the principal material direction.

(4) Find engineering strain in the principal material direction to stress in the principal material direction using Hooke’s Law for orthotropic material.

(5) Transform stress in the principal material direction to stress in the original arbitrary direction.

Convert Engineering Strain in the Arbitrary Direction to Tensorial

Strain

Rε εε ε

[ ] 11 0 00 1 0

10 0 2

Step 1

Transform Tensorial Strain in the Arbitrary Direction to Tensorial Strain

in the Principal Material Direction.

121 12 2

Tε εε ε

Step 2

Convert Tensorial Strain in the Principal Material Direction Back to the

Engineering Strain

[ ]1 1

12 121

Rε εε εγ γ

[ ]1 1

12 121

ε εγ γ

Step 3

Find the Stresses in the Principal Material Directions

Step 4

[ ]1 1

Qσ εσ ετ γ

Transform Back to the Arbitrary Direction

Tσ σσ στ τ

Step 5

All Operations in One Equation

[ ] [ ][ ][ ][ ]1 1x x

T Q R T Rσ εσ ετ γ

− −

Transformed stiffness matrix

[ ] [ ][ ][ ][ ]1 1Q T Q R T R

− − =

Reduces to

[ ] [ ][ ]1 TQ T Q T

− − =

The Transformed Stiffness Matrix

11 12 16

12 22 26

16 26 66

Q Q QQ Q Q Q

( )( )

( ) ( )( ) ( )( )

4 2 2 411 11 12 66 22

4 2 2 422 11 12 66 22

2 2 4 412 11 22 66 12

3 316 11 12 66 12 22 66

326 11 12 66 12

cos 2 2 sin cos sin

sin 2 2 sin cos cos

4 sin cos sin cos

2 sin cos 2 sin cos

2 sin cos

Q Q Q Q Q

Q Q Q Q Q Q Q

Q Q Q Q Q

θ θ θ θ

= + + +

= + − + +

= − − + − +

= − − + −( )( ) ( )

322 66

2 2 4 466 11 22 12 66 66

2 sin cos

2 2 sin cos sin cos

Q Q Q Q Q Q

θ θ θ θ

= + − − + +

Transformed Compliance Matrix

[ ][ ] [ ] [ ][ ]1 1x x

R T R S Tε σε σγ τ

− −

[ ][ ] [ ] [ ][ ]1 1S R T R S T− − =

[ ] [ ][ ]TS T S T =

Individual Coefficients In Transformed Compliance Matrix

11 12 16

12 22 26

16 26 66

S S SS S S S

( )( )

( ) ( )( ) ( )( )

4 4 2 211 11 22 12 66

4 4 2 222 11 22 12 66

2 2 4 412 11 22 66 12

3 316 11 12 66 22 12 66

326 11 12 66 22

cos sin 2 sin cos

sin cos 2 sin cos

sin cos cos sin

2 2 sin cos 2 2 sin cos

2 2 sin cos 2

S S S S S

S S S S S S S

S S S S S

θ θ θ θ

= + + +

= + − + +

= − − − − −

= − − − ( )( ) ( )

312 66

2 2 4 466 11 22 12 66 66

2 sin cos

2 2 2 4 sin cos cos sin

S S S S S S

θ θ θ θ

− −

= + − − + +

Invarient Forms Of Stiffness Coefficients

11 22 12 661

11 222

11 22 12 663

11 22 12 664

11 22 12 665

3 3 2 48

22 486 482 48

Q Q Q QU

+ + +=

+ − −=

+ + −=

+ − +=

11 1 2 3

22 1 2 3

12 4 3

16 2 3

26 2 3

66 5 3

cos2 cos4

sin 2 sin421

sin2 sin 42

Q U U U

= − +

= − −

= − +

Q 11U U cos 2 U cos 41 2 3

θ θ θ

Graphical Representation of InvarientComponents

Transformation Of Engineering Constants

2 21 2 12

cos sin cos sin

sin cos cos sin

2 cos sin 2 cos sin cos sin

ε ε θ ε θ γ θ θ

γ ε θ θ ε θ θ γ θ θ

= + −

= − + −

Stresses And Strains In The Principal Material Directions

1 12 21

2 21 12

σ ν σε

cos sin

σ σ θ

σ σ θτ σ θ θ

== −

Strains In Arbitrary Direction

4 4212

1 2 12 1

cos sin 1 1 2sin 2

4x x E E G Eθ θ ν

ε σ θ

= + + −

212 12

1 1 1 2 12

1 1 2 1 1sin 2

4y x E E E E Gν ν

ε σ θ

= − − + + −

212 12

1 2 12 1 1 2 12

1 1 1 2 1 1sin 2 cos

2xy x E E G E E E Gν ν

γ σ θ θ

= − + − − + + −

Young’s Modulus In Arbitrary Direction

4 4212

1 2 12 1

cos sin 1 1 2sin 2

E E G E

σε θ θ ν

+ + −

evaluated at 90θ +

4 4212

2 1 12 1

cos sin 1 1 2 sin 24

E E G E

σε θ θ ν θ

+ + −

( )2 4 412

1 2 1 12 12

12 2 4 1 1

2 sin 2 sin cosxyG

E E E G Gν

θ θ θ=

+ + − + +

( )4 4 212

1 1 2 12

1 1 1sin cos sin 2y

EE E E G

ε νν θ θ θ

= − = − − + −

Shear Modulus and Poisson’s Ratio in Arbitrary Direction

Considering only shear stress produces

Combining the equations for strain in both directions

Cross Coefficientsxσ xyγ xyτ

xε xm

γ = −

The stress can also produce shear strain , and shear stress can produce extensional strain , these can be related though a the cross coefficient

τε = −

σγ = −

ε = −

The cross coefficient mx and my

21 1 1 112 12

2 12 2 12

sin 2 cos 1 22x

E E E Em

E G E Gθ ν θ ν

= + − − + −

21 1 1 112 12

2 12 2 12

sin 2 sin 1 22y

E E E Em

E G E Gθ ν θ ν

= + − − + −

Transformed Engineering Compliance Matrix

x xx x

xy yy y

x yxy xy

mE E E

mmE E G

ε σν

ε σγ τ

− −

= − − − −

Off-axis Young’s Modulus and Shear Modulus for Carbon/epoxy

Lamina

0 10 20 30 40 50 60 70 80 90

Poisson's Ratiomxmy

Off-axis Poisson’s Ratio and Cross Coefficients for Carbon/epoxy Lamina

Restrictions On Elastic Constantsσε τγ

2(1 )E

1ν > −

Thermodynamic considerations require that Young’s modulus and shear modulus are positive values. The work of deformation, and are positive then

For isotropic materials the relation between the Young’s modulus and shear modulus is

For G and E to be positive .

x y zPK

φ ε ε ε= + + =

3(1 2 )E

2ν− < <

The volumetric strain, resulting from hydrostatic pressure P is

where K the bulk modulus is

For K and E to be positive

For all the elastic moduli to be positive then

Volume Strain Effects

Applying thermodynamic restraints to

For the plane stress condition Q11, Q22, Q66 >0 , then for example

and therefore. This is true if . From this it follows that

Relation Between The Elastic Constants For Orthotropic Materials

11 22 33 44 55 66, , , , , , 0C C C C C C >

12 211E

Qν ν

12 211 0ν ν− > 12 21 1ν ν <

Strength Of Orthotropic Lamina

For isotropic metals failure usually occurs by yielding and can be simply predicted by the maximum shear stress theory.

yieldσ σ τ− ≥ or I II yieldσ σ σ− ≥

σIσII

Strength Notation For Orthotropic Lamina

Fiber Direction

Comparison Of Positive And Negative Shear For Stress In The Principal Material

Directions

Positive Shear Negative Shear

Comparison Of Off-axis Positive And Negative Shear For Stress

Positive Shear Negative Shear

Biaxial Strength Theories

Maximum stress theory

X XY Y

σστ

′− < <′− < <′− < <

Failure occurs when any of the stress values exceeds the strength values

No interaction between applied stresses is considered

Composite Lamina Loaded In Arbitrary Direction X

xLoad Direction

sincos sin

σ σ θ

σ σ θτ σ θ θ

== −

Maximum Stress for Fibers in Arbitrary Direction

cos cos

sin sin

cos sin cos sin

σθ θ

σθ θ θ θ

′− < <

0 45 90

cos2θ

cos sin

Maximum Stress Theory Of Lamina Failure Plotted For Off-axis Loading

Maximum Strain Theory

1212 12

X XE EY Y

E ES S

′−< <

1 12 21

2 21 12

σ ν σε

τσσ

γεε

Hooke’s Law

Transforming To Arbitrary Off-axis Directions

2 21 12

2 22 21

cos sin

sin cos

cos sin

σε θ ν θ

σγ θ θ

2 2 2 212 12

cos sin cos sin

sin cos sin cos

cos sin cos sin

σθ ν θ θ ν θ

σθ θ θ θ

′− < <

− −′

− < <− −

′− < <

Strength Theories With Stress Interactions

The strength tensor

( ) ( ) ( ) 1ij i j ijk i j ki i F FFβ γα σ σ σ σ σσ + + + =K

where , , 1, 6i j k = K 1α β= =and

following coefficients are zero

4 5 6, 14 15 16, 24 25 26, 34 35 36, 44 45 46, , , , , , , , , ,F F F F F F F F F F F F F F F

1i i ij i jF Fσ σ σ+ =

The general polynomial

1 2 12 21 22 66, , , , ,F F F F F F

Non-zero coefficients

Tsai-Hill Failure Criterion

Using the Von Mises (distortion energy) criterion

1ij i jF σ σ =

2 2 211 1 22 2 12 1 2 66 62 1F F F Fσ σ σ σ σ+ + + =

Expanding

consider the case 1 Xσ = 2 0σ = 6 0σ =

211 1F X =

2 Yσ =then 1 0σ = 6 0σ =

222 1F Y =

6 Sσ = 1 2 0σ σ= =and

266 1F S =

apply a balanced biaxial stress, and

2122 1F X = −

Strength Coefficients

Coefficients

Tsai-Hill Criterion

22 261 2 1 2

2 2 2 2 1X Y X S

σσ σ σ σ+ − + =

4 42 2

2 2 2 2

1cos sin 1 1cos sin

X Y S X

σθ θ θ θ

= + + −

the failure stress, for off-axis loadingxσ

0 10 20 30 40 50 60 70 80 90

Fiber Orientation, θ

Failed

Safe Stress

Predicted Failure Stress

Tsai-Wu Failure Criterion2 2 2

6 6 1 1 2 2 12 1 2 11 1 22 2 66 62 1F F F F F F Fσ σ σ σ σ σ σ σ+ + + + + + =

21 11 1F X F X+ =

Stress in fiber direction is equal to the longitudinal tensile strength X

Compressive stress is equal to the compressive longitudinal strength, X’

21 11 1F X F X′ ′− + =

Coefficients are found

X X= −

= −′

Additional Coefficients

1 1 1 1 1 1 112

F B BB X X Y Y XX YY

= − −− + − + ′ ′ ′ ′

Strength Interaction CoefficientsApplying balanced biaxial stresses

F12 can be found empirically

*12 12 11 22F F F F=

F12 is between -0.5 and 0

Using *

12F = -0.5 the usual form of the Tsai-Wu criterion is

2 2 21 2 1 2 1 2 62

1 1 1 1 11 1 1 1 12XX YY XXYY SX X Y Y

σ σ σ σ σ σ σ + + + − + =− − ′ ′ ′ ′′ ′

Failure Envelopes Stress SpaceFailure envelope for isotropic brittle material

σΙΙ

Failure Envelopes for Isotropic Ductile Materials

Von Mises-Hencky

Tresca

σΙΙ

Failure Envelope For Orthotopic Material By The Maximum Stress Theory

Failure Envelope For Orthotopic Material By The Maximum Strain Theory

Tsai-Hill and Tsai-Wu Failure Envelope

rearranged in quadratic form

( ) ( )2 2 21 11 1 1 2 12 2 2 2 22 6 662 1 0F F F F F Fσ σ σ σ σ σ+ + + + + − =

Failure Ellipse In Strain Space

Using Hooke's law i ij jQσ ε=

1i i ij i jF Fσ σ σ+ =

In generalized failure criterion

1ij ik jl k l i ij jF Q Q F Qε ε ε+ =gives

ij ik jl klF Q Q H=Using the following notations and i ij iF Q H=

1ij i j i iH Hε ε ε+ =

Expanded Form2 2 2

1 1 2 2 11 1 12 1 2 22 2 66 62 1H H H H H Hε ε ε ε ε ε ε+ + + + + =

2 211 11 11 12 11 12 22 12

2 222 22 22 12 22 12 11 12

212 11 11 12 12 11 22 12 22 12 22

266 66 66

1 1 11 2 12

2 1 12 2 22

H F Q F Q Q F Q

H F Q Q F Q Q Q F Q Q

H F Q F QH FQ F Q

= + + +

= += +

Quadratic Form Plotted

( ) ( )2 21 11 1 1 2 12 2 2 2 22 6 662 1 0H H H H H Hε ε ε ε ε ε+ + + + + − =

FAILURE CRITERIA AS DESIGN TOOL

, ,x y sx y xy

X Y SR R R

σ σ τ= = =

Strength Ratios or Safety Factors:

Strength tensor in stress space in terms of R

( ) ( ) 2 1i i ij i jF R F Rσ σ σ+ =

In strain space

( ) ( ) 2 1i i ij i jH R H Rε ε ε+ =

b b acR

a+− + −

Solved Roots

b b acR

a−− − −

ij i j

== −

In stress space

In strain space

ij i j

== −

1Fij i jσ σ =

2 2 2 2

1 4 1 1 1

2 U X Y S− − −

1F Fi i ij i jσ σ σ+ =

X X−

Y Y−

XX ′

YY ′

111 22 2F F

1Fij i jσ σ =

X X−

Y Y−

XX ′

YY ′

2XX−

X X−

Y Y−

XX ′

YY ′

( )122

XX S X X X X Y Y

XX S XX

′ ′ ′− − − +−

′ ′

X X−

Y Y−

XX ′

YY ′

2XY−

1Fij i jσ σ =

X X−

Y Y−

XX ′

YY ′

*12 11 22F F F

Tsai-Wu

Tsai-Hill

Malmeister

Hoffman

Fischer

Ashkenazi

F66F12F22F11F2F1TensorTheory

Stress Interaction Strength Criteria

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