apuntes compositos i-presentaciones
TRANSCRIPT
(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
2
4
6
8
10
SPECIFIC STIFFNESS
SP
EC
IFIC
STR
EN
GTH
KEVLAR/EPOXY
BORON/EPOXY
GR/EPOXY (HS)
GR/EPOXY(HM)
GR/EPOXY(UHM)
E-GLASS/EPOXY
ALUMINUM ALLOYS
STEEL
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
YARN
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
10
20
30
40
50
0
FIBER ASPECT RATIO
FIBER REINFORCED
SOLID SOLUTIONAND PRECIPI-TATION HARDEN-ING
DISPERSIONSTRENGTHENING
CERMETS
COMPOSITE STRENGTH AFFECTED BY REINFORCEMENT SIZEM
ATR
IX S
TRE
NG
THE
NIN
G F
AC
TOR
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
5)ksi
Low(< 2%) Continuous Fiber
Metal MMCModerate(6 TO 16)
msi
High(10 TO150) ksi
High(20%)
Continuous andDiscontinuousFiber
Ceramic CMCHigh
(20 TO80)msi
High(20 to 80)
ksi
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
Dis
trib
uti
on
Surface Active Substance
WaterKerosene (Non-polar hydrocarbon)
Surface Chemistry Effects on Glass Fiber Strength
Diameter, µm
Str
eng
th L
oss
, % 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
Diff%
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
Ela
stic
Po
stac
tio
n
Relative Humidity, %
Elastic Post-Action
Time, min70
350
Def
orm
atio
n, m
m E-Glass
Soda-lime-silicate300
Elastic Post-Action vs Composition
Time, min.
580
400
70
Ela
stic
Po
st-a
ctio
n, µ
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
Mat
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
c
a
0.669 nm
0.3345 nm
0.142 nm
Crystal Structure of Graphite
Property Diamond(Cubic)
Graphite (Hexagonal)
a c
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
φ
Tensile axis1200
1000
800
600
400
200
0
Anlge between basal planes and tensile axis, φ
0 45 90
Effe
ctiv
e Y
oung
's m
odul
us, G
Pa
Young’s Modulus of Graphite
L a
cL
Idealized Graphite Fiber Structure
Fiber
axis
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
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
C C
H
H
HH
H Ethane
H
C C
H
HH
Ethylene
H
C C
H
HHN
Polyethylene
Olefin Chemistry
H
C C
H
H
H
C
C
H
H
H
C C
H
H
H
C C
H
H
C
C
N
N N
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
Hig
h s
tren
gth
car
bo
n f
iber
s
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
+ O
+ 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
CH2 CH
2
CH2
CH2 CH
2
CH CH
CH CH
CN C N
CN CN
+ O 2
CH C H CH
CH C H CH
O O O
CH CH
CH CH
CN CN
CN CN
+ H O2
PAN Conversion
C C C C C
C C C C C
C C C C C
C C C C C
C C C C C
C C C C C
H O HCN2 +
PAN Step II (Carbonization)
600
400
200 20
40
60
001200 1600 2000 2400 2800
Heat Treatment Temperature, C
Ten
sile
Mo
du
lus,
Msi
Ten
sile
Str
eng
th, K
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
FIBER
Mesophase Pitch Spinning
Precursor Yield(%)
Density E(GPa)
sTU
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
O
H
O O
H
O O O
- C - OH + - NH = - C - N + H O2
O O
Hcarboxolic acid amine amide water
H
Aramide Fibers
NH . . . O = C
Hydrogen Bond
C C C
C C
C C
N C N
O O
H HC C
C C
C 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
SHRED
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)
2
2
min
1
1 curvature
bendradius
2
f
fu
M cI
M EI
d ydx
ED
σ
ρ
ρ
ρ
ρσ
=
=
=
=
=
Minimum Bend Radius
Fiber s f
(GPa) D
(Microns)
E(GPa)
?min
(mm)
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
Mercury electrode
Tungsten filament
Reaction: 2BCl3+3H2→2B+6HCl
Boron Filament Processing
Tungsten Core
WB
W B2 5
Bulk amorphous boron
Skin
4WB
Boron Fiber Structure
FIBER
BORON
ST
RE
SS
_
+
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
µm
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
?
g/ccE
GPaE
msisTU
GPasTU
ksigf
%aR
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
0.20
Pich C P-100 10 2.15 690 100 2.20 325 0.3 -0.9
0.20
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= −
ff
c
vV
v=
mm
c
vV
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
(1 )
ff
c
mm
c
m f
wW
w
wW
w
W W
=
=
= −
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
fc m
c f m
ww wρ ρ ρ
= +
1c
f m
f m
W Wρ
ρ ρ
=+
Convert between volume fraction and weight fraction
f c
f f
V
Wρρ
=
Use the definition of weight fraction and express the weights in terms of volumes and densities
ff
c
f f ff f
c c c
wW
w
vW V
v
ρ ρρ ρ
=
= =
Voids in Composites
void actual theoreticalv v v= −
actual theoreticalvoid
actual
v vV
v−
=
Divide by actual composite volume gives:
Express volumes in terms of weights and densities
theoretical actualvoid
theoretical
Vρ ρ
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
P
P
c
c
L,1
T,2
S,3
MATRIX
FIBER
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
c c
AAA A
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=
f f
m m
E
E
σσ
=
f f
c c
E
E
σσ
=
Solving for the stress on the fiber
In terms of load
Consequences of ROM (Cont)
c ff
f f m m
E
E V E V
σσ =
+
f f f
m m m
P V EP V E
=
f
f m
f mc
m f
EP E
E VPE V
=+
and
1
10
100
1000
1 10 100
Ef/Em
Pf/P
m
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
Stre
ss
Fiber Fraction, Vf
0 1ε'fε'm
σfu
σmu σmu
σfu
fiber
matrix
Stre
ss
a) b)
Matrix contribution (1 )mu fVσ −
Fiber Contribution fu fVσ
Brittle Fiber Ductile Matrix With Different
Failure StrainsS
TR
ES
S
STRAIN
Vmin Vcrit0 1.0
FIBER FRACTION
σfu
σmu
εf
σm'
' (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'
'mu m
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 −
−−
= = =
49'49
49
3.60.02748
131Kevlarf
KevlarKevlarE
σε = = =
' 0.0310.04133
0.75epoxyf
epoxyepoxyE
σε = = =
49
49
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
WV
ρ
ρ−
−−
= = =
Volume fractions
4949
49
0.25(1.575)0.272
1.45Kevlar c
KevlarKevlar
WV
ρρ
= = =
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 =
GPa
GPa
Solution (page 4)The stresses when the Kevlar fails
GPa
The stresses when the Epoxy fails
When the epoxy fails
GPa
'1 1 49 57.67(0.02748) 1.585KevlarEσ ε= = ='1 2 49 22.37(0.02748) 0.615KevlarEσ ε= = = GPa
GPa
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
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
0 0.01 0.02 0.03 0.04 0.05 0.06
Strain
Str
ess,
GP
a
Solution (page 5)
Brittle matrix and ductile fiber with different failure strains
0 1.0
FIBER FRACTIONSTRAIN
ε'm
ST
RE
SS
Vtrans
σmu
fuσ
fσ '
matrix
fiber
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
'mu
transfu mu f
Vσ
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
2x'
x'
The force on the matrix, is transferred from the fiber by shear, , hence
md Vσ2 rdxπ
22 fm
Vd V rdx
rσ π τ
π= ⋅ ⋅
Model for calculating force on matrix
d x
r
FIBER
MATRIX
'
0 02muX
m
f
V rdx d
V
σσ
τ=∫ ∫
'
2m mu
f
V rX
Vσ
τ=
4x' 2x' <4x'
MA
TR
IX
ST
RE
SS
MA
TR
IX
STR
ES
S
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
STR
ESS
Slabing phenomenonin matrix
Slope = V f E
mu maxεε
f
Stress-strain diagram for brittle matrix-
ductile fiber composite
max 1 m mmu
f f
V EV E
ε ε
= − max 1
2m m
muf f
V EV E
ε ε
= −
2X’ slabs 4X’ slabs
Str
ess
Str
ess
Str
ess
Str
ess
Strain
Strain
Strain
Strain
Str
eng
thS
tren
gth
Str
eng
thS
tren
gth
0 Vf 1
0 Vf 1
0 Vf 1
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′−
=′+ −
minmu
fu mu f
Vσ
σ σ σ=
′+ −
minmu
fu mu f
Vσ
σ σ σ=
′+ −
minmu
fu mu f
Vσ
σ σ σ=
′+ −
fiber
fiber
fiber
fiber
matrix
mat
rix
mat
rix
matrix
Vmin
Vmin
Vmin
Vmin
minAbove : cu fu fV Vσ σ=
( )minBelow : 1cu mu f f fV V Vσ σ σ ′= − +
minAbove : cu fu fV Vσ σ=
( )minBelow : 1cu mu f f fV V Vσ σ σ ′= − +
minAbove : cu fu fV Vσ σ=
( )minBelow : 1cu mu f f fV V Vσ σ σ ′= − +
σf
σf
σf
σf
σmu
σmu
σmu
σmu
σ'm
σ'f
σ'f
σ'f
Stress-Strain Summary
0 1.0
FIBER FRACTIONSTRAIN
ε'm
STR
ES
S
σmu
fuσ
fuσ
fσ '
fσ '
matrix
fiber
Vmin
• 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
2X
X
Ductile fiber-brittle matrix
4X 2X <4X
MA
TR
IX
ST
RE
SS
MA
TR
IX
STR
ES
S
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
d x
r
FIBER
MATRIX
Force to Fracture Matrix
Force on matrix = total area for transfer x shear stress
22 fm
Vd V rdx
rσ π τ
π= ⋅ ⋅
Solving for dx in terms of dσ and integrating
0 02
2
muX m
f
m mu
f
V rdx dV
V rX V
σσ
τσ
τ
=
=
∫ ∫
Stress-Strain BehaviorCracking begins at εmu and ends at εmax when the last remaining slab splits
STRAIN
ST
RE
SS
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.
L,1
T,2
S,3
MATRIX
FIBER
Load
Slab Model
tftm/2 tm/2
matrix
reinforcement
Longitudinal
Transverse
Transverse Rule-of-Mixtures
c mf
c c m mf f
f mc mf
c c
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
( )( )
1
11
c mf
c mf
fc mf f
mT f
ff
mT f
P P P
V VE E E
VVE E E
σ σ σσσ σ
= == =
= + −
−= +
0 0.25 0.50 0.75 1.0
10
8
6
4
2
0
Fiber Volume Fraction , Vf
ET
/ E
m
Longitudinal(Parallel Connected)
Transverse(Serial Connected)
ROM Plot
( )1
1T
mmf f
f
EEE V VE
=+ −
Rearranging for plotting
1T
f m
mf
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
( )( )
( ) ( )( ) ( )
( ) ( )( ) ( )
21
2 22 1
2
2 2
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
ff
EK
ν=
− ( )2 1m
mm
EK
ν=
− ( )2 1f
ff
EG
ν=
+ ( )2 1m
mm
EG
ν=
+
where
The shear modulus
( ) ( )( )
( ) ( )( ) ( )
21
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
12 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
1
1fT
m f
VEE V
ξηη
+=
−
1f
m
f
m
EEEE
ηξ
−=
+( )2 a
bξ =
where
and
a
b
b
LOADLOAD
Cross Sectional Aspect Ratio
1.732ab
ξ =
Rectangular Cross Section Reinforcement
STRESSSTRESS
ba
For the case of then , hence and
STRESSSTRESS
a
b
0ξ = 2 0ab = 0a → b → ∞
Transverse Case 0ξ =
11
T
m f
EE Vη
=−
f m
f
E EE
η−
=
fT
m f m m f
EEE V E V E
=+
and
then
Results are identical to the serially connected constituent or slab model
Longitudinal Case
STRESSSTRESS b
a
ξ = ∞
For the case of and , then ξ = ∞a → ∞ 0b →
1
1fT
m f
VEE V
η
η
+ ∞=
−f m
m
E EE
η−
=∞
T f f m mE E V E V= +then
0.0
10.0
20.0
30.0
40.0
50.0
60.0
70.0
80.0
90.0
100.0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1Volume Fraction, V f
Yo
un
g's
Mo
du
lus
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
σσ
Tumu
S=
S Is the strength reduction factor
Assuming brittle fracture in the composite the transverse failure strain is
* TuT
TEσ
ε =
Estimates of Failure Strain
Analytical estimate:
ε επT mu
f m
f
V EE
*
/
= −
−
14
11 2
Empirical estimate:
( )ε εT mu fV* /= −1 1 3
Estimates of Strength
Analytical approach:
σ σπTu mu
T
m
f m
f
EE
V EE
= −
−
14
11 2/
Empirical approach:
( )1/31TTu mu f
m
EV
Eσ σ= −
Limitations to volume fraction
Square Array Close Packing
Random Regular Open
Packing Geometry
square array packing
max
2
2
40.785f
D
VD
π = =D
close packed array
Dmax
4
28 0.90734
f
D
VD
π
= =
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 =−=====
c
fm
c
ff w
wW
w
wW
89.1
)2.1(697.3107.1
)8.2(697.3595.2
11=
+=
+=
mc
m
fc
fth
ww
ww
ρρ
ρ
79.10.1636.1697.3
697.3
)(
=
−=
−= water
watercc
cact ww
wρρ
Acid Digestion Problem
Actual volumes
3.6972.065
1.79
2.595 1.0382.5
1.1020.918
1.2
cactual
actual
ff
f
mm
m
wv cc
wv cc
wv cc
ρ
ρ
ρ
= = =
= = =
= = =
1.0380.503
2.065
0.9180.444
2.065
1 1 0.503 0.444 0.053
ff
actual
mm
actual
v f m
vV
v
vV
v
V V V
= = =
= = =
= − − = − − =
Volume fractions
Acid Digestion Problem(page2)
Theoretical volume fractions
1.89
2.595 1.893.697 0.531
2.5
1 0.469
theoretical
f theoreticalf
f
m f
WV
V V
ρ
ρρ
=
= = =
= − =
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
Str
ess,
GP
a
0
1
2
Com
posi
te S
tren
gth,
GP
a
Rule of Mixtures Strength Problem(page2)
00325.0157
51.0
748.000761.0230
75.1
===′
=′=′===′
m
mum
mfff
fuf
E
EE
σε
εσσ
ε
σ’f
0.010
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
51.0
)1(
=−+
=′−+
=
=′+−
fmufu
mutransitionf
ffufffmu
VV
VVV
σσσσ
σσσ
b) What is the strength of the composite at this fiber fraction?
GPaVV
GPaGPaV
fffmu
transitionfucu
59.0)337(.748.0)337.1(51.0)1(
or
59.0)337.0(75.1
cu =+−=′+−=
===
σσσ
σσ
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
===
=
fu
cuf
ffucu
V
V
σσ
σσ
d) What fiber fraction is required to achieve a composite strength of 0.54 GPa?
126.051.0748.051.054.0
)1(cu
=−−
=−′−
=
′+−=
muf
mucuf
fffmu
V
VV
σσσσ
σσσ
Rule of Mixtures Strength Problem (page 5)
e) What is the composite strength at 20% fiber content?
GPa
VV fffmu
526.0)2.0(59.0)8.0(51.0
)1(
cu
cu
=+=
′+−=
σ
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
cf f
f m
g ccW W
ρ
ρ ρ
= = =− ++
Composite density
Fiber volume fraction
0.75(1.917)0.58
2.48f c
ff
WV
ρ
ρ= = =
Transverse StrengthProblem (page 2)
(1 )
(1 )
12
fT m
f
f
m
f
m
VE E
V
EEEE
ξη
η
η ξξ
+=
−
−= =
+
13.298TE GPa=
( )1 31 0.051
1.4
TTU mu f
mu
mu
TU
EV GPa
E
S
σ σ
σσ
= − =
= =
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
LS GS
GL
G
Sγγ
γ
θ
o
cos0 complete wetting
180 unwetting
GS LS LGγ γ γ θθ
θ
= +→
→
! 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
Fiber
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
x nd
md
Fibers unbonded
- Stiffness in bending is the moment of inertiaof each fiber x Ef x the number of fibers
Direct Effect of Interfacial Bonding
4
64U f
dD E mn=
π
Beam Stiffness with Bonded Fibers
/2 2
/ 2
2 2
/ 2 2
/2
/2 2
/ 2
3 4
4 41
14 4
14 4 12
nd
B c nd
nd
B f m nd
nd
B f m nd
B f m
D E X dA
d dnm nm
D E E X dAmd nd md nd
D E E X dA
mn dD E E
π π
π π
π π
−
−
−
=
= + − ⋅ ⋅ = + − = + −
∫
∫
∫
For the graphite fiber (P100)/epoxy matrix system:
Stiffness Ratio for Bonded & Unbonded Fibers
( )
6
6m
3 4
B
3 4
3 4
4
2
100 10 (P100 graphite fiber)
E 0.2 10 (epoxy)
D 0.000434 12
4 12
6448
43
f
f f
B f
fB
U f
B
U
E x
x
mn dE E
mn dD E
E mn dDD E mn d
D nD
π
π
π
π
=
=
= + =
= ⋅
=
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
( )
( )
* *
*
* *
**
1
Volume fraction of unbonded fibers
1
1
c f f m f f f
f
c f f m f
c c f f
fcf
c c
E E V E V E V
V
E E V E V
E E E V
EEV
E E
= + − −
=
= + −
= −
= −
**
**
1
1
fcf
c f f
fc
c f
EEV
E E V
VEE V
= −
= −
0.050.150.200.40
0 0.2 0.4 0.6 0.8
1.0
0.8
0.6
0.4
0.2
0
*
V f
V f
*E c
E c
Contribution of Unbonded Fibers
Interfacial Failure
Equilibrium forces on element of beam
dx
L dx
yM MB A
V
q
B A
c
f f
c c
ff
c
Vdx M MMyIE
EE MyE I
σ
σ
σ
σ
= −
=
=
= ⋅
Equilibrium Forces on Fiber
dx
y τ
PAPB
2fB fBP P rdxπ τ− = ⋅
Divide by cross section area of fiber
2 2
2fB fBP P rdxr r
π τπ π
− ⋅=
Gives stress on fiber
2f
dxr
τσ =
Stress on fiber in terms of bending moment
2f
c
E My dxE I r
τ=
usingM
dxV
=
Equilibrium Forces on Fiber(cont)
2f
c
E My ME I r V
τ=
Solving for τ
2f
c
VE yrE I
τ =
Dividing both sides byfσ
2f
c
ff
c
VE yrE I
E MyE I
τσ
=
or
2f
V rM
τσ
=
Ratio of Interfacial to Fiber stress
L
P
V=P/2
M=PL/4
f
f
f f
c c
fc
c
rL
rL
E
E
ErL E
τσ
τ σ
σσ
τ σ
=
=
=
=
Failure Under Compressive Loads
! Microbuckling! Transverse splitting - Poisson tensile strain! Shear failure
Microbuckling
Extensional Mode Shear Mode
( )
1 /2
23 1
f m fLU f
f
V E EV
Vσ
′ =
− 1
mLU
f
GV
σ ′ =−
Transverse Splitting
SPLITTING
( )( )1/31
T L LT
LU LTT
L
L TULU
LT
f f m m uLU
f f m m
E
E
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
0.100
1.000
10.000
100.000
0.0 0.2 0.4 0.6 0.8 1.0
Fiber Volume Fraction
Co
mp
ress
ive
Str
eng
th, m
psi
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
fiber
matrix
STRAIN
STR
ES
S τi
Elongation in Model System
All forces on fiber are transmitted from the matrix through fiber surface
LOAD LOAD
Stress on Fiber
2
rdx
CENTER-LINE
l
fσ σ+ d
ffσ
Sum forces in element dx
Forces consist of •End forces σfAc•Shear forces τiAs
2 2 2
/2
0
2
2
2fmax
fo
f i f f
i f
li
f
r rdx r d r
dx d r
d dxr
σ
σ
σ π τ π σ π σ π
τ σ
τσ
+ = +
=
=∫ ∫
Stress on Fiber (Cont.)
σ
σ
=
=max
0
largest stress developed in fiber
stress due to loading on free cross-sectionf
f
Evaluating the integral
maxi
f fol
rτ
σ σ= +
Stress due to loading on free cross-section can be neglected
max2 i
fl
dτ
σ =
Stress on Fiber ProfileIntegrating from fiber center to other end
/ 2
2fo
fmax
li
f ld dx
r
σ
σ
τσ =∫ ∫
Gives stress profile in fiber
i
LENGTH ALONG FIBER
SH
EA
R S
TR
ES
S
τ
l/ 2
FIB
ER
STR
ES
S
σmaxf
L
Ineffective LengthDiscontinuous but long fibers
Composite is loaded to cσ Where c cuσ σ<
maxf
f f cc
E
Eσ σ σ= =
LENGTH ALONG FIBER, X
2tl
l
FIB
ER
STR
ES
S
maxfσ
lt is called the ineffective length
f ct
c i
Er
Eστ
=l
FIB
ER
ST
RE
SS
l
LENGTH ALONG FIBER
fσmaxfσ
l t
If the composite contains fiber lengths , less than then the fibers cannot be loaded to their maximum potential hence
l tlmaxf fσ σ<
ff c
c
E
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
cuσ
cl
ffu cu
c
E
Eσ σ=
2fu
ci
dσ
τ=l
LENGTH ALONG FIBER
t
c
σc
== cu
=fσ
<
EE
f
c
EE
f
c
EE
f
cfσ σc
fσfu
σ σ
l
l
l
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=
z
θ
Matrix Displacement Around Fiber
w
DISTANCE FROM CENTER, zDIS
PL
AC
EM
EN
T, u
ufuR
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 =
d x
Rz
2r
τ
τeτ
eτ
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=
e
rz
τ τ=
e
m
rdudz zG
τ=
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
f
u Re
u rm
r dzdu
G zτ
=∫ ∫
lneR f
m
r Ru u
G rτ
− =
1ln ln
2 f
Rr V
π =
1 2ln ln
2 3 f
Rr V
π =
For a general arrangement
1ln ln
2f
f
PRr V
=
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
fm
f
E u u
Pr
V
τ
ν
−=
+
Calculating Stress Transfer
2f ed
dx r
σ τ= −
2
2 ( )
(1 ) ln
f m R f
fm
f
d E u udx P
rV
σ
ν
−= −
+
Calculating Stress Transfer (Cont)
Differentiating
( )2 2
12 2f
f f
d nE
dx r
σσ ε= −
( )
2 2
1 ln
m
ff m
f
EnP
EV
ν
=
+
1
cosh1
cosh2
f f
nxrE
nr
σ ε
= −
l1
1sinh
2
cosh2
f
e
nxnE
rnr
ετ
=
l
where
Fiber stress: Shear stress:
Shear Stress and Fiber Stress Distribution Along Fiber
DISTANCE ALONG FIBER
SH
EA
R S
TR
ES
S
l
l2
FIB
ER
ST
RE
SS
DISTANCE ALONG FIBER DISTANCE ALONG FIBER DISTANCE ALONG FIBER
FIB
ER
ST
RE
SS
FIB
ER
ST
RE
SS
FIB
ER
ST
RE
SS
STRAIN STRAIN STRAIN
ST
RE
SS
ST
RE
SS
ST
RE
SS
RIGID PERFECTLYPLASTIC MATRIX
ELASTIC FIBER
ELASTIC MATRIX
ELASTIC FIBER
ELASTIC-PLASTIC MATRIX
ELASTIC FIBER
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
2
2
if
f if
ld
ld
τσ
σ τσ
=
= =
l
FIB
ER
ST
RE
SS
fσ
Fiber Length Equal Ineffective Length
tl l=
max
2 2f c f
fc
E
E
σ σσ = =
FIB
ER
STR
ES
S σcEE
f
c
l t
Fiber Length > Ineffective Length
FIB
ER
ST
RE
SS
l
l2
t
maxfσ
( )
0
0
max max
max
2
12
l
f
f l
tf f t
f
tf f
dx
dx
ll l
lll
σσ
σ σσ
σ σ
=
+ − =
= −
∫∫
tl l>
Fiber Length >> Ineffective Length
50 tl l=
max
max
max max
12
1100
0.99
tf f
tf f
f f f
ll
ll
σ σ
σ σ
σ σ σ
= − = −
= ≈
Composite Strength with Average Fiber Stress
cl l< ( )1icu f mu f
lV V
dτ
σ σ= + −
cl l= ( )11
2cu fu f m fV Vσ σ σ ′= + −
cl l> ( )1 12
ccu fu f m f
lV V
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
1
tanh1f f
nrE
nr
σ ε
= −
l
l
Comparison of Elastic –Perfectly Plastic and Elastic-Elastic Cases
FIB
ER
ST
RE
SS
DISTANCE ALONG FIBER
maxfσ
( )max 1 1 tf f
lq
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
12
q ≥
ROM Strength Diagram
1.00
CO
MP
OS
ITE
ST
RE
NG
H
+ 1-( )VfV
VfV
1-( )VfV
1-( )VfV
VfV
VfV
Vmin Vcrit
σf
σf
σf
σ 'm
σ 'm
σ 'm
σmu
σmu
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=
11
2
1
2
T T fT
m T f
T
f
mT
f
m
VEE V
EEEE
ξ ηη
ξ
η
+=
−
=
−=
+
LOADLOAD d
l
Young’s Modulus of Short Fiber Composites
Longitudinal Modulus
1
1
2
1
2
L L fL
m L f
L
f
mL
f
m
VEE V
dE
EE
E d
ξ ηη
ξ
η
+=
−
=
−=
+
l
l
Empirical Values for ξ
102 40 f
aV
bξ = +
For EL102 40 f
lV
dξ = +
For ET
1.7321040 f
aV
bξ = +
For GLT
Correction for volume fraction becomes important for Vf>0.7
Random Orientation
5 38 8
2 18 8
12
random T L
random T L
randomrandom
random
E E E
G E E
EG
ν
= +
= +
= −
0
10
20
30S
TR
ES
S, k
si
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
80
60
40
20
00.2 0.60.4 0.8 1.00
ACTUAL MATRIX A
ACTUAL MATRIX B
ROM MATRIX AROM MATRIX B
ST
RE
NG
TH
, ksi
FIBER VOLUME FRACTION
Fracture Behavior of Short Fiber Composites
CRACK OPENING DISPLACEMENT
LOA
D0.29 V
f
EPOXY
Depends on:
Vf
Fiber Properties
Interfacial Properties
Strain Rate
Time
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
Y L,1
X
T,2
X
L,1
YT,2
θ
θ
Deformation in Orthotropic Lamina
Orthotropic lamina loaded in principal directions or isotropic lamina loaded in any direction
T
L
PP
a) Load in principal direction, L b) Shear load in LT direction
Deformation in Orthotropic Lamina
Orthotropic lamina loaded in an arbitrary direction
TL
P P
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=
x 3
x 2
x 1
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
E E
E E
E E
=
=
= −
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
CC
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
00
0 0
Q QQ Q
Q
σ εσ ετ γ
=
213
11 1133
CQ C
C= −
,
13 2312 12
33
C CQ C
C= −
,
223
22 2233
CQ C
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
00
0 0
S SS S
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
6666
1
SQS S S
SQS S S
SQ
S S S
QS
=−
=−
= −−
=
Inverting [S]
2211 2
11 22 12
1122 2
11 22 12
1212 2
11 22 12
6666
1
QS
Q Q QQ
SQ Q Q
QS
Q Q Q
SQ
=−
=−
= −−
=
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
00
0 0
S SS S
S
S SS SS
ε σε σγ τ
ε σ σε σ σγ τ
=
= += +=
expanded
Uniaxial Tensile Tests
2 12
1 11 1
0 0
S
σ τ
ε σ
= =
=
111
1 1
1S
Eεσ
= =1
1212
1
2
1122
ES
S
νσε
σε
−==
=
1 12
2 22 2
0 0
S
σ τ
ε σ
= =
=
222
2 2
1S
Eεσ
= =2
2112
2
1
2121
ES
S
νσε
σε
−==
=
2
1
21
12
EE
=νν
Shear Test
1266
12
12
126612
21
1
00
GS
S
==
=
==
τγ
τγ
σσ
Compliance And Stiffness In Terms Of Engineering Constants
111
222
12 2112
1 2
1112
1
1
1
SE
SE
SE E
SG
ν ν
=
=
= − = −
=
111
12 21
222
12 21
21 1 12 212
12 21 12 21
66 12
1
1
1 1
EQ
EQ
E EQ
Q G
ν ν
ν νν ν
ν ν ν ν
=−
=−
= =− −
=
12 1
21 2
EE
νν
=
Stress-strain Relation For Lamina in Any Direction
L
X
X
L
θ
σ
σ
[ ]1
2
12
x
y
xy
Tσ σσ στ τ
=
[ ]1
2
12
1 12 2
x
y
xy
Tε εε ε
γ γ
=
[ ]( )
2 2
2 2
2 2
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
[ ]( )
2 2
1 2 2
2 2
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
[ ]1
12
12
x
y
xy
Tσ σσ στ τ
−
=
Transforming strain from principal to arbitrary direction
[ ]1
12
12
1 12 2
x
y
xy
Tε εε ε
γ γ
−
=
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
[ ] 1
12
x x
y y
xyxy
Rε εε ε
γγ
−
=
[ ] 11 0 00 1 0
10 0 2
R−
=
Step 1
Transform Tensorial Strain in the Arbitrary Direction to Tensorial Strain
in the Principal Material Direction.
[ ]1
2
121 12 2
x
y
xy
Tε εε ε
γ γ
=
Step 2
Convert Tensorial Strain in the Principal Material Direction Back to the
Engineering Strain
[ ]1 1
2 2
12 121
2
Rε εε εγ γ
=
[ ]1 1
2 2
12 121
2
R
ε ε
ε εγ γ
=
Step 3
Find the Stresses in the Principal Material Directions
Step 4
[ ]1 1
2 2
12 12
Qσ εσ ετ γ
=
Transform Back to the Arbitrary Direction
[ ]1
12
12
x
y
xy
Tσ σσ στ τ
−
=
Step 5
All Operations in One Equation
[ ] [ ][ ][ ][ ]1 1x x
y y
xy xy
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
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 Q Q Q Q Q
Q Q Q Q Q
θ θ θ θ
θ θ θ θ
θ θ θ θ
θ θ 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 Q Q
θ θ
θ θ θ θ
+
= + − − + +
Transformed Compliance Matrix
[ ][ ] [ ] [ ][ ]1 1x x
y y
xy xy
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
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 S S S S S
S S S S S
θ θ θ θ
θ θ θ θ
θ θ θ θ
θ θ 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 S S
θ θ
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
Q QU
Q Q Q QU
Q Q Q QU
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
cos2 cos4
cos41
sin 2 sin421
sin2 sin 42
cos4
Q U U U
Q U U U
Q U U
Q U U
Q U U
Q U U
θ θ
θ θ
θ
θ θ
θ θ
θ
= + +
= − +
= −
= − −
= − +
= −
θ
=
Q 11U U cos 2 U cos 41 2 3
++
θ θ θ
θ θ
Graphical Representation of InvarientComponents
Transformation Of Engineering Constants
E
E
G12
1
ν12
21ν
2
( )
2 21 2 12
2 21 2 12
2 21 2 12
cos sin cos sin
sin cos cos sin
2 cos sin 2 cos sin cos sin
x
y
xy
ε ε θ ε θ γ θ θ
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
1 1
2 21 12
2 2
1212
12
E E
E E
G
σ ν σε
σ ν σε
τγ
= −
= −
=
21
22
12
cos
sin
cos sin
x
x
x
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
1
cos sin 1 1 2sin 2
4
xx
x
E
E E G E
σε θ θ ν
θ
= =
+ + −
evaluated at 90θ +
4 4212
2 1 12 1
1
cos sin 1 1 2 sin 24
yy
y
E
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
xy xx
EE E E G
ε νν θ θ θ
ε
= − = − − + −
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
1
x xxy
mEσ
γ = −
The stress can also produce shear strain , and shear stress can produce extensional strain , these can be related though a the cross coefficient
1
x xyx
m
E
τε = −
also
1
y yxy
m
E
σγ = −
1
y xyy
mEτ
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
1
1
1 1
1
1
1
xy x
x xx x
xy yy y
x yxy xy
yx
xy
mE E E
m
E E E
mmE E G
ν
ε σν
ε σγ τ
− −
= − − − −
Off-axis Young’s Modulus and Shear Modulus for Carbon/epoxy
Lamina
0
2
4
6
8
10
12
14
16
0 10 20 30 40 50 60 70 80 90
θ
En
gin
eeri
ng
Ela
stic
Co
nst
ants
, Mp
si
Ex
Ey
GLT
-0.5
0
0.5
1
1.5
2
0 10 20 30 40 50 60 70 80 90
θ
Poisson's Ratiomxmy
Off-axis Poisson’s Ratio and Cross Coefficients for Carbon/epoxy Lamina
Restrictions On Elastic Constantsσε τγ
2
0E
σ>
2
0Gτ
>
2(1 )E
Gν
=+
1ν > −
Thermodynamic considerations require that Young’s modulus and shear modulus are positive values. The work of deformation, and are positive then
and
For isotropic materials the relation between the Young’s modulus and shear modulus is
For G and E to be positive .
φ
x y zPK
φ ε ε ε= + + =
3(1 2 )E
Kν
=−
12
ν <
11
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
and
Relation Between The Elastic Constants For Orthotropic Materials
11 22 33 44 55 66, , , , , , 0C C C C C C >
111
12 211E
Qν ν
=−
12 211 0ν ν− > 12 21 1ν ν <
112
2
EE
ν>
221
1
EE
ν>
Strength Of Orthotropic Lamina
For isotropic metals failure usually occurs by yielding and can be simply predicted by the maximum shear stress theory.
2I II
yieldσ σ τ− ≥ or I II yieldσ σ σ− ≥
σIσII
τy
Strength Notation For Orthotropic Lamina
Y Y'
X
X'
S
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
1
2
12
X XY Y
S S
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
12
xLoad Direction
θ
21
22
12
cos
sincos sin
x
x
x
σ σ θ
σ σ θτ σ θ θ
=
== −
Maximum Stress for Fibers in Arbitrary Direction
2 2
2 2
cos cos
sin sin
cos sin cos sin
x
x
x
X X
Y Y
S S
σθ θ
σθ θ
σθ θ θ θ
′− < <
′− < <
′− < <
0 45 90
θ
σ x
X
cos2
X'
cos2θ
θ
S
cos sin
Y' Y
2sin
2sin
θ θ
θ θ
Maximum Stress Theory Of Lamina Failure Plotted For Off-axis Loading
Maximum Strain Theory
11 1
22 2
1212 12
X XE EY Y
E ES S
G G
ε
ε
γ
′−< <
′−< <
′−< <
1 12 21
1 1
2 21 12
2 2
1212
12
E E
E E
G
σ ν σε
σ ν σε
τγ
= −
= −
=
−
−
=
12
2
1
12
22
21
1
12
1
12
2
1
100
01
01
τσσ
ν
ν
γεε
G
EE
EE
Hooke’s Law
Transforming To Arbitrary Off-axis Directions
( )
( )
( )
2 21 12
1
2 22 21
2
1212
cos sin
sin cos
cos sin
x
x
x
E
E
G
σε θ ν θ
σε θ ν θ
σγ θ θ
= −
= −
= −
2 2 2 212 12
2 2 2 212 12
cos sin cos sin
sin cos sin cos
cos sin cos sin
x
x
x
X X
Y Y
S S
σθ ν θ θ ν θ
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 =
and
211 1F X =
2 Yσ =then 1 0σ = 6 0σ =
and
222 1F Y =
6 Sσ = 1 2 0σ σ= =and
266 1F S =
apply a balanced biaxial stress, and
2122 1F X = −
Strength Coefficients
11 2
22 2
66 2
12 2
1
1
1
12
FX
FY
FS
FX
=
=
=
= −
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
X Y S X
σθ θ θ θ
= + + −
the failure stress, for off-axis loadingxσ
0
10000
20000
30000
40000
50000
60000
70000
80000
90000
0 10 20 30 40 50 60 70 80 90
Fiber Orientation, θ
Str
ess,
σx
(psi
)
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
1
1 1F
X X= −
′
11
1F
XX=
′
21 1
FY Y
= −′
22
1F
YY=
′
66 2
1F
S=
Additional Coefficients
212 2
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
σΙ
σUTS
σUTS
σUTS
σUTS
Failure Envelopes for Isotropic Ductile Materials
Von Mises-Hencky
Tresca
σΙ
σΙΙ
Failure Envelope For Orthotopic Material By The Maximum Stress Theory
σ1
σ2
Y
Y'
XX'
Failure Envelope For Orthotopic Material By The Maximum Strain Theory
σ1
σ2
Y
Y'
XX'
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+ + + + + − =
σ1
σ2
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
2
2
H F Q F Q Q F Q
H F Q F Q Q F Q
H F Q Q F Q Q Q F Q Q
H F Q
H F Q F QH FQ F Q
= + +
= + +
= + + +
=
= += +
where
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+ + + + + − =
1ε
ε2
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+ =
2 42
b b acR
a+− + −
=
Solved Roots
2 42
b b acR
a−− − −
=
1
ij i j
i i
a F
b F
C
σ σ
σ
=
== −
In stress space
In strain space
1
ij i j
i i
a H
b H
C
ε ε
ε
=
== −
1Fij i jσ σ =
12X
12Y
2 2 2 2
1 4 1 1 1
2 U X Y S− − −
12S
1F Fi i ij i jσ σ σ+ =
1 1
X X−
′
1 1
Y Y−
′
1
XX ′
1
YY ′
111 22 2F F
S−
12S
1Fij i jσ σ =
12X
12Y 2
K
XX−
′
12S
1F Fi i ij i jσ σ σ+ =
1 1
X X−
′
1 1
Y Y−
′
1
XX ′
1
YY ′
1
2XX−
′
12S
1F Fi i ij i jσ σ σ+ =
1 1
X X−
′
1 1
Y Y−
′
1
XX ′
1
YY ′
( )122
XX S X X X X Y Y
XX S XX
′ ′ ′− − − +−
′ ′
12S
1F Fi i ij i jσ σ σ+ =
1 1
X X−
′
1 1
Y Y−
′
1
XX ′
1
YY ′
1
2XY−
12S
1Fij i jσ σ =
12X
12Y
122 X
−
12S
1F Fi i ij i jσ σ σ+ =
1 1
X X−
′
1 1
Y Y−
′
1
XX ′
1
YY ′
*12 11 22F F F
12S
Tsai-Wu
Tsai-Hill
Marin
Malmeister
Hoffman
Fischer
Cowin
Ashkenazi
F66F12F22F11F2F1TensorTheory
Stress Interaction Strength Criteria