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MSE 406: Thermal and Mechanical Behavior of Materials ©D.D. Johnson 2005

CompositesMany engineering components are composites

MSE 406: Thermal and Mechanical Behavior of Materials ©D.D. Johnson 2005

Composites

ISSUES TO ADDRESS...

• What are the classes and types of composites?

• Why are composites used instead of metals, ceramics, or polymers?

• How do we estimate composite stiffness & strength?

• What are some typical applications?

MSE 406: Thermal and Mechanical Behavior of Materials ©D.D. Johnson 2005

Classification of Composites• Composites: - Multiphase material w/significant proportions of ea. phase.• Matrix: - The continuous phase - Purpose is to: transfer stress to other phases protect phases from environment - Classification: MMC, CMC, PMC

• Dispersed phase: -Purpose: enhance matrix properties. MMC: increase σy, TS, creep resist. CMC: increase Kc PMC: increase E, σy, TS, creep resist. -Classification: Particle, fiber, structural

metal ceramic polymer

From D. Hull and T.W. Clyne, An Intro toComposite Materials, 2nd ed., CambridgeUniversity Press, New York, 1996, Fig. 3.6, p. 47.

MSE 406: Thermal and Mechanical Behavior of Materials ©D.D. Johnson 2005

Particle-reinforced

• Examples:

Adapted from Fig.10.10, Callister 6e.

Adapted from Fig.16.4, Callister 6e.

Adapted from Fig.16.5, Callister 6e.

COMPOSITE SURVEY: Particle-I

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MSE 406: Thermal and Mechanical Behavior of Materials ©D.D. Johnson 2005

• Elastic modulus, Ec, of composites: -- two approaches.

• Application to other properties: -- Electrical conductivity, σe: Replace E by σe. -- Thermal conductivity, k: Replace E by k.

Particle-reinforced

From Fig. 16.3,Callister 6e.

COMPOSITE SURVEY: Particle-II

MSE 406: Thermal and Mechanical Behavior of Materials ©D.D. Johnson 2005

• Aligned Continuous fibers

Fiber-reinforcedParticle-reinforced Structural

• Ex:

From W. Funk and E. Blank, “Creep deformationof Ni3Al-Mo in-situ composites", Metall. Trans. AVol. 19(4), pp. 987-998, 1988.

--Metal: γ'(Ni3Al)-α(Mo) by eutectic solidification.

--Glass w/SiC fibers formed by glass slurry Eglass = 76GPa; ESiC = 400GPa.

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

(a)

(b)

COMPOSITE SURVEY: Fiber-I

MSE 406: Thermal and Mechanical Behavior of Materials ©D.D. Johnson 2005

• Discontinuous, random 2D fibers

Fiber-reinforcedParticle-reinforced Structural

• Example: Carbon-Carbon --process: fiber/pitch, then burn out at up to 2500C. --uses: disk brakes, gas turbine exhaust flaps, nose cones.

• Other variations: --Discontinuous, random 3D --Discontinuous, 1D

fibers liein plane

view onto plane

C fibers:very stiffvery strongC matrix:less stiffless strong

(b)

(a)

COMPOSITE SURVEY: Fiber-II

MSE 406: Thermal and Mechanical Behavior of Materials ©D.D. Johnson 2005

Chapter 6: Elasticity of Composites

Stress-strain response depends on properties of• reinforcing and matrix materials (carbon, polymer, metal, ceramic)• volume fractions of reinforcing and matrix materials• orientation of fibre reinforcement (golf club, kevlar jacket)• size and dispersion of particle reinforcement (concrete)• absolute length of fibres, etc.

concentration size shape

distribution orientation

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MSE 406: Thermal and Mechanical Behavior of Materials ©D.D. Johnson 2005

Families of Composites: particle, fibre, structural reinforcements

Orientation dependence

Twisting,Bending

ceramics

MSE 406: Thermal and Mechanical Behavior of Materials ©D.D. Johnson 2005

Isostrain: Load & Reinforcements Aligned

Isoload: Load & Reinforcements Perpendicular(Isostress below)

Two simplest cases: Iso-load and Iso-strain

Strain or elongation of matrixand fibres are the same!

Load (Stress) across matrixand fibres is the same!

F

F

F

FEc= Eαα=1,N∑ ˜ V α

1E

c

=α=1,N∑

˜ V α

Eα

˜ V α= Vα

VTot

Volume fraction

MSE 406: Thermal and Mechanical Behavior of Materials ©D.D. Johnson 2005

Iso-strain Case in Ideal Composites

Isostrain Case:

Fc=Fm+Fr

εc=εm=εrstrain

forces

F

F

Composite Property: Pc= Pαα=1,N∑ ˜ V α

Properties include: elastic moduli, density, heat capacity, thermal expansion, specific heat, ...

*like law of mixtures

Load is distributed over matrix and fibers, so σcAc = σmAm + σfAf.

€

σ c =σm(Am /Ac)+σ f(Af /Ac)or σ c =σm

˜ V m +σ f˜ V f

*if the fibers are continuous,then volume fraction is easy.

For Elastic case:

€

σ c =εcEc =εmEm˜ V m +ε f Ef

˜ V f =εc(Em˜ V m +Ef

˜ V f )

MSE 406: Thermal and Mechanical Behavior of Materials ©D.D. Johnson 2005

Composite Property: Pc= Pαα=1,N∑ ˜ V α

density,

heat capacity,

thermal expansion,

*like law of mixtures

For Elastic case:

€

σ c =εcEc =εmEm˜ V m +ε f Ef

˜ V f =εc(Em˜ V m +Ef

˜ V f )

Consider Density, Heat Capacity, and Thermal Expansion

€

ρc = ραα = 1,N∑ ˜ V αN = 2

→ ρ1˜ V 1+ ρ2 ˜ V 2

€

Cc =C1 ρ1

˜ V 1+ C2 ρ2 ˜ V 2ρ1˜ V 1+ ρ2 ˜ V 2

€

αc =α1E1

˜ V 1+α2E2 ˜ V 2E1˜ V 1+ E2 ˜ V 2

How?

€

εc =σcEc

→αc =dεcdT

=d

dTσcEc

=

1Ec

dσcdT

=(dε1/dT )E1

˜ V 1+ (dε2 /dT )E2 ˜ V 2Ec

=α1E1

˜ V 1+α2E2 ˜ V 2E1

˜ V 1+ E2 ˜ V 2Need to assess the proper dependence of the properity to get Rule-of-Mixture correct.

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MSE 406: Thermal and Mechanical Behavior of Materials ©D.D. Johnson 2005

Iso-Load Case for Ideal CompositesIsoload Case:

Fc=Fm=Fr

εc=εm+εrstrain

forces

Composite Property1P

c

=α=1,N∑

˜ V α

Pα

Properties include: elastic moduli, density, heat capacity, thermal expansion, specific heat, ...

*like resistors in parallel.

Without de-bonding, loads are equal, therefore, strains must add, so

€

εc =εm˜ V m +ε f

˜ V f = σ

Em

˜ V m + σ

E f

˜ V f *if the fibers are continuousor planar, then area ofapplied stress is the same.elastic case

MSE 406: Thermal and Mechanical Behavior of Materials ©D.D. Johnson 2005

ISOSTRAIN Example

Suppose a polymer matrix (E= 2.5 GPa) has 33% fibrereinforcements of glass (E = 76 GPa).

What is Elastic Modulus?

Ec= ˜ V mEm+ ˜ V fEf =(1− ˜ V f)Em+ ˜ V fEf ≈˜ V fEf

= 26.7 GPA

* Stiffness of composite under isostrain is dominated by fibres.

~ 25 GPA

MSE 406: Thermal and Mechanical Behavior of Materials ©D.D. Johnson 2005

ISOLOAD Example

Suppose a polymer matrix (E= 2.5 GPa) has 33% fibrereinforcements of glass (E = 76 GPa).

What is Elastic Modulus? 1Ec

=˜ V mEm

+˜ V

fEf

EC=EmE

f˜ V f Em +(1− ˜ V f )Ef

≈ Em

(1− ˜ V f )

Rearrange:= 3.8 GPA

* Elastic modulus of composite under isoload conditionStrongly depends on stiffness of matrix, unlike isostraincase where stiffness dominates from fibres.

MSE 406: Thermal and Mechanical Behavior of Materials ©D.D. Johnson 2005

isostrain

isoload

•Particle reinforcements usually fall in between two extremes.

Modulus of Elasticity in Tungsten Particle Reinforced Copper

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MSE 406: Thermal and Mechanical Behavior of Materials ©D.D. Johnson 2005

Simplified Examples of Composites

Are these isostrain or isoload?

What are some real life examples?

MSE 406: Thermal and Mechanical Behavior of Materials ©D.D. Johnson 2005

• Critical fiber length for effective stiffening & strengthening:

Fiber-reinforcedParticle-reinforced Structural

fiber length > 15

σfdτc

fiber diameter

shear strength offiber-matrix interface

fiber strength in tension

• Ex: For fiberglass, fiber length > 15mm needed• Why? Longer fibers carry stress more efficiently!

fiber length > 15

σ fd

τc

Shorter, thicker fiber:

fiber length < 15

σ fd

τc

Longer, thinner fiber:

Poorer fiber efficiency Better fiber efficiency

Adapted from Fig.16.7, Callister 6e.

COMPOSITE SURVEY: Fiber-III

MSE 406: Thermal and Mechanical Behavior of Materials ©D.D. Johnson 2005

• Estimate of Ec and TS: --valid when

-- Elastic modulus in fiber direction:

--TS in fiber direction:

efficiency factor:--aligned 1D: K = 1 (anisotropic)--random 2D: K = 3/8 (2D isotropy)--random 3D: K = 1/5 (3D isotropy)

Fiber-reinforcedParticle-reinforced Structural

fiber length > 15

σ fd

τc

Ec = EmVm +KEfVf

(TS)c = (TS)mVm + (TS)f Vf (aligned 1D)

Values from Table 16.3, Callister 6e.

COMPOSITE SURVEY: Fiber-IV

MSE 406: Thermal and Mechanical Behavior of Materials ©D.D. Johnson 2005

Structural

• Stacked and bonded fiber-reinforced sheets -- stacking sequence: e.g., 0/90 -- benefit: balanced, in-plane stiffness

• Sandwich panels -- low density, honeycomb core -- benefit: small weight, large bending stiffness

Adapted fromFig. 16.16,Callister 6e.

Adapted from Fig. 16.17,Callister 6e.

COMPOSITE SURVEY: Structural

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MSE 406: Thermal and Mechanical Behavior of Materials ©D.D. Johnson 2005

• CMCs: Increased toughness • PMCs: Increased E/ρ

• MMCs: Increased creep resistance

Adapted from T.G. Nieh, "Creep ruptureof a silicon-carbide reinforcedaluminum composite", Metall. Trans. AVol. 15(1), pp. 139-146, 1984.

Composite Benefits

MSE 406: Thermal and Mechanical Behavior of Materials ©D.D. Johnson 2005

Laminate Composite (Ideal) Example

Gluing together these composite layerscomposed of epoxy matrix (Em= 5 GPa)with graphite fibres (Ef= 490 GPa andVf = 0.3). Central layer is oriented 900

from other two layers.

Case I - Load is applied parallel to fibres in outer two sheets.Case II - Load is applied parallel to fibres of central sheet.

What are effective elastic moduli in the two case?• First need to know how individual sheets respond, then average.1E⊥

= 0.3490GPa+

0.75GPa→E⊥ =7.1GPa For isoload case.

E||=0.3(490GPa)+0.7(5GPa)→E||=150.5GPa For isotrain case.

Case I: Elam=(2/3)(150.5 GPa) + (1/3)(7.1 GPa) = 102.7 GPa

Case II: Elam=(1/3)(150.5 GPa) + (2/3)(7.1 GPa) = 54.9 GPa

MSE 406: Thermal and Mechanical Behavior of Materials ©D.D. Johnson 2005

Mechanical Response of Laminate is Complex and NOT Ideal

3 Conditions required: consider top and bottom before laminated• strain compatibility- top and bottom must have same strain when glued.• stress-strain relations - need Hooke’s Law and Poisson effect.• equilibrium - forces and torques, or twisting and bending.

Isostrain for loadalong x-dir:

Poisson Effect andDisplacements in Δ:

• When glued together displacements have to be same! • Unequal displacements not allowed!So, top gets wider (σy

top > 0) and bottom gets narrower (σybott < 0).

Equilibrium: Fy = 0 = (σybot tbot + σy

top ttop)L. (t = thickness)

σxtop=EtopEbott

σxbott

Δytop =EtopEbott

Δybott

MSE 406: Thermal and Mechanical Behavior of Materials ©D.D. Johnson 2005

COMPATIBILITY: When glued, displacements have to be same!

As stress is applied, compatibility can be maintained, depending on thelaminate, only if materials twists.

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MSE 406: Thermal and Mechanical Behavior of Materials ©D.D. Johnson 2005

Symmetry of laminate composite dictates properties

Elastic constants are different for different symmetry laminates.

MSE 406: Thermal and Mechanical Behavior of Materials ©D.D. Johnson 2005

Orientation of layers dictates response to stresses

Want compressive stresses at end oflaminate so there are no tensile stresses tocause delamination - failure!

MSE 406: Thermal and Mechanical Behavior of Materials ©D.D. Johnson 2005

NO delamination - failure!

Apply in-pane Tensile Stress A B+90 +45+45 –45–45 +90–45 +90+45 –45+90 +45

Tensile -> delaminateCompressive

MSE 406: Thermal and Mechanical Behavior of Materials ©D.D. Johnson 2005

Why Laminate Composite is NOT Ideal

Depending on placement of load and the orientation of fibresinternal to sheet and the orientation of sheets relative to oneanother, the response is then very different.

Examples of orientations of laminated sheets that providedcompressive stresses at edges of composite and also tensilestresses there. >>>> Tensile stresses lead to delamination!

The stacking of composite sheets and their angular orientationcan be used to prevent “twisting” moments but allow “bending”moments. This is very useful for airplane wings, golf club shafts(to prevent slices or hooks), tennis rackets, etc., where poweror lift comes or is not reduced from bending.

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MSE 406: Thermal and Mechanical Behavior of Materials ©D.D. Johnson 2005

Thermal Stresses in Composites• Not just due to fabrication, rather also due to thermal expansion differencesbetween matrix and reinforcements αT

m and αTr.

• Thermal coatings, e.g.

• Material with most contraction (least) has positive (negative) residualstress. (For non-ceramics, you should consider plastic strain too.)

• Ceramic-oxide thermal layers, e.g. on gas turbine engines:• ceramic coating ZrO2-based (lower αT

r)• metal blade (NixCo1-x)CrAlY (higher αT

m)

•Failure by delamination without a good design of composite, i.e.compatibility maintained.

€

σ ≈|αTm –αT

r |ΔTE = ΔαTΔTEc

At T1 At T2

If compatible,composite willbend and rotate

MSE 406: Thermal and Mechanical Behavior of Materials ©D.D. Johnson 2005

• Composites are classified according to: -- the matrix material (CMC, MMC, PMC) -- the reinforcement geometry (particles, fibers, layers).• Composites enhance matrix properties: -- MMC: enhance σy, TS, creep performance -- CMC: enhance Kc

-- PMC: enhance E, σy, TS, creep performance• Particulate-reinforced: -- Elastic modulus can be estimated. -- Properties are isotropic.• Fiber-reinforced: -- Elastic modulus and TS can be estimated along fiber dir. -- Properties can be isotropic or anisotropic.• Structural: -- Based on build-up of sandwiches in layered form.

Summary