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Page 1: Phy351 ch 9

CHAPTER 9

COMPOSITES

PHY351

Page 2: Phy351 ch 9

INTRODUCTION

A composite material is a material system, a mixture or

combination of two or more micro or macro constituents

that differ in form and composition and do not form a

solution.

Properties of composite materials can be better-quality to

its individual components.

Examples:

- fiber reinforced plastics

- concrete

- asphalt

- wood etc. 2

Page 3: Phy351 ch 9

EXERCISE 1

a. Give 2 examples of natural composite.

b. Give 3 special properties of composite materials.

c. Give 2 applications of composite.

d. Distinguish between cement and concrete.

3

Page 4: Phy351 ch 9

FIBERS FOR REINFORCED PLASTIC COMPOSITE

MATERIALS

FIBER REINFORCED PLASTIC is a composite

materials consisting of a mixture of a matrix of a plastic

material such as a polyester or epoxy strengthened by

fibers of high strength such as:

glass

carbon

arimid

The fibers provide the high strength and stiffness and the

plastic matrix bonds the fibers together and supports

them.4

Page 5: Phy351 ch 9

GLASS FIBER FOR REINFORCING PLASTIC RESINS

GLASS fiber are used to reinforce plastic matrices to

form structural composites and molding compounds.

Glass fiber reinforced plastic composite materials have

high strength-weight ratio, good dimensional stability,

good temperature and corrosion resistance and low

cost.

The most important types of glass used to produce glass

fiber for composites are:

i. ‘S” Glass = ‘S’ (high-strength) glasses used for military and

aerospace application.

ii. ‘E” Glass = ‘E’ (electrical glasses) cheaper than S glass

5

Page 6: Phy351 ch 9

CARBON FIBERS FOR REINFORCED PLASTICS

CARBON fiber such as epoxy are characterized by having

a combination of light weight, very high strength and high

stiffness (modulus of elasticity).

Produced from polyacrylonitrile (PAN) and pitch.

Steps:

Stabilization: PAN fibers are stretched and oxidized in air at

about 2000C.

Carbonization: Stabilized carbon fibers are heated in inert

atmosphere at 1000-15000C which results in elimination of O,H

and N resulting in increase of strength.

Graphitization: Carried out at 18000C and increases modulus

of elasticity at the expense of strength

6

Page 7: Phy351 ch 9

ARAMID FIBERS FOR REINFORCING PLASTIC

RESINS

7

ARAMID fiber is the generic name for aromatic polyamide fibers.

Trade name is Kevlar. There are two commercial type:

Kevlar 29:- Low density, high strength, and used for ropes and

cables.

Kevlar 49:- Low density, high strength and modulus and used for

aerospace and auto applications.

Hydrogen bonds bond fiber together.

Used where resistance to fatigue, high strength and light weight is

important.

Page 8: Phy351 ch 9

COMPARISON OF MECHANICAL PROPERTIES

8

Carbon fibers provide best combination of properties.

Due to favorable properties, carbon and aramid fiber reinforced

composites have replaced steel and aluminum in aerospace applications.

Figure 12.7: Stress-strain behavior of various types of reinforcing fibers.

Page 9: Phy351 ch 9

9

Figure 12.8: Specific tensile strength (tensile strength to density) and specific

tensile modulus (tensile modulus to density) for various types of reinforcing

fibers.

Page 10: Phy351 ch 9

MATRIX MATERIALS FOR FIBER REINFORCED-

PLASTIC COMPOSITE MATERIALS

10

Two of the most important MATRIX plastic resins for fiber-reinforced plastics are:

• unsaturated polyester

• epoxy resins

The polyester resin are lower in cost but are usually not as strong as the epoxy resin.

Page 11: Phy351 ch 9

11

Fiberglass-reinforced POLYESTER resins:

Higher the wt% of glass, stronger the reinforced plastic is.

Nonparallel alignment of glass fibers reduces strength.

Carbon fiber reinforced EPOXY resins:

Carbon fiber contributes to rigidity and strength while epoxy matrix contributes to impact strength.

Polyimides, polyphenylene sulfides are also used.

Exceptional fatigue properties.

Carbon fiber epoxy material is laminated to meet strength requirements.

Page 12: Phy351 ch 9

PROPERTIES OF FIBER REINFORCED PLASTICS

12

Page 13: Phy351 ch 9

QUESTION 2

a. Cite the general difference in strengthening mechanism

between large-particle and dispersion-strengthened

particle-reinforced composites.

b. A undirectional Kevlar 49 fiber-epoxy composite contains

60% by volume of Kevlar 49 fibers and 40 % epoxy resin.

The density of the Kevlar 49 fibers is 1.48Mg/m3 and that of

the epoxy resin is 1.2 Mg/m3.

i. What are the weight percentages of Kevlar 49 and epoxy

resin in the composite material?

(Answer: 64.9%, 35.1%)

ii. What is the average density of the composite?

(Answer: 1.37 Mg/m3)13

Page 14: Phy351 ch 9

EQUATION FOR ELASTIC MODULUS OF LAMELLAR

COMPOSITE

14

Isostrain condition: Stress on composite causes uniform strain on all composite layers.

Pc = Pf + Pm

Known; σ = P/A

Therefore;

σcAc = σfAf + σmAm

Pc = Load on composite

Pf = Load on fibers

Pm = load on matrix

Page 15: Phy351 ch 9

Since length of layers are equal,

σcVc = σfVf + σmVm

Where;

Vf and Vm = volume fractions

Vc = 1

Therefore;

σc = σfVf + σmVm

15

Page 16: Phy351 ch 9

16

Since strains;

εc = εf = εm

Therefore;

Ec = EfVf + EmVm

m

mm

f

ff

c

c VV

(Rule of mixture of binary composites)

Page 17: Phy351 ch 9

Since σ = Eε and εf = εm

Pc = Pf + Pm

From above two equations, load on each of fiber and matrix

regions can be determined if values of Ef, Em, Vf, Vm and Pc

are known.

LOADS ON FIBER AND MATRIX REGIONS

17

mm

ff

mm

ff

mmm

fff

mm

ff

m

f

VE

VE

AE

AE

AE

AE

A

A

P

P

Page 18: Phy351 ch 9

QUESTION 3

a. The composite consists of a continuous glass-fober-

reinforced-epoxy resin produced by using 60% by volume of

E-glass fibers having a modulus of elasticity of Ef = 7.24 x

104 MPa and a tensile strength of 2.4 GPa and a harded

eposy resin with a modulus of Em = 3.1 x 103 MPa and

tensile strength of 0.06 GPa. Calculate the composite

i. A modulus of elasticity (Answer: 44.64 GPa)

ii. The tensile strength (Answer: 1.46 GPa)

iii. The fraction of the load carried by the fiber for the following

composite material stresses under isostrain conditions.

(Answer: 0.97)

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Page 19: Phy351 ch 9

ISOSTRESS CONDITION

19

Stress on the composite structure produces an equal stress

condition on all the layers.

σc = σf + σm

εc = εf + εm

Assuming no change in area and assuming unit length of the

composite

εc = εfVf + εmVm

Page 20: Phy351 ch 9

20

But;

Therefore;

m

m

f

f

c

cEEE

,,

m

m

f

f

c E

V

E

V

E

Page 21: Phy351 ch 9

ELASTIC MODULUS FOR ISOSTRESS CONDITION

21

We know that

m

m

f

f

c E

V

E

V

E

• Higher modulus values are obtained with isostrain loading for

equal volume of fibers

Page 22: Phy351 ch 9

22

Dividing by σ;

fmmf

mf

c

fm

fm

mf

mf

c

m

m

f

f

c

EVEV

EEE

EE

EV

EE

EV

E

E

V

E

V

E

1

1

Page 23: Phy351 ch 9

QUESTION 4

a. Calculate the modulus of elasticity for a composite material

consisting of 60% by volume of continuous E-glass fiber and 40%

epoxy resin for the matrix when stressed under isostress

conditions (i.e. the material is stresses perpendicular to the

continuous fiber). The modulus of elasticity of the E glass is 72.4

GPa and that of the epoxy resin is 3.1 GPa.

(Answer : 7.3 GPa)

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Page 24: Phy351 ch 9

WOOD

24

Wood is naturally occurring composite with polymeric material lignin and other organic compounds.

Nonhomogenous and highly anisotropic.

Consists of layers:

(a) Outer bark – provides protection

(b) Inner bark – moist and soft,

carries food

(c) Cambium layer – forms wood

and bark cells

(d) Sapwood – carries wood and sap.

(e) Heartwood – dead, dark and

provides strength

(f) Pith – Soft tissue at the center

(g) Wood rays

Page 25: Phy351 ch 9

PROPERTIES OF WOOD

25

Moisture content: Water occurs in wood as absorbed in fiber walls

or in cell fiber lumen.

150% for softwood and sapwood.

Mechanical strength: Compressive strength parallel to the grain is 10

times higher than that perpendicular to the grain.

Wood in green condition is weaker than kiln-dried wood.

Shrinkage: Green wood shrinks if dried.

Shrinkage is more in transverse direction.

Wood moisture content (wt%) =Wt of water in sample

Wt of dry wood sample

x 100

Page 26: Phy351 ch 9

QUESTION 5

A piece of wood containing moisture weighs 165.3g and

after oven drying to a constant weight, weighs 147.5g.

What is its percent moisture content?

(Answer: 12.1%)

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Page 27: Phy351 ch 9

REFERENCES

A.G. Guy (1972) Introduction to Material Science,

McGraw Hill.

J.F. Shackelford (2000). Introduction to Material Science

for Engineers, (5th Edition), Prentice Hall.

W.F. Smith (1996). Principle to Material Science and

Engineering, (3rd Edition), McGraw Hill.

W.D. Callister Jr. (1997) Material Science and

Engineering: An Introduction, (4th Edition) John Wiley.

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