This document discusses composites, which are materials made from two or more constituent materials with different physical or chemical properties. It describes different types of composites including fiber reinforced plastics, concrete, wood, and others. It focuses on fiber reinforced plastic composites, discussing different fiber types like glass, carbon, and aramid fibers and how they are produced and their mechanical properties. It also discusses matrix materials like polyester and epoxy resins and how they contribute to the properties of fiber reinforced composites. The properties and applications of different composites are summarized.

1. P HY 351
CHAPTER 9
COMPOSITES

2. 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

3. EXERCISE 1
a.
b.
c.
d.
Give 2 examples of natural composite.
Give 3 special properties of composite materials.
Give 2 applications of composite.
Distinguish between cement and concrete.
3

4. 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.
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5. 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.
ii.
‘S” Glass = ‘S’ (high-strength) glasses used for military and
aerospace application.
‘E” Glass = ‘E’ (electrical glasses) cheaper than S glass
5

6. 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

7. ARAMID FIBERS FOR REINFORCING PLASTIC
RESINS

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.
7

8. COMPARISON OF MECHANICAL PROPERTIES

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.
8
Figure 12.7: Stress-strain behavior of various types of reinforcing fibers.

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.
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10. MATRIX MATERIALS FOR FIBER REINFORCEDPLASTIC COMPOSITE MATERIALS

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.
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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.
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12. PROPERTIES OF FIBER REINFORCED PLASTICS
12

13. 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.
ii.
What are the weight percentages of Kevlar 49 and epoxy
resin in the composite material?
(Answer: 64.9%, 35.1%)
What is the average density of the composite?
(Answer: 1.37 Mg/m3)
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14. EQUATION FOR ELASTIC MODULUS OF LAMELLAR
COMPOSITE

Isostrain condition: Stress on composite causes uniform
strain on all composite layers.
Pc = P f + P m
Pc = Load on composite
Pf = Load on fibers
Pm = load on matrix
Known; σ = P/A
Therefore;
σcAc = σfAf + σmAm
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15. Since length of layers are equal,
σcVc = σfVf + σmVm
Where;
Vf and Vm = volume fractions
Vc = 1
Therefore;
σc = σfVf + σmVm
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16. Since strains;
εc = εf = εm
Therefore;
 c  f V f  mVm


c
f
m
Ec = EfVf + EmVm
(Rule of mixture of binary composites)
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17. LOADS ON FIBER AND MATRIX REGIONS

Since σ = Eε and εf = εm
 f Af
E f  f Af
E f Af
EfVf




Pm  m Am E m  m Am E m Am E mVm
Pf
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.
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18. QUESTION 3
The composite consists of a continuous glass-foberreinforced-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
a.
i.
ii.
iii.
A modulus of elasticity (Answer: 44.64 GPa)
The tensile strength (Answer: 1.46 GPa)
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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19. ISOSTRESS CONDITION

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
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20. But;
c 

Ec
, f 

Ef
, m 

Em
Therefore;

Ec

V f
Ef

Vm
Em
20

21. ELASTIC MODULUS FOR ISOSTRESS CONDITION

We know that

Ec

V f
Ef

Vm
Em
21
• Higher modulus values are obtained with isostrain loading for
equal volume of fibers

22. Dividing by σ;
V f Vm
1


Ec E f Em
V f E m Vm E f
1


Ec E f E m Em E f
Ec 
E f Em
V f E m  Vm E f
22

23. 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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24. WOOD
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





24

25. PROPERTIES OF WOOD

Moisture content: Water occurs in wood as absorbed in fiber walls
or in cell fiber lumen.
 150% for softwood and sapwood.
Wood moisture content (wt%) =
Wt of water in sample
x 100
Wt of dry wood sample

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.
25

26. 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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27. 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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