Lecture20-Nov30

8/10/2019 Lecture20-Nov30

http://slidepdf.com/reader/full/lecture20-nov30 1/21

Objectives/outcomes: You will learn the following:

• particle and fiber reinforcement.

• Various types of fibers

• Unidirectional

•

Rule of mixture.• Strength of fiber composites

MSE200Lecture 20(CH. 12.1-12.3)

Composite Materials

Instructor: Yuntian Zhu



Introduction

• A composite material is

• Properties of composite materials can be superior to itsindividual components.

• Examples:



Two types of composite materials

•

Classified according to the reinforcements – Particle reinforced composites

• Examples:

– Fiber reinforced composites

• Examples:



Glass Fibers for Reinforced Plastic Composite Materials

• Glass fiber reinforced plastic composite materials have

high strength-weight ratio, good dimensional stability,

good temperature and corrosion resistance and low

cost.

‘E’ Glass : 52-56% SiO2, + 12-16% Al2O3, 16-25%

CaO + 8-13% B2O3

Tensile strength = 3.44 GPa, E = 72.3 GPa

‘S” Glass : Used for military and aerospace

application.

65% SiO2 + 25% Al

2O

3 + 10% MgO

Tensile strength = 4.48 GPa, E = 85.4 GPa



Production of Glass Fibers

• Produced by drawing monofilaments from a furnace

and gathering them to form a strand.

Low cost and hencecommonly used.

http://www.google.com/search?q=glass+fiber+drawing&tbo=p&tbs=vid%3A1&source=vgc&hl=en&aq=f



Glass fiber products

http://video.google.com/videosearch?hl=en&q=glass%20fiber%20composite&gbv=2&ie=UTF-8&sa=N&tab=iv#hl=en&q=glass+fiber+composite&gbv=2&ie=UTF-8&sa=N&tab=iv&start=0



Carbon Fibers for Reinforced Plastics

• Light weight, very high strength and high stiffness.

• 7-10 micrometer in diameter.

•

Produced from polyacrylonitrile (PAN) and pitch.• Steps:

Stabilization: PAN fibers are stretched and oxidised in air

at about 2000C.

Carbonization: Stabilized carbon fibers are heated in inertatmosphere 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

• Tensile strength = 3.1-4.45 GPa, E = 193-241 GPa,

density = 1.7-2.1 g/cc.

The highest strength: 6.9 GPa, by Toray



Aramid Fibers (Kevlar) for Reinforcing Plastic Resins

• Aramid = aromatic polyamide fibers.

•

Trade name is Kevlar

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

Kevlar 49:- Low density, high strength and modulus and used foraerospace and auto applications.

•

Hydrogen bonds bond fiber together.

•

Used where resistance to fatigue, high

strength and light weight is important.



Kevlar fiber products

http://video.google.com/videosearch?q=kevlar&hl=en&emb=0&aq=f#



Comparison of Mechanical Properties

• Glass fibers are cheap, for cheap civilian products

•

Carbon fibers are strong but brittle, high strength structure

• Kevlar fibers are toughest, for body armor.



Fiber frontier: Carbon nanotube (CNT) fiber

Commercial

LANL CNT fiber



CNT fiber

Ribbons being pulled from array

Example of spun CNT fiber



Fiber Reinforced-Plastic Composite Materials

• Fiberglass-reinforced polyester resins:

Higher the wt% of glass, stronger the reinforcedplastic is.

Nonparallel alignment of glass fibers reducesstrength.

• Carbon fiber reinforced epoxy resins:

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

Polyimides, polyphenylene sulfides are also used.

Exceptional fatigue properties. Carbon fiber epoxy material is laminated to

meet strength requirements.



Properties of Fiber Reinforced Plastics

Fiberglass polyester

(Carbon fibers and epoxy)



Rule of Mixture (isostrain condition)

•

Stress on composite causes uniform strain on all composite layers.

Pc = Pf + Pm

= P/A

cAc = f Af + mAm

Pc = Load on composite

Pf = Load on fibers

Pm = load on matrix

Rule of mixture of binary composites

Ec = Ef Vf + EmVm

c = f Vf + mVm

c = f Vf + mVm



Isostress Condition

•

Stress on the composite structure produces an equal stress conditionon all the layers.

c = f = m

Assuming no change in area

and assuming unit length of the composite

c = f Vf + mVm

Therefore

m

m

f

f

c

c E E E

=== ,,

E c=

V f

E f +

V m

E m

1

E c=

V f

E f +V m

E m



Toughening Mechanisms in Composite Materials

• Toughening is due to fibers interfering with crack

propagation.

Crack deflection: Up on encountering reinforcement,

crack is deflected making propagation more

meandering.

Crack bridging: Fibers bridge the crack and help to

keep the cracks together. Fiber pullout: Friction caused by pulling out the fiber

from matrix results in higher toughness.



HW

•

Examples problems in Chapter 12: 12.1, 12.2, 12.3

Lecture20-Nov30

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Transcript of Lecture20-Nov30