Composites - MT4744
University of Moratuwa
Department of Materials Science and Engineering
Ashen Gunasekara
November 24, 2024
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Contents
1 Composites
1.1 What are Composites? . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Historical or Natural Examples of Composites . . . . . . . . . . . .
1.3 Other Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
5
5
6
2 Basic Fracture Mechanics
7
2.1 Stress Concentration Approach . . . . . . . . . . . . . . . . . . . . 7
2.2 Stress Intensity Approach . . . . . . . . . . . . . . . . . . . . . . . 9
2.3 Griffith’s Apporach . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.4 Weibull Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
3 Fiber Reinforced Composites
13
3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
3.2 Stress-Strain behavior of FRCs . . . . . . . . . . . . . . . . . . . . 13
3.3 Elastic Modulus of FRCs . . . . . . . . . . . . . . . . . . . . . . . . 15
3.3.1 Along the Fiber axis . . . . . . . . . . . . . . . . . . . . . . 15
3.3.2 Normal to the Fiber axis . . . . . . . . . . . . . . . . . . . . 16
3.3.3 Practice Questions . . . . . . . . . . . . . . . . . . . . . . . 16
3.3.4 Halpin-Tsai Equation . . . . . . . . . . . . . . . . . . . . . . 18
3.3.5 Bantrup Equation . . . . . . . . . . . . . . . . . . . . . . . . 18
3.4 Tensile Strength of FRCs . . . . . . . . . . . . . . . . . . . . . . . . 19
3.4.1 Along the fiber axis [Continuous Fibers] . . . . . . . . . . . 19
3.4.2 Along the fiber axis [Discontinuous Fibers] . . . . . . . . . . 21
3.5 Toughness of Composites . . . . . . . . . . . . . . . . . . . . . . . . 24
4 Glass Fibers
27
4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
4.2 Types of Glass Fibers . . . . . . . . . . . . . . . . . . . . . . . . . . 27
4.2.1 E-Glass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
4.2.2 S-Glass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
4.2.3 C-Glass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
4.2.4 A-Glass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
4.3 Properties of Glass Fibers . . . . . . . . . . . . . . . . . . . . . . . 28
4.4 Manufacturing of Glass Fibers . . . . . . . . . . . . . . . . . . . . . 28
3
4
CONTENTS
4.4.1
4.4.2
4.4.3
4.5
Continuous Filament Process . . . . . . . . . . . . . . . . .
Wool Processes . . . . . . . . . . . . . . . . . . . . . . . . .
Comparison of Continuous Filament Process and Wool Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
29
30
30
31
5 Boron Fibers
33
5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
5.2 Properties of Boron Fibers . . . . . . . . . . . . . . . . . . . . . . . 33
5.3 Manufacturing of Boron Fibers . . . . . . . . . . . . . . . . . . . . 33
5.4 Applications of Boron Fibers . . . . . . . . . . . . . . . . . . . . . . 34
5.5 Comparison with Glass Fibers . . . . . . . . . . . . . . . . . . . . . 34
5.6 Challenges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
6 Polymer Fibers
37
6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
6.2 Production Processes . . . . . . . . . . . . . . . . . . . . . . . . . . 37
6.2.1 Melt Spinning . . . . . . . . . . . . . . . . . . . . . . . . . . 37
6.2.2 Solution Spinning . . . . . . . . . . . . . . . . . . . . . . . . 38
6.3 Properties of Polymer Fibers . . . . . . . . . . . . . . . . . . . . . . 39
6.4 High-Performance Polymer Fibers . . . . . . . . . . . . . . . . . . . 40
6.5 Applications of Polymer Fibers . . . . . . . . . . . . . . . . . . . . 40
6.6 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
Chapter 1
Composites
1.1
What are Composites?
Mix of two or more constituent materials with significantly different
physical and/or chemical properties which remain separate and distinct
on a macroscopic level within the finished structure.
In their broadest form,
ˆ composites are materials consist of two or more constituents
ˆ the constituents are combined in such a way that they keep their individual
physical phases
ˆ the constituents are not soluble in each other or not to form a new chemical
compound
ˆ one constituent is called reinforcing phase and the one in which the reinforcing phase is embedded is called matrix
1.2
Historical or Natural Examples of Composites
Historical or natural examples of composites are abundant:
ˆ brick made of clay reinforced with straw
ˆ mud wall with bamboo shoots
ˆ concrete
ˆ concrete reinforced with steel rebar
ˆ granite consisting of quartz, mica and feldspar
ˆ wood (cellulose fibers in lignin matrix)
5
6
CHAPTER 1. COMPOSITES
1.3
Other Examples
Other examples include:
ˆ Fiber glass
ˆ Asphalt mix
ˆ Asbestos-Cement
ˆ Chip boards
ˆ Fiber boards
ˆ Cladded materials
ˆ Polymer Impregnated Concrete (PIC)
ˆ Plywood
Sandwich panels
Structural
Laminates
Randomly oriented
Discontinuous (short)
Composites
Aligned
Fiber-reinforced
Continuous (aligned)
Dispersion-Strengthened
Particle-reinforced
Large-Particle
Figure 1.1: Map of Composites
Chapter 2
Basic Fracture Mechanics
ˆ Fracture is the separation of a body into two or more pieces
ˆ Brittle fracture - Fracture that occurs without any significant plastic deformation
ˆ Ductile fracture - Fracture that involves significant amount of plastic deformation
ˆ Brittle materials undergo brittle fracture
ˆ Ductile materials also may fail in brittle manner under certain conditions
ˆ Fracture stress in brittle fracture depends on the size and the number of
cracks present
ˆ Fracture stress in brittle fracture (σf ) determined by different methods as
described below
ˆ Cracks can propagate under only tensile stress
2.1
Stress Concentration Approach
Consider an elastic body with an internal crack as shown
ˆ Crack is considered to be elliptical in shape (a - Major axis, b - Minor axis)
ˆ When b → 0 crack becomes a line(still an ellipse)
ˆ stress at edges are different though
2a
σtip = 1 +
·σ
b
What are the maximum and minimum possible values for σtip ?
7
8
CHAPTER 2. BASIC FRACTURE MECHANICS
Figure 2.1: Elastic Body with an internal crack
(
limb→0 σtip
3 · σ < σtip < ∞
limb→a σtip
= limb→0 1 + 2a
·σ =∞
b 2a
= limb→0 1 + b · σ = 3 · σ
(2.1)
Failure Occurs when σtip = σtheoretical . (Stress required to break bond between
two atoms, roughly σtheoretical = 0.1 · (Y oung ′ sM odulus)). That means at the
failure point,
2a
· σ = σtheoretical
σtip = 1 +
b
σtheoretical
σ=
1 + 2a
b
At failure point, σ = σf racture . In real materials,
2a
2a
a >>> b −→∴ 1 +
≈
b
b
b
σf racture =
· σtheoretical
2a
This is the proof that crack stress depend on the crack size. Larger the crack
(a) fracture strength will be lower. (Low Strength implies that fracture occurs at
lower stress). When the dimension ratio is getting near one we cant get proper
information from this equation.
CHAPTER 2. BASIC FRACTURE MECHANICS
2.2
9
Stress Intensity Approach
Figure 2.2: Elastic Body
The Stresses on this element will be as follows
Figure 2.3: Stresses on the Element

1 − sin θ sin 3θ 
σx
θ
2
2 K cos 2
 σy  = I√

· 1 + sin 2θ sin 3θ
2
2πr
τxy
sin 2θ sin 3θ
2

√
K1 is known as the stress intensity factor and defined as KI = σ πaf (a, α)
ˆ σ = Applied stress
ˆ a = Length of the Crack
ˆ f (a, α) = Function depend on the material geometry and loading system
For a Surface Crack f (a, α) = 1.12 and for a Internal Crack f (a, α) = 1
ˆ KI increases with the increase in the applied stress and reaches a maximum
known as KIC , at fracture
ˆ KIC is named as critical stress intensity factor or fracture toughness
10
CHAPTER 2. BASIC FRACTURE MECHANICS
KI
Surface Crack Strength
Internal Crack Strength
KIC
σ
Figure 2.4: Crack Strengths reaching to KIC
ˆ KIC is a material property – it does not depend on material geometry or
loading conditions
In General,
KIC
σf ∝ √
πa
Material tends to reach to failure faster when it is propagated by surface crack
strength. With Larger a(crack size) fracture strength will be lower.
2.3
Griffith’s Apporach
Griffith’s criterion is based on the premise that “a crack is unstable if the stored
energy released at fracture is greater than the creation of surface energy due to
the new surfaces”. According to Griffith’s energy criterion the total free energy of
a cracked body U is given by,
U = Ū − WR + US
Strain Energy of a non-cracked body
Ū =
σ2
/volume
2E
Energy released due to the formation of the crack
WR =
σ 2 πa2
[1 + ν][1 + k]
4E
CHAPTER 2. BASIC FRACTURE MECHANICS
11
Surface energy of the newly created surfaces,4a is for considering up and down
sides of the crack 2a for one side and γ0 is the surface energy per unit area.
US = 4aγ0 /thickness
σ2
σ 2 πa2
−
[1 + ν][1 + k] + 4aγ0
2E
4E
∂U
σ 2 πa
= 4γ0 −
[1 + ν][1 + k]
∂a
2E
U=
According to the Griffith Criterion,
∂U
> 0 −→ Crack is stable
∂a
∂U
< 0 −→ Crack is unstable (Crack propagates)
∂a
∂U
= 0 −→ Equilibrium state
∂a
At the equilibrium state we can say σ = σf ,
s
1
8Eγ0
σf =
=⇒ σf ∝ √
πa[1 + ν][1 + k]
a
For plain strain condition (εZ = 0) ,k = 3 − 4ν
s
2Eγ0
σf =
πa[1 − ν 2 ]
3−ν
1+ν
r
2Eγ0
σf =
πa
For plain stress condition (σZ = 0) , k =
12
CHAPTER 2. BASIC FRACTURE MECHANICS
σf
a
Figure 2.5: Exponential relationship between σf vs a
2.4
Weibull Analysis
ˆ Strength of brittle materials shows usually a wide dispersion of values
ˆ As such, a statistical approach is required
ˆ Weibull statistics leads to the following conclusion:
σf ∝
1
1
Vm
V is the volume of the material(This makes sense when considering small pieces of
glass which are stronger than a sheet of a glass but for a ductile material volume
does not matter), m is the weibull modulus(brittleness, lower the m higher the
brittleness, m is larger for ductile materials)
Chapter 3
Fiber Reinforced Composites
3.1
Introduction
The reinforcement is done with fibers in a FRC. But why fibers?. Following
relationships exist for the fracture stress of the brittle materials.
r
2Eγ0
σf =
πa
σf ∝
1
1
Vm
Imagine a cylindrical structure with diameter of 10mm, Then the largest possible crack can occur in that direction is 10mm. That means lesser the diameter,
possibility of having larger cracks reduces and with that volume is reduced resulting σf will be increased, Therefore fibers are stronger. Manufacture of fibers and
whiskers accomplished thisˆ Strength of SiC 450MPa
ˆ Strength of SiC fiber 20000MPa
Even though fibers are strong they cannot be used as they are. This problem is
overcome by using the ‘matrix’ to hold the fibers together. A material in which
fibers are incorporated in a matrix is named Fiber Reinforced Composite [FRC].
E.g.: GFRP – Glass Fiber Reinforced Plastic – commonly known as fiber glass.
3.2
Stress-Strain behavior of FRCs
In the study of σ − ε behavior following assumption is made: Fibers and matrix
are uniform and the fibers are firmly gripped by the matrix so that no slippage
can occur at the interface. Following notations are used in this study:
ˆ F - Load
13
14
CHAPTER 3. FIBER REINFORCED COMPOSITES
ˆ σ - Stress
ˆ E - Young’s Modulus
ˆ A - Cross sectional area
ˆ V - Volume fraction
ˆ e - extension
ˆ subscripts m, f and c refer to matrix fiber and composite respectively.
ˆ Vf =
fiber volume
= volume fraction of fibers
composite volume
ˆ Vm =
matrix volume
= volume fraction of matrix
composite volume
Vf + Vm = 1
Figure 3.1: σ − ε behavior of FRCs
CHAPTER 3. FIBER REINFORCED COMPOSITES
3.3
Elastic Modulus of FRCs
3.3.1
Along the Fiber axis
15
Figure 3.2: Strain along the longitudinal axis
Fc = Ff + Fm
σc Ac = σf Af + σm Am
Am
Af
σc = σf
+ σm
Ac
Ac
Af l
Am l
σc = σf
+ σm
Ac l
Ac l
σc = σf Vf + σm Vm
Stress equation
σc = σf Vf + σm (1 − Vf )
Ec εc = Ef εf Vf + Em εm (1 − Vf )
Strain Equation (young’s modulus)
Ec = Ef Vf + Em (1 − Vf )
(εc = εf = εm )
16
3.3.2
CHAPTER 3. FIBER REINFORCED COMPOSITES
Normal to the Fiber axis
Figure 3.3: Strain along the transverse axis
ec = ef + em
εc tc = εf tf + εm tm
tm
tf
εc = εf · + εm
tc
tc
εc = εf Vf + εm Vm
σc
σf
σm
=
Vf +
Vm
Ec
Ef
Em
1
1
1
=
Vf +
Vm
Ec
Ef
Em
(Thickness ratios = volume fractions)
(Since ε = Eσ )
(Assuming σc = σf = σm )
Young Modulus of Composite
Ec =
3.3.3
Ef Em
Em Vf + Ef Vm
Practice Questions
Question 01.
Ef = 400 GPa,
Em = 5 GPa,
Vf = 0.6,
Ecy =?,
Ecx =?
CHAPTER 3. FIBER REINFORCED COMPOSITES
17
Figure 3.4: Q1
Answer: Fibers are oriented only along y direction
Ecy = Ec = Ef Vf + Em (1 − Vf )
= 400 × 0.6 + 5(1 − 0.6)
= 242GP a
Ef Em
Em Vf + Ef Vm
400 × 5
=
5 × 0.6 + 400 × 0.4
= 12.27GP a
Ecx = Ec =
Question 02.
Ef = 400 GPa,
Em = 5 GPa,
Vf = 0.6,
Ecy =?,
Ecx =?,
Ecz =?
Figure 3.5: Q2
Answer: We cannot directly use derived equations because they are meant to be
uni-directionally aligned fibers. But here fibers are arranged in a bidirectional
manner, because of that for simplifying the problem we will separating the Top
half and Bottom half in a way only top half contains horizontally aligned fibers
18
CHAPTER 3. FIBER REINFORCED COMPOSITES
and bottom half contains only vertically aligned fibers without changing the Volume fraction of the fibers.
F
F
F = Fv + Fh
A
A
σc · A = σv · + σh ·
2
2
σv σh
σc =
+
2
2
Ev · ε Eh · ε
+
Ec · ε =
2
2
Ev Eh
+
Ec =
2
2
242 12.27
Ec =
+
2
2
Ec = 127.135GP a
Fibers are oriented along both x and y directions. But there are no fibers along z
direction. How can we approach this problem? The answer is 12.27GP a. Because
both fiber sets whatever the orientation, are parallel to the z direction. To z
direction, all fibers are at transverse orientation.
In practice the above equation does not give the correct elastic modulus in
transverse direction. Main reason may be the uneven distribution of the stress
between fibers and the matrix. Equations of empirical nature have been used for
the elastic modulus in transverse direction.
3.3.4
Halpin-Tsai Equation
1 + 2βVf
;
Ec = Em ·
1 − βVf
3.3.5
β=
Ef
−1
Em
Ef
+2
Em
Bantrup Equation
Ec =
′
Ef Em
;
′ V +E V
Em
f
f m
′
Em
=
Em
2
1 − νm
CHAPTER 3. FIBER REINFORCED COMPOSITES
3.4
19
Tensile Strength of FRCs
Tensile Strength is the Maximum Stress without Yielding or Fracturing, For Ductile Materials it’s the Yield Point, For Brittle Materials it’s the Fracture Point
3.4.1
Along the fiber axis [Continuous Fibers]
Stress on the Composite is given by
σc = σf Vf + σm (1 − Vf )
Following notations are used here:
ˆ σc,u - Composite fracture strength
ˆ σf,u - Fiber fracture strength
ˆ σm,u - Matrix fracture strength
Failure of a composite could happen in three different ways,
ˆ Both, fibers and the matrix fail at the same time: This case is very rare
unless both are made of the same material. Here, fibers and the matrix
reach their fracture stresses at the same time.In this case strength of the
composite is given by,
σc,u = σf,u Vf + σm,u (1 − Vf )
ˆ Fibers fail first, matrix unable to carry load, matrix fails: In this case
strength of the composite is given by,
′
σc,u = σf,u Vf + σm
(1 − Vf )
′
σm
- stress on matrix at the time when fibers reach their fracture stress
ˆ Fibers fail first, but matrix carries load, and matrix fails only then: In this
case strength of the composite is given by,
σc,u = σm,u (1 − Vf )
Strength of the composite( σc,u ) as a function of Fiber volume fraction (Vf ) is
shown in the graph below.
20
CHAPTER 3. FIBER REINFORCED COMPOSITES
σc,u
σf,u
σm,u
′
σm
Vf
Vmin
Vcrit
1
The Graph was done using the last two types of Failures, These two have two
different kind of σc,u − Vf behaviors.
ˆ Fibers fail first, matrix unable to carry load, matrix fails:
′
′
−→ y = mx + c
)Vf + σm
σc,u = (σf,u − σm
ˆ Fibers fail first, but matrix carries load, and matrix fails only then:
σc,u = −σm,u Vf + σm,u −→ y = −mx + c
In practice composite σc,u − Vf relationship is depicted by the red line, Its a hybrid
of both failures. Reason is second type is only possible if Vf > Vmin . Therefore
the strength of FRC is given by,
(
σm,u (1 − Vf )
if Vf < Vmin
σc,u =
′
σf,u Vf + σm (1 − Vf ) if Vf > Vmin
Value of Vmin will be given by the intersection point of the two graphs and shown
below.
′
σm,u − σm
Vmin =
′
σf,u + σm,u − σm
Manufacture of FRC will be meaning only if Vf > Vcrit (otherwise σc.u < σm,u ).
Therefore strength of an actual composite will be always given by
′
σc,u = σf,u Vf + σm
(1 − Vf )
Value of Vcrit could be found by replacing σc,u by σm,u and Vf by Vcrit in above
equation.
′
σm,u − σm
Vcrit =
′
σf,u − σm
CHAPTER 3. FIBER REINFORCED COMPOSITES
3.4.2
21
Along the fiber axis [Discontinuous Fibers]
We will be assuming:
ˆ All fibers are equal in length and the diameter.
ˆ Fibers are uniformly distributed
ˆ Bonding between the fibers and the matrix is good
When a stress is applied to the composite it is transferred from the matrix to the
fiber gradually. Therefore, the stress at the ends (of the fiber) is zero and increases
to a maximum as shown.
Figure 3.6: Stress along the composite
Figure 3.7: Strain along the composite
ˆ Strain on the composite in the fiber-less region is same as the strain in the
region where the stress is maximum on the fiber.
ˆ Maximum possible value for σmax is σf,u
22
CHAPTER 3. FIBER REINFORCED COMPOSITES
ˆ Only when σmax = σf,u fiber will be utilized to its full capacity
ˆ when σmax = σf,u , the dimension z will be taken as l2c and lc is called the
critical length of the fiber.
Figure 3.8: Critical length vs σf,u
ˆ Non-uniform stress distribution in the region BC of the fiber produces shear
stresses at the fiber-matrix interface
ˆ This is because the matrix tends to strain uniformly everywhere but fiber
restricts this due to the gradual increase of stress from its ends
ˆ l < lc → σf,u not reached
ˆ long fibers implies more strength capacity
ˆ As a result shear stress develops in that region as shown below
CHAPTER 3. FIBER REINFORCED COMPOSITES
23
Figure 3.9: Shear Stress and Tensile Stress
δx
σ + δσ
τ
σ
2r
τ
σ < σ + δσ → should deform but won’t due to the fact they are contained by the
shear forces (τ )
σ · πr2 + 2πr · τ δx = πr2 (σ + δσ)
2πrτ δx = πr2 δσ
δσ
2τ
=
r
δx
dσ
2τ
=
dx
r
2τ
We can graphically find the r from the gradient of σ − x graph
m=
σf,u
2τ
=
lc /2
r
24
CHAPTER 3. FIBER REINFORCED COMPOSITES
Then following equation obtained.
σf,u
lc
=
2τ
d
lc
is the critical aspect ratio. Higher the ratio more efficient the load transfer. (In
d
compressive stress condition aspect ratio less than 3 will not subjected to buckle,
This is important as when doing the compressive tests the sample should not be
subjected to buckling therefore the sample should have optimum aspect ratio).
Also we can obtain the equation for composite stress integrating under the stresslength curve. Discontinuous and aligned fiber composite with uniform distribution
of fibers in which l > lc , longitudinal strength
l−2z+l
× σf,u
2
′
· Vf + σm
(1 − Vf )
2
(l − z) × σf,u
′
(1 − Vf )
=
· Vf + σm
l
(l − l2c ) × σf,u
′
=
· Vf + σm
(1 − Vf )
l
lc
′
σf,u · Vf + σm
(1 − Vf )
= 1−
2l
σc,u =
If l > 15lc taken as continuous.
3.5
Toughness of Composites
ˆ Toughness is the strain energy stored in a material up to fracture
ˆ It is equal to the work of fracture - γF
ˆ In monolithic materials, γF is equal to the thermodynamic Surface energy
γS of the newly formed surfaces. Hence the units of γF will be J/m2
ˆ Toughness of a composte can be calculated using the γF values of the matrix
and fiber materials.
(γF )c = (γF )f × Vf + (γF )m × (1 − Vf )
Estimate the toughness of C/epoxy composite with 50% volume fraction of fibers
γF values of carbon and epoxy are 70J/m2 and 350 J/m2 respectively.
(γF )c = 70 × 0.5 + 350 × (1 − 0.5)
= 210J/m2
In practice, for a composite with continuous fibers the actual work of fracture will
be the same as the calculated value. But for a composite with discontinuous fibers
CHAPTER 3. FIBER REINFORCED COMPOSITES
25
actual value is much higher than the calculated value. In the above case it was
210 J/m2 . The reason for this is that the γF of such a composite contains ”other
work terms” in addition to γS .
γF = γd + γdf + γp + γS
ˆ γd = work done for debonding
ˆ γdf = work done for post debonding friction
ˆ γp = work done for pullout
ˆ γS = Surface energy
In a composite with discontinuous fibers under tensile stress, the matrix fractures
first, leading to the separation of the composite—this process is called pullout.
Before pullout occurs, the fiber must be de-bonded from the matrix. Once debonded, friction arises between the fiber and matrix due to their strain difference,
known as post-debonding friction. During pullout, additional friction acts
between the fiber and matrix. The work done against these forces, along with the
thermodynamic surface energy (which is higher than usual), contributes to the
composite’s increased toughness (work of fracture).
26
CHAPTER 3. FIBER REINFORCED COMPOSITES
Chapter 4
Glass Fibers
4.1
Introduction
Glass fibers are widely used non-metallic fibers primarily composed of silica (SiO2 ),
which forms the main component of glass. They are lightweight, strong, and
versatile, making them essential in various composite applications.
4.2
Types of Glass Fibers
Glass fibers are categorized into two main types:
ˆ General Purpose Fibers: These are widely used and include E-Glass
fibers, which conform to ASTM specifications.
ˆ Special Purpose Fibers: These include specialized fibers like S-Glass,
C-Glass, and A-Glass tailored for specific properties and applications.
4.2.1
E-Glass
E-Glass, or electrical glass, is the most common type of glass fiber. It offers
excellent electrical insulation and chemical resistance, making it suitable for a
wide range of applications.
4.2.2
S-Glass
S-Glass, or structural glass, has higher strength and stiffness compared to EGlass. It is often used in high-performance composites, such as those required in
aerospace applications.
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CHAPTER 4. GLASS FIBERS
4.2.3
C-Glass
C-Glass, or chemical glass, is known for its chemical resistance and is used in
environments exposed to harsh chemicals.
4.2.4
A-Glass
A-Glass, or alkali glass, is used where high alkali resistance is needed, such as in
certain glass containers.
4.3
Properties of Glass Fibers
Glass fibers exhibit the following key properties:
ˆ High Strength: Excellent tensile strength, suitable for reinforcing materials like plastics and composites.
ˆ Low Density: Lightweight, aiding in the reduction of overall weight in
composite materials.
ˆ High Modulus: High stiffness, enhancing the rigidity of composite structures.
ˆ Chemical Resistance: Resistance to chemical corrosion, depending on the
type of fiber.
ˆ Thermal Stability: Suitable for high-temperature applications.
ˆ Electrical Insulation: E-Glass provides excellent electrical insulation.
4.4
Manufacturing of Glass Fibers
Glass fibers are manufactured using the continuous filament process and Wool
Process.
Figure 4.1: General Process
CHAPTER 4. GLASS FIBERS
4.4.1
29
Continuous Filament Process
1. Weighing and Mixing: Ingredients are mixed to form a batch.
2. Melting: Glass is melted at approximately 1260°C.
3. Fiberization: Monofilaments are drawn from the molten glass and gathered
into strands.
4. Coating and Sizing: A thin homogeneous coating, known as sizing, is
applied to protect the fibers during handling and improve compatibility with
the matrix. Sizing consist of Coupling agents (nearly always contain an
organofunctional silane), Film formers, Additives or modifiers, Water.
(a) Prevents surface damage of fibers by forming protective film
(b) Binds individual filaments together
(c) Increases the compatibility of the fibers with the matrix
(d) Lubrication
5. Drying and Packaging: The fibers are dried and prepared for use in
various applications.
(a) Roving: Roving is a collection of parallel continuous ends of filaments.
Generally, rovings are made with fibers of diameter 9 or 13 µm
(b) Woven Roving: Rovings may be woven into a heavy, Coarse-weave
fabric for applications that require rapid thickness buildup over large
areas.
(c) Chopped-strand mat: Nonwoven material in which the fiberglass
strands from roving are chopped into 25-50-mm lengths, evenly distributed at random onto a horizontal plane, and bound together with
an appropriate chemical binder.
(d) Continuous-strand mat: Consists of un-chopped continuous strands
of fiberglass deposited and interlocked in a spiral fashion.
(e) Yarn: Combination of strands that can be woven suitably into textile materials. The continuous, individual strand as it comes from the
bushing represents the simplest form of textile fiberglass yarn
30
4.4.2
CHAPTER 4. GLASS FIBERS
Wool Processes
Figure 4.2: Wool Processing Methods
4.4.3
Comparison of Continuous Filament Process and Wool
Processes
Table 4.1: Comparison of Continuous Filament Process and Wool Processes
Aspect
Continuous-Filament
Wool Processes
Process
Typical Fiber Di- 4 to 30 microns
ameters
0.2 to 10 microns
Glass Types Used
E-glass, C-glass, R-glass, S- Soft alkali borosilicates,
glass, AR-glass, A-glass, D- mineral wool, modified slag,
glass, etc.
basalt, etc.
Primary Uses
Reinforcement, filtration, Thermal insulation, acousseparation, facers, thermal tic insulation, filtration,
insulation, fire blocking
separation
CHAPTER 4. GLASS FIBERS
4.5
31
Applications
Glass fibers are used in numerous industries, including:
ˆ Aerospace: Aircraft and spacecraft components.
ˆ Construction: Reinforcement for concrete and structural elements.
ˆ Marine: Boats and marine accessories.
ˆ Consumer Goods: Protective gear, industrial tools, and household items.
ˆ Electrical: Insulating rods, tubes, and components.
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CHAPTER 4. GLASS FIBERS
Chapter 5
Boron Fibers
5.1
Introduction
Boron fibers, also known as boron filaments, are amorphous elemental boron products. These fibers are highly valued for their exceptional mechanical properties
and are used in various high-performance applications, particularly as reinforcements in composite materials.
5.2
Properties of Boron Fibers
Boron fibers exhibit the following key properties:
ˆ High Strength: Boron fibers are significantly stronger than glass fibers.
ˆ High Stiffness: Exceptional modulus of elasticity, up to six times that of
glass fibers.
ˆ Low Density: Lightweight, making them ideal for aerospace and sporting
applications.
ˆ Thermal Resistance: High melting point of 2000°C.
ˆ Low Flexibility: These fibers are rigid and not flexible.
5.3
Manufacturing of Boron Fibers
Boron fibers are manufactured using the Chemical Vapor Deposition (CVD)
process:
1. A tungsten substrate with a diameter of 12 µm is introduced into a borosilicate glass reactor.
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34
CHAPTER 5. BORON FIBERS
2. The substrate is resistively heated to approximately 1300°C using a DC
power supply.
3. Boron trichloride and hydrogen gases are introduced, and boron is deposited
onto the heated substrate. 2 BCl3 (g) + H2 (g) −−→ 2 B(s) + 6 HCl(g)
4. The final fiber is composed of an amorphous boron outer layer and a fully
borided tungsten core.
5. To ensure quality, the temperature must be carefully controlled; temperatures exceeding 1350°C increase grain size and reduce filament strength.
Figure 5.1: Boron Fibers Manufacturing
5.4
Applications of Boron Fibers
Boron fibers are used in a variety of industries due to their strength and stiffness:
ˆ Aerospace: Aircraft empennage skins, space shuttle truss members, and
prefabricated aircraft repair patches.
ˆ Sporting Goods: Fishing rods, golf club shafts, skis, and bicycle frames.
ˆ Other Applications: Reinforcement in light alloys, particularly in environments requiring high compressive and tensile strength.
5.5
Comparison with Glass Fibers
ˆ Boron fibers have a modulus of elasticity six times higher than that of glass
fibers.
ˆ They also exhibit superior strength and stiffness but are heavier and more
expensive.
CHAPTER 5. BORON FIBERS
5.6
35
Challenges
ˆ Boron fibers require protection during contact with molten metals to avoid
chemical reactions that reduce strength. (SiC,B4 C)
ˆ High production costs limit their widespread use compared to other fibers
like carbon or glass.
36
CHAPTER 5. BORON FIBERS
Chapter 6
Polymer Fibers
6.1
Introduction
Polymer fibers are synthetic fibers derived from polymers. These fibers are lightweight,
flexible, and exhibit excellent mechanical and chemical properties, making them
highly versatile for industrial and commercial applications.
6.2
Production Processes
Polymer fibers can be categorized based on their production methods and properties:
ˆ Melt Spinning: Best suited for most thermoplastic polymers.
ˆ Solution Spinning: Includes dry spinning and wet spinning, used when
polymers need to be dissolved in a solvent.
6.2.1
Melt Spinning
Melt spinning is widely used for producing thermoplastic fibers:
1. The polymer is melted,extruded through a filter (Screen pack) to eliminate
small contaminants and extruded through a spinneret (die of multiple orifices) and solidified and stretched to the desired diameter.
2. Fibers such as nylon and PET are produced using this process.
3. High speed (275 to 1500 yds/min).
4. No solvents required.
5. No purification issues.
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CHAPTER 6. POLYMER FIBERS
Figure 6.1: Melt Spinning
6.2.2
Solution Spinning
Solution spinning is used when the polymer needs to be dissolved in a solvent:
ˆ Dry Spinning: The solution is extruded through a spinneret, and the
solvent is evaporated to form fibers.
ˆ Wet Spinning: The polymer solution is extruded into a coagulation bath,
where fibers are formed.
Dry Spinning
1. The polymer is dissolved in a volatile solvent.
2. The solution is pumped through a spinneret (die) with numerous holes (one
to thousands).
3. As the fibers exit the spinneret, air is used to evaporate the solvent so that
the fibers solidify and can be collected on a take-up wheel.
4. Stretching of the fibers provides for orientation of the polymer chains along
the fiber axis
CHAPTER 6. POLYMER FIBERS
39
Figure 6.2: Dry Spinning
Advantages of Wet Spinning:
ˆ Large tows can be handled.
Disadvantages:
ˆ Slow production speed (70-150 yds/min).
ˆ Requires solvent and chemical recovery.
6.3
Properties of Polymer Fibers
Polymer fibers exhibit the following key properties:
ˆ Lightweight: Makes them ideal for applications requiring high strength-toweight ratios.
ˆ Thermal Stability: Heat resistance up to high temperatures (e.g., aramid
fibers resist up to 800°F).
ˆ High Strength: Suitable for reinforcing materials in composites.
ˆ Chemical Resistance: Resistant to organic solvents and other chemicals.
ˆ Low Flammability: Some fibers, like Kevlar, are inherently flame-resistant.
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CHAPTER 6. POLYMER FIBERS
6.4
High-Performance Polymer Fibers
Aramid Fibers:
ˆ Short for ”aromatic polyamide.”
ˆ High strength, abrasion resistance, and heat resistance.
ˆ Common brands include Kevlar® and Nomex®.
® Properties:
Kevlar
ˆ Ballistic resistance: Absorbs and dissipates energy from high-velocity projectiles.
ˆ Cut and puncture resistance: Strong protective barrier.
ˆ Heat resistance: Protects against thermal hazards up to 800°F.
6.5
Applications of Polymer Fibers
Polymer fibers are widely used in:
ˆ Textiles: Clothing, ropes, and nets.
ˆ Industrial Applications: Belts, filters, and conveyor systems.
ˆ Composites: Reinforcements in aerospace, automotive, and marine industries.
ˆ Protective Gear: Bulletproof vests, helmets, and fire-resistant clothing.
6.6
Conclusion
Polymer fibers, with their remarkable properties and versatility, have revolutionized modern materials science. From high-performance applications to everyday
uses, these fibers continue to play a crucial role in various industries.