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Composites
4 Lectures for Paper B8 Prof. J.C. Tan
www.eng.ox.ac.uk/tan
Michaelmas Term 2014 1
Department of Engineering Science University of Oxford
Contents of the lectures
1. Introduction - engineering applications
2. Basics of two-component composites
3. Elasticity of uni-directional (UD) continuous fibre composites
4. Failure of UD continuous fibre composites
5. Laminated composites: elasticity, bending, twisting and failure
2
e1
e3
e2
eY
eX
eZ
MYXMX
MY
MXY
1. What are composites?
3
• Composite materials are physical mixtures of two or more constituent materials.
e.g. glass fibres mixed into a
polymer resin
Typically, the “particles” have at least two dimensions that are on a small scale (<< 1 mm), visible only under a microscope
• Usually one material forms a continuous matrix, in which particles/fibres of the other material – the “filler” or “reinforcement phase” – are embedded.
200 µm
Fibre–matrix interface
“Microstructure”
Handout P.3
Engineering applications (I)
4
• Advanced composites are used extensively in a wide range of high-tech engineering products
• e.g. in transportation (aircraft, boats, warship), marine structures and sports goods (rackets, rowing eights, fishing rods), amongst many others…
Engineering applications (II)
5
50% composites
Boeing 787 Dreamliner
High-performance composites
6
• The main reason is that they offer an excellent combination of high specific stiffness (E/ρ) and specific strength (σ*/ρ), and low mass (ρ)
• Composites are being developed based on matrices of metals (MMCs), ceramics (CMCs) and polymers (PMCs)
• But by far the most common composites employed in commercial applications/manufacturing are those based on polymer matrices (i.e. PMCs)
Typical fibre composite properties (65% fibre + 35% epoxy resin matrix)
7
Density
ρ kg/m3
Modulus
E GPa
Strength
σ* MPa
E/ρ
× 103
σ*/ρ
× 103
Carbon fibre (high
modulus)
1700 222 1630 133 977
Carbon fibre (high
strength)
1600 151 2080 93 1284
Kevlar 49 1400 82 1820 65 1300
E-glass 2100 50 1086 24 515
S-glass 2100 57 1358 27 644
Boron 2100 207 2210 97 1030
polyethylene 970 77 1700 79 1753
Some metals for comparison:
Mild steel 7900 210 450 27 57
Aluminium alloy 2800 70 450 25 161
Titanium 4500 110 960 24 213
“Specific” mechanical properties
Handout P.4
A good reinforcement material
8
1. It has high stiffness and strength 2. It has good particle shape and surface character for
effective mechanical coupling to the matrix 3. It preserves the desirable properties of the polymer
matrix (e.g. low shaping T, low conductivities)
The reinforcing action
Handout P.5
Particle aspect ratio
ad
= l Pinning effect
σ
σ
matrix
particle
What is the optimum size and shape for a reinforcing particle?
9
( )1 3
2 3 1 32 2/
/ /A a aV V
π −⎛ ⎞= +⎜ ⎟⎝ ⎠
Handout P.6-7
Surface-to-volume ratio of particles:
We want a reinforcement particles to have: (1) high surface area (2) strong bonding to the matrix
Thus we prefer reinforcing particles that are either a << 1 or a >> 1
Also prefer reinforcing particles that are small (i.e. low V), e.g. nanoparticles, exfoliated nano-sheets & nano-clays (1-nm thickness)
aspect ratio a0.01 0.1 1 10 100 1000
A/V in units of (2π/V)1/3
0
5
10
15
20
25
fibreplateletPlatelets Fibres
= l
d
a << 1 a >> 1
Minimum
The major reinforcing fibres
10
• Glass fibres • Carbon fibres • Aramid fibres
(e.g. Kevlar®, Nomex®, Technora®) Also emerging: • Boron fibres • Polyethylene fibres
Handout P.8-10
fibre axis
aromatic polyamide = aramid
Typical fibre properties
11
Density
ρ kg/m3
Axial tensile
modulus E1 GPa
Axial tensile
strength σ* MPa
E-glass 2540 76 1800
S-glass 2490 86 2500
Carbon (HM) 1860 340 2500
Carbon (HS) 1790 230 3200
Kevlar 49 1450 124 2800
Boron (on tungsten) 2600 400 3400
Polyethylene 970 117 2600
Handout P.10
Typical polymer matrix properties
12
Density ρ kg/m3
Tensile modulus E GPa
Tensile strength σ* MPa
Thermoset TS or
thermoplastic TP Epoxy resin 1300 2.4 60 TS
Thermoset polyester 1280 3.0 55 TS
PEEK (Polyether ether ketone)
1390 4.0 90 TP
• Note that whether a matrix is thermoset or a thermoplastic has major implications for manufacturing processes.
Handout P.11
Video: Advanced composites for engineering applications
13
Contents of the lectures
1. Introduction - engineering applications
2. Basics of two-component composites
3. Elasticity of uni-directional (UD) continuous fibre composites
4. Failure of UD continuous fibre composites
5. Laminated composites: elasticity, bending, twisting and failure
14
e1
e3
e2
eY
eX
eZ
MYXMX
MY
MXY
Composition
15
Matrix, mass mm , volume vm Filler, mass mf , volume vf
Total mass m and volume v
f mf m f 1v v,
v vϕ ϕ ϕ≡ ≡ = −
( )f f m mf f f m
f m1ρ ϕ ρ ϕ ρ= = ⋅ + ⋅ = + −m v m v m
v v v v v
“Rule of Mixtures” (RoM)
f m f m and in the absence of any holes m m m , v v v= + = +
Handout P.12
Composite density
Define: Volume Fractions
Symmetry
16
=x QX
1
1 0 00 1 00 0 1
−⎛ ⎞⎜ ⎟= = ⎜ ⎟⎜ ⎟⎝ ⎠
Q M 1
1 0 00 1 00 0 1
⎛ ⎞⎜ ⎟= = −⎜ ⎟⎜ ⎟−⎝ ⎠
Q R
Quantify symmetry in terms of orthogonal transformations, Q
Symmetry transformations if they leave the object for practical purposes unchanged
1 1
1 0 00 1 00 0 1
−⎛ ⎞⎜ ⎟= = − ≡⎜ ⎟⎜ ⎟−⎝ ⎠
Q M R C
The set of symmetry transformations = G (Symmetry Group)
• Most high-performance composites are anisotropic … • their non-scalar physical properties vary with direction,
and hence with the choice of reference axes.
Mirror symm. 2-fold Rotational symm.
e.g.
Centre of symm.
Handout P.13
..symmetry restriction on response
17
A e.g. state of ε
Agency Material Response
M R e.g. state of σ
AMR GGG ∩⊇The principle of “isotropy of space” imposes the restriction
superset intersection
Handout P.14
Basic mechanics
18
Handout P.15-17
e1e3
e2
σ11 σ11
Area A
Length L
σ11 =ϕ fσ f11 + 1−ϕ f( )σm11
( )11 f f11 f m111ε ϕ ε ϕ ε= + −
Assume L ≈ A d then:
We need “combining rules” for predicting the properties of the composite in terms of the proportions (φ) and properties of their constituents
RVE: Volume average stresses & strains in filler and matrix, both of which are homogeneous
Uniaxial loading
d = size of “microstructure”
>>
continuum
..the equations to solve in general
19
f f m m, ij ijkl kl ij ijkl klk l k l
σ ξ σ σ ξ σ= =∑∑ ∑∑
f11 f f f11 f
f22 f f f22 ff
f33 f f f33 f
1 - -1 - 1 -- - 1
TE
ε ν ν σ αε ν ν σ αε ν ν σ α
⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟= + Δ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠
( ) ( )f f f m f f f m1 1 for all ij ij ij ij ij ij, i , jσ ϕ σ ϕ σ ε ϕ ε ϕ ε= + − = + −
And similarly for an isotropic linear elastic matrix (m)…
Plus constitutive equations (σ-ε relations) of the constituents, e.g. isotropic linear thermo-elastic solid
Handout P.17-19
Continuum stresses — “Applied stresses” onto bulk material
Volume- averaged stresses (overbar) Coupling
coefficients (4-D arrays)
γ f23
γ f31
γ f12
⎛
⎝
⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟
= 1Gf
1 0 00 1 00 0 1
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟
τ f23
τ f31
τ f12
⎛
⎝
⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟
Examples Sheet (on P.77)
20
Today’s Homework:
try Question 1
Contents of the lectures
1. Introduction - engineering applications
2. Basics of two-component composites
3. Elasticity of uni-directional (UD) continuous fibre composites
4. Failure of UD continuous fibre composites
5. Laminated composites: elasticity, bending, twisting and failure
21
e1
e3
e2
eY
eX
eZ
MYXMX
MY
MXY
UD continuous fibre composites
22
e1
e3
e2
• The filler has the form of continuous cylindrical fibres aligned perfectly (along e1),
• .. but located randomly in the plane perpendicular to their axes (i.e. isotropic in 2-3 plane).
• Useful when a high degree of reinforcement is required in just one direction
• But more usually they act as building blocks of laminated composites (see later §5), used most widely in engineering composite structures.
e.g. • glass fibres • carbon fibres
Handout P.20
Symmetry in UD composites
23
*Because the fibres are not arranged randomly, this composite will be anisotropic in its physical properties
But it does possess symmetry, as may be expressed in terms of its symmetry group G, which includes
1 1 2
1 0 0 1 0 0 1 0 00 cos sin , 0 1 0 , 0 1 00 sin cos 0 0 1 0 0 1
ω ωω ω
−⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − = = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
U M M
e1
e3
e2
Handout P.21
U1
M1 M2
isotropic in 2-3 plane
Elastic constants of UD composites
24
Contracted notation for linear elasticity
1 111 11
2 222 22
3 333 33
4 423 23
5 531 31
6 612 12
,
σ εσ εσ εσ εσ εσ ετ γτ γτ γτ γτ γτ γ
⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟
≡ ≡⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠
6
1
ε σ=
=∑p pq qqS
11 12 12
12 22 23
12 23 2244 22 23
44
66
66
0 0 00 0 00 0 0
, where 2( )0 0 0 0 00 0 0 0 00 0 0 0 0
S S SS S SS S S
S S SS
SS
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟
= = −⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
S
6 x 1 column
matrices
Compliance matrix S linking strain to stress:
5 independent elastic constants, instead of the usual 2 for isotropic materials
Infinitesimal strain (<5%)
Handout P.22
6 x 6 square matrix
SYMM.
6 x 6 6 x 1 6 x 1
…in terms of moduli (E, G) and Poisson’s ratios (ν)
25
Handout P.23
12 212 23 23
1 22 (1 ), and E G
E Eν νν= + =
With isotropy in the 2-3 plane
1 1 11 21 2 21 2
1 1 112 1 2 23 2
1 1 112 1 23 2 2
123
112
112
0 0 0
0 0 0
0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
E E E
E E E
E E E
G
G
G
ν νν νν ν
− − −
− − −
− − −
−
−
−
⎛ ⎞− −⎜ ⎟− −⎜ ⎟
⎜ ⎟− −⎜ ⎟= ⎜ ⎟
⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
S
SYMM.
Coupling coefficients for UD composite
26
f1 f11 f12 f12 1
f2 f21 f22 f23 2
f3 f21 f23 f22 3
f4 f44 4
f5 f66 5
f6 f66 6
0 0 00 0 00 0 0
0 0 0 0 00 0 0 0 00 0 0 0 0
σ ξ ξ ξ σσ ξ ξ ξ σσ ξ ξ ξ στ ξ ττ ξ ττ ξ τ
⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟
=⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
f fm
f
11
pppp
ϕ ξξ
ϕ−
=− ( )
f fm
f1pq
pqϕ ξ
ξϕ
= −−
From symmetry
From equilibrium
Volume average stresses (induced @ microstruct-ural level of RVE = Response)
fibres
Continuum stresses (applied stresses = Agency)
Stress amplification associated with ξfij (i.e. coupling coefficients)
Handout P.24-25
(p ≠ q)
SYMM.
Axial Tension (1-direction)
27
( ) ( )1 1f1 f11 f f21 f 2 f 3 f21 f f11 f f 21
f f2 ,
E Eσ σε ξ ν ξ ε ε ξ ν ξ ν ξ= − = = − −
( ) ( )
( ) ( )
1m1 f f11 f m f21
f m
1m2 m3 f f21 m f f11 m f f21
f m
1 2 , 1
11
E
E
σε ϕ ξ ϕ ν ξϕ
σε ε ϕ ξ ν ϕ ξ ν ϕ ξϕ
= − +−
⎡ ⎤= = − + − −⎣ ⎦ −
f1 m1 1ε ε ε= =But ...
Handout P.25-28
• Because of coupling term ξf21, volume average tensile/compressive stresses may be induced in fibres and matrix in directions e2 and e3
• … there is transverse constraint on the f and m in these directions
Fibres:
Matrix:
Strong axial constraint on displacements of fibres and matrix
e1
e3
e2σ1
σ1 Axial coupling coeff.
Axial-transverse Coupling coeff.
Young’s modulus E1 and Poisson’s ratio ν12
28
f 21
f110ξ
ξ→ ( )
ff11
f f f m1E
E Eξ
ϕ ϕ=
+ −Hence…
Note the Rule of Mixtures! (series model = equal strain)
• The effect of transverse constraint (ξf21) is much smaller than that of axial constraint (ξf11)
Axial coupling coefficient
Axial tensile modulus
For anisotropic fibres such as carbon or aramid in an isotropic polymer matrix:
( )1 f f f m1E E Eϕ ϕ= + − ( )212 f f f m
11εν ϕν ϕ ν
ε= − = + −
Poisson’s ratio
…take the limit
E1 =ϕ fEf1 + 1−ϕ f( )Em
Transverse Tension
29
Applied stress
Note the complex stress state in matrix
From Hull and Clyne
fibre
matrix
σ
σ
View down the 2-3 plane (isotropic
e1
e3
e2
σ
σ
2
3
1
2
3
Transverse tension (2-direction)
30
( ) ( )
( )
2 2f1 f12 f f22 f32 f2 f22 f f12 f32
f f
2f3 f32 f f12 f22
f
, ,
E E
E
σ σε ξ ν ξ ξ ε ξ ν ξ ξ
σε ξ ν ξ ξ
⎡ ⎤ ⎡ ⎤= − + = − +⎣ ⎦ ⎣ ⎦
⎡ ⎤= − +⎣ ⎦
( ) ( )
( ) ( )
( ) ( )
2m1 f f12 m f f22 f f32
f m
2m2 f f22 m f f12 f32
f m
2m3 f f32 m f f22 f f12
f m
1 , 1
1 ,1
1 .1
E
E
E
σε ϕ ξ ν ϕ ξ ϕ ξϕσε ϕ ξ ν ϕ ξ ξϕ
σε ϕ ξ ν ϕ ξ ϕ ξϕ
⎡ ⎤= − − − −⎣ ⎦ −
⎡ ⎤= − + +⎣ ⎦ −
⎡ ⎤= − − − −⎣ ⎦ −
• We now need to model the case where a tensile stress is applied in the transverse direction of e2 only (or e3 only)
• There are 3 coupling coefficients involved: ξf12, ξf22, ξf32 Volume average strains for fibres:
.. and for matrix:
Handout P.29
e1
e3
e2
σ2
σ2
Find the transverse modulus E2
31
f32
f120ξ
ξ→ f1 m1 1ε ε ε= =
…gives an equation for the inverse modulus:
…take the limit Since transverse constraint can be neglected in comparison to axial constraint
We have no means of getting at ξf22 without doing a full stress analysis!
Handout P.30
1E2
=ε2σ 2
=ϕ fξf22
Ef+
1−ϕ fξf22( )Em
−ϕ f 1−ϕ f( )EfEm
ϕ fEf + 1−ϕ f( )Em
ν fEf
−νmEm
⎛
⎝⎜⎞
⎠⎟ν fξf22
Ef−νmEm
1−ϕ fξf22( )1−ϕ f( )
⎛
⎝⎜
⎞
⎠⎟
..but we may employ approximations
32
f22 1ξ =
( ) ( )( )
2f f f f mf f m
2 f m f f f m f m
1 111E E
E E E E E E Eϕ ϕ ϕϕ ν ν
ϕ ϕ− − ⎛ ⎞
= + − −⎜ ⎟+ − ⎝ ⎠
1. With
we get
2. With
we get
and f12 0ξ =
f22 1ξ =
The Inverse Rule of Mixtures
( )ff
2 f m
11E E E
ϕϕ −= +
assume no stress amplification
under-estimated since ξf22 >1 in reality
assume no axial constraint
Handout P.31
under-estimate even more than assumption #1 above
(parallel model = equal stress)
(3.26)
Alternative, empirical approach
33
3. The “Halpin-Tsai” (H-T) equations for property p
If p = E2 then it is found ζ ~1
f f mm
f f m
1 / 1 where =1- /
p pp pp p
ζηϕ ηηϕ ζ
⎛ ⎞+ −= ⎜ ⎟ +⎝ ⎠
Handout P.32
Comparison of the models
34
à Assume no stress amplification
Assume no stress amp. & no axial constraint
à Empirical approach
φ beyond 70% not achievable in practice
Handout P.33
A typical UD composite of glass fibres in epoxy resin
f22 1ξ =
Transverse modulus
Halpin-Tsai (H-T) model against some experimental data
35
Inverse RoM (equal stress = Lower Bound)
RoM series model (equal strain = Upper Bound)
H-T equation ζ ~1
From Hull and Clyne
Handout P.34
e1
e3
e2
Shear modulus parallel to fibres: G12 or G13
36
6 f f66 f f66
12 6 f m
11G G G
γ ϕ ξ ϕ ξτ
−= = +
Again the same options:
1. Assume something about the coupling, e.g. ξ66 =1 gives
2. Apply Halpin-Tsai (H-T) equation with ζ = 1
f f m12 m
f f m
1 / 1 where 1- / 1
G GG GG G
ηϕ ηηϕ
⎛ ⎞ ⎛ ⎞+ −= =⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠
Handout P.35
f f
12 f m
11G G G
ϕ ϕ−= +
Composite sheared in the 1-2 plane
Degree of stress amplification in the fibres ξf66
e1
e3
e2Inverse RoM
τ12
τ12
Fibre volume fraction φf
0.0 0.2 0.4 0.6 0.8
She
ar m
odul
us G
12 G
Pa
0
1
2
3
4
5
6S
tress amplification ξ
f66
0.0
0.5
1.0
1.5
2.0
Inverserule ofmixtures
H-T equation
Comparison of inverse rule of mixtures and H-T equation
37
Handout P.37
(lower bound)
Thermal expansion
38
Handout P.37
σf1
σf2
σf3
τ f4
τ f5
τ f6
⎛
⎝
⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟
=
ξf1T
ξf2T
ξf2T
000
⎛
⎝
⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟
ΔT
( )f
m T f Tf
( 1..3)1p p pϕξ ξ
ϕ= − =
−
A temperature rise ΔT gives rise to internal stresses due to thermal expansion mismatch between fibre and matrix…
where…
Stress-temperature coupling factors ξfpT (p = 1..6) relate the induced mean stress in the fibres to the temperature rise of the composite ΔT
stress amplification factors in the matrix from equilibrium
Due to symmetry, only two independent coupling coefficients ξf1T and ξf2T
… and hence to internal strains
39
( ) ( )f1 f1T f f2T f f f2 f3 f f2T f f1T f ff f
2 , 1T TE EE E
ε ξ ν ξ α ε ε ν ξ ν ξ αΔ Δ⎡ ⎤= − + = = − − +⎣ ⎦
f1 m1ε ε=f2T
f1T0ξ
ξ→
Then take the limit…
We get…
( )f f f f m mf11
1
1E ET E
ϕ α ϕ αεα+ −
= =Δ
( ) ( ) ( )2 f f f f m m 1 121 1 1α ϕ α ν ϕ α ν αν= + + − + −
Handout P.38-39
Axial coefficient of thermal expansion (CTE)
Transverse CTE
Axial strain Transverse strains
Anisotropy of thermal expansion
40 Fibre volume fraction φf
0.0 0.2 0.4 0.6 0.8 1.0
Thermal expansion coefficient x 106
K-1
01020304050607080
α2
α1
Handout P.40
• These calculations of CTEs fit experimental data rather well
• It can be seen that: α1 > α2
Why?
e.g. UD glass/epoxy composite
Examples Sheet (P.77)
41
Today’s Homework
try Questions 2,3,4
Contents of the lectures
1. Introduction - engineering applications
2. Basics of two-component composites
3. Elasticity of uni-directional (UD) continuous fibre composites
4. Failure of UD continuous fibre composites
5. Laminated composites: elasticity, bending, twisting and failure
42
e1
e3
e2
eY
eX
eZ
MYXMX
MY
MXY
4. Failure of UD Fibre Composites
43
Handout P.41
asterisk “ * ” denotes failure
In most polymer matrix composite materials, both the fibres and the matrix are relatively brittle: they fracture in tension at small strains ε* after little or no detectable plastic deformation
Fibre type Failure strain ε*
Glass 0.02-0.03
Carbon (HM) ~0.005
Carbon (HS) ~0.01
Kevlar 49 ~0.025
Matrix type Failure strain ε*
Epoxy resin 0.02-0.05
Thermoset polyester ~0.02
ε* < ~5%
…but fibre pull-out increases toughness
44
Handout P.42
crack
fibre
matrix
• What’s the toughening mechanism? - When a crack grows in the matrix and approaches a fibre, the
interface is sheared. - If the interface is weak, it fails in shear and the crack turns and runs
along the fibre, - ..until it finds a relatively weak spot on the fibre where that breaks,
and so on until the composite is completely broken.
• Weak fibre-matrix interfaces act as crack-stoppers
Fracture surface under a Scanning Electron Microscope (SEM)
45
A glass/polyester composite with
weak interfaces σ
σ • Pull-out of partially-debonded fibres from a fracture plane absorbs energy.
• Hence composite materials (especially those with weak interfaces) can be tougher than their constituent materials.
From Hull and Clyne
Calculation of tensile strength σ1* parallel to fibres (1-direction)
46
f m* *ε ε<
m f* *ε ε<
2 cases to consider:
Case I
Case II
Handout P.42
the fibres are more brittle than the matrix
the matrix is more brittle than the fibres
where ε* denotes failure strain
Case I: εf* < εm* e.g. high modulus carbon fibres (~1%)
in epoxy resin matrix (2~5%)
47
Strain
Stress
εm*εf*
σm’σm*
σf*
Matrix
Fibres
Stress carried by matrix at the failure strain of fibres (denoted by ’ )
Handout P.42
(relatively more brittle)
…again two possible cases
48
Strain
Stress
εm*εf*
(1−φ σf m) *
φσf f*
Matrix
Fibres
Composite
Strain
Stress
εm*εf*
φσf f*
Matrix
Fibres
Composite
(1−φ σf m) ’
σ1∗
( )1 f f f m* * 1 'σ ϕ σ ϕ σ= + −( )1 f m* 1 *σ ϕ σ= −
Handout P.43-44
Tensile strength at failure - the matrix is carrying all the stress alone!
ϕ <<f 1 At higher φf
Which one wins?
49
Fibre volume fraction φf0 1
Axial tensile strength σ1*
0
50
100
150
200
250
300
350
σm*σm'
σf*
σ 1*=φ fσ f*+
(1-φ f)σ m'
σ1*=φfσm*
φfmin
The tensile strength of the composite is the larger of the two candidate strengths given by equations in the previous slide
critical fibre volume fraction (e.g. ~3%)
Handout P.45
( )1f m
* 1*
σ ϕ σ= −
Case II: εf* > εm* e.g. glass fibres (2~3%) in a
thermoset polyester matrix (~2%)
50
Strain
Stress
εm* εf*
σf’
σm*
σf*
Matrix
Fibres
(relatively more brittle)
Stress carried by fibres at the failure strain of matrix (denoted by ’ )
Handout P.46
…and again two cases
51
Strain
Stress
εm* εf*
(1−φ σf m) *
φσf f’
Matrix
Fibres
Compositeσ1*
1 f f f m* ' (1 ) *σ ϕ σ ϕ σ= + −
Strain
Stress
εm* εf*(1−φ σf m) * Matrix
Fibres
Compositeφσf f*
1 f f* *σ ϕ σ=
f 1ϕ <<
Handout P.46-47
At higher φf
Which one wins this time?
52
Fibre volume fraction φf0 1
Axial tensile strength σ1*
0
50
100
150
200
250
300
350
σm*
σf'
σf*
σ1*=φfσf'+(1-φf)σm*
σ 1*=φ fσ f*
Handout P.47
Other modes of failure
53
σ2T*
Axial compression
Transverse tension
Axial shear
σ2T*
Handout P.48-50
σ1C*
τ12*
e1
eX
eY
θσ
Failure under off-axis stress
54
Handout P.50
e.g. off-axis tension
We need an anisotropic failure criterion
e.g. “Maximum Stress” failure criterion
1211 11 22 22
1T 1C 2T 2C 12MAX , , , , 1
* * * * *τσ σ σ σ
σ σ σ σ τ⎛ ⎞− − ≥⎜ ⎟⎝ ⎠
σ σ
• The idea is that the various failure modes outlined in the previous slide are assumed to be separate and non-interacting
• Failure will occur when the following condition is first satisfied:
Failure criterion in example P.53
55
2 2
1 2 12
cos sin sin cosMAX , , 1* * *
σ θ σ θ σ θ θσ σ τ
⎛ ⎞≥⎜ ⎟
⎝ ⎠
Angle θ degrees
0 10 20 30 40 50 60 70 80 90
Tens
ile s
treng
th σ*
0
200
400
600
800
A B C
A = axial tensile failureB = axial shear failure
C = transverse tensile failure
axial tension
axialshear transverse
tension
Failure modes
Data for 50% glass fibres in polyester resin
Handout P.53
T T
Contents of the lectures
1. Introduction - engineering applications
2. Basics of two-component composites
3. Elasticity of uni-directional (UD) continuous fibre composites
4. Failure of UD continuous fibre composites
5. Laminated composites: elasticity, bending, twisting and failure
56
e1
e3
e2
eY
eX
eZ
MYXMX
MY
MXY
Lamina or “ply”
5. Laminated Composites
57
Handout P.54
• UD fibre composites can offer high values of specific strength and stiffness parallel to the fibres,
• ..but such composites are highly anisotropic with poor transverse mechanical properties
• Hence we create a laminate of UD composites to withstand biaxial and triaxial stress states.
e.g. a 3-layer laminate with stacking sequence of [0/90/0]
Global co-ordinate axes with O at centre of laminate: eX eY eZ Local co-ordinate axes within a single lamina e1 e2 e3
O
Stacking sequence notation (I)
58
[ ]1 2 3/ / / nθ θ θ θ⋅ ⋅ ⋅ ⋅ ⋅
[ ] [ ] [ ]2 20 / 90 /0 0/90 /90/ 0 , 0/90/0 [0 / 90 /0 /0 / 90/0]≡ ≡
[ ] [ ]0 / 30 / 0 0 / 30 / 30 /0± ≡ + −
[ ] [ ] [ ]2s0 / 90 0/90 /90/ 0 0 /90 / 0≡ =
General stacking sequence:
Multiple adjacent plies
Balancing plies
Symmetrically arranged plies
Handout P.55
Stacking sequence notation (II)
59
These abbreviations can be combined, to enable complex stacking sequences to be expressed compactly
Handout P.56
0 / ±30⎡⎣ ⎤⎦2s≡ 0 / +30 / −30⎡⎣ ⎤⎦2s
≡ 0 / +30 / −30 / 0 / +30 / −30⎡⎣ ⎤⎦s
≡ 0 / +30 / −30 / 0 / +30 / −30 / −30 / +30 / 0 / −30 / +30 / 0⎡⎣ ⎤⎦
Eqn.(5.5) Mirror plane (Mz)
≡ A laminate stack of 12 plies (laminae)
Symmetry of laminates
60
• “Symmetric” laminates exhibit mirror symmetry in the mid-plane (Mz), i.e.
!
• “Balanced” laminates have equal numbers of
+θ and -θ plies
Handout P.56-57
[ ]1 2 3 s/ / ... mθ θ θ θLaminate has 2m of plies and cannot undergo out-of-plane deformation: i.e. it cannot bend or twist
No tension—shear coupling in the plane of eX and eY, when in-plane loads or displacements are applied parallel or perpendicular to these axes Balanced, symmetric plies are
preferred in practice.
Laminate elasticity – lamina (ply) stresses
61
Laminates mostly experience plane stress conditions (h << 1 mm). The strain-stress relation is then
[ ]
21
1 2
12
1 2
12
1 0
1 0
10 0
E E
SE E
G
ν
ν
⎡ ⎤−⎢ ⎥⎢ ⎥⎢ ⎥−= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
And we recall….
Handout P.58
“Reduced compliance matrix” [S] of the lamina
ε1
ε2
γ 6
⎛
⎝
⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟
=
S11 S12 0
S21 S22 0
0 0 S66
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
σ1
σ 2
τ6
⎛
⎝
⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟
3 x 3
e1
e3
e2
τ6 = τ12
σ1 σ2
σ3 = 0
RECAP à Compliance Matrix [S]
62
1 111 11
2 222 22
3 333 33
4 423 23
5 531 31
6 612 12
,
σ εσ εσ εσ εσ εσ ετ γτ γτ γτ γτ γτ γ
⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟
≡ ≡⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠
11 12 12
12 22 23
12 23 2244 22 23
44
66
66
0 0 00 0 00 0 0
, where 2( )0 0 0 0 00 0 0 0 00 0 0 0 0
S S SS S SS S S
S S SS
SS
⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟
= = −⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
S
Recap — Handout P.22-23
6 x 6
Plane stress (3 x 1)
6 x 1 1 1 1
1 21 2 21 21 1 1
12 1 2 23 21 1 1
12 1 23 2 2123
112
112
0 0 0
0 0 0
0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
E E E
E E E
E E E
G
G
G
ν νν νν ν
− − −
− − −
− − −
−
−
−
⎛ ⎞− −⎜ ⎟− −⎜ ⎟
⎜ ⎟− −⎜ ⎟= ⎜ ⎟
⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
S
Reduced compliance matrix (3 x 3 sub-matrix)
6 x 1
In-plane strains (3 x 1)
The reduced (plane stress) stiffness matrix [Q] of a single lamina
63
Inverting the 3x3 [S] sub-matrix we obtain:
( ) ( )
( ) ( )
1 21 1
12 21 12 21
1 12 2 2
12 21 12 21
12
01 1
[ ] [ ] 01 1
0 0
E E
E EQ S
G
νν ν ν ν
νν ν ν ν
−
⎡ ⎤⎢ ⎥− −⎢ ⎥⎢ ⎥
= = ⎢ ⎥− −⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
Handout P.58
Note: 1. [Q] is the stiffness of a single lamina w.r.t. local axes e1 & e2
2. Applicable only under plane stress conditions
In a single lamina (one ply)…
64
• The stresses are related to the in-plane strains by
[ ]1 1
2 2
6 6
Qσ εσ ετ γ
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟=⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
• In general, local axes e1, e2 of lamina will be inclined to the global axes eX, eY, by an angle θ
Handout P.59
e1
eX
eY
θσRecap:
Handout P.50 σ σ
3 x 3 3 x 1 3 x 1
Rotation of axes of [Q] via Mohr’s circle
65
1 XX 1 XX
2 YY 2 YY
6 XY 6 XY
[ ] , [ ]T Tσ ε
σ σ ε εσ σ ε ετ τ γ γ
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟= =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
2 2 2 2
2 2 2 2
2 2 2 2
2
[ ] 2 , [ ]
( ) 2 2 ( )
c s cs c s cs
T s c cs T s c cs
cs cs c s cs cs c sσ ε
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥
= − = −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− − − −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
XX 1 1 XX1 1 1
YY 2 2 YY
XY 6 6 XY
[ ] [ ] [ ] [ ] [ ][ ]T T Q T Q Tσ σ σ ε
σ σ ε εσ σ ε ετ τ γ γ
− − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟= = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Hence..
• To calculate the response of the whole laminate,
• first we need to establish σ-ε relation for each layer,
• .. expressed w.r.t. the global reference axes along X, Y, Z Global
(≡ Laminate) Local
(≡ Lamina)
c ≡ cosθ, s ≡ sinθ
Handout P.60-61
Global Global Local Local
Transformation matrices to convert stress and strain:
Q⎡⎣ ⎤⎦
Global (≡ Laminate)
Local (≡ Lamina)
result… Laminate Stiffness Matrix [Q]
66
XX XX1
YY YY
XY XY
[ ] where [ ] [ ] [ ][ ]Q Q T Q Tσ ε
σ εσ ετ γ
−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
4 4 2 211 11 22 12 662( )Q Q c Q s Q Q c s= + + +
3 316 11 12 66 12 22 66( 2 ) ( 2 )Q Q Q Q c s Q Q Q cs= − − + − +
4 4 2 266 66 11 22 12 66( ) ( 2 2 )Q Q c s Q Q Q Q c s= + + + − −
Handout P.60
The lamina stress-strain relation, expressed with respect to the global (laminate) reference axes (eX, eY, eZ)
=
Q11 Q12 Q16
Q12 Q22 Q26
Q16 Q26 Q66
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥ 3 x 3 Examples of some terms in [Q]:
Eqn.(5.15)
Laminate forces N (unit load: N/m)
67
eXeY
eZ
NXY NXY NYYNXX
Note the definitions of the unit loads (forces per unit width of composite), hence: e.g.
/ 2
XX XX/ 2
/ 2 / 2 / 2
11 XX 12 YY 16 XY/ 2 / 2 / 2
d
d d d .
h
hh h h
h h h
N z
Q z Q z Q z
σ
ε ε γ
−
− − −
=
= + +
∫
∫ ∫ ∫
Handout P.61
In a composite laminate, the stresses usually vary throughout the thickness, because of the differing elastic constants of the layers
in-plane strains are constant in through-thickness direction…
mid-plane
h
n laminae
Summing over the n laminae…
68
( )
( )
( )
XX 11 1 XX1
12 1 YY1
16 1 XY1
( )
( )
( ) .
n
k k kk
n
k k kk
n
k k kk
N Q h h
Q h h
Q h h
ε
ε
γ
−=
−=
−=
= −
+ −
+ −
∑
∑
∑
Handout P.62
hk
eZ
eXk
1
mid-plane
..the integrals can be replaced by summations over the n layers
Seen edgewise: (down eY)
..giving the laminate stiffness matrix [A]
69
Handout P.63
where…
( )11( )
n
ij ij k k kk
A Q h h −=
= −∑
• Repeat the step in previous slide to obtain similar expressions for the other unit loads NYY and NXY
• ... can be combined in the form of a matrix equation relating the unit loads {N} to the laminate in-plane strains {ε}:
NXX
NYY
NXY
⎛
⎝
⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟
=
A11 A12 A16
A12 A22 A26
A16 A26 A66
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
εXX
εYY
γ XY
⎛
⎝
⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟
the thickness of the k-th lamina
Steps in laminate calculations…
70
A typical problem in the mechanics of fibre composite laminates would be: a laminate is to be subjected to certain in-plane strains εXX, εYY, γXY, predict the unit loads induced.
Handout P.63
1. For each lamina, obtain the elastic constants:
2. For each lamina calculate matrix [Q] 3. For each lamina rotate axes to obtain
4. For the whole laminate calculate [A]
5. Multiply the strain column by [A] to obtain the unit loads {N}
1 2 12 21 12, , , , E E Gν ν
[ ]Q
Laminate example #1
71
Laminate example 1: Calculate the stiffness matrix [A] for a [+45/-45]s
laminate constructed from 4 identical laminae each with thickness 0.1 mm
and with elastic constants:
1 2 12 1255 GPa, 16 GPa, 0.28, 7.6 GPa.E E Gν= = = =
Handout P.65
Examples Sheet P.77
72
Today’s Homework:
try Questions 5,6,7
Contents of the lectures
1. Introduction - engineering applications
2. Basics of two-component composites
3. Elasticity of uni-directional (UD) continuous fibre composites
4. Failure of UD continuous fibre composites
5. Laminated composites: elasticity, bending, twisting and failure
73
e1
e3
e2
eY
eX
eZ
MYXMX
MY
MXY
Laminate bending and twisting
74
eY
eX
eZ
MYXMX
MY
MXY
We define unit moments as shown
Handout P.67
To retain the usual sign convention (sagging bending moments are positive); we view the laminate from underneath (+eZ pointing down).
Bending moments per unit width of laminate
Twisting moments per unit width of laminate
Special case: bending in the X-Z plane
75
MX MX
h
eZ
eX
XX X
YY
XY
00
kz
εεγ
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟=⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
/ 2 / 22
X XX 11 X/ 2 / 2
d dh h
h h
M z z Q z zkσ− −
= =∫ ∫
For curvature kX (=1/R) the strains:
Hence the bending moment in the XZ plane:
Handout P.67
Curvature in XZ plane
Plane strain
Not twisting
A wide plate A special case:
z
The general case of bending/twisting
76
XX X
YY Y
XY XY
kz kk
εεγ
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟=⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
( )
( )
( )
3 3X 11 1 X
1
3 312 1 Y
1
3 316 1 XY
1
1 ( )3
1 ( )3
1 ( ) .3
n
k k kk
n
k k kk
n
k k kk
M Q h h k
Q h h k
Q h h k
−=
−=
−=
= −
+ −
+ −
∑
∑
∑
Handout P.68
..and the bending moment in the XZ plane becomes
As before, we recognise that the elements of [Q] are piecewise constant through the thickness of the laminate, resulting in summations over the n laminae as follows
twisting curvature
curvature in YZ plane curvature in XZ plane
The bending/twisting stiffness matrix for the laminate [D]
77
X 11 12 16 X
Y 12 22 26 Y
XY 16 26 66 XY
M D D D kM D D D kM D D D k
⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟=⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
where
( )3 31
1
1 ( )3
n
ij ij k k kk
D Q h h −=
= −∑
Handout P.69
When this step is repeated for the YZ plane moment MY and twisting moment MXY (=MYX), the complete set of bending moments may be expressed as
The case of the symmetric laminate
78
eZ
eX
Stress )σ (XX z
StrainεXX( )z
MX MX
Answer : there is no net axial force
Handout P.70
Does curvature or twisting of the laminate lead to in-plane resultant loads NX, NY, NXY in addition?
.. a symmetric laminate does not show tension-bending/torsion coupling!
The full stiffness matrix of the laminate
79
[ ]0 0 00 0 00 0 0
B⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦
Where…
..for a symmetric laminate with 2m of plies
Handout P.71
Now assemble together all the 6 equations for in-plane forces and bending moments acting on the lamina, in terms of the in-plane strains and curvatures
Matrices [A] and [D] are sub-matrices from within the total stiffness matrix of the laminate
A
D
Failure of laminates – 2 modes
80
• Intra-lamina failure i.e. takes place within
laminae, for example:
• Inter-lamina failure i.e. takes place between
laminae, for example:
eXeY
eZ
eXeY
eZ
Handout P.71
VS.
..usually under in-plane stresses ..under out-of-plane stresses
Prediction of intra-lamina failure
81
1. Determine strain ratios εXX :εYY :εXY (wrt global axes) if necessary.
2. Rotate these strains to find strains ε1 :ε2 :ε6 (wrt local axes) in each lamina.
3. Find stresses σ1 , σ2 , σ6 in each lamina via {σ}=[Q]{ε}. 4. Apply “maximum stress failure criterion” (on P.51) to
predict first failure in each lamina and hence in the laminate
Handout P.72
Laminate example #2
82
Laminate example 2: The laminate in Laminate Example 1 has the following lamina strengths:
1T 1C 2T 2C 12* 1500 MPa, * 750 MPa, * 30 MPa, * 150 MPa, * 50 MPa.σ σ σ σ τ= = = = =
The laminate is bent in the XZ plane. Where does the first intra-lamina
failure occur? How does first failure occur? At what curvature does first
failure occur?
Handout P.73
When?
B8 Classes:
83
Week 8 sign up via WebLearn
PPT slides: www.eng.ox.ac.uk/tan (under teaching page)
Extras
84
..the other Poisson’s ratio ν21
85
• Easily obtained from symmetry of the compliance matrix S
221 12
1
EE
ν ν=
• Also we recall that…
Handout P.35
( )12 f f f m1ν ϕ ν ϕ ν= + −c.f. Handout P.23
1 1 11 21 2 21 2
1 1 112 1 2 23 2
1 1 112 1 23 2 2
123
112
112
0 0 0
0 0 0
0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
E E E
E E E
E E E
G
G
G
ν νν νν ν
− − −
− − −
− − −
−
−
−
⎛ ⎞− −⎜ ⎟− −⎜ ⎟
⎜ ⎟− −⎜ ⎟= ⎜ ⎟
⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠
S
SYMM.
Eqn.(3.5) on P.23
RoM
The case of a symmetric laminate
86
Handout P.64
eZ
eX
Uniform extensionof each layer
Stress distribution )σ (XX z
Note: there is no bending moment!
In a symmetric laminate, there is no coupling between uniform in-plane strains and bending moments
Mirror plane Mz
Recommended