Micromechanics- Classical Laminate Theory 1

Question 1

Laminate approach to manufacturing composites

Answer 1

Combine fibres/matrix into thin layers.
Consolidate several of these layers (laminae/plies) together to obtain desired thickness.
Many laminae consolidated together is a laminate

Question 2

Three levels at which to consider the material

Answer 2

Component level (fibres and matrix)
Lamina level
Laminate level

Question 3

Assumption for component level

Answer 3

All individual components (fibres and matrix) of the composite are isotropic. Valid for matrix (as generally isotropic polymers) and glass but not for carbon

Question 4

Shear modulus of fibres

Answer 4

Gf=Ef/2(1+νf)
All f subscript meaning fibre
Ef is tensile modulus
νf is Poisson’s ratio

Question 5

Shear modulus of matrix

Answer 5

Gm=Em/2(1+νm)
All m subscript mean matrix
Em is tensile modulus
νm is Poisson’s ratio

Question 6

Mass, volumes and fractions notation

Answer 6

Mass: m=mf+mm
Mass fractions: Mf+Mm=1
Volume: V=Vf+Vm+Vv (sub v means voids)
Volume fractions: ff+fm+fv=1

Question 7

Density of composite formula

Answer 7

ρ=m/V=(ρf)(ff)+(ρm)(fm)

Question 8

Void volume fraction formula in databook

Answer 8

fv=1-ρ((Mf/ρf)+(Mm/ρm))

Question 9

What void volume fraction does aerospace require?

Answer 9

Less than 1%

Question 10

Directions in a unidirectional lamina

Answer 10

All fibres oriented in the same direction. This direction is 1 (longitudinal). Perpendicular to this in the plane is 2 (transverse). Perpendicular to both and out if plane is 3

Question 11

Assumption for strain when longitudinal stress applied to unidirectional lamina

Answer 11

Equal longitudinal strain for the fibre, matrix and lamina
ε1=ε1f=ε1m

Question 12

How do longitudinal stresses in fibres compare to matrix?

Answer 12

σ1f much greater than σ1m

Question 13

Formula for longitudinal stress in lamina

Answer 13

σ1=ffσ1f+fmσ1m
σ1=(ffEf+fmEm)ε1

Question 14

Formula for longitudinal modulus in lamina in databook

Answer 14

E1=ffEf+(1-ff)Em
Known as longitudinal rule of mixtures
Example of equal strain or Voigt model

Question 15

Assumptions for rule of mixtures equations

Answer 15

Composite is unidirectional
Perfect adhesion between fibres and matrix
Fibres uniformly distributed within the matrix
Fibres have uniform properties in any given direction
Each fibre has same properties as any other
Matrix is isotropic and contains no voids
There are no residual stresses in the composite
The fibres and matrix both behave as linear elastic materials (ok for low strains)

Question 16

Assumption for transverse applied stress

Answer 16

Equal transverse stress so
σ2=σ2f=σ2m

Question 17

Transverse rule of mixtures and why is inaccurate

Answer 17

1/E2=ff/Ef + (1-ff)/Em
Example of equal stress or Reuss model
Inaccurate and rarely used as there is non-uniform strain distribution in the matrix (larger strains near fibres)
Ultimately the fibres are bearing more load than expected

Question 18

Halpin-Tsai equation

Answer 18

E2=Em(1+ξηff)/(1-ηff)
For transverse stress
Where η=(Ef-Em)/(Ef+ξEm)
And ξ=1 unless told otherwise
Not fully rigorous and based in experimental data but gives good results

Question 19

Modulus vs ff graphs for rule of mixtures and Halpin-Tsai

Answer 19

Longitudinal rule of mixtures is linear from Em to Ef (0-1 ff).
Transverse rule of mixtures starts and ends at same point but otherwise much lower with steep curve at end.
Halpin-Tsai is like transverse ROM but always a but above

Question 20

Why is unidirectional composite stiffer parallel to fibres than perpendicular to fibres?

Answer 20

Parallel (1) has matrix and fibres working in parallel so there is load sharing. Perpendicular (2) has matrix and fibres working in series so there is deformation sharing. Effect has very little to do with any anisotropy in fibres (like carbon)

Question 21

Formula for lamina shear modulus in databook

Answer 21

G12=G13=Gm(1+ξηff)/(1-ηff)
Where η=(Gf-Gm)/(Gf+ξGm)
And ξ=1
In reality G12 and G13 not same as wont be perfect adhesion between laminae

Question 22

Formula for shear modulus for shear response in plane normal to the fibre direction in databook

Answer 22

G23=E2/2(1+ν23)

Question 23

The 3 different lamina Poisson’s ratios

Answer 23

Major: stress along 1, equal applied strains but unequal Poisson contractions, ν12=ν13
Minor: stress along 2, unequal applied strains but equal Poisson contractions, ν21=ν31
Other: stress along 3, unequal applied strains and unequal Poisson contractions, ν23=ν32

Question 24

Lamina major Poisson’s ratio

Answer 24

ν12. Determined using longitudinal rule of mixtures.
ν12=ffνf+(1-ff)νm (databook)
Quantifies contraction in transverse direction 2 in response to extension in longitudinal direction 1

Question 25

Lamina minor Poisson’s ratio

Answer 25

ν21. Transverse anisotropy allows minor to be calculated using
ν12/E1=ν21/E2 (databook)
Quantifies contraction in longitudinal direction 1 in response to extension in the in-plane transverse direction 2
Expected to be small

Question 26

The third Poisson’s ratio

Answer 26

ν23. Accounts for contraction in other transverse direction 3 in response to extension in the in-plane transverse direction 2.
Expected to be large.
ν23=1-ν21-E2/3B (databook)
B is bulk modulus of composite

Question 27

Obtaining bulk modulus of composite

Answer 27

σh=ΔfBf=ΔmBm
Where σh is hydrostatic stress, Δ is overall volume change
σh=(σ1+σ2+σ3)/3
Δ=ffΔf+(1-ff)Δm
Bf=Ef/3(1-2νf), Bm=Em/3(1-2νm)
So overall: B=(ff/Bf + (1-ff)/Bm)^-1

Question 28

Graph of three Poisson’s ratios vs ff

Answer 28

ν12 is linear decrease from νm to νf over 0 to 1.
Others start and end at same points.
ν21 quickly goes below νf and stays low until near 1 when it goes back up to νf.
ν23 quickly rises from νm then gradually comes down then steeper at end down to νf

Question 29

Formulae for bulk modulus of fibres or matrix

Answer 29

Fibres:
Bf=Ef/3(1-2νf)
Matrix:
Bm=Em/3(1-2νm)